Wiley's Solomons, Fryhle, Synder Organic Chemistry for NEET and other Medical Entrance Examinations 9789354248894, 9789354248788

Table of contents :
Cover
Title Page
Copyright
Preface to the Adapted Version
Preface to the Original Edition
About the Authors
About the Adapting Author
Contents
1 Basic Principles of Organic Chemistry
PART I: BONDING AND MOLECULAR STRUCTURE
1.1 Life and the Chemistry of Carbon Compounds
1.1A Development of the Science of Organic Chemistry
1.2 Chemical Bonds: The Octet Rule
1.2A Ionic Bonds
1.2B Covalent Bonds and Lewis Structures
1.2C How to Write Lewis Structures
1.3 Isomers: Different Compounds that Have the Same Molecular Formula
1.4 How to Write and Interpret Structural Formulas
1.4A More About Dash Structural Formulas
1.4B Condensed Structural Formulas
1.4C Bond-Line Formulas
1.4D Three-Dimensional Formulas
1.5 Resonance Theory
1.5A The Use of Curved Arrows: How to Write Resonance Structures
1.5B Rules for Writing Resonance Structures
1.5C How Do We Decide When One Resonance Structure Contributes More to the Hybrid Than Another?
1.6 The Structure of Methane and Ethane: sp3 Hybridization
1.6A The Structure of Methane
1.6B The Structure of Ethane
1.7 The Structure of Ethene (Ethylene): sp2 Hybridization
1.7A Restricted Rotation and the Double Bond
1.7B Cis–Trans Isomerism
1.8 The Structure of Ethyne (Acetylene): sp Hybridization
1.8A Bond Lengths of Ethyne, Ethene, and Ethane
1.8B A Summary of Important Concepts
1.9 How to Predict Molecular Geometry: The Valence Shell Electron Pair Repulsion Model
1.9A Methane
1.9B Ammonia
1.9C Water
1.9D Carbon Dioxide
Part II: Families of Carbon Compounds
1.10 Classification of Organic Compounds
1.10A Acyclic or Open Chain Compounds
1.10B Alicyclic or Closed-Chain or Ring Compounds
1.10C Aromatic Compounds
1.10D Homologous Series
1.11 Hydrocarbons: Representative Alkanes, Alkenes, Alkynes, and Aromatic Compounds
1.11A Alkanes
1.11B Alkenes
1.11C Alkynes
1.11D Benzene: A Representative Aromatic Hydrocarbon
1.12 Functional Groups
1.12A Alkyl Groups and the Symbol R
1.12B Phenyl and Benzyl Groups
1.13 Alkyl Halides or Haloalkanes
1.14 Alcohols and Phenols
1.15 Ethers
1.16 Amines
1.17 Aldehydes and Ketones
1.18 Carboxylic Acids, Esters, and Amides
1.18A Carboxylic Acids
1.18B Esters
1.18C Amides
1.19 Nitriles
1.20 Summary of Important Families of Organic Compounds
1.20A Functional Groups in Biologically Important Compounds
1.21 Nomenclature of Organic Compounds
1.21A The IUPAC System of Nomenclature
1.21B IUPAC Nomenclature of Alkanes
1.21C Nomenclature of Organic Compounds having Functional Group(s)
1.21D Nomenclature of Substituted Benzene Compounds
1.22 Polar Covalent Bonds
1.22A Polar and Nonpolar Molecules
1.22B Dipole Moments in Alkenes
1.23 Physical Properties and Molecular Structure
1.23A Ionic Compounds: Ion–Ion Forces
1.23B Intermolecular Forces (van der Waals Forces)
1.23C Boiling Points
1.23D Solubilities
Part III: An Introduction to Organic Reactions and their Mechanisms
1.24 Acid–Base Reactions
1.24A Brønsted–Lowry Acids and Bases
1.24B Lewis Acids and Bases
1.25 How to Use Curved Arrows in Illustrating Reactions
1.26 Heterolysis of Bonds to Carbon: Carbocations and Carbanions
1.26A Electrophiles and Nucleophiles
1.27 Homolytic Fission of Bonds
1.28 Electron Displacement Effects in Organic Compounds
1.28A Inductive Effect
1.28B Resonance Structure and Effect
1.28C Electromeric Effect (E Effect)
1.28D Hyperconjugation
1.29 The Strength of Brønsted–Lowry Acids and Bases: Ka and pKa
1.29A Predicting the Strength of Bases
1.29B How to Predict the Outcome of Acid–Base Reactions
1.30 Relationships between Structure and Acidity
1.30A The Effect of Hybridization
1.30B Inductive Effects
1.31 Acidity: Carboxylic Acids versus Alcohols
1.31A The Effect of Delocalization
1.31B The Inductive Effect
1.31C Summary and a Comparison of Conjugate Acid–Base Strengths
1.31D Inductive Effects of Other Groups
1.32 The Effect of the Solvent on Acidity
1.33 A Mechanism for an Organic Reaction
Solved Examples
Solved Previous Years’ NEET Questions
Additional Objective Questions
Answer Key
Hints and Explanations
2 Isomerism
2.1 Types of Isomerism
2.1A Structural (Constitutional) Isomerism
2.1B Stereoisomerism
2.2 Geometrical Isomerism
2.2A E- and Z-Nomenclature
2.2B Geometrical Isomerism in Cyclic Structures
2.3 Chirality and Stereochemistry
2.3A The Biological Significance of Chirality
2.4 Enantiomers and Chiral Molecules
2.5 Molecules having One Chirality Center are Chiral
2.5A Tetrahedral versus Trigonal Stereogenic Centers
2.6 How to Test for Chirality: Planes of Symmetry
2.6A Plane of Symmetry
2.6B Center of Symmetry
2.6C Axis of Symmetry
2.6D Alternating or Improper Axis of Symmetry (Sn or σ)
2.7 Naming Enantiomers: The R,S-System
2.7A How to Assign (R) and (S) Configurations
2.8 Properties of Enantiomers: Optical Activity
2.8A Plane-Polarized Light
2.8B The Polarimeter
2.8C Specific Rotation
2.8D Racemic Forms
2.9 The Synthesis of Chiral Molecules
2.9A Racemic Mixture
2.9B Stereoselective Syntheses
2.10 Molecules with More than One Chirality Center
2.10A How to Draw Stereoisomers for Molecules having More than One Chirality Center
2.10B Meso Compounds
2.10C How to Name Compounds with More than One Chirality Center
2.11 Fischer Projection Formulas
2.11A How to Draw and Use Fischer Projections
2.12 Stereoisomerism of Cyclic Compounds
2.12A Cyclohexane Derivatives
2.13 Relating Configurations through Reactions in Which No Bonds to the Chirality Center are Broken
2.14 Chiral Molecules that do not Possess a Chirality Center
2.15 Conformational Isomerism
2.15A Conformations
2.15B Sawhorse Projections
2.15C Newman Projections
Solved Examples
Solved Previous Years’ NEET Questions
Additional Objective Questions
Answer Key
Hints and Explanations
3 Purification and Characterization of Organic Compounds
3.1 Methods of Purification of Organic Compounds
3.1A Sublimation
3.1B Crystallization
3.1C Distillation
3.1D Differential Extraction
3.1E Chromatography
3.2 Qualitative Analysis of Organic Compounds
3.2A Preliminary Tests
3.2B Detection of Carbon and Hydrogen
3.2C Detection of Other Elements
3.3 Quantitative Analysis
3.3A Carbon and Hydrogen
3.3B Nitrogen
3.3C Halogens
3.3D Sulphur
3.3E Phosphorus
3.3F Oxygen
3.4 Determination of Empirical Formula of the Organic Compound
Solved Examples
Solved Previous Years’ NEET Questions
Additional Objective Questions
Answer Key
Hints and Explanations
4 Alkanes and Cycloalkanes
4.1 Introduction to Alkanes and Cycloalkanes
4.1A Sources of Alkanes: Petroleum
4.2 Shapes of Alkanes
4.3 How to Name Alkanes and Alkyl Groups: The IUPAC System
4.3A How to Name Unbranched Alkyl Groups
4.3B How to Name Branched-Chain Alkanes
4.3C How to Name Branched Alkyl Groups
4.3D How to Classify Hydrogen Atoms
4.4 How to Name Cycloalkanes
4.4A How to Name Monocyclic Cycloalkanes
4.4B How to Name Bicyclic Cycloalkanes
4.5 Physical Properties of Alkanes and Cycloalkanes
4.6 Isomerism
4.7 Synthesis of Alkanes and Cycloalkanes
4.7A Hydrogenation of Alkenes and Alkynes
4.7B From Alkyl Halides
4.7C From Carboxylic Acids
4.8 How to Gain Structural Information from Molecular Formulas and the Index of Hydrogen Deficiency
4.8A Compounds Containing Halogens, Oxygen, or Nitrogen
4.9 Chemical Reactions of Alkanes
4.9A Reactions of Alkanes with Halogens
4.9B Chlorination of Methane: Mechanism of Reaction
4.9C Halogenation of Higher Alkanes
4.9D Other Substitution Reactions
4.9E Combustion of Alkanes
4.9F Controlled (Catalytic) Oxidation
4.9G Isomerization
4.9H Aromatization
4.9I Pyrolysis
4.10 Conformations
4.10A Newman Projections and How to Draw Them
4.10B Conformational Analysis of Ethane
4.10C Conformational Analysis of Butane
4.10D Stereoisomers and Conformational Stereoisomers
4.11 The Relative Stabilities of Cycloalkanes: Ring Strain
4.11A Cyclopropane
4.12 Conformations of Cyclohexane: The Chair and the Boat
4.12A Conformations of Higher Cycloalkanes
4.13 Disubstituted Cycloalkanes: Cis–Trans Isomerism
Solved Examples
Solved Previous Years’ NEET Questions
Additional Objective Questions
Answer Key
Hints and Explanations
5 Alkenes and Alkynes
Part I: Alkenes
5.1 Structure of the Double Bond
5.2 Isomerism
5.3 How to Name Alkenes and Cycloalkenes
5.4 The (E )–(Z) System for Designating Alkene Diastereomers
5.4A How to Use the (E )–(Z ) System
5.5 Relative Stabilities of Alkenes
5.5A Heat of Reaction
5.5B Overall Relative Stabilities of Alkenes
5.6 Cycloalkenes
5.7 Synthesis of Alkenes by Hydrogenation of Alkynes
5.7A Syn Addition of Hydrogen: Synthesis of cis-Alkenes
5.7B Anti Addition of Hydrogen: Synthesis of trans-Alkenes
5.8 Synthesis of Alkenes via Elimination Reactions
5.8A Dehydrohalogenation of Alkyl Halides
5.8B Dehalogenation of Vicinal Dihalides
5.8C Acid-Catalyzed Dehydration of Alcohols
5.9 Carbocation Stability and the Occurrence of Molecular Rearrangements
5.9A Rearrangements During Dehydration of Secondary Alcohols
5.9B Rearrangement After Dehydration of a Primary Alcohol
5.10 Hydrogenation of Alkenes
5.10A Hydrogenation: The Function of the Catalyst
5.11 Addition Reactions of Alkenes
5.11A How to Understand Additions to Alkenes
5.11B Electrophilic Addition of Hydrogen Halides to Alkenes: Mechanism and Markovnikov’s Rule
5.11C The Anti-Markovnikov Addition of Hydrogen Bromide Radical Addition to Alkenes
5.11D Addition of Water to Alkenes: Acid-Catalyzed Hydration
5.11E Electrophilic Addition of Bromine and Chlorine to Alkenes
5.11F Addition of Sulphuric Acid
5.12 Oxidation of Alkenes: Syn 1,2-Dihydroxylation
5.12A Oxidative Cleavage of Alkenes
5.12B Cleavage with Ozone
Part II: Alkynes
5.13 Structure of the Triple Bond
5.13A Bond Lengths of Ethyne, Ethene and Ethane
5.14 Isomerism
5.15 How to Name Alkynes
5.16 The Acidity of Terminal Alkynes
5.17 Methods of Preparation
5.17A Synthesis of Alkynes by Elimination Reactions
5.17B Synthesis of Higher Alkynes by Carbon–Carbon Bond Formation in Terminal Alkynes
5.17C Other Methods of Preparation
5.18 Chemical Reactivity
5.18A Electrophilic Addition Reactions
5.18B Oxidation of Alkynes
5.18C Polymerization
Part III: Conjugated Unsaturated Systems
5.19 The Stability of the Allyl Radical
5.19A Molecular Orbital Description of the Allyl Radical
5.19B Resonance Description of the Allyl Radical
5.19C Resonance Theory Revisited
5.20 The Allyl Cation
5.21 Alkadienes and Polyunsaturated Hydrocarbons
5.22 1,3-Butadiene: Electron Delocalization
5.22A Bond Lengths of 1,3-Butadiene
5.22B Conformations of 1,3-Butadiene
5.22C Molecular Orbitals of 1,3-Butadiene
5.23 The Stability of Conjugated Dienes
5.24 Allylic Substitution and Allylic Radicals
5.24A Allylic Chlorination (High Temperature)
5.24B Allylic Bromination with N-Bromosuccinimide (Low Concentration of Br2)
5.25 Electrophilic Attack on Conjugated Dienes: 1,4-Addition
5.25A Kinetic Control versus Thermodynamic Control of a Chemical Reaction
5.26 The Diels–Alder Reaction: A 1,4-Cycloaddition Reaction of Dienes
Solved Examples
Solved Previous Years’ NEET Questions
Additional Objective Questions
Answer Key
Hints and Explanations
6 Aromatic Compounds and their Reactions
6.1 The Discovery of Benzene
6.2 Nomenclature of Benzene Derivatives
6.3 Structure of Benzene
6.3A Reactions of Benzene
6.3B The Kekulé Structure for Benzene
6.4 The Thermodynamic Stability of Benzene
6.5 Modern Theories of the Structure of Benzene
6.5A The Resonance Explanation of the Structure of Benzene
6.5B The Molecular Orbital Explanation of the Structure of Benzene
6.6 Hückel’s Rule: The 4n + 2π Electron Rule
6.6A How to Diagram the Relative Energies of π Molecular Orbitals in Monocyclic Systems Based on Hückel’s Rule
6.6B The Annulenes
6.6C Aromatic Ions
6.6D Aromatic, Antiaromatic, and Nonaromatic Compounds
6.7 Other Aromatic Compounds
6.7A Benzenoid Aromatic Compounds
6.7B Nonbenzenoid Aromatic Compounds
6.7C Fullerenes
6.7D Heterocyclic Aromatic Compounds
6.7E Aromatic Compounds in Biochemistry
6.8 Preparation of Arenes
6.9 Physical Properties of Arenes
6.10 Electrophilic Aromatic Substitution Reactions
6.11 A General Mechanism for Electrophilic Aromatic Substitution
6.12 Halogenation of Benzene
6.13 Nitration of Benzene
6.14 Sulfonation of Benzene
6.15 Friedel–Crafts Reactions
6.15A Friedel–Crafts Alkylation
6.15B Friedel–Crafts Acylation
6.15C Limitations of Friedel–Crafts Reactions
6.15D Synthetic Applications of Friedel–Crafts Acylations
6.16 Substituents Can Affect Both the Reactivity of the Ring and the Orientation of the Incoming Group
6.16A How Do Substituents Affect Reactivity?
6.16B Ortho–Para-Directing Groups and Meta-Directing Groups
6.16C Electron-Donating and Electron-Withdrawing Substituents
6.16D Groups: Ortho–Para Directors
6.16E Deactivating Groups: Meta Directors
6.16F Halo Substituents: Deactivating Ortho–Para Directors
6.16G Classification of Substituents
6.17 How Substituents Affect Electrophilic Aromatic Substitution: A Closer Look
6.17A Reactivity: The Effect of Electron-Releasing and Electron-Withdrawing Groups
6.17B Inductive and Resonance Effects: Theory of Orientation
6.17C Meta-Directing Groups
6.17D Ortho–Para-Directing Groups
6.17E Ortho–Para Direction and Reactivity of Alkylbenzenes
6.17F Summary of Substituent Effects on Orientation and Reactivity
6.18 Reactions of the Side Chain of Alkylbenzenes
6.18A Benzylic Radicals and Cations
6.18B Benzylic Halogenation of the Side Chain
6.19 Oxidation Reactions
6.19A Oxidation of the Benzene Ring
6.19B Oxidation of the Side Chain
6.20 Addition Reactions
6.21 Reduction of Aromatic Compounds
6.21A The Birch Reduction
6.22 Synthetic Applications
6.22A Use of Protecting and Blocking Groups
6.22B Orientation in Disubstituted Benzenes
Solved Examples
Solved Previous Years’ NEET Questions
Additional Objective Questions
Answer Key
Hints and Explanations
7 Alkyl and Aryl Halides
7.1 Classification
7.1A Classification Based on Hybridization of Carbon Atom
Part I: Alkyl Halides (Haloalkanes)
7.2 Nomenclature
7.3 Physical Properties of Alkyl Halides
7.3A Melting and Boiling Points
7.3B Density
7.3C Solubility
7.4 Preparation of Alkyl Halides
7.4A From Alcohols
7.4B From Hydrocarbons
7.4C Halogen Exchange
7.5 Nucleophilic Substitution Reactions of Alkyl Halides
7.5A Nucleophiles and Leaving Groups
7.5B Substitution Nucleophilic Bimolecular (SN2)
7.5C Substitution Nucleophilic Unimolecular (SN1)
7.5D Factors Affecting the Rates of SN1 and SN2 Reactions
7.5E Organic Synthesis: Functional Group Transformations Using SN2 Reactions
7.6 Elimination Reactions of Alkyl Halides
7.6A Dehydrohalogenation
7.6B The E2 Reaction
7.6C The E1 Reaction
7.6D How to Determine Whether Substitution or Elimination is Favored
7.7 Overall Summary–Substitution and Elimination Reactions
7.8 Reaction of Alkyl Halides with Metals
7.9 Reduction of Alkyl Halides to Hydrocarbons
7.10 Polyhalogen Compounds
7.10A Dichloromethane (Methylene Chloride CH2Cl2)
7.10B Trichlormethane (Chloroform CHCl3 )
7.10C Triiodomethane (Iodoform CH3l)
7.10D Tetrachloromethane (Carbon Tetrachloride CCl4 )
7.10E Freons
Part II: Aryl Halides (Haloarenes)
7.11 Nomenclature
7.12 Preparation of Aryl Halides
7.12A By Electrophilic Substitution
7.12B From Diazonium Salts
7.13 Nucleophilic Aromatic Substitution in Aryl Halides
7.13A Nucleophilic Aromatic Substitution by Addition–Elimination: The SNAr Mechanism
7.13B Nucleophilic Aromatic Substitution through an Elimination–Addition Mechanism: Benzyne
7.13C Phenylation
7.14 Electrophilic Aromatic Substitution Reaction
7.15 Reaction with Metals
Solved Examples
Solved Previous Years’ NEET Questions
Additional Objective Questions
Answer Key
Hints and Explanations
8 Alcohols, Phenols and Ethers
8.1 Structure and Classification
8.1A Monohydric, Dihydric, Trihydric or Polyhydric Alcohols and Phenols
8.1B Classification of Ethers
8.2 Structures of Functional Groups
8.2A Alcohols
8.2B Phenols
8.2C Ethers
8.3 Nomenclature
8.3A Nomenclature of Alcohols
8.3B Nomenclature of Phenols
8.3C Nomenclature of Ethers
8.4 Physical Properties
8.4A Physical Properties of Alcohols
8.4B Physical Properties of Phenols
8.4C Physical Properties of Ethers
8.5 Synthesis of Alcohols
8.5A Alcohols from Alkenes
8.5B Alcohols by Reduction of Carbonyl Compounds
8.5C Alcohols from Grignard Reagents
8.5D By Substitution Reactions on Alkyl Halides
8.6 Synthesis of Phenols
8.6A Laboratory Synthesis
8.6B Industrial Syntheses
8.7 Synthesis of Ethers
8.7A Ethers by Intermolecular Dehydration of Alcohols
8.7B The Williamson Ether Synthesis
8.7C Synthesis of Ethers by Alkoxymercuration–Demercuration
8.8 Reactions of Alcohols
8.8A Reactions Involving Cleavage of O—H Bond (Alcohols as Acids)
8.8B Reactions Involving Cleavage of Carbon–Oxygen (C—O) Bond
8.8C Reactions Involving Alcohol Molecule as a Whole
8.8D Oxidation of Alcohols
8.9 Reactions of Phenols as Acids
8.9A Strength of Phenols as Acids
8.9B Distinguishing and Separating Phenols from Alcohols and Carboxylic Acids
8.9C Reactions of the Benzene Ring of Phenols
8.9D Other Reactions of Phenols
8.10 Reactions of Ethers
8.10A Cleavage of Ethers
8.10B Electrophilic Substitution
8.10C The Claisen Rearrangement
8.11 Distinction Between 1°, 2°, 3° Alcohols
8.11A Lucas Test
8.11B Victor Meyer Test
8.11C Iodoform Test
8.12 Some Commercially Important Alcohols and Ethers
8.12A Methanol
8.12B Ethanol
8.12C Ethylene and Propylene Glycols
8.12D Diethyl Ether
Solved Examples
Solved Previous Years’ NEET Questions
Additional Objective Questions
Answer Key
Hints and Explanations
9 Aldehydes and Ketones
9.1 Nomenclature of Aldehydes and Ketones
9.1A Isomerism in Aldehydes and Ketones
9.2 Structure of the Carbonyl Group
9.2A Comparison between C=O (Carbonyl) and C=C (Alkene) Double Bond
9.3 Physical Properties
9.4 Synthesis of Aldehydes
9.4A Aldehydes by Oxidation of 1° Alcohols
9.4B Aldehydes by Catalytic Dehydrogenation of Alcohols
9.4C Aldehydes by Ozonolysis of Alkenes
9.4D Aldehydes from Alkynes (Hydration and Hydroboration)
9.4E Aldehydes by Reduction of Acyl Chlorides, Esters, and Nitriles
9.4F Aromatic Aldehydes from Hydrocarbons
9.5 Synthesis of Ketones
9.5A Ketones from Alkenes, Arenes, and 2° Alcohols
9.5B Ketones from Nitriles
9.5C Ketones from Acyl Chlorides
9.6 Miscellaneous Methods for Preparation of Aldehydes and Ketones
9.6A Aqueous Alkali Hydrolysis of Geminal Dihalides
9.6B Distillation of Calcium Salts of Acids
9.6C Catalytic Decomposition of Carboxylic Acids
9.7 Nucleophilic Addition to the Carbon–Oxygen Double Bond
9.7A Reversibility of Nucleophilic Additions to the Carbon–Oxygen Double Bond
9.7B Relative Reactivity: Aldehydes versus Ketones
9.7C Addition Products Can Undergo Further Reactions
9.8 The Addition of Alcohols: Hemiacetals and Acetals
9.8A Hemiacetals
9.8B Acetals
9.8C Acetals Are Used as Protecting Groups
9.8D Thioacetals
9.9 The Addition of Hydrogen Cyanide: Cyanohydrins
9.10 The Addition of Sodium Hydrogen Sulfite
9.11 Addition of Grignard Reagents
9.12 The Addition of Primary and Secondary Amines
9.12A Imines
9.12B Oximes and Hydrazones
9.12C Enamines
9.13 The Addition of Ylides: The Wittig Reaction
9.14 Reduction of Aldehydes and Ketones
9.14A Reduction to Alcohols
9.14B Reduction to Hydrocarbons
9.15 Oxidation of Aldehydes and Ketones
9.15A The Baeyer–Villiger Oxidation
9.15B Oxidation Reactions Based Tests for Aldehydes and Ketones
9.16 The Acidity of the α Hydrogens of Carbonyl Compounds: Enolate Anions
9.17 Keto and Enol Tautomers
9.18 Reactions via Enols and Enolates
9.18A Racemization
9.18B Halogenation at the a Carbon
9.18C The Haloform Reaction
9.19 Lithium Enolates
9.20 Aldol Reactions: Addition of Enolates and Enols to Aldehydes and Ketones
9.20A Aldol Addition Reactions
9.20B The Retro-Aldol Reaction
9.20C Aldol Condensation Reactions: Dehydration of the Aldol Addition Product
9.20D Acid-Catalyzed Aldol Condensations
9.20E Synthetic Applications of Aldol Reactions
9.21 Crossed Aldol Condensations
9.21A Crossed Aldol Condensations Using Weak Bases
9.21B Crossed Aldol Condensations Using Strong Bases: Lithium Enolates and Directed Aldol Reactions
9.22 Cyclizations via Aldol Condensations
9.23 Other Important Reactions
9.23A Cannizaro Reaction
9.23B Benzilic Acid Rearrangement
9.23C Benzoin Condensation
9.23D Perkin Reaction
9.24 Summary of Aldehyde and Ketone Reactions
9.24A Aldehyde and Ketone Addition Reactions
9.24B Enolate Chemistry
9.24C Other Important Reactions
Solved Examples
Solved Previous Years’ NEET Questions
Additional Objective Questions
Answer Key
Hints and Explanations
10 Carboxylic Acids and Their Derivatives
10.1 Structure of the Carboxyl Group
10.2 Nomenclature and Physical Properties
10.2A Carboxylic Acids
10.2B Carboxylate Salts
10.2C Dicarboxylic Acids
10.2D Esters
10.2E Carboxylic Anhydrides
10.2F Acyl Chlorides
10.2G Amides
10.2H Nitriles
10.3 Preparation of Carboxylic Acids
10.4 Chemical Properties of Carboxylic Acids
10.4A Reactions Involving Cleavage of O—H Bond
10.4B Reactions Involving Cleavage of C—OH Bond
10.4C Reactions Involving—COOH Group
10.4D Substitution Reactions in the Hydrocarbon Part
10.5 Acyl Substitution: Nucleophilic Addition–Elimination at the Acyl Carbon
10.5A Relative Reactivity of Acyl Compounds
10.5B Synthesis of Acid Derivatives
10.6 Acyl Chlorides
10.6A Synthesis of Acyl Chlorides
10.6B Reactions of Acyl Chlorides
10.7 Carboxylic Acid Anhydrides
10.7A Synthesis of Carboxylic Acid Anhydrides
10.7B Reactions of Carboxylic Acid Anhydrides
10.8 Esters
10.8A Synthesis of Esters: Esterification
10.8B Base-Promoted Hydrolysis of Esters: Saponification
10.8C Lactones
10.9 Amides
10.9A Synthesis of Amides
10.9B Amides from Acyl Chlorides
10.9C Amides from Carboxylic Anhydrides
10.9D Amides from Esters
10.9E Amides from Carboxylic Acids and Ammonium Carboxylates
10.9F Hydrolysis of Amides
10.9G Nitriles from the Dehydration of Amides
10.9H Hydrolysis of Nitriles
10.9I Lactams
10.10 Decarboxylation of Carboxylic Acids
10.11 Summary of the Reactions of Carboxylic Acids and Their Derivatives
Solved Examples
Solved Previous Years’ NEET Questions
Additional Objective Questions
Answer Key
Hints and Explanations
11 Organic Compounds Containing Nitrogen
Part I: Amines
11.1 Nomenclature of Amines
11.1A Aliphatic Amines
11.1B Arylamines
11.1C Isomerism in Amines
11.2 Physical Properties and Structure of Amines
11.2A Physical Properties
11.2B Structure of Amines
11.3 Preparation of Amines
11.3A Through Nucleophilic Substitution Reactions
11.3B Preparation of Aromatic Amines through Reduction of Nitro Compounds
11.3C Preparation of Primary, Secondary, and Tertiary Amines through Reductive Amination
11.3D Preparation of Primary, Secondary, or Tertiary Amines through Reduction of Nitriles, Oximes, and Amides
11.3E Preparation of Primary Amines through the Hofmann and Curtius Rearrangements
11.4 Basicity of Amines: Amine Salts
11.4A Basicity of Arylamines
11.4B Amines versus Amides
11.4C Aminium Salts and Quaternary Ammonium Salts
11.4D Solubility of Amines in Aqueous Acids
11.5 Reactions of Amines
11.5A Oxidation of Amines
11.5B Reaction with Grignard Reagents
11.5C Carbylamine Reaction
11.5D Hofmann’s Mustard Oil Reaction
11.6 Reactions of Amines with Nitrous Acid
11.6A Reactions of Primary Aliphatic Amines with Nitrous Acid
11.6B Reactions of Secondary Amines with Nitrous Acid
11.6C Reactions of Tertiary Amines with Nitrous Acid
11.7 Reactions of Amines with Sulfonyl Chlorides: Test for Amines
11.7A The Hinsberg Test
11.8 Eliminations Involving Ammonium Compounds
11.8A The Hofmann Elimination
11.8B The Cope Elimination
Part II: Arenediazonium Salts
11.9 Preparation of Arenediazonium Salts
11.10 Replacement Reactions of Arenediazonium Salts
11.10A Syntheses Using Diazonium Salts
11.10B The Sandmeyer Reaction: Replacement of the Diazonium Group by —Cl, —Br, or —CN
11.10C Replacement by —I
11.10D Replacement by —F
11.10E Replacement by —OH
11.10F Replacement by Hydrogen: Deamination by Diazotization
11.11 Coupling Reactions of Arenediazonium Salts
Part III: Cyanides and Isocyanides
11.12 Preparation and Properties of Cyanides and Isocyanides
11.12A Methods of Preparation
11.12B Physical Properties
11.12C Chemical Properties
11.13 Summary of Reactions of Organic Compounds Containing Nitrogen
11.13A Preparation of Amines
11.13B Reactions of Amines
11.13C Preparation of Cyanides and Isocyanides
11.13D Reactions of Cyanides and Isocyanides
Solved Examples
Solved Previous Years’ NEET Questions
Additional Objective Questions
Answer Key
Hints and Explanations
12 Biomolecules
Part I: Carbohydrates
12.1 Classification of Carbohydrates
12.2 Monosaccharides
12.2A Classification of Monosaccharides
12.2B d and l Designations of Monosaccharides
12.2C Glucose
12.2D Fructose
12.2E Reactions of Monosaccharides
12.2F Other Reactions of Monosaccharides
12.3 Disaccharides
12.3A Sucrose
12.3B Maltose
12.3C Cellobiose
12.3D Lactose
12.4 Polysaccharides
12.4A Starch
12.4B Glycogen
12.4C Cellulose
12.4D Cellulose Derivatives
12.5 Other Biologically Important Sugars
12.6 Sugars that Contain Nitrogen
12.6A Glycosylamines
12.6B Amino Sugars
Part II: Amino Acids and Proteins
12.7 Amino Acids
12.7A Structures and Names
12.7B Essential Amino Acids
12.7C Amino Acids as Dipolar Ions
12.8 Polypeptides and Proteins
12.8A Primary Structure of Polypeptides and Proteins
12.8B Examples of Polypeptide and Protein Primary Structure
12.9 Secondary, Tertiary and Quaternary Structures of Proteins
12.9A Secondary Structure
12.9B Tertiary Structure
12.9C Quaternary Structure
12.9D Hemoglobin: A Conjugated Protein
12.10 Denaturation of Proteins
12.11 Introduction to Enzymes
12.12 Hormones
12.13 Vitamins
12.13A Classification of Vitamins
Part III: Nucleic Acids
12.14 Nucleotides and Nucleosides
12.15 Deoxyribonucleic Acid: DNA
12.15A Primary Structure
12.15B Secondary Structure
12.15C Replication of DNA
12.16 RNA and Protein Synthesis
Part IV: Lipids
12.17 Fatty Acids and Triacylglycerols
12.17A Hydrogenation of Triacylglycerols
12.17B Biological Functions of Triacylglycerols
12.17C Saponification of Triacylglycerols
12.17D Reactions of the Carboxyl Group of Fatty Acids
12.17E Reactions of the Alkenyl Chain of Unsaturated Fatty Acids
12.18 Terpenes and Terpenoids
12.19 Phospholipids and Cell Membranes
12.19A Phosphatides
12.20 Waxes
Solved Examples
Solved Previous Years’ NEET Questions
Additional Objective Questions
Answer Key
Hints and Explanations
13 Polymers
13.1 Some Terms Related to Polymers
13.2 Classification of Polymers
13.2A Classification Based on Source
13.2B Classification Based on Structure of Polymers
13.2C Classification Based on Mode of Polymerization
13.2D Classification Based on Molecular Forces
13.3 Addition Polymerization or Chain-Growth Polymerization
13.3A Free Radical Mechanism
13.3B Ionic Polymerization
13.4 Preparation of Some Important Addition Polymers
13.5 Condensation Polymerization or Step-Growth Polymerization
13.5A Copolymerization
13.6 Natural Rubber
13.6A Vulcanization of Rubber
13.7 Synthetic Rubbers
13.7A Preparation of Synthetic Rubbers
13.8 Molecular Mass of Polymers
13.9 Biodegradable Polymers
13.10 Polymers of Commercial Importance
Solved Examples
Solved Previous Years’ NEET Questions
Additional Objective Questions
Answer Key
Hints and Explanations
14 Chemistry in Everyday Life
14.1 Drugs and Their Classification
14.1A Classification of Drugs
14.2 Drug–Target Interaction
14.2A Enzymes as Drug Targets
14.2B Receptors as Drug Targets
14.3 Therapeutic Action of Different Classes of Drugs
14.3A Antacids
14.3B Antihistamines
14.3C Neurologically Active Drugs
14.3D Antimicrobials
14.3E Antifertility Drugs
14.4 Chemicals in Food
14.4A Artificial Sweetening Agents
14.4B Food Preservatives
14.5 Cleansing Agents
14.5A Soaps
14.5B Synthetic Detergents
Solved Examples
Solved Previous Years’ NEET Questions
Additional Objective Questions
Answer Key
Hints and Explanations
NEET–2020 CHEMISTRY PAPER
Index
Back Cover

Citation preview

HIGHLIGHTS OF THE BOOK Content reorganized as per NEET syllabus – additional chapters for complete coverage.

l

A detailed, relevant, lucid and up to date coverage of the concepts as per the NEET curriculum enriched with concept map, summary and review tools.

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Solved Examples to learn about application of concepts to problem solving.

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Chapter-wise Solved Questions from NEET previous years’ papers.

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NCERT Exemplar Problems included at the end of each chapter.

l

Additional Objective Questions for practice as per medical entrance exam requirement. v Single choice correct type ü

Exercise 1: Easy to Moderate

ü

Exercise 2: Moderate to Hard

ABOUT THE BOOK Organic Chemistry for NEET and other Medical Entrance Examinations is an adaptation of the classic book Solomons' Organic Chemistry, which is widely used by the students all over the world. This adapted version provides more concise content that is structured as per the syllabus requirement. It is a definitive text offering for students preparing for NEET-UG, which replaced the All India Pre Medical test (AIPMT) for admission to MBBS and dental courses across the country. The book derives advantage from the time-tested content of the original book and offers assessment as per the entrance examination requirement. It aims to enhance the learning experience and focus on the problem solving skills. The premise is to encourage self-study in students, help develop understanding of the concepts and build an aptitude to apply these to problem solving. About Maestro Series l

Idea: World-class content developed by “Master teachers” adapted to the needs of medical aspirants.

l

Process: Original books customized for preparation of medical entrance after in-depth study of syllabus and relative weightage of topics.

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Adapter: Adaptations carried out by teachers' expert in coaching students for preparation of entrance exam.

l

Two-fold advantage:

v Assertion-Reason Type l

Answer Key provided for all objective questions.

l

Hints and Explanations to selected questions available at the end of each chapter.

l

Includes NEET 2020 Chemistry solved paper.

v Conceptual strength provided by authoritative yet precise content as per syllabus requirement. v Assessment as per medical entrance through thought provoking end-of chapter exercises as per the pattern of NEET (previously AIPMT) and AIIMS question papers.

GET FREE RESOURCES Scan the given QR code or Visit www.wileyindia.com/testprep-resources

Wiley India Pvt. Ltd. Customer Care +91 120 6291100 [email protected] www.wileyindia.com www.wiley.com

ISBN 978-93-5424-878-8

2022 Solomons Fryhle Snyder

ORGANIC CHEMISTRY for NEET

l

CHOUHAN

9 789354 248788

Solomons|Fryhle|Snyder

ORGANIC CHEMISTRY

NEET

for

and other MEDICAL ENTRANCE EXAMINATIONS

M S CHOUHAN

Solomons|Fryhle|Snyder

ORGANIC CHEMISTRY

NEET

for

M S CHOUHAN

and other MEDICAL ENTRANCE EXAMINATIONS

Solomons|Fryhle|Snyder

ORGANIC CHEMISTRY for NEET and other Medical Entrance Examinations Author: M S Chouhan Published by Wiley India Pvt. Ltd., 4436/7, Ansari Road, Daryaganj, New Delhi-110002. Printed at: Yash Printographics First Edition: 2018 Edition: 2021 ISBN: 978-93-5424-878-8 ISBN: 978-93-5424-889-4 (ebk) Copyright © 2021 by Wiley India Pvt. Ltd. All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or scanning without the written permission of the publisher. This book is an adaptation of Organic Chemistry, 11/e by T. W. Graham Solomons, Craig B. Fryhle and Scott A. Snyder (ISBN: 978-81-265-5684-7). All rights remain with their respective holders. Limits of Liability: While the publisher and the author have used their best efforts in preparing this book, Wiley and the author make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and specifically disclaim any implied warranties of merchantability or fitness for any particular purpose. There are no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or extended by sales representatives or written sales materials. Disclaimer: The contents of this book have been checked for accuracy. Since deviations cannot be precluded entirely, Wiley or its author cannot guarantee full agreement. As the book is intended for educational purpose, Wiley or its author shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in the book. This publication is designed to provide accurate and authoritative information with regard to the subject matter covered. It is sold on the understanding that the Publisher is not engaged in rendering professional services. Trademarks: All brand names and product names used in this book are trademarks, registered trademarks or trade names of their respective holders. Wiley is not associated with any product or vendor mentioned in this book. Other Wiley Editorial Offices: John Wiley & Sons, Inc. 111 River Street, Hoboken, NJ 07030, USA Wiley-VCH Verlag GmbH, Pappellaee 3, D-69469 Weinheim, Germany John Wiley & Sons Australia Ltd, 42 McDougall Street, Milton, Queensland 4064, Australia John Wiley & Sons (Asia) Pte Ltd, 1 Fusionpolis Walk #07-01 Solaris, South Tower Singapore 138628 John Wiley & Sons Canada Ltd, 22 Worcester Road, Etobicoke, Ontario, Canada, M9W ILI www.wileyindia.com

Preface to the Original Edition “It’s Organic Chemistry!” That’s what we want students to exclaim after they become acquainted with our subject. Our lives revolve around organic chemistry, whether we all realize it or not. When we understand organic chemistry, we see how life itself would be impossible without it, how the quality of our lives depends upon it, and how examples of organic chemistry leap out at us from every direction. That’s why we can envision students enthusiastically exclaiming “It’s organic chemistry!” when, perhaps, they explain to a friend or family member how one central theme—organic chemistry—pervades our existence. We want to help students experience the excitement of seeing the world through an organic lens, and how the unifying and simplifying nature of organic chemistry helps make many things in nature comprehensible. Our book makes it possible for students to learn organic chemistry well and to see the marvelous ways that organic chemistry touches our lives on a daily basis. Our book helps students develop their skills in critical thinking, problem solving, and analysis—skills that are so important in today’s world, no matter what career paths they choose. The richness of organic chemistry lends itself to solutions for our time, from the fields of health care, to energy, sustainability, and the environment. After all, it’s organic chemistry! Guided by these goals, and by wanting to make our book even more accessible to students than it has ever been before, we have brought many changes to this edition.

New To This Edition With this edition we bring Scott Snyder on board as a co-author. We’re very excited to have Scott join our team. Scott brings a rich resource of new perspectives to the book, particularly in the arena of complex molecule synthesis. Scott has infused new examples and applications of exciting chemistry that help achieve our goals. In addition to adding his perspectives to the presentation of core chemistry throughout the book, Scott’s work is manifest in most of this edition’s chapter openers and in all of the chapter closers, couched in a new feature called “Why do these topics matter?”. “Why do these topics matter?”  is a new feature that bookends each chapter with a teaser in the opener and a captivating example of organic chemistry in the closer. The chapter opener seeks to whet the student’s appetite both for the core chemistry in that chapter as well as a prize that comes at the end of the chapter in the form of a “Why do these topics matter?” vignette. These new closers consist of fascinating nuggets of organic chemistry that stem from research relating to medical, environmental, and other aspects of organic chemistry in the world around

iv

Preface to the Original Edition

us, as well as the history of the science. They show the rich relevance of what students have learned to applications that have direct bearing on our lives and wellbeing. For example, in Chapter 6, the opener talks about the some of the benefits and drawbacks of making substitutions in a recipe, and then compares such changes to the nucleophilic displacement reactions that similarly allow chemists to change molecules and their properties. The closer then shows how exactly such reactivity has enabled scientists to convert simple table sugar into the artificial sweetener Splenda which is 600 times as sweet, but has no calories! Laying the foundation earlier  Certain tools are absolutely key to success in organic chemistry. Among them is the ability to draw structural formulas quickly and correctly. In this edition, we help students learn these skills even sooner than ever before by moving coverage of structural formulas and the use curved arrows earlier in the text (Section 3.5). We have woven together instruction about Lewis structures, covalent bonds, and dash structural formulas, so that students build their skills in these areas as a coherent unit, using organic examples that include alkanes, alkenes, alkynes, and alkyl halides. One could say that it’s a “use organic to teach organic” approach. Getting to the heart of the matter quicker Acid-base chemistry, and electrophiles and nucleophiles are at the heart of organic chemistry. Students cannot master the subject if they do not have a firm and early grasp of these topics. In this edition, we cut to the chase with these topics earlier in Chapter 3 than ever before, providing a streamlined and highly efficient route to student mastery of these critical concepts. Improving a core area: substitution reactions  All organic instructors know how important it is for their students to have a solid understanding of substitution reactions. This is one reason our text has proven its lasting value. In this edition we have even further enhanced the presentation of substitution reactions in several ways, including a revised introduction of SN1 reactions (Section 6.10) through the classic hydrolysis experiments of Hughes, and a newly organized presentation of solvent effects on the rate of substitution reactions. Striking a strong balance of synthetic methods Students need to learn methods of organic synthesis that are useful, as environmentally friendly as possible, and that are placed in the best overall contextual framework. In this edition we incorporate the Swern oxidation (Section 11.4B), long held as a useful ­oxidation method and one that provides a less toxic alternative to chromate o ­ xidations in some cases. We also restore coverage of the Wolff-Kishner reduction (Section 15.8C) and the Baeyer-Villiger oxidation (Section 15.12), two methods whose importance has been proven by the test of time. The chemistry of radical ­reactions has also been refocused and streamlined by reducing thermochemistry content and by centralizing the coverage of allylic and benzylic radical substitutions (including NBS reactions) in one chapter (Sections 9.7 and 9.8), instead of distributing it between two, as before. The addition of sulfuric acid to alkenes and the Kolbe reaction have been deleted from the text, since these have little practical use in the laboratory. Maintaining an eye for clarity  With every edition we improve the presentation of topics, reactions, and diagrams where the opportunity arises. This edition includes improved mechanism depictions for aromatic ­sulfonation and thionyl chloride substitution.

Preface to the Original Edition

Showing how things work A mechanistic understanding of organic chemistry is key to student success in organic chemistry. Mechanisms have always been central to the book, and in this edition the authors have added a mechanistic framework for the Swern and chromate alcohol oxidations (Section 11.4) by presenting elimination of the carbinol hydrogen and a leaving group from oxygen as the common theme.

Traditional Pedagogical Strengths Solved Problems  Knowing “where to begin” to solve organic chemistry problems is one of the greatest challenges faced by today’s students. By modeling problem solving strategies, students begin to understand the patterns inherent in organic chemistry and learn to apply that knowledge to new situations. In this edition we have added even more Solved Problems. They are usually paired with a related Practice Problem. Practice Problems  Students need ample opportunities to practice and apply their new found strategies for solving organic chemistry problems. We’ve added to our rich array of in-text Practice Problems to provide students with even more opportunities to check their progress as they study. If they can work the practice problem, they should move on. If not, they should review the ­preceding presentation. Solved Problem 11.2 Which reagents would you use to accomplish the following transformations? O OH

(a)

OH

(b)

O OH

H

(c) (d)

Answer (a) To oxidize a primary alcohol to a carboxylic acid, use (1) potassium permanganate in aqueous base, followed by (2) H3O+, or use chromic acid (H2CrO4). (b) To reduce a carboxylic acid to a primary alcohol, use LiAlH4. (c) To oxidize a primary alcohol to an aldehyde, use the Swern oxidation or pyridinium chlorochromate (PCC). (d) To reduce an aldehyde to a primary alcohol, use NaBH4 (preferably) or LiAlH4.

Practice Problem 11.3 Show how each of the following transformations could be accomplished: O (a)

OH

?

H

(b)

OH

?

O OH

End-of-Chapter Problems As athletes and musicians know, practice makes perfect. The same is true with organic chemistry. The End of Chapter problems, categorized by topic, provide essential practice for students and help them build mastery of both concepts and skills presented throughout the chapter. Many of the End of Chapter problems are cast in a visual format using structures, equations, and schemes. A Mechanism for the Reaction Understanding mechanisms and the ability to recognize patterns among them is a key component in determining student success in organic chemistry. We provide A Mechanism for the Reaction boxes that show step-by-step details about how reactions take place so that students have the tools to understand rather than memorize organic reactions.

SOLVED PROBLEMS model problem solving strategies.

PRACTICE PROBLEMS provides opportunities to check progress.

v

vi

Preface to the Original Edition End-of-chapter problems are grouped and labeled by topic. ­Students and instructors can more easily select problems for specific ­purposes.

7.28 Give the IUPAC names for each of the following:

CH3 D H

(d)

(a)

7.34 Outline a synthesis of phenylethyne from each of the following:

Br

(b)

(e)

Cl

Br

(b)

Br O

(f)

(c)

Br

(a)

H Cl

(c)

(d)

7.29 Without consulting tables, arrange the following compounds in order of decreasing acidity: Pentane

1-Pentene

1-Pentyne

1-Pentanol

Synthesis 7.30 Outline a synthesis of propene from each of the following: (a) Propyl chloride (b) Isopropyl chloride (c) Propyl alcohol (d) Isopropyl alcohol

A MECHANISM FOR THE REACTION Stepped out ­reactions with just the right amount of detail provides the tools for students to understand rather than memorize r­ eaction mechanisms.

Dehydrohalogenation and Dehydration 7.35 Write a three-dimensional representation for the transition state structure leading to formation of 2-methyl2-butene from reaction of 2-bromo-2-methylbutane with sodium ethoxide. 7.36 When trans-2-methylcyclohexanol (see the following reaction) is subjected to acid-catalyzed dehydration, the major product is 1-methylcyclohexene:

A MECHANISM FOR THE REACTION

Hydrogen Atom Abstraction

General Reaction X

+

H R

Reactive Alkane radical intermediate

X H

+ R Alkyl radical intermediate (reacts further)

Specific Example Cl

+

H CH3

Chlorine Methane atom (a radical)

Cl H + CH3 Methyl radical intermediate (reacts further)

Key Ideas as Bullet Points  The amount of content covered in organic chemistry can be overwhelming to students. To help students focus on the most essential topics, key ideas are emphasized as bullet points in every section. In preparing bullet points, we have distilled appropriate concepts into simple declarative statements that convey core ideas accurately and clearly. No topic is ever presented as a bullet point if its integrity would be diminished by oversimplification, however. “How to” Sections Students need to master important skills to support their conceptual ­learning. “How to” Sections throughout the text give step-by-step instructions to guide students in performing important tasks, such as using curved arrows, drawing chair conformations, p ­ lanning a Grignard synthesis, determining formal charges and writing Lewis structures. The Chemistry of . . . . Virtually every instructor has the goal of ­showing students how organic chemistry relates to their field of study and to their everyday life experience. The authors assist their colleagues in this goal by providing boxes titled “The Chemistry of . . .” that provide interesting and targeted examples that engage the student with chapter content. Summary and Review Tools  At the end of each chapter, Summary and Review Tools provide visually oriented roadmaps and frameworks that students can use to help organize and assimilate concepts as they study and review chapter content. Intended to accommodate diverse learning styles, these include Synthetic Connections, Concept Maps, Thematic

Preface to the Original Edition

[S U M M A R Y

A N Dand Review R E Tools V I E W Summary

T O O L S

SUMMARY AND REVIEW TOOLS Visually oriented study tools accommodate diverse learning styles.

]

Synthetic Connections of Alkynes and Alkenes: II

R1

H

X

C

C

H

X

R2

H

HX

R2

(2nd molar equivalent)

C

R1 C H

R2 O

O

C H

C

X R1 (If R1 in alkyne is H then addition is Markovnikov)

C

C

(2) Me2S

(1) Na, EtNH2 (2) NH4Cl

H

H *(1) KMnO4, HO , heat

R1 C

C (Z)

H2, catalyst (Pt, Pd, Rh, Ni)

CH2I2, Zn(Cu) (or other methods)

R2

H

R1

H

H

H

C

C

H

H

C (E)

H

H

R1

X

X

C

C

X

X

R2

(1) KMnO4, HO , heat (2) H3O or (1) O3 (2) HOAc

R2 (1) X2 (2) NaNH2 (2 equiv.), heat

H2, Lindlar’s or P-2 catalyst

R2

C

X

R1

(1st molar equivalent) R1

R1

(2nd molar equivalent)

X2

HX

(1) X2 (2) NaNH2 (2 equiv.), heat

*(1) O3

X2

C

R1

(1st molar equivalent)

• Hydrohalogenation • Halogenation • Hydrogenation • Carbene Addition • Ozonolysis • Potassium Permanganate Cleavage • Alkyne Synthesis by Double E2

R2

X

C

R2

OH

R1 C

(2) H3O

O

HO

O

C R2

CH2I2, Zn(Cu) (or other methods)

R2

R1

H

H

R2

*If R1 is H, then alkene ozonolysis generates formaldehyde and KMnO4 generates CO2. If the alkene is disubstituted at one end, a ketone results from cleavage by either ozonolysis or KMnO4.

[C O N C E P T

M A P

]

Conformers (Section 5.8) are Molecules that differ only by rotation about sigma (s s ) bonds can have Ring strain (Section 5.10)

Different potential energies of conformers can be represented by

among conformers is a function of Torsional strain (Section 5.8)

are a plot of

is caused by

Dihedral angle vs. potential energy

Angle strain (Section 5.10)

and loss of

result in 60

120 240 300 360 Degrees of Rotation u

Deviation from ideal bond angles

involves

Steric hindrance (Section 5.8)

420

is caused by

Hyperconjugative stabilization (Section 5.8)

Repulsive dispersion forces

E

0

is caused by

and

Conformer potential energy diagrams (Section 5.9)

Favorable overlap of occupied with unoccupied orbitals

can be represented by Newman projection formulas (Section 5.8)

of cyclohexane can be represented by

Chair conformational structures (Section 5.11)

or

Boat

have

can be used to show

Axial positions Eclipsed conformations

and

Staggered conformations

and twist-boat

with substituted groups (G) can be and G

G

Anti

or G

Gauche

G

Equatorial positions conformational structures

Mechanism Review Summaries, and the detailed Mechanism for the Reaction boxes already mentioned. We also provide Helpful Hints and richly annotated illustrations throughout the text.

Coverage Throughout the book, we have streamlined or reduced content to match the modern practice of organic chemistry, and we have provided new

vii

viii

Preface to the Original Edition

coverage of current reactions, while maintaining our commitment to an appropriate level and breadth of coverage. Chapters on carbonyl chemistry that are organized to emphasize mechanistic themes of nucleophilic addition, acyl substitution, and reactivity at the a-carbon.

●

Organization—An Emphasis on the Fundamentals So much of organic chemistry makes sense and can be generalized if students master and apply a few fundamental concepts. Therein lays the beauty of organic chemistry. If students learn the essential principles, they will see that memorization is not needed to succeed. Most important is for students to have a solid understanding of structure—of hybridization and geometry, steric hindrance, electronegativity, polarity, formal charges, and resonance—so that they can make intuitive sense of mechanisms. It is with these topics that we begin in Chapter 1. In Chapter 2 we introduce the families of functional groups— so that students have a platform on which to apply these concepts. We begin our study of mechanisms in the context of acid-base chemistry in Chapter 3. Acid-base reactions are fundamental to organic reactions, and they lend themselves to introducing several important topics that students need early in the course: (1) curved arrow notation for illustrating mechanisms, (2) the relationship between free-energy changes and equilibrium constants, and (3) the importance of inductive and resonance effects and of solvent effects. In Chapter 3 we present the first of many “A Mechanism for the Reaction” boxes, using an example that embodies both Brønsted-Lowry and Lewis acid-base principles. All throughout the book, we use boxes like these to show the details of key reaction mechanisms. All of the Mechanism for the Reaction boxes are listed in the Table of Contents so that students can easily refer to them when desired. A central theme of our approach is to emphasize the relationship between structure and reactivity. This is why we choose an organization that combines the most useful features of a functional group approach with one based on reaction mechanisms. Our philosophy is to emphasize mechanisms and fundamental principles, while giving students the anchor points of functional groups to apply their mechanistic ­knowledge and intuition. The structural aspects of our approach show students what organic chemistry is. Mechanistic aspects of our approach show students how it works. And wherever an opportunity arises, we show them what it does in living systems and the physical world around us. In summary, our writing reflects the commitment we have as teachers to do the best we can to help students learn organic chemistry and to see how they can apply their knowledge to improve our world. The enduring features of our book have proven over the years to help ­students learn organic chemistry.

Preface to the Adapted Version This book is dedicated to the lotus feet of Revered Guruji SHRI KESARAM JI MAHARAJ It is a matter of great pleasure for me to present this book before medical aspirants. This book was written because I felt the need for a textbook that corresponds closely to the course that I teach. Although the book is organized along tried-and- true functional-group lines, it contains some unique features that have served me well in both my teaching and ­understanding of Organic Chemistry. My four major concerns in both the initial writing and the revision of this text were readability, presentation, organization, and accuracy. In the current scenario of stiff competition especially for NEET, one must be clear that almost all the sincere applicants are well equipped with the facts of subject; yet, the winner is the one who knows how to use these facts with accuracy and efficiency. As an experienced teacher, I would like to suggest students three golden rules to score high in Organic Chemistry: 1. Do not lag behind in schedule. 2. Work out a number of problems of different types. 3. Revise through short notes. I hope that the present book will cater to the needs of NEET aspirants. As a matter of fact, they will enjoy the present venture and I would feel rewarded if this book is helpful for the students and teachers in real terms. All efforts have been made to make the book error-free; however, a few misprints may inadvertently creep. I acknowledge the blessing and support of my mother Smt. Raj Kanwar, father Shri B.S. Chouhan, brother Dr. V. S. Chouhan, wife Ms. Meena Chouhan and my children. They inspired me all the time during the preparation of this book. The support and valuable suggestions from Mr Jitendra Chandwani (JC Sir) are highly acknowledged. I also pay my sincere thanks to all the esteemed members of Wiley India and specially Ms. Anjali Chadha, Ms. Seema Sajwan and Mr. Rakesh Poddar in bringing out this book in such a nice form. In the end, constructive criticism and valuable suggestions from the readers are most welcome to make the book more useful. M.S.Chouhan (MSC Sir) Email: [email protected] Mobile: 9828025625

About the Authors T. W. Graham Solomons  did his undergraduate work at The Citadel and received his doctorate in organic chemistry in 1959 from Duke University where he worked with C. K. Bradsher. Following this he was a Sloan Foundation Postdoctoral Fellow at the University of Rochester where he worked with V. Boekelheide. In 1960 he became a charter member of the faculty of the University of South Florida and became Professor of Chemistry in 1973. In 1992 he was made Professor Emeritus. In 1994 he was a visiting professor with the Faculté des Sciences Pharmaceutiques et Biologiques, Université René Descartes (Paris V). He is a member of Sigma Xi, Phi Lambda Upsilon, and Sigma Pi Sigma. He has received research grants from the Research Corporation and the American Chemical Society Petroleum Research Fund. For several years he was director of an NSF-sponsored Undergraduate Research Participation Program at USF. His research interests have been in the areas of heterocyclic chemistry and unusual aromatic compounds. He has published papers in the Journal of the American Chemical Society, the Journal of Organic Chemistry, and the Journal of Heterocyclic Chemistry. He has received several awards for distinguished teaching. His organic chemistry textbooks have been widely used for 30 years and have been translated into French, Japanese, Chinese, Korean, Malaysian, Arabic, Portuguese, Spanish, Turkish, and Italian. He and his wife Judith have a daughter who is a building conservator and a son who is a research biochemist. Craig Barton Fryhle  is Chair and Professor of Chemistry at Pacific Lutheran University. He earned his B.A. degree from Gettysburg College and Ph.D. from Brown University. His experiences at these institutions shaped his dedication to mentoring undergraduate students in chemistry and the liberal arts, which is a passion that burns strongly for him. His research interests have been in areas relating to the shikimic acid pathway, including molecular modeling and NMR spectrometry of substrates and analogues, as well as structure and reactivity studies of shikimate pathway enzymes using isotopic labeling and mass spectrometry. He has mentored many students in undergraduate research, a number of who have later earned their Ph.D. degrees and gone on to academic or industrial positions. He has participated in workshops on fostering undergraduate participation in research, and has been an invited participant in efforts by the National Science Foundation to enhance undergraduate research in chemistry. He has received research and instrumentation grants from the National Science Foundation, the M J. Murdock Charitable Trust, and other private foundations. His work in chemical education, in addition to textbook coauthorship, involves incorporation of student-led teaching in the classroom and technology-based strategies in organic chemistry. He has also developed e­ xperiments for undergraduate students in organic laboratory and instrumental analysis courses. He has been a volunteer with the hands-on science program in Seattle public schools, and Chair of the Puget Sound Section of the  American Chemical Society. His passion for climbing has led to

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ascents of high peaks in several parts of the world. He resides in Seattle with his wife, where both enjoy following the lives of their two daughters as they unfold in new places. Scott A. Snyder  is Associate Professor of Chemistry at Columbia University. He grew up in the suburbs of Buffalo NY and was an undergraduate at Williams College, where he graduated summa cum laude in 1999, before pursuing his doctoral studies at The Scripps Research Institute under the tutelege of K. C. Nicolaou as an NSF, Pfizer, and Bristol-Myers-Squibb predoctoral fellow. While there, he ­co-authored the ­graduate textbook Classics in Total Synthesis II with his doctoral mentor. Scott was then an NIH postdoctoral fellow in the laboratories of E. J. Corey at Harvard University before assuming his current position in 2006. His research interests lie in the arena of natural products total synthesis, especially in the realm of unique polyphenols and halogenated materials, and to date he has trained more than 60 students at the high school, undergraduate, graduate, and postdoctoral levels and co-authored more than 40 research and review articles. Scott has received a number of awards and honors, including a Camille and Henry Dreyfus New Faculty Award, Amgen New Faculty and Young Investigator Awards, Eli Lilly New Faculty and Grantee Awards, a Bristol-Myers Squibb Unrestricted Grant Award, an NSF CAREER Award, an Alfred P. Sloan Foundation Fellowship, a DuPont Young Professor Award, and an Arthur C. Cope Scholar Award from the American Chemical Society. He has also received recognition for his teaching through a Cottrell Scholar Award from the Research Corporation for Science Advancement and a Columbia Presidential Teaching Award. He is a member of the international advisory board for The Chemical Record and the editorial board of Chirality. He lives north of New York City with his wife Cathy where he enjoys gardening, cooking, and watching movies.

About the Adapting Author Mahendra Singh Chouhan (MSC Sir) is a renowned name in the realm of Organic Chemistry. Though a Chemical engineer from Mumbai University, he has great passion for the subject that led him to impart guidance to medical and engineering aspirants on a regular basis. His in-depth knowledge and vast teaching experience has helped innumerable students to achieve their dream of excelling in those areas. He has launched a website to extend the benefits of his expertise beyond the geographical barriers to all those who dare to dream and seek—www.winvall.com and www.shivwin.com. The website provides expert guidance in all areas of the subject in a most skillful manner. There are quizzes, challenging questions, notes, e-books, videos, etc. This website is a complete guide in itself for organic chemistry and has been designed keeping in mind CBSE syllabus and various other syllabi. Highly recommended for high flyers!

Contents Preface to the Original Edition Preface to the Adapted Version About the Authors About the Adapting Author 1  Basic Principles of Organic Chemistry

iii ix xi xiii 1

Part I:  Bonding and Molecular Structure 1 1.1 Life and the Chemistry of Carbon Compounds 2 1.1A Development of the Science of Organic Chemistry2 1.2 Chemical Bonds: The Octet Rule 3 1.2A Ionic Bonds 3 1.2B Covalent Bonds and Lewis Structures 4 1.2C How to Write Lewis Structures 5 1.3 Isomers: Different Compounds that Have the Same Molecular Formula 6 1.4 How to Write and Interpret ­Structural Formulas 7 1.4A More About Dash Structural Formulas 7 1.4B Condensed Structural Formulas 8 1.4C Bond-Line Formulas 8 1.4D Three-Dimensional Formulas 9 1.5 Resonance Theory 11 1.5A The Use of Curved Arrows: How to Write Resonance Structures 13 1.5B Rules for Writing Resonance Structures 13 1.5C How Do We Decide When One Resonance Structure Contributes More to the Hybrid Than Another? 14 1.6 The Structure of Methane and Ethane: sp3 Hybridization 14 1.6A The Structure of Methane 15 1.6B The Structure of Ethane 17 1.7 The Structure of Ethene (Ethylene): sp2 ­Hybridization 18 1.7A Restricted Rotation and the Double Bond 20 1.7B Cis–Trans Isomerism 20 1.8 The Structure of Ethyne (Acetylene): sp Hybridization 21 1.8A Bond Lengths of Ethyne, Ethene, and Ethane 23 1.8B A Summary of Important Concepts 24 1.9 How to Predict Molecular Geometry: The Valence Shell Electron Pair Repulsion Model 24 1.9A Methane 24 1.9B Ammonia 25 1.9C Water 25 1.9D Carbon Dioxide 25 Part II:  Families of Carbon Compounds 1.10 Classification of Organic Compounds 1.10A  Acyclic or Open Chain Compounds 1.10B  Alicyclic or Closed-Chain or Ring Compounds

26 26 27 27

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1.10C  Aromatic Compounds 1.10D  Homologous Series 1.11 Hydrocarbons: Representative Alkanes, Alkenes, Alkynes, and Aromatic Compounds 1.11A Alkanes 1.11B Alkenes 1.11C Alkynes 1.11D  Benzene: A Representative Aromatic Hydrocarbon 1.12 Functional Groups 1.12A  Alkyl Groups and the Symbol R 1.12B  Phenyl and Benzyl Groups 1.13 Alkyl Halides or Haloalkanes 1.14 Alcohols and Phenols 1.15 Ethers 1.16 Amines 1.17 Aldehydes and Ketones 1.18 Carboxylic Acids, Esters, and Amides 1.18A  Carboxylic Acids 1.18B Esters 1.18C Amides 1.19 Nitriles 1.20 Summary of Important Families of Organic Compounds 1.20A  Functional Groups in Biologically Important Compounds 1.21 Nomenclature of Organic Compounds 1.21A  The IUPAC System of Nomenclature 1.21B  IUPAC Nomenclature of Alkanes 1.21C  Nomenclature of Organic Compounds having Functional Group(s) 1.21D  Nomenclature of Substituted Benzene Compounds 1.22 Polar Covalent Bonds 1.22A  Polar and Nonpolar Molecules 1.22B  Dipole Moments in Alkenes 1.23 Physical Properties and Molecular Structure 1.23A  Ionic Compounds: Ion–Ion Forces 1.23B  Intermolecular Forces (van der Waals Forces) 1.23C  Boiling Points 1.23D Solubilities

27 28 28 29 29 29 30 31 31 32 32 33 35 35 36 37 37 38 38 39 39 39 40 41 42 45 45 47 48 49 50 50 50 52 53

Part III:  An Introduction to Organic Reactions and their Mechanisms 54 1.24 Acid–Base Reactions 55 1.24A  Brønsted–Lowry Acids and Bases 55 1.24B  Lewis Acids and Bases 56 1.25 How to Use Curved Arrows in ­Illustrating Reactions 56 1.26 Heterolysis of Bonds to Carbon: Carbocations and Carbanions 57 1.26A  Electrophiles and Nucleophiles 59 1.27 Homolytic Fission of Bonds 60 1.28 Electron Displacement Effects in Organic Compounds 61 1.28A  Inductive Effect 61 1.28B  Resonance Structure and Effect 62 1.28C  Electromeric Effect (E Effect) 66 1.28D Hyperconjugation 66 1.29 The Strength of Brønsted–Lowry Acids and Bases: Ka and pKa68 1.29A  Predicting the Strength of Bases 69 1.29B  How to Predict the Outcome of Acid–Base Reactions 70 1.30 Relationships between Structure and Acidity 71

Contents





1.30A  The Effect of Hybridization 1.30B  Inductive Effects 1.31 Acidity: Carboxylic Acids versus Alcohols 1.31A  The Effect of Delocalization 1.31B  The Inductive Effect 1.31C  Summary and a Comparison of Conjugate Acid–Base Strengths 1.31D  Inductive Effects of Other Groups 1.32 The Effect of the Solvent on Acidity 1.33 A Mechanism for an Organic Reaction Solved Examples Solved Previous Years’ NEET Questions Additional Objective Questions Answer Key Hints and Explanations

2  Isomerism







72 73 73 74 75 75 75 76 76 78 85 88 95 96

103

2.1 Types of Isomerism 103 2.1A Structural (Constitutional) Isomerism 103 2.1B Stereoisomerism 106 2.2 Geometrical Isomerism 107 2.2A E- and Z-Nomenclature 109 2.2B Geometrical Isomerism in Cyclic Structures 110 2.3 Chirality and Stereochemistry 110 2.3A The Biological Significance of Chirality 111 2.4 Enantiomers and Chiral Molecules 112 2.5 Molecules having One Chirality Center are Chiral 114 2.5A Tetrahedral versus Trigonal Stereogenic Centers 115 2.6 How to Test for Chirality: Planes of Symmetry 115 2.6A Plane of Symmetry 116 2.6B Center of Symmetry 117 2.6C Axis of Symmetry 118 2.6D Alternating or Improper Axis of Symmetry (Sn or σ) 119 2.7 Naming Enantiomers: The R,S-System119 2.7A How to Assign (R) and (S) Configurations 120 2.8 Properties of Enantiomers: Optical Activity 122 2.8A Plane-Polarized Light 123 2.8B The Polarimeter 124 2.8C Specific Rotation 124 2.8D Racemic Forms 126 2.9 The Synthesis of Chiral Molecules 127 2.9A Racemic Mixture 127 2.9B Stereoselective Syntheses 128 2.10 Molecules with More than One Chirality Center 128 2.10A How to Draw Stereoisomers for Molecules having More than One Chirality Center 129 2.10B  Meso Compounds 130 2.10C How to Name Compounds with More than One Chirality Center 131 2.11 Fischer Projection Formulas 132 2.11A  How to Draw and Use Fischer Projections 132 2.12 Stereoisomerism of Cyclic Compounds 133 2.12A  Cyclohexane Derivatives 133 2.13 Relating Configurations through Reactions in Which No Bonds to the Chirality Center are Broken 135

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2.14 Chiral Molecules that do not Possess a Chirality Center 2.15 Conformational Isomerism 2.15A Conformations 2.15B  Sawhorse Projections 2.15C  Newman Projections Solved Examples Solved Previous Years’ NEET Questions Additional Objective Questions Answer Key Hints and Explanations

3  Purification and Characterization of Organic Compounds







3.1 Methods of Purification of Organic Compounds 3.1A Sublimation 3.1B Crystallization 3.1C Distillation 3.1D Differential Extraction 3.1E Chromatography 3.2 Qualitative Analysis of Organic Compounds 3.2A Preliminary Tests 3.2B Detection of Carbon and Hydrogen 3.2C Detection of Other Elements 3.3 Quantitative Analysis 3.3A Carbon and Hydrogen 3.3B Nitrogen 3.3C Halogens 3.3D Sulphur 3.3E Phosphorus 3.3F Oxygen 3.4 Determination of Empirical Formula of the Organic Compound Solved Examples Solved Previous Years’ NEET Questions Additional Objective Questions Answer Key Hints and Explanations

4  Alkanes and Cycloalkanes

4.1 Introduction to Alkanes and Cycloalkanes 4.1A Sources of Alkanes: Petroleum 4.2 Shapes of Alkanes 4.3 How to Name Alkanes and Alkyl Groups: The IUPAC System 4.3A How to Name Unbranched Alkyl Groups 4.3B How to Name Branched-Chain Alkanes 4.3C How to Name Branched Alkyl Groups 4.3D How to Classify Hydrogen Atoms 4.4 How to Name Cycloalkanes 4.4A How to Name Monocyclic Cycloalkanes 4.4B How to Name Bicyclic Cycloalkanes 4.5 Physical Properties of Alkanes and Cycloalkanes 4.6 Isomerism 4.7 Synthesis of Alkanes and Cycloalkanes 4.7A Hydrogenation of Alkenes and Alkynes

136 136 136 137 137 138 141 144 149 150

155 155 155 156 156 159 160 163 163 164 164 166 167 167 169 170 170 171 171 172 175 175 177 178

179 179 179 181 182 183 183 184 186 186 186 187 188 190 190 190

Contents







4.7B From Alkyl Halides 4.7C From Carboxylic Acids 4.8 How to Gain Structural Information from Molecular Formulas and the Index of Hydrogen Deficiency 4.8A Compounds Containing Halogens, Oxygen, or Nitrogen 4.9 Chemical Reactions of Alkanes 4.9A Reactions of Alkanes with Halogens 4.9B Chlorination of Methane: Mechanism of Reaction 4.9C Halogenation of Higher Alkanes 4.9D Other Substitution Reactions 4.9E Combustion of Alkanes 4.9F Controlled (Catalytic) Oxidation 4.9G Isomerization 4.9H Aromatization 4.9I Pyrolysis 4.10 Conformations 4.10A  Newman Projections and How to Draw Them 4.10B  Conformational Analysis of Ethane 4.10C  Conformational Analysis of Butane 4.10D  Stereoisomers and Conformational Stereoisomers 4.11 The Relative Stabilities of Cycloalkanes: Ring Strain 4.11A Cyclopropane 4.12 Conformations of Cyclohexane: The Chair and the Boat 4.12A  Conformations of Higher Cycloalkanes 4.13 Disubstituted Cycloalkanes: Cis–Trans Isomerism Solved Examples Solved Previous Years’ NEET Questions Additional Objective Questions Answer Key Hints and Explanations

5  Alkenes and Alkynes

191 194 195 196 197 197 199 201 204 204 205 205 205 206 206 206 207 208 210 210 210 211 213 213 214 217 218 222 222

227

Part I: Alkenes 228 5.1 Structure of the Double Bond 228 5.2 Isomerism 230 5.3 How to Name Alkenes and Cycloalkenes 231 5.4 The (E )–(Z ) System for Designating Alkene Diastereomers 233 5.4A How to Use the (E )–(Z ) System 233 5.5 Relative Stabilities of Alkenes 234 5.5A Heat of Reaction 234 5.5B Overall Relative Stabilities of Alkenes 235 5.6 Cycloalkenes 236 5.7 Synthesis of Alkenes by Hydrogenation of Alkynes 236 5.7A Syn Addition of Hydrogen: Synthesis of cis-Alkenes 236 5.7B Anti Addition of Hydrogen: Synthesis of trans-Alkenes 237 5.8 Synthesis of Alkenes via Elimination Reactions 237 5.8A Dehydrohalogenation of Alkyl Halides 237 5.8B Dehalogenation of Vicinal Dihalides 240 5.8C Acid-Catalyzed Dehydration of Alcohols 240 5.9 Carbocation Stability and the Occurrence of Molecular Rearrangements244 5.9A Rearrangements During Dehydration of Secondary Alcohols 244 5.9B Rearrangement After Dehydration of a Primary Alcohol 246

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5.10 Hydrogenation of Alkenes 5.10A  Hydrogenation: The Function of the Catalyst 5.11 Addition Reactions of Alkenes 5.11A  How to Understand Additions to Alkenes 5.11B Electrophilic Addition of Hydrogen Halides to Alkenes: Mechanism and Markovnikov’s Rule 5.11C The Anti-Markovnikov Addition of Hydrogen Bromide Radical Addition to Alkenes 5.11D  Addition of Water to Alkenes: Acid-Catalyzed Hydration 5.11E Electrophilic Addition of Bromine and Chlorine to Alkenes 5.11F  Addition of Sulphuric Acid 5.12 Oxidation of Alkenes: Syn 1,2-Dihydroxylation 5.12A  Oxidative Cleavage of Alkenes 5.12B  Cleavage with Ozone

247 248 249 249 250 254 257 258 262 263 264 264

Part II: Alkynes 5.13 Structure of the Triple Bond 5.13A  Bond Lengths of Ethyne, Ethene and Ethane 5.14 Isomerism 5.15 How to Name Alkynes 5.16 The Acidity of Terminal Alkynes 5.17 Methods of Preparation 5.17A  Synthesis of Alkynes by Elimination Reactions 5.17B Synthesis of Higher Alkynes by Carbon–Carbon Bond Formation in Terminal Alkynes 5.17C  Other Methods of Preparation 5.18 Chemical Reactivity 5.18A  Electrophilic Addition Reactions 5.18B  Oxidation of Alkynes 5.18C Polymerization

265 266 266 267 268 269 270 270

Part III: Conjugated Unsaturated Systems 5.19 The Stability of the Allyl Radical 5.19A  Molecular Orbital Description of the Allyl Radical 5.19B  Resonance Description of the Allyl Radical 5.19C  Resonance Theory Revisited 5.20 The Allyl Cation 5.21 Alkadienes and Polyunsaturated Hydrocarbons 5.22 1,3-Butadiene: Electron Delocalization 5.22A  Bond Lengths of 1,3-Butadiene 5.22B  Conformations of 1,3-Butadiene 5.22C  Molecular Orbitals of 1,3-Butadiene 5.23 The Stability of Conjugated Dienes 5.24 Allylic Substitution and Allylic Radicals 5.24A  Allylic Chlorination (High Temperature) 5.24B Allylic Bromination with N-Bromosuccinimide (Low Concentration of Br2) 5.25 Electrophilic Attack on Conjugated Dienes: 1,4-Addition 5.25A Kinetic Control versus Thermodynamic Control of a Chemical Reaction 5.26 The Diels–Alder Reaction: A 1,4-Cycloaddition ­Reaction of Dienes Solved Examples Solved Previous Years’ NEET Questions Additional Objective Questions Answer Key Hints and Explanations

276 277 277 277 278 279 279 281 281 281 282 282 283 284 285 286 287 288 290 295 300 307 308

271 272 273 273 275 276

Contents

6  Aromatic Compounds and their Reactions











6.1 The Discovery of Benzene 6.2 Nomenclature of Benzene Derivatives 6.3 Structure of Benzene 6.3A Reactions of Benzene 6.3B The Kekulé Structure for Benzene 6.4 The Thermodynamic Stability of Benzene 6.5 Modern Theories of the Structure of Benzene 6.5A The Resonance Explanation of the Structure of Benzene 6.5B The Molecular Orbital Explanation of the Structure of Benzene 6.6 Hückel’s Rule: The 4n + 2π Electron Rule 6.6A How to Diagram the Relative Energies of π Molecular Orbitals in Monocyclic Systems Based on Hückel’s Rule 6.6B The Annulenes 6.6C Aromatic Ions 6.6D Aromatic, Antiaromatic, and Nonaromatic Compounds 6.7 Other Aromatic Compounds 6.7A Benzenoid Aromatic Compounds 6.7B Nonbenzenoid Aromatic Compounds 6.7C Fullerenes 6.7D Heterocyclic Aromatic Compounds 6.7E Aromatic Compounds in Biochemistry 6.8 Preparation of Arenes 6.9 Physical Properties of Arenes 6.10 Electrophilic Aromatic Substitution Reactions 6.11 A General Mechanism for Electrophilic Aromatic Substitution 6.12 Halogenation of Benzene 6.13 Nitration of Benzene 6.14 Sulfonation of Benzene 6.15 Friedel–Crafts Reactions 6.15A  Friedel–Crafts Alkylation 6.15B  Friedel–Crafts Acylation 6.15C  Limitations of Friedel–Crafts Reactions 6.15D  Synthetic Applications of Friedel–Crafts Acylations 6.16 Substituents Can Affect Both the Reactivity of the Ring and the Orientation of the Incoming Group 6.16A  How Do Substituents Affect Reactivity? 6.16B  Ortho–Para-Directing Groups and Meta-Directing Groups 6.16C  Electron-Donating and Electron-Withdrawing Substituents 6.16D  Groups: Ortho–Para Directors 6.16E  Deactivating Groups: Meta Directors 6.16F  Halo Substituents: Deactivating Ortho–Para Directors 6.16G  Classification of Substituents 6.17 How Substituents Affect Electrophilic Aromatic Substitution: A Closer Look 6.17A  Reactivity: The Effect of Electron-Releasing and Electron-Withdrawing Groups 6.17B  Inductive and Resonance Effects: Theory of Orientation 6.17C  Meta-Directing Groups 6.17D  Ortho–Para-Directing Groups 6.17E  Ortho–Para Direction and Reactivity of Alkylbenzenes 6.17F  Summary of Substituent Effects on Orientation and Reactivity 6.18 Reactions of the Side Chain of Alkylbenzenes 6.18A  Benzylic Radicals and Cations 6.18B  Benzylic Halogenation of the Side Chain

317 317 318 320 320 321 322 323 324 324 325 326 327 328 329 331 331 332 332 333 335 335 336 336 337 339 341 342 343 343 344 346 348 350 350 350 351 351 353 353 354 354 354 355 357 358 361 362 362 363 364

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6.19 Oxidation Reactions 6.19A  Oxidation of the Benzene Ring 6.19B  Oxidation of the Side Chain 6.20 Addition Reactions 6.21 Reduction of Aromatic Compounds 6.21A  The Birch Reduction 6.22 Synthetic Applications 6.22A  Use of Protecting and Blocking Groups 6.22B  Orientation in Disubstituted Benzenes Solved Examples Solved Previous Years’ NEET Questions Additional Objective Questions Answer Key Hints and Explanations

7  Alkyl and Aryl Halides

365 365 365 366 366 367 368 368 370 370 376 380 387 387

395

7.1 Classification 7.1A Classification Based on Hybridization of Carbon Atom

396

Part I:  Alkyl Halides (Haloalkanes) 7.2 Nomenclature 7.3 Physical Properties of Alkyl Halides 7.3A Melting and Boiling Points 7.3B Density 7.3C Solubility 7.4 Preparation of Alkyl Halides 7.4A From Alcohols 7.4B From Hydrocarbons 7.4C Halogen Exchange 7.5 Nucleophilic Substitution Reactions of Alkyl Halides 7.5A Nucleophiles and Leaving Groups 7.5B Substitution Nucleophilic Bimolecular (SN2) 7.5C Substitution Nucleophilic Unimolecular (SN1) 7.5D Factors Affecting the Rates of SN1 and SN2 Reactions 7.5E Organic Synthesis: Functional Group ­Transformations Using SN2 Reactions 7.6 Elimination Reactions of Alkyl Halides 7.6A Dehydrohalogenation 7.6B The E2 Reaction 7.6C The E1 Reaction 7.6D How to Determine Whether ­Substitution or Elimination is Favored 7.7 Overall Summary–Substitution and Elimination Reactions 7.8 Reaction of Alkyl Halides with Metals 7.9 Reduction of Alkyl Halides to Hydrocarbons 7.10 Polyhalogen Compounds 7.10A  Dichloromethane (Methylene Chloride CH2Cl2) 7.10B  Trichlormethane (Chloroform CHCl3 ) 7.10C  Triiodomethane (Iodoform CH3l) 7.10D  Tetrachloromethane (Carbon Tetrachloride CCl4 ) 7.10E Freons

397 398 400 400 401 401 401 401 403 407 407 408 410 413 416

396

424 425 426 427 428 430 432 434 435 435 435 435 436 436 436

Contents

Part II: Aryl Halides (Haloarenes) 7.11 Nomenclature 7.12 Preparation of Aryl Halides 7.12A  By Electrophilic Substitution 7.12B  From Diazonium Salts 7.13 Nucleophilic Aromatic Substitution in Aryl Halides 7.13A Nucleophilic Aromatic Substitution by Addition–Elimination: The SNAr Mechanism 7.13B Nucleophilic Aromatic Substitution through an Elimination–Addition Mechanism: Benzyne 7.13C Phenylation 7.14 Electrophilic Aromatic Substitution Reaction 7.15 Reaction with Metals Solved Examples Solved Previous Years’ NEET Questions Additional Objective Questions Answer Key Hints and Explanations

8  Alcohols, Phenols and Ethers

8.1 Structure and Classification 8.1A Monohydric, Dihydric, Trihydric or Polyhydric Alcohols and Phenols 8.1B Classification of Ethers 8.2 Structures of Functional Groups 8.2A Alcohols 8.2B Phenols 8.2C Ethers 8.3 Nomenclature 8.3A Nomenclature of Alcohols 8.3B Nomenclature of Phenols 8.3C Nomenclature of Ethers 8.4 Physical Properties 8.4A Physical Properties of Alcohols 8.4B Physical Properties of Phenols 8.4C Physical Properties of Ethers 8.5 Synthesis of Alcohols 8.5A Alcohols from Alkenes 8.5B Alcohols by Reduction of Carbonyl Compounds 8.5C Alcohols from Grignard Reagents 8.5D By Substitution Reactions on Alkyl Halides 8.6 Synthesis of Phenols 8.6A Laboratory Synthesis 8.6B Industrial Syntheses 8.7 Synthesis of Ethers 8.7A Ethers by Intermolecular Dehydration of Alcohols 8.7B The Williamson Ether Synthesis 8.7C Synthesis of Ethers by Alkoxymercuration–Demercuration 8.8 Reactions of Alcohols 8.8A Reactions Involving Cleavage of O—H Bond (Alcohols as Acids) 8.8B Reactions Involving Cleavage of Carbon–Oxygen (C—O) Bond 8.8C Reactions Involving Alcohol Molecule as a Whole 8.8D Oxidation of Alcohols

437 437 438 438 439 440 441 442 443 446 448 449 454 458 468 469

475 475 476 478 479 479 479 479 480 480 481 482 482 482 484 484 486 486 488 491 493 493 493 494 495 495 496 498 498 499 501 502 504

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8.9 Reactions of Phenols as Acids 8.9A Strength of Phenols as Acids 8.9B Distinguishing and Separating Phenols from Alcohols and Carboxylic Acids 8.9C Reactions of the Benzene Ring of Phenols 8.9D Other Reactions of Phenols 8.10 Reactions of Ethers 8.10A  Cleavage of Ethers 8.10B  Electrophilic Substitution 8.10C  The Claisen Rearrangement 8.11 Distinction Between 1°, 2°, 3° Alcohols 8.11A  Lucas Test 8.11B  Victor Meyer Test 8.11C  Iodoform Test 8.12 Some Commercially Important Alcohols and Ethers 8.12A Methanol 8.12B Ethanol 8.12C  Ethylene and Propylene Glycols 8.12D  Diethyl Ether Solved Examples Solved Previous Years’ NEET Questions Additional Objective Questions Answer Key Hints and Explanations

9  Aldehydes and Ketones







9.1 Nomenclature of Aldehydes and Ketones 9.1A Isomerism in Aldehydes and Ketones 9.2 Structure of the Carbonyl Group 9.2A Comparison between C   O (Carbonyl) and C   C (Alkene) Double Bond 9.3 Physical Properties 9.4 Synthesis of Aldehydes 9.4A Aldehydes by Oxidation of 1° Alcohols 9.4B Aldehydes by Catalytic Dehydrogenation of Alcohols 9.4C Aldehydes by Ozonolysis of Alkenes 9.4D Aldehydes from Alkynes (Hydration and Hydroboration) 9.4E Aldehydes by Reduction of Acyl Chlorides, Esters, and Nitriles 9.4F Aromatic Aldehydes from Hydrocarbons 9.5 Synthesis of Ketones 9.5A Ketones from Alkenes, Arenes, and 2° Alcohols 9.5B Ketones from Nitriles 9.5C Ketones from Acyl Chlorides 9.6 Miscellaneous Methods for Preparation of Aldehydes and Ketones 9.6A Aqueous Alkali Hydrolysis of Geminal Dihalides 9.6B Distillation of Calcium Salts of Acids 9.6C Catalytic Decomposition of Carboxylic Acids 9.7 Nucleophilic Addition to the Carbon–Oxygen Double Bond 9.7A Reversibility of Nucleophilic Additions to the Carbon–Oxygen Double Bond 9.7B Relative Reactivity: Aldehydes versus Ketones 9.7C Addition Products Can Undergo Further Reactions

506 506 507 508 509 512 512 514 516 516 516 517 517 518 518 518 518 518 520 524 527 538 539

551 552 553 554 555 555 556 556 557 557 558 558 562 563 563 564 565 565 565 565 566 566 568 568 569

Contents

















9.8 The Addition of Alcohols: Hemiacetals and Acetals 9.8A Hemiacetals 9.8B Acetals 9.8C Acetals Are Used as Protecting Groups 9.8D Thioacetals 9.9 The Addition of Hydrogen Cyanide: Cyanohydrins 9.10 The Addition of Sodium Hydrogen Sulfite 9.11 Addition of Grignard Reagents 9.12 The Addition of Primary and Secondary Amines 9.12A Imines 9.12B  Oximes and Hydrazones 9.12C Enamines 9.13 The Addition of Ylides: The Wittig Reaction 9.14 Reduction of Aldehydes and Ketones 9.14A  Reduction to Alcohols 9.14B  Reduction to Hydrocarbons 9.15 Oxidation of Aldehydes and Ketones 9.15A  The Baeyer–Villiger Oxidation 9.15B Oxidation Reactions Based Tests for Aldehydes and Ketones 9.16 The Acidity of the α Hydrogens of Carbonyl Compounds: Enolate Anions 9.17 Keto and Enol Tautomers 9.18 Reactions via Enols and Enolates 9.18A Racemization 9.18B  Halogenation at the a Carbon 9.18C  The Haloform Reaction 9.19 Lithium Enolates 9.20 Aldol Reactions: Addition of Enolates and Enols to Aldehydes and Ketones 9.20A  Aldol Addition Reactions 9.20B  The Retro-Aldol Reaction 9.20C Aldol Condensation Reactions: Dehydration of the Aldol Addition Product 9.20D  Acid-Catalyzed Aldol Condensations 9.20E  Synthetic Applications of Aldol Reactions 9.21 Crossed Aldol Condensations 9.21A  Crossed Aldol Condensations Using Weak Bases 9.21B Crossed Aldol Condensations Using Strong Bases: Lithium Enolates and Directed Aldol Reactions 9.22 Cyclizations via Aldol Condensations 9.23 Other Important Reactions 9.23A  Cannizaro Reaction 9.23B  Benzilic Acid Rearrangement 9.23C  Benzoin Condensation 9.23D  Perkin Reaction 9.24 Summary of Aldehyde and Ketone Reactions 9.24A  Aldehyde and Ketone Addition Reactions 9.24B  Enolate Chemistry 9.24C  Other Important Reactions Solved Examples Solved Previous Years’ NEET Questions Additional Objective Questions Answer Key Hints and Explanations

569 569 572 573 574 574 576 576 577 577 578 579 580 581 581 582 583 584 585 586 587 588 588 589 591 592 592 593 594 594 594 595 596 597 598 599 600 600 602 603 603 603 603 605 606 607 611 618 628 628

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10  Carboxylic Acids and Their Derivatives

10.1 10.2

Structure of the Carboxyl Group Nomenclature and Physical Properties 10.2A Carboxylic Acids 10.2B Carboxylate Salts 10.2C Dicarboxylic Acids 10.2D Esters 10.2E Carboxylic Anhydrides 10.2F Acyl Chlorides 10.2G Amides 10.2H Nitriles 10.3 Preparation of Carboxylic Acids 10.4 Chemical Properties of Carboxylic Acids 10.4A Reactions Involving Cleavage of O—H Bond 10.4B Reactions Involving Cleavage of C—OH Bond 10.4C Reactions Involving—COOH Group 10.4D Substitution Reactions in the Hydrocarbon Part 10.5 Acyl Substitution: Nucleophilic Addition–Elimination at the Acyl Carbon 10.5A Relative Reactivity of Acyl Compounds 10.5B Synthesis of Acid Derivatives 10.6 Acyl Chlorides 10.6A Synthesis of Acyl Chlorides 10.6B Reactions of Acyl Chlorides 10.7 Carboxylic Acid Anhydrides 10.7A Synthesis of Carboxylic Acid Anhydrides 10.7B Reactions of Carboxylic Acid Anhydrides 10.8 Esters 10.8A Synthesis of Esters: Esterification 10.8B Base-Promoted Hydrolysis of Esters: Saponification 10.8C Lactones 10.9 Amides 10.9A Synthesis of Amides 10.9B Amides from Acyl Chlorides 10.9C Amides from Carboxylic Anhydrides 10.9D Amides from Esters 10.9E Amides from Carboxylic Acids and Ammonium Carboxylates 10.9F Hydrolysis of Amides 10.9G Nitriles from the Dehydration of Amides 10.9H Hydrolysis of Nitriles 10.9I Lactams 10.10 Decarboxylation of Carboxylic Acids 10.11 Summary of the Reactions of Carboxylic Acids and Their Derivatives Solved Examples Solved Previous Years’ NEET Questions Additional Objective Questions Answer Key Hints and Explanations

639 640 640 640 643 643 643 644 645 645 646 646 650 650 652 653 654 654 656 656 656 656 658 658 658 659 659 659 662 664 665 665 665 665 666 666 667 669 669 670 670 671 675 679 682 689 690

Contents

11  Organic Compounds Containing Nitrogen

697

Part I: Amines 11.1 Nomenclature of Amines 11.1A Aliphatic Amines 11.1B Arylamines 11.1C Isomerism in Amines 11.2 Physical Properties and Structure of Amines 11.2A Physical Properties 11.2B Structure of Amines 11.3 Preparation of Amines 11.3A Through Nucleophilic Substitution Reactions 11.3B Preparation of Aromatic Amines through Reduction of Nitro Compounds 11.3C Preparation of Primary, Secondary, and Tertiary Amines through Reductive Amination 11.3D Preparation of Primary, Secondary, or Tertiary Amines through Reduction of Nitriles, Oximes, and Amides 11.3E Preparation of Primary Amines through the Hofmann and Curtius Rearrangements 11.4 Basicity of Amines: Amine Salts 11.4A Basicity of Arylamines 11.4B Amines versus Amides 11.4C Aminium Salts and Quaternary Ammonium Salts 11.4D Solubility of Amines in Aqueous Acids 11.5 Reactions of Amines 11.5A Oxidation of Amines 11.5B Reaction with Grignard Reagents 11.5C Carbylamine Reaction 11.5D Hofmann’s Mustard Oil Reaction 11.6 Reactions of Amines with Nitrous Acid 11.6A Reactions of Primary Aliphatic Amines with Nitrous Acid 11.6B Reactions of Secondary Amines with Nitrous Acid 11.6C Reactions of Tertiary Amines with Nitrous Acid 11.7 Reactions of Amines with Sulfonyl Chlorides: Test for Amines 11.7A The Hinsberg Test 11.8 Eliminations Involving Ammonium Compounds 11.8A The Hofmann Elimination 11.8B The Cope Elimination

697 698 698 699 699 700 700 700 701 701

Part II:  Arenediazonium Salts 11.9 Preparation of Arenediazonium Salts 11.10 Replacement Reactions of Arenediazonium Salts 11.10A Syntheses Using Diazonium Salts 11.10B The Sandmeyer Reaction: Replacement of the Diazonium Group by —Cl, —Br, or —CN 11.10C Replacement by —I 11.10D Replacement by —F 11.10E Replacement by —OH 11.10F Replacement by Hydrogen: Deamination by Diazotization 11.11 Coupling Reactions of Arenediazonium Salts

718 718 719 719

703 704 705 706 707 708 709 709 710 711 713 713 713 714 714 714 715 715 715 716 717 717 718

720 720 721 721 721 722

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Part III:  Cyanides and Isocyanides 11.12 Preparation and Properties of Cyanides and Isocyanides 11.12A Methods of Preparation 11.12B Physical Properties 11.12C Chemical Properties 11.13 Summary of Reactions of Organic Compounds Containing Nitrogen 11.13A Preparation of Amines 11.13B Reactions of Amines 11.13C Preparation of Cyanides and Isocyanides 11.13D Reactions of Cyanides and Isocyanides Solved Examples Solved Previous Years’ NEET Questions Additional Objective Questions Answer Key Hints and Explanations

12  Biomolecules

724 724 724 724 725 726 726 728 729 729 730 735 741 750 751

757

Part I: Carbohydrates 12.1 Classification of Carbohydrates 12.2 Monosaccharides 12.2A Classification of Monosaccharides 12.2B d and l Designations of Monosaccharides 12.2C Glucose 12.2D Fructose 12.2E Reactions of Monosaccharides 12.2F Other Reactions of Monosaccharides 12.3 Disaccharides 12.3A Sucrose 12.3B Maltose 12.3C Cellobiose 12.3D Lactose 12.4 Polysaccharides 12.4A Starch 12.4B Glycogen 12.4C Cellulose 12.4D Cellulose Derivatives 12.5 Other Biologically Important Sugars 12.6 Sugars that Contain Nitrogen 12.6A Glycosylamines 12.6B Amino Sugars

757 758 758 758 759 760 764 765 768 771 771 772 772 775 775 776 777 777 778 778 779 779 779

Part II:  Amino Acids and Proteins 12.7 Amino Acids 12.7A Structures and Names 12.7B Essential Amino Acids 12.7C Amino Acids as Dipolar Ions 12.8 Polypeptides and Proteins 12.8A Primary Structure of Polypeptides and Proteins 12.8B Examples of Polypeptide and Protein Primary Structure 12.9 Secondary, Tertiary and Quaternary Structures of Proteins 12.9A Secondary Structure 12.9B Tertiary Structure 12.9C Quaternary Structure 12.9D Hemoglobin: A Conjugated Protein

779 781 781 781 784 786 787 787 788 788 791 791 792

Contents



12.10 Denaturation of Proteins 12.11 Introduction to Enzymes 12.12 Hormones 12.13 Vitamins 12.13A Classification of Vitamins

792 793 796 797 797

Part III:  Nucleic Acids 12.14 Nucleotides and Nucleosides 12.15 Deoxyribonucleic Acid: DNA 12.15A Primary Structure 12.15B Secondary Structure 12.15C Replication of DNA 12.16 RNA and Protein Synthesis

798 800 802 802 803 803 804

Part IV: Lipids 12.17 Fatty Acids and Triacylglycerols 12.17A Hydrogenation of Triacylglycerols 12.17B Biological Functions of Triacylglycerols 12.17C Saponification of Triacylglycerols 12.17D Reactions of the Carboxyl Group of Fatty Acids 12.17E Reactions of the Alkenyl Chain of Unsaturated Fatty Acids 12.18 Terpenes and Terpenoids 12.19 Phospholipids and Cell Membranes 12.19A Phosphatides 12.20 Waxes Solved Examples Solved Previous Years’ NEET Questions Additional Objective Questions Answer Key Hints and Explanations

806 806 807 807 809 810

13  Polymers



13.1 13.2

Some Terms Related to Polymers Classification of Polymers 13.2A Classification Based on Source 13.2B Classification Based on Structure of Polymers 13.2C Classification Based on Mode of Polymerization 13.2D Classification Based on Molecular Forces 13.3 Addition Polymerization or Chain-Growth Polymerization 13.3A Free Radical Mechanism 13.3B Ionic Polymerization 13.4 Preparation of Some Important Addition Polymers 13.5 Condensation Polymerization or Step-Growth Polymerization 13.5A Copolymerization 13.6 Natural Rubber 13.6A Vulcanization of Rubber 13.7 Synthetic Rubbers 13.7A Preparation of Synthetic Rubbers 13.8 Molecular Mass of Polymers 13.9 Biodegradable Polymers 13.10 Polymers of Commercial Importance Solved Examples Solved Previous Years’ NEET Questions

810 811 812 813 814 815 819 822 828 829

833 833 834 834 834 835 836 838 838 839 840 842 846 847 847 848 848 849 849 850 852 854

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Contents

Additional Objective Questions Answer Key Hints and Explanations

14  Chemistry in Everyday Life 14.1

Drugs and Their Classification 14.1A Classification of Drugs 14.2 Drug–Target Interaction 14.2A Enzymes as Drug Targets 14.2B Receptors as Drug Targets 14.3 Therapeutic Action of Different Classes of Drugs 14.3A Antacids 14.3B Antihistamines 14.3C Neurologically Active Drugs 14.3D Antimicrobials 14.3E Antifertility Drugs 14.4 Chemicals in Food 14.4A Artificial Sweetening Agents 14.4B Food Preservatives 14.5 Cleansing Agents 14.5A Soaps 14.5B Synthetic Detergents Solved Examples Solved Previous Years’ NEET Questions Additional Objective Questions Answer Key Hints and Explanations

NEET Paper Index

Organic Chemistry_FM.indd 30

857 862 862

865 865 866 866 866 868 868 868 869 869 871 873 874 874 875 876 876 878 881 882 883 888 888

P1 I1

07-12-2017 12:25:29

Basic Principles of Organic Chemistry

1

C H A P T E R OU TLIN E Part I: Bonding and Molecular Structure 1.1 Life and the Chemistry of Carbon Compound 1.2 Chemical Bonds: The Octet Rule 1.3 Isomers: Different Compounds that Have the Same Molecular Formula 1.4 How to Write and Interpret Structural Formulas 1.5 Resonance Theory 1.6 The Structure of Methane and Ethane: sp3 Hybridization 1.7 The Structure of Ethene (Ethylene): sp2 Hybridization 1.8 The Structure of Ethyne (Acetylene): sp Hybridization 1.9 How to Predict Molecular Geometry: The Valence Shell Electron Pair Repulsion Model Part II: Families of Carbon Compounds 1.10 Classification of Organic Compounds 1.11 Hydrocarbons: Representative Alkanes, Alkenes, Alkynes, and Aromatic Compounds 1.12 Functional Groups 1.13 Alkyl Halides or Haloalkanes 1.14 Alcohols and Phenols 1.15 Ethers 1.16 Amines

1.17 Aldehydes and Ketones 1.18 Carboxylic Acids, Esters, and Amides 1.19 Nitriles 1.20 Summary of Important Families of Organic Compounds 1.21 Nomenclature of Organic Compounds 1.22 Polar Covalent Bonds 1.23 Physical Properties and Molecular Structure

PHOTO CREDITS: clothing: © Sandra van der Steen/iStockphoto; inks: © Andrey Kuzman/iStockphoto; drugs: © cogal/iStockphoto

Part III: An Introduction to Organic Reactions and their Mechanisms 1.24 Acid–Base Reactions 1.25 How to Use Curved Arrows in Illustrating Reactions 1.26 Heterolysis of Bonds to Carbon: Carbocations and Carbanions 1.27 Homolytic Fission of Bonds 1.28 Electron Displacement Effects in Organic Compounds 1.29 The Strength of Brønsted–Lowry Acids and Bases: Ka and pKa 1.30 Relationships between Structure and Acidity 1.31 Acidity: Carboxylic Acids versus Alcohols 1.32 The Effect of the Solvent on Acidity 1.33 A Mechanism for an Organic Reaction

PART I: BONDING AND MOLECULAR STRUCTURE Organic chemistry plays a role in all aspects of our lives, from the clothing we wear, to the pixels of our television and computer screens, to preservatives in food, to the inks that color the pages of this book. If you take the time to understand organic chemistry, to learn its overall logic, then you will truly have the power to change society. Indeed, organic chemistry provides the power to synthesize new drugs, to engineer molecules that can make computer processors run more quickly, and to design ways to knock the calories out of

2

Chapter 1 | Basic Principles of Organic Chemistry

sugar while still making food taste deliciously sweet. It can explain biochemical processes like aging, neural functioning, and cardiac arrest, and show how we can prolong and improve life. It can do almost anything.

1.1

LIFE AND THE CHEMISTRY OF CARBON COMPOUNDS

Organic chemistry is the chemistry of compounds that contain the element carbon. If a compound does not contain the element carbon, it is said to be inorganic. Look for a moment at the periodic table inside the front cover of this book. More than a hundred elements are listed there. The question that comes to mind is this: why should an entire field of chemistry be based on the chemistry of compounds that contain this one element, carbon? There are several reasons, the primary one being this: carbon compounds are central to the structure of living organisms and therefore to the existence of life on Earth. We exist because of carbon compounds. What is it about carbon that makes it the element that nature has chosen for living organisms? There are two important reasons: carbon atoms can form strong bonds to other carbon atoms to form rings and chains of carbon atoms, and carbon atoms can also form strong bonds to elements such as hydrogen, nitrogen, oxygen, and sulfur. Because of these bond-forming properties, carbon can be the basis for the huge diversity of compounds necessary for the emergence of living organisms. From time to time, writers of science fiction have speculated about the possibility of life on other planets being based on the compounds of another element—for example, silicon, the element most like carbon. However, the bonds that silicon atoms form to each other are not nearly as strong as those formed by carbon, and therefore it is very unlikely that silicon could be the basis for anything equivalent to life as we know it.

1.1A Development of the Science of Organic Chemistry The science of organic chemistry began to flower with the demise of a nineteenth century theory called vitalism. According to vitalism, organic compounds were only those that came from living organisms, and only living things could synthesize organic compounds through intervention of a vital force. Inorganic compounds were considered those compounds that came from nonliving sources. Friedrich Wöhler, however, discovered in 1828 that an organic compound called urea (a constituent of urine) could be made by evaporating an aqueous solution of the inorganic compound ammonium cyanate. With this discovery, the synthesis of an organic compound, began the evolution of organic chemistry as a scientific discipline. O heat

NH4NCO

H 2N

Ammonium cyanate

C

NH2

Urea

Despite the demise of vitalism in science, the word “organic” is still used today by some people to mean “coming from living organisms” as in the terms “organic vitamins” and “organic fertilizers.” The commonly used term “organic food” means that the food was grown without the use of synthetic fertilizers and pesticides. An “organic vitamin” means to these people that the vitamin was isolated from a natural source and not synthesized by a chemist. While there are sound arguments to be made against using food contaminated with certain pesticides, while there may be environmental benefits to be obtained from organic farming, and while “natural” vitamins may contain beneficial substances not present in synthetic vitamins, it is impossible to argue that pure “natural” vitamin C, for example, is healthier than pure “synthetic” vitamin C, since the two substances are identical in all respects. In science today, the study of compounds from living organisms is called natural products chemistry. OH O

O C C

HO

CH

CH9CH2OH

C OH Vitamin C

1.2 | Chemical Bonds: The Octet Rule

1.2

CHEMICAL BONDS: THE OCTET RULE

The first explanations of the nature of chemical bonds were advanced by G. N. Lewis (of the University of California, Berkeley) and W. Kössel (of the University of Munich) in 1916. Two major types of chemical bonds were proposed: 1. Ionic (or electrovalent) bonds are formed by the transfer of one or more electrons from one atom to another to create ions. 2. Covalent bonds result when atoms share electrons. The central idea in their work on bonding is that atoms without the electronic configuration of a noble gas generally react to produce such a configuration because these configurations are known to be highly stable. For all of the noble gases except helium, this means achieving an octet of electrons in the valence shell. ● ●

The valence shell is the outermost shell of electrons in an atom. The tendency for an atom to achieve a configuration where its valence shell contains eight electrons is called the octet rule.

The concepts and explanations that arise from the original propositions of Lewis and Kössel are satisfactory for explanations of many of the problems we deal with in organic chemistry today. For this reason we shall review these two types of bonds in more modern terms.

1.2A Ionic Bonds Atoms may gain or lose electrons and form charged particles called ions. ●

An ionic bond is an attractive force between oppositely charged ions.

One source of such ions is a reaction between atoms of widely differing electronegativities (Table 1.1). For example, the formation of an ionic bond in the reaction between lithium and fluorine atoms. ● ●

Electronegativity is a measure of the ability of an atom to attract electrons. Electronegativity increases as we go across a horizontal row of the periodic table from left to right and it increases as we go up a vertical column (Table 1.1). Table 1.1 Electronegativities of some of the elements Increasing electronegativity H 2.1 Li 1.0

Be 1.5

B 2.0

C 2.5

N 3.0

O 3.5

F 4.0

Na 0.9

Mg 1.2

Al 1.5

Si 1.8

P 2.1

S 2.5

Cl 3.0

K 0.8

Increasing electronegativity

Br 2.8

We will use electronegativity frequently as a tool for understanding the properties and reactivity of organic molecules. Ionic substances, because of their strong internal electrostatic forces, are usually very high melting solids, often having melting points above 1000 °C. In polar solvents, such as water, the ions are solvated (see Section 2.13D), and such solutions usually conduct an electric current. ●

Ionic compounds, often called salts, form only when atoms of very different electronegativities transfer electrons to become ions.

3

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Chapter 1 | Basic Principles of Organic Chemistry

1.2B Covalent Bonds and Lewis Structures When two or more atoms of the same or similar electronegativities react, a complete transfer of electrons does not occur. In these instances the atoms achieve noble gas configurations by sharing electrons. ●

●

Covalent bonds form by sharing of electrons between atoms of similar electronegativities to achieve the configuration of a noble gas. Molecules are composed of atoms joined exclusively or predominantly by covalent bonds.

Molecules may be represented by electron-dot formulas or, more conveniently, by formulas where each pair of electrons shared by two atoms is represented by a line. ●

A dash structural formula has lines that show bonding electron pairs and includes elemental symbols for the atoms in a molecule.

Some examples are shown here: 1. Hydrogen, being in group IA of the periodic table, has one valence electron. Two hydrogen atoms share electrons to form a hydrogen molecule, H2. H2

H 1 H

HH

H

usually written

H

2. Because chlorine is in group VIIA, its atoms have seven valence electrons. Two chlorine atoms can share electrons (one electron from each) to form a molecule of Cl2. Cl2

Cl 1 Cl

Cl Cl

usually written

Cl

Cl

3. And a carbon atom (group IVA) with four valence electrons can share each of these electrons with four hydrogen atoms to form a molecule of methane, CH4. CH4

C

1

4H

H H C H H

H usually written

H

C

H

H

Two carbon atoms can use one electron pair between them to form a carbon–carbon single bond while also bonding hydrogen atoms or other groups to achieve an octet of valence electrons. Consider the example of ethane below.

C2H6

H H H C C H H H

and as a dash formula

H

H

H

C

C

H

H

H

Ethane

These formulas are often called Lewis structures; in writing them we show all of the valence electrons. Unshared electron pairs are shown as dots, and in dash structural formulas, bonding electron pairs are shown as lines. 4. Atoms can share two or more pairs of electrons to form multiple covalent bonds. For example, two nitrogen atoms possessing five valence electrons each (because nitrogen is in group VA) can share electrons to form a triple bond between them. N2

N N

and as a dash formula

N; N

Carbon atoms can also share more than one electron pair with another atom to form a multiple covalent bond. Consider the examples of a carbon–carbon double bond in ethene (ethylene) and a carbon–carbon triple bond in ethyne (acetylene).

1.2 | Chemical Bonds: The Octet Rule

H

C2H4

H C C

H

H

H

and as a dash formula

H C

C

H

H Ethene

C2H2

H C C H

and as a dash formula

H

C

C

H

Ethyne

5. Ions, themselves, may contain covalent bonds. Consider, as an example, the ammonium ion. H

H1 HNH H

1

NH4

and as a dash formula

H

1

N

H

H

1.2C How to Write Lewis Structures Several simple rules allow us to draw proper Lewis structures: 1. Lewis structures show the connections between atoms in a molecule or ion using only the valence electrons of the atoms involved. Valence electrons are those of an atom’s outermost shell. 2. For main group elements, the number of valence electrons a neutral atom brings to a Lewis structure is the same as its group number in the periodic table. Carbon, for example, is in group IVA and has four valence electrons; the halogens (e.g., fluorine) are in group VIIA and each has seven valence electrons; hydrogen is in group IA and has one valence electron. 3. If the structure we are drawing is a negative ion (an anion), we add one electron for each negative charge to the original count of valence electrons. If the structure is a positive ion (a cation), we subtract one electron for each positive charge. 4. In drawing Lewis structures we try to give each atom the electron configuration of a noble gas. To do so, we draw structures where atoms share electrons to form covalent bonds or transfer electrons to form ions. (a) Hydrogen forms one covalent bond by sharing its electron with an electron of another atom so that it can have two valence electrons, the same number as in the noble gas helium. (b) Carbon forms four covalent bonds by sharing its four valence electrons with four valence electrons from other atoms, so that it can have eight electrons (the same as the electron configuration of neon, satisfying the octet rule). (c) To achieve an octet of valence electrons, elements such as nitrogen, oxygen, and the halogens typically share only some of their valence electrons through covalent bonding, leaving others as unshared electron pairs. 5. If necessary, we use multiple bonds to satisfy the octet rule (i.e., give atoms the noble gas configuration). The carbonate ion (CO32−) illustrates this: 2

O O

C

O

The organic molecules ethene (C2H4) and ethyne (C2H2), as mentioned earlier, have a double and triple bond, respectively: H

H C

H

C

and H

H

C

C

H

5

6

Chapter 1 | Basic Principles of Organic Chemistry

6. Before we can write some Lewis structures, we must know how the atoms are connected to each other. Consider nitric acid, for example. Even though the formula for nitric acid is often written HNO3, the hydrogen is actually connected to an oxygen, not to the nitrogen. The structure is HONO2 and not HNO3. Thus, the correct Lewis structure is: O H

O

and not

N

H

O

1.3

N

O

O

O

ISOMERS: DIFFERENT COMPOUNDS THAT HAVE THE SAME MOLECULAR FORMULA

Now that we have had an introduction to Lewis structures, it is time to discuss isomers. ●

Isomers are compounds that have the same molecular formula but different structures.

We will learn about several kinds of isomers during the course of our study. For now, let us consider a type called constitutional isomers. ●

Constitutional isomers are different compounds that have the same molecular formula but differ in the sequence in which their atoms are bonded—that is, their connectivity.

Acetone, used in nail polish remover and as a paint solvent, and propylene oxide, used with seaweed extracts to make food-grade thickeners and foam stabilizers for beer (among other applications) are isomers. Both of these compounds have the molecular formula C3H6O and therefore the same molecular weight. Yet acetone and propylene oxide have distinctly different boiling points and chemical reactivity that, as a result, lend themselves to distinctly different practical applications. Their shared molecular formula simply gives us no basis for understanding the differences between them. We must, therefore, move to a consideration of their structural formulas. On examining the structures of acetone and propylene oxide several key aspects are clearly different (Fig. 1.1). Acetone contains a double bond between the oxygen atom and the central carbon atom. Propylene oxide does not contain a double bond, but has three atoms joined in a ring. The connectivity of the atoms is clearly different in acetone and propylene oxide. Their structures have the same molecular formula but a different constitution. They are constitutional isomers.* ●

Constitutional isomers usually have different physical properties (e.g., melting point, boiling point, and density) and different chemical properties (reactivity).

Acetone

H

H

O

H

C

C

C

H

H

Propylene oxide

H H

H

O

C

C

C

H

H

H

H

Figure 1.1 Chemical formulas show the different structures of acetone and propylene oxide. *An older term for isomers of this type was structural isomers. The International Union of Pure and Applied Chemistry (IUPAC) now recommends that use of the term “structural” when applied to constitutional isomers be abandoned.

1.4 | How to Write and Interpret Structural Formulas

1.4

7

HOW TO WRITE AND INTERPRET STRUCTURAL FORMULAS

Organic chemists use a variety of formats to write structural formulas. We have already used electron-dot formulas and dash formulas in previous sections. Two other important types of formulas are condensed formulas and bond-line (skeletal) formulas. Examples of these four types of structural formulas are shown in Fig. 1.2 using propyl alcohol as an example. H H H H C C C O H H H H

H

H

H

H

C

C

C

H

H

H

Electron-dot formula (b)

Ball-and-stick model (a)

O

H

OH

CH3CH2CH2OH

Dash formula (c)

Condensed formula (d)

Bond-line formula (e)

Figure 1.2 Structural formulas for propyl alcohol. Although electron-dot formulas account explicitly for all of the valence electrons in a molecule, they are tedious and time-consuming to write. Dash, condensed, and bond-line formulas are therefore used more often. Generally it is best to draw unshared electron pairs in chemical formulas, though sometimes they are omitted if we are not considering the chemical properties or reactivity of a compound. When we write chemical reactions, however, we shall see that it is necessary to include the unshared electron pairs when they participate in a reaction. It is a good idea, therefore, to be in the habit of writing unshared electrons pairs.

1.4A More About Dash Structural Formulas ●

Dash structural formulas have lines that show bonding electron pairs, and include elemental symbols for all of the atoms in a molecule.

If we look at the ball-and-stick model for propyl alcohol given in Fig. 1.2a and compare it with the electron-dot, dash, and condensed formulas in Figs. 1.2b–d we find that the chain of atoms is straight in those formulas. In the model, which corresponds more accurately to the actual shape of the molecule, the chain of atoms is not at all straight. Also of importance is this: Atoms joined by single bonds can rotate relatively freely with respect to one another. (We shall discuss the reason for this in Section 1.6B.) This relatively free rotation means that the chain of atoms in propyl alcohol can assume a variety of arrangements like these: H H

H

H

H C

C H

O

C H

H

H H

or

O

H

H C

C H

H

C H

H

H or

H

H

O C

C H

H

C

H

H

H

Equivalent dash formulas for propyl alcohol

It also means that all of the structural formulas above are equivalent and all represent propyl alcohol. Dash structural formulas such as these indicate the way in which the atoms are attached to each other and are not representations of the actual shapes of the molecule. (Propyl alcohol does not have 90° bond angles. It has tetrahedral bond angles.) Dash structural formulas show what is called the connectivity of the atoms. Constitutional isomers (Section 1.3) have different connectivities and, therefore, must have different structural formulas.

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Chapter 1 | Basic Principles of Organic Chemistry

Consider the compound called isopropyl alcohol, whose formula we might write in a variety of ways: H H

O

H

H

H

H

H

O9H

H 9 C 9 C 9 C 9 H or H 9 C 9 C 9 C 9 H or H 9 C 9 C 9 H H

H

H

H

O

H

H

H

H9C9H H

Equivalent dash formulas for isopropyl alcohol

Isopropyl alcohol is a constitutional isomer (Section 1.6A) of propyl alcohol because its atoms are connected in a different order and both compounds have the same molecular formula, C3H8O. In isopropyl alcohol the OH group is attached to the central carbon; in propyl alcohol it is attached to an end carbon.

1.4B Condensed Structural Formulas Condensed structural formulas are somewhat faster to write than dash formulas and, when we become familiar with them, they will impart all the information that is contained in the dash structure. In condensed formulas all of the hydrogen atoms that are attached to a particular carbon are usually written immediately after the carbon. In fully condensed formulas, all of the atoms that are attached to the carbon are usually written immediately after that carbon, listing hydrogens first. For example,

H

H

H

H

H

C

C

C

C

H

Cl

H

H

H

CH3CHCH2CH3 or CH3CHClCH2CH3 Cl

Dash formula

Condensed formulas

The condensed formula for isopropyl alcohol can be written in four different ways:

H

H

H

H

C

C

C

H

O

H

H Dash formula

H

CH3CHCH3

CH3CH(OH)CH3

or

OH CH3CHOHCH3

or

(CH3)2CHOH

Condensed formulas

1.4C Bond-Line Formulas The most common type of structural formula used by organic chemists, and the fastest to draw, is the bondline formula. (Some chemists call these skeletal formulas.) The formula in Fig. 1.3e is a bond-line formula for propyl alcohol. The sooner you master the use of bond-line formulas, the more quickly you will be able to draw molecules when you take notes and work problems. And, lacking all of the symbols that are explicitly shown in dash and condensed structural formulas, bond-line formulas allow you to more quickly interpret molecular connectivity and compare one molecular formula with another. How to Draw Bond-Line Formulas We apply the following rules when we draw bond-line formulas: ● ●

Each line represents a bond. Each bend in a line or terminus of a line represents a carbon atom, unless another group is shown explicitly.

1.4 | How to Write and Interpret Structural Formulas ● ●

●

●

●

No Cs are written for carbon atoms, except optionally for CH3 groups at the end of a chain or branch. No Hs are shown for hydrogen atoms, unless they are needed to give a three-dimensional perspective, in which case we use dashed or solid wedges (as explained in the next section). The number of hydrogen atoms bonded to each carbon is inferred by assuming that as many hydrogen atoms are present as needed to fill the valence shell of the carbon, unless a charge is indicated. When an atom other than carbon or hydrogen is present, the symbol for that element is written at the appropriate location (i.e., in place of a bend or at the terminus of the line leading to the atom). Hydrogen atoms bonded to atoms other than carbon (e.g., oxygen or nitrogen) are written explicitly.

Consider the following examples of molecules depicted by bond-line formulas. Bond-line formulas

CH3 CH2 CH3CHClCH2CH3  CH CH3 

Cl

Cl CH3 CH2 CH CH3 

CH3CH(CH3)CH2CH3 

CH3

(CH3)2NCH2CH3 

CH3 CH2 N CH3 

N

CH3

Bond-line formulas are easy to draw for molecules with multiple bonds and for cyclic molecules, as well. The following are some examples. CH2 CH2

H 2C

and

5

H2C9CH2 H2C9CH2



CH3 CH CH3 C CH2  CH3 CH2 " CHCH2OH  OH CH3

CH2

C

C

CH3 5

1.4D Three-Dimensional Formulas None of the formulas that we have described so far convey any information about how the atoms of a molecule are arranged in space. Molecules exist in three dimensions. We can depict three-dimensional geometry in molecules using bonds represented by dashed wedges, solid wedges, and lines. ● ● ●

A dashed wedge ( ) represents a bond that projects behind the plane of the paper. A solid wedge ( ) represents a bond that projects out of the plane of the paper. An ordinary line (}) represents a bond that lies in the plane of the paper.

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For example, the four C } H bonds of methane (CH4) are oriented toward the corners of a regular tetrahedron, with the carbon in the center and an approximately 109° angle between each C } H bond, as was originally postulated by J. H. van’t Hoff and L. A. Le Bel in 1874. Figure l.3 shows the tetrahedral structure of methane. We will discuss the physical basis for the geometries of carbon when it has only single bonds, a double bond, or a triple bond in Sections 1.6–1.8. For now, let us consider some guidelines for representing these bonding patterns in three dimensions using dashed and solid wedge bonds. H 109°28´ C

H

H Methane

H

Figure 1.3 The tetrahedral structure of methane. In general for carbon atoms that have only single bonds: ●

●

A carbon atom with four single bonds has tetrahedral geometry (Section 1.6) and can be drawn with two bonds in the plane of the paper separated by approximately 109°, one bond behind the plane using a dashed wedge, and one bond in front of the plane using a solid wedge. The dashed wedge and solid wedge bonds in tetrahedral geometry nearly eclipse each other when drawn in proper three-dimensional perspective.

For carbon atoms with a double or a triple bond: ●

●

A carbon atom with a double bond has trigonal planar geometry (Section 1.7) and can be depicted with bonds that are all in the plane of the paper and separated by 120°. A carbon atom with a triple bond has linear geometry (Section 1.8) and can be depicted with its bonds in the plane of the paper and separated by a 180° angle.

Last, when drawing three-dimensional formulas for molecules: ●

Draw as many carbon atoms in the plane of the paper as possible using ordinary lines, then use dashed or solid wedge bonds for substituent groups or hydrogen atoms that are needed to show three dimensions.

Some examples of three-dimensional formulas are shown below. H

C

H C H

H H

HH C

or

H

H

H etc.

C H

H

H

H C H

H Br

or

Ethane

OH

C

H H

or

H C H

H

etc.

Bromomethane

Br H

Br

Br

H

H OH HO

Br H

H

Br

Examples of bond-line formulas that include three-dimensional representations

An example involving trigonal planar geometry

An example involving linear geometry

The carbon chains are shown in the plane of the paper. The dashed and solid wedge bonds nearly eclipse each other.

Bonds to the carbon with the double bond are in the plane of the paper and separated by 120°.

Bonds to the carbon with the triple bond are in the plane of the paper and separated by 180°.

1.5 | Resonance Theory

1.5

RESONANCE THEORY

Often more than one equivalent Lewis structure can be written for a molecule or ion. Consider, for example, the carbonate ion (CO32−). We can write three different but equivalent structures, 1–3: O 

C

O

O O





1

O

C



O C

O

O



2

O



3

Notice two important features of these structures. First, each atom has the noble gas configuration. Second, and this is especially important, we can convert one structure into any other by changing only the positions of the electrons. We do not need to change the relative positions of the atomic nuclei. For example, if we move the electron pairs in the manner indicated by the curved arrows in structure 1, we change structure 1 into structure 2: O 2

O

C

O O

2

becomes

2

O

1

2

C

O

2

In a similar way we can change structure 2 into structure 3: O 2

O

C 2

2

O O

becomes

O

C

2

O

2

3

Structures 1–3, although not identical on paper, are equivalent. None of them alone, however, fits important data about the carbonate ion. Note: Curved arrows show movement of electron pairs, not atoms. The tail of the arrow begins at the current position of the electron pair. The head of the arrow points to the location where the electron pair will be in the next structure. X-ray studies have shown that carbon–oxygen double bonds are shorter than single bonds. The same kind of study of the carbonate ion shows, however, that all of its carbon–oxygen bonds are of equal length. One is not shorter than the others as would be expected from representations 1, 2, and 3. Clearly none of the three structures agrees with this evidence. In each structure, 1–3, one carbon–oxygen bond is a double bond and the other two are single bonds. None of the structures, therefore, is correct. How, then, should we represent the carbonate ion? One way is through a theory called resonance theory. This theory states that whenever a molecule or ion can be represented by two or more Lewis structures that differ only in the positions of the electrons, two things will be true: 1. None of these structures, which we call resonance structures or resonance contributors, will be a realistic representation for the molecule or ion. None will be in complete accord with the physical or chemical properties of the substance. 2. The actual molecule or ion will be better represented by a hybrid (average) of these structures. ●

Resonance structures, then, are not real structures for the actual molecule or ion; they exist only on paper. As such, they can never be isolated. No single contributor adequately represents the molecule

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or ion. In resonance theory we view the carbonate ion, which is, of course, a real entity, as having a structure that is a hybrid of the three hypothetical resonance structures. What would a hybrid of structures 1–3 be like? Look at the structures and look especially at a particular carbon–oxygen bond, say, the one at the top. This carbon–oxygen bond is a double bond in one structure (1) and a single bond in the other two (2 and 3). The actual carbon–oxygen bond, since it is a hybrid, must be something in between a double bond and a single bond. Because the carbon–oxygen bond is a single bond in two of the structures and a double bond in only one, it must be more like a single bond than a double bond. It must be like a one and one-third bond. We could call it a partial double bond. And, of course, what we have just said about any one carbon–oxygen bond will be equally true of the other two. Thus all of the carbon–oxygen bonds of the carbonate ion are partial double bonds, and all are equivalent. All of them should be the same length, and this is exactly what experiments tell us. The bonds are all 1.28 Å long, a distance which is intermediate between that of a carbon–oxygen single bond (1.43 Å) and that of a carbon–oxygen double bond (1.20 Å). One angstrom equals 1 × 10−10 meter. ●

One other important point: by convention, when we draw resonance structures, we connect them by double-headed arrows ( ) to indicate clearly that they are hypothetical, not real. For the carbonate ion we write them this way: O 

O

O

C

O





O

C



O O

O

C



O



We should not let these arrows, or the word “resonance,” mislead us into thinking that the carbonate ion fluctuates between one structure and another. These structures individually do not represent reality and exist only on paper; therefore, the carbonate ion cannot fluctuate among them because it is a hybrid of them. ●

Resonance structures do not represent an equilibrium.

In an equilibrium between two or more species, it is quite correct to think of different structures and moving (or fluctuating) atoms, but not in the case of resonance (as in the carbonate ion). Here the atoms do not move, and the “structures” exist only on paper. An equilibrium is indicated by and resonance by . Note: Each type of arrow in organic chemistry (e.g., , , and ) has a specific meaning. It is important that you use each type of arrow only for the purpose for which it is defined. How can we write the structure of the carbonate ion in a way that will indicate its actual structure? We may do two things: we may write all of the resonance structures as we have just done and let the reader mentally fashion the hybrid, or we may write a non-Lewis structure that attempts to represent the hybrid. For the carbonate ion we might do the following: O 

O



C

Hybrid

O O







O

C

O O





O

C



O O

O

C



O



Contributing resonance structures

The bonds in the structure on the left are indicated by a combination of a solid line and a dashed line. This is to indicate that the bonds are something in between a single bond and a double bond. As a rule, we use a solid line whenever a bond appears in all structures, and a dashed line when a bond exists in one or more but not all. We also place a δ− (read partial minus) beside each oxygen to indicate that something less than a full negative charge resides on each oxygen atom. In this instance, each oxygen atom has two-thirds of a full negative charge.

1.5 | Resonance Theory

1.5A The Use of Curved Arrows: How to Write Resonance Structures As we have mentioned earlier, curved arrows are often used in writing resonance structures, and as we shall see in Section 1.25 they are essential in writing reaction mechanisms. Let us now point out several important things to remember about their use. ● ●

●

●

Curved arrows are used to show the movement of both bonding and unshared electrons. A double-barbed curved arrow ( ) shows the movement of two electrons (an electron pair). [Later, we will see that a single-barbed arrow ( ) can be used to show the movement of a single electron.] A curved arrow should originate precisely at the location of the relevant electrons in the initial formula and point precisely to where those electrons will be drawn in the new formula. A new formula should be drawn to show the result of the electron shift(s). All formulas should be proper Lewis structures and should include formal charges as appropriate. The maximum number of valence electrons should not be exceeded for any atom in a formula.

1.5B Rules for Writing Resonance Structures 1. Resonance structures exist only on paper. Although they have no real existence of their own, resonance structures are useful because they allow us to describe molecules and ions for which a single Lewis structure is inadequate. We write two or more Lewis structures, calling them resonance structures or resonance contributors. We connect these structures by double-headed arrows (↔), and we say that the real molecule or ion is a hybrid of all of them. 2. We are only allowed to move electrons in writing resonance structures. The positions of the nuclei of the atoms must remain the same in all of the structures. Structure 3 is not a resonance structure of 1 or 2, for example, because in order to form it we would have to move a hydrogen atom and this is not permitted: H H

C H

H

H C



C H

1

H

C

H

C H

H

H

H

These are resonance structures.

C

C 2

H C H

H

H

C 

H

H C

C H 3

C

H

H

This is not a resonance structure of 1 or 2. It is an isomer because a hydrogen atom has been moved.

Generally speaking, when we move electrons, we move only those of multiple bonds (as in the example above) and those of nonbonding electron pairs. 3. All of the structures must be proper Lewis structures. We should not write structures in which carbon has five bonds, for example:

H H



C H

O H

This is not a proper resonance structure for methanol because carbon has five bonds. Elements of the first major row of the periodic table cannot have more than eight electrons in their valence shell.

4. The energy of the resonance hybrid is lower than the energy of any contributing structure. Resonance stabilizes a molecule or ion. This is especially true when the resonance structures are equivalent.

13

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Chapter 1 | Basic Principles of Organic Chemistry

Chemists call this stabilization resonance stabilization. If the resonance structures are equivalent, then the resonance stabilization is large. In Chapter 6 we shall find that benzene is highly resonance stabilized because it is a hybrid of the two equivalent forms that follow: or Resonance structures for benzene

Representation of hybrid

5. The more stable a structure is (when taken by itself), the greater is its contribution to the hybrid.

1.5C How Do We Decide When One Resonance Structure Contributes More to the Hybrid Than Another? The following rules will help us: 1. The more covalent bonds a structure has, the more stable it is. Consider the resonance structures for formaldehyde below. (Formaldehyde is a chemical used to preserve biological specimens.) Structure A has more covalent bonds, and therefore makes a larger contribution to the hybrid. In other words, the hybrid is more like structure A than structure B. Four covalent bonds

O H

C

O H

SS

H

C 



H

A

B

More stable

Less stable

Three covalent bonds

Resonance structures for formaldehyde

These structures also illustrate two other considerations: 2. Charge separation decreases stability. It takes energy to separate opposite charges, and therefore a structure with separated charges is less stable. Structure B for formaldehyde has separated plus and minus charges; therefore, on this basis, too, it is the less stable contributor and makes a smaller contribution to the hybrid. 3. Structures in which all the atoms have a complete valence shell of electrons (i.e., the noble gas structure) are more stable. Look again at structure B. The carbon atom has only six electrons around it, whereas in A it has eight. On this basis we can conclude that A is more stable and makes a larger contribution.

1.6

THE STRUCTURE OF METHANE AND ETHANE: sp 3 HYBRIDIZATION

The s and p orbitals used in the quantum mechanical description of the carbon atom, given in Section 1.10, were based on calculations for hydrogen atoms. These simple s and p orbitals do not, when taken alone, provide a satisfactory model for the tetravalent–tetrahedral carbon of methane. However, a satisfactory model of methane’s structure that is based on quantum mechanics can be obtained through an approach called orbital hybridization. Orbital hybridization, in its simplest terms, is nothing more than a mathematical approach that involves the combining of individual wave functions for s and p orbitals to obtain wave functions for new orbitals. The new orbitals have, in varying proportions, the properties of the original orbitals taken separately. These new orbitals are called hybrid atomic orbitals.

1.6 | The Structure of Methane and Ethane: sp3 Hybridization

According to quantum mechanics, the electronic configuration of a carbon atom in its lowest energy state—called the ground state—is that given here: C 1s 2s 2px 2py 2pz Ground state of a carbon atom

The valence electrons of a carbon atom (those used in bonding) are those of the outer level, that is, the 2s and 2p electrons.

1.6A The Structure of Methane Hybrid atomic orbitals that account for the structure of methane can be derived from carbon’s second-shell s and p orbitals as follows (Fig. 1.4): ●

●

●

Wave functions for the 2s, 2px, 2py, and 2pz orbitals of ground state carbon are mixed to form four new and equivalent 2sp3 hybrid orbitals. The designation sp3 signifies that the hybrid orbital has one part s orbital character and three parts p orbital character. The mathematical result is that the four 2sp3 orbitals are oriented at angles of 109.5° with respect to each other. This is precisely the orientation of the four hydrogen atoms of methane. Each H } C } H bond angle is 109.5°.

c (+)

c (+)

c (–) c (–)

c (+)

2s Orbital

c (–) c (+)

2py Orbital 2px Orbital

2pz Orbital Hybridization

109.5°

c (+)

c (+) – 109.5°

– – –

109.5°

Four sp3 hybrid orbitals.

109.5°

c (+)

c (+) 109.5°

Figure 1.4 Hybridization of pure atomic orbitals of a carbon atom to produce sp3 hybrid orbitals. If, in our imagination, we visualize the hypothetical formation of methane from an sp3-hybridized carbon atom and four hydrogen atoms, the process might be like that shown in Fig. 1.5. For simplicity

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Chapter 1 | Basic Principles of Organic Chemistry

H H H

109.5°

H

C

C H

H H

H

Methane, CH4

Figure 1.5 The hypothetical formation of methane from an sp3-hybridized carbon atom and four hydrogen atoms. In orbital hybridization we combine orbitals, not electrons. The electrons can then be placed in the hybrid orbitals as necessary for bond formation, but always in accordance with the Pauli principle of no more than two electrons (with opposite spin) in each orbital. In this illustration we have placed one electron in each of the hybrid carbon orbitals. In addition, we have shown only the bonding molecular orbital of each C } H bond because these are the orbitals that contain the electrons in the lowest energy state of the molecule. we show only the formation of the bonding molecular orbital for each carbon–hydrogen bond. We see that an sp3-hybridized carbon gives a tetrahedral structure for methane, and one with four equivalent C } H bonds. In addition to accounting properly for the shape of methane, the orbital hybridization model also explains the very strong bonds that ψ (–) C ψ (+) are formed between carbon and hydrogen. To see how this is so, con3 sider the shape of an individual sp orbital shown in Fig. 1.6. Because an sp3 orbital has the character of a p orbital, the positive lobe of an Figure 1.6 The shape of an sp3 orbital is large and extends relatively far from the carbon nucleus. sp3 orbital. It is the positive lobe of an sp3 orbital that overlaps with the positive 1s orbital of hydrogen to form the bonding molecular orbital of a carbon–hydrogen bond (Fig. 1.7). Because the positive lobe of the sp3 orbital is large and is extended into space, the overlap between it and the 1s orbital of hydrogen is also large, and the resulting carbon–hydrogen bond is quite strong.

c (–) C

c (+)

sp 3 Orbital

c (+) H 1s Orbital

c (–) C c (+) H Carbon–hydrogen bond (bonding MO)

Figure 1.7 Formation of a C } H bond. The bond formed from the overlap of an sp3 orbital and a 1s orbital is an example of a sigma (s) bond (Fig. 1.8). ●

●

A sigma (s) bond has a circularly symmetrical orbital cross section when viewed along the bond between two atoms. All purely single bonds are sigma bonds.

From this point on we shall often show only the bonding molecular orbitals because they are the ones that contain the electrons when the molecule is in its lowest energy state. Consideration of antibonding orbitals is important when a molecule absorbs light and in explaining certain reactions. We shall point out these instances later.

1.6 | The Structure of Methane and Ethane: sp3 Hybridization Circular cross section

Bond axis

H

C

C or

H

Bond axis

Figure 1.8 A s (sigma) bond.

1.6B The Structure of Ethane The bond angles at the carbon atoms of ethane, and of all alkanes, are also tetrahedral like those in methane. A satisfactory model for ethane can be provided by sp3-hybridized carbon atoms. Figure 1.9 shows how we might imagine the bonding molecular orbitals of an ethane molecule being constructed from two sp3-hybridized carbon atoms and six hydrogen atoms. The carbon–carbon bond of ethane is a sigma bond with cylindrical symmetry, formed by two overlapping sp3 orbitals. (The carbon–hydrogen bonds are also sigma bonds. They are formed from overlapping carbon sp3 orbitals and hydrogen s orbitals.) ●

Rotation of groups joined by a single bond does not usually require a large amount of energy. Consequently, groups joined by single bonds rotate relatively freely with respect to one another.

+

C

+

C

sp 3 Carbon

6

H

sp 3 Carbon

H

H Sigma bonds

H

H

C

C A sigma bond H

H

Figure 1.9 The hypothetical formation of the bonding molecular orbitals of ethane from two sp3hybridized carbon atoms and six hydrogen atoms. All of the bonds are sigma bonds. (Antibonding sigma molecular orbitals—called s* orbitals—are formed in each instance as well, but for simplicity these are not shown.)

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Chapter 1 | Basic Principles of Organic Chemistry

THE STRUCTURE OF ETHENE (ETHYLENE): sp 2 HYBRIDIZATION

1.7

The carbon atoms of many of the molecules that we have considered so far have used their four valence electrons to form four single covalent (sigma) bonds to four other atoms. We find, however, that many important organic compounds exist in which carbon atoms share more than two electrons with another atom. In molecules of these compounds some bonds that are formed are multiple covalent bonds. When two carbon atoms share two pairs of electrons, for example, the result is a carbon–carbon double bond: C

C

or

C

C

Hydrocarbons whose molecules contain a carbon–carbon double bond are called alkenes. Ethene (C2H4) and propene (C3H6) are both alkenes. Ethene is also called ethylene, and propene is sometimes called propylene. H

H C

H

C

H

H C

C

H3C

H Ethene

H

Propene

In ethene the only carbon–carbon bond is a double bond. Propene has one carbon–carbon single bond and one carbon–carbon double bond. The spatial arrangement of the atoms of alkenes is different from that of alkanes. The six atoms of ethene are coplanar, and the arrangement of atoms around each carbon atom is triangular (Fig. 1.10). ●

Carbon–carbon double bonds are comprised of sp2-hybridized carbon atoms.

The mathematical mixing of orbitals that furnish the sp2 orbitals for our model can be visualized in the way shown in Fig. 1.11. The 2s orbital is mathematically mixed (or hybridized) with two of the 2p orbitals. (The hybridization procedure applies only to the orbitals, not to the electrons.) One 2p orbital is left unhybridized. One electron is then placed in each of the sp2 hybrid orbitals and one electron remains in the 2p orbital.

H

H C

H

C

~121°

2p

Figure 1.10 The structure and bond angles of ethene. The plane of the atoms is perpendicular to the paper. The dashed wedge bonds project behind the plane of the paper, and the solid wedge bonds project in front of the paper.

~118°

H

Ground state

sp2-Hybridized state

Excited state 2p

2p 2sp2

Energy

18

2s

2s

1s

1s Promotion of electron

1s Hybridization

Figure 1.11 A process for deriving sp2-hybridized carbon atoms.

1.7 | The Structure of Ethene (Ethylene): sp2 Hybridization

The three sp2 orbitals that result from hybridization are directed toward the corners of a regular triangle (with angles of 120° between them). The carbon p orbital that is not hybridized is perpendicular to the plane of the triangle formed by the hybrid sp2 orbitals (Fig. 1.12). In our model for ethene (Fig. 1.13) we see the following: ●

●

y sp2 Orbital sp2 Orbital p Orbital

z

x

Two sp -hybridized carbon atoms form a sigma (s) bond sp2 Orbital between them by overlap of one sp2 orbital from each carbon. The remaining carbon sp2 orbitals form s bonds to four hydrogens through overlap with the hydrogen 1s orbitals. These five s bonds account for 10 of the Figure 1.12 An sp2-hybridized 12 valence electrons contributed by  the two carbons and carbon atom. four hydrogens, and comprise the s-bond framework of the molecule. The remaining two bonding electrons are each located in an unhybridized p orbital of each carbon. Sideways overlap of these p orbitals and sharing of the two electrons between the carbons leads to a pi (p) bond. The overlap of these orbitals is shown schematically in Fig. 1.14. 2

The bond angles that we would predict on the basis of sp2-hybridized carbon atoms (120° all around) are quite close to the bond angles that are actually found (Fig. 1.10). p Orbitals

Overlap

H

H

s Bonds

H

C

s Bond overlap

C

s Bonds

H

Figure 1.13 A model for the bonding molecular orbitals of ethene formed from two sp2-hybridized carbon atoms and four hydrogen atoms. We can better visualize how these p orbitals interact with each other if we view a structure showing calculated molecular orbitals for ethene (Fig. 1.14). We see that the parallel p orbitals overlap above and below the plane of the s framework.

H H

C

C

Bond

H H

Figure 1.14 A wedge–dashed wedge formula for the sigma bonds in ethene and a schematic depiction of the overlapping of adjacent p orbitals that form the p bond.

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Chapter 1 | Basic Principles of Organic Chemistry

Note the difference in shape of the bonding molecular orbital of a p bond as contrasted to that of a s bond. A s bond has cylindrical symmetry about a line connecting the two bonded nuclei. A p bond has a nodal plane passing through the two bonded nuclei and between the p molecular orbital lobes. ●

●

When two p atomic orbitals combine to form a p bond, two p molecular orbital molecular orbitals form: one is a bonding molecular orbital and the other is an antibonding molecular orbital. To summarize, a carbon–carbon double bond consists of one s bond and one p bond.

The s bond results from two sp2 orbitals overlapping end to end and is symmetrical about an axis linking the two carbon atoms. The p bond results from a sideways overlap of two p orbitals; it has a nodal plane like a p orbital. In the ground state the electrons of the p bond are located between the two carbon atoms but generally above and below the plane of the s bond framework.

1.7A Restricted Rotation and the Double Bond The s–p model for the carbon–carbon double bond also accounts for an important property of the double bond: ●

There is a large energy barrier to rotation associated with groups joined by a double bond.

Maximum overlap between the p orbitals of a p bond occurs when the axes of the p orbitals are exactly parallel. Rotating one carbon of the double bond 90° (Fig. 1.15) breaks the p bond, for then the axes of the p orbitals are perpendicular and there is no net overlap between them. Estimates based on thermochemical calculations indicate that the strength of the p bond is 264 kJ mol−1. This, then, is the barrier to rotation of the double bond. It is markedly higher than the rotational barrier of groups joined by carbon– carbon single bonds (13–26 kJ mol−1). While groups joined by single bonds rotate relatively freely at room temperature, those joined by double bonds do not.

C

C

C

Rotate 90º

C

Figure 1.15 A stylized depiction of how rotation of a carbon atom of a double bond through an angle of 90° results in breaking of the p bond.

1.7B Cis–Trans Isomerism Restricted rotation of groups joined by a double bond causes a new type of isomerism that we illustrate with the two dichloroethenes written as the following structures: Cl

Cl C

H

H C

H

cis-1,2-Dichloroethene ●

Cl

C H

C Cl

trans-1,2-Dichloroethene

These two compounds are isomers; they are different compounds that have the same molecular formula.

We can tell that they are different compounds by trying to place a model of one compound on a model of the other so that all parts coincide, that is, to try to superpose one on the other. We find that it cannot be done. Had one been superposable on the other, all parts of one model would correspond in three

21

1.8 | The Structure of Ethyne (Acetylene): sp Hybridization

dimensions exactly with the other model. (The notion of superposition is different from simply superimposing one thing on another. The latter means only to lay one on the other without the necessary condition that all parts coincide.) ●

We indicate that they are different isomers by attaching the prefix cis or trans to their names (cis, Latin: on this side; trans, Latin: across).

cis-1,2-Dichloroethene and trans-1,2-dichloroethene are not constitutional isomers because the connectivity of the atoms is the same in each. The two compounds differ only in the arrangement of their atoms in space. Isomers of this kind are classified formally as stereoisomers, but often they are called simply cis–trans isomers. (We shall study stereoisomerism in detail in Chapters 4 and 5.) The structural requirements for cis–trans isomerism will become clear if we consider a few additional examples. 1,1-Dichloroethene and 1,1,2-trichloroethene do not show this type of isomerism. Cl C

H

Cl

H

Cl

Cl

C

Cl

C

1,1-Dichloroethene (no cis-trans isomerism)

C H

1,1,2-Trichloroethene (no cis-trans isomerism)

1,2-Difluoroethene and 1,2-dichloro-1,2-difluoroethene do exist as cis–trans isomers. Notice that we designate the isomer with two identical groups on the same side as being cis: F

F C

F

C

H

C H

F

trans-1,2-Difluoroethene

F

F C

C

H

cis-1,2-Difluoroethene

F

H

C

C

Cl

Cl C

Cl

Cl

cis-1,2-Dichloro-1,2-difluoroethene

F

trans-1,2-Dichloro-1,2-difluoroethene

Clearly, then, cis–trans isomerism of this type is not possible if one carbon atom of the double bond bears two identical groups.

1.8

THE STRUCTURE OF ETHYNE (ACETYLENE): sp HYBRIDIZATION

Hydrocarbons in which two carbon atoms share three pairs of electrons between them, and are thus bonded by a triple bond, are called alkynes. The two simplest alkynes are ethyne and propyne. H

C

C

H

CH3

C

Ethyne (acetylene) (C2H2)

C

H

Propyne (C3H4)

Ethyne, a compound that is also called acetylene, consists of a linear arrangement of atoms. The H } C bond angles of ethyne molecules are 180°: H

C

C

180° 180°

H

C

Chapter 1 | Basic Principles of Organic Chemistry Ground state

Excited state

sp-Hybridized state 2p

2p

2p Energy

22

2sp 2s

2s

1s

1s

1s

Promotion of electron

Hybridization

Figure 1.16 A process for deriving sp-hybridized carbon atoms. We can account for the structure of ethyne on the basis of orbital hybridization as we did for ethane and ethene. In our model for ethane (Section 1.12B) we saw that the carbon orbitals are sp3 hybridized, and in our model for ethene (Section 1.13) we saw that they are sp2 hybridized. In our model for ethyne we shall see that the carbon atoms are sp hybridized. The mathematical process for obtaining the sp hybrid orbitals of ethyne can be visualized in the following way (Fig. 1.16). ● ●

The 2s orbital and one 2p orbital of carbon are hybridized to form two sp orbitals. The remaining two 2p orbitals are not hybridized.

Calculations show that the sp hybrid orbitals have their large positive lobes oriented at an angle of 180° with respect to each other. The two 2p orbitals that were not hybridized are each perpendicular to the axis that passes through the center of the two sp orbitals (Fig. 1.17). We place one electron in each orbital. We envision the bonding molecular orbitals of ethyne being formed in the following way (Fig. 1.18). ●

Two carbon atoms overlap sp orbitals to form a sigma bond between them (this is one bond of the triple bond). The remaining two sp orbitals at each carbon atom overlap with s orbitals from hydrogen atoms to produce two sigma C } H bonds.

p Orbitals

sp Orbital

C

sp Orbital

Figure 1.17 An sp-hybridized carbon atom.

Bond

Bond

H C

C H

Bond

Figure 1.18 Formation of the bonding molecular orbitals of ethyne from two sp-hybridized carbon atoms and two hydrogen atoms. (Antibonding orbitals are formed as well, but these have been omitted for simplicity.)

1.8 | The Structure of Ethyne (Acetylene): sp Hybridization

H

C

C

Figure 1.19 (a) The structure of ethyne (acetylene) showing the sigma-bond framework and a schematic depiction of the two pairs of p orbitals that overlap to form the two p bonds in ethyne. (b) A structure of ethyne showing calculated p molecular orbitals. Two pairs of p molecular orbital lobes are present, one pair for each p bond. The hydrogen atoms of ethyne (white spheres) can be seen at each end of the structure (the carbon atoms are hidden by the molecular orbitals).

H

π Bond (a)

●

●

(b)

The two p orbitals on each carbon atom also overlap side to side to form two p bonds. These are the other two bonds of the triple bond. The carbon–carbon triple bond consists of two p bonds and one s bond.

Structures for ethyne based on calculated molecular orbitals and electron density are shown in Fig. 1.19. Circular symmetry exists along the length of a triple bond (Fig. 1.19b). As a result, there is no restriction of rotation for groups joined by a triple bond (as compared with alkenes), and if rotation would occur, no new compound would form.

1.8A Bond Lengths of Ethyne, Ethene, and Ethane The carbon–carbon triple bond of ethyne is shorter than the carbon–carbon double bond of ethene, which in turn is shorter than the carbon–carbon single bond of ethane. The reason is that bond lengths are affected by the hybridization states of the carbon atoms involved. ●

●

The greater the s orbital character in one or both atoms, the shorter is the bond. This is because s orbitals are spherical and have more electron density closer to the nucleus than do p orbitals. The greater the p orbital character in one or both atoms, the longer is the bond. This is because p orbitals are lobe-shaped with electron density extending away from the nucleus.

In terms of hybrid orbitals, an sp hybrid orbital has 50% s character and 50% p character. An sp2 hybrid orbital has 33% s character and 67% p character. An sp3 hybrid orbital has 25% s character and 75% p character. The overall trend, therefore, is as follows: ●

Bonds involving sp hybrids are shorter than those involving sp2 hybrids, which are shorter than those involving sp3 hybrids. This trend holds true for both C } C and C } H bonds.

The bond lengths and bond angles of ethyne, ethene, and ethane are summarized in Fig. 1.20.

H H C

1188

H

H

1808

H

C

109.58

1218

1.20 Å

1.10 Å 1.09 Å

1.06 Å H

1.54 Å

1.34 Å

C

H

H

C

C

C

H H

Figure 1.20 Bond angles and bond lengths of ethyne, ethene, and ethane.

H H

23

24

Chapter 1 | Basic Principles of Organic Chemistry

1.8B A Summary of Important Concepts 1. The energy of electrons in a bonding molecular orbital is less than the energy of the electrons in their separate atomic orbitals. The energy of electrons in an antibonding orbital is greater than that of electrons in their separate atomic orbitals. 2. The number of molecular orbitals always equals the number of atomic orbitals from which they are formed. Combining two atomic orbitals will always yield two molecular orbitals—one bonding and one antibonding. 3. Hybrid atomic orbitals are obtained by mixing (hybridizing) the wave functions for orbitals of different types (i.e., s and p orbitals) but from the same atom. 4. Hybridizing three p orbitals with one s orbital yields four sp3 orbitals. Atoms that are sp3 hybridized direct the axes of their four sp3 orbitals toward the corners of a tetrahedron. The carbon of methane is sp3 hybridized and tetrahedral. 5. Hybridizing two p orbitals with one s orbital yields three sp2 orbitals. Atoms that are sp2 hybridized point the axes of their three sp2 orbitals toward the corners of an equilateral triangle. The carbon atoms of ethene are sp2 hybridized and trigonal planar. 6. Hybridizing one p orbital with one s orbital yields two sp orbitals. Atoms that are sp hybridized orient the axes of their two sp orbitals in opposite directions (at an angle of 180°). The carbon atoms of ethyne are sp hybridized and ethyne is a linear molecule. 7. A sigma (s) bond (a type of single bond) is one in which the electron density has circular symmetry when viewed along the bond axis. In general, the skeletons of organic molecules are constructed of atoms linked by sigma bonds. 8. A pi (p) bond, part of double and triple carbon–carbon bonds, is one in which the electron densities of two adjacent parallel p orbitals overlap sideways to form a bonding pi molecular orbital.

1.9

HOW TO PREDICT MOLECULAR GEOMETRY: THE VALENCE SHELL ELECTRON PAIR REPULSION MODEL

We can predict the arrangement of atoms in molecules and ions on the basis of a relatively simple idea called the valence shell electron pair repulsion (VSEPR) model. We apply the VSEPR model in the following way: 1. We consider molecules (or ions) in which the central atom is covalently bonded to two or more atoms or groups. 2. We consider all of the valence electron pairs of the central atom—both those that are shared in covalent bonds, called bonding pairs, and those that are unshared, called nonbonding pairs or unshared pairs or lone pairs. 3. Because electron pairs repel each other, the electron pairs of the valence H shell tend to stay as far apart as possible. The repulsion between nonbonding pairs is generally greater than that between bonding pairs. 4. We arrive at the geometry of the molecule by considering all of the electron pairs, bonding and nonbonding, but we describe the shape of C the molecule or ion by referring to the positions of the nuclei (or atoms) H H and not by the positions of the electron pairs. H

1.9A Methane The valence shell of methane contains four pairs of bonding electrons. Only a tetrahedral orientation will allow four pairs of electrons to have equal and maximum possible separation from each other (Fig. 1.21). Any other

Figure 1.21 A tetrahedral shape for methane allows the maximum separation of the four bonding electron pairs.

1.9 | How to Predict Molecular Geometry: The Valence Shell Electron Pair Repulsion Model

orientation, for example, a square planar arrangement, places some electron pairs closer together than others. Thus, methane has a tetrahedral shape. The bond angles for any atom that has a regular tetrahedral structure are 109.5°. A representation of these angles in methane is shown in Fig. 1.22.

H

1.9B Ammonia

H

The shape of a molecule of ammonia (NH3) is a trigonal pyramid. There are three bonding pairs of electrons and one nonbonding pair. The bond angles in a molecule of ammonia are 107°, a value very close to the tetrahedral angle (109.5°). We can write a general tetrahedral structure for the electron pairs of ammonia by placing the nonbonding pair at one corner (Fig. 1.23). A tetrahedral arrangement of the electron pairs explains the trigonal pyramidal arrangement of the four atoms. The bond angles are 107° (not 109.5°) because the nonbonding pair occupies more space than the bonding pairs.

C

109.5° 109.5° H

109.5°

H 109.5°

Figure 1.22 The bond angles of methane are 109.5°.

1.9C Water A molecule of water has an angular or bent shape. The H } O } H bond angle in a molecule of water is 104.5°, an angle that is also quite close to the 109.5° bond angles of methane. We can write a general tetrahedral structure for the electron pairs of a molecule of water if we place the two bonding pairs of electrons and the two nonbonding electron pairs at the corners of the tetrahedron. Such a structure is shown in Fig. 1.24. A tetrahedral arrangement of the electron pairs accounts for the angular arrangement of the three atoms. The bond angle is less than 109.5° because the nonbonding pairs are effectively “larger” than the bonding pairs and, therefore, the structure is not perfectly tetrahedral.

1.9D Carbon Dioxide The VSEPR method can also be used to predict the shapes of molecules containing multiple bonds if we assume that all of the electrons of a multiple bond act as though they were a single unit and, therefore, are located in the region of space between the two atoms joined by a multiple bond.

N H

H

107°

H

O H H

107°

N

H

H

H

Figure 1.23 The tetrahedral arrangement of the electron pairs of an ammonia molecule that results when the nonbonding electron pair is considered to occupy one corner. This arrangement of electron pairs explains the trigonal pyramidal shape of the NH3 molecule.

107°

O H

H 105°

25

Figure 1.24 An approximately tetrahedral arrangement of the electron pairs of a molecule of water that results when the pairs of nonbonding electrons are considered to occupy corners. This arrangement accounts for the angular shape of the H2O molecule.

26

Chapter 1 | Basic Principles of Organic Chemistry

This principle can be illustrated with the structure of a molecule of carbon dioxide (CO2). The central carbon atom of carbon dioxide is bonded to each oxygen atom by a double bond. Carbon dioxide is known to have a linear shape; the bond angle is 180°. 180° C O O

or

O

C

O

The four electrons of each double bond act as a single unit and are maximally separated from each other.

Such a structure is consistent with a maximum separation of the two groups of four bonding electrons. The nonbonding pairs associated with the oxygen atoms have no effect on the shape. The shapes of several simple molecules and ions as predicted by VSEPR theory are shown in Table 1.2. In this table we have also included the hybridization state of the central atom. Table 1.2 Shapes of molecules and ions from VSEPR theory Number of Electron Pairs at Central Atom Bonding

Nonbonding

Total

Hybridization State of Central Atom

2

0

2

sp

Linear

BeH2

3

0

3

sp

Trigonal planar

BF3, CH3

4

0

4

sp3

Tetrahedral

CH4, NH4

3

1

4

∼sp3

Trigonal pyramidal

NH3, CH3

2

2

4

∼sp3

Angular

H2O

2

Shape of Molecule or Iona

Examples +

+

−

Referring to positions of atoms and excluding nonbonding pairs.

a

PART II: FAMILIES OF CARBON COMPOUNDS 1.10 CLASSIFICATION OF ORGANIC COMPOUNDS The need to classify organic compounds arises from the fact that a large number of organic compounds exist and their number is constantly increasing. Their classification is done on the basis of structure and is facilitated by close relation between structure and properties. The broad classification of organic compounds can be depicted as shown in Fig. 1.25. Organic compounds

Acyclic (Open chain or aliphatic)

Cyclic (Closed chain or ring compounds)

Carbocyclic or Homocyclic

Alicyclic

Benzenoid compounds

Aromatic

Non-benzenoid compounds

Figure 1.25 Classification of organic compounds.

Heterocyclic

Non-aromatic

Aromatic

1.10 | Classification of Organic Compounds

1.10A Acyclic or Open Chain Compounds Acyclic or aliphatic compounds consist of continuous sequence of carbon atoms called a straight chain or branched chain. The straight chain means only that no carbon atom is bonded to more than two other carbons.

Straight chain

Branched chain

An organic compound in whose molecules the carbon atoms do not all occur one after another in a continuous sequence are called branched chains. Branched chains are also very common. Isooctane, for example, has a main chain of five carbon atoms (in black) carrying three CH3 branches. CH3 CH3 CH3CCH2CHCH3 CH3 Isooctane

1.10B Alicyclic or Closed-Chain or Ring Compounds Homocyclic compounds are those in which the carbon atoms are joined in the form of a ring. For example, H2C

H2C

CH2 5

C H2 Cyclopropane

CH2 CH2 5

H2C

C H2 Cyclopentane

A compound whose molecules have a ring containing an element other than carbon are called heterocyclic compounds. These compounds exhibit properties similar to aliphatic compounds.

1.10C Aromatic Compounds Benzene and all substances with structures and chemical properties resembling benzene are classified as aromatic compounds. The word aromatic originally referred to the rather pleasant odor many of these substances possess, but this meaning has been dropped. Like aliphatic compounds, aromatic compounds can have a hetero atom (other than carbon) in the ring and these are known as heterocyclic aromatic compounds. Benzenoid Aromatic Compounds Representatives of class benzenoid aromatic compounds are benzene and its derivatives as well as polycyclic aromatic hydrocarbons (PAH), illustrated as follows. CH3

NO2

8

1

7 6 5

Benzene

Toluene

Nitrobenzene

4

Naphthalene C10H8

8 2

7

3

6

9

1 2 3

5

10

4

Anthracene C14H10

27

28

Chapter 1 | Basic Principles of Organic Chemistry

Benzenoid polycyclic aromatic hydrocarbons consist of molecules having two or more benzene rings fused together. Non-Benzenoid Compounds The cyclopentadienyl anion, the cycloheptatrienyl cation, trans-15, 16-dimethyldihydropyrene and azulene, which do not contain benzene ring are classified as non-benzenoid aromatic compounds.

Azulene

Heterocyclic Aromatic Compounds Heterocyclic aromatic compounds containing nitrogen, oxygen, or sulphur are by far the most common. The four important examples are as follows: 4 5 6

N 1

3

4

2

5

Pyridine

N1

3

4

2

5

H Pyrrole

O 1 Furan

3

4

2

5

3 S 1

2

Thiophene

1.10D Homologous Series In the unbranched alkanes, each alkane differs from the preceding alkane by one } CH2} group. For example, butane is CH3(CH2)2CH3 and pentane is CH3(CH2)3CH3. A series of compounds like this, where each member differs from the next member by a constant unit (} CH2), is called a homologous series. This characteristic is also observed in alkenes, alkynes, haloalkanes, amines, etc. Members of a homologous series containing a characteristic functional group are called homologues. Some characteristic features of homologous series are: 1. Each homologous series can be represented by a general formula and a CH2 group gets added in the successive formulas. For example, the general formula for alkenes is CnH2n and the first three members are C2H4, C3H6, C4H8. 2. All the members of the same homologous series have the same functional group. For example, all monohydric alcohols can be represented by CnH2n+1OH. 3. The members of a homologous series can be prepared by same general methods. 4. The physical properties of the members of a homologous series show a regular trend with increasing molecular mass. 5. The chemical properties of members of a homologous series are similar but may vary somewhat in the first few members of the series.

1.11 ●

HYDROCARBONS: REPRESENTATIVE ALKANES, ALKENES, ALKYNES, AND AROMATIC COMPOUNDS

Hydrocarbons are compounds that contain only carbon and hydrogen atoms.

Methane (CH4) and ethane (C2H6) are hydrocarbons, for example. They also belong to a subgroup of compounds called alkanes. ●

Alkanes are hydrocarbons that do not have multiple bonds between carbon atoms, and we can indicate this in the family name and in names for specific compounds by the -ane ending.

Propane (an alkane)

Other hydrocarbons may contain double or triple bonds between their carbon atoms. ●

Alkenes contain at least one carbon–carbon double bond, and this is indicated in the family name and in names for specific compounds by an -ene ending.

Propene (an alkene)

1.11 | Hydrocarbons: Representative Alkanes, Alkenes, Alkynes, and Aromatic Compounds ●

●

Alkynes contain at least one carbon–carbon triple bond, and this is indicated in the family name and in names for specific compounds by an -yne ending. Aromatic compounds contain a special type of ring, the most common example of which is a benzene ring. There is no special ending for the general family of aromatic compounds.

Propyne (an alkyne)

Benzene (an aromatic compound)

We shall introduce representative examples of each of these classes of hydrocarbons in the following sections. Generally speaking, compounds such as alkanes, whose molecules contain only single bonds, are referred to as saturated compounds because these compounds contain the maximum number of hydrogen atoms that the carbon compound can possess. Compounds with multiple bonds, such as alkenes, alkynes, and aromatic hydrocarbons, are called unsaturated compounds because they possess fewer than the maximum number of hydrogen atoms, and they are capable of reacting with hydrogen under the proper conditions.

1.11A Alkanes The primary sources of alkanes are natural gas and petroleum. The smaller alkanes (methane through butane) are gases under ambient conditions. Methane is the principal component of natural gas. Higher molecular weight alkanes are obtained largely by refining petroleum. Methane, the simplest alkane, was one major component of the early atmosphere of this planet. Methane is still found in Earth’s atmosphere, but no longer in appreciable amounts. It is, however, a major component of Methane the atmospheres of Jupiter, Saturn, Uranus, and Neptune. Some living organisms produce methane from carbon dioxide and hydrogen. These very primitive creatures, called methanogens, may be Earth’s oldest organisms, and they may represent a separate form of evolutionary development. Methanogens can survive only in an anaerobic (i.e., oxygen-free) environment. They have been found in ocean trenches, in mud, in sewage, and in cow’s stomachs.

1.11B Alkenes Ethene and propene, the two simplest alkenes, are among the most important industrial chemicals. Ethene is used as a starting material for the synthesis of many industrial compounds, including ethanol, ethylene oxide, ethanal, and the polymer polyethylene. Propene is used in making the polymer polypropylene, and, in addition to other uses, propene is the starting material for a synthesis of acetone and cumene. Ethene also occurs in nature as a plant hormone. It is produced naturally by fruits such as tomatoes and bananas and is involved in the ripening process of these fruits. Much use is now made of ethene in the commercial fruit industry to bring about the ripening of tomatoes and bananas picked green because the green fruits are less susceptible to damage during shipping.

Ethene

1.11C Alkynes The simplest alkyne is ethyne (also called acetylene). Alkynes occur in nature and can be synthesized in the laboratory. Two examples of alkynes among thousands that have a biosynthetic origin are capillin, an antifungal agent, and dactylyne, a marine natural product that is an inhibitor of pentobarbital metabolism. Ethinyl estradiol is a synthetic alkyne whose estrogen-like properties have found use in oral contraceptives.

Ethyne

29

30

Chapter 1 | Basic Principles of Organic Chemistry

1.11D Benzene: A Representative Aromatic Hydrocarbon In Chapter 6 we shall study in detail a group of unsaturated cyclic hydrocarbons known as aromatic compounds. The compound known as benzene is the prototypical aromatic compound. Benzene can be written as a six-membered ring with alternating single and double bonds, called a Kekulé structure after August Kekulé, who first conceived of this representation: H H

H

C C

C

C

C C

H Benzene

or H

H Kekulé structure for benzene

Bond-line representation of Kekulé structure

Even though the Kekulé structure is frequently used for benzene compounds, there is much evidence that this representation is inadequate and incorrect. For example, if benzene had alternating single and double bonds as the Kekulé structure indicates, we would expect the lengths of the carbon–carbon bonds around the ring to be alternately longer and shorter, as we typically find with carbon–carbon single and double bonds. In fact, the carbon–carbon bonds of benzene are all the same length (1.39 Å), a value in between that of a carbon–carbon single bond and a carbon–carbon double bond. There are two ways of dealing with this problem: with resonance theory or with molecular orbital theory. If we use resonance theory, we visualize benzene as being represented by either of two equivalent Kekulé structures:

Two contributing Kekulé structures for benzene

A representation of the resonance hybrid

Based on the principles of resonance theory we recognize that benzene cannot be represented adequately by either structure, but that, instead, it should be visualized as a hybrid of the two structures. We represent this hybrid by a hexagon with a circle in the middle. Resonance theory, therefore, solves the problem we encountered in understanding how all of the carbon–carbon bonds are the same length. According to resonance theory, the bonds are not alternating single and double bonds, they are a resonance hybrid of the two. Any bond that is a single bond in the first contributor is a double bond in the second, and vice versa. ●

All of the carbon–carbon bonds in benzene are one and one-half bonds, have a bond length in between that of a single bond and a double bond, and have bond angles of 1208.

In the molecular orbital explanation, which we shall describe in much more depth in Chapter 6, we begin by recognizing that the carbon atoms of the benzene ring are sp2 hybridized. Therefore, each carbon has a p orbital that has one lobe above the plane of the ring and one lobe below, as shown on the next page in the schematic and calculated p orbital representations. H

H C

H

C

1.09 Å

H

C 1208

C

1.39 Å

C

1208

C 1208

H

H

1.12 | Functional Groups

H

H

H

H

H

H Calculated p orbital shapes in benzene

Schematic representation of benzene p orbitals

Calculated benzene molecular orbital resulting from favorable overlap of p orbitals above and below plane of benzene ring

The lobes of each p orbital above and below the ring overlap with the lobes of p orbitals on the atoms to either side of it. This kind of overlap of p orbitals leads to a set of bonding molecular orbitals that encompass all of the carbon atoms of the ring, as shown in the calculated molecular orbital. Therefore, the six electrons associated with these p orbitals (one electron from each orbital) are delocalized about all six carbon atoms of the ring. This delocalization of electrons explains how all the carbon–carbon bonds are equivalent and have the same length.

1.12 FUNCTIONAL GROUPS ●

Functional groups are common and specific arrangements of atoms that impart predictable reactivity and properties to a molecule.

The functional group of an alkene, for example, is its carbon–carbon double bond. When we study the reactions of alkenes in greater detail in Chapter 8, we shall find that most of the chemical reactions of alkenes are the chemical reactions of the carbon–carbon double bond. The functional group of an alkyne is its carbon–carbon triple bond. Alkanes do not have a functional group. Their molecules have carbon–carbon single bonds and carbon–hydrogen bonds, but these bonds are present in molecules of almost all organic compounds, and C } C and C } H bonds are, in general, much less reactive than common functional groups. We shall introduce other common functional groups and their properties in Sections 1.12–1.19. Table 1.3 (Section 1.20) summarizes the most important functional groups. First, however, let us introduce some common alkyl groups, which are specific groups of carbon and hydrogen atoms that are not part of functional groups.

1.12A Alkyl Groups and the Symbol R Alkyl groups are the groups that we identify for purposes of naming compounds. They are groups that would be obtained by removing a hydrogen atom from an alkane: Alkane

Alkyl Group

Abbreviation

CH3 — H

H3C —

Me-

Methane

Methyl

CH3CH2 — H

CH3CH2 —

Ethane

Ethyl

CH3CH2CH2 — H

CH3CH2CH2 —

Propane

Propyl

CH3CH2CH2CH2 — H

CH3CH2CH2CH2 —

Butane

Butyl

Et-

Pr-

Bu-

Bond-line

31

32

Chapter 1 | Basic Principles of Organic Chemistry

While only one alkyl group can be derived from methane or ethane (the methyl and ethyl groups, respectively), two groups can be derived from propane. Removal of a hydrogen from one of the end carbon atoms gives a group that is called the propyl group; removal of a hydrogen from the middle carbon atom gives a group that is called the isopropyl group. The names and structures of branched alkyl groups derived from butane and other hydrocarbons are discussed in Section 1.21. We can simplify much of our future discussion if, at this point, we introduce a symbol that is widely used in designating general structures of organic molecules: the symbol R. R is used as a general symbol to represent any alkyl group. For example, R might be a methyl group, an ethyl group, a propyl group, or an isopropyl group: CH3 CH3CH2 CH3CH2CH2 CH3CHCH3

Methyl Ethyl Propyl Isopropyl

These and others can be designated by R.

Thus, the general formula for an alkane is R } H.

1.12B Phenyl and Benzyl Groups When a benzene ring is attached to some other group of atoms in a molecule, it is called a phenyl group, and it is represented in several ways: or or f

or Ar

or C6H5

or Ph

(if ring substituents are present)

Ways of representing a phenyl group

The combination of a phenyl group and a methylene group (} CH2} ) is called a benzyl group: CH2 or

or C6H5CH2

or Bn

Ways of representing a benzyl group

1.13 ALKYL HALIDES OR HALOALKANES Alkyl halides are compounds in which a halogen atom (fluorine, chlorine, bromine, or iodine) replaces a hydrogen atom of an alkane. For example, CH3Cl and CH3CH2Br are alkyl halides. Alkyl halides are also called haloalkanes. The generic formula for an alkyl halide is R X where X = fluorine, chlorine, bromine, or iodine. Alkyl halides are classified as being primary (1°), secondary (2°), or tertiary (3°). This classification is based on the carbon atom to which the halogen is directly attached. If the carbon atom that bears the halogen is directly attached to only one other carbon, the carbon atom is said to be a primary carbon atom and the alkyl halide is classified as a primary alkyl halide. If the carbon that bears the halogen is itself directly attached to two other carbon atoms, then the carbon is a secondary carbon and the alkyl halide is a secondary alkyl halide. If the carbon that bears the halogen is directly attached to three other

1.14 | Alcohols and Phenols

carbon atoms, then the carbon is a tertiary carbon and the alkyl halide is a tertiary alkyl halide. Examples of primary, secondary, and tertiary alkyl halides are the following: 18 Carbon

H

H

H

C

C

H

Cl

or

28 Carbon

Cl

H

H

H

H

C

C

C

H

H

Cl

38 Carbon

CH3 H

or

H

A 18 alkyl chloride

CH3

Cl

C

Cl or

CH3

A 28 alkyl chloride

Cl

A 38 alkyl chloride

An alkenyl halide is a compound with a halogen atom bonded to an alkene carbon. In older nomenclature such compounds were sometimes referred to as vinyl halides. An aryl halide is a compound with a halogen atom bonded to an aromatic ring such as a benzene ring. Br

Cl

An alkenyl chloride

A phenyl bromide

1.14 ALCOHOLS AND PHENOLS Methyl alcohol (also called methanol) has the structural formula CH3OH and is the simplest member of a family of organic compounds known as alcohols. The characteristic functional group of this family is the hydroxyl (} OH) group attached to an sp3-hybridized carbon atom. Another example of an alcohol is ethyl alcohol, CH3CH2OH (also called ethanol). C

O

H

This is the functional group of an alcohol.

Alcohols may be viewed structurally in two ways: (1) as hydroxyl derivatives of alkanes and (2) as alkyl derivatives of water. Ethyl alcohol, for example, can be seen as an ethane molecule in which one hydrogen has been replaced by a hydroxyl group or as a water molecule in which one hydrogen has been replaced by an ethyl group: Ethyl group

CH3CH2 CH3CH3

109.5°

H

Ethane

O

Hydroxyl group Ethyl alcohol (ethanol)

H 104.5°

O

H Water

As with alkyl halides, alcohols are classified into three groups: primary (1°), secondary (2°), and tertiary (3°) alcohols. This classification is based on the degree of substitution of the carbon to which the

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Chapter 1 | Basic Principles of Organic Chemistry

hydroxyl group is directly attached. If the carbon has only one other carbon attached to it, the carbon is said to be a primary carbon and the alcohol is a primary alcohol:

H

H

H

C

C

H

H

18 Carbon O

H

OH

OH

OH

or

Ethyl alcohol (a 18 alcohol)

Geraniol (a 18 alcohol)

Benzyl alcohol (a 18 alcohol)

If the carbon atom that bears the hydroxyl group also has two other carbon atoms attached to it, this carbon is called a secondary carbon, and the alcohol is a secondary alcohol: 28 Carbon

H

H

H

H

C

C

C

H

O

H

H

or

OH

OH

H Menthol (a 28 alcohol found in peppermint oil)

Isopropyl alcohol (a 28 alcohol)

If the carbon atom that bears the hydroxyl group has three other carbons attached to it, this carbon is called a tertiary carbon, and the alcohol is a tertiary alcohol: H H

C

H H

H 38 Carbon H

C

C

C

H

O

H

H

or

OH

H tert-Butyl alcohol (a 38 alcohol)

When a hydroxyl group is bonded to a benzene ring the combination of the ring and the hydroxyl is called a phenol. Phenols differ significantly from alcohols in terms of their relative acidity, and thus they are considered a distinct functional group. OH

Thymol (a phenol found in thyme)

1.16 | Amines

1.15 ETHERS Ethers have the general formula R} O} R or R} O} R′, where } R′ may be an alkyl (or phenyl) group different from R. Ethers can be thought of as derivatives of water in which both hydrogen atoms have been replaced by alkyl groups. The bond angle at the oxygen atom of an ether is only slightly larger than that of water: R9

R or

O R

CH3

O

1108

R

CH3

General formula for an ether

C

The functional group of an ether

Dimethyl ether (a typical ether)

CH2

H2C

C

O

O

O

O

Ethylene oxide

Tetrahydrofuran (THF)

Two cyclic ethers

1.16 AMINES Just as alcohols and ethers may be considered as organic derivatives of water, amines may be considered as organic derivatives of ammonia: H—N—H

R—N—H H2N

NH2

H

H

Ammonia

An amine

Amphetamine (a dangerous stimulant)

NH2

Putrescine (found in decaying meat)

Amines are classified as primary, secondary, or tertiary amines. This classification is based on the number of organic groups that are attached to the nitrogen atom: R—N—H

R—N—H

R—N—R0

H

R9

R9

A primary (18) amine

A secondary (28) amine

A tertiary (38) amine

Notice that this is quite different from the way alcohols and alkyl halides are classified. Isopropylamine, for example, is a primary amine even though its } NH2 group is attached to a secondary carbon atom. It is a primary amine because only one organic group is attached to the nitrogen atom:

H

H

H

H

C

C

C

H

NH2 H

H or NH2

N H

Isopropylamine (a 18 amine)

Piperidine (a cyclic 28 amine)

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Chapter 1 | Basic Principles of Organic Chemistry

Amines are like ammonia (Section 1.9B) in having a trigonal pyramidal shape. The C } N} C bond angles of trimethylamine are 108.7°, a value very close to the H } C} H bond angles of methane. Thus, for all practical purposes, the nitrogen atom of an amine can be considered to be sp3 hybridized with the unshared electron pair occupying one orbital (see below). This means that the unshared pair is relatively exposed, and as we shall see this is important because it is involved in almost all of the reactions of amines.

N

H 3C

CH3

CH3

Bond angle = 108.7° Trimethylamine

1.17 ALDEHYDES AND KETONES Aldehydes and ketones both contain the carbonyl group—a group in which a carbon atom has a double bond to oxygen: O C The carbonyl group

The carbonyl group of an aldehyde is bonded to one hydrogen atom and one carbon atom (except for formaldehyde, which is the only aldehyde bearing two hydrogen atoms). The carbonyl group of a ketone is bonded to two carbon atoms. Using R, we can designate the general formulas for aldehydes and ketones as follows:

ALDEHYDES O R

or

KETONES O

RCHO R

H

(R = H in formaldehyde)

or

RCOR

R

(where R9 is an alkyl group that may be the same or different from R)

Some specific examples of aldehydes and ketones are the following:

H

O

O

O H

C6H5

H

Formaldehyde

Acetaldehyde

O

O

H

Benzaldehyde

Acetone

Ethyl methyl ketone

O

O H

trans-Cinnamaldehyde (present in cinnamon)

Carvone (from spearmint)

1.18 | Carboxylic Acids, Esters, and Amides

Aldehydes and ketones have a trigonal planar arrangement of groups around the carbonyl carbon atom. The carbon atom is sp2 hybridized. In formaldehyde, for example, the bond angles are as follows: O 121°

H

121°

C 118°

H

1.18 CARBOXYLIC ACIDS, ESTERS, AND AMIDES Carboxylic acids, esters, and amides all contain a carbonyl group that is bonded to an oxygen or nitrogen atom. As we shall learn in later chapters, all of these functional groups are interconvertible by appropriately chosen reactions.

1.18A Carboxylic Acids Carboxylic acids have a carbonyl group bonded to a hydroxyl group, and they have the general formula O R

C

O O

. The functional group,

H O

R

C

C

O

, is called the carboxyl group (carbonyl + hydroxyl):

H

O

O O

H

or

or R

OH

RCO2H

C

O

H

or

CO2H or

COOH

The carboxyl group

A carboxylic acid

Examples of carboxylic acids are formic acid, acetic acid, and benzoic acid: O H

C

O O

H

or

or

OH

H

HCO2H

Formic acid

O CH3

C

O O

H

or OH

or CH3CO2H

Acetic acid

O C

O OH

or

OH

or

C6H5CO2H

Benzoic acid

Formic acid is an irritating liquid produced by ants. (The sting of the ant is caused, in part, by formic acid being injected under the skin. Formic is the Latin word for ant.) Acetic acid, the substance responsible for the sour taste of vinegar, is produced when certain bacteria act on the ethyl alcohol of wine and cause the ethyl alcohol to be oxidized by air.

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Chapter 1 | Basic Principles of Organic Chemistry

1.18B Esters Esters have the general formula RCO2R9 (or RCOOR9), where a carbonyl group is bonded to an alkoxyl (} OR) group: O

O R

C

OR9 or RCO2R9

or R

R9

O

General formula for an ester

O CH3

C

O OCH2CH3

or

or

O

CH3CO2CH2CH3

Ethyl acetate is an important solvent.

O O Pentyl butanoate has the odor of apricots and pears.

Esters can be made from a carboxylic acid and an alcohol through the acid-catalyzed loss of a molecule of water. For example: O

O CH3

C

OH

1 HOCH2CH3

Acetic acid

acid-catalyzed

CH3

Ethyl alcohol

C

OCH2CH3

1 H2O

Ethyl acetate

1.18C Amides Amides have the formulas RCONH2, RCONHR′, or RCONR′R″ where a carbonyl group is bonded to a nitrogen atom bearing hydrogen and/or alkyl groups. General formulas and some specific examples are shown below. O

O R

N

H

R

O N

H

R9

R

N

H

An unsubstituted amide

R9

R0

An N-substituted amide

An N,N-disubstituted amide

General formulas for amides

O

O

O N

H

H Acetamide

N

Me

N

H N-Methylacetamide

Me

Me N,N-Dimethylacetamide

Specific examples of amides

N- and N,N-indicate that the substituents are attached to the nitrogen atom.

1.20 | Summary of Important Families of Organic Compounds

1.19 NITRILES A nitrile has the formula R} C N: (or R} CN). The carbon and the nitrogen of a nitrile are sp hybridized. In IUPAC systematic nomenclature, acyclic nitriles are named by adding the suffix -nitrile to the name of the corresponding hydrocarbon. The carbon atom of the } C N group is assigned number 1. The name acetonitrile is an acceptable common name for CH3CN, and acrylonitrile is an acceptable common name for CH2 CHCN: 2

CH3

1

C

4

N

3

2

CH3CH2CH2

Ethanenitrile (acetonitrile)

1

C

3

N

Butanenitrile

2

5

1

CN

Propenenitrile (acrylonitrile)

3

4

2

1

N

4-Pentenenitrile

Cyclic nitriles are named by adding the suffix -carbonitrile to the name of the ring system to which the } CN group is attached. Benzonitrile is an acceptable common name for C6H5CN: C

N

Benzenecarbonitrile (benzonitrile)

C

N

Cyclohexanecarbonitrile

1.20 SUMMARY OF IMPORTANT FAMILIES OF ORGANIC COMPOUNDS A summary of the important families of organic compounds is given in Table 1.3. You should learn to identify these common functional groups as they appear in other, more complicated molecules.

1.20A Functional Groups in Biologically Important Compounds Many of the functional groups we have listed in Table 1.3 are central to the compounds of living organisms. A typical sugar, for example, is glucose. Glucose contains several alcohol hydroxyl groups ( } OH) and in one of its forms contains an aldehyde group. Fats and oils contain ester groups, and proteins contain amide groups. See if you can identify alcohol, aldehyde, ester, and amide groups in the following examples. O OH

O

OH

HO HO

O

O

n

O

n

OH

O n

Glucose

O A typical fat

O etc.

1R

N

N H

O

H

O

3R

N

N 2R

H

Part of a protein

O

H

O

N 4R

H

etc.

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Chapter 1 | Basic Principles of Organic Chemistry

Table 1.3 Important families of organic compounds Family Alkane Functional C } H group and C} C bonds General formula

RH

Alkene C

C

RCH

CH2

RCH

CHR

R2C

CHR

R2C

CR2 CH2

Alkyne

Aromatic

} C C}

Aromatic ring

ArH RC

CH

RC

CR

HC

CH

Haloalkane

C

X

Alcohol

C

OH

RX

ROH

CH3CH2Cl

CH3CH2OH

Phenol Ether OH

ArOH

C

O

C

ROR

Specific example

CH3CH3

CH2

IUPAC name

Ethane

Ethene

Ethyne

Benzene

Chloroethane

Ethanol

Phenol

Methoxymethane

Common namea

Ethane

Ethylene

Acetylene

Benzene

Ethyl chloride

Ethyl alcohol

Phenol

Dimethyl ether

Amine

Aldehyde

Ketone

Carboxylic Acid Ester

Functional }C }N group

General formula

RNH2 R2NH R3N

O

O C } }H

C} }C } C}

O

O

RCH

RCH

O

O

C } } OH

C } } O}C }

O

O

RCOH

CH3OCH3

Amide

Nitrile

O

C

N

C} } N}

O

RCOR

OH

RCN

RCNH2 O

RCNHR O RCNR R O

Specific example

CH3NH2

IUPAC name

Methanamine Ethanal

Common name

Methylamine

CH3CH

O

O

CH3CCH3

CH3COH

CH3COCH3

CH3CNH2

Propanone

Ethanoic acid

Methyl ethanoate

Ethanamide

Acetic acid

Methyl acetate Acetamide

Acetaldehyde Acetone

O

O

CH3C

N

Ethanenitrile Acetonitrile

These names are also accepted by the IUPAC.

a

1.21 NOMENCLATURE OF ORGANIC COMPOUNDS In the early development of organic chemistry, each new compound was given a name, usually by the person who had isolated or synthesized it. Names were not systematic, but often conveyed some

1.21 | Nomenclature of Organic Compounds

Table 1.4 Compounds and their common names Compound

Common Name

CH4

Methane

C2H4

Ethylene

C2H2

Acetylene

CH2Cl2

Methylene chloride

CH3COOH

Acetic acid

CH3HC

Acetaldehyde

O

CH3COCH3

Acetone

C6H5OH

Phenol

C6H5COOH

Benzoic acid

C6H5NH2

Aniline

C6H5OCH3

Anisole

CH3OCH3

Dimethyl ether

information – usually about the origin of the substance. A single compound was often known by several names. For example, the active ingredient in alcoholic beverages has been called alcohol, ethyl alcohol, methyl carbinol, grain alcohol, spirit and ethanol. A meeting in Geneva in 1892 initiated the development of an international system for naming compounds. In its present form, the method recommended by the International Union of Pure and Applied Chemistry is systematic, generally unambiguous and internationally accepted. It is called the IUPAC System. Despite the existence of the official IUPAC System, a great many well-established common, or trivial, names and abbreviations (e.g., TNT and DDT) are used because of their brevity or convenience. It is thus necessary to have knowledge of both the IUPAC system and many common names. Common names of some organic compounds are given in Table 1.4.

1.21A The IUPAC System of Nomenclature The name we give to any compound with a chain of carbon atoms consists of three parts – a prefix, an infix (a modifying element inserted into a word) and a suffix. Each part provides specific information about the structural formula of the compound. 1. The prefix shows the number of carbon atoms in the parent chain. Prefixes that show the presence of 1–20 carbon atoms in a chain are given in Table 1.5.

Table 1.5 Prefixes used in the IUPAC system to show the presence of 1–20 carbons in an unbranched chain Prefix

Number of Carbon Atoms

Prefix

Number of Carbon Atoms

meth-

1

undec-

11

eth-

2

dodec-

12

prop-

3

tridec-

13

but-

4

tetradec-

14

pent-

5

pentadec-

15

hex-

6

hexadec-

16

hept-

7

heptadec-

17

oct-

8

octadec-

18

non-

9

nonadec-

19

dec-

10

eicos-

20

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Chapter 1 | Basic Principles of Organic Chemistry

The first four prefixes listed in Table 1.5 were chosen by the IUPAC because they were well established in the language of organic chemistry. 2. The infix shows the nature of the carbon–carbon bonds in the parent chain. Infix

Nature of Carbon–Carbon Bonds in the Parent Chain

-an

All single bonds

-en-

One or more double bonds

-yn

One or more triple bond

3. The suffix shows the class of the compound to which the substance belongs. Suffix

Class of Compound

-e

Hydrocarbon

-ol

Alcohol

-al

Aldehyde

-one

Ketone

-oic acid

Carboxylic acid

The hydrocarbons containing carbon–carbon single bonds saturated with hydrogen are called alkanes while those containing double and triple bonds are unsaturated hydrocarbons called alkenes and alkynes, respectively.

1.21B IUPAC Nomenclature of Alkanes Straight-Chain Hydrocarbons For naming alkanes, the parent chain hydrocarbon is identified, the prefix is fixed indicating the number of carbon atoms in the chain and the suffix –ane is added to it. The IUPAC names of some straight-chain (saturated) hydrocarbons are given in Table 1.6. These alkanes are homologues as they differ from each other only by a } CH2 group. Table 1.6 Unbranched alkanes Name

Number of Carbon Atoms

Structure

Methane

1

CH4

Ethane

2

CH3CH3

Propane

3

CH3(CH2)CH3

Butane

4

CH3(CH2)2CH3

Pentane

5

CH3(CH2)3CH3

Hexane

6

CH3(CH2)4CH3

Heptane

7

CH3(CH2)5CH3

Octane

8

CH3(CH2)6CH3

Nonane

9

CH3(CH2)7CH3

Decane

10

CH3(CH2)8CH3

Undecane

11

CH3(CH2)9CH3

Dodecane

12

CH3(CH2)10CH3

Tridecane

13

CH3(CH2)11CH3

Tetradecane

14

CH3(CH2)12CH3

Pentadecane

15

CH3(CH2)13CH3

Hexadecane

16

CH3(CH2)14CH3

1.21 | Nomenclature of Organic Compounds

Branched-Chain Hydrocarbons The parent chain is branched and it bears small chains of carbon atoms called alkyl groups which are formed by removing hydrogen from alkane. Alkyl groups have the general formula CnH2n+1 (one less hydrogen atom than the corresponding alkane). The missing H atom may have been detached from any carbon in the alkane. The name of the group is formed from the name of the corresponding alkane by simply dropping -ane and substituting a -yl ending. The letter R is often used in formulas to mean any of the many possible alkyl groups: R = CnH2n+1

Any alkyl group

Alkyl groups are the groups that we identify for purposes of naming compounds: Alkane

Abbreviation

Alkyl group

CH4

CH3

Methane

Me

Methyl group

CH3CH3

CH3CH2

Ethane

or C2H5

Et

Ethyl group

CH3CH2CH3 Propane

CH3CH2CH2

Pr

Propyl group CH3

CH3CH2CH3 Propane

CH3CHCH3 or CH3CH

i-Pr

Isopropyl group

The names and formulas of selected alkyl groups up to and including four carbon atoms are given in Table 1.7. Three prefixes commonly indicate structural information. They are iso-, sec- (for secondary) and tert- or t(for tertiary). Structural unit called the neopentyl group is also found in some alkyl groups. Table 1.7 Names and formulas of selected alkyl groups Formula

Name

CH3 }

Methyl

Formula

CH3

Name Isopropyl

CH3CH CH3CH2 }

Ethyl

CH3CH2CH2 }

Propyl

Isobutyl

CH3 CH3CHCH2

CH3CH2CH2CH2 }

Butyl

CH3(CH2)3CH2 }

Pentyl

CH3(CH2)4CH2 }

Hexyl

CH3

sec-Butyl (secondary butyl)

CH3CH2CH CH3(CH2)5CH2 }

Heptyl

CH3(CH2)6CH2 }

Octyl

CH3

tert-Butyl or t-butyl (tertiary butyl)

CH3C CH3

(Continued)

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Chapter 1 | Basic Principles of Organic Chemistry

Table 1.7 (Continued) Formula CH3(CH2)7CH2–

Name

Formula

Nonyl

H3C CH3(CH2)8CH2–

Name Neopentyl

CH3 CH2 CH3

Decyl

According to the IUPAC System, a few relatively simple rules are all that are needed to name a great many alkanes. In later sections, these rules will be extended to cover other classes of compounds, but advanced texts or references must be consulted for the complete system. Rules for Naming Branched-Chain Alkanes The rules for nomenclature of branched-chain alkanes are briefly stated here. They would be discussed in greater detail in Chapter 4. 1. Select the longest continuous chain (parent or root chain) of carbon atoms as the parent compound, and consider all alkyl groups attached to it as branch chains that have replaced hydrogen atoms of the parent hydrocarbon. If two chains of equal length are present, use the chain that has the larger number of substituents attached to it as the parent compound. The name of the alkane consists of the name of the parent compound prefixed by the names of the branched-chain alkyl groups attached to it. 2. Number the carbon atoms in the parent carbon chain from one end to the other, starting from the end closest to the first carbon atom that has a branch chain: CH3 CH3CH2CHCH2CH2CH3 1 2 3 4 5 6

CH3 (not CH3CH2CHCH2CH2CH3) 6 5 4 3 2 1

3. Name each branched-chain alkyl group and designate its position on the parent carbon chain by a number (e.g., 3-methyl means a methyl group attached to C3). 4. When the same alkyl-group branch chain occurs more than once, indicate this by a prefix (di-, tri-, tetra-, etc.) written in front of the alkyl-group name (e.g., dimethyl indicates two methyl groups). The numbers indicating the positions of these alkyl groups are separated by a comma, followed by a hyphen, and placed in front of the name (e.g., 2,3-dimethyl). 5. When several different alkyl groups are attached to the parent compound, list them in alphabetical order and the lower number is given to the one coming first alphabetically (e.g., ethyl before methyl as in 3-ethyl-4-methyloctane). Note: Adding numerical prefixes (di-, tri-, tetra-, etc.) to the alkyl name does not change the alphabetical order. The prefixes iso- and neo- are considered to be part of fundamental name of the alkyl group while sec- and tert- are not. The use of iso and other common prefixes used in naming of alkyl groups is permitted by IUPAC as long as these are not further substituted. 6. In case the alkyl group attached to the parent hydrocarbon is branched, the carbon atom that attaches to the root alkane is numbered 1. The name of such branched chain is placed in parenthesis while writing the name of the compound. When more than one alkyl groups are attached to the parent hydrocarbon, then selection of substituent alkyl group and naming of compound is done based on the following: (a) If the two chains are of the same length, then the chain with greater number of side chains is selected. (b) After selection of the alkyl chain, the numbering is done from the end closer to the substituent. 7. Saturated cyclic hydrocarbons with only one ring are named by attaching the prefix cyclo- to the names of the alkanes possessing the same number of carbon atoms. For the side chains present, the same rules used for naming of alkyl groups in straight chain compounds are applicable.

1.21 | Nomenclature of Organic Compounds

Unsaturated Hydrocarbons The following additional rules are followed for naming of hydrocarbons with multiple (double or triple) bonds. 1. The multiple bond must be a part of the parent chain, regardless of the fact that it corresponds to the longest chain or not. 2. The numbering of the parent chain is done in a manner that the double or the triple bond gets the lowest number. In case both double and triple bond are present, the parent chain is numbered so as to assign lowest number to the double bond. 3. If the compound contains only one double or triple bond, its locant or positional number is placed before the suffix indicating the name. For example, pent-2-ene. In case both double and triple bonds are present in the compound, the positional number is written before each suffix, the compound is named as a derivative of alkyne and the suffix -e is removed from -ene. For example, pent-3-en-1-yne. 4. In case all the double and triple bonds present in the compound cannot be included in the longest chain, those not part of the chain are indicated by prefixes of their names, such as methylene, ethlyidene, vinyl, ethynyl, etc.

1.21C Nomenclature of Organic Compounds having Functional Group(s) The naming of different classes of compounds is based on the following guidelines: 1. For naming of different classes of compounds, the functional group is identified first and this determines the appropriate suffix for the name. 2. The longest chain containing the functional group is identified and numbered such that the carbon bearing the functional group gets the lowest possible number. 3. In case of polyfunctional compounds: (a) In case more than one functional group of the same type are present on the molecule, their number is indicated in the name of the compound as di-, tri, etc. before the suffix for the functional group. Note that for such compounds, the full name of parent alkane is written before the class suffix. In case of compounds containing more than one double bond, the term –ene from the name is dropped and replaced with diene or triene. (b) In case more than one different functional groups are present in a compound, based on priority, one of the functional group is chosen as the main or principal functional group and the remaining are named as substituents on the appropriate carbon number using suitable prefixes. The order of decreasing priority for some of the functional groups is COOH,

SO3H,

COOR,

COCl,

CONH2,

CN,

HC

O, >C

O,

OH,

NH2, >C

CR

R 38 > (most stable)

C1

H >R

C1

H 28

H >

H >H

18

C1

H > Methyl (least stable)

This order of stability of carbocations can be explained on the basis of inductive effect and hyperconjugation, which are discussed later. ●

Carbanions are electron rich. They are anions and have an unshared electron pair. Carbanions, therefore, are Lewis bases and react accordingly.

The carbon atom bearing the negative charge in the carbanion is in sp3 hybridized state. It is bonded to three other atoms, and the unshared electron pair occupies the apex of the tetrahedron (Fig. 1.32). The relative order of stability of carbanions is

CH3

H

CH3

CH3

C2 > CH3

C2 > CH3

C2

H 28

CH3 38

H 18 (Most stable) >

>

CH3 CH3 CH3

C2 sp3

hybridized carbon

Figure 1.32 A wedgedashed, wedge-line representation of the tert-butyl anion.

1.26 | Heterolysis of Bonds to Carbon: Carbocations and Carbanions

The stability order can be explained with the help of +I effect of the alkyl groups. Greater the number of alkyl groups attached to the C bearing the negative charge, greater will be the electrondonating effect, and hence the negative charge will be intensified. Therefore, tertiary carbanion carrying three alkyl groups is least stable. The carbanions are stabilized by the presence of electronwithdrawing groups. If we compare the carbanions of an alkane, alkene and alkyne, the stability of the carbanion having maximum s character is the maximum. As the s character decreases, the stability decreases. CH

2

sp

C > CH2

2

sp2

CH > CH3

2

sp3

CH2

This is because more the s character more will be the electronegativity of that C; and hence it will be easier for it to carry a negative change.

1.26A Electrophiles and Nucleophiles Because carbocations are electron-seeking reagents chemists call them electrophiles (meaning electron-loving). ●

●

Electrophiles are reagents that seek electrons so as to achieve a stable shell of electrons like that of a noble gas. All Lewis acids are electrophiles. By accepting an electron pair from a Lewis base, a carbocation fills its valence shell. C





Carbocation Lewis acid and electrophile

●

C

B

B

Lewis base

Carbon atoms that are electron poor because of bond polarity, but are not carbocations, can also be electrophiles. They can react with the electron-rich centers of Lewis bases in reactions such as the following:

B



Lewis base

+



C

O



B

C

O



Lewis acid (electrophile)

Carbanions are Lewis bases. Carbanions seek a proton or some other positive center to which they can donate their electron pair and thereby neutralize their negative charge. When a Lewis base seeks a positive center other than a proton, especially that of a carbon atom, chemists call it a nucleophile (meaning nucleus loving; the nucleo- part of the name comes from nucleus, the positive center of an atom). ●

A nucleophile is a Lewis base that seeks a positive center such as a positively charged carbon atom.

Since electrophiles are also Lewis acids (electron pair acceptors) and nucleophiles are Lewis bases (electron pair donors), why do chemists have two terms for them? The answer is that Lewis acid and Lewis base are terms that are used generally, but when one or the other reacts to form a bond to a carbon atom, we usually call it an electrophile or a nucleophile.

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Chapter 1 | Basic Principles of Organic Chemistry



Nu



C

+

Nucleophile



C



Nu



C

O

C

Nu

Electrophile



Nu

+

Electrophile

O

Nucleophile

1.27 HOMOLYTIC FISSION OF BONDS Another broad category of reactions has mechanisms that involve homolysis (Greek: homo, the same, + lysis) of covalent bonds with the production of intermediates possessing unpaired electrons called radicals (or free radicals): The shift of a single electron is represented by a fish-hook arrow ( ) or single-barbed curved arrows to show the movement of a single electron. For example, in the molecule X–Y where two electrons are shared between atoms X and Y; homolytic cleavage leads to the formation of X⋅ and Y⋅ free radicals possessing unpaired electrons. X1Y

hv

X 1Y Free radicals

For example, when chlorine gas is exposed to sunlight, two chlorine free radicals are formed. Cl2

hv

Cl 1 Cl

Free radical is a planar species in which the C atom bearing the odd electron is sp2 hybridized. The carbon atom lies in the center of the triangle and the three bonds are directed towards the three corners of planar trigonal geometry. The odd electron lies in the unhybridized p orbital, perpendicular to the plane.

2C

1208

In some free radicals, the central carbon atom may be sp3 hybridized with trigonal pyramidal geometry. The odd electron lies in the sp3 hybrid orbital.

C 1098289

The relative stability of alkyl free radicals follows the order: (CH3)3C > (CH3)2 C H > CH3 38

28

C H2 > C H3 > Vinyl 18

1.28 | Electron Displacement Effects in Organic Compounds

The relative stability can be explained on the basis of ease of formation. Lower the homolytic bond dissociation energy, greater is the ease of formation of free radicals. The bond dissociation energy decreases in the order: .CH3 > 1° > 2° > 3°. Therefore ease of formation is of the order: 3° > 2° > 1° > .CH3. Greater the stability of the radical, more rapidly or easily it will be formed. Therefore stability of free radicals also follows same order. Energy must be supplied to cause homolysis of covalent bonds, and this is usually done in two ways – by heating or by irradiation with light. For example, compounds with an oxygen–oxygen single bond called peroxides, undergo homolysis readily when heated because the oxygen–oxygen bond is weak. The products are two radicals called alkoxyl radicals: OO

R

R

Dialkyl peroxide

Heat

O

2R

Alkoxyl radicals

Almost all small radicals are short-lived, highly reactive species. When they collide with other molecules, they tend to react in a way that leads to pairing of their unpaired electron. One way they can do this is by abstracting an atom from another molecule. For example, a halogen atom may abstract a hydrogen atom from an alkane. This hydrogen abstraction gives the halogen atom an electron (from the hydrogen atom) to pair with its unpaired electron. Notice, however, that the other product of this abstraction is . another radical, in this case, an alkyl radical, R . General Reaction X 1 HR



X H1 R

Alkane

Alkyl radical

Cl 1 H CH3

Cl H 1 CH3

Specific Example

Methane

Methyl radical

This behavior is characteristic of radical reactions. Consider another example, one that shows another way in which radicals can react. They can combine with a compound containing a multiple bond to produce a new, larger radical. R

R C

C

Alkene

C

C

New radical

1.28 ELECTRON DISPLACEMENT EFFECTS IN ORGANIC COMPOUNDS The bonding pairs of the covalent bond in organic compounds undergo electronic displacements on their own or under the influence of an attacking agent. These displacements may be temporary or permanent in nature and are described as follows.

1.28A Inductive Effect In a covalent single bond between non-identical atoms, the bonding pair (electron pair) forming the bond is not shared absolutely evenly between the two atoms. It is always attracted towards the more

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electronegative atom of the two. For example, in an alkyl halide

X when X = Cl, the electron density

C

is greater near Cl than C because Cl is more electronegative than C. This is represented as d1

C

X

d2

or

C

X

Although the bonding pair is permanently displaced towards the more electronegative atom, yet it remains in the same valence shell. If the carbon atom that is bonded to the halogen is itself further bonded to carbon atoms, the effect is transmitted further down the chain as shown: dddd1

C4

ddd1

dd1

C3

d1

C2

d2

C1

X

Here, X leaves C1 slightly electron deficient, therefore C1 pulls more than its share of electrons from the s bond joining it to C2, and this continues down the chain. The effect of Cl on C2 is less than the effect of Cl on C1 and the extent of partial charge keeps on decreasing away from the polar bond. This effect is known as inductive effect. It decreases with the increase in distance and is not noticeable beyond C2. Some characteristics of inductive effect are listed as follows. 1. The effect is said to be −I if a group pulls the electrons away from the carbon atom. Such a group is called an electron-withdrawing group. Electron-withdrawing groups in the order of decreasing −I effect are shown below. 1

2 2 2 2 2 2 2 2 R3N > NO2 2 > CN > COOH > F > Cl > Br > I > OAr > OH > Phenyl

2. The effect is said to be +I if a group pushes the electrons towards the carbon atom. Such a group is called electron-releasing group. These electron-releasing groups in order of decreasing +I effect are shown: O2 > COO2 > (CH3)C > (CH3)2 CH > CH3CH2CH2 > CH3CH2 > CH3

3. Inductive effect causes development of a dipole moment in the molecule, which increases with increase in inductive effect. For example, CH3

Cl > CH3

F > CH3

Br > CH3

I

as the order of inductive effect is F > Cl > Br > I. 4. The bond length of the organic molecules decreases with increase in inductive effect. For example, CH3

Cl < CH3

F < CH3

Br < CH3

I

as the inductive effect increases in the order F > Cl > Br > I.

1.28B Resonance Structure and Effect One problem with Lewis structures is that they impose an artificial location on the electrons. As a result, more than one equivalent Lewis structure can be written for many molecules and ions. Consider, for example, the carbonate ion ( CO2− 3 ). We can write three different but equivalent structures, 1–3: O

O2

C 2

O

C O2

1

O2

2

O

C O

2

O

O2 3

1.28 | Electron Displacement Effects in Organic Compounds

Notice two important features of these structures. First, each atom has the noble gas configuration. Second, and this is especially important, we can convert one structure into any other by changing only the positions of the electrons. We do not need to change the relative positions of the atomic nuclei. For example, if we move the electron pairs in the manner indicated by the curved arrows in structure 1, we change structure 1 into structure 2: O

O2

C 2

O

O2

becomes

C 2

O

1

O 2

In a similar way, we can change structure 2 into structure 3: O2 C 2

O

O

O2 becomes

2

C O

O2 3

Structures 1–3, although not identical on paper, are equivalent. None of them alone, however, fits the important data about the carbonate ion. X-ray studies have shown that carbon–oxygen double bonds are shorter than single bonds. The same kind of study of the carbonate ion shows, however, that all of its carbon–oxygen bonds are of equal length. One is not shorter than the others as would be expected from the representations 1, 2 and 3. Clearly none of the three structures agrees with this evidence. In each structure, 1–3, one carbon–oxygen bond is a double bond and the other two are single bonds. None of the structures, therefore, is correct. One way to represent structure of carbonate ion is through a theory called resonance theory. This theory states that whenever a molecule or ion can be represented by two or more Lewis structures that differ only in the positions of the electrons, two things will be true: 1. None of these structures, which we call resonance structures or resonance contributors, will be a correct representation for the molecule or ion. None will be in complete accord with the physical or chemical properties of the substance. 2. The actual molecule or ion will be better represented by a hybrid (average) of these structures. Resonance structures, then, are not structures for the actual molecule or ion; they exist only on paper. As such, they can never be isolated. No single contributor adequately represents the molecule or ion. In resonance theory we view the carbonate ion, which is, of course, a real entity, as having a structure that is a hybrid of these three hypothetical resonance structures. In the three structural representations of carbonate ion, carbon–oxygen bond is a double bond in one structure (1) and a single bond in the other two (2 and 3). The actual carbon–oxygen bond, since it is a hybrid, must be something in between a double bond and a single bond. Since the carbon–oxygen bond is a single bond in two of the structures and a double bond in only one, it must be more like a single bond than a double bond. It must be like a one and one-third bond. We could call it a partial double bond. Thus all of the carbon–oxygen bonds of the carbonate ion are partial double bonds, and all are equivalent. All of them should be of the same length, and this is exactly what experiments tell us. The bonds are all 128 pm long, a distance which is intermediate between that of a carbon–oxygen single bond (143 pm) and that of a carbon–oxygen double bond (120 pm). By convention, when we draw resonance structures, we connect them by double-headed arrows to indicate clearly that they are hypothetical, not real. For the carbonate ion we write them this way: O

O

C 2

O

2

O2

C O

2

2

O

C O

O

O2

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Note: By the arrows between the structure, we should not be misled to think that the carbonate ion fluctuates between one structure and another. The carbonate ion cannot fluctuate among them because it is a hybrid of them. It is also important to distinguish between resonance and equilibrium. In equilibrium between two, or more, species, it is quite correct to think of different structures and moving (or fluctuating) atoms, but not in the case of resonance (as in the carbonate ion). Here the atoms do not move, and the structures exist only on paper. An equilibrium is indicated by and resonance by ↔. For the carbonate ion we might write a non-Lewis structure representing the hybrid of resonance structures. 2

O

O3 2

C

C 2 32

O

O

2

O 32

2

O

2

O2

C O2

2O

C O

O

O2

The bonds in the structure on the left are indicated by a combination of a solid line and a dashed line. This is to indicate that the bonds are something in between a single bond and a double bond. As a rule, we use a solid line whenever a bond appears in all structures, and a dashed line when a bond exists in one or more but not all. Although they have no real existence of their own, resonance structures are useful because they allow us to describe molecules and ions for which a single Lewis structure is inadequate. We write two or more Lewis structures, calling them resonance structures or resonance contributors. We connect these structures by double-headed arrows (↔), and we say that the real molecule or ion is a hybrid of all of them. The rules applied for writing resonance structures are: 1. The resonance structures should differ only in the position of electrons and not in the position of atoms or nuclei. 2. All the resonance structures should have the same number of unpaired electrons. 3. In case of atoms present in the second period of the periodic table, those resonance structures which violate the octet rule should not be considered. For example, CH2

1

CH

NH3

1

CH2

I

CH

NH3

II

Structure (II) cannot be considered as a resonance structure of (I) since it violates the octet rule because N in this structure has 10 electrons which is not feasible in absence of d orbitals. 4. The energy of the actual molecule is lower than the energy that might be estimated for any contributing structure. Chemists often call this kind of stabilization resonance stabilization. 5. Equivalent resonance structures make equal contributions to the hybrid, and a system described by them has a large resonance stabilization. The following cation, for example, is more stable than either contributing resonance structure taken separately: CH2

1

1

CH CH2 CH2 CH CH2 4 5 Contributing resonance structures

1 21

CH2

CH

1

CH2 2 1

Resonance hybrid

In Chapter 6 we shall find that benzene is highly resonance stabilized because it is a hybrid of the two equivalent forms that follow: or Resonance structures for benzene

Representation of hybrid

1.28 | Electron Displacement Effects in Organic Compounds

6. The more stable a structure is (when taken by itself), the greater is its contribution to the hybrid. 7. The following rules will help us in making decisions about the relative stabilities of resonance structures. The more covalent bonds a structure has, the more stable it is. This is because forming a covalent bond lowers the energy of atoms. This means that of the following structures for 1,3-butadiene, 6 is by far the most stable and makes by far the largest contribution because it contains one more bond. CH2

CH

CH 6

1

CH2

CH2

CH

CH 7

2

2

CH2

CH2

CH

CH 8

1

CH2

8. Structures in which all of the atoms have a complete valence shell of electrons (i.e., the noble gas structure) are especially stable and make large contributions to the hybrid. This means, for example, that 10 makes a larger stabilizing contribution to the cation below than 9 because all of the atoms of 10 have a complete valence shell. 1

CH2

O 9

CH3

CH2

1

O CH3 10

9. Charge separation decreases stability. Separating opposite charges requires energy. Therefore, structures in which opposite charges are separated have greater energy (lower stability) than those that have no charge separation. In the following two structures for vinyl chloride, structure 11 makes a larger contribution because it does not have separated charges. CH2

CH 11

2

Cl

CH2

CH 12

Cl

1

10. Resonance contributors with negative charge on highly electronegative atoms are more stable than ones with negative charge on less or non-electronegative atoms. Conversely, resonance contributors with positive charge on highly electronegative atoms are less stable than ones with positive charge on non-electronegative atoms. Resonance Effect Electron redistributions take place in unsaturated, especially in conjugated systems via their p orbitals. This generates centers of high- and low-electron densities. This phenomenon is called resonance or mesomeric effect (R or M effect). The resonance effect produces polarity in the molecule by the interaction of two p bonds or between a p bond and a lone pair. It is a permanent effect and is of two types. 1. Negative resonance effect (−R effect): Groups which tend to withdraw electrons from the conjugated system possess −R or −M effect, for example, } NO2, C N, } C O, etc. CH2

CH

C

1

O

CH2

CH

H O

C

O2 (2M effect)

H O

2

O

O2

2

2

2. Positive resonance effect (+R effect): Groups which tend to donate electrons to the conjugated system possess +R or +M effect, for example, } Cl, } Br, } I, } NH2, } OH, } OCH3, etc.

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CH2

CH

H

2

Cl

CH2

H

Cl1 (1M effect)

CH

H O1

O

H O1

O1 2

2

2

Note: Curved arrows used for drawing resonance structures do not represent the motion of electrons; they are only tools that assist in drawing of resonance structures with ease. These tools treat electrons as if they were moving, even though electrons are not actually moving. It is essential that the head and tail of every arrow is dawn in precise and proper location the tail where the electrons are coming from and head where the electrons are going. Two rules that must be followed when drawing curved arrows for resonance structures are: 1. Avoid breaking of single bonds. 2. Do not exceed an octet for second row elements.

1.28C Electromeric Effect (E Effect) This effect involves the displacement of p electrons in a multiple bond towards one of the bonded atoms at the demand of an attacking species. It is a temporary displacement and a time-variable effect and is nullified when attacking reagent is removed from the reaction domain. The electron shift is represented by a curved arrow. There are two types of effects: 1. Positive electromeric effect (+E effect): If the transference of p electrons occurs towards the atom where attacking species attaches, it is called +E effect. H C

C 1 H1

C1

C

(1E effect)

Attacking species

2. Negative electromeric effect (−E effect): If the transference of p electrons occurs away from the atom where the attacking group attaches, it is called −E effect. C

2

O 1 CN Attacking species

●

C

2

O (2E effect) CN

When inductive and electromeric effects are both operative in opposite directions, the electromeric effect predominates.

1.28D Hyperconjugation Electron delocalization (via orbital overlap) from a filled bonding orbital to an adjacent unfilled orbital or to an atom of an unsaturated system is called hyperconjugation. It involves conjugation of s electrons of

1.28 | Electron Displacement Effects in Organic Compounds

a single C } H bond and the unshared p orbital or the p electrons of the adjacent multiple bond. Hyperconjugation generally H Orbitals overlap here has a stabilizing permanent effect and is also known as Baker Nathan effect. To understand hyperconjugation between electrons of Vacant p orbital 2e2 C } H bond and an atom with unshared p orbital, consider the case of a carbocation. Here, the unfilled orbital is the vacant p C C1 orbital of the carbocation, and the filled orbitals are C } H or H C } C s bonds at the carbons adjacent to the p orbital of the carbocation. Sharing of electron density from adjacent C } H H or C } C s bonds with the carbocation p orbital delocalizes the Figure 1.33 Schematic representation positive charge. Any time a charge can be dispersed or delo- of hyperconjugation. calized, a system will be stabilized. Figure 1.33 shows a stylized representation of hyperconjugation between a s bonding orbital and an adjacent carbocation p orbital. Electron density from one of the carbon–hydrogen s bonds of the methyl group flows into the vacant p orbital of the carbocation because the orbitals can partly overlap. Shifting electron density in this way makes the sp2 hybridized carbon of the carbocation somewhat less positive, and the hydrogens of the methyl group assume some of the positive charge. Delocalization (dispersal) of the charge in this way leads to greater stability. This interaction of a bond orbital with a p orbital is called hyperconjugation. Tertiary carbocations have three carbons with C } H bonds (or depending on the specific example, C } C bonds instead of C } H) adjacent to the carbocation that can overlap partially with the vacant p orbital. Secondary carbocations have only two adjacent carbons with C } H or C } C bonds to overlap with the carbocation; hence, the possibility for hyperconjugation is less and the secondary carbocation is less stable. Primary carbocations have only one adjacent carbon from which to derive hyperconjugative stabilization, and so they are even less stable. A methyl carbocation has no possibility for hyperconjugation, and it is the least stable of all in this series. The following are specific examples: δ1

CH3 δ 1CH

3

Cδ 1 CH3

δ1

is more stable than

CH3 δ 1CH

3

Cδ 1 H

is more stable than

is more H C stable H than H

δ 1CH

3

δ1

H C1 H

δ1

tert-Butyl cation (3°) (most stable)

Isopropyl cation (2°)

Ethyl cation (1°)

Methyl cation (least stable)

Thus, the relative stability of carbocations is 3° > 2° > 1° > methyl. Hyperconjugation is also observed in alkenes and alkylarenes. It is called s–p conjugation or no-bond resonance because in the various canonical forms, no bond exists between C and H atom of the alkyl group directly attached to the multiplebonded C atoms. This may be depicted diagrammatically as p electrons or adjacent non-bonding p orbitals may also be involved in hyperconjugation. σ2π hyperconjugation π

H

H H

C

2 3 sp sp 2 σ

H C

C H

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Chapter 1 | Basic Principles of Organic Chemistry

For example in CH3 } CH

CH2, the hyperconjugation may be represented as

H H

C

H CH

CH2

H

1

C

H

H 2

CH

1

CH2

H

C

H I

H CH

2

CH2

H

C

CH

2

CH2

H1

H II

III

IV

1.29 THE STRENGTH OF BRØNSTED–LOWRY ACIDS AND BASES: Ka AND pKa Many organic reactions involve the transfer of a proton by an acid–base reaction. An important consideration, therefore, is the relative strengths of compounds that could potentially act as Brønsted–Lowry acids or bases in a reaction. In contrast to the strong acids, such as HCl and H2SO4, acetic acid is a much weaker acid. When acetic acid dissolves in water, the following reaction does not proceed to completion: O CH3

O

C

OH

 H 2O

CH3

C

O

 H 3O 

Experiments show that in a 0.1 M solution of acetic acid at 25 °C only about 1% of the acetic acid molecules ionize by transferring their protons to water. Therefore, acetic acid is a weak acid. The acid strength is characterized in terms of acidity constant (Ka ) or pKa values. Using a generalized hypothetical weak acid (HA), the reaction in water is HA

H3O1 1

H2O

1

A2

and the expression for the acidity constant is Ka

[H3O1][A2] [HA]

Because the concentrations of the products of the reaction are written in the numerator and the concentration of the undissociated acid in the denominator, a large value of Ka means the acid is a strong acid and a small value of Ka means the acid is a weak acid. If the Ka is greater than 10, the acid will be, for all practical purposes, completely dissociated in water at concentrations less than 0.01 M. Chemists usually express the acidity constant, Ka, as its negative logarithm, pKa: pKa = −log Ka For acetic acid the pKa is 4.75: pKa = −log(1.76 × 10−5) = −(−4.75) = 4.75 ●

The larger the value of the pKa, the weaker is the acid.

For example, acetic acid with pKa = 4.75 is a weaker acid than trifluoroacetic acid with pKa = 0 (Ka = 1). Hydrochloric acid with pKa = −7 (Ka = 107) is a far stronger acid than trifluoroacetic acid. (It is understood that a positive pKa is larger than a negative pKa.) CH3CO2H  CF3CO2H HCl pKa 4.75 Weak acid

pKa 0

pKa 7 Very strong acid

Increasing acid strength

1.29 | The Strength of Brønsted–Lowry Acids and Bases: Ka and pKa

Table 1.12 lists pKa values for a selection of acids relative to water as the base. Table 1.12 Relative strength of selected acids and their conjugate bases Acid Strongest acid

Approximate pKa < −12

HSbF6

−10

HI

I

Weakest base

−

−9

HBr HCl C6H5SO3H

−9 −7 −6.5

Br− Cl− C6H5SO3−

−3.8

(CH3)2O

+

+

HSO4−

(CH3)2C

−2.5

CH3OH

+

−1.74

H2O

−1.4 0.18 3.2 4.21 4.63 4.75 6.35 9.0

NO3− CF3CO2− F− C6H5CO2− C6H5NH2

OH

CH3OH2 H3O HNO3 CF3CO2H HF C6H5CO2H C6H5NH3+ CH3CO2H H2CO3 CH3COCH2COCH3

O

CH3CO2− HCO3− − CH3COCHCOCH3

NH4+

9.2

NH3

C6H5OH

9.9

C6H5O−

HCO3−

10.2

CO32−

CH3NH3+

10.6

CH3NH2

H2O

15.7

HO−

CH3CH2OH

16

CH3CH2O−

(CH3)3COH

18

(CH3)3CO−

CH3COCH3

19.2

−

HC CH C6H5NH2 H2 (i-Pr)2NH NH3

25 31 35 36 38

HC C− C6H5NH− H− (i-Pr)2N− − NH2

CH2

44

CH2

50

CH3CH2−

CH2

CH3CH3

Increasing base strength

−2.9

+

(CH3)2C

Increasing acid strength

SbF6−

H2SO4

(CH3)2OH

Weakest acid

Conjugate base

CH2COCH3

CH− Strongest base

1.29A Predicting the Strength of Bases In our discussion so far we have dealt only with the strengths of acids. Arising as a natural corollary to this is a principle that allows us to estimate the base strength. Simply stated, the principle is this: ●

The stronger the acid, the weaker will be its conjugate base.

We can, therefore, relate the strength of a base to the pKa of its conjugate acid. ●

The larger the pKa of the conjugate acid, the stronger is the base.

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Chapter 1 | Basic Principles of Organic Chemistry

Consider the following as examples: Increasing base strength

Cl

CH3CO2

HO

Very weak base pKa of conjugate acid (HCl)  7

Weak base pKa of conjugate acid (CH3CO2H)  4.75

Strong base pKa of conjugate acid (H2O)  15.7

We see that the hydroxide ion is the strongest in this series of three bases because its conjugate acid, water, is the weakest acid. We know that water is the weakest acid because it has the largest pKa.

1.29B How to Predict the Outcome of Acid–Base Reactions By learning the relative scale of acidity of common acids now, you will be able to predict whether or not an acid–base reaction will occur as written. ●

The general principle to apply is this: acid–base reactions always favor the formation of the weaker acid and the weaker base.

The reason for this is that the outcome of an acid–base reaction is determined by the position of an equilibrium. Acid–base reactions are said, therefore, to be under equilibrium control, and reactions under equilibrium control always favor the formation of the most stable (lowest potential energy) species. The weaker acid and weaker base are more stable (lower in potential energy) than the stronger acid and stronger base. Using this principle, we can predict that a carboxylic acid (RCO2H) will react with aqueous NaOH in the following way because the reaction will lead to the formation of the weaker acid (H2O) and weaker base (RCO2−): O R

C

O O

H  Na  O

Stronger acid pKa  3–5

H

R

Stronger base

C

O



Na  H

Weaker base

O

H

Weaker acid pKa  15.7

Because there is a large difference in the value of the pKa of the two acids, the position of equilibrium will greatly favor the formation of the products. In instances like these we commonly show the reaction with a one-way arrow even though the reaction is an equilibrium. Water Solubility as the Result of Salt Formation Although acetic acid and other carboxylic acids containing fewer than five carbon atoms are soluble in water, many other carboxylic acids of higher molecular weight are not appreciably soluble in water. Because of their acidity, however, water-insoluble carboxylic acids dissolve in aqueous sodium hydroxide; they do so by reacting to form water-soluble sodium salts: O C

O O

H  Na  O

C

H

Insoluble in water

O  Na  H

O

H

Soluble in water (due to its polarity as a salt)

We can also predict that an amine will react with aqueous hydrochloric acid in the following way: H

H R

N

1

H Stronger base

H

O

1

H

Cl2

H Stronger acid pKa 5 –1.74

R

N 1 H Cl2 H

Weaker acid pKa 5 9 –10

1

O

H

H Weaker base

1.30 | Relationships between Structure and Acidity

While methylamine and most amines of low molecular weight are very soluble in water, amines with higher molecular weights, such as aniline (C6H5NH2), have limited water solubility. However, these water-insoluble amines dissolve readily in hydrochloric acid because the acid–base reactions convert them into soluble salts: H

H C6H5

N



H

H

O  H Cl

C6H5

H

N  H Cl H

Water insoluble



O

H

H

Water-soluble salt

1.30 RELATIONSHIPS BETWEEN STRUCTURE AND ACIDITY The strength of a Brønsted–Lowry acid depends on the extent to which a proton can be separated from it and transferred to a base. Removing the proton involves breaking a bond to the proton, and it involves making the conjugate base more electrically negative. ●

Bond strength to the proton decreases as we move down the column, increasing its acidity.

This phenomenon is mainly due to decreasing effectiveness of orbital overlap between the hydrogen 1s orbital and the orbitals of successively larger elements in the column. The less effective the orbital overlap, the weaker is the bond, and the stronger is the acid. The acidities of the hydrogen halides furnish an example: pKa 3.2 H 9 F 7 H9 Cl

Group VIIA

9 H9 Br 10 H 9 I

A c i d i t y i n c r e a s e s

Comparing the hydrogen halides with each other, H } F is the weakest acid and H—I is the strongest. This follows from the fact that the H } F bond is by far the strongest and the H } I bond is the weakest. ●

Acidity increases from left to right when we compare compounds in a given row of the periodic table.

Bond strengths vary somewhat, but the predominant factor becomes the electronegativity of the atom bonded to the hydrogen. The electronegativity of the atom in question affects acidity in two related ways: (1) it affects the polarity of the bond to the proton and (2) it affects the relative stability of the anion (conjugate base) that forms when the proton is lost. We can see an example of this effect when we compare the acidities of the compounds CH4, NH3, H2O, and HF. These compounds are all hydrides of first-row elements, and electronegativity increases across a row of the periodic table from left to right: Electronegativity increases

C

N

O

F

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Chapter 1 | Basic Principles of Organic Chemistry

Because fluorine is the most electronegative, the bond in H } F is most polarized, and the proton in H } F is the most positive. Therefore, H } F loses a proton most readily and is the most acidic in this series: Acidity increases

H3C

d2

H

H2N

pKa 5 48

d1

d2

H

HO

pKa 5 38

d1

d2

H

F

pKa 5 15.7

d1

H

pKa 5 3.2

The above compounds directly illustrate this trend based on electronegativity and increasing polarization of the bonds to hydrogen. Almost no positive charge is evident at the hydrogens of methane. Very little positive charge is present at the hydrogens of ammonia. This is consistent with the weak electronegativity of both carbon and nitrogen and hence with the behavior of methane and ammonia as exceedingly weak acids (pKa values of 48 and 38, respectively). Water shows significant positive charge at its hydrogens (pKa more than 20 units lower than ammonia), and hydrogen fluoride clearly has the highest amount of positive charge at its hydrogen (pKa of 3.2), resulting in strong acidity. Because H } F is the strongest acid in this series, its conjugate base, the fluoride ion (F−), will be the weakest base. Fluorine is the most electronegative atom and it accommodates the negative charge most readily: Basicity increases

CH3

H2N

HO

F

The methanide ion (CH3−) is the least stable anion of the four, because carbon being the least electronegative element is least able to accept the negative charge. The methanide ion, therefore, is the strongest base in this series.

1.30A The Effect of Hybridization ●

An alkyne hydrogen is weakly acid. Alkene and alkane hydrogens are essentially not acidic.

The pKa values for ethyne, ethene, and ethane illustrate this trend. H H

C

C

H

C H

Ethyne pKa  25

H C H

Ethene pKa  44

H

H H

C H

C

H H

Ethane pKa  50

We can explain this order of acidities on the basis of the hybridization state of carbon in each compound. Electrons of 2s orbitals have lower energy than those of 2p orbitals because electrons in 2s orbitals tend, on the average, to be much closer to the nucleus than electrons in 2p orbitals. (Consider the shapes of the orbitals: 2s orbitals are spherical and centered on the nucleus; 2p orbitals have lobes on either side of the nucleus and are extended into space.) ●

With hybrid orbitals, having more s character means that the electrons of the anion will, on the average, be lower in energy, and the anion will be more stable.

The sp orbitals of the C } H bonds of ethyne have 50% s character (because they arise from the combination of one s orbital and one p orbital), those of the sp2 orbitals of ethene have 33.3% s character, while those of the sp3 orbitals of ethane have only 25% s character. This means, in effect, that the sp carbon atoms of ethyne act as if they were more electronegative than the sp2 carbon atoms of ethene and the sp3 carbon atoms of ethane. (Remember: electronegativity measures an atom’s ability to hold bonding electrons close to its nucleus, and having electrons closer to the nucleus makes it more stable.) ●

An sp carbon atom is effectively more electronegative than an sp2 carbon, which in turn is more electronegative than an sp3 carbon.

1.31 | Acidity: Carboxylic Acids Versus Alcohols

In summary, the order of relative acidities of ethyne, ethene, and ethane parallels the effective electronegativity of the carbon atom in each compound: Relative Acidity of the Hydrocarbons

HC

H2 C

CH

CH2

H3C

CH3

As expected based on the properties of acid–base conjugate pairs, an sp3 carbanion is the strongest base in a series based on carbon hybridization, and an sp carbanion (an alkynide) is the weakest base. This trend is illustrated here with the conjugate bases of ethane, ethene, and ethyne. Relative Basicity of the Carbanions

H3C

CH2

2

H2C

CH

2

HC

C

2

1.30B Inductive Effects The carbon–carbon bond of ethane is completely nonpolar because at each end of the bond there are two identical methyl groups: CH3 } CH3 Ethane The C — C bond is nonpolar.

This is not the case with the carbon–carbon bond of ethyl fluoride, however: d1

d1

d2

CH3

CH2

F

2

1

One end of the bond, the one nearer the fluorine atom, is more negative than the other. This polarization of the carbon–carbon bond results from an intrinsic electron-attracting ability of the fluorine (because of its electronegativity) that is transmitted through space and through the bonds of the molecule. Chemists call this kind of effect an inductive effect. ●

Inductive effects are electronic effects transmitted through bonds. The inductive effect of a group can be electron donating or electron withdrawing. Inductive effects weaken as the distance from the group increases.

In the case of ethyl fluoride, the positive charge that the fluorine imparts to C1 is greater than that imparted to C2 because the fluorine is closer to C1.

1.31 ACIDITY: CARBOXYLIC ACIDS VERSUS ALCOHOLS Carboxylic acids are weak acids, typically having pKa values in the range of 3–5. Alcohols, by comparison, have pKa values in the range of 15–18, and essentially do not give up a proton unless exposed to a very strong base. To understand the reasons for this difference, let’s consider acetic acid and ethanol as representative examples of simple carboxylic acids and alcohols. O CH3

C

OH

Acetic acid pKa 5 4.75

CH3CH2

OH

Ethanol pKa 5 16

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Consider first the structural changes that occur if both acetic acid and ethanol act as acids by donating a proton to water. Acetic Acid Acting as an Acid

O H3C

C

O

H

O



C

O

H 3C

H

Acetic acid

H

O

H

Water



H! O





H

Acetate ion

Hydronium ion

Ethanol Acting as an Acid

H

H CH3CH2!O!H

O



CH3CH2!O



H!O



H Ethanol



H Ethoxide ion

Water

Hydronium ion

What we need to focus on is the relative stability of the conjugate bases derived from a carboxylic acid and an alcohol. This is because the smaller free-energy change for ionization of a carboxylic acid (e.g., acetic acid) as compared to an alcohol (e.g., ethanol) has been attributed to greater stabilization of the negative charge in the carboxylate ion as compared to an alkoxide ion. Greater stabilization of the carboxylate ion appears to arise from two factors: (a) delocalization of charge (as depicted by resonance structures for the carboxylate ion (1.31 A), and (b) an inductive electron-withdrawing effect (1.28 A).

1.31A The Effect of Delocalization Delocalization of the negative charge is possible in a carboxylate anion, but it is not possible in an alkoxide ion. We can show how delocalization is possible in carboxylate ions by writing resonance structures for the acetate ion. Two Resonance Structures that can be Written for Acetate Anion

O

O H3C

C

O



CH3

C



O

Resonance stabilization in acetate ion (The structures are equivalent and there is no requirement for charge separation.)

The two resonance structures we drew above distributed the negative charge to both oxygen atoms of the carboxylate group, thereby stabilizing the charge. This is a delocalization effect (by resonance). In contrast, no resonance structures are possible for an alkoxide ion, such as ethoxide. (You may wish to review the rules we have given in Section l.5 for writing proper resonance structures.) CH3

CH2

O

No resonance stabilization

H

 H2O

CH3

CH2



O



H3O

No resonance stabilization

No resonance structures can be drawn for either ethanol or ethoxide anion.

A rule to keep in mind is that charge delocalization is always a stabilizing factor, and because of charge stabilization, the energy difference for formation of a carboxylate ion from a carboxylic acid is less

1.31 | Acidity: Carboxylic Acids Versus Alcohols

than the energy difference for formation of an alkoxide ion from an alcohol. Since the energy difference for ionization of a carboxylic acid is less than for an alcohol, the carboxylic acid is a stronger acid.

1.31B The Inductive Effect We have already shown how the negative charge in a carboxylate ion can be delocalized over two oxygen atoms by resonance. However, the electronegativity of these oxygen atoms further helps to stabilize the charge, by what is called an inductive electron-withdrawing effect. A carboxylate ion has two oxygen atoms whose combined electronegativity stabilizes the charge more than in an alkoxide ion, which has only a single electronegative oxygen atom. In turn, this lowers the energy barrier to forming the carboxylate ion, making a carboxylic acid a stronger acid than an alcohol. Negative charge in the acetate anion is evenly distributed over the two oxygen atoms, whereas in ethoxide the negative charge is localized on its sole oxygen atom. It is also reasonable to expect that a carboxylic acid would be a stronger acid than an alcohol when considering each as a neutral molecule (i.e., prior to loss of a proton), because both functional groups have a highly polarized O } H bond, which in turn weakens the bond to the hydrogen atom. However, the significant electron-withdrawing effect of the carbonyl group in acetic acid and the absence of an adjacent electron-withdrawing group in ethanol make the carboxylic acid hydrogen much more acidic than the alcohol hydrogen. O CH3

C

O

H

CH3

Acetic acid (stronger acid)

CH2

O

H

Ethanol (weaker acid)

1.31C Summary and a Comparison of Conjugate Acid–Base Strengths In summary, the greater acidity of a carboxylic acid is predominantly due to the ability of its conjugate base (a carboxylate ion) to stabilize a negative charge better than an alkoxide ion, the conjugate base of an alcohol. In other words, the conjugate base of a carboxylic acid is a weaker base than the conjugate base of an alcohol. Therefore, since there is an inverse strength relationship between an acid and its conjugate base, a carboxylic acid is a stronger acid than an alcohol. ●

The more stable a conjugate base is, the stronger the corresponding acid.

1.31D Inductive Effects of Other Groups The acid-strengthening effect of other electron-attracting groups (other than the carbonyl group) can be shown by comparing the acidities of acetic acid and chloroacetic acid: O CH3

C

O Cl O

CH2

H

C

O

H

pKa  2.86

pKa 4.75

This is an example of a substituent effect. The greater acidity of chloroacetic acid can be attributed, in part, to the extra electron-attracting inductive effect of the electronegative chlorine atom. By adding its inductive effect to that of the carbonyl group and the oxygen, it makes the hydroxyl proton of chloroacetic acid even more positive than that of acetic acid. It also stabilizes the chloroacetate ion that is formed when the proton is lost by dispersing its negative charge: O

O Cl CH2

C

O

H

Cl



 H2O

CH2

C

O



 H3O

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Dispersal of charge always makes a species more stable, and, as we have seen now in several instances, any factor that stabilizes the conjugate base of an acid increases the strength of the acid.

1.32 THE EFFECT OF THE SOLVENT ON ACIDITY In the absence of a solvent (i.e., in the gas phase), most acids are far weaker than they are in solution. In the gas phase, for example, acetic acid is estimated to have a pKa of about 130 (a Ka of ∼ 10−130)! The reason is this: when an acetic acid molecule donates a proton to a water molecule in the gas phase, the ions that are formed are oppositely charged particles and the particles must become separated: O CH3

O

C

OH



H 2O

CH3

C

O



H 3O 

In the absence of a solvent, separation is difficult. In solution, solvent molecules surround the ions, insulating them from one another, stabilizing them, and making it far easier to separate them than in the gas phase. In a solvent such as water, called a protic solvent, solvation by hydrogen bonding is important. ●

A protic solvent is one that has a hydrogen atom attached to a strongly electronegative element such as oxygen or nitrogen.

A protic solvent, therefore, can form hydrogen bonds to the unshared electron pairs of an acid and its conjugate base, but they may not stabilize both equally. ●

The stability of a conjugate base is enhanced if it is solvated to a greater extent than the corresponding acid.

Relative acidity cannot be predicted solely on the basis of solvation, however. Steric factors affecting solvation, and the relative order or disorder of the solvent molecules (entropic parameters), can enhance or decrease acidity.

1.33 A MECHANISM FOR AN ORGANIC REACTION Virtually all organic reactions fall into one of the four categories: Substitution, addition elimination or rearrangement. 1. Substitutions are the characteristic reactions of saturated compounds such as alkanes and alkyl halides and of aromatic compounds (even though they are unsaturated). In a substitution, one group replaces another. For example, chloromethane reacts with sodium hydroxide to produce methyl alcohol and sodium chloride: H3C

Cl 1 Na1OH2

H2O

H3C

OH 1 Na1Cl2

In this reaction, a hydroxide ion from sodium hydroxide replaces the chlorine of methyl chloride. We shall study this reaction in detail in Chapter 7. 2. Additions are characteristic of compounds with multiple bonds. Ethene, for example, reacts with bromine by an addition. In an addition reaction, all parts of the adding reagent appear in the product; two molecules become one. H

H C

H

C

1 Br H

Br

CCl4

H

H

H

C

C

Br Br

H

1.33 | A Mechanism for an Organic Reaction

3. Eliminations are the opposite of additions. In an elimination, one molecule loses the elements of another small molecule. Elimination reactions give us a method for preparing compounds with double and triple bonds. In Chapter 7, for example, we shall study an important elimination called dehydrohalogenation, a reaction that is used to prepare alkenes. In dehydrohalogenation, as the word suggests, the elements of a hydrogen halide are eliminated. An alkyl halide becomes an alkene: H H

H

C

C

H

Br

H

H

KOH

H C

(2HBr)

C

H

H

4. In a rearrangement reaction, a molecule undergoes a reorganization of its constituent parts. For example, heating the following alkene with a strong acid causes the formation of another isomeric alkene: H H3C

H C

C

C H3C

acid cat.

H

H3C

CH3 C

C

H3C

CH3

CH3

In this rearrangement, not only have the positions of the double bond and a hydrogen atom changed, but a methyl group has moved from one carbon to another. Let us consider now one mechanism as an example, dissolving tert-butyl alcohol in concentrated (concd) aqueous hydrochloric acid soon results in the formation of tert-butyl chloride. The reaction is a substitution reaction: CH3 C

H3C

CH3 OH



H

CH3



O

H  Cl



H2O

H3C

Cl  2 H2O

CH3

H

tert-Butyl alcohol (soluble in H2O)

C

Concd HCl

tert-Butyl chloride (insoluble in H2O)

Considerable experimental evidence, indicates that the reaction occurs in the following way.

Reaction of tert-Butyl Alcohol with Concentrated Aqueous HCl

MECHANISM Step 1

CH3 H3C

C CH3

CH3 H O

H  H

O



H

H

H3C

C

O H 

CH3

O

H

H

tert-Butyloxonium ion tert-Butyl alcohol acts as a base and accepts a proton from the hydronium ion. (Chloride anions are spectators in this step of the reaction.)

The products are a protonated alcohol and water (the conjugate acid and base).

(continues on next page)

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Chapter 1 | Basic Principles of Organic Chemistry

Step 2

CH3 H3C

H

C

O

H

CH3 

H

H3C

C

H

O



CH3

CH3

Carbocation The bond between the carbon and oxygen of the tert-butyloxonium ion breaks heterolytically, leading to the formation of a carbocation and a molecule of water.

Step 3

CH3

CH3 H3C

C

Cl





H3C

C

CH3

Cl

CH3 tert-Butyl chloride

The carbocation, acting as a Lewis acid, accepts an electron pair from a chloride ion to become the product.

SOLVED EXAMPLES 1. There are two constitutional isomers with the formula C2H6O. Write structural formulas for these isomers.

3. Write each of the following condensed structural formulas as a bond-line formula:

Solution Based on the fact that a carbon atom can form four covalent bonds, oxygen can form two, and hydrogen only one, we can arrive at the following constitutional isomers H H

C

H O

H

C

H

H

H

H

H

C

C

H

H

Dimethyl ether

O

OH

B

C

OH

OH D

(c) (CH3)2C

CHCH2CH3

(f) CH2 C(CH2CH3)2

2. Which of the following compounds are constitutional isomers of one another?

A

(b) (CH3)2CHCH2CH2OH

(e) CH3CH2CH(OH)CH2CH3

Ethanol

OH

(a) (CH3)2CHCH2CH3

(d) CH3CH2CH2CH2CH3

H

It should be noted that these two isomers are clearly different in their physical properties. At room temperature and 1 atm pressure, dimethyl ether is a gas. Ethanol is a liquid.

O

isomers of each other. A, C, and E also have the same molecular formula (C3H6O) and are constitutional isomers of one another.

(g)

O CH3CCH2CH2CH2CH3

(h) CH3CHClCH2CH(CH3)2 Solution CH3

(a) CH3

E

Solution First determine the molecular formula for each compound. You will then see that B and D have the same molecular formula (C4H8O) but have different connectivities. They are, therefore, constitutional

CH

CH2

=

CH3

CH3

(b) CH3

(c)

CH

CH2

CH3 CH3

CH2

= OH

H C

C

CH2

CH3

=

OH

Solved Examples (d)

CH2

CH3

CH2

(e) CH3

CH2

CH2

CH2

CH

H

=

CH3

(c)

CH3

=

CH2

(f) CH2 C

(d)

CH3

CH2

O

(g)

C

CH3

CH2

CH2

Cl

(h) CH3

CH

CH2

CH3 =

=

CH

O

CH3

(a) and (d) are constitutional isomers with the molecular formula C5H12. (b) and (e) are constitutional isomers with the molecular formula C5H12O. (c) and (f) are constitutional isomers with the molecular formula C6H12. 5. Write the resonance structure that would result from moving the electrons as the curved arrows indicate. Include formal charges if needed. C

? O

−

C

C

H

?

H

?

H H

+

N

C H

(d)

C

C

N

?

O

O2

C

C H

(b)

2

C H

CH3

The resonance structure on the left would be the major contributor. There are two contributing factors; first, the resonance structure on the left has one more formal bond, the C O bond, compared to the one on the right. Second, in the resonance structure on the left there is no formal charge separation as there is the one on the right. 7. Indicate the number of bonds and lone pairs of electrons on each of the following atoms: (a) A neutral carbon (b) A positively charged oxygen (c) A negatively charged nitrogen (d) A positively charged carbon

Four bonds and zero lone pairs. Three bonds and one lone pair. Two bonds and two lone pairs. Three bonds and zero lone pairs.

Br Cl Cl

and Cl and

H

1

ClCH2CH(CH3)2

(d) O

H

C H

C H

Cl

C

Cl H

C H

2

Br

H

(c) H C Cl and

O H

2

N

O2 C1 H3C CH3

H

Solution

H

C

H

(b)

H

(a)

C

Solution

(a)

H −

N

8. Consider each pair of structural formulas that follow and state whether the two formulas represent the same compound, whether they represent different compounds that are constitutional isomers of each other, or whether they represent different compounds that are not isomeric.

H

(c)

C

6. Which resonance structure shown below for acetone would contribute more to the overall structure of acetone? Explain.

(a) (b) (c) (d)

H

H

H C

Solution

O

(b)

H

Cl

Solution

H

N

H

H 2

H

1

C

O

4. Which molecules in Solved Example 3 form sets of constitutional isomers?

(a)

H

H3C

CH3 CH2

1

C

H

=

CH3

H

N

H

OH

OH

H

Cl

H

F F

and

F F

CH3

(e) CH3 C CH2Cl and CH3

Cl

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Chapter 1 | Basic Principles of Organic Chemistry

(f) CH2 CHCH2CH3 (g) (h)

O

and

O

arrangement of atoms or groups within the molecule. Based in this, following are the structural isomers for C6H14:

and

CH3

CH3CH2

and

CH2CH3

H3C

9. Which of the following pairs of compounds are positional isomers?

CH

CH2

O

CH2

CH3

H3C

CH3

CH2

CH

CH2

CH2

C

CH

CH2

H3C

CH2

C

C

CH2

CH3

H3C

CH3

CH

CH3

O

CH

CH3

CH3 CH3

2,2-Dimethyl butane

2,3-Dimethyl butane

11. For the following write all possible resonance structures. Be sure to include formal charges where appropriate. 2

(c)

O

O

(b) Br

(3) CH3 CH2 CH2 C CH3 and O CH2

CH3

(a)

CH3

CH3

3-Methyl pentane

CH3

O CHO

CH2

CH3

(2) CH3 CH2 CH2 C CH3 and

CH3

CH2

2-Methyl pentane

(1) CH3 CH2 CH2 CH2 CHO and

CH3

CH2

CH3

Different compounds, not isomeric Same compound Same compound Same compound Same compound Constitutional isomers Different compounds, not isomeric Same compound

CH3

CH2

n-Hexane

Solution (a) (b) (c) (d) (e) (f) (g) (h)

CH2

O

(d)

O

N

1

Solution 2

(a)

2

(4) CH3 CH2 C CH2 CH3 and

O

2

O

O2

O

O CH3 CH3

1

CH

CH2

CHO

Br

O 1

Br

O

1

O

Br

1

O

(b)

Solution (3) Compounds that differ in the position of the substituent or the functional group on the parent carbon chain are known as the positional isomers. O

Br

2

O

O CH2

C

O2 1

1

CH2

CH3 are positional isomers.

O

(3) 5

(2) 4

(4) 6

Solution (3) Structural isomers refers to the compounds that have same molecular formula but different

2O

2

10. The number of structural isomers for C6H14 is (1) 3

2

1

(c)

Thus, CH3 CH2 CH2 C CH3 and CH3

1

1

O

1

N

O

N

2

(d) O

1

O

1

C

2

N

C

2

N

Solved Examples 12. In allene (C3H4), the type(s) of hybridization of the carbon atoms is (are) (1) sp and sp3 (2) sp2 and sp 2 (3) only sp (4) sp2 and sp3 Solution

Solution (3) 2-Butene exhibits geometrical isomerism. In rest of the compounds, two identical groups are attached to the doubly bonded carbon atom, so their geometrical isomerism is not possible.

H H

π

sp 2 sp sp 2

H

13. Which of the following compounds will exhibit geometrical isomerism? (1) 3-Phenyl-1-butene (3) 1,1-Diphenyl-1-propane (2) 2-Phenyl-1-butene (4) 1-Phenyl-2-butene Solution

Ph-H2C H

C

C

(cis)

H 3C

C

C

will

CH3

H3C

cis

exhibit

geometrical

Ph-H2C

H

H

C

For a CO2 molecule to have a zero dipole moment, the bond moments of the two carbon–oxygen bonds must cancel each other. This can happen only if molecules of carbon dioxide are linear. C

O

(trans)

m

CH3

0D

18. Write bond-line formulas for three esters with the formula C5H10O2. Solution O

O

O CH3

5s and 1p

H

H F N

3s and 1p

N

F Cl H

C

C

N

H

5s and 1p

H

2s and 2p

15. Which one the following does not have sp2 hybridized carbon? (1) Acetone (3) Acetonitrile (2) Acetamide (4) Acetic acid Solution

1

CH3

others

19. Arrange the following compounds according to their expected boiling points, with the lowest boiling point first, and explain your answer. Notice that the compounds have similar molecular weights. O OH

Cl C

O

O

(2) The number of s and p bonds in the given molecule are: H

H

Solution

O

C

CH3

17. Although molecules of CO2 have polar bonds (oxygen is more electronegative than carbon), carbon dioxide has no dipole moment. What can you conclude about the geometry of a carbon dioxide molecule?

C

Solution

C

C

H

14. Which of the following molecules has two sigma (s) and two pi (p) bonds? (1) C2H4 (3) C2H2Cl2 (2) N2F2 (4) HCN

H

C

trans

O

CH3

H

H

π

Cσ CσC

(4) 1-Phenyl-2-butene isomerism.

H

H

(2) In allene molecule (C3H4), the central carbon atom is sp hybridized and other terminal carbon atoms are sp2 hybridized.

sp3

sp

(3) Acetonitrile ( CH3 C N) has no double bonds and hence no sp2 hybridized carbon. 16. The alkene that exhibits geometrical isomerism is (1) propene. (3) 2-butene. (2) 2-methylpropene. (4) 2-methyl-2-butane.

Diethyl ether

sec-Butyl alcohol

Pentane

Solution Pentane < Diethyl ether < sec-Butyl alcohol Increasing boiling point

Pentane has no polar groups and has only dispersion forces holding its molecules together. It would have the lowest boiling point. Diethyl ether has the polar ether group that provides dipole–dipole forces which are greater than dispersion forces, meaning it would have a higher boiling point than pentane. sec-Butyl alcohol has an } OH group that can form strong hydrogen bonds; therefore, it would have the highest boiling point.

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Chapter 1 | Basic Principles of Organic Chemistry 20. Identify the functional group in the below compound which are circled.

O

(c)

S H3C N H H3C

O

O

OH

(a)

H3C

H3C

O

CH3

(c)

alkene

N H ether

ester

O

O

H

alcohol

OH

O H3C

H3C

O ketone

OH CH3

alcohol

21. Write structural formulas for each of the following: (a) Three ethers with the formula C4H10O. (b) Three primary alcohols with the formula C4H8O. (c) A secondary alcohol with the formula C3H6O. (d) A tertiary alcohol with the formula C4H8O. (e) Two esters with the formula C3H6O2. Solution (a) Me

O

O

O

Me OH

OH

OH

because its molecules can form

hydrogen bonds to each other through its O H group.

Solution S

OH

OH because its molecules can form hydrogen bonds to each other through its O H group.

(b) HO

OH

H3C

or

OH

Solution

H

O

H3C

O

(d)

or

OH because its molecules can form hydrogen bonds to each other.

(d)

OH same as reason (c).

23. Explain why 1-pentanol has a solubility of 2.7g per 100 mL of water whereas methanol is completely miscible in water. Solution 1-Pentanol contains five carbons while methanol has only one carbon. The } OH group in 1-pentanol does confer some water solubility (like-dissolves-like) however; with the hydrocarbon portion of the molecule being rather large (C5H11 } ) the molecule looks mostly like an alkane, which would have no water solubility. Methanol does not look as much like an alkane; half alkane and half water. In methanol the } OH group (aids water solubility) is a larger contributor to the overall structure. 24. Hydrogen fluoride has a dipole moment of 1.83 D; its boiling point is 19.34°C. Ethyl fluoride (CH3CH2F) has an almost identical dipole moment and has a larger molecular weight, yet its boiling point is −37.7°C. Explain. Solution

(b)

OH

OH OH

(c)

OH

(d) O

(e) H

O O

O

Me

22. Which compound in each of the following pairs would have the higher boiling point? Explain your answers. (a)

OH

or

(b)

OH

or

O HO

OH

The attractive forces between hydrogen fluoride molecules are the very strong dipole-dipole attractions that we call hydrogen bonds. The partial positive charge of a hydrogen fluoride molecule is relatively exposed because it resides on the hydrogen nucleus. By contrast, the positive charge of an ethyl fluoride molecule is buried in the ethyl group and is shielded by the surrounding electrons. Thus the positive end of one hydrogen fluoride molecule can approach the negative end of another hydrogen fluoride molecule much more closely, with the result that the attractive force between them is much stronger. 25. Why does one expect the cis isomer of an alkene to have a higher boiling point than the trans isomer?

Solved Examples Solution

Solution The cis isomer is polar while the trans isomer is non-polar. The intermolecular attractive forces are therefore greater in the case of the cis isomer, and thus its boiling point should be the higher of the two. 26. Explain why the boiling point of 2,2-dimethylbutane (49.7°C) is lower than the boiling point of hexane (69°C) despite the fact that they have the same molecular formula, C6H14. Solution

(3) 6

Cl 1

2 3

5

Br

4

According to the order of precedence of groups the substituents are ordered alphabetically and the position is mentioned by the first carbon of the double bond. 29. The IUPAC name of the following compound is

Hexane is a linear, unbranched alkane. Because of the lack of branching, the hexane chains can get close to one another. The intermolecular attractive forces, although rather weak, can reinforce each other resulting in the chains being held together more closely. The energy supplied as heat, first goes to getting the hexane chain apart or away from one another and then into the vapor phase. 2,2-dimethylbutane is a branched hexane constitutional isomer. Because of the high degree of branching the chains cannot get close to one another, and thus are not held together as tightly as in hexane. When energy is supplied as heat, it does not have to initially get the chains apart from one another, since they are not held so closely together, resulting in the lower boiling point. 27. Designate the Lewis acid and Lewis base in each of the following reactions: Cl

(a) CH3CH2

Cl 1 AlCl3

CH3CH2

1

Al2 Cl

Cl

(1) (2) (3) (4)

3-ethyl-4, 4-dimethylheptane. 1, 1-diethyl-2, 2-dimethylpentane. 4, 4-dimethyl-5, 5-diethylpentane. 5, 5-diethyl-4, 4-diemthylpentane.

Solution (1) Naming of given organic compound is done by following step wise process: (i) Selection of longest chain (ii) Numbering is done from that end having lowest set of locants for substituents. (iii) Writing the name by using the below formula. Name of substituent with position in alphabetical order + Root word + Suffix (nature of functional group) 6

Cl F

(b) CH3

OH 1 BF3

CH3

7

Lewis base

Cl CH3CH2

Cl1 Al2 Cl

Lewis acid

Cl F

(b) CH3 Lewis base

OH 1 BF3

CH3

Lewis acid

1

O

B2 F

H

F

28. The IUPAC name of the compound shown below is Cl

1

30. The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is (1) } COOH, } SO3H, } CONH2, } CHO (2) } SO3H, } COOH, } CONH2, } CHO (3) } CHO, } COOH, } SO3H, } CONH2 (4) } CONH2, } CHO, } SO3H, } COOH Solution (1) According to latest IUPAC system priority order of various functional group is } COOH > SO3H > } CONH2 > CHO 31. The shape of a carboanion is (1) linear. (3) pyramidal. (2) planar. (4) tetrahedral.

Br

(1) (2) (3) (4)

3

3-Ethyl-4, 4-dimethylheptane

F

Solution Cl 1 AlCl3

2

O1 B2 F H

(a) CH3CH2

5

4

2-bromo-6-chlorocyclohex-1-ene 6-bromo-2-chlorocyclohexene 3-bromo-1-chlorocyclohexene 1-bromo-3-chlorocyclohexene

Solution (3) The carbon atom bearing the negative charge in the caboanion is in sp3 hybridized state. It is bonded to three other atoms and the unshared

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Chapter 1 | Basic Principles of Organic Chemistry release of proton. As the distance between Cl− and COOH increases, the dispersal of negative charge of the corresponding carboxylate ion becomes less pronounced, and hence acidic strength decreases. The pKa values of the compounds are:

electron pair occupies the apex of the tetrahedron. Thus, the shape of carboanion is pyramidal. R R C2 R sp 3 hybridized carbon

32. The increasing order of stability of the following free radicals is (1) (CH3)2 CH < (CH3)3 C < (C6H5)2 CH < (C6H5)3 C

Compound

pKa

HCOOH

3.75

CH3

CH2

CH

COOH

2.86

(3) (C6H5)2 CH < (C6H5)3 C < (CH3)3 C < (CH3)2 CH

CH2

CH2

Cl CH2

COOH

4.52

(4) (CH3)2 CH < (CH3)3 C < (C6H5)3 C < (C6H5)2 CH

Cl

(2) (C6H5)3 C < (C6H5)2 CH < (CH3)3 C < (CH3)2 CH

CH3COOH

Solution (1) The relative stability of free radicals is measured through the homolytic bond dissociation energies of R } H bonds. Greater is the value of bond dissociation energy the lesser is the stability. Tertiary C } H bond is easier to break than secondary C } H bond. Therefore, the order is 3° > 2° > 1° > methyl. 2 2

2

35. Supply the curved arrows necessary for the following reactions: O

(a)

2

C

2

2

2

(1) C6H5CH2 > CCl3 > (CH3)3C > (CH3)2CH 2

2

2

2

2

2

C

H

H

2

2

O

1 H

O

O

O 2

C H

O

CH3

1 O

H

H

2

H

C

O

O

CH3

Solution

2

(3) CCl3 > C6H5CH2 > (CH3)2CH > (CH3)3C 2

1 O

2

(2) (CH3)2CH > CCl3 > C6H5CH2 > (CH3)3C 2

H

H

(b)

2

O

2

2

(CH3)3C,CCl3,(CH3)2CH,C6H5CH2 in order of their decreasing stability:

O 2

H

33. Arrange the carbanions, (CH3)3C,CCl3,(CH3)2CH,C6H5CH2 2 2

4.76

O

2

(4) (CH3)3C > (CH3)2CH > C6H5CH2 > CCl3 H3C

Solution (3) Because in CCl3−, Cl− being electronegative stabilizes the carbanion, in C6H5CH2−, the presence of phenyl group stabilizes it due to resonance and in case of secondary and tertiary, the order of reactivity is 2° > 3°.

O

O

(a)

H3C

H Na1 H2 O

(b)

O2Na1

H 1 Na1OH2

1 H2

O2Na1 1 H2O

CH22 2CCl 3

.

Resonance

. CH3

CH3

CH3

CH . CH3

C

2

2

CH3

1I Effect

34. The strongest acid amongst the following compounds is (1) HCOOH (2) CH3CH2CH(Cl)COOH (3) ClCH2CH2CH2COOH (4) CH3COOH

36. In the anion HCOO− the two carbon–oxygen bonds are found to be of equal length. What is the reason for it? (1) Electronic orbitals of carbon atom are hybridized. (2) The C O bond is weaker than the C } O bond. (3) The anion HCOO– has two resonating structures. (4) The anion is obtained by removal of a proton from the acid molecule. Solution (3) The anion HCOO− can be represented as following resonating structures.

Solution (2) Cl– being an electron withdrawing group withdraws electron density, thereby facilitating the

O H

C

O2 O2

H

C

O

Solved Previous Years’ NEET Questions gets finally attached, the effect is called +E effect. For example, addition of acids to alkenes.

37. Addition of acids to alkenes is an example of (1) +I effect (3) +E effect (2) −I effect (4) −E effect

C

Solution

C

1

1 H1

C

C H

(3) If the electrons of the p bond are transferred to that atom of the double bond to which the reagent

SOLVED PREVIOUS YEARS’ NEET QUESTIONS 1. Base strength of the following compounds is in the order of (I) H3C

CH22

(II) H2C

CH2

(III) H

(1) (2) (3) (4)

sp, sp2, sp3 and sp2 sp, sp3, sp2 and sp3 sp3, sp2, sp2 and sp sp, sp2, sp2 and sp3 (AIPMT 2009)

C2

C

(1) I > II > III (2) II > I > III

(3) III > II > I (4) I > III > II (AIPMT 2008)

Solution (1) The protons of ethyne are more acidic than those of ethene, which in turn are more acidic than those of ethane. The stronger the acid, the weaker will be its conjugate base. Relative acidity of hydrocarbons CH > H2C

HC

CH2 > H3C

CH22 > H2C CH

CH

5

6

CH2

4

3

sp; C3

sp3; C5

C2

CH

C 2

sp2 and

4. Which of the following compounds will exhibit cis-trans (geometrical) isomerism? (1) 2-Butenol (2) 2-Butene (3) Butanol (4) 2-Butyne (AIPMT 2009) Solution

CH22 > HC

2. In the hydrocarbon CH3

(2) In the compound, C2 C6 sp3.

CH3

Relative basicity of carbanions H3C

Solution

(2) For geometrical isomerism, the two atoms or groups attached to the double-bonded carbon should be different.

1

The states of hybridization of carbons 1, 3 and 5 are in the following sequence: (1) sp3, sp2, sp, (3) sp, sp3, sp2 2 3 (2) sp , sp, sp (4) sp, sp2, sp3

H

CH3

CH3

C C H

(AIPMT 2008)

H C C

CH3

cis-2-Butene

H

CH3

trans-2-Butene

Solution 5. Which of the following species is not electrophilic in nature? 1 (1) BH3 (3) NO2

(3) The hybridization is 6

CH3

5

CH

4

CH

sp2

3

2

CH2

C

sp3

1

CH sp

1

3. The state of hybridization of C2, C3, C5 and C6 of the hydrocarbon CH3 7

CH3

6

C

CH3 5

CH

4

CH

3

CH

CH3

is in the following sequence:

2

C

1

CH

(2) H3O

1

(4) C l (AIPMT MAINS 2010)

Solution (2) Among the following given species, H3O+ has lone pair of electrons for donation, thus it is not electrophilic in nature.

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86

Chapter 1 | Basic Principles of Organic Chemistry 6. The correct order of increasing bond length of C } H, C } O, C } C and C C is (1) C } H < C } O < C } C < C C (2) C } H < C C < C } O < C } C (3) C } C < C C < C } O < C } H (4) C } O < C } H < C } C < C C (AIPMT 2011)

9. Which of the following organic compounds has same hybridization as its combustion product (CO2)? (1) Ethane (3) Ethene (2) Ethyne (4) Ethanol (AIPMT 2014) Solution (2) The combustion reaction is

Solution (2) C } H has the lowest bond length due to the smallest size of hydrogen followed by C C bond length. In C } O, because of greater electronegativity of oxygen atom, the bond strength of C } O is stronger than that of C } C, consequently, bond length of C } O is shorter than that of C } C. Hence the correct order is C } H < C C < C } O < C } C. 7. Considering the state of hybridization of carbon atoms, find out the molecule among the following which is linear? (1) CH3 } CH2 } CH2 } CH3 (2) CH3 } CH CH } CH3 (3) CH3 } C C } CH3 (4) CH2 CH } CH2 } C CH (AIPMT PRE 2011) Solution

C2H2 1

C

C

(1)

OH COOH

H CH3

H 3C H 2C

H CH3

(1) 8 (2) 12

(3) 16 (4) 4 (AIPMT 2015)

Solution (1) There are four p-bonds in the given structure as shown below. H H

CH3 π CH3

H

π

CH3

π

π CH2

(3) COOH

H CH3

Each p-bond contains two electrons; therefore, total number of p-bond electrons is eight. 11. Consider the following compounds: CH3 CH3

C

Ph C

Ph

C

Ph

CH3

CH3 H

OH

(2)

H H

H 3C

sp sp

OH

COOH

(4)

COOH

OH

(NEET 2013)

CH

10. The total number of p-bond electrons in the following structure is

CH3

8. Structure of the compound whose IUPAC name is 3-ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid is

2CO2 1 H2O

The hybridization of both CO2 and CH is sp.

(3) sp hybridized carbon atoms consist of linear arrangements of atoms. The bond angle between them is 180°. CH3

5 O 2 2

(I)

(II)

(III)

Hyperconjugation occurs in (1) II only. (3) I and III. (2) III only. (4) I only. (AIPMT 2015)

Solution

Solution

(2)

OH 3 6

2

1

COOH

4 5

3-Ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid

(2) When s-electrons of a C } H bond are in conjugation with an adjacent p-bond, then this type of conjugation is called hyperconjugation. The presence of α-hydrogen at an alkene, alkyl free radical is the main condition for hyperconjugation.

Solved Previous Years’ NEET Questions 12. In which of the following compounds, the C } Cl bond ionization shall give most stable carbonium ion? H3C

(1)

C

H3C

Cl

(3)

CH3 H CH

(2)

H

C O2NH2C H

H3C H (4) C H3C

Cl

Cl

(1) 3 and 4 (2) 2 and 4

(NEET-II 2016) Solution

Cl

(AIPMT 2015)

(3) In pyrrole, the four p-electrons are contributed by the carbon atoms of the pyrrole ring and two by sp2 hybridized nitrogen to complete the aromatic sextet. The possible resonance structures are as follows: 4

Solution

5

(1) The order of stability of carbonium ions is as follows: H3C H3C

1

C

(3) 2 and 5 (4) 2 and 3

H3C

1

CH2 .

.

H3C

CH3

2

2

N H

N1 H

(II)

2

(III)

C1 O2NH2C

2

Tertiary carbocation is most stable followed secondary carbocation whose stability is equivalent to benzylic carbocation. Primary carbocation is least stable.

(RE-AIPMT 2015)

2

N1 H (V)

N1 H (IV)

From the resonance structures, we can see that the maximum electron density is at position (2) and (5) in the ring as resonating structures (III) and (IV) are more stable than (II) and (V) so are the major contributors.

13. The number of structural isomers possible from the molecular formula C3H9N is (1) 2 (3) 4 (2) 3 (4) 5

16. Which one is the correct order of acidity? (1)

Solution (3) There are four structural isomers possible for molecule C3H9N. The structures are CH3

N1 H

(I)

H CH1 .

3

CH2 CH3

CH2

NH2,

CH3

CH2

CH

CH3

CH3

N

NH2

NH

CH

CH

CH3

C

−

is present in which of the following orbitals? (1) sp2 (3) 2p (2) sp (4) sp3 (NEET-I 2016) Solution (2) In the carbanion, the carbon has one s bond, two p bonds and one lone pair. Therefore, carbon is sp hybridized. 15. In pyrrole 4 5

2

N1 H

the electron density is maximum on

CH2 > CH3

CH3

C

CH > CH3

CH3

CH3 > CH2

CH2 > CH3

C

CH > CH

CH

CH2 > CH3

C

CH > CH

CH

(4) CH2

CH2 > CH3

CH

(NEET 2017) Solution (1) The hydrogen bonded to the carbon of a terminal alkyne, called an acetylenic hydrogen atom, is considerably more acidic than those bonded to carbons of an alkene or alkane. Hybridization of terminal carbon is shown as: CH

3

CH > CH2

CH2 > CH3

(3)

CH3

14. The pair of electron in the given carbanion, CH 3C

C

(2)

CH3

CH3

CH > CH2

CH > CH3

sp

CH > CH3

C

sp

CH > CH2

sp 2

CH2 > CH3

sp 3

CH3

Greater the s orbital character in carbon hybridization, greater is its acidity. Being the most electronegative, the sp-hybridized carbon atom of ethyne polarizes its C } H bonds to the greatest extent, causing its hydrogens to be most

87

88

Chapter 1 | Basic Principles of Organic Chemistry positive. Therefore, ethyne donates a proton to a base more readily. Alkyl being an electron donating group reduces the acidity of alkynes.

(4) electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from a nucleophile.

17. The correct statement regarding electrophile is (1) electrophile is negatively charged species and can form a bond by accepting a pair of electrons from another electrophile. (2) electrophiles are generally neutral species and can form a bond by accepting a pair of electrons from a nucleophile. (3) electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile.

(NEET 2017) Solution (3) Electrophiles are reagents that seek electrons so as to achieve a stable shell of electrons like that of a noble gas. They can be neutral (BF3 and AlCl3) or positively charged ( CH +3 ) species and can form a bond by accepting a pair of electrons from a nucleophile.

ADDITIONAL OBJECTIVE QUESTIONS NCERT Exemplar Questions

6. What is the correct order of decreasing stability of the following cations?

1. Which of the following is the correct IUPAC name? (1) 3-Ethyl-4, 4-dimethylheptane (2) 4,4-Dimethyl-3-ethylheptane (3) 5-Ethyl-4, 4-dimethylheptane (4) 4,4-Bis(methyl)-3-ethylheptane O

2. The is (1) (2) (3) (4)

O

IUPAC name for CH3 CH2 C CH2 C OH

Cl NO2

is CH3

(1) (2) (3) (4)

CH (I)

CH3

CH3

CH OCH3 (II)

CH3

1

CH

CH2 (III)

OCH3

1

(1) II > I > III (2) II > III > I

(3) III > I > II (4) I > II > III

7. Correct IUPAC name for CH3 CH CH CH3 is ___________. C2H5 C2H5 (1) 2- ethyl-3-methylpentane (2) 3,4- dimethylhexane (3) 2-sec-butylbutane (4) 2, 3-dimethylbutane

1-hydroxypentane-1,4-dione. 1,4-dioxopentanol. 1-carboxybutan-3-one. 3-oxopentanoic acid.

3. The IUPAC name for

1

CH3

1-chloro-2-nitro-4-methylbenzene. 1-chloro-4-methyl-2-nitrobenzene. 2-chloro-1-nitro-5-methylbenzene. m-nitro-p-chlorotoluene.

4. Electronegativity of carbon atoms depends upon their state of hybridization. In which of the following compounds, the carbon marked with asterisk is most electronegative? (1) CH3 } CH2 } *CH2 } CH3 (2) CH3 } *CH CH } CH3 (3) CH3 } CH2 } C *CH (4) CH3 } CH2 } CH *CH2 5. In which of the following, functional group isomerism is not possible? (1) Alcohols (3) Alkyl halides (2) Aldehydes (4) Cyanides

8. In which of the following compounds, the carbon marked with asterisk is expected to have greatest positive charge? (1) *CH3 } CH2 } Cl (2) *CH3 } CH2 } Mg+Cl– (3) *CH3 } CH2 } Br (4) *CH3 } CH2 } CH3 9. Ionic species are stabilized by the dispersal of charge. Which of the following carboxylate ion is the most stable? O

(1) CH3 C

O2 O

(2) Cl CH2 C

O2

O

(3) F CH2 C (4)

F F

O2

O CH

C

O2

Additional Objective Questions 10. Electrophilic addition reactions proceed in two steps. The first step involves the addition of an electrophile. Name the type of intermediate formed in the first step of the following addition reaction. H3C } HC (1) 2° Carbanion (2) 1° Carbocation

4

CH2 + H+ →? (3) 2° Carbocation (4) 1° Carbanion

11. Covalent bond can undergo fission in two different ways. The correct representation involving a heterolytic fission of CH3 } Br is Br

CH3 1 Br2

(2) CH3

Br

CH3 1 Br2

(3) CH3

Br

2

CH3 1

(4) CH3

Br

CH3 1 Br

5

3

(1) sp2, sp2 (2) sp2, sp

(3) sp2, sp3 (4) sp, sp2

5. Which of the following is a set of constitutional isomers? Br Br

1

C

C

(3) H1

(2) H1

C

C

(4) All of these are possible

C

C

Exercise 1 1. How many sigma 1s-2sp3 bonds are there in ethane? (1) 7 (3) 5 (2) 6 (4) 3 2. Which is the correct bond-line formula for the following structure? H

H

C

C

C

C

H

H

H

H

(1)

(3)

(2)

(4)

Br

Br

(1) H1

H

(III)

(I)

Br1

12. The addition of HCl to an alkene proceeds in two steps. The first step is the attack of H+ ion to C C portion which can be shown as

H

H

C

C

(II)

(1)

(2)

N

(3)

O

H N

O

O

N

(4)

O

N

7. Which of the following is the Lewis structure for CH3CH2O2H? H

H

C

C

H

H

H

H

(2) H

C

O

O

H

O

H

H O

C

H

O

H

C

C

O

H

H

H

(3) H

(4) H

C H

H

H

H

CH3CH2CH(OH)CHCH2CH3 CH3CH2CH(OH)CH(CH2CH3)2 CH3CH2(OH)CH(CH2CH3)CH3 CH3CH2CH2(OH)CH(CH2CH3)CH2CH3

(3) I, II, and III (4) I, III, and IV

6. Which compound is not a constitutional isomer of the others?

H

OH

(IV)

(1) I and II (2) II and III

(1) H

3. Which is the correct condensed formula for the following structure?

(1) (2) (3) (4)

6

7

1

2

1

(1) CH3

H

4. Identify the atomic orbitals involved in the C } 2 and C } 3 sigma bond (indicated by an arrow) in the following molecule:

O

C

O

H

H

8. The IUPAC name of neopentane is (1) 2-methylbutane. (2) 2,2-dimethylpropane. (3) 2-methylpropane. (4) 2,2-dimethylbutane.

89

90

Chapter 1 | Basic Principles of Organic Chemistry 9. The correct IUPAC name of the following compound is

17. Which of the following compounds are classified correctly? O

O

C

(1) (2) (3) (4)

CH3

4-methyl -3-ethylhexane 3-ethyl -4-methylhexane 3, 4-ethylmethylhexane 4-ethyl-3-methylhexane

H

ketone

(I)

(II)

(III)

H

(3) Benzyl, hexyl (4) Benzyl, heptyl

Cl

1o alcohol and 2o alkyl chloride ether and 2o alcohol 1o alkyl chloride and 1o alcohol 1o alkyl chloride and 2o alcohol

14. Which of following compound can show geometrical isomerism? (3) CH

CCl

(4) (Cl)(Br)C

CH(I)

15. The carbon atom has the correct orbital hybridization indicated in which of the following structures?

sp 2

(II) CH2

O CH2

sp

(IV) CH N (V) O

sp

C

(III) CH4 (3) I, II, III (4) I, IV, V

16. cis-trans isomerism is possible only in the case of (1) CH2 CBr2 (3) BrCH CHBr (2) CH2

CHBr

(V)

(3) I, III, V (4) I, III, IV

18. What is the correct structure for the aldehyde, which has the formula C4H8O? OH

(4) Br2C

(2) CH3 C CH2 CH3 O

(3) CH3 CH2 CH2 CH (4) CH2

CH

CH2

O

CH3

OH O

(1) (2) (3) (4)

primary amino and carboxyl primary amino and ketone secondary amino and carboxyl secondary amino and ester

20. The order of stability of the following carbocations is 1

CH2 H2C

CH

1

CH2;

H 3C

(I)

O

sp 2

(1) II, IV, V (2) II, III, IV

(IV)

NH2

13. The general formula CnH2nO2 could be for open chain (1) diketones. (3) diols. (2) carboxylic acids. (4) dialdehydes.

(I) CH2

carboxylic acid

19. The following molecule contains the _________ and __________ functional groups.

OH

sp

OH

amine

O

12. What functional group(s) is(are) present in the following compound?

(2) CH3CH2CH2Cl

C CH3

(1) CH3 CH CH CH2

Ph

CHCl

CH3

N

CH3

O

(1) CH2

CH3

aldehyde

(1) III, IV, V (2) II, III, IV

11. What alkyl groups make up the following ketone?

(1) (2) (3) (4)

C CH3

O

1-Cyclopropylcyclobutane 1,1′-Dicyclobutane 1-Cyclobutane-1-cyclopropane None of these

(1) Phenyl, pentyl (2) Hexyl, phenyl

OH

alcohol

10. The correct IUPAC name of the following compound is

(1) (2) (3) (4)

CH3

CHBr

CH2

1

CH2;

(II)

(1) II > III > I (2) I > II > III

(III)

(3) III > I > II (4) III > II > I

21. Which if the following reactions involves a nucleophile? (I) CH3COOH 1 HO2 CH3COO2 1 H2O 2

(II) CH3COCH3 1 CN 1

(III) C6H5 1 CH3CO

(CH3)2C(CN) 1 (OH) C6H5COCH3

Additional Objective Questions (1) I and II (2) I and III

(3) III only (4) II and III

CH3

22. Which statement about contributing structures is false? (1) All contributing structures must have the same number of valence electrons. (2) All contributing structures must obey the rules of covalent bonding. (3) The position of nuclei may change. (4) Third period atoms may have up to 18 electrons around them. 23. Which of the structures below is not expected to contribute to the CO2 resonance hybrid? 1

(1) O

C

2

1

(2) O

2

(3) O

O

C

2

(4) O

O

C

O

11

2

C

H

F

(2)

CH3

F

H

(1) H C

(2) H C

C

(IV)

H

(4) H

(1) I and II (2) III, IV and V

26. Which of the following are correct resonance structures of structure I? O N

H

N

O

(1) I and II (2) II, III and IV

H

H

C

C

C

C

H

H

(IV)

H

H

H

C

C

C

C

H

H

H

H

H

C

C

C

C

H

H

H

(3) NH +4

(2) H2O

(4) HCl

30. Arrange the following species in the order of increasing acidity (weakest to strongest). (II) H3O+

N

1

(III) NH +4

(3) III, II, I (4) I, III, II NH2

H

O H

H HH

H

(1) CH3COOH

(1)

2

1

C

31. Which of the following structures are Lewis bases?

2

N

C

1

(III)

2

(II)

C

H

(1) II, III, I (2) I, II, III O

N

H

C

(I) H2O

H

O

(I)

H

29. Which of these has the lowest numerical value of pKa and is therefore the strongest acid?

(3) III and V (4) IV and V

I

BF3

H H H

(III)

1

C

(V) CH2Cl2

OH

CH3

H H H

O

(II) C5H12

H

H

(3) H

OH

CH2

H H H

25. Which of the following are polar molecules?

(I) CCl4

O

(3) Brønsted acid (4) Brønsted base

H

H

1

CH2

CH3 1 BF3

28. Which of the following is the best, correct Lewis structure for CH2C(CH3)CH2CH3?

F

H

CH2

(1) Lewis acid (2) Lewis base

(4) H

O

2

O

24. Which molecule would you expect to have no dipole moment (i.e., µ = 0 (4))? (1) CHF3 (3) :NF3 F

CH2

2

(3)

O

2

H

(3) I, II and IV (4) all of them

27. What is the role of diethyl ether in the following reaction?

(2)

(4) All of these

32. The compounds ethane, ethene, and ethyne exhibit this order of increasing acidity (1) Ethyne < ethene < ethane (2) Ethene < ethyne < ethane (3) Ethane < ethyne < ethene (4) Ethane < ethene < ethyne

91

92

Chapter 1 | Basic Principles of Organic Chemistry 33. Which of the following is the most acidic? H (1)

40. The inductive effect (1) decreases with increase of distance. (2) increases with increase of distance. (3) indicates the transfer of p pair of electron from less electronegative atom to more electronegative atom in a molecule. (4) shows the transfer of lone pair of electrons.

H

(2) H H

H

(3)

H

(4) All of these are equally acidic 34. Which of the following are acid-base reactions according to Brønsted–Lowry theory? O

(I) CH3

C

42. Which of these shows incorrect representation of inductive effect?

O 2

O

1 NH3

1 CH3

CH3

C

41. Electrophilic reagents are (1) electron pair donors. (2) Lewis acids. (3) odd electron molecules. (4) None of these.

OH

CH3 O

1 CH3

(II) AlCl3 1

Cl

NH2 1 H

(IV) CH3

1 NH3 1 H

(1) I, III (2) I, II, II, IV

AlCl2 4

2

(III) CH3

NH2

CH3

Cl O

2

CH3

1 NH3 1 Cl

(1)

O

37. The solid alkane CH3(CH2)18CH3 is expected to exhibit the greatest solubility in which of the following solvents? (1) CCl4 (3) H2O (2) CH3OH (4) CH3NH2 38. Which sets are pairs of acceptable resonance contributing structures? 2 O CH3

CH3

O

(2) CH3

C

CH3

NH2 1 H2O

36. Which of the following has the highest boiling point? (1) Pentane (3) Pentyne (2) Pentene (4) Pentyl alcohol

C

(3)

(3) I, II, III (4) I, III, IV

O

CH3

CH2

2

O

2

35. Which of the following compounds would you expect to have the highest melting point? (1) n-Butyl alcohol (3) sec-Butyl alcohol (2) Isobutyl alcohol (4) tert-Butyl alcohol

(1) CH3

C

C 1

CH3

O

H

C

CH3

(3) Both of these. (4) None of these. 39. Due to the presence of an unpaired electron, free radicals are (1) chemically reactive. (2) chemically inactive. (3) anions. (4) cations.

C

OH

(2)

(4) CH3

C

CH2

CH3

O

43. The reaction between electrophile and nucleophile involves (1) an ionic bond. (2) a covalent bond. (3) a coordinate covalent bond. (4) None of these. 44. What type of reaction is shown below? Br

HBr/CCl4

(1) (2) (3) (4)

Addition Substitution Elimination Rearrangement

Exercise 2 1. The C4–C5 carbon-carbon bond in the following molecule results from the overlap of which orbitals (in the order C4–C5)? O 7

(1) sp–sp2 (2) sp–sp3

6

5

4

3

2

1

(3) sp2–sp2 (4) sp3–sp2

Additional Objective Questions 2. Which molecule has the shortest carbon-carbon single bond?

7. Which of the following pairs are NOT resonance structures? 1

(3)

(1)

2

(1) H3C O N O and H3C O N O 1

2

(2) O C O and O C O (2)

(4)

(3) H3C

3. Which of the following represent a pair of constitutional isomers? (1) and (2) CH3CH

CH2

and

CH2

CHCH3

N

Br

(3) H

and

1

H

(II) CH3CH2CH2CH2CH2CH2CH

(1)

O

(3) 1

(2) O

CH2

Which structures can exist as cis-trans isomers? (1) I and II (3) I and IV (2) I and III (4) II and III

N H3CO H

5. Which molecule among the following does not have a dipole moment?

(IV)

CF3OH

F

O +

F

(4)

CI F

6. Which of the following species is a resonance form of the species in the box? P(CH3)2 1

P(CH3)2

1 P(CH ) 3 2

P(CH3)2 1

(4) N H

(1) (2) (3) (4)

V > III > II > IV > I II > V > III > I > IV II > III > V > IV > I III > I > V > II > IV

(1) F3C

N H

O

H

O

(3) F3C

H

O O

H

(4) F3C

O

H

O

N H P(CH3)2

1

CO2H (III)

(2) F3C

(3) N1 H

(2)

CI

10. Which of the following is the most acidic?

N H

(1)

H

(V)

(II)

F

H

O

(I)

F

F

O

1

(4)

9. Rank the bold-faced hydrogens for the following compounds from most acidic to least acidic.

CHCH2CH2CH2CH2CH2CH3

(3)

1

O

1

CHCH2CH2CH2CH3

(2)

2

O

4. Consider the following: (I) CH3CH2CH2CH CHCH2CH2CH3

F

O

Br

H

(1)

O

N

8. Which of the following species is not a resonance form of the following species?

H

(4) More than one of these

(III) CH3CH2CH

1

O and H3C

(4) Each of these pairs represents resonance structures.

Br

Br

(IV) CH2

O

11. Which acid-base reaction would not take place as written? (1) CH3Li + CH3CH2OH → CH4 + CH3CH2OLi (2) H2C CH2 + NaOH → H2C CHNa + H2O (3) CH3C CNa + H2O → CH3C CH + NaOH (4) (CH3)2CHOH + NaH → (CH3)2CHONa + H2

93

94

Chapter 1 | Basic Principles of Organic Chemistry 12. Which of the following can be applied to explain relative order of stability of carbocations? (1) Resonance (2) Inductive effect (3) Hyperconjugation (4) All of these 13. Which of the following is not a conjugate acid – conjugate base pair (in that order)? (1) H3PO4, H2PO4− (2) HBF4, BF4− (3) CH3CH2OH, CH3CH2O− (4) HPO4−, H2PO4− 14. Which of the acids below would have the strongest conjugate base? (1) CH3CH2OH pKa = 18 (2) CH3CO2H pKa = 4.75 (3) ClCH2CO2H pKa = 2.81 (4) Cl2CHCO2H pKa = 1.29

18. Which of the following are pairs of contributing structures? (I) CH3

O CH3 2

CH2

O 2

(III) CH2 CH CH2 2

N

C

O

CH2 N

(1) II, IV (2) I, II, III

OH

1

2

(II) CH2 O

(IV)

CH2

CH3

C

CH O

CH2

2

(3) III, IV (4) II, III, IV

19. In which of the following pairs, A is more stable than B? A

B

(4) Ph3C

(CH3)3C

(1)

15. What is the role of water in the following reaction? NH3 1 H3O1

(1) (2) (3) (4)

(2)

H2O 1 NH14

Acid Base Conjugate acid Conjugate base

(3)

16. Identify the conjugate acids in the following reactions. H CH3

H

O

N

H 1 CH3

C

CH3

O

H 1 H

1

CH3 O

N

H

Cl

O H 1 CH3

2

O

H (I)

20. The correct order of stability of the following free radicals is

(II) 1

CH3

C

O

2

H 1 Cl

H (III) CH3

CH2

O

1

H 1 Na H

2

CH3

CH2

2

1

O Na 1 H

(V)

(1) I, IV, VI (2) I, III, VI

O CH3O

(I)

CH3

O2

(II)

(1) II > III > IV > I (3) III > II > IV > I

Br

(IV)

(1) III > I > II > IV (2) I > II > IV > III

(3) IV > II > I > III (4) I > III > II > IV

Exercise 3 In the following set of questions, two statements “Assertion” and “Reason” are given. Choose the correct answer from the following options:

(III)

O2

(II)

H

(3) II, IV, V (4) I, III, V

O2

(III)

(VI)

17. The order of decreasing basicity of following anions is

CH3C

(I) (IV)

O2 (IV)

(2) I > IV > II > III (4) III > II > I > IV

(1) Assertion and Reason are true and the Reason is the correct explanation of the Assertion. (2) Assertion and Reason are true but the Reason is not a correct explanation of the Assertion. (3) Assertion is true but the Reason is false. (4) Assertion and Reason both are false.

Answer Key 1. Assertion: Carbon possesses property of catenation. Reason: Carbon atom forms double as well as triple bond during catenation. 2. Assertion: sp2–sp2 overlapping is more efficient than sp3–sp3. Reason: sp3 orbital has higher electrons density than sp2 orbital. 3. Assertion: The order of reactivity of carbocation ions is 2° > 3° > 1°. Reason: Carbon atom in carbocation ions is in sp3 state of hybridization. 4. Assertion: Boiling point of cis isomers is higher than trans isomers.

5. Assertion: Heterolytic fission involves the breaking of a covalent bond in such a way that both the electrons of the shared pair are carried away by one of the atoms. Reason: Heterolytic fission occurs readily in polar covalent bonds. 6. Assertion: Free radicals are short lived and highly reactive. Reason: Free radicals are highly unstable. 7. Assertion: Carbon–oxygen bonds are of equal length in carbonate ion. Reason: Bond length decreases with the multiplicity of bond between two atoms.

Reason: Dipole moment of cis isomers is higher than trans isomers.

ANSWER KEY NCERT Exemplar 1. (2)

2. (4)

3. (2)

4. (3)

5. (3)

6. (1)

7. (2)

8. (1)

9. (4)

10. (3)

11. (2)

12. (2)

3. (2)

4. (4)

5. (4)

Exercise 1 1. (2)

2. (2)

6. (4)

7. (1)

8. (2)

9. (2)

10. (1)

11. (3)

12. (4)

13. (2)

14. (4)

15. (1)

16. (3)

17. (1)

18. (3)

19. (1)

20. (3)

21. (1)

22. (3)

23. (4)

24. (2)

25. (3)

26. (3)

27. (2)

28. (4)

29. (4)

30. (4)

31. (4)

32. (4)

33. (1)

34. (4)

35. (4)

36. (4)

37. (1)

38. (1)

39. (1)

40. (1)

41. (2)

42. (4)

43. (3)

44. (1)

1. (4)

2. (4)

3. (3)

4. (2)

5. (4)

6. (3)

7. (3)

8. (1)

9. (1)

10. (4)

11. (2)

12. (4)

13. (4)

14. (1)

15. (4)

16. (2)

17. (3)

18. (3)

19. (4)

20. (1)

1. (2)

2. (3)

3. (4)

4. (1)

5. (2)

6. (2)

7. (2)

Exercise 2

Exercise 3

95

96

Chapter 1 | Basic Principles of Organic Chemistry

HINTS AND EXPLANATIONS Exercise 1 2. (2) Option (1): Greater number of carbon atoms. Option (2): Correct. Option (3): CH3 on carbon 3 is missing. Option (4): Lesser number of carbon atoms. 3. (2) Option (1): CH2CH3 on carbon 4 is missing. Option (2): Correct Option (3): OH and CH2CH3 are on the wrong carbons and a CH2CH3 is missing. Option (4): Too many hydrogen and too many bonds on carbon #3. 8. (2) The structure is of neopentane is CH3 3

2

H3C

1

C

CH3

CH3

The IUPAC name is 2,2-dimethylpropane. 9. (2) The IUPAC name of the compound is 3-ethyl-4-methylhexane 2 1

3

5

13. (2) This is a general formula for carboxylic acids. For example, for n = 2, CH3COOH 15. (1) Carbon atoms in (II), (IV) and (V) have the correct orbital hybridization indicated. 16. (3) Geometrical isomerism is not observed in a molecule if either of the doubly bounded atoms has two similar groups. 18. (3) Aldehyde has functional group } CHO. 20. (3) Stability is directly proportional to extent of delocalization of positive charge. Thus, order of stability is CH12 > CH2

CH

1 CH2 > CH3

CH2

1 CH2

21. (1) The reactions I and II involve nucleophiles, OH− being the nucleophile in the (I) and CN− being the ambident nucleophile in II. 24. (2) In the following molecule, net dipole moment cancels out. F H

C

C

H F

26. (3) Resonance structures involve the movement of electrons. You cannot move atoms and bonds or violate the octet rule when drawing resonance structures. 28. (4) In option (1): Connectivity is wrong. In a condensed structural formula, atoms in parenthesis are attached to the previous atom. In option (2): Connectivity is correct, but the number of electrons is incorrect. The first two carbons in this drawing have open octets. In option (3): Connectivity and electrons are correct. However, this structure shows charge separation and open octets. It is not the best Lewis Structure. 31. (4) Structures (1), (2) and (3) all have lone pairs that can donate electrons and act as Lewis Bases (electron pair donors). 33. (1) Alkynes are sp hybridized. Alkenes are sp2 hybridized. Alkanes are sp3 hybridized. The order of acidity is sp > sp2 > sp3.

6

4

25. (3) Molecule with a net dipole moment is a polar molecule. Molecules (I), (II) and (III) consist of more than two atoms, have polar bonds, but have no dipole moment. In molecules (IV) and (V) have net dipole moment hence, they are polar.

36. (4) For boiling point, the hydrogen bonding of the alcohol is the most powerful intermolecular force. 37. (1) A general rule for solubility is that “like dissolves like” in terms of comparable polarities, therefore, nonpolar solids are soluble in nonpolar solvents. 38. (1) For (1): A pair of resonance contributing structures. They differ only in the distribution of valence electrons. For (2): Not a pair of resonance contributing structures. They differ in the arrangement of their atoms. Oxygen is bonded to a hydrogen atom in the Lewis structure on the right, but the other structure contains no such bond. 39. (1) Free radicals are highly reactive due to the presence of unpaired electron. R

H

R 1 H (where R is an alkyl group)

42. (4) Inductive effect may be defined as the induction of polarity in an otherwise covalent bond, due to incomplete shifting of the electron pair between the two atoms that have different electronegativity. It is permanent effect.

Hints and Explanations It is represented by the symbol → It propagates through the carbon chain. CH3

C

CH2

CH3

O

This is the incorrect representation of inductive effect correct representation is CH3

C

CH2

CH3

10. (4) Carboxylic acids are more acidic than corresponding alcohols due to greater stabilization of the carboxylate ion due to resonance and inductive electron-withdrawing effect. Fluorine being electron withdrawing, stabilizes the negative change on the carboxylate ion. The inductive effect decreases with increase in distance from the charge, so CF3COOH is more acidic than CF3CH2COOH. 11. (2) CH2=CH2 is not an acidic compound hence, the reaction (2) would not take place.

O

43. (3) A reaction between a nucleophile and electrophile involves formation of a coordinate covalent bond. 44. (1) Addition reactions are observed when multiple bonds are replaced by another functional group

12. (4) Stability of carbocations can be applied to resonance, inductive effect and hyperconjugation effect. (1) Resonance: For example consider allyl carbocation 1

CH2

Exercise 2 2. (4) C2 — C3 single bond would be the shortest bond among the given compounds. 1

sp

sp

2

3

4

6. (3) P(CH3)2

P(CH3)2 ⊕

⊕

N

N

H

H

CH2

CH

1

CH2

(2) Inductive effect: The order of stability of carbocation can also be explained on the basis of positive inductive effect of alkyl group. It is clear that lesser the positive formal charge on the carbon atom of the carbocation due to +IE of alkyl group, greater is the stability of carbocation. (3) Hyperconjugation: The stability of carbocations can be explained on the basis of number of hyperconjugating structures. As the number of hyperconjugating structure increases, the stability of carbocation increases.

16. (2) The molecule or ion that forms when a base accepts a proton is called the conjugate acid. 17. (3) All given anions are basic in nature because they contain negative charge. Now this strength is enhanced by electron donating groups and decreases by electron withdrawing groups.

N H

7. (3) In resonance structures, position of atom does not change, thus, compounds in option (3) are not the correct representation of resonating structures. 8. (1) ⊕

O

CH2

14. (1) The molecule or ion that forms when an acid loses its proton is called the conjugate base of that acid. The larger the value of the pKa, the weaker is the acid and stronger is its conjugate base. ⊕ P(CH3)2

O

CH

Since –COCH3 and –Br groups are electron withdrawing, they decrease the basic strength. –CH3 and –OCH3 are electron donating groups, so they increase the basic strength. 19. (4) Stability of radicals can be compared on the basis of electronic effect and angle strain. In option (1): B is more stable than a due to resonance.

⊕

⊕

< O A

B

In option (2): 3 radical is more stable as compared to 2o. o

O

⊕


I > II > IV.

Part I: Bonding and Molecular Structure

CONCEPT MAP Organic Molecules have

can be predicted by

VSEPR Theory

can be predicted by

Three-dimensional shape

Quantum mechanics utilizes

requires creation of must be

Proper Lewis structures

show all

show all Valence electrons

Wave functions

Resonance structures

are used to generate

show all

Atomic orbitals

Formal charges

include all

of 2nd row elements consist of

are averaged in the

Bonding and nonbonding electrons

y

Resonance hybrid

One 2s

repel each other to achieve

and three 2p orbitals y y

Maximum separation in 3-D space

z

Linear geometry

is present in

are at each triplebonded carbon of

π Bond

σ Bond

H

C

y

x

z

x

z may become

Alkynes of two groups* of electrons leads to

x

x z

C

Two sp hybrid and two p orbitals p Orbitals

sp Orbital

C

sp Orbital

H π Bond

C

C

Alkenes of three groups* of electrons leads to

Trigonal planar geometry

is present in

H

Overlap H

C

C

sp2 Orbital z

sp2 Orbital

x sp2 Orbital

C

are at each singlebonded carbon of

H 109.5°

H

C H H

* A single bond, a double bond, a triple bond, and a nonbonding electron pair each represent a single ‘group’ of electrons.

y

H

Alkanes is present in Tetrahedral geometry

Three sp2 hybrid and one p orbital

p Orbital

H

C

of four groups* of electrons leads to

are at each doublebonded carbon of

Four sp3 hybrid orbitals 109.5°

c (+)

c (+) –

109.5°

c (+)

– – –

109.5° 109.5° c (+)

109.5°

C

Part II: Families of Carbon Compounds

CONCEPT MAP Functional Groups

help us organize knowledge about

that are most common in organic compounds are:

C C

Alkenes

C C

Alkynes

are hydrocarbons

help us predict Aromatics

Physical Properties

Reactions

R X

include

Alkyl halides

R O R´

Ethers

R C N

Nitriles

R OH mp, bp, and solubility

R

Alcohols

N H

Amines (there can be one, two, or three alkyl groups)

R´ O

are strongly influenced by

R

R

Hydrophobic

C

N

Amides, N-substituted (can also be R´ N,N-disubstituted or unsubstituted)

O Dipole–dipole forces

include

R

Hydrogen bonds

R

Polar molecules are Hydrophilic

have

C

Esters

OR´

O

predominate in

are

OH

H

include

Nonpolar molecules

Carboxylic acids

O

Intermolecular (van der Waals) forces

Dispersion (London) forces

C

C

R´

Polar covalent bonds result from differences in

Electronegativity

contain Heteroatoms are

Atoms that have unshared electron pairs and covalent bonds

Aldehydes (R´= H), ketones

CONCEPT MAP

Part III: An Introduction to Organic Reactions and Their Mechanisms

Curved-arrow notation is used to show Acids

Reaction mechanisms

can be

Brønsted–Lowry acids are

are a subcategory of

Proton donors contain/have Small or negative pKa values and large Ka values are associated with Strong acids have Weak conjugate bases

Bases

often involve

Lewis acids

Lewis bases

are Electron pair acceptors are Electrophiles can be Carbocations

can be

are Electron pair donors

are a subcategory of

Brønsted–Lowry bases are Proton acceptors

are

contain/have

Nucleophiles

Large and positive pKa values and small Ka values

can be

can be Carbanions

are associated with Strong bases have Weak conjugate acids

2

Isomerism C H A P T E R OU TLIN E 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

Types of Isomerism Geometrical Isomerism Chirality and Stereochemistry Enantiomers and Chiral Molecules Molecules Having One Chirality Center are Chiral How to Test for Chirality: Planes of Symmetry Naming Enantiomers: The R,S-System Properties of Enantiomers: Optical Activity

2.9 The Synthesis of Chiral Molecules 2.10 Molecules with More than One Chirality Center 2.11 Fischer Projection Formulas 2.12 Stereoisomerism of Cyclic Compounds 2.13 Relating Configurations through Reactions in which No Bonds to the Chirality Center are Broken 2.14 Chiral Molecules That Do Not Possess a Chirality Center 2.15 Conformational Isomerism

The phenomenon of two or more compounds having the same molecular formula, but different structural arrangements, is called isomerism. The individual compounds are called isomers. The term isomers comes from the Greek words isos and meros, meaning “made of same parts”. Thus, isomers are compounds that are constructed from the same atoms (hence have same molecular formula) but have different properties. These are compounds with the same molecular formula but different structural formulas. Isomerism is common among organic compounds and is another reason for the large number of known compounds. There are 3 isomers of pentane, 5 isomers of hexane, 9 isomers of heptane, 18 isomers of octane, 35 isomers of nonane and 75 isomers of decane. The phenomenon of isomerism is a compelling reason for using structural formulas.

2.1

TYPES OF ISOMERISM

The types of isomerism observed in organic compounds are illustrated in Fig. 2.1 and described as follows.

2.1A Structural (Constitutional) Isomerism Structural isomers are compounds that have the same molecular formula, but different structural formulas. By “different structural formulas”, we mean that these compounds differ in the kinds of bonds they have (single, double or triple) or in their connectivity (the order of attachment among their atoms). These are also known as constitutional isomers. Chain Isomerism For the molecular formulas CH4, C2H6 and C3H8, only one order of attachment of atoms is possible. For the molecular formula C4H10, two orders of attachment of atoms are possible. In one of these, named butane, the four carbons are bonded in a chain; in the other, named

104

Chapter 2 | Isomerism

ISOMERISM Structural Isomerism

Stereoisomerism

Isomers with different structures

Isomers with the same structure and bond sequence but different orientation of atoms in space

Chain Isomerism Owing to arrangement of carbon skeleton

Geometric Isomerism Involves a double bond, usually C C, that does not allow free rotation

Position Isomerism Owing to position of functional group

Optical Isomerism Involves an atom, usually carbon, bonded to four different atoms or groups of atoms

Functional Group Isomerism Owing to different functional groups Metamerism Owing to different alkyl chains on either side of functional group

Figure 2.1 Isomerism in organic compounds. 2-methylpropane, three carbons are bonded in a chain, with the fourth carbon as a branch on the middle carbon of the chain. CH3 CH3CH2CH2CH3

and

Butane (C4H10)

CH3CHCH3 Isobutane (C4H10)

Butane and isobutane are constitutional isomers (Table 2.1). Position Isomerism Compounds that differ in the position of the substituent or the functional group on the parent carbon chain are known as position isomers and the phenomenon is known as positions isomerism. For example, CH3CH2CH2Cl

and

1-Chloropropane (C3H7Cl)

CH3CHCH3 Cl 2-Chloropropane (C3H7Cl)

Table 2.1 Constitutional isomers of some compounds Molecular Formula

Constitutional Isomers CH3

C4H10

CH3CH2CH2CH3

and CH3CHCH3

Butane CH3CH2CH2Cl C3H7Cl

1-Chloropropane

C2H6O

CH3CH2OH Ethanol

Isobutane

and

CH3CHCH3 Cl 2-Chloropropane

and

CH3OCH3 Dimethyl ether

2.1 | Types of Isomerism

Functional Group Isomerism Compounds that have the same molecular formula but have different functional groups are called functional group isomers and the phenomenon is termed as functional group isomerism. For example, CH3CH2OH and

CH3OCH3

Ethanol

Dimethyl ether

(C2H6O)

(C2H6O)

Metamerism Compounds that have different alkyl chains attached to the functional group are said to exihibit metamerism. For example, the compound with molecular formula C5H10O can exist as following metamers: O CH3

CH2

CH3

CH2

CH

CH2

CH2

CH3

C

C H

O H

CH3

CH2

CH2

CH3

CH2

C

C

CH3

CH2

CH3

O

O

Ring-Chain Isomerism Compounds that have the same molecular formula but possess open-chain or cyclic structures are called ring-chain isomers and the phenomenon is called ring-chain isomerism. The classes of organic compounds that exhibit this type of isomerism are alkenes and cyclic compounds (e.g., propene and cyclopropane); alkynes and cycloalkenes (e.g., propyne and cyclopropene) and unsaturated alcohols and cyclic ethers (e.g., prop-2-en-ol and oxetane). Tautomerism A single compound that can exist in two readily interconvertible structures that differ in the relative position of at least one atom (generally hydrogen) is said to exhibit tautomerism. For example, carbonyl compounds and their corresponding enols that are in rapid equilibrium, are tautomers, obtained by simultaneous shift of H atom and a double bond at 1 and 3 positions. Tautomerism can be considered as a special case of functional isomerism, where equilibrium exists between an enol and a ketone. The enol and ketone are said to be constitutional isomers that rapidly interconvert via the migration of a proton. The essential condition for an aldehyde or ketone to exhibit keto-enol tautomerism is the presence of at least one α- hydrogen atom. So whereas acetophenone, proprionaldehyde, butan-2-one exihibit keto-enol tautomerism due to the presence of α- hydrogen atom; benzaldehyde and benzophenone do not show the same due to absence of α- hydrogen atom. For most of the aldehydes and ketones, both the forms exist in equilibrium but the percentage of enol form is usually insignificant. H

O

O

C

C

H3C

CH3

keto form

H3C

CH2

enol form

Similarly, ethylacetoacetate also exists as a mixture of keto and enol forms in equilibrium with each other, but at ordinary temperatures the keto form is almost 99% (because of its greater stability). O

OH

O O

keto form (93%)

O O

enol form (7%)

The major reason for the instability of enols is that the C O bond of a carbonyl group is stronger than the C C bond of an enol. In the presence of acids and esters, the additional instability of enols results

105

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Chapter 2 | Isomerism

from the loss of the stabilizing resonance interaction between the carboxylate oxygen and the carbonyl p -electrons, which are present in the carbonyl forms. However, in some compounds such as 1,3-dicarbonyl compounds, the amount of enol form is much higher. For example, in the case of acetylacetone, the enol form has greater stability than expected because of intramolecular hydrogen bonding leading to a six-membered cyclic transition state O

O

O

keto form (24%)

H

O

enol form (76%)

The enol content increases in non-polar aprotic solvents (e.g., hexane, benzene) and decreases in polar protic or aprotic solvents (e.g., water, acetone). The keto-enol tautomerism can be both acid and base catalyzed. In alkaline solutions, hydroxide ion extracts a proton from the α- carbon of the keto tautomer; thus forming an enolate ion. Protonation on oxygen forms the enol tautomer, whereas protonation on the α- carbon reforms the keto tautomer.

O RCH 2

C

H

O

2

H

O R

RCH

r.d.s

C

OH RCH

R

Enolate ion

H

C

R 1 HO2

Enol tautomer

HO

Keto tautomer

In an acidic solution, the carbonyl oxygen of the keto tautomer is protonated and water removes a proton from the α- carbon, forming the enol. 1

O RCH2

C

OH R

H1 2H1

RCH2

Keto tautomer

C

OH R

r.d.s

RCH

C

R 1 H3O1

Enol tautomer

H H2O

2.1B Stereoisomerism Stereoisomers have the same molecular formula and the same connectivity of atoms in their molecules, but different three-dimensional orientations of their atoms in space. The one example of stereoisomers we have seen so far is that of cis–trans isomers in cycloalkanes and alkenes: CH3

CH3 and

CH3 cis-1,2-Dimethylcyclohexane

CH3 trans-1,2-Dimethylcyclohexane

H3C H

C

C

CH3 H

H3 C and

cis-2-Butene

H

H C

C CH3

trans-2-Butene

The stereoisomers are classified as geometrical isomers or optical isomers. ●

The existence of isomers whose molecules have identical atomic organizations but different geometries are called geometrical isomers or cis–trans isomers.

2.2 | Geometrical Isomerism ●

Stereoisomers other than geometrical (cis–trans) isomers and that include substances that can rotate the plane of plane-polarized light are called optical isomers.

Stereoisomers can be subdivided into two general categories: those that are enantiomers of each other, and those that are diastereomers of each other. ●

Enantiomers are stereoisomers whose molecules are nonsuperposable mirror images of each other.

All other stereoisomers are diastereomers. ●

Diastereomers are stereoisomers whose molecules are not mirror images of each other.

The alkene isomers cis- and trans-1,2-dichloroethene shown here are stereoisomers that are diastereomers. Cl

H

Cl

Cl

H

cis-1,2-Dichloroethene (C2H2Cl2)

H

H Cis and trans alkene isomers are diastereomers.

Cl

trans-1,2-Dichloroethene (C2H2Cl2)

By examining the structural formulas for cis- and trans-1,2-dichloroethene, we see that they have the same molecular formula (C2H2Cl2) and the same connectivity (both compounds have two central carbon atoms joined by a double bond, and both compounds have one chlorine and one hydrogen atom attached to each carbon atom). But, their atoms have a different arrangement in space that is not interconvertible from one to another (due to the large barrier to rotation of the carbon–carbon double bond), making them stereoisomers. Furthermore, they are stereoisomers that are not mirror images of each other; therefore they are diastereomers and not enantiomers. Cis and trans isomers of cycloalkanes furnish us with another example of stereoisomers that are diastereomers. Consider the following two compounds:

Me Me H

Me

H

cis-1,2-Dimethylcyclopentane (C7H14)

H

H

Cis and trans cycloalkane isomers are diastereomers.

Me

trans-1,2-Dimethylcyclopentane (C7H14)

These two compounds have the same molecular formula (C7H14), the same sequence of connections for their atoms, but different arrangements of their atoms in space. In one compound both methyl groups are bonded to the same face of the ring, while in the other compound the two methyl groups are bonded to opposite faces of the ring. Furthermore, the positions of the methyl groups cannot be interconverted by conformational changes. Therefore, these compounds are stereoisomers, and because they are stereoisomers that are not mirror images of each other, they can be further classified as diastereomers.

2.2

GEOMETRICAL ISOMERISM

The phenomenon of geometrical isomerism is a general one and is observed in any class of compound containing carbon–carbon double bond. Because of restricted rotation about a carbon–carbon double bond, any compound in which each carbon of the double bond has two different groups bonded to it shows cis–trans isomerism. If two identical groups are on the same side of the double bond, the compound can be designated cis; if they are on opposite sides it can be designated trans:

107

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Chapter 2 | Isomerism

H

H C

H

C

H3C

C CH3

cis-2-Butene

CH3

C

H3C

H

trans-2-Butene

Consider, for example, 2-butene. In cis-2-butene, the two methyl groups are on the same side of the double bond; in trans-2-butene, the two methyl groups are on opposite sides of the double bond. These two compounds cannot be converted into one another at room temperature because of the restricted rotation about the double bond. They are different compounds, with different physical and chemical properties. It takes approximately 264 kJ (63 kcal mol−1) to break the p bond of ethylene, that is, to rotate one carbon by 90° with respect to the other where there is zero overlap between the 2p orbitals of adjacent carbons. This energy is considerably greater than the thermal energy available at room temperature; consequently, rotation about a carbon–carbon double bond does not occur under normal conditions. The two important and necessary conditions for a molecule to exhibit geometrical isomerism are: 1. The molecule must contain a carbon–carbon double bond. 2. Each carbon atom of the double bond must have different atoms or groups attached to it. For example, compounds of the type abC = Cab, abC = Cdc, abC = Cad show this type of isomerism. The following show the cases where abC = Cab and abC = Cad. CH3 H

C

C

CH3 H

CH3 H

(cis) CH3 H

C

C

(cis)

C

H

C

CH3

(trans) Cl

CH3

H

H

C

H

C

Cl

(trans)

Cis alkenes are less stable than their trans isomers because of non-bonded interaction strain between alkyl substituents on the same side of the double bond in the cis isomer. Geometric isomers have different physical properties, such as melting points, boiling points, refractive indices, solubility’s, densities and dipole moment. The cis and trans isomers may be distinguished on the basis of their physical properties as follows. 1. From boiling and melting points: The boiling point of the cis isomer is higher than the trans isomer. The boiling point of isomeric compounds depends upon the dipole–dipole interactions. Since cis isomers usually have a higher value of dipole moment than the corresponding trans isomer, therefore their boiling point is more. For example, for a pair of geometric isomers cis- and trans-butene, the cis isomers have higher boiling points due to higher polarity. However, due to its lower symmetry, it does not fit into the crystalline lattice as well as the trans isomer and thus has a lower melting point. H3C

H C

H3C

C

C H3C

H C

H

H

CH3

cis-Butene

trans-Butene

μ 5 0.33 D bp 5 4° C mp 5 2139° C

μ 50 bp 5 1° C mp 5 2106° C

2.2 | Geometrical Isomerism

The melting point and density of a trans isomer is higher than the cis isomer due to the symmetrical nature and more close packing in trans isomer. 2. From dipole moments: Cis isomers have higher dipole moments than trans isomers which may have zero dipole moments. For example, the dipole moment of cis-maleic acid is quite high while that of fumaric acid is zero because dipole moments of C}COOH bonds cancel out the effects of each other as these are in opposite directions. H

C

COOH

H

C

COOH

H

C

COOH

HOOC

C

H

μ 5 7.22 D Maleic acid (cis)

μ 50D Fumaric acid (trans)

Similarly in 1,2-dichloroethene, the dipole moment of C}Cl bond is cancelled out in the trans isomer. H

H

Cl

H

Cl

C

C

C

C Cl

H

Cl

μ 5 1.85 D (Cis)

μ 50D (Trans)

2.2A E- and Z-Nomenclature The alkenes showing geometrical isomerism that we have considered so far are of the general formula abC = Cba. However, geometrical isomerism is also shown by alkenes of the type abC = Cad or abC = Ccd. An E and Z system of nomenclature is used to denote geometrical isomers of this type. Consider the following alkene as an example: Br

Cl C

C

H

F A

It is impossible to decide whether A is cis or trans since no two groups are the same. A system that works in all cases is based on the priorities of groups in the Cahn-Ingold-Prelog convention (discussed in Section 2.7). This system, called the (E)-(Z) system applies to alkene diastereomers of all types. In the (E)-(Z) system, we examine the two groups attached to one carbon atom of the double bond and decide which has higher priority. Then we repeat that operation at the other carbon atom: Higher priority

Cl

F

F

C C

Higher priority

Br

Cl C

Higher priority

Cl . F

C H

(Z )-2-Bromo-1-chloro1-fluoroethene

Higher priority

Br

H

Br . H

(E )-2-Bromo-1-chloro1-fluoroethene

We take the group of higher priority on one carbon atom and compare it with the group of higher priority on the other carbon atom. If the two groups of higher priority are on the same side of the double bond,

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Chapter 2 | Isomerism

the alkene is designated (Z) (from the German word zusammen, meaning together). If the two groups of higher priority are on opposite sides of the double bond, the alkene is designated (E) (from the German word entgegen, meaning opposite). The following examples illustrate this: H3C

C

C

H

CH3 H

(Z )-2-Butene or (Z )-but2-ene (cis-2-butene) Cl

Cl C

H

C Br

H3C

CH3 . H

H C

C

H

CH3

(E)-2-Butene or (E)-but2-ene (trans-2-butene) Cl

Cl . H Br . Cl

Br C

H

C Cl

(Z )-1-Bromo-1,2dichloroethene

(E)-1-Bromo-1,2dichloroethene

2.2B Geometrical Isomerism in Cyclic Structures In cyclic compounds, presence of ring systems restricts the rotation about a carbon-carbon single bond. These compounds can thus exist as geometrical isomers, provided there are at least two other groups besides hydrogen on the ring. These groups are present on different carbon atoms of the ring. CH3

CH3

H

CH3 H3C

CH3

H

CH3

H

1,1-Dimethyl cyclohexane (no geometrical isomers possible)

2.3

Trans-1,2-dimethyl cyclohexane

H

Cis-1,2-dimethyl cyclohexane

CHIRALITY AND STEREOCHEMISTRY

Chirality is a phenomenon that pervades the universe. How can we know whether a particular object is chiral or achiral (not chiral)? ●

We can tell if an object has chirality by examining the object and its mirror image.

Every object has a mirror image. Many objects are achiral. By this we mean that the object and its mirror image are identical—that is, the object and its mirror image are superposable one on the other.* Superposable means that one can, in one’s mind’s eye, place one object on the other so that all parts of each coincide. Simple geometrical objects such as a sphere or a cube are achiral. So is an object like a water glass. ●

A chiral object is one that cannot be superposed on its mirror image.

*To be superposable is different than to be superimposable. Any two objects can be superimposed simply by putting one object on top of the other, whether or not the objects are the same. To superpose two objects (as in the property of superposition) means, on the other hand, that all parts of each object must coincide. The condition of superposability must be met for two things to be identical.

Photo by Michael Watson for John Wiley & Sons, Inc.

Photo by Michael Watson for John Wiley & Sons, Inc.

2.3 | Chirality and Stereochemistry

Figure 2.2 The mirror image of a right hand is a left hand.

Figure 2.3 Left and right hands are not superposable.

Each of our hands is chiral. When you view your right hand in a mirror, the image that you see in the mirror is a left hand (Fig. 2.2). However, as we see in Fig. 2.3, your left hand and your right hand are not identical because they are not superposable. Your hands are chiral. In fact, the word chiral comes from the Greek word cheir meaning hand. An object such as a mug may or may not be chiral. If it has no markings on it, it is achiral. If the mug has a logo or image on one side, it is chiral.

2.3A The Biological Significance of Chirality The human body is structurally chiral, with the heart lying to the left of center and the liver to the right. Helical seashells are chiral and most are spiral, such as a right-handed screw. Many plants show chirality in the way they wind around supporting structures. DNA is a chiral molecule. The double helical form of DNA turns in a right-handed way. Chirality in molecules, however, involves more than the fact that some molecules adopt left- or righthanded conformations. All but one of the 20 amino acids that make up naturally occurring proteins are chiral, and all of these are classified as being left-handed. The molecules of natural sugars are almost all classified as being right-handed. In fact, most of the molecules of life are chiral, and most are found in only one mirror image form. The binding specificity for a chiral molecule (like a hand) at a chiral receptor site (a glove) is only favorable in one way. If either the molecule or the biological receptor site had the wrong handedness, the natural physiological response (e.g., neural impulse, reaction catalysis) would not occur. A diagram showing how only one amino acid in a pair of enantiomers can interact in an optimal way with a hypothetical binding site (e.g., in an enzyme) is shown in Fig. 2.4. Because of the chirality center of the amino acid, three-point binding can occur with proper alignment for only one of the two enantiomers. R C H +NH 3

R CO2–

–O

2C

C H +NH

3

Figure 2.4 Only one of the two amino acid enantiomers shown (the left-hand one) can achieve three-point binding with the hypothetical binding site (e.g., in an enzyme). Chirality has tremendous importance in our daily lives. Most pharmaceuticals are chiral. Usually only one mirror-image form of a drug provides the desired effect. The other mirror-image form is often

111

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Chapter 2 | Isomerism

inactive or, at best, less active. In some cases the other mirror-image form of a drug actually has severe side effects or toxicity. For several years before 1963 the drug thalidomide was used to alleviate the symptoms of morning sickness in pregnant women. In 1963 it was discovered that thalidomide was the cause of horrible birth defects in many children born subsequent to the use of the drug. O O

N N O

H

O

Thalidomide

Even later, evidence began to appear indicating that whereas one of the thalidomide enantiomers (the right-handed molecule) has the intended effect of curing morning sickness, the other enantiomer, which was also present in the drug (in an equal amount), may be the cause of the birth defects. Our senses of taste and smell also depend on chirality. As we shall see, one mirror-image form of a chiral molecule may have a certain odor or taste while its mirror image smells and tastes completely different. One enantiomeric form of a compound called limonene is primarily responsible for the odor of oranges and the other enantiomer for the odor of lemons.

(+)-Limonene (the enantiomer of limonene found in oranges)

(–)-Limonene (the enantiomer of limonene found in lemons)

One enantiomer of a compound called carvone is the essence of caraway, and the other is the essence of spearmint.

2.4

ENANTIOMERS AND CHIRAL MOLECULES

Enantiomers occur only with compounds whose molecules are chiral. ●

A chiral molecule is one that is not superposable on its mirror image.

The trans isomer of 1,2-dimethylcyclopentane is chiral because it is not superposable on its mirror image, as the following formulas illustrate.

Me

H

H

Me

H

Me

Me

H

Mirror images of trans-1,2-dimethylcyclopentane They are not superposable and therefore are enantiomers.

113

2.4 | Enantiomers and Chiral Molecules

Enantiomers do not exist for achiral molecules. ●

An achiral molecule is superposable on its mirror image.

The cis and trans isomers of 1,2-dichloroethene are both achiral because each isomer is superposable on its mirror image, as the following formulas illustrate. Cl

H

H

Cl

Cl

H

H

Cl

Cl

H

H

Cl

H

Cl

Cl

H

cis-1,2-Dichloroethene mirror images

trans-1,2-Dichloroethene mirror images

The mirror images of the cis isomer are superposable on each other (try rotating one by 180° to see that it is identical to the other), and therefore the cis formulas both represent the same, achiral molecule. The same analysis is true for the trans isomer.

● ●

Enantiomers only occur with compounds whose molecules are chiral. A chiral molecule and its mirror image are called a pair of enantiomers. The relationship between them is enantiomeric.

The universal test for chirality of a molecule, or any object, is the nonsuperposability of the molecule or object on its mirror image. We encounter chiral and achiral objects throughout our daily life. Shoes are chiral, for example, whereas most socks are achiral. The chirality of molecules can be demonstrated with relatively simple compounds. Consider, for example, 2-butanol:

OH 2-Butanol

Until now, we have presented the formula for 2-butanol as though it represented only one compound and we have not mentioned that molecules of 2-butanol are chiral. Because they are, there are actually two different 2-butanols and these two 2-butanols are enantiomers. We can understand this if we examine the drawings and models in Fig. 2.5.

H

H HO

OH

HO

H OH

H

HO II

H

OH I

HO II

(a)

H

CH2

CH2

CH3

CH3

CH3 I

CHCH 3 3

CH3 II

I (b)

(b)

CH2

CH2

CH3

CH3

CH3

CH2

CH CH 3 3

H

H HO

II CH2 CH3

H OH H

OH

I

I

CH2 CH3

CH

CH3 2 CH3 CH3

II (c)

(c)

Figure 2.5 (a) Three-dimensional drawings of the 2-butanol enantiomers I and II. (b) Models of the 2-butanol enantiomers. (c) An unsuccessful attempt to superpose models of I and II. If model I is held before a mirror, model II is seen in the mirror and vice versa. Models I and II are not superposable on each other; therefore, they represent different, but isomeric, molecules. Because models I and II are nonsuperposable mirror images of each other, the molecules that they represent are enantiomers.

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Chapter 2 | Isomerism

2.5 ● ●

MOLECULES HAVING ONE CHIRALITY CENTER ARE CHIRAL A chirality center is a tetrahedral carbon atom that is bonded to four different groups. A molecule that contains one chirality center is chiral and can exist as a pair of enantiomers.

Molecules with more than one chirality center can also exist as enantiomers, but only if the molecule is not superposable on its mirror image. (We shall discuss that situation later in the chapter.) For now we will focus on molecules having a single chirality center. Chirality centers are often designated with an asterisk (*). The chi(hydrogen) rality center in 2-butanol is C2 (Figure 2.6). The four different groups H attached to C2 are a hydroxyl group, a hydrogen atom, a methyl group, 1 2& 3 4 * and an ethyl group. (It is important to note that chirality is a property (methyl) CH39C9CH 2CH3 (ethyl) & of a molecule as a whole, and that a chirality center is a structural OH feature that can cause a molecule to be chiral.) (hydroxyl) An ability to find chirality centers in structural formulas will help us Figure 2.6 The tetrahedral recognize molecules that are chiral, and that can exist as enantiomers. carbon atom of 2-butanol that ● The presence of a single chirality center in a molecule guarantees bears four different groups. that the molecule is chiral and that enantiomeric forms are [By convention, chirality possible. centers are often designated Figure 2.7 demonstrates that enantiomeric compounds can exist whenwith an asterisk (*).] ever a molecule contains a single chirality center. ●

An important property of enantiomers with a single chirality center is that interchanging any two groups at the chirality center converts one enantiomer into the other. X

X

X

X

W W

Y

Y

Y

Y

W

Z

Z III

Z

Mirror (a)

III

IV

W

Z

IV

Mirror (a)

Mirror

X X

III Y Y

III (rotated) (rotated)

Y Z

W (b)

Y

W

Z W Z

ZW

X

IV IV

Y X

X

X

Mirror

III

W

III

W

Y Z

X

X

IV IV

Z Y

Z

W W

Y

Z (c)

Figure 2.7 A demonstration of chirality of a generalized molecule containing one chirality center. (c) (b) III is rotated and (a) The four different groups around(b)the carbon atom in III and IV are arbitrary. placed in front of a mirror. III and IV are found to be related as an object and its mirror image. (c) III and IV are not superposable; therefore, the molecules that they represent are chiral and are enantiomers.

2.6 | How to Test for Chirality: Planes of Symmetry

In Fig. 2.5b, it is easy to see that interchanging the methyl and ethyl groups converts one enantiomer into the other. You should now convince yourself that interchanging any other two groups has the same result. ●

Any atom at which an interchange of groups produces a stereoisomer is called a stereogenic center. (If the atom is a carbon atom, it is usually called a stereogenic carbon.)

When we discuss interchanging groups like this, we must take care to notice that what we are describing is something we do to a molecular model or something we do on paper. An interchange of groups in a real molecule, if it can be done, requires breaking covalent bonds, and this is something that requires a large input of energy. This means that enantiomers such as the 2-butanol enantiomers do not interconvert spontaneously. The chirality center of 2-butanol is one example of a stereogenic center, but there are stereogenic centers that are not chirality centers. The carbon atoms of cis-1,2-dichloroethene and of trans-1,2dichloroethene are stereogenic centers because an interchange of groups at either carbon atom produces the other stereoisomer. The carbon atoms of cis-1,2-dichloroethene and trans-1,2-dichloroethene are not chirality centers, however, because they do not have four different groups attached to them. ●

If all of the tetrahedral atoms in a molecule have two or more groups attached that are the same, the molecule does not have a chirality center. The molecule is superposable on its mirror image and is an achiral molecule.

An example of a molecule of this type is 2-propanol; carbon atoms 1 and 3 bear three identical hydrogen atoms and the central atom bears two identical methyl groups. If we write three-dimensional formulas for 2-propanol, we find (Fig. 2.8) that one structure can be superposed on its mirror image.

HO

H

H

H3C

CH3 V

OH

H3C Mirror

CH3 VI

(a) Superposable

HO HO H3C H3C

H H CH3 CH3 VI

(b) therefore

Not enantiomers

Figure 2.8 (a) 2-Propanol ( V ) and its mirror image ( VI ). (b) When either one is rotated, the two structures are superposable and so do not represent enantiomers. They represent two molecules of the same compound. 2-Propanol does not have a chirality center.

Thus, we would not predict the existence of enantiomeric forms of 2-propanol, and experimentally only one form of 2-propanol has ever been found.

2.5A Tetrahedral versus Trigonal Stereogenic Centers It is important to clarify the difference between stereogenic centers, in general, and a chirality center, which is one type of stereogenic center. The chirality center in 2-butanol is a tetrahedral stereogenic center. The carbon atoms of cis- and trans-1,2-dichloroethene are also stereogenic centers, but they are trigonal stereogenic centers. They are not chirality centers. An interchange of groups at the alkene carbons of either 1,2-dichloroethene isomer produces a stereoisomer (a molecule with the same connectivity but a different arrangement of atoms in space), but it does not produce a nonsuperposable mirror image. A chirality center, on the other hand, is one that must have the possibility of nonsuperposable mirror images. ● ●

2.6

Chirality centers are tetrahedral stereogenic centers. Cis and trans alkene isomers contain trigonal stereogenic centers.

HOW TO TEST FOR CHIRALITY: PLANES OF SYMMETRY

The ultimate way to test for molecular chirality is to construct models of the molecule and its mirror image and then determine whether they are superposable. If the two models are superposable, the molecule that they represent is achiral. If the models are not superposable, then the molecules that they represent are

115

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Chapter 2 | Isomerism

chiral. We can apply this test with actual models, as we have just described, or we can apply it by drawing three-dimensional structures and attempting to superpose them in our minds. There are other aids, however, that will assist us in recognizing chiral molecules. We have mentioned one already: the presence of a single chirality center. Other aids are based on the absence of certain symmetry elements in the molecule.

2.6A Plane of Symmetry ● ●

A molecule will not be chiral if it possesses a plane of symmetry (σ). A plane of symmetry (also called a mirror plane) is defined as an imaginary plane that bisects a molecule in such a way that the two halves of the molecule are mirror images of each other.

The plane may pass through atoms, between atoms, or both. For example, 2-chloropropane has a plane of symmetry (Fig. 2.9a), whereas 2-chlorobutane does not (Fig. 2.9b). ●

All molecules with a plane of symmetry in their most symmetric conformation are achiral.

Plane of symmetry

Cl

CH3

Cl

CH3

CH3

C2H5

Cl H3C

H

H

CH3

H

Achiral

Chiral

(a)

(b)

Figure 2.9 (a) 2-Chloropropane has a plane of symmetry and is achiral. (b) 2-Chlorobutane does not possess a plane of symmetry and is chiral. Other examples include: 1. Cis 2-Butene σ xz (Molecular plane) H

H

CH3

CH3

or

σ H

H

CH3

σ yz

CH3

2. Trans 2-Butene σ xz (Molecular plane) CH3

H

H

CH3

CH3

or

H

C H

CH3 σ yz

(Not possible)

2.6 | How to Test for Chirality: Planes of Symmetry

3. 2, 3,4-Trichloropentane

117

CH3 H

Cl

H

Cl

H

Cl

σ

CH3

4. 2, 4-Dibromopentane

CH3 H

Br σ

CH2 Br

H CH3

5. 1, 2, 3, 4, 5, 6-Hexachlorocyclohexane Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl Cl

Cl Cl

Cl Cl

Cl Cl

Cl Cl

Cl

Cl

Cl

Cl

Cl Cl

Cl Cl

Cl Cl

Cl Cl

Cl Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Each of the above structures possess plane of symmetry and therefore are optically inactive. However, the below structure has no plane of symmetry and so is optically active. Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl No plane of symmetry

2.6B Center of Symmetry The center of symmetry (COS) is an imaginary point through which a line is drawn in a direction from an atom or group to the same atom or group placed at the same distance just opposite to the imaginary center. This rule is applicable for each atom of the molecule and the operation is applicable only for threedimensional formula and not Fischer projection formula. Some examples are listed as follows. 1. Benzene molecule: COS present

H H

H

H

H H

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Chapter 2 | Isomerism

2. O

O

NH

CH3

(H)

H CH3

H NH

NH (H) CH3

CH3 NH

O (Trans) COS present

O (Cis) COS absent

3. COS present Cl C

CH3

C

CH3

Cl

4. COS not visible COOH H

OH H

OH COOH

5.

CH3

C

H

H

C

CH3

H H3C

(Trans) COS present

C

H

C

CH3

(Cis) COS absent

2.6C Axis of Symmetry The axis of symmetry (AOS) is an imaginary axis around which by rotating through a minimum angle of rotation (360°/n), the same compound is obtained. For example, 1. C2

C3

C4

C5

C6

21 H

H

2. If there are two or more possible AOS with the same value of n, then the preference is given to the axis that passes through more number of atoms.

Cl

Cl

C2

2.7 | Naming Enantiomers: The R,S-System

3. AOS absent H

F C

C

H

Cl

4. Some compounds with C2 axis of symmetry Br H

C2

H C

C

H

C2

H

Br

2.6D Alternating or Improper Axis of Symmetry (Sn or σ) The symmetry element alternating axis of symmetry involves rotation followed by reflection, that is, rotation by an angle of 360°/n followed by reflection in a plane perpendicular to the axis of rotation. The resultant molecule is identical to the original molecule. Hence, this operation is also called rotational reflection axis of symmetry. The following expression is true for Sn symmetry: S1 = σ ; S2 = i; S4 = for cyclobutane A

CH3

H CH3

H CH3

H CH3

CH3

H Rotation by 90°

H

H

CH3

H CH3

H H

H

B

2.7

H

CH3 CH3

CH3 Reflection in a plane perperdicular to axis

CH3 CH3

H

Identical

NAMING ENANTIOMERS: THE R,S-SYSTEM

The two enantiomers of 2-butanol are the following: H

OH I

H

HO II

If we name these two enantiomers using only the IUPAC system of nomenclature that we have learned so far, both enantiomers will have the same name: 2-butanol (or sec-butyl alcohol). This is undesirable because each compound must have its own distinct name. Moreover, the name that is given a compound should allow a chemist who is familiar with the rules of nomenclature to write the structure of the compound from its name alone. Given the name 2-butanol, a chemist could write either structure I or structure II.

119

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Chapter 2 | Isomerism

Three chemists, R. S. Cahn (England), C. K. Ingold (England), and V. Prelog (Switzerland), devised a system of nomenclature that, when added to the IUPAC system, solves both of these problems. This system, called the R,S-system or the Cahn–Ingold–Prelog system, is part of the IUPAC rules. According to this system, one enantiomer of 2-butanol should be designated (R)-2-butanol and the other enantiomer should be designated (S )-2-butanol. [(R ) and (S ) are from the Latin words rectus and sinister, meaning right and left, respectively.] These molecules are said to have opposite configurations at C2.

2.7A How to Assign (R) and (S) Configurations We assign (R ) and (S ) configurations on the basis of the following procedure. 1. Each of the four groups attached to the chirality center is assigned a priority or preference a, b, c, or d. Priority is first assigned on the basis of the atomic number of the atom that is directly attached to the chirality center. The group with the lowest atomic number is given the lowest priority, d; the group with next higher atomic number is given the next higher priority, c; and so on. (In the case of isotopes, the isotope of greatest atomic mass has highest priority.) We can illustrate the application of the rule with the following 2-butanol enantiomer: (a) HO H3C (b or c)

(d) H C

CH2CH3 (b or c)

One of the 2-butanol enantiomers

Oxygen has the highest atomic number of the four atoms attached to the chirality center and is assigned the highest priority, a. Hydrogen has the lowest atomic number and is assigned the lowest priority, d. A priority cannot be assigned for the methyl group and the ethyl group by this approach because the atom that is directly attached to the chirality center is a carbon atom in both groups. 2. When a priority cannot be assigned on the basis of the atomic number of the atoms that are directly attached to the chirality center, then the next set of atoms in the unassigned groups is examined. This process is continued until a decision can be made. We assign a priority at the first point of difference.* When we examine the methyl group of the 2-butanol enantiomer above, we find that the next set of atoms bonded to the carbon consists of three hydrogen atoms (H, H, H). In the ethyl group the next set of atoms bonded to the carbon consists of one carbon atom and two hydrogen atoms (C, H, H). Carbon has a higher atomic number than hydrogen, so we assign the ethyl group the higher priority, b, and the methyl group the lower priority, c, since (C, H, H) > (H, H, H): (a) HO H H (H, H, H)

C C (c) H

(d ) H

H

C H C (b) H H H (C, H, H)

3. We now rotate the formula (or model) so that the group with lowest priority (d ) is directed away from us: *The rules for a branched chain require that we follow the chain with the highest priority atoms.

2.7 | Naming Enantiomers: The R,S-System (a) OH (a)

OH

(d ) (c)

H (d )

(b)

Me

Et

CH3 (c)

Newman projection

CH2CH3 (b)

Viewer

One of the 2-butanol enantiomers

Then we trace a path from a to b to c. If, as we do this, the direction of our finger (or pencil) is clockwise, the enantiomer is designated (R ). If the direction is counterclockwise, the enantiomer is designated (S ). On this basis the 2-butanol enantiomer II is (R )-2-butanol: (a) OH

(a)

OH

(d ) (c)

(b)

Me

Et

Newman projection

HO C CH3

H = CH2CH3

H (d) CH3 (c)

CH2CH3 (b)

Viewer

Arrows are clockwise.

(R)-2-Butanol

The first three rules of the Cahn–Ingold–Prelog system allow us to make an (R ) or (S ) designation for most compounds containing single bonds. For compounds containing multiple bonds one other rule is necessary: 4. Groups containing double or triple bonds are assigned priorities as if both atoms were duplicated or triplicated—that is, (Y) (C) C

Y

as if it were

C

Y

and

C

Y

as if it were

C

(Y) (C)

Y

(Y) (C)

where the symbols in parentheses are duplicate or triplicate representations of the atoms at the other end of the multiple bond. Thus, the vinyl group, } CH CH2, is of higher priority than the isopropyl group, } CH(CH3)2. That is,

9 CH

CH2

is treated as though it were

H

H

9C9C9H (C) (C)

which has higher priority than

H

H

9 C 99 C9H H H9 C 9H H

121

122

Chapter 2 | Isomerism

because at the second set of atoms out, the vinyl group (see the following structure) is C, H, H, whereas the isopropyl group along either branch is H, H, H. (At the first set of atoms both groups are the same: C, C, H.) H

H

9 C9 C 9 H

H 

H

9 C 99 C9H

(C) (C)

H H9C9H H

C, H, H Vinyl group



H, H, H Isopropyl group

Other rules exist for more complicated structures, but we shall not study them here.

2.8

PROPERTIES OF ENANTIOMERS: OPTICAL ACTIVITY

The molecules of enantiomers are not superposable and, on this basis alone, we have concluded that enantiomers are different compounds. How are they different? Do enantiomers resemble constitutional isomers and diastereomers in having different melting and boiling points? The answer is no. ●

Pure enantiomers have identical melting and boiling points.

Do pure enantiomers have different indexes of refraction, different solubilities in common solvents, different infrared spectra, and different rates of reaction with achiral reagents? The answer to each of these questions is also no. Many of these properties (e.g., boiling points, melting points, and solubilities) are dependent on the magnitude of the intermolecular forces operating between the molecules, and for molecules that are mirror images of each other these forces will be identical. We can see an example of this if we examine Table 2.2, where boiling points of the 2-butanol enantiomers are listed. Mixtures of the enantiomers of a compound have different properties than pure samples of each, however. The data in Table 2.2 illustrate this for tartaric acid. The natural isomer, (+)-tartaric acid, has a melting point of 168–170 °C, as does its unnatural enantiomer, (−)-tartaric acid. An equal mixture tartaric acid enantiomers, (+/−)-tartaric acid, has a melting point of 210–212 °C, however. ●

Enantiomers show different behavior only when they interact with other chiral substances, including their own enantiomer.

This is evident in the melting point data below. Enantiomers also show different rates of reaction toward other chiral molecules—that is, toward reagents that consist of a single enantiomer or an excess of a single enantiomer. And, enantiomers show different solubilities in solvents that consist of a single enantiomer or an excess of a single enantiomer.

Table 2.2 Physical properties of 2-butanol and tartaric acid enantiomers Compound

Boiling Point (bp) or Melting Point (mp)

(R )-2-Butanol

99.5 °C (bp)

(S )-2-Butanol

99.5 °C (bp)

(+)-(R,R )-Tartaric acid

168–170 °C (mp)

(−)-(S,S )-Tartaric acid

168–170 °C (mp)

(+/−)-Tartaric acid

210–212 °C (mp)

2.8 | Properties of Enantiomers: Optical Activity

One easily observable way in which enantiomers differ is in their behavior toward plane-polarized light. ●

●

●

When a beam of plane-polarized light passes through an enantiomer, the plane of polarization rotates. Separate enantiomers rotate the plane of plane-polarized light equal amounts but in opposite directions. Separate enantiomers are said to be optically active compounds. Because of their effect on planepolarized light.

In order to understand this behavior of enantiomers, we need to understand the nature of plane-polarized light. We also need to understand how an instrument called a polarimeter operates.

2.8A Plane-Polarized Light Light is an electromagnetic phenomenon. A beam of light consists of two mutually perpendicular oscillating fields: an oscillating electric field and an oscillating magnetic field (Fig. 2.10).

Electric field Electric wave

Magnetic field

Magnetic wave

Direction of motion of the light beam

Figure 2.10 The oscillating electric and magnetic fields of a beam of ordinary light in one plane. The waves depicted here occur in all possible planes in ordinary light.

If we were to view a beam of ordinary light from one end, and if we could actually see the planes in which the electrical oscillations were occurring, we would find that oscillations of the electric field were occurring in all possible planes perpendicular to the direction of propagation (Fig. 2.11). (The same would be true of the magnetic field.) When ordinary light is passed through a polarizer, the polarizer interacts with the electric field so that the electric field of the light that emerges from the polarizer (and the magnetic field perpendicular to it) is oscillating only in one plane. Such light is called plane-polarized light (Fig. 2.12a). If the plane-polarized beam encounters a filter with perpendicular polarization, the light is blocked (Fig. 2.12b). This phenomenon can readily be demonstrated with lenses from a pair of polarizing sunglasses or a sheet of polarizing film (Fig. 2.12c).

Figure 2.11 Oscillation of the electric field of ordinary light occurs in all possible planes perpendicular to the direction of propagation.

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Chapter 2 | Isomerism

(a)

(b)

(c)

Photo by Michael Watson for John Wiley & Sons, Inc.

124

Figure 2.12 (a) Ordinary light passing through the first polarizing filter emerges with an electric wave oscillating in only one plane (and a perpendicular magnetic wave plane not shown). When the second filter is aligned with its polarizing direction the same as the first filter, as shown, the plane-polarized light can pass through. (b) If the second filter is turned 90°, the plane-polarized light is blocked. (c) Two polarizing sunglass lenses oriented perpendicular to each other block the light beam.

2.8B The Polarimeter ●

The device that is used for measuring the effect of optically active compounds on plane-polarized light is a polarimeter.

A sketch of a polarimeter is shown in Fig. 2.13. The principal working parts of a polarimeter are (1) a light source (usually a sodium lamp), (2) a polarizer, (3) a cell for holding the optically active substance (or solution) in the light beam, (4) an analyzer, and (5) a scale for measuring the angle (in degrees) that the plane of polarized light has been rotated. The analyzer of a polarimeter (Fig. 2.13) is nothing more than another polarizer. If the cell of the polarimeter is empty or if an optically inactive substance is present, the axes of the plane-polarized light and the analyzer will be exactly parallel when the instrument reads 0°, and the observer will detect the maximum amount of light passing through. If, by contrast, the cell contains an optically active substance, a solution of one enantiomer, for example, the plane of polarization of the light will be rotated as it passes through the cell. In order to detect the maximum brightness of light, the observer will have to rotate the axis of the analyzer in either a clockwise or counterclockwise direction. If the analyzer is rotated in a clockwise direction, the rotation, α (measured in degrees), is said to be positive (+). If the rotation is counterclockwise, the rotation is said to be negative (−). A substance that rotates plane-polarized light in the clockwise direction is also said to be dextrorotatory, and one that rotates plane-polarized light in a counterclockwise direction is said to be levorotatory (Latin: dexter, right, and laevus, left).

2.8C Specific Rotation ●

The number of degrees that the plane of polarization is rotated as the light passes through a solution of an enantiomer depends on the number of chiral molecules that it encounters.

To normalize optical rotation data relative to experimental variables such as tube length and the concentration of the enantiomer, chemists calculate a quantity called the specific rotation, [α], by the following equation: [α ] =

α c⋅l

where [α] = the specific rotation α = the observed rotation c  = the concentration of the solution in grams per milliliter of solution (or density in g mL−1 for neat liquids) l  = the length of the cell in decimeters (1 dm = 10 cm)

2.8 | Properties of Enantiomers: Optical Activity +

Analyzer (can be rotated)

0° +90°

–

Observed angle of rotation

–90°

As the arrows indicate, the optically active substance in solution in the cell is causing the plane of the polarized light to rotate.

180°

Degree scale (fixed)

The plane of polarization of the emerging light is at a different angle than that of the entering polarized light.

Polarimeter sample cell

Polarizer (fixed) Light source • Polarizer and analyzer are parallel. • No optically active substance is present. • Polarized light can get through analyzer.

(a)

• Polarizer and analyzer are perpendicular. • No optically active substance is present. • No polarized light can emerge from analyzer.

(b) –40°

• Substance in cell between polarizer and analyzer is optically active. • Analyzer has been rotated to the left (from observer’s point of view) to permit rotated polarized light through (substance Analyzer Observer is levorotatory in this example).

(c)

Polarizer

Figure 2.13 The principal working parts of a polarimeter and the measurement of optical rotation. (Reprinted with permission of John Wiley & Sons, Inc. from Holum, J. R., Organic Chemistry: A Brief Course, (Reprinted with permission of John Wiley & Sons, Inc. from Holum, J. R., Organic Chemistry: A Brief Course, p. 316. Copyright 1975.) p. 316. Copyright 1975.)

The specific rotation also depends on the temperature and the wavelength of light that is employed. Specific rotations are reported so as to incorporate these quantities as well. A specific rotation might be given as follows: [α ]25 D = +3.12 This means that the D line of a sodium lamp (λ = 589.6 nm) was used for the light, that a temperature of 25 °C was maintained, and that a sample containing 1.00 g mL−1 of the optically active substance, in a 1 dm tube, produced a rotation of 3.12° in a clockwise direction.* The specific rotations of (R )-2-butanol and (S )-2-butanol are given here: H

OH

(R)-2-Butanol []25 D  13.52

HO

H

(S )-2-Butanol []25 D  13.52

*The magnitude of rotation is dependent on the solvent used when solutions are measured. This is the reason the solvent is specified when a rotation is reported in the chemical literature.

125

126

Chapter 2 | Isomerism ●

The direction of rotation of plane-polarized light is often incorporated into the names of optically active compounds.

The following two sets of enantiomers show how this is done: H

CH3 OH

(R)-()-2-Methyl-1-butanol []25 D  5.756

H

H

CH3

HO

(S )-()-2-Methyl-1-butanol []25 D  5.756

CH3 Cl

H Cl

(R)-()-1-Chloro-2-methylbutane []25 D  1.64

CH3

(S)-()-1-Chloro-2-methylbutane []25 D  1.64

The previous compounds also illustrate an important principle: ●

No obvious correlation exists between the (R) and (S) configurations of enantiomers and the direction [(+) or (−)] in which they rotate plane-polarized light.

(R)-(+)-2-Methyl-1-butanol and (R)-(−)-1-chloro-2-methylbutane have the same configuration; that is, they have the same general arrangement of their atoms in space. They have, however, an opposite effect on the direction of rotation of the plane of plane-polarized light: H

CH3 OH

Same configuration

(R)-()-2-Methyl-1-butanol

H

CH3 Cl

(R)-()-1-Chloro-2-methylbutane

These same compounds also illustrate a second important principle: ●

No necessary correlation exists between the (R) and (S) designation and the direction of rotation of plane-polarized light.

(R)-2-Methyl-1-butanol is dextrorotatory (+), and (R)-1-chloro-2-methylbutane is levorotatory (−). A method based on the measurement of optical rotation at many different wavelengths, called optical rotatory dispersion, has been used to correlate configurations of chiral molecules. A discussion of the technique of optical rotatory dispersion, however, is beyond the scope of this text.

2.8D Racemic Forms A sample that consists exclusively or predominantly of one enantiomer causes a net rotation of plane-polarized light. Figure 2.14a depicts a plane of polarized light as it encounters a molecule of (R)-2-butanol, causing the plane of polarization to rotate slightly in one direction. (For the remaining purposes of our discussion we shall limit our description of polarized light to the resultant plane, neglecting consideration of the circularly-polarized components from which plane-polarized light arises.) Each additional molecule of (R)-2-butanol that the beam encounters would cause further rotation in the same direction. If, on the other hand, the mixture contained molecules of (S )-2-butanol, each molecule of that enantiomer would cause the plane of polarization to rotate in the opposite direction (Fig. 2.14b). If the (R) and (S ) enantiomers were present in equal amounts, there would be no net rotation of the plane of polarized light. ●

An equimolar mixture of two enantiomers is called a racemic mixture (or racemate or racemic form). A racemic mixture causes no net rotation of plane-polarized light.

2.9 | The Synthesis of Chiral Molecules

CH3 H

CH3

C

C C2H5

(R)–2–butanol

H

HO

OH

C2H5

Rotation

There is net (S)–2–butanol Equal and rotation if (if present) opposite rotation (R)–2–butanol by the enantiomer is present predominantly or exclusively. (b) (c)

(a)

Figure 2.14 (a) A beam of plane-polarized light encounters a molecule of (R )-2-butanol, a chiral molecule. This encounter produces a slight rotation of the plane of polarization. (b) Exact cancellation of this rotation occurs if a molecule of (S)-2-butanol is encountered. (c) Net rotation of the plane of polarization occurs if (R)-2-butanol is present predominantly or exclusively.

In a racemic mixture the effect of each molecule of one enantiomer on the circularly-polarized beam cancels the effect of molecules of the other enantiomer, resulting in no net optical activity. The racemic form of a sample is often designated as being (±). A racemic mixture of (R)-(−)-2-butanol and (S)-(+)-2-butanol might be indicated as (±)-2-butanol

2.9

or

(±)-CH3CH2CHOHCH3

THE SYNTHESIS OF CHIRAL MOLECULES

2.9A Racemic Mixture Reactions carried out with achiral reactants can often lead to chiral products. In the absence of any chiral influence from a catalyst, reagent, or solvent, the outcome of such a reaction is a racemic mixture. In other words, the chiral product is obtained as a 50:50 mixture of enantiomers. An example is the synthesis of 2-butanol by the nickel-catalyzed hydrogenation of butanone. In this reaction the hydrogen molecule adds across the carbon–oxygen double bond in much the same way that it adds to a carbon–carbon double bond. CH3CH2CCH3  H9H O Butanone (achiral molecules)

Ni

* ()-CH3CH2CHCH3 OH

Hydrogen (achiral molecules)

()-2-Butanol [chiral molecules but 50:50 mixture (R ) and (S )]

Figure 2.15 illustrates the stereochemical aspects of this reaction. Because butanone is achiral, there is no difference in presentation of either face of the molecule to the surface of the metal catalyst. The two faces of the trigonal planar carbonyl group interact with the metal surface with equal probability. Transfer of the hydrogen atoms from the metal to the carbonyl group produces a chirality center at carbon 2. Since there has been no chiral influence in the reaction pathway, the product is obtained as a racemic mixture of the two enantiomers, (R)-(−)-2-butanol and (S)-(+)-2-butanol. We shall see that when reactions like this are carried out in the presence of a chiral influence, such as an enzyme or chiral catalyst, the result is usually not a racemic mixture.

127

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Chapter 2 | Isomerism

O

C

H

H

CH2CH3 CH3

CH3CH2 CH3

C

O

H

H

(a)

(b) OH

HO C

CH2CH3 CH3

CH3CH2 CH3

H

H HO

C

H OH

H

(R )-(–)-(2)-Butanol (50%)

(S)-(+)-(2)-Butanol (50%)

Figure 2.15 The reaction of butanone with hydrogen in the presence of a nickel catalyst. The reaction rate by path (a) is equal to that by path (b). (R-15)-(−)-2Butanol and (S)-(+)-2-butanol are produced in equal amounts, as a racemate.

2.9B Stereoselective Syntheses Stereoselective reactions are reactions that lead to a preferential formation of one stereoisomer over other stereoisomers that could possibly be formed. ●

●

If a reaction produces preferentially one enantiomer over its mirror image, the reaction is said to be an enantioselective reaction. If a reaction leads preferentially to one diastereomer over others that are possible, the reaction is said to be an diastereoselective reaction.

For a reaction to be either enantioselective or diastereoselective, a chiral reagent, catalyst, or solvent must assert an influence on the course of the reaction.

2.10 MOLECULES WITH MORE THAN ONE CHIRALITY CENTER So far we have mainly considered chiral molecules that contain only one chirality center. Many organic molecules, especially those important in biology, contain more than one chirality center. Let us consider 2,3-dibromopentane, shown here in a two-dimensional bond-line formula. 2,3-Dibromopentane has two chirality centers: Br *

*

Br 2,3-Dibromopentane

A useful rule gives the maximum number of stereoisomers: ●

In compounds whose stereoisomerism is due to chirality centers, the total number of stereoisomers will not exceed 2n, where n is equal to the number of chirality centers.

For 2,3-dibromopentane we should not expect more than four stereoisomers (22 = 4). Our next task is to write three-dimensional bond-line formulas for the possible stereoisomers.

2.10 | Molecules with More than One Chirality Center

2.10A How to Draw Stereoisomers for Molecules Having More Than One Chirality Center Using 2,3-dibromopentane as an example, the following sequence explains how we can draw all of the possible isomers for a molecule that contains more than one chirality center. Remember that in the case of 2,3-dibromopentane we expect a maximum of four possible isomers because there are two chirality centers (2n, where n is the number of chirality centers). 1. Start by drawing the portion of the carbon skeleton that contains the chirality centers in such a way that as many of the chirality centers are placed in the plane of the paper as possible, and as symmetrically as possible. In the case of 2,3-dibromopentane, we simply begin by drawing the bond between C2 and C3, since these are the only chirality centers. 2. Next we add the remaining groups that are bonded at the chirality centers in such a way as to maximize the symmetry between the chirality centers. In this case we start by drawing the two bromine atoms so that they project either both outward or both inward relative to the plane of the paper, and we add the hydrogen atoms at each chirality center. Drawing the bromine atoms outward results in formula 1, shown below. Even though there are eclipsing interactions in this conformation, and it is almost certainly not the most stable conformation for the molecule, we draw it this way so as to maximize the possibility of finding symmetry in the molecule. Br

H H

Br

1

3. To draw the enantiomer of the first stereoisomer, we simply draw its mirror image, either side-by-side or top and bottom, by imagining a mirror between them. The result is formula 2. Br

H H

Br

Br

1

H H

Br

2 Mirror

4. To draw another stereoisomer, we interchange two groups at any one of the chirality centers. By doing so we invert the R,S configuration at that chirality center. ●

All of the possible stereoisomers for a compound can be drawn by successively interchanging two groups at each chirality center.

If we interchange the bromine and hydrogen atoms at C2 in formula 1 for 2,3-dibromopentane, the result is formula 3. Then to generate the enantiomer of 3, we simply draw its mirror image, and the result is 4.

Br

H Br

3

H

H

Mirror

Br H

Br

4

5. Next we examine the relationship between all of the possible pairings of formulas to determine which are pairs of enantiomers, which are diastereomers, and, for special cases like we shall see in Section 2.10B, which formulas are actually identical due to an internal plane of symmetry.

129

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Chapter 2 | Isomerism

Since structures 1 and 2 are not superposable, they represent different compounds. Since structures 1 and 2 differ only in the arrangement of their atoms in space, they represent stereoisomers. Structures 1 and 2 are also mirror images of each other; thus 1 and 2 represent a pair of enantiomers. Structures 3 and 4 correspond to another pair of enantiomers. Structures 1–4 are all different, so there are, in total, four stereoisomers of 2,3-dibromopentane. At this point you should convince yourself that there are no other stereoisomers by writing other structural formulas. You will find that rotation about the single bonds, or of the entire structure, or of any other arrangement of the atoms will cause the structure to become superposable with one of the structures that we have written here. Better yet, using different colored balls, make molecular models as you work this out. The compounds represented by structures 1–4 are all optically active compounds. Any one of them, if placed separately in a polarimeter, would show optical activity. The compounds represented by structures 1 and 2 are enantiomers. The compounds represented by structures 3 and 4 are also enantiomers. But what is the isomeric relation between the compounds represented by 1 and 3? We can answer this question by observing that 1 and 3 are stereoisomers and that they are not mirror images of each other. They are, therefore, diastereomers. ●

Diastereomers have different physical properties—different melting points and boiling points, different solubilities, and so forth.

2.10B Meso Compounds A structure with two chirality centers does not always have four possible stereoisomers. Sometimes there are only three. As we shall see: ●

Some molecules are achiral even though they contain chirality centers.

To understand this, let us write stereochemical formulas for 2,3-dibromobutane. We begin in the same way as we did before. We write formulas for one stereoisomer and for its mirror image:

Br

H Br

H

H

A

Br H

Br

B

Structures A and B are nonsuperposable and represent a pair of enantiomers. When we write the new structure C (see below) and its mirror image D, however, the situation is different. The two structures are superposable. This means that C and D do not represent a pair of enantiomers. Formulas C and D represent identical orientations of the same compound:

Br

H

C

H

Br

Br

H H

Br

D

The molecule represented by structure C (or D) is not chiral even though it contains two chirality centers. ● ●

If a molecule has an internal plane of symmetry it is achiral. A meso compound is an achiral molecule that contains chirality centers and has an internal plane of symmetry. Meso compounds are not optically active.

2.10 | Molecules with More than One Chirality Center

Another test for molecular chirality is to construct a model (or write the structure) of the molecule and then test whether or not the model (or structure) is superposable on its mirror image. If it is, the molecule is achiral. If it is not, the molecule is chiral. We have already carried out this test with structure C and found that it is achiral. We can also demonstrate that C is achiral in another way. Figure 2.16 shows that structure C has an internal plane of symmetry.

Br

H

H

Br

Figure 2.16 The plane of symmetry of meso-2,3dibromobutane. This plane divides the molecule into halves that are mirror images of each other.

2.10C How to Name Compounds with More Than One Chirality Center 1. If a compound has more than one chirality center, we analyze each center separately and decide whether it is (R ) or (S ). 2. Then, using numbers, we tell which designation refers to which carbon atom. Consider stereoisomer A of 2,3-dibromobutane: H Br

Br

2

H

3

1

4

A 2,3-Dibromobutane

When this formula is rotated so that the group of lowest priority attached to C2 is directed away from the viewer, it resembles the following: (b) CHBrCH3

(d) H

CH3

(c) Viewer

Br (a) (R) Configuration

The order of progression from the group of highest priority to that of next highest priority (from } Br, to } CHBrCH3, to } CH3) is clockwise. Therefore, C2 has the (R) configuration. When we repeat this procedure with C3, we find that C3 also has the (R) configuration: (b) CHBrCH3

(d) H

CH3

(c) Viewer

Br (a) (R) Configuration

Compound A, therefore, is (2R,3R)-2,3-dibromobutane.

131

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Chapter 2 | Isomerism

2.11 FISCHER PROJECTION FORMULAS So far in writing structures for chiral molecules we have only used formulas that show three dimensions with solid and dashed wedges. The reason is that formulas with solid and dashed wedges unambiguously show three dimensions, and they can be manipulated on paper in any way that we wish so long as we do not break bonds. Their use, moreover, teaches us to see molecules (in our mind’s eye) in three dimensions, and this ability will serve us well. Chemists, however, sometimes use formulas called Fischer projections to show three dimensions in chiral molecules such as acyclic carbohydrates. Fischer projection formulas are useful in cases where there are chirality centers at several adjacent carbon atoms, as is often the case in carbohydrates. Use of Fischer projection formulas requires rigid adherence to certain conventions, however. Used carelessly, these projection formulas can easily lead to incorrect conclusions.

2.11A How to Draw and Use Fischer Projections Let us consider how we would relate a three-dimensional formula for 2,3-dibromobutane using solid and dashed wedges to the corresponding Fischer projection formula. 1. The carbon chain in a Fischer projection is always drawn from top to bottom, rather than side to side as is often the case with bond-line formulas. We consider the molecule in a conformation that has eclipsing interactions between the groups at each carbon. For 2,3-dibromobutane we turn the bond-line formula so that the carbon chain runs up and down and we orient it so that groups attached to the main carbon chain project out of the plane like a bow tie. The carbon–carbon bonds of the chain, therefore, either lie in the plane of the paper or project behind it. CH3 Br

H Br

Br H  H

H

C C

Br

CH3 A

A

2. From the vertical formula with the groups at each carbon eclipsed we “project” all of the bonds onto the paper, replacing all solid and dashed wedges with ordinary lines. The vertical line of the formula now represents the carbon chain, each point of intersection between the vertical line and a horizontal line represents a carbon atom in the chain, and we interpret the horizontal lines as bonds that project out toward us. Doing this with the vertical, eclipsed form of 2,3-dibromobutane leads to the Fischer projection shown here. CH3 Br

C

CH3

H 

H

C CH3 A

Br

Br

H

H

Br CH3 A

Fischer projection formula

3. To test the superposability of two structures represented by Fischer projections we are allowed to rotate them in the plane of the paper by 180°, but by no other angle. We must always keep the Fischer projection formulas in the plane of the paper, and we are not allowed to flip them over. If we flip a

2.12 | Stereoisomerism of Cyclic Compounds

Fischer projection over, the horizontal bonds project behind the plane instead of in front, and every configuration would be misrepresented as the opposite of what was intended. CH3

CH3

Br

H

H

Br CH3

rotate 180° in plane

CH3

Br

H

H

Br

H

Br

Br

H

CH3

A Same structure

CH3

A

CH3 H

Br

Br

H

B Not the same (Flipping the projection formula over sideways creates the projection formula for the enantiomer of A.)

CH3 Not the same (Flipping the projection formula over end for end creates the projection formula for the enantiomer of A.)

2.12 STEREOISOMERISM OF CYCLIC COMPOUNDS Cyclopentane derivatives offer a convenient starting point for a discussion of the stereoisomerism of cyclic compounds. For example, 1,2-dimethylcyclopentane has two chirality centers and exists in three stereoisomeric forms 5, 6, and 7: H Me Me

Me H

H

H

Me

Enantiomers 5

H

H

Me Me Meso compound 7

6

The trans compound exists as a pair of enantiomers 5 and 6. cis-1,2-Dimethylcyclopentane (7) is a meso compound. It has a plane of symmetry that is perpendicular to the plane of the ring:

H

H

Me Me Plane of symmetry 7

2.12A Cyclohexane Derivatives 1,4-Dimethylcyclohexanes If we examine a formula of 1,4-dimethylcyclohexane, we find that it does not contain any chirality centers. However, it does have two stereogenic centers. 1,4-Dimethylcyclohexane can exist as cis–trans isomers and the cis and trans forms (Fig. 2.17) are diastereomers. Neither compound is chiral and, therefore, neither is optically active. Notice that both the cis and trans forms of 1,4-dimethylcyclohexane have a plane of symmetry.

133

134

Chapter 2 | Isomerism Plane of symmetry

Plane of symmetry

Me

Me Me

H H

Me

Me

Me

or

Me

H

Me

or

cis-1,4Dimethylcyclohexane

H trans-1,4Dimethylcyclohexane

Figure 2.17 The cis and trans forms of 1,4-dimethylcyclohexane are diastereomers of each other. Both compounds are achiral, as the internal plane of symmetry (blue) shows for each. 1,3-Dimethylcyclohexanes 1,3-Dimethylcyclohexane has two chirality centers; we can, therefore, expect as many as four stereoisomers (22 = 4). In reality there are only three. cis-1,3-Dimethylcyclohexane has a plane of symmetry (Fig. 2.18) and is achiral. Plane of symmetry

Me

Me

Me or

H

Me

Figure 2.18 cis-1,3-Dimethylcyclohexane has a plane of symmetry, and is therefore achiral.

H

trans-1,3-Dimethylcyclohexane does not have a plane of symmetry and exists as a pair of enantiomers (Fig. 2.19). You may want to make models of the trans-1,3-dimethylcyclohexane enantiomers. Having done so, convince yourself that they cannot be superposed as they stand and that they cannot be superposed after one enantiomer has undergone a ring flip. Me

Me H

H Me

Me

Me

H

H Me

(a)

(b)

(no plane of symmetry) (c)

Figure 2.19 trans-1,3-Dimethylcyclohexane does not have a plane of symmetry and exists as a pair of enantiomers. (c) A simplified representation of (b). 1,2-Dimethylcyclohexanes 1,2-Dimethylcyclohexane also has two chirality centers, and again we might expect as many as four stereoisomers. Indeed there are four, but we find that we can isolate only three stereoisomers. trans-1,2-Dimethylcyclohexane (Fig. 2.20) exists as a pair of enantiomers. Its molecules do not have a plane of symmetry. Me Me (a)

Me Me (b)

Figure 2.20 trans-1,2-Dimethylcyclohexane has no plane of symmetry and exists as a pair of enantiomers (a and b).

2.13 | Relating Configurations through Reactions in which No Bonds to the Chirality Center are Broken

cis-1,2-Dimethylcyclohexane, shown in Fig. 2.21, presents a somewhat more complex situation. If we consider the two conformational structures (c) and (d), we find that these two mirror-image structures are not identical. Neither has a plane of symmetry and each is a chiral molecule, but they are interconvertible by a ring flip. Therefore, although the two structures represent enantiomers, they cannot be separated because they rapidly interconvert even at low temperature. They simply represent different conformations of the same compound. Therefore, structures (c) and (d) are not configurational stereoisomers; they are conformational stereoisomers. H

H H

H

Me (c)

2.13

●

Me

Me

Me

(d)

Figure 2.21 cis-1,2-Dimethylcyclohexane exists as two rapidly interconverting chair conformations (c) and (d ).

RELATING CONFIGURATIONS THROUGH REACTIONS IN WHICH NO BONDS TO THE CHIRALITY CENTER ARE BROKEN

A reaction is said to proceed with retention of configuration at a chirality center if no bonds to the chirality center are broken. This is true even if the R,S designation for the chirality center changes because the relative priorities of groups around it changes as a result of the reaction.

First consider an example that occurs with retention of configuration and that also retains the same R,S designation in the product as in the reactant. Such is the case when (S )-(−)-2-methyl-l-butanol reacts with hydrochloric acid to form (S )-(+)-l-chloro-2-methylbutane. Note that none of the bonds at the chirality center are broken. Same configuration

CH3

H OH  H

CH3

heat

Cl

(S)-()-2-Methyl-1-butanol []25  5.756 D

H Cl  H OH

(S)-()-1-Chloro-2-methylbutane []25  1.64 D

This example also reminds us that the sign of optical rotation is not directly correlated with the R,S configuration of a chirality center, since the sign of rotation changes but the R,S configuration does not. Next consider the reaction of (R )-1-bromo-2-butanol with zinc and acid to form (S )-2-butanol. At this point we do not need to know how this reaction takes place, except to observe that none of the bonds to the chirality center are broken. H

Zn, H (ZnBr2)

OH Br

retention of configuration

(R)-1-Bromo-2-butanol

H

OH H

(S)-2-Butanol

This reaction takes place with retention of configuration because no bonds to the chirality center are broken, but the R,S configuration changes because the relative priorities of groups bonded at the chirality center changes due to substitution of hydrogen for bromine.

135

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Chapter 2 | Isomerism

2.14

CHIRAL MOLECULES THAT DO NOT POSSESS A CHIRALITY CENTER

A molecule is chiral if it is not superposable on its mirror image. The presence of a tetrahedral atom with four different groups is only one type of chirality center, however. While most of the chiral molecules we shall encounter have chirality centers, there are other structural attributes that can confer chirality on a molecule. For example, there are compounds that have such large rotational barriers between conformers that individual conformational isomers can be separated and purified, and some of these conformational isomers are stereoisomers. Allenes are compounds that exhibit stereoisomerism. Allenes are molecules that contain the following double-bond sequence: C

C

C

The planes of the p bonds of allenes are perpendicular to each other: R R C

C

C R

R

This geometry of the p bonds causes the groups attached to the end carbon atoms to lie in perpendicular planes, and, because of this, allenes with different substituents on the end carbon atoms are chiral (Fig. 2.22). (Allenes do not show cis–trans isomerism.) H C Cl

C

C

H Cl

H Cl

H C

C

C Cl

Mirror

Figure 2.22 Enantiomeric forms of 1,3-dichloroallene. These two molecules are nonsuperposable mirror images of each other and are therefore chiral. They do not possess a tetrahedral atom with four different groups, however.

2.15 CONFORMATIONAL ISOMERISM The different spatial arrangements that a molecule can have due to free rotation about carbon–carbon single bond are known as conformations. Groups bonded by only a sigma (σ) bond (i.e., by a single bond) can undergo rotation about that bond with respect to each other. The isomers that differ in their conformations are called conformational isomers or conformers, and the phenomenon is known as conformational isomerism. An analysis of the energy changes associated with a molecule undergoing rotation about single bonds is called conformational analysis. This type of isomerism mostly occurs in alkanes and cycloalkanes and their substituted derivatives.

2.15A Conformations Conformations of a molecule are three-dimensional arrangements that differ only by rotation around a single bond. Each increment of rotation, however small, produces a change of conformation. In particular, the atoms remain connected in the same order during conformational change, with bonds being neither formed nor broken. For example, in n-butane, H

CH3

H

H

CH3

CH3

Rotation

H

H CH3

H

H H

2.15 | Conformational Isomerism

Thus, a molecule can assume an infinite number of conformations because of rotation around its sigma bonds. However, certain conformations are more stable than others. Isolation of conformers in non-rigid systems is not feasible because the potential energy barrier for the inter-conversion is too low. The different conformations of a molecule: ● ● ● ●

have the same molecular formula. have the same atomic connections. have different three-dimensional shapes. are interchanged by the rotation of single bonds.

Around the C}C bond, there can be an infinite number of conformations. However, there is a torsional strain or weak repulsive interaction between adjacent bonds causing an energy barrier of 1–20 kJ mol−1. This hinders the rotation of the C}C bond. Among the infinite number of conformations possible, the two extreme cases are when hydrogen atoms on two carbon atoms are the closest and the most far apart. The former is known as eclipsed conformation and the latter as staggered confirmation. The intermediate positions of hydrogen on the two carbons are called skew conformations. When we do conformational analysis, we will find that certain types of structural formulas are especially convenient to use. One of these types is called a Newman projection formula and another type is a sawhorse formula (Fig. 2.23).

Newman projection formula

Sawhorse formula

Figure 2.23 Conformational analysis using Newman and Sawhorse formulas.

2.15B Sawhorse Projections To draw a sawhorse projection, the molecule is viewed along the molecular axis and carbon–carbon bond is represented by an elongated straight line. The carbon in the front is shown at the lower end of the line and that at the back is represented by the upper end of the line. The line is tilted slightly towards left or right side for clear depiction of both the carbons. Each carbon end has three lines attached to it, at an angle 120°, to represent the attached hydrogen atoms. The sawhorse projections of staggered and eclipsed forms of ethane are shown in Fig. 2.24

(a) Staggered

(b) Eclipsed

Figure 2.24 Sawhorse projections.

2.15C Newman Projections In this projection, the molecule is viewed along the bond joining the carbon atoms. For example, in ethane molecule, the carbon atoms C1 and C2 are represented as superimposed circles, only one circle being drawn as shown in Fig. 2.25. The center of the circle represents the front carbon C1 while the circumference represents the back carbon atom C2, the line joining the two carbon atoms is not visible. The remaining bonds on each H carbon atom are shown by small straight lines at angles C2 H of 120° joined to the center and to the circumference. The H H different conformations are characterized by the angles H H C2 formed between the bonds on the front carbon atom C1 C1 and the bonds on the rear carbon atom, which are called H H H H H dihedral angles. H The conformational isomerism in alkanes is discussed in detail in Chapter 4. Figure 2.25 Newman projections.

137

138

Chapter 2 | Isomerism

SOLVED EXAMPLES 1. Which of the following are chiral and, therefore, capable of existing as enantiomers? (a) 2-Butanol (b) 1,1-Dibromopropane (c) 2-Chloro-2-methylpropane (d) 3-Bromopentane (e) 2-Propanol (f) 2-Bromopentane (g) 1-Fluoro-2-ethylpentane Solution (a) It is a chiral compound, therefore, can exist as enantiomers. (b) 1,1-Dibromopropane-No (c) 2-Chloro-2-methylpropane-No (d) 3-Bromopentane-No (e) 2-Propanol-No (f) 2-Bromopentane-Yes (g) 1-Fluoro-2-ethylpentane-Yes 2. Which of the following is not chiral? (1) 2-Butanol (2) 2,3-Dibromopentane (3) 3-Bromopentane (4) 2-Hydroxypropanoic acid

H C

CH2

(a) (b) (c) (d) (e) (f)

} Cl > } SH > } OH > } H } CH2Br > } CH2Cl > } CH2OH > } CH3 } OH > } CHO > } CH3 > } H } C(CH3)3 > } CH CH2 > } CH(CH3)2 > } H } OCH3 > } N(CH3)2 > } CH3 > } H } OPO3H2 > } OH > } CHO > } H

5. Assign absolute configurations (R or S) to each of the following compounds. Cl

H

H

(b)

CH3

Br

3. Indicate with an asterisk all chiral centers in the following molecules

(a) The order of priority is

} Cl > } CH2CH3 > } CH3 > } H. The group of lowest priority, H, points away from you. Reading the groups in the order 1, 2, 3 occurs in the counterclockwise direction, so the configuration is S. 1

OH OH

(c) HO

OH

(d)

CH3

N

OH

*

*

(c) HO

O

2

OH

N CH3

(d)

* *

OH R

6. The chirality of the following compound is Br

NH2 *

N

1

O

OH

(b)

H

3

Solution (a)

3

(b) The order of priority is } OH > } CH CH > } CH2 } CH2 > } H With hydrogen, the group of lowest priority, pointing away from you, reading the groups in the order 1, 2, 3 occurs in the clockwise direction, so the configuration is R.

NH2 N

H

2

OH

O

(b)

Cl

S

O

(a)

OH

Solution

(3) 3-Bromopentane is an achiral compound. H2C

Solution

(a)

Solution

H3C

4. List the substituents in each of the following sets in order of priority, from highest to lowest: (a) } Cl, } OH, } SH, } H (b) } CH3, } CH2Br, } CH2Cl, } CH2OH (c) } H, } OH, } CHO, } CH3 (d) } CH(CH3)2, } C(CH3)3, } H, } CH CH2 (e) } H, } N(CH3)2, } OCH3, } CH3 (f) } OH, } OPO3H2, } H, } CHO

C

OH

H3C

(1) E (2) R

H Cl

(3) S (4) Z

Solved Examples Solution

Br1 H4 C H33C Cl2

(2) Each of four groups attached to the stereogenic carbon is assigned a priority which is first done on the basis of atomic number. Therefore, the priority of atoms are Br > Cl > CH3 > H

The molecule is designated as (R) due to its clockwise rotation.

7. Write three-dimensional formulas for all of the stereoisomers of each of the following compounds. Label pairs of enantiomers and label meso compounds. Cl

F

F

(a)

Cl

(c) Cl Cl

(e)

F

Br OH HO2C

(b) OH

CO2H

(f)

(d) OH

OH

Cl

OH Tartaric acid

Solution H

(a)

H

Cl

Cl

Cl

Cl

(1)

H

H

H

Cl

(2)

H HO OH H

Meso compound

HO

H H

(1)

OH

H HO H

(2)

Meso compound F

F

(c)

F

F

H

F

(2)

H

HO

(1)

H H (2)

Enantiomers

H (3)

Enantiomers

H OH Cl

Cl Cl

(1)

(d)

F

Cl Cl

H

H

H

Cl Cl

OH

(3)

Enantiomers H

H (3)

Enantiomers

(b)

Cl

Meso compound

Cl

H OH H

Cl

HO

(3)

H Cl (4)

Enantiomers

H

139

140

Chapter 2 | Isomerism H

(e) H

F

F

Br

Br

(1)

H

F

H

H (2)

H

Br

Br

F

H (4)

Enantiomers HO HO2C

OH

HO HO2C

H

HO

(1)

H CO2H

CO2H

(f)

H

(3)

Enantiomers H OH HO2C

H

H

(2)

H CO2H

OH (3)

Enantiomers

Meso compound

8. Which of the following pairs of compounds are enantiomers? CH3

CH3

HO

H

and

H

OH

HO

H

CH3

CH3

CH3

CH3

CH3

OH

HO

H

HO

and

H

HO

CH3

H

H

CH3

and

H

OH

H

OH CH3 CH3

H

HO

(4)

CH3 OH

H

(3)

CH3

H

(2)

H

HO

OH

H

(1)

CH3

OH

and

CH3

HO

H

HO

H CH3

Solution (1) Enantiomers are stereoisomers whose molecules are non-superimposable mirror images of each other. CH3 H

CH3 OH

HO

HO

H

H OH

H

CH3

CH3

9. How many stereoisomers exist for 1,3-cyclohexanediol? Solution 1,3-Cyclohexanediol has two stereocenters, and, according to the 2n rule, a maximum of 22 = 4 stereoisomers is possible. The trans isomer of this compound exists as a pair of enantiomers. The cis isomer has a plane of symmetry and is a meso compound. Therefore, although the 2n rule predicts a maximum of four stereoisomers for 1,3-cyclohexanediol, only three exist—one pair of enantiomers and one meso compound: plane of symmetry

OH

OH cis-1,3-Cyclohexanediol (a meso compound)

OH

HO

OH

OH

trans-1,3-Cyclohexanediol (a pair of enantiomers)

Solved Previous Years’ NEET Questions 10. Tell whether the compounds of each pair are enantiomers, diastereomers, constitutional isomers, or not isomeric. CH2OH

(a)

(c)

CH2OH

H

OH

HO

H

H

OH

HO

H

CHO HO H

H

OH

H

OH

H

OH

H

OH

OH

H

OH CH2CH3

CH2OH

CHO

CHO

CH2CH3

OH

CH2OH

(d)

H

C(O)CH3

H

H

CH2Br

CH2Br

(b)

C(O)CH3

CHO OH

HO

HO

H

H OH

H CH2OH

C(O)CH3

Solution (a) Enantiomers (b) Diastereomers

(c) Not isomeric (d) Not isomeric

SOLVED PREVIOUS YEARS’ NEET QUESTIONS 1. If there is no rotation of plane polarized light by a compound in a specific solvent, thought to be chiral, it may mean that (1) the compound may be a racemic mixture. (2) the compound is certainly a chiral. (3) the compound is certainly meso. (4) there is no compound in the solvent.

Solution (4) The R configuration is shown by the below compound 2

C2H5 R 3

1

CH3

Cl

(AIPMT 2007) Solution

4

(3) Optical activity is measured by the degree of rotation of plane-polarized light upon passage through a chiral medium. Molecules that are achiral cause no rotation of the plane of polarized light. Meso compounds, because they are achiral, are optically inactive. 2. CH3 } CHCl } CH2 } CH3 has a chiral center, which one of the following represents its R configurations? CH3

(1) H

C

C2H5

(3) H

Cl

C2H5

C H

CH3

Cl

3. How many stereoisomers’ does the molecule have? CH3CH

C

CH3

Solution (2) The molecule will have two geometrical isomers and four stereoisomers. H3C

(AIPMT 2007)

C

C

H

H

CH2

C*

trans H H3C

H

(3) 6 (4) 8 (AIPMT 2008)

C2H5

(4) Cl

CHCH2CHBrCH3

(1) 2 (2) 4

H

Cl

C2H5

(2) H3C

C

H

C

C cis

Chiral carbon CH3

Two stereoisomers

Br H

H

CH2

C* Br

Chiral carbon CH3

Two stereoisomers

141

142

Chapter 2 | Isomerism 4. Which of the following acids does not exhibit optical isomerism? (1) Maleic acid (3) Lactic acid (2) α-Amino acids (4) Tartaric acid

H

H

σ

σ

C

O σ

σ

σ

σ

C

σ π

H

C

σ

σ

σ π

σ

C

H

H

O

H

σ

O

σ

H

C

σ

σ

σ

C

σ

σ

H

H

H

(AIPMT PRE 2012) Solution

7. Given:

(1) The mirror image of maleic acid is exactly the same, so it exhibits no optical isomerism. H H

C

COOH

5. The order of stability of the following tautomeric compounds is

CH2

C

O CH2

C

O CH3

CH3

C

(I)

O CH2

CH3

C

(II) OH CH3

CH3 CH3

C

O

O

O

(I)

(II)

(III)

CH

CH3 CH3

Which of the given compounds can exhibit tautomerism? (1) I and III (2) II and III (3) I, II and III (4) I and II (AIPMT 2015)

O CH3

C

(III)

(1) I > II > III (2) III > II > I

CH3

COOH

C

Maleic acid

OH

CH3

(3) II > I > III (4) II > III > I (NEET 2013)

Solution (4) The keto and enol forms of carbonyl compounds are constitutional isomers. They are easily interconverted in the presence of traces of acids or bases. CH3

CH3

CH3

CH3

Solution H H

(2) The compound (III) is the most stable due to intermolecular hydrogen bonding and because it is a conjugate diene. Compound (II) is not a conjugated diene, but involves more acidic hydrogen in tautomerism. Compound (I) has the least number of acidic hydrogens for tautomerism.

O

OH CH3 CH3

6. The enolic form of ethyl acetoacetate as below has:

CH3 CH3

H H3C

(1) (2) (3) (4)

H C

O

C

C

OH

OC2H5

H3C

H2 C

O

O

C

C

O

OC2H5

16 sigma bonds and 1 p bond 9 sigma bonds and 2 p bonds 9 sigma bonds and 1 p bond 18 sigma bonds and 2 p bonds

8. Two possible stereostructures of CH3CHOH⋅COOH, which are optically active, are called (1) enantiomers. (2) mesomers. (3) diastereomers. (4) atropisomers. (RE-AIPMT 2015)

(AIPMT 2015) Solution (4) There are two p bonds and eighteen σ bonds in the given structure.

OH

Solution (1) There is one chiral center in the molecule. Thus, two optically active enantiomers are possible. The structures are

Solved Previous Years’ NEET Questions Solution COOH

COOH

H

OH

HO

H

CH3

CH3

9. The correct statement regarding a carbonyl compound with a hydrogen atom on its alpha-carbon is (1) a carbonyl compound with a hydrogen atom on its alpha-carbon never equilibrates with its corresponding enol. (2) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as aldehyde-ketone equilibration. (3) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as carbonylation. (4) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as keto-enol tautomerism. (NEET-I 2016)

(2) Substituted biphenyl compounds can show optical activity even in the absence of chiral carbon because of restricted rotation about carbon-carbon single bond. Steric hindrance arises due to presence of bulkier groups at ortho positions of benzene rings, the biphenyl system becomes non-planar and hence optically active. Br Br

I

Restricted rotation around bond

●

●

Biphenyl systems in option (2) are optically active because the ortho positions in both the benzene rings are occupied by groups, thus restricting the free rotation around the single bond. neither of the rings have plane of symmetry and center of symmetry.

11. Which among the given molecules can exhibit tautomerism?

Solution

O

(4) A carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as keto-enol tautomerism. H R

I

C

C

H

O

OH

(Keto form)

(Enol form)

R′

R

CH

C

R′

I CH3

(4) I

I

Both I and III Both I and II Both II and III III only

Solution

CH3

(NEET-I 2016)

OH

O

I

Br Br

(III)

(4) Keto-enol tautomerism is only possible for compound III.

(3)

(2)

(II)

(NEET-II 2016)

I

(1)

O

Ph Ph (I)

(1) (2) (3) (4)

10. Which of the following biphenyls is optically active? O2N

O

H H Keto form

Enol form

In compound I, the hydrogen atoms at the bridge head cannot participate in keto-enol tautomerism. In compound II, one of the α-positions is at the bridge head and the other does not contain α-hydrogen.

143

144

Chapter 2 | Isomerism

ADDITIONAL OBJECTIVE QUESTIONS NCERT Exemplar Questions 1. In which of the following molecules carbon atom marked with asterisk (*) is asymmetric? H I

H

C* CI

D I

C* CH 3 C2H5 (III)

OH

Br (I)

H

C* CI

H

Br (II)

(1) I, II, III, IV (2) I, II, III

C* CH 3 C2H5 (IV)

(3) II, III, IV (4) I, III, IV

2. Which of the following structures is enantiomeric with the molecule (A) given below?

4. Which of the following compound is not chiral? (1) 1-Chloropentane (2) 3-Chloro-2-methylpentane (3) 1-Chloro-2-methylpentane (4) 2-Chloropentane 5. Which of the following is true about any (R)-enantiomer? (1) It is dextrorotatory. (2) It is levorotatory. (3) It is an equal mixture of (+) and (−). (4) It is the mirror image of the (S)-enantiomer. 6. How many stereogenic centers are there in the following compound? O

HO

H H5C2

3. Which of the following is a comparatively insignificant factor affecting the magnitude of specific optical rotation? (1) Concentration of the substance of interest (2) Purity of the sample (3) Temperature of the measurement (4) Length of the sample tube

CH3

C

O

Br

O

(A) H

H

(1)

C2H5

C

H3C

(3)

H3C

C2H5

Br CH3

(2)

Br

H

C C2H5

Br

C

Br

(4)

H5C2

H

C

(1) 8 (2) 5

(3) 6 (4) 7

7. An alkane which can exhibit optical activity is (1) neopentane. (3) 3-methylpentane. (2) isopentane. (4) 3-methylhexane. 8. Which molecule among the following is achiral?

CH3

3. Molecules whose mirror image is non-superimposable over them are known as chiral. Which of the following molecules is chiral in nature? (1) 2-Bromobutane (3) 2-Bromopropane (2) 1-Bromobutane (4) 2-Bromopropan-2-ol

Exercise 1 1. Enantiomers are (1) molecules that have a mirror image. (2) molecules that have at least one stereogenic center. (3) non-superposable molecules. (4) non-superposable molecules that are mirror images of each other. 2. Which of the following is not true of enantiomers? They have the same (1) boiling point. (3) specific rotation. (2) melting point. (4) density.

H

(1)

H

Cl

Br

Br

H

(3)

Cl

Br

Cl

Br

Cl

Br

Cl

(2)

H

Br

H

Cl

(4) None of these

Br

9. (R)-2-Chlorobutane is represented by H

CH3

(1) H3C

Cl

(3) H3C

CH2CH3

CH2CH3 CH3

(2) Cl

Cl

CH2CH3 CH3

CH2CH3

(4) H3C

Cl H

Additional Objective Questions 10. Which structure represents (S)-2-bromobutane? CH3

(1) H

(3) H3C

Br

H Br

(1) (2) (3) (4)

CH3

(4) None of these

H CH2CH3

11. Which statement is not true for a meso compound? (1) The specific rotation is 0°. (2) There are one or more planes of symmetry. (3) A single molecule is identical to its mirror image. (4) The stereochemical labels, (R) and (S), must be identical for each stereogenic center. 12. An achiral molecule is one that is _________________ upon its mirror image. (1) superposable. (2) nonsuperposable. (3) superposable to some extent. (4) None of these. 13. To be superposable, when one object is placed on top of another, _________________. (1) some part of each object must coincide. (2) all parts of each object must coincide. (3) coinciding of each object is not a necessary requirement. (4) None of these.

Cl Cl H Cl (1) Diastereomers (2) Enantiomers

21. Which of the following is the enantiomer of the given substance (A)? H H Br

16. Hexane and 3-methylpentane are examples of (1) enantiomers. (3) diastereomers. (2) stereoisomers. (4) constitutional isomers. 17. Among the following four structures I to IV, it is true that,

CH (I)

C3H7

CH3

CH3 (A) CH3

H Br H

CH3 Br H H

CH3 H

H Br (III)

(II)

I II III It does not have a non-superposable enantiomer.

22. Five alcohols can be drawn for formula C4H10O. How many of these are optically active? (1) 1 (3) 3 (2) 2 (4) 4

(3) 6 (4) 7

CH3

H (3) Meso compounds (4) Identical compounds

20. Which of the following will have mesoisomer also? (1) 2-Chlorobutane (2) 2-Hydroxypropanoic acid (3) 2,3-dichloropentane (4) 2,3-dichlorobutane

(1) (2) (3) (4)

O

C 3H 7

19. How are the following compounds related? H3C Cl CH3

(I)

O

CH (IV)

18. Which compound would show optical activity? (1) cis-1,4-Dimethylcyclohexane (2) trans-1,4-Dimethylcyclohexane (3) cis-1,4-Dimethylcycloheptane (4) trans-1,4-Dimethylcycloheptane

15. How many chiral centers are present in the following compound?

C2H5

CH3

all four are chiral compounds. only I and II are chiral compounds. only III is a chiral compound. only II and IV are chiral compounds.

14. Molecule that contains stereogenic carbons but is nonetheless achiral is referred to as a(an) __________. (1) diastereoisomer. (3) meso compound. (2) enantiomer. (4) identical compound.

(1) 4 (2) 5

CH3

C1 H (III)

CH2CH3

CH2CH3

(2) Br

H

H

O

CH3

C

CH (II)

23. Amongst the following compounds, the optically active compound having lowest molecular mass is (1)

H3C CH3

(3) H

CH3 C 2H 5 CH3

C2H5

(2) H3C

CH

(4) H3C

CH3

145

146

Chapter 2 | Isomerism 24. Which of the following is a meso compound? OH CH3

H

(1) HO

CH2

H

H3C H

(2)

30. Out of the following, the alkene that exhibits optical isomerism is (1) 2-methyl-2-pentene. (2) 3-methyl-2-pentene. (3) 4-methyl-1-pentene. (4) 3-methyl-1-pentene. 31. How many pairs of enantiomers are possible for the following compound?

OH CH3

H3C HO

OH

H

CH3

H

CH2OH

(3) H

OH

(4) H

H

CH2OH

H

CH2OH

HO

25. Rank the following substituents in order of increasing priority. H OH H

OCH3

O

H (I)

H (II)

(1) III, II, IV, I (2) I, IV, II, III

(III)

(IV)

(3) IV, I, III, II (4) I, II, III, IV

(1) 16 (2) 32

33. Which of the following pairs contain identical compounds?

27. Which types of isomerism is shown by 2,3-dichlorobutane? (1) Diastereomers (3) Geometric (2) Optical (4) Structural

CHO H

(1)

F

F (I)

H

CH3

Br

Br

H

H

OH

H

Br

H

CH3

Br

CHO

CHO H

H

OH

HO

H CH2OH

CHO

(3)

CH } CH(OH) } CH3 is

(4)

CHO OH

H

(3) II, III and IV (4) III and IV

(3) 4 (4) 6

OH

H

and

CH2OH

(IV)

29. The number of stereoisomers possible for a compound of the molecular formula (1) 3 (2) 2

OH CH2OH

H

CH3 } CH

H

(V)

Cl

(1) I, II and V (2) I, III and IV

H

HO

and

CH2OH

HO

(2)

H3C

(III)

(II)

CHO OH

28. Which of the following compounds does not possess a plane of symmetry? H

(3) 64 (4) 128

32. The optically inactive compound from the following is (1) 2-chloropropanal. (2) 2-chloropentane. (3) 2-chlorobutane. (4) 2-chloro-2-methylbutane.

26. CH3CHBrCH2CHClCH3 is the generalized representation of what number of stereoisomers? (1) 3 (3) 5 (2) 4 (4) 6

H

H

OH

and

HO

H

HO

H

CH2OH

CH2OH

CH2OH

CHO

H

OH

H

OH CHO

and

HO

H

HO

H CH2OH

Additional Objective Questions

Exercise 2

6. Which of the following molecules is a meso compound?

1. How many stereoisomers are possible for the following compound? N

H N O

(1) 0 (2) 2

H

F

H

H

H

CH3

F

CH3

OH F

F

H

H

F

H

CH3

F

CH3

(I)

S N

H

H

O

O

(II)

(1) I and II (2) IV and V

(3) 4 (4) 8

2. What kind of isomerism is possible in the following organic compound?

(3) II and III (4) None of the above

Br

CH3

CH3

CH3

(1) (2) (3) (4)

Optical Geometrical Both (1) and (2) None of these

3. How many chiral carbons are there in the following compound? H3C

Br

CH3

8. Which pair of structures are enantiomers? CH3

CH3

COOH H OH

HOOC

CH3

H3C

Br

(II)

Br

H Cl

Br

(III)

4. Which of the following statements about stereoisomers are true? (I) Enantiomers and diastereomers have the same physical properties. (II) 50/50 mixtures of R and S enantiomers are called racemic mixtures. (III) Meso isomers rotate the plane of plane polarized light. (IV) Dextrorotatory compounds rotate plane polarized light to the right. (1) I, II (3) II, IV (2) II, III (4) III, IV 5. The compounds below are:

Cl

F

Br

Br

constitutional isomers. enantiomers. diastereomers. identical.

(1) I, II (2) II, III

(3) I, III (4) I, II, III

CHO

(1) HO

H2N H

CH2OH

(3)

HN2

H

H

Ph

Ph COOH

SH

(4) H2N

H H

F CH3

H CH2CH3

HO

9. Which of the following molecules is expected to rotate the plane of plane-polarized light?

(2)

Cl

CH3

CH3

CH3 H

HO

CH2CH3 H

Cl

CH2CH3

(3) 8 (4) 9

CH3 H

HO

CH2CH3

HO

(1) 6 (2) 7

Br

the same melting point. different melting points. equal but opposite optical rotations. More than one of the above options.

(I)

CH3

(V)

7. The compounds whose structures are shown below, would have

CH3

(1) (2) (3) (4)

(IV)

CH3

H3C

(1) (2) (3) (4)

(III)

10. Which of the following compounds contain stereocenters? (I) 1-Chloropentane (III) 3-Chloropentane (II) 2-Chloropentane (IV) 1,2-Dichloropentane (1) I, II (2) III, IV

(3) I, III (4) II, IV

147

148

Chapter 2 | Isomerism 16. The stereochemical relationship between the following molecules is

11. I and II are F

I

F

(I)

(1) (2) (3) (4)

Br

(II)

CH3

Br

OH

H

H

H

H

CH3 H HO HO

CH3

OH H H

CH3

H H HO

CH3

(I)

OH H OH

OH OH H CH3

(III)

HO HO H

(1) I and II (2) II and III

CH3

H CH2CH2CH3

(III)

(V)

CH2CH2CH3 CH3

H3C

Cl H

(II)

(IV)

(1) I, II and III, IV (2) I, II

(3) III, IV (4) IV, V

18. Of the compounds which correspond to the general name “dichlorocyclobutane”, how many are optically active? (1) 0 (3) 2 (2) 1 (4) 3

H

H

(1) Constitutional isomers (3) Diastereomers (2) Enantiomers (4) Different molecules 20. Which of the following compounds (I–IV) represent enantiomers?

(3) III and IV (4) III and V

CH2OH

14. Rank the following substituents in order of increasing priority.

(1) (2) (3) (4)

H

H H OH

(IV)

(II)

Cl

19. Indicate the relationship of the pair of molecules shown below.

(V)

CH3

CH3

CH2Cl

CH2CH2CH3

(I)

CH3

CH3

H3C

CH2CH2CH3

13. Which structures among the following represent the same compound? OH OH OH

Cl

Cl

(3) 2 (4) 4

CH3

H

CH2CH2CH3 COOH

(1) Zero (2) 3

H HO H

H3C

CH3

HOOC H

(3) Enantiomers (4) Constitutional isomers

17. Pairs of enantiomers are

12. How many meso forms are possible for the following organic compound? CH3 OH

I

(1) Identical (2) Diastereomers

constitutional isomers. enantiomers. identical. diastereomers.

H H H

Br

OH

Cl

CH3

H

(I)

(II)

(III)

(IV)

I, III, II, IV II, I, III, IV III, I, II, IV IV, III, I, II

15. How many stereoisomers are possible for 2,3butanediol? (1) 1 (3) 3 (2) 2 (4) 4

CH3

H

OH

HO

H

H

H

HO

CH3

CH3

H

(I)

(III)

H

CH3

HO

CH3

H

H

CH3

H3C

OH (II)

OH H OH (IV)

Answer Key (1) I and II (2) II and III

Exercise 3

(3) III and IV (4) II and IV

In the following set of questions, two statements “Assertion” and “Reason” are given. Choose the correct answer from the following options:

21. Which molecule has a plane of symmetry? F Cl

H

H

H Cl (I)

H

H

CH3

H

(II)

(1) I (2) II

(1) Assertion and Reason are true and the Reason is the correct explanation of the Assertion. (2) Assertion and Reason are true but the Reason is not a correct explanation of the Assertion. (3) Assertion is true but the Reason is false. (4) Assertion and Reason both are false.

H

Cl

CH3 CH3 (III)

(3) III (4) More than one of these

22. Which of the following is a meso compound? H H OH

3. Assertion: Diastereomers have different physical properties.

OH

Reason: They are non-superposable mirror images.

H

4. Assertion: Chirality is the essential condition for a molecule to be optically active.

H

OH H

(2)

Reason: Propadiene has a plane of symmetry. H

(3)

H

Reason: All molecules with chirality centers can rotate the plane-polarized light in clockwise or anti-clockwise direction.

OH

(4)

OH H

OH

Br

23. What is the molecular formula for the alkane of smallest molecular weight which possesses a stereogenic center? (1) C4H10 (3) C6H14 (2) C5H12 (4) C7H16 24. The absolute configuration of:

Br

5. Assertion:

are stereoisomers. Br

Br

Reason: Stereoisomers have different orientation of groups in space. 6. Assertion: A racemic mixture causes no net rotation of plane-polarized light. Reason: Racemic mixture is a mixture of meso structures.

CO2H H

OH

H

Cl

7. Assertion: Meso-tartaric acid is optically inactive. Reason: Meso-tartaric acid contains two asymmetrical carbon atoms. 8. Assertion: Every optically active substance rotates the plane of polarized light towards right.

CH3

(1) (2R, 3R) (2) (2R, 3S)

Reason: All chiral molecules have chiral centers. 2. Assertion: Propadiene is optically active.

OH OH

(1)

1. Assertion: Molecules that are not superimposable on their mirror images are chiral.

Reason: Optically active stereoisomers must possess dissymmetric structures.

(3) (2S, 3R) (4) (2S, 3S)

ANSWER KEY NCERT Exemplar 1. (2)

2. (1)

3. (1)

1. (4)

2. (3)

3. (3)

4. (1)

5. (4)

6. (1)

7. (4)

8. (1)

9. (3)

10. (1)

Exercise 1

149

150

Chapter 2 | Isomerism 11. (4)

12. (1)

13. (2)

14. (3)

15. (3)

16. (4)

17. (2)

18. (4)

19. (1)

20. (4)

21. (4)

22. (2)

23. (3)

24. (1)

25. (2)

26. (2)

27. (2)

28. (4)

29. (3)

30. (4)

31. (2)

32. (4)

33. (4)

2. (3)

3. (3)

4. (3)

5. (4)

Exercise 2 1. (1) 6. (4)

7. (4)

8. (1)

9. (1)

10. (4)

11. (4)

12. (4)

13. (4)

14. (4)

15. (3)

16. (1)

17. (3)

18. (3)

19. (3)

20. (2)

21. (4)

22. (4)

23. (4)

24. (3)

1. (3)

2. (4)

3. (2)

4. (4)

6. (4)

7. (2)

8. (4)

Exercise 3 5. (4)

HINTS AND EXPLANATIONS Exercise 1 4. (1) 1-Chloropentane. This is because it does not have a chiral or asymmetric carbon atom. 6. (1)

17. (2) Because in structures I and II, there is an asymmetric center present, that is, carbon atom having four different substituent attached. 18. (4) Trans-1,4-dimethylcycloheptane does not possess any symmetry element, hence is optical active.

O

HO * O * *

O * *

*

*

*

7. (4)

19. (1) The stereoisomers that are optically active isomers but not mirror images are called diastereoisomers or diastereomers. 20. (4) 2,3-Dichlorobutane has a plane of symmetry that divides the molecule into halves that are mirror images of each other.

*

CH3 H

15. (3) The carbon atom that is attached by four different group of atoms is called chiral carbon and represents chiral center. There are 6 chiral centers present in the given compound.

Cl Plane of symmetry Cl

H CH3

O

H O

22. (2) H3C

C*

CH3 CH2

CH3

H3C

OH

C

OH

CH3

H

16. (4) Both hexane and 3-methylpentane have molecular formula C6H14, therefore, are the examples of constitutional isomers.

H3C

C CH3

CH2OH

CH3

CH2

CH2

CH2OH

Hints and Explanations 23. (3) Among the given compounds, there is only one optically active alkane. CH3

Cl CH3

CH *

CH3

CHO

CH2

CH2

C2H5

2-Chloropentane

2-Chloropropanal

Chiral center

CH3

Four different substituents attached (chiral carbon). 26. (2)

CH3 Br

Cl

*

*

Stereocenters = 2, therefore stereoisomers 2 = 2 = 4.

CH3

CH3

H

Cl

H

Cl

Cl

H

H

Cl

Cl

H

H

Cl

CH3

CH3

H

H

H

C

C

C*

OH

CH3CHCH2CH

3-Methyl-2-pentene (No chiral carbon)

4-Methyl-1-pentene (No chiral carbon) CH3CH2CH

H H2C

CH

C*

Geometrical isomers due to double bond *

H3C

C

3-Methyl-1-pentene (One chiral carbon)

Chiral carbon CH3

CH3

3. (3) There are 8 chiral carbon atoms present. H3C

CH3

CH3

CH3

CH3

2-Methyl-2-pentene (No chiral carbon)

CH3

2-Chloro-2-methylbutane

2. (3) The compound shows both optical and geometrical isomerism. Optical isomerism is shown by those compounds that contain at least one chiral carbon and geometrical isomerism shown by the compound containing double bond.

CH3

CH2CH3

H3C

HO

31. (2) There are five chiral carbons in the following molecules.

* H

H

* *

CH3

Cl

CH2

CH3

CH3

C

1. (1) There is no chiral center in the molecule, therefore, no stereoisomers are possible.

30. (4) The alkene that has a chiral center exhibits optical isomerism. CH2CH3

CH2

33. (4) In option (1) – Diastereomers Option (2) – mirror images, so enantiomers Option (3) – mirror images, so enantiomers Option (4) – rotating anyone compound by 180° in plane of paper produces second structure, so, both structure are identical.

CH3

C

CH3

CH3

Exercise 2

CH3

29. (3) The compound has one chiral center. Therefore, 2n = 21 = 2 optically active compounds are possible. Also, two geometrical isomers (cis and trans) are possible. So, total of four isomers are possible.

CH3CH2

* CH

2-Chlorobutane 2

27. (2) Optical isomerism is shown by 2,3-dichlorobutane. CH3

CH2

Cl n

CH3

CH3

Cl

*

H

* CH

9. (1) The compound has a chiral carbon atom.

OH *

CHO

(1)

HO

*

H

Chiral carbon (A carbon attached to four different substituents)

* H CH2OH

HO

Hence, the number of enantiomers are = 2n = 25 = 32. 32. (4) Among the given compounds, 2-chloro-2methylbutane does not contain any chiral center; hence is optically inactive.

(2) SH

Plane of symmetry (which divides the molecule into two equal halves)

Achiral

151

152

Chapter 2 | Isomerism H2N

(3)

16. (1) On 180° rotation, the configuration of both the molecules is identical, that is, 2S, 3S. Hence, the molecules are identical.

NH2

H

H Ph

Ph

(1)

Plane of symmetry Achiral

I

COOH

(4) H2N

(3) 3 4 (2)

H Two identical groups attached H

H

*

CH3

H3C

*

Cl

Cl

Cl

2-Chloropentane (II)

H 3C

H

H

CH3

HO

H

H

OH

Br

H

Br

H

H

H

OH

H

H

HO H3C

CH3 COOH

COOH

(I)

2S, 3S

R

Cl

Cl S

R

S Cl

Cl (1R, 2R )-dichlorocyclobutane

(1S, 2S )-dichlorocyclobutane

19. (3) There is two stereocenters, so disastereomers are not eliminated. The molecules have the same formula and connectivity so different molecules and constitutional isomers are eliminated. The two molecules are not mirror images of each other and not superposable; therefore, enantiomers are eliminated. The first molecule is (2R, 3S) and the second is (2R, 3R). So the molecules are disatereomers of each other. 23. (4) The following alkane will possess a stereogenic center.

(III)

* COOH H

CH3

COOH H 3C

H

H

OH

Br

H

Br

H

H

H

CH3 COOH

3-Methylhexane

H

HO HO

H

24. (3) 3 3

OH

H3C

H COOH

COOH H 4

(IV)

All these four molecules possess symmetry; so, all are meso compounds.

1 2S

4

1

4

H (II)

I

(1)

18. (3) Two of the dichlorocyclobutane isomers are optically active: (1R, 2R)-1,2-dichlorocyclobutane and (1S, 2S)-1,2-dichlorocyclobutane.

12. (4) Meso compounds are the inactive compound because the rotation by upper half of molecule is internally compensated by rotation of the lower half.

COOH

2 2 (4)

1,2-Dichloropentane (IV)

COOH

H

(4) (3) 3 4

H

Br

(1)

2S, 3S

10. (4) The following compounds contain stereocenters. H3C

(3) 1

(2) (3) 2

(4)

Achiral

Br

(4)

H

2 2

OH

2 2

Cl 3

CH3

1

4

1 3R 3

Achiral molecules

Alkene E (trans) or Z (cis) isomers have

have Meso compounds

include

have

Enantiomers

include

No optical activity

have the

Stereoisomers

are (except for alkene diastereomers)

Optical activity

is called

Rotation of plane-polarized light

Nonsuperposable mirror images

have

Chiral molecules

are

Resolution

is necessary for

A diastereomeric relationship

have

Stereoisomers that are not mirror images

exist among compounds with

Separation of enantiomers

is the

Tetrahedral atoms w/ four different groups

Alkenes capable of cis–trans isomerism

include

Chirality centers

Different physical properties

have

are

are

contain

include

Diastereomers

Same connectivity but different orientation of groups in space

cause

Identical physical properties (except optical rotation)

An internal plane of symmetry

Superposable mirror images

have

have

is a 50/50 mixture of

Nonsuperposable mirror images

A racemic mixture (racemate, racemic form)

Different connectivity

have

Constitutional isomers

can be subdivided into

Isomers

C E P T M A P ]MAP [ C O N CONCEPT

Purification and Characterization of Organic Compounds

3

C H A P T E R OU TLIN E 3.1 Methods of Purification of Organic Compounds 3.2 Qualitative Analysis of Organic Compounds

3.3 Quantitative Analysis 3.4 Determination of Empirical Formula of the Organic Compound

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Characterization of a substance almost always requires a pure sample of the material. This is particularly difficult to obtain in case of organic compounds because most organic reactions generate several products. The desired product can be obtained from the product mixture as the pure or major component by means of a number of separation techniques. This chapter describes the various separation techniques used in the laboratory in detail. The next challenge after separation of a pure component is its identification. There are millions of organic compounds recorded in chemical literature, so identifying one certain compound from this vast array appears like a formidable task. However, determination of the physical properties of a molecule, the functional groups present and the reactions the molecule undergoes has allowed the chemist to establish a systematic identification scheme for organic compounds. This will also be discussed in the chapter.

3.1

METHODS OF PURIFICATION OF ORGANIC COMPOUNDS

Organic compounds isolated from natural sources or synthesized in the laboratory are invariably impure because they are contaminated with other materials present in the source or in the reaction mixture. These are purified using some fundamental purification techniques such as sublimation, crystallization, distillation, steam distillation, differential extraction with solvents and chromatography. The techniques or the combination of techniques used are determined by the nature of the compound and the kind of impurities present in it. The purity of the compound is determined by measuring its melting or boiling points; refractive index is also used. A sharp melting point for a solid indicates that the compound is pure; whereas impure substances melt gradually over a range of temperature. Similarly, a pure liquid will have sharp and constant boiling point at any given pressure and will boil-off without leaving any residue.

3.1A Sublimation The phenomenon of a solid directly passing to the vapor state without first melting (passing to liquid state) is known as sublimation. This technique is used to separate volatile substances from non-volatile impurities. The substance to be purified is placed in a porcelain dish and heated gently with a small flame. The vapors of the pure substance are condensed on a cooled surface such as an inverted glass funnel or a flask placed in cold water. Iodine is generally purified by this method. Some organic compounds that can be heated and separated from the impurity by separating its vapor using inverted funnel method are benzoic acid, salicylic acid, camphor, anthracene, naphthalene, etc.

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3.1B Crystallization Solid organic compounds are purified by crystallization from a suitable solvent making use of difference in solubilities of the compound and impurities in the solvent. Choosing the correct solvent system is critical to a successful crystallization. Ideally the solvent should be such that the compound to be purified is very soluble in the hot solvent, but nearly insoluble in the cold solvent. Further, the impurities should either be very soluble in the solvent at all temperatures or not soluble at all at any temperature. The solvent should be low boiling so that its traces can be easily removed from the purified compound. The following are the essential steps in crystallization: 1. Select a suitable solvent. 2. Dissolve the material to be purified in the minimum amount of warm solvent. 3. Once the material is fully dissolved, filter the heated solution and bring to the point of saturation by evaporating a part of the solvent. 4. Cool the warm saturated solution to reduce the solubility of the solute which will generally cause the material to precipitate out. 5. Isolate the pure solid by filtration and remove the traces of solvent. 6. The cycle may be repeated for recrystallization to further purify the compound. The process of separation of different components of a mixture by repeated crystallization is called fractional crystallization. It is carried out separate two or more compounds which have different solubility in the solvent. The crystals of less soluble compounds separate out first on cooling the hot saturated solution and separated from the mother liquor. The process is repeated a number of times by again heating the solution and cooling till the component separate out completely. For example, pure benzoic acid can be obtained from a mixture of benzoic acid and anthracene by fractional crystallization because it is sparingly soluble in cold water and highly soluble in cold water.

3.1C Distillation Distillation is a process that is used to separate: 1. Substances that have different boiling points: These vaporize at different temperatures and are collected separately after condensation. 2. Volatile substances from non-volatile impurities: The volatile substance vaporizes on heating and is obtained after condensation while non-volatile impurities are left in the distillation flask. The impure compound or the mixture of components is taken in a round bottom flask fitted with a thermometer. The side-tube of the flask is connected to a water or air condenser (for liquids that boil above 130 °C), the other end of which is fitted into the mouth of a dry and clean receiving flask. The round bottom flask is heated in a water or sand bath or over an asbestos covered wire-gauze. The vapors of the low boiling component rise first, pass through the condenser and are collected in the receiver flask. The liquid with higher boiling point can be collected in a separate flask on heating to a higher temperature (Fig. 3.1). During the distillation process, the following points of caution should be observed: 1. The flask should not be more than one-half full. 2. The bulb of the thermometer should be near the side tube of the distillation flask and not dipping in the solution. 3. Add pieces of porous plate in the flask before heating to eliminate the risk of bumping. 4. Regulate the heating during distillation by removing the burner. 5. Distillation of volatile organic compounds such as ether should be carried out in heated water bath rather than on direct flame. Fractional Distillation The process of fractional distillation is used when the boiling points of the components to be separated differ by less than 30–40 °C and a fairly complete separation is desired. A fractionating

3.1 | Methods of Purification of Organic Compounds

Thermometer To sink Condenser Round bottom flask Water inlet

Liquid substance

Figure 3.1 Mixtures such as hexane–toluene, chloroform– aniline can be separated by simple distillation.

column fitted over the mouth of receiving flask is used to accomplish this separation and the vapors are passed through it for condensation. The vapors of higher boiling component condense before the vapors of lower boiling fraction. However, the vapors are richer in lower boiling component of the binary mixture, and they rise in the fractionating column and condense to form a liquid richer in lower boiling component. This will however, not be pure and only somewhat enriched if the difference in boiling point of two components is small. If the condensate is vaporized a second time, it will be further enriched in the lower boiling component. The trick for separating such binary mixtures is to repeat the cycle a number of times. Each cycle is one theoretical plate. A number of column designs are available which achieve different number of theoretically plates (Fig. 3.2).

Thermometer To sink Condenser Adaptor Fractionating column with packing Water inlet

Liquid to be distilled

Oil bath

Distilled liquid

Figure 3.2 Schematic representation of fractional distillation technique.

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Most of the distillation columns are designed so that the fractionation efficiency is achieved by having a large surface area in contact with the vapor phase. This can be accomplished by packing the column with wire gauze or glass beads (Fig. 3.3). These provide surfaces for exchange of heat between ascending vapors and descending condensed liquid. This technique is used in the petroleum industry for fractional distillation of crude oil. Distillation Under Reduced Pressure Liquids which boil at very high temperatures or which decompose below their boiling points or undergo chemical changes on heating are separated by carrying out distillation at reduced pressure. The pressure of the mixture in the round bottom flask is reduced using a water or vacuum pump to make it fall below the atmospheric pressure. The reduction in pressure makes the liquids boil below their normal boiling points. Distillation under reduced pressure may be carried out with or without heating and is also known as low temperature distillation. The technique is used for separation of glycerol from spent lye in the soap industry. It is used for removal of solvent from materials that are sensitive to heat (Fig. 3.4).

Glass beads

(a)

(b)

Figure 3.3 (a) Simple packed column; (b) Bubble plate column.

Stopcock Manometer

Capillary to regulate the air

To sink

Vacuum pump

Water inlet

Distilled liquid

Figure 3.4 Schematic representation for distillation under reduced pressure. Steam Distillation This technique is used for purifying substances that are steam volatile and immiscible with water. The steam is generated in a boiling flask placed and heated over asbestos covered wire gauze. The round bottom distillation flask is fitted with an inlet tube for steam and another U-shaped tube which serves as a vapor outlet. The inlet tube is connected to the steam generator and the vapor outlet tube to a water condenser. The flask is titled at an angle of about 45° to check the splashing of the contents in the flask into the condenser. The steam passes through the compound to be distilled and the mixture of organic compound vapors and steam condense and are collected in the receiver flask. The liquid in the distillation flask boils when the sum of the vapor pressures of the organic compound (p1) and stem (p2) become equal to the atmospheric pressure (p). Thus, the organic compound would boil at a lower temperature than its boiling point. Being immiscible with water, the organic compound can be separated using a separating funnel. The technique is used in separation of mixture such as aniline and water (Fig. 3.5).

3.1 | Methods of Purification of Organic Compounds

Safety tube To sink

Condensed water vapors Water Compound to be Water distilled inlet

Water Distilled liquid

Figure 3.5 Schematic representation for steam distillation technique.

3.1D Differential Extraction Solvent extraction is frequently used in the laboratory to separate out the desired compound out of the mixture or from impurities depending upon their differential solubility in various solvents. The solubility characteristics of a given compound govern its distribution between two phases of immiscible solvents (in which the material has been dissolved) when these phases are mixed thoroughly. Thus, an organic compound can be separated from an aqueous mixture by shaking with an organic solvent in a separating funnel. The mixture forms two distinct layers which are separated and the organic compound is obtained by evaporation of solvent (Fig. 3.6). Multiple or continuous extractions are carried out to separate out

Organic solvent Organic compound in aqueous layer

Organic compound in solvent layer Aqueous layer

Figure 3.6 Formation of two layers by differential extraction. compounds that are not very soluble in the organic solvent. The same solvent is repeatedly used for extraction and it becomes enriched in the organic compound with each extraction. In liquid–liquid extraction, the desired properties in chosen solvent for extraction are: 1. It must be immiscible with the solution solvent. 2. It should readily separate from the desired component after extraction by evaporation, that is, it should be low boiling. 3. It should not react chemically with any of the component in the aqueous mixture being extracted.

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3.1E Chromatography Chromatography is the most powerful of techniques for separating mixtures and isolating pure organic substances either in solid or liquid state. Chromatography may be defined as a method of separating a mixture of components into individual components by equilibrium distribution between the two phases, one stationary and the other mobile. The various methods of chromatography (also see Table 3.1) are categorized on the basis of the phases involved as: 1. Solid–liquid chromatography. 2. Liquid–liquid chromatography (paper). 3. Vapor phase chromatography (gas–liquid or simply gas chromatography). Table 3.1 Different types of chromatographic techniques Stationary Phase

Moving Phase

Type

Alumina

Liquid solvents

Column chromatography

Silica gel

Liquid solvent

Thin layer chromatography

Magnesium oxide

Liquid

Column chromatography

Paper

Liquid solvent

Paper chromatography

Alumina

Gas

Gas chromatography

The word chromatography is derived from the Greek word chromatos. Tswett discovered this technique in 1903 while studying ways to separate mixtures of natural plant pigments. The zones corresponding to different components of the mixture were detected by simply observing the visual adsorption bands, and hence the name. The principle on which the separation of components is achieved is differential solubility of components of the mixture in the two phases involved. The process involves adsorption of the mixture on a stationary phase (solid or liquid) and elution of the components using mobile phases (liquid or gas). On the basis of the principle involved in separation, chromatography can be classified as adsorption chromatography and partition chromatography. Adsorption Chromatography It is based on the differences in the rate at which the component of the mixture moves through a stationary phase under the influence of moving phase. It involves the following steps: 1. 2. 3. 4.

Adsorption or retention of a substance on the stationary phase. Separation of adsorbed substances by the moving phase. Recovery of the separated substances by a continuous flow of the mobile phase (elution). Qualitative and quantitative analysis of the eluted substances.

The two main chromatographic techniques based on differential adsorption on a solid phase are: Column chromatography and thin layer chromatography. 1. Column chromatography: This technique is used extensively in inorganic chemistry for separation of mixtures obtained from natural extracts or purification of compounds synthesized in the laboratory. This method involves use of a glass or plastic column that is filled with particulate material such as silica or alumina. The sample to be analyzed is dissolved in minimum amount of solvent and loaded on the column. The separation is carried out by passing the solvent through the column at atmospheric pressure. The solvent elutes different components present in the sample based on adsorption or partition. A molecule (component) that is strongly adsorbed on the stationary phase moves slowly down the column, whereas the molecule that is weakly adsorbed will move at a faster rate. Thus, a complex mixture can be resolved into separate bands of materials which are eluted with solvents of different polarity (Fig. 3.7). The polarity of the sample is increased stepwise and the eluant is collected as small fractions, each of which is concentrated and checked by TLC for purity. The apparatus used is simple but the technique involves use of large volumes of solvent.

3.1 | Methods of Purification of Organic Compounds

1. Mixture to be separated is dissolved in the mobile phase

2. Mobile phase is added throughout the process

3. Components separate Stationary phase

Chromatography column

4. Each component is collected as it reaches the bottom of the column

Figure 3.7 Steps involved in column chromatography. 2. Thin layer chromatography (TLC): This technique is particularly useful in rapid analysis of purity of samples and is used as a preparative technique for obtaining small amounts of materials with high purity. The method involves the use of a thin layer (2–20 mg) of particulate adsorbent such as silica or alumina on an inert sheet of glass or plastic, known as TLC plates. The steps involved are: (a) The sample to be analyzed/separated is dissolved in minimum amount of solvent and applied on the TLC plate with the tip of a thin capillary tube; about 2 cm above the end of the plate. (b) The plate is dipped in a covered glass jar containing the solvent (mobile phase). The solvent is allowed to travel up the plate with the sample spotted on the sorbent just above the solvent. When the solvent front nears the top of the plate, it is removed from the jar and the position of solvent is marked on the plate. (c) The separation of mixture into different spots occurs by the same mechanism as in column chromatography. Different components of the mixture travel up to different distances with the eluant depending upon the extent of adsorption on the adsorbent (Fig. 3.8). Note the Solvent tank

TLC plate

Figure 3.8 Different components travelling different distances.

Solvent

Time zero

Solvent

After ten minutes

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difference that in column chromatography the solvent (eluant) descends while it ascends in case of TLC. (d) The TLC properties of compounds are reported in terms of retardation factor or as Rf values. The Rf value is the distance traveled by the substance divided by the distance traveled by the solvent front (Fig. 3.9). Rf =

Distance moved by the substance from base line (x ) Distance moved by the solvent from the base line ( y )

(e) The colored spots are visible on the TLC plate as such, while colorless spots are developed by placing the plate in iodine-vapor chamber. Iodine forms complex with most of the organic compounds that appear as brown spots. The sample spots can also be detected with handheld UV lamps. The TLC plates are prepared with UV-activated fluorescent indicator mixed in the silica. The samples quench the fluorescence induced by the UV lamp and appear as dark spots. The presence of some compounds on the TLC plates is also indicated by spraying with suitable reagents. This chromatographic technique offers the following advantages:

Solvent front

Spot y x Base line

Figure 3.9 Rf value calculated from x and y components.

1. A wide range of compounds can be detected using different indicators and reactive agents such as iodine vapor, sulphuric acid, etc. 2. When non-destructive indicators like fluorescent/UV lamps are used, the separated components can be scrapped off the plate and studied by other analytical techniques. 3. The plates used are cheap, reusable and easily disposable. 4. Separation of larger amounts of samples can be carried out by using preparative TLC plates. Partition Chromatography In partition chromatography, continuous differential partitioning of the components takes place between mobile and stationary phases. This was the first analytical chromatographic technique developed using paper. It consists of a solvent moving along filter or blotting paper. The interaction between the components of the sample, the solvent and the paper, results in separation of the components. In the modern form, chromatography is used in partition chromatography. It is carried out using a special quality of paper called chromatography paper. Here the cellulose (of the paper) serves as an inert support, and the water adsorbed from air onto the hydroxyl groups of the cellulose becomes the stationary phase. The steps involved are as follows: 1. The mixture to be separated is dissolved in minimum amount of suitable solvent and applied as spots near the base of the chromatography sheet, which is then suspended in a solvent or mixture of solvents. The paper is sometimes positioned to hang down in a container holding the paper and the elution solvent. 2. The solvent acts as the mobile phase and rises up the paper and flows over the spot. The paper selectively retains different components in accordance with their partition coefficient in the two liquid phases. The paper developed after chromatography is called chromatogram. 3. The spots corresponding to the different components are visible at different heights either under visible or UV light or by use of suitable spray reagents (Fig. 3.10).

3.2 | Qualitative Analysis of Organic Compounds

4. Paper chromatography is widely used for separation of colored compounds, highly polar compounds and polyfunctional species such as sugars and amino acids. It can work on very small sample sizes but the time taken for the paper chromatograms to elute (3–4 h) is a disadvantage.

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Chromatography paper Spot Base line

3 Applications of Chromatography Different types of 4 Solvent 1 chromatography techniques find diverse applications in the following: Figure 3.10 Appearance of spots in partition chromatography. 1. Chemical industry: Column chromatography is used for separation of the required component from the mixture of compounds obtained after synthesis. Thin layer chromatography is useful for monitoring large-scale column chromatography. Analyses of the fractions can guide decisions on the solvent– elution sequence. 2. Pharmaceutical industry: Chromatography is used to prepare large amounts of pure component or for purification of the component from trace impurities associated with it. It is used for separation of chiral compounds to obtain the pharmaceutically active optical isomer. 3. Food industry: Chromatographic techniques are used for quality control in food industry. It can be used to separate additives, flavors, etc. and to determine their presence and use in specified amounts. Chromatography can also be used to detect presence of contaminants and other harmful growths such as that of molds and bacteria. 4. Environment-testing laboratories: The presence and quantity of pollutants in the air and drinking water is determined using chromatographic techniques. 5. Diagnostic techniques: The presence of certain drugs and the marker compounds for medical diagnosis in blood and urine are determined using chromatographic techniques.

3.2

QUALITATIVE ANALYSIS OF ORGANIC COMPOUNDS

The objective of qualitative analysis of organic compounds is to identify the compounds by conducting a number of screening tests and placing them into one of the specific classes. The systematic approach to analysis for identification of a compound involves: 1. Preliminary tests to determine the physical nature of the compound. 2. Determination of solubility characteristics. 3. Chemical tests to help identify presence of elements carbon, hydrogen, oxygen, nitrogen, phosphorus, sulphur and halogens. 4. Classification tests to identify the nature of functional group present. 5. Instrumental analysis to support the findings of chemical tests. We will now learn about some chemical tests carried out for qualitative analysis.

3.2A Preliminary Tests These tests, conducted with little expenditure of time and sample material, can offer valuable clues as to which class a given compound belongs. These are further classified as nonchemical for physical for determination of physical state and ignition tests. Physical Tests If the compound is a solid, the melting point is determined and a narrow range (1-2 °C) is a good indication that the material is pure. If a broad range is observed, the compound may be recrystallized

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from a suitable solvent. Boiling point is similarly determined for a liquid sample and a narrow range indicates that the material is pure. Refractive index can also provide valuable information for identification purposes. Since most of the organic compounds are colorless, the presence of color may provide some clue about the nature of the compound. It generally indicates the presence of a chromophore (extended conjugation) in the compound. However, presence of slight amount of impurity may impart color to a compound which is colorless in its pure form. Detection of a compound’s odor can also sometime be of assistance since the vast majority of organic compounds are odorless. However, some common organic compounds or classes have characteristic odors which when detected can be used as indication for their presence. For example, esters have fruity smell, aliphatic amines have fishy smell, benzaldehyde has an almond odor and most of the common solvents also have characteristic odors. Ignition Tests This test is carried out by placing 1–2 mg of the sample on a spatula, followed by heating and ignition in a micro-burner flame. As the sample is heated, the following observations should be made: 1. Any melting or sublimation taking place. 2. Color of the flame. 3. Nature of combustion – rapid, almost instantaneous (indicates high hydrogen content) or explosive (indicates presence of nitrogen or nitrogen-oxygen containing groups, such as nitro). 4. Nature of residue if present after ignition: ● Black residue that disappears on further heating is that of carbon. ● A residue that swells during heating indicates the presence of a carbohydrate. Some important observations related nature and color of flame in ignition tests are summarized in Table 3.2. Table 3.2 Ignition test observations Type of Compound

Observation

Example

Lower aliphatic compounds

Yellow, almost non-smoky flame

Hexane

Higher aliphatic, unsaturated or aromatic compounds

Yellow sooty flame

Toluene

Oxygen containing organic compounds

Clear bluish flame

Ethanol

Polyhalogen compounds

Generally do not ignite until burner flame is applied directly to the substance

Chloroform

3.2B Detection of Carbon and Hydrogen All organic compounds generally contain carbon and hydrogen. Their presence in the compound is detected by heating it with copper (II) oxide. Carbon in the compound is oxidized to carbon dioxide and its formation is detected by passing the gas through limewater which turns milky due to precipitation of calcium carbonate.  → 2Cu + CO C + 2CuO  2

CO2 + Ca(OH)2 → CaCO3 + H2O

Hydrogen present in the compound is oxidized to water and its presence is indicated when it turns anhydrous copper sulphate blue.  → 2Cu + H O 2H + CuO  2

5H2O + CuSO4 → CuSO4 ⋅ 5H2O White

Blue

3.2C Detection of Other Elements Other than carbon, hydrogen and oxygen, the elements that are most often present in organic compounds are nitrogen, sulphur, phosphorus and halogens. To detect the presence of these elements, the

3.2 | Qualitative Analysis of Organic Compounds

organic compound is generally fused with metallic sodium. This reaction converts these heteroatoms to water-soluble inorganic compounds and their presence can be determined by inorganic qualitative analysis tests.  → NaCN Na + C + N   → Na S 2Na + S  2

 → NaX Na + X 

(where X = Cl, Br or I)

A small piece of sodium metals is heated and mixed with the organic compound to be analyzed. The test tube is heated till red hot and then broken in a beaker containing distilled water. The extract so obtained is called sodium fusion extract and the tests carried out using this is known as Lassaigne’s test. Tests for Nitrogen The tests used and steps involved are listed as follows. 1. The sodium fusion extract is boiled with iron (II) sulphate and acidified with sulphuric acid. Appearance of blue coloration confirms the presence of nitrogen. Sodium cyanide formed in Lassaigne’s test first reacts with FeSO4 to form sodium hexacyanoferrate(II). 6CN− + Fe2 + → Na 4 [Fe(CN)6 ]

On addition of sulphuric acid, some part of Fe2+ is oxidized to Fe3+ which reacts with sodium hexacyanoferrate(II) to form iron(III) hexacyanoferrate(II) which is Prussian blue in color. HO 3Na 4 [Fe(CN)6 ] + 4Fe3 +  → Fe4 [Fe(CN)6 ]3 ⋅ x H2O + 12Nacl 2

Prussian blue

2. In the presence of halogens and sulphur, nitrogen is detected by another method. A mixture of p-nitrobenzaldehyde (1.5%) and o-dinitrobenzene (1.7%) in 2-methoxyethanol and NaOH in water (25%) is added to the sodium fusion extract. Formation of deep purple color confirms the presence of CN- in the extract, and hence the presence of nitrogen in the compound. In absence of CN-, yellow or tan coloration is obtained. 3. Soda lime test: If on heating a small amount of organic compound in a dry test tube with soda lime (NaOH + CaO) evolves ammonia (identified by smell or turning of blue litmus paper red), it indicates the presence of nitrogen in the compound. Test for Sulphur The tests used and the steps involved are listed as follows. 1. Acidify sodium fusion extract with a few drops of acetic acid and add 1% lead acetate solution. The formation of a black precipitate indicates the presence of sulphur in the compound. Na 2S + Pb(CH3COO)2 → PbS + 2CH3COONa Black ppt.

2. Add a few drops of water to the sodium fusion extract and then add dilute (2%) aqueous sodium nitroprusside solution. The formation of a deep violet color is a positive test for sulphur. Na 2S + Na 2Fe(CN)5NO → Na 4 [Fe(CN)5NOS] Blue − violet complex

3. If both nitrogen and sulphur are present, sodium thiocyanate is formed in the sodium fusion extract. This on reaction with ferric ions gives blood-red coloration instead of Prussian blue coloration formed with cyanide ions. Fe3 + + SCN− → [Fe(SCN)(H2O)5 ]2 + Blood red

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4. If the fusion extract is prepared with an excess of sodium, the thiocyanate decomposes to form sodium cyanide and sodium sulphide which give their usual tests. NaSCN + 2Na → NaCN + Na 2S

Tests for Halogens The tests used and the steps involved are listed as follows. 1. The sodium fusion extract is acidified with nitric acid and then treated with aqueous silver nitrate solution (0.1 M). The formation of a heavy curdy precipitate is a positive test for the presence of halogens, and appearance of faint turbidity is a negative test for halogens. The halogen present is further identified from the color of the precipitate and its solubility in dilute ammonium hydroxide. NaX + AgNO3 →

AgX Thick precipitate

+ NaNO3

where X = Cl, Br or I. AgCl forms white precipitate soluble in ammonium hydroxide. AgBr forms yellowish precipitate sparingly soluble in ammonium hydroxide. AgI forms yellow precipitate insoluble in ammonium hydroxide. Presence of fluorine is not detected by this test as AgF is soluble in water. Once the presence of halogens is confirmed, their identity can be further established by layer test. Acidify the sodium fusion extract with nitric acid and add 1% aqueous KMnO4 solution and shake well. Next add oxalic acid (10–15 mg) to decolorize the excess purple permanganate, followed by methylene chloride. Shake well and allow the layers to separate. Observe the color of the methylene chloride layer to identify the halogen: (a) Colorless methylene chloride layer indicates presence of chlorine. (b) Brown color of the methylene chloride layer indicates presence of bromine. (c) Violet color of the methylene chloride layer indicates the presence of iodine. Note: In the presence of nitrogen or sulphur, the sodium fusion extract is boiled with concentrated nitric acid first to decompose sodium cyanide or sulphide during Lassaigne’s test which would otherwise interfere with silver nitrate test. 2. Beilstein’s test: In this test organic compounds that contain chlorine, bromine or iodine and hydrogen decompose on ignition in presence of copper oxide to yield the corresponding hydrogen halides. These hydrogen halides react to form volatile cupric halides that impart green or blue–green color to nonluminous flame. It is a sensitive test but some nitrogen-containing compounds and some carboxylic acids also give positive results. Test for Phosphorus The compound is heated with sodium peroxide which oxidizes the phosphorus present in the compound to phosphate. Then the reagent (NH4)2MoO4 (ammonium molybdate) is added in the presence of strongly acidic solution of HNO3 which reacts with HPO24 ion to produce a canary yellow precipitate of ammonium phosphomolybdate. Na 3PO4 + 3HNO 3 → H3PO4 + 3NaNO3 H3PO4 + 12(NH4 )2MoO4 + 21HNO3 → (NH4 )3PO4 ⋅ 12MoO3 + 21NH4NO3 + 12H2O Canary yellow ppt.

Note: Presence of arsenic in the organic compound also gives similar reaction under boiling conditions. To distinguish between phosphorus and arsenic, the filtrate is heated with silver nitrate. Appearance of a brown precipitate confirms the presence of arsenic.

3.3

QUANTITATIVE ANALYSIS

Once the identity of elements present in the organic compound is established by qualitative analysis, their amounts present are determined by quantitative analysis. The laboratory methods for estimation of

3.3 | Quantitative Analysis

carbon, hydrogen, nitrogen, phosphorus, sulphur, oxygen and halogens are described in the following sections. Though these methods continue to be used in laboratories, they are increasingly being replaced by instrumental techniques, such as an elemental analyzer for CHN and other automatic experimental techniques. These techniques require micro-quantities of samples and require a very short time for the analysis.

3.3A Carbon and Hydrogen The amount of carbon and hydrogen present in the organic compound is determined by heating a known amount of sample in a combustion tube (also known as Liebig’s combustion tube) in presence of excess oxygen and copper(II) oxide. Carbon is oxidized to carbon dioxide and hydrogen to water. The general reaction is y  Cx Hy + (x + y )O2 → x CO2 +   H2O 2

The amounts formed are determined by passing these through previously weighed U-tubes containing concentrated solution of potassium hydroxide and anhydrous calcium chloride, respectively, and determining the increase in their mass. Note: If nitrogen, sulphur and halogens are also present in the organic compound, the apparatus is modified to include a copper spiral for decomposing oxides of nitrogen formed. Silver gauze is added to retain sulphur and halogens to prevent their absorption in the potassium hydroxide tube. Calculation From the mass of H2O and CO2 obtained, we can calculate the percentage of C and H in the organic compound as follows. Let the mass of organic compound taken be m g, and the amount of CO2 and H2O formed be m1g and m2g, respectively. The molar masses of CO2 and H2O are 44 and 18, respectively. The amount of C in 44 g of CO2 = 12 g; therefore, the amount in m1g of CO2 = Percentage of carbon =

12 × m1. Thus, 144

12 × m1 × 100 % 144 × m

Similarly, the amount of H in 18 g of H2O = 2 g; therefore, the amount in m2g of H 2 O = Percentage of hydrogen =

2 × m2. Thus, 18

2 × m 2 × 100 % 18 × m

3.3B Nitrogen The nitrogen present in an organic compound can be estimated by Dumas’ method or Kjeldahl’s method. The first method involves conversion to nitrogen (N2) gas and the second method involves its conversion to ammonia followed by titration against a standard acid. Dumas’ Method In this method a known amount of organic compound containing nitrogen is placed in a combustion tube with copper oxide and heated in an atmosphere of carbon dioxide. Carbon and hydrogen are oxidized to carbon dioxide and water, respectively, and free nitrogen is obtained. The general reaction is  2x + y  y  z   2x + y  Cx Hy Nz +   CuO → x CO2 +  2  H2O +  2  N2 +  2  Cu 2        

Oxides of nitrogen if formed are converted back to free nitrogen by passing over a copper gauze. The mixture of gases liberated is collected over aqueous solution of potassium hydroxide. The carbon dioxide formed is absorbed in the solution and nitrogen is collected in the upper part of the graduated tube and its volume measured (Fig. 3.11).

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CuO1 organic compound

Coarse CuO

CuO gauze

CO2

Reduced copper gauze

Nitrogen

KOH solution

Furnace

Mercury seal

Nitrometer

Figure 3.11 Schematic representation of Dumas method. Calculation: From the volume of N2 collected, we can estimate the percentage of nitrogen in the organic compound as follows: Let the mass of organic compound taken = m g, volume of N2 collected = V1 mL, room temperature = T1 K. Then the volume of gas collected at STP V =

pV 1 1 × 273 760 × T1

where p1 is the pressure of nitrogen which is different from the atmospheric pressure at which the gas is collected. It can be obtained from the relation Corrected pressure(p1) = Atmospheric pressure – Vapor pressure of water at T (Aqueous tension) Using the relation that 22400 mL of N2 at STP weighs 28 g, we can get the weight of V mL of N2 gas as =

28 × V g. Thus, 22400 Percentage of nitrogen =

28 × V × 100 % 22400 × m

Kjeldahl’s Method In this method, the organic compound containing nitrogen is heated with concentrated sulphuric acid in presence of CuSO4 which converts the nitrogen in the compound to ammonium sulphate. Organic compound + H2SO4 → (NH4 )2SO4

The solution containing ammonium sulphate is heated with excess sodium hydroxide to liberate ammonia which is absorbed in an excess of standard solution of sulphuric acid (Fig. 3.12). (NH4 )2SO4 + 2NaOH → Na 2SO4 + 2NH3 + 2H2O 2NH3 + H2SO4 → (NH4 )2SO4

Calculation: The amount of nitrogen in the compound is estimated by measuring the amount of sulphuric acid consumed in the reaction. The amount of sulphuric acid left after the absorption of ammonia is estimated by titrating it with standard alkali solution.

3.3 | Quantitative Analysis

Kjeldahl’s trap Water out Organic compound 1 Conc.H2SO4 1 CuSO4 Kjeldahl’s flask

Water inlet Contents of Kjeldahl’s flask after digestion 1NaOH Known volume of standard acid

Figure 3.12 Schematic representation of Kjeldahl’s method. Let the mass of organic compound taken = m g, volume of H2SO4 of molarity M taken = V mL and volume of NaOH of molarity M used for titration of excess acid = V1 mL. We know that for solution of same molarity, V1 mL of NaOH = V1/2 mL H2SO4. Therefore, the volume of H2SO4 left unconsumed after absorption of ammonia = (V - V1/2) mL. Again if molarity of NH3 solution is M, then (V - V1/2) mL of H2SO4 = 2(V - V1/2) mL of NH3. Using the fact that 1000 mL of 1 M solution of ammonia will contain 17 g of ammonia or 14 g of nitrogen, we can calculate the weight of nitrogen in 2(V - V1/2) mL of ammonia solution of molarity M as 14 × M × 2(V − V1 / 2) g 1000

Therefore, Percentage of nitrogen =

14 × M × 2(V − V1 / 2) 100 % × 1000 m

Note: Kjeldahl method cannot be applied for estimating nitrogen in organic compounds containing nitro and azo groups or heterocyclic compounds containing nitrogen, as the nitrogen present in them does not get converted to ammonium sulphate under the test conditions.

3.3C Halogens The estimation of halogens in organic compounds is done using Carius method. A known amount of organic compound is heated in a furnace with fuming nitric acid in the presence of silver nitrate in a hard glass tube known as Carius tube. The halogen present gets converted into silver halide which is filtered, washed, dried and weighed. The carbon and hydrogen present in the compound are oxidized to carbon dioxide and water, respectively (Fig. 3.13). Calculation: The amount of halogen present in the organic compound can be calculated from the amount of silver halide formed as follows.

Organic substance

Carius tube

Iron tube

HNO3 1 AgNO3

Furnace

Figure 3.13 Carius tube.

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Let the mass of organic compound taken = m g, mass of silver halide formed = m1 g. Now, molecular mass of AgX contains 1 atomic mass of X. Therefore, Mass of halogen (X) in m1 g of AgX =

Atomic mass of X × m1 g Molecular mass of AgX

Percentage of halogen =

Atomic mass of X × m1 × 100 % Molecular mass of AgX × m

3.3D Sulphur Sulphur is estimated using a slight modification in Carius method. A known mass of the organic compound is heated in a Carius tube with fuming nitric acid or sodium peroxide. The sulphur in the organic compound is oxidized to sulphuric acid, which is precipitated as barium sulphate adding aqueous solution of barium chloride in excess. The precipitate of barium sulphate is filtered, washed, dried and weighed and used for estimation of sulphur. Calculation The amount of sulphur present in the organic compound can be estimated from the amount of barium sulphate formed as follows. Let the mass of organic compound taken = m g Mass of barium sulphate formed = m1 g. Since 1 mol or 233 g of BaSO4 contains 32 g of sulphur, therefore, the amount of sulphur in m1 g of BaSO4 =

32 × m1 g. Hence, 233

Percentage of sulphur =

32 × m1 × 100 % 233 × m

3.3E Phosphorus The amount of phosphorus in an organic compound is also determined using Carius method. A known amount of organic compound is heated with fuming nitric acid in Carius tube and the phosphorus present in the compound gets converted to phosphoric acid. The phosphoric acid formed can be precipitated using two different reagents: 1. It can be treated with magnesia mixture (MgCl2, NH4Cl and NH3) which precipitates out magnesium ammonium phosphate. The precipitate is ignited to give magnesium pyrophosphate (Mg2P2O7) which is weighed and used for estimation of the amount of phosphorus in the organic compound. 2. It can be precipitated as ammonium phosphomolybdate, (NH3)2PO4⋅12MoO3, using ammonia and ammonium molybdate. The precipitate is filtered, washed, dried and weighed and then used for estimation of phosphorus in the compound. Calculations The amount of phosphorus present can be calculated for the two methods as follows. For the first method: Let the mass of organic compound taken = m g and mass of magnesium pyrophosphate formed = m1 g. Since 1 mol or 222 g of Mg2P2O7 contains 62 g of phosphorus, therefore, Amount of phosphorus in m1 g of Mg2P2O7 =

62 × m1 g . Thus, 222

Percentage of phosphorus =

62 × m1 × 100 % 222 × m

For the second method: Let the mass of organic compound taken = m g, mass of ammonium phosphomolybdate formed = m1 g. Since 1 mol or 1877 g of (NH3)2PO4⋅12MoO3 contains 31 g of phosphorus, therefore,

3.4 | Determination of Empirical Formula of the Organic Compound

Amount of phosphorus in m1 g of (NH3)2PO4⋅12MoO3 =

31× m1 g. Thus, 1877

Percentage of phosphorus =

31× m1 × 100 % 1877 × m

3.3F Oxygen The estimation of oxygen in a compound is done indirectly. It is usually found by adding the percentages of all other elements determined and subtracting the sum from total percentage composition (i.e., 100). If the sum of percentages of all the elements in the compound becomes hundred, it implies that the compound does not contain oxygen. If the sum is less than hundred, then the percentage of oxygen in the compound = 100 – Sum of percentages of all other elements present. The amount of oxygen in the compound can also be determined directly by decomposing a known amount of organic compound by heating it in a stream of nitrogen gas. The gaseous products obtained are passed over red-hot coke which converts the free oxygen liberated to carbon monoxide. The mixture of gases is then passed through warm iodine pentoxide (I2O5) which leads to formation of carbon dioxide and iodine. Decomposition Organic compound  → O2 + other gases ∆

High temperature 2C + O2  → 2CO

I2O5 + 5CO → I2 + 5CO2

(3.1) (3.2)

Calculation The amount of oxygen present in the organic compound can be estimated from the amount of carbon dioxide formed as follows. Let the mass of organic compound taken = m g, mass of carbon dioxide formed = m1 g. From Eqs. (3.1) and (3.2), we can see that each mole of oxygen liberated will produce two moles of carbon dioxide or 88 g of CO2 is obtained from 32 g. Therefore, the amount of m1 g of CO2 is obtained from 32 × m1 g of O2 88 Percentage of oxygen =

32 × m1 × 100 % 88 × m

The amount of oxygen in the compound can also be estimated from the amount of iodine formed as follows. Let the mass of organic compound taken = m g, mass of iodine formed = m1 g. From Eqs. (3.1) and (3.2), we can see that each mole of oxygen liberated will produce 2/5 mol of iodine or (2 × 127)/5 g of I2 is obtained from 32 g. Therefore, amount of m1 g of I2 is obtained from 5 × 32 × m1 g of O2 2 × 127 Percentage of O2 ==

3.4

5 × 32 × m1 % 2 × 127 × m

DETERMINATION OF EMPIRICAL FORMULA OF THE ORGANIC COMPOUND

The empirical formula of a compound represents the simplest mole ratio in which the various elements are present in the organic compound. Molecular formula shows the actual number of each atom present

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in the molecule whereas an empirical formula shows the ratio of the elements in the molecule. Once the percentage composition of all the elements in the compound is obtained by elemental analysis, the empirical formula of the compound can be obtained by the following steps: 1. Divide the percentages of all the elements by their respective atomic masses. 2. Divide the numbers thus obtained by the lowest number to obtain quotients that will give the simplest ratio in which the elements exist in the compound. 3. If the quotients are not whole number, then multiply with a suitable common factor.

SOLVED EXAMPLES 1. Liebig’s test is used to estimate (1) H (3) Both (1) and (2) (2) C (4) N

(3) In Liebig’s test, the amount of carbon and hydrogen present in the organic compound is determined by heating a known amount of sample in a combustion tube (also known as Liebig’s combustion tube) in the presence of excess oxygen and copper(II) oxide. Carbon is oxidized to carbon dioxide and hydrogen to water. 2. In Kjeldahl’s method, CuSO4 acts as (1) oxidizing agent. (2) reducing agent. (3) hydrolyzing agent. (4) catalytic agent. Solution (4) In Kjeldahl’s method, the organic compound containing nitrogen is heated with concentrated sulphuric acid in presence of CuSO4, which converts the nitrogen in the compound to ammonium sulphate. 3. Match the organic compounds in Column-I with the Lassaigne’s test results in Column-II appropriately:

(a) Aniline

Column-II (p) Red color with FeCl3

(b) Benzene (q) Violet color with sodium sulphonic acid nitroprusside (c) Thiourea

(1) (2) (3) (4)

(4) Lassaigne’s test results (a) Aniline

Solution

Column-I

Solution

(r) Blue color with hot and acidic solution of FeSO4

(a) – (q); (b) – (p); (c) – (r) (a) – (r); (b) – (q); (c) – (p) (a) – (q); (b) – (r); (c) – (p) (a) – (r); (b) – (p); (c) – (q)

(r) Blue color with hot and acidic solution of FeSO4

(b) Benzene (p) Red color with FeCl3 sulphonic acid (c) Thiourea

(q) Violet color with sodium nitroprusside

4. The suitable technique of separation of the components from a mixture of calcium sulphate and camphor is (1) sublimation. (2) distillation. (3) crystallization. (4) chromatography. Solution (1) Sublimation. Camphor sublimes on heating and can be collected as the sublimate. 5. The presence of halogen, in an organic compound, is detected by (1) iodoform test. (3) Beilstein’s test. (2) silver nitrate test. (4) Millon’s test. Solution (3) In Beilstein test, a (clean) copper wire is heated in the non-luminous flame of the Bunsen burner until it ceases to impart any green or bluish green color to the flame. The heated end of the wire is dipped into the organic compound and is again introduced into the Bunsen flame. The appearance of a bluish green or green flame due to the formation of volatile cupric halides shows the presence of halogens in the given organic compound. 6. In Kjeldahl’s method, 29.5 mg of an organic compound containing nitrogen was digested and the evolved ammonia was absorbed in 20 ml of 0.1 M

Solved Examples HCl solution. The excess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is (1) 29.5 (3) 47.4 (2) 59.0 (4) 23.7 Solution (4) Weight of substance taken = 29.5 × 10-3 g Volume of acid taken = 20 mL of 0.1 N HCl Now, 15 mL of 0.1 N NaOH = 15 mL of 0.1 N HCl Therefore, volume of acid used = 20 - 15 = 5 mL of 0.1 N HCl % Nitrogen = =

1.4 × Volume × Normality 00 × 10 Weight of substance taken

Solution (4) Percentage of Br Atomic mass of Br × Weight of AgBr × 100 = Molecular mass of AgBr × Weight of organic substance 80 × 141× 100 = = 24% 188 × 250

8. 1.4 g of an organic compound was digested according to Kjeldahl’s method and the ammonia evolved was absorbed in 60 mL of M/10 H2SO4 solution. The excess sulphuric acid required 20 mL of M/10 NaOH solution for neutralization. The percentage of nitrogen in the compound is (1) 3 (3) 10 (2) 5 (4) 24 Solution (3)

=

Solution The percentage of carbon is given by 12 Mass of carbon dioxide formed (m1) × × 100% Mass of compound taken (m ) 144

Given that m = 0.210 g and amount of carbon dioxide formed (m1) = 0.307 g. Substituting the values in the equation, we get Percentage of carbon =

1.4 × 5 × 10 −3 L × 1 gL−1 × 100 = 23.73% 29.5 × 10 −3 g

7. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is (atomic mass Ag = 108; Br = 80) (1) 36 (3) 60 (2) 48 (4) 24

% of Nitrogen =

calculate the percentage of carbon and hydrogen in the organic compound.

1.4 × m.eq of H2SO4 used to neutralize NH3 Mass of the compound 1 1  1.4 ×  60 × × 2 − 20 ×  10 10  = 10%  1 .4

9. If 0.210 g of an organic compound on complete combustion gave 0.307 g of CO2 and 0.127 g of H2O,

=

12 × m1 × 100 % 144 × m 12 × 0.307 × 100 = 39.87% 144 × 0.210

The percentage of hydrogen in the compound is given by Mass of water formed (m 2 ) 2 × × 100% 18 Mass of compound taken (m )

Given that m = 0.210 g and amount of water formed (m2) = 0.127 g. Substituting the values in the equation, we get 2 × m 2 × 100 % 18 × m 2 × 0.127 × 100 = = 6.72% 18 × 0.210

Percentage of hydrogen =

10. A organic compound (0.1840 g) containing nitrogen, when analyzed by Dumas’ method gave 30.0 mL of nitrogen collected at 287 K and 758 mm Hg pressure. Calculate the percentage of nitrogen in the organic compound. (Aqueous tension at 287 K is 14 mm Hg.) Solution Given that volume of nitrogen collected at 287 K and 758 mm Hg pressure is 30 mL. Taking into account the aqueous tension of water, the actual pressure = 758 - 14 = 744 mm Hg. Therefore, volume of nitrogen at STP is 273 × 744 × 30 = 27.935 mL 287 × 760

We know that 22,400 mL of N2 at STP weighs 28 g. Therefore, 27.945 mL of N2 weighs =

28 × 27.935 = 3.49 g 22400

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Chapter 3 | Purification and Characterization of Organic Compounds Therefore, Percentage of N = =

Mass of N2 × 100 Mass of organic sample 0.0349 g × 100 = 18.98% 0.1840 g

11. In estimation of nitrogen in an organic compound (0.35 g) by Kjeldahl’s method, the ammonia evolved was neutralized by passing through 60 mL of 0.2 M H2SO4. The excess of acid required 22.0 mL of 0.2 M NaOH for complete neutralization. Determine the percentage of nitrogen in the organic compound. Solution Volume of 0.2 N H2SO4 taken (V) = 60 mL. Volume of acid used to neutralize ammonia solution (V1) V1 = Total volume of the acid − Volume of acid neutralized by NaOH = 60 − 22 = 38 mL

We know that for neutralization, (V - V1/2) mL of H2SO4 of molarity M = 2(V - V1/2) mL of NH3 solution with molarity M. Therefore, 38 mL of 0.2 M H2SO4 = 76 mL of 0.2M NH3 Also, 1000 mL of 1 M NH3 solution contains 17 g NH3 or 14 g of N. Therefore, amount of nitrogen in 76 mL of NH3 is 14 × 76 = 0.1064 g of N 1000

Therefore, Percentage of N = =

Mass of Nitrogen × 100 Mass of organic compound 0.1064 g × 100 = 30.4% 0.35 g

12. An organic compound (0.8 g) containing bromine was heated with concentrated nitric acid and silver nitrate in a Carius tube. The amount of silver bromide formed was 0.76 g. Determine the amount of bromine in the compound. Solution The percentage of bromine in a organic compound when estimated as silver bromide is given by 80 Mass of silver bromide formed (m1) × 100 % × 188 Mass of compound taken (m)

Given that mass of silver bromide formed (m1) = 0.76 g and the amount of organic compound taken (m) = 0.8 g. Substituting the values in the equation, we get 80 × m1 × 100 % 188 × m 80 × 0.76 × 100 = = 4 0 .4 % 188 × 0.8

Percentage of halogen =

13. Analysis of 0.3 g of an organic compound containing sulphur gave 0.4 g of BaSO4 when analyzed using Carius tube. Determine the percentage of sulphur in the compound. Solution The percentage of sulphur in a compound when estimated as barium sulphate is given by 32 Mass of barium sulphate formed (m1) × 100 × 233 Mass of compound taken (m )

Given that mass of organic compound (m) = 0.3 g and mass of BaSO4 formed (m1) = 0.4 g. Substituting in the equation, we get 32 × m1 × 100 % 233 × m 32 × 0.4 × 100 = = 1 8 .3 % 233 × 0.3

Percentage of sulphur =

14. Analysis of an organic compound (0.36 g) containing phosphorus gave 0.66 g of Mg2P2O7 when treated with concentrated nitric acid followed by magnesia mixture. Calculate the amount of phosphorus present in the compound. Solution The percentage of phosphorus in an organic compound when estimated as magnesium pyrophosphate is given by 62 Mass of Mg2P2O7 formed (m1) × 100 % × 222 Mass of compound taken (m )

Given that mass of compound taken (m) = 0.36 g and amount of Mg2P2O7 formed (m1) = 0.66  g. Substituting the values in the equation, we get 62 × m1 × 100 % 222 × m 62 × 0.66 × 100 = = 51.20% 36 222 × 0.3

Percentage of phosphorus =

Additional Objective Questions

SOLVED PREVIOUS YEARS’ NEET QUESTIONS 1. In Dumas’ method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is (1) 18.20 (3) 15.76 (2) 16.76 (4) 17.36 (AIPMT 2015) Solution (2) Given: Mass of organic compound = 0.25 g; Volume of moist nitrogen (V1) = 40 mL at 725 mm; Pressure of nitrogen (p1) = 725 – 25 = 700 mm We know that

p1V1 p2V2 = T1 T2 V2 =

p1V1 T2 × T1 p2

700 × 40 × 273 300 × 760 7644000 = = 33.52 mL 228000

2. The Lassaigne’s extract is boiled with conc. HNO3 while testing for halogens. By doing so it, (1) increases the concentration of NO3- ions. (2) decomposes Na2S and NaCN, if formed. (3) helps in the precipitation of AgCl. (4) increases the solubility product of AgCl. (AIPMT PRE 2011) Solution (2) In the presence of nitrogen or sulphur, the sodium fusion extract is boiled with concentrated nitric acid first to decompose sodium cyanide or sulphide formed during Lassaigne’s test which would otherwise interfere with silver nitrate test. 3. The most suitable method of separation of 1:1 mixture of ortho and para-nitrophenols is (1) chromatography. (3) steam distillation. (2) crystallization. (4) sublimation. (NEET 2017)

=

22,400 mL of nitrogen at STP weighs = 28 g Therefore, 33.52 mL of nitrogen at STP weighs =

28 × 33.52 938.56 = = 0.0418 g 22400 22400

Hence, % of nitrogen in organic compound = 0.0418 × 100 = 16.76 % 0.25

Solution (3) The ortho and para isomers can be separated by steam distillation. Steam distillation is used for purifying substances that are steam volatile and immiscible with water. Ortho-nitrophenol is steam volatile due to intramolecular hydrogen bonding while para-nitrophenol is less volatile due to intermolecular hydrogen bonding which cause association of molecules.

ADDITIONAL OBJECTIVE QUESTIONS NCERT Exemplar Questions 1. The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapor in vapor phase. A suitable method for the extraction of these oils from the flowers is (1) distillation. (2) crystallization.

(3) distillation under reduced pressure. (4) steam distillation. 2. During hearing of a court case, the judge suspected that some changes in the documents had been carried out. He asked the forensic department to check the ink used at two different places. According to you which technique can give the best results? (1) Column chromatography (2) Solvent extraction (3) Distillation (4) Thin layer chromatography

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Chapter 3 | Purification and Characterization of Organic Compounds 3. The principle involved in paper chromatography is (1) adsorption. (3) solubility. (2) partition. (4) volatility.

OH

(1)

Exercise 1 1. The organic mixture is dissolved in a solvent in which two components have different solubility. The process is (1) sublimation. (2) distillation. (3) fractional crystallization. (4) simple crystallization. 2. Chromatography is a valuable method for the separation, isolation, purification and identification of the constituents of a mixture and it is based on general principles of (1) phase rule. (2) phase distribution. (3) interphase separation. (4) phase contact. 3. Which of the following is not commonly used solvent for crystallization? (1) Water (3) Phenol (2) Alcohol (4) Acetone 4. A mixture of camphor and benzoic acid can be separated by (1) chemical method. (2) sublimation. (3) fractional distillation. (4) extraction with a solvent. 5. Which of the following methods is not used for purification of an organic compound? (1) Simple crystallization (3) Sublimation (2) Vaporization (4) Steam distillation 6. Anthracene is purified by (1) filtration. (2) distillation.

10. Which of the following cannot be purified by steam distillation?

(3) crystallization. (4) sublimation.

7. Acetone and methanol in a mixture can be separated by (1) vacuum distillation. (2) fractional distillation. (3) steam distillation. (4) None of the above. 8. Two volatile liquids A and B differ in their boiling points by 15 K. The process that can be used to separate them is (1) fractional distillation. (2) steam distillation. (3) fractional crystallization. (4) simple distillation. 9. In Kjeldahl’s method, nitrogen present is estimated as (1) N2 (3) NO2 (2) NH3 (4) N2O

OH CHO

(3)

COOH

(2)

NO2 OH

(4)

COCH3

11. An organic substance can be separated from its aqueous solution by (1) distillation. (2) steam distillation. (3) solvent extraction. (4) fractional distillation.

Exercise 2 1. Simple distillation can be used to separate a mixture of (1) benzene (b.p. 80 °C) and toluene (b.p. 110 °C). (2) ether (b.p. 35 °C) and toluene (b.p. 110 °C). (3) ethanol (b.p. 78 °C) and water (b.p. 100 °C). (4) None of the above 2. Which of the following reagents is useful in separating aniline from a mixture of aniline and nitrobenzene? (1) Aqueous NaOH (3) Aqueous NaHCO3 (2) H2O (4) Aqueous HCl 3. In steam distillation, the vapors pressure of the volatile organic compound is (1) equal to atmospheric pressure. (2) less than atmospheric pressure. (3) more than atmospheric pressure. (4) None of the above 4. In the Lassaigne’s test, sulphur present in the organic compound first changes into (1) CS2 (3) Na2SO4 (2) Na2S (4) Na2SO3 5. In Lassaigne’s test for sulphur in an organic compound, the purple coloration with sodium nitroprusside is formed is due to (1) Na4[Fe(CN)5NOS] (2) Na3[Fe(CN)5S] (3) Na2[Fe(CN)5NOS] (4) Na3[Fe(CN)6] 6. On complete combustion, 0.3 g of an organic compound gave 0.2 g of CO2 and 0.1 g of water. What is the percentage of carbon in the compound? (1) 18.18 (3) 28.36 (2) 21.95 (4) 46.24

Answer Key 7. On complete combustion, 5 g of an organic compound gave 2 g of water. What is the percentage of hydrogen in the compound? (1) 21.69 (3) 4.44 (2) 40.44 (4) 2.69 8. An organic compound is found to have the formula C5H10ONCl. The percentage of nitrogen present in it is (1) 21.36% (3) 44.05% (2) 10.3% (4) 20.6%

(1) Assertion and Reason are true and the Reason is the correct explanation of the Assertion. (2) Assertion and Reason are true but the Reason is not a correct explanation of the Assertion. (3) Assertion is true but the Reason is false. (4) Assertion and Reason both are false.

9. An alkane has a C/H ratio of 5.1428 by mass. Its molecular formula is (1) C5H12 (3) C8H18 (2) C6H14 (4) C7H10 10. The empirical formula of compound is CH2. The mass of one mole of the compound is 42 g. Therefore, its structural formula is (1) CH3CH2CH3 (2) CH3 } CH CH2 (3) CH2 CH-CH CH2 (4) CH3 } C

CH

11. Empirical formula of a hydrocarbon containing 80% carbon and 20% hydrogen is (1) CH (3) CH3 (2) CH2 (4) CH4

1. Assertion: During test for nitrogen with Lassaigne’s extract, on adding FeCl3 solution sometimes a red precipitate is obtained. Reason: Sulphur is also present. 2. Assertion: A mixture of plant pigments can be separated by chromatography. Reason: Chromatography is used for the separation of colored substances into individual components. 3. Assertion: Essential oils are purified by steam distillation. Reason: The compounds which decompose at their boiling points can be purified by steam distillation. 4. Assertion: Non-volatile impurities are purified by simple distillation. Reason: Vaporization of a liquid by heating and subsequent condensation of vapors.

12. The percentage of N2 in urea is about (1) 18.05% (3) 46.66% (2) 28.29% (4) 85.56%

5. Assertion: Carbon tetrachloride does not give white precipitate with AgNO3.

Exercise 3

6. Assertion: Glycerol is purified by distillation under reduced pressure.

Reason: It is a covalent compound and AgNO3 is ionic.

In the following set of questions, two statements “Assertion” and “Reason” are given. Choose the correct answer from the following options:

Reason: Organic compounds in liquid state are purified by distillation.

ANSWER KEY NCERT Exemplar 1. (4)

2. (4)

3. (2)

1. (3)

2. (3)

3. (3)

4. (2)

5. (2)

6. (4)

7. (2)

8. (1)

9. (2)

10. (2)

2. (4)

3. (2)

4. (2)

5. (1)

6. (1)

7. (3)

8. (2)

9. (2)

10. (2)

11. (3)

12. (3)

3. (3)

4. (1)

5. (1)

Exercise 1

11. (3)

Exercise 2 1. (2)

Exercise 3 1. (1) 6. (2)

2. (2)

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HINTS AND EXPLANATIONS Exercise 1 1. (3) Fractional crystallization is the basic principle. 3. (3) Phenol is not commonly used as a solvent for crystallization. Other common solvents used are H2O, alcohol, ether, CHCl3, CCl4, CH3COCH3, benzene and ether, etc. 4. (2) Sublimation process can be used for separation of camphor and benzoic acid. This technique is used to separate volatile substances like camphor from non-volatile impurities like benzoic acid. 5. (2) Vaporization is not used for purification of an organic compound.

4. (2) Sulphur present in organic compounds first gives Na2S in Lassaigne’s test. 5. (1) Na 2S + Na 2 [Fe(CN)5 NO] → Na 4 [Fe(CN)5 NOS] Purple coloration

12 × Weight of carbon dioxide × 100 12 × Weight of organic compound 12 × 0.2 = × 100 = 18.18% 44 × 0.3

6. (1) % Carbon =

7. (3) % of Hydrogen = =

6. (4) Anthracene is purified by sublimation. 7. (2) Both can be separated by fractional distillation because they have different boiling points. 8. (1) This method is used when the boiling points of the two liquids of the mixture are very close to one another (i.e., differ by 10 K or so). It involves repeated distillations and condensations.

11. (3) Solvent extraction is the process of removing an organic substance from its solution in water by another immiscible solvent such as ether, chloroform, CCl4, etc.

Exercise 2 1. (2) Simple distillation can be used for separating two liquids if they differ in boiling point by 30 to 50°C. 2. (4) Aniline forms aniline hydrochloride with a aqueous HCl. 3. (2) At boiling point, pmixture = pcompound + psteam

2×2 × 100 = 4.44% 18 × 5

8. (2) Molecular mass of C5H10ONCl = 135.5. Percentage of N =

14 × 100 = 10.3% 135.5

9. (2) The C/H ratio by mass in C6H14 is

9. (2) Nitrogen is estimated as NH3. 10. (2) Steam distillation is used purify nitrobenzene, bromobenzene, ortho-nitrophenol, salicyaldehyde, o-hydroxyacetophenone, essential oils, turpentine oil, etc. The compounds which have intermolecular hydrogen bonding have higher boiling points and cannot be separated by this process, for example, benzoic acid.

2 × Weight of H2O formed × 100 18 × Weight of organic compound

6 × 12 72 = = 5.1428 14 × 1 14

10. (2) Empirical mass = 12 + 2 = 14 42 1

= = 42 Molecular mass

42 14

Therefore, number of units = = 3 11. (3) Relative number of Simplest Atomic moles ratio Element Percentage mass C

80

12

6.6

1

H

20

1

20

3

Empirical formula is CH3. 12. (3) The structure of urea is NH2CONH2. Its molecular mass is 60. N2 =

28 × 100 = 46.66% 60

Basic Principles of Alkanes and Organic Chemistry Cycloalkanes

4

C H A P T E R OU TLIN E 4.8 How to Gain Structural Information from Molecular Formulas and the Index of Hydrogen Deficiency 4.9 Chemical Reactions of Alkanes 4.10 Conformations 4.11 The Relative Stabilities of Cycloalkanes: Ring Strain 4.12 Conformations of Cyclohexane: The Chair and the Boat 4.13 Disubstituted Cycloalkanes: Cis–Trans Isomerism

Charles D. Winters/Photo Researchers, Inc.

4.1 Introduction to Alkanes and Cycloalkanes 4.2 Shapes of Alkanes 4.3 How to Name Alkanes and Alkyl Groups: The IUPAC System 4.4 How to Name Cycloalkanes 4.5 Physical Properties of Alkanes and Cycloalkanes 4.6 Isomerism 4.7 Synthesis of Alkanes and Cycloalkanes

Hydrocarbons are compounds composed entirely of carbon and hydrogen atoms bonded to each other by covalent bonds. Fossil fuels – natural gas, petroleum and coal – are the principal sources of hydrocarbons. Natural gas, found in the upper strata during drilling of oil wells, is primarily methane with small amounts of ethane, propane and butane. Petroleum is a mixture of hydrocarbons from which petrol, kerosene, fuel oil, lubricating oil, paraffin wax and petrolatum (semi-solid mixture of hydrocarbons) are separated. Petrol and compressed natural gas (CNG) are widely used as automobile fuels.

4.1

INTRODUCTION TO ALKANES AND CYCLOALKANES

The family of organic compounds called hydrocarbons can be divided into several groups on the basis of the type of bond that exists between the individual carbon atoms. Those hydrocarbons in which all of the carbon–carbon bonds are single bonds are called alkanes, those hydrocarbons that contain a carbon– carbon double bond are called alkenes, and those with a carbon–carbon triple bond are called alkynes. Cycloalkanes are alkanes in which all or some of the carbon atoms are arranged in a ring. ●

Alkanes have the general formula CnH2n12; cycloalkanes containing a single ring have two fewer hydrogen atoms and thus have the general formula CnH2n.

4.1A Sources of Alkanes: Petroleum The primary source of alkanes is petroleum. Petroleum is a complex mixture of organic compounds, most of which are alkanes and aromatic compounds (cf. Chapter 14). It also contains small amounts of oxygen-, nitrogen-, and sulfur-containing compounds. Some of the molecules in petroleum are clearly of biological origin. Most scientists believe that petroleum originated with accumulation of dead microorganisms that settled to the bottom of the sea and that were entombed in sedimentary rock. These microbial remains eventually were transformed into oil by the heat radiating from Earth’s core.

180

Chapter 4 | Alkanes and Cycloalkanes

The Chemistry of...

T

he first step in refining petroleum is distillation; the object here is to separate the petroleum into fractions based on the volatility of its components. Complete separation into fractions containing individual compounds is economically impractical and virtually impossible technically. More than 500 different compounds are contained in the petroleum distillates boiling below 200 °C, and many have almost the same boiling points. Thus the fractions taken contain mixtures of alkanes of similar boiling points (see the table below). Mixtures of alkanes, fortunately, are perfectly suitable for uses as fuels, solvents, and lubricants, the primary uses of petroleum. The demand for gasoline is much greater than that supplied by the gasoline fraction of petroleum. Important processes in the petroleum industry, therefore, are concerned with converting hydrocarbons from other fractions into gasoline. When a mixture of alkanes from the gas oil fraction (C12 and higher) is heated at very high temperatures (∼500 °C) in the presence of a variety of catalysts, the molecules break apart and rearrange to smaller, more highly branched hydrocarbons containing 5–10 carbon atoms. This process is called catalytic cracking. Cracking can also be done in the absence of a catalyst—called thermal cracking—but in this process the products tend to have unbranched chains, and alkanes with unbranched chains have a very low “octane rating.”

PETROLEUM REFINING The highly branched compound 2,2,4trimethylpentane (called isooctane in the petroleum industry) burns very smoothly (without knocking) in internal combustion engines and is used as one of the standards by which the octane rating of gasolines is established. According to this scale, 2,2,4-trimethylpentane has an octane rating of 100. Heptane, CH3(CH2)5CH3, a compound that produces much knocking when it is burned in an internal combustion engine, is given an octane rating of 0. Mixtures of 2,2,4-trimethylpentane and heptane are used as standards for octane ratings between 0 and 100. A gasoline, for example, that has the same characteristics in an engine as a mixture of 87% 2,2,4-trimethylpentane and 13% heptane would be rated as 87-octane gasoline. CH3 CH3

C

CH3 CH2

CH

CH3

CH3 or

2,2,4-Trimethylpentane (“isooctane”)

Typical fractions obtained by distillation of petroleum Boiling range of fraction (°C)

Number of carbon atoms per molecule

Use

Below 20

C1}C4

Natural gas, bottled gas, petrochemicals

20–60

C5}C6

Petroleum ether, solvents

60–100

C6}C7

Ligroin, solvents

40–200

C5}C10

Gasoline (straight-run gasoline)

175–325

C12}C18

Kerosene and jet fuel

250–400

C12 and higher

Gas oil, fuel oil, and diesel oil

Nonvolatile liquids

C20 and higher

Refined mineral oil, lubricating oil, and grease

Nonvolatile solids

C20 and higher

Paraffin wax, asphalt, and tar

Adapted with permission of John Wiley & Sons, Inc., from Holum, J. R., General, Organic, and Biological Chemistry, Ninth Edition, p. 213. Copyright 1995.

4.2 | Shapes of Alkanes

Hydrocarbons are also found in outer space. Asteroids and comets contain a variety of organic compounds. Methane and other hydrocarbons are found in the atmospheres of Jupiter, Saturn, and Uranus. Saturn’s moon Titan has a solid form of methane–water ice at its surface and an atmosphere rich in methane. Whether of terrestrial or celestial origin, we need to understand the properties of alkanes. We begin with a consideration of their shapes and how we name them.

4.2

SHAPES OF ALKANES

A general tetrahedral orientation of groups—and thus sp3 hybridization—is the rule for the carbon atoms of all alkanes and cycloalkanes. We can represent the shapes of alkanes as shown as follows: Propane

Butane

Pentane

CH3CH2CH3

CH3CH2CH2CH3

CH3CH2CH2CH2CH3

or

or

or

Butane and pentane are examples of alkanes that are sometimes called “straight-chain” alkanes. One glance at three-dimensional models, however, shows that because of their tetrahedral carbon atoms the chains are zigzagged and not at all straight. Indeed, the structures depicted above are the straightest possible arrangements of the chains because rotations about the carbon–carbon single bonds produce arrangements that are even less straight. A better description is unbranched. This means that each carbon atom within the chain is bonded to no more than two other carbon atoms and that unbranched alkanes contain only primary and secondary carbon atoms. Primary, secondary, and tertiary carbon atoms were defined in Section 1.13. Isobutane, isopentane, and neopentane are examples of branched-chain alkanes. In neopentane the central carbon atom is bonded to four carbon atoms. Isobutane

Isopentane

CH3CHCH3

CH3CHCH2CH3

CH3 or

CH3 or

Neopentane CH3 CH3CCH3 CH3 or

Butane and isobutane have the same molecular formula: C4H10. The two compounds have their atoms connected in a different order and are, therefore, constitutional isomers (Section 1.3). Pentane, isopentane, and neopentane are also constitutional isomers. They, too, have the same molecular formula (C5H12) but have different structures. Constitutional isomers, as stated earlier, have different physical properties. The differences may not always be large, but constitutional isomers are always found to have different melting points, boiling points, densities, indexes of refraction, and so forth. Table 4.1 gives some of the physical properties of the C6H14 isomers, of which there are only five. Note that the number of constitutional isomers that is possible increases dramatically as the number of carbon atoms in the alkane increases. Prior to the development near the end of the nineteenth century of a formal system for naming organic compounds, many organic compounds had already been discovered or synthesized. Early chemists named these compounds, often on the basis of the source of the compound. Acetic acid (systematically called ethanoic acid) is an example; it was obtained by distilling vinegar, and it got its name from the Latin word for vinegar, acetum. Formic acid (systematically called methanoic acid) had been obtained by the distillation of the bodies of ants, so it got the name from the Latin word for ants, formicae. Many of these older names for compounds, called common or trivial names, are still in wide use today. Today, chemists use a systematic nomenclature developed and updated by the International Union of Pure and Applied Chemistry (IUPAC). Underlying the IUPAC system is a fundamental principle: each different compound should have a different and unambiguous name.* *The complete IUPAC rules for nomenclature can be found through links at the IUPAC website.

181

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Chapter 4 | Alkanes and Cycloalkanes

Table 4.1 Physical constants of the hexane isomers Molecular formula

Condensed structural formula

Bond-line formula

mp (°C)

bp (°C)a (1 atm)

Density (g mL21) at 20 °C

Index of refractionb (nD 20 °C)

C6H14

CH3CH2CH2CH2CH2CH3

295

68.7

0.6594

1.3748

C6H14

CH3CHCH2CH2CH3

2153.7

60.3

0.6532

1.3714

2118

63.3

0.6643

1.3765

2128.8

58

0.6616

1.3750

298

49.7

0.6492

1.3688

CH3 CH3CH2CHCH2CH3

C6H14

CH3 C6H14

CH3CH—CHCH CH3 CH3 CH3

C6H14

CH3— C — CH2CH CH3 Unless otherwise indicated, all boiling points given in this book are at 1 atm or 760 torr. The index of refraction is a measure of the ability of the alkane to bend (refract) light rays. The values reported are for light of the D line of the sodium spectrum (nD).

a

b

4.3

HOW TO NAME ALKANES AND ALKYL GROUPS: THE IUPAC SYSTEM

We have learnt about IUPAC system of nomenclature for organic compounds in Chapter 1 and its use in naming hydrocarbons and organic compounds containing functional groups. The rules for naming of straight and branched chained alkanes using in the IUPAC system have also been briefly discussed. We will now discuss the nomenclature of unbranched and branched chain alkanes and alkyl groups in detail. The names for several of the unbranched alkanes are listed in Table 4.2. The ending for all of the names of alkanes is -ane. The stems of the names of most of the alkanes (above C4) are of Greek and Table 4.2 The unbranched alkanes Name

Number of carbon atoms

Structure

Name

Number of carbon atoms

Structure

Methane

1

CH4

Undecane

11

CH3(CH2)9CH3

Ethane

2

CH3CH3

Dodecane

12

CH3(CH2)10CH3

Propane

3

CH3CH2CH3

Tridecane

13

CH3(CH2)11CH3

Butane

4

CH3(CH2)2CH3

Tetradecane

14

CH3(CH2)12CH3

Pentane

5

CH3(CH2)3CH3

Pentadecane

15

CH3(CH2)13CH3

Hexane

6

CH3(CH2)4CH3

Hexadecane

16

CH3(CH2)14CH3

Heptane

7

CH3(CH2)5CH3

Heptadecane

17

CH3(CH2)15CH3

Octane

8

CH3(CH2)6CH3

Octadecane

18

CH3(CH2)16CH3

Nonane

9

CH3(CH2)7CH3

Nonadecane

19

CH3(CH2)17CH3

Decane

10

CH3(CH2)8CH3

Eicosane

20

CH3(CH2)18CH3

4.3 | How to Name Alkanes and Alkyl Groups: The IUPAC System

Latin origin. Learning the stems is like learning to count in organic chemistry. Thus, one, two, three, four, and five become meth-, eth-, prop-, but-, and pent-.

4.3A How to Name Unbranched Alkyl Groups If we remove one hydrogen atom from an alkane, we obtain what is called an alkyl group. These alkyl groups have names that end in -yl. When the alkane is unbranched, and the hydrogen atom that is removed is a terminal hydrogen atom, the names are straightforward: CH3 } H

CH3CH2 } H

CH3CH2CH2 } H

CH3CH2CH2CH2 } H

Methane

Ethane

Propane

Butane

CH3 }

CH3CH2 }

CH3CH2CH2 }

CH3CH2CH2CH2 }

Methyl

Ethyl

Propyl

Butyl

Me-

Et-

Pr-

Bu-

4.3B How to Name Branched-Chain Alkanes Branched-chain alkanes are named according to the following rules: 1. Locate the longest continuous chain of carbon atoms; this chain determines the parent name for the alkane. We designate the following compound, for example, as a hexane because the longest continuous chain contains six carbon atoms: CH3CH2CH2CH2CHCH3 ––

or

CH3

Longest chain

The longest continuous chain may not always be obvious from the way the formula is written. Notice, for example, that the following alkane is designated as a heptane because the longest chain contains seven carbon atoms:

2. Number the longest chain beginning with the end of the chain nearer the substituent. Applying this rule, we number the two alkanes that we illustrated previously in the following way:

6

5

3

4

2

1

7

6

5

4

Substituent 3 2

Substituent

1

3. Use the numbers obtained by application of rule 2 to designate the location of the substituent group. The parent name is placed last, and the substituent group, preceded by the number designating its location on the chain, is placed first. Numbers are separated from words by a hyphen. Our two examples are 2-methylhexane and 3-methylheptane, respectively: 6

5

4

3

2

1

Substituent Locant

2-Methylhexane

7

Longest chain

6

5

4

3 2 1

3-Methylheptane

183

184

Chapter 4 | Alkanes and Cycloalkanes

4. When two or more substituents are present, give each substituent a number corresponding to its location on the longest chain. For example, we designate the following compound as 4-ethyl-2methylhexane:

4-Ethyl-2-methylhexane

The substituent groups should be listed alphabetically (i.e., ethyl before methyl).* In deciding on alphabetical order, disregard multiplying prefixes such as “di” and “tri.” 5. When two substituents are present on the same carbon atom, use that number twice:

3-Ethyl-3-methylhexane

6. When two or more substituents are identical, indicate this by the use of the prefixes di-, tri-, tetra-, and so on. Then make certain that each and every substituent has a number. Commas are used to separate numbers from each other:

2,3-Dimethylbutane

2,2,4,4-Tetramethylpentane

2,3,4-Trimethylpentane

Application of these six rules allows us to name most of the alkanes that we shall encounter. Two other rules, however, may be required occasionally: 7. When two chains of equal length compete for selection as the parent chain, choose the chain with the greater number of substituents: 7

5

6

3

4

1

2

2,3,5-Trimethyl-4-propylheptane (four substituents)

8. When branching first occurs at an equal distance from either end of the longest chain, choose the name that gives the lower number at the first point of difference: 6

5

4

3

2

1

2,3,5-Trimethylhexane (not 2,4,5-trimethylhexane)

4.3C How to Name Branched Alkyl Groups In Section 4.3A you learned the names for the unbranched alkyl groups such as methyl, ethyl, propyl, and butyl, groups derived by removing a terminal hydrogen from an alkane. For alkanes with more than *Some handbooks also list the groups in order of increasing size or complexity (i.e., methyl before ethyl). An alphabetical listing, however, is now by far the most widely used system.

4.3 | How to Name Alkanes and Alkyl Groups: The IUPAC System

two carbon atoms, more than one derived group is possible. Two groups can be derived from propane, for example; the propyl group is derived by removal of a terminal hydrogen, and the 1-methylethyl or isopropyl group is derived by removal of a hydrogen from the central carbon: Three-Carbon Groups CH3CH2CH2 Propane

CH3CHCH3

Propyl

Isopropyl

Pr-

i-Pr-

––

CH3CH2CH2––

1-Methylethyl is the systematic name for this group; isopropyl is a common name. Systematic nomenclature for alkyl groups is similar to that for branched-chain alkanes, with the provision that numbering always begins at the point where the group is attached to the main chain. There are four C4 groups. Four-Carbon Groups CH3CH2CH2CH3 Butane

––

CH3

CH3CHCH2––

CH3CH2CHCH3

(CH3)3C––

Butyl

Isobutyl

sec-Butyl

tert-Butyl (or t-Bu)

––

CH3CH2CH2CH2––

The following examples show how the names of these groups are employed:

4-(1-Methylethyl)heptane or 4-isopropylheptane

4-(1,1-Dimethylethyl)octane or 4-tert-butyloctane

The common names isopropyl, isobutyl, sec-butyl, and tert-butyl are approved by the IUPAC for the unsubstituted groups, and they are still very frequently used. You should learn these groups so well that you can recognize them any way they are written. In deciding on alphabetical order for these groups you should disregard structure-defining prefixes that are written in italics and separated from the name by a hyphen. Thus tert-butyl precedes ethyl, but ethyl precedes isobutyl.*

*The abbreviations i-, s-, and t- are sometimes used for iso-, sec-, and tert-, respectively.

185

186

Chapter 4 | Alkanes and Cycloalkanes

There is one five-carbon group with an IUPAC approved common name that you should also know: the 2,2-dimethylpropyl group, commonly called the neopentyl group. CH3 CH39 C9 CH29 CH3 2,2-Dimethylpropyl or neopentyl group

4.3D How to Classify Hydrogen Atoms The hydrogen atoms of an alkane are classified on the basis of the carbon atom to which they are attached. A hydrogen atom attached to a primary carbon atom is a primary (1°) hydrogen atom, and so forth. The following compound, 2-methylbutane, has primary, secondary (2°), and tertiary (3°) hydrogen atoms: 1º Hydrogen atoms

CH3 CH3

CH

CH2

3º Hydrogen atom

CH3 2º Hydrogen atoms

On the other hand, 2,2-dimethylpropane, a compound that is often called neopentane, has only primary hydrogen atoms: CH3 H3 C

C

CH3

CH3

2,2-Dimethylpropane (neopentane)

4.4

HOW TO NAME CYCLOALKANES

4.4A How to Name Monocyclic Cycloalkanes Cycloalkanes are named by adding “cyclo” before the parent name. 1. Cycloalkanes with one ring and no substituents: Count the number of carbon atoms in the ring, then add “cyclo” to the beginning of the name of the alkane with that number of carbons. For example, cyclopropane has three carbons and cyclopentane has five carbons. CH2

H2C

CH2

H2C =

C H2 Cyclopropane

=

CH2

H2C C H2

Cyclopentane

4.4 | How to Name Cycloalkanes

2. Cycloalkanes with one ring and one substituent: Add the name of the substituent to the beginning of the parent name. For example, cyclohexane with an attached isopropyl group is isopropylcyclohexane. For compounds with only one substituent, it is not necessary to specify a number (locant) for the carbon bearing the substituent. Cl

Isopropylcyclohexane

Chlorocyclopentane

3. Cycloalkanes with one ring and two or more substituents: For a ring with two substituents, begin by numbering the carbons in the ring, starting at the carbon with the substituent that is first in the alphabet and number in the direction that gives the next substituent the lower number possible. When there are three or more substituents, begin at the substituent that leads to the lowest set of numbers (locants). The substituents are listed in alphabetical order, not according to the number of their carbon atom. 1

OH

2

3 2

1

3 4

1

2

2-Methylcyclohexanol

1-Ethyl-3-methylcyclohexane (not 1-ethyl-5-methylcyclohexane)

Cl

4-Chloro-2-ethyl-1-methylcyclohexane (not 1-chloro-3-ethyl-4-methylcyclohexane)

4. When a single ring system is attached to a single chain with a greater number of carbon atoms, or when more than one ring system is attached to a single chain, then it is appropriate to name the compounds as cycloalkylalkanes. For example.

1-Cyclobutylpentane

1,3-Dicyclohexylpropane

4.4B How to Name Bicyclic Cycloalkanes 1. We name compounds containing two fused or bridged rings as bicycloalkanes and we use the name of the alkane corresponding to the total number of carbon atoms in the rings as the parent name. The following compound, for example, contains seven carbon atoms and is, therefore, a bicycloheptane. The carbon atoms common to both rings are called bridgeheads, and each bond, or each chain of atoms connecting the bridgehead atoms, is called a bridge. One-carbon bridge

H C

CH2

H2C Two-carbon bridge

Bridgehead

CH2 CH2

H2C C H

Two-carbon = bridge

Bridgehead A bicycloheptane

=

187

188

Chapter 4 | Alkanes and Cycloalkanes

2. We then interpose an expression in brackets within the name that denotes the number of carbon atoms in each bridge (in order of decreasing length). Fused rings have zero carbons in their bridge. For example, Bridged

Fused

H C CH2

H2C CH2

H C =

CH2

H2C

CH2

H2C

=

C H

C H Bicyclo[2.2.1]heptane (also called norbornane)

Bicyclo[1.1.0]butane

3. In bicycloalkanes with substituents, we number the bridged ring system beginning at one bridgehead, proceeding first along the longest bridge to the other bridgehead, then along the next longest bridge back to the first bridgehead; the shortest bridge is numbered last. Bridged 7

1

5

9

2 8

6

Fused

3 4

8-Methylbicyclo[3.2.1]octane

4.5

1

2 3

8 7

6

4 5

8-Methylbicyclo[4.3.0]nonane

PHYSICAL PROPERTIES OF ALKANES AND CYCLOALKANES

If we examine the unbranched alkanes in Table 4.2, we notice that each alkane differs from the preceding alkane by one } CH2 } group. Butane, for example, is CH3(CH2)2CH3 and pentane is CH3(CH2)3CH3. A series of compounds like this, where each member differs from the next member by a constant unit, is called a homologous series. Members of a homologous series are called homologues. At room temperature (25 °C) and 1 atm pressure the first four members of the homologous series of unbranched alkanes are gases (Fig. 4.1), the C5 } C17 unbranched alkanes (pentane to heptadecane) are liquids, and the unbranched alkanes with 18 and more carbon atoms are solids. Boiling Points The boiling points of the unbranched alkanes show a regular increase with increasing molecular weight (Fig. 4.1a) in the homologous series of straight-chain alkanes. Branching of the alkane chain, however, lowers the boiling point. The hexane isomers in Table 4.1 examplify this trend. With unbranched alkanes, as molecular weight increases, so too do molecular size and, even more importantly, molecular surface area. With increasing surface area, the dispersion forces between molecules increase; therefore, more energy (a higher temperature) is required to separate molecules from one another and produce boiling. Chain branching, on the other hand, makes a molecule more compact, reducing its surface area and with it the strength of the dispersion forces operating between it and adjacent molecules; this has the effect of lowering the boiling point. Figure 4.2 illustrates this for two C8 isomers. Melting Points The unbranched alkanes do not show the same smooth increase in melting points with increasing molecular weight (gray line in Fig. 4.1b) that they show in their boiling points. There is an alternation as one progresses from an unbranched alkane with an even number of carbon atoms to the next one with an odd number of carbon atoms. If, however, the even- and odd-numbered alkanes are plotted on separate curves (white and gray lines in Fig. 4.1b), there is a smooth increase in melting point with increasing molecular weight.

189

4.5 | Physical Properties of Alkanes and Cycloalkanes 300 0 Melting point, °C

Boiling point, °C

200 Cycloalkanes

100

Alkanes 0

–100 –200

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Number of carbon atoms (a)

–50 –100 –150 –200

1

3

5 7 9 11 Number of carbon atoms (b)

13

15

Figure 4.1 (a) Boiling points of unbranched alkanes (in gray) and cycloalkanes (in white). (b) Melting points of unbranched alkanes. &+ &+ &+

&+&+&+&+&+&+&+&+

&

&

&+

&+ &+ D 2FWDQHESƒ&

E 7HWUDPHWK\OEXWDQHESƒ&

Figure 4.2 Chain-branching decreases the contact surface area between molecules, as for the branched C8 isomer in (b), lessening the dispersion forces between them and leading to a lower boiling point than for the unbranched C8 isomer (a). X-ray diffraction studies, which provide information about molecular structure, have revealed the reason for this apparent anomaly. Alkane chains with an even number of carbon atoms pack more closely in the crystalline state. As a result, attractive forces between individual chains are greater and melting points are higher. Cycloalkanes also have much higher melting points than their open-chain counterparts (Table 4.3). Table 4.3 Physical constants of cycloalkanes Number of carbon atoms

Name

bp (°C) (1 atm)

mp (°C)

Density at 20 °C (g mL21)

Refractive index (n)

3

Cyclopropane

233

2126.6

—

—

4

Cyclobutane

13

290

—

1.4260

5

Cyclopentane

49

294

0.751

1.4064

6

Cyclohexane

81

6.5

0.779

1.4266

7

Cycloheptane

118.5

212

0.811

1.4449

8

Cyclooctane

149

13.5

0.834

—

Density As a class, the alkanes and cycloalkanes are the least dense of all groups of organic compounds. All alkanes and cycloalkanes have densities considerably less than 1.00 g mL21 (the density of water at 4 °C). As a result, petroleum (a mixture of hydrocarbons rich in alkanes) floats on water. Solubility Alkanes and cycloalkanes are almost totally insoluble in water because of their very low polarity and their inability to form hydrogen bonds. Liquid alkanes and cycloalkanes are soluble in one another, and they generally dissolve in solvents of low polarity. Good solvents for them are benzene, carbon tetrachloride, chloroform, and other hydrocarbons.

190

Chapter 4 | Alkanes and Cycloalkanes

4.6

ISOMERISM

For the molecular formulas CH4, C2H6 and C3H8, only one order of attachment of atoms is possible. For the molecular formula C4H10, two orders of attachment of atoms are possible. In one of these, named butane, the four carbons are bonded in a chain; in the other, named 2-methylpropane, three carbons are bonded in a chain, with the fourth carbon as a branch on the middle carbon of the chain.

CH3 CH3CH2CH2CH3

CH3CHCH3

Butane (boiling point = −0.5 °C)

2-Methylpropane (boiling point = −11.6 °C)

Butane and 2-methylpropane are structural isomers; they are different compounds and have different physical and chemical properties. Their boiling points, for example, differ by approximately 11 °C. Structural isomers are compounds that have the same molecular formula, but different structural formulas. By “different structural formulas”, we mean that these compounds differ in the kinds of bonds they have (single, double or triple) or in their connectivity (the order of attachment among their atoms). Structural isomers are also known as constitutional isomers. These structural isomers which differ in chain of carbon atoms are called chain isomers. The carbon atom which is attached to none or only one carbon atom is called primary (1°), the one attached to two, three and four are called secondary (2°), tertiary (3°) and quaternary (4°) carbon atoms.

4.7

SYNTHESIS OF ALKANES AND CYCLOALKANES

4.7A Hydrogenation of Alkenes and Alkynes A chemical synthesis may require, at some point, the conversion of a carbon–carbon double or triple bond to a single bond. This conversion is easily accomplished by a reaction called hydrogenation. There are several reaction conditions that can be used to carry out hydrogenation, but among the common ways is use of hydrogen gas and a solid metal catalyst such as platinum, palladium, or nickel. ●

Alkenes and alkynes react with hydrogen in the presence of metal catalysts such as nickel, palladium, and platinum to produce alkanes.

The general reaction is one in which the atoms of the hydrogen molecule add to each atom of the carbon–carbon double or triple bond of the alkene or alkyne. This converts the alkene or alkyne to an alkane: General Reaction C C



Alkene

H

Pt, Pd, or Ni

C

H

C

H

solvent, pressure

C

H

C

Alkane

 2 H2

Alkyne

Pt

H

C

H

solvent, pressure

H

C

H

Alkane

4.7 | Synthesis of Alkanes and Cycloalkanes

The reaction is usually carried out by dissolving the alkene or alkyne in a solvent such as ethyl alcohol (C2H5OH), adding the metal catalyst, and then exposing the mixture to hydrogen gas under pressure in a special apparatus. One molar equivalent of hydrogen is required to reduce an alkene to an alkane. Two molar equivalents are required to reduce an alkyne. Specific Examples CH3

CH3 C"CH C 2  H2 CH3

Ni

CH39C9CH2

EtOH (25 °C, 50 atm)

H

2-Methylpropene

Isobutane Pd

 H2

EtOH (25 °C, 1 atm)

Cyclohexene

O

Cyclohexane

O

Pd

2 H2



H

ethyl acetate

Cyclononyn-6-one

Cyclononanone

4.7B From Alkyl Halides Alkyl halides can be converted into corresponding alkanes either by replacement of the halide group by hydrogen atom (reduction) or alkyl group (coupling reaction). Reduction of Alkyl Halides Most alkyl halides (except fluorides) react with zinc and aqueous acid to produce an alkane. The general reaction is as follows: General Reaction R

X 1 Zn 1 HX

R

H 1 ZnX2

R

H

or R

X

Zn, HX (2ZnX2)

Specific Examples HBr

2 CH3CH2CHCH3

Zn

Br sec-Butyl bromide (2-bromobutane)

H Butane

CH3 2 CH3CHCH2CH2

2 CH3CH2CHCH3 1 ZnBr2

CH3 Br

Isopentyl bromide (1-bromo-3-methylbutane)

HBr Zn

2 CH3CHCH2CH2

H 1 ZnBr2

Isopentane (2-methylbutane)

In these reactions, zinc atoms transfer electrons to the carbon atom of the alkyl halide. Therefore, the reaction is a reduction of the alkyl halide. Zinc is a good reducing agent because it has two electrons in an orbital far from the nucleus which are readily donated to an electron acceptor. The mechanism for the

191

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Chapter 4 | Alkanes and Cycloalkanes

reaction is complex because the reaction takes place in a separate phase at or near the surface of the zinc metal. It is possible that an alkyl zinc halide forms first and then reacts with the acid to produce the alkane: δ1

Zn 1 R Reducing agent

δ2

2

R Zn21 X

X

HX

2

R H 1 Zn21 1 2 X

Alkylzinc halide

2

Alkane

The other reducing agents that can be used for the reduction of alkyl halides are: 1. Catalytic hydrogenolysis: The reaction of alkyl halide with excess hydrogen in the presence of a catalyst leads to formation of alkanes. Pd-C catalyst CH3CH2Br + H2  → CH3CH3 + HBr or RaneyNi

2. Reduction with HI and red phosphorus: Alkyl iodides react when heated with concentrated hydrogen iodide in the presence of red phosphorus to give corresponding alkanes. Red P CH3CH2I + conc. HI  → CH3CH3 + I2

Red phosphorus is used to remove the iodine formed so that it does not react again with alkane to form alkyl iodide. 3. Reduction with metal hydrides: Primary and secondary halides are reduced by lithium aluminium hydride to form the corresponding alkanes, while tertiary halides undergo dehydrohalogenation. Sodium boron hydride preferentially reduces secondary and tertiary halides but not primary. CH3CH2Br + H− → CH3CH3 + Br −

By Grignard Reagent These reagents are usually prepared by the reaction of an organic halide and magnesium metal (turnings) in an ether solvent: RX 1 Mg ArX 1 Mg

Et2O Et2O

RMgX ArMgX

Grignard reagents

Grignard reagents are seldom isolated but are used for further reactions in ether solution. Two examples are shown here: CH3I 1 Mg

Et2O 35°C

CH3MgI Methylmagnesium iodide (95%)

C6H5Br 1 Mg

Et2O 35°C

C6H5MgBr Phenylmagnesium bromide (95%)

The carbon–magnesium bond is polar due to difference in electronegativity, hence Grignard reagent readily reacts with compounds containing active hydrogen, such as alcohols, acids, amines, water, ammonia, etc. to form alkanes. For example, CH3CH2MgBr + CH3OH → CH3CH3 + Mg(OCH3 )Br

4.7 | Synthesis of Alkanes and Cycloalkanes

By Wurtz Reaction This involves the interaction of two molecules of an alkyl halide (preferably bromide or iodide) with metallic sodium in the presence of dry ether to form symmetrical alkanes containing double the number of carbon atoms presence in the alkyl halide. General Reaction R X 1 2Na 1 X Alkyl halide

R

Dry ether

R R 1 2NaX Alkane

Specific Examples CH3 Br 1 2Na 1Br Bromomethane CH3CH2 I 1 2Na 1 I Iodoethane

CH2CH3

CH3

Dry ether

CH3

Dry ether

CH3CH3 1 2NaBr Ethane CH3CH2CH2CH3 1 2NaI Butane

Thus, Wurtz reaction is a convenient method for the preparation of symmetrical alkanes (R–R), that is, alkanes containing even number of carbon atoms. However, if two different alkyl halides are used, a mixture of three alkanes is actually obtained. For example, CH3

I 1 2Na 1 I

Iodomethane CH3

CH2CH3

Dry ether

Iodoethane

I 1 2Na 1 I

CH3

Propane Dry ether

Iodomethane CH3CH2

CH3CH2CH3 1 2NaI

I 1 2Na 1 I

CH3CH3 1 2NaI Ethane

CH2CH3

Dry ether

CH3CH2CH2CH3 1 2NaI

Iodoethane

Butane

The boiling points of these alkanes are very close and hence cannot be separated by fractional distillation. That is why Wurtz reaction is useful only for the preparation of symmetrical alkanes and not for the preparation of unsymmetrical alkanes, that is, alkanes containing odd number of carbon atoms. By Corey-House Synthesis A highly versatile method for the synthesis of alkanes and other hydrocarbons from organic halides has been developed by E.J. Corey, G.H. Posner, G.M. Whitesides and H.O. House and is commonly known as Corey–House synthesis. The overall synthesis provides a way for coupling the alkyl groups of two alkyl halides to produce an alkane: R

X 1 R9

X

several steps (22 X)

R

R9

It overcomes the limitation of Wurtz reaction for synthesis of unsymmetrical alkanes (containing odd number of carbon atoms) in poor yields and can be used to prepare both symmetrical and unsymmetrical alkanes in good yield. In order to accomplish this coupling, we must transform one alkyl halide into a lithium dialkylcuprate (R2CuLi). This transformation requires two steps. First, the alkyl halide is treated with lithium metal in an ether solvent to convert the alkyl halide into an alkyllithium, RLi: R

X 1 2 Li

diethyl ether

RLi 1 LiX Alkyllithium

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Chapter 4 | Alkanes and Cycloalkanes

Then the alkyllithium is treated with cuprous iodide (CuI). This converts it to the lithium dialkylcuprate: 2 RLi

1

CuI

R2CuLi

Alkyllithium

1

LiI

Lithium dialkylcuprate

When the lithium dialkylcuprate is treated with the second alkyl halide (R′}X), coupling takes place between one alkyl group of the lithium dialkylcuprate and the alkyl group of the alkyl halide, R′}X: R2CuLi

R9

1

X

R

Lithium Alkyl halide dialkylcuprate

R9 1 RCu 1 LiX

Alkane

For the last step to give a good yield of the alkane, the alkyl halide R′}X must be a methyl halide, a primary alkyl halide, or a secondary cycloalkyl halide. The alkyl groups of the lithium dialkylcuprate may be methyl, 1°, 2°, or 3°. Moreover, the two alkyl groups being coupled need not be different. The overall scheme for this alkane synthesis is shown here: CuI

RLi An alkyllithium

R2CuLi A lithium dialkylcuprate

R′X

R

R′ 1 RCu 1 LiX

Li Et2O

R X Any alkyl halide

R′ X A methyl, 1° alkyl, or 2° cycloalkyl halide

These are the organic starting materials. The R groups need not be different.

and R′

Consider as examples the synthesis of hexane from methyl iodide and pentyl iodide and the synthesis of nonane from butyl bromide and pentyl bromide: CH3I

Li Et2O

CH3Li

Cul

(CH3)2CuLi

CH3CH2CH2CH2CH2I

CH3

CH2CH2CH2CH2CH3 Hexane (98%)

CH3CH2CH2CH2Br

Li Et2O

(CH3CH2CH2CH2)2CuLi

CH3CH2CH2CH2Li CH3CH2CH2CH2CH2Br

CuI

CH3CH2CH2CH2

CH2CH2CH2CH2CH3

Nonane (98%)

4.7C From Carboxylic Acids 1. Decarboxylation: When sodium salts of the carboxylic acids are heated with soda lime (NaOH + CaO), alkanes are obtained. The reaction involves removal of carbon dioxide and is known as decarboxylation. The product alkane has one carbon less than the original acid. CaO CH3CH2COO−Na + + NaOH  → CH3CH2 + Na 2CO3 Sodium propanoate Propane

4.8 | How to Gain Structural Information from Molecular Formulas and the Index of Hydrogen Deficiency

2. Kolbe’s electrolytic method: This reaction is used to prepare some alkanes and alkynes by electrolysis of aqueous solution of sodium or potassium salt of suitable acids. For example, ethane is produced when an aqueous solution of potassium acetate is electrolyzed. 2CH3COOK → 3CH3COO− + 2K + Potassium acetate 2H2O  2OH− + 2H+

(ionization)

(ionization)

Oxidation half-cell reaction at the anode: 2CH3COO− − 2e − → CH3 − CH3 + 2CO2 Ethane

Reduction half-cell reaction at the cathode: 2H+ + 2e − → H2 Both K+ and H+ are present but H+ ions preferentially discharged due to their lower discharge potential. Note: Methane cannot be prepared by this method because it contains only one carbon atom.

4.8

HOW TO GAIN STRUCTURAL INFORMATION FROM MOLECULAR FORMULAS AND THE INDEX OF HYDROGEN DEFICIENCY

A chemist working with an unknown compound can obtain considerable information about its structure from the compound’s molecular formula and its index of hydrogen deficiency (IHD). ●

The index of hydrogen deficiency (IHD)* is defined as the difference in the number of pairs of hydrogen atoms between the compound under study and an acyclic alkane having the same number of carbons.

Saturated acyclic hydrocarbons have the general molecular formula CnH2n22. Each double bond or ring reduces the number of hydrogen atoms by two as compared with the formula for a saturated compound. Thus each ring or double bond provides one unit of hydrogen deficiency. For example, 1-hexene and cyclohexane have the same molecular formula (C6H12) and they are constitutional isomers.

1-Hexene (C6H12)

Cyclohexane (C6H12)

Both 1-hexene and cyclohexane (C6H12) have an index of hydrogen deficiency equal to 1 (meaning one pair of hydrogen atoms), because the corresponding acyclic alkane is hexane (C6H14). C6H14 5 formula of corresponding alkane (hexane) C6H12 5 formula of compound (1-hexene or cyclohexane) H2 5 difference 5 1 pair of hydrogen atoms Index of hydrogen deficiency 5 1 Alkynes and alkadienes (alkenes with two double bonds) have the general formula CnH2n22. Alkenynes (hydrocarbons with one double bond and one triple bond) and alkatrienes (alkenes with three double bonds) have the general formula CnH2n24, and so forth.

1,3-Butadiene IHD = 2

But-1-en-3-yne IHD = 3

1,3,5-Hexatriene IHD = 3

*Some organic chemists refer to the index of hydrogen deficiency as the “degree of unsaturation” or “the number of double-bond equivalencies.”

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The index of hydrogen deficiency is easily determined by comparing the molecular formula of a given compound with the formula for its hydrogenation product. ●

●

●

Each double bond consumes one molar equivalent of hydrogen and counts for one unit of hydrogen deficiency. Each triple bond consumes two molar equivalents of hydrogen and counts for two units of hydrogen deficiency. Rings are not affected by hydrogenation, but each ring still counts for one unit of hydrogen deficiency.

Hydrogenation, therefore, allows us to distinguish between rings and double or triple bonds. Consider again two compounds with the molecular formula C6H12: 1-hexene and cyclohexane. 1-Hexene reacts with one molar equivalent of hydrogen to yield hexane; under the same conditions cyclohexane does not react:  H2  H2

Pt 25 °C Pt 25 °C

no reaction

Or consider another example. Cyclohexene and 1,3-hexadiene have the same molecular formula (C6H10). Both compounds react with hydrogen in the presence of a catalyst, but cyclohexene, because it has a ring and only one double bond, reacts with only one molar equivalent. 1,3-Hexadiene adds two molar equivalents:



H2



2 H2

Pt 25 °C

Cyclohexene Pt 25 °C

1,3-Hexadiene

4.8A Compounds Containing Halogens, Oxygen, or Nitrogen Calculating the index of hydrogen deficiency (IHD) for compounds other than hydrocarbons is relatively easy. For compounds containing halogen atoms, we simply count the halogen atoms as though they were hydrogen atoms. Consider a compound with the formula C4H6Cl2. To calculate the IHD, we change the two chlorine atoms to hydrogen atoms, considering the formula as though it were C4H8. This formula has two hydrogen atoms fewer than the formula for a saturated alkane (C4H10), and this tells us that the compound has IHD 5 1. It could, therefore, have either one ring or one double bond. [We can tell which it has from a hydrogenation experiment: If the compound adds one molar equivalent of hydrogen (H2) on catalytic hydrogenation at room temperature, then it must have a double bond; if it does not add hydrogen, then it must have a ring.] For compounds containing oxygen, we simply ignore the oxygen atoms and calculate the IHD from the remainder of the formula. Consider as an example a compound with the formula C4H8O. For the purposes of our calculation we consider the compound to be simply C4H8 and we calculate IHD 5 1. Again, this means that the compound contains either a ring or a double bond. Some structural possibilities for this compound are shown next. Notice that the double bond may be present as a carbon–oxygen double bond:

4.9 | Chemical Reactions of Alkanes O OH

OH O

O

O

and so on

H

For compounds containing nitrogen atoms we subtract one hydrogen for each nitrogen atom, and then we ignore the nitrogen atoms. For example, we treat a compound with the formula C4H9N as though it were C4H8, and again we get IHD 5 1. Some structural possibilities are the following: NH NH2

NH2 H

H

N

N

NH

4.9

and so on

CHEMICAL REACTIONS OF ALKANES

Alkanes, as a class, are characterized by a general inertness to many chemical reagents. Carbon–carbon and carbon–hydrogen bonds are quite strong; they do not break unless alkanes are heated to very high temperatures. Because carbon and hydrogen atoms have nearly the same electronegativity, the carbon– hydrogen bonds of alkanes are only slightly polarized. As a consequence, they are generally unaffected by most bases. Molecules of alkanes have no unshared electrons to offer as sites for attack by acids. This low reactivity of alkanes toward many reagents accounts for the fact that alkanes were originally called paraffins (parum affinis, Latin: little affinity). The term paraffin, however, was probably not an appropriate one. We all know that alkanes react vigorously with oxygen when an appropriate mixture is ignited. This combustion occurs, for example, in the cylinders of automobiles, in furnaces, and, more gently, with paraffin candles. When heated, alkanes also react with chlorine and bromine, and they react explosively with fluorine. These reactions are discussed as follows.

4.9A Reactions of Alkanes with Halogens ●

Alkanes react with molecular halogens to produce alkyl halides by a substitution reaction called radical halogenation.

A general reaction showing formation of a monohaloalkane by radical halogenation is shown below. It is called radical halogenation because, as we shall see, the mechanism involves species with unpaired electrons called radicals. This reaction is not a nucleophilic substitution reaction. R ●

H 1 X2

R

X 1 HX

A halogen atom replaces one or more of the hydrogen atoms of the alkane, and the corresponding hydrogen halide is formed as a by-product.

Only fluorine, chlorine, and bromine react this way with alkanes. Iodine is essentially unreactive due to unfavorable reaction energetics. Multiple Halogen Substitution One complicating factor of alkane halogenations is that multiple substitutions almost always occur unless we use an excess of the alkane. The following example illustrates this phenomenon. If we mix an equimolar ratio of methane and chlorine (both substances are gases at room

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Chapter 4 | Alkanes and Cycloalkanes

temperature) and then either heat the mixture or irradiate it with light of the appropriate wavelength, a reaction begins to occur vigorously and ultimately produces the following mixture of products: H H

C

H H 1 Cl2

H Methane

heat or light

Chlorine

H

C

Cl Cl 1 H

H

C

Cl Cl 1 H

H

C

Cl Cl 1 Cl

Cl

C

Cl

1

H

Cl

Cl

Chloromethane Dichloromethane Trichloromethane Tetrachloromethane Hydrogen chloride (The sum of the number of moles of each chlorinated methane produced equals the number of moles of methane that reacted.)

To understand the formation of this mixture, we need to consider how the concentration of reactants and products changes as the reaction proceeds. At the outset, the only compounds that are present in the mixture are chlorine and methane, and the only reaction that can take place is one that produces chloromethane and hydrogen chloride: H H9C 9H

H Cl2



H 9C9C l

H



H9C l

H

As the reaction progresses, however, the concentration of chloromethane in the mixture increases, and a second substitution reaction begins to occur. Chloromethane reacts with chlorine to produce dichloromethane: H H9C 9Cl

Cl 

Cl2

H 9C9 C l

H



H9C l

H

The dichloromethane produced can then react to form trichloromethane, and trichloromethane, as it accumulates in the mixture, can react with chlorine to produce tetrachloromethane. Each time a substitution of — Cl for — H takes place, a molecule of H — Cl is produced. Lack of Chlorine Selectivity Chlorination of most higher alkanes gives a mixture of isomeric monochlorinated products as well as more highly halogenated compounds. ●

Chlorine is relatively unselective; it does not discriminate greatly among the different types of hydrogen atoms (primary, secondary, and tertiary) in an alkane. An example is the light-promoted chlorination of isobutane: Cl2 light

Cl



 Cl

Isobutane

●

●

Isobutyl chloride (48%)

Polychlorinated products



HCl

(23%)

tert-Butyl chloride (29%)

Because alkane chlorinations usually yield a complex mixture of products, they are not useful as synthetic methods when the goal is preparation of a specific alkyl chloride. An exception is the halogenation of an alkane (or cycloalkane) whose hydrogen atoms are all equivalent (i.e., homotopic). [Homotopic hydrogen atoms are defined as those that on replacement by some other group (e.g., chlorine) yield the same compound.]

4.9 | Chemical Reactions of Alkanes

Neopentane, for example, can form only one monohalogenation product, and the use of a large excess of neopentane minimizes polychlorination:

heat

 Cl2 Neopentane (excess)

●

Cl  HCl

or light

Neopentyl chloride

Bromine is generally less reactive toward alkanes than chlorine, and bromine is more selective in the site of attack when it does react.

We shall examine the selectivity of bromination further in Section 4.9C.

4.9B Chlorination of Methane: Mechanism of Reaction The reaction of methane with chlorine (in the gas phase) provides a good example for studying the mechanism of radical halogenation. CH4 1 Cl2

CH3Cl 1 HCl (1 CH2Cl2, CHCl3, and CCl4)

Several experimental observations help in understanding the mechanism of this reaction: 1. The reaction is promoted by heat or light. At room temperature methane and chlorine do not react at a perceptible rate as long as the mixture is kept away from light. Methane and chlorine do react, however, at room temperature if the gaseous reaction mixture is irradiated with UV light at a wavelength absorbed by Cl2, and they react in the dark if the gaseous mixture is heated to temperatures greater than 100 °C. 2. The light-promoted reaction is highly efficient. A relatively small number of light photons permits the formation of relatively large amounts of chlorinated product. A mechanism that is consistent with these observations has several steps, shown below. The first step involves the dissociation of a chlorine molecule, by heat or light, into two chlorine atoms. The second step involves hydrogen abstraction by a chlorine atom.

MECHANISM Reaction

Radical Chlorination of Methane

heat

CH4  Cl2

or light

CH3Cl  HCl

Mechanism Chain Initiation Step 1: Halogen dissociation

Cl

Cl

Under the influence of heat or light a molecule of chlorine dissociates; each atom takes one of the bonding electrons.

heat or light

Cl  Cl This step produces two highly reactive chlorine atoms.

(continues on next page)

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Chapter 4 | Alkanes and Cycloalkanes

H

Chain Propagation Step 2: Hydrogen abstraction → Cl

 H

C

H H

Cl H 

This step produces a molecule of hydrogen chloride and a methyl radical.

H

H H

C



Cl

Cl

H

H

C Cl



Cl

H

A methyl radical abstracts a chlorine atom from a chlorine molecule.

Chain Termination

This step produces a molecule of chloromethane and a chlorine atom. The chlorine atom can now cause a repetition of step 2.

H

H H

H

H

H A chlorine atom abstracts a hydrogen atom from a methane molecule.

Step 3: Halogen abstraction

C

C



Cl

H

H

C Cl H

Coupling of any two radicals depletes the supply of reactive intermediates and terminates the chain. Several pairings are possible for radical coupling termination steps (see the text).

In step 3 the highly reactive methyl radical reacts with a chlorine molecule by abstracting a chlorine atom. This results in the formation of a molecule of chloromethane (one of the ultimate products of the reaction) and a chlorine atom. The latter product is particularly significant, for the chlorine atom formed in step 3 can attack another methane molecule and cause a repetition of step 2. Then, step 3 is repeated, and so forth, for hundreds or thousands of times. (With each repetition of step 3 a molecule of chloromethane is produced.) ●

This type of sequential, stepwise mechanism, in which each step generates the reactive intermediate that causes the next cycle of the reaction to occur, is called a chain reaction.

Step 1 is called the chain-initiating step. In the chain-initiating step radicals are created. Steps 2 and 3 are called chain-propagating steps. In chain-propagating steps one radical generates another. Chain Initiation: creation of radicals Step 1

Cl2

light or heat

2 Cl

Chain Propagation: reaction and regeneration of radicals Step 2

CH4 1 Cl ?

Step 3

? CH3 1 Cl2

? CH3 1 H

Cl

CH3Cl 1 Cl ?

4.9 | Chemical Reactions of Alkanes

The chain nature of the reaction accounts for the observation that the light-promoted reaction is highly efficient. The presence of a relatively few atoms of chlorine at any given moment is all that is needed to cause the formation of many thousands of molecules of chloromethane. What causes the chain reaction to terminate? Why does one photon of light not promote the chlorination of all of the methane molecules present? We know that this does not happen because we find that, at low temperatures, continuous irradiation is required or the reaction slows and stops. The answer to these questions is the existence of chain-terminating steps: steps that happen infrequently but occur often enough to use up one or both of the reactive intermediates. The continuous replacement of intermediates used up by chain-terminating steps requires continuous irradiation. Plausible chain-terminating steps are as follows. Chain Termination: consumption of radicals (e.g., by coupling) H

H H

C

Cl



H

C Cl

H

H H H

H H H

C

C



H

H

C C

H H Cl

(Ethane by-product)

H

H H Cl



Cl Cl

Our radical mechanism also explains how the reaction of methane with chlorine produces the more highly halogenated products, CH2Cl2, CHCl3, and CCl4 (as well as additional HCl). As the reaction progresses, chloromethane (CH3Cl) accumulates in the mixture and its hydrogen atoms, too, are susceptible to abstraction by chlorine. Thus chloromethyl radicals are produced that lead to dichloromethane (CH2Cl2). Side Reactions: multihalogenated by-product formation Cl Step 2

Cl

 H C

Cl H

H Cl

Cl

Cl H

 Cl Cl

C H

H

H

H Step 3

C



H

C Cl  Cl H

(Dichloromethane)

Then step 2 is repeated, then step 3 is repeated, and so on. Each repetition of step 2 yields a molecule of HCl, and each repetition of step 3 yields a molecule of CH2Cl2.

4.9C Halogenation of Higher Alkanes Higher alkanes react with halogens by the same kind of chain mechanism as those that we have just seen. Ethane, for example, reacts with chlorine to produce chloroethane (ethyl chloride). The mechanism is as follows:

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Chapter 4 | Alkanes and Cycloalkanes

Radical Halogenation of Ethane

MECHANISM Chain Initiation

Cl2

Step 1

light or heat

2 Cl

Chain Propagation Step 2

CH3CH2 H  Cl

Step 3

CH3CH2  Cl Cl

CH3CH2  H Cl CH3CH2 Cl  Cl

Chain propagation continues with steps 2, 3, 2, 3, and so on.

Chain Termination

CH3CH2 CH3CH2



Cl

CH3CH2 Cl

 CH2CH3

CH3CH2 CH2CH3

Cl

Cl Cl



Cl

Chlorination is not Selective Chlorination of most alkanes whose molecules contain more than two carbon atoms gives a mixture of isomeric monochloro products (as well as more highly chlorinated compounds). Several examples follow. The percentages given are based on the total amount of monochloro products formed in each reaction. The ratios of products that we obtain from chlorination reactions of higher alkanes are not identical to what we would expect if all the hydrogen atoms of the alkane were equally reactive. We find that there is a correlation between reactivity of different hydrogen atoms and the type of hydrogen atom (1°, 2°, or 3°) being replaced. The tertiary hydrogen atoms of an alkane are most reactive, secondary hydrogen atoms are next most reactive, and primary hydrogen atoms are the least reactive. These examples show the nonselectivity of chlorination. Cl2 light, 25 °C

Propane



1-Chloropropane (45%) Cl2 light 25 °C

2-Methylpropane

Cl

Cl

Cl

2-Chloropropane (55%)

 Cl

1-Chloro-2-methyl propane (63%)

2-Chloro-2-methyl propane (37%)

4.9 | Chemical Reactions of Alkanes

Cl2



Cl

300 °C

2-Methylbutane

Cl

1-Chloro-2-methylbutane (30%)



2-Chloro-2-methyl butane (22%)



Cl

Cl 2-Chloro-3-methylbutane (33%)

1-Chloro-3-methyl butane (15%)

We can account for the relative reactivities of the primary, secondary, and tertiary hydrogen atoms in a chlorination reaction on the basis of the homolytic bond dissociation energies. Of the three types, breaking a tertiary C— H bond requires the least energy, and breaking a primary C— H bond requires the most. Since the step in which the C— H bond is broken (i.e., the hydrogen atom–abstraction step) determines the location or orientation of the chlorination, we would expect the Eact for abstracting a tertiary hydrogen atom to be least and the Eact for abstracting a primary hydrogen atom to be greatest. Thus tertiary hydrogen atoms should be most reactive, secondary hydrogen atoms should be the next most reactive, and primary hydrogen atoms should be the least reactive. The differences in the rates with which primary, secondary, and tertiary hydrogen atoms are replaced by chlorine are not large, however. ●

Chlorine does not discriminate among the different types of hydrogen atoms in a way that makes chlorination of higher alkanes a generally useful laboratory synthesis.

Selectivity of Bromine Bromine shows a much greater ability to discriminate among the different types of hydrogen atoms. ●

●

Bromine is less reactive than chlorine toward alkanes in general but bromine is more selective in the site of attack. Bromination is selective for substitution where the most stable radical intermediate can be formed.

The reaction of 2-methylpropane and bromine, for example, gives almost exclusive replacement of the tertiary hydrogen atom: Br2 hn, 127 °C

Br

 Br (99%)

(trace)

A very different result is obtained when 2-methylpropane reacts with chlorine: Cl2 h, 25 °C

Cl

 Cl (37%)

(63%)

Fluorine, being much more reactive than chlorine, is even less selective than chlorine. Because the energy of activation for the abstraction of any type of hydrogen by a fluorine atom is low, there is very little difference in the rate at which a 1°, 2°, or 3° hydrogen reacts with fluorine. Reactions of alkanes with fluorine give (almost) the distribution of products that we would expect if all of the hydrogens of the alkane were equally reactive.

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Chapter 4 | Alkanes and Cycloalkanes

4.9D Other Substitution Reactions In other substitution reactions, one or more of hydrogen atoms of alkanes may be replaced by nitro or sulphonic acid groups. Lower alkanes do not exhibit these reactions. Sulphonation The replacement of hydrogen atom in an alkane by sulphonic acid (−SO3H) group is known as sulphonation. Higher alkanes, starting from hexane, undergo sulphonation on heating with oleum (H2SO4 + SO3) at 400 °C. The reaction proceeds through free radical mechanism. SO

3 C7H15 − H + HOSO3H  → C7H15 − SO3H + H2O

Nitration In this substitution reaction, the hydrogen atom of alkane is replaced by nitro (−NO2) group. Alkanes undergo nitration in vapor phase when treated with fuming nitric acid at a temperature of 450−500 °C. The reaction proceeds through free radical mechanism. ∆ C2H5 − H + HO − NO2  → C2H5 − NO2 + H2O

4.9E Combustion of Alkanes When alkanes react with oxygen (e.g., in oil and gas furnaces and in internal combustion engines) a complex series of reactions takes place, ultimately converting the alkane to carbon dioxide and water. Although our understanding of the detailed mechanism of combustion is incomplete, we do know that the important reactions occur by radical chain mechanisms with chain-initiating and chain-propagating steps such as the following reactions: RH  O2 R

OO

R 

OOH

R  O2

R

OO

 R

R

OOH  R

H

Initiating Propagating

One product of the second chain-propagating step is R— OOH, called an alkyl hydroperoxide. The oxygen–oxygen bond of an alkyl hydroperoxide is quite weak, and it can break and produce radicals that can initiate other chains: RO

OH

RO ? 1 ? OH

Since the alkanes can produce a large amount of heat, these are used as fuels. The following is the general combustion equation for any alkane:  3n + 1 CnH2n + 2 +   O2 → nCO2 + (n + 1)H2O  2 

Consider, as an example, the combustion of butane and isobutane: CH3CH2CH2CH3 1 612 O2 (C4H10)

4 CO2 1 5 H2O

ΔH8 5 22877 kJ mol21

1 612 O2

4 CO2 1 5 H2O

ΔH8 5 22868 kJ mol21

CH3CHCH3 CH3 (C4H10)

Incomplete combustion (in the presence of insufficient air or oxygen) of alkanes leads to the formation of carbon black which is subsequently used in the manufacture of inks and pigments. Incomplete CH4 (g) + O2 (g)  combustion → C(s) + 2H2O(l)

Combustion of alkanes in limited supply of air or oxygen also produces carbon monoxide and water 2CH4 + 3O2 → 2CO + 4H2O

4.9 | Chemical Reactions of Alkanes

When methane is heated with steam in presence of nickel as a catalyst, carbon monoxide and hydrogen are obtained. The reaction is used for commercial product of hydrogen gas. Heat CH4 + H2O  → CO + 3H2

4.9F Controlled (Catalytic) Oxidation Alkanes when heated with limited supply of air or oxygen in presence of different catalysts leads to the formation of different oxidation products. 1. Formation of alcohols: When a mixture of methanol and oxygen are heated (573 K) at high pressure (100 atm) and passed through copper tube, methanol is obtained. 2CH4 + O2 → 2CH3OH

2. Formation of aldehydes: When a mixture of methanol and oxygen under pressure is passed over heated molybdenum oxide, methanol is obtained. CH4 + O2 → HCHO + 2H2O

3. Formation of carboxylic acids: Higher alkanes on oxidation with silver oxide give carboxylic acids. CH3 − CH3 + O2 → CH3COOH + 2H2O

4.9G Isomerization Straight-chain alkanes when heated with anhydrous aluminium chloride and hydrogen chloride gas at high temperature and pressure undergo isomerization to yield branched alkanes. For example, CH3CH2CH2CH2CH3

anhyd. AlCl3/HCl Heat, pressure

CH3CH2CHCH2 CH3

CH3(CH2)4CH3

anhyd. AlCl3/HCl

CH3CH

n-Hexane

(CH2)2

CH3 1 CH3CH2

CH3

CH

CH2

CH3

CH3

2-Methylpentane

3-Methylpentane

The isomerization of n-alkanes may lead to formation of a number of branched products, but only some are formed in substantial amounts and only these major products are normally indicated in the reaction.

4.9H Aromatization Straight-chain alkanes containing six to eight carbon atoms can be converted into corresponding aromatic compounds when heated to 500 °C at 10–20 atm pressure in the presence of oxides of vanadium, molybdenum or chromium supported over alumina. The reaction is known as aromatization. For example, aromatization of n-hexane yields benzene. CH3 CH2 CH3

Cr2O3, V2O5, Mo2O3/Al2O3

CH2 CH2

5008 C, 10−20 atm

CH2 n-Hexane

23H2

Cyclohexane

Benzene

Aromatization of n-heptane yields toluene. CH3 CH3

CH2 CH2 CH3

Cr2O3, V2O5, Mo2O3/Al2O3

CH2 CH2

5008 C, 10−20 atm

CH2

CH3 23H2

Methylcyclohexane

Toluene

205

206

Chapter 4 | Alkanes and Cycloalkanes

4.9I Pyrolysis Pyrolysis or cracking is a process by which the carbon–carbon bonds and carbon–hydrogen bonds of higher alkanes are broken by action of heat to yield lower alkanes, alkenes and some hydrogen. The reaction proceeds by a free radical mechanism The reaction can take place either on heating to high temperature (thermal cracking) or in the presence of a catalysts (catalytic cracking). For example, Heat C6H14  → C6H12 + H2 + C4H8 + C2H6 + C3H6 + C2H4 + CH4 Pressure Hexane

Hexene

Butene

Ethane

Propene

Ethene

Methane

Heat C16H34  → C8H18 + C8H16 Catalyst Alkane

Alkane

Alkene

The lower alkenes obtained by cracking processes, when separated and purified, are an important raw material for synthesis of other aliphatic compounds. Two modifications to the cracking process are hydrocracking and steam cracking. In the former, smaller hydrocarbons are obtained when the reaction is carried out in the presence of a catalyst and hydrogen, under high pressure but much lower temperature. In the latter, hydrocarbons are diluted with steam and heated for a fraction of second to very high temperatures. The mixture is then quenched by cooling and lower alkenes are obtained. The most important use of pyrolysis or cracking is in conversion of crude oil into fractions suitable for use as fuel. The preparation of petrol gas or oil gas from petrol and kerosene, respectively, is achieved by pyrolysis in presence of a catalyst. Dodecane, a constituent of kerosene, on heating in presence of platinum, palladium or nickel as catalyst, gives a mixture of heptanes and pentane along with other products. Catalyst C12H26  → C7H16 + C5H10 + Other products

4.10 CONFORMATIONS Two groups bonded by only a single bond can undergo rotation about that bond with respect to each other. ● ● ●

The temporary molecular shapes that result from such a rotation are called conformations of the molecule. Each possible structure is called a conformer. An analysis of the energy changes that occur as a molecule undergoes rotations about single bonds is called a conformational analysis.

4.10A Newman Projections and How to Draw Them When we do conformational analysis, we will find that certain types of structural formulas are especially convenient to use. One of these types is called a Newman projection formula and another type is a sawhorse formula. Sawhorse formulas are much like dash–wedge three-dimensional formulas we have used so far. In conformational analyses, we will make substantial use of Newman projections.

Newman projection formula

Sawhorse formula

To write a Newman projection formula: ●

We imagine ourselves taking a view from one atom (usually a carbon) directly along a selected bond axis to the next atom (also usually a carbon atom).

207

4.10 | Conformations

●

The front carbon and its other bonds are represented as

●

The back carbon and its bonds are represented as

.

.

In Figs. 4.3a, b we show ball-and-stick models and a Newman projection formula for the staggered conformation of ethane. The staggered conformation of a molecule is that conformation where the dihedral angle between the bonds at each of the carbon–carbon bonds is 180° and where atoms or groups bonded to carbons at each end of a carbon–carbon bond are as far apart as possible. The 180° dihedral angle in the staggered conformation of ethane is indicated in Fig. 4.3b. The dihedral angle (f) between these hydrogens is 1808. H H H

f = 608

f

H

H

H

H H

H

H

(a)

H f

H

(b)

(c)

Figure 4.3 (a) The staggered conformation of ethane. (b) The Newman projection formula for the staggered conformation. (c) The dihedral angle between these hydrogen atoms is 60°. The eclipsed conformation of ethane is shown in Fig. 4.4 using ball-and-stick models and a Newman projection. In an eclipsed conformation the atoms bonded to carbons at each end of a carbon–carbon bond are directly opposed to one another. The dihedral angle between them is 0°.

4.10B Conformational Analysis of Ethane Now let us consider a conformational analysis of ethane. Clearly, infinitesimally small changes in the dihedral angle between C — H bonds at each end of ethane could lead to an infinite number of conformations, including, of course, the staggered and eclipsed conformations. These different conformations are not all of equal stability, however, and it is known that the staggered conformation of ethane is the most stable conformation (i.e., it is the conformation of lowest potential energy). The explanation for greater stability of the staggered conformation relates mainly to steric replusion between bonding pairs of electrons. In the eclipsed conformation electron clouds from the C — H bonds are closer and repel one another. The staggered conformation allows the maximum possible separation of the electron pairs in the C — H bonds. In addition, there is a phenomenon called hyperconjugation that involves favorable overlap between filled and unfilled sigma orbitals in the staggered conformation. Hyperconjugation helps to stabilize the staggered conformation. The more important factor, however, is the minimization of steric

The dihedral angle () between these hydrogens is 0.  H H H H

(a)

(b)

H H

Figure 4.4 (a) The eclipsed conformation of ethane. (b) The Newman projection formula for the eclipsed conformation.

Chapter 4 | Alkanes and Cycloalkanes

repulsions in the staggered form. In later chapters we shall explain hyperconjugation further and the role it plays in relative stability of reactive species called carbocations. ● The energy difference between the conformations of ethane can be represented graphically in a potential energy diagram, as shown in Figure 4.5.

H H H H

Potential energy

208

Eclipsed

H H

12 kJ mol–1

H

H

H

H

H

H

H

H

H

H

H Staggered

H Staggered Rotation

Figure 4.5 Potential energy changes that accompany rotation of groups about the carbon–carbon bond of ethane.

In ethane the energy difference between the staggered and eclipsed conformations is about 12 kJ mol21. This small barrier to rotation is called the torsional barrier of the single bond. Because of this barrier, some molecules will wag back and forth with their atoms in staggered or nearly staggered conformations, while others with slightly more energy will rotate through an eclipsed conformation to another staggered conformation. At any given moment, unless the temperature is extremely low (2250 °C), most ethane molecules will have enough energy to undergo bond rotation from one conformation to another. What does all this mean about ethane? We can answer this question in two different ways. If we consider a single molecule of ethane, we can say, for example, that it will spend most of its time in the lowest energy, staggered conformation, or in a conformation very close to being staggered. Many times every second, however, it will acquire enough energy through collisions with other molecules to surmount the torsional barrier and it will rotate through an eclipsed conformation. If we speak in terms of a large number of ethane molecules (a more realistic situation), we can say that at any given moment most of the molecules will be in staggered or nearly staggered conformations.

4.10C Conformational Analysis of Butane Now let us consider rotation about the C2—C3 bond of butane. The barriers to rotation about the C2—C3 bond in butane are larger than for rotation about the C—C bond in ethane, but still not large enough to prevent the rotations that lead to all possible butane conformers. ●

The factors involved in barriers to bond rotation are together called torsional strain and include the repulsive interactions called steric hindrance between electron clouds of the bonded groups.

In butane, torsional strain results from steric hindrance between the terminal methyl groups and hydrogen atoms at C-2 and C-3 and from steric hindrance directly between the two methyl groups. These interactions result in six important conformers of butane, shown as I–VI below.

Butane

H CH3

H H H3C

H CH3

CH3 H

H

H H H3C

CH3

I An anti conformation

H H

CH3

H

H

H H

H H

II An eclipsed conformation

H3C CH3

H3C CH3

CH3

H3C

H H

H H

H

CH3

H

H

H CH3 H H

H

H CH3

VI An eclipsed conformation

V A gauche conformation

IV An eclipsed conformation

III A gauche conformation

H

III II | Conformations gauche An eclipsed 4.10 A conformation conformation

I An anti conformation

H

H H

H CH3

CH3

The anti conformation not H (I) doesCH 3 have torsional strain from steric hindrance because the groups are staggered and the methyl groups are far apart. The anti conformation is the most stable. The methyl H H H H groups conformations III enough to each other that the dispersion forces H H CHclose H in the gauche H H and V are 3 between them are repulsive; the electron clouds of the two groups are so close that they repel each other. H This repulsion causes the gauche conformations to have approximately 3.8 kJ mol21 more energy than the VI V IV anti conformation. An eclipsed A gauche An eclipsed The eclipsed conformations (II, IV, and VI) represent energy maxima in the potential energy diagram conformation conformation conformation (Fig. 4.6). Eclipsed conformations II and VI have repulsive dispersion forces arising from the eclipsed methyl groups and hydrogen atoms. Eclipsed conformation IV has the greatest energy of all because of the added large repulsive dispersion forces between the eclipsed methyl groups as compared to II and VI.

H3C CH3 H CH 3 H H3C

H H

H CH 3

H H Eclipsed IV

H H

H H

Potential energy

Eclipsed II

H CH3 Eclipsed VI

19 kJ mol–1 16 kJ mol

–1

16 kJ mol–1 3.8 kJ

CH3 H H

3.8 kJ

CH3 H

H3C

H

H

CH3 Anti I

0°

mol–1

60°

mol–1

CH3 H

H

H

H

H

H

Gauche III

Gauche V

120°

180°

240°

CH3 CH3

H

H

H

H H CH3 Anti I

300°

360°

Rotation

Figure 4.6 Energy changes that arise from rotation about the C2 — C3 bond of butane. Although the barriers to rotation in a butane molecule are larger than those of an ethane molecule they are still far too small to permit isolation of the gauche and anti conformations at normal temperatures. Only at extremely low temperatures would the molecules have insufficient energies to surmount these barriers. We have learnt that dispersion forces can be attractive. Here, however, we find that they can also be repulsive, leading to steric hindrance. Whether dispersion interactions lead to attraction or to repulsion depends on the distance that separates the two groups. As two nonpolar groups are brought closer and closer together, the first effect is one in which a momentarily unsymmetrical distribution of electrons in one group induces an opposite polarity in the other. The opposite charges induced in those portions of the two groups that are in closest proximity lead to attraction between them. This attraction increases to a

209

210

Chapter 4 | Alkanes and Cycloalkanes

maximum as the internuclear distance of the two groups decreases. The internuclear distance at which the attractive force is at a maximum is equal to the sum of what are called the van der Waals radii of the two groups. The van der Waals radius of a group is, in effect, a measure of its size. If the two groups are brought still closer—closer than the sum of their van der Waals radii—their electron clouds begin to penetrate each other, and strong electron–electron repulsion occurs.

4.10D Stereoisomers and Conformational Stereoisomers Gauche conformers III and V of butane are examples of stereoisomers. ●

●

Stereoisomers have the same molecular formula and connectivity but different arrangements of atoms in three-dimensional space. Conformational stereoisomers are related to one another by bond rotations.

Conformational analysis is but one of the ways in which we will consider the three-dimensional shapes and stereochemistry of molecules. We shall see that there are other types of stereoisomers that cannot be interconverted simply by rotations about single bonds. Among these are cis–trans cycloalkane isomers and others that we shall consider in Chapter 5.

4.11 THE RELATIVE STABILITIES OF CYCLOALKANES: RING STRAIN Cycloalkanes do not all have the same relative stability. Experiments have shown that cyclohexane is the most stable cycloalkane and that, in comparison, cyclopropane and cyclobutane are much less stable. This difference in relative stability is due to ring strain, which comprises angle strain and torsional strain. Angle strain is the result of deviation from ideal bond angles caused by inherent structural constraints (such as ring size). ● Torsional strain is the result of repulsive dispersion forces that cannot be relieved due to restricted conformational mobility. ●

4.11A Cyclopropane The carbon atoms of alkanes are sp3 hybridized. The normal tetrahedral bond angle of an sp3-hybridized atom is 109.5°. In cyclopropane (a molecule with the shape of a regular triangle), the internal angles must be 60° and therefore they must depart from this ideal value by a very large amount—by 49.5°: HH C H H

C

60°

C

H H

Angle strain exists in a cyclopropane ring because the sp3 orbitals comprising the carbon–carbon s bonds cannot overlap as effectively (Fig. 4.7a) as they do in alkanes (where perfect end-on overlap is possible). The carbon–carbon bonds of cyclopropane are often described as being “bent.” Orbital overlap is less effective. (The orbitals used for these bonds are not purely sp3; they contain more p character.) The carbon–carbon bonds of cyclopropane are weaker, and as a result the molecule has greater potential energy. While angle strain accounts for most of the ring strain in cyclopropane, it does not account for it all. Because the ring is (of necessity) planar, the C } H bonds of the ring are all eclipsed (Figs. 4.7b, c), and the molecule has torsional strain from repulsive dispersion forces as well.

4.12 | Conformations of Cyclohexane: The Chair and the Boat

211

H H C

H H

H

C

C

H (a)

H

C C

H

H

H

1.089 Å (b)

1.510 Å H C 115°

H H H H

H

CH2 (c)

(d )

Figure 4.7 (a) Orbital overlap in the carbon–carbon bonds of cyclopropane cannot occur perfectly end-on. This leads to weaker “bent” bonds and to angle strain. (b) Bond distances and angles in cyclopropane. (c) A Newman projection formula as viewed along one carbon–carbon bond shows the eclipsed hydrogens. (Viewing along either of the other two bonds would show the same picture.) (d) Ball-and-stick model of cyclopropane.

4.12 CONFORMATIONS OF CYCLOHEXANE: THE CHAIR AND THE BOAT Cyclohexane is more stable than the other cycloalkanes we have discussed, and it has several conformations that are important for us to consider. ● ●

The most stable conformation of cyclohexane is the chair conformation. There is no angle or torsional strain in the chair form of cyclohexane.

In a chair conformation (Fig. 4.8), all of the carbon–carbon bond angles are 109.5°, and are thereby free of angle strain. The chair conformation is free of torsional strain, as well. When viewed along any carbon–carbon bond (viewing the structure from an end, Fig. 4.9), the bonds are seen to be perfectly staggered. Moreover, the hydrogen atoms at opposite corners of the cyclohexane ring are maximally separated. By partial rotations about the carbon–carbon single bonds of the ring, the chair conformation can assume another shape called the boat conformation (Fig. 4.10). ● The boat conformation has no angle strain, but it does have torsional strain. ●

When a model of the boat conformation is viewed down carbon–carbon bond axes along either side (Fig. 4.11a), the C } H bonds at those carbon atoms are found to be eclipsed, causing torsional strain. Additionally, two of the hydrogen atoms on C1 and C4 are close enough to each other to cause van der Waals repulsion (Fig. 4.11b). This latter effect has been called the “flagpole” interaction of the boat

H H H (a)

(b)

H

H

H

H

H

H H H

H

(c)

Figure 4.8 Representations of the chair conformation of cyclohexane: (a) tube format; (b) balland-stick format; (c) line drawing; (d ) space-filling model of cyclohexane. Notice that there are two orientations for the hydrogen substituents—those that project obviously up or down and those that lie around the perimeter of the ring in more subtle up or down orientations.

(d)

212

Chapter 4 | Alkanes and Cycloalkanes

conformation. Torsional strain and flagpole interactions cause the boat conformation to have considerably higher energy than the chair conformation. H H

H

H

4

H

CH2 6

5

2

CH2

H

H

3

H

H

1

H

H

H

(a)

(b)

Figure 4.9 (a) A Newman projection of the chair conformation of cyclohexane. (Comparisons with an actual molecular model will make this formulation clearer and will show that similar staggered arrangements are seen when other carbon–carbon bonds are chosen for sighting.) (b) Illustration of large separation between hydrogen atoms at opposite corners of the ring (designated C1 and C4) when the ring is in the chair conformation.

H H H H

H

H

H

H

H H

H H

H

HH H H

H

H

H H

H H

H

(a)

(b)

Figure 4.10 (a) The boat conformation of cyclohexane is formed by “flipping” one end of the chair form up (or down). This flip requires only rotations about carbon–carbon single bonds. (b) Ball-and-stick model of the boat conformation. Although it is more stable, the chair conformation is much more rigid than the boat conformation. The boat conformation is quite flexible. By flexing to a new form—the twist conformation (Fig. 4.12)—the boat conformation can relieve some of its torsional strain and, at the same time, reduce the flagpole interactions. ●

The twist boat conformation of cyclohexane has a lower energy than the pure boat conformation, but is not as stable as the chair conformation. HH

HH

CH2 CH2

H H

H H

(a)

H H

H

1

4

H

(a)

(b)

Figure 4.11 (a) Illustration of the eclipsed conformation of the boat conformation of cyclohexane. (b) Flagpole interaction of the C1 and C4 hydrogen atoms of the boat conformation. The C1–C4 flagpole interaction is also readily apparent in Fig. 4.13c.

(b)

Figure 4.12 (a) Tube model and (b) line drawing of the twist conformation of cyclohexane.

4.13 | Disubstituted Cycloalkanes: Cis–Trans Isomerism

The stability gained by flexing is insufficient, however, to cause the twist conformation to be more stable than the chair conformation. The chair conformation is estimated to be lower in energy than the twist conformation by approximately 23 kJ mol21. The energy barriers between the chair, boat, and twist conformations of cyclohexane are low enough (Fig. 4.13) to make separation of the conformers impossible at room temperature. At room temperature the thermal energies of the molecules are great enough to cause approximately 1 million interconversions to occur each second. ●

Because of the greater stability of the chair, more than 99% of the molecules are estimated to be in a chair conformation at any given moment. Halfchair

Halfchair

Relative energy, kJ mol–1

40

Boat

45.2 kJ

20

mol–1

Twist boat

Twist boat

30 kJ mol–1 23 kJ mol–1

Chair

Chair

0

Chair

Halfchair

Twist boat

Boat

Twist boat

Halfchair

Chair

Figure 4.13 The relative energies of the various conformations of cyclohexane. The positions of maximum energy are conformations called half-chair conformations, in which the carbon atoms of one end of the ring have become coplanar.

4.12A Conformations of Higher Cycloalkanes Cycloheptane, cyclooctane, and cyclononane and other higher cycloalkanes also exist in nonplanar conformations. The small instabilities of these higher cycloalkanes appear to be caused primarily by torsional strain and repulsive dispersion forces between hydrogen atoms across rings, called transannular strain. The nonplanar conformations of these rings, however, are essentially free of angle strain.

4.13 DISUBSTITUTED CYCLOALKANES: CIS–TRANS ISOMERISM The presence of two substituents on different carbons of a cycloalkane allows for the possibility of cis–trans isomerism similar to the kind we saw for alkenes in Chapter 1 Section 1.7B. These cis–trans isomers are also stereoisomers because they differ from each other only in the arrangement of their atoms in space. Consider 1,2-dimethylcyclopropane (Fig. 4.14) as an example.

213

214

Chapter 4 | Alkanes and Cycloalkanes

The planarity of the cyclopropane ring makes the cis–trans isomerism obvious. In the first structure the methyl groups are on the same side of the ring; therefore, they are cis. In the second structure, they are on opposite sides of the ring; they are trans. Cis and trans isomers such as these cannot be interconverted without breaking carbon–carbon bonds. They will have different physical properties (boiling points, melting points, and so on). As a result, they can be separated, placed in separate bottles, and kept indefinitely. H

H

H

CH3

CH3

CH3

CH3

H

cis-1,2-Dimethylcyclopropane

trans-1,2-Dimethylcyclopropane

Figure 4.14 The cis- and trans-1,2-dimethylcyclopropane isomers.

SOLVED EXAMPLES 1. Give systematic IUPAC names for each of the following: (a)

(b)

2. Write the structure and give the IUPAC systematic name of an alkane or cycloalkane with the formulas (a) C8H18 that has only primary hydrogen atoms, (b) C6H12 that has only secondary hydrogen atoms, (c) C6H12 that has only primary and secondary hydrogen atoms, and (d) C8H14 that has 12 secondary and 2 tertiary hydrogen atoms. Solution H3C

(a) (c)

CH3

C H3C

CH3 C

CH3

CH3

CH3

(c)

1,1Dimethylcyclobutane

2,2,3,3Tetramethylbutane

(d)

(b)

(d) Cyclohexane

Solution (a) (b) (c) (d)

4-Ethyl-3,5,7,8-tetramethyldecane 2,5-Dimethyl-3-iso-propylhexane 4-Ethyl-2,3-dimethylheptane 4-(1-Methylethyl)octane

CH3

Bicyclo[2.2.2]octane

3. (a) Why are alkanes inert? (b) Why the (C } C) bond rather than (C } H) bond breaks when alkanes are pyrolyzed? (c) Why the combustion of alkanes does not occur at moderate temperature, although it is an exothermic process?

Solved Examples (b) These three alkanes are constitutional isomers with the molecular formula C8H18. Their relative boiling points depend on the degree of branching. 2,2,4-Trimethylpentane, the most highly branched isomer, has the smallest surface area and the lowest boiling point. Octane, the unbranched isomer, has the largest surface area and the highest boiling point.

Solution (a) Alkanes do not have reactive sites and therefore are inert. A reactive site in a molecule must have: (i) one or more unshared pair of electrons. (ii) a polar bond. (iii) an electron deficient atom. (iv) an atom with an expandable octet. However, alkanes have none of them. (b) The (C } C) bond has a lower bond energy (∆H + 347 kJ mol−1) than (C } H) bond (∆H + 415 kJ mol−1). (c) Since the energy of activation (∆Hact) is very high, the reaction is very slow at room temperature.

the greater the branching, the lower is the surface area, causing a decrease in the dispersion forces and a decrease in boiling point

4. Write the structures of three pentenes that would all yield pentane on hydrogenation.

2,2,4-Trimethylpentane (bp 99 8C)

2-Methylheptane (bp 118 8C)

Solution Octane (bp 125 8C) 1-Pentene

cis-2-Pentene

trans-2-Pentene

5. Specify the missing compounds and/or reagents in each of the following syntheses:

7. How many different staggered conformations are there for 2-methylpropane? How many different eclipsed conformations are there? Solution

? (a) trans-5-Methyl-2-hexene  → 2-Methylhexane

2-Methylpropane has only one staggered conformation and one eclipsed conformation, as shown in the Newman projections.

?

(b) Solution

H

(a) H2; Pd, Pt or Ni catalyst, pressure (b) H2; Pd, Pt or Ni catalyst, pressure 6. Arrange the alkanes in each set in order of increasing boiling point: (a) Butane, decane, and hexane (b) 2-Methylheptane, octane, and 2,2,4-trimethylpentane Solution (a) All of the compounds are unbranched alkanes. As the number of carbon atoms in the chain increases, the dispersion forces among molecules increase, and the boiling points increase. Decane has the highest boiling point, butane the lowest. Butane (bp 20.5 °C)

Decane (bp 174 °C)

Hexane (bp 69 °C) in unbranched hydrocarbons, the longer the chain length, the higher is the surface area. This results in an increase in the dispersion forces and an increase in boiling point

CH3

H

H CH3

H CH3

H

H Staggered

H Eclipsed

H CH3

8. Draw three staggered conformations of 1-fluoro2-methylbutane as sighted down the C1–C2 bond. Determine the most stable and least stable of these conformations. Solution H

F

H

CH2CH3 H

CH3

H F

H

CH2CH3

H CH3 Most Stable

H

H

CH2CH3

H

F CH3 Least Stable

The conformer on the right is the least stable because the fluorine, methyl and ethyl groups are bunched together resulting in the largest steric strain. The conformer in the middle has a methyl and fluorine close together, but that interaction has less strain than having an ethyl and fluorine close together as found in the conformer on the left.

215

216

Chapter 4 | Alkanes and Cycloalkanes 9. What generalizations can you make about the densities of alkanes relative to that of water?

Solution (a)

Solution

CH3CH3

In general, the density of an alkane is less than that of water. However, as the molecular weight of an alkane increases, its density also tends to increase.

Ethane

10. Consider the compounds given below: (I)

(II)

Cl2

Which of the following statements is not correct about these compounds? (1) Heat of combustion order is I > II > III (2) Order of stability of compounds is III > II > I (3) The average combustion energy per CH2 unit is equal (4) The number of carbon atoms in each compound is the same.

Cl2

CH4

(b)

CH3CH2CH2CH3 Butane

CH3Cl

hν

Na, ether

CH3CH3

Wurtz reaction

Ethane

13. In alkanes, except methane, there is chain termination reaction called disproportionation. What are products when ethyl free radical is involved? Why is this reaction not possible with methyl free radical? Solution Disproportionation is hydrogenation, that is, reduction of one species at the cost of the same which is oxidized.

Solution

⋅C2H5 + ⋅ C2H5 → C2H6 + C2H4

(3) All the three compounds contain the same number of carbon atoms, but compound I contains a three-membered ring, which contains maximum angle strain. All carbon atoms in I are sp3 hybridized, it implies C } C } C bond angle should be 109°28´ but geometry does not allow this angle. According to geometry, this angle should be 60°, whereas it deviates to 49°28´ in reality. It implies that three-membered ring is most unstable and five-membered ring is the most stable out of given options. The heat of combustion is inversely proportional to the stability of the compound, if the number of carbon atoms is the same in each case. Therefore, compound I being most unstable has the maximum heat of combustion per mole and per unit of CH2. 11. Which branched chain isomer of the hydrocarbon with molecular mass 72 u gives only one isomer of monosubstituted alky halide? (1) Tertiary butyl chloride (3) Isohexane (2) Neopentane (4) Neohexane

⋅CH3 + ⋅ CH3 → CH4 + CH2

Since CH2 (methylene) does not exist. Hence, this reaction is not possible with .CH3. 14. How many chiral compounds are possible on monochlorination of 2-methylbutane? (1) 8 (3) 4 (2) 2 (4) 6 Solution (2) The reaction involved is 1

CH3

2

CH

3

4

CH2

CH3

CH3 2-Methylbutane Cl2/hν or heat

Cl CH3

CH

CH2

CH2Cl

1

CH3

CH3

Solution 1 H3C

CH3 C

CH3 CH3

Cl2 /hv

CH3 Neopentane All hydrogens are equivalent Molecular mass 5 C5H12 5 12 3 5 1 12 5 72 u

CH3

C

*

CH

CH2

CH2Cl CH2Cl

CH3 Monosubstituted alkyl halide

12. How will you convert? (a) Ethane to butane (b) Methane to ethane

C

CH2

CH3

CH3

(Achiral)

(2) The reaction involved is CH3

Wurtz reaction

Ethyl chloride

Methane

(III)

2Na, ether

CH3CH2Cl

hν

(Chiral)

(Achiral)

CH3

1

CH3

CH

*

CH

CH3

CH3 Cl (Chiral)

Out of all the four isomers formed only two compounds are optically active. 15. What is the total number of products obtained on monochlorination of 2,2,5-trimethyl hexane in the presence of light and chlorine? (1) 5 (3) 8 (2) 7 (4) 10

Solved Previous Years’ NEET Questions Solution (3) In this chlorination, there are five different possibilities to produce five constitutional isomers, out of which three contain chiral carbon centers. Since no stereospecific reagent is used in this reaction, optical isomers are also produced and racemic mixture is produced in the case of enantiomers. Hence, a total of seven chlorination products are obtained. CH3

CH3

CH2Cl

CH3CCH2CH2CHCH3 1 Cl2 CH3

D

CH3

CH3

CH3CCH2CH2CHCH3 1 CH3C CH3

CH3 * CHCH2CHCH3 1

CH3 Cl 1 its enantiomer

CH3

CH3 CH3 CH3 CH3 CH3 * CH3CCH2CHCHCH3 1 CH3CCH2CH2CCH3 1 CH3CCH2CH2CHCH2Cl 1 HCl * CH3 Cl CH3 Cl CH3 1 its enantiomer 1 its enantiomer

SOLVED PREVIOUS YEARS’ NEET QUESTIONS 1. Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by (1) cracking. (2) distillation under reduced pressure. (3) hydrolysis. (4) oxidation. (AIPMT PRE 2010) Solution (1) In case of cracking higher hydrocarbons are broken into smaller alkanes. A process used in the petroleum industry for breaking down the molecules of larger alkanes into mixture of gaseous hydrocarbons. Cracking may be accomplished with heat (thermal cracking), or with a catalyst (catalytic cracking).

are minimum because both the bulky CH5 groups are 180° apart. This conformation corresponds to minimum energy and is hence most stable. 3. Which of the following conformers for ethylene glycol is most stable?

(1) H H

(1)

H

CH3

CH3

H

CH3 H

(3)

CH3 CH3

(2)

H H

(4) H H CH3

H H

H

OH H HO H

H

(4)

H H

H H

OH

H OH

OH

OH H

H H

(AIPMT MAINS 2010) Solution (3) Among the given conformations, the most stable

H H

H

(3)

H H

(2)

2. In the following the most stable conformation of n-butane is H

OH OH

H3C H CH3

H

is

H

CH3 H

(1) Anti-conformation of butane is the most stable conformation as the torsional and steric strains

OH

H

OH

due to intermolecular H-bonding.

H

4. The structure of isobutyl group in an organic compound is

(AIPMT PRE 2010) Solution

H

(1)

CH3 CH3

CH

CH22

(2) CH3 CH CH2

CH3

217

218

Chapter 4 | Alkanes and Cycloalkanes (3) CH3 CH2 CH2

6. The compound that will react most readily with gaseous bromine has the formula (1) C2H2 (3) C2H4 (2) C4H10 (4) C3H6 (NEET-II 2016)

CH2

CH3

(4) CH3 C CH3

(NEET 2013) Solution (1) The isobutyl group is

CH3 CH3

CH

CH2

5. The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is (1) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain. (2) The eclipsed conformation of ethane is more stable than staggered conformation, because eclipsed conformation has no torsional strain. (3) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain. (4) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain. (NEET-I 2016) Solution (4) Staggered conformation of ethane is more stable.

(4) Given that the reaction involves gaseous bromine, suggests that the reaction is taking place at higher temperature, as bromine is a liquid at room temperature conditions. Thus the bromination reaction would take place by free radical mechanism which is possible with butane (C4H10) and propene (C3H6). The reaction that results in more stable free radical will take place faster. CH3

CH

CH2

CH2

CH

CH2

CH3

CH

CH2

Propene CH3

CH2

CH2

CH3

CH3

Butane

Gaseous bromine will react faster with propene because the resulting allylic free radical is more stable due to conjugation. 7. With respect to the conformers of ethane, which of the following statements is true? (1) Bond angle changes but bond length remains same. (2) Both bond angle and bond length change. (3) Both bond angle and bond length remain the same. (4) Bond angle remains the same but bond length changes. (NEET 2017) Solution

H H H H

Solution

H H H

Eclipsed form (Torsional strain)

H

H

H H Staggered form (More stable) H

(3) Any three-dimensional arrangement of atoms that results from rotation about a single bond is called a conformation. In conformation of ethane, there is no change in bond length and bond angle. Change occurs only in dihedral or torsional angle.

ADDITIONAL OBJECTIVE QUESTIONS NCERT Exemplar Questions 1. Arrange the following in decreasing order of their boiling points. (I) n-Butane (II) 2-Methylbutane (III) n-Pentane (IV) 2,2-Dimethylpropane

(1) (2) (3) (4)

I > II > III > IV II > III > IV > I IV > III > II > I III > II > IV > I

2. Arrange the halogens F2, Cl2, Br2, I2, in order of their increasing reactivity with alkanes. (1) I2 < Br2 < Cl2 < F2 (3) F2 < Cl2 < Br2 < I2 (2) Br2 < Cl2 < F2 < I2 (4) Br2 < I2 < Cl2 < F2

Additional Objective Questions 3. The correct IUPAC name of the following alkane is H3C

CH2

CH

CH2

CH2

(1) (2) (3) (4)

CH2

CH3

CH2

CH CH3

CH

CH3

CH3

7. Which of the following molecules has the most negative heat of combustion in kcal/mole? (1) Methane (3) Propane (2) Ethane (4) Butane

3,6-diethyl-2-methyloctane. 5-isopropyl-3-ethyloctane. 3-ethyl-5-isopropyloctane. 3-isopropyl-6-ethyloctane.

4. Which of the following reactions of methane is incomplete combustion? Cu/523K/100 atm (1) 2CH4 + O2  → 2CH3OH Mo O (2) CH4 + O2  → HCHO + H2O 2

(4) CH4 + 2O2 → CO2(g) + 2H2O(l)

Exercise 1

10. Which is a conformer of pentane?

1. Which is the correct IUPAC name for the following cycloalkane?

2,4-Dimethyl-1-ethylcyclopentane 1,3-Dimethyl-5-ethylcyclopentane 1-Ethyl-2,4-dimethylcyclopentane 1-Ethyl-3,5-dimethylcyclopentane

2. The IUPAC name for

(1) (2) (3) (4)

is

3. Which of these is the common name for the 1,1-dimethylpropyl group? (1) tert-Butyl (3) Isopentyl (2) tert-Pentyl (4) Neopentyl 4. How many secondary hydrogens are there in the following alkane? CH3 CH3

(1)

(2)

6-ethyl-3,4-dimethylheptane. 2-ethyl-4,5-dimethylheptane. 3,4,6-trimethyloctane. 3,5,6-trimethyloctane.

(1) 1 (2) 2

8. Which of the following alkane is not liquid at room temperature? (1) C5H12 (3) C10H22 (2) C17H36 (4) C4H10 9. Conformation arises due to rotation around (1) carbon–carbon double bond. (2) carbon–carbon triple bond. (3) carbon–carbon single bond. (4) All of these.

3

(3) CH4 + O2 → C(s) + 2H2O(l)

(1) (2) (3) (4)

6. What is the simplest alkane, that is, the one with the smallest molecular weight, which possesses primary, secondary and tertiary carbon atoms? (1) 2-Methylpropane (3) 2-Methylpentane (2) 2-Methylbutane (4) 3-Methylpentane

CHCH2CH2CH3

(3) 4 (4) 9

5. How many compounds with the formula C7H16 (heptanes) contain a single tertiary carbon atom? (1) 2 (3) 4 (2) 3 (4) 5

CH3

H H

CH3

CH3 H

CH3

H

CH3

H H

CH3

H

(3)

H

H CH3

(4)

H H

CH3

H CH2CH3

H

H CH2CH3

11. The most stable conformation of 1,2-dibromoethane is

(1)

Br

H

Br

H H

(3)

H Br

H

H HH

(2)

H H

Br Br

H

(4)

H

H

H H

Br H

Br Br

12. Increasing order of stability among the three main conformations (i.e., eclipse, anti, gauche) of 2fluoroethanol is (1) eclipse, anti, gauche. (2) anti, gauche, eclipse. (3) eclipse, gauche, anti. (4) gauche, eclipse, anti. 13. Pick out the alkane which differs from the other members of the group. (1) 2,2-Dimethylpropane (2) Pentane (3) 3-Methylbutane (4) 2,2-Dimethylbutane

219

220

Chapter 4 | Alkanes and Cycloalkanes 14. The chlorination of an alkane involves (1) Cl⋅ free radicals.

(3) Cl– species.

(2) Cl+ species.

(4) CH4 free radicals.

⋅

15. What is the term for the process of forming ethene (an unsaturated hydrocarbon) from ethane (a saturated hydrocarbon)? CH3

(1) (2) (3) (4)

CH3

CH2

CH2 1 H2

Combustion Fractional distillation Thermal cracking Catalytic reformation CH3

16. CH3 CH CH2 CH3

Cl2 UV light

23. Which of these molecules is not expected to be obtained as a product of the high temperature chlorination of methane? (1) CCl4 (3) CH2Cl2 (2) HCCl3 (4) CH2 = CH2 24. Which of the following reactions would have the smallest energy of activation?

?

(1)

(A)

1 Br ?

1 HBr

major

The compound A is

(2)

1 Br ?

(3)

1 Br ?

(4)

1 Br ?

1 HBr

?

CH3

CH3

(1) CH3 CH CH CH3 (3) CH3 CH CH2 CH2Cl Cl CH3

CH2Cl

(2) CH3 C CH2 CH3 (4) CH3 C CH2 CH3 Cl

H

17. Isomerization in alkane may be brought about by using (1) Al2O3 (3) AlC13 and HCl (2) Fe2O3 (4) concentrated H2SO4 18. Which are the stoichiometric coefficients that complete the following equation? 1 x O2

(1) (2) (3) (4)

22. The major product obtained in the photocatalyzed bromination of 2-methylbutane is (1) l-bromo-2-methylbutane (2) l-bromo-3-methylbutane (3) 2-bromo-3-methylbutane (4) 2-bromo-2-methylbutane

y CO2 1 z H2O

1 HBr

? ?

1 HBr

25. An alkane C7H16 is produced by the reaction of lithium di(3-pentyl)cuprate with ethyl bromide. The structural formula of the product is (1) 3-ethylpentane. (3) 3-methylhexane. (2) 2-methylpentane. (4) 2-methylhexane.

Exercise 2 1. The correct IUPAC name for the following compound is

x = 9 ½, y = 6, z = 7 x = 18, y = 12, z = 14 x = 1, y = 1, z = 2 x = 6, y = 6, z = 4

19. Which of the following compounds cannot be prepared singly by the Wurtz reaction? (1) C2H6 (3) CH3CH2CH2CH3 (2) (CH3)2CHCH3 (4) All can be prepared 20. A chain reaction is one that (1) involves a series of steps. (2) involves two steps of equal activation energy. (3) can be initiated by light. (4) involves a series of steps, each of which generates a reactive intermediate that brings about the next step. 21. Radical halogenation reactions using ___ are the most ___ and often lead to multiple products. While radical halogenation reactions using ___ are the most ___ and produce primarily the major product. (1) bromine, selective, chlorine, reactive (2) bromine, reactive, chlorine, selective (3) chlorine, selective, bromine, reactive (4) chlorine, reactive, bromine, selective

(1) (2) (3) (4)

3-ethyl-5-isobutyl-3-methylnonane 5-(1-Methylpropyl)nonane 4-tert-butyl-2-methylheptane 5-ethyl-3-methylheptane

2. Which of the below is 8-methylbicyclo[4.3.0]nonane? (1)

(3)

(2)

(4)

3. Which conformation of 2-methylbutane is the most stable? CH3 H

(1) H

CH3

H H CH3

(2)

CH3

H

H CH3

CH3

Additional Objective Questions

(3)

CH3

CH3

H

CH3

(4)

H

H

CH3

H

H

CH3 CH3

H

4. Which compound has the highest boiling point? (1) 2-Methylpentane (3) 2,2-Dimethylbutane (2) 2,3-Dimethylbutane (4) Methane 5. The most stable conformation for 1,2-ethanediol (ethylene glycol) is shown below. It is the most stable conformation because H H H

OH H

OH

(1) this corresponds to an anti-conformation. (2) in general, gauche conformations possess the minimum energy. (3) it is stabilized by intramolecular hydrogen bonding. (4) it is a staggered conformation. 6. Which cycloalkane has the greatest ring strain? (1) Cyclopropane (3) Cyclopentane (2) Cyclobutane (4) Cyclohexane 7. Which alkene would yield 3-methylpentane when subjected to catalytic hydrogenation? (1)

(3)

(2)

(4)

8. What product is formed when 5-chlorocyclohex3-en-1-ol is reacted with Pd/C and H2? (1) 5-Chlorocyclohexanol (3) 5-Chloro-3-hexanol (2) 3-Chlorocyclohexanol (4) 2-Chloro-4-hexanol 9. What is the index of hydrogen deficiency (or degree of unsaturation) of benzene (C6H6)? (1) 2 (3) 4 (2) 3 (4) 5 10. Which of the following gas-phase reactions is a possible chain-terminating step in the light-initiated chlorination of methane? (1) Cl } Cl → 2Cl· (2) Cl· + CH4 → CH3· + H } Cl (3) CH3· + CH3· → CH3 } CH3 (4) CH3· + Cl } Cl → CH3Cl + Cl· 11. Methane cannot be prepared by (1) Corey–House synthesis. (2) Wurtz reaction. (3) Fittig reaction. (4) All of these. 12. The reactivity of hydrogen atom in an alkane toward substitution by bromine atom is (1) 1° H > 2° H > 3° H

(3) 1° H > 2° H > 3° H

(2) 1° H > 2° H > 3° H

(4) 1° H > 2° H > 3° H

13. Isopropyl bromide on Wurtz reaction gives (1) hexane. (3) 2,3-dimethylbutane. (2) propane. (4) neohexane. 14. On halogenation, an alkane (C5H12) gives only one monohalogenated product. The alkane is (1) n-pentane. (3) 2,2-dimethyl propane. (2) 2-methyl butane. (4) cyclopentane. 15. In the presence of light, ethane (1 mol) reacts with chlorine (1 mol) to form which product(s)? (I) CH2ClCHCl2 (III) CH3CH2Cl (II) CH3CHCl2 (IV) ClCH2CH2Cl (1) I, III, IV (2) II, IV

(3) II, III, IV (4) I, II, III, IV

16. For the following reaction, hv Na/ether C5H12 + Cl2  → C5H11Cl  →(C) (A ) (B)

the structures of (A), (B) and (C) are CH3

CH3

(1) CH3 C CH3

CH3

C

CH3

CH3

CH3 CH3

C

CH2Cl

CH3 CH2

CH2

C

CH3

CH3

CH3

CH3

CH3

(2) CH3 CH CH2CH3

CH3

C

CH2CH3

Cl CH3CH3 CH3

CH2

C

C

CH2CH3

CH3CH3

(3) Both (1) and (2) (4) None of these 17. Which of the following reactions would not produce saturated hydrocarbon as a major product? ether (1) CH3CH2Br + Na  → Zn(Hg)/HCl, ∆ (2) CH3COCH3  → N H /KOH (3) CH3CHO  → 2

4

LiAlH (4) CH3COOCH3  → 4

Exercise 3 In the following set of questions, two statements “Assertion” and “Reason” are given. Choose the correct answer from the following options: (1) Assertion and Reason are true and the Reason is the correct explanation of the Assertion. (2) Assertion and Reason are true but the Reason is not a correct explanation of the Assertion. (3) Assertion is true but the Reason is false. (4) Assertion and Reason both are false.

221

222

Chapter 4 | Alkanes and Cycloalkanes Reason: The conformation in which the bond pairs of two central atoms are very far from one another is called staggered form. 5. Assertion: Neopentane forms only one monosubstituted compound. Reason: Neopentane has high bond energy. 6. Assertion: CH4 does not react with Cl2 in dark. Reason: Chlorination of CH4 takes place in sunlight. 7. Assertion: Corey–House reaction can be used to prepare both symmetrical and unsymmetrical alkanes. Reason: The reaction involves the interaction between lithium dialkyl copper with an alkyl halide both of which may contain even or odd number of carbon atoms.

1. Assertion: Alkanes can have an infinite number of conformations. Reason: In configurational isomerism, the isomers are distinct individual substances. 2. Assertion: Cyclobutane is less stable than cyclopentane. Reason: The bond angles in cyclobutane and cyclopentane are 90° and 108° respectively and angle strain decreases. 3. Assertion: Melting point of neopentane is higher than that of n-pentane but the boiling point of n-pentane is higher than that of neopentane. Reason: Melting point depends upon packing of molecules in the crystal lattice while boiling point depends only upon surface area of the molecules. 4. Assertion: Staggered form is less stable than the eclipsed form.

ANSWER KEY NCERT Exemplar 1. (4)

2. (1)

3. (1)

4. (3)

2. (3)

3. (2)

4. (3)

5. (3)

Exercise 1 1. (3) 6. (2)

7. (4)

8. (4)

9. (3)

10. (4)

11. (3)

12. (1)

13. (4)

14. (1)

15. (3)

16. (2)

17. (3)

18. (1)

19. (2)

20. (4)

21. (4)

22. (4)

23. (4)

24. (3)

25. (1)

1. (2)

2. (3)

3. (2)

4. (1)

5. (3)

6. (1)

7. (2)

8. (2)

9. (3)

10. (3)

11. (4)

12. (3)

13. (3)

14. (3)

15. (4)

16. (1)

17. (4)

3. (1)

4. (1)

5. (3)

Exercise 2

Exercise 3 1. (1)

2. (2)

6. (4)

7. (1)

HINTS AND EXPLANATIONS Exercise 1 5. (4) There are four following compounds with formula C7H16 that contains a single 3° carbon atom.

2-Methylhexane

3-Methylhexane

3-Ethylpentane

3,3-Dimethylpentane

11. (3) The anti-conformation does not have torsional strain from steric hindrance because the groups are staggered and the bromine groups are far apart. Hence, the anti-conformation is the most stable.

Hints and Explanations The bromine groups in the gauche conformations are close enough to each other that the dispersion forces between them are repulsive; the electron clouds of the two groups are so close that they repel each other. Eclipsed conformation has the greatest energy of all because of the added large repulsive dispersion forces between the eclipsed bromine groups. H

Br Br

H H

H H

H

Br

Br

H

H

H

H

Gauche

CH3

CH3

UV light

C

CH2

CH3

AlCl − HCl CH3 − CH2 − CH2 − CH3  → (CH3 )2 − CH − CH3 n -Butane Isobutane 3

19. (2) Unsymmetrical alkanes cannot be prepared by Wurtz reaction. For their preparation, Wurtz–Fittig reaction is required.

Br

Anti

CH2

17. (3)

Br

H

CH

CH3 Cl2

Cl

Br Br

H

2

CH3

Eclipsed

Gauche

H

16. (2) CH3

HH

H

Compounds (1), (2), and (3) contain five carbons and these occur in the form of isomers; compound (4) contains six carbon atoms. So (4) is different from the other compounds.

20. (4) Chain reaction involves sequential, stepwise mechanism, in which each step generates the reactive intermediate that causes the next cycle of the reaction to occur.

1

12. (1) H2C CH2OH (2-Fluoroethanol)

21. (4) Chlorine is the most reactive halogen and bromine is the most selective.

F

22. (4) The reaction is

HO F

OH

H

H H

H

H

H

H

H Eclipsed

H

OH

CH3

F

H

H

F

H

Anti

Gauche

CH3

CH

CH3 CH2

Br2

CH3

hv

CH3

C

CH2

CH3

Br 2-Bromo-2-methylbutane (major)

23. (4) The products obtained from chlorination of methane are

Stability increases

Stability depends on repulsion. Since gauche conformations have lesser repulsion, it is more stable than eclipse conformations. In anti-conformation, two electronegative atoms are present in opposite sides; hence, it is most stable. Generally the anti-conformation is the most stable due to least steric strain and repulsion. However, in one of the gauche conformations of 2-haloethanols, hydrogen bonding between OH and X groups makes the conformation more stable.

H H

C

H H 1 Cl2

H Methane

heat or light

Chlorine

H

C H

C

C

Cl

H Dichloromethane

Cl Cl 1 Cl

Cl

13. (4)

Cl 1 H

Chloromethane

Cl 1 H

Cl

C

Cl 1 H

Cl

Cl

Trichloromethane

Tetrachloromethane

Hydrogen chloride

CH3 CH3

C

CH3 CH3

CH3 2,2-Dimethylpropane

CH2

CH2

CH2

CH3

Pentane

25. (1) (C5H11)2 CuLi 1 C2H5Br CH3

CH3 CH3

CH2

CH

2-Methylbutane

CH3 CH3

CH2

24. (3) Reaction given in option (3) involves the smallest energy of activation as leads to the formation of most stable tertiary radical.

C

H3C CH3

CH3 2,2-Dimethylbutane

CH2

CH

CH2

CH3 1 CuC5H11 1 LiBr

C2H5 3-Ethylpentane

This is Corey–House reaction.

223

224

Chapter 4 | Alkanes and Cycloalkanes

Exercise 2

10. (3) Combination of two methyl radicals is one of the chain-terminating steps in the chlorination of methane.

2. (3) The structure of the compound 8-methylbicyclo[4.3.0]nonane is 9 1 2 8 7

6

12. (3) H3C

CH

4

13. (3) 2H3C

5

Cl

OH

2,3-Dimethylbutane

14. (3)

16. (1) CH3

CH3 C

CH3 Cl2

CH3

2HCl

H3C

C

CH2Cl

CH3

CH2

CH3

Pd/C

CH3 C

CH3 CH3

CH3 Neopentane (A) Cl

Cl2 hv

CH3

C

CH3 CH2

CH2

C

CH3

CH3 (C)

17. (4) Major products of the reactions are as follows: ether CH3CH2Br + Na   → CH3CH2CH2CH3 Zn(Hg)/HCl, ∆ CH3COCH3  → CH3CH2CH3 N H /KOH

Rings are not affected by hydrogenation, but each ring still counts for one unit of hydrogen deficiency.

Wurtz reaction

CH3

3-Chlorocyclohexanol

1 3H2

Na/ether

CH2Cl

CH3 CH3

9. (3) The index of hydrogen deficiency of benzene is four.

C

CH3 Neopentyl chloride (B)

OH

H2

5-Chlorocyclohex3-en-1ol

CH(CH3)2 1 2NaBr

CH3

3-Methylpentane

8. (2)

Dry ether

2,2-Dimethylpropane

CH3 CH

CH3CH2CH2Cl

CH3 1 2Na

(CH3)2CH

CH3

Pt H2

CH2

? CH3CH2CH2

Isopropyl bromide

3-Methylpentane

CH3

CH(Br)

H3C CH2

H2Cl Peroxide Cl?

CH3 CH2

CH3

3

4. (1) 2-Methylpentane has highest boiling point because it has the least branched chain structure as compared to 2, 3-dimethylbutane and 2, 2-dimethylbutane. Therefore, it has largest surface area and, hence the highest boiling point. 7. (2) CH2

CH2

2 4 → CH3CH3 CH3CHO 

LiAlH

4 → CH3CH2OH CH3COOCH3 

CONCEPT MAP ]

[C O N C E P T

M A P

Conformers are Molecules that differ only by rotation about sigma ( s ) bonds can have Ring strain

Different potential energies of conformers can be represented by

among conformers is a function of

is caused by

and Conformer potential energy diagrams

Torsional strain is caused by

are a plot of Dihedral angle vs. potential energy

Angle strain

and loss of

Repulsive dispersion forces

Hyperconjugative stabilization

result in

E

0

60

120 240 300 360 Degrees of Rotation u

Deviation from ideal bond angles

involves Favorable overlap of occupied with unoccupied orbitals

Steric hindrance

420

is caused by

can be represented by Newman projection formulas

of cyclohexane can be represented by

Chair conformational structures

or

Boat

have

can be used to show

Axial positions Eclipsed conformations

and

Staggered conformations

and twist-boat

with substituted groups (G) can be and G

G

Anti

or G

Gauche

G

Equatorial positions conformational structures

[C O N C E P T

CONCEPT MAP

CONCEPT MAP

M A P

]

Mechanism Review of Radical Reactions

Radical Halogenation of Alkanes X

heat or light

X

If X = Br, hydrogen abstraction is selective. If X = Cl, hydrogen abstraction is not selective.

C

H

Chain initiation Radical Polymerization

Chain propagation

O

O H

C·

X

X

X

C

R

X  X·

C

O

O

C

R

Coupling The substitution product

X· C

X·  X·

X

C

C

Some possible chain-terminating steps Anti-Markovnikov Addition of HBr to Alkenes RO

OR

RO·  H

heat

R Chain initiation

ROH  Br·

Br

Addition of the bromine radical to the alkene occurs so as to form the more stable carbon radical intermediate (The alkene reactant shown is meant to indicate any alkene where a difference exists in the extent of alkyl substitution at the initial alkene carbons.) R C

RO·  RO·

C

C

n Chain propagation

R

·C R Br·

C

C

C

HBr

R

C

C

 Br·

Br H Br Coupling The anti-Markovnikov addition product C

C

C

Br Br Br R R Br Some possible chain-terminating steps

Disproportionatio

Possible cha

5

Alkenes and Alkynes C H A P T E R OU TLIN E Part I: Alkenes 5.1 Structure of the Double Bond 5.2 Isomerism 5.3 How to Name Alkenes and Cycloalkenes 5.4 The (E )–(Z ) System for Designating Alkene Diastereomers 5.5 Relative Stabilities of Alkenes 5.6 Cycloalkenes 5.7 Synthesis of Alkenes by Hydrogenation of Alkynes 5.8 Synthesis of Alkenes via Elimination Reactions 5.9 Carbocation Stability and the Occurrence of Molecular Rearrangements 5.10 Hydrogenation of Alkenes 5.11 Addition Reactions of Alkenes 5.12 Oxidation of Alkenes: Syn 1,2-Dihydroxylation Part II: Alkynes 5.13 Structure of the Triple Bond

5.14 5.15 5.16 5.17 5.18

Isomerism How to Name Alkynes The Acidity of Terminal Alkynes Methods of Preparation Chemical Reactivity

Part III: Conjugated Unsaturated Systems 5.19 The Stability of the Allyl Radical 5.20 The Allyl Cation 5.21 Alkadienes and Polyunsaturated Hydrocarbons 5.22 1,3-Butadiene: Electron Delocalization 5.23 The Stability of Conjugated Dienes 5.24 Allylic Substitution and Allylic Radicals 5.25 Electrophilic Attack on Conjugated Dienes: 1,4-Addition 5.26 The Diels–Alder Reaction: A 1,4-Cycloaddition Reaction of Dienes

PHOTO CREDIT: Media Bakery

Alkenes are hydrocarbons whose molecules contain a carbon–carbon double bond. An old name for this family of compounds that is still often used is the name olefins. Ethene  (ethylene), the simplest olefin (alkene), was called olefiant gas (Latin: oleum, oil + facere, to make) because gaseous ethene (C2H4) reacts with chlorine to form C2H4Cl2, a liquid (oil). Hydrocarbons whose molecules contain the carbon–carbon triple bond are called alkynes. The common name for this family is acetylenes, after the simplest member, HC CH: H

H C

H

C

C H

H Ethene

CH3 C

H C

C

H

H

H Propene

Ethyne

Alkenes and alkynes have physical properties similar to those of corresponding alkanes. Alkenes and alkynes up to four carbons (except 2-butyne) are gases at room temperature. Being relatively nonpolar themselves, alkenes and alkynes dissolve in nonpolar solvents or in solvents of low polarity. Alkenes and alkynes are only very slightly soluble in water (with alkynes being slightly more soluble than alkenes). The densities of alkenes and alkynes are lower than that of water.

228

Chapter 5 | Alkenes and Alkynes

PART I: ALKENES 5.1

STRUCTURE OF THE DOUBLE BOND

The carbon atoms of many of the molecules that we have considered so far have used their four valence electrons to form four single covalent (σ) bonds to four other atoms. We find, however, that many important organic compounds exist in which carbon atoms share more than two electrons with another atom. In molecules of these compounds, some bonds that are formed are multiple covalent bonds. When two carbon atoms share two pairs of electrons, for example, the result is a carbon–carbon double bond: C C or

C

C

Hydrocarbons whose molecules contain a carbon–carbon double bond are called alkenes. Ethene (C2H4) and propene (C3H6) are both alkenes. (Ethene is also called ethylene, and propene is sometimes called propylene.) H

H C

H

H

C

H

C H

H3C

Ethene

C H

Propene

In ethane, the only carbon–carbon bond is a double bond. Propene has one carbon–carbon single bond and one carbon–carbon double bond. The spatial arrangement of the atoms of alkenes is different from that of alkanes. The six atoms of ethene are coplanar, and the arrangement of atoms around each carbon atom is triangular (Fig. 5.1). H

H C

H

C

121°

118°

H

Figure 5.1 The structure and bond angles of ethene. The plane of the atoms is perpendicular to the paper. The dashed wedge bonds project behind the plane of the paper, and the solid wedge bonds project in front of the paper. A satisfactory model for the carbon–carbon double bond can be based on sp2 hybridized carbon atoms. The mathematical mixing of orbitals that furnish the sp2 orbitals for our model can be visualized in the way shown in Fig. 5.2. The 2s orbital is mathematically mixed (or hybridized) with two of the 2p orbitals. (The hybridization procedure applies only to the orbitals, not to the electrons.) One 2p orbital is left unhybridized. One electron is then placed in each of the sp2 hybrid orbitals and one electron remains in the 2p orbital. The three sp2 orbitals that result from hybridization are directed towards the corners of a regular triangle (with angles of 120° between them). The carbon p orbital that is not hybridized is perpendicular to the plane of the triangle formed by the hybrid sp2 orbitals (Fig. 5.3). In our model for ethane, as shown in Fig. 5.4(a), we see that two sp2 hybridized carbon atoms form a σ bond between them by the overlap of one sp2 orbital from each. The remaining sp2 orbitals of the carbon atoms form σ bonds to four hydrogen atoms through overlap with the 1s orbitals of the hydrogen atoms. These five bonds account for 10 of the 12 bonding electrons of ethene, and they are called the σ-bond framework. The bond angles that we would predict on the basis of sp2 hybridized

5.1 | Structure of the Double Bond

Ground state 2p

sp 2-Hybridized state

Excited state 2p

2p

Energy

2sp 2 2s

2s

1s

1s

1s

Promotion of electron

Hybridization

Figure 5.2 A process for obtaining sp2 hybridized carbon atoms.

y carbon atoms (120° all around) are quite close to the bond angles that are actually found. The remaining two bonding electrons in our model are located in the sp 2 Orbital p orbitals of each carbon atom. The overlap of these z sp 2 Orbital p orbitals is shown schematically in Fig. 5.4 (b). We see that the parallel p orbitals overlap above and below the plane of the σ framework. This sideways overlap of x the p orbitals results in a new type of covalent bond, p Orbital known as a pi (π) bond. A π bond has cylindrical symmetry about a line connecting the two bonded sp 2 Orbital nuclei. A π bond has a nodal plane passing through the two bonded nuclei and between the σ molecular orbital lobes. The strength of the carbon–carbon double bond is more (bond enthalpy 681 kJ mol−1) than that Figure 5.3 An sp2 hybridized carbon atom. of the carbon–carbon single bond (bond enthalpy 384 kJ mol−1). The bond length of carbon–carbon double bond (134 pm) is shorter than that of single bond (154 pm). The electrons of the π bond are loosely held due to weak sideways overlap of the

p Orbitals Overlap

H

H

σ Bonds

C

H

C

σ Bonds

H

H C

C

H

H

H

σ Bond overlap (a)

π Bond (b)

Figure 5.4 Schematic depiction of: (a) Bonding molecular orbitals of ethene formed from two sp2 hybridized carbon atoms and four hydrogen atoms. (b) The σ bonds and the overlapping of adjacent π orbitals.

229

230

Chapter 5 | Alkenes and Alkynes

orbitals. As a result of this, alkenes behave as a source of loosely held mobile electrons and are easily attacked by reagents that are looking for electrons (electrophiles or electrophilic reagents). The presence of weaker π bond makes alkenes less stable and they are readily converted to alkanes on reaction with electrophilic reagents. Note: There is a large energy barrier to rotation associated with groups joined by a double bond. While groups joined by single bonds rotate relatively freely at room temperature, those joined by double bonds do not. This is because the barrier to rotation of the double bond (264 kJ mol−1) is markedly higher than the rotational barrier of groups joined by carbon–carbon single bonds (13 to 26 kJ mol−1).

5.2

ISOMERISM

Both structural and geometric isomerism are exhibited by alkenes. 1. Structural isomerism: Higher alkenes (number of carbon atoms > 3), like higher alkanes also exhibit structural isomerism. They can exist as more than one structure which differ in the position of the double bond or position of the side chain. These are known as position and chain isomers, respectively. For example, the various straight-chain structural isomers possible for alkene with molecular formula C5H10 are: (a) Position isomers: CH2

CHCH2CH2CH3 1-Pentene

CH3CH CHCH2CH3 2-Pentene

(b) Chain isomers: CH3 CH2

CH3

CCH2CH3

CH3C

2-Methyl-1-butene

CH3

CHCH3

CH2

2-Methyl-2-butene

CHCHCH3

3-Methyl-1-butene

Note: The compound with molecular formula C5H10 can also exist as cycloalkanes, cyclopentane, methyl cyclobutane, dimethyl cyclopropane, etc. 2. Geometric isomerism: Because of restricted rotation about a carbon–carbon double bond, an alkene in which each carbon of the double bond has two different groups bonded to it shows cis– trans isomerism. These isomers are called geometric isomers. If two identical groups are on the same side of the double bond, the compound can be designated cis; if they are on opposite sides it can be designated trans: H

H C

H3C

H

C

C CH3

cis-2-Butene mp 2139°C, bp 4°C

H3C

C

CH3 H

trans-2-Butene mp 2106°C, bp 1°C

Cis alkenes are less stable than their trans isomers because of non-bonded interaction strain between alkyl substituents on the same side of the double bond in the cis isomer. Alkenes are weakly polar but due to presence of weakly held π electrons, their dipole moments are larger than that of alkanes. The bond joining the alkyl group to the doubly bonded carbon has a small dipole moment, which is not cancelled by the corresponding polarity in the opposite direction. In case of

5.3 | How to Name Alkenes and Cycloalkenes

cis isomers, the alkyl groups are present on the same side of the double bond, so a net dipole moment is observed. In case of trans isomer, the dipole moment on one side is cancelled by the dipole moment on the other side of the carbon–carbon double bond, so net dipole moment is zero. CH3

H C

CH3CH2

C H H Propene µ 5 0.35 D

H C

C H H 1-Butene µ 5 0.37 D

For a pair of geometric isomers, for example, cis- and trans-butene, the cis isomers have higher boiling points due to higher polarity. However, due to its lower symmetry, it does not fit into the crystalline lattice as well as the trans isomer and thus has a lower melting point. H3C

H C

H3C

C H3C

H C C

H

cis-Butene µ 5 0.33 D bp 5 550 8C mp 5 401 8C

H

CH3

trans-Butene µ5 0 bp 5 547 8C mp 5 440 8C

Geometric isomers of alkenes can be distinguished on the basis of different physical properties and identified after determination of configuration.

5.3

HOW TO NAME ALKENES AND CYCLOALKENES

Many older names for alkenes are still in common use. Propene is often called propylene, and 2-methylpropene frequently bears the name isobutylene: CH3 CH2

CH2

CH3CH

CH2

C CH3

IUPAC: Common:

Ethene Ethylene

Propene Propylene

CH2

2-Methylpropene Isobutylene

231

232

Chapter 5 | Alkenes and Alkynes

The IUPAC rules for naming alkenes are similar in many respects to those for naming alkanes: 1. Determine the parent name by selecting the longest chain that contains the double bond and change the ending of the name of the alkane of identical length from -ane to -ene. Thus, if the longest chain contains five carbon atoms, the parent name for the alkene is pentene; if it contains six carbon atoms, the parent name is hexene, and so on. 2. Number the chain so as to include both carbon atoms of the double bond, and begin numbering at the end of the chain nearer the double bond. Designate the location of the double bond by using the number of the first atom of the double bond as a prefix. The locant for the alkene suffix may precede the parent name or be placed immediately before the suffix. We will show examples of both styles: 1

CH2

2

3

4

CHCH2CH3

CH3CH

1-Butene (not 3-butene)

CHCH2CH2CH3

2-Hexene (not 4-hexene)

3. Indicate the locations of the substituent groups by the numbers of the carbon atoms to which they are attached: 4 2

1

4

3

1

2-Methyl-2-butene or 2-methylbut-2-ene

CH3 CH3CH “ CHCH2 C9CH3 1

2

3

4

5

6

2

6

5

3

2,5-Dimethyl-2-hexene or 2,5-dimethylhex-2-ene 4

3

2

1

CH3CH “ CHCH2 Cl

CH3 5,5-Dimethyl-2-hexene or 5,5-dimethylhex-2-ene

1-Chloro-2-butene or 1-chlorobut-2-ene

4. Number substituted cycloalkenes in the way that gives the carbon atoms of the double bond the 1 and 2 positions and that also gives the substituent groups the lower numbers at the first point of difference. With substituted cycloalkenes it is not necessary to specify the position of the double bond since it will always begin with C1 and C2. The two examples shown here illustrate the application of these rules: 1

1 5

2

4

3

6

2

5

3 4

1-Methylcyclopentene (not 2-methylcyclopentene)

3,5-Dimethylcyclohexene (not 4,6-dimethylcyclohexene)

5. Name compounds containing a double bond and an alcohol group as alkenols (or cycloalkenols) and give the alcohol carbon the lower number: OH 1

OH

2 5

4

3

2

1

4-Methyl-3-penten-2-ol or 4-methylpent-3-en-2-ol

3

2-Methyl-2-cyclohexen-1-ol or 2-methylcyclohex-2-en-1-ol

5.4 | The (E )–(Z ) System for Designating Alkene Diastereomers

6. Two frequently encountered alkenyl groups are the vinyl group and the allyl group:

The vinyl group

The allyl group

Using substitutive nomenclature, the vinyl and allyl groups are called ethenyl and prop-2-en-1-yl, respectively. The following examples illustrate how these names are employed: OH

Cl

Br Bromoethene or vinyl bromide (common)

Ethenylcyclopropane or vinylcyclopropane

3-Chloropropene or allyl chloride (common)

3-(Prop-2-en-1-yl)cyclohexan-1-ol or 3-allylcyclohexanol

7. If two identical or substantial groups are on the same side of the double bond, the compound can be designated cis; if they are on opposite sides it can be designated trans:

Cl

Cl

Cl

Cl cis-1,2-Dichloroethene

trans-1,2-Dichloroethene

THE (E )–(Z ) SYSTEM FOR DESIGNATING 7.2 5.4 ALKENE DIASTEREOMERS

The terms cis and trans to designate the stereochemistry of alkene diastereomers (cis–trans isomers) are unambiguous, only when applied to disubstituted alkenes. If the alkene is trisubstituted or tetrasubstituted, the terms cis and trans are either ambiguous or do not apply at all. Consider the following alkene as an example: Br

Cl C

H

C A

F

It is impossible to decide whether A is cis or trans since no two groups are the same. A system that works in all cases is based on the priorities of groups in the Cahn–Ingold–Prelog convention. This system, called the (E )–(Z ) system, applies to alkene diastereomers of all types.

5.4A How to Use the (E )–(Z ) System 1. Examine the two groups attached to one carbon atom of the double bond and decide which has higher priority. 2. Repeat that operation at the other carbon atom:

233

234

Chapter 5 | Alkenes and Alkynes

Cl

Higher priority

C C

Br

Higher priority

F

F

H

Br

Higher priority

C C

Cl

Cl  F

Higher priority

H

Br  H

(E)-2-Bromo-1-chloro-1fluoroethene

(Z)-2-Bromo-1-chloro-1fluoroethene

3. Compare the group of higher priority on one carbon atom with the group of higher priority on the other carbon atom. If the two groups of higher priority are on the same side of the double bond, the alkene is designated (Z ) (from the German word zusammen, meaning together). If the two groups of higher priority are on opposite sides of the double bond, the alkene is designated (E ) (from the German word entgegen, meaning opposite). The following isomers provide another example. H3C

H3C

CH3 C

CH3 . H

C

H

C

H

H

(Z)-2-Butene or (Z)-but-2-ene (cis-2-butene)

H C

CH3

(E)-2-Butene or (E)-but-2-ene (trans-2-butene)

7.3 5.5 RELATIVE STABILITIES OF ALKENES Cis and trans isomers of alkenes do not have the same stability. ●

Strain caused by crowding of two alkyl groups on the same side of a double bond makes cis isomers generally less stable than trans isomers (Fig. 5.5).

Figure 5.5 Cis and trans alkene isomers. The cis isomer is less stable due to greater strain from crowding by the adjacent alkyl groups. This effect can be measured quantitatively by comparing thermodynamic data from experiments involving alkenes with related structures, as we shall see later.

5.5A Heat of Reaction The addition of hydrogen to an alkene (hydrogenation, Sections 4.7A and 5.10A) is an exothermic reaction; the enthalpy change involved is called the heat of reaction or, in this specific case, the heat of hydrogenation. C

C H

H

Pt

C

C

H

H

H °  –120 kJ mol1

5.5 | Relative Stabilities of Alkenes

+ H2 + H2

7 kJ mol–1

+ H2

Enthalpy

5 kJ mol–1

∆H  = –120 kJ mol–1 ∆H  = –115 kJ mol–1

∆H  = –127 kJ mol–1

Figure 5.6 An energy diagram for platinum-catalyzed hydrogenation of the three butene isomers. The order of stability based on the differences in their heats of hydrogenation is trans-2-butene > cis-2-butene > 1-butene.

We can gain a quantitative measure of relative alkene stabilities by comparing the heats of hydrogenation for a family of alkenes that all become the same alkane product on hydrogenation. The results of such an experiment involving platinum-catalyzed hydrogenation of three butene isomers are shown in Fig. 5.6. All three isomers yield the same product—butane—but the heat of reaction is different in each case. On conversion to butane, 1-butene liberates the most heat (127 kJ mol−1), followed by cis-2-butene (120 kJ mol−1), with trans-2-butene producing the least heat (115 kJ mol−1). These data indicate that the trans isomer is more stable than the cis isomer, since less energy is released when the trans isomer is converted to butane. Furthermore, it shows that the terminal alkene, 1-butene, is less stable than either of the disubstituted alkenes, since its reaction is the most exothermic.

5.5B Overall Relative Stabilities of Alkenes Studies of numerous alkenes reveal a pattern of stabilities that is related to the number of alkyl groups attached to the carbon atoms of the double bond. ●

The greater the number of attached alkyl groups (i.e., the more highly substituted the carbon atoms of the double bond), the greater is the alkene’s stability.

This order of stabilities can be given in general terms as follows:* Relative Stabilities of Alkenes R

R

R

R

R

Tetrasubstituted

R

R



R

H

Trisubstituted

R

H

R

H

R

H





R



H

R

Disubstituted

R

H



H

H

H

H

H

H



H

H

Monosubstituted Unsubstituted

*This order of stabilities may seem contradictory when compared with the explanation given for the relative stabilities of cis and trans isomers. Although a detailed explanation of the trend given here is beyond our scope, the relative stabilities of substituted alkenes can be rationalized. Part of the explanation can be given in terms of the electron-releasing effect of alkyl groups, an effect that satisfies the electron-withdrawing properties of the sp2-hybridized carbon atoms of the double bond.

235

236

Chapter 5 | Alkenes and Alkynes

5.6

CYCLOALKENES

The rings of cycloalkenes containing five carbon atoms or fewer exist only in the cis form (Fig. 5.7). The introduction of a trans double bond into rings this small would, if it were possible, introduce greater strain than the bonds of the ring atoms could accommodate. (handheld molecular models.) trans-Cyclohexene might resemble the structure shown in Fig. 5.8. There is evidence that it can be formed as a very reactive short-lived intermediate in some chemical reactions, but it is not isolable as a stable molecule.

Cyclopropene

Cyclobutene

Cyclopentene

Cyclohexene

Figure 5.8 Hypothetical transcyclohexene. This molecule is apparently too strained to exist at room temperature.

Figure 5.7 cis-Cycloalkenes.

trans-Cycloheptene has been observed spectroscopically, but it is a substance with a very short lifetime and has not been isolated. trans-Cyclooctene (Fig. 5.9) has been isolated, however. Here the ring is large enough to accommodate the geometry required by a trans double bond and still be stable at room temperature. trans-Cyclooctene is chiral and exists as a pair of enantiomers.

cis-Cyclooctene

trans-Cyclooctene

Figure 5.9 The cis and trans forms of cyclooctene.

SYNTHESIS OF ALKENES BY 5.7 5.7

HYDROGENATION OF ALKYNES

Depending on the conditions and the catalyst employed, one or two molar equivalents of hydrogen will add to a carbon–carbon triple bond. When a platinum catalyst is used, the alkyne generally reacts with two molar equivalents of hydrogen to give an alkane: CH3C

Pt, H

2 CCH3 99: [CH3CH"CHCH3]

Pt, H

2 99: CH3CH2CH2CH3

However, hydrogenation of an alkyne to an alkene can be accomplished through the use of special catalysts or reagents. Moreover, these special methods allow the preparation of either (E)- or (Z)-alkenes from disubstituted alkynes.

5.7A Syn Addition of Hydrogen: Synthesis of cis-Alkenes A heterogeneous catalyst that permits hydrogenation of an alkyne to an alkene is the nickel boride compound called P-2 catalyst. The P-2 catalyst can be prepared by the reduction of nickel acetate with sodium borohydride:

( ) O

Ni OCCH3

2

NaBH4 EtOH

Ni2B P-2

5.8 | Synthesis of Alkenes via Elimination Reactions ●

Hydrogenation of alkynes in the presence of P-2 catalyst causes syn addition of hydrogen. The alkene formed from an internal alkyne has the (Z) or cis configuration.

The hydrogenation of 3-hexyne illustrates this method. The reaction takes place on the surface of the catalyst accounting for the syn addition: Syn addition of hydrogen to an alkyne

H2/Ni2B (P-2) (syn addition)

3-Hexyne

(Z )-3-Hexene (cis-3-hexene) (97%)

Other specially conditioned catalysts can be used to prepare cis-alkenes from disubstituted alkynes. Metallic palladium deposited on calcium carbonate can be used in this way after it has been conditioned with lead acetate and quinoline. This special catalyst is known as Lindlar’s catalyst: R

C

C

R

H2, Pd/CaCO3 (Lindlar's catalyst) quinoline (syn addition)

R

R C

H

C H

5.7B Anti Addition of Hydrogen: Synthesis of trans-Alkenes ●

Anti addition of hydrogen to the triple bond of alkynes occurs when they are treated with lithium or sodium metal in ammonia or ethylamine at low temperatures.

This reaction, called a dissolving metal reduction, takes place in solution and produces an (E)- or trans-alkene. The mechanism involves radicals, which are molecules that have unpaired electrons (see Chapter 10). Anti addition of hydrogen to an alkyne.

(1) Li, EtNH2, –78 °C (2) NH4Cl

4-Octyne

(anti addition)

(E)-4-Octene (trans-4-octene) (52%)

7.4 5.8 SYNTHESIS OF ALKENES VIA ELIMINATION REACTIONS Elimination reactions are the most important means for synthesizing alkenes. In this chapter we shall study two methods for alkene synthesis based on elimination reactions: dehydrohalogenation of alkyl halides and dehydration of alcohols.

5.8A Dehydrohalogenation of Alkyl Halides HH

H C H ●

H

base

C

HX

X

H

H

H

H

The best reaction conditions to use when synthesizing an alkene by dehydrohalogenation are those that promote an E2 mechanism.

237

238

Chapter 5 | Alkenes and Alkynes

In an E2 mechanism, a base removes a β hydrogen from the β carbon, as the double bond forms and a leaving group departs from the α carbon. B



H 

C

E2



C

C

C



B H



X

X

Reaction conditions that favor elimination by an E1 mechanism should be avoided because the results can be too variable. The carbocation intermediate that accompanies an E1 reaction can undergo rearrangement of the carbon skeleton, as we shall see in Section 5.9, and it can also undergo substitution by an SN1 mechanism, which competes strongly with formation of products by an E1 path. How to Favor an E2 Mechanism 1. Use a secondary or tertiary alkyl halide if possible. Why? Because steric hindrance in the substrate will inhibit substitution. 2. When a synthesis must begin with a primary alkyl halide, use a bulky base. Why? Because the steric bulk of the base will inhibit substitution. 3. Use a high concentration of a strong and nonpolarizable base such as an alkoxide. Why? Because a weak and polarizable base would not drive the reaction toward a bimolecular reaction, thereby allowing unimolecular processes (such as SN1 or E1 reactions) to compete. 4. Sodium ethoxide in ethanol (EtONa/EtOH) and potassium tert-butoxide in tert-butyl alcohol (t-BuOK/t-BuOH) are bases typically used to promote E2 reactions. Why? Because they meet criterion 3 above. Note that in each case the alkoxide base is dissolved in its corresponding alcohol. (Potassium hydroxide dissolved in ethanol or tert-butyl alcohol is also sometimes used, in which case the active base includes both the alkoxide and hydroxide species present at equilibrium.) 5. Use elevated temperature because heat generally favors elimination over substitution. Why? Because elimination reactions are entropically favored over substitution reactions (because the products are greater in number than the reactants). Hence ΔS ° in the Gibbs free-energy equation, ΔG ° = ΔH ° − TΔS ° is significant, and ΔS ° will be increased by higher temperature since T is a coefficient, leading to a more negative (favorable) ΔG °. Zaitsev’s Rule: Formation of the More Substituted Alkene is Favored with a Small Base We have seen that in some dehydrohalogenation reactions only a single elimination product was possible. For example:

EtONa EtOH, 55 °C

Br

(79%)

Br

EtONa EtOH, 55 °C

(91%)

( )15

Br

t-BuOK t-BuOH, 40 °C

( )15 (85%)

5.8 | Synthesis of Alkenes via Elimination Reactions

Dehydrohalogenation of many alkyl halides, however, yields more than one product. For example, dehydrohalogenation of 2-bromo-2-methylbutane can yield two products: 2-methyl-2-butene and 2-methyl-1-butene, as shown here by pathways (a) and (b), respectively: B (a)

(a)



(b)

B 

Br



 H

B 

Br



H 2-Methyl-2-butene

H (b)

Br 2-Bromo-2-methylbutane ●

 H

2-Methyl-1-butene

If we use a small base such as ethoxide or hydroxide, the major product of the reaction will be the more highly substituted alkene (which is also the more stable alkene).

Br

EtONa EtOH, 70 °C



2-Methyl-2-butene (69%)

2-Methyl-1-butene (31%)

Trisubstituted: more stable

Disubstituted: less stable

2-Methyl-2-butene is a trisubstituted alkene (three methyl groups are attached to carbon atoms of the double bond), whereas 2-methyl-1-butene is only disubstituted. 2-Methyl-2-butene is the major product. ●

Whenever an elimination occurs to give the more stable, more highly substituted alkene, chemists say that the elimination follows Zaitsev’s rule, named for the nineteenth-century Russian chemist A. N. Zaitsev (1841–1910) who formulated it. (Zaitsev’s name is also transliterated as Zaitzev, Saytzeff, Saytseff, or Saytzev.)

Formation of the Less Substituted Alkene Using a Bulky Base ●

Carrying out dehydrohalogenations with a bulky base such as potassium tert-butoxide (t-BuOK) in tert-butyl alcohol (t-BuOH) favors the formation of the less substituted alkene: A bulky base

Br

t-BuOK t-BuOH, 75 °C



2-Methyl-2-butene (27.5%)

2-Methyl-1-butene (72.5%)

More substituted but formed more slowly.

Less substituted but formed faster.

The reasons for this behavior are related in part to the steric bulk of the base and to the fact that in tert-butyl alcohol the base is associated with solvent molecules and thus made even larger. The large tert-butoxide ion appears to have difficulty removing one of the internal (2°) hydrogen atoms because of greater crowding at that site in the transition state. It removes one of the more exposed (1°) hydrogen atoms of the methyl group instead. ●

When an elimination yields the less substituted alkene, we say that it follows the Hofmann rule.

239

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Chapter 5 | Alkenes and Alkynes

5.8B Dehalogenation of Vicinal Dihalides Vicinal (Latin: vicinalis, neighboring) dihalides contain halogen atoms on adjacent carbon atoms. These undergo dehalogenation, that is, loss of halogen, in the presence of zinc metal to form an alkene. CH2Br

CH2Br 1 Zn

CH2

CH2 1 ZnBr2

CH3CHBr

CH2Br 1 Zn

CH3CH

CH2 1 ZnBr2

The dehalogenation reaction is restricted by the fact that the dihalides are themselves prepared by halogenations of alkenes. The reaction, however, is used when protection of a double bond is required in synthesis of some compounds. The double bond is converted into vicinal halide and then regenerated at the end of reaction by dehalogenation. Similarly, impure alkenes may be purified by this reaction.

5.8C Acid-Catalyzed Dehydration of Alcohols ●

Most alcohols undergo dehydration (lose a molecule of water) to form an alkene when heated with a strong acid. C

C

H

OH

HA heat

C

C

H2O



The reaction is an elimination and is favored at higher temperatures. The most commonly used acids in the laboratory are Brønsted acids—proton donors such as sulfuric acid and phosphoric acid. Lewis acids such as alumina (Al2O3) are often used in industrial, gas-phase dehydrations. 1. The temperature and concentration of acid required to dehydrate an alcohol depend on the structure of the alcohol substrate. (a) Primary alcohols are the most difficult to dehydrate. Dehydration of ethanol, for example, requires concentrated sulfuric acid and a temperature of 180 °C: H

H

H

H9C9C9H H

concd H2SO4

H C

180 °C

C

H

OH



H 2O

H Ethene

Ethanol (a 1° alcohol)

(b) Secondary alcohols usually dehydrate under milder conditions. Cyclohexanol, for example, dehydrates in 85% phosphoric acid at 165–170 °C: OH

85% H3PO4

 H2O

165–170 °C

Cyclohexanol

Cyclohexene (80%)

(c) Tertiary alcohols are usually so easily dehydrated that relatively mild conditions can be used. tert-Butyl alcohol, for example, dehydrates in 20% aqueous sulfuric acid at a temperature of 85 °C: CH3 CH39C9OH CH3 tert-Butyl alcohol

CH2 20% H2SO4 85 °C

CH3

C

CH3

 H 2O

2-Methylpropene (84%)

5.8 | Synthesis of Alkenes via Elimination Reactions ●

The relative ease with which alcohols undergo dehydration is 3 ° > 2 ° > 1 °. R

H

R

R9 C9OH



R9 C9OH

R



R9 C9OH

H

3° Alcohol

H

2° Alcohol

1° Alcohol

This behavior, as we shall related to the relative stabilities of carbocations. 2. Some primary and secondary alcohols also undergo rearrangements of their carbon skeletons during dehydration. Such a rearrangement occurs in the dehydration of 3,3-dimethyl-2-butanol: CH3 CH3 9C99 CH9 CH3 CH3

H 3C

85% H3PO4

C"C

80 °C

3,3-Dimethyl-2-butanol

C ! CHCH3

 CH3

H3C

OH

CH3

H3C

CH3

H2C

2,3-Dimethyl-2-butene (80%)

2,3-Dimethyl-1-butene (20%)

Notice that the carbon skeleton of the reactant is C

C

C9C9C9C while that of the products is

C C!C C

C

C

The carbon skeleton has rearranged

We shall see in Section 5.9 that this reaction involves the migration of a methyl group from one carbon to the next so as to form a more stable carbocation. (Rearrangements to carbocations of approximately equal energy may also be possible with some substrates.) Mechanism for Dehydration of Secondary and Tertiary Alcohols: An E1 Reaction Explanations for these observations can be based on a stepwise mechanism originally proposed by F. Whitmore (of Pennsylvania State University). The mechanism is an E1 reaction in which the substrate is a protonated alcohol. Consider the dehydration of tert-butyl alcohol as an example: CH3

Step 1

CH3

C

H O

H  H

O

CH3 H CH3



H

CH3



C

O

H  H

O H

CH3 Protonated alcohol

Protonation of the alcohol

In this step, an acid–base reaction, a proton is rapidly transferred from the acid to one of the unshared electron pairs of the alcohol. In dilute sulfuric acid the acid is a hydronium ion; in concentrated sulfuric acid the initial proton donor is sulfuric acid itself. This step is characteristic of all reactions of an alcohol with a strong acid. The presence of the positive charge on the oxygen of the protonated alcohol weakens all bonds to oxygen, including the carbon–oxygen bond, and in step 2 the carbon–oxygen bond breaks. The leaving group is a molecule of water: CH3 H

Step 2

H

CH3 

CH39C9O9H CH3 Departure of a water molecule

CH3

C

CH3

A carbocation



O9H

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Chapter 5 | Alkenes and Alkynes

The carbon–oxygen bond breaks heterolytically. The bonding electrons depart with the water molecule and leave behind a carbocation. The carbocation is, of course, highly reactive because the central carbon atom has only six electrons in its valence level, not eight. Finally, in step 3, a water molecule removes a proton from the β carbon of the carbocation by the process shown below. The result is the formation of a hydronium ion and an alkene. The electron pair left behind when a proton is removed becomes the second bond of the double bond of the alkene. H H9C9H

Step 3

CH3

C

CH3



H

CH2

O9H

C

CH3

H  H9O9 H



CH3

2-Methylpropene Removal of a b hydrogen

Acid-Catalyzed Dehydration of Secondary or Tertiary Alcohols: An E1 Reaction

MECHANISM Step 1

R C

C

H

R

O

H



2° or 3° Alcohol (R may be H)

H

A

fast

Acid catalyst (typically sulfuric or phosphoric acid)

R

H

C

C

O

H

R



H

A



Protonated alcohol

Conjugate base

The alcohol accepts a proton from the acid in a fast step.

Step 2

R C

C

H

R

H O

H

R 

H

slow (rate determining)

C

C

O



H

R

H

The protonated alcohol loses a molecule of water to become a carbocation. This step is slow and rate determining.

Step 3

R C H

C R

R 

A

fast

C

C



H

A

R Alkene

The carbocation loses a proton to a base. In this step, the base may be another molecule of the alcohol, water, or the conjugate base of the acid. The proton transfer results in the formation of the alkene. Note that the overall role of the acid is catalytic (it is used in the reaction and regenerated).

5.8 | Synthesis of Alkenes via Elimination Reactions

Carbocation Stability and the Transition State The order of stability of carbocations is tertiary > secondary > primary > methyl: R R9C

H 

R9C

R 3°

H 

R9C

R 

2°

H 

H9C



Methyl

H 1°



(most stable)

H (least stable)

In the dehydration of secondary and tertiary alcohols the slowest step is formation of the carbocation as shown in step 2 of the “A Mechanism for the Reaction” box above. The first and third steps involve simple acid–base proton transfers, which occur very rapidly. The second step involves loss of the protonated hydroxyl as a leaving group, a highly endergonic process, and hence it is the rate-determining step. Because step 2 is the rate-determining step, it is this step that determines the overall reactivity of alcohols toward dehydration. With that in mind, we can now understand why tertiary alcohols are the most easily dehydrated. ●

The transition state that leads to the tertiary carbocation is lowest in free energy because it resembles the carbocation that is lowest in energy.

By contrast, the transition state that leads to the primary carbocation occurs at highest free energy because it resembles the carbocation that is highest in energy. In each instance, moreover, the same factor stabilizes the transition state that stabilizes the carbocation itself: delocalization of the charge. We can understand this if we examine the process by which the transition state is formed: H C

‡

H 

O

H

C

Protonated alcohol



O



H

H

C

Transition state

O



H

Carbocation

The oxygen atom of the protonated alcohol bears a full positive charge. As the transition state develops, this oxygen atom begins to separate from the carbon atom to which it is attached. The carbon atom begins to develop a partial positive charge because it is losing the electrons that bonded it to the oxygen atom. This developing positive charge is most effectively delocalized in the transition state leading to a tertiary carbocation because three alkyl groups are present to contribute electron density by hyperconjugation to the developing carbocation. The positive charge is less effectively delocalized in the transition state leading to a secondary carbocation (two electron-releasing groups) and is least effectively delocalized in the transition state leading to a primary carbocation (one electron-releasing group). For this reason the dehydration of a primary alcohol proceeds through a different mechanism —an E2 mechanism. d1

R

Hyperconjugative stabilization

d1 R

C R

d1

d1

H O

R

H

d1

d1

Transition state leading to 3° carbocation (most stable)

d1

R

C H

d1

H O

H H

d1

Transition state leading to 2° carbocation

d1

R

C H

H d1

O

H

d1

Transition state leading to 1° carbocation (least stable)

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Chapter 5 | Alkenes and Alkynes

A Mechanism for Dehydration of Primary Alcohols: An E2 Reaction Dehydration of primary alcohols apparently proceeds through an E2 mechanism because the primary carbocation required for dehydration by an E1 mechanism is relatively unstable. The first step in dehydration of a primary alcohol is protonation, just as in the E1 mechanism. Then, with the protonated hydroxyl as a good leaving group, a Lewis base in the reaction mixture removes a β hydrogen simultaneously with formation of the alkene double bond and departure of the protonated hydroxyl group (water).

MECHANISM

Dehydration of a Primary Alcohol: An E2 Reaction

H C

C

H

H

O

H



H

A

fast

C H

Primary alcohol

H

H

C

O



H

H Protonated alcohol

Acid catalyst (typically sulfuric or phosphoric acid)

A



Conjugate base

The alcohol accepts a proton from the acid in a fast step.

H A



C

C

H

H

H O H

H

H

slow

C

(rate determining)

C

Alkene



H

A



O

H

H

A base removes a hydrogen from the  carbon as the double bond forms and the protonated hydroxyl group departs. The base may be another molecule of the alcohol or the conjugate base of the acid.

CARBOCATION STABILITY AND THE OCCURRENCE 7.7 5.9 OF MOLECULAR REARRANGEMENTS

With an understanding of carbocation stability and its effect on transition states, we can now proceed to explain the rearrangements of carbon skeletons that occur in some alcohol dehydrations.

5.9A Rearrangements During Dehydration of Secondary Alcohols Consider again the rearrangement that occurs when 3,3-dimethyl-2-butanol is dehydrated: CH3

H3C

CH39C99CH9CH3 CH3

OH

3,3-Dimethyl-2-butanol

85% H3PO4

CH3 C

heat

H3C

C

C

 CH3

2,3-Dimethyl-2-butene (major product)

CH3

H3C

CH

CH3

H2C 2,3-Dimethyl-1-butene (minor product)

5.9 | Carbocation Stability and the Occurrence of Molecular Rearrangements

The first step of this dehydration is the formation of the protonated alcohol in the usual way: Step 1 CH3

H

C

CH3

CH3 

CH

CH3 O

H

O

CH3

CH3



H

H

C

CH9CH3

 H2O

CH3 OH2 

Protonated alcohol

Protonation of the alcohol

In the second step the protonated alcohol loses water and a secondary carbocation forms: Step 2 CH3

CH3

C

CH3

CH

CH3

CH9CH3

C

CH3

CH3 OH2





H2O

CH3



A 2° carbocation

Departure of a water molecule

Now the rearrangement occurs. The less stable, secondary carbocation rearranges to a more stable tertiary carbocation: Step 3 CH3

CH3

CH3

CHCH3

C

‡

CH3







CHCH3

C

CH3

H 3C



C

CHCH3

H 3C

CH3

CH3

2° Carbocation (less stable)

Transition state

3° Carbocation (more stable)

Rearrangement by migration of a methyl group

The rearrangement occurs through the migration of an alkyl group (methyl) from the carbon atom adjacent to the one with the positive charge. The methyl group migrates with its pair of electrons (called a methanide shift). In the transition state the shifting methyl is partially bonded to both carbon atoms by the pair of electrons with which it migrates. It never leaves the carbon skeleton. After the migration is complete, the carbon atom that the methyl anion left has become a carbocation, and the positive charge on the carbon atom to which it migrated has been neutralized. Because a group migrates from one carbon to an adjacent one, this kind of rearrangement is also called a 1, 2 shift. The final step of the reaction is the removal of a proton from the new carbocation (by a Lewis base in the reaction mixture) and the formation of an alkene. This step, however, can occur in two ways: Step 4 H 2C

A

(a)



(a)

H

CH2

or (b)



C

H 3C

H C

Less stable alkene

CH3

(minor product)

CH3

 HA

CH3 CH3 Removal of a b hydrogen

CH9CH3

C

H3C (b)

CH3 C

H 3C

C CH3

(major product)

More stable alkene

245

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Chapter 5 | Alkenes and Alkynes

The more favored product is dictated by the stability of the alkene being formed. The conditions for the reaction (heat and acid) allow equilibrium to be achieved between the two forms of the alkene, and the more stable alkene is the major product because it has lower potential energy. Such a reaction is said to be under equilibrium or thermodynamic control. Path (b) leads to the highly stable tetrasubstituted alkene and this is the path followed by most of the carbocations. Path (a), on the other hand, leads to a less stable, disubstituted alkene, and because its potential energy is higher, it is the minor product of the reaction. ●

Formation of the more stable alkene is the general rule in acid-catalyzed dehydration of alcohols (Zaitsev’s rule).

Studies of many reactions involving carbocations show that rearrangements like those just described are general phenomena. They occur almost invariably when an alkanide shift or hydride shift can lead to a more stable carbocation. The following are examples: An alkanide shift

CH3 CH3

C

+

CH

CH3 CH3

CH3

+

C

CH

CH3

CH3

CH

H3C

CH3

H +

C

CH

CH3

H 3C

CH3

3° Carbocation

2° Carbocation

+

C

CH3

CH3

A hydride shift

H

2° Carbocation

3° Carbocation

Rearrangements of carbocations can also lead to a change in ring size, as the following example shows: Ring expansion

OH CH3 CH



CH3 CH

CH3

CH3

HA, heat (H2O)

2° Carbocation

CH3 CHCH3



CH3

CH3 HA





3° Carbocation

H

CH3 

CH3

A

Ring expansion by migration is especially favorable if relief in ring strain occurs.

5.9B Rearrangement After Dehydration of a Primary Alcohol Rearrangements also accompany the dehydration of primary alcohols. Since a primary carbocation is unlikely to be formed during dehydration of a primary alcohol, the alkene that is produced initially from a primary alcohol arises by an E2 mechanism, as described. However, an alkene can accept a proton to generate a carbocation in a process that is essentially the reverse of the deprotonation step in the E1 mechanism for dehydration of an alcohol. When a terminal alkene does this by using its π electrons to bond a proton at the terminal carbon, a carbocation forms at the second carbon of the chain.* This carbocation, since it is internal to the chain, will be secondary or tertiary, depending on the specific substrate. Various processes that you have already learned can now occur from this carbocation: (1) a different β hydrogen *The carbocation could also form directly from the primary alcohol by a hydride shift from its β carbon to the terminal carbon as the protonated hydroxyl group departs: R C H

H

H

C

C

R

H

1

H

O

R C

H

H

H 1

C

C

R

H

H

1

H2O

5.10 | Hydrogenation of Alkenes

may be removed, leading to a more stable alkene than the initially formed terminal alkene; (2) a hydride or alkanide rearrangement may occur leading to a yet more stable carbocation (e.g., moving from a 2° to a 3° carbocation) or to a carbocation of approximately equal stability, after which the elimination may be completed; or (3) a nucleophile may attack any of these carbocations to form a substitution product. Under the high-temperature conditions for alcohol dehydration the principal products will be alkenes rather than substitution products.

Formation of a Rearranged Alkene During Dehydration of a Primary Alcohol

MECHANISM R

H

H

C

C

C

H

R

H

R

C

O

H

H

1

A

H E2

Primary alcohol (R may be H)

H

H C

C

O

1

H

H

1

A

H

R

Initial alkene

The primary alcohol initially undergoes acid-catalyzed dehydration by an E2 mechanism.

R

C H

C

C

H

A

H

R

R

C

H

H protonation

H

C C R

H



A

H

The  electrons of the initial alkene can then be used to form a bond with a proton at the terminal carbon, forming a secondary or tertiary carbocation.*

A

R

C 

H

C R



R H C

H

C H

deprotonation

H

C R

C

H



H

A

H

Final alkene A different b hydrogen can be removed from the carbocation, so as to form a more highly substituted alkene than the initial alkene. This deprotonation step is the same as the usual completion of an E1 elimination. (This carbocation could experience other fates, such as further rearrangement before elimination or substitution by an SN1 process.)

7.11 5.10 HYDROGENATION OF ALKENES ●

Alkenes react with hydrogen in the presence of a variety of metal catalysts to add one hydrogen atom to each carbon atom of the double bond.

Hydrogenation reactions that involve insoluble platinum, palladium, or nickel catalysts proceed by heterogeneous catalysis because the catalyst is not soluble in the reaction mixture. Hydrogenation reactions that

247

248

Chapter 5 | Alkenes and Alkynes

involve soluble catalysts occur by homogeneous catalysis. Typical homogeneous hydrogenation catalysts include rhodium and ruthenium complexes that bear various phosphorus and other ligands. One of the most well-known homogeneous hydrogenation catalysts is Wilkinson’s catalyst, tris(triphenylphosphine) rhodium chloride, Rh[(C6H5)3P]3Cl. The following are some examples of hydrogenation reactions under heterogeneous and homogeneous catalysis:  H2  H2

Pt, Pd, or Ni 25 C Rh[(C6H 5)3P]3Cl

Catalytic hydrogenation reactions, like those shown above, are a type of addition reaction (versus substitution or elimination), and they are also a type of reduction. This leads to a distinction between compounds that are saturated versus those that are unsaturated. ●

●

Compounds containing only carbon–carbon single bonds (alkanes and others) are said to be saturated compounds because they contain the maximum number of hydrogen atoms that a given formula can possess. Compounds containing carbon–carbon multiple bonds (alkenes, alkynes, and aromatic compounds) are said to be unsaturated compounds because they contain fewer than the maximum number of hydrogen atoms possible for a given formula.

5.10A Hydrogenation: The Function of the Catalyst Unsaturated compounds can be reduced to saturated compounds by catalytic hydrogenation. Hydrogenation of an alkene is an exothermic reaction (DH °≅ −120 kJ mol−1): R

CH

CH

hydrogenation

R 1 H2

R

CH2

CH2

R 1 heat

Although the process is exothermic, there is usually a high free energy of activation for uncatalyzed alkene hydrogenation, and therefore, the uncatalyzed reaction does not take place at room temperature. However, hydrogenation will take place readily at room temperature in the presence of a catalyst because the catalyst provides a new pathway for the reaction that involves lower free energy of activation. Heterogeneous hydrogenation catalysts typically involve finely divided platinum, palladium, nickel, or rhodium deposited on the surface of powdered carbon (charcoal). Hydrogen gas introduced into the atmosphere of the reaction vessel adsorbs to the metal by a chemical reaction where unpaired electrons on the surface of the metal pair with the electrons of hydrogen and bind the hydrogen to the surface. The collision of an alkene with the surface bearing adsorbed hydrogen causes adsorption of the alkene as well. A stepwise transfer of hydrogen atoms takes place, and this produces an alkane before the organic molecule leaves the catalyst surface. As a consequence, both hydrogen atoms usually add from the same side of the molecule. This mode of addition is called a syn addition: C

Pt

C 

H

C H

H

C H

Catalytic hydrogenation is a syn addition.

Syn and Anti Additions An addition that places the parts of the adding reagent on the same side (or face) of the reactant is called syn addition. We have just seen that the platinum-catalyzed addition of hydrogen (X = Y = H) is a syn addition: C

C



X

Y

C X

C Y

Syn addition

5.11 | Addition Reactions of Alkenes

The opposite of a syn addition is an anti addition. An anti addition places the parts of the adding reagent on opposite faces of the reactant. Y C

C

X



Y

C

C

Anti addition

X

5.11 ADDITION REACTIONS OF ALKENES The alkenes are much more reactive than the corresponding alkanes. This greater reactivity is due to the carbon–carbon double bonds. In organic chemistry, a reaction in which two substances join to produce one compound is called an addition reaction. The most characteristic reaction of alkenes is addition to the carbon–carbon double bond in such a way that the π bond is broken and, in its place, σ bonds are formed to two new atoms or groups of atoms. We have already studied one addition reaction of alkenes—hydrogenation—in which a hydrogen atom is added at each end of a double (or triple) bond. Now we shall study other alkene addition reactions that do not involve the same mechanism as hydrogenation. We can depict this type of reaction generally, using E for an electrophilic portion of a reagent and Nu for a nucleophilic portion, as follows. C

C E

Nu

addition

E

C

C

Nu

Some specific reactions of this type include addition of hydrogen halides, sulfuric acid, water (in the presence of an acid catalyst), and halogens.

C

C

H

X

H

OH

HA (cat.)

H

C

C

X

H

C

C

OH

X

C

C

X

Alkyl halide

Alcohol

Alkene

X

X

Dihaloalkane

5.11A How to Understand Additions to Alkenes Two characteristics of the double bond help us understand why these addition reactions occur: 1. An addition reaction results in the conversion of one π bond and one σ bond into two σ bonds. The result of this change is usually energetically favorable. The energy released in making two σ bonds exceeds that needed to break one σ bond and one π bond (because π bonds are weaker), and, therefore, addition reactions are usually exothermic: C

C E

 bond

Nu

 bond

Bonds broken

C

C

E

Nu

2 bonds Bonds formed

249

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Chapter 5 | Alkenes and Alkynes

2. The electrons of the π bond are exposed. Because the π bond results from overlapping p orbitals, the π electrons lie above and below the plane of the double bond.

The electron pair of the p bond is distributed throughout both lobes of the p molecular orbital.

Electrophilic Addition ●

Electrons in the π bond of alkenes react with electrophiles.

●

Electrophiles are electron-seeking reagents. They have the property of being electrophilic.

Electrophiles are Lewis Acids Electrophiles are molecules or ions that can accept an electron pair. Nucleophiles are molecules or ions that can furnish an electron pair (i.e., Lewis bases). Any reaction of an electrophile also involves a nucleophile. In the protonation of an alkene the electrophile is the proton donated by an acid; the nucleophile is the alkene: H X9 H

C



Electrophile

C

C

C  X



Nucleophile

In the next step, the reaction of the carbocation with a halide ion, the carbocation is the electrophile and the halide ion is the nucleophile: H C

C

Electrophile



X



H

X

C

C

Nucleophile

5.11B Electrophilic Addition of Hydrogen Halides to Alkenes: Mechanism and Markovnikov’s Rule Hydrogen halides (HI, HBr, HCl, and HF) add to the double bond of alkenes: C

C

 H X

C

C

H

X

These additions are sometimes carried out by dissolving the hydrogen halide in a solvent, such as acetic acid or CH2Cl2, or by bubbling the gaseous hydrogen halide directly into the alkene and using the alkene itself as the solvent. HF is prepared as polyhydrogen fluoride in pyridine. ●

The order of reactivity of the hydrogen halides in alkene addition is HI . HBr . HCl . HF

Unless the alkene is highly substituted, HCl reacts so slowly that the reaction is not one that is useful as a preparative method.

5.11 | Addition Reactions of Alkenes

The addition of HX to an unsymmetrical alkene could conceivably occur in two ways. In practice, however, one product usually predominates. The addition of HBr to propene, for example, could conceivably lead to either 1-bromopropane or 2-bromopropane. The main product, however, is 2-bromopropane:





little Br

H Br



Br 2-Bromopropane

1-Bromopropane

When 2-methylpropene reacts with HBr, the main product is 2-bromo-2-methylpropane, not 1-bromo2-methylpropane: H Br





2-Bromo-2-methylpropane

2-Methylpropene

Br

little

Br



1-Bromo-2-methylpropane

Consideration of many examples like this led the Russian chemist Vladimir Markovnikov in 1870 to formulate what is now known as Markovnikov’s rule. ●

One way to state Markovnikov’s rule is to say that in the addition of HX to an alkene, the hydrogen atom adds to the carbon atom of the double bond that already has the greater number of hydrogen atoms.*

The addition of HBr to propene is an illustration: Alkene carbon atom with the greater number of hydrogen atoms

CH2

H

CHCH3

CH2

CHCH3

H

Br

Markovnikov addition product

Br

Reactions that illustrate Markovnikov’s rule are said to be Markovnikov additions. A mechanism for addition of a hydrogen halide to an alkene involves the following two steps:

MECHANISM

Addition of a Hydrogen Halide to an Alkene

Step 1

C C

C C

 

H H

slow slow

X X

C C

H H C C

 



X X

The p electrons of the alkene form a bond with a proton The from p electrons of thea alkene form aand bond with aion. proton HX to form carbocation a halide from HX to form a carbocation and a halide ion.

Step 2  X

X

 

C C

H H C C

fast fast

C C X X

H H C C

The halide ion reacts with the carbocation by The halide ion reacts the carbocation by donating an electron pair;with the result is an alkyl halide. donating an electron pair; the result is an alkyl halide.

*In his original publication, Markovnikov described the rule in terms of the point of attachment of the halogen atom, stating that “if an unsymmetrical alkene combines with a hydrogen halide, the halide ion adds to the carbon atom with the fewer hydrogen atoms.”

251

252

Chapter 5 | Alkenes and Alkynes

The important step—because it is the rate-determining step—is step 1. In step 1 the alkene donates a pair of electrons to the proton of the hydrogen halide and forms a carbocation. This step is highly endergonic and has a high free energy of activation. Consequently, it takes place slowly. In step 2 the highly reactive carbocation stabilizes itself by combining with a halide ion. This exergonic step has a very low free energy of activation and takes place very rapidly. Theoretical Explanation of Markovnikov’s Rule If the alkene that undergoes addition of a hydrogen halide is an unsymmetrical alkene such as propene, then step 1 could conceivably lead to two different carbocations: H X 9 H  CH3CH



CH3CH

CH2

CH2  X

1 Carbocation (less stable)

CH3CH



CH2  H 9 X

CH3CH

CH2

H  X

2 Carbocation (more stable)

These two carbocations are not of equal stability, however. The secondary carbocation is more stable, and it is the greater stability of the secondary carbocation that accounts for the correct prediction of the overall addition by Markovnikov’s rule. In the addition of HBr to propene, for example, the reaction takes the following course: 

CH3CH2CH2

Br

1-Bromopropane (little formed)

1°

CH3CH

CH2

HBr slow 

CH3CHCH3 2°

CH3CH2CH2Br

Br fast

CH3CHCH3 Br 2-Bromopropane (main product)

Step 1

Step 2

The chief product of the reaction is 2-bromopropane because the more stable secondary carbocation is formed preferentially in the first step. ● ●

●

The more stable carbocation predominates because it is formed faster. The reaction leading to the secondary carbocation (and ultimately to 2-bromopropane) has the lower free energy of activation. This is reasonable because its transition state resembles the more stable carbocation. The reaction leading to the primary carbocation (and ultimately to 1-bromopropane) has a higher free energy of activation because its transition state resembles a less stable primary carbocation. This second reaction is much slower and does not compete appreciably with the first reaction.

The reaction of HBr with 2-methylpropene produces only 2-bromo-2-methylpropane, for the same reason regarding carbocations stability. Here, in the first step (i.e., the attachment of the proton) the choice is even more pronounced—between a tertiary carbocation and a primary carbocation. Thus, 1-bromo-2-methylpropane is not obtained as a product of the reaction because its formation would require

5.11 | Addition Reactions of Alkenes

the formation of a primary carbocation. Such a reaction would have a much higher free energy of activation than that leading to a tertiary carbocation. ●

Rearrangements invariably occur when the carbocation initially formed by addition of HX to an alkene can rearrange to a more stable one.

MECHANISM

Addition of HBr to 2-Methylpropene

This reaction takes place:

H3C

CH3

H3C C

C9CH29H

CH2

H3C

H

CH3



H3C

Br

Br



C

Major

CH3 product

Br

3° Carbocation (more stable carbocation)

2-Bromo-2-methylpropane

This reaction does not occur to any appreciable extent:

CH3

H3C

C

CH3

CH2

H3C

Br

C H

H

CH3 

CH3

CH2

Br

CH

CH2

Br

Little formed



1° Carbocation (less stable carbocation)

1-Bromo-2-methylpropane

General Statement of Markovnikov’s Rule With this understanding of the mechanism for the ionic addition of hydrogen halides to alkenes, we can now generalize about how electrophiles add to alkenes. ●

●

General statement of Markovnikov’s rule: In the ionic addition of an unsymmetrical reagent to a double bond, the positive portion of the adding reagent attaches itself to a carbon atom of the double bond so as to yield the more stable carbocation as an intermediate. Addition of the electrophile determines the overall orientation of the addition, because it occurs first (before the addition of the nucleophilic portion of the adding reagent).

Notice that this formulation of Markovnikov’s rule allows us to predict the outcome of the addition of a reagent such as ICl. Because of the greater electronegativity of chlorine, the positive portion of this molecule is iodine. The addition of ICl to 2-methylpropene takes place in the following way and produces 2-chloro-1-iodo-2-methylpropane: H3C C H3C

CH2 





I

Cl

2-Methylpropene

CH3

H3C C CH2 H3C

CH3

I Cl



C

CH2

I

Cl 2-Chloro-1-iodo2-methylpropane

253

254

Chapter 5 | Alkenes and Alkynes

Regioselective Reactions Chemists describe reactions like the Markovnikov additions of hydrogen halides to alkenes as being regioselective. Regio comes from the Latin word regionem meaning direction. ●

When a reaction that can potentially yield two or more constitutional isomers actually produces only one (or a predominance of one), the reaction is said to be a regioselective reaction.

The addition of HX to an unsymmetrical alkene such as propene could conceivably yield two constitutional isomers, for example. As we have seen, however, the reaction yields only one, and therefore it is regioselective. An Exception to Markovnikov’s Rule The exception to Markovnikov’s rule concerns the addition of HBr to alkenes when the addition is carried out in the presence of peroxides (i.e., compounds with the general formula ROOR). ●

When alkenes are treated with HBr in the presence of peroxides, an anti-Markovnikov addition occurs in the sense that the hydrogen atom becomes attached to the carbon atom with the fewer hydrogen atoms.

With propene, for example, the addition takes place as follows: CH3 CH ●

CH2 1 HBr

ROOR

CH3CH2CH2Br

This anti-Markovnikov addition occurs only when HBr is used in the presence of peroxides and does not occur significantly with HF, HCl, and HI even when peroxides are present. This is known as peroxide effect or kharasch effect.

5.11C The Anti-Markovnikov Addition of Hydrogen Bromide Radical Addition to Alkenes Before 1933, the orientation of the addition of hydrogen bromide to alkenes was the subject of much confusion. At times addition occurred in accordance with Markovnikov’s rule; at other times it occurred in just the opposite manner. Many instances were reported where, under what seemed to be the same experimental conditions, Markovnikov additions were obtained in one laboratory and anti-Markovnikov additions in another. At times even the same chemist would obtain different results using the same conditions but on different occasions. The mystery was solved in 1933 by the research of M. S. Kharasch and F. R. Mayo (of the University of Chicago). The explanatory factor turned out to be organic peroxides present in the alkenes—peroxides that were formed by the action of atmospheric oxygen on the alkenes. R

O

O

R

An organic peroxide ●

R

O

O

H

An organic hydroperoxide

When alkenes containing peroxides or hydroperoxides react with hydrogen bromide, antiMarkovnikov addition of HBr occurs.

For example, in the presence of peroxides propene yields 1-bromopropane. In the absence of peroxides, or in the presence of compounds that “trap” radicals, normal Markovnikov addition occurs. Br

HBr, ROOR (peroxides present)

Anti-Markovnikov addition

Br HBr (peroxides absent)

Markovnikov addition

5.11 | Addition Reactions of Alkenes ●

Hydrogen bromide is the only hydrogen halide that gives anti-Markovnikov addition when peroxides are present.

Hydrogen fluoride, hydrogen chloride, and hydrogen iodide do not give anti-Markovnikov addition even when peroxides are present. The mechanism for anti-Markovnikov addition of hydrogen bromide is a radical chain reaction initiated by peroxides.

MECHANISM

Anti-Markovnikov Addition of HBr

Chain Initiation

R

Step 1

O

O

R

heat

2R

O

Heat brings about homolytic cleavage of the weak oxygen–oxygen bond.

R

Step 2

O

 H

Br

R

O H 

Br

The alkoxyl radical abstracts a hydrogen atom from HBr, producing a bromine radical.

Chain Propagation

Br

Step 3

 H2C

CH

CH3

Br CH2

CH

CH3

2° Radical A bromine radical adds to the double bond to produce the more stable 2° alkyl radical.

Br

Step 4

CH2

CH

CH3  H

Br

Br

CH2

CH H

CH3  Br

1-Bromopropane The alkyl radical abstracts a hydrogen atom from HBr. This leads to the product and regenerates a bromine radical. Then repetitions of steps 3 and 4 lead to a chain reaction.

Step 1 is the simple homolytic cleavage of the peroxide molecule to produce two alkoxyl radicals. The oxygen–oxygen bond of peroxides is weak, and such reactions are known to occur readily: R

OO

R

Peroxide

2R

O

DH8

1150 kJ mol21

Alkoxyl radical

Step 2 of the mechanism, abstraction of a hydrogen atom by the radical, is exothermic and has a low energy of activation: R

O 1 H Br

R

O H 1 Br

DH8

296 kJ mol21

Eact is low

255

256

Chapter 5 | Alkenes and Alkynes

Step 3 of the mechanism determines the final orientation of bromine in the product. It occurs as it does because a more stable secondary radical is produced and because attack at the primary carbon atom is less hindered. Had the bromine attacked propene at the secondary carbon atom, a less stable, primary radical would have been the result,

Br  CH2

CHCH3

CH2CHCH3 Br 1° Radical (less stable)

and attack at the secondary carbon atom would have been more hindered. Step 4 of the mechanism is simply the abstraction of a hydrogen atom from hydrogen bromide by the radical produced in step 3. This hydrogen atom abstraction produces a bromine atom (which, of course, is a radical due to its unpaired electron) that can bring about step 3 again; then step 4 occurs again—a chain reaction. Summary of Markovnikov versus Anti-Markovnikov Addition of HBr to Alkenes We can now see the contrast between the two ways that HBr can add to an alkene. In the absence of peroxides, the reagent that attacks the double bond first is a proton. Because a proton is small, steric effects are unimportant. It attaches itself to a carbon atom by an ionic mechanism so as to form the more stable carbocation. The result is Markovnikov addition. Polar, protic solvents favor this process. Ionic Addition Br –

H ¬Br

+

Br

H Markovnikov product

Addition to form the more stable carbocation

In the presence of peroxides, the reagent that attacks the double bond first is the larger bromine atom. It attaches itself to the less hindered carbon atom by a radical mechanism, so as to form the more stable radical intermediate. The result is anti-Markovnikov addition. Nonpolar solvents are preferable for reactions involving radicals. Radical Addition H ¬Br Br

H Br

Br

Addition to form the more stable alkyl radical



Br

Anti-Markovnikov product

Stereochemistry of the Ionic Addition to an Alkene Consider the following addition of HX to 1-butene and notice that the reaction leads to the formation of a 2-halobutane that contains a chirality center: CH3CH2CH "CH2  HX

CH3CH2*CHCH3 X

5.11 | Addition Reactions of Alkenes

The product, therefore, can exist as a pair of enantiomers. The question now arises as to how these enantiomers are formed. Is one enantiomer formed in greater amount than the other? The answer is no; the carbocation that is formed in the first step of the addition (see the following scheme) is trigonal planar and is achiral (a model will show that it has a plane of symmetry). When the halide ion reacts with this achiral carbocation in the second step, reaction is equally likely at either face. The reactions leading to the two enantiomers occur at the same rate, and the enantiomers, therefore, are produced in equal amounts as a racemic mixture.

5.11D Addition of Water to Alkenes: Acid-Catalyzed Hydration The acid-catalyzed addition of water to the double bond of an alkene (hydration of an alkene) is a method for the preparation of low-molecular-weight alcohols. This reaction has its greatest utility in large-scale industrial processes. The acids most commonly used to catalyze the hydration of alkenes are dilute aqueous solutions of sulfuric acid and phosphoric acid. These reactions, too, are usually regioselective, and the addition of water to the double bond follows Markovnikov’s rule. In general, the reaction takes the form that follows: C

C

 HOH

H3O

C

C

H

OH

An example is the hydration of 2-methylpropene:  HOH

H3O

H

25 °C

2-Methylpropene (isobutylene)

HO 2-Methyl-2-propanol (tert-butyl alcohol)

Because the reactions follow Markovnikov’s rule, acid-catalyzed hydrations of alkenes do not yield primary alcohols except in the special case of the hydration of ethene: CH2

CH2 1 HOH

H3PO4 300 °C

CH3CH2OH

Mechanism The mechanism for the hydration of an alkene is simply the reverse of the mechanism for the dehydration of an alcohol. We can illustrate this by giving the mechanism for the hydration of 2-methylpropene and by comparing it with the mechanism for the dehydration of 2-methyl-2-propanol given in Section 7.7A.

MECHANISM

Acid-Catalyzed Hydration of an Alkene

CH2

Step 1

H 3C

C

CH3

H  H

O

slow

H

CH2 CH3

C



H

H 

O

H

CH3

The alkene donates an electron pair to a proton to form the more stable 3° carbocation.

H

CH3 H 3C

C

CH3



O

CH3 H H

fast

CH3

C CH3

O

(mechanism continues on the next page)



The carbocation reacts with a molecule of water to form a protonated alcohol.

H

257

CH2

258

H

 H C H C CH 3 3 Chapter 5 | Alkenes and Alkynes

O

H

C

CH3



H

H

CH2

slow



O

H

CH3

The alkene donates an electron pair to a proton to form the more stable 3° carbocation.

CH3

Step 2

H3C

C

H

CH3



CH3 H

O

fast

H

CH3

C

O

CH3

H



The carbocation reacts with a molecule of water to form a protonated alcohol.

Step 3

CH3 H CH3

H

O H 

C

O

CH3 fast

H

CH3

C

CH3

H O

H  H

O

H



CH3

A transfer of a proton to a molecule of water leads to the product.

The rate-determining step in the hydration mechanism is step 1: the formation of the carbocation. It is this step, too, that accounts for the Markovnikov addition of water to the double bond. The reaction produces 2-methyl-2-propanol because step 1 leads to the formation of the more stable tertiary (3 °) cation rather than the much less stable primary (1 °) cation: CH2 C H3C

 CH3



H H

O

very

H

CH39C

slow



H

CH2 H 

O

For all practical purposes this reaction does not take place because it produces a 1 carbocation.

H

CH3

The reactions whereby alkenes are hydrated or alcohols are dehydrated are reactions in which the ultimate product is governed by the position of an equilibrium. Therefore, in the dehydration of an alcohol it is best to use a concentrated acid so that the concentration of water is low. (The water can be removed as it is formed, and it helps to use a high temperature.) In the hydration of an alkene it is best to use dilute acid so that the concentration of water is high. It also usually helps to use a lower temperature. Rearrangements ● One complication associated with alkene hydrations is the occurrence of rearrangements. Because the reaction involves the formation of a carbocation in the first step, the carbocation formed initially invariably rearranges to a more stable one (or possibly to an isoenergetic one) if such a rearrangement is possible. An illustration is the formation of 2,3-dimethyl-2-butanol as the major product when 3,3-dimethyl-1-butene is hydrated: cat. H2SO4 H2O

3,3-Dimethyl-1-butene

HO 2,3-Dimethyl-2-butanol (major product)

5.11E Electrophilic Addition of Bromine and Chlorine to Alkenes Alkenes react rapidly with bromine and chlorine in nonnucleophilic solvents to form vicinal dihalides. An example is the addition of chlorine to ethene. H

H C

Cl2

C

H

H Ethene

H

Cl

H

C

C

H

Cl

H

1,2-Dichloroethane

5.11 | Addition Reactions of Alkenes

This addition is a useful industrial process because 1,2-dichloroethane can be used as a solvent and can be used to make vinyl chloride, the starting material for poly(vinyl chloride). Cl H

H

C

C

H

Cl

H E2 base (–HCl)

H

H C

polymerization (see Section 10.10)

C

H

Cl

H

H

C

C

H

Cl n

1,2-Dichloroethane

Vinyl chloride

Poly(vinyl chloride)

Other examples of the addition of halogens to a double bond are the following: Cl

Cl2

Cl trans-2-Butene

meso-1,2-Dichlorobutane

Br

Br2

Br Br

Br Cyclohexene

trans-1,2-Dibromocyclohexane (racemic)

These two examples show an aspect of these additions that we shall address later when we examine a mechanism for the reaction: the addition of halogens is an anti addition to the double bond. When bromine is used for this reaction, it can serve as a test for the presence of carbon–carbon multiple bonds. If we add bromine to an alkene (or alkyne, see Section 8.17), the red-brown color of the bromine disappears almost instantly as long as the alkene (or alkyne) is present in excess: Rapid decolorization of Br2 is a positive test for alkenes and alkynes.

Br C

C



Br2

room temperature in the dark

C

C Br

An alkene (colorless)

vic-Dibromide (a colorless compound)

Bromine (red-brown)

This behavior contrasts markedly with that of alkanes. Alkanes do not react appreciably with bromine or chlorine at room temperature and in the absence of light. When alkanes do react under those conditions, however, it is by substitution rather than addition and by a mechanism involving radicals that we shall discuss in Chapter 10: R9 H  Alkane (colorless)

Br2

room temperature in the dark

No appreciable reaction

Bromine (red-brown)

Mechanism of Halogen Addition A possible mechanism for the addition of a bromine or chlorine to an alkene is one that involves the formation of a carbocation. Br C

C

+ Br ¬ Br

C

+

C

-

+ Br

Br

Br

C

C

Although this mechanism is similar to ones we have studied earlier, such as the addition of H } X to an alkene, it does not explain an important fact. As we have just seen the addition of bromine or chlorine to an alkene is an anti addition.

259

260

Chapter 5 | Alkenes and Alkynes

The addition of bromine to cyclopentene, for example, produces trans-1,2-dibromocyclopentane, not cis-1,2-dibromocyclopentane. + Br ¬ Br

Br

anti

Br

Br

not

addition

Br trans-1,2-Dibromocyclopentane (as a racemic mixture)

cis-1,2-Dibromocyclohexane (a meso compound)

A mechanism that explains anti addition is one in which a bromine molecule transfers a bromine atom to the alkene to form a cyclic bromonium ion and a bromide ion, as shown in step 1 of “A Mechanism for the Reaction” that follows. The cyclic bromonium ion causes net anti addition, as follows. In step 2, a bromide ion attacks the back side of either carbon 1 or carbon 2 of the bromonium ion (an SN2 process) to open the ring and produce the trans-1,2-dibromide. Attack occurs from the side opposite the bromine of the bromonium ion because attack from this direction is unhindered. Attack at the other carbon of the cyclic bromonium ion produces the enantiomer.

MECHANISM

Addition of Bromine to an Alkene

Step 2

Step 1 C

C



Br



Br

C

C



Br



C

Br

Br





Bromonium ion

Bromide ion

As a bromine molecule approaches an alkene, the electron density of the alkene p bond repels electron density in the closer bromine, polarizing the bromine molecule and making the closer bromine atom electrophilic. The alkene donates a pair of electrons to the closer bromine, causing displacement of the distant bromine atom. As this occurs, the newly bonded bromine atom, due to its size and polarizability, donates an electron pair to the carbon that would otherwise be a carbocation, thereby stabilizing the positive charge by delocalization. The result is a bridged bromonium ion intermediate.

Br C



Bromonium ion

Br



C

C



enantiomer

Br Bromide ion

vic-Dibromide

A bromide anion attacks at the back side of one carbon (or the other) of the bromonium ion in an SN2 reaction, causing the ring to open and resulting in the formation of a vic-dibromide.

The mechanisms for addition of Cl2 and I2 to alkenes are similar to that for Br2, involving formation and ring opening of their respective halonium ions. The anti addition of a halogen to an alkene provides us with an example of what is called a stereospecific reaction. ●

A stereospecific reaction is one where a particular stereoisomer of the starting material yields a specific stereoisomeric form of the product.

Consider the reactions of cis- and trans-2-butene with bromine shown below. When trans-2-butene adds bromine, the product is the meso compound, (2R,3S)-2,3-dibromobutane. When cis-2-butene

5.11 | Addition Reactions of Alkenes

adds bromine, the product is a racemic mixture of (2R,3R)-2,3-dibromobutane and (2S,3S)2,3-dibromobutane: Reaction 1 H

CH3

CH3

H

Br2

CH3 H

Br H

‚

Br

CH3

Br

Br

H CH3

H CH3

(2R,3S)-2,3-Dibromobutane (a meso compound)

trans-2-Butene

Reaction 2 H

H

CH3

CH3

Br2

H

Br H

Br

CH3



H CH3

Br

Br

CH3

cis-2-Butene

H CH3

(2R,3R)

(2S,3S)

(a pair of enantiomers)

Halohydrin Formation ●

When the halogenation of an alkene is carried out in aqueous solution, rather than in a nonnucleophilic solvent, the major product is a halohydrin (also called a halo alcohol) instead of a vicdihalide.

Molecules of water react with the halonium ion intermediate as the predominant nucleophile because they are in high concentration (as the solvent). The result is formation of a halohydrin as the major product. If the halogen is bromine, it is called a bromohydrin, and if chlorine, a chlorohydrin. X

OH C

 X2  H2O

C

C

C

C



X X  Cl or Br

C

 HX

X

Halohydrin (major)

vic-Dihalide (minor)

Halohydrin formation can be described by the following mechanism.

MECHANISM

Step 1

Halohydrin Formation from an Alkene

C

C X



X



C

C



X



X 

Halonium ion

Halide ion

This step is the same as for halogen addition to an alkene (see Section 8.11A).

O

H 

O

Steps 2 and 3

C

C X 



O H

H

C X

H

H

H

(continues on next page)

H

O

C

C X

C



H

O H H

261

Step 1

262

C

C

C

X



Chapter 5 | Alkenes and Alkynes X



C

X





X 

Halonium ion

Halide ion

This step is the same as for halogen addition to an alkene (see Section 8.11A).

Steps 2 and 3

C

C

O



X

H

C

H

O

O H

H

H

O C

C

C



H

X

X

H

H

O H H



Halonium ion

Protonated halohydrin

Here, however, a water molecule acts as the nucleophile and attacks a carbon of the ring, causing the formation of a protonated halohydrin.

Halohydrin

The protonated halohydrin loses a proton (it is transferred to a molecule of water). This step produces the halohydrin and hydronium ion.

The first step is the same as that for halogen addition. In the second step, however, the two mechanisms differ. In halohydrin formation, water acts as the nucleophile and attacks one carbon atom of the halonium ion. The three-membered ring opens, and a protonated halohydrin is produced. Loss of a proton then leads to the formation of the halohydrin itself. ●

If the alkene is unsymmetrical, the halogen ends up on the carbon atom with the greater number of hydrogen atoms.

Bonding in the intermediate bromonium ion is unsymmetrical. The more highly substituted carbon atom bears the greater positive charge because it resembles the more stable carbocation. Consequently, water attacks this carbon atom preferentially. The greater positive charge on the tertiary carbon permits a pathway with a lower free energy of activation even though attack at the primary carbon atom is less hindered: This bromonium ion is bridged asymmetrically because the 3° carbon can accommodate more positive charge than the 1° carbon.

H H 

Br2

O

HO

OH2

OH2

Br

Br



H3O

 Br (73%)

5.11F Addition of Sulphuric Acid When alkenes are treated with cold concentrated sulphuric acid, they dissolve because they react by addition to form alkyl hydrogen sulphates. The mechanism is similar to that for the addition of HX. In the first step of this reaction the alkene donates a pair of electrons to a proton from sulphuric acid to form a carbocation; in the second step the carbocation reacts with a hydrogen sulphate ion to form an alkyl hydrogen sulphate: H

O C

C

1H

O

S

O

H

1

C

C

O 2 1 O

O Alkene

Sulfuric acid

S

O

HO3SO

H

C

C

H

O Carbocation

Hydrogen sulfate ion

Alkyl hydrogen sulfate

Soluble in sulfuric acid

5.12 | Oxidation of Alkenes: Syn 1,2-Dihydroxylation

The addition of sulphuric acid is also regioselective and it follows Markovnikov’s rule. Propene, for example, reacts to yield isopropyl hydrogen sulphate rather than propyl hydrogen sulphate: H

H

H C

H3C

CH2 H

CH2

C H3C

OSO3H

1

H3C

H

C

2

CH3

OSO3H

OSO3H

28 Carbocation Isopropyl hydrogen (more stable carbocation) sulfate

5.12 OXIDATION OF ALKENES: SYN 1,2-DIHYDROXYLATION Alkenes undergo a number of reactions in which the carbon–carbon double bond is oxidized. ●

1,2-Dihydroxylation is an important oxidative addition reaction of alkenes.

Osmium tetroxide is widely used to synthesize 1,2-diols (the products of 1,2-dihydroxylation, sometimes also called glycols). Potassium permanganate can also be used, although because it is a stronger oxidizing agent it is prone to cleave the diol through further oxidation. 1. OsO4, pyridine 2. NaHSO3 /H2O

OH OH 1,2-Propanediol (propylene glycol)

Propene

CH2

HO

CH2  KMnO4

H2O, cold

Ethene

●

HO

OH

1,2-Ethanediol (ethylene glycol)

The mechanism for the formation of a 1,2-diol by osmium tetroxide involves a cyclic intermediate that results in syn addition of the oxygen atoms (see below).

After formation of the cyclic intermediate with osmium, cleavage at the oxygen–metal bonds takes place without altering the stereochemistry of the two new C } O bonds. C

C

C

pyridine

O

 O

O Os

O O

Os O

O

NaHSO3 H 2O

C

C

C

OH

OH

HO  OH

O

Os

An osmate ester

O

O

The syn stereochemistry of this dihydroxylation can readily be observed by the reaction of cyclopentene with osmium tetroxide. The product is cis-1,2-cyclopentanediol.

+ OsO4

NaHSO3 25 °C, pyridine

H2O

O

O Os

O

O

HO

OH

cis -1,2-Cyclopentanediol (a meso compound)

263

264

Chapter 5 | Alkenes and Alkynes

5.12A Oxidative Cleavage of Alkenes Alkenes can be oxidatively cleaved using potassium permanganate or ozone (as well as by other reagents). Potassium permanganate (KMnO4) is used when strong oxidation is needed. Ozone (O3) is used when mild oxidation is desired. Cleavage with Hot Basic Potassium Permanganate ●

Treatment with hot basic potassium permanganate oxidatively cleaves the double bond of an alkene.

Cleavage is believed to occur via a cyclic intermediate similar to the one formed with osmium tetroxide and intermediate formation of a 1,2-diol. ● ● ●

Alkenes with monosubstituted carbon atoms are oxidatively cleaved to salts of carboxylic acids. Disubstituted alkene carbons are oxidatively cleaved to ketones. Unsubstituted alkene carbons are oxidized to carbon dioxide.

The following examples illustrate the results of potassium permanganate cleavage of alkenes with different substitution patterns. In the case where the product is a carboxylate salt, an acidification step is required to obtain the carboxylic acid. O

O CH3CH

KMnO4, HO2, H2O

CHCH3

heat

2 C H3C O2

(cis or trans)

CH2

H3O1

2 H3C

Acetate ion 1. KMnO4, HO2 heat 2. H3O1

O 1

C OH

Acetic acid

O

C

O 1 H2O

One of the uses of potassium permanganate, other than for oxidative cleavage, is as a chemical test for unsaturation in an unknown compound. ●

If an alkene is present (or an alkyne, Section 5.18B), the purple color of a potassium permanganate solution is discharged and a brown precipitate of manganese dioxide (MnO2) forms as the oxidation takes place.

The oxidative cleavage of alkenes has also been used to establish the location of the double bond in an alkene chain or ring. The reasoning process requires us to think backward much as we do with retrosynthetic analysis. Here we are required to work backward from the products to the reactant that might have led to those products. We can see how this might be done with the following example.

5.12B Cleavage with Ozone ●

The most useful method for cleaving alkenes is to use ozone (O3).

Ozonolysis consists of bubbling ozone into a very cold (−78 °C) solution of the alkene in CH2Cl2, followed by treatment of the solution with dimethyl sulfide (or zinc and acetic acid). The overall result is as follows: R C R9

1. O3 , CH2Cl2 , 2 78 °C

C

C

2. Me2S

H

R0

R

R0

R9

O 1

O

C

H

The reaction is useful as a synthetic tool, as well as a method for determining the location of a double bond in an alkene by reasoning backward from the structures of the products. ●

The overall process (above) results in alkene cleavage at the double bond, with each carbon of the double bond becoming doubly bonded to an oxygen atom.

5.12 | Oxidation of Alkenes: Syn 1,2-Dihydroxylation

The following examples illustrate the results for each type of alkene carbon. O

O

1. O3 , CH 2Cl2 ,278 °C

1

2. Me2S

2-Methyl-2-butene

Acetone

H

Acetaldehyde

O 1. O3 , CH2Cl2 , 2 78 °C

3-Methyl-1-butene

O

2. Me2S

1

H

H

Isobutyraldehyde Formaldehyde

The mechanism of ozone addition to alkenes begins with formation of unstable compounds called initial ozonides (sometimes called molozonides). The process occurs vigorously and leads to spontaneous (and sometimes noisy) rearrangement to compounds known as ozonides. The rearrangement is believed to occur with dissociation of the initial ozonide into reactive fragments that recombine to yield the ozonide. Ozonides are very unstable compounds, and low-molecular-weight ozonides often explode violently.

MECHANISM

Ozonolysis of an Alkene

C

C



O

O

C O

O

O

Ozone adds to the alkene to form an initial ozonide.

O

O

O

O





O

The initial ozonide fragments.

O C



O

C



Initial ozonide



C

C

C

O

C

C

O

Me2S

O

Ozonide The fragments recombine to form the ozonide.

C

O  O

C

Aldehydes and/or ketones

 Me2SO Dimethyl sulfoxide

Combustion Alkenes, like alkanes, on heating in the presence of air or oxygen undergo complete oxidation to CO2 and H2O. The reaction is accompanied by evolution of large amount of heat. CnH2n 1

3n O 2 2

nCO2 1 nH2O 1 Heat

PART II: ALKYNES The common name for this family is ethynes (acetylenes), after the first member, HC CH which is often used as oxy-acetylene (in combination with oxygen) and is used for welding purposes The general formula of alkynes is CnH2n−2. H

C C H Ethyne

265

Chapter 5 | Alkenes and Alkynes

5.13 STRUCTURE OF THE TRIPLE BOND The functional group of an alkyne is a carbon–carbon triple bond. The simplest alkyne is ethyne, C2H2. Ethyne is a linear molecule; all of its bond angles are 180° (Fig. 5.10).

H

C C H Acetylene

Side view

End view

Figure 5.10 Shape of ethyne (C2H2). The mathematical process for obtaining the sp hybrid orbitals of ethyne can be visualized in the following way (Fig. 5.11). The 2s orbital and one 2p orbital of carbon are hybridized to form two sp orbitals. Ground state

Energy

266

Excited state

2p

2p

2s

2s

1s

1s

sp-Hybridized state 2p 2sp

Promotion of electron

1s Hybridization

Figure 5.11 A process for obtaining sp hybridized carbon atoms. The remaining two 2p orbitals are not hybridized. p Orbitals Calculations show that the sp hybrid orbitals have their large positive lobes oriented at an angle of 180° with respect to each other. The two 2p orbitals that were not hybridized are each perpendicular to the axis sp Orbital C that passes through the center of the two sp orbitals (Fig. 5.12). sp Orbital We envision the bonding molecular orbitals of ethyne being formed in the following way (Fig. 5.13). Two carbon atoms overlap sp orbitals to form a σ bond between them (this is one bond of the triple bond). The remaining two sp orbitals at each carbon Figure 5.12 An sp-hybridized carbon atom. atom overlap with s orbitals from hydrogen atoms to produce two σ C } H bonds. The two p orbitals on each carbon atom also overlap from side to side to form two π bonds. These are the other two bonds of the triple bond. Thus we see that the carbon–carbon triple bond consists of two π bonds and one σ bond.

5.13A Bond Lengths of Ethyne, Ethene and Ethane The carbon–carbon triple bond is shorter than the carbon–carbon double bond, and the carbon–carbon double bond is shorter than the carbon–carbon single bond. The carbon–hydrogen bonds of ethyne are also shorter than those of ethene, and the carbon–hydrogen bonds of ethene are shorter than those of ethane. This illustrates a general principle: The shortest C } H bonds are associated with those carbon

5.14 | Isomerism

π Bond σ Bond H C C H π Bond

Figure 5.13 Formation of the bonding molecular orbitals of ethyne from two sp-hybridized carbon atoms and two hydrogen atoms.

orbitals with the greatest s character. The sp orbitals of ethyne, 50% s (and 50% p) in character, form the shortest C } H bonds. The sp3 orbitals of ethane, 25% s (and 75% p) in character, form the longest C } H bonds. The differences in bond lengths and bond angles of ethyne, ethene and ethane are summarized in Fig. 5.14.

H

C

H

180°

118° C

1.20Å

H H 121°

1.34Å

C 1.06Å

1.09Å H

C

H 109.5°

1.54Å 1.10Å H

C

H

H

H

C

H

H

Figure 5.14 Bond angles and bond lengths of ethyne, ethene and ethane.

5.14 ISOMERISM Alkynes, having a linear shape across the carbon–carbon triple bond, do not exist as cis and trans isomers. For alkynes, ethyne and propyne, only one structure is possible. But in higher alkynes such as butyne, pentyne, etc., where by changing the position of the triple bond, different structures can be obtained are called position isomers. For example, but-1-yne and but-2-yne are position isomers. CH3

CH2 C CH But-1-yne

CH3

C C CH3 But-2-yne

267

268

Chapter 5 | Alkenes and Alkynes

Another type of isomerism observed in alkynes with five or more carbons is chain isomerism that arises due to difference in placement of the carbon side chains. For example, the possible isomers for C5H8 are CH3 CH3

CH2 CH2 C Pent-1-yne (I)

CH

CH3

C C CH2 Pent-2-yne (II)

CH

C

CH

CH3 3-Methyl but-1-yne

CH3

(III)

Here, I and II are positions isomers while II and III are chain isomers. Alkynes also show functional isomerism with dienes, which are compounds containing two double bonds (discussed later in the chapter). For example, CH3

C

CH2

CH

CH2

But-1-yne

CH

CH

CH2

Buta-1,3-diene

Alkynes can also show ring chain isomerism with cycloalkenes. For example, propyne and cyclopropene are ring chain isomers.

5.15 HOW TO NAME ALKYNES Alkynes are named in much the same way as alkenes. Unbranched alkynes, for example, are named by replacing the -ane of the name of the corresponding alkane with the ending -yne. The chain is numbered to give the carbon atoms of the triple bond the lower possible numbers. The lower number of the two carbon atoms of the triple bond is used to designate the location of the triple bond. The IUPAC names of three unbranched alkynes are shown here: H

C

C

H

Ethyne or acetylene*

1-Penten-4-yne† or pent-1-en-4-yne

2-Pentyne

The locations of substituent groups of branched alkynes and substituted alkynes are also indicated with numbers. An—OH group has priority over the triple bond when numbering the chain of an alkynol: Cl

2

3

1

3-Chloropropyne

5 6

4

3

2 1

5-Methyl-1-hexyne or 5-methylhex-1-yne

4

3

Cl

2

1

1-Chloro-2-butyne or 1-chlorobut-2-yne

5

4

2

1 4

3

3-Butyn-1-ol or but-3-yn-1-ol

OH

1

3

4,4-Dimethyl-1-pentyne or 4,4-dimethylpent-1-yne

*The name acetylene is retained by the IUPAC system for the compound HC

OH

2

1

2

3

4

5

2-Methyl-4-pentyn-2-ol or 2-methylpent-4-yn-2-ol CH and is used frequently.

†When double and triple bonds are present, the direction of numbering is chosen so as to give the lowest overall set of locants. In the face of equivalent options, then preference is given to assigning lowest numbers to the double bonds.

5.16 | The Acidity of Terminal Alkynes

Monosubstituted acetylenes or 1-alkynes are called terminal alkynes, and the hydrogen attached to the carbon of the triple bond is called the acetylenic hydrogen atom. Acetylenic hydrogen

R

C

C

H

A terminal alkyne

When named as a substituent, the HC C } group is called the ethynyl group. The anion obtained when the acetylenic hydrogen is removed is known as an alkynide ion or an acetylide ion. As we shall see in Section 5.17B, these ions are useful in synthesis: R

C or

R

C

–

CH3C or

–

An alkynide ion (an acetylide ion)

C

–

–

The propynide ion

7.8 5.16 THE ACIDITY OF TERMINAL ALKYNES The hydrogen bonded to the carbon of a terminal alkyne, called an acetylenic hydrogen atom, is considerably more acidic than those bonded to carbons of an alkene or alkane. The pKa values for ethyne, ethene, and ethane illustrate this point: A terminal alkyne is ~1020 times more acidic than an alkene or alkane.

H

H H

C

C

C

H H

pKa = 25

C

H H

pKa = 44

H

H

C

C

H

H

H

pKa = 50

The order of basicity of their anions is opposite that of their relative acidity: Relative Basicity CH3CH2: 2 . CH2

CH:2 . HC

C:2

If we include in our comparison hydrogen compounds of other first-row elements of the periodic table, we can write the following orders of relative acidities and basicities. This comparison is useful as we consider what bases and solvents to use with terminal alkynes. Relative Acidity Most acidic

H

Least acidic

OH . H

pKa 15.7

OR . H

16–17

C 25

CR . H

N H2 . H 38

CH

CH2 . H 44

CH2 CH3 50

Relative Basicity Least basic 2

OH ,

2

OR , 2 C

Most basic

CR , 2 N H2 , 2 CH

CH2 , 2 CH2 CH3

We see from the order just given that while terminal alkynes are more acidic than ammonia, they are less acidic than alcohols and are less acidic than water.

269

270

Chapter 5 | Alkenes and Alkynes

7.9 5.17 METHODS OF PREPARATION 5.17A Synthesis of Alkynes by Elimination Reactions ●

Alkynes can be synthesized from alkenes via compounds called vicinal dihalides.

A vicinal dihalide (abbreviated vic-dihalide) is a compound bearing the halogens on adjacent carbons (vicinus, Latin: adjacent). Vicinal dihalides are also called 1,2-dihalides. A vicinal dibromide, for example, can be synthesized by addition of bromine to an alkene (Section 5.11C). The vic-dibromide can then be subjected to a double dehydrohalogenation reaction with a strong base to yield an alkyne. Vicinal dihalide formation

RCH

H

CHR  Br2

C

X

X

A vic–dihalide

Double dehydrohalogenation

H

R9C9C9R

C

2 NaNH2

R9C

C 9 R  2 NH3  2 NaBr

Br Br A vic-dibromide

The dehydrohalogenations occur in two steps, the first yielding a bromoalkene, and the second, the alkyne.

Dehydrohalogenation of vic-Dibromides to Form Alkynes

MECHANISM Reaction H

H 

R9C9C9R  2 NH2

R9C

C 9 R  2 NH3  2 Br

Br Br

Mechanism Step 1

H

N



R



H

H

H

C

C

C

R

C

Br

Br Br

Amide ion

H

R

vic-Dibromide



H

R

N

H

Bromoalkene

Ammonia

Bromide ion

H

R C Br



H

The strongly basic amide ion brings about an E2 reaction.

Step 2

Br



C

 R



N

H

R

C

C

H

Bromoalkene Amide ion A second E2 reaction produces the alkyne.

R



H

N

H



Br



H Alkyne

Ammonia

Bromide ion

5.17 | Methods of Preparation ●

Geminal dihalides can also be converted to alkynes by dehydrohalogenation.

X

A geminal dihalide (abbreviated gem-dihalide) has two halogen atoms bonded to the same carbon (geminus, Latin: twins). Ketones can be converted to gem-dichlorides by reaction with phosphorus pentachloride, and the gem-dichlorides can be used to synthesize alkynes. O

Cl

C

C

PCl5

CH3

Cyclohexyl methyl ketone

C

C

X A gem–dihalide

(1) 3 equiv. NaNH2 ,

CH3

Cl

0 °C (POCl3)

C

C

CH

mineral oil, heat (2) HA

A gem-dichloride (70–80%)

Cyclohexylacetylene (46%)

5.17B Synthesis of Higher Alkynes by Carbon–Carbon Bond Formation in Terminal Alkynes ●

The acetylenic proton of ethyne or any terminal alkyne (pKa 25) can be removed with a strong base such as sodium amide (NaNH2). The result is an alkynide anion. C

C

H 1 NaNH2

CH3 C

C

H 1 NaNH2

H

●

C

C 2 Na 1 1 NH3

CH3 C

C 2 Na 1 1 NH3

H

liq. NH3 liq. NH3

Alkynide anions are useful nucleophiles for carbon–carbon bond forming reactions with primary alkyl halides or other primary substrates.

The following are general and specific examples of carbon–carbon bond formation by alkylation of an alkynide anion with a primary alkyl halide. General Example R

C

C



Na  RCH2

Sodium alkynide

Br

R

Primary alkyl halide

C

C

CH2R  NaBr

Mono- or disubstituted acetylene

(R or R′ or both may be hydrogen.)

Specific Example CH3CH2C

C



Na  CH3CH2

Br

liq. NH3 6h

CH3CH2C

CCH2CH3



NaBr

3-Hexyne (75%)

The alkynide anion acts as a nucleophile and displaces the halide ion from the primary alkyl halide. R RC

C

C



Na Sodium alkynide

H

Br

H 1° Alkyl halide

nucleophilic substitution S N2

RC

C

CH2R 

NaBr

271

272

Chapter 5 | Alkenes and Alkynes ●

Primary alkyl halides should be used in the alkylation of alkynide anions, so as to avoid competition by elimination.

Use of a secondary or tertiary substrate causes E2 elimination instead of substitution because the alkynide anion is a strong base as well as a good nucleophile. R RC

C

H

C

H

C

H

Br

CH  RCH

RC

E2

CHR  Br

R

2° Alkyl halide

5.17C Other Methods of Preparation 1. From calcium carbide: Ethyne (acetylene) is prepared on industrial scale by the reaction of calcium carbide with water. At one time, this reaction was the best known source for obtaining ethyne used in oxy-acetylene flame. Now, it is also obtained from processing oil. Calcium carbide is made commercially by strongly heating lime and coke: Quicklime can be obtained by heating limestone (CaCO3). CaC2 1 CO; DH 5 1466 kJ mol21

CaO 1 3C

Calcium carbide reacts exothermically with water, liberating ethyne or acetylene and so they are also called acetylides. CaC2 1 H2O

Ca(OH)2 1 HC

CH

2. From carbon and hydrogen: Ethyne is obtained by passing a stream of hydrogen gas between two carbon electrodes and striking an electric arc between them. 2C 1 H2

Electric arc

HC

CH

3. From haloforms: When chloroform or iodoform are heated in presence of silver powder, they undergo dehalogenation to form ethyne. CHCl3 1 6Ag 1 CHCl3

D

HC

CH 1 6AgCl

CHl3 1 6Ag 1 CHl3

D

HC

CH 1 6AgI

4. From tetrahalides: Tetrahaloalkanes can be dehalogenated using zinc dust to yield alkynes. Br Br R

C

C

R

1 2Zn

Methanol D

RC

CR

Br Br

5. From electrolysis: Ethyne is obtained by electrolysis of aqueous solution of potassium fumerate or maleate. CH COOK CH COOK

1 2H2O

Electrolysis

CH 1 2CO2 1 H2 CH 1 KOH

5.18 | Chemical Reactivity

5.18 CHEMICAL REACTIVITY Like a double bond, the triple bond is very reactive towards a number of electrophilic reagents due to the presence of loosely held π electrons. The addition reactions with reagents such as halogens, hydrogen, etc., proceed in a manner similar to alkenes, except that two molecules of the reagent are consumed by the triple bond instead of one in the double bond. Besides addition reactions, alkynes undergo reactions due to the acidic nature of the hydrogen atom attached to the triple bond as well as oxidation and polymerization.

5.18A Electrophilic Addition Reactions Addition of halogens, halogen hydrides and dihydrogen to alkynes proceeds through electrophilic addition mechanism, via the formation of a carbocation intermediate as in case of alkenes. However, the intermediate in this case is a vinylic cation. H C

C

1H

Z

C

C 1

1Z

H

Z

C

C

Vinylic cation

The reactivity of alkynes and the addition products formed depends on the stability of the carbocation intermediate. The addition of the protic acid to alkynes occurs at the same rate as alkenes and addition to unsymmetrical alkynes takes place according to Markovnikov’s rule. Addition of Dihydrogen As discussed under preparation of alkenes, depending on the conditions and the catalyst employed, one or two molar equivalents of hydrogen will add to a carbon–carbon triple bond. When a platinum catalyst is used, the alkyne generally reacts with two molar equivalents of hydrogen to give an alkane: CH3C

CCH3

Pt H2

[CH3CH

CHCH3]

Pt H2

CH3CH2CH2CH3

Addition of Bromine and Chlorine to Alkynes ● ●

Alkynes show the same kind of addition reactions with chlorine and bromine that alkenes do. With alkynes the addition may occur once or twice, depending on the number of molar equivalents of halogen we employ: Br Br

Br C

C

Br2

C

Br2

C

Br

Tetrabromoalkane

Cl Cl

Cl C

C

C

Br Br

Dibromoalkene Cl2

C

C

C

Cl

Cl2

C

C

Cl Cl

Dichloroalkene

Tetrachloroalkane

It is usually possible to prepare a dihaloalkene by simply adding one molar equivalent of the halogen: OH

Br OH

Br2 (1 mol) 0 C

Br ●

Addition of one molar equivalent of chlorine or bromine to an alkyne generally results in anti addition and yields a trans-dihaloalkene.

273

274

Chapter 5 | Alkenes and Alkynes

Addition of bromine to acetylenedicarboxylic acid, for example, gives the trans isomer in 70% yield: HO2C

C

C

CO2H

Br

HO2C

Br2

C

C

(1 mol)

Br

Acetylenedicarboxylic acid

CO2H (70%)

Addition of Hydrogen Halides to Alkynes ●

●

Alkynes react with one molar equivalent of hydrogen chloride or hydrogen bromide to form haloalkenes, and with two molar equivalents to form geminal dihalides. Both additions are regioselective and follow Markovnikov’s rule: H C

C

HX

C

HX

C X

Haloalkene

H

X

C

C

H

X

gem-Dihalide

The hydrogen atom of the hydrogen halide becomes attached to the carbon atom that has the greater number of hydrogen atoms. 1-Hexyne, for example, reacts slowly with one molar equivalent of hydrogen bromide to yield 2-bromo-1-hexene and with two molar equivalents to yield 2,2-dibromohexane: HBr

HBr

Br

Br 2-Bromo-1-hexene

Br

2,2-Dibromohexane

The addition of HBr to an alkyne can be facilitated by using acetyl bromide (CH3COBr) and alumina instead of aqueous HBr. Acetyl bromide acts as an HBr precursor by reacting with the alumina to generate HBr. For example, 1-heptyne can be converted to 2-bromo-1-heptene in good yield using this method: Br

‘‘HBr’’ CH3COBr/alumina CH2Cl2

(82%)

Anti-Markovnikov addition of hydrogen bromide to alkynes occurs when peroxides are present in the reaction mixture. These reactions take place through a free-radical mechanism (Section 10.10): HBr

99999:

Br

peroxides

H

(E ) and (Z )

(74%)

Addition of Hypohalous Acids Alkynes add two moleceules of hypohalous acid (Cl2 or Br2 + H2O) to form dihalocarbonyl compound. The reaction proceeds in two steps: CH

δ1

CH 1 Br

δ2

OH

CH

[Br

CH

δ1

Br

Br CH

CH

O

Br 2,2-Dibromoethanal

2H2O

Br

CH

CH

Br

OH

δ2

OH

unstable

OH

CH]

5.18 | Chemical Reactivity

Addition of Water In the presence of concentrated sulphuric acid and mercuric salts, alkynes undergo the addition of water in a reaction that follows Markovnikov’s rule; that is, it corresponds to the addition of H to the less substituted carbon of the triple bond and –OH to the more substituted carbon, as illustrated by the hydration of propyne. OH CH3C

CH 1 H2O

Propyne

H2SO4

CH3C

HgSO4

O CH3CCH3

CH2

Propen-2-ol (an enol)

Propanone (acetone)

The initial product of hydration of an alkyne is an enol, a compound containing a hydroxyl group bonded to a carbon of a carbon–carbon double bond. The name “enol” is derived from the fact that the compound is both an alkene (-en-) and an alcohol (-ol). Other Addition Reactions 1. Addition of hydrogen cyanide: Alkynes react with hydrogen cyanide in the presence of barium cyanide as catalyst to form vinyl cyanide or acrylonitrile, which is further used for polymer polyacrylonitrile. CH

Ba(CN)2

CH 1 HCN

CH2

CN

CH

2. Addition of alcohols: Alkynes add a molecule of alcohol to form vinyl ethers. The reaction is catalyzed by the presence of alkali. CH

CH 1 ROH

KOH

CH2

D

CH

OR

3. Addition of carboxylic acids: Alkynes add a molecule of carboxylic acid to give vinyl esters in presence of Lewis acid as a catalyst. CH

CH 1 RCOOH

Hg21

CH2

D

CH

OCOR

5.18B Oxidation of Alkynes Oxidative Cleavage of Alkynes Treating alkynes with ozone followed by acetic acid, or with basic potassium permanganate followed by acid, leads to cleavage at the carbon–carbon triple bond. The products are carboxylic acids: R

C

C

R′

(1) O3 (2) HOAc

RCO2H 1 R′CO2H

or R

C

C

R′

(1) KMnO4, HO2 (2) H3O2

RCO2H 1 R′CO2H

Combustion Complete combustion of alkynes in the presence of air or oxygen leads to formation of carbon dioxide and water. The reaction is highly exothermic and large amount of heat is liberated. 2CH

CH 1 5O2

4CO2 1 2H2O

Under normal conditions, ethyne burns in oxygen with a sooty yellow flame. However, when it is burnt with oxygen under high pressure, it burns with a blue flame producing a high temperature of 3000 K. The heat produced by the flame is used for welding metals and is known as oxy-acetylene flame.

275

276

Chapter 5 | Alkenes and Alkynes

5.18C Polymerization Linear Polymerization Under suitable conditions, gaseous ethyne undergoes polymerization to yield polyethyne (polyacetylene). The polymer is linear with high molecular weight and can be represented by the repeating unit [ } CH CH } CH CH } ]n. Polyethyne is the first known example of an organic polymer capable of conducting electricity. H C

C

H

H C

C

H

H C

H

C

C

H

C

H

H C

C

H

H C

C n

H

By itself, polyethyne is limited in its application because it poorly conducts electricity. However, when prepared as a cation or anion by a process called doping, it can conduct electricity almost as well as copper wire. Both cationic and anionic polymers are stabilized by resonance and conduct electricity efficiently. 2 an

d n Ad ctro e l e Re an move ele ctr on

Polyethyne

Resonance stabilized

1

Resonance stabilized

Polyethyne is used in the packaging materials for computer parts, due to its ability to dissipate static charges that can damage the sensitive circuits. Thin films of polyactetylene are used as electrodes in batteries in place of metal conductors as they are lighter, cheaper and good conductors. Polyethyne is limited in its application due to its sensitivity to air and moisture. However, some conducting polymers based on it show much better conducting properties and wider applications. For example, poly(p-phenylene vinylene) or PPV is a conducting polymer used in light-emitting diode (LED) displays. Cyclic Polymerization This polymerization of alkynes is a route for converting aliphatic hydrocarbons to aromatic compounds. When gaseous ethyne is passed through red hot iron tube, three molecules undergo cyclic polymerization to form benzene, which can be further used for preparation of a number of derivatives. 3 HC

CH

Catalyst

PART III: CONJUGATED UNSATURATED SYSTEMS At its essence, a conjugated system involves at least one atom with a p orbital adjacent to at least one π bond. The adjacent atom with the p orbital can be part of another π bond, as in 1,3-butadiene, or a radical, cationic, or anionic reaction intermediate. If an example derives specifically from a propenyl group, the common name for this group is allyl. In general when we are considering a radical, cation, or anion that is adjacent to one or more π bonds in a molecule other than propene, the adjacent position is called allylic. The formula for butadiene, resonance hybrids for the allyl radical and an allylic carbocation as shown as follows. +

1,3-butadiene (a conjugated diene)

The allyl radical

+

An allylic carbocation

Radical substitution at an allylic position is especially favorable because the intermediate radical is a part of the conjugated system.

5.19 | The Stability of the Allyl Radical

5.19 THE STABILITY OF THE ALLYL RADICAL An explanation of the stability of the allyl radical can be approached in two ways: in terms of molecular orbital theory and in terms of resonance theory. As we shall see soon, both approaches give us equivalent descriptions of the allyl radical.

5.19A Molecular Orbital Description of the Allyl Radical As an allylic hydrogen atom is abstracted from propene (see the following diagram), the sp3-hybridized carbon atom of the methyl group changes its hybridization state to sp2. The p orbital of this new sp2hybridized carbon atom overlaps with the p orbital of the central carbon atom. ●

●

●

In the allyl radical three p orbitals overlap to form a set of π molecular orbitals that encompass all three carbon atoms. The new p orbital of the allyl radical is said to be conjugated with those of the double bond, and the allyl radical is said to be a conjugated unsaturated system. The unpaired electron of the allyl radical and the two electrons of the π bond are delocalized over all three carbon atoms.

Delocalization of the unpaired electron accounts for the greater stability of the allyl radical when compared to primary, secondary, and tertiary radicals. Although some delocalization occurs in primary, secondary, and tertiary radicals, delocalization is not as effective because it occurs only through hyperconjugation with σ bonds. We can illustrate the picture of the allyl radical given by molecular orbital theory with the following structure: H H

1

1 2

C

2

C

H

3

C 21

H

H

We indicate with dashed lines that both carbon–carbon bonds are partial double bonds. This accommodates one of the things that molecular orbital theory tells us: that there is a π bond encompassing all three atoms. We also place the symbol 21 beside the C1 and C3 atoms. This presentation denotes a second thing molecular orbital theory tells us: that electron density from the unpaired electron is equal in the vicinity of C1 and C3. Finally, implicit in the molecular orbital picture of the allyl radical is this: the two ends of the allyl radical are equivalent. This aspect of the molecular orbital description is also implicit in the formula just given.

5.19B Resonance Description of the Allyl Radical One structure that we can write for the allyl radical is A: A

However, we might just as well have written the equivalent structure, B: B

In writing structure B, we do not mean to imply that we have simply taken structure A and turned it over. We have not moved the nuclei. What we have done is move the electrons in the following way:

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Resonance theory (Section 1.5) tells us that whenever we can write two structures for a chemical entity that differ only in the positions of the electrons, the entity cannot be represented by either structure alone but is a hybrid of both. We can represent the hybrid in two ways. We can write both structures A and B and connect them with a double-headed arrow, the special arrow we use to indicate that they are resonance structures: A

B

Or we can write a single structure, C, that blends the features of both resonance structures: d

C

d

We see, then, that resonance theory gives us exactly the same picture of the allyl radical that we obtained from molecular orbital theory. Structure C describes the carbon–carbon bonds of the allyl radical as partial double bonds. The resonance structures A and B also tell us that the unpaired electron is associated only with the C1 and C3 atoms. We indicate this in structure C by placing a δ beside C1 and C3. Because resonance structures A and B are equivalent, the electron density from the unpaired electron is shared equally by C1 and C3. Another rule in resonance theory is the following: ●

Whenever equivalent resonance structures can be written for a chemical species, the chemical species is much more stable than any single resonance structure would indicate.

If we were to examine either A or B alone, we might decide incorrectly that it resembled a primary radical. Thus, we might estimate the stability of the allyl radical as approximately that of a primary radical. In doing so, we would greatly underestimate the stability of the allyl radical. Resonance theory tells us, however, that since A and B are equivalent resonance structures, the allyl radical should be much more stable than either, that is, much more stable than a primary radical. This correlates with what experiments have shown to be true: the allyl radical is even more stable than a tertiary radical. We should point out that a structure that would indicate an unpaired electron on the central carbon of the allyl system, as shown here, is not a proper resonance structure because resonance theory dictates that all resonance structures must have the same number of unpaired electrons. This structure shows three unpaired electrons, whereas the other resonance structures for the allyl radical have only one unpaired electron. An incorrect resonance structure

5.19C Resonance Theory Revisited We had an introduction to resonance theory in Section 1.5 and an initial presentation of some rules for writing resonance structures. In view of further discussion on stability o carbocations, radicals and conjugated systems, these rules can now be reviewed and expanded as well as the ways for estimating the relative contribution a given structure will make to the overall hybrid. How to Write Proper Resonance Structure Some key points are listed as follows: 1. 2. 3. 4. 5. 6.

Resonance structures exist only on paper. In writing resonance structures, we are only allowed to move electrons. All the contributing structures must be proper Lewis structures. All resonance structures must have the same number of unpaired electrons. All atoms that are part of the delocalized π-electron system must lie in a plane or be nearly planar. The resonating energy of the actual molecule is lower than the energy that might be estimated for any contributing structures.

5.21 | Alkadienes and Polyunsaturated Hydrocarbons

7. Equivalent resonance structures make equal contributions to the hybrid, a system described by them has a larger resonance stabilization. 8. The more stable a structure is (when taken by itself), the greater is its contribution to the hybrid. How to Estimate the Relative Stability of Contributing Resonance Structures The following rules will help make decision about relative stability of the contributing resonance structures. 1. The more covalent bonds a structure has, the more stable it is. 2. Structures in which all the atoms have a complete valence shell of electrons (i.e., the noble gas structure) are especially stable and make large contributions to the hybrid. 3. Charge separation reduces stability.

5.20 THE ALLYL CATION Carbocations can be allylic as well. ●

The allyl (propenyl) cation

(



) is even more stable than a secondary carbocation and is almost

as stable as a tertiary carbocation. In general terms, the relative order of stabilities of carbocations is that given here. Relative order of carbocation stability 

 Substituted allylic 



3°





Allyl











2°





1°

 



Vinyl

The bonding π molecular orbital of the allyl cation, like that of the allyl radical contains two spinpaired electrons. The nonbonding π molecular orbital of the allyl cation, however, is empty. Resonance theory depicts the allyl cation as a hybrid of structures D and E represented here: 1

2

3

1



2 

D

3

E

Because D and E are equivalent resonance structures, resonance theory predicts that the allyl cation should be unusually stable. Since the positive charge is located on C3 in D and on C1 in E, resonance theory also tells us that the positive charge should be delocalized over both carbon atoms. Carbon atom 2 carries none of the positive charge. The hybrid structure F includes charge and bond features of both D and E: 1 d

5.21

2 F

3 d

ALKADIENES AND POLYUNSATURATED HYDROCARBONS

Many hydrocarbons are known that contain more than one double or triple bond. A hydrocarbon that contains two double bonds is called an alkadiene; one that contains three double bonds is called an alkatriene,

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and so on. Colloquially, these compounds are often referred to simply as dienes or trienes. A hydrocarbon with two triple bonds is called an alkadiyne, and a hydrocarbon with a double and triple bond is called an alkenyne. The following examples of polyunsaturated hydrocarbons illustrate how specific compounds are named. Recall from IUPAC rules that the numerical locants for double and triple bonds can be placed at the beginning of the name or immediately preceding the respective suffix. We provide examples of both styles. 1

2

3

CH2"C"CH2 1,2-Propadiene (allene, or propa-1,2-diene)

1

2

3

4

5

2

1

4

3

1,3-Butadiene (buta-1,3-diene) 2

6

3

2

1

6

2

6

1

2

5

(3Z)-Penta-1,3-diene (cis-penta-1,3-diene)

(2Z,4E)-Hexa-2,4-diene (cis,trans-hexa-2,4-diene) 1

3

5

4

1

(2E,4E)-2,4-Hexadiene (trans,trans-2,4-hexadiene)

4

5

3

4

3

2

1

Pent-1-en-4-yne

4

5

6

7

8

(2E,4E,6E)-Octa-2,4,6-triene (trans,trans,trans-octa-2,4,6-triene) 3

3

2

4

1

4 5

5

6

1,3-Cyclohexadiene

1,4-Cyclohexadiene

The multiple bonds of polyunsaturated compounds are classified as being cumulated, conjugated, or isolated. ●

The double bonds of a 1,2-diene (such as 1,2-propadiene, also called allene) are said to be cumulated because one carbon (the central carbon) participates in two double bonds.

Hydrocarbons whose molecules have cumulated double bonds are called cumulenes. The name allene is also used as a class name for molecules with two cumulated double bonds: H Cumulated double bonds

C H

C

C

H

C

H

Allene

C

C

A cumulated diene

An example of a conjugated diene is 1,3-butadiene. ●

In conjugated polyenes the double and single bonds alternate along the chain: C

Conjugated double bonds

C C

1,3-Butadiene

C

A conjugated diene

(2E,4E,6E)-Octa-2,4,6-triene is an example of a conjugated alkatriene. ●

If one or more saturated carbon atoms intervene between the double bonds of an alkadiene, the double bonds are said to be isolated.

An example of an isolated diene is 1,4-pentadiene:

Isolated double bonds

C

C

C

C

(CH2)n An isolated diene (n 0)

1,4-Pentadiene

5.22 | 1,3-Butadiene: Electron Delocalization

Cumulated dienes have had some commercial importance, and cumulated double bonds are occasionally found in naturally occurring molecules. In general, cumulated dienes are less stable than isolated dienes. The double bonds of isolated dienes behave just as their name suggests—as isolated “enes.” They undergo all of the reactions of alkenes, and, except for the fact that they are capable of reacting twice, their behavior is not unusual. Conjugated dienes are far more interesting because we find that their double bonds interact with each other. This interaction leads to unexpected properties and reactions. We shall therefore consider the chemistry of conjugated dienes in detail.

5.22 1,3-BUTADIENE: ELECTRON DELOCALIZATION 5.22A Bond Lengths of 1,3-Butadiene The carbon–carbon bond lengths of 1,3-butadiene have been determined and are shown here: 1.34 Å

Double bonds are shorter than single bonds.

2

4

1 3

1.47 Å

The C1 } C2 bond and the C3 } C4 bond are (within experimental error) the same length as the carbon–carbon double bond of ethene. The central bond of 1,3-butadiene (1.47 Å), however, is considerably shorter than the single bond of ethane (1.54 Å). This should not be surprising. All of the carbon atoms of 1,3-butadiene are sp2 hybridized and, as a result, the central bond of butadiene results from overlapping sp2 orbitals. And, as we know, a sigma bond that is sp3–sp3 is longer. There is, in fact, a steady decrease in bond length of carbon–carbon single bonds as the hybridization state of the bonded atoms changes from sp3 to sp (Table 5.1). Table 5.1 Carbon–carbon single-bond lengths and hybridization state Compound

Hybridization State

Bond Length (Å)

H3C } CH3

sp –sp

1.54

CH2

CH } CH3

sp –sp

1.50

CH } CH

sp –sp

2

1.47

C } CH3

sp–sp3

1.46

HC

C } CH

2

sp–sp

1.43

HC

C}C

sp–sp

1.37

CH2 HC

3

3

2

3

2

CH2

CH2 CH

5.22B Conformations of 1,3-Butadiene There are two possible planar conformations of 1,3-butadiene: the s-cis and the s-trans conformations. 2

3

s-cis Conformation of 1,3-butadiene

rotate about C29C3

2

3

s-trans Conformation of 1,3-butadiene

These are not true cis and trans forms since the s-cis and s-trans conformations of 1,3-butadiene can be interconverted through rotation about the single bond (hence the prefix s). The s-trans conformation is

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the predominant one at room temperature. We shall see that the s-cis conformation of 1,3-butadiene and other 1,3-conjugated alkenes is necessary for the Diels–Alder reaction (Section 5.26).

5.22C Molecular Orbitals of 1,3-Butadiene The central carbon atoms of 1,3-butadiene are close enough for overlap to occur between the p orbitals of C2 and C3. This overlap is not as great as that between the orbitals of C1 and C2 (or those of C3 and C4). The C2 } C3 orbital overlap, however, gives the central bond partial double-bond character and allows the four π electrons of 1,3-butadiene to be delocalized over all four atoms. The four p orbitals of 1,3-butadiene combine to form a set of four π molecular orbitals. ●

●

●

Two of the π molecular orbitals of 1,3-butadiene are bonding molecular orbitals. In the ground state these orbitals hold the four π electrons with two spin-paired electrons in each. The other two π molecular orbitals are antibonding molecular orbitals. In the ground state these orbitals are unoccupied. The delocalized bonding that we have just described for 1,3-butadiene is characteristic of all conjugated polyenes.

5.23 THE STABILITY OF CONJUGATED DIENES ●

Conjugated alkadienes are thermodynamically more stable than isomeric isolated alkadienes.

Two examples of this extra stability of conjugated dienes can be seen in an analysis of the heats of hydrogenation given in Table 5.2. Table 5.2 Heats of hydrogenation of alkenes and alkadienes Compound 1-Butene

H2 (mol)

DH ° (kJ mol−1)

1

−127

1-Pentene

1

−126

trans-2-Pentene

1

−115

1,3-Butadiene

2

−239

trans-1,3-Pentadiene

2

−226

1,4-Pentadiene

2

−254

1,5-Hexadiene

2

−253

In itself, 1,3-butadiene cannot be compared directly with an isolated diene of the same chain length. However, a comparison can be made between the heat of hydrogenation of 1,3-butadiene and that obtained when two molar equivalents of 1-butene are hydrogenated: H (kJ mol1)

 2 H2 9: 2

2

2  (127)  254

1-Butene

 2 H2 9: 1,3-Butadiene

Difference

 239 15 kJ mol1

Because 1-butene has a monosubstituted double bond like those in 1,3-butadiene, we might expect hydrogenation of 1,3-butadiene to liberate the same amount of heat (254  kJ  mol−1) as two molar

5.24 | Allylic Substitution and Allylic Radicals

equivalents of 1-butene. We find, however, that 1,3-butadiene liberates only 239 kJ mol−1, 15 kJ mol−1 less than expected. We conclude, therefore, that conjugation imparts some extra stability to the conjugated system (Fig. 5.15).

2

+ 2 H2

H 

Difference 15 kJ mol–1

∆H ° = –254 kJ mol–1

+ 2 H2

∆H ° = –239 kJ mol–1

Figure 5.15 Heats of hydrogenation of 2 mol of 1-butene and 1 mol of 1,3-butadiene.

An assessment of the stabilization that conjugation provides trans-1,3-pentadiene can be made by comparing the heat of hydrogenation of trans-1,3-pentadiene to the sum of the heats of hydrogenation of 1-pentene and trans-2-pentene. This way we are comparing double bonds of comparable types:

1-Pentene

trans-2-Pentene

trans-1,3-Pentadiene

H  126 kJ mol1

H  115 kJ mol1 Sum  241 kJ mol1 H  226 kJ mol1 Difference  15 kJ mol1

We see from these calculations that conjugation affords trans-1,3-pentadiene an extra stability of 15 kJ mol−1, a value that is equivalent, to two significant figures, to the one we obtained for 1,3-butadiene (15 kJ mol−1). When calculations like these are carried out for other conjugated dienes, similar results are obtained; conjugated dienes are found to be more stable than isolated dienes. The question, then, is this: what is the source of the extra stability associated with conjugated dienes? There are two factors that contribute. The extra stability of conjugated dienes arises in part from the stronger central bond that they contain and, in part, from the additional delocalization of the π electrons that occurs in conjugated dienes.

5.24 ALLYLIC SUBSTITUTION AND ALLYLIC RADICALS ●

An atom or group that is bonded to an sp3-hybridized carbon adjacent to an alkene double bond is called an allylic group. The group is said to be bonded at the allylic position.

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The following are some examples. The hydrogen atoms bonded to the highlighted carbons below are allylic hydrogens.

The chlorine and bromine atoms below are bonded at allylic positions.

CI Br

Allylic hydrogens are especially reactive in radical substitution reactions. We can synthesize allylic halides by substitution of allylic hydrogens. For example, when propene reacts with bromine or chlorine at high temperatures or under radical conditions where the concentration of the halogen is small, the result is allylic substitution. X

high temperature

X2



and low concentration of X2

 HX

Propene At high temperature (or in the presence of a radical initiator) and low concentration of X2 a substitution reaction occurs.

On the other hand, when propene reacts with bromine or chlorine at low temperatures, an addition reaction of the type we studied in Section 5.11E occurs. X



X2

At low temperature an addition reaction occurs.

low temperature CCl4

X

To bias the reaction toward allylic substitution we need to use reaction conditions that favor formation of radicals and that provide a low but steady concentration of halogen.

5.24A Allylic Chlorination (High Temperature) Propene undergoes allylic chlorination when propene and chlorine react in the gas phase at 400 °C. 

Cl

400 °C

Cl2

gas phase



HCl

3-Chloropropene (allyl chloride)

Propene

The mechanism for allylic substitution is the same as the chain mechanism for alkane halogenations that we saw earlier in the chapter. In the chain-initiating step, the chlorine molecule dissociates into chlorine atoms. Chain-Initiating Step Cl

Cl

2 Cl

In the first chain-propagating step the chlorine atom abstracts one of the allylic hydrogen atoms. The radical that is produced in this step is called an allylic radical.

5.24 | Allylic Substitution and Allylic Radicals

First Chain-Propagating Step H 

 H

Cl

Cl

An allylic radical

In the second chain-propagating step the allyl radical reacts with a molecule of chlorine. Second Chain-Propagating Step  Cl

Cl

Cl



Cl

Allyl chloride

This step results in the formation of a molecule of allyl chloride (2-chloro-1-propene) and a chlorine atom. The chlorine atom then brings about a repetition of the first chain-propagating step. The chain reaction continues until the usual chain-terminating steps consume the radicals.

5.24B Allylic Bromination with N-Bromosuccinimide (Low Concentration of Br2) Propene undergoes allylic bromination when it is treated with N-bromosuccinimide (NBS) in the presence of peroxides or light: Br 

O

H

N

O

Br

light or ROOR

N-Bromosuccinimide (NBS)



N

O

3-Bromopropene (allyl bromide)

O

Succinimide

The reaction is initiated by the formation of a small amount of Br? (possibly formed by dissociation of the N } Br bond of the NBS). The main propagation steps for this reaction are the same as for allylic chlorination: H

Br

 Br



Br

Br



HBr



Br

N-Bromosuccinimide is a solid that provides a constant but very low concentration of bromine in the reaction mixture. It does this by reacting very rapidly with the HBr formed in the substitution reaction. Each molecule of HBr is replaced by one molecule of Br2. Br O

N

H O

 HBr

O

N

O

 Br2

Under these conditions, that is, in a nonpolar solvent and with a very low concentration of bromine, very little bromine adds to the double bond; it reacts by substitution and replaces an allylic hydrogen atom instead. The following reaction with cyclohexene is another example of allylic bromination with NBS: Br NBS, ROOR ∆

82–87% ●

In general, NBS is a good reagent to use for allylic bromination.

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5.25

ELECTROPHILIC ATTACK ON CONJUGATED DIENES: 1,4-ADDITION

Not only are conjugated dienes somewhat more stable than nonconjugated dienes, they also display special behavior when they react with electrophilic reagents. ●

Conjugated dienes undergo both 1,2- and 1,4-addition through an allylic intermediate that is common to both.

For example, 1,3-butadiene reacts with one molar equivalent of hydrogen chloride to produce two products, 3-chloro-1-butene and 1-chloro-2-butene: 1,2-Addition

1,4-Addition

Cl HCl 25 °C

1,3-Butadiene

H



3-Chloro-1-butene (78%)

H

Cl

1-Chloro-2-butene (22%, primarily E)

If only the first product (3-chloro-1-butene) were formed, we would not be particularly surprised. We would conclude that hydrogen chloride had added to one double bond of 1,3-butadiene in the usual way. It is the second product, 1-chloro-2-butene, that is initially surprising. Its double bond is between the central atoms, and the elements of hydrogen chloride have added to the C1 and C4 atoms. To understand how both 1,2- and 1,4-addition products result from reaction of 1,3-butadiene with HCl, consider the following mechanism. Step 1 1

Cl

H

1

1

1

Cl

2

An allylic cation equivalent to 1

Step 2 (a)

(a) 1

1

(b)

Cl

1

1

Cl 1,2-Addition

2

(b)

Cl

1,4-Addition

In step 1 a proton adds to one of the terminal carbon atoms of 1,3-butadiene to form, as usual, the more stable carbocation, in this case a resonance-stabilized allylic cation. Addition to one of the inner carbon atoms would have produced a much less stable primary cation, one that could not be stabilized by resonance: H  Cl 

H

Cl



Addition of the electrophile in this fashion does not lead to an allylic (resonance-stabilized) carbocation.

5.25 | Electrophilic Attack on Conjugated Dienes: 1,4-Addition

In step 2 a chloride ion forms a bond to one of the carbon atoms of the allylic cation that bears a partial positive charge. Reaction at one carbon atom results in the 1,2-addition product; reaction at the other gives the 1,4-addition product. Note that the designations 1,2 and 1,4 only coincidentally relate to the IUPAC numbering of carbon atoms in this example. ●

Chemists typically use 1,2 and 1,4 to refer to modes of addition to any conjugated diene system, regardless of where the conjugated double bonds are in the overall molecule.

Thus, addition reactions of 2,4-hexadiene would still involve references to 1,2 and 1,4 modes of addition. 1,3-Butadiene shows 1,4-addition reactions with electrophilic reagents other than hydrogen chloride. Two examples are shown here, the addition of hydrogen bromide (in the absence of peroxides) and the addition of bromine: Br HBr



40 °C

20%

80%

Br Br2 –15 °C

Br



Br

54%

Br

Br 46%

Reactions of this type are quite general with other conjugated dienes. Conjugated trienes often show 1,6-addition. An example is the 1,6-addition of bromine to 1,3,5-cyclooctatriene: Br2 CHCl3

Br

Br 68%

5.25A Kinetic Control versus Thermodynamic Control of a Chemical Reaction The addition of hydrogen bromide to 1,3-butadiene allows the illustration of another important aspect of reactivity—the way temperature affects product distribution in a reaction that can take multiple paths. In general: ●

●

The favored products in a reaction at lower temperature are those formed by the pathway having the smallest energy of activation barrier. In this case the reaction is said to be under kinetic (or rate) control, and the predominant products are called the kinetic products. The favored products at higher temperature in a reversible reaction are those that are most stable. In this case the reaction is said to be under thermodynamic (or equilibrium) control, and the predominant products are called the thermodynamic (or equilibrium) products.

Let’s consider specific reaction conditions for the ionic addition of hydrogen bromide to 1,3-butadiene. Case 1. When 1,3-butadiene and hydrogen bromide react at low temperature (−80 °C), the major product is formed by 1,2-addition. We obtain 80% of the 1,2-product and 20% of the 1,4-product. Case 2. When 1,3-butadiene and hydrogen bromide react at high temperature (40 °C), the major product is formed by 1,4-addition. We obtain about 20% of the 1,2-product and about 80% of the 1,4-product.

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Case 3. When the product mixture from the low temperature reaction is warmed to the higher temperature, the product distribution becomes the same as when the reaction was carried out at high temperature—that is, the 1,4-product predominates. We summarize these scenarios here: 80 °C

Br  80%

 HBr 40 °C

Br 20%

40 °C

Br 

Br

20%

80%

Furthermore, when a pure sample of 3-bromo-1-butene (the predominant product at low temperature) is subjected to the high temperature reaction conditions, an equilibrium mixture results in which the 1,4-addition product predominates. Br 40 °C, HBr

Br 1,2-Addition product

1,4-Addition product

Because this equilibrium favors the 1,4-addition product, that product must be more stable.

5.26

THE DIELS–ALDER REACTION: A 1,4-CYCLOADDITION REACTION OF DIENES

In 1928 two German chemists, Otto Diels and Kurt Alder, developed a 1,4-cycloaddition reaction of dienes that has since come to bear their names. The reaction proved to be one of such great versatility and synthetic utility that Diels and Alder were awarded the Nobel Prize in Chemistry in 1950. An example of the Diels–Alder reaction is the reaction that takes place when 1,3-butadiene and maleic anhydride are heated together at 100 °C. The product is obtained in quantitative yield: O 

O O

1,3-Butadiene (diene)

●

Maleic anhydride (dienophile)

O

benzene, 100 °C

O O

Adduct (100%)

In general terms, the Diels–Alder reaction is one between a conjugated diene (a 4π-electron system) and a compound containing a double bond (a 2π-electron system) called a dienophile (diene + philia, Greek: to love). The product of a Diels–Alder reaction is often called an adduct.

5.26 | The Diels–Alder Reaction: A 1,4-Cycloaddition Reaction of Dienes

In the Diels–Alder reaction, two new σ bonds are formed at the expense of two π bonds of the diene and dienophile. The adduct contains a new six-membered ring with a double bond. Since σ bonds are usually stronger than π bonds, formation of the adduct is usually favored energetically, but most Diels–Alder reactions are reversible. We can account for all of the bond changes in a Diels–Alder reaction like that above by using curved arrows in the following way: O

O O

O O Diene

O

Dienophile

Adduct

The Diels–Alder reaction is an example of a pericyclic reaction. Pericyclic reactions are concerted reactions that take place in one step through a cyclic transition state in which symmetry characteristics of molecular orbitals control the course of the reaction. The simplest example of a Diels–Alder reaction is the one that takes place between 1,3-butadiene and ethene. This reaction, however, takes place much more slowly than the reaction of butadiene with maleic anhydride and also must be carried out under pressure:

sealed tube, 200 °C

20%

In general, the dienophile reacts with a conjugated diene by 1,4-addition to form a six-membered ring. The process is called a [4 + 2] cycloaddition, named according to the number of atoms from each reactant that join to form the ring, and it is brought about by heat (a thermal reaction). Any position on either side of the diene or dienophile can be substituted. Some representative electron-withdrawing groups that can be part of the dienophile are shown below as Z and Z9.

z

z +

heat

z or

+

z

heat

z9

z9

Where Z and Z’ can be CHO, COR, CO2H, CO2R, CN, Ar, CO-O-CO, or halogen, as well as others.

Pericyclic reactions in which two alkenes combine in the following way are also known. +

light

These are called [2 + 2] cycloadditions and require light energy (they are photochemical reactions).

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Chapter 5 | Alkenes and Alkynes

SOLVED EXAMPLES 1. Each of the following names is incorrect. Give the correct name and explain your reasoning. (a) cis-1-Butene (b) 2-Ethyl-5-methyl-3-hexene (c) (E)-3-Pentene Solution

(c) 2-Methyl-3-hexyne (d) 1-Chloro-2-methylcyclohexene 4. Consider the two alkenes 2-methyl-1-pentene and 2-methyl-2-pentene and decide which would be more stable. Solution

(a) Correct name should be 1-butene. 1-Butene is a terminal alkene with identical substituents on the terminal carbon. In this type of alkene cis/ trans isomerism is not possible. (b) Correct name should be 2,5-dimethyl-3-heptene. The longest continuous carbon chain containing the C C double bond is not necessarily the longest linear that one observes or visualizes. (c) Correct name should be (E)-2-pentene. The chain needs to be numbered in such a way to give the C C the lowest possible number. 2. Write a structural formula for each of the following: (a) (Z)-3-Heptene (b) 1-Methylcyclohexene (c) 1-Methyl-5-ethyl-1,3-cyclohexadiene (d) 2,3-Dimethyl-1,3-butadiene Solution

2-Methyl-1-pentene (disubstituted, less stable)

2-Methyl-2-pentene (trisubstituted, more stable)

2-Methyl-2-pentene has three substituents on its double bond, whereas 2-methyl-1-pentene has two, and therefore 2-methyl-2-pentene is the more stable. 5. Write structural formulas for all the products that would be obtained when each of the following alkyl halides is heated with sodium ethoxide in ethanol. When more than one product results, you should indicate which would be the major product and which would be the minor product(s). You may neglect cis–trans isomerism of the products when answering this question.

CH3

(a)

(a)

(c)

(c)

CH2CH3 CH3

(b)

Br

H3C

(b)

H3C

Solution

(d)

Br

(d)

3. Give the IUPAC names for each of the following:

1

(a) major

(a)

minor 1

(b)

(b)

Br

major (1stereoisomer)

minor

Br

(c)

(d)

Br

Br

1

(c) Cl

major

Solution (a) Z-5,6-Dimethyl-oct-4-ene-1-yne (b) E-2,3-Dibromo-4,7,8-trimethyl-non-4-ene

(d) only product

minor

Solved Examples (c)

6. Predict the products of the following reactions: Br

Br

NaOMe

(a)

ONa

MeOH

I

only product

OH

(d)

K 12 OC(CH3)3

Br

(CH3)3COH

(b)

291

ONa

1

OH

CH2CH3

major

Solution

minor

9. What reagent is needed for the following reaction? (b)

(a)

CH2CH3

(1) (2) (3) (4)

7. The alkene formed as a major product in the below elimination reaction is Me 1

Me

Δ

Solution

N

OH2

(4) The product is a cis double bond. H2/Lindlar’s catalyst and H2/Ni2B can both be used to turn an alkyne into a cis double bond.

Et n-Bu Me

(1) Me

10. The correct order of reactivity of the following alkenes to acid-catalyzed hydration is:

(3)

Me

(2) CH2 CH2

(4)

Solution (2) From Hofmann’s rule, the alkene having least alkylated double bond is the major product. 8. Starting with an appropriate alkyl halide and base, outline syntheses that would yield each of the following alkenes as the major (or only) product: (a)

(c)

(b)

(d)

H2/Lindlar’s catalyst (i) Li in NH3, (ii) NH4Cl H2/Ni2B Both (1) and (3)

(1) (CH3)2C

CH2 > H3CHC

CH2 > H2C

CH2

(2) H3CHC

CH2 > (CH3)2C

CH2 > H2C

CH2

(3) H2C

CH2 > (CH3)2C

CH2 > H3CHC

CH2

(4) H2C

CH2 > H3CHC

CH2 > (CH3)2C

CH2

Solution (1) Acid catalyzed hydration is the addition of water molecule in the presence of acid. It occurs through carbocation formation in the first step which is also slow step, so, more stable carbocation formation indicates faster reaction. H2C

CH2

H3CHC

CH2

Solution (H3C)2C

(a)

1

H1

CH2

H2CCH3 (18 carbocation, least stable) 1

H1

CH3CHCH3 (28 carbocation, moderately stable)

H1

1

(CH3)2CCH3 (38 carbocation, most stable)

Thus the order of reactivity is Br

ONa

only product

OH

ONa OH

CH2 > H3CHC

CH2 > H2C

CH2

11. Arrange the following alcohols in order of their reactivity toward acid-catalyzed dehydration (with the most reactive first):

(b) Br

(CH3)3 C

only product

1-Pentanol

2-Methyl-2-butanol

3-Methyl-2butanol

292

Chapter 5 | Alkenes and Alkynes (c) Br2 in CCl4 (d) KMnO4, OH−, heat, then H3O+

Solution CH3

CH3

Solution

CH3CCH2CH3 . CH3CHCHCH3 . CH3CH2CH2CH2CH2OH OH 3°

OH 2°

(c)

OH

(b)

OH

OH

OH

(b)

Br

(d)

Br O OH

1 CO2

15. Give the structure of the products that you would expect from the reaction of 1-hexyne with: (a) One molar equivalent of Br2 (b) Two molar equivalents of HBr (c) One molar equivalent of HBr (d) H2 (in excess)/Pt Solution

Solution

Br

(a) OH

(c)

(a)

12. Give the products that would be formed when each of the following alcohols is subjected to acidcatalyzed dehydration. If more than one product would be formed, designate the alkene that would be the major product. (Neglect cis–trans isomerism.) (a)

Br

I

1°

(a) Br HA

1

heat (2H2O)

major (1 stereoisomer)

1

Br

minor

(b)

minor

Br Br

(b) OH

(c) H

HA heat (2H2O) rearrangement

1 H major

(c) HA

OH

1

heat (2H2O)

major

(d)

minor

minor

13. What is the major product of the following reaction?

16. The hydrocarbon which can react with sodium in liquid ammonia is (1) CH3CH2CH2C CCH2CH2CH3 (2) CH3CH2C CH (3) CH3CH CHCH3 (4) CH3CH2C CCH2CH3 Solution

85% H3PO4

OH

(2) Terminal alkyne is acidic, hence they will react with Na in presence of liquid ammonia.

Δ

CH3

(1)

(3)

(2)

(4)

Solution (3) Upon formation of the secondary carbocation a methyl group will migrate to create a tertiary carbocation. Then the Zaitsev product will form. 14. Write structural formulas for the products that form when 1-butene reacts with each of the following reagents: (a) HI (b) Br2 in H2O

CH2

C

CH 1 Na

NH3

CH3

CH2

C

C2Na1

17. Give the structure of the products you would expect from the reaction (if any) of 2-butyne with: (a) Two molar equivalents of HBr (b) One molar equivalent of Br2 (c) Two molar equivalents of H2, Pt (d) H2 (in excess), Pt Solution Br Br

(a)

(c) Br

(b)

(d) Br

Solved Examples 18. Which of the following reactions will yield 2,2dibromopropane? (1) CH3

C

CH3(CH2)2C

CHBr 1 HBr

(3) CH

CH 1 2HBr

(4) CH3

CH

CH2 1 HBr

(a)

(Markovnikov’s)

CH 1 HBr

CH3

addition

C

HBr

(b)

OH

CH2

(c)

Br

CH3

(Markovnikov’s addition)

C

OH HBr (no peroxides)

(b)

19. Show how 1-butyne could be synthesized from each of the following (a) 1-Butene (b) 1-Chloro-1-butene (c) 1,1-Dichlorobutane

Solution (3) Anti-Markovnikov addition of hydrogen bromide to alkenes occurs when peroxides are present in the reaction mixture. The reaction takes place through a free radical mechanism.

Br 3 NaNH2

CCl4

mineral oil, heat

Br NH4Cl

2 NaNH2

CH3

2

Na1

CH3CH

NH4Cl

mineral oil, heat

Cl

3 NaNH2

CH2

Propene

HBr Peroxides

O3

CHCH3

The compound B is (1) CH3CH2CHO (2) CH3COCH3

(c) Cl

CH

CH3

CH2

CH2

Br

n-Propyl bromide

23. In the following sequence of reactions, the alkene forms the compound B

(b) Cl

Cl OH

22. Reaction of HBr with propene in the presence of peroxide gives (1) 3-bromopropane. (3) n-propyl bromide. (2) allyl bromide. (4) isopropyl bromide.

Solution Br2

Br

Cl2, H2O

(c)

CH3

Br 2,2-Dibromopropane

Na1

Cl HO

H3O1, H2O

(a)

Br

2

C

CH3CH2CH2 H trans-2-Hexene

Br

(a)

CH3 C

Solution

(1) The reaction is C

H

Li/NH3

21. Starting with 2-methylpropene (isobutylene) and using any other needed reagents, outline a synthesis of each of the following:

Solution

CH3

CH3

2-Hexyne

CH 1 2HBr

(2) CH3CH

C

2

Na1

NH4Cl

H2O

(3) CH3CH2COCH3 (4) CH3CHO

(4) This is an example of reductive ozonolysis H3C

CH

CH

CH3

O3

O H3C

CH

CH

O

O

CH3 (Ozonide) (A)

H2O/Zn

Solution (2) An anti-addition of hydrogen atoms to the triple bond occurs when alkynes are reduced with lithium or sodium metal in ammonia at low temperature.

B.

Zn

Solution

mineral oil, heat

20. 2-Hexyne gives trans-2-hexene on treatment with (1) Pt/H2 (3) Pd/BaSO4 (2) Li/NH3 (4) LiAlH4

A

O

(2 mol)

C H3C (B)

H

293

294

Chapter 5 | Alkenes and Alkynes 24. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene is (1) ethane. (3) 1-butene. (2) propene. (4) 2-butene.

Solution OK (2 equiv.)

Br

(a) Br

OH, heat

Solution (4) The reaction is CH3

CH

CH

CH3

O3 Zn/H2O

2-Butene (Symmetrical alkene)

Molecular mass (24 1 16 1 4 5 44 u)

CHCH2CH3

(cis or trans)

CH3 CH3

(2) CH3CH2C

CCH3

(cis or trans)

CH2CH3 CH3

(3) CH2

(c)

Cl

concd H2SO4 heat

OK OH, heat

CH3 C

OH

2CH3CHO

25. An alkene adds hydrogen in the presence of a catalyst to give 3,4-dimethylhexane. Ozonolysis of the alkene followed by treatment with zinc and acetic acid gives a single organic product. The structure of the alkene is: (1) CH3CH

(b)

27. Explain the difference between a conjugated diene, an isolated diene, and a cumulated diene. Solution The difference between these three types of dienes is the positional relationship of the two C C double bonds. In a conjugated diene there are p-orbitals on four adjacent carbon atoms. The two C C double bonds interact directly. In an isolated diene the array of p-orbitals is interrupted by an sp3 hybridized carbon. The two C C double bonds do not interact directly. In a cumulated diene the two C C double bonds emanate from the same central carbon. 28. Provide the reagents necessary to transform 2,3-dimethyl-1,3-butadiene into each of the following compounds

CCH2CHCH2CH3 CH3 CH2

(c)

(a)

(4) CH3CH2CCHCH2CH3 CH3

Br O

Solution CH3

(2) CH3CH2C

CH3 H2

CCH3

Pd/C

CH2CH3

(b) O

CH3CH2CHCHCH3 CH2CH3

Solution

3,4-Dimethylhexane Ozonolysis

1. O3 2. Zn/HOAC

(a)

H2, Pd/C

CH3 2 CH3CH2

C

O

(b)

1. O3 2. Me2S

26. Provide the reagents needed to synthesize 1,3-butadiene starting from (a) 1,4-Dibromobutane (c) (b)

OH

Cl

O

O

(c)

Br2, hv

Br

Solved Previous Years’ NEET Questions 29. Predict the products of the following reactions.

Solution

HBr 215 °C

(a)

Br

HBr

(a)

215 8C

Br

HBr 40 °C

(b)

HBr

(b)

1

40 8C

Br

Major

30. Provide a mechanism that explains formation of the following products. OH

Cl

HCI (conc)

1 Cl

Solution OH

Cl

HCI (conc)

1 Cl b

H1

a

1

OH2 2H2O

or

1 a

1

OH2

31. Predict the products of the following reactions

Solution NO2

NO2

(a)

(a)

D

1

b Cl2

NO2 D

1

(b) O

O

CH3

(b)

D

1

O CH3

CH3 D

1

O

O

O

SOLVED PREVIOUS YEARS’ NEET QUESTIONS 1. Predict the product C obtained in the following reaction of butyne-1 CH3

CH2

C

I

(1) CH3CH2 C CH3 Cl

CH 1 HCl

B

HI

I

(2) CH3 CH CH2CH2I (4) CH3 CH2 CH CH2Cl

C I

(3) CH3 CH2 CH2 C H Cl

Cl

(AIPMT 2007)

Solution (1) The reaction will take place according to Markovnikov’s rule. It states that in the addition

295

296

Chapter 5 | Alkenes and Alkynes of HX to an alkene, the hydrogen atom adds to the carbon atom of the double bond that already has the greater number of hydrogen atoms. The reaction is CH3

CH2

C

CH 1 HCl

CH3

CH2

C

(B)

HI

C

CH3 CH3

CH

CH

CH

CH3

CH3

Cl (C)

CH3

2. Which of the compounds with molecular formula C5H10 yields acetone on ozonolysis? (1) 2-Methyl-1-butene ( 3 ) 3-Methyl-l-butene (2) 2-Methyl-2-butene ( 4 ) C y c lo p e n t ane

C

CH2

(AIPMT 2007)

CH

CH

(2) Ozonolysis of 2-methyl-2-butene, gives acetone and acetaldehyde. The reaction involved is

(1) CH3

CH3

C

CH

O3

CH3

C

CH3

CH3

O

C O

O CH3

C

CH3 1 CH3

Acetone

CH2

CH3

3° Carbocation (More stable) A

CH

CH

CH3

CH

CH2

CH2Br

CH3

CH3

Br

2-Methyl-2-butene H2O 2H2O2

1

C

CH3 Br

(2) CH3

H

CH3

Hydride shift

CH3

A (predominantly) is

O

CH

CH3

CH2 1 HBr

Solution

CH3

Br2

CH3

Br (A) 2-Bromo-2-methylbutane

4. H3C

1

CH3 CH3 2° Carbocation (Less stable)

CH2

CH3 H Br 3-Methylbutene

I CH2

(3) 3-Methylbutene on reaction with hydrogen bromide yields 2-bromo-2-methylbutane via formation of more stable 3° carbocation through hydride shift.

CH2

Cl

CH3

Solution

(3) CH3

CH2CH3

CH3

O C

C

H

(4) CH3

Acetaldehyde

CH

CH

Br

CH3

CH3

(AIPMT 2008) 3. H3C

CH

CH

CH2 1 HBr

A

Solution

CH3

(3) 3-Methylbutene on reaction with hydrogen bromide yields 2-bromo-2-methylbutane via formation of more stable 3° carbocation through hydride shift.

A (predominantly) is (1) CH3

(2) CH3

CH

CH

CH3

Br

CH

CH2

CH3

CH3

CH3

CH

CH

CH3 H Br 3-Methyl butene

CH2Br

CH3 Br C

CH

CH

Br

CH3

C

CH2

CH3

(AIPMT 2008)

1

CH3 CH3 2° Carbocation (Less stable) CH3

Br (A) 2-Bromo-2-methylbutane

CH2CH3

CH3

(4) CH3

CH

CH3

CH3

(3) CH3

CH2

Br2

CH

CH3 CH3

1

C

CH2

H2 Shift

CH3

3° Carbocation (More stable)

5. The IUPAC name of the compound having the formula CH C } CH CH2 is (1) 1-butene-3-yne. (3) 1-butyn-3-ene. (2) 3-butene-1-yne. (4) but-1-yne-3-ene. (AIPMT 2009)

Solved Previous Years’ NEET Questions Solution

Solution

(1) In numbering the carbon atoms, double bond is preferred over the triple bond. Therefore, the IUPAC name of the compound is 1-butene-3-yne.

(3) The reactions involved are

1

2

CH2

3

CH

4

C

CH3 CH3

CH

CH

CH

CH3

OH

6. The IUPAC name of the compound CH3CH CHC CH is (1) pent-3-en-1-yne (2) pent-2-en-4-yne (3) pent-1-yn-3-ene (4) pent-4-yn-2-ene

Heat H1

CH3 C

CH3

CH3 1 CH3

CH

CH3CH

3

2

CH

CH3

CH3 CH2

CH

CH

CH3

Br

CH3

(D)

Br

1

C

C

CH3

CH3

CH3

(1) The correct name of the compound is pent-3-en-1-yne.

CH2

(Minor) (B)

HBr

Solution

4

C

(Major) (A)

(AIPMT MAINS 2010)

5

CH2

(C)

CH

8. The correct IUPAC name of the compound is

7. In the following reactions, (a) CH3 CH3

CH

CH

CH3

H1/heat

OH

(b) A

A

HBr, dark in absence of peroxide

C

B

1

(Major product)

(Minor product)

(AIPMT PRE 2011)

(Minor product)

Solution

The major product (A) and (C) are respectively CH3

(1) CH2

C

(2) This is the name based on longest chain including functional group, along with lowest sum rule.

CH3 CH2

CH3 and CH3

3-(1-ethyl propyl) hex-1-ene. 4-ethyl-3-propyl hex-1-ene. 3-ethyl-4-ethenyl heptane. 3-ethyl-4-propyl hex-5-ene.

D

1

(Major product)

(1) (2) (3) (4)

C

CH2

1

CH3

2 3

Br CH3

(2) CH2

C

CH2

CH3 and CH2

CH3 C

CH

CH2

9. In the following reaction CH3

CH

CH3 and CH3

CH3 C

6

4-Ethyl-3-propyl hex-1-ene

CH3

C

CH3 CH2

CH3

CH3 CH

CH3 and CH3

CH

H3C

C

CH

CH2

H2O/H1

CH3

Br

(4) CH2

5

CH3

Br

(3) CH3

4

A Major product

1

B Minor product

The major product is CH

CH3

Br

(AIPMT PRE 2011)

CH3

CH3

(1) H3C C CH CH3

(2) CH2 C CH2 CH3

OH CH3

OH

CH3

297

298

Chapter 5 | Alkenes and Alkynes CH3

(3) H3C

C

Solution

CH3 CH

(4) H3C C CH2 CH2

CH3

CH3 OH

CH3

(4) The reaction takes place via the formation of benzyl intermediate which is stable due to the dispersal of positive charge in the ring. The reaction is

OH

(AIPMT PRE 2012)

Br

Solution

CH

(1) The reaction involved is CH3 CH3

C

C

CH3 H1

CH2

CH3

CH3

C

1

CH3

2H1

C

CH3

CH

(Minor product)

CH3 OHC

CH3

1

C

CH

CH3

OH CH3

H2O

CH3

2H1

C

CH3

CH3

CH

CH3

(3° carbocation) More stable

C C H H2

C H2

C

O

(1)

H3C H3C

CH3

(3)

CH3 CH3

H3C

(2)

(4)

CH3

(AIPMT MAINS 2012)

CH2

Solution

(1) 1-Butyne reacts with sodium amide and liberates ammonia gas and sodium butynide precipitate. as it contains acidic hydrogen, while 2-butyne does not react with sodium amide due to the lack of acidic hydrogen.

(4) The reaction is

CH 1 NaNH2

CH3

1-Butyne CH3

(AIPMT 2015)

Solution

C

C

CH2

C

CH3

CH3

1. O3, CH2Cl2 2788C

CH3 1 NaNH2

O O O

C2Na1 1 NH3

2. Zn/HOAC

No reaction

2-Butyne

OHC

11. The reaction of C6H5CH duces (1) C6H5CH2 CH CH3

CH3

CH3

Sodium butynide C

(3)

CHCH3 with HBr proCH

CH3

CH3

H3C

(Major product)

10. Which of the following reagents will be able to distinguish between 1-butyne and 2-butyne? (1) NaNH2 (3) O2 (2) HCl (4) Br2

CH3

CH2

12. A single compound of the following structure is obtainable from ozonolysis of which of the following cyclic compounds?

1,2-methyl shift

CH3

CH

CH3

CH3 OH

(2° carbocation) less stable

CH3 1 HBr

CH3

H2O

CH3

CH

CH

CHCH3

CH CH2

CH2

CH3 C

O

CH3

13. Given: H3C

CH3

H3C

CH2

H2C

CH2

Br

Br

(2) C6H5CH2CH2CH2Br

(4) C6H5CH CH2 CH3 Br

(AIPMT 2015)

CH3

CH3

(I)

(II)

CH2 (III)

The enthalpy of the hydrogenation of these compounds will be in the order as: (1) III > II > I (2) II > III > I (3) II > I > III (4) I > II > III (AIPMT 2015)

Solved Previous Years’ NEET Questions Solution

Solution

(1) The quantity of heat evolved when one mole of unsaturated compound is hydrogenated is called heat of hydrogenation. The differences in heat of hydrogenation arise due to difference in stability. Greater the stability of alkene, lesser is the heat of hydrogenation and greater the number of alkyl group attached to the doubly bonded carbon atom, more stable the alkene. Hence, the correct order of enthalpy of hydrogenation is I < II < III.

(4) Heating 3,3-dimethyl-2-butene in the presence of strong acid gives 2,3-dimethyl-2-butene. CH3 H3C

C

CH3

CH

H1

CH2

CH3

C

CH3

1

CH

CH2

CH3 Alkyl shift

CH3

14. In the reaction with HCl, an alkene reacts in accordance with the Markovnikov’s rule, to give a product 1-chloro-1-methylcyclohexane. The possible alkene is

C

C

2H1

CH3

1

CH3

C

CH3CH3

CH

CH3

CH3CH3

2,3-Dimethyl-2-butene

3° Carbocation (More stable)

16. Which of the following is not the product of dehy-

CH2

(1)

(3) (1) and (2)

CH3

dration of

OH

?

CH3

(4)

(2)

(1)

(3)

(2)

(4)

(RE-AIPMT 2015) Solution (3) The first step involves the formation of a carbocation, next step is the attack of nucleophile. CH3

CH3

CH3

1

H1

CH2

Solution Cl

Cl2

CH3

CH3

1

H1

(RE-AIPMT 2015)

(4) The possible products are

Cl

Cl2

2H1

15. 2,3-Dimethyl-2-butene can be prepared by heating which of the following compounds with a strong acid? (1) (CH3)2C

CH

CH2

(2) (CH3)2CH CH2 CH (3) (CH3)2CH CH CH

1

H1

OH

1

1

17. In the reaction

CH3

H

CH2

C

C

H

1. NaNH2/liq.NH3 2. CH3CH2Br

X

1. NaNH2/liq.NH3 2. CH3CH2Br

Y,

X and Y are CH2

(1) (2) (3) (4)

CH3

(4) (CH3)3C CH CH2 (RE-AIPMT 2015)

X = 1-Butyne; Y = 3-Hexyne X = 2-Butyne; Y = 3-Hexyne X = 2-Butyne; Y = 2-Hexyne X = 1-Butyne; Y = 2-Hexyne (NEET-I 2016)

299

300

Chapter 5 | Alkenes and Alkynes Solution

(3) A : H3C

C

(1) The reaction is HC

CH

NaNH2 liq.NH3

21

HC

CH2

B : H 3C

OH

CNa

H3C

CH2

Br

HC

C CH2 (X) 1-Butyne

CH3

(4) A : H3C

C

H3C

CH2 C C 3-Hexyne (Y)

CH2

CH3

CH2

Br

H3C

CH2

C

B : H 3C

CH2

C

H2O, H2SO4

CH

(1) A : H3C

HgSO4

C

CH2

OH

(2) A : H3C

C

C

(NEET 2017)

21

CNa

B : H3C

C

(3) In the presence of concentrated sulphuric acid and mercuric salts, alkynes undergo the addition of water in a reaction that follows Markovnikov’s rule that is, it corresponds to the addition of H to the less substituted carbon of the triple bond and –OH to the more substituted carbon.

Product (B) CH2

OH

SO4 CH3

CH3

Solution

Intermediate (A) B : H3C

C O

SO4

18. Predict the correct intermediate and product in the following reaction:

H3C

CH3

O

NaNH2 liq.NH3 H3C

C

CH3C

CH 1 H2O

H2SO4

Propyne

CH

CH3C

HgSO4

O CH2

Propen-2-ol (an enol) (A)

O

CH3CCH3 Propanone (acetone) (B)

ADDITIONAL OBJECTIVE QUESTIONS NCERT Exemplar Questions 1. The addition of HBr to 1-butene gives a mixture of products A, B and C Br

(A) H5C2

C

H (A) C2H5

(B)

C2H5 C CH3

H5C2

Br (B)

CH3

CH3

CH2

CH2

CH2

Br

H CH3

C

CH2Br

CH3

(III) CH3 } CH2 } CH2 } Br

The mixture consists of (1) A and B as major and C as minor products. (2) B as major, A and C as minor products. (3) B as minor, A and C as major products. (4) A and B as minor and C as major products. 2. Arrange the following hydrogen halides in order of their decreasing reactivity with propene. (1) HCl > HBr > HI (3) HI > HBr > HCl (2) HBr > HI > HCl (4) HCl > HI > HBr 3. Arrange the following carbanions in order of their decreasing stability. (II) H } C

(4) III > I > II

(II) CH3 } CH2 } Br

CH3

H3C } C

(2) II > I > III

4. Arrange the following alkyl halides in decreasing order of the rate of β-elimination reaction with alcoholic KOH. (I)

Br (C) CH(B) } CH2 } CH2 } CH2 } Br 3

(I)

(3) III > II > I

(C)

C H

(1) I > II > III

C– C–

(III) CH3

CH2 2

(1) I > II > III

(3) II > III > I

(2) III > II > I

(4) I > III > II

Exercise 1 1. What is the IUPAC name for the following structure? CH3 CH3

C

CH

CH2

CH3

(1) (2) (3) (4)

3,3-Dimethyl-4-pentene 3-Methyl-3-ethyl-1-butene Isopropylpentene 3,3-Dimethyl-1-butene

Additional Objective Questions 2. How many Z-isomers are possible for an alkene with the formula C4H7Cl? (1) 1 (3) 3 (2) 2 (4) 4

(3) combine with a nucleophile. (4) do all of the above. 8. Which of the following carbocations would not be likely to undergo rearrangement?

3. Which structure is Z-2-bromo-3-methyl-2-pentene? CH3

(1) CH3CHCHCH3

CH2CH3

(1)

C

1

CH3

(2) CH3CHCCH3 1

CH C

CH3 C

CH3

(4)

CH3

(1)

(3)

CH2CH3

(2)

(4)

C

Br

10. Upon catalytic hydrogenation, a compound C6H6 absorbs four moles of hydrogen. Select a structure for C6H6.

Br C

(4) CH3CHCH1 2

9. Which alkene would liberate the most heat per mole when subjected to catalytic hydrogenation?

Br

(3)

CH3

CH3

CH2CH3

C

H

1

CH3

H3C H

(3) CH3CCH2CH3

CH3

C

Br

(2)

CH3

C

CH3

CH2CH3

4. The hydrocarbon with seven carbon atoms containing a neopentyl and a vinyl group is (1) 2,2-dimethyl-4-pentene. (2) 4,4-dimethylpentene. (3) isopropyl-2-butene. (4) 2,2-dimethyl-3-pentene. 5. What is the correct order of stability (most stable to least stable) for alkenes? (1) Tetrasubstituted > cis-disubstituted > transdisubstituted > monosubstituted (2) Trisubstituted > terminal disubstituted > transdisubstituted > monosubstituted (3) Tetrasubstituted > trans-disubstituted > cisdisubstituted > trisubstituted (4) Monosubstituted > cis-disubstituted > transdisubstituted > trisubstituted 6. Rank the following cycloalkenes in order of increasing stability.

(I)

(IV)

(1) I, II (2) III

(II)

(III)

(V)

(3) II, III (4) IV, V

11. Which of the following pairs of compounds have the same general formula CnH2n? (1) Alkenes and cycloalkynes (2) Alkenes and cycloalkenes (3) Alkenes and alkynes (4) Alkenes and cycloalkanes 12. Which of these compounds belongs to the class of substances commonly known as halohydrins? (1) BrCH2CH2Cl

(3) ICH2CH2OH

(2) ClCH2CO2H

(4) FCH2CH2NH2

13. Which set of reagents will produce an electrophilic addition to the π bond of an alkene? (I)

(1) II < III II > III > IV > V. 5. Complete each acid–base reaction and name the salt formed: (a)

or A

(IV)

O2N

B

(b)

COOH 1 NaOH OH COOH

1 NaHCO3

675

676

Chapter 10 | Carboxylic Acids and Their Derivatives Solution (a) Sodium hydroxide (NaOH) typically reacts to yield carbonic acid, which subsequently decomposes to give CO2 and H2O.

Solution (a) O OH

1. KMnO4, HO2, heat

COOH + NaOH

2. H3O1

Butanoic acid

1 CO2 O

(b)

OH

1. KMnO4, HO2, heat 2. H3O1

COO−Na+ + H2O

O

Sodium butanoate

(c)

(b) The given carboxylic acid is converted to its sodium salt. Carbonic acid is formed (not shown)  which decomposes to carbon dioxide and water:

2. H3O1

O

(d)

H

OH

Solution + H2O +CO2

COO−Na+ Sodium 2-hydroxypropanoate (Sodium lactate)

(a)

(b)

(c) F

O

(c)

OH

OH

O ONa

O

O

O OH 1 HCl

6. Rank the following according to acid strength; weakest to strongest.

(d)

OH F

O OH
ester > acid anhydride > amide. (2) acyl chloride > acid anhydride > ester > amide. (3) ester > acyl chloride > amide > acid anhydride. (4) acid anhydride > amide > ester > acyl chloride.

COCl

The correct decreasing order of their reactivity towards hydrolysis is (1) (b) > (d) > (a) > (c) (3) (a) > (b) > (c) > (d) (2) (b) > (d) > (c) > (a) (4) (d) > (b) > (a) > (c) (AIPMT 2007) Solution

O C

O NO2 > Cl

O Cl

Solution (2) Acyl chlorides are the most reactive toward nucleophilic addition–elimination, and amides are the least reactive. In general, the overall order of reactivity is O

(1) The rate of hydrolysis enhances with the increase in magnitude of positive charge on carbonyl group. Electron withdrawing groups like } NO2 and } CHO increase the positive charge and electron releasing groups like } CH3 decrease the negative charge on carbonyl carbon. Amongst } NO2 and } CHO groups nitro has more } I effect than } CHO. Therefore, the reactivity towards hydrolysis is

Cl

(AIPMT 2008)

C

CHO >

C

O > Cl

C

CH3

2. Which of the following presents the correct order of the acidity in the given compounds? (1) FCH2COOH > ClCH2COOH > BrCH2COOH > CH3COOH (2) CH3COOH > BrCH2COOH > ClCH2COOH > FCH2COOH (3) FCH2COOH > CH3COOH > BrCH2COOH > ClCH2COOH (4) BrCH2COOH > ClCH2COOH > FCH2COOH > CH3COOH (AIPMT 2007) Solution (1) Carboxylic acids having electron withdrawing groups are stronger than unsubstituted acids. Electronegativity decreases in the order F > Cl > Br. Because fluorine is the most electronegative, the O } H bond is most polarized, and the proton in O } H is most positive. Therefore,

R

O >R

C

O >R

C

O R9

Acyl chloride

>R

C

Cl

O C

OR9

NH2

C

O Acid anhydride

Ester

Amide

The general order of reactivity of acid derivatives can be explained by taking into account the basicity of the leaving groups. When acyl chlorides react, the leaving group is a chloride ion. When acid anhydrides react, the leaving group is a carboxylic acid or a carboxylate ion. When esters react, the leaving group is an alcohol, and when amides react, the leaving group is an amine (or ammonia). Of all of these bases, chloride ions are the weakest bases and acyl chlorides are the most reactive acyl compounds. Amines (or ammonia) are the strongest bases and so amides are the least reactive acyl compounds. 4. Propionic acid with Br2 /P yields a dibromo product. Its structure would be (having restricted amount of Br2/P) Br

Br

(1) CH3

C

COOH

C

CH2COOH

Br

Br

(2) CH2Br

(3) H

CHBr

COOH (4) CH2Br

CH2

COBr

(AIPMT 2009) Solution (1) In the presence of a small amount of phosphorus, aliphatic carboxylic acids react smoothly with chlorine or bromine to yield a compound in which α-hydrogen is replaced by halogen. This is called Hell–Volhard–Zelinsky reaction.

679

680

Chapter 10 | Carboxylic Acids and Their Derivatives 7. Match the compounds given in List-I and List-II and select the suitable option using the code given below

Br Br2/P

CH3CH2COOH

CH3CH

COOH

Br2/P

List-I

Br CH3

C

COOH

Br

5. Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is (3) CH3COCl

(2) CH3COOCOCH3

(4) CH3COOCH3 (AIPMT PRE 2010)

Solution (3) Cl has a strong electron withdrawing nature and is the best leaving group, so CH3COCl is the most reactive.

CH3

C

O2 Z

C

CH3

Nu2

O Z

Nu 1

Z2

Nu

6. In a set of reaction, ethyl benzene yields a product D CH2CH3

KMnO4 KOH

B

Br2

C

FeBr3

(i) (ii) (iii) (iv)

Phenolphthalein Benzoin condensation Oil of wintergreen Fries rearrangement

(1) (2) (3) (4)

(a) (ii) (iv) (iv) (ii)

(b) (i) (i) (ii) (iii)

(c) (iv) (iii) (iii) (iv)

(d) (iii) (ii) (i) (i) (AIPMT MAINS 2011)

Solution

C

CH3

Benzaldehyde Phthalic anhydride Phenyl benzoate Methyl salicylate Code:

(1) CH3CONH2

O

(a) (b) (c) (d)

List-II

C2H5OH H1

(1) Option (a)-(ii): Benzoin condensation: This reaction involves an aromatic aldehyde, Ph } CHO, and is similar to Cannizzaro reaction. O

D

C

H

KCN

‘D’ would be Br

(3)

O

OH

O

Br

O

Br

CH2COOC2H5

CH

Option (b)-(i): The reaction is

COOC2H5

(1)

C

O H1

COOH

(2)

(4)

CH2

CH Br

OCH2CH3

OH

Solution

OH

OH

OH

Option (c)-(iv): Fries rearrangement: It is one of the methods for the preparation of acyl phenols.

COOH CH2CH3 Oxidation

O

B Halogenation

COOH

O

C

C6H5

OH

AICI3 D

Esterification

D

O

Phenolpthalein

(3) The reactions involved are

COOC2H5

H

1

(AIPMT PRE 2010)

Br

O

H

COOC2H5

2H2O

O C

OH C6H5 1 C

C

Br

O

C6H5

Solved Previous Years’ NEET Questions Option (d)-(iii): Methyl salicylate is known as oil of wintergreen.

OCOCH3

(4)

O

H3CO

C

OCH3

(RE-AIPMT 2015)

HO

Solution 8. The correct order of decreasing acid strength of trichloroacetic acid a, trifluoroacetic acid b, acetic acid c and formic acid d is (1) b > a > d > c (3) a > b > c > d (2) b > d > c > a (4) a > c > b > d

(3) Esters undergo base-promoted hydrolysis known as saponification. It involves nucleophilic addition-elimination at the acyl carbon, thus, the presence of electron withdrawing group increases the rate of hydrolysis. The reaction involved is

(AIPMT PRE 2012) Solution

O CH3

C

(1) CF3COOH > CCl3COOH > HCOOH > CH3COOH Acidic character increases with the electronegativity of the group attached to the carbon. As F is more electronegative than Cl, so CF3COOH is more acidic than CCl3COOH; whereas in CH3COOH, the CH3 group being an electron donating group decreases the acidic character.

O

NO2 2 OH

O2 CH3

CH3

Br

A

H3O1

The end product (C) is (1) acetone. (2) methane.

B

Ether

C

CH3

(3) acetaldehyde. (4) ethyl alcohol.

Solution

CH3Br

CH3CN Methyl nitrile (A)

CH3COOH Acetic acid (B)

OCOCH3

OCOCH3

(2) Cl

COOH

(3) O2N

COOH

NO2

COOH O

O II

Ether

C2H5OH Ethyl alcohol (C)

(1) II > III > I (2) III > II > I

III

(3) II > I > III (4) I > II > III

Solution (1) The presence of electron withdrawing groups increases the acidic strength of carboxylic acids due to a combination of inductive and mesomeric effects. Compounds II and III are more acidic than carboxylic acid I due to − I effect of oxygen present in the ring. However, II is more acidic than III as the inductive effect reduces with increase in distance. COOH

OCOCH3

2 OH 1 O

(NEET-II 2016) LiAlH4

10. Which one of the following esters gets hydrolyzed most easily under alkaline conditions? (1)

C

11. The correct order of strengths of the carboxylic acids is

I

(4) The sequence of reactions is H3O1

NO2

O

LiAlH4

(AIPMT PRE 2012)

KCN

O

OH

9. In the following sequence of reactions KCN

C

. O (2I effect is more) (II)

COOH O

(2I effect is less) (III)

COOH

. (1I effect) (I)

681

682

Chapter 10 | Carboxylic Acids and Their Derivatives

ADDITIONAL OBJECTIVE QUESTIONS NCERT Exemplar Questions 1. The correct order of increasing acidic strength is (1) phenol < ethanol < chloroacetic acid < acetic acid. (2) ethanol < phenol < chloroacetic acid < acetic acid. (3) ethanol < phenol < acetic acid < chloroacetic acid. (4) chloroacetic acid < acetic acid < phenol < ethanol.

O

C

of

of

O OH

O

O

1 2

KO

OH O

(3)

OH

1 2

KO O O OH

(4)

O

of

KOH solution?

2 1

OK

3. Which structures are correctly named? O

O

(1)

CHO

C

21 OK 1

C

KO

21 12 OK 1 KO

21 OK

OH

(4)

OH methacrylic acid

4. The only carboxylic acid derivative with two carbonyl groups is the (1) anhydride. (3) acid chloride (2) amide. (4) acetate. 5. What is the IUPAC name for

O C

O

O acrylic acid

O

(3) 12

HO

CH2OH

21 OK

21 OK 1

OH

malonic acid O

(2)

O

(3) OH

OH formic acid

O

(4)

KO

of

CHO is treated with concentrated aqueous

(2)

(1)

O

3. Which product is formed when the compound

12

1 2

Ph can be prepared by the

(1) phenol and benzoic acid in the presence NaOH. (2) phenol and benzoyl chloride in the presence pyridine. (3) phenol and benzoyl chloride in the presence ZnCl2. (4) phenol and benzaldehyde in the presence palladium.

(1) KO

2-Oxohexanoic acid 5-Oxohexanoic acid Methyl butyroxo ketone 4-Ketopentanoic acid

2. Which is the structure for potassium hydrogen oxalate?

(2)

O

2. Compound Ph reaction of

(1) (2) (3) (4)

(1) (2) (3) (4)

O?

O

2,3-Dimethylbutyl acetate 2,3-Dimethyl-4-oxoethanal 2,3-Dimethylbutyl methanoate 2,3-Dimethylbutyl methylate

6. The correct structure for ethyl 3-methylbutanoate is

Exercise 1

O

1. Which is the IUPAC name for the following compound? O

(1)

O

(3)

O O

O

(2) OH

O

(4)

O O O

Additional Objective Questions 7. Which functional groups are correctly named? O O

O

O

O

HC

CH3 N

CH3CH2COCH3 amide (I)

ester (III) O

O

O

anhydride (II)

CH3 amide (V)

O

lactone (IV)

(1) I, II, III, IV (2) I, III, IV, V

(3) III, IV, V (4) I, II, IV, V

8. In which of these species are all the carbon-oxygen bonds of equal length? (1) Diethyl carbonate (2) Methyl butanoate (3) Lithium acetate (4) Propionic anhydride 9. Which is the correct structure for Z-3-hexenedioic acid? O

(1) OH

OH

O OH

O

(3)

O OH O

(4) OH

OH O

10. Which of the following best represents the structure of the carboxylate ion? Oδ2

Oδ1

(1) R

(3) R

C Oδ2

C Oδ1

Oδ2

(2) R

Oδ2

14. Carboxylic acids do not give the characteristic properties of (1)

C

(3) alkyl group.

O group

(2) } COOH group.

(4) None of these.

15. Formic acid is not a representative member of the carboxylic acids because (1) it is the first member of the series. (2) it does not contain in alkyl group. (3) it is a gas. (4) it contains an aldehydic group while the other acids not have the aldehydic acid group.

18. The correct order of the boiling point of comparable molecular weight of acid and its derivatives is (1) Primary amides > carboxylic acids > nitriles > esters (2) Carboxylic acids > primary amides > nitriles > esters (3) Carboxylic acids > nitriles > primary amides > esters (4) Amides > carboxylic acids > esters > nitriles 19. Which is the correct order of increasing boiling point of the following compounds (lowest first)? O

(4) None of the above.

C

13. Monocarboxylic acids show functional isomerism with (1) esters. (3) ethers. (2) alcohols. (4) aldehydes.

17. Carboxylic acids and amides have in general higher boiling points than esters and anhydrides because of which property? (1) Dispersion forces (2) Resonance stabilization (3) Conjugated functional groups (4) Hydrogen bonding

O HO

12. Which of the following is not the functional isomer of monocarboxylic acid? (1) Amide (2) Ester (3) Hydroxy aldehyde (4) Hydroxy ketone

16. Acetic acid exists in a dimer state in benzene due to (1) condensation reaction. (2) hydrogen bonding. (3) presence of carbonyl group. (4) presence of α-hydrogen.

O

(2) HO

11. What is the main reason for the fact that carboxylic acids can undergo ionization? (1) Absence of alpha-hydrogen (2) Resonance stabilization of the carboxylate ion (3) High reactivity of alpha-hydrogen (4) Hydrogen bonding

H3CCOCH3 (I)

O H3CCCI (II)

O

O

H3CCNH2 (III)

H3CCOH (IV)

683

684

Chapter 10 | Carboxylic Acids and Their Derivatives (1) IV, I, III, II (2) I, III, IV, II

(3) II, I, IV, III (4) IV, III, II, I

O

(3)

OH

O OH

25. In which of the following sequences are the compounds listed in order of decreasing acidity? (1) CH3COOH > H2O > PhOH > HC CH > NH3 CH > NH3

(3) I, IV, II, III (4) II, III, IV, I

(3) CH3COOH > PhOH > H2O > HC

CH > NH3

(4) H2O > CH3COOH > PhOH > HC

CH > NH3

CH3 O

OH

26. Which of the following would be the strongest acid?

(I)

(1)

(III) O

O

OH

CH3

(II)

(IV)

(1) II, I, III, IV (2) I, II, III, IV

(3) III, I, II, IV (4) IV, I, II, III

22. Which compound would be the strongest acid? CHCl2CH2CH2CO2H ClCH2CHClCH2CO2H CH3CCl2CH2CO2H CH3CH2CCl2CO2H

23. Among the following acids which has the lowest pKa value? (1) CH3 COOH (2) HCOOH (3) (CH3)2CH } COOH (4) CH3CH2COOH 24. Which reactions proceed nearly to completion as written?

CO2H

1 NaOH

1 Na2CO3

(1) BrCH2CH2CH2COOH (2) ClCH2CH2CH2COOH (3) Cl2CHCH2CH2COOH (4) ICHBrCH2CH2COOH 28. In which of the following sequences are the compounds listed in order of increasing acidity? (1) NH3 < HC CH < CH3CH2OH < H2O < CH3COOH (2) CH3CH2OH < NH3 < H2O < HC CH < CH3COOH (3) CH3COOH < CH3CH2OH < H2O < NH3 < HC CH (4) H2O < CH3COOH < CH3CH2OH < HC CH < NH3 29. Which of the following would be the weakest acid? OCH3

CO2H

1 H2O 1 CO2

CO2H

(3) H3CO CO2H

O2Na1

CO2H

27. Which of the following acids would have the largest value for pKa?

1 H2O

O

Cl

Cl

O2Na1

OH

OH

(4)

Cl

(1)

O

CO2H

(3) Cl

(2)

(2)

Cl

CO2H

O2K1

H

O

OH

(2) PhOH > CH3COOH > H2O > HC

21. Arrange the compounds in order of increasing boiling point (lowest first).

O

O2Na1

(IV)

(II)

(1)

O2Na1

OH

O

O2K1

(1) (2) (3) (4)

O

1

O

C

1

OH (III)

(I)

(1) II, III, I, IV (2) IV, I, III, II

O

O

(4)

O

1 H2SO3 O2Na1

OH

20. Arrange the compounds in order of increasing solubility in water (least soluble first). O

O

1 NaHSO4

(2)

OCH3 CO2H

(4) H3CO

OCH3

Additional Objective Questions 30. The following reaction is fastest when z is which group? O C

O 2 Z 1 OH

(1)

C

2 O 1 HZ

(3) CH3O

O

(2) CH3

(4)

31. Fruity smell is given by (1) esters. (2) alcohols.

O CH3CO

(3) chloroform. (4) acid anhydrides.

32. } OH group in alcohol is neutral, while it is acidic in carboxylic acid because (1) alcohol has alkyl group with +I effect. (2) carboxylic acid is an electrovalent compound. (3) alcohol is a covalent compound. (4) in carboxylic acid, } OH group is joined to the electron withdrawing carbonyl group. 33. Choose the reagent(s) that would bring about the following reaction: CH3CH2CH2COCl

CH3CH2CH2CHO

(1) H2/Ni

(3) LiAl(OC(CH3)3)3H

(2) Li/liq NH3

(4) NaBH4, CH3OH

34. HCOOH reacts with concentrated H2SO4 to produce (1) CO (3) NO (2) CO2 (4) NO2 35. An acetic acid reacts with isotopically labelled methanol to produce (1) methyl acetate having the labelled oxygen. (2) water having all the labelled oxygen. (3) both methyl acetate and water contain isotopic oxygen. (4) no esterification. 36. What is the expected product, A, of the following reaction sequence? Cl

1. NaCN 2. 70% H2SO4, reflux

A 1 NH41

(1) HCO2CH2C6H5

(3) C6H5CH2OSO3H

(2) C6H5CH2COOH

(4) C6H5CHClCOOH

37. An acid chloride is prepared from the related carboxylic acid by reaction with which of these? (1) HCl (3) SOCl2 (2) Cl2 (4) HOCl 38. g- and d-hydroxy acids can be esterified intramolecularly to form compounds known as which of these? (1) Anhydrides (3) Lactones (2) Cycloalkenes (4) Lactams

39. Which reagent would serve as the basis for a simple chemical test to distinguish between benzoic acid and benzamide? (1) Cold dilute NaOH (2) Cold dilute NaHCO3 (3) Cold conc. H2SO4 (4) More than one of these 40. Which reagent would best serve as the basis for a simple chemical test to distinguish between C6H5CH CHCOOH and C6H5CH CHCH3? (1) Conc. H2SO4 (3) CrO3/H2SO4 (2) Br2/CCl4 (4) NaHCO3/H2O 41. Which compound is prepared by reaction of benzoic acid with ammonia and water? O

O C

(1)

NH2

O2

(3)

NH1 3 O C

(2)

O2NH1 4

NH2

(4)

42. What is the order of decreasing reactivity toward nucleophilic acyl substitution for these carboxylic acid derivatives (most reactive first)? O

O

O

O O

H3CCOCH3

H3CCCl

H3CCNH2

H3CCOCCH3

(I)

(II)

(III)

(IV)

(1) II, IV, I, III (2) III, IV, I, II

(3) IV, I, II, III (4) I, II, III, IV

43. What is the reactant of the following reaction sequence? ?

OH

1. Mg/ether 2. CO2

O

3. H1

(1) HCO2CH2C6H5 (2) C6H5CH2COOH

(3) C6H5CH2Cl (4) C6H5CHClCOOH

Exercise 2 1. Which compound would be most reactive toward nucleophilic acyl addition-elimination? (1) CH3CO2Na (3) (CH3CO)2O (2) CH3COCl (4) CH3CONH2 2. Which reactions yield the same carboxylic acid? O

(I)

(II)

CH2

CH

CH2CH2Br

K2Cr2O7 H2SO4 Mg diethyl ether

CO2

H3O1/H2O

685

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Chapter 10 | Carboxylic Acids and Their Derivatives

(III)

KMnO4

CH2CH3

KOH/H2O

(1) I, II, IV (2) I, III, IV

O

(2)

(4)

8. In a set of the given reactions, acetic acid yielded a product C.

H2SO4

(3) II, III, IV (4) I, II, III

CH3COOH 1 PCl5

3. In the reaction below, product P is

A

C6H6 Anhy. AlCl3

(1) CH3CH(OH)C2H5

(2) CH3COC6H5 OH

COOH

(4)

O C6H5

C

C6H5

4. In the presence of a small amount of phosphorous, aliphatic carboxylic acids react with chlorine or bromine to yield a compound in which α-hydrogen has been replaced by halogen. This reaction is known as (1) Wolff–Kishner reaction (2) Etard reaction (3) Hell–Volhard–Zelinsky reaction (4) Rosenmund reaction 5. Which conditions are the best for the following transformation? HO

OH

(1) LiAlH4

(3) Ag(NH3)2OH / NH4OH

(2) NaBH4

(4) Pyridine Cr3O4

6. The } OH group of an alcohol or the carboxylic acid can be replaced by } Cl, using (1) (2) (3) (4)

hypochlorous acid. chlorine. hydrochloric acid. phosphorous pentachloride.

O

(a) CH3C

O

O

(b) CH3C

O

NO2

O

OCH3

O

CH3

O

(c) CH3C O

(d) CH3C

12. What would be the final product of this reaction sequence? CO2H

1. PCl3 2. NH3

O

3. P4O10

(1)

CN

(3)

NH2

1. DIBAL-H, 278°C

O

2. H3O1

(1)

(OH)C6H5

11. Consider the esters given below. Which will show the lowest rate of hydrolysis?

7. What would be the final organic product of the following reaction?

O

C

10. When propionic acid is treated with aqueous sodium bicarbonate, CO2 is liberated. The ‘C’ of CO2 comes from (1) methyl group. (3) methylene group. (2) carboxylic group. (4) bicarbonate.

O

O OH

(4) CH3

9. The formation of ester from acetyl chloride and alcohol is an example of (1) electrophilic addition. (2) nucleophilic addition. (3) nucleophilic substitution. (4) electrophilic substitution.

(3)

(2)

C

ether

C2H5

CHO

(1)

C2H5MgBr

(3) CH3CH(OH)C6H5

P

2. H3O1

O

B

Product C would be

MgBr 1. CO2

O

K2Cr2O7

CH2OH

(IV)

OH

H3O1/H2O

OH

(3)

(2)

HO OH

2

NH2

(4)

1

CO2 NH4

Additional Objective Questions 13. Predict the major organic product in the following reaction, HCl

* BaCO3

X (gas)

COOH

(3) *

COOH

*

H3O1

Et2O

* COOH

(1) (2)

CH3CH2CH2MgBr

(1) (2) (3) (4)

Succinic acid Salicyclic acid 1, 4-Benzene dicarboxylic acid Methyl benzoic acid

18. What is the ultimate product of this sequence of reactions? O

COOH

(4)

Cl

*

14. Maleic acid and fumaric acid are two unsaturated dicarboxylic acids.Which of the following statements is incorrect? (1) The boiling point of fumaric acid is greater than maleic acid. (2) Dehydration product of both acids is maleic anhydride. (3) Maleic acid dehydrates at lower temperature than fumaric acid. (4) The pKa1 of maleic acid is greater than fumaric acid, whereas pKa2 of fumaric acid is greater than maleic acid.

CH3COOH

A

Benzene Anhy, AlCl3

B

HCN

C

H2O

(2)

O H N

(2)

(3)

Cl

N H O

(4) O

O

19. Consider the derivatives of carbonic acid H2CO3 given below. O

O

ClCCl (I)

CH3OCOCH3 (IV) O

O

C6H5CH2OCCI

H2NCNH2

(II)

(V)

O O

COOH

CH3

(CH3)3COCOCOC(CH3)3

COOH

The correct order of decreasing reactivity in nucleophilic addition – eliminations is (1) I > II > III > IV > V (2) V > IV > III > II > I (3) V > I > II > III > IV (4) III > I > II > IV > V

CH2

(III)

C

CH3

CN C

CH3

20. When CH2 CH } COOH is reduced with LiAlH4, the compound obtained will be (1) CH3 } CH2 } COOH

OH OH

(4)

N H

D

OH

(3)

O

H N

OH C

CH3NH2

(1 eq.)

O

The structure of D would be

(1)

CH3CH2OH

O

(1)

15. In a set of reactions, acetic acid yielded a product D SOCl2

Cl

CH2

C

CH3

CN

16. Self-condensation of two moles of ethyl acetate in presence of sodium ethoxide yields (1) ethyl butyrate. (2) acetoacetic ester. (3) methyl acetoacetate. (4) ethyl propionate. 17. Which of the following acids is isomeric with phthalic acid?

(2) CH2

CH } CH2OH

(3) CH3 } CH2 } CH2OH (4) CH3 } CH2 } CHO 21. Which of the following reactions would serve as a synthesis for 2,2-dimethylpropanoic acid? (1)

(2)

Br

1. Mg, Et2O 2. CO2 3. H3O1

OH 1. KMnO4, OH2, heat 2. H3O1

687

688

Chapter 10 | Carboxylic Acids and Their Derivatives (3)

25. The compound formed as a result of oxidation of ethylbenzene by KMnO4 is (1) benzyl alcohol. (3) acetophenone. (2) benzophenone. (4) benzoic acid.

1. CN2

Br

2. OH2, H2O, heat 3. H3O1

(4) (1) and (2)

26. Which of the following products is not possible in the reaction given below?

22. Consider the reaction sequence below: OCH3

(C2H5COO)2 Ca 1 (C6H5COO)2 Ca Succinic anhydride AlCl3

A

Clemmensen’s reduction

X

(1) C2H5COC2H5 (2) C2H5COC6H5

distillation

(3) C6H5COC6H5 (4) C6H5CH2COCH3

27. 1 mol of benzene-1,2-dicarboxylic acid reacts with 2 mol of thionyl chloride. What is the product formed?

The structure of product X is OCH3 OH

O

OCH3

OH

(c)

(a)

Dry

SOCI2

OH OH

O OH

O

H3CO

O

H3CO

(b)

(d)

(1)

23. What would be the final product, F, of the following sequence of reactions? CO2H

PBr3

1. Mg, Et2O

2. H2O

2. CO2

Br

(2)

CO2H

(2) Br

CO2H

(4)

24. Identify the product(s) of the following reaction. O CO2H

O

(1)

O

(3)

(4)

1 CO2

O 2

Cl

O

O

28. Consider the two reactions given below to prepare the carboxylic acid (CH3)3C } COOH. Cl

(I)

Cl

(II)

Mg

1. CO2

Et2O

2. H3O1

NaCN

1. HO2, H2O

(80%)

2. H3O1

Select the correct statement about the outcome of reactions. (1) Both reactions produce desired acid in good yield. (2) Only I produces desired acid in good yield. (3) Only II produces desired acid in good yield. (4) None of them is a good method to prepare the acid. 29. What is the reactant of the following reaction sequence?

OH

(2)

heat

Cl

(4)

O

F

CO2H

(3)

O

O

3. H3O1

(1)

OH O

OH

1. LAH, Et2O

Cl

(3)

O

CO2H

CO2H

?

1. LAH, Et2O

(1) 1 CO

(2)

PBr3

2. H2O

CO2H

2. CO2 3. H3O1

Br

Br

1. Mg, Et2O

CO2H

(3) (4)

CO2H

CO2H

Answer Key (1) C6H5CH2OH + KMnO4/OH−/H2O, heat; then H3O+

30. The following compound is prepared by Fischer esterification of which combination of reagents?

(2) C6H5CH3 + KMnO4/OH−/H2O, heat; then H3O+ (3) C6H6 + CO2, high pressure

O

(4) C6H5COCH3 + I2/OH−/H2O; then H3O+

OCH3 OCH3

Exercise 3

O

(1) (2) (3) (4)

In the following set of questions, two statements “Assertion” and “Reason” are given. Choose the correct answer from the following options:

1,2-Dihydroxybenzene and methanol Salicylic acid and methanol Phthalic acid and methanol Benzoic acid and methanol

31. Which of the following reaction produces benzoic acid? Br

(a)

Mg

1. CO2

Et2O

2. H3O1

(1) Assertion and Reason are true and the Reason is the correct explanation of the Assertion. (2) Assertion and Reason are true but the Reason is not a correct explanation of the Assertion. (3) Assertion is true but the Reason is false. (4) Assertion and Reason both are false. 1. Assertion: Carboxylic acids are acidic in nature and readily give up a proton to form a carboxylate ion. Reason: The carboxylate ion formed is stabilized by resonance.

1. KMnO4, HO2

(b)

2. Assertion: First four aliphatic monocarboxylic acids are colorless.

heat 2. H3O1

Br

(c)

NaCN

Reason: Carboxylic acids with more than five carbon atoms are insoluble in water.

1. HO2, H2O

3. Assertion: Chloroacetic acid is a stronger acid than acetic acid.

2. H3O1

Reason: In aliphatic carboxylic acids, the presence of strong electron-withdrawing groups on the hydrocarbon chain increases the acidity.

1. O3, CH3CO2H

(d)

2. H2O2

32. Which of the following would serve as syntheses of (CH3)3CCO2H? (1)

(2)

(3)

O

1. Cl2 /OH2 (excess) 2. H3O1

Br

Br

Reason: Carboxylic acids exist as cyclic dimers in solid, liquid and even in vapor state. 5. Assertion: Fluorine is more electronegative than chlorine, but p-fluorobenzoic acid is a weaker acid than p-chlorobenzoic acid.

1. CN2 2. H3O1 (heat)

Reason: F has a stronger +R effect than Cl due to similar sizes of p orbitals of F and Cl atoms.

1. Mg, Et2O

6. Assertion: Pure acetic acid is converted into ice like solid called glacial acetic acid.

2. CO2 3. H3O1

Reason: Acetic acid is stronger than HCOOH.

(4) (1) and (2) 33. Which of these combinations will not produce benzoic acid?

ANSWER KEY NCERT Exemplar 1. (3)

4. Assertion: Carboxylic acids do not give characteristic reaction of carbonyl group.

2. (2)

3. (2)

7. Assertion: Carboxylic acids have higher boiling points than alkanes. Reason: Carboxylic acids are resonance hybrids.

689

690

Chapter 10 | Carboxylic Acids and Their Derivatives

Exercise 1 1. (2)

2. (3)

3. (3)

4. (1)

5. (3)

6. (4)

7. (3)

8. (3)

9. (2)

10. (2)

11. (2)

12. (1)

13. (1)

14. (1)

15. (4)

16. (2)

17. (4)

18. (1)

19. (3)

20. (2)

21. (4)

22. (4)

23. (2)

24. (1)

25. (3)

26. (3)

27. (1)

28. (1)

29. (3)

30. (4)

31. (1)

32. (4)

33. (3)

34. (1)

35. (1)

36. (2)

37. (3)

38. (3)

39. (4)

40. (4)

41. (2)

42. (1)

43. (3)

1. (2)

2. (2)

3. (2)

4. (3)

5. (2)

6. (4)

7. (4)

8. (4)

9. (3)

10. (4)

11. (3)

12. (1)

13. (1)

14. (4)

15. (1)

16. (2)

17. (3)

18. (1)

19. (1)

20. (2)

21. (4)

22. (3)

23. (4)

24. (2)

25. (4)

26. (4)

27. (4)

28. (2)

29. (4)

30. (3)

31. (1)

32. (4)

33. (3)

1. (1)

2. (3)

3. (1)

4. (2)

5. (1)

6. (3)

7. (2)

Exercise 2

Exercise 3

HINTS AND EXPLANATIONS Exercise 1

18. (1) Boiling point of amides is the maximum due to dipole–dipole interactions and intermolecular hydrogen bonding.

8. (3) In lithium acetate, both the C } O bonds will have equal length due to resonance. 2

O Li

O CH3

C

2

O Li 1

CH3

C

R

1

R

O Dipole-dipole interactions

10. (2) The best representation of the structure is Oδ2 R

C Oδ2

R R

1

N O2

R C

Dipole-dipole C N 16. (2) Acetic acid exists as a dimer due to hydrogen 1 interactions R bonding. The structure of the dimer is O H-O O-H O

O2

O2

R R R

1

N

R C

O2 R

O2

O2

R R

C

N 1

R C

N 1

H

R

O2

R Intermolecular H-bond

H C

N 1

1

H

O2

R Intermolecular H-bond

R C

N

R

Carboxylic acid also have strong hydrogen bonding due to which their boiling point are more than nitriles and esters but lower than primary amides.

C

Hints and Explanations group (i.e., alkyl group) doesn’t assist the release of proton easily.

Intermolecular H-bond O R

OH C

C OH

24. (1) Reaction (1) will proceed nearly to completion as sodium hydroxide is a strong base.

R

O

Nitriles contain strong dipole-dipole interactions, so, their boiling points are more than esters but lower than acids. δ2 δ2

Dipole-dipole interactions

R

N

C

C

N

R

δ1 δ2

19. (3) Amides with nitrogen atoms bearing one or two hydrogen atoms are able to form strong hydrogen bonds to each other. Therefore, they have higher boiling point as compare to acids. Esters are polar compounds, but, lacking a hydrogen attached to oxygen, their molecules cannot form strong hydrogen bonds to each other. As a result, esters have boiling points that are lower than those of acids. Acyl chloride has boiling points in the same range as esters of comparable molecular weight. Hence, the increasing order of boiling point is CH3COCl

,

CH3COOCH3 ,

CH3COOH ,

25. (3) Phenol is a weak acid when compared with a carboxylic acid such as acetic acid and are stronger acid than that of water. Ammonia is the weakest acid among the given compounds. Hence, the correct order is CH3COOH . PhOH . H2O . HC

26. (3) Presence of three electron withdrawing groups makes the compound (3) strongest acid. 27. (1) The acid strength is characterized in terms of acidity constant (Ka ) or pKa values. The larger the value of the pKa, the weaker is the acid. Among the given compounds, in option (1), there is only one electron withdrawing group –Br is present, thus, it will be the weakest acid and therefore, have highest pKa value. 30. (4) Due to resonance stabilization, the reaction is O

fastest when Z is CH3CO

CH3CONH2

20. (2) The solubility of a carboxylic acid in water decreases as its molecular weight increases due to the increased hydrophobic interaction of hydrocarbon part. Carboxylic acids are polar compounds but they consist of two regions of different polarity a polar hydrophilic carboxyl group and, except for formic acid, a non-polar hydrophobic hydrocarbon chain. The hydrophilic carboxyl group increases water solubility; the hydrophobic hydrocarbon chain decreases water solubility. Hence, the increasing order of solubility in water IV > I > III > II. 21. (4) Carboxylic acids generally have boiling points higher than other types of organic compounds of comparable molecular weights, such as alcohols, aldehydes and ketones. The higher boiling points of carboxylic acids result from their polarity and from the fact that they form very strong intermolecular hydrogen bonds which are not broken completely even in the vapor phase. Also, alcohols have higher boiling point than that of ketone due to formation of strong intermolecular hydrogen bonding. Ketone molecules are associated by dipole-dipole interaction. Hence, the correct order is IV < I < II pKa2 (fumaric acid).

Cl

2. NH3

3. P4O10

O

NH2

15. (1) The reaction is O

13. (1)

C

O 21

OMgBr HCl * CH3CH2CH2MgBr * BaCO3 CO2 CH3CH2CH2 Et2O

C *

H3

SOCl2

CH3COOH

O

(B) HCN

CH2 HOOC

C

O O HO Fumaric

(C)

CH3

C

O

CH2CH3 1 CH3

C

O

OH O Maleic acid

2008C

O

CH2CH3

O 2C2H5OH

CH3

Dehydration of both acids produces maleic anhydride but maleic acid’s dehydration occurs readily.

C

C2H5ONa

CH2

C

O

CH2CH3

O O Acetoacetic ester

17. (1)

O O 1 H 2O O Maleic anhydride

Being a trans-dicarboxylic acid, fumaric acid must undergo isomerization to maleic acid first. This isomerization requires a higher temperature.

CH3

16. (2) The acetoacetate ester is produced mainly by self-condensation of ethyl acetate in the presence of sodium ethoxide.

Being in cis form, maleic acid contains intramolecular hydrogen bonding due to which its boiling point is less than fumaric acid.

OH

C

NC

(D)

O

O

HO

OH H2O

OH

Maleic acid

Anhy, AlCl3

(A)

14. (4) Maleic acid and fumaric acid are cis and trans forms of butenedioic acid, and this structural feature is the cause of their different physical and chemical properties.

O

Benzene

Cl

O1

* CH3CH2CH2COOH

HO OH

C

CH3

CH3

Br

CN

1. NaCN

reflux 2. 70% H2SO4

COOH

1 NH41

693

694

Chapter 10 | Carboxylic Acids and Their Derivatives 19. (1) Nucleophilic addition-elimination is the characteristic reactions of acid derivatives. The two step mechanism involves attack of nucleophile at carbonyl carbon and elimination of weal base from carbonyl carbon.

25. (4) Alkyl chains on aromatic ring can be oxidized to } COOH by oxidizing agent KMnO4. CH2CH3 KMnO4

If more and more electronegative atom is attached to the carbonyl carbon atom then it produces is helpful in both steps. In this way acyl chloride is most reactive and amides are least reactive. In these compounds more than one leaving group is attached with carbonyl carbon atom, so, the effects are added. Thus the correct order of rate is I > II > III > IV > V. 20. (2) LiAlH4 reduces carboxylic acid to a primary alcohol. CH

CH2

LiAlH4

COOH

OMe

CH2

O

Anisole

HO

1. AlCl3

MeO

HO

4

(A)

2 3

Haworth reaction

26. (4) On dry distillation calcium salts of acid anhydrides produce ketones as shown below. Calcium carbonate is also produced in the reaction. Dry

(C2H5COO)2 Ca

distillation

(C6H5COO)2 Ca

distillation

MeO

3 4

7-Methoxy-1-tetralone

4-(p-Methoxybenzyl) butanoic acid

C6H5COC2H5 1 CaCO3

O

O OH

SOCI2

Cl

OH

(2mp)

Cl

O

2

3

Dry distillation

27. (4) The reaction is

1

6

4

C6H5COC6H5 1 CaCO3

Therefore, the compound C6H5CH2COCH3 cannot be formed in the reaction.

O

7

C2H5COC2H5 1 CaCO3

Dry

(C6H5COO)2 Ca 1 (C2H5COO)2 Ca

(Clemmensen reduction)

8

Benzoic acid

Hence ethyl benzene on oxidation with KMnO4 gives benzoic acid.

3

O 4-(p-Methoxyphenyl)4-oxobutanoic acid

Friedel−Crafts acylation (Intramolecular)

1

2

2. H3

Zn(Hg)/HCl

O

Ethybenzene

1

O1

O Succinic anhydride

H1

CH2OH

MeO O

1

CH

COOH

O

28. (2) CN− being a strong base gives elimination reaction instead substitution, so, nitrite is not produced in the second reaction and carboxylic acid is not produced in good yield. O Cl

OCH3

Mg

MgCl

Et2O

1. CO2 2. H3O1

OH

29. (4) (X)

1. LAH, Et2O

21. (4) The following reactions would serve as the synthesis of 2,2-dimethylpropanoic acid.

COOH

OH

2. H2O

PBr3

1. Mg, Et2O 2. CO2 3. H3O1

2 1

OMgBr Br

1. Mg Et2O

MgBr

2. CO2

C

COOH

O 3. H3O1

31. (1) Only reaction in first option produces benzoic acid.

OH C OH 1. KMnO4, OH2, D

COO K

2 1

2. H3O1

Br

O

O Br COOH

Mg Et2O

MgBr

1. CO2 2. H3O1

OH

Hints and Explanations The alkyl group on benzene ring can be oxidized to carboxylic acid group if it contains benzylic hydrogen atom. In the reaction in option (2), benzene ring does not bear benzylic hydrogen atom, so, no oxidation occurs in this case. In the reaction in option ((3), aryl halide cannot give SN2 reactions, so, no nitrile is possible which on hydrolysis can produce carboxylic acid. In the reaction in option (4), ozonolysis oxidizes double and triple bonds of carbon only, so, it produces acetic acid instead benzoic acid.

O

1. O3, CH3CO2H

H3C

2. H2O2

OH

32. (4) O

1. Cl2/OH2

O OH

2. H3O1

1 CHCl3 O

Br 1. CN2

CN

2. H3O1

OH

695

Organic Compounds Containing Nitrogen

11 5

C H A P T E R OU TLIN E Part I: Amines 11.1 Nomenclature of Amines 11.2 Physical Properties and Structure of Amines 11.3 Preparation of Amines 11.4 Basicity of Amines: Amine Salts 11.5 Reactions of Amines 11.6 Reactions of Amines with Nitrous Acid 11.7 Reactions of Amines with Sulfonyl Chlorides: Test for Amines 11.8 Eliminations Involving Ammonium Compounds

Part II: Arenediazonium Salts 11.9 Preparation of Arenediazonium Salts 11.10 Replacement Reactions of Arenediazonium Salts 11.11 Coupling Reactions of Arenediazonium Salts

SO23 Na1

OH

N

N

Orange II

Part III: Cyanides and Isocyanides 11.12 Preparation and Properties of Cyanides and Isocyanides 11.13 Summary of Reactions of Organic Compounds Containing Nitrogen

Carbon, hydrogen and oxygen are the three most common elements in organic compounds. Due to the wide distribution of nitrogen-containing compounds in the biological world, it is the fourth most common component of organic compounds. Amines and diazonium salts are the most common classes of organic compounds containing nitrogen. Alkyl (aryl) cyanides and isocyanides form another group of organic compounds containing nitrogen. Amine groups are present in many synthetic and naturally occurring drugs. They may be useful as antidepressants, antihistamines, antibiotics, analgesics, diuretics, tranquilizers, among others. Aliphatic amines of low molecular weight are used as reagents and solvents in laboratories and industries and in manufacture of drugs, dyes, disinfectants and insecticides. Aromatic amines find use in synthesis of various organic compounds in form of arene diazonium salts, which act as intermediates, in the manufacture of dyes and drugs, rubber and plastic materials in industries.

PART I: AMINES Amines are considered as derivatives of ammonia in which one or more hydrogen atoms are replaced by the corresponding number of alkyl or aryl groups.

698

Chapter 11 | Organic Compounds Containing Nitrogen

11.1 NOMENCLATURE OF AMINES 11.1A Aliphatic Amines In common nomenclature most primary amines are named as alkylamines. In systematic nomenclature (names in parentheses below) they are named by adding the suffix -amine to the name of the chain or ring system to which the NH2 group is attached with replacement of the final -e. Amines are classified as being primary (1°), secondary (2°), or tertiary (3°) on the basis of the number of organic groups attached to the nitrogen. Primary Amines NH2

NH2

CH3NH2 Methylamine (methanamine)

Ethylamine (ethanamine)

NH2

Isobutylamine (2-methyl-1-propanamine)

Cyclohexylamine (cyclohexanamine)

Most secondary and tertiary amines are named in the same general way. In common nomenclature we either designate the organic groups individually if they are different or use the prefixes di- or tri- if they are the same. In systematic nomenclature we use the locant N to designate substituents attached to a nitrogen atom. Secondary Amines H

H

N

N

Ethylmethylamine (N-methylethanamine)

Diethylamine (N-ethylethanamine)

Tertiary Amines N

N

Triethylamine Ethylmethylpropylamine (N,N-diethylethanamine) (N-ethyl-N-methyl-1-propanamine)

In the IUPAC system, the substituent } NH2 is called the amino group. We often use this system for naming amines containing an OH group or a CO2H group: O H2N

OH

2-Aminoethanol

H2N

OH

3-Aminopropanoic acid

Secondary and tertiary amines are commonly named as N-substituted primary amines. For unsymmetrical amines, the largest group is taken as the parent amine; then the smaller group or groups bonded to nitrogen are named. Finally, their location is indicated by the prefix N (indicating that they are attached to nitrogen): CH3 N CH3 N, N-Dimethylcyclopentanamine

11.1 | Nomenclature of Amines

11.1B Arylamines Some common arylamines have the following names: NH2

NHCH3

Aniline (benzenamine)

NH2

NH2

CH3

N-Methylaniline (N-methylbenzenamine)

OCH3

p-Toluidine (4-methylbenzenamine)

p-Anisidine (4-methoxybenzenamine)

The simple derivatives of aniline are named with the prefixes o-, m- and p-, or numbers to locate substituents. Several derivatives of aniline have common names that are still widely used, such as toluidine (methyl-substituted aniline) and anisidine (methoxy substituted aniline).

11.1C Isomerism in Amines The amines can exhibit the following types of isomerism. 1. Functional isomerism: Primary, secondary and tertiary amines are functional isomers of each other. For example, CH3CH2CH2NH2

CH3CH2NHCH3

(CH3)3N

1-Propanamine (Propylamine)

N-Methylethanamine (Ethylmethyl amine)

N, N-Dimethylmethaminine (Trimethyl amine)

2. Position isomerism: These isomers differ in the position of the amino group. For example, o-, m- and p-toluidine. NH2

NH2

NH2

CH3 CH3

CH3

3. Chain isomerism: These isomers differ in the alkyl chain part of the amine. For example, CH3CH2CH2CH2NH2

CH3

n-Butyl amine

CH

CH2

NH2

CH3 Isobutyl amine

CH3 CH3

CH2

CH

NH2

CH3

CH3

C

NH2

CH3

sec-Butyl amine

tert-Butyl amine

4. Metamerism: This type of isomerism is shown by secondary amines. For example, CH3CH2CH2NH

CH3

C2H5NHC2H5

Methyl-n-propylamine

Diethyl amine

699

700

Chapter 11 | Organic Compounds Containing Nitrogen

11.2 PHYSICAL PROPERTIES AND STRUCTURE OF AMINES 11.2A Physical Properties Amines are moderately polar substances; they have boiling points that are higher than those of alkanes but generally lower than those of alcohols of comparable molecular weight. Molecules of primary and secondary amines can form strong hydrogen bonds to each other and to water. Molecules of tertiary amines cannot form hydrogen bonds to each other, but they can form hydrogen bonds to molecules of water or other hydroxylic  solvents. As a result, tertiary amines generally boil at lower temperatures than primary and secondary amines of comparable molecular weight, but all low-molecular-weight amines are very water soluble. Table 11.1 lists the physical properties of some common amines. Table 11.1 Physical properties of amines Name

Water Solubility (25 °C) (g 100 mL−1)

pKa (aminium ion)

Structure

mp (°C)

bp (°C)

Methylamine

CH3NH2

−94

−6

Very soluble

10.64

Ethylamine

CH3CH2NH2

−81

17

Very soluble

10.75

Isopropylamine

(CH3)2CHNH2

−101

33

Very soluble

10.73

Cyclohexylamine

Cyclo-C6H11NH2

−18

134

Slightly soluble

10.64

Benzylamine

C6H5CH2NH2

10

185

Slightly soluble

9.30

Aniline

C6H5NH2

−6

184

3.7

4.58

Primary Amines

4-Methylaniline

4-CH3C6H4NH2

44

200

Slightly soluble

5.08

4-Nitroaniline

4-NO2C6H4NH2

148

332

Insoluble

1.00

Dimethylamine

(CH3)2NH

−92

7

Very soluble

10.72

Diethylamine

(CH3CH2)2NH

−48

56

Very soluble

10.98

Diphenylamine

(C6H5)2NH

53

302

Insoluble

0.80

3

Very soluble

9.70

Secondary Amines

Tertiary Amines Trimethylamine

(CH3)3N

−117

Triethylamine

(CH3CH2)3N

−115

90

14

10.76

N,N-Dimethylaniline

C6H5N(CH3)2

3

194

Slightly soluble

5.06

11.2B Structure of Amines The nitrogen atom of most amines is like that of ammonia; it is approximately sp3 hybridized. The three alkyl groups (or hydrogen atoms) occupy corners of a tetrahedron; the sp3 orbital containing the unshared electron pair is directed toward the other corner. We describe the shape of the amine by the location of the atoms as being trigonal pyramidal. However, if we were to consider the unshared electron pair as being a group we would describe the geometry of the amine as being tetrahedral. The electrostatic potential map for the van der Waals surface of trimethylamine indicates localization of negative charge where the nonbonding electrons are found on the nitrogen:

N

R9 R0

R-

Structure of an amine.

11.3 | Preparation of Amines

The bond angles are what one would expect of a tetrahedral structure; they are very close to 109.5°. The bond angles for trimethylamine, for example, are 108°. If the alkyl groups of a tertiary amine are all different, the amine will be chiral. There will be two enantiomeric forms of the tertiary amine, and, theoretically, we ought to be able to resolve (separate) these enantiomers. In practice, however, resolution is usually impossible because the enantiomers interconvert rapidly:

R–

R¿

R¿

N

N

R‡

R–

R‡

Interconversion of amine enantiomers

This interconversion occurs through what is called a pyramidal or nitrogen inversion. The barrier to the interconversion is about 25 kJ mol−1 for most simple amines, low enough to occur readily at room temperature. In the transition state for the inversion, the nitrogen atom becomes sp2 hybridized with the unshared electron pair occupying a p orbital. Ammonium salts cannot undergo nitrogen inversion because they do not have an unshared pair. Therefore, those quaternary ammonium salts with four different groups are chiral and can be resolved into separate (relatively stable) enantiomers: R

R N



R

R

R X



N

R R

R X

Quaternary ammonium salts such as these can be resolved.

11.3 PREPARATION OF AMINES In this section we discuss a variety of ways to synthesize amines. Some of these methods will be new to you, while others are methods you have studied earlier in the context of related functional groups and reactions.

11.3A Through Nucleophilic Substitution Reactions Alkylation of Ammonia Salts of primary amines can be prepared from ammonia and alkyl halides by nucleophilic substitution reactions. Subsequent treatment of the resulting aminium salts with a base gives primary amines: NH3  R ●

X

R



NH3 X

HO

RNH2

This method is of very limited synthetic application because multiple alkylations occur.

When ethyl bromide reacts with ammonia, for example, the ethylaminium bromide that is produced initially can react with ammonia to liberate ethylamine. Ethylamine can then compete with ammonia and react with ethyl bromide to give diethylaminium bromide. Repetitions of alkylation and proton transfer reactions ultimately produce some tertiary amines and even some quaternary ammonium salts if the alkyl halide is present in excess.

701

702

Chapter 11 | Organic Compounds Containing Nitrogen

MECHANISM

Alkylation of NH3

NH 3



Br



NH 3



Br

H  N H



NH 2

NH 3





NH 4

H H NH 2

N+

Br





Br, etc.

H

Multiple alkylations can be minimized by using a large excess of ammonia. An example of this technique can be seen in the synthesis of alanine from 2-bromopropanoic acid: O

O

O – NH4

OH  NH3 Br

NH2

(1 mol)

(70 mol)

Alanine (65–70%)

This method cannot be used for the preparation of arylamines since aryl halides are less reactive towards nucleophilic substitution reaction. Alkylation of Azide Ion and Reduction A much better method for preparing a primary amine from an alkyl halide is first to convert the alkyl halide to an alkyl azide (R } N3) by a nucleophilic substitution reaction, then reduce the azide to a primary amine with lithium aluminum hydride. R9X 







N" N " N

SN2 (X )

Azide ion (a good nucleophile)







R9N"N"N

LiAlH4

RNH2

Alkyl azide

Caution: Alkyl azides are explosive, and low-molecular-weight alkyl azides should not be isolated but should be kept in solution. Sodium azide is used in automotive airbags. Amination of Alcohols The reaction of an alcohol with ammonia when passed over alumina (Al2O3) heated to 723 K leads to formation of a mixture of primary, secondary and tertiary aliphatic amines. For example, CH3CH2OH 1 NH3 → CH3CH2NH2 1 H2O CH3CH2OH 1 CH3CH2NH2 → (CH3CH2)2NH 1 H2O CH3CH2OH 1 (CH3CH2)2NH → (CH3CH2)3N 1 H2O

Higher concentration of alcohols leads to formation of secondary and tertiary amines while the use of ammonia in excess yields mainly primary amine. The Gabriel Synthesis Potassium phthalimide (see the following reaction) can also be used to prepare primary amines by a method known as the Gabriel synthesis. This synthesis also avoids the complications of multiple alkylations that occur when alkyl halides are treated with ammonia:

11.3 | Preparation of Amines

O N H

Step 1

Step 2

O

KOH

R



N K

X

NH 2NH 2

N R

(KX)

O

O

Step 3

O

EtOH, reflux (several steps)

O

Phthalimide

N-Alkylphthalimide

O

O NHNH 2 N R

(several

N

steps)

N

H

O

O

H 

R

NH2

H

Phthalazine-1,4-dione

Primary amine

Phthalimide is quite acidic (pKa = 9); it can be converted to potassium phthalimide by  potassium hydroxide (step 1). The phthalimide anion is a strong nucleophile and (in step 2) it reacts with an alkyl halide by an SN2 mechanism to give an N-alkylphthalimide. At this point, the N-alkylphthalimide can be hydrolyzed with aqueous acid or base, but the hydrolysis is often difficult. It is often more convenient to treat the N-alkylphthalimide with hydrazine (NH2NH2) in refluxing ethanol (step 3) to give a primary amine and phthalazine-1,4-dione. Syntheses of amines using the Gabriel synthesis are, as we might expect, restricted to the use of methyl, primary, and secondary alkyl halides. The use of tertiary halides leads almost exclusively to eliminations. Alkylation of Tertiary Amines Multiple alkylations are not a problem when tertiary amines are alkylated with methyl or primary halides. Reactions such as the following take place in good yield: R3N

 RCH 2

Br

SN2



R3N

CH2R  Br

11.3B Preparation of Aromatic Amines through Reduction of Nitro Compounds The most widely used method for preparing aromatic amines involves nitration of the ring and subsequent reduction of the nitro group to an amino group: Ar

H

HNO3 H2SO4

Ar

NO2

[H]

Ar

NH2

The ring nitration is applicable to a wide variety of aromatic compounds. Reduction of the nitro group can also be carried out in a number of ways. The most frequently used methods employ catalytic hydrogenation, or treatment of the nitro compound with acid and iron. Zinc, tin, or a metal salt such as SnCl2 can also be used. Overall, this is a 6e− reduction. General Reaction Ar

NO2

H2, catalyst or 1. Fe, HCl 2. HO

Ar

NH2

Specific Example NO2

NH2

1. Fe, HCl 2. HO2

97%

703

704

Chapter 11 | Organic Compounds Containing Nitrogen

11.3C Preparation of Primary, Secondary, and Tertiary Amines through Reductive Amination Aldehydes and ketones can be converted to amines through catalytic or chemical reduction in the presence of ammonia or an amine. Primary, secondary, and tertiary amines can be prepared this way: H NH 3

1 Amine

[H]

R

R H

O

RNH 2

Aldehyde or ketone

H

R¿ R–

N

2 Amine

[H]

R

H

N

R R‡

RRNH

H

R¿ R–

N

3 Amine

[H]

R

H

R¿

This process, called reductive amination of the aldehyde or ketone (or reductive alkylation of the amine), appears to proceed through the following general mechanism (illustrated with a 1° amine).

MECHANISM

Reductive Amination

O R

R

Aldehyde or ketone

 H 2N

R

HO

two steps

R

1 Amine

NHR

N

(H 2O)

R

R

Hemiaminal

R

R Imine

[H]

NHR R

H

R

2 Amine

When ammonia or a primary amine is used, there are two possible pathways to the product: via an amino alcohol that is similar to a hemiacetal and is called a hemiaminal or via an imine. When secondary amines are used, an imine cannot form, and, therefore, the pathway is through the hemiaminal or through an iminium ion: R R



N

R R

Iminium ion

11.3 | Preparation of Amines

The reducing agents employed include hydrogen and a catalyst (such as nickel) or NaBH3CN or LiBH3CN (sodium or lithium cyanoborohydride). The latter two reducing agents are similar to NaBH4 and are especially effective in reductive aminations. Three specific examples of reductive amination follow: O H

NH2

NH3, H2, Ni

90 atm 40– 70 °C

Benzylamine (89%)

Benzaldehyde

O H

1. CH3CH2NH2

N

2. LiBH 3CN

H

Benzaldehyde

N-Benzylethanamine (89%) 1. (CH3)2NH

O

2. NaBH3CN

Cyclohexanone

CH3 N CH3

N,N -Dimethylcyclohexanamine (52–54%)

11.3D Preparation of Primary, Secondary, or Tertiary Amines through Reduction of Nitriles, Oximes, and Amides Nitriles, oximes, and amides can be reduced to amines. Reduction of a nitrile or an oxime yields a primary amine; reduction of an amide can yield a primary, secondary, or tertiary amine:

R

C

N

[H]

Nitrile

RCH

NOH

[H]

Oxime

RCH2NH2 18 Amine

Nitriles can be prepared from alkyl halides and 2CN or from aldehydes and ketones as cyanohydrins.

RCH2NH2

Oximes can be prepared from aldehydes and ketones.

18 Amine

O R

R9

C N

[H]

RCH2N

R9

R0

R0

Amides can be prepared from acid chlorides, acid anhydrides, and esters.

38 Amine

Amide

(In the last example, if R9 = H and R0 = H, the product is a 1° amine; if only R9 = H, the product is a 2° amine.) All of these reductions can be carried out with hydrogen and a catalyst or with LiAlH4. Oximes are also conveniently reduced with sodium in ethanol. Specific examples follow: OH N

Na EtOH

NH2 50–60%

705

706

Chapter 11 | Organic Compounds Containing Nitrogen

NH2

CN

2 H2, Raney Ni 140 °C

2-Phenylethanenitrile (phenylacetonitrile)

2-Phenylethanamine (71%)

CH3

CH3

N

N 1. LiAlH4 2. H2O

O N-Methylacetanilide

N-Ethyl-N -methylaniline

Reduction of an amide is the last step in a useful procedure for monoalkylation of an amine. The process begins with acylation of the amine using an acyl chloride or acid anhydride; then the amide is reduced with lithium aluminum hydride. For example, O

O

NH2

Cl base

N

1. LiAlH4

N

H

2. H2O

H

Benzylamine

Benzylethylamine

11.3E Preparation of Primary Amines through the Hofmann and Curtius Rearrangements Hofmann Rearrangement Amides with no substituent on the nitrogen react with solutions of bromine or chlorine in sodium hydroxide to yield amines through loss of their carbonyl carbon by a reaction known as the Hofmann rearrangement or Hofmann degradation: O R

C

NH2

H2O

 Br2  4 NaOH

R

NH2  2 NaBr  Na2CO3  2 H2O

From this equation we can see that the carbonyl carbon atom of the amide is lost (as CO32−) and that the R group of the amide becomes attached to the nitrogen of the amine. Primary amines made this way are not contaminated by 2° or 3° amines. The mechanism for this interesting reaction is shown in the following scheme. An examination of the first two steps of this mechanism shows that, initially, two hydrogen atoms must be present on the nitrogen of the amide for the reaction to occur. Consequently, the Hofmann rearrangement is limited to amides of the type RCONH2. Studies of the Hofmann rearrangement of optically active amides in which the chirality center is directly attached to the carbonyl group have shown that these reactions occur with retention of configuration. Thus, the R group migrates to nitrogen with its electrons, but without inversion. Curtius Rearrangement The Curtius rearrangement is a rearrangement that occurs with acyl azides and yields a primary amine with loss of the acyl carbon. It resembles the Hofmann rearrangement in that an R group migrates from the acyl carbon to the nitrogen atom as the leaving group departs. In this instance the leaving group is N2 (the best of all possible leaving groups since it is highly stable, is virtually nonbasic, and being a gas, removes itself from the medium). Acyl azides are easily prepared by allowing acyl chlorides to react with sodium azide. Heating the acyl azide brings about the rearrangement; afterward, adding water causes hydrolysis and decarboxylation of the isocyanate: O R

C

O NaN3

Cl

Acyl chloride

(NaCl)

R

C

N





N

Acyl azide

N

heat (–N2)

R

N

C

O

Isocyanate

H2O

R

NH2  CO2

Amine

11.4 | Basicity of Amines: Amine Salts

MECHANISM

The Hofmann Rearrangement

O

O

C

R



N

H

OH

C

R

H

O N



Br

Br

C

R

N

H

Br  Br 

H

 H2O Amide

N-Bromo amide

Base-promoted N-bromination of the amide occurs.

O R

C

O N

Br



R

OH

H

C

N  Br

Base removes a proton from the nitrogen to give a bromo amide anion.

R

N

C

O



R

N

C

R

Isocyanate

R

N

C

N

O C



O

H

OH

Transfer of a proton leads to a carbamate ion

OH

O

The R¬ group migrates to the nitrogen as a bromide ion departs. This produces an isocyanate.

H

O

C

Isocyanate

OH OH

N

 H2O

N-Bromo amide



R

(Br )

R

NH2  CO2  HO

Amine

HCO3

O

The isocyanate undergoes hydrolysis and decarboxylation to produce the amine.

11.4 BASICITY OF AMINES: AMINE SALTS

●

Amines are relatively weak bases. Most are stronger bases than water but are far weaker bases than hydroxide ions, alkoxide ions, and alkanide anions.

A convenient way to compare the base strengths of amines is to compare the pKa values of their conjugate acids, the corresponding alkylaminium ions. 1

RNH3 1 H2O Ka 5

RNH2 1 H3O1 [RNH2][H3O1] [RNH31]

pKa 5 2log Ka

707

708

Chapter 11 | Organic Compounds Containing Nitrogen

The equilibrium for an amine that is relatively more basic will lie more toward the left in the above chemical equation than for an amine that is less basic. ●

The aminium ion of a more basic amine will have a larger pKa than the aminium ion of a less basic amine.

When we compare aminium ion acidities in terms of this equilibrium, we see that most primary alkylaminium ions (RNH3+) are less acidic than ammonium ion (NH4+). In other words, primary alkylamines (RNH2) are more basic than ammonia (NH3): H

H



H N H

CH3

H Aminium ion pKa

H



N H

CH3CH2

H

9.26



N H H

10.64

10.75

We can account for this on the basis of the electron-releasing ability of an alkyl group. An alkyl group releases electrons, and it stabilizes the alkylaminium ion that results from the acid–base reaction by dispersing its positive charge. It stabilizes the alkylaminium ion to a greater extent than it stabilizes the amine: H R

N

H



H

OH

R

N

H



H





OH

H By releasing electrons, R stabilizes the alkylaminium ion through dispersal of charge.

11.4A Basicity of Arylamines ●

Aromatic amines are much weaker bases than alkylamines.

Considering amine basicity from the perspective of aminium ion acidity, when we examine the pKa values of the conjugate acids of aromatic amines (e.g., aniline and 4-methylaniline) in Table 11.1, we see that they are much weaker bases than the nonaromatic amine, cyclohexylamine: H

H 

H 

H N H

H N H



H N H

CH3 Aminium ion pKa

10.64

4.58

5.08

We can account for this effect, in part, on the basis of resonance contributions to the overall hybrid of an arylamine. For aniline, the following contributors are important: NH2

NH2







NH2

NH2

NH2 





1

2

3

4

5

Structures 1 and 2 are the Kekulé structures that contribute to any benzene derivative. Structures 3–5, however, delocalize the unshared electron pair of the nitrogen over the ortho and para positions of the ring. This delocalization of the electron pair makes it less available to a proton, and delocalization of the electron pair stabilizes aniline.

11.4 | Basicity of Amines: Amine Salts

Another important effect in explaining the lower basicity of aromatic amines is the electronwithdrawing effect of a phenyl group. Because the carbon atoms of a phenyl group are sp2 hybridized, they are more electronegative (and therefore more electron withdrawing) than the sp3-hybridized carbon atoms of alkyl groups.

11.4B Amines versus Amides ●

Amides are far less basic than amines (even less basic than arylamines). The pKa of the conjugate acid of a typical amide is about zero.

The lower basicity of amides when compared to amines can be understood in terms of resonance and inductive effects. An amide is stabilized by resonance involving the nonbonding pair of electrons on the nitrogen atom. However, an amide protonated on its nitrogen atom lacks this type of resonance stabilization. This is shown in the following resonance structures: Amide O

O R

NH2

R



O



NH2





R

NH2

Larger resonance stabilization

N-Protonated Amide O

O 

R

NH3

R





Smaller resonance stabilization



NH3

However, a more important factor accounting for amides being weaker bases than amines is the powerful electron-withdrawing effect of the carbonyl group of the amide. Comparing the following equilibria, the reaction with the amide lies more to the left than the corresponding reaction with an amine. This is consistent with the amine being a stronger base than an amide. O R

O NH2

RNH2

 H2O



R

NH3 

 H2O

RNH3

 HO  HO

The nitrogen atoms of amides are so weakly basic that when an amide accepts a proton, it does so on its oxygen atom instead. Protonation on the oxygen atom occurs even though oxygen atoms (because of their greater electronegativity) are typically less basic than nitrogen atoms. Notice, however, that if an amide accepts a proton on its oxygen atom, resonance stabilization involving the nonbonding electron pair of the nitrogen atom is possible: 

OH

OH R

NH2

R



OH NH2

R



NH2

11.4C Aminium Salts and Quaternary Ammonium Salts When primary, secondary, and tertiary amines act as bases and react with acids, they form compounds called aminium salts. In an aminium salt the positively charged nitrogen atom is attached to at least one hydrogen atom: H N

H H



HCl



H2 O

N H Cl H Ethylaminium chloride (an aminium salt)

709

710

Chapter 11 | Organic Compounds Containing Nitrogen

H

H



N



HBr

N

H2O

Br

H Diethylaminium bromide



N



HI

N

H2O

I

H

Triethylaminium iodide

When the central nitrogen atom of a compound is positively charged but is not attached to a hydrogen atom, the compound is called a quaternary ammonium salt. For example,



Br

N

Tetraethylammonium bromide (a quaternary ammonium salt)

Quaternary ammonium halides—because they do not have an unshared electron pair on the nitrogen atom—cannot act as bases. Quaternary ammonium hydroxides, however, are strong bases. As solids, or in solution, they consist entirely of quaternary ammonium cations (R4N+) and hydroxide ions (HO−); they are, therefore, strong bases—as strong as sodium or potassium hydroxide. Quaternary ammonium hydroxides react with acids to form quaternary ammonium salts: 1

1

(CH3)4NOH 2 1 HCl

(CH3)4NCl2 1 H2O

11.4D Solubility of Amines in Aqueous Acids ●

Almost all alkylaminium chloride, bromide, iodide, and sulfate salts are soluble in water. Thus, primary, secondary, or tertiary amines that are not soluble in water will dissolve in dilute aqueous HCl, HBr, HI, and H2SO4.

Solubility in dilute acid provides a convenient chemical method for distinguishing amines from nonbasic compounds that are insoluble in water. Solubility in dilute acid also gives us a useful method for separating amines from nonbasic compounds that are insoluble in water. The amine can be extracted into aqueous acid (dilute HCl) and then recovered by making the aqueous solution basic and extracting the amine into ether or CH2Cl2. H N



H X (or H2SO4)

N

Water-insoluble amine

X (or HSO4) Water-soluble aminium salt

Because amides are far less basic than amines, water-insoluble amides do not dissolve in dilute aqueous HCl, HBr, HI, or H2SO4: O R

NH2

Water-insoluble amide (not soluble in aqueous acids)

11.5 | Reactions of Amines

11.5 REACTIONS OF AMINES We have encountered a number of important reactions of amines in earlier sections. In Section 11.3 we saw reactions in which primary, secondary, and tertiary amines act as bases. In Section 11.4 we saw their reactions as nucleophiles in alkylation reactions, and in Chapter 10 as nucleophiles in acylation reactions. In Chapter 6 we saw that an amino group on an aromatic ring acts as a powerful activating group and as an ortho–para director. The feature of amines that underlies all of these reactions and that forms a basis for our understanding of most of the chemistry of amines is the ability of nitrogen to share an electron pair: Acid–Base Reactions N

H  H

A

N

A



An amine acting as a base

Alkylation

CH2R N

 R

CH2

Br

 Br

N

An amine acting as a nucleophile in an alkylation reaction

Acylation O N

H

 R

C

O

O 

Cl

N

C

R

H

Cl

C

N

(HCl)

R

A primary or secondary amine acting as a nucleophile in an acylation reaction

The reaction is shown by both alkyl and aryl amines to form acyl derivatives or substituted amides. The substituted aryl amides are called anilides and acylation proceeds in the presence of a base. CH3NH2 1 CH3COCl

CH3NHCOCH3 1 HCl

N–Methyl acetamide pyridine

C6H5NH2 1 CH3COCl

C6H5NHCOCH3 1 HCl Acetanilide

NaOH

C6H5NH2 1 C6H5COCl

C6H5NHCOC6H5 1 HCl (Schotten−Baumann reaction) Benzanilide

In the preceding examples the amine acts as a nucleophile by donating its electron pair to an electrophilic reagent. In the following example, resonance contributions involving the nitrogen electron pair make carbon atoms nucleophilic: Electrophilic Aromatic Substitution H

N

H

H E



N

A

H

A H

H



N

H E

E H

N

HA

H

H

H



N

H

N

H

HA

A

E



A

H

E

E

The amino group acting as an activating group and as an ortho–para director in electrophilic aromatic substitution

711

712

Chapter 11 | Organic Compounds Containing Nitrogen

Some specific examples of electrophilic aromatic substitution are listed as follows: 1. Bromination: At room temperature, bromine water reacts with aniline to produce a white precipitate of 2, 4, 6-tribromoaniline. NH2

NH2 Br2/H2O

3 Br2

1

Br

Br 3 HBr

1 Br

For preparing monosubstituted aniline derivative, the –NH2 group is protected by acetic anhydride by acylation. The substituted amide is finally hydrolyzed to substituted amine. O NH2

H

N

C

O H

CH3

(CH3CO)2O

Br2

Pyridine

CH3COOH

Aniline

N

C

NH2 1

CH3 OH2 or H1

2 3

N-Phenylethanamide (Acetanilide)

Br

Br

(Major)

4-Bromoaniline

The activating effect of –NHCOCH3 is less than that of –NH2 group as the lone pair of electrons on nitrogen in the former interacts with oxygen due to resonance, making it inadequately available for the benzene ring. 2

O N

C

O 1

CH3

N

CH3

C

2. Nitration: In the presence of nitric and sulphuric acids, aniline forms meta-directing anilinium ion leading to the formation of all the three derivatives. NH2

NH2 288K HNO3, H2SO4

NH2

NH2

1

NO2

1 NO2

NO2 (47%)

(51%)

(2%)

The –NH2 group can be protected by acetylation to produce p-nitroaniline as the major product. NH2

NHCOCH3 (CH3CO)2O

NHCOCH3 2

288K HNO3, H2SO4

Pyridine

NO2 Acetanilide

p-Nitroacetanilide

NH2

OH or H1

NO2 p-Nitroaniline

3. Sulphonation: Aniline forms anilinium hydrogen sulphate with concentrated sulphuric acid which on  further heating with sulphuric acid at 453–473 K form sulphanilic acid (p-aminobenzene sulphonic acid).

11.5 | Reactions of Amines 1

NH2

2

NH3HSO4 H2SO4

1

NH2

NH3

SO3H

SO23

453 2 473K

Anilinium hydrogensulphate

Sulphanilic acid

Zwitter ion

4. Friedel–Crafts reaction: Aniline forms salt with aluminium chloride, a Lewis acid used as a catalyst and hence it does not undergo Friedel–Crafts reaction (alkylation and acylation). The nitrogen of aniline acquires positive charge, acts as a strong deactivating group and does not promote further reaction.

11.5A Oxidation of Amines Primary and secondary aliphatic amines are subject to oxidation, although in most instances useful products are not obtained. Complicated side reactions often occur, causing the formation of complex mixtures. Tertiary amines can be oxidized cleanly to tertiary amine oxides. This transformation can be brought about by using hydrogen peroxide or a peroxy acid: O

R 3N

H2O2 or RCOOH



R 3N

O

A tertiary amine oxide

Tertiary amine oxides undergo a useful elimination reaction to be discussed in Section 11.8B. Arylamines are very easily oxidized by a variety of reagents, including the oxygen in air. Oxidation is not confined to the amino group but also occurs in the ring. (The amino group through its electrondonating ability makes the ring electron rich and hence especially susceptible to oxidation.) The oxidation of other functional groups on an aromatic ring cannot usually be accomplished when an amino group is present on the ring, because oxidation of the ring takes place first.

11.5B Reaction with Grignard Reagents Both aliphatic and aromatic primary and secondary amines react with Grignard reagents to form alkanes corresponding to the alkyl group present in the Grignard reagent. RNH2 1 CH3MgBr → CH4 1 RNHMgBr R2NH 1 CH3Mg I → CH4 1 R2NMgI

11.5C Carbylamine Reaction Primary amines when heated with chloroform and ethanolic potassium hydroxide form pungent isocyanides or carbylamines. This reaction is exclusive to primary amines and is used as a test to distinguish it from secondary and tertiary amines. R 2 NH2 1 CHCl3 1 3KOH

Heat

R 2 NC 1 3KCl 1 3H2O

713

714

Chapter 11 | Organic Compounds Containing Nitrogen

11.5D Hofmann’s Mustard Oil Reaction The reaction of a primary amines (e.g., aniline) with alcoholic carbon disulfide followed by heating with excess of mercuric chloride form isothiocyanates having pungent smell similar to mustard oil. C6H5NH2 1 S 5 C 5 S

HgCl2

C6H5NCS 1 2HCI 1 HgS

11.6 REACTIONS OF AMINES WITH NITROUS ACID Nitrous acid (HO } N O) is a weak, unstable acid. It is always prepared in situ, usually by treating sodium nitrite (NaNO2) with an aqueous solution of a strong acid: HCl(aq) 1 NaNO2(aq) → HONO(aq) 1 NaCl(aq) H2SO4 1 2 NaNO2(aq) → 2 HONO(aq) 1 Na2SO4(aq)

Nitrous acid reacts with all classes of amines. The products that we obtain from these reactions depend on whether the amine is primary, secondary, or tertiary and whether the amine is aliphatic or aromatic.

11.6A Reactions of Primary Aliphatic Amines with Nitrous Acid Primary aliphatic amines react with nitrous acid through a reaction called diazotization to yield highly unstable aliphatic diazonium salts. Even at low temperatures, aliphatic diazonium salts decompose spontaneously by losing nitrogen to form carbocations. The carbocations go on to produce mixtures of alkenes, alcohols, and alkyl halides by removal of a proton, reaction with H2O, and reaction with X−: General Reaction R9NH2  NaNO2  2 HX 1 Aliphatic amine

(HONO) H2O

[ R9N # N 

X

]

 NaX  2 H2O

Aliphatic diazonium salt (highly unstable) N2 (i.e., N#N )

R  X

Alkenes, alcohols, alkyl halides ●

Diazotizations of primary aliphatic amines are of little synthetic importance because they yield such a complex mixture of products.

Diazotizations of primary aliphatic amines are used in some analytical procedures, however, because the evolution of nitrogen is quantitative. They can also be used to generate and thus study the behavior of carbocations in water, acetic acid, and other solvents. The most important reaction of amines with nitrous acid, by far is the reaction of primary aryl amines to give arene diazonium salts (Section 11.9). Primary arylamines can be converted to aryl halides, nitriles and phenols via aryl diazonius ions.

11.7 | Reactions of Amines with Sulfonyl Chlorides: Test for Amines

11.6B Reactions of Secondary Amines with Nitrous Acid Secondary amines—both aryl and alkyl—react with nitrous acid to yield N-nitrosoamines. N-Nitrosoamines usually separate from the reaction mixture as oily yellow liquids: Specific Examples (CH3)2NH

(HONO) H 2O

 HCl  NaNO2

Dimethylamine

(CH3)2N

N

O

N-Nitrosodimethylamine (a yellow oil)

N

H

 HCl

N

 NaNO2

CH3

(HONO) H 2O

N-Methylaniline

O

N CH3

N-Nitroso-N-methylaniline (87–93%) (a yellow oil)

11.6C Reactions of Tertiary Amines with Nitrous Acid When a tertiary aliphatic amine is mixed with nitrous acid, an equilibrium is established among the tertiary amine, its salt, and an N-nitrosoammonium compound: 2 R3N

HX

NaNO2

R3NH X

Tertiary aliphatic amine

R3N

N

OX

N-Nitrosoammonium compound

A mine salt

Although N-nitrosoammonium compounds are stable at low temperatures, at higher temperatures and in aqueous acid they decompose to produce aldehydes or ketones. These reactions are of little synthetic importance, however. Tertiary arylamines react with nitrous acid to form C-nitroso aromatic compounds. Nitrosation takes place almost exclusively at the para position if it is open and, if not, at the ortho position. The reaction is another example of electrophilic aromatic substitution. Specific Example H3C

H3C

N

 HCl

 NaNO2

H3C

H2O, 8 °C

N

N

O

H3C p-Nitroso-N,N-dimethylaniline (80–90%)

REACTIONS OF AMINES WITH SULFONYL

11.7 CHLORIDES: TEST FOR AMINES

Primary and secondary amines react with sulfonyl chlorides to form sulfonamides: O

O R

N

H

H 18 Amine

O

O 1

CI

R

S

Ar

Sulfonyl chloride

2HCl

N

S

Ar

H N-Substituted sulfonamide

O

O R

N

H

R 28 Amine

CI

R

O

O 1

S

Ar

2HCl

N

S

Ar

R N,N-Disubstituted sulfonamide

715

716

Chapter 11 | Organic Compounds Containing Nitrogen

When heated with aqueous acid, sulfonamides are hydrolyzed to amines: O

O R

N

S

(1) H3O, heat (2) HO

Ar

R

N



H

O



R

R

O

O 

S

Ar

This hydrolysis is much slower, however, than hydrolysis of carboxamides.

11.7A The Hinsberg Test ●

Sulfonamide formation is the basis for a chemical test, called the Hinsberg test, that can be used to demonstrate whether an amine is primary, secondary, or tertiary.

A Hinsberg test involves two steps. First, a mixture containing a small amount of the amine and benzenesulfonyl chloride is shaken with excess potassium hydroxide. Next, after allowing time for a reaction to take place, the mixture is acidified. Each type of amine—primary, secondary, or tertiary—gives a different set of visible results after each of these two stages of the test. Primary amines react with benzenesulfonyl chloride to form N-substituted benzenesulfonamides. These, in turn, undergo acid–base reactions with the excess potassium hydroxide to form water-soluble potassium salts. (These reactions take place because the hydrogen attached to nitrogen is made acidic by the strongly electron-withdrawing } SO2 } group.) At this stage our test tube contains a clear solution. Acidification of this solution will, in the next stage, cause the water-insoluble N-substituted sulfonamide to precipitate: R

N

O

O

H

CI



O

O

S

R

H

HO

N

(HCl)

H

S Acidic hydrogen

1 Amine KOH

O

O R

N

S

HCl

K O O R  S N

H Water insoluble (precipitate)

Water-soluble salt (clear solution)

Secondary amines react with benzenesulfonyl chloride in aqueous potassium hydroxide to form insoluble N,N-disubstituted sulfonamides that precipitate after the first stage. N,N-Disubstituted sulfonamides do not dissolve in aqueous potassium hydroxide because they do not have an acidic hydrogen. Acidification of the mixture obtained from a secondary amine produces no visible result; the nonbasic N,N-disubstituted sulfonamide remains as a precipitate and no new precipitate forms: R

N R

H

O

O 

CI

S

HO

(HCl)

O

O R

N

S

R Water insoluble (precipitate)

11.8 | Eliminations Involving Ammonium Compounds

If the amine is a tertiary amine and if it is water insoluble, no apparent change will take place in the mixture as we shake it with benzenesulfonyl chloride and aqueous KOH. When we acidify the mixture, the tertiary amine dissolves because it forms a water-soluble salt.

11.8 ELIMINATIONS INVOLVING AMMONIUM COMPOUNDS 11.8A The Hofmann Elimination All of the eliminations that we have described so far have involved electrically neutral substrates. However, eliminations are known in which the substrate bears a positive charge. One of the most important of these is the E2-type elimination that takes place when a quaternary ammonium hydroxide is heated. The products are an alkene, water, and a tertiary amine: HO



H 

HOH



heat

NR3 A quaternary ammonium hydroxide

NR3



An alkene

A tertiary amine

This reaction was discovered in 1851 by August W. von Hofmann and has since come to bear his name. Quaternary ammonium hydroxides can be prepared from quaternary ammonium halides in aqueous solution through the use of silver oxide or an ion exchange resin: 

NMe3 X 

R



Ag2O

R

H2O

A quaternary ammonium halide

NMe3 HO



2 AgX

A quaternary ammonium hydroxide

Silver halide precipitates from the solution and can be removed by filtration. The quaternary ammonium hydroxide can then be obtained by evaporation of the water. Although most eliminations involving neutral substrates tend to follow the Zaitsev rule, eliminations with charged substrates tend to follow what is called the Hofmann rule and yield mainly the least substituted alkene. We can see an example of this behavior if we compare the following reactions: EtONa EtOH, 25 °C

Br

+

NMe3

 NaBr  EtOH

 75%

– OH 150 °C

25%

 NMe3  H2O

 5%

95%

The precise mechanistic reasons for these differences are complex and are not yet fully understood. One possible explanation is that the transition states of elimination reactions with charged substrates have considerable carbanionic character. Therefore, these transition states show little resemblance to the final alkene product and are not stabilized appreciably by a developing double bond: 



HO

H

HO

H



+NMe

3

Carbanion-like transition state (gives Hofmann orientation)

Br   Alkene-like transition state (gives Zaitsev orientation)

With a charged substrate, the base attacks the most acidic hydrogen instead. A primary hydrogen atom is more acidic because its carbon atom bears only one electron-releasing group.

717

718

Chapter 11 | Organic Compounds Containing Nitrogen

11.8B The Cope Elimination Tertiary amine oxides undergo the elimination of a dialkylhydroxylamine when they are heated. The  reaction is called the Cope elimination, it is a syn elimination and proceeds through a cyclic transition state. R

Me



N H

O

H

Me

R

150 °C

A tertiary amine oxide



Me O N Me

An alkene

N,N-Dimethylhydroxylamine

Tertiary amine oxides are easily prepared by treating tertiary amines with hydrogen peroxide (Section 11.5A). The Cope elimination is useful synthetically. Consider the following synthesis of methylenecyclohexane: H O N



160 °C

CH3 CH3

(CH 3)2NOH

98%

PART II: ARENEDIAZONIUM SALTS Diazonium salts are almost always prepared by diazotizing primary aromatic amines. Primary arylamines can be synthesized through reduction of nitro compounds that are readily available through direct nitration reactions.

11.9 PREPARATION OF ARENEDIAZONIUM SALTS ●

Primary arylamines react with nitrous acid to give arenediazonium salts.

Even though arenediazonium salts are unstable, they are still far more stable than aliphatic diazonium salts; they do not decompose at an appreciable rate in solution when the temperature of the reaction mixture is kept below 5 °C: Ar

NH2

Primary ary lamine

NaNO2

2 HX

Ar

N

N X

A renediazonium salt (stable if kept below 5 °C)

NaX

2 H2O

Diazotization of a primary amine takes place through a series of steps. In the presence of strong acid, nitrous acid dissociates to produce +NO ions. These ions then react with the nitrogen of the amine to form an unstable N-nitrosoaminium ion as an intermediate. This intermediate then loses a proton to form an N-nitrosoamine, which, in turn, tautomerizes to a diazohydroxide in a reaction that is similar to keto–enol tautomerization. Then, in the presence of acid, the diazohydroxide loses water to form the diazonium ion. ●

Diazotization reactions of primary arylamines are of considerable synthetic importance because 1 the diazonium group, N N can be replaced by a variety of other functional groups.

11.10 | Replacement Reactions of Arenediazonium Salts

MECHANISM 1

O 1 H3O

N

HO

Diazotization

1 A

H Ar

N

H 1

1N

O

N

N

H

N

O 1 H2O

O

2H3O1

Ar

2HA

O1 H

H

Ar

1HA

1N

O

O

H

N-Nitrosoaminium ion

N

N

N

H

N

2 H2O 1

OH2

1

Ar

18 Arylamine (or alkylamine)

Ar

1

H2O

2

N

H

A

N-Nitrosoamine

N

1HA

OH

2

A

Ar

2HA

Diazohydroxide

Ar

1

N

N

N

OH2

Ar

N

1

1

N

N 1 H2O

Diazonium ion

11.10 REPLACEMENT REACTIONS OF ARENEDIAZONIUM SALTS ●

Arenediazonium salts are highly useful intermediates in the synthesis of aromatic compounds, because the diazonium group can be replaced by any one of a number of other atoms or groups, including } F, } Cl, } Br, } I, } CN, } OH, and } H.

11.10A Syntheses Using Diazonium Salts Most arenediazonium salts are unstable at temperatures above 5–10 °C, and many explode when dry. Fortunately, however, most of the replacement reactions of diazonium salts do not require their isolation. We simply add another reagent (CuCl, CuBr, KI, etc.) to the mixture, gently warm the solution, and the replacement (accompanied by the evolution of nitrogen) takes place: Cu2O, Cu 2, H 2O CuCl CuBr

Ar

NH 2

HONO 0–5 °C

Ar



N2

Arenediazonium salt

CuCN KI (1) HBF 4 (2) heat H 3PO2, H 2O

Ar

OH

Ar

Cl

Ar

Br

Ar

CN

Ar

I

Ar

F

Ar

H

Only in the replacement of the diazonium group by } F need we isolate a diazonium salt. We do this by adding HBF4 to the mixture, causing the sparingly soluble and reasonably stable arenediazonium fluoroborate, ArN2+ BF4−, to precipitate.

719

720

Chapter 11 | Organic Compounds Containing Nitrogen

11.10B The Sandmeyer Reaction: Replacement of the Diazonium Group by } Cl, } Br, or } CN Arenediazonium salts react with cuprous chloride, cuprous bromide, and cuprous cyanide to give products in which the diazonium group has been replaced by } Cl, } Br, and } CN, respectively. These reactions are known generally as Sandmeyer reactions. Several specific examples follow. The mechanisms of these replacement reactions are not fully understood; the reactions appear to be radical in nature, not ionic. CH3

CH3 NH 2

1

CH3

N2

HCl, NaNO2

Cl 2

Cl CuCl

o-Toluidine

o-Chlorotoluene (74–79% overall) 1

NH 2

N2 Br 2

HBr, NaNO2

Br CuBr

1 N2

100 °C

H2O (0–10 °C)

Cl

Cl

m-Chloroaniline

NO2

NO2 NH 2

1 N2

15–60 °C

H2O (0–5 °C)

Cl

m-Bromochlorobenzene (70% overall)

NO2

1

N2 Cl2

HCl, NaNO2

CN CuCN

H2O 25 °C

1 N2

90–100 °C

o-Nitroaniline

o-Nitrobenzonitrile (65% overall)

In a modification of Sandmeyer reaction, called Gattermann reaction, benzenediazonium chloride is treated with copper powder and a halogen acid (instead of cuprous halide dissolved in the corresponding halogen acid) to form aryl halide. 1

N

NCl2

Benzenediazonium chloride 1

N

Cu/HCl Heat

Cl 1 N2

Chlorobenzene

NCl2

Benzenediazonium chloride

Cu/HBr Heat

Br 1 N2

Bromobenzene

11.10C Replacement by } I Arenediazonium salts react with potassium iodide to give products in which the diazonium group has been replaced by } I. An example is the synthesis of p-iodonitrobenzene: NO2

NO2 H 2SO4 , NaNO2

NO2 KI



H2O 0–5 °C

NH 2 p-Nitroaniline

N2 HSO4

I p-Iodonitrobenzene (81% overall)

N2

11.10 | Replacement Reactions of Arenediazonium Salts

11.10D Replacement by } F The diazonium group can be replaced by fluorine by treating the diazonium salt with fluoroboric acid (HBF4). The diazonium fluoroborate that precipitates is isolated, dried, and heated until decomposition occurs. An aryl fluoride is produced: CH3

CH3

CH3

(1) HONO, H 

NH 2



heat

(2) HBF 4

N2 BF4

m-Toluidine

N2



BF3

F

m-Toluenediazonium fluoroborate (79%)

m-Fluorotoluene (69%)

11.10E Replacement by } OH The diazonium group can be replaced by a hydroxyl group by adding cuprous oxide to a dilute solution of the diazonium salt containing a large excess of cupric nitrate: H3C

N2 HSO4

Cu2O Cu 2, H2O

H3C

OH

p-Toluenediazonium hydrogen sulfate

p-Cresol (93%)

This variation of the Sandmeyer reaction (developed by T. Cohen, University of Pittsburgh) is a much simpler and safer procedure than an older method for phenol preparation, which required heating the diazonium salt with concentrated aqueous acid.

11.10F Replacement by Hydrogen: Deamination by Diazotization Arenediazonium salts react with hypophosphorous acid (H3PO2) to yield products in which the diazonium group has been replaced by } H. Since we usually begin a synthesis using diazonium salts by nitrating an aromatic compound, that is, replacing } H by } NO2 and then by } NH2, it may seem strange that we would ever want to replace a diazonium group by } H. However, replacement of the diazonium group by } H can be a useful reaction. We can introduce an amino group into an aromatic ring to influence the orientation of a subsequent reaction. Later we can remove the amino group (i.e., carry out a deamination) by diazotizing it and treating the diazonium salt with H3PO2. We can see an example of the usefulness of a deamination reaction in the following synthesis of m-bromotoluene. CH3

CH3

O 2

CH3 H 2SO4, NaNO2

(1) Br 2

O

(2) HO, H2O heat

NH 2 p -Toluidine

H2O 0–5 °C

Br NH 2

HN

65% (from p-toluidine)

O CH3

CH3 H3PO2

Br N2

H2O 25 °C

 Br

m-Bromotoluene (85% from 2-bromo-4methylaniline)

N2

721

722

Chapter 11 | Organic Compounds Containing Nitrogen

We cannot prepare m-bromotoluene by direct bromination of toluene or by a Friedel–Crafts alkylation of bromobenzene because both reactions give o- and p-bromotoluene. (Both CH3 } and Br } are ortho– para directors.) However, if we begin with p-toluidine (prepared by nitrating toluene, separating the para isomer, and reducing the nitro group), we can carry out the sequence of reactions shown and obtain m-bromotoluene in good yield. The first step, synthesis of the N-acetyl derivative of p-toluidine, is done to reduce the activating effect of the amino group. (Otherwise both ortho positions would be brominated.) Later, the acetyl group is removed by hydrolysis.

11.11 COUPLING REACTIONS OF ARENEDIAZONIUM SALTS Arenediazonium ions are weak electrophiles; they react with highly reactive aromatic compounds—with phenols and tertiary arylamines—to yield azo compounds. This electrophilic aromatic substitution is often called a diazo coupling reaction. General Reaction 

N N X

Q

Q

N

Q

 NR2 or



N

OH

X

H

HX

Q N

N

An azo compound

Specific Examples

OH N

N2

Cl

OH



Benzenediazonium chloride

Phenol

NaOH, H2O, 0 °C

N

p-(Phenylazo)phenol (orange solid)

N(CH 3)2 O

N21

Cl2

Benzenediazonium chloride

1

N(CH 3)2 N,N-Dimethylaniline

CH3CO2 Na1 0 °C, H2O

N

N

N,N-Dimethyl-p-(phenylazo)aniline (yellow solid)

Couplings between arenediazonium cations and phenols take place most rapidly in slightly alkaline solution. Under these conditions an appreciable amount of the phenol is present as a phenoxide ion, ArO−, and phenoxide ions are even more reactive toward electrophilic substitution than are phenols themselves. If the solution is too alkaline (pH > 10), however, the arenediazonium salt itself reacts with hydroxide ion to form a relatively unreactive diazohydroxide or diazotate ion:

11.11 | Coupling Reactions of Arenediazonium Salts

O

OH HO HA

Phenol (couples slowly)

Ar



N

HO

N

Ar

HA

Arenediazonium ion (couples)

Phenoxide ion (couples rapidly)

N

N

HO

OH

Ar

HA

Diazohydroxide (does not couple)

N

N

O

Diazotate ion (does not couple)

Couplings between arenediazonium cations and amines take place most rapidly in slightly acidic solutions (pH 5–7). Under these conditions the concentration of the arenediazonium cation is at a maximum; at the same time an excessive amount of the amine has not been converted to an unreactive aminium salt: 

NR2

HNR2 HA HO

Amine (couples)

Aminium salt (does not couple)

If the pH of the solution is lower than 5, the rate of amine coupling is low. With phenols and aniline derivatives, coupling takes place almost exclusively at the para position if it is open. If it is not, coupling takes place at the ortho position. 

HO

OH

N2 Cl

N



NaOH H 2O



N

CH3

CH3 4-Methylphenol (p-cresol)

4-Methyl-2-(phenylazo)phenol

Azo compounds are usually intensely colored because the azo (diazenediyl) linkage, } N N } , brings the two aromatic rings into conjugation. This gives an extended system of delocalized π electrons and allows absorption of light in the visible region. Azo compounds, because of their intense colors and because they can be synthesized from relatively inexpensive compounds, are used extensively as dyes. Azo dyes almost always contain one or more } SO3−Na+ groups to confer water solubility on the dye and assist in binding the dye to the surfaces of polar fibers (wool, cotton, or nylon). Many dyes are made by coupling reactions of naphthylamines and naphthols. Orange II, a dye introduced in 1876, is made from 2-naphthol: SO3 Na

OH N

N

Orange II

723

724

Chapter 11 | Organic Compounds Containing Nitrogen

PART III: CYANIDES AND ISOCYANIDES Alkyl cyanides (R } C N) and alkyl isocyanides (R } N C) are isomeric compounds that are organic derivatives of hydrocyanic acid. Cyanide ion is ambident in nature and can form covalent bonds either from carbon or nitrogen atom.

PREPARATION AND PROPERTIES

11.12 OF CYANIDES AND ISOCYANIDES 11.12A Methods of Preparation The important methods of preparation are discussed as follows. From Alkyl Halides RX 1 KCN

RCN 1 KCl

CH3CH2Cl 1 KCN

CH3CH2CN 1 KCl Ethyl cyanide (Propanenitrile)

CH3CH2I 1 AgCN

CH3CH2NC 1 AgI Ethyl isocyanide

From Primary Amines (Carbylamine Reaction) R2NH2 1 CHCl3 1 3KOH(alc.)

Heat

CH3CH2NH2 1 CHCl3 1 3KOH(alc.) C6H5NH2 1 CHCl3 1 3KOH(alc.)

R2NC 1 3KCl 1 3H2O Heat

Heat

CH3CH22NC 1 3KCl 1 3H2O

C6H52NC 1 3KCl 1 3H2O

From Arenediazonium Salts 1

ArN2 X2 1 CuCN/KCN

ArCN 1 N2 (Sandmeyer’s reaction)

From Aldoximes CH3CH 5 NOH

(CH3CO)2O

CH3CN 1 H2O

From Grignard Reagent CH3CH2MgBr 1 ClCN → CH3CH2CN 1 Mg(Br)Cl

11.12B Physical Properties Alkyl nitriles generally have a pleasant smell whereas isonitriles have unpleasant smell and are mostly poisonous. The lower members of alkyl nitriles and isonitriles are colourless liquids but higher members are solids. The cyanide and ioscyanide functional groups are highly polar in nature, with strong intermolecular forces. The nitriles have higher boiling points than the isomeric isonitriles because R−CN have dipole moment than R−NC. The lower members of alkyl cyanides are soluble in water as they form hydrogen bond with water. The solubility deceases with increase in size of the alkyl group (hydrophobic). Alkyl isocyanides are however, insoluble in water because the lone pair on nitrogen is involved in bonding and not available for formation of hydrogen bond.

11.12 | Preparation and Properties of Cyanides and Isocyanides

11.12C Chemical Properties The characteristic reactions of cyanides and isocyanides are hydrolysis and reduction. Hydrolysis Alkyl cyanides are hydrolyzed in presence of dilute acid or bases to give carboxylic acids and ammonia. Alkyl isocyanides, on the other hand are hydrolyzed only in the presence dilute acids to form primary amines and formic acid. H

RCN 1 H2O

H1 H2O

RCONH2

Alkyl cyanide

Acid amide 2

H

RCN 1 H2O

2

OH

RCONH2

Alkyl cyanide

RCOOH 1 NH3 Carboxylic acid

Acid amide

RCOO 1 NH3 2

Carboxylate ion

O 1

H

RNC 1 H2O

1

R 2 NH 2 C 2 H

H / H2O

Isocyanide

RNH2 1 HCOOH

Primary amine Formic acid

Reduction Alkyl cyanides can be completely reduced either catalytic hydrogenation or by lithium aluminium hydride to give the corresponding primary amine. When the reduction is carried out in presence of sodium and alcohol, it is called Mendius reaction. Pt or Ni

RCN 1 2H2 Alkylcyanide LiAIH4

RCN

RCH2NH2

reduction

Cyanide

RCH2NH2 Primary amine

Primary amine

Isocyanides on complete reduction give secondary amines, that is, N-methyl amines. Pt or Ni or Na/C2H5OH

RNC 1 2 H2 Isocyanide

RNHCH3

N – Methyl alkylamine

On reducing the solution of an alkyl cyanide in ether with hydrogen chloride gas and stannous chloride at room temperature, a precipitate of amine hydrochloride is formed which on hydrolysis with boiling water gives aldehydes. This partial reduction reaction is known as Stephen reduction. SnCI2 1 2HCI RCN 1 2[H] 1 HCI

Ether 290–295 K

Alkyl cyanide

SnCI4 1 2[H]

RCH

Boiling

NH . HCI

RCHO 1 NH4CI

water

Amine hydrochloride

Alderhyde

Isomerism When alkyl isocyanides are heated for a long time, they rearrange to form cyanides. R2N

C

D

Isocyanide

R2C

N

Cyanide

Reaction with Grignard Reagent Alkyl cyanides react with Grignard reagent to form ketones. R9 RC

N 1 RMgX

ether

Alkyl cyanide Grignard reagent

1

3R 2 C 2 N Mg Br4 Amine salt

H / H2O 2Mg

(OH)X

2NH 3

R R9

C

O

Ketone

725

726

Chapter 11 | Organic Compounds Containing Nitrogen

Addition Reactions Alkyl isocyanates form addition products on reaction with sulphur, halogens and ozone and are oxidized by HgO. R

N1

C2 1 S

R

Alkyl isocyanide

R

NC 1 CI2

R

Alkyl isocyanide

R

C

S

N

CCI2

Alkyl iminocarbonyl chloride

NC 1 O3

R

Alkyl isocyanide

R

N

Alkyl isothiocyanate

N

C

O 1 O3

Alkyl isocyanate

NC 1 2HgO

R

Alkyl isocyanide

N

C

O 1 Hg2O

Alkyl isocyanate

SUMMARY OF REACTIONS OF ORGANIC

11.13 COMPOUNDS CONTAINING NITROGEN 11.13A Preparation of Amines 1. Gabriel synthesis

N H

O

O

O (1) KOH (2) R

NH 2NH 2

N R

X

R

ethanol, reflux

NH 2 

O

O

N N O

2. By reduction of alkyl azides R

NaN3

Br

R

ethanol

N



N



N

Na/alcohol

R

or LiAlH4

NH2

3. By amination of alkyl halides R

Br  NH3

RNH3 Br  R2NH2 Br  R3NH Br  R4N Br HO

RNH2  R2NH  R3N  R4N OH (A mixture of products results.)

(R = a 1° alkyl group)

4. By amination of alcohols R

OH 1 NH3

RNH2 1 H2O

R

OH 1 RNH2

R2NH 1 H2O

R

OH 1 R2NH

R3N 1 H2O

5. By reduction of nitroarenes Ar

NO2

H2, catalyst or (1) Fe/HCl (2) NaOH

Ar

NH2

H

H

11.13 | Summary of Reactions of Organic Compounds Containing Nitrogen

6. By reductive amination H NH 3

1 Amine

[H]

R

R

H

H

O

2 Amine

[H]

Aldehyde or ketone

R¿ R–

N

RNH 2

R

H

N

R

H

R‡ RRNH

R¿ R–

N

3 Amine

[H]

R

H

R¿

7. By reduction of nitriles, oximes, and amides H 1. LiAlH4, Et2O

CN

R

N

R

2. H2O

OH R

R

R

R

N

O R

1. LiAlH4, Et2O

H

N

2. H2O

R9

N

R

1. LiAlH4, Et2O

N

R

2. H2O

O R

N

H 18 Amine

R9

28 Amine

H

H

R9

1 Amine

H

H O

18 Amine

H

NH2

Na/ethanol

R

N

1. LiAlH4, Et2O

R

2. H2O

N

R9

38 Amine

R0

R0

8. Through the Hofmann and Curtius rearrangements Hofmann Rearrangement O H R

Br2, HO

N

R

NH2  CO32

N

C

H

Curtius Rearrangement O

O R

Cl

NaN3 (NaCl)

R

N3

(N2) heat

R

O

H2O

R

NH2  CO2

727

728

Chapter 11 | Organic Compounds Containing Nitrogen

11.13B Reactions of Amines 1. As bases H R

N

R  H

A



R

N

R A 

R R (R, R, and/or R may be alkyl, H, or Ar)

2. Conversion to sulfonamides O

O R

N

H

R

1. ArSO2Cl, HO2

N

2. HCl

H

S

R

Ar

H

N

ArSO2Cl, HO2

R9

H

O

O

R

N

S

Ar

R9

3. Conversion to amides O

O

R

N

H

R0

O

Cl

base

R0

H

N

N

1 Cl

2

H

R0

R

N

H

R0

2

O

O

R0

H

H O

R

R

O N

R

1

H

O

Cl

base

R0

R

N

R9

1 Cl2

R9

4. Hofmann and Cope eliminations Hofmann Elimination H 

HO 

NR3

 H 2O 

heat

NR3

Cope Elimination  Me N Me

H

O–

syn elimination heat



Me

N

Me

OH

5. Diazotization of 1° arylamines and replacement of, or coupling with, the diazonium group Cu 2O, Cu 2, H 2O CuCl CuBr

Ar

NH 2

HONO 0–5 °C

Ar



N2

CuCN KI (1) HBF4 (2) heat H 3PO2 , H 2O

Ar

OH

Ar

Cl

Ar

Br

Ar

CN

Ar

I

Ar

F

Ar

H

R0

OH

11.13 | Summary of Reactions of Organic Compounds Containing Nitrogen

Q Q



N2

N

N

Q  NR2 or OH

11.13C Preparation of Cyanides and Isocyanides 1. From alkyl halides RX 1 KCN → RCN 1 KCl

2. From primary amines (Carbylamine reaction) Heat

R 2 NH2 1 CHCl3 1 3KOH(alc.)

R 2 NC 1 3KCl 1 3H2O

3. From arenediazonium salts 1

ArN2 X2 1 CuCN/KCN

ArCN 1 N2 (Sandmeyer’s reaction)

11.13D Reactions of Cyanides and Isocyanides 1. Hydrolysis H1

RCN 1 H2O

2

OH

RCN 1 H2O

H1 H2O

RCONH2

RCOOH 1 NH3

2

OH

RCONH2

RCOO 1 NH3 2

O RNC 1 H2O

1

1

H

R 2 NH 2 C 2 H

H /H2O

RNH2 1 HCOOH

2. Reduction Pt or Ni

RCN 1 2H2 RCN

LiAIH4

RCN 1 2[H] 1 HCI

Ether 290–295 K

RCH2NH2

reduction

RNC 1 2H2 RCH

RCH2NH2

Pt or Ni or Na/C2H5OH

NH . HCI

Boiling water

RNHCH3 RCHO 1 NH4CI (Stephen reduction)

3. Isomerism R2N

C

D

R2C

N

4. With Grignard reagent R9 RC

N 1 RMgX

ether

1

R 2 C 2 N MgBr

H /H2O 2Mg

(OH)X

2NH3

R R9

C

O

729

730

Chapter 11 | Organic Compounds Containing Nitrogen

5. Addition reactions R 2 N1

C2 1 S

R2N

R 2 NC 1 CI2

R2N

R 2 NC 1 O3

R2N

R 2 NC 1 2HgO

C CCI2

C

R2N

S

O 1 O3 C

O 1 Hg2O

SOLVED EXAMPLES 1. Classify the following amines as primary, secondary or tertiary. (a) CH3NHCH3 (b) CH3CH2N(CH3)2 NH

(c)

3. Arrange the following isomeric amines in order of increasing boiling points. Provide an explanation for your answer. (1) n-Butyl amine (2) Diethyl amine (3) Ethyl dimethyl amine Solution

(d) CH3

N

CH2CH2CH2CH3

N

CH2CH3

Solution (a) Secondary (b) Tertiary

(c) Secondary (d) Tertiary

2. Provide the structural formula for each amine. (a) 1-Hexanamine (b) 2-Methyl-1-propanamine (c) 1,4-Butanediamine (d) trans-4-Methylcyclohexanamine (e) 1-Phenyl-2-Propanamine (f) N-Ethyl-N-methylaniline (g) Benzylamine Solution (a) CH3CH2CH2CH2CH2CH2 (b) CH3CH

CH2

NH2

NH2

CH3

(c) NH2

CH2CH2CH2CH2

NH2

H3C

NH2

1

(e)

3 2

NH2

(f)

CH3

(g)

CH3

N

NH2

NH2

,

The main difference between these isomeric amines is their degree; primary, secondary or tertiary. N,Ndimethyl ethyl amine has the lowest boiling point (36–38°C) since it is the most branched structure, which does not allow for the molecules to get close to one another. Also, being a tertiary amine it can only function as an H-acceptor. Diethyl amine has the next highest boiling point (55.5°C) because the branching is less, meaning molecules can get close to one another and being a secondary amine it can function as both an H-bond donor and acceptor. Butyl amine has the highest boiling point (77°C) since this molecule has no branching and therefore adjacent molecules can get very close to one another, which strengthens the intermolecular attractive forces. Being a primary amine, it can function as both H-bond donor and acceptor. 4. Account for the fact that 1-butanamine has a lower boiling point than 1-butanol. Solution

CH3

(d)

N H

,

1-Butanol has a higher boiling point than 1-butanamine because an O−H—O hydrogen bond is stronger than an N−H—N hydrogen bond because of the higher polarity of the OH bond. NH2 bp 78°C 1-Butanamine

OH bp 117°C 1-Butanol

Solved Examples The intermolecular forces are greater in 1-butanol compared to 1-butanamine, hence the higher boiling point of 1-butanol. 5. What factors must one take into account when estimating the basicity of amines?

NH2 O2N

NH2

6. Select the stronger acid from each pair of ions. (a) O2N

NH31 or CH3 A

NH31

NH

1

or

N

(d)

7. Each of the following compounds contains two nitrogens. Indicate which one would get protonated first when treated with HCl. NH2 (c)

NH2

(b) NH2

O2N

NH2 N

(d)

N H

Solution (a) The lone pair of electrons on N1 is localized while N2 electron pair is delocalized in the ring. 1

NH2

1

NH3Cl2

HCl

1

N H

8. Amines can act as nucleophiles. For each of the following molecules, circle the most likely atom that would be attacked by the nitrogen of an amine: (a)

(a) When comparing the strengths of acids, examine the conjugate bases. (A) is the stronger acid because its conjugate base, p-nitroaniline, is more stable, as a result of the inductively withdrawing nitro group, than the conjugate base of (B), p-methylaniline. (b) (C) is the stronger acid because its conjugate base, pyridine, is more stable, as a result of the sp2-hybridized nitrogen atom, than the conjugate base of (D), cyclohexanamine. The nitrogen atom of cyclohexanamine is sp3-hybridized.

N

NH3Cl2 H N

HCl

N H

NH31

H2N

1

N

NH2

2

(b)

(a)

1

N

O

B

Solution

H 2N

NH3Cl2

HCl

The factors one must take into account when estimating the basicity of amines are as follows: (1) electron delocalization through extended π bonding (resonance), (2) inductive effects, (3) solvation, (4) steric inhibition of resonance, (5) steric effects towards solvation, (6) s character, and (7) field effects.

2

1

O2N

(c) N1 is more basic than N2.

Solution

(b)

NH2 HCl

O OCH3

(c) Cl

Br

Solution O

(a) O

(b)

OCH3

(c) Cl

Br

In each of these molecules, the most likely atom that would be attacked by the nitrogen of an amine is the one that has the greatest δ+ charge (the most electrophilic). However, realize that according to kinetic molecular theory, not every attack (collision) is successful. Although the indicated carbon in (c) is more likely to be attacked than is the carbon atom bearing the Br, the latter is more likely to yield a successful collision due to −Br being a better leaving group than −Cl. 9. The reaction sequence below shows how a methyl group on a benzene ring can be replaced by an amino group. Supply the missing reagents and intermediates. CH3

2. H3O1

H2N

(b) Nitro is an electron withdrawing group, thus, it reduces the electron density on nitrogen. Nitro reduces electron density on ortho and para position greater than that in meta position.

1. KMnO4, HO2, Δ

A B

O NH2

E

D

C

Cl

731

732

Chapter 11 | Organic Compounds Containing Nitrogen Solution

12. For the following reaction, provide the intermediates A and B.

CO2H A=

Br NaCN

B = SO2Cl C = NH3

A

1. CH3Li

O

1. NH3 2. LiBH3CN

NH2

D=

B

2. H2O

E = Br2/HO2

H2N

10. The starting material and reactants needed to complete the following reaction are? NH2 Cl

Amphetamine

Solution O

O

CN

A=

B=

OH

13. Provide the major organic product from each of the following reactions.

Solution NO2

NH2

Cl

Cl

1. NaCN 2. LiAIH4

Cl

(a)

3. H3O1

H2/Pt or Ni

O

NH2

O OH

OH

(b)

11. How will you carry out the following conversions? (a) Elthanoic acid to diethylamine (b) Benzene to aniline (c) n-Propylchloride to n-proply amine by ammonolysis (d) Acetaldehyde to ethyl amine

O 1 N(CH3)2

(c)

NH2

(a) CH COOH 3

SOCl2

(a)

CH3COCI

Ethanoic acid

base

CH3CH2NH2

(b)

O CH3

C

NH

CH2

N H

CH3

(c)

LiAIH4

1

N

CH3 CH2 NH CH2 CH3

NH2 Fe/HCl

14. Outline a synthesis of 4-methylpentanamine using the Gabriel synthesis.

Aniline

(c) CH2CH2CH2Cl 1 NH3 n-Propyl chloride

O Acetaldehyde

O

O

CH3CH2CH2NH2

(Excess)

N

H

1. KOH 2.

n-Propyl amine Δ 2H2O

I

Solution

Benzene

H 1 NH3

2

CH3

NO2 HNO3/H2SO4 Δ

CH3 CH3

Diethylamine

(d) CH3C

CH3I

Solution

Solution

(b)

NaBH3CN

N Br

O

O CH3

CH

NH

H2 /Ni

CH3CH2NH2 Ethylamine

NH2NH2 EtOH, heat

O N

NH2 1

N O

H

H

Solved Examples 15. Show how you might utilize the reduction of an amide, oxime, or nitrile to carry out each of the following transformations: O OH

(a)

N H

Solution

Br

(b)

NH2

O

(c)

(a) Benzylamine dissolves in dilute HCl at room temperature

N

OH

NH2

3

1 H3O1 1 Cl2

NH2

O

(d)

16. Write equations for simple chemical tests that would distinguish between (a) Benzylamine and benzamide (b) Allylamine and propylamine (c) p-Toluidine and N-methylaniline (d) Triethylamine and diethylamine

25°C

Solution O

(a) C6H5

1

O OH

SOCl2

C6H5

NH3Cl2 Cl NH2

Benzamide does not dissolve O

O

NH2

N H

C6H5

1 H3O1 1 Cl2

25°C

No reaction

(b) Allylamine reacts with bromine

LiAIH4

Br C6H5

NH2

N H N

Br NaCN

(b)

NH2 O

(c)

O

No reaction if the mixture is not heated or irradiated

1 Br2

(c) These can be distinguished on the basis of reaction with benzene sulfonyl chloride. NH2 1 C6H5SO2Cl

H3C

SOCl2

OH

1 Br2

Propylamine does not: NH2

LiAlH4

Cl

NH2

Br

H N

KOH H2O

O

K1 NSO2C6H5

H3C N

Soluble H3O1 LiAIH4

NHSO2C6H5

H3C N

Precipitate

HO

(d)

O

NH2OH

KOH

N

NSO2C6H5

NHCH3 1 C6H5SO2Cl H O 2

Precipitate Na/

NH2

OH

CH3

H3O1

Precipitate remains

733

734

Chapter 11 | Organic Compounds Containing Nitrogen (d) These can be distinguished on the basis of raction with benzene sulfonyl chloride. 3N

1 C6H5SO2Cl

KOH H2O

CH3

(c)

No reaction

Cl

Cl2 hv

(excess)

CH3

H3O1

N1

(CH3)3N

1

CH3

3NH

Cl2 CH3

Soluble 2NH

KOH H2O

1 C6H5SO2Cl

2NSO2C6H5

Precipitate

O

CH3

(d)

OH

1. KMnO4, HO2 2. H3O1

O2N

O2N

H3O1

Δ NaOH, CaO

Precipitate remains

Fe

17. Using a different method for each part, but taking care in each case to select a good method, show how each of the following transformations might be accomplished:

HCl

NH2

NO2

HNO3, H2SO4

NH2

(a)

NO2 CH3O

CH3O NH2 NH2

(b) CH3O

CH3O

18. Outline a synthesis of butter yellow from benzene and N, N-dimethylaniline.

CH3

CH3

N1

(c)

CH2Cl2

N(CH3)2

CH3

(excess)

N CH3

N

NH2

(d) O2N

Butter yellow

O 2N

Solution

Solution NO2

HNO3

(a)

NO2

NH2

H2SO4

CH3O

HNO3

CH3O (1 ortho; separate)

H2SO4

1. Fe, HCl 2. OH2

NH2

(0–5 °C)

Fe HCl

N12 Cl2

CH3O O

O Cl AlCl3

(b)

H2SO4/NaNO2

CH3O

CH3O

pH 5–7

NH2

N(CH3)2

NH3, H2 Ni

CH3O

N

N

N(CH3)2

Solved Previous Years’ NEET Questions 19. How would you prepare (a) p-toludine, (b) m-chlorotoluene, (c) m-bromotoluene, (d) m-iodotoluene, (e) m-tolunitrile (mCH3C6H4CN) and (f) m-toluic acid from toluene? CH3

CHO

CHO

Solution NO2

HNO3

PCC

(a)

20. Show how you might prepare each of the following compounds from benzene: (a) m-Chloroaniline (b) m-Bromoaniline

H2SO4

NO2

(a)

NO2

HNO3

Cl2

H2SO4

Fe

Fe HCl

CH3

Cl

CHO NH2

1. Fe, HCI, heat

2. HO2

NH2

NH2

OH2

NH2 NH2

Cl

CH3 (b)

NO2

CuCI

Br2

(b)

FeBr3

CI

(c)

CH3 HONO (0–5 °C) HCI

NH2

Br

[From part (a)]

CH3 CH3

NO2

1. Fe, HCI, heat

CuBr

2. HO2

NH2

Br CH3 1 N2 CI2

(d) KI

Br I CH3

(e)

CuCN

CN CH3 (f)

H3O1 H2O

CO2H

SOLVED PREVIOUS YEARS’ NEET QUESTIONS 1. Which one of the following on reduction with lithium aluminium hydride yields a secondary amine? (1) Methyl cyanide (2) Nitroethane (3) Methylisocyanide (4) Acetamide (AIPMT 2007)

Solution (3) Alkyl isocyanides on reduction in the presence of LiAlH4 yield secondary amines containing methyl as one of the alkyl groups. R

N

C 1 4[H]

LiAiH4

R

NH

CH3

2° amine

735

736

Chapter 11 | Organic Compounds Containing Nitrogen 2. In a reaction of aniline, a colored product C was obtained. NH2

NaNO2 HCl

N

B

CH3

(3)

N

CH3 CH3

Cold

C

NO2

NHCH3

A

NO

(4)

The structure of C would be

NHCH3 1

CH3

(1)

N

N

NO

N

(AIPMT 2009)

CH3

Solution N

(2)

N

CH2

N

(2) Secondary amines-both aryl and alkyl react with nitrous acid to yield N-nitrosoamines. It usually separates from the reaction mixture as oily yellow liquids.

CH3 CH3

(3)

CH3 N

H

N

N CH3

NH

(4)

NH

N

N-Methyl aniline

CH3

(HONO) H2O

CH3

N

(AIPMT 2008)

CH3 N-Nitroso-N-methylaniline (87–93%) (a yellow oil)

(1) The reaction is A

N21Cl2 B (Diazonium salt)

Weak electrophile attacks on activated rings to undergo diazo coupling. N

1

N1

N

4. Acetamide is treated with the following reagents separately. Which one of these would yield methylamine? (1) soda lime (3) PCl5 (2) hot conc. H2SO4 (4) NaOH–Br2 (AIPMT PRE 2010)

CH3

Solution

CH3

N

N

N

C

CH3 CH3

(4) Methylamine can be obtained using Hofmannbromamide reaction. CH3

C

Br2

NH2 4NaOH

CH3NH2 1 K2CO3 1 2KBr 1 2H2O

O

5. Aniline in a set of the following reactions yielded a colored product ‘Y’.

3. Predict the product NHCH3 1 NaNO2 1 HCl

NH2 NaNO2/HCl

N

N,N-Dimethylaniline

X

(273–278 K)

OH

The structure of ‘Y’ would be

CH3

CH3

(1) HN

CH3

(2)

O

N

Solution

(1)

1 HCl 1 NaNO2

N

N

O

(2) H3C

CH3 NH

N

NH

N

NH2

Y

Solved Previous Years’ NEET Questions CH3

CH3

(3) HN

N

N

CH3

(4)

N

(c) CH3CH2COOCH3

(iii) Gives white ppt. with ammoniacal AgNO3

(d) CH3CH(OH)CH3

(iv) With Lucas reagent cloudiness appears after 5 min.

NH

N

N CH3

(AIPMT PRE 2010) Solution (4) Primary arylamines react with nitrous acid to give arenediazonium ions which are weak electrophiles; they react with highly reactive compounds like N,N-dimethylaniline to yield an azo compound. NH2

CH3

NaNO2/HCl

2 N1 2 Cl 1

(273–278K)

N

CH3 N,N-dimethylaniline

(X)

Coupling reaction

CH3 N

N

Options: (a) (iii) (ii) (iv) (ii)

(1) (2) (3) (4)

(b) (ii) (iii) (ii) (i)

(d) (iv) (iv) (i) (iii)

(AIPMT MAINS 2010) Solution (2) (a) → (ii): Primary amines when heated with chloroform and ethanolic potassium hydroxide form pungent isocyanide or carbylamines. CH3(CH2)3NH2 1 CHCl3 1 2KOH Heat

N CH3

(c) (i) (i) (iii) (iv)

CH3(CH2)3NC 1 3KCl 1 3H2O

(Y)

6. Which of the following statements about primary amines is ‘False’? (1) Alkyl amines react with nitrous acid to produce alcohols. (2) Aryl amines react with nitrous acid to produce phenols. (3) Alkyl amines are stronger bases than ammonia. (4) Alkyl amines are stronger bases than arylamines.

(b) → (iii): Ammoniacal AgNO3 is Tollens’ reagent. It reacts with terminal alkynes to form silver acetylide which precipitates. CH3C

CH 1 AgNO3(ammoniacal)

CH3C

(AIPMT PRE 2010) Solution (2) Primary arylamines react with nitrous acid to give arenediazonium salts. Ar

NH2 1 NaNO2 1 2HX

C2Ag1↓ 1HNO3 White ppt

(c) → (i): Esters undergoes base promoted hydrolysis known as saponification to produce an alcohol and the sodium salt of the acid. CH3CH2COOCH3 1 NaOH

Primary arylamine

Ar

1

N

CH3CH2COONa 1 CH3OH N X 1 NaX 1 2H2O

Arenediazonium salt (Stable if kept below 5°C)

7. Match the compounds given in List-I with their characteristic reactions given in List-II. Select the correct option. List-I List-II Compounds Reactions (a) CH3CH2CH2CH2NH2 (i) Alkaline hydrolysis (b) CH3C CH

Heat

(ii) With KOH (alcohol) and CHCl3 produces bad smell

(d) → (iv): Secondary alcohols form halide with Lucas reagent (HCl + ZnCl2) and produce turbidity in solution after 5 minutes. CH3CH(OH)CH3 1 conc. HCl Anhy. ZnCl2

CH3

CH

CH3 1 H2O

Cl Cloudiness appears in 5 minutes

737

738

Chapter 11 | Organic Compounds Containing Nitrogen 8. What is the product obtained in the following reaction? NO2

Zn NH4Cl

NH2

N

(1)

(3)

N

10. An organic compound ‘A’ on treatment with NH3 gives ‘B’ which on heating gives ‘C’. ‘C’ when treated with Br2 in the presence of KOH produces ethylamine. Compound A is (1) CH3CH2COOH

(3) CH3CH2CH2COOH

(2) CH3COOH

(4) CH3 CHCOOH CH3

O2

NHOH

(2)

N

(4)

(AIPMT MAINS 2011)

N

Solution

1

(AIPMT PRE 2011)

(1) The reaction involved is

Solution (2) The product formed is NO2

9. In a set of reactions m-bromobenzoic acid gave a product D. Identify the product D. COOH

A

B

NH3

C

Br

CONH2

(1)

NaOH Br2

D

Δ(2H2O)

2 1

CH3CH2COONH4 (B) KOH 1 Br2

CH3CH2CONH2 (C)

NHOH Zn NH4Cl

SOCl2

NH3

CH3CH2COOH (A) Propanoic acid

CH3CH2NH2

Hofmann bromamide degradation reaction

Ethyl amine

11. An organic compound (C3H9N) (A), when treated with nitrous acid, gave an alcohol and N2 gas was evolved. (A) on warming with CHCl3 and caustic potash gave (C) which on reduction gave isopropylmethylamine. Predict the structure of (A). (1)

CH3

CH

CH3

(3) CH3 N CH3

NH2

CH3

COOH

(2) CH3CH2

(3)

CH3 (4) CH3CH2CH2

NH

NH2

(AIPMT MAINS 2012) Br

NH2 SO2NH2

(2)

Solution

NH2

(1)

(4)

CH3

Br

Br

(AIPMT PRE 2011)

CH

HNO3

NH2

CH3

OH 1 N2 ↑

CH CH3

CH3 (A) Isopropyl amine

Isopropyl alcohol

Solution Carbylamine test

(4) The reaction involved is O COOH

C

O Cl

SOCl2

(A)

Br

C

NH2

NH3

(B)

Br

(C)

Br

NaOH/Br2 Hoffmann bromamide reaction

CH3

CHCl3/KOH

(A) CH

N

C

Reduction

CH3 (C) Isopropyl isocyanide (C)

CH3

CH

NO2

NO2 A

Br (D)

Br

N21Cl2

CH3

Isopropyl methylamine

12. In the reaction NH2

NH

CH3

Br

739

Solved Previous Years’ NEET Questions A is (1) HgSO4/H2SO4 (2) Cu2Cl2

(3) H3PO2 and H2O (4) H+/H2O (NEET 2013)

Solution (3)

stable (due to resonance stabilization) than aliphatic diazonium salts; they do not decompose at an appreciable rate in solution when the temperature of the reaction mixture is kept below 5°C. The resonating structures of arenediazonium salts are 1

NO2

N

NO2

1

N

N

N

2

N

N

1

2

N

N

1

2

1

H3PO2

Br

H2O

Br

1

N21Cl2

1

N

13. In the following reaction, the product (A) is 1

N

NCl2

N

N

H1

(A) Yellow dye

15. The electrolytic reduction of nitrobenzene in strongly acidic medium produces (1) azoxybenzene. (3) aniline. (2) azobenzene. (4) p-aminophenol. (AIPMT 2015)

NH

NH2

(2)

N

Solution

N

(4) The electrolytic reduction of nitrobenzene under highly acidic condition converts it into p-aminophenol by the rearrangement of the phenylhydroxylamine intermediate as shown in the reaction below.

NH2

(3)

N

N

(4)

N

N

NO2

NH2

Solution

1

N

N

NH2

(A)

14. Which of the following will be most stable diazonium salt RN+2X−? (1) CH3N+2X− (3) CH3CH2N+2X− (4) C6H5CH2N+2X− (AIPMT 2014)

2e2 2H1

NH2

OH

NH2

H2

NH2

H1

(4) Arenediazonium ions are weak electrophiles; they react with highly reactive aromatic compounds to yield azo compound. This electrophilic aromatic substitution is often called diazo coupling reaction. N2 Cl2 1 H

NHOH Al2 4H1

(AIPMT 2014)

(2) C6H5N+2X−

2

NH2

1

(1)

N

16. Method by which aniline cannot be prepared is (1) reduction of nitrobenzene with H2/Pd in ethanol. (2) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution. (3) hydrolysis of phenyl isocyanide with acidic solution. (4) degradation of benzamide with bromine in alkaline solution. (RE-AIPMT 2015)

Solution

Solution

(2) Primary arylamines react with nitrous acid to give arenediazonium salts. Even though arenediazonium salts are unstable, they are still far more

(2) The reactions are as follows Option (1) Nitrobenzene can be reduced to aniline by hydrogenation (H2/Pd/C in ethanol).

740

Chapter 11 | Organic Compounds Containing Nitrogen NO2

pair of electrons on nitrogen over ortho and para positions of the ring. The delocalized electron pair is less available to a proton.

NH2 H2/Pd Ethanol

NH2

Option (2) Aryl amines cannot be prepared via this method since aryl halides  do not  undergo simple nucleophilic substitution. O

Cl

R Arylamine (less basic)

O

N2K1 1

N

O

O NaOH(aq.) OH N

Alkyl amine (more basic)

18. A given nitrogen-containing aromatic compound A reacts with Sn/HCl, followed by HNO2 to give an unstable compound B. B, on treatment with phenol, forms a beautiful colored compound C with the molecular formula C12H10N2O. The structure of compound A is NO2

CONH2

(3)

(1) CN

O

Option (3) Phenyl isocyanide on acidic solution produces aniline. NC

NH2

NH2

(2)

(4) (NEET-II 2016)

NH2

Solution

H1 H2O

(1) The given reaction sequence can be represented as follows.

Option (4) When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, degradation of amide takes place leading to the formation of primary amine.  CONH2

NH2 Br2

NaOH(aq.)

NO2

1

NH2

N2Cl2 HNO2

Sn 1 HCl Redution

A

Aniline

B Benzene diazonium chloride

1 CO2

17. The correct statement regarding the basicity of arylamines is (1) arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring π electron system. (2) arylamines are generally more basic than alkylamines because the nitrogen lone-pair electrons are not delocalized by interaction with the aromatic ring π electron system. (3) arylamines are generally more basic than alkylamines because of aryl group. (4) arylamines are generally more basic than alkylamines, because the nitrogen atom in arylamines is sp-hybridized. (NEET-I 2016)

Phenol

OH

N

N

p-Hydroxy azobenzene (red color azo dye)

19. The correct increasing order of basic strength for the following compound is NH2

(I)

(1) III < I < II (2) III < II < I

NO2

NH2

(II)

CH3

NH2

(III)

(3) II < I < III (4) II < III < I (NEET 2017)

Solution

Solution

(1) Arylamines are generally less basic than alkylamines. It is due to delocalization of the lone

(3) Aryl amines are basic in nature due to the presence of lone pair of electrons on nitrogen atom.

Additional Objective Questions Among the given compounds, (III) is the most basic as electron donating alkyl group increases the electron density on the nitrogen atom. In compound (II) electron density decreases due to the presence of electron withdrawing nitro group, thus is less basic than aniline (I). Hence, the increasing order of basic strength is (II) < (I) < (III).

Solution (1) Hofmann hypobromamide reaction is used for the conversion of an amide into a primary amine with one carbon less. The overall reaction is as follows: O CH3

20. Which of the following reactions is appropriate for converting acetamide to methanamine? (1) Hofmann hypobromamide reaction (2) Stephens reaction (3) Gabriels phthalimide synthesis (4) Carbylamine reaction (NEET 2017)

C

NH2 1 Br2 1 4KOH

Acetamide

CH3

NH2 1 2KBr 1 K2CO3 1 2H2O

Methanamine

ADDITIONAL OBJECTIVE QUESTIONS NCERT Exemplar Questions 1. Which of the following is a 3° amine? (1) 1-Methylcyclohexylamine (2) Triethylamine (3) tert-Butylamine (4) N-Methylaniline 2. The correct IUPAC name for CH2 = CHCH2NHCH3 is (1) allylmethylamine. (2) 2-amino-4-pentene. (3) 4-aminopent-1-ene. (4) N-methylprop-2-en-1-amine. 3. Amongst the following, the strongest base in aqueous medium is (1) CH3NH2 (3) (CH3)2NH (2) NCCH2NH2 (4) C6H5NHCH3 4. Which of the following is the weakest Brönsted base? NH2 NH2

(1)

(2)

(3)

N

H

(4) CH3NH2

5. Benzylamine may be alkylated as shown in the following equation: C6H5CH2NH2 1 R

X

C6H5CH2NHR

Which of the following alkyl halides is best suited for this reaction through SN1 mechanism? (1) CH3Br (3) C6H5CH2Br (2) C6H5Br (4) C2H5Br

6. Which of the following reagents would not be a good choice for reducing an aryl nitro compound to an amine? (1) H2 (excess)/Pt (3) Fe and HCl (2) LiAlH4 in ether (4) Sn and HCl 7. In order to prepare a 1° amine from an alkyl halide with simultaneous addition of one CH2 group in the carbon chain, the reagent used as source of nitrogen is (1) Sodium amide, NaNH2 (2) Sodium azide, NaN3 (3) Potassium cyanide, KCN (4) Potassium phthalimide, C6H4(CO)2N−K+ 8. The source of nitrogen in Gabriel synthesis of amines is (1) Sodium azide, NaN3 (2) Sodium nitrite, NaNO2 (3) Potassium cyanide, KCN (4) Potassium phthalimide, C6H4(CO)2N−K+ 9. Amongst the given set of reactants, the most appropriate for preparing 2° amine is (1) 2° R—Br + NH3 (2) 2° R—Br + NaCN followed by H2/Pt (3) 1° R—NH2 + RCHO followed by H2/Pt (4) 1° R—Br (2 mol) + potassium phthalimide followed by H3O+/heat 10. The best reagent for converting 2-phenylpropanamide into 2-phenylpropanamine is (1) excess H2. (2) Br2 in aqueous NaOH. (3) iodine in the presence of red phosphorus. (4) LiAlH4 in ether. 11. The best reagent for converting, 2-phenylpropanamide into 1-phenylethanamine is (1) excess H2/Pt. (3) NaBH4/methanol. (2) NaOH/Br2. (4) LiAlH4/ether.

741

742

Chapter 11 | Organic Compounds Containing Nitrogen 12. Hofmann bromamide degradation reaction is shown by (1) ArNH2 (2) ArCONH2 (3) ArNO2 (4) ArCH2NH2 13. The correct increasing order of basic strength for the following compounds is NH2

NH2

NH2

(I)

NO2 (II)

CH3 (III)

(1) II < III < I (2) III < I < II

(3) III < II < I (4) II < I < III

14. Methylamine reacts with HNO2 to form (1) CH3 O N O (2) CH3 O CH3 (3) CH3OH (4) CH3CHO 15. The gas evolved when methylamine reacts with nitrous acid is (1) NH3 (3) H2 (2) N2 (4) C2H6 16. In the nitration of benzene using a mixture of conc. H2SO4 and conc. HNO3, the species which initiates the reaction is (1) NO2 (3) NO2+ (2) NO+ (4) NO2− 17. Reduction of aromatic nitro compounds using Fe and HCl gives (1) aromatic oxime. (2) aromatic hydrocarbon. (3) aromatic primary amine. (4) aromatic amide. 18. The most reactive amine towards dilute hydrochloric acid is (1) CH3 NH2

(3)

H3C H3C

N

CH3

NH2

(2)

H3C H3C

NH

(4)

19. Acid anhydrides on reaction with primary amines give (1) Amide. (3) secondary amine. (2) imide. (4) Imine. Cu/HCl

2 20. The reaction ArN1 2 Cl named as (1) Sandmeyer reaction. (2) Gatterman reaction. (3) Claisen reaction. (4) Carbylamine reaction.

ArCl 1 N2 1 CuCl is

21. Best method for preparing primary amines from alkyl halides without changing the number of carbon atoms in the chain is (1) Hofmann bromamide reaction. (2) Gabriel phthalimide synthesis. (3) Sandmeyer reaction. (4) Reaction with NH3. 22. Which of the following compound will not undergo azo coupling reaction with benzene diazonium chloride? (1) Aniline (3) Anisole (2) Phenol (4) Nitrobenzene 23. Which of the following compounds is the weakest Brönsted base? NH2

(1)

OH

(3) NH2

(2)

OH

(4)

24. Among the following amines, the strongest Brönsted base is H

NH2

(1)

(3)

N

H

(2) NH3

(4)

N

25. The correct decreasing order of basic strength of the following species is _______. H2O, NH3, OH−, NH2− (1) NH2− > OH − > NH3 > H2O (2) OH − > NH2− > H2O > NH3 (3) NH3 > H2O > NH2− > OH − (4) H2O > NH3 > OH − > NH2− 26. Which of the following should be most volatile? CH3CH2

(I) CH3CH2CH2NH2

(III)

(II) (CH3)3N (1) II (2) IV

(IV) CH3CH2CH3 (3) I (4) III

NH CH3

27. Which of the following methods of preparation of amines will not give same number of carbon atoms in the chain of amines as in the reactant? (1) Reaction of nitrite with LiAlH4. (2) Reaction of amide with LiAlH4 followed by treatment with water. (3) Heating alkyl halide with potassium salt of phthalimide followed by hydrolysis. (4) Treatment of amide with bromine in aqueous solution of sodium hydroxide.

Additional Objective Questions

Exercise 1 1. Trimethylamine has (1) planar geometry. (2) trigonal bipyramidal geometry. (3) pyramidal shape. (4) octahedral geometry.

(3)

N

(4)

N

8. Procaine is a local anesthetic drug. What is the classification of the amines I and II?

2. What type of amine is N-methyl-2-methyl-3hexanamine? (1) Primary (3) Tertiary (2) Secondary (4) Quaternary 3. Which is the classification of the following compound? 1

N

2

Cl

N

O

I O

II NH2

Procaine

(1) (2) (3) (4)

I: primary aliphatic; II: primary aromatic I: tertiary aliphatic, II: primary aromatic I: tertiary aliphatic, II: tertiary aromatic I: tertiary aromatic, II: tertiary aliphatic

9. o-Phenylenediamine has the structure (1) 1° (2) 2°

(3) 3° (4) 4°

NH2

(1)

4. Which is a correct common name for the following substance?

NH2 CH(CH3)2

(2)

NH2

N

(1) (2) (3) (4)

Ethylethylisobutylamine Diethylisobutylamine sec-Butyldiethylamine Ethylethyl-sec-butylamine

5. Which is the IUPAC name for the following structure? CH3 H CH3

C

N

CH3

CH3

(1) (2) (3) (4)

N-Methyl-tert-butylamine 2,N-Methyl-1,1-dimethylethylamine 2,N-Dimethyl-2-propanamine tert-Butyl methyl amine

CH(NH2)2

(3)

(4) None of these.

10. Which of these is properly termed a “quaternary ammonium salt”? (1) (CH3)3CCH2CH2NH3+ Cl− (2) (CH3CH2CH(CH3)CH2)2NH2+ Cl− (3) (CH3CH2CH2)3NH+ Cl− (4) (CH3CH2CH2)4N+ Cl− 11. How many isomeric amines with that formula C7H9N contain a benzene ring? (1) two (3) four (2) three (4) five 12. Which of the following compounds would be the weakest base? O NH2

(1)

NH2

(3)

6. p-Anisidine can be represented by the formula (1) H3CO

NH2

NH2

(2)

NH2

(4) O2N

(2) NO2

OCH3

(3) NH2

NH2

(4) NH2

OC2H5

7. What is the structure of N,N-diethylpentanamine? (1)

H N

(2)

N

13. Arrange the amines in order of increasing boiling point (lowest first). N

H N

(I)

(III)

NH2 (II)

(1) II, III, I, IV (2) III, I, IV, II

NH13 Cl2 (IV)

(3) IV, II, III, I (4) IV, III, I, II

743

744

Chapter 11 | Organic Compounds Containing Nitrogen 14. Arrange the amines in order of increasing boiling point (lowest first). NH2

NH2

NH2

(1)

NH2

F F (I)

F (IV)

(III)

(II)

(1) I, III, II, IV (2) II, IV, I, III

1. KOH 2. ethyl bromide

NH

3. NH2NH2, ethanol reflux

O O

H N Cl

(1)

N

(III)

(I)

O

H N

O

H N

Cl Cl (II)

(2)

(IV)

(1) III, I, IV, II (2) II, I, III, IV

(3) I, III, IV, II (4) IV, II, III, I

16. Which of the following groups does not decrease the basic strength of aniline? (1) OCH3 (3) CN (2) NO2 (4) halogen 17. Aromatic nitriles (ArCN) are not prepared by reaction (1) ArX + KCN (3) ArCONH2 + P2O5 (2) ArN2+ + CuCN (4) ArCONH2 + SOCl2 18. Arrange the following in order of increasing strength of the hydrogen bonds (weakest first). (I) H2NH}OH2 (III) H2O}HOH (II) H3N}HNH2 (IV) H3N}HOH (1) I, II, III, IV (3) II, IV, I, III (2) II, I, IV, III (4) I, IV, II, III 19. The correct order of increasing basic nature for bases NH3, CH3NH2 and (CH3)2NH is (1) CH3NH2 < NH3 < (CH3)2NH (2) (CH3)2NH < NH3 < CH3NH2 (3) NH3 < CH3NH2 < (CH3)2NH (4) CH3NH2 < (CH3)2NH < NH3

N

1 NH3

1 NH3

(3)

NH2

(4)

N H

1 NH3

1 NH3

(4)

24. Which is the product of the following reaction? NO2 Fe, HCl ethanol/water

CH3

(1) HOOC

O

1. NH3 2. Br2, NaOH, H2O

COOH NH31Cl2

(2) CH3

CH3

CH3

CH3

(3)

21. What is the product of the following reaction? Cl

CH3 NO2

l

O2N

NH2

23. Electrolytic reduction of nitrobenzene in weakly acidic medium gives (1) aniline. (3) N-phenyl hydroxylamine. (2) p-hydroxyaniline. (4) nitrosobenzene.

(3)

(1)

NH

O

20. Which of these is the strongest acid?

(2)

NH2

O

15. Arrange the amines in order of increasing solubility in water (least first).

Cl

(4)

22. What is the product of the following reaction?

(3) IV, II, III, I (4) II, III, IV, I

H N

O

NH2

(2)

NH2

(3)

NH2

Cl

NO2

Cl

(4) CH3

CH3

Additional Objective Questions 25. Trinitrotoluene (TNT) reacts with large quantities of hydrogen in the presence of a nickel catalyst. Which reaction product is formed? O2N

NO2

31. Consider the synthesis below. What is reagent ‘Z’? NO2 1. HONO, 0–5°C 2. Z

H2 /Ni H2O

NH2

NO2 O2N

H2N

NO2

(1)

NH2

(3) NH2

NO2 H2N

(1) CuCl (2) CuCl2

(3) NaCl (4) KCl

32. Which reactions will proceed predominantly to products as written? 1

(I) CH3NH2 1 H2O

2

CH3NH3 1 OH

(II) CH3NH2 1 CH3 C OH

(4)

O

NH2

1

CH3NH3 1 CH3

26. Ethyl isocyanide on hydrolysis in acidic medium generates (1) ethylamine salt and methanoic acid (2) propanoic acid and ammonium salt (3) ethanoic acid and ammonium salt (4) methylamine salt and ethanoic acid

CH3 2 CH3NH 1 CH3

28. Which reagent would serve as the basis for a simple chemical test that would distinguish between the pair of compounds listed below?

CH3NH2 1

and N H

AgNO3 in H2O Dilute NaHCO3 Dilute NaOH C6H5SO2Cl/OH−, then H3O+

29. Which one of the following methods is neither meant  for the synthesis nor for separation of amines? (1) Hinsberg method (3) Wurtz reaction (2) Hofmann method (4) Curtius reaction 30. The overall conversion RBr → RCH2NH2 can be accomplished by successive application of which of these sets of reagents? (1) Mg, ether; then NH3 (2) NaN3; then LiAlH4, ether (3) NaCN; then LiAlH4, ether (4) H2C = O; then NH3

O2

CH3

(IV) CH3NH2 1

N

C

(III) CH3NH2 1 CH3 N CH3

27. In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of amine produced are (1) four moles of NaOH and one mole of Br2. (2) one mole of NaOH and one mole of Br2. (3) four moles of NaOH and two moles of Br2. (4) two moles of NaOH and two moles of Br2.

(1) (2) (3) (4)

Cl

O

NH2

(2)

NO2

N1

CH3

H

1

(1) I, II (2) III, IV

1

NH3

NH2

(3) II, IV (4) I, III

33. Conversion of benzene diazonium chloride to chlorobenzene is an example of which of the following reactions? (1) Claisen (3) Sandmeyer (2) Friedel–Crafts (4) Wurtz 34. Aniline is reacted with bromine water and the resulting product is treated with an aqueous solution of sodium nitrite in presence of dilute HCl. The compound so formed is converted into tetrafluoroborate which is subsequently heated dry. The final product is (1) p-bromofluorobenzene. (2) p-bromoaniline. (3) 2,4,6-tribromofluorobenzene. (4) 1,3,5-tribromobenzene. 35. During the preparation of arenediazonium salts, the excess of nitrous acid, if any, is destroyed by adding (1) aq. NaOH (3) aq. NH2CONH2 (2) aq. Na2CO3 (4) aq. KI

745

746

Chapter 11 | Organic Compounds Containing Nitrogen 36. Which of the following diazonium salt is the most stable? (1) p-Nitrobenzenediazonium chloride (2) 2, 4-Dinitrobenzenediazonium chloride (3) 2, 4, 6-Trinitrobenzenediazonium chloride (4) p-Methoxybenzenediazonium chloride

4. By heating which mixture, propane nitrile will be obtained? (1) Ethyl alcohol + KCN (2) Propyl alcohol + KCN (3) Ethyl chloride + KCN (4) Propyl chloride + KCN

37. When the process ArNH2 → ArY is carried out via an intermediate diazonium salt, this salt is isolated only in the case in which Y is which of these groups? (1) 2F (3) 2Br (2) 2Cl (4) 2I

5. What is the product of the following reaction?

Exercise 2 1. Arrange the following amines in the order of increasing basicity. NH2

(1)

NH2 ,

,

OCH3

NO2

NH2

NH2

(2)

NH2

,

,

CH3NH2

,

CH3NH2

NH2 ,

OCH3

NO2

NH2

NH2

(3) CH3NH2 ,

,

NH2

NH2

(4)

NH2 ,

NO2

,

,

CH3NH2

OCH3

(1)

H N

(3)

NH

(2)

N

(4)

N

6. Which of the following is the correct statement about coupling reaction of diazonium salts? (1) In coupling reactions, diazonium ion works as a nucleophile. (2) Coupling takes place almost exclusively at the ortho position with aniline and phenol. (3) Coupling with phenol and aniline is most rapid in slightly acidic medium. (4) Coupling with phenol is most rapid in slightly basic medium. 7. In order to distinguish between C2H5NH2 and C6H5NH2 which of the following reagents is useful? (1) Hinsberg reagent (3) CHCl3/KOH (2) 2-Naphthol (4) NaOH

(I) (1 mol) benzyl bromide 1 (1 mol) NH3

(III) Phthalimide

NH3 2. 4.

SOCl2

R

1.

R

CH2CONH2

R

CH2OH

5.

LAH

3.

R

CH2NH2

NH2NH2 C2H5OH

9. What would be the product of the following reaction sequence? Cl

1.

O

, AlCl3

2. (CH3)2NH 3. LiBH3CN

CH2COCl R

benzyl bromide

Et2O

(3) III (4) II and III

3. A sequential reaction may be performed as represented below: CH2CO2H

KOH

(1) I (2) II

2. Considering the basic strength of amines in aqueous solution, which one has the smallest pKb value? (1) (CH3)2NH (3) (CH3)3N (2) CH3NH2 (4) C6H5NH2

R

2. H2O

O

(II) (1 mol) benzyl bromide 1 (1 mol) NaN3

NH2

NO2

1. LiAlH4, Et2O

8. Which of the following can be used to prepare benzylamine (pure)?

, OCH3

N

N2

(1)

O

(3)

N2

CO2H

The appropriate reagent for step (3) is (1) NaBr (3) HBr (2) Bromine + alkali (4) P2O5

OH

(2)

O N

(4) N

Additional Objective Questions 14. In a set of reactions propionic acid yielded a compound D.

10. In the chemical reaction. NH2 NaNO2

HBF4

A

HCl, 5°C

CH3CH2COOH B

the compounds A and B, respectively are (1) nitrobenzene and chlorobenzene. (2) nitrobenzene and fluorobenzene. (3) phenol and benzene. (4) benzene diazonium chloride and fluorobenzene.

SOCl2

NH3

B

O

NH

1. LAH 2. H2O

O O

(3)

(2)

(CH3)2NH

CH3NH2 H2, Ni

?

(3)

2. H2O

H N O

16. Ethanamine is treated with nitrous acid at ordinary temperature, the products will be (1) ethanol only. (2) ethanol, acetic acid, N2 and H2O. (3) acetic acid, ethane and H2O. (4) ethanol, ethene, ethyl chloride and N2.

(1) CH3

Zn(Hg), HCl

1 H2N

NaBH3CN

N

13. The final product C, obtained in this reaction, would be

N

(3) (CH3)2N

NH N

NH2

NH2 A

Br2 CH3COOH

B

H2O H1

CH3 NH2

Br

CH3

CH3

COOH3

NHCOCH3

CH3

(1)

2. Ag2O, H2O 3. heat

N OH

(2) CH3

HNO2

(3)

N

Br

(4)

CuCN

1. CH3I

N

Br

A

B C

H2 Ni

The structure of the product D would be (1) C6H5CH2OH (3) C6H5NHOH (2) C6H5CH2NH2 (4) C6H5NHCH2CH3

(3)

(2)

NaNO2

D C

NHCH3

N

19. What would be the product of the following reaction sequence?

NH2 COOH3

N

18. Aniline in a set of reactions yielded a product D HCl

AC2O

NH2

N

(2) (CH3)2N

(4) CH3NH

(4) None of these

(1)

OH

17. Aniline when diazotized in cold and then treated with dimethylaniline gives a coloured product. Its structure would be

1. LiAlH4, ether

N H O

(4) H2N

NH

H2, Ni

O

(2)

NH2

OH

(1 mol)

(4) (1) and (2)

(1)

(3)

CH3I

12. Which combination of reactants will not produce H N

O

NH

(1)

(2)

D

15. What is the final product from the following reaction sequence?

H N

(1)

Br2

The structure of D would be (1) CH3CH2CH2NH2 (3) CH3CH2NHCH3 (2) CH3CH2CONH2 (4) CH3CH2NH2

11. Identify the best method(s) to prepare.

NH2 (1 mol)

KOH

C

(4) N

N

747

748

Chapter 11 | Organic Compounds Containing Nitrogen 20. The reagent B in the following reaction is Br

Br NaNO2/HCl 0248C

NH2

(1) (2) (3) (4)

A

B

Br

Cu powder and HBr Cuprous bromide Either (1) or (2) Neither (1) nor (2)

OH

26. Fluorobenzene (C6H5F) can be synthesized in the laboratory (1) by heating phenol with HF and KF. (2) from aniline by diazotization followed by heating the diazonium salt with HBF4. (3) by direct fluorination of benzene with F2 gas. (4) by reacting bromobenzene with NaF solution.

1. SOCl2 2. NaN3 3. heat

O

NH2

4. H2O

NH2

(3) O

NH2

(2) 22.

1

N

(4)

NH2

27. Which of these alkyl halides can be used to prepare amines using Gabriel pthalimide synthesis? (1) Vinyl bromide (2) 1-Bromo-3-methylpentane (3) Bromobenzene (4) 2-Bromo-2,3-dimethylbutane 28. In the chemical reaction,

Δ

Me

CH3

(3) CH3CH2COOH (4) CH3COOH

21. What is the product of the following reaction?

(1)

25. An organic compound (A) upon reacting with NH3 gives (B). On heating (B) gives (C). (C) in presence of KOH react with Br2 to give CH3CH2NH2. (A) is (1) CH3CH2CH2COOH (2) H3C CH COOH

CH3CH2NH2 1 CHCl3 1 3KOH

Me Et

HO2n-Bu

The alkene formed as a major product in the above elimination reaction is Me

the compound (A) and (B) are, respectively, (1) C2H5CN and 3KCl (2) CH3CH2CONH2 and 3KCl (3) C2H5NC and K2CO3 (4) C2H5NC and 3KCl 29. What is the product of the following reaction?

(3)

(1) Me

NH2 1. HCl, NaNO2, H2O

Me

(2) CH2 5 CH2

2. CuCl

(4) NO2

23. The reaction of chloroform with alcoholic KOH and p-toluidine forms (1) H3C

CN

(2) H3C

N2CI

(A) 1 (B) 1 3H2O,

(1)

Cl

(3) N 1N

(3) H3C

NHCHCI2

(4) H3C

NC

(2)

24. What is the chief product of the Hofmann elimination reaction applied to the compound shown? N 1

(1)

(3)

(2)

(4)

OH2

Cu

Cl2

(4)

30. The final product formed when methyl amine is treated with NaNO2 and HCl is (1) Diazomethane (3) Methyl cyanide (2) Methyl alcohol (4) Nitromethane 31. The test to distinguish primary, secondary and tertiary amines is (1) Sandmeyer’s reaction. (2) Carbylamine reaction. (3) Mustard oil test. (4) C6H5SO2Cl.

Additional Objective Questions 32. Which of the following amines is least basic? NH2

(1)

(3) N NO2 NH

(2)

NH2

(4)

33. Consider the following ions. 2

1

(I) Me2N

N

(II) O2N

N

N

1

N 1

(III) CH3O

N 1

(IV) CH3

N

In the following set of questions, two statements “Assertion” and “Reason” are given. Choose the correct answer from the following options:

N N

34. What is the product of the following reaction? NH2 1. HCl, NaNO2, H2O 2.

OH

OH

N

(3)

(1) Assertion and Reason are true and the Reason is the correct explanation of the Assertion. (2) Assertion and Reason are true but the Reason is not a correct explanation of the Assertion. (3) Assertion is true but the Reason is false. (4) Assertion and Reason both are false. 1. Assertion: Carbylamine reaction involves the reaction between primary amine and chloroform in the presence of alkali. Reason: In carbylamine reaction, NH2 group changes to NC group. 2. Assertion: Alkyl cyanides and alkyl isocyanides have much higher boiling points than corresponding alkyl halides. Reason: Cyanides and isocyanides are much more polar than alkyl halides.

HO

N N

36. In the diazotization of aniline with sodium nitrite and hydrochloric acid, the excess of hydrochloric acid is used primarily to (1) suppress the concentration of free aniline. (2) suppress the hydrolysis to phenol. (3) ensure a stochiometric amount of nitrous acid. (4) neutralize the base liberated.

Exercise 3

The reactivity of these ions towards azo-coupling reaction under similar conditions is (1) I < IV < II < III (3) III < I < II < IV (2) I < III < IV < II (4) III < I < IV < II

(1)

35. Nitration of aniline is carried out after acylation because (1) acylation deactivates the NH2 group. (2) oxidation can be prevented. (3) o- and p-products are obtained in good yield. (4) all of these.

N

3. Assertion: Aniline does not undergo Friedel–Crafts reaction. Reason: Friedel–Crafts reaction is an electrophilic substitution reaction. 4. Assertion: Aniline reacts with bromine water to form 2,4,6-tribromoaniline. Reason: Aniline is resonance stabilized. 5. Assertion: The order of basicity among the following is CH3CH2NH2 > NH3 > C6H5NH2. Reason: Electron releasing groups increase the basicity of amines while electron withdrawing groups decrease the basicity.

OH HO

N

(2)

N

N

(4)

N

6. Assertion: n-Propylamine has a higher boiling point than trimethylamine. Reason: Among n-propylamine molecules there is hydrogen bonding, but there is no hydrogen bonding among trimethylamine molecules. 7. Assertion: All the amines, except tertiary amines are capable of forming intermolecular hydrogen bonds.

749

750

Chapter 11 | Organic Compounds Containing Nitrogen Reason: Molecules of CH3CH2CH2NH2 form hydrogen bonds while (CH3)3N molecules are incapable of forming hydrogen bonds.

Reason: Tertiary amines have larger molecules and surface area. 8. Assertion: Aromatic amines are less basic than alkyl amines. Reason: The π electrons on the aromatic ring decrease the basic character. 9. Assertion: (CH3)3N boils at 276 K, while CH3CH2CH2NH2 boils at 322 K though both have same molecular mass.

10. Assertion: In strongly acidic solutions, aniline becomes less reactive towards electrophilic reagents. Reason: The amino group being completely protonated in strongly acidic solution, the lone pair of electrons on nitrogen is no longer available for resonance.

ANSWER KEY NCERT Exemplar Questions 1. (2)

2. (4)

3. (3)

4. (1)

5. (3)

6. (2)

7. (3)

8. (4)

9. (3)

10. (4)

11. (2)

12. (2)

13. (4)

14. (3)

15. (2)

16. (3)

17. (3)

18. (2)

19. (1)

20. (2)

21. (2)

22. (4)

23. (3)

24. (4)

25. (1)

26. (2)

27. (4)

Exercise 1 1. (3)

2. (2)

3. (4)

4. (3)

5. (3)

6. (1)

7. (2)

8. (2)

9. (1)

10. (4)

11. (4)

12. (1)

13. (1)

14. (4)

15. (2)

16. (1)

17. (1)

18. (2)

19. (3)

20. (2)

21. (1)

22. (3)

23. (3)

24. (2)

25. (2)

26. (1)

27. (1)

28. (4)

29. (3)

30. (3)

31. (1)

32. (3)

33. (3)

34. (3)

35. (4)

36. (1)

37. (1)

3. (2)

4. (3)

5. (2)

Exercise 2 1. (4)

2. (1)

6. (4)

7. (2)

8. (4)

9. (3)

10. (4)

11. (3)

12. (4)

13. (3)

14. (4)

15. (1)

16. (4)

17. (2)

18. (1)

19. (1)

20. (3)

21. (1)

22. (4)

23. (4)

24. (1)

25. (3)

26. (2)

27. (2)

28. (4)

29. (3)

30. (2)

31. (4) 36. (1)

32. (2)

33. (2)

34. (3)

35. (4)

1. (1)

2. (1)

3. (2)

4. (2)

5. (2)

6. (1)

7. (3)

8. (3)

9. (1)

10. (1)

Exercise 3

Hints and Explanations

HINTS AND EXPLANATIONS Exercise 1 1. (3) In triethylamine, the hybridization is sp3, but because of lone pair the shape gets distorted to pyramidal. sp3 N H3C

CH3

CH3

6. (1) The compound is 4-methoxyaniline or panisidine. 2

3 4

H3CO

1 5

6

9. (1) The compound is 2-aminoaniline or o-phenylenediamine. 5

1

16. (1) This is because electron releasing group due to their +R effect increases the electron density and thus increases the basic character, whereas electron withdrawing groups such as CN, NO2, halogens decrease the basic character. 18. (2) O H … O hydrogen bond is stronger than an N H … N hydrogen bond as a result of the higher polarity of the OH bond.

NH2

7. (2) N,N-diethylpentanamine is a 5 carbon structure with a secondary amine on carbon 1. The amine on carbon 1 has two ethyl groups attached to the nitrogen.

6

15. (2) The solubility of a compound is influenced by the strength of the intermolecular forces. In compound II, two electron-withdrawing groups are present which reduces the “availability” of the lone pair on nitrogen hence, reduces its ability to form hydrogen bonding with water, thus, it would be least soluble. The correct order of increasing solubility is II < I < III < IV.

19. (3) More is the number of alkyl groups present, more is the + I effect, more is the electron density and stronger is the base. 21. (3) Hofmann degradation turns amides into primary amines. Cl

NH2

NH2

1. NH3

O

O

2

4 3

2. Br2, NaOH, H2O

NH2

11. (3) The isomeric amines are as follows: CH2NH2

CH3

CH3

NH2

CH3 NH2

22. (3) Gabriel Synthesis produces primary amines. NH2

23. (3)

NO2

NH2

NOH 1 4H

12. (1) Benzamide would be the weakest base among the given compounds as lone pair on nitrogen gets delocalized with carbonyl group and less available for donation. O

reduction (in weak acidic medium)

N-Phenyl hydroxyl amine

26. (1) The reaction involved is C2H5NC

O2 NH2

Electrolytic

C2H2NH2 1 HCOOH Ethyl amine Methanoic acid

1

NH2

13. (1) Molecules of primary and secondary amines can form strong hydrogen bonds to each other and to water. Molecules of tertiary amines cannot form hydrogen bonds to each other, but they can form hydrogen bonds to molecules of water. As a result, tertiary amines generally boil at lower temperatures than primary and secondary amines of comparable molecular weight. Hence, the order of increasing boiling point is II < III (CH3)2NH > CH3NH2 > NH3. However, solvation of conjugate acids occurs by hydrogen bonding which depends on the number of hydrogen atoms attached with nitrogen atom. The order of solvation of conjugate base is (CH3)3N < (CH3)2NH < CH3NH2 < NH3. In this way, both factors are against each other and the order of resultant basic strength becomes (CH3)2NH > CH3NH2 > (CH3)3N > NH3

In slightly basic conditions, phenol is in deprotonated form, due to which it is more reactive. However, on increasing basicity, hydroxide ions oxidize diazonium ions and coupling would not occur. Slightly acidic conditions favour diazonium ion’s concentration, but below pH = 5 aniline gets protonated and becomes unreactive. 7. (2) Both are primary amines, therefore, 2-naphthol is used to differentiate (as one is aromatic and forms a colored dye) NH2 1 NaNO2 1 HCl

OH 1

N

NCl21

2-Naphthol dil. NaOH

OH N

N

Hence the amine that has minimum pKb (maximum basic) is (CH3)2NH. 5. (2) Tertiary amides are reduced by lithium aluminum hydride to form tertiary amines.

Phenylazo-2-naphthol (Orange-red dye)

Hints and Explanations 8. (4)

753

13. (3) The reaction involved is Br

CH2

I.

NH2

NH2

CH2

NH3

NHCOCH3 Ac2O

CH2COOH

CH3

CH3 (A)

2

CH3 (B) H2O

N

CH2

1

Br

Br2

NH

CH2

1

NHCOCH3

H1

NH2 Br

3

1

NBr

CH2

1

2

CH3

4

Br

CH2

II.

CH2

NaN3

(C) 1

N

2

N

N

14. (4) The reaction is CH3CH2COOH

SOCl2

CH3CH2CONH2

(B)

LiAlH4

CH2

NH3

CH3CH2COCl

(C) Hofmann bromamide degradation

NH2

KOH/Br2

CH3CH2NH2 (D)

III. O

16. (4)

O NH

KOH

CH3

N

2

O

CH2

CH2

NH2

2N2

NH2

N

ErOH

CH3

10. (4) 1

F

N2Cl2 NaNO2

HBF4

1 N2 1 BF3 1 HCl

HCl, 58C

(A) Benzenediazonium chloride

(B)

H N CH3I (1 mol)

(1 mol)

N

N

CH3

CH2

OH

2Cl2

CH3

CH2

Cl

2H1

CH2

CH2

NH2 Aniline

NaNO2 1 HCl 0258C

1

N2Cl2 1

Benzenediazonium chloride 08C H2O

11. (1) NH2

1

CH2

1

CH2

17. (2) Arenediazonium ions are weak electrophiles; they react with highly reactive aromatic compounds – with phenols and tertiary arylamines – to yield azo compounds. This electrophilic aromatic substitution is often called as diazo coupling reaction.

CH2

O

NH2

CH3

2H1

Br

O NH2

HNO2

H2O

O

CH2

NH2

N

N

N(CH3)2

N, N-Dimethyl aniline O CH3

C

O2Na1

N(CH3)2

N,N-Dimethyl-p-(phenylazo) aniline (yellow solid)

754

Chapter 11 | Organic Compounds Containing Nitrogen 18. (1) The reaction is NH2

23. (4) The reaction involved is carbylamines reaction.

1Cl2

N2

CH2NH2

CN

NaNO2

NH2

H2/Ni

HCl

C

1 CHCl3 1 3KOH

CuCN

(A)

N

(B)

1 3KCl 1 3H2O

(C) CH3 HNO2

CH2

CH3

25. (3) The reaction can be elucidated as O

O

OH CH3CH2

C

OH1NH3

CH3CH2

(A)

O

NaNO2/HCl 0248C

N2Cl

Br Cu2Br2

12

N2Cl

Br

The diazonium salt can also undergo Gattermann reaction with Cu powder and HBr to give 1,3-dibromobenzene. Br

Br

HBr

N2Cl

Br

21. (1) This reaction is an example of Curtius rearrangement. The carboxylic acid is converted to acid chloride and then azide, which then loses N2 followed by CO2 to yield primary amine with one carbon atom less than the parent acid. 1. SOCl2 2. NaN3

OH

3. heat

NH2

4. H2O

O

22. (2) This is an example of Hofmann elimination reaction Me

Me D

1 2

N

n-Bu

Me OH2 Et

1 Me

CH3CH2

C

NH2

(C)

N2X

2

1

N2BF4 HBF4

HONO

F Heat

(0258C)

27. (2) Gabriel synthesis is used to prepare primary amines from haloalkanes. The reaction proceeds by SN2 mechanism, so primary haloalkanes or those which can undergo SN2 reaction are used to obtain amine in good yield. As the reaction takes place by SN2 mechanism, the back side of carbon–halogen bond should be least hindered; otherwise elimination reaction will take place to produce alkene. From the given options, only 1-bromo-3-methyl pentane is a primary halide and would undergo SN2 mechanism. Hence, it can be used in Gabriel phthalimide synthesis.

Cu powder

12

1

NH2

The diazonium salt will then undergo Sandmeyer reaction with cuprous bromide to give 1,3-dibromobenzene. Br

Hofmann degradation

26. (2) The reaction involved is called Balz–Schiemann reaction.

12

(A)

Br2/KOH

CH3CH2NH2

Br

NH2

(2H2O)

D

20. (3) NaNO2/HCl will first convert 3-bromoaniline into diazonium salt.

21

ONH4

(B)

(D)

Br

C

28. (4) Primary amines react with chloroform and alkali metal hydroxide to give carbylamine (isonitrile) reaction. CH3CH2NH2 1 CHCl3 1 3KOH C2H5

NC 1 3KCl 1 3H2O

29. (3) Aniline reacts with nitrous acid to produce a diazonium salt. With the addition of CuCl, the Sandmeyer reaction occurs turning the diazonium salt into a chloride. 1

NH2

N2

Cl

Et N

1. HCl, NaNO2

n-Bu

H2O 0258C

2. CuCl

Hints and Explanations 30. (2) Methyl diazonium ion is unstable hence converted into alcohol in presence of H2O. CH3

NH2

HNO2 (NaNO21HCl)

CH3

H2O

1

N2

(2N2)

CH3

OH

31. (4) Hinsberg test is used to distinguish between primary, secondary and tertiary amines. The reagent used is benzenesulphonyl chloride (C6H5SO2Cl). O C6H5SO2Cl 1 H

N

C2H5

C6H5

H (18 amine)

Since –NO2 is most electron withdrawing, compound (II) is least stable. In contrast, –NMe2 is most electron donating, and so compound (I) is the most stable. In this way, the stability of diazonium ion decreases in the order I > III > IV > II. Therefore, reactivity will increase in the reverse order, that is, I < III < IV < II. 34. (3) NH2

S

N

O

H

NH2

C2H5 1 HCl 1 3Br2(aq)

1 3HBr

Soluble in alkali Br 2,4,6-Tribromoaniline

O C6H5SO2Cl 1 H

N

C2H5

C2H5 (28 amine)

Br

Br

H5C6

S

N

C2H5 1 HCl

O

C2H5

NaNO2 /HCl

Insoluble in alkali

No reaction is observed with 3° amine. 32. (2) The basicity of amines depends on the lone pair available on nitrogen atom. In benzyl amine, lone pair is easily available since this lone pair is not delocalized over benzene ring. Also the nitrogen atom is in sp3 hybridized state, so it is the most basic, even more basic than pyridine in which nitrogen atom is sp2 hybridized. Increasing s-character in hybridization decreases basicity. From the given options, however, p-nitroaniline is a weak base because nitrogen’s lone pair is delocalized in the benzene ring. In pyrrole, nitrogen’s lone pair is involved in aromaticity and least available for protonation. This effect makes it least basic among the given compounds. 33. (2) Diazonium salt contains positive charge on nitrogen which is connected with benzene ring. This implies that it is electron deficient. Its stability can be increased by any group which can supply electron density to this azo linkage.

1

N

F Br

273−278 K

Br

Br

HBF4

NCl2 Br

Heat

Br

Br 2,4,6-Tribromofluorobenzene

35. (3) NH2

CH3COCl

NHCOCH3

H2SO4 HNO3

NO2 O2N

NHCOCH3 1

NHCOCH3

36. (1) Excess of HCl suppresses concentration of free aniline, so that reactions shifts in the forward direction.

755

Biomolecules

12 2

C H A P T E R OU TLIN E Part I: Carbohydrates 12.1 12.2 12.3 12.4 12.5

Classification of Carbohydrates Monosaccharides Disaccharides Polysaccharides Other Biologically Important Sugars 12.6 Sugars That Contain Nitrogen

12.11 Introduction to Enzymes 12.12 Hormones 12.13 Vitamins Part III: Nucleic Acids 12.14 Nucleotides and Nucleosides 12.15 Deoxyribonucleic Acid: DNA 12.16 RNA and Protein Synthesis

Part II: Amino Acids and Proteins

Part IV: Lipids

12.7 Amino Acids 12.8 Polypeptides and Proteins 12.9 Secondary, Tertiary and Quaternary Structures of Proteins 12.10 Denaturation of Proteins

12.17 Fatty Acids and Triacylglycerols 12.18 Terpenes and Terpenoids 12.19 Phospholipids and Cell Membranes 12.20 Waxes

PHOTO CREDITS: Ingram Publishing/Getty Images, Inc. illustration credit: PDB ID: 1A4K, http://www.pdb.org. Romesberg, F. E., Spiller, B., Schultz, P. G., Stevens, R. C. Immunological origins of binding and catalysis in a Diels–Alderase antibody. Science 279, pp. 1929–1933, 1998.

Chemical substances present in living organisms range in complexity from water and simple salts to DNA (deoxyribonucleic acid) molecules containing tens of thousands of atoms. Four of the chemical elements – hydrogen, carbon, nitrogen and oxygen – make up approximately 95% of the mass of living matter. Small amounts of sulphur, phosphorus, calcium, sodium, potassium, chlorine, magnesium and iron, together with trace amounts of many other elements such as copper, manganese, zinc, cobalt and iodine, are also found in living organisms. The chemical reactions in living organisms involve chemical compounds, called biomolecules found in living organisms. These include chemicals that are composed of mainly carbon, hydrogen, oxygen, nitrogen, sulphur and phosphorus. These molecules play an important role in vital functions of living organisms and are called the building blocks of life. The four major classes of biomolecules upon which all life depends are carbohydrates, lipids, proteins and nucleic acids. Of these, carbohydrates and proteins form an important part of our diet. Each kind of living organism has an amazing ability to select and synthesize a large portion of the many complicated molecules needed for its existence. Vitamins are required in very small quantities but these play a significant role in the proper functioning of the body.

PART I: CARBOHYDRATES Carbohydrates are the most abundant organic constituents of plants. They not only serve as an important source of chemical energy for living organisms (sugars and starches are important in this respect), but also in plants and in some animals they serve as important constituents of supporting tissues (this is the primary function of the cellulose found in wood, cotton, and flax, for example). We encounter carbohydrates at almost every turn of our daily lives. The paper on which this book is printed is largely cellulose; so, too, is the cotton of our clothes and the wood of our houses. The flour from which we make bread is mainly starch, and starch is also a major constituent of many other foodstuffs, such as potatoes, rice, beans, corn, and peas. Carbohydrates are central to metabolism, and they are important for cell recognition.

758

Chapter 12 | Biomolecules

The group of compounds known as carbohydrates received their general name because of early observations that they often have the formula Cx(H2O)y—that is, they appear to be “hydrates of carbon” as noted in the chapter opener. They are also characterized by the functional groups that they contain. ●

Carbohydrates are usually defined as polyhydroxy aldehydes and ketones or substances that hydrolyze to yield polyhydroxy aldehydes and ketones. They exist primarily in their hemiacetal or acetal forms.

12.1 CLASSIFICATION OF CARBOHYDRATES Carbohydrates are mainly classified on the basis of their hydrolytic behavior. The simplest carbohydrates, those that cannot be hydrolyzed into simpler carbohydrates, are called monosaccharides. On a molecular basis, carbohydrates that undergo hydrolysis to produce only 2 molecules of monosaccharide are called disaccharides; those that yield 3 molecules of monosaccharide are called trisaccharides; and so on. (Carbohydrates that hydrolyze to yield 2–10 molecules of monosaccharide are sometimes called oligosaccharides.) Carbohydrates that yield a large number of molecules of monosaccharides (>10) are known as polysaccharides. Maltose and sucrose are examples of disaccharides. On hydrolysis, 1 mol of maltose yields 2 mol of the monosaccharide glucose; sucrose undergoes hydrolysis to yield 1 mol of glucose and 1 mol of the monosaccharide fructose. Starch and cellulose are examples of polysaccharides; both are glucose polymers. Hydrolysis of either yields a large number of glucose units. The following shows these hydrolyses in a schematic way: O

O O

OH

H3O

1 mol of maltose A disaccharide

O

O

(

O O

)

OH

O

OH +

O

1 mol of sucrose A disaccharide

2

2 mol of glucose A monosaccharide

H3O

O

O

OH

O

1 mol of glucose  1 mol of fructose Monosaccharides

O O

OH

H3O

O n

OH

m

1 mol of starch or 1 mol of cellulose Polysaccharides

many moles of glucose Monosaccharides

Carbohydrates can also be classified as either reducing or non-reducing, depending on their reaction with Fehling’s solution and Tollen’s reagent. All monosaccharides, whether containing an aldehydic or ketonic group reduce Fehling’s solution and Tollen’s reagent, so are called reducing sugars. Disaccharides in which aldehydic or ketonic groups are bonded are called non-reducing sugars (e.g., sucrose) while those disaccharides in which these groups are free are called reducing sugars (e.g., maltose, lactose).

12.2 MONOSACCHARIDES 12.2A Classification of Monosaccharides Monosaccharides are classified according to (1) the number of carbon atoms present in the molecule and (2) whether they contain an aldehyde or keto group. Thus, a monosaccharide containing three carbon atoms is called a triose; one containing four carbon atoms is called a tetrose; one containing five carbon

12.2 | Monosaccharides

atoms is a pentose; and one containing six carbon atoms is a hexose. A monosaccharide containing an aldehyde group is called an aldose; one containing a keto group is called a ketose. These two classifications are frequently combined. A C4 aldose, for example, is called an aldotetrose; a C5 ketose is called a ketopentose.

O (H

12.2B

D

and

L

C C

O

CH2OH

H OH)n

(H

C

CH2OH

H

C

O

C

O

H

C

OH

H

C

OH

C

OH)n

H

C

OH

H

C

OH

CH2OH

CH2OH

An aldose

A ketose

CH2OH

CH2OH

An aldotetrose (C4)

A ketopentose (C5)

Designations of Monosaccharides

The simplest monosaccharides are the compounds glyceraldehyde and dihydroxyacetone (see the following structures). Of these two compounds, only glyceraldehyde contains a chirality center. CHO H *C

CH2OH

OH

C

O

CH2OH

CH2OH

Glyceraldehyde (an aldotriose)

Dihydroxyacetone (a ketotriose)

Glyceraldehyde exists, therefore, in two enantiomeric forms that are known to have the absolute configurations shown here: 2

+

& &

2

+

2+

&+2+

DQG

 *O\FHUDOGHK\GH

+2

& &

+

+

&+2+

 *O\FHUDOGHK\GH

We have learnt that, according to the Cahn–Ingold–Prelog convention, (+)-glyceraldehyde should be designated (R)-(+)-glyceraldehyde and (−)-glyceraldehyde should be designated (S)-(−)-glyceraldehyde. Early in the twentieth century, before the absolute configurations of any organic compounds were known, another system of stereochemical designations was introduced. According to this system (first suggested by M. A. Rosanoff of New York University in 1906), (+)-glyceraldehyde is designated d-(+)-glyceraldehyde and (−)-glyceraldehyde is designated l-(−)-glyceraldehyde. These two compounds, moreover, serve as configurational standards for all monosaccharides. A monosaccharide whose highest numbered chirality center (the penultimate carbon) has the same configuration as d-(+)-glyceraldehyde is designated as a d sugar; one whose highest numbered chirality center has the same configuration as l-glyceraldehyde is designated as an l sugar. By convention, acyclic forms of monosaccharides are drawn vertically with the aldehyde or keto group at or nearest the top. When drawn in this way, d sugars have the }OH on their penultimate carbon on the right: 1 2 3

CHO

H *C H *C H *C 4 5

OH OH OH

CH2OH

CH2OH

2

C

H *C

3 4

1

OH

H *C OH HO *C H 5

highest numbered chirality center

6

A D-aldopentose

O

CH2OH

An L-ketohexose

759

760

Chapter 12 | Biomolecules

The D and L nomenclature designations are like (R) and (S) designations in that they are not necessarily related to the optical rotations of the sugars to which they are applied. Thus, one may encounter other sugars that are d-(+) or d-(−) and ones that are l-(+) or l-(−). The d–l system of stereochemical designations is thoroughly entrenched in the literature of carbohydrate chemistry, and even though it has the disadvantage of specifying the configuration of only one chirality center—that of the highest numbered chirality center—we shall employ the d–l system in our designations of carbohydrates.

12.2C Glucose Glucose (C6H12O6) is the most important of the monosaccharides. It is the key sugar of the body and is carried by the bloodstream to all body parts. The concentration of glucose in the blood is normally 80–100 mg per 100 mL of blood. Because glucose is the most abundant carbohydrate in the blood, it is also sometimes known as blood sugar. Glucose requires no digestion and therefore may be given intravenously to patients who cannot take food by mouth. Among the common sugars, glucose is of intermediate sweetness. Glucose can be prepared by any of the following methods: 1. From sucrose: In the laboratory, it is prepared from cane sugar by its acid hydrolysis (using dilute hydrochloric acid) in the presence of alcohol. Fructose is more soluble in alcohol than glucose; hence when cooled, glucose separates out by crystallization. The solution is then filtered to separate the two. C12 H22 O11 1 H2 O

H1

Cane sugar (Sucrose)

C6 H12 O6 1 C6 H12 O6 Glucose

Fructose

2. From starch: Commercially, glucose is obtained by hydrolysis of starch (from potato or corn) in the presence of dilute hydrochloric or sulphuric acid at 393 K under 2–3 atm pressure. (C6 H12 O5 )n 1 nH2 O Starch

H1 Δ

nC6 H12 O6 Glucose

Cyclic Structure of Glucose Fischer represented the structure of d-(+)-glucose with the cross formulation (1) in Fig. 12.1. This type of formulation is now called a Fischer projection and is still useful for carbohydrates. In Fischer projections, by convention, horizontal lines project out toward the reader and vertical lines project behind the plane of the page. When we use Fischer projections, however, we must not (in our mind’s eye) remove them from the plane of the page in order to test their superposability and we must not rotate them by 90 °. In terms of more familiar formulations, the Fischer projection translates into formulas 6 and 7. In IUPAC nomenclature and with the Cahn–Ingold–Prelog system of stereochemical designations, the open-chain form of d-(+)-glucose is (2R,3S,4R,5R)-2,3,4,5,6-pentahydroxyhexanal. Although many of the properties of d-(+)-glucose can be explained in terms of an open-chain structure (1, 2, or 3), a considerable body of evidence indicates that the open-chain structure exists, primarily, in equilibrium with two cyclic forms. These can be represented by structures 4 and 5 or 6 and 7. The cyclic forms of d-(+)-glucose are hemiacetals formed by an intramolecular reaction of the }OH group at C5 with the aldehyde group (Fig. 12.2). Cyclization creates a new chirality center at C1, and this chirality center explains how two cyclic forms are possible. These two cyclic forms are diastereomers that differ only in the configuration of C1. ●

In carbohydrate chemistry diastereomers differing only at the hemiacetal or acetal carbon are called anomers, and the hemiacetal or acetal carbon atom is called the anomeric carbon atom.

Structures 4 and 5 for the glucose anomers are called Haworth formulas and, although they do not give an accurate picture of the shape of the six-membered ring, they have many practical uses. Figure 12.2 demonstrates

12.2 | Monosaccharides

CHO H

CHO H

OH

HO

OH

HO

H

H

OH

H

OH

CHO

H

H

OH

H

OH

CH2OH

H

C

OH

HO

C

H

H

C

OH

H

C

OH

CH2OH

CH2OH

Fischer projection formula

Circle-and-line formula

Wedge–line– dashed wedge formula

1

2

3

HOCH2

HOCH2 H HO

O H OH

H

H

OH

O

H

H

H

+ HO

OH

OH

H

H

OH

OH H

Haworth formulas 4

5

OH HO

OH O +

HO

HO

OH

HO

O HO

OH

HO

a-D-(+)-Glucopyranose

b-D-(+)-Glucopyranose

6

7

Figure 12.1 Formulas 1–3 are used for the open-chain structure of D-(+)-glucose. Formulas 4–7 are used for the two cyclic hemiacetal forms of D-(+)-glucose. how the representation of each chirality center of the open-chain form can be correlated with its representation in the Haworth formula. Each glucose anomer is designated as an α anomer or a β anomer depending on the location of the }OH group of C1. When we draw the cyclic forms of a d sugar in the orientation shown in Figs. 12.1 or 12.2, the α anomer has the }OH trans to the }CH2OH group and the β anomer has the }OH cis to the }CH2OH group. Studies of the structures of the cyclic hemiacetal forms of d-(+)-glucose using X-ray analysis have demonstrated that the actual conformations of the rings are the chair forms represented by conformational formulas 6 and 7 in Fig. 12.1. It is especially interesting to notice that in the β anomer of d-glucose all of the large substituents, }OH and }CH2OH, are equatorial. In the α anomer, the only bulky axial substituent is the }OH at C1.

761

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H

O 1C 2

H

C

3

HO

C

4

H

C

5

H

C

OH H OH OH

6

CH2OH

Glucose

(plane projection formula) When a model of this is made it will coil as follows:

H 5

H 4

C

6

CH2OH

OH

C

1

OH

HO

C

3

H

CH

O

2

C

OH H If the group attached to C4 is pivoted as the arrows indicate, we have the structure below.

6

CH2OH H C HO

C

CH2OH

O H

H OH

5

H

C

C

H

OH

C

* OH

H 4

C

C

O

C

H

H 1

CH

O

C

C

O

H OH

H

C

C

C

OH

H

OH

OH 3

CH2OH

H

H

HO

This OH group adds across the C O to close a ring of six atoms and make a cyclic hemiacetal.

H 2

HO

* OH C

a-D-(+)-Glucopyranose

Open-chain form of D-glucose

b-D-(+)-Glucopyranose

(Starred ¬OH is the hemiacetal ¬OH, which in a-glucose is on the opposite side of the ring from the ¬CH2OH group at C5.)

(The proton transfer step occurs between separate molecules. It is not intramolecular or concerted.)

(Starred ¬OH is the hemiacetal ¬OH, which in b-glucose is on the same side of the ring as the ¬CH2OH group at C5.)

Figure 12.2 Haworth formulas for the cyclic hemiacetal forms of D-(+)-glucose and their relation to the PHOTO CREDIT:polyhydroxy (Reprinted with permission of John Wiley and Sons, Inc. from Holum, J. R., Organic Chemistry: A Brief Course, p. 316. Copyright 1975.) open-chain aldehyde structure.

H

12.2 | Monosaccharides

It is convenient at times to represent the cyclic structures of a monosaccharide without specifying whether the configuration of the anomeric carbon atom is α or β. When we do this, we shall use formulas such as the following: CH2OH

OH

O

H

H OH

HO

H

H The symbol

OH

HO HO

O HO

OH

OH indicates  or  (three-dimensional view not specified).

Not all carbohydrates exist in equilibrium with six-membered hemiacetal rings; in several instances the ring is five membered. (Even glucose exists, to a small extent, in equilibrium with five-membered hemiacetal rings.) Because of this variation, a system of nomenclature has been introduced to allow designation of the ring size. ●

If the monosaccharide ring is six membered, the compound is called a pyranose; if the ring is five membered, the compound is designated as a furanose.

These names come from the names of the oxygen heterocycles pyran and furan + ose:

O

O

A pyran

Furan

Thus, the full name of compound 4 (or 6) is α-d-(+)-glucopyranose, while that of 5 (or 7) is β-d-(+)-glucopyranose. Mutarotation Part of the evidence for the cyclic hemiacetal structure for d-(+)-glucose comes from experiments in which both α and β forms have been isolated. Ordinary d-(+)-glucose has a melting point of 146 °C. However, when d-(+)-glucose is crystallized by evaporating an aqueous solution kept above 98 °C, a second form of d-(+)-glucose with a melting point of 150 °C can be obtained. When the optical rotations of these two forms are measured, they are found to be significantly different, but when an aqueous solution of either form is allowed to stand, its rotation changes. The specific rotation of one form decreases and the rotation of the other increases, until both solutions show the same value. A solution of ordinary d-(+)-glucose (mp 146 °C) has an initial specific rotation of +112, but, ultimately, the specific rotation of this solution falls to +52.7. A solution of the second form of d-(+)-glucose (mp 150 °C) has an initial specific rotation of +18.7, but, slowly, the specific rotation of this solution rises to +52.7. ●

This change in specific rotation toward an equilibrium value is called mutarotation.

The explanation for this mutarotation lies in the existence of an equilibrium between the open-chain form of d-(+)-glucose and the α and β forms of the cyclic hemiacetals: O OH HO HO

O HO

OH

 -D -()-Glucopyranose (mp 146 C; [] D25  112)

H HO H H

H OH H OH OH OH

Open-chain form of D-()-glucose

OH HO HO

O

OH

HO

 -D-()-Glucopyranose (mp 150 C; [ ] D25  18.7)

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Chapter 12 | Biomolecules

X-ray analysis has confirmed that ordinary d-(+)-glucose has the α configuration at the anomeric carbon atom and that the higher melting form has the β configuration.

12.2D Fructose Fructose (C6H12O6), also known as levulose, is a ketohexose and occurs in fruit juices, honey, and (along with glucose) as a constituent of the disaccharide sucrose. Fructose is the major constituent of the polysaccharide inulin, a starch-like substance present in many plants, such as dahlia tubers, chicory roots, etc. Note: Fructose is the sweetest of all the sugars, being more than twice as sweet as glucose. This sweetness accounts for the sweet taste of honey; the enzyme invertase, which is present in bees, splits sucrose into glucose and fructose. Fructose is metabolized directly but is also readily converted to glucose in the liver. Structure of Fructose Fructose has an open-chain structure as shown: CH2OH C HO

O H

H

OH

H

OH CH2OH

D-Fructose

It belongs to the D-series and is a laevorotatory compound. It can be written as D-(−)-fructose. It also exists as a five-membered cyclic furanose ring which is formed by addition of –OH at C5 to the group. 1

2

C

OH

HO

3

H

H

4

OH

H

5

HOH2C

2

O

CH2OH

HO

3

H

H

4

OH

H

5

6

O

6

CH2OH

α-D-(–)-Fructofuranose

1

C

HO

CH2OH

β-D-(–)-Fructofuranose

The Haworth structures of two anomers of fructose are as follows: 6 5

H

1

O

HOH2C

H 4

OH

CH2OH 2

OH OH 3

H

α-D-(–)-Fructofuranose

6

O

HOH2C

5

H

H 4

OH

OH 2

OH CH OH 2 3 1

H

β-D-(–)-Fructofuranose

C

O

12.2 | Monosaccharides

12.2E Reactions of Monosaccharides Glycoside Formation When a small amount of gaseous hydrogen chloride is passed into a solution of d-(+)-glucose in methanol, a reaction takes place that results in the formation of anomeric methyl acetals: OH HO HO

OH O

OH

CH3OH, HCl (HOH)

HO

D-()-Glucose

●

HO HO

OH O

HO HO



HO

OCH3

Methyl  -D-glucopyranoside (mp 165 C; [] D25  158)

O

OCH3

HO

Methyl  -D-glucopyranoside (mp 107 C; [] D25  33)

Carbohydrate acetals are generally called glycosides, and an acetal of glucose is called a glucoside. (Acetals of mannose are mannosides, acetals of fructose are fructosides, and so on.)

The methyl d-glucosides have been shown to have six-membered rings so they are properly named methyl α-d-glucopyranoside and methyl β-d-glucopyranoside. Oxidation Reactions of Monosaccharides A number of oxidizing agents are used to identify functional groups of carbohydrates, in elucidating their structures, and for syntheses. The most important are (1) Benedict’s or Tollens’ reagents, (2) bromine water, (3) nitric acid, and (4) periodic acid. Each of these reagents produces a different and usually specific effect when it is allowed to react with a monosaccharide. We shall now examine what these effects are. Benedict’s or Tollens’ Reagents: Reducing Sugars Benedict’s reagent (an alkaline solution containing − a cupric citrate complex ion) and Tollens’ solution [Ag+(NH3)2Ο H] oxidize and thus give positive tests with aldoses and ketoses. The tests are positive even though aldoses and ketoses exist primarily as cyclic hemiacetals. Since the solutions of cupric tartrates and citrates are blue, the appearance of a brick-red precipitate is a vivid and unmistakable indication of a positive test. O Cu21 (complex)

1

(H

C C

H OH)n

CH2OH

CH2OH

or (H

C

O

C

OH)n

Cu2O

1 oxidation products

CH2OH Benedict’s solution (blue) ●

Aldose

Ketose

(brick-red reduction product)

Sugars that give positive tests with Tollens’ or Benedict’s solutions are known as reducing sugars, and all carbohydrates that contain a hemiacetal group give positive tests.

In aqueous solution the hemiacetal form of sugars exists in equilibrium with relatively small, but not insignificant, concentrations of noncyclic aldehydes or α-hydroxy ketones. It is the latter two that undergo the oxidation, perturbing the equilibrium to produce more aldehyde or α-hydroxy ketone, which then undergoes oxidation until one reactant is exhausted. ●

Carbohydrates that contain only acetal groups do not give positive tests with Benedict’s or Tollens’ solutions, and they are called nonreducing sugars.

Acetals do not exist in equilibrium with aldehydes or α-hydroxy ketones in the basic aqueous media of the test reagents.

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Chapter 12 | Biomolecules

Reducing Sugar

Nonreducing Sugar Alkyl group or another sugar

C

O

O

H

C

O

C C

O

R

C R

C

Hemiacetal (R  H or CH2OH) (gives positive Tollens’ or Benedict’s test)

R

Acetal (R  H or CH2OH) (does not give a positive Tollens’ or Benedict’s test)

Bromine Water: The Synthesis of Aldonic Acids Monosaccharides do not undergo isomerization and fragmentation reactions in mildly acidic solution. Thus, a useful oxidizing reagent for preparative purposes is bromine in water (pH 6.0). ●

Bromine water is a general reagent that selectively oxidizes the }CHO group to a }CO2H group, thus converting an aldose to an aldonic acid: CHO (H

C

OH)n

CO2H Br2 H2O

(H

CH2OH

C

OH)n

CH2OH

Aldose

Aldonic acid

Nitric Acid Oxidation: Aldaric Acids ●

Dilute nitric acid—a stronger oxidizing agent than bromine water—oxidizes both the }CHO group and the terminal }CH2OH group of an aldose to }CO2H groups, forming dicarboxylic acids are known as aldaric acids: CHO (H

C

OH)n

CO2H HNO3

(H

CH2OH

C

OH)n

CO2H

Aldaric acid

Aldose

Periodate Oxidations: Oxidative Cleavage of Polyhydroxy Compounds ●

Compounds that have hydroxyl groups on adjacent atoms undergo oxidative cleavage when they are treated with aqueous periodic acid (HIO4). The reaction breaks carbon–carbon bonds and produces carbonyl compounds (aldehydes, ketones, or acids).

The stoichiometry of oxidative cleavage by periodic acid is C

OH

C

OH

O 

HIO4

2

C



HIO3



H2O

Notice in these periodate oxidations that for every C}C bond broken, a C}O bond is formed at each carbon.

12.2 | Monosaccharides

1. When three or more }CHOH groups are contiguous, the internal ones are obtained as formic acid. Periodate oxidation of glycerol, for example, gives two molar equivalents of formaldehyde and one molar equivalent of formic acid: O H

H H

C

OH

H

C

OH

H

C

OH

C

(formaldehyde)

H

 O 

2 IO4

H

C

(formic acid)

OH

 O

H Glycerol

H

C

(formaldehyde)

H

2. Oxidative cleavage also takes place when an }OH group is adjacent to the carbonyl group of an aldehyde or ketone (but not that of an acid or an ester). Glyceraldehyde yields two molar equivalents of formic acid and one molar equivalent of formaldehyde, while dihydroxyacetone gives two molar equivalents of formaldehyde and one molar equivalent of carbon dioxide: O H O

C C

OH

H

C

OH

(formic acid)

OH

 O

H

H

C



2 IO4

H

C

(formic acid)

OH

 O

H Glyceraldehyde

H

C

(formaldehyde)

H

O H H

H

C

OH

C

O

C

OH

H 

2 IO4

O

C 

(formaldehyde)

H

C

O

(carbon dioxide)



O

H Dihydroxyacetone

H

C

(formaldehyde)

H

3. Periodic acid does not cleave compounds in which the hydroxyl groups are separated by an intervening }CH2} group, nor those in which a hydroxyl group is adjacent to an ether or acetal function: CH2OH CH2 CH2OH

CH2OCH3 

IO4

no cleavage

H

C

OH

CH2R



IO4

no cleavage

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Chapter 12 | Biomolecules

Reduction of Monosaccharides: Alditols ●

Aldoses (and ketoses) can be reduced with sodium borohydride to give compounds called alditols: CH2OH

CHO (H

C

OH)n

NaBH4

(H

or H2, Pt

CH2OH

C

OH)n

CH2OH Alditol

Aldose

Reduction of d-glucose, for example, yields d-glucitol: H OH HO HO

O

OH

HO

O

H HO H H

OH H OH OH OH

NaBH4

H HO H H

OH OH H OH OH OH

D-Glucitol (or D-sorbitol)

Reactions of Monosaccharides with  Phenylhydrazine: Osazones The aldehyde group of an aldose reacts with such carbonyl reagents as hydroxylamine and phenylhydrazine. With hydroxylamine, the product is the expected oxime. With enough phenylhydrazine, however, three molar equivalents of phenylhydrazine are consumed and a second phenylhydrazone group is introduced at C2. The product is called a phenylosazone. Phenylosazones crystallize readily (unlike sugars) and are useful derivatives for identifying sugars. O

C

H

H

C

OH

(H

C

OH)n  3 C6H5NHNH2

CH2OH Aldose

H

(H

C

NNHC6H5

C

NNHC6H5

C

OH)n  C6H5NH2  NH3  H2O

CH2OH Phenylosazone

12.2F Other Reactions of Monosaccharides Some other important reactions of monosaccharides are described as follows: Enolization, Tautomerization, and Isomerization Dissolving monosaccharides in aqueous base causes them to undergo a series of enolizations and keto–enol tautomerizations that lead to isomerizations. For example, if a solution of  d-glucose containing calcium hydroxide is allowed to stand for several days, a number of products can be isolated, including d-fructose and d-mannose (Fig. 12.3). This type of reaction is called the Lobry de Bruyn–Alberda van Ekenstein transformation after the two Dutch chemists who discovered it in 1895. Formation of Ethers ●

Hydroxyl groups of sugars can be converted to ethers using a base and an alkyl halide by a version of the Williamson ether synthesis.

Benzyl ethers are commonly used to protect hydroxyl groups in sugars. Benzyl halides are easily introduced because they are highly reactive in SN2 reactions. Sodium or potassium hydride is typically used as

12.2 | Monosaccharides

H

C

H

H

O OH

HO



HO

H

H

OH

H

OH

OH

C

HO

H2O

H

O

C

C

H

HO

H

HO

HO

H

HO

H

OH

H

OH

H

OH

H

OH

H

OH

OH

OH

OH

Enolate ion

H

CH2OH O

OH

C

OH

HO

tautomerization

D-Mannose

HO

C

H

H2O

O

H

H2O

HO

OH

C

OH

D-Glucose (open-chain form)

C

H

O

H

OH

Carbonyl group isomerized to C-2 relative to D-glucose

C

H

H

OH

H

OH

H

OH

H

OH

OH

Figure 12.3 Monosaccharides undergo isomerizations via enolates and enediols when placed in aqueous base. Here we show how D-glucose isomerizes to D-mannose and to D-fructose.

OH Enediol

D-Fructose

Epimerized at C-2 relative to D-glucose

the base in an aprotic solvent such as DMF or DMSO. The benzyl groups can later be easily removed by hydrogenolysis using a palladium catalyst. Benzyl Ether Formation OBn

OH O

HO HO

HO

C6H5CH2Br

O

BnO BnO

NaH in DMF, heat

BnO

OMe

OMe

Bn  C6H5CH2

Benzyl Ether Cleavage OBn BnO BnO

OH O

BnO

H2, Pd

OMe

HO HO

O HO



4 C6H5CH3

OMe

Methyl ethers can also be prepared. The pentamethyl derivative of glucopyranose, for example, can be synthesized by treating methyl glucoside with excess dimethyl sulfate in aqueous sodium hydroxide. Sodium hydroxide is a competent base in this case because the hydroxyl groups of monosaccharides are

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more acidic than those of ordinary alcohols due to the many electronegative atoms in the sugar, all of which exert electron-withdrawing inductive effects on nearby hydroxyl groups. In aqueous NaOH the hydroxyl groups are all converted to alkoxide ions, and each of these, in turn, reacts with dimethyl sulfate in an SN2 reaction to yield a methyl ether. The process is called exhaustive methylation: OH HO HO

OH O

OH

HO

O

HO HO

OCH3

OCH3

O

CH3

OSO3CH3

Methyl glucoside

OH

OCH3 O

HO HO

repeated

OCH3

O

CH3O CH3O

methylations

OCH3

OCH3 OCH3

Pentamethyl derivative

Conversion to Esters Treating a monosaccharide with excess acetic anhydride and a weak base (such as pyridine or sodium acetate) converts all of the hydroxyl groups, including the anomeric hydroxyl, to ester groups. If the reaction is carried out at a low temperature (e.g., 0 °C), the reaction occurs stereospecifically; the α anomer gives the α-acetate and the β anomer gives the β-acetate. Acetate esters are common protecting groups for carbohydrate hydroxyls. O OH HO HO

O

O HO

O

O

O

O pyridine, 0 C

O

OH

O O

O

O O

O O

Conversion to Cyclic Acetals We have learnt that aldehydes and ketones react with open-chain 1,2-diols to produce cyclic acetals: O OH

HO



1,2-Diol

HA

O

O



HOH

Cyclic acetal

If the 1,2-diol is attached to a ring, as in a monosaccharide, formation of the cyclic acetals occurs only when the vicinal hydroxyl groups are cis to each other. For example, α-d-galactopyranose reacts with acetone in the following way: OH OH HO

O

O H2SO4

HO

OH

O O

OH O

 2 H2O

O O

Cyclic acetals are commonly used to protect vicinal cis hydroxyl groups of a sugar while reactions are carried out on other parts of the molecule. When acetals such as these are formed from acetone, they are called acetonides.

12.3 | Disaccharides

12.3 DISACCHARIDES 12.3A Sucrose Ordinary table sugar is a disaccharide called sucrose. Sucrose, the most widely occurring disaccharide, is found in all photosynthetic plants and is obtained commercially from sugarcane or sugar beets. Sucrose has the structure shown in Fig. 12.4. 6

HO 5

From D-glucose

4

HO

OH

1

O 1

O

H

3

HO

O

2

OH

From

5

2

OH

3

6

OH

D-fructose

4

OH

 - Glucosidic linkage

 - Fructosidic linkage

OH HO

O HO HO

HO

O O

HO OH

OH

Figure 12.4 Two representations of the formula for (+)-sucrose (α-D-glucopyranosyl β-D-fructofuranoside).

The structure of sucrose is based on the following evidence: 1. Sucrose has the molecular formula C12H22O11. 2. Acid-catalyzed hydrolysis of 1 mol of sucrose yields 1 mol of d-glucose and 1 mol of d-fructose. 6

HO 5

H

O H 4

OH

OH

HO

2

3 1

OH

H

Fructose (as a  -furanose)

3. Sucrose is a nonreducing sugar; it gives negative tests with Benedict’s and Tollens’ solutions. Sucrose does not form an osazone and does not undergo mutarotation. These facts mean that neither the glucose nor the fructose portion of sucrose has a hemiacetal group. Thus, the two hexoses must have a glycosidic linkage that involves C1 of glucose and C2 of fructose, for only in this way will both carbonyl groups be present as full acetals (i.e., as glycosides). 4. The stereochemistry of the glycosidic linkages can be inferred from experiments done with enzymes. Sucrose is hydrolyzed by an α-glucosidase obtained from yeast but not by β-glucosidase enzymes. This hydrolysis indicates an α configuration at the glucoside portion. Sucrose is also hydrolyzed by sucrase,

771

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Chapter 12 | Biomolecules

an enzyme known to hydrolyze β-fructofuranosides but not α-fructofuranosides. This hydrolysis indicates a β configuration at the fructoside portion. 5. Methylation of sucrose gives an octamethyl derivative that, on hydrolysis, gives 2,3,4,6-tetra-O-methyld-glucose and 1,3,4,6-tetra-O-methyl-d-fructose. The identities of these two products demonstrate that the glucose portion is a pyranoside and that the fructose portion is a furanoside.

12.3B Maltose When starch is hydrolyzed by the enzyme diastase, one product is a disaccharide known as maltose (Fig. 12.5). The structure of maltose was deduced based on the following evidence: 1. When 1 mol of maltose is subjected to acid-catalyzed hydrolysis, it yields 2 mol of d-(+)-glucose. 2. Unlike sucrose, maltose is a reducing sugar; it gives positive tests with Fehling’s, Benedict’s, and Tollens’ solutions. Maltose also reacts with phenylhydrazine to form a monophenylosazone (i.e., it incorporates two molecules of phenylhydrazine). 3. Maltose exists in two anomeric forms: α-(+)-maltose, [α]d25 = +168, and β-(+)-maltose, [α]d25 = +112. The maltose anomers undergo mutarotation to yield an equilibrium mixture, [α]d25 = +136. Facts 2 and 3 demonstrate that one of the glucose residues of maltose is present in a hemiacetal form; the other, therefore, must be present as a glucoside. The configuration of this glucosidic linkage can be inferred as α, because maltose is hydrolyzed by α-glucosidase enzymes and not by β-glucosidase enzymes.

OH

6

H 4

HO

5

O 1

H OH

H

H

OH

O

H OH

O

2

H

5

4

H

3

OH

6

H

3

OH 1

H

2

OH

H

 - Glucosidic linkage

OH HO

O HO

HO

OH O HO

O OH HO

Figure 12.5 Two representations of the structure of the β anomer of (+)-maltose, 4-O-(α-D-glucopyranosyl)-β-Dglucopyranose.

4. Maltose reacts with bromine water to form a monocarboxylic acid, maltonic acid (Fig. 12.6a). This fact, too, is consistent with the presence of only one hemiacetal group. 5. Methylation of maltonic acid followed by hydrolysis gives 2,3,4,6-tetra-O-methyl-d-glucose and 2,3,5,6-tetra-O-methyl-d-gluconic acid. That the first product has a free } OH at C5 indicates that the nonreducing glucose portion is present as a pyranoside; that the second product, 2,3,5,6-tetraO-methyl-d-gluconic acid, has a free } OH at C4 indicates that this position was involved in a glycosidic linkage with the nonreducing glucose. Only the size of the reducing glucose ring needs to be determined. 6. Methylation of maltose itself, followed by hydrolysis (Fig. 12.6b), gives 2,3,4,6-tetra-O-methyl-dglucose and 2,3,6-tri-O-methyl-d-glucose. The free } OH at C5 in the latter product indicates that it must have been involved in the oxide ring and that the reducing glucose is present as a pyranose.

12.3C Cellobiose Partial hydrolysis of cellulose gives the disaccharide cellobiose (C12H22O11) (Fig. 12.7). Cellobiose resembles maltose in every respect except one: the configuration of its glycosidic linkage. Cellobiose, like maltose, is a reducing sugar that, on acid-catalyzed hydrolysis, yields two molar equivalents of d-glucose. Cellobiose also undergoes mutarotation and forms a monophenylosazone. Methylation studies show that C1 of one glucose unit is connected in glycosidic linkage with C4 of the other and that both rings are six membered. Unlike maltose, however, cellobiose is hydrolyzed by β-glucosidase enzymes and not by α-glucosidase enzymes: This indicates that the glycosidic linkage in cellobiose is β (Fig. 12.7).

12.3 | Disaccharides

HO 5

H

HO

6

O 1

H OH

4

HO

H

H

O

2

OH

H

6

O 1

H OH

4

H

3

5

OH

H

3

2

OH

H

Maltose

(a)

(b)

(1) CH3OH, H (2) (CH3)2SO4, HO

Br2/H2O

HO

HO H HO

O

H

H OH

H

H

OH

CH3O OH

H

O

H

H OH

H CO2H

H

OH

CH3O

CH3O O

H

H OCH3 H

H OCH3 H

O

OCH3

H

O

H

H

OCH3

OCH3

Maltonic acid (CH3)2SO4 OH

CH3O

CH3O H CH3O

H3O

O H OCH3 H

H

O

OCH3

H

OCH3 H OCH3 H CO2CH3

H

CH3O

CH3O

H OH  HO

OCH3

H

2,3,5,6-Tetra-O-methyl5 D-gluconic acid H 4

H

5

H

HO 5

4

HO

6

O

H OH 3

H

1

H

O

4

O

H OH

H

6

O

H OH 3

H

H

6

3

HO

H

H

HO

OCH3

H

1

HO HO

2

OH

OCH3

O

4

6 of to maltonic acid 5 maltose O followed byOH methylation and H hydrolysis. (b)1 Methylation H OH subsequent and hydrolysis of H maltose itself. 3 2

OH

H

H

2

OH OH O OH

O

O HO

OH OH

2

OH

Figure 12.7OHTwo representations of the β anomer of cellobiose, 4-O-(β-D-glucopyranosyl)OH β-D-glucopyranose. O

HO HO

OH

O

O HO

OH OH

OH

2,3,6-Tri-Omethyl-D-glucose (as a pyranose)

OH

OH H

1

H

O H OCH3 H

HO Figure 12.6 (a) Oxidation

 - Glycosidic linkage

HO  - Glycosidic linkage

HO

OCH3 H OCH3 H CO2H H

2,3,4,6-Tetra-O-methylD-glucose (as a pyranose)

OH 

OCH3

H

CH3O O H OCH3 H

H

2,3,4,6-Tetra-Omethyl-D-glucose (as a pyranose)

H3O

H

O H OCH3 H

H

OCH3

H

CH3O

CH3O

CH3O

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The Chemistry of...

ARTIFICIAL SWEETENERS (HOW SWEET IT IS)

S

ucrose (table sugar) and fructose are the most common natural sweeteners. We all know, however, that they add to our calorie intake and promote tooth decay. For these reasons, many people find artificial sweeteners to be an attractive alternative to the natural and calorie-contributing counterparts. Perhaps the most successful and widely used artificial sweetener is aspartame, the methyl ester of a dipeptide formed from phenylalanine and aspartic acid. Aspartame is roughly 100 times as sweet as sucrose. It undergoes slow hydrolysis in solution, however, which limits its shelf life in products such as soft drinks. It also cannot be used for baking because it decomposes with heat. Furthermore, people with a genetic condition known as phenylketonuria cannot use aspartame because their metabolism causes a buildup of phenylpyruvic acid derived from aspartame. Accumulation of phenylpyruvic acid is harmful, especially to infants. Alitame, on the other hand, is a compound related to aspartame, but with improved properties. It is more stable than aspartame and roughly 2000 times as sweet as sucrose. H

O N

CO2H

H

O N

H NH2 OCH3 H

O Aspartame

H

CO2H

O

Cl OH O Cl HO

HO

O O

HO

Sucralose

S

Alitame

Sucralose is a trichloro derivative of sucrose that is an artificial sweetener. Like aspartame, it is also approved for use by the U.S. Food and Drug Administration (FDA). Sucralose is 600 times sweeter than sucrose and has many properties desirable in an artificial sweetener. Sucralose looks and tastes like sugar, is stable at the temperatures used for cooking and

Cl

OH

Cyclamate and saccharin, used as their sodium or calcium salts, were popular sweeteners at one time. A common formulation involved a 10:1 mixture of cyclamate and saccharin that proved sweeter than either compound individually. Tests showed, however, that this mixture produced tumors in animals, and the FDA subsequently banned it. Certain exclusions to the regulations nevertheless allow continued use of saccharin in some products. H N

SO2 N H

SO3H

O

Cyclamate

H NH2 H N CH3

baking, and it does not cause tooth decay or provide calories.

Saccharin

Many other compounds have potential as artificial sweeteners. For example, l sugars are also sweet, and they presumably would provide either zero or very few calories because our enzymes have evolved to selectively metabolize their enantiomers instead, the d sugars. OH

OH O

HO OH

OH

L-Glucose

Much of the research on sweeteners involves probing the structure of sweetness receptor sites. One model proposed for a sweetness (continues on next page)

12.4 | Polysaccharides

receptor incorporates eight binding interactions that involve hydrogen bonding as well as van der Waals forces. Sucronic acid is a synthetic compound designed on the basis of this model. Sucronic acid is reported to be 200,000 times as sweet as sucrose.

H CH2

HO2C

N

N C

CN

N

H

Sucronic acid

12.3D Lactose Lactose (Fig. 12.8) is a disaccharide present in the milk of humans, cows, and almost all other mammals. Lactose is a reducing sugar that hydrolyzes to yield d-glucose and d-galactose; the glycosidic linkage is β.  -Glycosidic linkage

H

HO From D -galactose

HO H

OH

O H OH

H

H

OH

O

O

H

H

H

OH

HO OH

From

D -glucose

H

OH O

HO

OH

H OH

O

O OH

HO

OH OH

Figure 12.8 Two representations of the β anomer of lactose, 4-O-(β-D-galactopyranosyl)-β-D-glucopyranose.

12.4 POLYSACCHARIDES ●

Polysaccharides, also known as glycans, consist of monosaccharides joined together by glycosidic linkages.

Polysaccharides that are polymers of a single monosaccharide are called homopolysaccharides; those made up of more than one type of monosaccharide are called heteropolysaccharides. Homopolysaccharides are also classified on the basis of their monosaccharide units. A homopolysaccharide consisting of glucose monomeric units is called a glucan; one consisting of galactose units is a galactan, and so on. Three important polysaccharides, all of which are glucans, are starch, glycogen, and cellulose. ●

Starch is the principal food reserve of plants, glycogen functions as a carbohydrate reserve for animals, and cellulose serves as structural material in plants.

As we examine the structures of these three polysaccharides, we shall be able to see how each is especially suited for its function.

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12.4A Starch Starch occurs as microscopic granules in the roots, tubers, and seeds of plants. Corn, potatoes, wheat, and rice are important commercial sources of starch. Heating starch with water causes the granules to swell and produce a colloidal suspension from which two major components can be isolated. One fraction is called amylose and the other amylopectin. Most starches yield 10–20% amylose and 80–90% amylopectin. ●

Amylose typically consists of more than 1000 d-glucopyranoside units connected in α linkages between C1 of one unit and C4 of the next (Fig. 12.9).

 (1

HO H

4 ) Glucosidic linkage

HO

6 5

O

H 4 OH 3

H

H

6

HH 4

1

OH

O

2

O

5 3

OH

H

O

2

OH

H n

H 1

Figure 12.9 Partial structure of amylose, an unbranched polymer of D-glucose connected in α(1 → 4) glycosidic linkages.

n

500

Thus, in the ring size of its glucose units and in the configuration of the glycosidic linkages between them, amylose resembles maltose. Chains of d-glucose units with α-glycosidic linkages such as those of amylose tend to assume a helical arrangement. This arrangement results in a compact shape for the amylose molecule even though its molecular weight is quite large (150,000–600,000). ●

Amylopectin has a structure similar to that of amylose [i.e., α(1 → 4) links], except that in amylopectin the chains are branched. Branching takes place between C6 of one glucose unit and C1 of another and occurs at intervals of 20–25 glucose units (Fig. 12.10).

Physical measurements indicate that amylopectin has a molecular weight of 1–6 million; thus amylopectin consists of hundreds of interconnecting chains of 20–25 glucose units each.

HO

HO

Branch

H O

Main chain

H O

H OH H

O H OH

O

HH O

H OH H

HO

HO

O

OH

H

H

OH

5

HH

H

4

O

OH

H OH 3

H

HO

O

HH O

6

OH

H

H

OH

O

H 1

H

2

OH

 (1 : 6) branch point

O 5

HH O

4

O

OH 3

H

 (1 : 4)

Figure 12.10 Partial structure of amylopectin.

HO

6

H

O

HH 1

2

OH

O

OH

H

H

OH

H

O

12.4 | Polysaccharides

12.4B Glycogen ●

Glycogen has a structure very much like that of amylopectin; however, in glycogen the chains are much more highly branched.

Methylation and hydrolysis of glycogen indicate that there is one end group for every 10–12 glucose units; branches may occur as often as every 6 units. Glycogen has a very high molecular weight. Studies of glycogens isolated under conditions that minimize the likelihood of hydrolysis indicate molecular weights as high as 100 million. The size and structure of glycogen beautifully suit its function as a reserve carbohydrate for animals. First, its size makes it too large to diffuse across cell membranes; thus, glycogen remains inside the cell, where it is needed as an energy source. Second, because glycogen incorporates tens of thousands of glucose units in a single molecule, it solves an important osmotic problem for the cell. Amylopectin presumably serves a similar function in plants. The fact that amylopectin is less highly branched than glycogen is, however, not a serious disadvantage. Plants have a much lower metabolic rate than animals—and plants, of course, do not require sudden bursts of energy. Animals store energy as fats (triacylglycerols) as well as glycogen. Fats, because they are more highly reduced, are capable of furnishing much more energy. The metabolism of a typical fatty acid, for example, liberates more than twice as much energy per carbon as glucose or glycogen.

12.4C Cellulose When we examine the structure of cellulose, we find another example of a polysaccharide in which nature has arranged monomeric glucose units in a manner that suits its function. ●

Cellulose contains d-glucopyranoside units linked in (1 → 4) fashion in very long unbranched chains. Unlike starch and glycogen, however, the linkages in cellulose are β-glycosidic linkages (Fig. 12.11).

(1

HO

H 4

3

H

H

6 5

O

O

H OH

H

H

4

O

H OH

HO

H

6 5

4)

3

H

1

2

O H

1

2

OH

OH

n

Figure 12.11 A portion of a cellulose chain. The glycosidic linkages are β(1 → 4). The β-glycosidic linkages of cellulose make cellulose chains essentially linear; they do not tend to coil into helical structures as do glucose polymers when linked in an α(1 → 4) manner. The linear arrangement of β-linked glucose units in cellulose presents a uniform distribution of } OH groups on the outside of each chain. When two or more cellulose chains make contact, the hydroxyl groups are ideally situated to “zip” the chains together by forming hydrogen bonds. Zipping many cellulose chains together in this way gives a highly insoluble, rigid, and fibrous polymer that is ideal as cell-wall material for plants. This special property of cellulose chains, we should emphasize, is not just a result of β(1 → 4) glycosidic linkages; it is also a consequence of the precise stereochemistry of d-glucose at each chirality center. There is another interesting and important fact about cellulose: the digestive enzymes of humans cannot attack its β(1 → 4) linkages. Hence, cellulose cannot serve as a food source for humans, as can starch. Cows and termites, however, can use cellulose (of grass and wood) as a food source because symbiotic bacteria in their digestive systems furnish β-glucosidase enzymes.

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12.4D Cellulose Derivatives A number of derivatives of cellulose are used commercially. Most of these are compounds in which two or three of the free hydroxyl groups of each glucose unit have been converted to an ester or an ether. This conversion substantially alters the physical properties of the material, making it more soluble in organic solvents and allowing it to be made into fibers and films. Treating cellulose with acetic anhydride produces the triacetate known as “Arnel” or “acetate,” used widely in the textile industry. Cellulose trinitrate, also called “gun cotton” or nitrocellulose, is used in explosives. Rayon is made by treating cellulose (from cotton or wood pulp) with carbon disulfide in a basic solution. This reaction converts cellulose to a soluble xanthate: S Cellulose

OH



NaOH

CS2

cellulose

O

C

S Na

Cellulose xanthate

The solution of cellulose xanthate is then passed through a small orifice or slit into an acidic solution. This operation regenerates the } OH groups of cellulose, causing it to precipitate as a fiber or a sheet: S Cellulose

O

C

S Na

H3 O

cellulose

OH

Rayon or cellophane

The fibers are rayon; the sheets, after softening with glycerol, are cellophane.

12.5 OTHER BIOLOGICALLY IMPORTANT SUGARS Monosaccharide derivatives in which the } CH2OH group at C6 has been specifically oxidized to a carboxyl group are called uronic acids. Their names are based on the monosaccharide from which they are derived. For example, specific oxidation of C6 of glucose to a carboxyl group converts glucose to glucuronic acid. In the same way, specific oxidation of C6 of galactose would yield galacturonic acid: HO

CO2H

H HO OH or H H

O HO

OH OH

HO

CO2H

OH H OH OH

H HO OH or HO

O

HO OH

H OH

CO2H

D-Glucuronic

●

O

acid

O OH H H OH CO2H

D-Galacturonic

acid

Monosaccharides in which an } OH group has been replaced by } H are known as deoxy sugars.

The most important deoxy sugar, because it occurs in DNA, is deoxyribose. Other deoxy sugars that occur widely in polysaccharides are l-rhamnose and l-fucose: HO

O

OH

HO

O CH3

O

OH

CH3 HO

OH -2-Deoxy-D-ribose

OH

OH

-L-Rhamnose (6-deoxy-L-mannose)

OH

HO

HO -L-Fucose (6-deoxy-L-galactose)

12.6 | Sugars That Contain Nitrogen

12.6 SUGARS THAT CONTAIN NITROGEN 12.6A Glycosylamines A sugar in which an amino group replaces the anomeric } OH is called a glycosylamine. Examples are β-d-glucopyranosylamine and adenosine: NH2 N HO

OH O

HO HO

NH2

HO

H

H

HO

-D-Glucopyranosylamine

N

O H

N N

H

OH Adenosine

Adenosine is an example of a glycosylamine that is also called a nucleoside. ●

Nucleosides are glycosylamines in which the amino component is a pyrimidine or a purine and in which the sugar component is either d-ribose or 2-deoxy-d-ribose (i.e., d-ribose minus the oxygen at the 2 position).

Nucleosides are the important components of RNA (ribonucleic acid) and DNA (deoxyribonucleic acid). We shall describe their properties in detail in later sections.

12.6B Amino Sugars ●

A sugar in which an amino group replaces a nonanomeric } OH group is called an amino sugar.

D-Glucosamine

is an example of an amino sugar. In many instances the amino group is acetylated as in N-acetyl-D-glucosamine. N-Acetylmuramic acid is an important component of bacterial cell walls. HO H HO

HO O H OH H

H

OH H

NH2

-D-Glucosamine

H HO

HO O H OH H

H

OH H

NHCOCH3

-N-Acetyl-D -glucosamine (NAG)

H HO

O H OR H

H

CH 3

OH R H

H CO2H

NHCOCH3

-N-Acetylmuramic acid (NAM)

d-Glucosamine can be obtained by hydrolysis of chitin, a polysaccharide found in the shells of lobsters and crabs and in the external skeletons of insects and spiders.

PART II: AMINO ACIDS AND PROTEINS Of the three groups of biopolymers, polysaccharides, proteins, and nucleic acids, proteins have the most diverse functions. As enzymes and hormones, proteins catalyze and regulate the reactions that occur in the

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body; as muscles and tendons they provide the body with the means for movement; as skin and hair they give it an outer covering; as hemoglobin molecules they transfer all-important oxygen to its most remote corners; as antibodies they provide it with a means of protection against disease; and in combination with other substances in bone they provide it with structural support. Given such diversity of functions, we should not be surprised to find that proteins come in all sizes and shapes. By the standard of most of the molecules we have studied, even small proteins have very high molecular weights. Lysozyme, an enzyme, is a relatively small protein and yet its molecular weight is 14,600. The molecular weights of most proteins are much larger. Their shapes cover a range from the globular proteins such as lysozyme and hemoglobin to the helical coils of α-keratin (hair, nails, and wool) and the pleated sheets of silk fibroin. And yet, in spite of such diversity of size, shape, and function, all proteins have common features that allow us to deduce their structures and understand their properties. Later in this chapter we shall see how this is done. ●

Proteins are polyamides, and their monomeric units are composed of about 20 different α-amino acids: O α

HO

NH2

R An α-amino acid R is a side chain at the α carbon that determines the identity of the amino acid (Table 12.1).



R2

O NH R1

R4

O NH

NH O

R3

O NH

NH O

R5



A portion of a protein molecule Amide (peptide) linkages are shaded. R1 to R5 may be any of the possible side chains. ●

The exact sequence of the different α-amino acids along the protein chain is called the primary structure of the protein.

A protein’s primary structure, as its name suggests, is of fundamental importance. For the protein to carry out its particular function, the primary structure must be correct. We shall see later that when the primary structure is correct, the protein’s polyamide chain folds in particular ways to give it the shape it needs for its particular task. ●

●

●

Folding of the polyamide chain gives rise to higher levels of complexity called the secondary and tertiary structures of the protein. Quaternary structure results when a protein contains an aggregate of more than one polyamide chain. Hydrolysis of proteins with acid or base yields a mixture of amino acids.

Although hydrolysis of naturally occurring proteins may yield as many as 22 different amino acids, the amino acids have an important structural feature in common: with the exception of glycine (whose molecules are achiral), almost all naturally occurring amino acids have the l configuration at the α carbon.* That is, they have the same relative configuration as l-glyceraldehyde: *Some d-amino acids have been obtained from the material comprising the cell walls of bacteria and by hydrolysis of certain antibiotics.

12.7 | Amino Acids

O

O

R

OH

H

HO

NH2

OH

An L-a-amino acid [usually an (S)-a-amino acid]

L-Glyceraldehyde

[(S)-glyceraldehyde]

CO2H H2N

CHO

H

HO

H CH2OH

R

Fischer projections for an L-a-amino acid and L-glyceraldehyde

12.7 AMINO ACIDS 12.7A Structures and Names ●

The 22 α-amino acids that can be obtained from proteins can be subdivided into three different groups on the basis of the structures of their side chains, R. These are given in Table 12.1.

Only 20 of the 22 α-amino acids in Table 12.1 are actually used by cells when they synthesize proteins. Two amino acids are synthesized after the polyamide chain is intact. Hydroxyproline (present mainly in collagen) is synthesized by oxidation of proline, and cystine (present in most proteins) is synthesized from cysteine. The conversion of cysteine to cystine requires additional comment. The } SH group of cysteine makes cysteine a thiol. One property of thiols is that they can be converted to disulfides by mild oxidizing agents. This conversion, moreover, can be reversed by mild reducing agents: 2R

S

H

O H

R

Thiol

S

S

R

Disulfide Disulfide linkage

O

O

2 HS

OH NH2 Cysteine

O H

O S

HO

S

OH NH2

NH2 Cystine

We shall see later how the disulfide linkage between cysteine units in a protein chain contributes to the overall structure and shape of the protein.

12.7B Essential Amino Acids Amino acids can be synthesized by all living organisms, plants and animals. Many higher animals, however, are deficient in their ability to synthesize all of the amino acids they need for their proteins. Thus, these higher animals require certain amino acids as a part of their diet. For adult humans there are eight essential amino acids; these are identified in Table 12.1 by a footnote.

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Table 12.1 L-Amino acids found in proteins Structure

Name

pKa pKa pKa 2 1 3 Abbreviationsa α -CO2H α -NH3+ R group

pI

Glycine

G or Gly

2.3

9.6

6.0

Alanine

A or Ala

2.3

9.7

6.0

Valineb

V or Val

2.3

9.6

6.0

Leucineb

L or Leu

2.4

9.6

6.0

Isoleucineb

I or Ile

2.4

9.7

6.1

Phenylalanineb

F or Phe

1.8

9.1

5.5

Tyrosine

Y or Tyr

2.2

9.1

Tryptophanb

W or Trp

2.4

9.4

5.9

Serine

S or Ser

2.2

9.2

5.7

Threonineb

T or Thr

2.6

10.4

6.5

Neutral Amino Acids

O H2N

OH O OH

NH2 O OH NH2 O OH NH2 O OH NH2 O OH NH2 O

10.1

5.7

OH NH2

HO

O OH NH2

N H O OH

HO NH2 O

OH

HO NH2

(Continued)

12.7 | Amino Acids

Table 12.1 (Continued) Structure

O

Name

pKa pKa pKa 2 1 3 Abbreviationsa α -CO2H α -NH3+ R group

pI

Proline

P or Pro

2.0

10.6

6.3

4-Hydroxyproline O or Hyp (cis and trans)

1.9

9.7

6.3

Cysteine

C or Cys

1.7

10.8

Cystine

Cys-Cys

1.6 2.3

7.9 9.9

5.1

Methionineb

M or Met

2.3

9.2

5.8

Asparagine

N or Asn

2.0

8.8

5.4

Glutamine

Q or Gln

2.2

9.1

5.7

Aspartic acid

D or Asp

2.1

9.8

3.9

3.0

Glutamic acid

E or Glu

2.2

9.7

4.3

3.2

OH NH O OH HO

NH O

8.3

5.0

OH

HS NH2 NH2 HO

O S

S

OH NH2

O O MeS

OH NH2 O

H2N

OH O

NH2

O

O

H2N

OH NH2

Side Chains Containing an Acidic (Carboxyl) Group

O HO

OH O O

NH2 O OH

HO NH2

(Continued)

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Table 12.1 (Continued) Structure

Name

pKa pKa pKa 1 2 3 Abbreviationsa α -CO2H α -NH3+ R group

pI

Lysineb

K or Lys

2.2

9.0

10.5c

9.8

Arginine

R or Arg

2.2

9.0

12.5c

10.8

Histidine

H or His

1.8

9.2

6.0c

7.6

Side Chains Containing a Basic Group

O H2N

OH NH2 NH

H2N

H N

O OH

N H

NH2 O OH NH2

N

Single-letter abbreviations are now the most commonly used form in current biochemical literature. An essential amino acid. c pKa is of protonated amine of R group. a

b

12.7C Amino Acids as Dipolar Ions ● ●

●

Amino acids contain both a basic group (} NH2) and an acidic group (} CO2H). In the dry solid state, amino acids exist as dipolar ions, a form in which the carboxyl group is present as a carboxylate ion, } CO2−, and the amino group is present as an aminium ion, } NH3+. (Dipolar ions are also called zwitterions.) In aqueous solution, an equilibrium exists between the dipolar ion and the anionic and cationic forms of an amino acid. O R

O HO



OH +

NH3

Cationic form (predominant in strongly acidic solutions, e.g., at pH 0)

H3O

R

O HO



O– +

NH3

Dipolar ion

H3O

R

O– NH2

Anionic form (predominant in strongly basic solutions, e.g., at pH 14)

The predominant form of the amino acid present in a solution depends on the pH of the solution and on the nature of the amino acid. In strongly acidic solutions all amino acids are present primarily as cations; in strongly basic solutions they are present as anions. ●

The isoelectric point (pI ) is the pH at which the concentration of the dipolar ion is at its maximum and the concentrations of the anions and cations are equal.

Each amino acid has a particular isoelectric point. These are given in Table 12.1. Proteins have isoelectric points as well. Let us consider first an amino acid with a side chain that contains neither acidic nor basic groups—an amino acid, for example, such as alanine. If alanine is dissolved in a strongly acidic solution (e.g., pH 0), it is present in mainly a net cationic form. In this state the amino group is protonated (bears a formal +1 charge) and the carboxylic acid group

12.7 | Amino Acids

is neutral (has no formal charge). As is typical of α-amino acids, the pKa for the carboxylic acid hydrogen of alanine is considerably lower (2.3) than the pKa of an ordinary carboxylic acid (e.g., propanoic acid, pKa 4.89): O

O OH

+

OH

NH3

Cationic form of alanine pKa1 5 2.3

Propanoic acid pKa 5 4.89

The reason for this enhanced acidity of the carboxyl group in an α-amino acid is the inductive effect of the neighboring aminium cation, which helps to stabilize the carboxylate anion formed when it loses a proton. Loss of a proton from the carboxyl group in a cationic α-amino acid leaves the molecule electrically neutral (in the form of a dipolar ion). This equilibrium is shown in the red-shaded portion of the equation below. The protonated amino group of an α-amino acid is also acidic, but less so than the carboxylic acid group. The pKa of the aminium group in alanine is 9.7. The equilibrium for loss of an aminium proton is shown in the blue-shaded portion of the equation below. The carboxylic acid proton is always lost before a proton from the aminium group in an α-amino acid. O

O HO

2

OH +

NH3

O O–

H3O1

+

Cationic form (pKa1 5 2.3)

NH3

Dipolar ion (pKa2 5 9.7)

HO

2

H3O1

O– NH2 Anionic form

The state of an α-amino acid at any given pH is governed by a combination of two equilibria, as shown in the above equation for alanine. The isoelectric point (pI ) of an amino acid such as alanine is the average of pKa1 and pKa2: pl =

1 ( 2 .3 + 9 .7 ) = 6 . 0 2

(isoelectric point of alanine)

When a base is added to a solution of the net cationic form of alanine (initially at pH 0, for example), the first proton removed is the carboxylic acid proton, as we have said. In the case of alanine, when a pH of 2.3 is reached, the carboxylic acid proton will have been removed from half of the molecules. This pH represents the pKa of the alanine carboxylic acid proton, as can be demonstrated using the Henderson–Hasselbalch equation. ●

The Henderson–Hasselbalch equation shows that for an acid (HA) and its conjugate base (A−) when [HA] = [A−], then pH = pKa. pK a = pH + log

[HA ] [A − ]

Henderson–Hasselbalch equation

Therefore, when the acid is half neutralized, [HA ] = [ A − ], log

[HA ] = 0, and thus pH = pK a [A − ]

As more base is added to this solution, alanine reaches its isoelectric point (pI ), the pH at which all of alanine’s carboxylic acid protons have been removed but not its aminium protons. The molecules are therefore electrically neutral (in their dipolar ion or zwitter ionic form) because the carboxylate group carries a −1 charge and the aminium group a +1 charge. The pI for alanine is 6.0.

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Chapter 12 | Biomolecules

Now, as we continue to add the base, protons from the aminium ions will begin to be removed, until at pH 9.7 half of the aminium groups will have lost a proton. This pH represents the pKa of the aminium group. Finally, as more base is added, the remaining aminium protons will be lost until all of the alanine molecules have lost their aminium protons. At this point (e.g., pH 14) the molecules carry a net anionic charge from their carboxylate group. The amino groups are now electrically neutral.

12.8 POLYPEPTIDES AND PROTEINS Amino acids are polymerized in living systems by enzymes that form amide linkages from the amino group of one amino acid to the carboxyl group of another. ●

A molecule formed by joining amino acids together is called a peptide, and the amide linkages in them are called peptide bonds or peptide linkages. Each amino acid in the peptide is called an amino acid residue.

Peptides that contain 2, 3, a few (3–10), or many amino acids are called dipeptides, tripeptides, oligopeptides, and polypeptides, respectively. Proteins are polypeptides consisting of one or more polypeptide chains. O H3N

O

O

+

O 

H3N

+

O

R

H3

[H2O]

R

N+

O

N R

R

H

O

A dipeptide

Polypeptides are linear polymers. One end of a polypeptide chain terminates in an amino acid residue that has a free } NH3+ group; the other terminates in an amino acid residue with a free } CO2− group. These two groups are called the N-terminal and the C-terminal residues, respectively:

( )

O H3N

+

R

N

R

N-Terminal residue ●

O

H

H

O

N

n

O R

C-Terminal residue

By convention, we write peptide and protein structures with the N-terminal amino acid residue on the left and the C-terminal residue on the right: O

O

H3N

+

H3N

O

N H

+

H

O

Glycyl valine (GV)

O

N O

Valyl glycine (VG)

The tripeptide glycylvalylphenylalanine has the following structural formula: H

O

H3N

+

N

N H

O

Glycylvalylphenylalanine (GVF)

O O

12.8 | Polypeptides and Proteins

It becomes a significant task to write a full structural formula for a polypeptide chain that contains any more than a few amino acid residues. In this situation, use of the one-letter abbreviations (Table 12.1) is the norm for showing the sequence of amino acids. Very short peptide sequences are sometimes still represented with the three-letter abbreviations (Table 12.1). When a protein or polypeptide is refluxed with 6 M hydrochloric acid for 24 h, hydrolysis of all the amide linkages usually takes place, liberating its constitutent amino acids as a mixture. Chromatographic separation and quantitative analysis of the resulting mixture can then be used to determine which amino acids composed the intact polypeptide and their relative amounts.

12.8A Primary Structure of Polypeptides and Proteins The sequence of amino acid residues in a polypeptide or protein is called its primary structure. A simple peptide composed of three amino acids (a tripeptide) can have 6 different amino acid sequences; a tetrapeptide can have as many as 24 different sequences. For a protein composed of 20 different amino acids in a single chain of 100 residues, there are 2100 = 1.27 × 10130 possible peptide sequences, a number much greater than the number of atoms estimated to be in the universe (9 × 1078)! Clearly, one of the most important things to determine about a protein is the sequence of its amino acids. Fortunately, there are a variety of methods available to determine the sequence of amino acids in a polypeptide. These include terminal residue analysis techniques used to identify the N-(Edman degradation and Sanger analysis) and C-(throughout of digestive enzymes carboxypeptidases) terminal amino acids.

12.8B Examples of Polypeptide and Protein Primary Structure ●

The covalent structure of a protein or polypeptide is called its primary structure (Fig. 12.12).

N-Terminal end

C-Terminal end

Hydrogen

Oxygen

Nitrogen

R group

Carbon

Peptide bond

Figure 12.12 A representation of the primary structure of a tetrapeptide. Insulin, a hormone secreted by the pancreas, regulates glucose metabolism. Insulin deficiency in humans is the major problem in diabetes mellitus. The amino acid sequence of bovine insulin shown below was determined by Sanger in 1953 after 10 years of work. Bovine insulin has a total of 51 amino acid residues in two polypeptide chains, called the A and B chains. These chains are joined by two disulfide linkages. The A chain contains an additional disulfide linkage between cysteine residues at positions 6 and 11. A Chain GIVEQCCASVCSLYQLENYCN B Chain FVNQHLCGSHLVEALYLVCGERGFFYTPKA

Human insulin differs from bovine insulin at only three amino acid residues: Threonine replaces alanine once in the A chain (residue 8) and once in the B chain (residue 30), and isoleucine replaces valine once in the A chain (residue 10). Insulins from most mammals have similar structures.

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12.9 SECONDARY, TERTIARY AND QUATERNARY STRUCTURES OF PROTEINS The structure and shape of proteins is understood in terms of four levels of organization: primary, secondary, tertiary, and quaternary structure. The amide and disulfide linkages constitute the covalent or primary structure of proteins. Of equal importance in understanding how proteins function is knowledge of the way in which the peptide chains are arranged in three dimensions. The secondary and tertiary structures of proteins are involved here.

12.9A Secondary Structure ● ●

The secondary structure of a protein is defined by the local conformation of its polypeptide backbone. Secondary structures are specified in terms of regular folding patterns called α helices, β sheets, and coil or loop conformations.

To understand how these interactions occur, let us look first at what X-ray crystallographic analysis has revealed about the geometry at the peptide bond itself. ●

Peptide bonds tend to assume a geometry such that six atoms of the amide linkage are coplanar (Fig. 12.13). O H

1.24 123.5°

120.5°

1

1.5

C

Peptide bond

C

R

C

122°

116°

1.3

3

N

119.5°

.46

Figure 12.13 The geometry and bond lengths (in angstroms, Å) of the peptide linkage. The six enclosed atoms tend to be coplanar and assume a “transoid” arrangement. (Reprinted with

111°

1

118.5°

1.0

H

R

Amide plane

H

permission of John Wiley & Sons, Inc., from Voet, D. and Voet, J. G., Biochemistry, Second Edition. © 1995 Voet, D. and Voet, J. G.).

trans-Peptide group PHOTO CREDIT: (Reprinted with permission of John Wiley and Sons, Inc. from Voet, D. and Voet, J. G., Biochemistry, Second Edition. © 1995 Voet, D. and Voet, J. G.)

The carbon–nitrogen bond of the amide linkage is unusually short, indicating that resonance contributions of the type shown here are important: O N

O

C



N

Hindered rotation d

N

O

C



Contributing resonance structures for an amide

d

C

Hybrid resonance structure for an amide

Free rotation

●

●

The amide carbon–nitrogen bond, consequently, has considerable double-bond character (∼ 40%), and rotations of groups about this bond are severely hindered. Rotations of groups attached to the amide nitrogen and the carbonyl carbon are relatively free, however, and these rotations allow peptide chains to form different conformations.

12.9 | Secondary, Tertiary and Quaternary Structures of Proteins

A transoid arrangement of groups around the relatively rigid amide bond would cause the side-chain R groups to alternate from side to side of a single fully extended peptide chain:

0DLQFKDLQ

6LGHFKDLQ

Calculations show that such a polypeptide chain would have a repeat distance (i.e., distance between alternating units) of 7.2 Å. Fully extended polypeptide chains could hypothetically form a flat-sheet structure, with each alternating amino acid in each chain forming two hydrogen bonds with an amino acid in the adjacent chain: H

O

N

C C

H

R C

N

C

R

H

H

O

R

H

O

H

crowding

C N

O

H

O

N

C C

C H

R

N

R

H

H

R

H

O

crowding

C

N

C

C

H

C

C

N

O

H

Hypothetical flat-sheet structure (not formed because of steric hindrance)

However, this structure does not exist in naturally occurring proteins because of the crowding that would exist between R groups. If such a structure did exist, it would have the same repeat distance as the fully extended peptide chain, that is, 7.2 Å. ●

Many proteins incorporate a β sheet or β configuration (Fig. 12.14).

In a β sheet structure, slight bond rotations from one planar amide group to the next relieve the steric strain from small- and medium-sized R groups. This allows amide groups on adjacent polypeptide segments to form hydrogen bonds between the chains (see Fig. 12.14). The β sheet structure has a repeat distance of 7.0 Å between amide groups in a chain. The predominant secondary structure in silk fibroin (48% glycine and 38% serine and alanine residues) is the β sheet. ●

The α helix is also a very important secondary structure in proteins (Fig. 12.15).

The α helix of a polypeptide is right-handed with 3.6 amino acid residues per turn. Each amide group in the chain has a hydrogen bond to an amide group at a distance of three amino acid residues in either direction, and the R groups all extend away from the axis of the helix. The repeat distance of the α helix is 5.4 Å. The α-helical structure is found in many proteins; it is the predominant structure of the polypeptide chains of fibrous proteins such as myosin, the protein of muscle, and of α-keratin, the protein of hair, unstretched wool, and nails. Helices and pleated sheets account for only about one-half of the structure of the average globular protein. The remaining polypeptide segments have what is called a coil or loop conformation. These

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7.0 Å PHOTO CREDIT: (Illustration, Irving Geis. Image from the Irving Geis Collection, HHMI. Rights owned by Howard Hughes Medical Institute. Not to be reproduced without permission.)

Figure 12.14 The β sheet or β configuration of a protein. (Illustration, Irving Geis. Image from the Irving Geis Collection, HHMI. Rights owned by Howard Hughes Medical Institute. Not to be reproduced without permission.)

nonrepetitive structures are not random; they are just more difficult to describe. Globular proteins also have stretches, called reverse turns or β bends, where the polypeptide chain abruptly changes direction. These often connect successive strands of β sheets and almost always occur at the surface of proteins. ●

The locations of the side chains of amino acids of globular proteins are usually those that we would expect from their polarities:

1. Residues with nonpolar, hydrophobic side chains, such as valine, leucine, isoleucine, methionine, and phenylalanine, are almost always found in the interior of the protein, out of contact with the aqueous solvent. (These hydrophobic interactions are largely responsible for the tertiary structure of proteins that we discuss in the next section.) 2. Side chains of polar residues with positive or negative charges, such as arginine, lysine, aspartic acid, and glutamic acid, are usually on the surface of the protein in contact with the aqueous solvent. 3. Uncharged polar side chains, such as those of serine, threonine, asparagine, glutamine, tyrosine, and tryptophan, are most often found on the surface, but some of these are found in the interior as well. When they are found in the interior, they are virtually all hydrogen bonded to other similar residues. Hydrogen bonding apparently helps neutralize the polarity of these groups. Certain peptide chains assume what is called a random coil arrangement, a structure that is flexible, changing, and statistically random. Synthetic polylysine, for example, exists as a random coil and does not normally form an α helix. At pH 7, the ε-amino groups of the lysine residues are positively charged, and, as a result, repulsive forces between them are so large that they overcome any stabilization that would be gained through hydrogen bond formation of an α helix. At pH 12, however, the ε-amino groups are uncharged and polylysine spontaneously forms an α helix.

Figure 12.15 representation PHOTO CREDIT:A (Illustration, Irving Geis. Image α from the Irvingstructure Geis Collection,of HHMI. of the -helical a Rights owned by Howard Hughes Medical Institute. polypeptide. Hydrogen bonds Not to be reproduced without permission.) are denoted by dashed lines. (Illustration, Irving Geis. Image from the Irving Geis Collection, HHMI. Rights owned by Howard Hughes Medical Institute. Not to be reproduced without permission.)

12.9 | Secondary, Tertiary and Quaternary Structures of Proteins

The presence of proline or hydroxyproline residues in polypeptide chains produces another striking effect: because the nitrogen atoms of these amino acids are part of five-membered rings, the groups attached by the nitrogen–α carbon bond cannot rotate enough to allow an α-helical structure. Wherever proline or hydroxyproline occur in a peptide chain, their presence causes a kink or bend and interrupts the α helix.

12.9B Tertiary Structure ●

The tertiary structure of a protein is the overall three-dimensional shape that arises from all of the secondary structures of its polypeptide chain.

Proteins typically have either globular or fibrous tertiary structures. These tertiary structures do not occur randomly. Under the proper environmental conditions the tertiary structure of a protein occurs in one particular way—a way that is characteristic of that particular protein and one that is often highly important to its function. The sequence of amino acids (primary structure) ultimately determines which folding pattern is selected, so both secondary and tertiary structures depend on primary structure. Various forces involved in stabilizing tertiary structures are hydrogen bonding, van der Waals and electrostatic forces of attraction, including the disulfide bonds of the primary structure. ●

One characteristic of most proteins is that the folding takes place in such a way as to expose the maximum number of polar (hydrophilic) groups to the aqueous environment and enclose a maximum number of nonpolar (hydrophobic) groups within its interior.

The overall three-dimensional shape of a protein is referred to as its tertiary structure. In Fig. 12.16 the tertiary structure of a protein is shown, with the blue arrows representing β-sheets, red coils representing α-helices, and the yellow rope representing other twists and turns of the polypeptide chain.

Figure 12.16 Tertiary structure. [Source: PDB ID: 1AV5, http://www.pdb.org. Lima, C.D., Klein, M.G., Hendrickson, W.A. Structure-based analysis of catalysis and substrate definition in the HIT protein family. Science 278, pp. 286 (1997). Berman, H.M., Westbrook, J., Feng, Z., Gilliland, G., Bhat, T.N., Weissig, H., Shindyalov, I.N., Bourne, P.E. The Protein Data Bank. Nucleic Acids Research, 28, pp. 235–242 (2000).]

The soluble globular proteins tend to be much more highly folded than fibrous proteins. Myoglobin an example of a globular protein. However, fibrous proteins also have a tertiary structure; the α-helical strands of α-keratin, for example, are wound together into a “superhelix.” The superhelix makes one complete turn for each 35 turns of the α helix. The tertiary structure does not end here, however. Even the superhelices can be wound together to give a ropelike structure of seven strands.

12.9C Quaternary Structure Many proteins exist as stable and ordered noncovalent aggregates of more than one polypeptide chain. The overall structure of a protein having multiple subunits is called its quaternary structure. The quaternary

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structure of hemoglobin, for example, involves four subunits. Some proteins require more than one polypeptide chain to be biologically active. Quaternary structure refers to the arrangement of these chains. Human insulin consists of two polypeptide chains: the A chain has 21 amino acid residues and the B chain has 30 residues and these are linked by disulphide bonds. Disulphide bond B A

Figure 12.17 Quaternary structure. [Source: Hua, QX., Gozani, S.N., Chance, R.E., Hoffmann, J.A., Frank, B.H., Weiss, M.A., The Protein Data Bank. Nat. Struct. Biol. 2(2), 129–38 (1995). http://www.pdb.org

12.9D Hemoglobin: A Conjugated Protein Some proteins, called conjugated proteins, contain as a part of their structure a nonprotein group called a prosthetic group. An example is the oxygen-carrying protein hemoglobin. Each of the four polypeptide chains of hemoglobin is bound to a prosthetic group called heme (Fig. 12.18). The four polypeptide chains of hemoglobin are wound in such a way as to give hemoglobin a roughly spherical shape (Fig. 12.19). Moreover, each heme group lies in a crevice with the hydrophobic vinyl groups of its porphyrin structure surrounded by hydrophobic side chains of amino acid residues. The two propanoate side chains of heme lie near positively charged amino groups of lysine and arginine residues. O

Fe

D -subunit

HO

D -subunit N

N

E -subunit

Fe N

N

Fe

HO O

Figure 12.18 The structure of heme, the prosthetic group of hemoglobin. Heme has a structure derived from the heterocyclic ring, porphyrin. The iron of heme is in the ferrous (2+) oxidation state.

E -subunit

Fe

Figure 12.19 Hemoglobin. The four subunits are marked along with iron atoms. (PDB ID: IOUU, http://www.pdb.org. Tame, J. R., Wilson, J. C., Weber, R. E. The crystal structures of trout Hb I in the deoxy and carbonmonoxy forms. J. Mol. Biol. Volume 259, Issue 4, pp. 749–760, 1996.)

12.10 DENATURATION OF PROTEINS The functions and properties of a protein arise from a combination of its secondary, tertiary, and quaternary structures that give the protein its particular shape and conformation. The protein with a unique three-dimensional structure found in a living system showing biological activity is called a native protein.

12.11 | Introduction to Enzymes

Denaturation is any change in protein conformation caused by disruption of the non-covalent forces and disulphide bonds responsible for maintaining secondary, tertiary and quaternary structures (Fig. 12.20). Therefore, denaturation causes disruption of secondary and tertiary structures, but primary structure is maintained. A loss of biological activity normally accompanies denaturation, and this process is reversible only if the changes that take place are minor. Denaturation can be caused by a variety of factors, including changes in temperature or pH, agitation and the use of detergents or soaps. An increase in temperature is associated with an increase in kinetic energy, and this increased motion can be enough to disrupt hydrogen bonds and other non-covalent interactions. Varying pH affects protein shape, in part, because the charges on amino acid side chains involved in salt bridges may disappear. Heat, for example, breaks apart hydrogen bonds, so boiling a protein destroys its α-helical and β-pleated sheet structures. The polypeptide chains of globular proteins unfold when heated; the unraveled proteins can then bond strongly to each other and precipitate or coagulate. This is what happens when an egg is boiled and the “liquid” white of the egg is turned into a “solid.”

Denaturing agent Native protein (active form)

Denatured protein (inactive form)

Figure 12.20 Denaturation.

12.11 INTRODUCTION TO ENZYMES ●

The reactions of cellular metabolism are mediated by remarkable biological catalysts called enzymes.

Enzymes have the ability to bring about vast increases in the rates of reactions; in most instances, the rates of enzyme-catalyzed reactions are faster than those of uncatalyzed reactions by factors of 106–1012. For living organisms, rate enhancements of this magnitude are important because they permit reactions to take place at reasonable rates, even under the mild conditions that exist in living cells (i.e., approximately neutral pH and a temperature of about 35 °C). ●

Enzymes show remarkable specificity for their substrates and for formation of specific products.

The specificity of enzymes is far greater than that shown by most chemical catalysts. In the enzymatic synthesis of proteins, for example, polypeptides consisting of well over 1000 amino acid residues are synthesized virtually without error. It was Emil Fischer’s discovery, in 1894,

Carbonic anhydrase PHOTO CREDIT: (PDB IDis CA2, Carbonic anhydrase an enhttp://www.pdb.org. Eriksson, Jones, zyme that catalyzes the Liljas, Proteins: Structure, Function foland Genetics, Volume 4,H Issue lowing reaction: O4,+ CO2 2 1988,   pp. H 274–282.) CO . (PDB ID CA2,

  2 3 http://www.pdb.org. Eriksson, Jones, Liljas, Proteins: Structure, Function and Genetics, Volume 4, Issue 4, 1988, pp. 274–282.)

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of the ability of enzymes to distinguish between α- and β-glycosidic linkages that led him to formulate his lock-and-key hypothesis for enzyme specificity. ●

●

●

According to the lock-and-key hypothesis, the specificity of an enzyme (the lock) and its substrate (the key) comes from their geometrically complementary shapes. In an enzyme-catalyzed reaction, the enzyme and the substrate combine to form an enzyme–substrate complex. Formation of the enzyme–substrate complex often induces a conformational change in the enzyme called an induced fit that allows it to bind the substrate more effectively.

Binding of the substrate can cause certain of its bonds to become strained, and therefore more easily broken. The product of the reaction usually has a different shape from the substrate, and this altered shape, or in some instances the intervention of another molecule, causes the complex to dissociate. The enzyme can then accept another molecule of the substrate, and the whole process is repeated:     Enzyme + substrate    enzyme–substrate complex    enzyme + product ●

The place where a substrate binds to an enzyme and where the reaction takes place is called the active site.

The noncovalent forces that bind the substrate to the active site are the same forces that account for the conformations of proteins: dispersion forces, electrostatic forces, hydrogen bonding, and hydrophobic interactions. The amino acids located in the active site are arranged so that they can interact specifically with the substrate. ●

Reactions catalyzed by enzymes are stereospecific because enzymes are chiral.

The specificity of enzymes arises in the way enzymes bind their substrates. An α-glycosidase will only bind the α stereoisomeric form of a glycoside, not the β form. Enzymes that metabolize sugars bind only d sugars; enzymes that synthesize most proteins bind only l amino acids; and so on. Although enzymes catalyze reactions stereospecifically, they often vary considerably in what is called their geometric specificity. By geometric specificity, we mean a specificity that is related to the identities of the chemical groups of the substrates. Some enzymes will accept only one compound as their substrate. Others, however, will accept a range of compounds with similar groups. Carboxypeptidase A, for example, will hydrolyze the C-terminal peptide from all polypeptides as long as the penultimate residue is not arginine, lysine, or proline and as long as the next preceding residue is not proline. Chymotrypsin, a digestive enzyme that catalyzes the hydrolysis of peptide bonds, will also catalyze the hydrolysis of esters. O

O R

N

R 

H2O

chymotrypsin –

R

O





H3N

R

H Peptide

O

O R

O

R  H2O

chymotrypsin

R

OH  HO

R

Ester

●

A compound that can negatively alter the activity of an enzyme is called an inhibitor. A competitive inhibitor is a compound that competes directly with the substrate for the active site.

12.11 | Introduction to Enzymes

The Chemistry of...

SOME CATALYTIC ANTIBODIES

A

reactions, allowed antibodies to be generated against these molecules (called haptens), and then isolated the resulting antibodies. The antibodies thus produced are catalysts when actual substrate molecules are provided. The following are examples of haptens used as transition state analogs to elicit catalytic antibodies for a Claisen rearrangement, hydrolysis of a carbonate, and a Diels–Alder reaction. The reaction catalyzed by the antibody generated from each hapten is shown as well.

ntibodies are chemical warriors of the immune system. Each antibody is a protein produced specifically in response to an invading chemical species (e.g., molecules on the surface of a virus or pollen grain). The purpose of antibodies is to bind with these foreign agents and cause their removal from the organism. The binding of each antibody with its target (the antigen) is usually highly specific. One way that catalytic antibodies have been produced is by prompting an immune response to a chemical species resembling the transition state for a reaction. According to this idea, if an antibody is created that preferentially binds with a stable molecule that has a transition state-like structure, other molecules that are capable of reaction through this transition state should, in principle, react faster as a result of binding with the antibody. (By facilitating association of the reactants and favoring formation of the transition state structure, the antibody acts in a way similar to an enzyme.) In stunning fashion, precisely this strategy has worked to generate catalytic antibodies for certain Diels–Alder reactions, Claisen rearrangements, and ester hydrolyses. Chemists have synthesized stable molecules that resemble transition states for these

A hapten related to(PDB theID: Diels–Alder adduct from PHOTO CREDIT: 1A4K, http://www.pdb.org. Romesberg, F. E., Spiller, B., Schultz, P. G., Stevens,within R. C. a cyclohexadiene and maleimide, bound Immunological origins of binding and catalysis in a Diels–Alderase catalytic antibody. (PDB ID: 1A4K, Diels–Alderase antibody. Science 279, pp. 1929–1933, 1998.)

http://www.pdb.org. Romesberg, F. E., Spiller, B., Schultz, P. G., Stevens, R. C. Immunological origins of binding and catalysis in a Diels–Alderase antibody. Science 279, pp. 1929–1933, 1998.)

Claisen Rearrangement O2C



O O

R

CO2

Hapten

N H N2

OR O2 C

CO2

‡



CO2

O2C



O

O CO2



O OH

CO2 OH

OH

Transition state

(continues on next page)

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Chapter 12 | Biomolecules Carbonate Hydrolysis O

Hapten

O2N

2

P

O

O

O

OH

O O2N

O

C

O OCH3

O2 N

O

1 HO2

OH C

‡

OCH3

Transition state O2 O2N

O

OH C

OCH3

O2 1 CO2 1 CH3OH

O2N

Diels–Alder Reaction O C

Hapten

N

CH3 CH3

Fe H N

CO2H

‡

NH

N

1 O

O O

CH2 Diene

CH3 CH3

H H RO2CNH

CON(CH3)2

COOH Dienophile

Transition state

O C

CH3

N

CH3 O

N H

CH2

COOH

O Exo product

This marriage of enzymology and immunology, resulting in chemical offspring, is just one area of exciting research at the interface of chemistry and biology.

12.12 HORMONES Hormones function as chemical messengers in the body. They are produced in the endocrine glands and travel to the organs and tissues in the blood stream. Chemically, hormones maybe steroids (e.g., estrogen and androgen); polypeptides (e.g., insulin and endorphins) or amino acid derivatives (e.g., epinephrine and norepinephrine). Hormones affect a number of functions in the body including growth and development; metabolism; sexual functions and reproduction. They also affect moods. They react slowly but are needed only in very

12.13 | Vitamins

small quantities to cause big changes at the cellular level and in the body. Some important hormones with their functions are: 1. Insulin and glucagon: These regulate the level of glucose in the blood. Insulin is released when the glucose level rises in the blood and glucagon increases the release of glucose when its level falls in the blood. 2. Epinephrine and norepinephrine: These mediate response to external stimuli. 3. Thyroxine: This is produced in the thyroid gland and controls metabolism. Low levels of thyroxine cause hypothyroidism which results in obesity and general tiredness, while high levels cause hyperthyroidism. The condition can be controlled by maintaining iodine level in the blood by intake of “iodized salt”. 4. Adrenaline and noradrenaline: These two hormones are secreted in the medulla of the adrenal gland. Released into the blood stream, adrenaline causes an increase in blood pressure, a strengthening of the heart rate, and a widening of the passages of the lungs. Noradrenaline also causes increase in blood pressure and it is involved in the transmission of impulses from one end of the nerve fiber to the next. These are also called epinephrine and norepinephrine, respectively. 5. Steroid hormones: These regulate tissue growth and reproductive processes and are secreted by gonads and adrenal cortex. Androgens (testosterone and androsterone) are male sex hormones produced in the testes. They control development of secondary male characteristics such as facial hair, deep voice, etc. Estrogens (estradiol and estrone) and progestin are female sex hormones produced in the ovaries. Estrogens are responsible for development of secondary female characteristics and regulating menstrual cycle. Progestin prepares the uterus for implantation of the fertilized egg. Hormones released by the cortex of adrenal glands are called adrenocortical hormones and they are characterized by presence of a hydroxyl or a carbonyl group. Examples include cortisone, cortisol, glucocorticoids and mineralocorticoids. Glucocorticoids control carbohydrate metabolism, inflammatory reaction and reactions to stress, whereas mineralocorticoids control excretion of water and salts by the kidneys. Disruption in the function of adrenal cortex causes hormonal imbalance and results in Addison’s disease which is characterized by hypoglycemia, weakness and increases stress level.

12.13 VITAMINS The term vitamine originates from vital + amine as the early compounds identified carried amino groups. Not all vitamins carry amino groups so instead of vitamine it is now known as vitamin. They are organic compounds required in small amounts in the diets of animals in order to ensure healthy growth and reproduction. They are essential nutrients because the animal cannot synthesize such compounds in amounts adequate for its daily needs. The absence of a vitamin in the diet, or its poor absorption from the digestive tract, usually produces a disease with characteristic symptoms. Plants and some bacteria (present in the gut) also produce vitamins which can be used by our bodies.

12.13A Classification of Vitamins English alphabets are used to denote the vitamins such as A, B, C, D, E, etc. Some vitamins, such as vitamin B are also subdivided into subgroups such as B1, B2, B3, etc. Vitamins can be classified according to their solubility into two groups as fat soluble or water soluble. 1. Fat soluble vitamins: Vitamins A, D, E and K are fat soluble and water insoluble. They are stored in adipose tissue and liver. 2. Water soluble vitamins: Vitamins of B group and vitamin C are water soluble. They cannot be stored in the body (except vitamin B12) and hence their intake should be regular. A disease caused by chronic or long-term deficiency of a vitamin is called avitaminosis. It may also be caused by a defect in metabolic conversion of that vitamin. These are named by the same letter as the vitamin. The condition caused by over-retention of fat-soluble vitamins in the body is called hypervitaminosis. Table 12.2 lists the vitamins, their sources and the diseases caused by their deficiencies.

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Table 12.2 Sources of vitamins and deficiency diseases Vitamin

Sources

Functions

Deficiency Diseases

Vitamin A

Carrots, green leafy vegetables, fish, liver oil, milk, butter

Promotes growth and vision and increases resistance to diseases.

Night blindness, Xerophthalmia

Vitamin B1 (thiamine)

Milk, green vegetables, cereal grains, yeast

Helps convert sugar and starches Beri beri, polyneuritis into energy; promotes digestion, strong heart muscle, child growth.

Vitamin B2 (riboflavin)

Milk, liver, kidney, egg white, yeasts

Aids in releasing energy to body cells; enables utilization of fats, proteins and sugars.

Dermatitis, impaired growth and reproduction, cheilosis, digestive disorders, burning skin sensation

Vitamin B3 (nicotinic acid)

Meat products

Promotes energy metabolism, proper digestion, cell respiration and healthy nervous system.

Pellagra, dermatitis, black tongue (in dogs)

Vitamin B6 (pyridoxine)

Milk, egg yolk, cereal grains, grams, yeast

Aids metabolism of proteins, carbohydrates and fats; aids chemical balance between blood and tissue; prevents water retention; builds haemoglobin.

Neurological disorders, dermatitis, convulsions

Vitamin B12

Curd, meat, fish, egg

Promotes energy metabolism, helps in for formation of red blood cells; builds nucleic acid; helps nervous system.

Pernicious anaemia

Vitamin C (ascorbic acid)

Citrus fruits and green leafy vegetables

Increases resistance of the body towards diseases, improves iron absorption and aids formation of collagen to hold the cells together and for healthy teeth, gums and blood vessels.

Scurvy

Vitamin D

Sunlight, fish oil, egg yolk

Promotes absorption and use of calcium and phosphate for healthy bones and teeth.

Rickets and osteomalacia

Vitamin E

Sunflower oil, wheat germ oil

Protects red blood cells and helps prevent destruction of vitamins A and C by destructive oxidation.

Fragility of RBCs, muscular weakness

Vitamin K

Green leafy vegetables, fish meal

Helps in normal blood clotting and synthesis of proteins found in plasma, bone and kidneys.

Increased time for blood clotting

PART III: NUCLEIC ACIDS Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are molecules that carry genetic information in cells. DNA is the molecular archive of instructions for protein synthesis. RNA molecules transcribe and translate the information from DNA for the mechanics of protein synthesis. The storage of genetic information, its passage from generation to generation, and the use of genetic information to create the working parts of the cell all depend on the molecular structures of DNA and RNA. For these reasons, we shall focus our attention on the structures and properties of these nucleic acids and of their components, nucleotides and nucleosides.

12.13 | Vitamins

DNA is a biological polymer composed of two molecular strands held together by hydrogen bonds. Its overall structure is that of a twisted ladder with a backbone of alternating sugar and phosphate units and rungs made of hydrogen-bonded pairs of heterocyclic amine bases (Fig. 12.21). DNA molecules are very long polymers. If the DNA from a single human cell were extracted and laid straight end-to-end, it would be roughly a meter long. To package DNA into the microscopic container of a cell’s nucleus, however, it is supercoiled and bundled into the 23 pairs of chromosomes with which we are familiar from electron micrographs.

Coiled DNA (a) Chromosome

(b) DNA double helix

Guanine

(c) The four bases

Thymine Cytosine Adenine

Sugar–phosphate backbone of DNA

Figure 12.21 The basics of genetics. Each cell in the human body (except red blood cells) contains PHOTO CREDITS: (Science and Technology Review, November 1996, “The Human 23 pairs of chromosomes.Genome Chromosomes are inherited: each parent contributes one chromosome per Project,” https://www.llnl.gov/str/Ashworth.html. Credit must be given to Linda pair to their children. (a) Each chromosome is made up of a tightly coiled strand Ashworth, the University of California, Lawrence Livermore National Laboratory, and theof DNA. The structure Department of Energy underfamiliar whose auspices the work was performed, whenIfthis information of DNA in its uncoiled state reveals (b) the double-helix shape. we picture DNA as a or a reproduction of it is used.) twisted ladder, the sides, made of sugar and phosphate molecules, are connected by (c) rungs made of heterocyclic amine bases. DNA has four, and only four, bases—adenine (A), thymine (T), guanine (G), and cytosine (C)—that form interlocking pairs. The order of the bases along the length of the ladder is called the DNA sequence. Within the overall sequence are genes, which encode the structure of proteins. (Science and Technology Review, November 1996, “The Human Genome Project,” https://www.llnl. gov/str/Ashworth.html. Credit must be given to Linda Ashworth, the University of California, Lawrence Livermore National Laboratory, and the Department of Energy under whose auspices the work was performed, when this information or a reproduction of it is used.)

Four types of heterocyclic bases are involved in the rungs of the DNA ladder, and it is the sequence of these bases that carries the information for protein synthesis. Human DNA consists of approximately 3 billion base pairs. In an effort that marks a milestone in the history of science, a working draft of the sequence of the 3 billion base pairs in the human genome was announced in 2000. A final version was announced in 2003, the 50th anniversary of the structure determination of DNA by Watson and Crick. ● ●

Each section of DNA that codes for a given protein is called a gene. The set of all genetic information coded by DNA in an organism is its genome.

There are approximately 30,000–35,000 genes in the human genome. The set of all proteins encoded within the genome of an organism and expressed at any given time is called its proteome.

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12.14 NUCLEOTIDES AND NUCLEOSIDES Mild degradations of nucleic acids yield monomeric units called nucleotides. A general formula for a nucleotide and the specific structure of one called adenylic acid are shown in Fig. 12.22.

Heterocyclic base

O HO

P

O 5

OH

N O

3

N 5

O  -N-Glycosidic linkage

HO

P OH

1

4

NH2

7

2

N1

8

O 5

9

O

N 4 N

2

3

1

4 3

OH OH

6

2

OH OH

A nucleotide

Adenylic acid

(a)

(b)

Figure 12.22 (a) General structure of a nucleotide obtained from RNA. The heterocyclic base is a purine or pyrimidine. In nucleotides obtained from DNA, the sugar component is 2 ′-deoxy-d-ribose; that is, the } OH at position 2 ′ is replaced by } H. The phosphate group of the nucleotide is shown attached at C5′; it may instead be attached at C3′. In DNA and RNA a phosphodiester linkage joins C5′ of one nucleotide to C3′ of another. The heterocyclic base is always attached through a β-N-glycosidic linkage at C1′. (b) Adenylic acid, a typical nucleotide. Complete hydrolysis of a nucleotide furnishes: 1. A heterocyclic base from either the purine or pyrimidine family. 2. A five-carbon monosaccharide that is either d-ribose or 2-deoxy-d-ribose. 3. A phosphate ion. The central portion of the nucleotide is the monosaccharide, and it is always present as a five-membered ring, that is, as a furanoside. The heterocyclic base of a nucleotide is attached through an N-glycosidic linkage to C1′ of the ribose or deoxyribose unit, and this linkage is always β. The phosphate group of a nucleotide is present as a phosphate ester and may be attached at C5′ or C3′. (In nucleotides, the carbon atoms of the monosaccharide portion are designated with primed numbers, i.e., 1′, 2′, 3′, etc.) Removal of the phosphate group of a nucleotide converts it to a compound known as a nucleoside (Section 12.6A). The nucleosides that can be obtained from DNA all contain 2-deoxy-d-ribose as their sugar component and one of four heterocyclic bases: adenine, guanine, cytosine, or thymine: NH2 N N

N N

H

O N N

N N

H

Adenine (A)

Guanine (G) Purines

H NH2

NH2

O H3C

N N H

N N

O

Cytosine (C)

H O

H

Thymine (T) Pyrimidines

The nucleosides obtained from RNA contain d-ribose as their sugar component and adenine, guanine, cytosine, or uracil as their heterocyclic base.

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12.14 | Nucleotides and Nucleosides

O N

Uracil replaces thymine in an RNA nucleoside (or nucleotide). Some nucleosides obtained from specialized forms of RNA may also contain other, but similar, purines and pyrimidines.

N

H O

H Uracil (a pyrimidine)

The heterocyclic bases obtained from nucleosides are capable of existing in more than one tautomeric form. The forms that we have shown are the predominant forms that the bases assume when they are present in nucleic acids. Adenine

Guanine

NH2 N

HO

O N

N

N

O

Cytosine

HO

N

HO

O

N

N

H3C

N HO

HO

2′-Deoxyadenosine

O

NH2

H NH2

N

Thymine

O

N

HO

2′-Deoxyguanosine

O

HO

H

N O

N

O

HO

2′-Deoxycytidine

2′-Deoxythymidine

Figure 12.23 Nucleosides that can be obtained from DNA. DNA is 2′-deoxy at the position where the blue shaded box is shown. RNA (see Fig. 12.24) has hydroxyl groups at that location. RNA has a hydrogen where there is a methyl group in thymine, which in RNA makes the base uracil (and the nucleoside uridine). The names and structures of the nucleosides found in DNA are shown in Fig. 12.23; those found in RNA are given in Fig. 12.24. Adenine

Guanine

NH2 N HO

HO

N

N

N

O

O

HO

N

O

N

Uracil

O

NH2 N O

HO

N

OH

Cytidine

NH2

Guanosine

Cytosine

HO

N

H

OH

HO

OH

Adenosine

N

N O

HO

O

HO

N

OH

Uridine

H O

Figure 12.24 Nucleosides that can be obtained from RNA. DNA (see Fig. 12.23) has hydrogen atoms where the red hydroxyl groups of ribose are shown (DNA is 2′-deoxy with respect to its ribose moiety).

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12.15 DEOXYRIBONUCLEIC ACID: DNA 12.15A Primary Structure Nucleotides bear the same relation to a nucleic acid that amino acids do to a protein: they are its monomeric units. The connecting links in proteins are amide groups; in nucleic acids they are phosphate ester linkages. Phosphate esters link the 3′-OH of one ribose (or deoxyribose) with the 5′-OH of another. This makes the nucleic acid a long unbranched chain with a “backbone” of sugar and phosphate units with NH2

5 end

N

O

N

O

5

N N

Adenine

pentose

O

3

H3C

O



P

O

O

N

O

5

H

N

O

phosphate

O Thymine

pentose

N P

O

N

O

5

H

N

O O

base

O

3



base

NH2

N

phosphate

Guanine

O

O

base

N

O 

pentose

NH2

3

P

O

O

O

5

N

O

Cytosine

phosphate pentose

base

3

O O

P

O

3 end

O

Figure 12.25 A segment of one DNA chain showing how phosphate ester groups link the 3′- and 5′-OH groups of deoxyribose units. RNA has a similar structure with two exceptions: a hydroxyl replaces a hydrogen atom at the 2′ position of each ribose unit and uracil replaces thymine. heterocyclic bases protruding from the chain at regular intervals (Fig. 12.25). We would indicate the direction of the bases in Fig. 12.25 in the following way: 5′ ← A—T—G—C → 3′ It is, as we shall see, the base sequence along the chain of DNA that contains the encoded genetic information. The sequence of bases can be determined using enzymatic methods and chromatography.

12.15 | Deoxyribonucleic Acid: DNA

12.15B Secondary Structure Of prime importance to Watson and Crick’s proposal was an earlier observation (made in the late 1940s) by Erwin Chargaff that certain regularities can be seen in the percentages of heterocyclic bases obtained from the DNA of a variety of species. Chargaff pointed out that for all species examined: 1. The total mole percentage of purines is approximately equal to that of the pyrimidines, that is, (%G + %A)/(%C + %T) ≅ 1. 2. The mole percentage of adenine is nearly equal to that of thymine (i.e., %A/%T ≅ 1), and the mole percentage of guanine is nearly equal to that of cytosine (i.e., %G/%C ≅ 1). Chargaff also noted that the ratio which varies from species to species is the ratio (%A  + %T)/ (%G + %C). He noted, moreover, that whereas this ratio is characteristic of the DNA of a given species, it is the same for DNA obtained from different tissues of the same animal and does not vary appreciably with the age or conditions of growth of individual organisms within the same species. Watson and Crick also had X-ray data that gave them the bond lengths and angles of the purine and pyrimidine rings of model compounds. In addition, they had data from Franklin and Wilkins that indicated a repeat distance of 34 Å in DNA. It was the now-classic proposal of James Watson and Francis Crick (made in 1953 and verified shortly thereafter through the X-ray analysis by Maurice Wilkins) that gave a model for the secondary structure of DNA. The secondary structure of DNA is especially important because it enables us to understand how genetic information is preserved, how it can be passed on during the process of cell division, and how it can be transcribed to provide a template for protein synthesis. Reasoning from these data, Watson and Crick proposed a double helix as a model for the secondary structure of DNA. According to this model, two nucleic acid chains are held together by hydrogen bonds between base pairs on opposite strands. This double chain is wound into a helix with both chains sharing the same axis. The base pairs are on the inside of the helix, and the sugar–phosphate backbone is on the outside. The pitch of the helix is such that 10 successive nucleotide pairs give rise to one complete turn in 34 Å (the repeat distance). The exterior width of the spiral is about 20 Å, and the internal distance between 1′ positions of ribose units on opposite chains is about 11 Å. Using molecular-scale models, Watson and Crick observed that the internal distance of the double helix is such that it allows only a purine–pyrimidine type of hydrogen bonding between base pairs. Purine– purine base pairs do not occur because they would be too large to fit, and pyrimidine–pyrimidine base pairs do not occur because they would be too far apart to form effective hydrogen bonds. Specific base pairing also means that the two chains of DNA are complementary. Wherever adenine appears in one chain, thymine must appear opposite it in the other; wherever cytosine appears in one chain, guanine must appear in the other (Fig. 12.26). Notice that while the sugar–phosphate backbone of DNA is completely regular, the sequence of heterocyclic base pairs along the backbone can assume many different permutations. This is important because it is the precise sequence of base pairs that carries the genetic information. Notice, too, that one chain of the double strand is the complement of the other. If one knows the sequence of bases along one chain, one can write down the sequence along the other, because A always pairs with T and G always pairs with C. It is this complementarity of the two strands that explains how a DNA molecule replicates itself at the time of cell division and thereby passes on the genetic information to each of the two daughter cells.

12.15C Replication of DNA Just prior to cell division the double strand of DNA begins to unwind. Complementary strands are formed along each chain (Fig. 12.27). Each chain acts, in effect, as a template for the formation of its complement. When unwinding and replication are complete, there are two identical DNA molecules where only one had existed before. These two molecules can then be passed on, one to each daughter cell.

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C G

20 Å P T

S P S

P

S

C G

S

34 Å

S

P S S

C

P

S T P

C

G

C S P P

S

S

S

S

G

S P

Base

C

P CS S Sugar

P

Phosphate

P G G

S

P

G A

P S C

S

A

P S G

G

T

G

T P

C

3.4 Å

A

P

C

P

P

Major groove

G S

S

S

P

S

Minor S groove P P A S T S

S

G C

G P

S C P

P

A

S C

P S

P Purine

Pyrimidine

Figure 12.26 Diagram of the DNA double helix showing complementary base pairing. The arrows indicate the 3′ → 5′ direction.

12.16 RNA AND PROTEIN SYNTHESIS Soon after the Watson–Crick hypothesis was published, scientists began to extend it to yield what Crick called “the central dogma of molecular genetics.” This dogma stated that genetic information flows as follows: DNA → RNA → protein The synthesis of protein is, of course, all important to a cell’s function because proteins (as enzymes) catalyze its reactions. Even the very primitive cells of bacteria require as many as 3000 different enzymes. This means that the DNA molecules of these cells must contain a corresponding number of genes to direct the synthesis of these proteins. A gene is that segment of the DNA molecule that contains the information necessary to direct the synthesis of one protein (or one polypeptide).

12.16 | RNA and Protein Synthesis

G

S P

P S S

S

S

C P

S

S A T P P S P P S C G S

P

A

T C

A P S P P A S T

S S

G

C C

S

P

S

T P

P

S

G

G

S

P

C

P

P

S

S G

P

P

S

S

P

S A S T P P S P P S C G S P P S P–P–P–S A A T S P T P S–P–P–P A A S T S–P T P –P–P–P–P–S S P C G G S–P –P–P P C P S G C C S–P–P–P S P P S G C G S S P P P S S P P P P S A S T A S S T P P S S C C G G P P S S S P P C G C G S S P P T A S S T AP S P P P S S P P P P S S C G G C S P P S S T A T A S P P S S P P G C G C S S

S S

S S

S S

Figure 12.27 Replication of DNA. The double strand unwinds from one end and complementary strands are formed along each chain.

Protein synthesis takes place primarily in that part of the cell called the cytoplasm. Protein synthesis requires that two major processes take place; the first occurs in the cell nucleus, the second in the cytoplasm. The first is transcription, a process in which the genetic message is transcribed onto a form of RNA called messenger RNA (mRNA). The second process involves two other forms of RNA, called ribosomal RNA (rRNA) and transfer RNA (tRNA).

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PART IV: LIPIDS Lipids are compounds of biological origin that dissolve in nonpolar solvents, such as chloroform and diethyl ether. The name lipid comes from the Greek word lipos, for fat. Unlike carbohydrates and proteins, which are defined in terms of their structures, lipids are defined by the physical operation that we use to isolate them. Not surprisingly, then, lipids include a variety of structural types. Examples are the following: O O

R

CH3

O O

OH R′

OH

O O

CH(CH3)2 R″

A fat or oil (a triacylglycer ol)

Menthol (a terpenoid)

Vitamin A (a terpenoid)

O O

R O

O

CH3

CH3

O O–

H

CH3

R′

O P O

H

H

+

N(CH3)3

H H

HO

A lecithin (a phosphatide)

Cholesterol (a steroid)

12.17 FATTY ACIDS AND TRIACYLGLYCEROLS O

Only a small portion of the total lipid fraction obtained by extraction with a nonpolar solvent consists of long-chain carboxylic acids. Most of the carboxylic acids of biological origin are found as esters of glycerol, that is, as triacylglycerols (Fig. 12.28).* Triacylglycerols are the oils of plants and the fats of animal origin. They include such common substances as peanut oil, soybean oil, corn oil, sunflower oil, butter, lard, and tallow. ●

Triacylglycerols that are liquids at room temperature are generally called oils; those that are solids are called fats.

Triacylglycerols can be simple triacylglycerols in which all three acyl groups are the same. More commonly, however, the triacylglycerol is a mixed triacylglycerol in which the acyl groups are different.

O OH OH OH

O

O O

O

R R R

Glycerol

A triacylglycerol

(a)

(b)

Figure 12.28 (a) Glycerol. (b) A triacylglycerol. The groups R, R′, and R″ are usually long-chain alkyl groups. R, R′, and R″ may also contain one or more carbon–carbon double bonds. In a triacylglycerol R, R′, and R″ may all be different.

*In the older literature triacylglycerols were referred to as triglycerides, or simply as glycerides. In IUPAC nomenclature, because they are esters of glycerol, they should be named as glyceryl trialkanoates, glyceryl trialkenoates, and so on.

12.17 | Fatty Acids and Triacylglycerols ●

Hydrolysis of a fat or oil produces a mixture of fatty acids: O

O O O O

O O

R R′ R″

A fat or oil ●

R

OH 1. HO− in H2O, heat 2. H3O+

OH OH

+

R′

O O

R″

Glycerol

OH OH OH

Fatty acids

Most natural fatty acids have unbranched chains and, because they are synthesized from two-carbon units, they have an even number of carbon atoms.

Table 12.3 lists some of the most common fatty acids. Notice that in the unsaturated fatty acids in Table 12.3 the double bonds are all cis. Many naturally occurring fatty acids contain two or three double bonds. The fats or oils that these come from are called polyunsaturated fats or oils. The first double bond of an unsaturated fatty acid commonly occurs between C9 and C10; the remaining double bonds tend to begin with C12 and C15 (as in linoleic acid and linolenic acid). The double bonds, therefore, are not conjugated. Triple bonds rarely occur in fatty acids. The carbon chains of saturated fatty acids can adopt many conformations but tend to be fully extended because this minimizes steric repulsions between neighboring methylene groups. ●

●

Saturated fatty acids pack efficiently into crystals, and because dispersion force attractions are large, they have relatively high melting points. The melting points increase with increasing molecular weight. The cis configuration of the double bond of an unsaturated fatty acid puts a rigid bend in the carbon chain that interferes with crystal packing, causing reduced dispersion force attractions between molecules. Unsaturated fatty acids, consequently, have lower melting points.

Fatty acids known as omega-3 fatty acids are those where the third to last carbon in the chain is part of carbon-carbon double bond. Long-chain omega-3 fatty acids incorporated in the diet are believed to have beneficial effects on health. Figure 12.29 shows how the introduction of a single cis double bond affects the shape of a triacylglycerol and how catalytic hydrogenation can be used to convert an unsaturated triacylglycerol into a saturated one.

12.17A Hydrogenation of Triacylglycerols Solid commercial cooking fats are manufactured by partial hydrogenation of vegetable oils. The result is the familiar “partially hydrogenated fat” present in so many prepared foods. Complete hydrogenation of the oil is avoided because a completely saturated triacylglycerol is very hard and brittle. Typically, the vegetable oil is hydrogenated until a semisolid of appealing consistency is obtained. One commercial advantage of partial hydrogenation is to give the fat a longer shelf life. Polyunsaturated oils tend to react by autoxidation, causing them to become rancid. One problem with partial hydrogenation, however, is that the catalyst isomerizes some of the unreacted double bonds from the natural cis arrangement to the unnatural trans arrangement, and there is accumulating evidence that trans fats are associated with an increased risk of cardiovascular disease.

12.17B Biological Functions of Triacylglycerols The primary function of triacylglycerols in animals is as an energy reserve. When triacylglycerols are converted to carbon dioxide and water by biochemical reactions (i.e., when triacylglycerols are metabolized),

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Table 12.3 Common fatty acids mp (°C) Saturated Carboxylic Acids 54

O OH Myristic acid (tetradecanoic acid)

63

O OH Palmitic acid (hexadecanoic acid)

70

O OH Stearic acid (octadecanoic acid)

Unsaturated Carboxylic Acids 32

O OH Palmitoleic acid (cis-9-hexadecenoic acid)

4

O OH

Oleic acid (cis-9-octadecenoic acid)

−5

O OH

Linoleic acid (cis,cis-9,12-octadecadienoic acid)

−11

O Linolenic acid (cis,cis,cis-9,12,15-octadecatrienoic acid)





6

OH

−44

CO2H

DHA, an omega-3 fatty acid [(4Z,7Z,10Z,13Z,16Z,19Z)-4,7,10,13,16,19-docosahexaenoic acid]

D

D

CO2H

4

Arachidonic acid, an omega-6 fatty acid [(5Z,8Z,11Z,14Z)-5,8,11,14-eicosatetraenoic acid]

−49

12.17 | Fatty Acids and Triacylglycerols O O O O O O An unsaturated fat

H2, Ni

O

Figure 12.29 Two typical triacylglycerols, one unsaturated and one saturated. The cis double bond of the unsaturated triacylglycerol interferes with efficient crystal packing and causes an unsaturated fat to have a lower melting point. Hydrogenation of the double bond causes an unsaturated triacylglycerol to become saturated.

O O O O O A saturated fat

they yield more than twice as many kilocalories per gram as do carbohydrates or proteins. This is largely because of the high proportion of carbon–hydrogen bonds per molecule. In animals, specialized cells called adipocytes (fat cells) synthesize and store triacylglycerols. The tissue containing these cells, adipose tissue, is most abundant in the abdominal cavity and in the subcutaneous layer. Men have a fat content of about 21%, women about 26%. This fat content is sufficient to enable us to survive starvation for 2–3 months. By contrast, glycogen, our carbohydrate reserve, can provide only one day’s energy need. All of the saturated triacylglycerols of the body, and some of the unsaturated ones, can be synthesized from carbohydrates and proteins. Certain polyunsaturated fatty acids, however, are essential in the diets of higher animals. The amount of fat in the diet, especially the proportion of saturated fat, has been a health concern for many years. There is compelling evidence that too much saturated fat in the diet is a factor in the development of heart disease and cancer.

12.17C Saponification of Triacylglycerols ●

Saponification is the alkaline hydrolysis of triacylglycerols, leading to glycerol and a mixture of salts of long-chain carboxylic acids: O O

O R

OH

O O

O R9

O O

O2 Na1

R

R0

3NaOH H2O

OH

1

O2 Na1

R9 O

OH Glycerol

R0

O2 Na1

Sodium carboxylates “soap”

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Chapter 12 | Biomolecules

These salts of long-chain carboxylic acids are soaps, and this saponification reaction is the way most soaps are manufactured. The action of soaps and detergents is discussed in detail in Chapter 14.

12.17D Reactions of the Carboxyl Group of Fatty Acids Fatty acids, as we might expect, undergo reactions typical of carboxylic acids. They react with LiAlH4 to form alcohols, with alcohols and mineral acid to form esters, with bromine and phosphorus to form α-halo acids, and with thionyl chloride to form acyl chlorides: R

1. LiAlH4 2. H2O

OH

Long-chain alcohol

O R O R

OCH3

CH3OH, HA

Methyl ester

OH

O

Fatty acid Br2, P

R

OH Br

α-Halo acid SOCl2 pyridine

O R

Cl

Long-chain acyl chloride

12.17E Reactions of the Alkenyl Chain of Unsaturated Fatty Acids The double bonds of the carbon chains of fatty acids undergo characteristic alkene addition reactions: H H2, Ni

( )

m

Br

( )

( )

n

( )

m

HO

OH

1. OsO4 2. aq. NaHSO3

OH

m

Br

( )

OH

O

( )

n

OH

O

( )

n

H HBr

m

( )

OH

O

( )

n

O

O

( )

n

Br Br2

H

m

OH

+

Br

( )

n

H

O

( )

m

OH

12.18 | Terpenes and Terpenoids

12.18 TERPENES AND TERPENOIDS People have isolated organic compounds from plants since antiquity. By gently heating or by steam distilling certain plant materials, one can obtain mixtures of odoriferous compounds known as essential oils. These compounds have had a variety of uses, particularly in early medicine and in the making of perfumes. As the science of organic chemistry developed, chemists separated the various components of these mixtures and determined their molecular formulas and, later, their structural formulas. Even today these natural products offer challenging problems for chemists interested in structure determination and synthesis. Research in this area has also given us important information about the ways the plants themselves synthesize these compounds. ●

●

Hydrocarbons known generally as terpenes and oxygen-containing compounds called terpenoids are the most important constituents of essential oils. Most terpenes have skeletons of 10, 15, 20, or 30 carbon atoms and are classified in the following way: Number of Carbon Atoms 10

●

Class Monoterpenes

15

Sesquiterpenes

20

Diterpenes

30

Triterpenes

One can view terpenes as being built up from two or more C5 units known as isoprene units. Isoprene is 2-methyl-1,3-butadiene.

Isoprene and the isoprene unit can be represented in various ways: C C

C

2-Methyl-1,3-butadiene (isoprene)

C

C

or

An isoprene unit

Many terpenes also have isoprene units linked in rings, and others (terpenoids) contain oxygen:

Limonene (from oil of lemon or orange)

β -Pinene (from oil of turpentine)

CH 3 OH OH Geraniol (from roses and other flowers)

Menthol (from peppermint)

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Chapter 12 | Biomolecules

The carotenes are tetraterpenes. They can be thought of as two diterpenes linked in tail-to-tail fashion:

α -Carotene

β -Carotene

γ -Carotene

The carotenes are present in almost all green plants. In animals, all three carotenes serve as precursors for vitamin A, for they all can be converted to vitamin A by enzymes in the liver.

OH

Vitamin A

Natural rubber can be viewed as a 1,4-addition polymer of isoprene. In fact, pyrolysis degrades natural rubber to isoprene. Pyrolysis (Greek: pyros, a fire, + lysis) is the heating of a substance in the absence of air until it decomposes. The isoprene units of natural rubber are all linked in a head-to-tail fashion, and all of the double bonds are cis:

etc.

etc.

Natural rubber (cis-1,4-polyisoprene)

12.19 PHOSPHOLIPIDS AND CELL MEMBRANES Another large class of lipids are those called phospholipids. Most phospholipids are structurally derived from a glycerol derivative known as a phosphatidic acid. In a phosphatidic acid, two hydroxyl groups of glycerol are joined in ester linkages to fatty acids and one terminal hydroxyl group is joined in an ester linkage to phosphoric acid:

12.19 | Phospholipids and Cell Membranes

O O

R

From fatty acids

O O

R′

O O

P

OH

From phosphoric acid

OH A phosphatidic acid (a diacylglyceryl phosphate)

12.19A Phosphatides In phosphatides, the phosphate group of a phosphatidic acid is bound through another phosphate ester linkage to one of the following nitrogen-containing compounds: O

+

N(CH3)3 X –

HO

HO

NH2

O–

HO +

Choline

2-Aminoethanol (ethanolamine)

NH3

L-Serine

The most important phosphatides are the lecithins, cephalins, phosphatidylserines, and plasmalogens (a phosphatidyl derivative). Their general structures are shown in Table 12.4. Table 12.4 Phosphatides Lecithins

Cephalins O

O O

O

O

O

O

R

O

R′

O

+

N(CH3)3

O−

(from choline) R is saturated and R′ is unsaturated.

P O−

O P O O−

OR O

O O–

O +

NH3

(from L-serine) R is saturated and R′ is unsaturated.

R is CH CH(CH 2)nCH3 (This linkage is that of an α, β-unsaturated ether.)

R′

O O

O O

+

NH3

Plasmalogens

R R′

O

R′

(from 2-aminoethanol)

Phosphatidylserines O O

R

O

P O

O

O

O P O

+

NH3

O− (from 2-aminoethanol) or

OCH 2CH 2N+ (CH3)3 (from choline) R′ is an unsaturated fatty acid.

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Polar group

Nonpolar group O CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2

COCH2 O

CH3CH2CH2CH2CH2CH2CH2CH2CH

CHCH2CH2CH2CH2CH2CH2CH2

COCH O +

CH2OPOCH2CH2N(CH3)3 O–

(a)

(b)

Figure 12.30 (a) Polar and nonpolar sections of a phosphatide. (b) A phosphatide micelle or lipid bilayer. Phosphatides resemble soaps and detergents in that they are molecules having both polar and nonpolar groups (Fig. 12.30a). Like soaps and detergents, too, phosphatides “dissolve” in aqueous media by forming micelles. There is evidence that in biological systems the preferred micelles consist of threedimensional arrays of “stacked” bimolecular micelles (Fig. 12.30b) that are better described as lipid bilayers. On hydrolysis, sphingomyelins yield sphingosine, choline, phosphoric acid, and a C24 fatty acid called lignoceric acid. In a sphingomyelin this last component is bound to the } NH2 group of sphingosine. The sphingolipids do not yield glycerol when they are hydrolyzed. The cerebroside shown in Fig. 23.10 is an example of a glycolipid. Glycolipids have a polar group that is contributed by a carbohydrate. They do not yield phosphoric acid or choline when they are hydrolyzed. The sphingolipids, together with proteins and polysaccharides, make up myelin, the protective coating that encloses nerve fibers or axons. The axons of nerve cells carry electrical nerve impulses. Myelin has a function relative to the axon similar to that of the insulation on an ordinary electric wire (see the chapter opening vignette).

12.20 WAXES Most waxes are esters of long-chain fatty acids and long-chain alcohols. Waxes are found as protective coatings on the skin, fur, and feathers of animals and on the leaves and fruits of plants. Several esters isolated from waxes are the following:

DD

O

14

O

DD

14

Cetyl palmitate (from spermaceti)

DD

n

O O

DD

m

n = 24 or 26; m = 28 or 30 (from beeswax)

HO

DD

n

O O

DD

n = 16–28; m = 30 or 32 (from carnauba wax)

m

Solved Examples

SOLVED EXAMPLES Solution

1. Define each of the following: (a) Aldohexose (b) Mutarotation (c) Glycoside (d) Polysaccharides (e) D-sugars

Sucrose is a disaccharide consisting of D-glucose and D-fructose. These two monosaccharides are linked by a glycosidic bond that is hydrolyzed under acidic conditions. The resulting 1:1 mixture of D-glucose and D-fructose has a sweeter taste than sucrose alone. 4. Which disaccharides are reduced by NaBH4? (a) Sucrose (b) Lactose (c) Maltose

Solution (a) Aldohexose is a monosaccharide with six carbon atoms and aldehyde group in the chain structure. (b) Mutarotation occurs by a reversible ring opening of each anomer to the open-chain aldehyde, followed by reclosure. (c) Glycoside is a carbohydrate acetal, obtained in the reaction of monosaccharide hemiacetal with an alcohol under acidic conditions. (d) Polysaccharides are carbohydrates in which tens, hundreds, or even thousands of simple sugars are linked together through glycoside bonds. (e) D-sugars have the hydroxyl group at the lowest chiral carbon atom on the right in Fisher projection.

Solution In order for a sugar to be reduced by NaBH4, which is capable of reducing a carbonyl group to an alcohol, the sugar must contain at least one anomeric carbon that is in equilibrium with the open-chain form. For this to be the case, the sugar must have a hemiacetal. Maltose and lactose both contain a monosaccharide that is a hemiacetal, but sucrose does not. Both of the anomeric carbons in sucrose are glycosides (acetals). 5. What is the product of the following reaction? CHO

2. The structure below is a(n) _______. Ring A is a(n) _______. Ring B is a(n) _______. Ring C is a(n) _______. HO H

HO

H H

(1) (2) (3) (4)

H

O H

O H

OH

OH H OH

O H OH H

OH H OH O HO

H O

H OH OH H

OH Br2/H2O

H OH CH2OH

(1)

CO2H

(3)

CHO

H

OH

H

OH

H

OH

H

OH

HO H

Solution

3. In making candy or syrups from sugar, sucrose is boiled in water with a little acid, such as lemon juice. Why does the product mixture taste sweeter than the starting sucrose solution?

H

H

trisaccharide, pentose, furanose, alpha. trisaccharide, hexose, pyranose, beta. tetrasaccharide, pentose, furanose, alpha. disaccharide, hexose, pyranose, beta.

(2) The given structure on hydrolysis yields three molecules of monosaccharide, therefore, known as trisaccharide. Ring A contains six carbon atoms thus, is a hexos while ring B contains five carbon atoms therefore is a pyranose or pentose. Ring C is a β anomer as it has the } OH cis to the } CH2OH group.

OH

HO

H

OH

H

H

HO

OH

H

CH2OH

(2)

CH2OH

H OH CO2H

(4)

CO2H

H

OH

H

OH

H

OH

H

OH

HO H

H OH CH2OH

HO H

H OH CO2H

815

816

Chapter 12 | Biomolecules Solution

CHO

(1) Bromine water selectively oxidizes the } CHO group to a } CO2H group, converting the aldose to aldonic acid. CHO

CO2H

H

OH

H

OH

HO

Br2/H2O

H

H

H

OH

H

OH

HO

OH

H

H

CH2OH

CO2H

H

OH

H

OH

HO

HNO3

H

H

OH

OH

HO

H

O

(1) (2) (3) (4)

(3)

CHO H

OH

H

OH

H

OH

H

HO

OH

H

CH2OH

(2)

CH2OH

H

CO2H

(4)

CO2H

OH

H

OH

H

OH

H

OH

H

H OH CH2OH

OH

3 formic acids, 1 carbon dioxide, 2 formaldehyde 4 formic acids, 2 formaldehydes 4 formic acids, 1 carbon dioxide, 1 formaldehyde 4 formic acids, 2 carbon dioxides

HO H

(3) This is an example of periodate oxidative cleavage reaction. H C

O

C

OH

C

O

HO

C

H

H

C

OH

H

OH

H

HO

HlO4

Solution

OH

H

H

CH2OH HNO3

H

HO

OH

OH

H

OH

CO2H

H

CHO

OH

CH2OH

(1)

HO

CO2H

HO

H

H

OH

7. What are the products of the following reaction?

CHO

H

H

H

6. What is the product of the following reaction?

OH

OH

CH2OH

CH2OH

H

H

H OH CO2H

O 1

(4) This reaction involves oxidation of monosaccharides to dicarboxylic acids called aldaric acids.

CO2

1

(Carbon dioxide)

CH2OH

O C 1 3H OH (Formic acid)

8. What is the product of the following reaction? CHO H

OH

H

OH

HO

Solution

C 2H H (Formaldehyde)

5 IO2 4

H

H OH CH2OH

NaBH4

Solved Examples (1)

(3)

CO2H H

OH

H

OH

H

OH

H

OH

HO

H

H

HO

OH

CH2OH

H H

H

H

CH2OH

(2)

OH

(4)

H

O

OH

OH H

CH3

H

OH CH2

O

OH

H

OH

H

OH

HO

H

H H

OH

OH CH2OH HO H OH

OH

O

H

H

CH3

H

OH CH2

O H

H H

(2) This reaction involves the reduction of monosaccharides to alditols.

H

OH

H

OH

H

OH

H

OH

HO H

H OH

HO H

(1)

Solution (2) Sucrose, the most widely occurring disaccharide, is found in all photosynthetic plants and is obtained commercially from sugarcane or sugar beets. Lactose is a disaccharide present in the milk of humans, cows, and almost all other mammals. The chains of glycogen are much more highly branched and animals store energy as fats (triacylglycerols) as well as glycogen.

CH2OH HO H OH

OH

OCH3 H

CH2OH

H

O

H H

O

H H OH

(2)

CH2OH

9. The disaccharide _______ is found in table sugar. _______ is the sugar found in milk. _______ is the polysaccharide consisting of highly branched polymers of glucose that human and other animals use to store sugars in the body. (1) fructose, maltose, glycosaminoglycans (2) sucrose, lactose, glycogen (3) sucrose, maltose, starch (4) fructose, lactose, cellulose

H

HO

H OH

O

HO

OH HOCH2

HOCH2

CH2OH

H

11. Which of the following carbohydrates does not undergo mutarotation?

CH2OH

NaBH4

OCH3

OH

OH

Solution

CHO

H OH

H

OH

CO2H

CH2OH

CH2OH HO

There are three glycosidic bonds in the molecule as shown by the arrow.

H

H

OH

O

Solution

CO2H H

H

H OH

H

OH

H H

OH

H

OH CH2OH HO

CO2H

H

HO

10. How many glycosidic bonds exist in the following polysaccharide?

CHO

H OH

OH CH2OH

H

(3) HO

O H OH

H

H

OH

OH CH3

CH2OH H

(4) HO

O H OH

H

H

OH

OCH2CH3 H

Solution (4) For a carbohydrate to be able to undergo mutarotation, the anomeric carbon needs to be

817

818

Chapter 12 | Biomolecules a hemiacetal. Unlike glycosides (acetals of sugars), which are much more stable, hemiacetals are in equilibrium with their open-chain forms. 12. The following structure is a(n) _______. It contains the following amino acids, _______, _______ and _______.

1

H 3N

H

H

O

H N

H2N

1

H2N

O

Solution (4) Peptide is the name given to a short polymer of amino acids. A molecule containing 2 amino acids joined by an amide bond is called a dipeptide. Those containing 4 amino acids are called tetrapeptides. Serine OH O

H N

H2N

O Alanine

H N H

H

O

H N

OH

H

O

H

CO22 1

H2N

H

H

OH

H

(With cystine, both chirality centres are α-carbon atoms; thus, according to the problem, both must have the L-configuration, and no isomers of this type can be written.) (c) Diastereomers 14. (a) Describe three (3) functions of proteins. (b) Write a general formula for an α-amino acid. (c) What is the stereochemistry of all natural α-amino acids and how did this correlation come about? (d) What is the isoelectric point, pI? Solution (a) The functions for proteins are as follows: (i) To provide strength to cells and tissues. (ii) Act as hormones. (iii) Help in movement of molecule across membrane. 2

COO

O NH2

Phenylalanine

HO

HO

H

tripeptide, alanine, phenylalanine, arginine tripeptide, alanine, phenylalanine, glutamine tetrapeptide, alanine, threonine, glutamic acid tetrapeptide, alanine, serine, glutamine

H

H

and

NH2

(1) (2) (3) (4)

1

CH3

OH H

O

H3N

CH3 CO22

O

H N

N H

H

O

H

and

OH

OH

H

CO22

CO22

Glutamine

1

(b) H3N

13. (a) Which amino acids in Table 12.1 have more than one chirality center? (b) Write Fischer projections for the isomers of each of these amino acids that would have the L configuration at the α carbon. (c) What kind of isomers have you drawn in each case?

H

R α-amino acid

(c) The stereochemistry of all natural α-amino acids is the L-form. This correlation came about b7y comparison to L-glyceraldehyde in carbohydrates to L-alanine. O

Solution (a) Isoleucine, threonine, hydroxyproline, and cystine. CO22

CO22 1

1

H3N

H

(b) CH3

H

H3N

and

2

CH

HO

COO

H

1

H3N

H

H

H

CH3

CH2

CH2

CH3

CH3

CH2OH L-glyceraldehyde

CH3 L-alanine

(d) The isoelectric point, pI, is the pH at which amino acid exists only as a dipolar ion with zero net charge.

Solved Previous Years’ NEET Questions Solution

15. Draw zwitterion forms of these amino acids: (a) glycine (b) proline (c) tryptophan

(2)

Solution

O HO O P

O

1

(a) H3N

N H

1

O

N H

O

2

2

(b)

2

O

1

N H

1

O

H

NH3

O

N H O 2

2

O (c)

1

2

N H

H

NH3

1

O

NH3

59

O

O

N O

49

β -N-Glycosidic linkage

19

29 39 OH OH A nucleotide

O

O

1

O

1

2

O

O

2 OH

2

H

H3N

Heterocyclic base

18. _______ are formed by removing a _______ from a nucleotide. (1) purines, deoxyribose ring (2) nucleic acids, ribose ring (3) DNA, nucleic acid (4) nucleosides, phosphate Solution

N H

16. The isoelectric point of histidine is 7.64. Toward which electrode does histidine migrate during paper electrophoresis at pH 7.0? Solution If the pH is identical to the pI of the amino acid, the amino acid would have no net charge. When the pH is lower than the pI of amino acid, the amino acid is positively charged and will migrate towards the negative electrode. 17. _______ contain heterocyclic bases attached to a _______ that has a _______ attached to one of the OH’s. (1) nucleosides, 5 carbon monosaccharide, phosphate ion (2) nucleotides, 5 carbon monosaccharide, phosphate ion (3) nucleosides, 6 carbon monosaccharide, phosphate ion (4) nucleotides, 6 carbon monosaccharide, phosphate ion

(4) Removal of the phosphate group of a nucleotide converts it to a compound known as a nucleoside. 19. Fats are _______ that are _______ at room temperature. In general, oils are more _______ than the corresponding fat. (1) fatty acids, liquids, saturated (2) fatty acids, solids, unsaturated (3) triacylglycerols, liquids, saturated (4) triacylglycerols, solids, unsaturated Solution (4) Triacylglycerols that are liquids at room temperature are generally called oils; those that are solids are called fats. 20. If a fatty acid has 1 or more double bonds it is called a(n) _______ lipid. (1) saturated (3) monounsaturated (2) unsaturated (4) polyunsaturated Solution (2) Unsaturated fatty acids have one or more than one double bonds.

SOLVED PREVIOUS YEARS’ NEET QUESTIONS 1. RNA and DNA are chiral molecules, their chirality is due to (1) D-sugar component. (2) L-sugar component. (3) chiral bases. (4) chiral phosphate ester units. (AIPMT 2007)

Solution (1) Nucleic acids (DNA and RNA) are long-chain polymers of nucleotides and are also known as polynucleotides. Mild degradations of nucleic acids yield monomeric units called nucleotides. It consists of a heterocyclic base, a fivecarbon monosaccharide and a phosphate ion. The sugar moiety of RNA is β-D-ribose and

819

820

Chapter 12 | Biomolecules that of DNA is β-D-2-deoxyribose and these cause the chirality. 5

HOH2C H

H

3

1

2

OH

HOH2C

OH

O

4

H

5

4

H

H

OH

H

H

3

1

2

OH

β-D-Ribose

OH

O

H

H

β-D-2-Deoxyribose

2. In DNA, the complementary bases are (1) uracil and adenine; cytosine and guanine. (2) adenine and thymine; guanine and cytosine. (3) adenine and thymine; guanine and uracil. (4) adenine and guanine; thymine and cytosine. (AIPMT 2008) Solution (2) Complementary bases in DNA are adenine and thymine and guanine and cytosine. 3. Which one of the following is an amine hormone? (1) Progesterone (2) Thyroxine (3) Oxypurin (4) Insulin (AIPMT 2008) Solution (2) Thyroxine is 3, 5, 3′, 5′-tetraidothyronine, which is basically an amine hormone. I HO

I

O OH

O I

H

NH2 I

5. Which of the statements about ‘Denaturation’ given below are correct? Statements (I) Denaturation of proteins causes loss of secondary and tertiary structures of the protein. (II) Denaturation leads to the conversion of double strand of DNA into single strand. (III) Denaturation affects primary structure which gets distorted. Options: (1) (I), (II) and (III) (2) (II) and (III) (3) (I) and (III) (4) (I) and (II) (AIPMT MAINS 2011) Solution (4) During denaturation, primary structure remains intact while secondary and tertiary structures of proteins are destroyed. 6. Which one of the following, statements is incorrect about enzyme catalysis? (1) Enzymes are mostly proteinous in nature. (2) Enzyme action is specific. (3) Enzymes are denatured by ultraviolet rays at high temperature. (4) Enzymes are least reactive at optimum temperature. (AIPMT PRE 2012) Solution (4) Enzymes are most active at optimum temperature. At this temperature, the rate of an enzyme reaction becomes maximum. 7. D-(+)-glucose reacts with hydroxyl amine and yields an oxime. The structure of the oxime would be (1)

Amine group

4. The segment of DNA which acts as the instructional manual for the synthesis of the protein (1) nucleoside. (3) ribose. (2) nucleotide. (4) gene. (AIPMT 2009)

CH

NOH

H

C

OH

HO

C

HO H

(3)

CH HO

C

H

H

H

C

OH

C

H

HO

C

H

C

OH

H

C

OH

CH2OH

(2)

NOH

CH

Solution

HO

C

(1) DNA carries all of the information for physical characteristics and is the genetic material in most of the organisms. The information is determined by proteins. Each protein is encoded by a gene. The order of nucleotides within a gene specifies the order and types of amino acids that make a protein.

HO

C

H

C

H

C

NOH

CH2OH

(4)

CH

NOH

H

H

C

OH

H

HO

C

H

OH

H

C

OH

OH

H

C

OH

CH2OH

CH2OH

(AIPMT 2014)

Solved Previous Years’ NEET Questions Solution (4) Glucose reacts with hydroxylamine to form an oxime. CH

CH

H

C

OH

HO

C

H

H

C

OH

H

C

OH

1 NH2OH

NOH

H

C

OH

HO

C

H

H

C

OH

H

C

OH

2H2O

CH2OH

10. The correct statement regarding RNA and DNA, respectively is (1) the sugar component in RNA is arabinose and the sugar component in DNA is 2′-deoxyribose. (2) the sugar component in RNA is ribose and the sugar component in DNA is 2′-deoxyribose. (3) the sugar component in RNA is arabinose and the sugar component in DNA is ribose. (4) the sugar component in RNA is 2′-deoxyribose and the sugar component in DNA is arabinose.

CH2OH

D-glucose

Oxime

8. Which of the following hormones is produced under the condition of stress which stimulates glycogenolysis in the liver of human being? (1) Thyroxin (3) Adrenaline (2) Insulin (4) Estradiol (AIPMT 2014) Solution (3) In stressful situations and during exercise, impulses from the hypothalamus stimulate sympathetic preganglionic neurons, which in turn stimulate the chromaffin cells to secrete epinephrine and norepinephrine, also called adrenaline and  noradrenaline,  respectively. In response to these hormones, liver cells perform glycogenolysis (breakdown of glycogen to glucose), and adipose tissue cells perform lipolysis (breakdown of triglycerides to fatty acids and glycerol). 9. In a protein molecule various amino acids are linked together by (1) α-glycosidic bond. (3) peptide bond. (2) β-glycosidic bond. (4) dative bond. (NEET-I 2016) Solution (3) In biopolymer protein, monomers (amino acids) are joined to each other by peptide bonds. This bond is formed when the carboxyl group of one amino acid reacts with the amino group of the other, and releases a molecule of water (H2O). H2NCH

COOH 1 H2N 2H2O

R H2N

CH

CH

C

R

O

COOH

R

NH CH R

Peptide bond

COOH

(NEET-I 2016) Solution (2) The sugar component of RNA is β-D-ribose and that of DNA is β-D-2-deoxyribose. HOCH2 H

OH

O

H

H

OH

OH

HOCH2

H

H

β -D-ribose Sugar component of RNA

OH

O

H

H

OH

H

H

β -D-deoxyribose Sugar component of DNA

11. The central dogma of molecular genetics states that the genetic information flows from (1) DNA → Carbohydrates → Proteins (2) DNA → RNA → Proteins (3) DNA → RNA → Carbohydrates (4) Amino acids → Proteins → DNA (NEET-II 2016) Solution (2) DNA

Transcription

RNA

Translation

Proteins

12. Which of the following statements is not correct? (1) Ovalbumin is a simple food reserve in egg-white. (2) Blood proteins thrombin and fibrinogen are involved in blood clotting. (3) Denaturation makes the proteins more active. (4) Insulin maintains sugar level in the blood of a human body. (NEET 2017) Solution (3) The three-dimensional shapes of proteins are dependent on the temperature. If the environment changes, a protein may lose its characteristic shape (secondary, tertiary and quaternary structure). This process is called denaturation. Denatured proteins are no longer functional that is, they become inactive.

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Chapter 12 | Biomolecules

ADDITIONAL OBJECTIVE QUESTIONS NCERT Exemplar Questions

(4)

1. Glycogen is a branched chain polymer of α-Dglucose units in which chain is formed by C1—C4 glycosidic linkage whereas branching occurs by the formation of C1 } C6 glycosidic linkage. Structure of glycogen is similar to _______. (1) Amylose (2) Amylopectin (3) Cellulose (4) Glucose 2. Which of the following polymer is stored in the liver of animals? (1) Amylose (2) Cellulose (3) Amylopectin (4) Glycogen 3. Sucrose (cane sugar) is a disaccharide. One molecule of sucrose on hydrolysis gives _______. (1) 2 molecules of glucose (2) 2 molecules of glucose + 1 molecule of fructose (3) 1 molecule of glucose + 1 molecule of fructose (4) 2 molecules of fructose

H HO H H

(2) H HO

OH

HO

H

H

HO

H

OH

H

OH

CHO HO

(1)

H

H

H

OH

HO

H

H

OH

HO

H

CH2OH

(3)

H

OH

HO H

CH2OH

H

OH

HO H

H OH

H

HO O

H HO H

OH H OH

H CH2OH

CH2OH

O

HO

H

H CH2OH

CH2OH O

H OH

H

H

OH

H

H O

O H OH

H

H

OH

H

O

H

H OH

H

H

OH

HO

O

HOH2C H

O

OH

H OH

OH

CH2OH

H

CH2OH O

H OH

H

H

OH

HO O H

CH2OH H

H

OH

O

OH

CH2OH

(4) (3)

H

CH2OH

HO

OH

H

H

CH2OH

H

H

HO

6. In disaccharides, if the reducing groups of monosaccharides, that is, aldehydic or ketonic groups are bonded, these are non-reducing sugars. Which of the following disaccharide is a non-reducing sugar?

OH

CHO

H

H

5. Proteins are found to have two different types of secondary structures viz. α-helix and β-pleated sheet structure. α-helix structure of protein is stabilized by (1) peptide bonds. (2) van der Waals forces. (3) hydrogen bonds. (4) dipole-dipole interactions.

(2)

CH2OH

O

HO

CH2OH

HO

CH2OH

OH

OH

H

OH

H

H

H

CHO

CHO

OH

HO

4. Which of the following pairs represents anomers? (1)

H

O H OH

H

H

OH

OH H

CH2OH O

H OH

H

H

OH

H O H

O H OH

H

H

OH

7. Which of the following acids is a vitamin? (1) Aspartic acid (3) Adipic acid (2) Ascorbic acid (4) Saccharic acid

OH H

Additional Objective Questions 8. Dinucleotide is obtained by joining two nucleotides together by phosphodiester linkage. Between which carbon atoms of pentose sugars of nucleotides are these linkages present? (1) 5′ and 3′ (3) 5′ and 5′ (2) 1′ and 5′ (4) 3′ and 3′ 9. Nucleic acids are the polymers of _______. (1) Nucleosides (3) Bases (2) Nucleotides (4) Sugars 10. Which of the following statements is not true about glucose? (1) It is an aldohexose. (2) On heating with HI it forms n-hexane. (3) It is present in furanose form. (4) It does not give 2,4-DNP test. 11. Each polypeptide in a protein has amino acids linked with each other in a specific sequence. This sequence of amino acids is said to be _______. (1) primary structure of proteins. (2) secondary structure of proteins. (3) tertiary structure of proteins. (4) quaternary structure of proteins.

(1) I and II (2) II and III

16. Which of the following reactions of glucose can be explained only by its cyclic structure? (1) Glucose forms pentaacetate. (2) Glucose reacts with hydroxylamine to form an oxime. (3) Pentaacetate of glucose does not react with hydroxylamine. (4) Glucose is oxidized by nitric acid to gluconic acid. 17. Optical rotations of some compounds along with their structures are given below. Which of them have D configuration? CHO

(I) H

CH2OH (I)

(II)

CHO H HO H H

14. Which of the following bases is not present in DNA? (1) Adenine (3) Cytosine (2) Thymine (4) Uracil

(1) rotation (II)

H

OH

H

OH

HO H

(III) O

H OH

(II)

HO H HO H

HO

H

HO

H

HO

H OH

H CH2OH

HO H H

OH H OH

H CH2OH

O

H OH OH CH2OH (III)

O

(1) I, II, III (2) II, III

(3) I, II (4) III

18. Structure of a disaccharide formed by glucose and fructose is given below. Identify anomeric carbon atoms in monosaccharide units. f CH2OH e O

CH2OH

H

O

C

(2) rotation

H

H

CH2OH

(III)

15. Three cyclic structures of monosaccharides are given below. Which of these are anomers? (I)

OH H OH OH CH2OH

13. Which of the following B group vitamins can be stored in our body? (1) Vitamin B1 (3) Vitamin B6 (4) Vitamin B12

OH

(1) rotation

12. DNA and RNA contain four bases each. Which of the following bases is not present in RNA? (1) Adenine (3) Thymine (2) Uracil (4) Cytosine

(2) Vitamin B2

(3) I and III (4) III is anomer of I and II

H H d OH HO c H

(1) (2) (3) (4)

H b OH

H a

a HOH2C O

b

O H c OH

H HO d

e CH2OH f

H

‘a’ carbon of glucose and ‘a’ carbon of fructose. ‘a’ carbon of glucose and ‘e’ carbon of fructose. ‘a’ carbon of glucose and ‘b’ carbon of fructose. ‘f ’ carbon of glucose and ‘f ’ carbon of fructose.

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Chapter 12 | Biomolecules 19. Three structures are given below in which two glucose units are linked. Which of these linkages between glucose units are between C1 and C4 and which linkages are between C1 and C6? CH2OH H HO

CH2OH O

H OH

H

H

OH

(A)

H

O H

O H OH

H

H

OH

OH H

(I) CH2OH H OH

O

H

H OH

H

H

OH

OH

O

H OH

H

H

OH

OH

O H OH

CH2OH (C) O H

HO

H

OH H

OH

O H OH

H

H

OH

4. Invert sugar is (1) cellobiose. (2) maltose.

(3) sucrose. (4) lactose.

5. Cellobiose is a (1) monosaccharide. (2) disaccharide.

(3) polymer. (4) trisaccharide

(3) 6 carbon atoms. (4) 8 carbon atoms.

8. Which of the following sugars is the most sweet? (1) Glucose (3) Maltose (2) Fructose (4) Sucrose

H

CH2OH

H

3. D-Galactose contains how many carbon atoms? (1) 4 (2) 5 (3) 6 (4) 8

7. D-ribose contains (1) 4 carbon atoms. (2) 5 carbon atoms.

O (B)

(II)

H

2. Which of the following is a ketohexose? (1) D-Glucose (3) D-Mannose (2) D-Fructose (4) D-Ribose

6. The change in optical rotation with time of freshly prepared solutions of sugar, is known as (1) specific rotation. (3) rotatory activity. (2) inversion. (4) mutarotation.

CH2 H

(3) Fructose is the sweetest of all sugars. (4) They do not conjugate with lipids.

H

(III)

(1) (A) is between C1 and C4, (B) and (C) are between C1 and C6 (2) (A) and (B) are between C1 and C4, (C) is between C1 and C6 (3) (A) and (C) are between C1 and C4, (B) is between C1 and C6 (4) (A) and (C) are between C1 and C6, (B) is between C1 and C4

Exercise 1 1. What is not true for carbohydrates? (1) General formula is CnH2nOn. (2) Glucose is the most common monomer of carbohydrates.

9. Dihydroxy ketone (CH2OH·CO·CH2OH) has the general formula of carbohydrate but not included in this class because (1) it does not contain polyhydroxy group. (2) it does not contain aldehyde group. (3) it is not optically active. (4) all of the above. 10. What can be said, correctly, about a monosaccharide, the name of which is preceded only by (+)? (1) The compound is the α-anomer. (2) The compound exists in the pyranose form. (3) The compound is dextrorotatory. (4) The compound has the same stereochemistry at the penultimate carbon as D-(+)-glucose. 11. The relation between glucose and galactose is that they are (1) metamers. (3) anomers. (2) epimers. (4) homomers. 12. Which among the following is a reducing sugar with an α-glycosidic linkage? (1) Sucrose (3) Lactose (2) Maltose (4) Cellobiose 13. An amino acid contains an amino group attached to (1) α-carbon atom. (3) γ-carbon atom. (2) β-carbon atom. (4) δ-carbon atom. 14. Which functional group participates in disulphide bond formation in proteins? (1) Thioether (3) Thioester (2) Thiol (4) Thiolactone 15. The sugar present in DNA is (1) glucose. (3) ribose. (2) deoxyribose. (4) fructose.

Additional Objective Questions 16. The purine present in RNA is (1) adenine. (3) uracil. (2) cytosine. (4) thymine.

24. The vitamins which are soluble in water are (1) A and B. (3) C and D. (2) B and C. (4) A and D.

17. The function of DNA is (1) to synthesize RNA. (2) to synthesize proteins. (3) to carry hereditary characteristics. (4) All of these

25. Chemical name of vitamin A is (1) Thiamine. (3) Retinol. (2) Ascorbic acid. (4) Nicotinamide 26. Which of the following are esters of glycerol? (1) Lipids (3) Cholesterol (2) Glycolipids (4) Sterol

18. Zwitterion is a (1) neutral species. (2) singly charged species. (3) doubly charged species. (4) multifunctional species. 19. Which of the following structures represents the peptide chain? O

H

(1)

N

C

N

O

H

C

NH C

N

C

C

Exercise 2

NH

1. Which of these is α-D-glucopyranose?

H

(2)

C

C

H

O

N C

C

CH2OH

(1) OH

H N

C

H

O

C N C

C

(4)

N C

O

(3)

OH

OH

OH

OH

CH2OH

CH2OH O

O O

H

C C N C H

C

(2)

N C C

C

O

20. The pH at which the concentration of the dipolar ion (zwitterion) form of an amino acid is at a maximum and the cationic and anionic forms are at equal concentrations is termed the (1) end point. (3) neutral point. (2) equivalence point. (4) isoelectric point. 21. Which of these amino acids contains a hydrophobic side chain? (1) Lysine (3) Methionine (2) Serine (4) Arginine 22. The primary structure of a protein refers to its (1) sequence of amino acid residues. (2) disulfide bonds. (3) helical structure. (4) hydrogen bonding. 23. Which attractive force is responsible for maintaining the tertiary structure of a protein? (1) Hydrogen bonds (2) van der Waals forces (3) Hydrophobic interactions (4) All of these

OH

OH

OH

OH

OH

O H

CH2OH

O

O

(3)

27. Which of the following statements regarding triacylglycerols is true? (1) They form micelles when mixed with water. (2) They are liquid if they are unsaturated. (3) They are solid if they have alkyne bonds. (4) They can be used to wash dirty dishes.

(4) OH

OH

OH

OH

OCH3 OH

OH

2. A glycoside is a compound which contains the structural features of the following classes of organic compounds (1) aldehydes and alcohols. (2) acetals and alcohols. (3) hemiacetals and alcohols. (4) ketones and alcohols. 3. Which of these is a glycoside? CH2OH

CH2OH

O

(1) OH

O

(2)

OH

OH OH

(3) OH

CH2OH O

OH

(4)

OH

OH OH

OH

OH

CH2OH O

OH

OH

OH OH

OCH3

825

826

Chapter 12 | Biomolecules 4. Refer to the structures below. Which are L-sugars? CHO OH

HO

H

OH

H

H

OH

HO

CH2OH

(3)

H

OH

OH

HO

H

H

OH

HO

H

H

OH

HO

H

CH2OH

8. Which reagent would be used for the following transformation? HO

CH2OH

(II)

?

O

OH

(2)

O OCH3

HO HO HO

(1) CH3I, KOH

H

H

O OH

HO

OH

HO

HO

HO HO

(V)

CHO H

OH OH

7. Glucose when heated with CH3OH in the presence of dry HCl gas, α- and β-methyl glucosides are formed. This is because it contains (1) an aldehydic group. (3) five hydroxyl groups. (2) } CH2OH group. (4) a ring structure.

CHO

H

CH2OH

OH OH

(IV)

CHO

OCH3

(4)

OH

H CH2OH

(I)

HOH2C O

O

CHO

H

HO

CH2OCH3

(3) (CH3)2SO4, NaOH

O

(4) CH3OH, HCl

O

CH2OH

9. Which of these is a non-reducing monosaccharide?

(III)

(1) II and IV (2) I, II, and III

CH2OH

(3) I and V (4) IV and V (1)

5. An aldaric acid is represented by: (1)

(3)

CHO

OH

CO2H

HO

H

HO

H

HO

H

HO

H OH

H

OH

H

OH

CH2OH

CO2H

CO2H

CH2OH

H

HO

H

HO

H

HO

H

OH

H

OH

H

OH

H

OH

(2)

(1) OH

OH

OH OCH3

OH CH2OH O

(4) OH

OH

OCH3

H

OH

CO2H

HO

OH H

H ?

OH

HO

H

H

OH

H

OH

H

OH

H

OH

CH2OH I OH

OH

OCH3

OH

CHO

O

(2)

OH

11. Which reagent would be used for the following transformation?

CH2OH OH

OH

OH

10. The sugar present in honey is (1) sucrose. (3) fructose. (2) glucose. (4) maltose.

6. Which of these reacts with dilute HCl to produce methanol? O

OH

OH

CH2OH

CH2OH

OH

OH

H

CH2OH

OH

O

H

HO

(3)

CH2OH

OH

(4)

O

OH

H

(2)

CH2OH

O

(1) [Ag(NH3)]2+ (2) HNO3

CH2OH ll

(3) Br2/H2O (4) (1) and (3)

Additional Objective Questions 12. Configuration of mannose and glucose differ at C2 position, they are termed as (1) epimers. (3) mesomers. (2) anomers. (4) racimers.

20. The number of amino acids commonly found in protein is (1) 10 (3) 20 (2) 15 (4) 25

13. In which of the following groups, all the compounds are polysaccharides? (1) Sucrose, glucose and fructose (2) Maltose, galactose and fructose (3) Glucogen, sucrose and maltose (4) Glycogen, cellulose and starch

21. What might be concluded upon determining that an unknown amino acid has its isoelectric point near pH 10? (1) It must have a hydrophobic side chain. (2) It must have a hydrophilic side chain. (3) Its side chain must contain more basic groups than acidic functions. (4) Its side chain must contain an acidic group.

14. Glucose and Fructose can be distinguished by (1) Lucas test. (2) Ninhydrin test. (3) Benedict reagent. (4) All of these. 15. Antibodies are (a) carbohydrates. (b) proteins.

(c) lipids. (d) enzymes

16. Which statement is incorrect about peptide bond? (1) C } N bond length in proteins is longer than usual bond length of C } N bond. (2) Spectroscopic analysis shows planar structure of } CO } NH } group. (3) C } N bond length in proteins is smaller than usual bond length of C } N bond. (4) None of these. 17. Disulfide bonds in proteins (1) result from an oxidation of thiols. (2) help to maintain the shape of proteins. (3) can be broken by reduction. (4) All of the above 18. Which of these amino acids cannot be described as an L amino acid? CO2H H

H2N

H HO2C

(I)

NH2 (IV)

OH

H HO2C

NH2 (II)

H2N

HO2C

NH2 H (V)

CO2H H H (III)

(1) I (2) II, IV and V

(3) I and III (4) II and IV

19. Which of these natural amino acids contains an } OH group? (1) Serine (3) Tyrosine (2) Threonine (4) All of these

22. The predominant form of aspartic acid in water at pH 1 would be (1) HO2C

(2)

CO2H 1NH 3

CO2H 2O

2C

1NH 3

CO22

(3) HO2C

1NH 3

CO2H

(4)

2O

2C

NH2

23. At intermediate pH values of about 6.0, an amino acid behaves as a dipolar ion or zwitterion. On decreasing and increasing the pH values, the amino acid (1) becomes basic and acidic, respectively. (2) becomes acidic and basic, respectively. (3) remains in the state of a neutral molecule. (4) loses its optical activity with the exception of glycine. 24. All of the following are sulphur-containing amino acids found in proteins, except (1) cysteine. (3) cystine. (2) threonine. (4) methionine. 25. Consider the double helix structure for DNA. The base pairs are (1) part of the backbone structure. (2) inside the helix. (3) outside the helix. (4) None of these. 26. Peptides are formed by joining of amino acids through amide linkage. Which of the following statement is not true in this respect? (1) Amide groups do not contribute in the hydrogen bonding interactions. (2) π-resonance stabilizes the amide linkage. (3) Amide groups are more resistant to hydrolysis than the similar ester groups. (4) Stable conformations of peptides are restricted to those having planar amide groups. 27. Which of the following statement about DNA is not correct? (1) It has a double helix structure. (2) It undergoes replication. (3) The two stands in DNA molecule are exactly similar. (4) It contains the pentose sugar, 2-dexoyribose.

827

828

Chapter 12 | Biomolecules (3) Assertion is true but the Reason is false. (4) Assertion and Reason both are false.

28. The best source of vitamin A is (1) oranges. (3) carrots. (2) beans. (4) wheat.

1. Assertion: In acidic medium, an amino acid migrates to the cathode.

29. Riboflavin is also called (1) Vitamin B12. (3) Vitamin E. (2) Vitamin B2. (4) Vitamin D.

Reason: Zwitterion of the amino acid changes to cation in an acidic medium and migrates to the cathode.

30. The vitamin which is neither soluble in fat nor in water is (1) biotin. (3) thiamine. (2) phylloquinone. (4) pyridoxine.

2. Assertion: Fructose does not contain an aldehyde group but still reduces Tollens’ reagent. Reason: In presence of a base, fructose undergoes a rearrangement to form glucose and mannose.

31. The vitamins absorbed from intestine along with fats are (1) A, D. (3) A, C. (2) A, B. (4) D, B.

3. Assertion: Any disruption in the secondary and tertiary structures of proteins is known as denaturation. Reason: Abrupt changes in temperature and pH cause changes in primary structure of proteins.

32. The deficiency of vitamin C causes (1) scurvy. (3) pyorrhea. (2) rickets. (4) pernicious anaemia.

4. Assertion: A solution of sucrose in water is dextrorotatory but on hydrolysis in presence of little hydrochloric acid, it becomes laevorotatory.

33. Which of the following compounds show aromatic properties? (1) Valine (3) Serine (2) Leucine (4) Tyrosine

Reason: Sucrose on hydrolysis gives unequal amounts of glucose and fructose as a result of which change in sign of rotation is observed.

34. An analysis of a food sample involves smearing a small amount of the food sample onto newsprint. This is done to test for the presence of (a) starch. (c) protein. (b) carbohydrates. (d) fat.

5. Assertion: Cellulose is not digested by human beings. Reason: Cellulose is a polymer of β-D-glucose. 6. Assertion: All enzymes are proteins but all proteins are not enzymes.

35. Which of the following statements regarding lipids is not true? (1) Lipids are soluble in non-polar organic solvents. (2) All lipids have the same functional groups. (3) Lipids include waxes, steroids, and triacylglycerols. (4) Lipids have little in common except their solubility.

Reason: Enzymes are biocatalysts and possess a stable configuration having an active site pocket. 7. Assertion: Insulin is a globular protein. Reason: Globular proteins are water soluble. 8. Assertion: Glycosides are hydrolyzed in acidic conditions.

Exercise 3 In the following set of questions, two statements “Assertion” and “Reason” are given. Choose the correct answer from the following options: (1) Assertion and Reason are true and the Reason is the correct explanation of the Assertion. (2) Assertion and Reason are true but the Reason is not a correct explanation of the Assertion.

Reason: Glycosides are acetals. 9. Assertion: Sucrose is a non-reducing sugar. Reason: It has glycosidic linkage 10. Assertion: Maltose is a reducing sugar which gives two moles of D-glucose on hydrolysis. Reason: Maltose has a 1,4-β-glycosidic linkage.

ANSWER KEY NCERT Exemplar 1. (2)

2. (4)

3. (3)

4. (3)

5. (3)

6. (2)

7. (2)

8. (1)

9. (3)

10. (3)

11. (1)

12. (3)

13. (4)

14. (4)

15. (1)

16. (3)

17. (1)

18. (3)

19. (4)

Hints and Explanations

Exercise 1 1. (4)

2. (2)

3. (3)

4. (3)

5. (2)

6. (4)

7. (2)

8. (2)

9. (3)

10. (3)

11. (2)

12. (2)

13. (1)

14. (3)

15. (2)

16. (1)

17. (4)

18. (3)

19. (3)

20. (4)

21. (3)

22. (1)

23. (4)

24. (2)

25. (3)

26. (1)

27. (2)

3. (4)

4. (4)

5. (3)

Exercise 2 1. (1)

2. (2)

6. (4)

7. (4)

8. (4)

9. (4)

10. (3)

11. (4)

12. (1)

13. (4)

14. (3)

15. (2)

16. (1)

17. (4)

18. (2)

19. (4)

20. (3)

21. (3)

22. (1)

23. (2)

24. (2)

25. (2)

26. (1)

27. (3)

28. (3)

29. (2)

30. (1)

31. (1)

32. (1)

33. (4)

34. (4)

35. (2)

1. (1)

2. (1)

3. (3)

4. (3)

5. (2)

6. (1)

7. (1)

8. (1)

9. (2)

10. (3)

Exercise 3

HINTS AND EXPLANATIONS Exercise 1

8. (2) Fructose is the sweetest naturally occurring sugar.

1. (4) Carbohydrates conjugate with lipids as they form the main substrate for digestion. 3. (3) D-galactose contains 6 carbon atoms. 4. (3) Invert sugar is obtained by the hydrolysis of sugar, and it is called invert because the optical rotation changes from dextrorotatory to laevorotatory. 5. (2) Cellobiose is a disaccharide since it contains two glucose molecules linked by a β(1 → 4′) bond. CH2OH H 4

OH

5

CH2OH O

H OH 3

H

H 2

H 1

H 49 OH

O

H

39

OH

H

7. (2) Ribose contains 5 carbon atoms 1

H

C

H

C

H

C

H

2 3 4

C

O OH OH OH

5

CH2OH D-Ribose

12. (2) Maltose is a reducing sugar with an α-glycosidic linkage. α-carbon

13. (1)

H2N

O

CH

α-amino group

C

OH

Carboxyl group

R

Side chain

O

59

11. (2) They are epimers that differ at C4.

H 29

OH

OH 19

H

14. (2) Thiol group participates in disulphide bond formation (SH) in proteins. (} SH) containing amino acids combine and eliminate H2 to form disulphide bond in protein. SH

HS

S

S

15. (2) Deoxyribose is the monosaccharide present in DNA. 5

O

HOH2C 4

H

H 3

OH

OH H 2

1

H

H

β -2-Deoxy-D(2)ribose

829

Chapter 12 | Biomolecules 16. (1) Adenine is the purine present in RNA; remaining are all pyrimidine bases.

12. (1) Epimers are compounds which differ in configuration at any one carbon.

17. (4) All of these. DNA is a hereditary material present in the nucleus and helps synthesize RNA, which helps in protein synthesis.

13. (4) Glucose and fructose are monosaccharides; sucrose and maltose are disaccharides; glycogen, cellulose and starch are polysaccharides.

18. (3) Zwitterion is a molecule with both positive and negative charges, hence is known as doubly charged species. 1

H3N

CH2

COO2

R

CHO HO HO 14. (2) H H

C

OH 1 H

HN

CH

COOH

R9

NH2

H H OH OH CO2H

16. (1) In peptide linkage, the C } N bond is shorter due to resonance than usual C } N bond. O

2H2O

O R

HNO3

15. (2) Antibodies are proteins found in animal cells.

O CH2

HO HO H H

CH2OH

19. (3) Peptide linkage is formed as follows: R

CO2H

H H OH OH

CH2 NH2

C

NH

CH

COOH

R9

Peptide linkage

20. (4) The isoelectric point (pI ) is the pH at which the concentration of the dipolar ion is at its maximum and the concentrations of the anions and cations are equal. 22. (1) The sequence of amino acid residues in a polypeptide or protein is called its primary structure. 24. (2) Vitamins B and C are water soluble. Vitamins A, D, E and K are fat soluble.

C

OH C

NH

N

18. (2) Almost all naturally occurring amino acids have the L configuration at the α- carbon. That is, they have the same relative configuration as L-glyceraldehyde. 20. (3) There are total of 26 amino acids, out of which 20 are found in proteins whereas the remaining 6 are found in special tissues. 23. (2) In the acidic medium, COO− of the zwitterion accepts a proton to form the cation (I); whereas in the basic medium, NH3− ion loses a proton to form the anion (II). 1

H3N

25. (3) Chemical name of vitamin A is retinol. Thiamine is vitamin B1; ascorbic acid is vitamin C and nicotinamide is vitamin B3 (form of niacin).

CH

COOH

OH2 H1

1

H3N

CH

R

R Zwitterion

26. (1) Lipids are esters of glycerol, for example, triglycerides and fatty acids.

H1

H2N

Exercise 2

8. (4) When a small amount of gaseous hydrogen chloride is passed into a solution of D-(+)-glucose in methanol, the reaction results in the formation of anomeric methyl acetals. 9. (4) Carbohydrates that contain only acetal groups do not give positive tests with Benedict’s or Tollens’ solutions, and they are called non-reducing sugars. 10. (3) Fructose is the sweetest of all the sugars, being more than twice as sweet as glucose. This sweetness accounts for the sweet taste of honey.

OH2

CH

COO2

R

3. (4) Carbohydrate acetals are generally called glycosides and an acetal of glucose is called a glucoside. 7. (4) This is observed due to the ring structure of glucose. A carbohydrate, often undergoes cyclization due to which carbonyl carbon transforms into new stereo units forming α- and β-forms.

COO2

(I)

(II)

24. (2) The structures of sulphur-containing cysteine, cystine and methionine are as follows: NH2

O HS

OH NH2 Cysteine

HO

O S

O

830

S

OH NH2

Cystine O MeS

OH NH2 Methionine

Hints and Explanations Threonine does not contain sulphur. O HO

OH NH2 Threonine

25. (2) In DNA double helix structure, sugar–phosphate backbone is on the outside of the helix whereas the base pairs (adenine, thymine, cytosine, guanine) are present inside the helix, stacking perpendicular to the helical axis.

30. (1) Vitamin H or B7 (also called biotin) is neither water nor fat soluble and is found in yeast, liver, kidney and milk. 31. (1) Intestine absorbs vitamins A and D along with fats and these vitamins as well as vitamins E and K are called fat soluble vitamins. Vitamins B and C are water soluble vitamins. 32. (1) The deficiency of vitamin C causes scurvy. It causes general weakness, anaemia, gum and skin problems. 33. (4) Tyrosine has an aromatic structure. O

26. (3) Pyrimidine bases present in DNA are cytosine and thymine, whereas purine bases are adenine and guanine. 27. (3) The two strands of DNA are never similar instead they are complementary to each other. This complimentary structure of both the strands keeps them affixed in a double helical shape. 28. (3) β-carotene present in carrots is metabolized into vitamin A.

OH HO

NH2

34. (4) An analysis of a food sample involves smearing a small amount of food sample onto newsprint. This is done to test for the presence of fat because fat is an oily substance, and will make the newsprint translucent.

831

13

Polymers

OH

C H A P T E R OU TLIN E

OH

etc.

13.1 Some Terms Related to Polymers 13.2 Classification of Polymers 13.3 Addition Polymerization or Chain-Growth Polymerization 13.4 Preparation of Some Important Addition Polymers 13.5 Condensation Polymerization or Step-Growth Polymerization

13.6 Natural Rubber 13.7 Synthetic Rubbers 13.8 Molecular Mass of Polymers 13.9 Biodegradable Polymers 13.10 Polymers of Commercial Importance

OH etc.

OH

OH

OH etc. Bakelite

Natural fibers such as silk, wool and cotton are polymers and have been used for thousands of years. However, in the past few years synthetic polymers such as polyolefins, polyesters, acrylics, nylons, etc., have replaced them extensively as clothing and protective materials. It may be added that biological materials such as proteins, deoxyribonucleic acid (DNA) and polysaccharides are also polymers. Many synthetic polymeric materials, known as biomaterials, are increasingly being employed within the human body either as artificial organs, bone cements, dental cements, ligaments, pacemakers or contact lenses because they are inert and accepted by the body. These include natural polymeric materials, which are derived from animals or plants such as cellulosics, chitin (or chitosan), dextran, agarose and collagen and synthetic polymeric materials such as polysiloxane, polyurethane, polymethyl methacrylate, polyacrylamide, polyester and polyethene oxides. It is being predicted that within a few years, polymers may replace many more conventional materials for various applications. The term polymer is derived from the Greek word polumerés, meaning “having many parts”. Polymer is a macromolecule with high molecular mass, formed by the repeated unit of several simple molecules called monomers. The process of linking the repeating units is called polymerization. The repeating units in a polymer are linked through covalent bonds. For example, polyethene (PE), polymethylmethacrylate (PMMA), polyvinyl chloride (PVC), etc. CH3 [ CH2

CH2 ]n

[ CH2

C] n COOCH3

[ CH2

CH] n Cl

Here, n represents the chain length of the polymer.

13.1 SOME TERMS RELATED TO POLYMERS 1. Monomers are simple molecules, which combine with each other to form polymers. Monomers are also called “building blocks” of polymers. To qualify as a monomer, a molecule should have at least two bonding sites (in the form of double or triple bonds or functional groups, such as } OH, } COOH, } NH2, } NCO, etc.).

etc. OH etc.

834

Chapter 13 | Polymers

2. Functionality is the total number of bonding sites or functional groups present in a monomer molecule. For example, ethene (bifunctional), methylmethacrylate (bifunctional), adipic acid (bifunctional), phenol (trifunctional), glycerol (trifunctional), etc. 3. Polymerization is the process of conversion of substances having low molecular weight (monomers) into substances having high molecular weight (polymers) with or without the elimination of byproducts such as HCl, H2O, NH3, etc. Polymerization reaction usually takes place in the presence of initiators. The following are the two kinds of polymerization reactions: (a) nCH2

CH2

Ethene

[ CH2

CH2 ]n

Polyethene H

(b) nNH2(CH2)6 NH2 1 nHOOC(CH2)4COOH Hexamethylene diamine

Polymerization

[N

Adipic acid

(CH2)6

H

O

N

C

O (CH2)4

C ]n

Nylon-6,6

4. Degree of polymerization (DP) is the number of repeating units present in a polymer. Here n is the degree of polymerization. When the value of n is very large, that is, in the range of hundreds or thousands, the polymers are called high polymers. For low values of n, that is, less than 10, the polymers are called oligomers in general; and more specifically as dimers (n = 2), trimers (n = 3), tetramers (n = 4), and so on.

13.2 CLASSIFICATION OF POLYMERS On the basis of their source, structure, mode of polymerization and molecular forces, the polymers are classified as follows:

13.2A Classification Based on Source 1. Natural polymers: The polymers obtained from natural sources such as plants and animals are called natural polymers. For example, wood, cellulose, jute, cotton, wool, silk, proteins, natural rubber, etc. 2. Semisynthetic polymers: These are polymers obtained from natural polymers by subjecting them to some chemical processes. For example, rubber is subjected to process of vulcanization to get vulcanized rubber. A large number of polymers derived from cellulose such as cellulose acetate and cellulose nitrate are also semisynthetic polymers. 3. Synthetic polymers: The polymers synthesized from simple molecules (monomers), are called synthetic polymers. These include plastics, synthetic fibers and synthetic rubbers. Some examples are PE, PVC, PMMA, polytetrafluoro ethene (PTFE), polycarbonate (PC), polyurethane, nylon, Buna-S rubber, etc.

13.2B Classification Based on Structure of Polymers 1. Linear polymers: These are polymers in which the molecules form long chains without branches or cross-linked structures. These linear chains may be held together by physical forces but no chemical interactions occur across the chains. For example, bifunctional ethene monomer polymerizes to yield linear structure of high-density polyethene.

13.2 | Classification of Polymers

(a) (b)

(c)

Figure 13.1 (a) Linear, (b) branched chain and (c) cross-linked polymers.

2. Branched chain polymers: These polymers contain molecules having a linear backbone with branches arising randomly from it. Branched chain polymer is formed when a bifunctional monomer is mixed and polymerized with a little amount of trifunctional monomer. For example, low-density polyethene (LDPE), polystyrene, PMMA, etc. 3. Cross-linked or network polymers: When the functionality of monomer is 3 and above, cross-linked three-dimensional network of polymer is formed. They are linked by strong covalent bonds. In fact, whenever a multifunctional monomer is polymerized, the polymer ultimately forms a network polymer. For example, bakelite, urea–formaldehyde, etc. The structures of linear, branched chain and cross-linked polymers are shown in Fig. 13.1.

13.2C Classification Based on Mode of Polymerization The polymerization reactions can be classified into two broad categories: (a) addition polymerization or chain-growth polymerization and (b) condensation polymerization or step-growth polymerization. 1. Addition polymers: The polymers formed by self-addition of several monomers having double and triple bonds without elimination of byproducts are called addition polymers. Examples of addition polymers are PE, PVC, PMMA, etc. H

H

H

H

H

H

C

C

C

C

C

C

H

H

H

CH3

H

C6H5

n

Polyethene

n

Polypropene

Polystyrene

H

H

H

H

H

H

C

C

C

C

C

C

H

Cl

H

OOCCH3

H

CN

n

Polyvinyl chloride

n

n

Polyvinyl acetate

n

Polyacrylonitrile

835

836

Chapter 13 | Polymers

Polymers formed from the same type of monomers are called homopolymers; for example, polyethene formed by ethene monomers. nCH2

Homopolymerization

CH2

[CH2

CH2]n2

Homopolymer

If two or more different repeating units (monomers) make up the polymer, it is known as a copolymer, for example, Buna-S. nCH2CH2

CH

CH2 1 nC6H5CH

CH2

Copolymerization

[CH2

CH

CH

CH2

CH2

CH

C6H5]n2

Copolymer

2. Condensation polymers: The polymers formed by intermolecular condensation reaction by the functional groups of monomers with continuous elimination of byproducts are called condensation polymers. For example, phenol–formaldehyde resin, polyester, epoxy resin, nylon-6,6, etc. nOH

(CH2)2

OH 1 nHOOC

Ethylene glycol

C6H4

[O

COOH

(CH2)2

Terephthalic acid

OOC

C6H4

CO ]n 1 2nH2O

Polyester (terylene)

13.2D Classification Based on Molecular Forces The polymeric chains are held together by intermolecular forces such as van der Waals forces. These forces determine the mechanical properties of a polymer. Further, the different uses and applications of polymers are based on mechanical properties such elasticity, toughness, tensile strength, etc. Based on the nature of intermolecular forces present, polymers may be classified into the following four categories. 1. Elastomers: The polymers which undergo very long elongation when pulled apart, and return to their original length on release are called elastomers. These are mainly coiled and long-chain polymers with no intermolecular forces except weak van der Waals forces which allow the polymer to be stretched. Sometimes a few cross-links are introduced between the chains, as in the case of vulcanized rubber, which help the polymer to retract to its original position. Some examples of elastomers are natural rubber, neoprene, Buna-S, butyl rubber, silicone rubber, etc. C6H5 CH2

C

CH

Cl Neoprene

CH2

CH2 n

CH

CH

CH2

CH2

CH n

Butadiene-styrene copolymer (Buna-S)

2. Fibers: These are long, thin and thread-like polymers, whose length is at least 100 times their diameter. They do not undergo stretching and deformation like elastomers but have high tensile strength and modulus. They are linked to each other by strong intermolecular forces like hydrogen bonding, which

13.2 | Classification of Polymers

837

hold the polymer chains together and make the structure of these polymers crystalline. For example, natural fibers such as jute, wood, silk, etc., and synthetic fibers such as nylon-6,6 and terylene. H N

(CH2)6

H

O

O

O

O

N

C(CH2)4

C

C

C

n

Nylon-6,6

O

(CH2)2

O

Polyethene terephthalate

n

3. Thermoplastic polymers: The polymers that soften on heating and which can be converted into any shape on cooling are called thermoplastics. The process of heating, reshaping and retaining the shape on cooling can be repeated several times without affecting their properties much. They are either linear or slightly branched long-chain molecules. Their intermolecular forces of attraction are intermediate between elastomers and fibers. For example, polythene, polystyrene, polyvinyls, etc.

CH2

Cl

H

CH

C

C

C

C

H

H

H

H

n

PVC

H

H

n

C

C

H

H

Polystyrene

4. Thermosetting polymers: Some cross-linked or heavily branched polymers undergo chemical changes and cross-linking on heating and become permanently hard, rigid and infusible on cooling. These are called thermosetting polymers and they do not soften on reheating; instead they undergo degradation. These are not reusable. For example, phenol–formaldehyde (bake-lite), urea–formaldehyde, epoxy resin, etc.

O O

H CH2

O

H CH2 n

Bakelite

HO

H

H

C

N

H

H2C C O

N

C N

H

H

N

C

CH2

C H2

N

H C

OH H

H

N

C

O

OH

H

H1

Urea-formaldehyde resin

The structures of thermoplastic and thermosetting polymers are illustrated in Fig. 13.2.

Monomer

Monomers

Linear thermoplastic polymer

Non-linear thermosetting polymer

Figure 13.2 Diagrams of thermoplastic and thermosetting polymer structures.

838

Chapter 13 | Polymers

13.3

ADDITION POLYMERIZATION OR CHAIN-GROWTH POLYMERIZATION

A polymerization reaction in which monomers containing one or more double bonds are linked to each other without the elimination of any byproducts, usually in the presence of free radical initiator, is called addition polymerization. Polymerization may also take place in the presence of ionic species. The monomers used in this reaction are unsaturated compounds such as alkenes, alkadienes, etc. Since in this polymerization the monomers are linked by a chain reaction, it is also called chain-growth polymerization and the polymers formed are called chain-growth polymers. Propylene (propene), for example, can be polymerized to form polypropylene. polymerization

n

Propylene

n

Polypropylene (PP)

The main features of addition (chain) polymerization are as follows: 1. 2. 3. 4. 5.

Only olefinic or vinyl compounds can undergo addition polymerization. There is no elimination of byproducts. Double bond provides required bonding sites. The addition of monomers takes place rapidly. The elemental composition of the polymer is the same as that of the monomer.

The addition reactions occur through radical, cationic, or anionic mechanisms depending on how they are initiated. The following examples illustrate these mechanisms. All of these reactions are chain reactions:

13.3A Free Radical Mechanism The general reaction of free radical mechanism is as follows: C

R

1 C

C

R C

C

C

C

R

C

C

C

C

C

etc.

Addition polymerization that takes place through free radical mechanism involves three distinct stages: initiation, propagation and termination. The three stages are described as follows for the polymerization of ethane to polyethene. 1. Chain initiation: This process involves two reactions. The first reaction involves the production of free radicals by the homolytic dissociation of an initiator, such as dibenzoyl peroxide (other examples are acetyl peroxide, tert-butyl peroxide, etc.) to yield a pair of radicals. A free radical is an atomic or molecular species having an odd or unpaired electron. They are highly active species. O C

O O

O

O Heat

C

O

C

O 1 O

Dibenzoyl peroxide O

O C Benzoyl free radical

O

Heat

1 C (R ) Phenyl free radical

O

C

13.3 | Addition Polymerization or Chain-Growth Polymerization

In the second reaction, the process of polymerization begins by the addition of benzoyl or phenyl free radical to the double bond of ethene and is called the chain initiating step. The addition of this radical to the first monomer molecule produces the chain initiating species. R 1 CH2

CH2

R

CH2

CH2

2. Chain propagation: In this step, the radical attacks another monomer to produce yet another free radical and the process continues to build new and bigger radicals that carry the reaction forward until termination occurs. R

CH2

CH2 1 CH2

CH2

R

CH2

CH2

CH2

CH2 R

nCH2

(CH2

CH2

CH2)n

CH2

CH2

3. Chain termination: At some point, the propagating polymer chain stops growing and terminates. Termination can happen by many methods, one of which is coupling or combination of two growing chains. R

(CH2

CH2)n

CH2

CH2 1 CH2 R

CH2 (CH2

(CH2

CH2)n

CH2)n

CH2

R CH2

CH2

CH2

(CH2)n

R

13.3B Ionic Polymerization Addition polymerization can also be initiated by ionic species instead of free radicals and this is known as ionic addition polymerization. It can be classified into two types – cationic and anionic, depending on the nature of ionic initiator involved. Cationic Polymerization In cationic addition polymerization, the initiator is a cation obtained either from protonic acids (HCl, HNO3, H2SO4) or lewis acids (AlCl3, BF3, etc.). In case of lewis acid, presence of water is essential as a cocatalyst. Some common examples of cationic polymerization are homopolymerization of isobutylene and copolymerization of isobutylene with 2-methyl propene. The various steps involved in the mechanism are explained as follows: 1. Initiation: The addition of cation to the double bond of the monomer leads to the formation of a carbonium ion (positively charged ionic species with trivalent carbon) and this initiates the polymerization. H1 1 CH2

CH

CH3

R

1

CH

R Carbonium ion

The presence of an electron releasing group facilitates the reaction by stabilizing the carbonium ion formed. 2. Propagation: In the next step, the carbonium ion formed adds to the double bond of another monomer, generating a bigger carbonuim ion and the chain propagates in a similar manner. CH3

1

CH 1 CH2

CH

R

R

CH3

CH

CH2

R

1

CH R

Chain propagation

CH3

1

CH ( CH2

CH2 )n CH2

CH

R

R

R

839

840

Chapter 13 | Polymers

3. Termination: The chain is terminated by the removal of H+. CH3

1

1

CH ( CH2

CH )n CH2

CH

R

R

R

2H

CH3

CH ( CH2

CH )n CH

R

R

CHR

Anionic Polymerization Anionic addition polymerization proceeds in the presence of anionic initiators such as alkali metal amides (e.g., sodium or potassium amide) or alkali metal alkyls (e.g., n-butyl lithium). Some common examples of anionic polymerization include polymerization of styrene, acrylonitrile and methyl methacrylate. The various steps involved in the mechanism are described as follows: 1. Initiation: The anionic initiator adds to the electron deficient site of the alkene monomer resulting in the formation of a carbanion (negatively charged species with trivalent carbon) and this initiates the polymerization. 12

NaNH2 1 CH2

CH

H2N

CH2

X

2

CH Na1 X

where X is an electron withdrawing group. The presence of an electron withdrawing group facilitates the reaction by stabilizing the carbanion formed. 2. Propagation: In the next step, the carbanion formed adds to the double bond of another monomer generating a bigger carbanion ion and the chain propagates in a similar manner. H2N

CH2

CH Na1 1 CH2

2

CH

X

X

H2N

CH2

CH

CH2

X

2

CH Na1 X

Chain propagation

H2N

CH2 ( CH2

2

CH )n CH Na1 X

X

3. Termination: The polymeric anionic chain is quite stable and continues to grow until terminated by addition of an acid. H2N

CH2 [ CH2

2

CH ]n CH X

X

H1

H2N

CH2 [ CH2

CH ]n CH2x X

13.4 PREPARATION OF SOME IMPORTANT ADDITION POLYMERS 1. Polythene: The two types of polyethenes are prepared as follows: (a) Low density polythene (LDP): Ethene is polymerized under high pressure (1000–2000 atm) and temperature (350–570 K) in the presence of dioxygen or the catalyst peroxide initiator. Free radical addition and H atom abstraction take place leading to a highly branched structure of LDP. LDP is tough, chemically inert, flexible and a poor conductor of electricity. It is used in electric wire insulation, toys, squeeze bottles, pipes, etc. (b) High density polythene (HDP): Ethene is polymerized in a hydrocarbon solvent in the presence of triethyl aluminium and titanium chloride (Ziegler–Natta catalyst) as catalysts and

13.4 | Preparation of Some Important Addition Polymers

at 6–7 atm pressure and 333–343 K temperature. The linear molecules are closely packed. Pipes, bottles, buckets, etc. are manufactured from HDP. It is chemically inert and also more tough and hard. 2. Polytetrafluoroethene (Teflon): Tetrafluoroethene (TFE) is heated with a free radical peroxide or ammonium persulphate catalyst at high pressure to give polytetrafluoroethene (PTFE). It is a highly crystalline polymer (93–98%). It has high melting temperature (Tm = 327°C), high temperature stability and low temperature flexibility. It has low coefficient of friction and remains slippery over a wide range of temperature (−40°C to 300°C). It is practically insoluble in any solvent and is not wetted by either oil or water. It has excellent electrical insulating properties. n CF2

Emulsion polymerization

CF2

Peroxide as initiator

TFE

[

CF2

CF2

]n

PTFE

Teflon is used for insulation of motors, generators, transformers, coils, capacitors, wires and cables. It is used as coating on food processing equipments such as bakery trays, frying pans because it is not wetted by oil or water. It is also used for coating on army weapons as an anticorrosive coat. It is also used as a dry lubricant since it remains slippery over a wide range of temperatures. 3. Polyacrylonitrile: Acrylonitrile in the presence of a peroxide catalyst forms polyacrylonitrile. H n

H

Free radical homopolymerization

C

C H

C

N

CH2

CH n

C N Polyacrylonitrile

Acrylonitrile

It is used in the manufacture of fibres such as orlon and acrilan and is used as a substitute of wool. Fiber obtained by homopolymerization of acrylonitrile is used for making hot gas filtration systems, outdoor awnings and sails for yachts. These fibers also find use in fiber reinforced concrete. Table 13.1 lists several other common chain-growth polymers. Table 13.1 Other common chain-growth polymers Monomer

Polymer

Cl

Names

)

)n

Polypropylene

)

)n

Poly(vinyl chloride), PVC

)n

Polyacrylonitrile, Orlon

Cl

)

CN

CN F

F

F

F

F

)

CO2 Me

)n F

F

)

Poly(tetrafluoroethene), Teflon

F

)n CO2 Me

Poly(methyl methacrylate), Lucite, Plexiglas, Perspex

841

842

Chapter 13 | Polymers

13.5

CONDENSATION POLYMERIZATION OR STEP-GROWTH POLYMERIZATION

A polymerization reaction in which bifunctional or polyfunctional monomers undergo intermolecular condensation leading to the formation of high molecular mass polymers with continuous elimination of byproducts, such as H2O, HCl, NH3, etc., is called condensation polymerization. Condensation polymerization reaction occurs in steps. In each step, a bifunctional species is formed which undergoes further condensation to produce a new functionalized species and the polymer chain grows with each step. Hence, it is also called step-growth polymerization and the polymers formed are called step-growth polymers. The reaction is normally catalyzed by acids or bases. Depending on their linkage units, some important polymers formed by condensation polymerization are as follows: 1. Polyamides: Silk and wool are two naturally occurring polymers that humans have used for centuries to fabricate articles of clothing. The search for a synthetic material with properties similar to those of silk led to the discovery of a family of synthetic polyamides called nylons. Polyamides are polymers possessing an amide linkage and are prepared by condensation polymerization of dicarboxylic acids with diamines or amino acids with their lactams as in nylon-6,6 and nylon-6, respectively. (a) Nylon-6,6: One of the most important nylons, called nylon-6,6, can be prepared from the sixcarbon dicarboxylic acid, adipic acid and the six-carbon diamine, hexamethylenediamine (hexane1,6-diamine) by heating them at 553 K under high pressure. In the commercial process these two compounds are allowed to react in equimolar proportions in order to produce a 1:1 salt, which on heating to a temperature of 270°C at a pressure of 17 atm causes a polymerization to take place. O

Water molecules are lost as condensation reactions occur between the salt to give the polyamide.

O2 and –NH3+ groups of

C

O HO

OH 1

H2N

NH2

O Adipic acid

Hexamethylenediamine O

2O

n

O

) O

)2

H3N

heat

1

NH3

(polymerization)

1:1 salt (nylon salt) H

O 2O

1

O2

N H

O

N

N O

)

)3

1

NH3

H n21

Nylon 6,6 (a polyamide)

The nylon-6,6 produced in this way has a molecular weight of about 10,000, has a melting point of about 250°C, and when molten can be spun into fibers from a melt. The fibers are then stretched to about four times their original length. This orients the linear polyamide molecules so that they

13.5 | Condensation Polymerization or Step-Growth Polymerization

are parallel to the fiber axis and allows hydrogen bonds to form between } NH } and } C O groups on adjacent chains. Called cold drawing, this stretching greatly increases the fibers’ strength. Nylon-6,6 fibers are used for making fabric for clothing, socks, sportswear, carpets, etc. They are also used for making combs and bristles of the brushes; as also used for making sports gear, fishing lines and other recreational equipment. (b) Nylon-6: It can be prepared by a ring-opening polymerization of ε-caprolactam in the presence of water at 533–543 K: O

O NH

O H2O

2OC

1

( CH2 (5 NH3 1

NH

( H2O) 250°C

NH

O

O

C ( CH2 (5 NH

C ( CH2 (5 NH

ε -Caprolactam (a cyclic amide)

O nC

Nylon 6

In this process, ε-caprolactam is allowed to react with water, converting some of it into ε-aminocaproic acid. Then, heating this mixture at 250°C drives off water as ε-caprolactam and ε-aminocaproic acid (6-aminohexanoic acid) react to produce the polyamide. Nylon-6 can also be converted into fibers by melt spinning. Nylon-6 is mainly used for making cords for tyres. It is also used for making ropes and fabrics. Nylons find extensive use because of their unique thermo-mechanical properties as described below: (i) Nylon fibers are linear structures in which the molecular chains are arranged parallel to each other and held together by hydrogen bonding. The strong intermolecular hydrogen bonding makes the structure of these fibers crystalline and imparts them high strength, elasticity and melting point. (ii) They are chemically stable and resistant to abrasion. (iii) The yarn is smooth, long-lasting and can be spun into fabric. (iv) The fabric is tough, lustrous, moisture resistant, easy to dye, retains color and can be set by heat or steam. 2. Polyesters: These are formed from condensation of dicarboxylic acids and diols. One of the most important polyesters is poly (ethene (ethylene) terephthalate), a polymer that is marketed under the names Dacron, Terylene and Mylar. O O O O

O O

O O

n

Poly ethylene ethene terephthalate or PET (Dacron, Terylene, or Mylar)

Poly(ethene terephthalate) can be obtained by a direct acid-catalyzed esterification of ethylene glycol and terephthalic acid at 420–460 K. The catalyst used in the reaction is zinc acetate and antimony trioxide.

843

844

Chapter 13 | Polymers

O

O OH

HO

1

HO

OH

O

O

HA heat

O

1 n H2O

O

Terephthalic acid

1,2-Ethanediol

n

Poly(ethylene terephthalate) (a polyester)

The poly(ethene terephthalate) thus produced melts at about 270°C. It can be melt-spun into fibers to produce crease-resistant Dacron or Terylene; it can also be made into a film, in which form it is marketed as Mylar. It is also used in making safety helmets. 3. Phenol–formaldehyde polymer (Bakelite and related polymers): One of the first synthetic polymers to be produced was a polymer (or resin) known as Bakelite. It is made by a condensation reaction between phenol and formaldehyde; the reaction can be catalyzed by either acids or bases. A large excess of phenol on heating with an aqueous solution of formaldehyde in the presence of acid catalyst first gives a linear polymer of low molecular weight called novalac resin. Novolacs (used in paints) are phenol–formaldehyde polymers made when the molar ratio of formaldehyde to phenol is less than one. OH

OH CH2 OH H

OH CH2 OH

OH

H

CH2

CH2

H1

Novolac

Further cross-linking of novolac polymer can be done by adding excess of formaldehyde and heating to give highly cross-linked product called Bakelite. The mechanism for base-catalyzed reaction is shown in Fig. 13.3. Reaction can take place at the ortho and para positions of phenol and o- and/or p-hydroxymethyl phenol derivatives are formed first. These react further with phenol to form compounds with rings joined through the –CH2 group. Bakelite is used for making electrical equipment parts such as switches, plugs, switchboards, cooker handles, welding tongs, electric iron parts and telephone parts. Due to its hardness, it is used for making golf ball and heads for typewriters. It is used as hot-setting adhesives for plywood, cement and for lamination of paper, cardboards and wood. Some important thermo-mechanical properties that are responsible for its use are as follows: (i) (ii) (iii) (iv)

Bakelite is a rigid, hard, scratch resistance material. It shows good electrical insulation property. It has thermal stability of up to 200°C and good dimensional stability. It is resistant to water and shows low flammability.

4. Melamine-formaldehyde (MF) polymer: Reaction of melamine with neutralized formaldehyde at about 80–100°C leads to the production of a mixture of water-soluble methylol melamines. On further H N

H2N N

NH2 N

(HOCH2)2N 1 (CH2O)

NH2 Melamine

N(CH2OH)2

N N

N

H

N H3O1

Hexamethylolmelamine (Resin)

CH2

N

N

N(CH2OH)2 Formaldehyde

N

N

N H

H

Cross-linked MF

13.5 | Condensation Polymerization or Step-Growth Polymerization

OH

O2

O CH2

OH2

O2

O

OH

H

CH2

H

B

O2

O2

O CH2

2:B

CH2 OH2

O2

H B :B2 CH2

H

O2

OH

O

CH2

2H1, 1H1

more CH2O O2 more

OH

OH

etc.

etc. OH

OH etc.

OH

OH

etc. OH

etc. Bakelite

Figure 13.3 Mechanism for base-catalyzed reaction.

heating, the methylol melamines condense and a point is reached where hydrophilic resin separates out on cooling. This resin on polymerization yields melamine–formaldehyde polymer commonly known as melamine. MF is hard, scratch resistant, fairly heat resistant and free from color with wide colorability. The principal application of melamine polymer is in the manufacture of unbreakable tableware. Cellulose-filled compounds find usage in outlet for trays, clock cases and radio cabinets. Mineral-filled powders are used in electrical applications and knobs and handles for kitchen utensils.

845

Chapter 13 | Polymers

5. Polyurethanes: A urethane is the product formed when an alcohol reacts with an isocyanate: O R

OH 1 O C N R′

R

C

O

N

H

R′ Alcohol

An isocyanate

A urethane (a carbamate)

The reaction takes place in the following way: O2

O R

OH 1 C

1

R

O

C

H

N

O R

N

C

1

O H

R′

R′

H

O

A

2

R

N

R′

2

O

C

N

H

R′

A

A urethane is also called a carbamate because formally it is an ester of an alcohol (ROH) and a carbamic acid (R′NHCO2H). Polyurethanes are usually made by allowing a diol to react with a diisocyanate. The diol is typically polyester with } CH2OH end groups. The diisocyanate is usually toluene 2,4-diisocyanate: HO

O

polymer

C

N

N C O

OH 1 CH3 Toluene 2,4diisocyanate H O

846

O

N

polymer

O

N

O

CH3

H A polyurethane

n

Polyurethane foams, as used in pillows and paddings, are made by adding small amounts of water to the reaction mixture during the polymerization with the diisocyanate. Some of the isocyanate groups react with water to produce carbon dioxide, and this gas acts as the foaming agent: R

N

C

O 1 H2O

R

NH2 1 CO2

13.5A Copolymerization The polymerization of two or more different monomers resulting in the formation of a polymer containing both monomers linked in one chain is called copolymerization. It can be considered as a type of addition polymerization. The polymers so formed are called copolymers. They are formed both by chain-growth and step-growth polymerization reactions. x CH2

CH

CH

Butadiene

CH2 1 nCH2

CH

Styrene

[ CH2

CH

CH

CH2 ]x [ CH2

CH ]n

Styrene butadiene rubber (SBR)

13.6 | Natural Rubber

Copolymers can have any of the following three structures. If the various repeating units occur randomly along the chain structure, the polymer is called a random copolymer. When repeating units of each kind appear in blocks, it is called a block copolymer. Graft copolymers are formed when chains of one kind are attached to the backbone of a different polymer. For example, in case of linear chains synthesized from repeat units A and B, the block, random and graft copolymers can be represented as A

A

A

A

A

A

B

B

B

B

B

B

A

A

Block copolymer

A

A

B

A

B

B

A

B

A

A

B

A

A

A

A

B

B

B

B

A

A

A

B B Graft copolymer

Random copolymer

Copolymers have properties different from homopolymers of the constituting monomers. For example, butadiene–styrene copolymer has high abrasion resistance and load-bearing capacity. It has low oxidation resistance and swells in oil and solvents, like natural rubber. It also vulcanizes to produce cold rubber, which has greater tensile strength and abrasion resistance. It is used in motor tyres, shoes soles, footwear components, insulation of wires and cables, carpet backing, gaskets and adhesives.

13.6 NATURAL RUBBER Natural rubber, the original elastomer, still plays an important role among elastomers. It is prepared from the milky juice (latex) which forms a colloidal dispersion of rubber in water, obtained from Hevea rubber trees or other sources like gutta-percha and balata. The rubber trees are found in India, Sri Lanka, Indonesia, Malaysia and South America. Natural rubber is a linear polymer of isoprene hence it is also called as polyisoprene. nCH2

CH

C

[ CH2

CH2

CH3

C

CH

CH2 [n

CH3

Chemically, natural rubber from Hevea trees is a linear polymer of isoprene (2-methyl-1,3-butadiene) also known as cis-1,4-polyisoprene and that tapped from other sources (gutta-percha and balata) is the trans isomer of polyisoprene. n

CH2 CH2 C

n

O

CH2 H

CH2 H C

n

Natural rubber

O

CH3 CH2 n n

Gutta-percha

Rubber (mainly cis-1,4 polyisoprene) consists of long, coiled-up polymer chains that are interlinked at a few points and held together by van der Waals forces. This imparts properties of flexibility and extensibility to rubber. On application of force, the molecules straighten out in the direction in which they are being pulled and stretch like a spring. They are restored to their normal random arrangement on release.

13.6A Vulcanization of Rubber The process of heating and mixing crude rubber with sulphur to a definite temperature for a specific time is known as vulcanization. At high temperature (>335 K), the natural rubber is soft and at low temperatures (< 283 K), it becomes brittle. It has high water absorption capacity, is soluble in non-polar solvents and is non-resistant to attack by oxidizing agents. The natural rubber is cured by vulcanization

847

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Chapter 13 | Polymers

which modifies it to become more durable with better mechanical properties. The sulphur combines chemically at the double bonds forming cross-links between different polymer chains. This cross-linking brings about stiffness in the rubber, and subsequently prevents intermolecular sliding of rubber polymer chains. CH3 CH2

CH2

C

C

CH

CH

CH3 CH2 1 S 1 CH2

CH3

C

CH3

CH

C

CH

CH2

CH2

CH2

CH2

CH3

C

CH

S

S

C

CH

CH2

CH2

CH3

CH3

C

CH

S

S

C

CH

CH2

CH2

CH3

The extent of stiffness of vulcanized rubber depends upon the amount of the sulphur added. For example, a rubber tyre contains 5% sulphur, whereas a battery case made from rubber may contain as much as 30% of sulphur. This highly vulcanized rubber is called ebonite. Important properties of vulcanized rubber responsible for its use are as follows: 1. 2. 3. 4. 5.

It has good tensile strength and extensibility. It has excellent resilience. It possesses low water absorption capacity. It has higher resistance to oxidation and to abrasion, wear and tear resistance. It is resistant to organic solvents such as petrol, benzene, CCl4 and fats and oils. However, it swells in these liquids. 6. It exhibits good electrical insulation property. 7. It has a wide useful temperature range from −40°C to 100°C. 8. It is very easy to tailor-make the properties of vulcanized rubber for the specific needs.

13.7 SYNTHETIC RUBBERS Synthetic rubbers are a substitute for natural rubber and have improved material properties to overcome the drawbacks associated with natural rubber. Synthetic rubber can get stretched to double its length, but can return back to normal once the stretching force is released. Synthetic rubbers are made from homopolymers of 1,3-butadiene derivatives or copolymers of 1,3-butadiene or its derivatives with another unsaturated monomer.

13.7A Preparation of Synthetic Rubbers Neoprene (Polychloroprene) Neoprene is prepared by polymerization of chloroprene in the presence of persulphate as initiator (free radical polymerization). nCH2

C

CH

CH2

Cl 2-Chloro-1,3-butadiene

[ CH2

C

CH

CH2 ]n

Cl Neoprene

Neoprene rubber has greater resistance to vegetable and mineral oils as compared to natural rubber. It is used for manufacture of conveyor belts, car fan belts, gaskets, hoses, mouse pads, shoe heels, etc. It is also used for corrosion resistance coatings, lining material in reaction vessels and tubes for carrying oils and chemicals. It also finds use as adhesive in tapes and highway joints, and in gloves.

13.9 | Biodegradable Polymers

Buna-N Buna-N rubber is a copolymer of acrylonitrile and 1,3-butadiene prepared in the presence of a peroxide catalyst. x CH2

CH

CH

Butadiene

CH2 1 nCH2

[ CH2

CH

CN Acrylonitrile

CH

CH

CH2 ]x [ CH2

CH ]n CN

Nitrile rubber (Buna-N)

Nitrile rubber has low swelling and low solubility. It has good tensile strength and is resistant to action of petrol, lubricating oils and organic solvents. It also possesses good heat resistance. It is used in fuel tanks, gasoline hoses, as an adhesive and in the form of latex for impregnating paper, leather and textiles. Synthetic rubbers have several advantages over natural rubber: 1. 2. 3. 4. 5.

It is produced using monomers derived from petrochemical raw materials. It is more economical than natural rubber. It is superior to natural rubber in certain properties, such as tensile strength and abrasion resistance. It can be tailor-made to the end properties and use. It has low temperature flexibility and high temperature stability.

13.8 MOLECULAR MASS OF POLYMERS Molecular weight is an extremely important variable in polymers because it relates directly to the physical properties of macromolecules. In general, higher the molecular weight, tougher the polymer; however, too high molecular weight will lead to processing difficulties. The molecular weight and its distribution determine the viscous and elastic properties of the molten polymer. A variety of methods are available for molecular weight determination, and they are applicable in different ranges of molecular weight. Since any polymer sample contains a mixture of chain lengths of different sizes, the molecular weight is always expressed as average. The number average molecular weight of a polymer is given by Mn =

∑ M i Ni Ni

where Ni is the number of molecules or the number of moles of those molecules having molecular weight Mi. The weight average molecular weight is given by Mw =

2 ∑w i M i ∑ NiM i = ∑w i ∑ NiM i

13.9 BIODEGRADABLE POLYMERS Biopolymers such as polysaccharides (starch and cellulose), proteins, nucleic acids, etc., undergo natural degradation by the action of naturally occurring microorganisms, such as bacteria, fungi and algae. Most of the synthetic polymers do not get degraded in this manner and thus pose many environmental problems. Therefore, some synthetic biodegradable polymers have now been designed and developed whose functional groups are similar to those of biopolymers and hence they undergo biodegradation. The biodegradation reactions of synthetic biodegradable polymers is same as that of natural polymers.

849

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Chapter 13 | Polymers

These are enzyme-catalyzed and proceed in aqueous media. Biodegradable polymers can be broadly classified as 1. Natural biodegradable polymers that have been modified with additives and fillers. 2. Synthetic biodegradable polymers, primarily polyesters. The major category of synthetic biodegradable polymers consists of aliphatic polyesters with a hydrolysable linkage along the polymer chain such as polylactic acid (PLA). Other widely available synthetic types include aliphatic/aromatic copolyesters. The following are some important examples: 1. Poly β -hydroxybutyrate – co-β -hydroxy valerate (PHBV): It is obtained by copolymerization of 3-hydroxybutanoic acid and 3-hydroxypentanoic acid. It can be degraded by the bacteria. It finds use in packaging orthopaedic devices and in controlled release of drugs. OH CH3

CH

OH CH2 COOH 1 CH3

3-Hydroxybutanoic acid

CH2

CH

CH2

COOH

3-Hydroxypentanoic acid O

CH CH2

C

CH3

O

O

CH

CH2

CH2CH3

C O n

PHBV

2. Nylon-2-nylon-6: It is a polyamide formed by copolymerization of glycine and amino caproic acid, and is biodegradable. [H2N

CH2

COOH] 1 [H2N

(CH2)5

COOH]

H2N

[CH2

CO

NH (CH2)5] n

COOH

Some other biodegradable polymers are polyglycollic acid (used for stitching wounds), polylactic acid, poly(e-caprolactone; PCL).

13.10 POLYMERS OF COMMERCIAL IMPORTANCE The polymers of commercial importance are as given in Table 13.2 Table 13.2 Some commercially important polymers Polymer

Monomer

Polypropene

Propene

Structure

Manufacture of pipes, toys, etc.

CH3 CH2

Polystyrene

Uses

Styrene

CH n C6H5

CH2

CH n

Manufacturing of household wares, such as combs, toys, buttons, radio and television parts, refrigerator parts, machine housings and high frequency electrical insulation material.

(Continued)

13.10 | Polymers of Commercial Importance

Table 13.2 (Continued) Polymer

Monomer

Polyvinyl chloride (PVC)

Vinyl chloride

Structure

Used in rigid pipes, flooring and vinyl siding, wire and cable insulations and in packaging, flooring, artificial leather, wall coverings and carpet backing, pipe fittings, electrical outlet boxes and parts for automotive bumpers.

Cl CH2

Urea(a) Urea formaldehyde (b) Formaldehyde resin

Glyptal

Uses

NH

(a) Ethylene glycol (b) Phthalic acid

OCH2

CO

CH n

NH

Used as adhesives for the particle board, plywood and furniture industries and as casting material in textile finishing, as foams to aid floral decoration and as artificial snow in cinema and television productions.

CH2 n

CH2OOC

Used in paint and lacquer manufacturing.

CO n

Bakelite

(a) Phenol (b) Formaldehyde

O H CH2

O H CH2 n

Neoprene

Nylon-6,6

(a) Chloroprene (b) Persulphate (as initiator)

(a) Adipic acid (b) Hexamethylenediamine

CH2

C

CH

CH2

Cl

H N

n

H O (CH2)6

N

C(CH2)4

Manufacture of switches, plugs, switchboards, cooker handles, welding tongs, electric iron parts and telephone parts.

Manufacture of conveyor belts, car fan belts, gaskets, hoses, mouse pads, shoe heels. Also used for corrosion resistance coatings, lining material in reaction vessels and tubes for carrying oils and chemicals. It also finds use as adhesive in tapes and highway joints, and in gloves.

Used for making fabric for clothing, socks, sportswear, C n carpets, etc. It is used for making combs and bristles of the brushes. It is also is used for making sports gear, fishing lines and other recreational equipment.

O

851

852

Chapter 13 | Polymers

SOLVED EXAMPLES 1. Identify the monomers required for the synthesis of each step-growth polymer: O

O

(a)

(a) N H

O

N H

O

OH

O O

NC

(d) CO2Me

O n

3. Monomer of this following polymer is

O

H N

CH3

N H

O

n

O

C

CH2

CH3

O

(d)

Cl Cl

(b) O

(c)

(c)

n

O

(b)

Solution

H N

N H

O

n

(1) 2-methyl propene. (2) styrene.

n

(3) propylene. (4) ethane.

Solution

Solution

(a)

(1) 2-Methylpropene is the monomer of the given polymer.

O

O

Cl

Cl 1 H2N

CH3

NH2

C

CH2

CH3

(b) O

4. Acrilan is a hard, horny and a high melting point material. Which of the following represent its structure?

O HO Cl 1

Cl

OH

(1)

(c) O Cl

CH2 CH CN

(2)

CH2

(d)

(4)

C COOC2H5

O Cl 1 NH2

NH2

O

n

CH2

CH Cl

n

n

Solution (1) The structure of acrolein is

2. Draw the structure of the alkene monomer used to make each chain-growth polymer: (a)

(c)

n

O

OH

O

Cl

Cl

n

O

CH2

CH

n

CN

The other name of acrolein is polyacrylonitrile (PAN). It is a chain-growth polymer of acrylonitrile CH2 CH } CN 5. Complete the mechanism for radical chain propagation begun below.

n

(b)

C COOC2H5

n

CH3

O

O

CH2

NH2

HN Cl 1 2

Cl

(3)

(d)

n

NC

CO2Me

In

Cl

Cl

Cl In

Cl

Solved Examples Solution

Cl

In

Polymerization

nNH2(CH2)6NH2 1 nHOOC(CH2)4COOH Hexamethylene diamine

Cl

Cl

In

Cl

6. (a) Can a copolymer be formed in both addition and condensation polymerization? Explain. (b) Can a homopolymer be formed in both addition and condensation polymerization? Explain.

H [N

Adipic acide

H O (CH2)6

N

C

9. The monomer of the polymer CH3

(a) Yes, copolymers can be formed both in addition and condensation, polymerization. For example, Buna-S is an addition copolymer of styrene and 1, 3-butadiene while nylon-6,6, bakelite and polyester are condensation copolymers. (b) Yes, homopolymers can be formed both in addition and condensation polymerization. For example, polythene, PVC, PMMA, PAN, neoprene, etc. are examples of addition homopolymers while nylon-6 is an example of condensation homopolymer. 7. Poly(butylene terephthalate) is a plastic material used in automotive ignition systems and has the formula:

O

O

O

C

C

CH2

C

1,4-Butanediol

CH3

CHCH3

(3) (CH3)2C

(2) CH3CH

CH2

(4) H2C

C

C(CH3)2 CH3 CH3

Solution (4) The steps involved in the formation of polymer are Step 1: Formation of carbocation in the presence of acid. H2C

C

CH3

H1

1

CH3

CH3

C

CH3 CH3

Step 2: Attack of carbocation on another molecule and so on.

H2C

C

CH3 CH3

1 CH3

1

C

(a) 1,4-Butanediol and terephthalic acid (b) Condensation polymer (c) The reaction involved is

OH 1 HO

C

(1) CH3CH

Solution

HO CH2CH2CH2CH2

CH3

1

CH2

CH3

O )n

(a) Suggest the monomers which might be used to synthesize this polymer. (b) Is this an addition polymer or condensation polymer? (c) Write the reaction.

C ]n

(CH2)4

Nylon-6,6

Solution

( CH2CH2CH2CH2

O

CH3 CH3

CH3 H2C

O

O

C

C

C

CH2

CH3 OH

Terephthalic acid Heat HA

Poly(butylene terephthalate) 1 H2O

8. [NH(CH2)6NHCO(CH2)4CO]n is a (1) copolymer. (3) thermosetting polymer. (2) addition polymer. (4) homopolymer.

1

C

CH3 CH3

10. Arrange the following alkenes towards their increasing order of reactivity in cationic polymerization: H 2C H 2C

CHCH3, H2C CHCO2CH3.

CHCl,

H2C

CHC6H5,

Solution The stability of carbocation increases in the order: CH3CH+COOCH3 < CH3CH+Cl < CH3CH+CH3 < CH3CH+C6H5

Solution

Therefore, the reactivity towards the cationic polymerization increases in the same order:

(1) [NH(CH2)6NHCO(CH2)4CO]n is a copolymer of hexamethylenediamine and adipic acid.

CH2 CHCOOCH3 < CH2 < CH2 CHC6H5

CHCl < CH2

CHCH3

853

854

Chapter 13 | Polymers

SOLVED PREVIOUS YEARS’ NEET QUESTIONS 1. Which one of the following polymers is prepared by condensation polymerization? (1) Styrene (3) Teflon (2) Nylon-6,6 (4) Rubber

Solution (3) The correct structure of neoprene is CH2

(AIPMT 2007) Solution (2) The polymers formed by intermolecular condensation reaction between the functional groups of monomers with continuous elimination of byproducts are called condensation polymers. Nylon-6,6 is prepared by condensation polymerization of adipic acid and hexamethylenediamine. nNH2(CH2)6NH2 1 nHOOC(CH2)4COOH Hexamethylene diamine H [N

(CH2)6

Polymerization

Adipic acid

H

O

N

C

CH

CH2 n

Cl

4. Which of the following structure represents neoprene polymer? CN

(1) ( CH2

CH )n Cl

(2) ( CH2

CH )n

(3) ( CH

CH2 )n

C6H5

(4) ( CH2

C

CH2 )n

CH

Cl

O (CH2)4

C

C ]n

(AIPMT PRE 2010)

Nylon-6,6

Solution

2. Which one of the following statements is not true? (1) Natural rubber is a 1, 4-polymer of isoprene. (2) In vulcanization, the formation of sulphur bridges between different chains makes rubber harder and stronger. (3) Natural rubber has the trans-configuration at every double bond. (4) Buna-S is a copolymer of butadiene and styrene.

(4) Neoprene is prepared by polymerization of chloroprene (2-chlorobuta-1,3-diene) in the presence of persulphate as initiator (free radical polymerization). nCH2

C

CH

CH2

CH2

C

CH

Cl Neoprene

Cl 2-Chloro-1,3-butadiene

CH2 n

(AIPMT 2008)

5. Of the following which one is classified as polyester polymer? (1) Nylon-6,6 (3) Bakelite (2) Terylene (4) Melamine

(3) Natural rubber is a polyisoprene, but not all bends have trans configuration.

(AIPMT PRE 2011)

Solution

3. Structures of some common polymers are given. Which one is not correctly presented? (1) Nylon-6, 6: [ NH(CH2)6NHCO(CH2)4 CO ]2 (2) Teflon: [ CF2 CF2 ]n (3) Neoprene: [ CH2 C

CH

CH2

CH2 ]n

Cl

(4) Terylene:

OC

COOCH2

CH2

O n

(AIPMT 2009)

Solution (2) Polyesters are polymers formed from a dicarboxylic acid and a diol. Among the given polymers, terylene is a polyester polymer. O

O

C

C

O

CH2

CH2

O n

6. Which one of the following is not a condensation polymer? (1) Melamine (3) Dacron (2) Glyptal (4) Neoprene (AIPMT PRE 2012)

855

Solved Previous Years’ NEET Questions

n CH2

Solution

Solution

(4) Neoprene is an addition polymer of chloroprene

(3) Polyamides are polymers possessing an amide linkage and are prepared by condensation polymerization of dicarboxylic acids with diamines or amino acids with their lactams as in nylon-6,6 and nylon-6, respectively.

C

CH

CH2

Polymerization

CH2

C

CH

CH2 n

CI Neoprene

CI Chloroprene

10. Which is the monomer of neoprene in the following?

7. Which of the following statements is false? (1) Artificial silk is derived from cellulose. (2) Nylon-6,6 is an example of elastomer. (3) The repeat unit in natural rubber is isoprene. (4) Both starch and cellulose are polymers of glucose.

(1) CH2

CH } C

(2) CH2 C CH (3) CH2 C CH

CH } CH

CH2 (NEET 2013)

8. Which one of the following sets of monomers forms the biodegradable polymer? (1) CH2 CH } CN and CH2 CH } CH CH2 (2) HN2 } CH2 } COOH and HN2 } (CH2)5 } COOH

HOOC

Solution (3) Neoprene is prepared by polymerization of chloroprene in the presence of persulphate as initiator (free radical polymerization). nCH2

CH

(1) CH2 and CH2

CH } CH

CH2

CH2

(AIPMT MAINS 2012)

C

CH

Cl Neoprene

CH2 n

C

CH

n

CH

H

(2) Nylon-2-nylon-6 is an alternating copolymer of glycine (NH2 } CH2 } COOH) and amino capric acid (NH2 } (CH2)5 } COOH) and is biodegradable.

CH2

Cl

Solution (3)

(CH2)8

N

H

O

O

N

C(CH2)4

C

OH

(4)

COOH 1 nNH2 ( CH2 )5 COOH

Glycine

CH2

Cl

CH2 (2)

CH2

CH2

11. Which one of the following is an example of a thermosetting polymer?

COOH

CH

C

Cl 2-Chloro-1,3-butadiene

(3) HO } CH2 } CH2 } OH and

nH2N

CH2

Cl

(4) CH2

(2) Nylon-6,6 is an example of a fiber. Fibers are the polymers which have strong intermolecular forces between the polymer chains.

(4)

CH2

CH3

(AIPMT PRE 2012) Solution

CH

n

OH CH2

CH2

n

Amino caproic acid H2O

(AIPMT 2014) Solution

C O

CH2

NH

C ( CH2 )5 NH O

Nylon-2-nylon-6

9. Nylon is an example of (1) polyester. (3) polyamide. (2) polysaccharide. (4) polythene. (NEET 2013)

(4) Phenol-formaldehyde (bakelite) is an example of thermosetting polymer. 12. Which of the following organic compounds polymerizes to form the polyester Dacron? (1) Propylene and para-HO } (C6H4) } OH. (2) Benzoic acid and ethanol. (3) Terephthalic acid and ethylene glycol. (4) Benzoic acid and para-HO } (C6H4) } OH. (AIPMT 2014)

856

Chapter 13 | Polymers Solution (3) The reaction involved in the preparation of Dacron is nHOH2C

CH2OH 1 nHOOC

Ethylene glycol

(3) alternate cis-and trans-configuration. (4) random cis-and trans-configuration. (NEET-I 2016) Solution

COOH

(1) Natural rubber is cis polyisoprene.

Tetraphthalic acid H2C

C

CH

CH2

Polymerization

CH3 OCH2

CH2

O

O

C

C

Isoprene CH3

n

CH2

Dacron

13. Biodegradable polymer which can be produced from glycine and aminocaproic acid is (1) PHBV. (3) Nylon 6, 6. (2) Buna-N. (4) Nylon-2-nylon 6. (AIPMT 2015)

C C

CH2

C

(1)

(2)

(3)

(RE-AIPMT 2015) (4)

Solution (3) Nylon-6 can be prepared by a ring-opening polymerization of ε-caprolactum in the presence of water at 553–543 K. O 1

OC (CH2)5 NH3 1

(2H2O) 2508C

ε-Caprolactan (a cyclic amide) O NH

O

66 H2 H2 C H C H C C

H2 H2 C H C H C C

CH3

6

COOH

6

CI

H (CH ) H2 2 6 C) N C( C 2C H2 O

NH

H2 H2 C H C H C C NH2

CH3

66

(NEET-II 2016) Solution (3)

NH H2O

H

O

14. Caprolactam is used for the manufacture of (1) terylene. (3) nylon-6. (2) nylon-6, 6. (4) teflon.

NH

NH2

NH2

O n

O

CH3

CH2

H2 H2 C H C H C C

NH (CH2)5 C

O

CH2

C C

16. Which one of the following structures represents nylon 6,6 polymer?

NH2

NH

CH2

cis-Polyisoprene

Solution (2) Nylon-2-nylon-6 is an alternating polyamide copolymer of glycine (H2N } CH2 } COOH) and aminocaproic acid [NH2(CH2)5COOH]. It is biodegradable polymer and its structure is

H

nNH2 (CH2 )6NH2 1 nHOOC(CH2 )4COOH Hexamethylene diamine

Adipic acid

Polymerization

O

C (CH2)5 NH

15. Natural rubber has (1) all cis-configuration. (2) all trans-configuration.

C (CH2)5 NH nC Nylon-6 H N

O

H O (CH2)6

N

C

(CH2)4

Nylon-6,6

C n

Additional Objective Questions

ADDITIONAL OBJECTIVE QUESTIONS NCERT Exemplar Questions

CH3

1. Which of the following polymers of glucose is stored by animals? (1) Cellulose (3) Amylopectin (2) Amylose (4) Glycogen 2. Which of the following is not a semisynthetic polymer? (1) cis-Polyisoprene (3) Cellulose acetate (2) Cellulose nitrate (4) Vulcanized rubber 3. The commercial name of polyacrylonitrile is ____________. (1) Dacron (3) PVC (2) Orlon (acrilan) (4) Bakelite

( CH2

7.

C

CH3

CH2 C )n

CH3

CH3 is a polymer having monomer

units ____________. H

(3)

(1)

H

(2)

(4)

8. Which of the following polymer can be formed by using the following monomer unit?

4. Which of the following polymer is biodegradable? (1) ( CH2 C

H

CH CH2 )n

H2C

Cl

H2C H 2C

CN

(2) ( CH2 CH

CH CH2

(1) Nylon 6, 6 (2) Nylon 2-nylon 6

CH )n

CH2

N

O C CH2 CH2

(3) Melamine polymer (4) Nylon-6

(3) ( O CH CH2 C O CH CH2 C )n CH3

O

H

CH2CH3

H O

(4) ( N (CH2)6

N C

O

O (CH2)4

1. Which of the following is a copolymer? (1) Natural rubber (3) Gutta-percha (2) Synthetic rubber (4) Saran

C )n

5. In which of the following polymers ethylene glycol is one of the monomer units? (1) ( OCH2 CH2OOC

Exercise 1

CO )n

2. Which of the following is a linear polymer? (1) Nylon (2) Bakelite (3) Low-density polyethene (4) Melamine–formaldehyde polymer

(2) ( CH2 CH2 )n

3. When chains of one kind are attached to the backbone of a different polymer, the copolymer is called (1) random copolymer. (3) graft copolymer. (2) block copolymer. (4) none of these.

(3) ( CH2 CH

4. An example of a condensation homopolymer is (1) bakelite. (2) melamine-formaldehyde resin. (3) alkyl resin. (4) Perlon or nylon-6.

CH CH2 CH

CH2 )n

(4) ( O CH CH2 C O CH CH2 C )n CH3

O

CH2CH3

O

6. Which of the following statements is not true about low density polythene? (1) Tough (2) Hard (3) Poor conductor of electricity (4) Highly branched structure

5. Which of the following is a branched polymer? (1) Low-density polyethene (2) Polyester (3) Nylon (4) PVC 6. The monomer of synthetic rubber is (1) butadiene. (3) 2-methyl-1,2-butadiene. (2) chloroprene. (4) 2-methyl-1,3-butadiene.

857

858

Chapter 13 | Polymers 7. Which of the following polymers has amide linkages? (1) Nylon (3) Terylene (2) Bakelite (4) PVC 8. Starch is the condensation polymer of (1) α-glucose. (3) α-fructose. (2) β-glucose. (4) β-fructose. 9. Chemical name of melamine is (1) 2,4-diamino-1,3,5-triazine. (2) 2-amino-1,3,5-triazine. (3) 1,3,5-triamino-2,4,6-triazine. (4) 2,4,6-triamino-1,3,5-triazine. 10. Which of the following is not a biopolymer? (1) Proteins (3) Cellulose (2) Nucleic acids (4) Neoprene 11. The monomer of natural rubber is (1) butadiene. (3) 2-methyl-1,2-butadiene. (2) chloroprene. (4) 2-methyl-1,3-butadiene. 12. Which of the following is a chain-growth polymer? (1) Starch (3) Polystyrene (2) Nucleic acid (4) Protein 13. Which of the following sets contain only addition homopolymers? (1) Polyethene, natural rubber, cellulose (2) Starch, nylon, polyester (3) Teflon, bakelite, Orlon (4) Neoprene, PVC, polyethene 14. Which of the following polymers is synthesized using a free radical polymerization technique? (1) Terylene (3) Nylon-6,6 (2) Melamine (4) Teflon 15. Which of the following statements does not explain fibers? (1) Fibers are made of linear long chains which permit side-by-side alignment. (2) Structures of fibers are controlled by enthalpy instead of entropy. (3) There exist strong intermolecular forces to prevent slipping between chains. (4) Nylon and rubbers are typical examples of fibers. 16. Benzoyl peroxide acts as an initiator for (1) cationic polymerization. (2) anionic polymerization. (3) free radical polymerization. (4) condensation polymerization. 17. Which of the following statements is incorrect about polymers? (1) Polymers do not carry any charge. (2) Polymers have high viscosity. (3) Polymers scatter light. (4) Polymers have low molecular weight. 18. What is characteristic of the process of chain-growth polymerization? (I) In the beginning of the polymerization process, there are many oligomers (an oligomer is made from a few monomers, its chain length is short).

(II) High macromolecular weights are only reached at the very end of the polymerization process. (III) The macromolecular weight of the polymer chains grows approximately linear during the time required for the polymerization process. (IV) Polymer chains grow by addition of monomers to the polymer. (1) I, II (3) III, IV (2) II, III (4) II, IV 19. The species which can best serve as an initiator for the cationic polymerization is: (1) LiAlH4 (3) AlCl3 (2) HNO3 (4) BuLi 20. Which of the following type of forces are present in nylon-6,6? (1) van der Waals forces of attraction (2) Hydrogen bonding (3) Three-dimensional network of bonds (4) Metallic bonding 21. Polyethene is a(an) (1) elastomer. (2) fiber.

(3) thermoplastic polymer. (4) thermosetting polymer.

22. Which terms are correctly matched to their definitions? (I) Plastic – a long chain molecule synthesized by linking together many single parts. (II) Polymer – can be molded when hot and retains its shape when cooled. (III) Thermoplastic – can be molded when hot and retains its shape when cooled. (IV) Thermosetting plastic – can be molded when first prepared, but hardens irreversibly when cooled. (1) III, IV (3) I, II (2) II, III (4) I, IV 23. What small molecules can be released during stepgrowth polymerization? (1) H2O (3) methanol (2) NH3 (4) all of these 24. The correct repeating structural unit of polystyrene is (1)

CH2

CH

CH

CH2

C6H5 C6H5

(2)

CH2

CH

CH2

C6H5

(3)

CH

CH2

C6H5 CH2

C6H5

(4)

CH2

CH

CH C6H5

CH C6H5

CH2

CH

CH

CH2

Additional Objective Questions 25. Number average molecular mass (Mn ) and weight average molecular mass (Mw ) of synthetic polymers are related as: (1) Mn < Mw

(3) Mn = Mw

(2) Mn > Mw

(4) None of these.

Exercise 2 1. Which of the following polymers are step-growth polymers? (I) nylon (III) Teflon (II) HDPE (IV) PET (1) I, II (3) I, IV (2) II, III (4) III, IV 2. Which one of the following class of compounds is obtained by polymerization of acetylene? (1) Poly-yne (3) Poly-ester (2) Poly-ene (4) Poly-amide 3. In bakelite, the rings are joined to each other through (1) } CH2 }

(3) } O } O

C

8. Among the following, the weakest interparticle forces of attraction are present in (1) thermosetting polymers. (2) thermoplastic polymers. (3) fibers. (4) elastomers. 9. Which of the following is not an addition polymer? (1) Polystyrene (3) Polypropylene (2) PVC (4) Nylon 10. N1 , N2 , N3 ,…, are the number of molecules with average molecular masses M1 , M2 , M3 ,… respectively, then average molecular mass is expressed as 2 (1) ∑ N i Mi

OH

(2)

(III) If two bifunctional monomers (e.g. diamine and dicarboxylic acid) are used, slight deviations from 1:1 stoichiometry will prevent high macromolecular weights at the end of the polymerization process. (IV) Polymer chains can grow individually to high macromolecular weight at any time during the polymerization process. (1) I and II (3) I, II and III (2) II and III (4) II and IV

H

(4)

H

C

O

O

(2) ∑ N i Mi

4. Which compound would react to form the polymer shown?

∑ Ni

(3) Cl

(2)

(4)

(4) None of these

⋅

11. Structures of some important polymers are given below. Which one of these represents Buna-S?

n

(1)

(3) Both (1) and (2)

⋅

∑ N i Mi

CH3 OH

O

5. Which is not a step in the mechanism of a chaingrowth polymerization? (1) Initiation (3) Proliferation (2) Propagation (4) Termination 6. Low-density polyethene is prepared by (1) free radical polymerization. (2) cationic polymerization. (3) anionic polymerization. (4) Ziegler–Natta polymerization. 7. What is characteristic of the process of step-growth polymerization? (I) In the beginning of the polymerization process, there are many oligomers (an oligomer is made from a few monomers, its chain length is short). (II) High macromolecular weights are only reached at the very end of the polymerization process.

(1) (

CH2

C

(2) (

CH2

CH

CH CH

CH2 CH2

)n CH

CH2

)n

CH2

)n

C6H5

(3) (

CH2

CH

CH

CH2

CH CN

CI

(4) (

CH2

C

CH

CH2

)n

12. The polymer used for making contact lenses for eyes is (1) polyethylacrylate. (2) nylon-6. (3) polymethylmethacrylate. (4) polyethene. 13. Which of the following is a biodegradable polymer? (1) Cellulose (2) Polyethene (3) Polyvinyl chloride (4) Nylon-6

859

860

Chapter 13 | Polymers 14. Natural silk is a (1) polypeptide. (2) polysaccharide. (3) polychloroprene. (4) polyacrylonitrile. 15. Homopolymers are made from (1) only one type of monomers. (2) two different types of monomers. (3) three different types of monomers. (4) several different types of monomers. 16. Extent of stiffness of vulcanized rubber depends upon (1) temperature of vulcanization. (2) time of vulcanization. (3) amount of sulphur. (4) All of these. 17. Which of the following is not a cellulose product? (1) Gun-cotton (3) Rayon (2) Celluloid (4) Dacron 18. Identify the addition polymer that would be produced from 2-chloro-2-butene:

(1)

CI

H

CI

H

CI

H

C

C

C

C

C

C

CH3 CH3 CH3 CH3 CH3 CH3

(2)

CI

CI

CI

CI

CI

CI

C

C

C

C

C

C

CH3 CH3 CH3 CH3 CH3 CH3

(3)

(4)

C

C

C

C

C

C

CH3 CH3 CH3 CH3 CH3 CH3 CH3 CI

CH3 CI

CH3 CI

C

C

C

C

CH3 H

C

CH3 H

C

CH3 H

19. Complete the following free-radical addition equation RO 1 CH2

CH

CH3

(1) ROCH2CH2CH2 (2) ROH 1 CH2 (3) ROCH

CH

(4) ROCH2CH

CH

CH2

CH2 1 H2 CH3

20. Soft drinks and baby feeding bottles are generally made up of (1) polyester. (2) polyurethane. (3) polystyrene. (4) polyamide.

21. Which statement below does not occur during the formation of an addition polymer? (1) Free radicals initiate the process. (2) Certain double bonds in monomers are replaced with single bonds. (3) Propagation involves a reaction between two free radicals. (4) Termination occurs when the free radicals are used up. 22. Which of the following statements about low density polythene is FALSE? (1) It is used in the manufacture of buckets, dustbins etc. (2) Its synthesis requires high pressure. (3) It is a poor conductor of electricity. (4) Its synthesis requires dioxygen or a peroxide initiator as a catalyst. 23. Artificial silk is a (1) polypeptide. (3) polysaccharide. (2) polyvinyl chloride. (4) polyethene. 24. Terylene is a condensation polymer of ethylene glycol and (1) benzoic acid. (3) salicylic acid. (2) phthalic acid. (4) terephthalic acid. 25. In a polymer sample, 30% of molecules have a molecular mass of 20,000, 40% have 30,000 and the rest 60,000. What is the weight average molecular mass of the polymer? (1) 40,300 (3) 43,333 (2) 30,600 (4) 33,353 26. Which of the following is a fully fluorinated polymer? (1) Neoprene (3) Thiokol (2) Teflon (4) PVC 27. Which of the following is not a synthetic rubber? (1) Buna-S (3) Teflon (2) Thiokol (4) Neoprene 28. A copolymer of acrylonitrile and 1,3-butadiene is called (1) Buna-N. (3) neoprene. (2) polystrene. (4) Buna-S. 29. The polymer, which has nitrogen is (1) PVC (3) nylon. (2) terylene. (4) teflon. 30. Which of the following monomers would be most reactive towards cationic polymerization? (1) (2) (3) (4)

CH2 CH2 CH2 CH2

CH } OCH3 CH } CN CH } NO2 CH } CHO

31. The molecular formula of hexamethylene diamine adipate (monomer of nylon-6,6) is (1) C12H22O2N2 (3) C12H26O4N2 (2) C10H26O4N2 (4) C12H24O3N2

Additional Objective Questions 32. Which pair of compounds would react to form the polymer shown under appropriate reaction conditions? O

O

O Cl

(3) O

H2NCH2CH2CH2CH2CH2CH2NH2 1 HOCCH2CH2CH2CH2COH

(4) CH3 2

Na O

2 1

O Na 1

C CH3 O H3C

C

H CH2Cl

1 H2NCH2CH2NH2

33. Determine the repeating unit for the following polymer.

CH2CH2

O

O

CH2CH2

(1)

O

CH2CH2

O

CH2CH2

O

O

O

CH2CH2

O n

(2)

O

CH2CH2

O n

O

O

O

CH2CH2

(1) Assertion and Reason are true and the Reason is the correct explanation of the Assertion. (2) Assertion and Reason are true but the Reason is not a correct explanation of the Assertion. (3) Assertion is true but the Reason is false. (4) Assertion and Reason both are false.

(2)

1

O

In the following set of questions, two statements “Assertion” and “Reason” are given. Choose the correct answer from the following options:

HOCCH2CH2CH2CH2COH 1 HOCH2CH2OH

O

CH2CH2

Exercise 3

O

C

O

n

(1)

Cl

CH2CH2

n

H

H2NCH2CH2CH2CH2CH2CH2NH2 1

O

n

(4)

CCH2CH2CH2CH2CNCH2CH2CH2CH2CH2CH2N

O

(3)

1. Assertion: 1,3-Butadiene is the monomer for natural rubber. Reason: Natural rubber is formed through anionic addition polymerization. 2. Assertion: Bakelite is used for making electrical switches, plugs and boards. Reason: Bakelite is hard, has high melting point and electrical insulation property. Interparticle forces of attraction in it are due to hydrogen bonding. 3. Assertion: Natural rubber is flexible and extensible whereas vulcanized rubber is stiffer and has good tensile strength. Reason: Vulcanization is the process that results in cross-linking of the double bonds in natural rubber with sulphur. 4. Assertion: Polyester is a copolymer. Reason: The repeating structural unit of polyester is derived from two types of monomer units, dicarboxylic acids and diols. 5. Assertion: Teflon has high thermal stability and chemical inertness. Reason: Teflon is a thermosetting polymer. 6. Assertion: Polypropylene is an addition polymer. Reason: Addition polymerization generally occurs among molecules which contain double bonds. 7. Assertion: Polybutadiene is an example of chain growth polymer. Reason: In chain-growth polymers, the reactive species are either free radicals or ions. 8. Assertion: Natural rubber is an elastomer. Reason: The intermolecular forces of attraction are due to dipole–dipole interactions.

861

862

Chapter 13 | Polymers

ANSWER KEY NCERT Exemplar 1. (4)

2. (1)

3. (2)

6. (3)

7. (1)

8. (4)

2. (1)

3. (3)

4. (3)

5. (1)

4. (4)

5. (1)

Exercise 1 1. (4) 6. (2)

7. (1)

8. (1)

9. (4)

10. (4)

11. (4)

12. (3)

13. (4)

14. (4)

15. (4)

16. (3)

17. (4)

18. (3)

19. (3)

20. (2)

21. (3)

22. (1)

23. (4)

24. (2)

25. (1)

1. (3)

2. (2)

3. (1)

4. (1)

5. (3)

6. (1)

7. (3)

8. (4)

9. (4)

10. (3)

11. (2)

12. (3)

13. (1)

14. (1)

15. (1)

16. (3)

17. (4)

18. (1)

19. (4)

20. (3)

21. (3)

22. (1)

23. (3)

24. (4)

25. (3)

26. (2)

27. (3)

28. (1)

29. (3)

30. (1)

31. (4)

32. (3)

33. (2)

1. (4)

2. (1)

3. (1)

4. (1)

5. (3)

6. (1)

7. (2)

8. (3)

Exercise 2

Exercise 3

HINTS AND EXPLANATIONS Exercise 1

3. (3) The structure is A

1. (4) Vinylidene dichloride is polymerized with vinyl chloride to prepare copolymer saran. 2. (1) Nylon is a linear polymer with repeating unit

A

C

A

A

B

B

B

B

B NH

A

(CH2)5

A

A

A

B Graft copolymer

4. (4)

O

H

Low-density polyethene is a branched chain polymer, while melamine–formaldehyde polymer Bakelite are cross-linked polymers.

and

H2C H2C H2C

N

C

O

CH2 CH2

Caprolactum (Monomer)

O 533−543 K H2O

C

H (CH2)5

N

Nylon-6 (Polymer)

Hints and Explanations 5. (1) Low-density polyethene is a branched polymer whereas nylon, PVC and polyesters are linear polymers. 6. (2) The monomer of neoprene rubber (or synthetic rubber) is chloroprene or 2-chloro-1,3-butadiene. Cl CH2

C

CH

CH2

Chloroprene (2-chloro-1,3-butadiene)

7. (1) Nylon has an amide linkage. C

NH

(CH2)5

O n

8. (1) Starch is a condensation polymer of α-glucose. 9. (4) Melamine is 2,4,6-triamino-1,3,5-triazine. N 6

2

N 5

4

16. (3) Benzoyl peroxide acts as an initiator in free radical polymerization. 19. (3) AlCl3 is a Lewis acid, so in the presence of traces of water it serves as an initiator in the cationic polymerization. 20. (2) In nylon-6,6, there is orientation of linear polyamide molecules so that they are parallel to the fiber axis. This allows hydrogen bonds to form between } NH } and } C O groups on adjacent chains. 21. (3) Polythene softens on heating and can be moulded into any desirable shape. 24. (2)

1

H2N

van der Waals forces. Polyamides (nylon), polyesters, polyacrylonitrile, polyurethanes and isotactic polypropylene are the typical examples of fibers. However, rubber (polymer of neoprene) is not a fiber because rubbers are not formed by straight chains.

NH2

N3

NH2

10. (4) Neoprene is a synthetic manmade polymer whereas proteins, nucleic acids and cellulose are natural biopolymers.

CH3

H

C

C

H

C6H5

n Polystyrene

25. (1) From the expressions of number average and weight average mass.

11. (4) Natural rubber is an addition polymer containing isoprene (2-methyl-1,3-butadiene) units. CH2

H

Mn =

∑ Mi N i ∑ Ni

Mw =

∑ wi Mi ∑ N i Mi2 = ∑ wi ∑ N i Mi

CH2 C

C H

12. (3) Polystyrene undergoes addition polymerization, so it is an addition or chain-growth polymer. 13. (4) Homopolymers are formed from only one repeating unit. Neoprene, PVC and polyethene are addition homopolymers formed by addition polymerization of chloroprene, vinyl chloride and ethene, respectively. 14. (4) Terylene, melamine and nylon-6,6 are condensation polymers. Teflon, also known as polytetrafluoroethylene is an addition polymer. Its monomer, tetrafluoroethylene, is an alkene derivative (CF2 CF2) that polymerizes by free radical vinyl polymerization. 15. (4) Fibers are long-chain polymers in which chains are linked by strong intermolecular forces. So, the structure is favored by enthalpy instead of entropy. In elastomers, entropy is the dominating factor for their structures. The strong intermolecular forces in fibers, that prevent slipping between chains are, in general, strong hydrogen bonds, dipole–dipole and

we have Mn < Mw .

Exercise 2 2. (2) nCH

CH

Polymerization

(CH

CH)n CH

CH

(Poly-ene)

3. (1) Bakelite is formed by crosslinking of novolac through –CH2– bonds. 6. (1) R? 1

C C

R C C? C

C

C

C?

C

R

where R

C

} C2H5.

C

C

863

864

Chapter 13 | Polymers 8. (4) Elastomers have very weak intermolecular forces which can be overcome easily and the polymer chains can be stretched by applying small stress and they can regain shape on removing stress. 9. (4) Nylon is a condensation polymer formed by reaction between hexamethylene diamine and adipic acid. 11. (2) The reaction involved the preparation of Buna-S rubber is CH

manufacturing of buckets, dustbin etc., but instead is used in manufacturing of toys, carrier bags, etc. 23. (3) Artificial silk is a cellulose derivative, hence a polysaccharide. It gives thread of ash on burning. 24. (4) nHOH2C

CH2OH 1 nHOOC

Ethylene glycol

COOH

Terephthalic acid

CH2 Zn(OCOCH3)2 1 Sb2O3 4202460 K

1 CH2

CH

( CH2

CH CH

CH2 CH2

CH

CH

CH2 )n

OCH2

CH2

O

O

O

C

C

C6H5

12. (3) The polymer used in hard contact lenses is polymethylmethacrylate (PMMA). 13. (1) Cellulose is a biopolymer and it is biodegradable.

1 (2n21)H2O n

Dacron

25. (4) The weight average molecular mass of the polymer sample is calculated as.

14. (1) Natural silk is made of protein fiber consisting of fibroin, held together by various peptide linkages.

Weight average molecular mass

15. (1) Homopolymers are formed from only one repeating unit. For example, polyethene, styrene, etc.

=

30 × (20000)2 + 40 × (30000)2 + 30 × (60000)2 30 × 20000 + 40 × (30000) + 30 × 60000 = 43333

16. (3) Rubber containing about 5% sulphur is used in making rubber tyres. Similarly, vulcanized rubber containing about 30% sulphur is employed for making cases of batteries, etc.

26. (2) The structure of Teflon is } (CF2 } CF2 })n so it is a fully fluorinated polymer.

17. (4) Dacron is made up of ethylene glycol and terephthalic acid.

28. (1)

18. (1) The addition polymerization reaction of 2-chloro2-butene is

CH3

C

CH

CH3

CI

H

C

C

CH3 CH3

CI

CH

CH3

ROCH2CH

nCH2

CH

CH

CH2 1 nCH2

CH

CN Acrylonitrile

1,3-Butadiene

Heat Peroxide

n

19. (4) The given reaction represents chain initiation step, wherein the free radical adds to the monomer molecule. RO 1 CH2

27. (3) Teflon } (CF2 } CF2 })n.

CH3

20. (3) Soft drinks and baby feeding bottles are generally made up of polystyrene because it is a transparent thermoplastic material and floats over water. 21. (3) In the chain propagation step, a free radical attacks another monomer to produce yet another free radical and the process continues. The reaction between two free radicals leads to termination of reaction. 22. (1) LDP (Low density polyethene) is obtained when ethene is polymerized under high pressure and temperature in the presence of dioxygen or peroxide initiator. It is chemically inert and flexible and also a poor conductor of electricity. It is not used in the

[ CH2

CH

CH

CH2

CH2

CH ]n CN

Nitrile rubber (Buna-N)

29. (3) Nylon is a polyamide and hence contains nitrogen. 30. (1) Cationic polymerization is the polymerization in which cations are the reacting species. So, carbocations are the intermediates in the polymerization. The compound in which carbocation produced is very stable, will react relatively faster than the others. The monomer given in the first option produces most stable carbocation because electron releasing group } OCH3 stabilize it by resonance. Hence, it is most reactive in cationic polymerization. 33. (2) Cellulose on acetylation with acetic anhydride, in presence of H2SO4 (conc.) gives cellulose diacetate which is used for making threads of acetate rayon.

Chemistry in Everyday Life

14

C H A P T E R OU TLIN E 14.1 Drugs and Their Classification 14.2 Drug–Target Interaction 14.3 Therapeutic Action of Different Classes of Drugs

14.4 Chemicals in Food 14.5 Cleansing Agents

H N

R O

H

N O

S

CH3 CH3 COOH

Penicillin

Chemical compounds are certainly central to our lives. Hydrocarbons (compounds composed only of carbon and hydrogen) are used as fuel to power our cars and to heat our homes. Our bodies are fueled with compounds obtained from the food we eat in the form of sugars (carbohydrates), fats and proteins. This food is made more palatable by flavorings, is wrapped in plastic and is kept from spoiling with preservatives. Our clothes are made of organic compounds, out of which some come from plant and animal sources (cotton and wool) or are synthetic (nylon and dacron). These fabrics are made colorful with dyes. The other applications of chemistry are paints and varnishes, dyes, soaps and detergents, cosmetics, etc. When we are ill, we take drugs that may also be organic: aspirin relieves headaches, codeine suppresses coughs, and diazepam (valium) calms nerves. The applications of chemistry include cures for many diseases that have plagued the human race for thousands of years. Eventually, it may lead to a cure or vaccine for AIDS. These are only a few examples of how we use chemicals daily. In modern times, chemistry is being applied to produce materials for space travel, fuels and rocket propellants, all sorts of building materials, synthetic fibers, microchips for computers, and explosives and chemical weapons. Living organisms – from microbes to humans – have a range of chemical substances varying in complexity from water and simple salts to DNA (deoxyribonucleic acid) molecules containing tens of thousands of atoms. Four of the chemical elements – hydrogen, carbon, nitrogen and oxygen – make up approximately 95% of the mass of living matter. Small amounts of sulphur, phosphorus, calcium, sodium, potassium, chlorine, magnesium and iron, together with trace amounts of many other elements such as copper, manganese, zinc, cobalt and iodine, are also found in living organisms. In this chapter, we will focus on applications of chemistry to drugs, food additives and cleansing agents.

14.1 DRUGS AND THEIR CLASSIFICATION Drugs are chemicals of low molecular masses (~100–500μ) whose intake by a living system has a physiological effect. If the drug helps in treatment of disease and reduces pain and suffering of the body, it is known as a medicine. This branch of science which deals with treatment of various diseases with suitable chemical substances is known as chemotherapy. Each drug has a specific dose for a particular organism in which it can be considered safe. Beyond this dose, the same drug may be toxic and behave like a poison.

866

Chapter 14 | Chemistry in Everyday Life

14.1A Classification of Drugs The drugs are classified on the following basis: 1. On the basis of pharmacological effect: This classification is based on the effect of the drug on biological systems. Some examples include analgesics which are used as pain relievers, antipyretics to bring down body temperature and antibiotics to restrict the growth of microorganisms and prevent spread of infection. 2. On the basis of drug action: This classification is based on the effect of the drug on particular biochemical process. For example, COX (cyclooxygensae) inhibitors affect the action of enzyme cyclooxygensae which causes inflammation and pain. Similarly, antihistamines inhibit the action of histamine by blocking it from attaching to histamine receptors. Histamines produce symptoms of allergic reaction, such as runny nose or watery eyes, by increasing vascular permeability that results in flow of fluids from capillaries into the tissues. 3. On the basis of chemical structure: Drugs with similar chemical structures have similar actions on biological systems, for example, sulphonamides. O S

H 2N

NHR

O Sulphonamide

4. On the basis of molecular targets: This classification is based on the nature of biomolecules that the drugs target, such as carbohydrates, lipids, proteins, etc. This classification based on nature of target molecules is important for medicinal chemists to understand and predict the drug–target interaction.

14.2 DRUG–TARGET INTERACTION Drugs, when administered to a living system, interact with the biomolecules. The important biomolecules in the body are lipids and carbohydrates which form structures of the cell membrane. The nucleic acids code the genetic information for the cell. Proteins that function as biological catalysts are called enzymes; those which are involved in receiving communication signals are called receptors; and those which transport polar molecules across cell membrane are called carriers. A drug when administered may target any one of these biomolecules and the therapeutic effect results from this interaction.

14.2A Enzymes as Drug Targets Catalytic Action of Enzymes Almost all enzymes are proteins. To understand the interaction between the enzyme and the drug, it is essential to understand the catalytic action of the enzyme. The enzymes catalyze chemical reactions involving the substrate(s) which are molecule(s) on which the enzyme acts. In any enzyme-catalyzed reaction, the role of an enzyme is two-fold. 1. The first role of the enzyme is to bind the substrate (reactant or drug) and hold the substrate molecule at a site where the chemical reaction with the reagent takes place. This site is called the active site. The enzyme and the substrate combine to form an enzyme–substrate complex. The non-covalent forces that bind the substrate to the active site are: van der Waals forces, electrostatic forces, hydrogen bonding and hydrophobic interactions. These are the same forces that account for the conformations of the proteins themselves. Formation of the enzyme–substrate complex often induces a conformational change in the enzyme that allows it to bind the substrate more effectively. This is called an induced fit. Binding the substrate also often causes certain bonds to become strained, which are therefore more easily broken. The product of the reaction usually has a different shape from the substrate, and this altered shape, or

14.2 | Drug–Target Interaction

in some instances the intervention of another molecule, causes the complex to dissociate. The enzyme can then accept another molecule of the substrate, and the whole process is repeated. 2. The enzyme provides functional groups to carry out the reaction. The amino acids located in the active site are arranged so that they can interact specifically with the substrate (Fig. 14.1). Substrate Active site Enzyme Enzyme binds to substrate

Enzyme releases products

Figure 14.1 Enzyme binding with substrate.

Drug–Enzyme Interaction The drugs interact with the enzymes and inhibit their activity in two ways – by either blocking the active site or by inhibiting the catalytic activity of the enzyme. Such drugs are known as enzyme inhibitors. In the first type of interaction, also known as competitive inhibition, compounds that may or may not be structurally related to the natural substrate combine reversibly with the enzyme. These drugs may combine with the enzyme in the following two ways: 1. The drug (enzyme inhibitor) and the substrate compete for the same active site (Fig. 14.2). The drug prevents the substrate from binding to the enzyme. Such drugs are known as competitive inhibitors. 2. Some drugs show allosteric inhibition, in which the interaction of an enzyme molecule with a drug is at a site other than the active site. Conformational change at the active site is observed after non-competitive inhibitor binds at the allosteric site. As a result, the substrate cannot bind to the active site anymore. Substrate

Inhibitor

Inhibitor blocks active site

Active site

Enzyme–inhibitor complex

Enzyme

Figure 14.2 Inhibitor (drug) and substrate compete for active site.

When the drug–enzyme interaction involves the formation of strong covalent bonds, the binding is irreversible and the enzyme is permanently blocked. This enzyme–inhibitor complex then gets degraded and a new enzyme is synthesized (Fig. 14.3). Allosteric site

Substrate site

1

Inhibitor

Inhibitor

1

Enzyme

Figure 14.3 Allosteric inhibition of enzyme.

Substrate

Enzyme

Substrate

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14.2B Receptors as Drug Targets A drug will bind with a biological receptor (a protein involved in signal communication) if the drug possesses a specific three-dimensional arrangement of functional groups. In most situations, the physiological response produced by a drug is attributed to the interaction between the drug and a biological receptor site. A receptor is a region within a biological macromolecule (cell membrane) that can serve as a pouch with its active site on the outer side of cell membrane within which the drug molecule can fit (Fig. 14.4).

Outer surface of cell membrane

Chemical messenger (Drug mimics the action)

Binding site

Binding site (Drug receptor binding)

Induced fit

Interior of cell (a)

Cell membrane Message

(b)

(c)

Figure 14.4 (a) Cell receptor receives chemical messenger. (b) The messenger binds to the receptor and modifies its shape. (c) Receptor regains shape on removal of receptor. Initially, this mechanism was considered to work much like a lock and key. That is, a drug molecule would function as a key, either fitting or not fitting into a particular receptor. Extensive research on drug–receptor interactions has forced us to modify this simple lock-and-key model. It is now understood that both the drug and the receptor are flexible, constantly changing their shapes. As such, drugs can bind to receptors with various levels of efficiency, with some drugs binding more strongly and other drugs binding more weakly. Inside the body, some chemicals act as messengers for transferring signals between neurons and between neurons and muscles. These are known as chemical messengers. The receptors show selectivity of chemical messengers and act by receiving them at their active sites, following which the shape of the receptor site changes leading to transmission of message without physical relocation of the receptor protein. The drug binding to receptor site and inhibiting its natural function is called an antagonist. The drug which mimics the effect of natural messenger thus helping in signal transmission is called an agonist. While an antagonist blocks message transmission, an agonist is useful when natural messenger is lacking.

14.3 THERAPEUTIC ACTION OF DIFFERENT CLASSES OF DRUGS Some specific classes of drugs are discussed as follows. The chemical compounds used as drugs in these classes are also given with their structures. However, these complex structures are only for understanding and not evaluation.

14.3A Antacids Most of us have experienced occasional heartburn, commonly referred to as acidity, especially after eating food. Heartburn is caused by the buildup of excessive amounts of stomach acid (primarily HCl). This acid is used to digest the food we eat, but it can often travel back into the esophagus, causing the burning sensation called as heartburn. The symptoms of heartburn can be treated by using a mild base to neutralize the excess hydrochloric acid. Many different antacids are available in the market and can be purchased over the counter. All these antacids function in a similar way. They are all mild bases that can neutralize

14.3 | Therapeutic Action of Different Classes of Drugs

HCl in a proton-transfer reaction. For example, sodium bicarbonate deprotonates HCl to form carbonic acid, which then quickly degrades into carbon dioxide and water: O HO

O

2

Na1

H

1

O

CI

HO

Sodium bicarbonate

OH

1 Na1

CI

2

Carbonic acid

CO2 1 H2O

Constant use of sodium bicarbonate makes stomach alkaline, leading to the production of more acid. Alternatively, metal hydroxides such as aluminium and magnesium hydroxides are used. Being insoluble, they do not raise pH above 7. Acidity can only be controlled by antacids. Hyperacidity may cause ulcers, and surgery is the only alternative.

14.3B Antihistamines Histamine is an aromatic compound that plays many roles in the biological processes of mammals. It is involved in the regulation of gastric acid secretion in the stomach. For treating hyperacidity, years of research led to the discovery of cimetidine (Tegamet) which was used to control the overproduction of gastric acid associated with acid reflux diseases and ulcers. It binds with histamine receptors in stomach wall and prevents histamine to interact with the receptors. Ranitidine (Zantac) soon overcame cimetidine (Tegamet) as the best-selling drug worldwide. Histamine also plays a role in vasodilation and in the mediation of allergic reactions. It contracts the smooth muscles in the bronchi and stomach and is responsible for allergic responses to pollen and nasal congestion. Antihistaminic drugs such as brompheniramine (Dimetapp) and terfenadine (Seldane) inhibit the action of histamine by competitively binding with the receptors. These drugs do not regulate the secretion of gastric acid because they work on receptors different from those of antacids. H N

H 2N

N

HN H

N S

H3C

H Histamine

N H

N

CN

N H

CH3

H 3C

H N

H N CHNO2

Cimetidine

HO N

S

O

N

CH3

CH3

HCl

Ranitidine

N

OH

N

CH3 H 3C Terfenadine

CH3

Br

CH3

CH3 Brompheniramine

14.3C Neurologically Active Drugs Tranquilizers These drugs reduce emotional instability, tension, fear, anxiety and induce sleep. Such drugs are called anxiolytic agents. They are also used in relieving stress, anxiety and irritability. They are used to treat mild mental diseases and form an essential component of sleeping pills. Their use

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creates a sense of tranquility and well-being. The mechanism of action of tranquillizers differs, depending upon the receptors targeted by them. For example, antidepressant drugs act on enzymes which catalyze the degradation of a neurotransmitter noradrenaline. When the levels of this neurotransmitter fall, its signal-sending activity is reduced and depression sets in. Use of drugs which inhibit the enzyme affecting its degradation reduces the rate of metabolism of this neurotransmitter. The receptor targeted by it is activated for a longer duration of time and the effect of depression reduced. Thus, these drugs act as antidepressants. Examples of antidepressant drugs include iproniazid and phenelzine. N

N H

O

N H

N H

CH3

NH2

CH3

Iproniazid

Phenelzine

Barbituric acid and its derivatives, known as barbiturates, serve as good tranquillizers. Some important drugs of this class are veronal, amytal, nembutal, luminal, etc. These are hypnotic and induce sleep. H O HN O

N

O NH

H

O

Barbituric acid

N O

H 3C

O

O C2H 5

CH

HN O

C 2H5

N H

Veronal

CH3

CH3

O

Amytal

Another commonly used tranquilizers is valium (diazepam) which is a sedative and muscle relaxant, and is used in the treatment of anxiety, insomnia and seizures. Equanil is another tranquillizer used in depression and hypertension. Serotonin is an antidepressant drug and its use induces a feeling of well-being. CH3 O

N

CH2CH2NH2

HO N

Cl

O

N H

H2N

C

CH3 O

CH2

C

O CH2

O

CH3 Valium

Serotonin

Equanil

Chlordiazepoxide and meprobamate are mild tranquilizers used for relieving tension. H N

Cl

N

C

C

1 N

C6H5

CH3 O

CH2 2 O

Chlordiazepoxide

H2N

C

CH3 O

CH2

C

O CH2

(CH2)2CH3 Meprobamate

O

C

NH2

C

NH2

14.3 | Therapeutic Action of Different Classes of Drugs

Analgesics Analgesic is a pain reliever which when taken does not result in the loss of consciousness. Its use does not cause mental confusion or any other disturbance in the nervous system. Analgesics are classified as follows: 1. Non-narcotic (non-addictive) analgesics: Aspirin is prepared from salicylic acid, a compound found in the bark of the willow tree that has been used for its medicinal properties for thousands of years. Aspirin plays role in blocking the synthesis of prostaglandins which have many biological functions, including stimulating inflammation and inducing fever. With a decreased concentration of prostaglandins, the onset of inflammation is slowed and fevers are reduced. These drugs, thus, relieve skeletal pain and bring down body temperature. Use of asprin also prevents platelet coagulation and hence formation of blood clots. It is useful in prevention of heart attack. Paracetamol, acetaminophen, naproxen are some other examples of non-narcotic analgesics. O COOH O

H 3C

CH3

N

H

CH3

H3C O HO

O OH Paracetamol

Aspirin

NH

OH

O

H3CO

Acetaminophen

Naproxen

2. Narcotic (addictive) analgesics: Morphine is a very potent analgesic that is known to act on the central nervous system as a depressant (causing sedation and slower respiratory function) and as a stimulant (relieving symptoms of anxiety and causing an overall state of euphoria). Since morphine is addictive, it is primarily used for the short-term treatment of acute pain after surgery or delivery, and for terminally ill patients suffering from extreme pain. Other examples of narcotic analgesics are heroin and codeine. Heroin exhibits stronger activity than morphine and is extremely addictive. Codeine shows less activity than morphine and is less addictive. Codeine is currently used as an analgesic and cough suppressant. OH H

CH3

N Morphine

O

H 3C

O

O

H

H

OH

HO

H N

CH3

Codeine

14.3D Antimicrobials The pathogenic microorganisms such as bacteria, viruses, fungi, etc., are common causative agents of most of the diseases in living beings. Antimicrobials are substances that either kill microbes (microbiocidal) or prevent their growth (microbiostatic). These can be antibacterial, antiviral, antifungal or antiparasitic depending on the type of microorganism they target. The three important classes of antimicrobials are antibiotics, antiseptics and disinfectants. Antibiotics The term antibiotic was coined by Selman Waksman in 1942. It is derived from Greek – anti meaning against and bios meaning life. Antibiotic is a substance produced by a microorganism but is antagonistic to the growth of other microorganisms. Now-a-days, many antibiotics originally produced by microorganisms are being prepared by synthetic procedures. They act at low concentrations and inhibit the growth or kill the microorganisms by disrupting their normal metabolic activities.

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The development of antibiotics began in Germany with Paul Ehrlich in the late 1880s. He investigated arsenic-based structures for the treatment of syphilis and developed arsphenamine (salvarsan) as a medicine. The drug was less toxic to human beings but more toxic towards bacteria causing syphilis. He received a Nobel Prize in Medicine for his discovery in 1908. Ehrlich, while working on azodyes, also noticed similarity in the structure of azodyes with salvarsan (the – As = As – linkage in salvarsan was similar to – N = N – linkage of azodyes). In 1932, Ehrlich prepared prontosil, the first effective antibacterial agent structurally similar to salvarsan. NH2 AS HO

OH

AS

N

HO

N

H 2N Salvarsan

Azoyde

Prontosil exhibited a very curious property; it was found to be totally inactive against bacteria in vitro (literally “in glass”, in bacterial cultures grown in glass dishes). Its antibacterial properties were only observed in vivo (literally “in life”, when administered to living creatures, such as mice and humans). These observations inspired much research on the activity of prontosil; and in 1935, it was found that prontosil is metabolized in the body to produce a compound called sulphanil-amide. This compound interferes with bacterial cell growth and is the actual active drug. In a glass dish, prontosil is not converted into sulphanilamide, explaining why the antibacterial properties were only observed in vivo. After this discovery, scientists began designing new potential drugs based on structural modifications to sulphanilamide rather than prontosil. NH2 H 2N

N

O N

S

O NH2

S

O

NH2

O

Prontosil

Sulphanilamide

However, research was still needed to combat bacterial infections. In 1929, Alexander Fleming made a serendipitous discovery that had a profound impact on the field of medicine. He was growing a colony of Staphylococcus bacteria in a Petri dish that was accidentally contaminated with spores of Penicillium notatum mould. Fleming noticed that the spores of the mould prevented the colony from growing, and he summarized that the mould was producing a compound with antibiotic properties. He called the compound penicillin and its isolation and purification for clinical trials took 13 years. Penicillin was initially used to treat the wounded soldiers as early as 1943, and shortly thereafter was used on the general population. It has been credited with saving millions of lives, and for his discovery, Fleming was the co-recipient of the 1945 Nobel Prize in Physiology or Medicine. Antibacterials can either kill (cidal) the microbes or inhibit (static) their growth. A few examples of bactericidal antibiotics are penicillin, ofloxacin, cephalosporin, vancomycin, aminoglycosides, etc.; whereas the bacteriostatic antibiotics include erythromycin, tetracycline, chloramphenicol, spectinomycin, etc. (Table 14.1). Table 14.1 Bactericidal vs. bacteriostatic Bactericidal (Kill bacteria)

Bacteriostatic (Stop growth of bacteria)

Cephalosporin

Erythromycin

Aminoglycosides

Spectinomycin

Penicillin

Tetracycline

Ofloxacin

Chloramphenicol

14.3 | Therapeutic Action of Different Classes of Drugs

Based on their spectrum of action (range of microorganisms affected), the antibiotics can be of following kinds: 1. A broad spectrum antibiotic is one which can be used to treat a wide range of infections caused by both Gram positive and Gram negative bacteria. For example, ampicillin, amoxycillin, vancomycin, ofloxacin, chloramphenicol (administered orally as these are rapidly absorbed from gastrointestinal tract). 2. A narrow spectrum antibiotic is only effective against Gram positive or Gram negative bacteria. For example, Penicillin G. 3. A limited spectrum antibiotic is one which is effective only against a single organism or disease. For example, Dysidazirine is toxic against cancer cells. H N

R

H

S

OH

CH3 O

N O

O2N

CH3

CH

CH

CH2

OH

NH

CO

CHCl2

COOH Penicillin

Chloramphenicol

Antiseptics and Disinfectants Antiseptics are chemical substances that prevent the growth of microorganisms or kill them but are not harmful when applied to human tissues. These compounds are applied on wounds, cuts, disease or ulcers to cut down the chances of infection or sepsis. They differ from antibiotics as they cannot be ingested. They are also added to powders, soaps, deodorants, etc. to reduce the odor arising from bacterial decomposition. Some common examples are furacine, soframicine, dettol (a mixture of chloroxylenol and terpineol), iodine, etc. Tincture of iodine (2–3% solution in ethanol–water) and iodoform both are used to apply on wounds. Dilute solution of boric acid is used in the eyes. Bithionol is added to soaps to impart antiseptic properties. A number of dyes are also used as antiseptics. CH3

OH

Cl

OH

OH

Cl

S H3C

CH3 Cl

Chloroxylenol

H3C

OH CH3

Terpineol

Cl

Cl Bithionol

Disinfectants are used on non-living substances, such as instruments, floors, etc. These are harmful to  human tissues and cannot be applied directly to the wounds. For example, chlorine (0.2–0.4 ppm in aqueous solution), sodium hypochlorite and very low concentration of sulphur dioxide are disinfectants. Phenol solution (0.2%) is an antiseptic while 1% solution is a disinfectant.

14.3E Antifertility Drugs Progesterone is a progestin that prepares the uterus for nurturing a fertilized egg during pregnancy. During pregnancy, ovulation is inhibited by the release of estrogens and progestins from the placenta and ovaries. This process is mimicked by most birth control formulations, which generally contain a mixture of a synthetic estrogen (e.g., ethynyl estradiol) and a synthetic progestin (e.g., norethindrone). This mixture

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of compounds inhibits ovulation in much the same way that the body naturally inhibits ovulation during pregnancy. H3C H

H 3C

OH CH H

H

O

CH

H

H H

OH

H

HO Norethindrone

Ethynylestradiol

14.4 CHEMICALS IN FOOD Chemicals are added to food, add color, flavor, nutritive value or act as preservatives. They can thus act as: 1. 2. 3. 4. 5. 6. 7.

nutritional or dietary supplements such as minerals, vitamins and amino acids. fat emulsifiers and stabilizing agents. food colors (e.g. azo dyes). food flavors and artificial sweeteners. antioxidants and preservatives. antistaling and bleaching agents. flour improvers.

14.4A Artificial Sweetening Agents Sucrose (table sugar) and fructose are the most common natural sweeteners. We all know, however, that they add to our calorie intake and promote tooth decay. For these reasons, many people find artificial sweeteners to be an attractive alternative to the natural and calorie-contributing counterparts (Table 14.2). Table 14.2 Comparison of sweetners of artificial sweetners with cane sugar Artificial sweetener

Sweetness value in comparison with cane sugar

Saccharin

550

Aspartame

100

Alitame

2000

Sucralose

600

Saccharin (o-sulphobenzimide), 550 times as sweet as cane sugar, was discovered in 1879. It was a popular sweetener at one time and used as sodium or calcium salt. Saccharin is inert, harmless and an alternative substitute for diabetic patients. A common formulation involved a 10:1 mixture of cyclamate and saccharin that proved sweeter than either compound individually. Tests showed, however, that this mixture produced tumors in animals, and its use was banned by some countries. H N

SO2 SO3H

Cyclamate

N

O Saccharin

H

14.4 | Chemicals in Food

Perhaps the most successful and widely used artificial sweetener these days is aspartame, the methyl ester of a dipeptide formed from phenylalanine and aspartic acid. Aspartame is roughly 100 times as sweet as sucrose. It undergoes slow hydrolysis in solution, which limits its shelf life in products such as soft drinks. It also cannot be used for cooking and baking because it decomposes with heat. Alitame, on the other hand, is a compound related to aspartame, but with improved properties. It is more stable than aspartame and roughly 2000 times as sweet as sucrose. H

O N

H

H NH2 OCH3

O

H

CO2H

N

H

O

CH3

Aspartame

H NH2 H N O

CO2H

S

Alitame

Sucralose is a trichloro derivative of sucrose, which is an artificial sweetener. Sucralose is 600 times sweeter than sucrose and has many properties desirable in an artificial sweetener. Sucralose looks and tastes like sugar, is stable at the temperatures used for cooking and baking, and it does not cause tooth decay or provide calories. Cl

6

CH2OH

O

4 5

HO

2 1

3

OH

6

O

CH2Cl O

5 4

HO

OH 3

2

CH2Cl 1

Sucralose

14.4B Food Preservatives Food preservatives increase the shelf life of foods by stopping or slowing down the spoilage caused by microbes. Some common examples of food preservatives are common salt and sugar, vinegar, alcohol and vegetable oils. Other preservatives are sodium benzoate, EDTA and salts of sorbic acid and propanoic acid. These chemical compounds, used in small quantities as preservatives, are metabolized by the body and do not accumulate and cause harmful side effects. Antioxidants Food products containing unsaturated fats and oils have a short shelf-life. These oils are generally a mixture of triglycerides that are made up of long alkyl chains containing double bonds, which are susceptible to autooxidation. The hyperoxides that are formed on oxidation contribute to the rancid smell that develops over a time in food products containing unsaturated oils. Antioxidants are, therefore, added to slow the autooxidation process and help in preservation of food.

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The antioxidants used are radical inhibitors that are more reactive towards oxygen than the materials they are protecting. The two most common radical inhibitors used as food preservatives are butylated hydroxyl toluene (BHT) and butylated hydroxyl anisole (BHA). Sometimes BHT and BHA are used in combination with citric or ascorbic acids for more effective food preservation. OH (CH3)3C

OH C(CH3)3

(CH3)3C

CH3

C(CH3)3

OCH3

BHT

BHA

Some examples of use of antioxidants in foods are as follows: 1. Use of BHA in butter to increase its storage life. 2. Use of sulphur dioxide and sulphites (such as sulphite, bisulphite and metabisulphite of sodium) in wines, beer, sugar syrups and preserved cut and dried fruits and vegetables.

14.5 CLEANSING AGENTS Detergents are surfactants that clean fabric or skin when in dilute solutions. They improve the cleansing properties of water by helping in removal of dirt and oil stuck on fabrics or skin. They are mainly of two types – synthetic detergents and soaps. From 1930, a number of new cleansing agents that were superior in many respects to ordinary soap began to appear in the market. Since they were both synthetic organic products and detergents, they were called synthetic detergents, or syndets. A soap is distinguished from a synthetic detergent on the basis of chemical composition and not on the basis of function or usage.

14.5A Soaps Soap has been used as a cleansing agent for at least 2000 years. As people discovered long ago, soap could be made by heating animal fat together by wood ashes, which contain alkaline substances. Most soaps are produced from fats and oils, which contain three ester moieties. Upon treatment with a strong base, such as sodium hydroxide, the ester moieties are hydrolyzed, giving glycerol and three soap molecules. Alkaline hydrolysis (i.e., saponification) of triacylglycerols produces glycerol and a mixture of salts of long-chain carboxylic acids: O O

O R

OH

O

O R

O O

O Na

R

O

R

3 NaOH H2O

OH



O Na

R O

OH Glycerol

R

O Na

Sodium carboxylates “soap”

The identity of the alkyl chains can vary depending on the source of the fat or oil, but the concept is the same for all soaps. Specifically, the three ester moieties are hydrolyzed under basic conditions to produce soap molecules. The process is called saponification, from the Latin word sapo (meaning soap). The

14.5 | Cleansing Agents

soap so formed is separated from glycerol by a process called salting out. Potassium soaps are softer than sodium soaps for the skin, although both are soluble in water. O C −

O

H

H

H

H

H

H

H

H

H

H

H

H

H

H

C

C

C

C

C

C

C

C

C

C

C

C

C

C

H

H

H

H

H

H

H

H

H

H

H

H

H

H

Polar group (hydrophilic)

H

Non-polar group (hydrophobic)

Figure 14.5 Polar and non-polar groups in a soap molecule. Soaps are compounds that have a polar group on one end of the molecule and a non-polar group on the other end (Fig. 14.5). The polar group represents the hydrophilic region of the molecule (literally, “loves water”), while the non-polar group represents the hydrophobic region of the molecule (literally, “afraid of water”). Oil molecules are surrounded by the hydrophobic tails of the soap molecules, forming a micelle. The surface of the micelle is composed of all of the polar groups, rendering the micelle water soluble. Soap micelles are usually spherical clusters of carboxylate anions that are dispersed throughout the aqueous phase. The carboxylate anions are packed together with their negatively charged (and thus, polar) carboxylate groups at the surface and with their non-polar hydrocarbon chains on the interior. The sodium ions are scattered throughout the aqueous phase as individual solvated ions (Fig. 14.6). 1D

1D 2

&

1D

2² 2 &

2²

2 &

1D

2²

1D

2

$TXHRXVSKDVH ²

2

&

2

&

1D

2

²

0LFHOOH LQWHULRU

2

1D

&

2² 2

&

2²

1D

Figure 14.6 A portion of a soap micelle showing its interface with the polar dispersing medium.

Micelle formation accounts for the fact that soaps dissolve in water. The non-polar (and thus hydrophobic) alkyl chains of the soap remain in a non-polar environment – in the interior of the micelle. The polar (and therefore hydrophilic) carboxylate groups are exposed to a polar environment – that of the aqueous phase. As the surfaces of the micelles are negatively charged, individual micelles repel each other and remain dispersed throughout the aqueous phase. Soaps serve their function as “dirt removers” in a similar way. Most dirt particles (e.g., on the skin) become surrounded by a layer of an oil or fat. Water molecules alone are unable to disperse these greasy globules because they are unable to penetrate the oily layer and separate the individual particles from each other or from the surface to which they are stuck. Soap solutions, however, are able to separate the individual particles because their hydrocarbon chains can “dissolve” in the oily layer (Fig. 14.7). As this happens, each individual particle develops an outer layer of carboxylate anions and presents the aqueous

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Chapter 14 | Chemistry in Everyday Life

phase with a much more compatible exterior—a polar surface. The individual globules now repel each other and thus become dispersed throughout the aqueous phase. Shortly thereafter, they make their way down the drain. 1D

1D 1D

1D

1D

1D

1D 1D

1D

1D

&

2

²2

2

&

&

2² 2

2²

² 2 2

&

&

2²

1D

2

1D

1D ²

2 2 2² 2² 2 2² 2 & & & &

²

&

&

2

1D

1D

2

&

2²

2²

2

1D

&

2

2²

&

$TXHRXVSKDVH

2

2

²

&

2 ²

2

1D

1D

2

1D

1D

1D

1D

2

878

2LORUGLUW 6XUIDFH

Figure 14.7 Dispersal of a hydrophobic material (e.g., oil, grease, or fat) by a soap. Types of Soaps There are a variety of soaps available commercially to suit the requirement of the consumers. These can be toilet soaps (made by using good quality fats and oils, removal of excess alkali and addition of color and perfumes) and transparent soaps (made by dissolving in alcohol and evaporating the excess solvent). There are soaps that float in water (made by beating tiny bubbles before hardening), medicated soaps (containing substances antispecties like bithionol), shaving soaps (containing glycerol which increases time of drying) and laundry soaps (containing fillers such as sodium carbonate or silicate). Some soaps are made into chips and granules; soap powders and scouring soaps are also available. Laundry soaps contains borax, Na2CO3, sodium rosinate and sodium silicate, etc. Soaps in Hard Water The usefulness of soap is diminished in the presence of water that contains high concentrations of calcium ions (Ca2+) or magnesium ions (Mg2+). When soap is used with such water, called hard water, a precipitate is formed as a result of the following ion-exchange reaction. O 2

R

O

2

Na1

Soap

1 Ca21

O R

O

Ca21 1 2Na1

2

2

Soap scum

The generation of a precipitate, often called soap scum, limits the usefulness of soap. Thus, hard water is not fit for cleaning and washing purposes.

14.5B Synthetic Detergents To circumvent the problem of formation of soap scum with hard water, chemists have developed synthetic detergents that do not form precipitates when used with hard water and can even be used in ice cold water. Synthetic detergents (Fig. 14.8) function in the same way as soaps; they have long nonpolar alkane chains

14.5 | Cleansing Agents

O S O– Na+ O A sodium alkanesulfonate O O

S O– Na+ O

A sodium alkyl sulfate O S O– Na+ O

Figure 14.8 Typical synthetic detergents.

A sodium alkylbenzenesulfonate (ABS)

with polar groups at the end. The polar groups of most synthetic detergents are sodium sulfonates or sodium sulfates. (At one time, extensive use was made of synthetic detergents with highly branched alkyl groups. These detergents proved to be nonbiodegradable, and their use was discontinued.) Synthetic detergents offer an advantage over soaps; they function well in “hard” water, that is, water containing Ca2+, Fe2+, Fe3+, and Mg2+ ions. Calcium, iron, and magnesium salts of alkanesulfonates and alkyl hydrogen sulfates are largely water soluble, and thus synthetic detergents remain in solution. Soaps, by contrast, form precipitates—the ring around the bathtub—when they are used in hard water. Some of the detergents generally used are sodium lauryl sulphate and sodium p-dodecylbenzene sulphonate:

CH3(CH2)10CH2OSO32Na1 Sodium lauryl sulphate

CH3(CH2)10CH2

1 SO2 3 Na

Sodium p-dodecylbenzene sulphonate

Sodium lauryl sulphate and sodium p-dodecylbenzene sulphonate act in water in much the same way as sodium palmitate does. Like the palmitate ion, the negative lauryl sulphate ion has a long hydrocarbon chain that is soluble in grease and a sulphate group that is attracted to water:

CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2 Non-polar hydrophobic end, grease soluble

OSO2 3

Polar hydrophilic end, water soluble

Synthetic detergents are classified into following three categories depending upon the charge present on them: 1. Anionic detergents: These have a negative ionic group and are sodium salts of sulphonated long-chain alcohols or hydrocarbons (sodium salts of alkylbenzenesulphonates). Like soap, synthetic detergents also contain both hydrophobic and hydrophilic regions, but the identity of the hydrophilic region has

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Chapter 14 | Chemistry in Everyday Life

been modified. Rather than being sodium salts of carboxylic acids, synthetic detergents are sodium salts of alkybenzenesulphonates. aq. NaOH CH3 (CH2 )10CH2OH + H2SO4 → CH3 (CH2 )10 CH2OSO3H + H2O  → CH3 (CH2 )10CH2SO3− Na Lauryl alcohol

Lauryl hydrogen sulphate

+

Sodium lauryl sulphate

aq. NaOH CH3 (CH2 )11 − C6H5 + H2SO4 → CH3 (CH2 )11 − C6H5OSO3H + H2O  → CH3 (CH2 )11 − C6H5SO3− Na Dodecylbenzene

Dodecylbenzenesulphonic acid

+

Sodium dodecylbenzene sulphonate

In the case of detergents, an ion-exchange reaction does not generate a precipitate because the calcium salt is water soluble, hence no soap scum is formed. Sodium lauryl sulphate is in fact a common ingredient found in many shampoo formulations. Anionic detergents are also used in toothpastes. 2. Cationic detergents: These have a long hydrocarbon chain and a positive charge. They are quaternary ammonium salts with a positive charge on the N atom and acetates, chlorides or bromides as anions. A representative structure is as follows: +

CH3(CH2)14CH2

N(CH3)3

Grease soluble, hydrophobic

Water soluble, hydrophilic

These are germicidal and expensive. Cetyltrimethylammonium bromide and trimethylhexadecylammonium chloride are well-known cationic detergents. The former is a common ingredient of hair conditioners. CH3 CH3(CH2)15

N

1

CH3 Br2 Cl2

CH3

Cetyltrimethyl ammonium bromide/chloride

3. Non-ionic detergents: These are molecular substances and do not contain an ionic group. The molecule of a non-ionic detergent contains a grease-soluble component and a water-soluble component. The structure of a representative non-ionic detergent is as follows: CH3(CH2)10CH2 Grease soluble, hydrophobic

O

(CH2CH2O)7

CH2CH2OH

Water soluble, hydrophilic

Some of these substances are especially useful in automatic washing machines because they have good detergent, but low sudsing properties. They have a polar part to provide required water solubility. They form less foam than ionic detergents and are mainly used by the dishwashing liquids. An example is detergent formed by reaction of stearic acid with polyethylene glycol. −H O CH3 (CH2 )16COOH + HO(CH2CH2O)n CH2CH2OH  → CH3 (CH2 )16 COO(CH2CH2O)n CH2CH2OH 2

Soaps, sodium salts of long-chain fatty acids, are readily degraded by microorganisms in septic/treatment tanks. They are “soft” and biodegradable and hence do not cause any water pollution. A number of years ago a serious environmental pollution problem arose in connection with use of synthetic detergents.

Solved Examples

Some of the early detergents, which contained highly branched chain hydrocarbons, had no counterparts in nature. Therefore, enzymes capable of degrading them did not exist, and the detergents were essentially non-biodegradable and broke down very, very slowly. Microorganisms generally degrade long-chain C atoms by first converting the terminal CH3– group to –COOH group. Then they consume rest of the chain, two carbons at a time by further oxidation. Branching of chain blocks the process and makes detergents “hard” and non-biodegradable. As a result, these detergents get accumulated in water supplies, where they cause severe pollution problems due to excessive foaming and other undesirable effects. This in turn severely affects aquatic flora and fauna. Detergent manufacturers, acting on the recommendations of chemists and biologists, changed from a branched-chain alkyl benzene to a straight-chain alkyl benzene raw material. Detergents that contain the straight-chain alkyl groups are biodegradable. The structural difference is as follows: SO3– Na+

CH3CHCH2CHCH2CHCH2CH2 CH3

CH3

CH3

CH3

Non-biodegradable detergent CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2

SO3– Na+

Biodegradable detergent

While formulating a detergent, the branching, therefore, is kept minimum. In 1966, linear alkyl sulphonates (LAS) were introduced which were “soft” and biodegradable.

SOLVED EXAMPLES 1. Pick the odd one amongst the following on the basis of their medicinal properties giving reasons: chloroxylenol, phenol, chloramphenicol, bithional. Solution Chloramphenicol. It is an antibiotic while the other three are used as antiseptic. 2. The correct statement in respect of protein haemoglobin is that it (1) maintains blood sugar level. (2) acts as an oxygen carrier in the blood. (3) forms antibodies and offers resistance to diseases. (4) functions as a catalyst for biological reactions Solution (2) Human haemoglobin (Hb) is the protein that carries O2 through the bloodstream to the muscles, and consists of four polypeptide chains—two α chains that contain 141 amino acids and two β chains that contain 146 amino acids. Oxygen and haemoglobin bind in an easily reversible reaction to form oxyhaemoglobin.

3. Why is BHA added to butter? Solution BHA is added to butter to increase its storage life from months to years. 4. What are preservatives? How are they different from antioxidants? Solution Food preservatives increase the shelf life of foods by stopping or slowing down the spoilage caused by microbes. Preservatives are used to prevent the spoilage of food due to microbial growth, whereas antioxidants are the chemical substances which prevent the oxidation of food containing oils and fats. 5. Which artificial sweetener contains chlorine? (1) Aspartame (3) Sucralose (2) Saccharin (4) Alitame Solution (3) Sucralose is an artificial sweetener that contains chlorine.

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Chapter 14 | Chemistry in Everyday Life 6. Position of non-polar and polar parts in micelle is (1) polar at the outer surface but non-polar at the inner surface. (2) polar at the inner surface but non-polar at the outer surface. (3) distributed over all the surface. (4) on the surface only.

reacts to a particular medication will vary. Adverse effect may also be caused by drug interaction. 9. Give one example of a cationic detergent, write its structure also. Solution Cetyltrimethyl ammonium bromide is a cationic detergent. CH3

Solution (1) Micelles are formed by hydrophilic and hydrophobic groups in which hydrophobic (polar) groups are projected on the outer surface while hydrophobic (non-polar) groups are projected on the inner surface. As the concentration increases, the non-polar parts move away from the solvent, approach each other and form a cluster and remain in the interior. (Refer Fig. 14.6) 7. Among the following organic acids, the acid present in rancid butter is (1) pyruvic acid. (3) butyric acid. (2) lactic acid. (4) acetic acid.

CH3(CH2)15

N

+

CH3 Br –

CH3

10. Which two substances are mainly used as antioxidants in wine and sugar syrup? Solution Sulphur dioxide (SO2) gas and sulphite ( SO32− ) solutions are mainly used as antioxidants in wine and sugar syrup respectively. 11. How does the hard water use affect the cleaning action of detergents?

Solution

Solution

(3) Rancidity is development of any disagreeable flavour in fat. When butter becomes rancid it breaks down into glycerol and fatty acid. Rancid smell is due to formation of butyric acid (CH3CH2CH2 COOH).

Synthetic detergents function in the same way as soaps; they have long nonpolar alkane chains with polar groups at the end. The polar groups of most synthetic detergents are sodium sulfonates or sodium sulfates. Synthetic detergents offer an advantage over soaps; they function well in “hard” water, that is, water containing Ca2+, Fe2+, Fe3+, and Mg2+ ions. Calcium, iron, and magnesium salts of alkanesulfonates and alkyl hydrogen sulfates are largely water soluble, and thus synthetic detergents remain in solution.

8. Why do drugs cause side effects? Solution Drugs cause some side effects because they affect different people in different ways. The way a person

SOLVED PREVIOUS YEARS’ NEET QUESTIONS 1. Which one of the following is employed as a tranquilizer? (1) Chlorpheninamine (2) Equanil (3) Naproxen (4) Tetracycline (AIPMT PRE 2010) Solution (2) Equanil is tranquilizer used in the depression and hypertension. 2. Which one of the following is employed as antithistamine? (1) Omeprazole (2) Chloramphenicol (3) Diphenyl hydramine (4) Norothindrone (AIPMT PRE 2011)

Solution (3) Diphenyl hydramine is used as antihistamine. 3. Chloroamphenicol is an (1) antifertility drug. (2) antihistaminic. (3) antiseptic and disinfectant. (4) antibiotic-broad spectrum. (AIPMT MAINS 2012) Solution (4) Chloramphenicol is a broad spectrum antibiotic, it is rapidly absorbed from the gastrointestinal tract and is given orally in case of typhoid, dysentery, acute fever and certain form of urinary infections.

Additional Objective Questions 4. Antiseptics and disinfectants either kill or prevent growth of microorganisms. Identify which of the following statements is not true? (1) A 0.2% solution of phenol is an antiseptic while 1% solution acts as a disinfectant. (2) Chlorine and iodine are used as strong disinfectants. (3) Dilute solutions of boric acid and hydrogen peroxide are strong antiseptics. (4) Disinfectants harm the living tissues. (NEET 2013) Solution (3) Dilute solutions of boric acid and H2O2 are mild antiseptics. 5. Bithionol is generally added to the soaps as an additive to function as a/an (1) dryer. (3) antiseptic. (2) buffering agent. (4) softener. (AIPMT 2015) Solution (3) Bithionol is added to the soaps as an additive to function as an antiseptic.

6. Which of the following is an analgesic? (1) Novalgin (2) Penicillin (3) Streptomycin (4) Chloromycetin (NEET-I 2016) Solution (1) Novalgin is an analgesic. It is used for the treatment of pain. Penicillin and streptomycin are antibiotics. Chloromycetin is used for the treat of various infections. 7. Mixture of chloroxylenol and terpineol acts as (1) antiseptic. (3) antibiotic. (2) antipyretic. (4) analgesic. (NEET 2017) Solution (1) Antiseptics are chemical substances that prevent the growth of microorganisms or kill them but are not harmful when applied to human tissues. Mixture of mixture of chloroxylenol and terpineol is known as dettol which is an example of antiseptic.

ADDITIONAL OBJECTIVE QUESTIONS NCERT Exemplar Questions 1. Which of the following statements is not correct? (1) Some antiseptics can be added to soaps. (2) Dilute solutions of some disinfectants can be used as antiseptic. (3) Disinfectants are antimicrobial drugs. (4) Antiseptic medicines can be ingested. 2. Which is the correct statement about birth control pills? (1) Contain estrogen only. (2) Contain progesterone only. (3) Contain a mixture of estrogen and progesterone derivatives. (4) Progesterone enhances ovulation. 3. Which statement about aspirin is not true? (1) Aspirin belongs to narcotic analgesics. (2) It is effective in relieving pain. (3) It has antiblood clotting action. (4) It is a neurologically active drug. 4. The most useful classification of drugs for medicinal chemists is _________. (1) on the basis of chemical structure. (2) on the basis of drug action. (3) on the basis of molecular targets. (4) on the basis of pharmacological effect.

5. Which of the following statements is correct? (1) Some tranquilizers function by inhibiting the enzymes which catalyse the degradation of noradrenaline. (2) Tranquilizers are narcotic drugs. (3) Tranquilizers are chemical compounds that do not affect the message transfer from nerve to receptor. (4) Tranquilizers are chemical compounds that can relieve pain and fever. 6. Salvarsan is arsenic containing drug which was first used for the treatment of ____________. (1) syphilis (3) meningitis (2) typhoid (4) dysentry 7. A narrow spectrum antibiotic is active against _______________. (1) gram positive or gram negative bacteria. (2) gram negative bacteria only. (3) single organism or one disease. (4) both gram positive and gram negative bacteria. 8. The compound that causes general antidepressant action on the central nervous system belongs to the class of _____________. (1) analgesics (3) narcotic analgesics (2) tranquilizers (4) antihistamines

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Chapter 14 | Chemistry in Everyday Life 9. Compound which is added to soap to impart antiseptic properties is __________. (1) sodium laurylsulphate (2) sodium dodecylbenzenesulphonate (3) rosin (4) bithional 10. Equanil is __________. (1) artificial sweetener (2) tranquilizer

(3) antihistamine (4) antifertility drug

11. Which of the following enhances leathering property of soap? (1) Sodium carbonate (2) Sodium rosinate (3) Sodium stearate (4) Trisodium phosphate 12. Glycerol is added to soap. It functions __________. (1) as a filler. (2) to increase leathering. (3) to prevent rapid drying. (4) to make soap granules. 13. Which of the following is an example of liquid dishwashing detergent? (1) CH3(CH2)10 (2) C9H19

CH2OSO32Na1 O

(3) CH3

( CH2

CH3(CH2)15

O

)5

CH2CH2OH

1 SO2 3 Na

CH3

(4)

CH2

N

1

CH3 Br2

CH3

14. Polyethyleneglycols are used in the preparation of which type of detergents? (1) Cationic detergents (2) Anionic detergents (3) Non-ionic detergents (4) Soaps 15. Which of the following is not a target molecule for drug function in body? (1) Carbohydrates (3) Vitamins (2) Lipids (4) Proteins 16. Which of the following statements is not true about enzyme inhibitors? (1) Inhibit the catalytic activity of the enzyme. (2) Prevent the binding of substrate. (3) Generally a strong covalent bond is formed between an inhibitor and an enzyme. (4) Inhibitors can be competitive or noncompetitive. 17. Which of the following chemicals can be added for sweetening of food items at cooking temperature and does not provide calories? (1) Sucrose (3) Aspartame (2) Glucose (4) Sucralose

18. Which of the following will not enhance nutritional value of food? (1) Minerals (3) Vitamins (2) Artificial sweeteners (4) Amino acids

Exercise 1 1. Which of the following is used as antipyretic? (1) Paracetamol (3) Chloramphenicol (2) Chloroquine (4) LSD 2. An ester used as medicine is (1) ethyl acetate. (3) methyl salicylate. (2) methyl acetate. (4) ethyl benzoate. 3. Paracetamol is (1) both antipyretic and analgesic. (2) only analgesic. (3) only antipyretic. (4) antimicrobial. 4. Antiseptic properties of iodoform are due to the decomposition of (1) iodoform itself. (2) free iodine. (3) methyl iodide. (4) None of these. 5. Aspirin is an acetylation product of (1) p-dihydroxybenzene. (2) o-hydroxybenzoic acid. (3) o-dihydroxybenzene. (4) m-hydroxybenzoic acid. 6. Oral contraceptive drugs contain (1) mestranol. (3) Both (1) and (2). (2) norethindrone. (4) None of these. 7. Penicillin was first discovered by (1) A. Fleming. (3) G. Thompson. (2) L. Pasteur. (4) A. Noble. 8. The substances which affect the central nervous system and induce sleep are called (1) antipyretics. (3) analgesics. (2) tranquillizers. (4) None of these. 9. Detergents are prepared by the action of H2SO4 followed by neutralization by starting with (1) cholesterol. (3) cyclohexanol. (2) lauryl alcohol. (4) p-nitrophenol. 10. Chemically heroin is (1) morphinediacetate. (2) morphinemonoacetate. (3) morphinedibenzoate. (4) morphinemonobenzoate. 11. Which of the following is not an antipyretic? (1) Aspirin (3) Barbituric acid (2) Paracetamol (4) Phenacetin 12. A substance which can act both as an antiseptic and disinfectant is (1) aspirin. (3) analgin. (2) phenol. (4) sodium pentothal.

Additional Objective Questions 13. Antiseptics are different from disinfectants as (1) antiseptics merely inhibit the growth and disinfectant kill the microorganisms. (2) antiseptics are used against microorganisms while disinfectants are used against insects. (3) antiseptics are used only over skin while disinfectants can be taken orally also. (4) antiseptics are used over living tissues while disinfectants cannot be used over living tissues. 14. Which of the following is not an analgesic? (1) Ibuprofen (3) Naproxen (2) Diclofenac sodium (4) Ofloxacin 15. Which of the following statements is not true? (1) Some disinfectants can be used as antiseptics at low concentration. (2) Sulphadiazine is a synthetic antibacterial. (3) Ampicillin is a natural antibiotic. (4) Aspirin is both analgesic and antipyretic. 16. Which of the following food additives is an antioxidant? (1) Butylated hydroxyanisole (2) Cyclamate (3) Sodium metabisulphite (4) Amaranth 17. Drug which helps to reduce anxiety and brings about calmness is a(an) (1) tranquilizer. (3) analgesic. (2) diuretic. (4) antihistamine. 18. Antiallergy drugs are (1) antimicrobials. (2) antihistamines.

(3) antivirals. (4) antifungals.

19. The following compound is used as COOH

(3) analgesic. (4) pesticide.

20. Which of the following is used as a local anaesthetic agent? (1) Diazepam (3) Mescaline (2) Procaine (4) Seconal 21. Which of the following represents an analgesic? (1) Equanil (3) Novalgin (2) Quinine (4) All of these 22. Barbiturates are used as (1) analgesics. (2) food preservatives.

25. Phenyl salicylate can be used as (1) antiseptic. (3) analgesic. (2) antipyretic. (4) disinfectant. 26. Select the incorrect statement regarding detergents: (1) They can be used well in acidic solutions. (2) They can be used with hard water without any problem. (3) They can be used on woolen garments too. (4) None of these. 27. Aspartame is one of the good artificial sweeteners whose use is limited to cold foods and soft drinks because (1) it has very low boiling point. (2) it gets dissociated at cooking temperature. (3) it is sweetener at low temperature only. (4) it is insoluble at higher temperatures. 28. Choose the correct statement. (1) Saccharin is 650 times sweeter than sugar. (2) Aspartame is 550 times sweeter than sugar. (3) Sucralose is 160 times sweeter than sugar. (4) Alitame is 2000 times sweeter than sugar. 29. Which of the following represents a synthetic detergent? (1) C15H31COOK (2) CH3[CH2]6COONa (3) C12H25

SO3Na

(4) All of these

OCOCH3

(1) antiseptic. (2) antibiotic.

24. Point out the wrong statement. (1) Penicillin was discovered by A. Fleming. (2) Phenacetin is a very important antibiotic. (3) Chloroquine is an antimalarial drug. (4) Ether is an anaesthetic.

(3) antipyretics. (4) tranquilizers.

23. Which of the following statement is not true about the drug barbital? (1) It causes addiction. (2) It is a non-hypnotic drug. (3) It is a tranquilizer. (4) It is used in sleeping pills.

30. Which of the following is an anionic detergent? (1) Trimethylstearyl ammonium chloride (2) Sodium p-dodecyl benzene sulphonate (3) Cetyltrimethyl ammonium chloride (4) None of these

Exercise 2 1. 2-Acetoxybenzoic acid is used as an (1) antimalarial. (3) antiseptic. (2) antidepressant. (4) antipyretic. 2. The antibiotic used for the treatment of typhoid is (1) penicillin. (2) chloramphenicol. (3) terramycin. (4) sulphadiazine. 3. An antibiotic contains nitro group attached to aromatic nucleus in its structure. It is (1) penicillin. (3) tetracycline. (2) streptomycin. (4) chloramphenicol.

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Chapter 14 | Chemistry in Everyday Life 4. A large number of antibiotics have been isolated from (1) bacteria Actinomycetes. (2) bacteria Staphylococcus. (3) bacteria Rhizobium. (4) acids. 5. Which of the following is not a surfactant? (1) CH3–(CH2)15–N+(CH3)3Br− (2) CH3–(CH2)14–CH2NH2 (3) CH3–(CH2)16–CH2OSO2Na+ (4) OHC–(CH2)14–CH2–COO−Na+ 6. Morphine is an (1) antiseptic. (2) analgesic.

7. Which of the following detergents can cause maximum pollution? CH2CH2CH(C4H9)(CH2)3CH(C4H9) CH2CH2SO3Na

(2) CH3(CH2)11

SO3Na SO3Na

(3)

OH

(1)

(4) Detergents are always pollution-free

Cl

(3) CONH2

CONH2

OH

Cl

(2)

(3) antibiotic. (4) anaesthetic.

(1) CH3CH(C4H9)

13. The correct structure of the drug paracetamol is

(4) COCH3

NHCOCH3

14. In the following sets of compounds, the one which contains only medicinal compounds is (1) Alizarin, phenacetin, morphine (2) Aspirin, genatin, violet, phenolphthalein (3) Boric acid, chloramphenicol, aspirin (4) 9-Oxdecanoic acid, boric acid, morphine 15. LSD (lysergic acid diethyl amide) is (1) sweetening agent. (2) synthetic fiber. (3) psychedelic drug. (4) antibiotic.

8. Which of the following is not an insecticide? (1) DDT (3) Carbaryl (Sevin) (2) BHC (4) Aspartame

16. Glycerol is not used in (1) explosives. (3) soaps.

9. Which of the following is a broad-spectrum antibiotic? (1) Streptomycin (3) Ampicillin (2) Penicillin (4) Chloramphenicol

17. Which of the following can possibly be used as analgesic without causing addiction and modification? (1) Morphine (2) Diazepam (3) Tetrahydrocational (4) N-Acetyl-p-aminophenol

10. Which of the following is an example of nonbiodegradable detergent? (1) CH3

(CH2)11

(2) CH3

(CH2)9CH

SO3Na SO3Na

CH3

(3)

CH3

CH3

CH3

CHCH2

CH

SO3Na

(4) CH3(CH2)10CH2OSO3Na 11. Various phenol derivatives, tincture of iodine (2−3% I2 in water/alcohol) and some dyes like methylene blue are (1) antiseptics. (3) analgesics. (2) disinfectants. (4) antipyretics. 12. Which set has different class of compounds? (1) Tranquilizers: Equanil, heroin, valium (2) Antiseptics: Bithional, Dettol, boric acid (3) Analgesics: Naproxen, morphine, aspirin (4) Bactericidal: Penicillin, aminoglycosides, ofloxacin

(2) cosmetics. (4) matches.

18. What chemical is added to washing powders to keep them dry? (1) Sodium perborate (3) Sodium sulphate (2) Sodium carbonate (4) None of these 19. Which of the following is used as an antidepressant drug? (1) Valium (3) Methadrine (2) Opium (4) None of these 20. Monosodium glutamate, a food additive is a (1) sweetener. (3) flavor enhancer. (2) flavoring agent. (4) antioxidant. 21. What should be the feature of detergent molecule structure so as to be biodegradable? (1) It should be saturated. (2) It should be unsaturated. (3) Branching should be maximum. (4) Branching should be minimum. 22. Identify the bactericidal antibiotic from the following: (1) Tetracycline (2) Ofloxacin (3) Erythromycin (4) Chloramphenicol

Additional Objective Questions 23. Which of the following is an anionic detergent? (1) CH3(CH2)16CH2OSO3Na (2) CH3(CH2)16N(CH3)3Cl− (3) CH3(CH2)16COO(CH2CH2O)nCH2CH2OH (4) C5H5SO3Na 24. Which of the following is not used as an antacid? (1) Magnesium hydroxide (2) Sodium carbonate (3) Sodium bicarbonate (4) Aluminium phosphate 25. Which of the following drugs is an analgesic? (1) Sulphaguanidine (2) Paludrine (3) Analgin (4) Iodex 26. Which of the following is an artificial edible color? (1) Saffron (3) Tetrazine (2) Carotene (4) Melamine 27. Substance used for the preservation of colored fruit juices is (1) benzene. (2) benzoic acid. (3) phenol. (4) sodium metabisulphite. 28. 2-Acetoxy benzoic acid is used as (1) antimalarial. (2) antidepressant. (3) antiseptic. (4) antipyretic. 29. Which substance is added to soaps to impart antiseptic properties? (1) Aspirin (2) Chloroxylenol (3) Bithional (4) Phenol 30. The drug, which is not a tranquillizer, is (1) ibuprofen. (3) luminal. (2) veronal. (4) seconal. 31. Which of the following does not behave as a surfactant? (1) Soap (2) Detergent (3) Phospholipid (4) Triglycerides 32. The drug used for prevention of heart attacks is (1) aspirin. (2) valium. (3) chloramphenicol. (4) cephalsporin. 33. An antibiotic effective in treatment of pneumonia, bronchitis, etc., is (1) penicillin. (2) patalin. (3) chloromycetin. (4) tetracycline.

34. The drug given during hypertension is (1) streptomycin. (3) equanil. (2) chloroxylenol. (4) aspirin. 35. The antibiotic used for curing tuberculosis is (1) penicillin. (3) tetracycline. (2) streptomycin. (4) chloromycetin. 36. Which of the following is an antihistamine drug? (1) Chlorpheniramine (2) Ciprofloxacin (3) Chloramphenicol (4) Chloroquine

Exercise 3 In the following set of questions, two statements “Assertion” and “Reason” are given. Choose the correct answer from the following options: (1) Assertion and Reason are true and the Reason is the correct explanation of the Assertion. (2) Assertion and Reason are true but the Reason is not a correct explanation of the Assertion. (3) Assertion is true but the Reason is false. (4) Assertion and Reason both are false. 1. Assertion: Aspirin can cause ulcers in the stomach. Reason: The ester group in aspirin gets hydrolyzed to give salicylic acid which causes ulcers in the stomach. 2. Assertion: Cimetidine is an antacid. Reason: It prevents the interaction of histamine, thereby producing less hydrochloric acid in the stomach. 3. Assertion: A micelle formed by sodium stearate in water has the carboxylic group at the surface of water. Reason: The addition of sodium stearate reduces the surface tension of water. 4. Assertion: Tetracycline is bacteriostatic antibiotic. Reason: It inhibits the growth of organisms. 5. Assertion: A 0.2% solution of phenol is an antiseptic while 1% solution is a disinfectant. Reason: Disinfectants kill microorganisms but are harmful to human tissues. 6. Assertion: A low level of noradrenaline in the body causes depression. Reason: Antidepressant drugs catalyze the degradation of noradrenaline. 7. Assertion: Ranitidine is used to treat hyperacidity while brompheniramine is used to treat hypersensitivity. Reason: Both the drugs are antihistamines.

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Chapter 14 | Chemistry in Everyday Life

ANSWER KEY NCERT Exemplar 1. (4)

2. (3)

3. (1)

4. (3)

5. (1)

6. (1)

7. (1)

8. (2)

9. (4)

10. (2)

11. (2)

12. (3)

13. (2)

14. (3)

15. (3)

16. (3)

17. (4)

18. (2)

2. (3)

3. (1)

4. (1)

5. (2)

Exercise 1 1. (1) 6. (3)

7. (1)

8. (2)

9. (2)

10. (1)

11. (3)

12. (2)

13. (4)

14. (4)

15. (3)

16. (1)

17. (1)

18. (2)

19. (3)

20. (2)

21. (3)

22. (4)

23. (2)

24. (2)

25. (1)

26. (4)

27. (2)

28. (4)

29. (3)

30. (2)

2. (2)

3. (4)

4. (1)

5. (2)

Exercise 2 1. (4) 6. (2)

7. (1)

8. (4)

9. (4)

10. (3)

11. (1)

12. (1)

13. (2)

14. (3)

15. (3)

16. (4)

17. (1)

18. (3)

19. (3)

20. (3)

21. (4)

22. (2)

23. (1)

24. (2)

25. (3)

26. (3)

27. (2)

28. (4)

29. (3)

30. (1)

31. (4)

32. (1)

33. (3)

34. (3)

35. (2)

1. (1)

2. (1)

3. (1)

4. (1)

5. (2)

6. (3)

7. (2)

36. (1)

Exercise 3

HINTS AND EXPLANATIONS Exercise 1 1. (1) Paracetamol is used as antipyretic drug to reduce fever.

5. (2) o-Hydroxybenzoic acid on acetylation gives aspirin. OH COOH

2. (3) Methyl salicylate (oil of winter green) is used to treat joint and muscular pains. 3. (1) Paracetamol is used to reduce fever and it relieves pain as well that is why it is used both as antipyretic and analgesic.

6. (3) They are progesterone derivatives which are used for the control of menstrual cycle and to suppress ovulation.

4. (1) Iodoform on decomposition produces iodine which comes into contact with the skin and thus imparts antiseptic properties.

7. (1) A. Fleming worked many years and first discovered penicillin having empirical formula C9H11O4SN2R.

Hints and Explanations 8. (2) Tranquilizers affect central nervous system as they are used for the mental relief and are sleep inducing. 9. (2) The process yields anionic detergents. C11H23CH2OH  → C11H23CH2OSO3H conc. H2SO4

Lauryl alcohol

Lauryl hydrogen sulphate

→ C11H23CH2OSO3−Na + NaOH

Sodium lauryl sulphate (anionic detergent)

10. (1) It is a derivative of morphine and therefore of opium. 11. (3) Aspirin and paracetemol are antipyretics. Phenacetin is a very moderate and safe antipyretic, whereas barbituric acid acts as an antipyretic tranquilizer. 12. (2) Phenol in smaller concentration (0.2%) acts as an antiseptic while in a 1% concentration of solution, it acts as disinfectant. 13. (4) Disinfectants are harmful to the human tissue.

compounds; also the free amino acids present in it undergo racemization at high temperatures producing significant amounts of unnatural D-type amino acid. 28. (4) Alitame has the maximum sweetness (2000 times that of sugar), followed by sucralose (650 times), saccharin (300 times) and aspartame (160 times). 30. (2) Sodium salts of sulphonated long-chain alcohols or hydrocarbons act as anionic detergents.

Exercise 2 1. (4) It is the IUPAC name for aspirin and is used as an antipyretic, that is, for reducing fever. It is also used as an analgesic. 2. (2) It is a specific antibiotic used for the treatment of typhoid. 3. (4) The structure of chloramphenicol is OH

14. (4) Ofloxacin is a chemotherapeutic antibiotic. 15. (3) Ampicillin is a semi-synthetic antibiotic. 16. (1) BHA acts in contact with O2, light and heat to retard food degradation. OH C(CH3)3

OCH3 BHA

17. (3) It acts as both antiseptic and disinfectant depending on low or high concentration. Phenol in smaller concentration (0.2%) acts on antiseptic while in 1% concentration of solution, it acts as disinfectant. 18. (2) Antihistamines are drugs used to reduce allergy caused by proteins.

O2N

CH

CH

CH2

OH

NH

CO

CHCl3

Chloramphenicol

4. (1) After the discovery of penicillin, the bacteria Actinomycetes were exploited to give a series of antibiotics. 5. (2) For a molecule to behave as an surfactant, it should have both hydrophobic as well as hydrophilic part. –NH2 is the hydrophilic part and the hydrocarbon chain is the hydrophobic part. 6. (2) Morphine produces analgesia and sleep. It is a very potent drug. 7. (1) Since it has a long chain, it undergoes slow biodegradation. 8. (4) Aspartame is an artificial sweetener.

19. (3) It is aspirin which is a non-narcotic analgesic used to prevent heart attacks and relieving skeletal pain. It is also used as antipyretic.

9. (4) Chloramphenicol is used for both gram positive and gram negative bacteria, hence is a broadspectrum antibiotic.

20. (2) Procaine is a local anesthetic drug of the aminoester group.

10. (3) It is an alkyl benzene sulphonate detergent and is non-biodegradable as it contains branched chains.

21. (3) Quinine is an antimalarial drug whereas equanil is used for treating anxiety and nervousness.

11. (1) Lesser concentration of phenol acts like antiseptic. Tincture of iodine in ethanol is used to treat wounds. Methylene blue is used as an antidote to cyanide and has veterinary use as antiseptic.

22. (4) Barbituric acid along with its 5,5-disubsituted derivatives are used as tranquilizers, that is, sleep inducing agents. 23. (2) Barbital is used as a hypnotic drug to reduce mental tension. 24. (2) Phenacetin is an antipyretic. 25. (1) Phenyl salicylate or salol is used as an antiseptic.

12. (1) Heroin is a narcotic drug. 13. (2) Paracetemol is a derivative of alcohol. 15. (3) LSD is used to alter thinking process, and hence is a psychedelic drug.

26. (4) Hard water inhibits the detergent’s action.

17. (1) Morphine is a very potent drug and its chronic use leads to addiction.

27. (2) Aspartame has the strong tendency to react with other food ingredients forming unique chemical

18. (3) Na2SO4 acts like silica gel and hence keeps washing powders free from moisture.

889

890

Chapter 14 | Chemistry in Everyday Life 19. (3) Also called as mood boosters, vitaline and cocaine. 20. (3) Monosodium glutamate is added to food to enhance its flavor. 21. (4) Branching enhances degradation process. 22. (2) Ofloxacin is used as a bactericidal antibiotic as it kills the microorganisms in the body.

30. (1) Ibuprofen is a moderate pain killer. 31. (4) For a molecule to behave as surfactant, it should have both hydrophobic and hydrophilic parts; triglycerides do not fulfill the criteria. 32. (1) Aspirin is used for the prevention of heart attacks because of its anti-blood clotting action. 33. (3) Chloromycetin is a broad range antibiotic.

23. (1) As it is sodium salt of sulphonated long chain − alcohol. SO3 group makes it anionic.

34. (3) Equanil is used for the treatment of symptoms of anxiety and nervousness.

24. (2) Sodium carbonate is used in washing powder.

35. (2) Streptomycin is the antibiotic widely used for the treatment of tuberculosis.

25. (3) Analgin is used for relieving pain and fever (analgesic). 26. (3) Tetrazine is a food preservative which imparts color to the food. 29. (3) Bithional retains its antiseptic property in all kinds of environment.

36. (1) Chlorpheniramine is an antihistamine drug. Ciprofloxacin and chloramphenicol are antibiotics, and chloroquine is used as an antimalarial drug.

2020

neet QUESTION PAPER

1. Reaction between benzaldehyde and acetophenone in presence of dilute NaOH is known as (1) (2) (3) (4)

6. Reaction between acetone and methylmagnesium chloride ­followed by hydrolysis will give (1) tert-butyl alcohol. (3) isopropyl alcohol.

cross Cannizzaro’s reaction. cross Aldol condensation. aldol condensation. cannizzaro’s reaction.

2. Measuring Zeta potential is useful in determining which property of colloidal solution? (1) Stability of the colloidal particles. (2) Size of the colloidal particles. (3) Viscosity. (4) Solubility. 3. A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of which of the following?

7. The following metal ion activates many enzymes, participates in the oxidation of glucose to produce ATP and with Na, is responsible for the transmission of nerve signals. (1) Calcium (3) Iron

(1) (2) (3) (4)

(2) Potassium (4) Copper

8. Which of the following is a basic amino acid? (1) Tyrosine (3) Serine

(2) Lysine (4) Alanine

9. Identify compound X in the following sequence of reactions. CH3

(1) −R effect of − CH3 groups. (2) Hyperconjugation. (3) −I effect of −CH3 groups. (4) + R effect of −CH3 groups.

CHO Cl2/hν

4. The correct option for free expansion of an ideal gas under ­adiabatic condition is q < 0, ΔT = 0 and w = 0 q > 0, ΔT > 0 and w > 0 q = 0, ΔT = 0 and w = 0 q = 0, ΔT < 0 and w > 0

(2) isobutyl alcohol. (4) sec-butyl alcohol.

X

H2 O 373 K

CHCl2

(1)

(2)

Cl

(3)

CCl3

CH2Cl

(4)

5. Match the following. Column I (Oxide) (a) CO (b) BaO (c) Al2O3 (d) Cl2O7

Column II (Nature) (i) Basic (ii) Neutral (iii) Acidic (iv) Amphoteric

Which of the following is correct option? (a) (b) (c) (d) (1) (iii) (iv) (i) (ii) (2) (iv) (iii) (ii) (i) (3) (i) (ii) (iii) (iv) (4) (ii) (i) (iv) (iii)

Solomons Organic Chemistry for NEET 2020.indd 1

10. Which of the following is the correct order of increasing field strength of ligands to form coordination compounds? (1) F− < SCN− < C2O24 − < CN− (2) CN− < C2O24 − < SCN− < F− (3) SCN− < F− C2O24 − < CN− (4) SCN− < F− < CN− < C2O24 − 11. Which of the following is a cationic detergent? (1) (2) (3) (4)

Cetyltrimethyl ammonium bromide. Sodium dodecylbenzene sulphonate. Sodium lauryl sulphate. Sodium stearate.

26-Mar-21 4:45:40 PM

P2 neet 2020 12. Which one of the followings has maximum number of atoms? (1) (2) (3) (4)

1 g of O2(g) [Atomic mass of O = 16] 1 g of Li(s) [Atomic mass of Li = 7] 1 g of Ag(s) [Atomic mass of Ag = 108] 1 g of Mg(s) [Atomic mass of Mg = 24]

(1) (2) (3) (4)

14. Which of the following amine will give the carbylamine test?

(1)

NHC2H5

(2)

NH2

(3)

8.314 J mol−1K−1 × 300 K × ln(3 × 1013) −8.314 J mol−1K−1 × 300 K × ln(4 × 1013) −8.314 J mol−1K−1 × 300 K × ln(2 × 1013) 8.314 J mol−1K−1 × 300 K × ln(2 × 1013)

21. For the reaction, 2Cl(g) → Cl2(g), the correct option is

IUPAC Official Name (i) Mendelevium (ii) Lawrencium (iii) Seaborgium (iv) Darmstadtium (2) (d), (iv) (4) (b), (ii)

N(CH3)2

If the equilibrium constant (KC) is 2 × 1013 at 300 K, the value of ΔrG° at the same temperature will be: (1) (2) (3) (4)

13. Identify the incorrect match. Name (a) Unnilunium (b) Unniltrium (c) Unnilhexium (d) Unununnium (1) (c), (iii) (3) (a), (i)



ΔrH < 0 and ΔrS > 0 ΔrH < 0 and ΔrS < 0 ΔrH > 0 and ΔrS > 0 ΔrH > 0 and ΔrS < 0

22. Find out the solubility of Ni(OH)2 in 0.1 M NaOH. Given that the ionic product of Ni(OH)2 is 2 × 10−15 (1) 1 × 10−13 M (3) 2 × 10−13 M

(2) 1 × 108 M (4) 2 × 10−8 M

23. On electrolysis of dil. sulphuric acid using platinum (Pt) ­electrode, the product obtained at anode will be

NHCH3

(4)

(1) H2S gas. (3) hydrogen gas.

(2) SO2 gas. (4) oxygen gas.

24. Which of the following is not correct about carbon monoxide? (1) The carboxyhaemoglobin (haemoglobin bound to CO) is less stable than oxyhaemoglobin. (2) It is produced due to incomplete combustion. (3) It forms carboxyhaemoglobin. (4) It reduces oxygen carrying ability of blood.

15. Paper chromatography is an example of (1) (2) (3) (4)

thin layer chromatography. column chromatography. adsorption chromatography. partition chromatography.

16. A mixture of N2 and Ar gases in a cylinder contains 7 g of N2 and 8 g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of N2 is [Use atomic masses (in g mol ): N = 14, Ar = 40] (1) 15 bar (2) 18 bar (3) 9 bar (4) 12 bar −1

17. The number of protons, neutrons and electrons in respectively, are (1) 71, 71 and 104 (3) 71, 104 and 71

175 71

Lu,

(2) 175, 104 and 71 (4) 104, 71 and 71

(2) 1000 s (4) 200 s

19. Identify a molecule which does not exist. (1) C2 (3) He2

(2) O2 (4) Li2

20. Hydrolysis of sucrose is given by the following reaction.

 Sucrose + H2O   Glucose + Fructose 

Solomons Organic Chemistry for NEET 2020.indd 2

(1) 3 (3) 1

(2) 4 (4) 2

26. Elimination reaction of 2-Bromopentane to form pent-2-ene is

18. The rate constant for a first order reaction is 4.606 × 10−3 s−1. The time required to reduce 2.0 g of the reactant to 0.2 g is (1) 500 s (3) 100 s

25. The number of Faradays(F) required to produce 20 g of calcium from molten CaCl2 (Atomic mass of Ca = 40 u) is

(a) (b) (c) (d) (1) (3)

β-Elimination reaction. Follows Zaitsev rule. Dehydrohalogenation reaction. Dehydration reaction. (b), (c), (d) (2) (a), (b), (d) (a), (b), (c) (4) (a), (c), (d)

27. What is the change in oxidation number of carbon in the following reaction? CH4(g) + 4Cl2 (g) → CCl4 (l) + 4HCl (g) (1) −4 to +4 (2) 0 to −4 (3) +4 to +4 (4) 0 to +4 28. Which of the following alkane cannot be made in good yield by Wurtz reaction? (1) n-Heptane (2) n-Butane (3) n-Hexane (4) 2,3-Dimethylbutane

26-Mar-21 4:45:42 PM

P3

Question paper 29. Sucrose on hydrolysis gives (1) (2) (3) (4)

(a) (1) (iii) (2) (i) (3) (iii) (4) (iii)

α-D-Glucose + β-D-Fructose α-D-Fructose + β-D-Fructose β-D-Glucose + α-D-Fructose α-D-Glucose + β-D-Glucose

30. Identify the incorrect statement. (1) Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals. (2) The oxidation states of chromium in CrO24 − and Cr2O72 − are not the same. (3) Cr2+ (d4) is a stronger reducing agent than Fe2+ (d6) in water. (4) The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and to form complexes.

36. The mixture which shows positive deviation from Raoult’s law is (1) (2) (3) (4)

Only MgCl2 NaCl, MgCl2 and CaCl2 Both MgCl2 and CaCl2 Only NaCl

OH

(1)

(a) CO2(g) is used as refrigerant for ice-cream and frozen food. (b) The structure of C60 contains twelve six carbon rings and twenty-five carbon rings. (c) ZSM-5, a type of zeolite, is used to convert alcohols into gasoline. (d) CO is colorless and odorless gas. (1) (b) and (c) only (2) (c) and (d) only (3) (a), (b) and (c) only (4) (a) and (c) only 33. An increase in the concentration of the reactants of a reaction leads to change in (2) collision frequency. (4) heat of reaction.

34. The calculated spin only magnetic moment of Cr2+ ion is (1) 5.92 BM (3) 3.87 BM

(2) 2.84 BM (4) 4.90 BM

35. Match the following and identify the correct option. Column I (a) CO(g) + H2(g)

Column II (i) Mg(HCO3)2 + Ca(HCO3)2 (b) Temporary hardness (ii) An electron deficient of water hydride (c) B2H6 (iii) Synthesis gas (d) H2O2 (iv) Non-planar structure

Solomons Organic Chemistry for NEET 2020.indd 3

I + C2H5I

(2)

+ C2H5OH

OH

(3)

32. Identify the correct statements from the following.

(1) threshold energy. (3) activation energy.

acetone + chloroform chloroethane + bromoethane ethanol + acetone benzene + toluene

37. Anisole on cleavage with HI gives

31. HCl was passed through a solution of CaCl2, MgCl2 and NaCl. Which of the following compound(s) crystallize(s)? (1) (2) (3) (4)

(b) (c) (d) (iv) (ii) (i) (iii) (ii) (iv) (i) (ii) (iv) (ii) (i) (iv)

I + CH3I

(4)

+ CH3OH

38. Urea reacts with water to form A which will decompose to form B. B when passed through Cu2+(aq), deep blue color solution C is formed. What is the formula of C from the following? (1) Cu(OH)2 (3) CuSO4

(2) CuCO3Cu(OH)2 (4) [Cu(NH3)4]2+

39. The freezing point depression constant (Kf) of benzene is 5.12 K kg mol−1. The freezing point depression for the s­ olution of molality 0.078 m containing a non-electrolyte solute in ­benzene is (rounded off upto two decimal places): (1) 0.40 K (3) 0.20 K

(2) 0.60 K (4) 0.80 K

40. Which of the following oxoacid of sulphur has − O−O− l­ inkage? (1) H2S2O8, peroxodisulphuric acid (2) H2S2O7, pyrosulphuric acid (3) H2SO3, sulphurous acid (4) H2SO4, sulphuric acid 41. Identify the correct statement from the following. (1) Vapor phase refining is carried out for Nickel by van Arkel method. (2) Pig iron can be moulded into a variety of shapes. (3) Wrought iron is impure iron with 4% carbon. (4) Blister copper has blistered appearance due to evolution of CO2. 42. Which of the following is a natural polymer? (1) Polybutadiene (2) Poly (butadiene-acrylonitrile) (3) Cis-1, 4-polyisoprene (4) Poly (butadiene-styrene)

26-Mar-21 4:45:43 PM

P4 neet 2020 43. An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is (1) (3)

4

4 × 288 pm (2) × 288 pm 3 2

3 2 × 288 pm (4) × 288 pm 4 4

CH

CH

CH3

(3)

CH2

CH2

CH3

(4)

44. An alkene on ozonolysis gives methanal as one of the product. Its structure is CH2

CH

45. Which of the following set of molecules will have zero dipole moment?

CH2

(1)

(1) Nitrogen trifluoride, beryllium difluoride, water, 1,3-dichlorobenzene. (2) Boron trifluoride, beryllium difluoride, carbon dioxide, 1,4-dichlorobenzene. (3) Ammonia, beryllium difluoride, water, 1,4-dichlorobenzene. (4) Boron trifluoride, hydrogen fluoride, carbon dioxide, 1,3-dichlorobenzene.

CH2CH2CH3

(2)

ANSWER KEY 1. (1) 11. (1) 21. (2) 31. (4) 41. (2)

2.  (1) 12. (2) 22. (3) 32. (2) 42. (3)

3.  (2) 13. (2) 23. (4) 33. (2) 43. (3)

4.  (3) 14. (3) 24. (1) 34. (4) 44. (1)

5.  (4) 15. (4) 25. (3) 35. (3) 45. (2)

6.  (1) 16. (1) 26. (3) 36. (3)

7.  (2) 17. (4) 27. (1) 37. (3)

8.  (3) 18. (1) 28. (1) 38. (4)

9.  (1) 19. (3) 29. (1) 39. (1)

10.  (3) 20. (3) 30. (2) 40. (1)

ANSWERS WITH EXPLANATION 1. The following reaction is an example cross Cannizzaro’s ­reaction. H C

O + CH3

C

d+

CH3

CH3



C

d+

tert-Butyl cation (3°) (most stable)

C+

CH3

H Isopropyl cation (2°)



Answer (2)

4. Under adiabatic condition, we know that Δq = 0. For free expan-

CH3

sion, w = 0. Therefore, ΔU = 0 and ΔT = 0. Answer (1)

Answer (1)

3. This order of stability of carbocations can be explained on the basis of hyperconjugation.

Solomons Organic Chemistry for NEET 2020.indd 4

> CH3

2. Measuring Zeta potential is useful in determining the stability of colloidal solution. 

C+

CH3

d+

O CH

d+

CH3

O

293 K/ dil. NaOH3

CH

d+



Answer (3)

5. The correct match is as follows: (a)→(ii): CO is neutral in nature. (b)→(i): BaO is basic in nature. (c)→(iv): Al2O3 is amphoteric in nature. (d)→(iiii): Cl2O7 is acidic in nature. 

Answer (4)

26-Mar-21 4:45:45 PM

Question paper 6. The reaction involved is O CH3

C

14. Primary amines when heated with chloroform and ethanolic potassium hydroxide form pungent isocyanides or carbylamines. This reaction is exclusive to primary amines and is used as a test to distinguish them from secondary and tertiary amines.

OMgCl −

+

CH3 + CH3MgCl

CH3

C

CH3

CH3

NH2

H2O

N + CHCl3

CH3 CH3

C

Answer (1)

7. Potassium ion activates many enzymes, participates in the ­oxidation of glucose to produce ATP and with Na, is responsible for the transmission of nerve signals. Answer (2)

8. Lysine is basic amino acid and considered essential, since the human body is not capable of synthesizing it. 

Answer (3)

9. The chlorination of alkyl chain in methyl benzene forms benzal chloride which on hydrolysis gives benzaldehyde. CH3

CHCl2

Cl2 /hn

15. Paper chromatography is an example of partition chromatography. It is widely used for separation of colored compounds, highly polar compounds and polyfunctional species such as sugars and amino acids. 

Answer (4)

16. The partial pressure of N2 can be calculated as:

Number of moles of N2 =

Mass 7 = = 0.25 mol Molar mass 28



Number of moles of Ar =

Mass 8 = = 0.2 mol Molar mass 40



Thus, partial pressure of N2 can be calculated as

CHO

No.of moles of N2

=



H2O

(X)

Benzaldehyde



Answer (1)

10. The arrangement of ligands in the order of their abilities to split the energies of the d-orbitals is called the spectrochemical series. The correct order is SCN− < F− < C2O24 − < CN− . 

Answer (3)

11. Cetyltrimethylammonium bromide is a well-known cationic detergent. The structure is as follows: CH3 CH3(CH2)15

N

+

Total no.of moles

=



Answer (1)



Answer (4)

18. For the first order of reaction, we have a 2.303 log  a−x t

a 2.303 log k a−x 2.303 t= × log 10 4.606 × 10−3 2.303 × 1= 500s t= 4.606 × 10−3



Answer (1)

19. He2 does not exist as the bond order of He2 is zero.

13. The correct match is:



Unnilunium (At. no. 101) → Mendelevium Unniltrium (At. no. 103) → Lawrencium Unnilhexium (At. no. 106) → Seaborgium Unununnium (At. no. 111) → Roentgenium 

20. ΔG can be calculated as

Solomons Organic Chemistry for NEET 2020.indd 5

(1)

t=

Answer (1)

Least the atomic mass greater will be the number of atoms. Hence, 1 g of Li(s) will have maximum number of atoms.  Answer (2)

k=

Substituting the values in Eq. (1), we get

CH3



0.25 × 27 =15 bar 0.45

17. In Lu, since the atomic number is 71, therefore, the number of protons = number of electrons = 71. Number of neutrons = 175 − 71 = 104



12. Number of atoms can be calculated as: 1 = × NA Atomic massof an element

× Total pressure

175 71



CH3 Br –



+ 3 KCl + 3H2O

Answer (3)

373 K

Toluene

Heat



tert -Butyl alcohol



+ 3 KOH

C

OH

CH3



P5

Answer (3) o

Answer (2)



ΔGo = − RT ln KC = −8.314 J mol−1K −1 × 300 K × ln(2 × 103 ) Answer (3)

26-Mar-21 4:45:48 PM

P6 neet 2020 21. The reaction 2Cl(g) → Cl2 (g) involves the formation bond. Hence, it will be an exothermic reaction ( ΔH = −ve) . Also, since number of molecules are getting decreased from reactant to product which implies decrease in randomness. Hence, ΔS < 0. 

Answer (2)

22. The reaction involved is as follows: Ni(OH)2 → Ni2 + (aq ) + 2OH− (aq )



We know 0.1 M NaOH = 0.1 M OH ions



2 × 10−15 = 2 × 10−13 M (0.1)2 Answer (3)

23. The reactions involved are as follows: H2SO4 (dil.) → 2H+ + SO24 −



Answer (2)

31. Solubility of NaCl is less than CaCl2 and MgCl2. Thus, when HCl is passed through the saturated solution of NaCl containing CaCl2 and MgCl2, NaCl only precipitates, because its Ionic product (Qsp) exceeds solubility product. Answer (4)

32. Option (A): Incorrect. Solid CO2 is used as refrigerant for ­ice-cream and frozen food.

Ksp = [ Ni2 + ][OH− ]2 2 × 10−15 = S × (0.1)2 ⇒ S =





−



30. The oxidation states of chromium in CrO2− and Cr2O72− is 4 same, that is, +6.

At anode: 2H2O → 4 H+ + 4e− + O2 

Answer (4)

24. CO has a strong affinity towards hemoglobin and forms a stable complex with it known as carboxyhemoglobin (CoHb). Oxyhemoglobin is less stable than CoHb. 

Answer (1)



Option (B): Incorrect. Fullerene consists of a fused system of five- and six-membered rings. It contains 20 six membered rings and 12 five membered rings.  Answer (2) 33. When the concentration of the reactant is increased, the n­ umber of molecules per unit volume increases, thereby increasing the collision frequency, which ultimately causes the increase in reaction rate. 

Answer (2)

34. The spin only magnetic moment can be calculated as

Cr2+ =[Ar]4s0 3d 4

25. CaCl2 → Ca2 + + 2Cl−

Atomic weight valency 40 = = 20 g 2 1 equivalent = 1 × 96500 C = 1 F 

Therefore,

Eq. weight =

26. CH3

CH2

CH2

CH

CH3

alc. KOH

μspin = 4( 4 + 2)

Answer (3) CHCH3

35. (a)→(iii): CO is made by blowing steam through red or white hot coke at 473–1273 K to obtain a mixture of CO and H2 known as water gas or synthesis gas. Water gas

Pent-2-ene

The reaction is an example β-elimination reaction where ­bromine gets eliminated from β-carbon leading to d­ ehydrohalogenation. It also follows Saytzeff ’s rule.  Answer (3) +4

−1

27. C H4 (g) + 4 Cl2 (g) → CCl4 (l) + 4 H Cl(g)

From the above reaction, we can see that oxidation state of ­carbon in CH4 is −4 and in CCl4 is +4.  Answer (1) 28. Wurtz reaction is not suitable for the preparation of alkanes containing odd number of carbon. Hence, n-heptane cannot prepared in good yield by Wurtz reaction. 

(b)→(i): Temporary hardness results in water that contains bicarbonates of calcium and magnesium. (c)→(ii): B2H6 is an electron deficient compound. is denser (d)→(iv): H2O2 has a skew (non-planar) structure.  Answer (3) 36. Ethanol and acetone shows positive deviation from Raoult’s law. It is due to because the interaction between ethanol-acetone is weaker than ethanol-ethanol and acetone-acetone. 

Answer (3)

37. The reaction involved is OCH3

Answer (1)

29. Sucrose on hydrolysis gives glucose and fructose. C H O

12 22 11 Cane sugar (Sucrose)

+ H2 O



Solomons Organic Chemistry for NEET 2020.indd 6

Answer (4)

heat C(s) + H2O(vap) ⎯Red ⎯⎯⎯ → CO + H2



0

= 24 = 4.9 BM



CH3CH2CH

Br 2-Bromopentane

−4

n( n + 2)

Cr =[Ar]4s1 3d 5

+

⎯→ ⎯CH H

HI

O +C H O

6 12 6 Glucose

OH

6 12 6 Fructose

Answer (1)

Anisole



+ CH3I Phenol

Answer (3)

26-Mar-21 4:45:52 PM

Question paper 38. The reaction involved is Urea + H2O

(NH4)2CO3 Ammonium carbonate (A)

Decompose

NH3 (B)

]2+

[Cu(NH3)4 (C)

Answer (4) ΔTf = Kf × m

Substituting a = 288 pm in above equation, we get



O

O



Answer (3)

O CH2

CH

CH2

CH2 O3

40. The structure of peroxidisulphuric acid is as follows: S

3 × 288 pm 4

44. The reaction involved is

Substituting Kf = 5.12 K kg mol−1 and m = 0.078 m in above equation, we get ΔTf = 5.12 × 0.078 = 0.40 K  Answer (1)

HO

3a 4

43. We know that for bcc 4 r = 3a ⇒ r =

r=

 39. We know

Cu2+(aq)

O

O

S

O

O

H2O

O

CH2CHO

Answer (1)



Option (4) Incorrect. During the process of extraction of copper from cuprous oxide, in the final process the molten Cu obtained is poured into large container and allowed to cool. During cooling, the dissolved SO2 comes up to the surface and forms blisters. The resulting copper is known as blister copper.  Answer (2)

+ HCHO Methanal



Answer (3)

Answer (1)

45. The following set of molecules will have zero net dipole moment:

42. Natural rubber from Hevea trees is a linear polymer of isoprene (2-methyl-1,3-butadiene) also known as cis-1,4-polyisoprene.

Solomons Organic Chemistry for NEET 2020.indd 7

CH2

Ozonolysis

OH

Option (3): Incorrect. Wrought iron is the purest form of iron which has the total impurity less than 0.5%.



CH

O

41. Option (1): Incorrect. Vapor phase refining is carried out for Nickel by Mond’s process.

P7



Molecule

Structure

BF3

Trigonal planar

BeF2

Linear

CO2

Linear

1,4-dichlorobenzene

Planar Answer (2)

26-Mar-21 4:45:55 PM

Solomons Organic Chemistry for NEET 2020.indd 8

26-Mar-21 4:45:55 PM

Index

A acetals 572–574 acetone 6 acetylenes 227 acetylenic hydrogen atom  269 acetyl group  553 achiral molecule  113, 115 acid–base reactions  55–56 acid-catalyzed esterifications  500 acid-catalyzed halogenation  590–591 acid-catalyzed hydration of alkenes  257–258, 486 acid catalyzed reaction  401 acidity effect of solvent on  76 relationship between structure and  71–73 activating groups  351 acylation 448 acyl chlorides  645, 656–658 reactions of  672–673 acyl groups  553 1,4-addition 286–288 adduct 288 adipocytes 809 alcohol dehydrations  244–247 alcohols 33–34 acid-catalyzed dehydration of  240–244, 502–504 as acids  499–500 addition of  569–574 from alkenes  486–488 catalytic dehydrogenation of  557 cleavage of O—H bond  499–501 dehydrogenation 504 distinction of  516–517 esterification of  500–501 functional group of  479 from Grignard reagents  491–493 hydroboration–oxidation reaction  487 hydrogen bonding of  482, 485 with hydrogen halides  502 intermolecular dehydration of  495–496 nomenclature of  480–481 oxidation of  504–506 oxymercuration–demercuration reaction  486 physical properties of  482–484 polarization of  498 protonation of  498–499 reactions of  498–506 reduction of aldehydes and ketones to  581–582 by reduction of carbonyl compounds  488–491 structure and classification  475–478 by substitution reactions on alkyl halides  493 aldehydes  36–37, 490 addition reactions  603–605 additions to  569–581

Index.indd 1

from alkynes  558 aromatic 562–563 carbonyl group of  551 by catalytic dehydrogenation of alcohols  557 elimination reaction in  569 hydration of  558 hydroboration of  558 from hydrocarbons  562–563 isomerism in  553–554 nomenclature of  552–554 nucleophilic addition to  566–569 oxidation of  556–557, 583–586 physical properties  555–556 reduction of  581–583 by reduction of acyl chlorides, esters, and nitriles  558–562 relative reactivity of  568–569 aldol additions and condensations  592–596, 606 acid catalyzed condensation  594–595 cyclizations via condensations  599 synthetic applications of  595–596 aliphatic ether  478 alkadiene 279–281 alkadiyne 280 alkanes  28–29, 181–184 chemical reactions of  197–206 IUPAC nomenclature of  42–45 synthesis of  190–195 alkanoyl 553 alkatriene 279 alkenes  18, 28–29, 228 acid-catalyzed hydration of  486 addition reactions of  249–263 alcohol from  486–488 aldehydes by ozonolysis of  557 alkyl halides from  405 diastereomers 233–234 dipole moments in  49 hydroboration–oxidation reaction  487 hydrogenation of  247–249 less substituted  239 naming 231–233 oxidation of  263–265 oxymercuration–demercuration of  486 relative stability of  234–236 synthesis via hydrogenation of alkynes  236–237 via elimination reactions  237–244 alkenyl halide  33 alkenyl halides  395 alkenyne 280 alkoxy group  482 alkoxyl radicals  61 alkoxymercuration–demercuration of ethers  498 alkyl alcohols  480 alkylation 448 alkylbenzenes 365

12-12-2017 17:59:28

I2

Index ortho–para direction and reactivity of  361–362 reactions of side chain of  362–365 alkyl groups  31–32, 182–186, 358–361 alkyl halides  395–398 elimination reaction of  425–432 nucleophilic substitution reactions  407–425 physical properties of  400–401 preparation of  401–407 reaction with metals  434 reduction to hydrocarbons  435 substitution and elimination reactions of simple  432–433 alkynes 29 acidity of terminal  269 alkyl halides from  405 methods of preparation  269–272 naming of  268–269 oxidation of  275 allylic group and allylic radicals  283–285 allylic halides  396 allylic substitution  406 allyl (propenyl) cation  279 allyl radical  277–279 alpha carbon atom  426 amides  38, 645–646, 665–669 reactions of  674 amines  35–36, 577 aliphatic 698 arylamines 699 basicity of  707–710 eliminations involving ammonium compounds  717–718 isomerism of  699 oxidation of  713 physical properties and structure of  700–701 preparation of  701–707, 726–727 reactions of  711–717, 728 amino acids  779–781 as dipolar ions  784–786 essential 781–784 amino groups  357 amino sugar  779 ammonia 25 ammonium cyanate  2 analgesics 871 anhydrous diethyl ether (Et2O) 490 aniline 319 annulenes 327–328 antacids 868–869 anti addition of hydrogen  237 antibiotics 871–873 antibonding molecular orbital  20 antifertility drugs  873–874 antihistaminic drugs  869 anti-Markovnikov addition  254–257, 406 anti-Markovnikov regioselectivity  488 antiseptics 873 arenediazonium salts coupling reactions of  722–723 preparation of  718–719 replacement reactions of  719–722 arenes 362 physical properties of  336 preparation of  335–336 aromatic compounds  27–28 reduction of  366–367 aromatic ether  478 aromatic ions  328–329 aromaticity 322 artificial sweetening agents  874–875

Index.indd 2

aryl halides  33, 395–397 from diazonium salts  439–440 double-bond character of  440–441 eectrophilic aromatic substitution reaction of 446–448 by electrophilic substitution  438–439 as insecticides  444 nomenclature 437 nucleophilic aromatic substitution in  440–446 nucleophilic aromatic substitution of  441–442 preparation of  438–440 reaction with metals  448–449 atomic number  120 autoxidation 519 axis of symmetry (AOS)  118–119 axons 814

B Baeyer–Villiger oxidation  584–585 Balz-Schiemann reaction  439 base-promoted halogenation  590 base strength  69–70 Beilstein’s test  166 benzaldehyde  552, 562 benzene  30–31, 317–318 derivatives 318–320 halogenation of  339–340 hydrogenation of  366 Kekulé structure for  321–322 modern theories of structure  323–325 nitration of  341 orientation in disubstituted  370 reactions of  320–321 sulfonation of  342–343 thermodynamic stability of  322–323 benzenediols 481 benzenoid aromatic compounds  27–28, 331 benzenoids 317 benzilic acid rearrangement  602 benzoin condensation  603 benzoquinone 511 benzyl 320 benzyl group  32 benzylic cation  363 benzylic halides  396 benzylic halogenation  364 benzylic hydrogen atoms  363 benzylic radical  363 benzylic substituent  363 benzyne 442–443 beta carbon atom  426 biomolecules 757 biphenyl ethers  445–446 Birch reduction  367 blocking group  369 blocking group, use of  368–369 boiling point  52–53 σ-bond framework  228 bonding molecular orbital  16 bond-line formula  8–9 branched-chain alkanes  44 branched-chain hydrocarbons  43–44 breathalyzer alcohol test  506 bromobenzene 442 bromohydrin 261 Brønsted–Lowry acid–base reactions  55, 68–71 1,3-butadiene 281–282

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Index C Cahn-Ingold-Prelog convention  109 Cahn–Ingold–Prelog system  120 calcium salts of carboxylic acids, dry distillation of  565–566 Cannizaro reaction  600–601 crossed 601–602 intramolecular 601 carbaldehyde 552 carbanion 57–60 carbocation 57–60 carbohydrates 757–758 classification of  758 carbon atom  10 carbon–carbon double bond  4 carbon–carbon single bond  4 carbon–carbon triple bond  4, 266–267 carbon compounds  2 carbon dioxide, shapes of molecules in  25–26 carbon tetrachloride  436 carbonyl compounds  554–555 carboxylate salts  643 carboxyl group nomenclature 640–646 physical properties of  640–646 structure 640 carboxylic acid anhydrides  658–659 reactions of  673 carboxylic acids  37, 73, 490, 566, 640–643 chemical properties of  650–654 decarboxylation of  670–671 derivatives 639 preparation of  646–650 reactions of  671–672 carboxylic anhydrides  644 carbylamine reaction  713 catalytic hydrogenation  248 cellobiose 772–773 cellulose 777–778 center of symmetry (COS)  117–118 CHClBrCF3 395 chemotherapy 865 chirality 110–112 test for  115–119 chirality center  114–115, 129–131 retention of configuration at  135 chiral molecules  112, 128, 136 chloramphenicol 395 chlorobenzene  442, 447, 494 chlorofluorocarbons (CFCs)  436 chloroform 435–436 chloroform (CHCl3) 395 chlorohydrin 261 chloromethane 493 chromatography 160–163 chromic acid (H2CrO4) oxidation  505–506 chromic oxide  563 chromyl chloride  562 cine substitution  443 cis–trans isomerism  20–21 Claisen rearrangement  516 cleansing agents  876–881 Clemmensen reduction  582 condensed structural formulas  8 conformational analysis  206–210 conformational isomerism  136–137 conformational stereoisomers  135, 210 conjugate acid  55

Index.indd 3

I3

conjugate acid–base strengths  75 conjugated dienes  282–283 constitutional isomers  6 Cope elimination  718 copolymerization 846–847 covalent bonds  3–5 cresols 481 crossed aldol reaction  596–599 crystallization 156 cumene hydroperoxide  495 Curtius rearrangement  706–707 curved arrows  56–57, 63 cyanides and isocyanides chemical properties  725 physical properties  724 preparation of  724, 729 reactions of  729–730 cyanohydrins 574–575 cyclic acetals  573 [4 + 2] cycloaddition  289 1,4-cycloaddition reaction of dienes  288–289 cycloalkanes 186–189 cis  236 disubstituted 213–214 naming 231–233 relative stability of  210–211 synthesis of  190–195 trans  236 cyclobutadiene 330 cycloheptatriene 329 cyclohexane 211–213 cyclooctatetraene  322, 330–331 cyclopropane 210–211

D Darzen’s process  403 dash structural formula  4, 7–8 DDT 444–445 deactivating groups  353 deamination reaction  721–722 debye 48 decyl alcohol  483–484 dehydrohalogenation 426 bases used in  426–427 mechanism of  427 delocalization 74–75 deoxyribonucleic acid (DNA)  798, 802–804 deoxyribose 778 deoxy sugars  778 dextrorotatory 124 dialkylcadmium 565 diastereomers 107 diazo coupling reaction  722 diazonium salts  718 dibromobenzenes 319 dicarboxylic acids  643 dichloromethane 435 Diels–Alder reaction  288–289 diene 288 dienophile 288 diethyl ether  518–519 differential extraction  159 1,2-Difluoroethene 21 dihalogen derivatives  399 1,2-dihydroxylation 263–265 diisobutylaluminum hydride (DIBAL-H)  559, 561 1,2-Dimethylcyclohexane 134–135

12-12-2017 17:59:28

I4

Index.indd 4

Index 1,3-Dimethylcyclohexane 134 1,4-Dimethylcyclohexanes 133–134 dimethyl ether  485 diols 481 1,4-dioxane 482 dipole–dipole forces  51 dipole moment  47 directed aldol reaction  598–599 disaccharides 771–775 disinfectants 873 dispersion forces  52 dissolving metal reduction  237 distillation 156–159 disubstituted benzenes  46 double bond  20 structure of  228–230 double bond  10 Dow process  494 drugs 865 classification of  866 drug–target interaction  866–867 receptors as targets  868

alkoxymercuration–demercuration of  498 alkyl aryl  513 benzyl alkyl  514 Claisen rearrangement in  516 cleavage of  512–514 concerted rearrangement in  516 electrophilic substitution of  514–515 in Friedel–Crafts reaction  515 functional group of  479–480 halogenation of  515 by intermolecular dehydration of alcohols  495–496 nitration of  515 nomenclature of  482 physical properties of  484–486 reactions of  512–516 structure and classification  478 Williamson ether synthesis  496–497 ethyl alcohol  479 ethylene glycol  518 ethyl groups  32 ethyne (acetylene), structure of  21–24 (E )–(Z ) system of nomenclature  109, 233–234

E

F

electromeric effect (E effect)  66 electronegativity 3 electronegativity 47 electron-withdrawing groups  442 electrophiles 59–60 electrophilic addition  250, 258–261, 273–275 electrophilic aromatic substitution reactions  336–339 electrophilic substitutions of bromobenzene  447 elimination–addition mechanism  442–443 elimination reactions  237–244 eliminations 77 1,2 eliminations  426 β eliminations  426 enamines 579–580 enantiomers  107, 112–113 naming 119–120 properties of  122–127 energy of electrons  24 enolate 586–587 enolate chemistry  605–606 enol forms  587 enols 478 enzymes 793–796 epimerization 589 epimers 589 equilibrium 12 E1 reaction  238, 241–242, 427–430 vs SN1 reaction  432 E2 reaction  244, 427–428 esterification  500–501, 659–663 esters  38, 643–644, 659–663 acid-catalyzed hydrolysis  660 from acyl chlorides  500 from carboxylic acid anhydrides  501 reactions of  673–674 Etard reaction  562 ethane conformational analysis of  207–208 structure of  17 ethanol 518–519 ethanoyl 553 ethene (ethylene), structure of  18–21 ethers 35

fatty acids  806–810 Fehling’s reagent  586 Fehling’s test  586 Finkelstein reaction  407 Fischer esterifications  500, 660 Fischer projections  132–133 Fittig reaction  449 fluorobenzene 439 food preservatives  875–876 formic acid  37 formyl group  553 Freon-11 435 Friedel–Crafts acylation  344–349, 564 Friedel–Crafts alkylation  343–344 of benzene  495 Friedel–Crafts reaction  448, 515 fructose 764 fullerenes 332–333 functional groups  31–32, 39, 48 functional group transformation  424–425

G Gabriel synthesis  702–703 galacturonic acid  778 gammaxene 366 Gattermann–Koch reaction  563 Gattermann reaction  439–440, 720 geminal dihalides, aqueous alkali hydrolysis of 565 gene  799, 804 genome 799 geometrical isomers  106–109 in cyclic structures  110 glucose 760–764 glucuronic acid  778 glycogen 777 glycolipid 814 glycols 263 glycosylamines 779 goitre 395 Grignard reagents  434, 491–493, 565, 576, 713 ground state  15

12-12-2017 17:59:28

Index H haloalkanes  32–33, 398–399 haloarenes  395, 446 haloform reaction  591–592 halogenation  403, 447 halogenation at alpha carbon  589–590 halohydrin 261 Hammond–Leffler postulate  355 hemiacetals 569–572 Henderson–Hasselbalch equation  785 heteroatoms 48 heterocyclic aromatic compounds  28, 335 heterocyclic compounds  333–334 heterogeneous catalysis  247 heterolysis of a bond  57–60 hindered (small) base/nucleophile  432 Hinsberg test  716 Hofmann rearrangement  706 Hofmann rule  239, 717 Hofmann’s mustard oil reaction  714 homogeneous catalysis  248 homologous series  28 homolysis 60–61 homolytic fission of bonds  60–61 hormones 796–797 Hückel’s rule  325–331 Hunsdiecker reaction  440 hybrid atomic orbitals  14, 24 hybridization 72–73 hybridization of carbon atom  396–398 hydrazones 578–579 hydride ion  490, 581 hydroboration 487–488 hydroboration–oxidation reaction  487 hydrocarbons 28–31 hydrogen abstraction  61 hydrogenation of alkenes  247–249 hydrogenation of alkynes  236–237 β-hydrogen atom  426 hydrogen atoms  186 hydrogen bond  51–52 hydrogen cyanide  574–575 hydrogen halides  501 hydrolysis  416, 572 hydrophilic 54 hydrophobic 54 hydrophobic effect  54 hydroquinone 511 hydroxide ion  423 hyperconjugation 68–68

I imines 577–578 index of hydrogen deficiency (IHD)  195–197 inductive effect  61–62, 73 inductive electron-withdrawing effect  75 intermolecular dehydration of alcohols  495–496 intermolecular forces  50–52 iodoform reaction  517 Iodoform test  517 ion–dipole forces  53 ionic bonds  3 ion–ion forces  50 isomerism  190, 230–231, 267–268 chain  103–104, 230, 554 cis–trans 213–214

Index.indd 5

I5

functional group  105, 553 geometric 230 metamerism 105 position  104, 230, 267, 553 ring-chain 105 structural (constitutional)  103–106, 230 isomers 6 isopropyl group  32 IUPAC system of nomenclature  41–45, 182, 398–399, 480

J Jones reagent  505

K Kekulé structure  30 keto–enol tautomers  587–588 keto forms  587 ketones  36–37, 490–492 from acyl chlorides  565 addition reactions  603–605 from alkenes, arenes, and alcohols  563–564 carbonyl group of  551 elimination reaction in  569 isomerism in  553–554 from nitriles  564–565 nomenclature of  552–554 nucleophilic addition to  566–569 oxidation of  583–586 physical properties  555–556 reduction of  581–583 relative reactivity of  568–569 stabilization of  569 Kolbe’s reaction  510

L lactams 670 lactones 664 lactose 775 leaving group  407, 422–423 negative charge at  422 leaving groups  409–410 levorotatory 124 Lewis acid–base theory  56 Lewis acids  250 Lewis structures  4–6 Lindlar’s catalyst  237 lipid bilayers  814 lipids 806 lithium aluminum hydride (LiAlH4)  489, 558–559, 582 lithium enolates  592 lithium reagents  492 lithium tri-tert-butoxy-aluminum hydride  559 lone pairs  24 Lucas test  516–517

M maltose 772 Markovnikov additions  405 Markovnikov regioselectivity  486 Markovnikov’s rule  250–253, 405–406 Meisenheimer intermediate  441 melting point  50 meso compounds  130–131 meta-directing groups  357–358

12-12-2017 17:59:28

I6

Index meta directors  353, 357–358 meta-nitro group  441 methane, structure of  15–17, 24–25 methanol  499, 518 methanolysis 416 methanoyl 553 methyl benzene (toluene)  562 methylene group  32 methyl groups  32 methylphenols 481 m-nitrochlorobenzene 441 molecular formula  6 molecules 4 monohalogen compounds  396 monohydric alcohols  477 monomers 833 monosaccharides 758–760 reactions of  765–770 monosubstituted benzenes  46 multiple covalent bonds  4

N naphthols 476 natural products chemistry  2 Newman projections  137, 206–207 nitration 448 nitriles  39, 646, 669–670 reactions of  674–675 nitrobenzoic acids  319 noble gas configuration  5 non-benzenoid aromatic compounds  28, 332 non-benzenoids 317 nonbonding pairs  24 nucleic acids  798–799 nucleophiles  59–60, 407 effect of concentration and strength of  418 negatively charged  409 neutral nucleophile  409 relative strengths of  419 nucleophilic addition–elimination of acyl (carbonyl) carbon atoms 654–656 nucleophilic aromatic substitution (SNAr) 441–442 nucleophilicity 418 vs basicity  419 nucleoside 800–801 nucleotides 800–801

O 2-octanol 412 octet rule  3 olefins 227 optical isomers  107 optically active compounds  123 orbital hybridization  14 organic chemistry, development of  2 organic compounds classification of  26–28, 40 detection tests  164–166 electron displacement effects in  61–68 empirical formula of  171–172 nomenclature of  40–47 purification of  155–163 qualitative analysis of  163–166 quantitative analysis of  166–171 organic food  2 organic halides  445

Index.indd 6

organic vitamin  2 organohalogens 395 organolithium reagents (RLi)  492 ortho–para-directing groups  351–353, 358–361 oxetane 482 oximes 578–579 oxirane 482 oxonium salts  512 oxymercuration–demercuration reaction  486 ozonides. 265 ozonolysis 264–265

P pair of enantiomers  113 partial racemization  415 Perkin reaction  603 peroxides 61 petroleum 179–181 phenanthrols 476 phenols  34, 319, 476–477 as acids  506–507 bromination of  508 carbon–oxygen bond in  479 coupling reaction with  511 distinguishing and separating from alcohols 507–508 electrophilic aromatic substitutions of  508 esterification of  509 functional group of  477 industrial synthesis of  494–495 in Kolbe’s reaction  510 laboratory synthesis of  493–494 nitration of  509 nomenclature of  481 oxidation of  511 physical properties of  484 reactions of the benzene ring of  508–509 reaction with zinc dust  511 in Reimer–Tiemann reaction  510 sulfonation of  509 in Williamson synthesis  497 phenylalanine hydroxylase  335 phenylation 443 phenyl group  32, 320 phenyl halides  395 phosphatides 813 phospholipids 812–814 phosphorus tribromide  501 pi bond  19, 24, 229 pi molecular orbital  20 plane of symmetry  116–117 plane-polarized light  123 polar covalent bonds  47–50 polarimeter 124 polar molecule  48 polybromodiphenyl ethers (PBDEs)  445–446 polychlorinated biphenyls (PCBs)  445–446 polycyclic aromatic hydrocarbons (PAH)  27, 331 polyfunctional compounds  45 polyhalogens  399, 435–436 polyhydroxy alcohols  476 polymerization  276, 833 addition 838–839 chain-growth 838–839 condensation 842–846 ionic 839–840 step-growth 842–846

12-12-2017 17:59:28

Index polymers 833 addition 840–841 biodegradable 849–850 classification of  834–837 of commercial importance  850–851 functionality 834 molecular mass of  849 polymerization reactions  834 polypeptides 786–787 polysaccharides 775–778 potassium permanganate (KMnO4) 506 potassium tert-butoxide (t-BuOK) 427 primary alcohol  34, 240, 477, 503–504 primary alkyl halide  32 primary carbon  32, 396, 477 primary halides  431 propylene 18 propylene glycol  518 propylene oxide  6 propyl group  32 proteins denaturation of  792–793 native 792 structure of  788–792 synthesis 804–805 proteome 799 protic solvent  76 pyridine 403 pyridinium chlorochromate (PCC)  506

R racemic mixture  126–127 racemization  415, 588–589 rate-determining step  413 rate-limiting step  413 rearrangement reaction  77 reductive amination  704 regioselective reaction  254 Reimer–Tiemann reaction  510 replacement nomenclature  482 resonance contributors  11, 63 resonance effect  65–66 resonance energy  323 resonance structures  11–14, 62–66 resonance theory  11–14 retro-aldol reaction  594 ribonucleic acid (RNA)  798 Riemmer–Tiemann reaction  563 Rosenmund reduction  560 rotational reflection axis of symmetry  119 R,S-system 120–122 rubber natural 847–848 synthetic 848–849

S salts 3 Sandmeyer’s reaction  439, 720 saponification 662–663 saturated compounds  29 sawhorse projection  137 secondary alcohol  240, 477, 503, 505 secondary alkyl halide  32, 396 secondary carbon  396, 477 secondary halides  431 1, 2 shift  245

Index.indd 7

I7

side chain  362 sigma bond  16, 24 SN1 reaction  402, 413–421 factors favoring  424 mechanism for  413–415 reactivity of organic substrates in  418 solvent effects in  419 stereochemistry of  415–416 vs E1 reaction  432 vs SN2 reaction  424 SN2 reaction  402, 410–412, 493 effect of protic solvents  421–422 factors favoring  424 functional group transformation using  424–425 mechanism for  410–412 polar aprotic solvents in  419–420 polar protic solvents in  420–421 relative basicity and polarizability of  432 solvent effects in  419 steric effects and relative rates in  417 substrate effects in  416–417 vs E2 reaction  430, 432 vs SN1 reaction  424 soaps 876–878 sodium alkoxides  427 sodium alkynides  493 sodium benzenesulfonate  494 sodium borohydride (NaBH4) 490 sodium hydrogen sulfite  576 solubilities 53–54 solvent effects  419 solvolysis reaction  409, 416 specific rotation  124–126 sp2 hybridization  18–21 sp3 hybridization  14–17 sp2 orbitals  18–19 starch 776 Stephen reaction  562 Stephen reduction  562 stereogenic carbon  115 stereogenic center  115 stereoisomerism of cyclic compounds  133–135 stereoisomers  21, 106–107, 210 stereoselective reactions  128 stereospecific reaction  260 steric effect  417 steric hindrance  417 straight-chain hydrocarbons  42 structural formulas bond-line formula  8–9 condensed structural formulas  8 dash structural formula  7–8 definition of  7 three-dimensional formula  9–10 structural isomers  6 sublimation 155 substituent effect  75–76 substituent group  350–351 on benzene ring  367 classification of substituents  354 electron-donating and electron-withdrawing substituents  351 electrophilic aromatic substitution, impact on  354–355 halo substituents  353 inductive effect of a substituent  355–356 resonance effect of a substituent  356–357 substitution reactions of aromatic rings  368 substitutions 76 substrate  55, 407

12-12-2017 17:59:28

I8

Index sucrose  771–772, 774–775 sulphonation 448 superposition 20–21 Swarts reaction  407 Swern oxidation  505 symmetrical ether  478 syn addition  248 syn stereoselectivity  488 synthetic detergents  878–881

T tautomerism  105–106, 587–588 terpenes 811–812 terpenoids 811–812 tert-butoxide ion  431 tert-butyl alcohol  483 tert-butyl chloride  493 tertiary alcohols  240, 477, 503, 505 tertiary alkyl halide  33, 396 tertiary carbon  33, 396, 477 tertiary substrate  431 tetrachloromethane 436 tetrahedral stereogenic centers  115 tetrahydrofuran (THF)  482 thioacetals 574 thionyl chloride  403, 501 three-dimensional formula  9–10 Tollens’ test  585 toluene 319 tranquilizers 869–870 transition state  410 trans-3-methylcyclopentanol 411 triacylglycerols 806–810 tri- and polysubstituted benzenes  46–47 trichlormethane 435–436 1,1,2-trichloroethene 21 triflate ion  423 trifluoromethyl group  357–358 trigonal planar  24, 60 trigonal pyramid  25 trigonal stereogenic centers  115 triiodomethane 436 triple bond  4, 10

Index.indd 8

U Ullmann reaction  449 unhindered (small) base/nucleophile  432 unsaturated compounds  29 unsaturated hydrocarbons  45 unshared pairs  24 unsymmetrical ethers  478 urea 2 uronic acids  778

V valence shell  3 valence shell electron pair repulsion (VSEPR) model  24–26 van der Waals forces  50–52 vicinal dihalides  240, 258, 270 Victor Meyer test  517 vinylic halides  397 vitamins 797–798

W water, shapes of molecules in  25 Watson–Crick hypothesis  804 waxes 814 Williamson ether synthesis  496–497 Wittig reaction  580–581 Wolff–Kishner reduction  349, 582–583 Wurtz–Fittig reaction  448–449 Wurtz reaction  434

X xylenes 319

Y ylide 580–581

Z Zaitsev’s rule  238–239, 246

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