Organic Chemistry Student Study Guide/Solution [4 ed.] 1308115040, 9781308115047

Table of contents :
Structure and Bonding
Acids and Bases
Introduction to Organic Molecules and Functional Groups
Alkanes
Stereochemistry
Understanding Organic Reactions
Alkyl Halides and Nucleophilic Substitution
Alkyl Halides and Elimination Reactions
Alcohols, Ethers, and Epoxides
Alkenes
Alkynes
Oxidation and Reduction
Mass Spectrometry and Infrared Spectroscopy
Nuclear Magnetic Resonance Spectroscopy
Radical Reactions
Conjugation, Resonance, and Dienes
Benzene and Aromatic Compounds
Reactions of Aromatic Compounds
Carboxylic Acids and the Acidity of the O-H Bond
Introduction to Carbonyl Chemistry
Aldehydes and Ketones—Nucleophilic Addition
Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution
Substitution Reactions of Carbonyl Compounds at the a Carbon
Carbonyl Condensation Reactions
Amines
Carbon-Carbon Bond-Forming Reactions in Organic Synthesis
Pericyclic Reactions
Carbohydrates
Amino Acids and Proteins
Lipids
Synthetic Polymers

Citation preview

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition University of Pretoria

http://create.mcgraw-hill.com Copyright 2014 by McGraw-Hill Education. All rights reserved. Printed in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without prior written permission of the publisher. This McGraw-Hill Create text may include materials submitted to McGraw-Hill for publication by the instructor of this course. The instructor is solely responsible for the editorial content of such materials. Instructors retain copyright of these additional materials. ISBN-10: 1308115040

ISBN-13: 9781308115047

Contents 1. Structure and Bonding 1 2. Acids and Bases 35 3. Introduction to Organic Molecules and Functional Groups 61 4. Alkanes 81 5. Stereochemistry 117 6. Understanding Organic Reactions 145 7. Alkyl Halides and Nucleophilic Substitution 167 8. Alkyl Halides and Elimination Reactions 201 9. Alcohols, Ethers, and Epoxides 233 10. Alkenes 267 11. Alkynes 299 12. Oxidation and Reduction 323 13. Mass Spectrometry and Infrared Spectroscopy 353 14. Nuclear Magnetic Resonance Spectroscopy 371 15. Radical Reactions 397 16. Conjugation, Resonance, and Dienes 425 17. Benzene and Aromatic Compounds 453 18. Reactions of Aromatic Compounds 477 19. Carboxylic Acids and the Acidity of the O-H Bond 517 20. Introduction to Carbonyl Chemistry 539 21. Aldehydes and Ketones—Nucleophilic Addition 573 22. Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 609 23. Substitution Reactions of Carbonyl Compounds at the a Carbon 645 24. Carbonyl Condensation Reactions 675 25. Amines 707 26. Carbon-Carbon Bond-Forming Reactions in Organic Synthesis 741 27. Pericyclic Reactions 765 28. Carbohydrates 789 29. Amino Acids and Proteins 825 30. Lipids 861 31. Synthetic Polymers 877

iii

Credits 1. Structure and Bonding: Chapter 1 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 1

2. Acids and Bases: Chapter 2 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 35

3. Introduction to Organic Molecules and Functional Groups: Chapter 3 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 61

4. Alkanes: Chapter 4 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 81

5. Stereochemistry: Chapter 5 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 117

6. Understanding Organic Reactions: Chapter 6 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 145

7. Alkyl Halides and Nucleophilic Substitution: Chapter 7 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 167

8. Alkyl Halides and Elimination Reactions: Chapter 8 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 201

9. Alcohols, Ethers, and Epoxides: Chapter 9 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 233

10. Alkenes: Chapter 10 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 267

11. Alkynes: Chapter 11 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 299

12. Oxidation and Reduction: Chapter 12 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 323

13. Mass Spectrometry and Infrared Spectroscopy: Chapter 13 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 353

14. Nuclear Magnetic Resonance Spectroscopy: Chapter 14 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 371

15. Radical Reactions: Chapter 15 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 397

16. Conjugation, Resonance, and Dienes: Chapter 16 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 425

17. Benzene and Aromatic Compounds: Chapter 17 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 453

18. Reactions of Aromatic Compounds: Chapter 18 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 477

iv

19. Carboxylic Acids and the Acidity of the O-H Bond: Chapter 19 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 517

20. Introduction to Carbonyl Chemistry: Chapter 20 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 539

21. Aldehydes and Ketones—Nucleophilic Addition: Chapter 21 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 573

22. Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution: Chapter 22 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 609

23. Substitution Reactions of Carbonyl Compounds at the a Carbon: Chapter 23 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 645

24. Carbonyl Condensation Reactions: Chapter 24 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 675

25. Amines: Chapter 25 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 707

26. Carbon-Carbon Bond-Forming Reactions in Organic Synthesis: Chapter 26 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 741

27. Pericyclic Reactions: Chapter 27 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 765

28. Carbohydrates: Chapter 28 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 789

29. Amino Acids and Proteins: Chapter 29 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 825

30. Lipids: Chapter 30 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 861

31. Synthetic Polymers: Chapter 31 from Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition by Smith, Berk, 2014 877

v

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

1

Structure and Bonding 1–1

Chapter 1 Structure and Bonding Chapter Review Important facts •

The general rule of bonding: Atoms strive to attain a complete outer shell of valence electrons (Section 1.2). H “wants” 2 electrons. Second-row elements “want” 8 electrons. nonbonded electron pair

H

C

N

O

X

Usual number of bonds in neutral atoms

1

4

3

2

1

Number of nonbonded electron pairs

0

0

1

2

3

X = F, Cl, Br, I

The sum (# of bonds + # of lone pairs) = 4 for all elements except H.

•

Formal charge (FC) is the difference between the number of valence electrons on an atom and the number of electrons it “owns” (Section 1.3C). See Sample Problem 1.3 for a stepwise example.

Definition: Examples:

formal charge

C

• C shares 8 electrons. • C "owns" 4 electrons. • FC = 0

•

=

number of valence electrons

–

number of electrons an atom "owns"

C

• C shares 6 electrons. • C "owns" 3 electrons. • FC = +1

C

• C shares 6 electrons. • C has 2 unshared electrons. • C "owns" 5 electrons. • FC = −1

Curved arrow notation shows the movement of an electron pair. The tail of the arrow always begins at an electron pair, either in a bond or a lone pair. The head points to where the electron pair “moves” (Section 1.6). Move an electron pair to O.

O H C N H

A

O H C N H

B

Use this electron pair to form a double bond.

•

Electrostatic potential plots are color-coded maps of electron density, indicating electron rich (red) and electron deficient (blue) regions (Section 1.12).

2

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 1–2

The importance of Lewis structures (Sections 1.3–1.5) A properly drawn Lewis structure shows the number of bonds and lone pairs present around each atom in a molecule. In a valid Lewis structure, each H has two electrons, and each second-row element has no more than eight. This is the first step needed to determine many properties of a molecule. Geometry

[linear, trigonal planar, or tetrahedral] (Section 1.7)

Hybridization

Lewis structure

Types of bonds

[sp, sp2, or sp3] (Section 1.9) [single, double, or triple] (Sections 1.3, 1.10)

Resonance (Section 1.6) The basic principles: • Resonance occurs when a compound cannot be represented by a single Lewis structure. • Two resonance structures differ only in the position of nonbonded electrons and π bonds. • The resonance hybrid is the only accurate representation for a resonance-stabilized compound. A hybrid is more stable than any single resonance structure because electron density is delocalized. O

O

O

CH3CH2 C

CH3CH2 C O

delocalized charges

CH3CH2 C O

δ−

O

δ−

delocalized π bonds

resonance structures

hybrid

The difference between resonance structures and isomers: • Two isomers differ in the arrangement of both atoms and electrons. • Resonance structures differ only in the arrangement of electrons. O

O

CH3 C

O

CH3CH2 C O CH3

isomers

CH3CH2 C O H

O H

resonance structures

Geometry and hybridization The number of groups around an atom determines both its geometry (Section 1.7) and hybridization (Section 1.9). Number of groups 2 3 4

Geometry linear trigonal planar tetrahedral

Bond angle (o) 180 120 109.5

Hybridization sp sp2 sp3

Examples BeH2, HC≡CH BF3, CH2=CH2 CH4, NH3, H2O

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

3

Structure and Bonding 1–3

Drawing organic molecules (Section 1.8) •

Shorthand methods are used to abbreviate the structure of organic molecules. CH3 H =

C

C

H

H

CH3

skeletal structure

•

CH3

CH3 C

CH3

(CH3)2CHCH2C(CH3)3

=

isooctane

condensed structure

A carbon bonded to four atoms is tetrahedral in shape. The best way to represent a tetrahedron is to draw two bonds in the plane, one in front, and one behind. Four equivalent drawings for CH4

H

H

H

HH

C

C

C

H H

H H

H

H

H

H C H

HH

Each drawing has two solid lines, one wedge, and one dashed line.

Bond length •

Bond length decreases across a row and increases down a column of the periodic table (Section 1.7A). C H

>

N H

>

O H

H F

Increasing bond length

•

H Cl


−OH > −NH2 > −CH3

d. –COOH > –CHO > –CH2OH > –H

b. −(CH2)3CH3 > −CH2CH2CH3 > −CH2CH3 > −CH3

e. –Cl > –SH > –OH > –CH3

c. −NH2 > −CH2NHCH3 > −CH2NH2 > −CH3

f. –C≡CH > –CH=CH2 > –CH(CH3)2 > –CH2CH3

5.46 Use the rules in Answer 5.13 to assign R or S to each stereogenic center. 1

1 I

a.

H

c.

C

C

H CH3

CH3CH2

2

T

CH3

switch H and CH3

C

CH3 D

H D

T

2

3

3

counterclockwise It looks like an S isomer, but we must reverse the answer, S to R.

counterclockwise S isomer

R isomer 1

2

NH2

b.

CH3

Cl

C

d.

CH2CH3

3 H

Br ICH2

2

Cl switch H and Br

C

H ICH2

H

C

1 Br

3

clockwise, but H in front S isomer

counterclockwise It looks like an S isomer, but we must reverse the answer, S to R.

R isomer CH3

e.

CH(CH3)2

C

C CH3 SH

H HO

HOOC

f.

C

CH3

C

NH2 H

H

g.

H

CH3

HO

H

S, R

R, R

S

5.47 a. (3R)-3-methylhexane

c. (3R,5S,6R)-5-ethyl-3,6-dimethylnonane CH3 H R

CH3 H

CH3 H

b. (4R,5S)-4,5-diethyloctane 4R

R

H CH2CH3

d. (3S,6S)-6-isopropyl-3-methyldecane H CH(CH3)2

H CH2CH3

CH3CH2 H

S

S

5S

H CH3

S

Cl

h. Cl

S S

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Stereochemistry 5–19

5.48

4

9

1

3

a.

6

6R

H

5

b.

S

2

4R 5

7

6R 3

3R

c.

10

9

(4R,6R)-4-ethyl-6-methyldecane

(3S)-3-methylhexane

5S

1 7

1

(3R,5S,6R)-5-isobutyl-3,6-dimethylnonane

5.49 Two enantiomers of the amino acid leucine. COOH C (CH3)2CHCH2

COOH

H NH2

H H2N

S isomer naturally occurring

C CH2CH(CH3)2

R isomer

5.50 S S CH3

COOH

a. OH

L-dopa

a. 1R,2S C1

OH

NH Cl

b.

H NH2

HO

CH3CH2O2C N

N H

c. ketamine enalapril

O

S

O

CO2H

S

S

5.51 OH

d. NHCH3

S

R NHCH3

e.

enantiomer of (–)-ephedrine

C2

ephedrine b. 1S,2S C1

OH

OH NHCH3

R

R NHCH3

diastereomer of (–)-ephedrine

C2

pseudoephedrine c. Ephedrine and pseudoephedrine are diastereomers (one stereogenic center is the same; one is different).

5.52 OH C C H

NH2 H N

a. HO

O

S

b.

O

c.

O O

N O

amoxicillin

COOH

O

N

O

norethindrone

CH3

O

heroin

136

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 5–20

5.53 OH

O a. CH3CH(OH)CH(OH)CH2CH3

c.

b. CH3CH2CH2CH(CH3)2

OH

HO OH

HO

0 stereogenic centers 2 stereogenic centers 22 = 4 possible stereoisomers

4 stereogenic centers 24 = 16 possible stereoisomers

5.54 a. CH3CH(OH)CH(OH)CH2CH3 CH3

CH2CH3

C

CH3CH2

C

H HO

CH3

C H OH

C

H HO

A

CH3

CH2CH3

C H OH

HO

C

CH3

C H OH

H

B

CH3CH2 C

H HO

C

H

OH

D

Pairs of enantiomers: A and B, C and D. Pairs of diastereomers: A and C, A and D, B and C, B and D. b. CH3CH(OH)CH2CH2CH(OH)CH3 OH

OH

OH

OH

OH

OH

A

C

B

OH

OH

identical meso compound

Pair of enantiomers: A and B. Pairs of diastereomers: A and C, B and C. c. CH3CH(Cl)CH2CH(Br)CH3 Cl

Br

Br

A

Cl

Cl

B

Br

Br

C

Cl

D

Pairs of enantiomers: A and B, C and D. Pairs of diastereomers: A and C, A and D, B and C, B and D. d. CH3CH(Br)CH(Br)CH(Br)CH3 Br

Br

Br

Br

Br

Br

A identical

Br

Br

Br

Br

Br

Br

Br

Br

Br

B

C

D

meso compound Pair of enantiomers: B and C. Pairs of diastereomers: A and B, A and C, A and D, B and D, C and D.

Br

Br

Br

identical meso compound

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

137

Stereochemistry 5–21

5.55 HOCH2

a.

C

CH3

CH3

H OH

H HO

C

CH2OH

C

H HO

CH3

C

C H OH

HO

H I

I H

H I

I H

c.

OH

HO

CH3

d.

C

H HO

OH H

diastereomer

I H

diastereomer H 2N

H 2N

or

HO

enantiomer

CH2OH

C

H OH

I H

H 2N

or

C

H

enantiomer NH2

CH3

diastereomer

enantiomer

b.

CH2OH

HO

diastereomer

CH3

diastereomer

CH3

CH3

or CH2CH3

CH3CH2

CH3CH2

CH3CH2

diastereomer

enantiomer

diastereomer

5.56 CH3

CH3

CH3

CH3

a. CH3

CH3

A

identical

CH3

CH3

B

C

meso compound Pair of enantiomers: B and C. Pairs of diastereomers: A and B, A and C.

b.

CH3

CH3

CH3

CH3

CH3

A

CH3

CH3

CH3

B

identical

identical

Pair of diastereomers: A and B. Meso compounds: A and B.

c.

Cl

Cl

Br

Cl

Br

Br

B

A

C

Cl Br

D

Pairs of enantiomers: A and B, C and D. Pairs of diastereomers: A and C, A and D, B and C, B and D.

5.57

achiral

achiral

chiral

chiral

achiral

achiral

138

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 5–22

5.58 Explain each statement. All molecules have a mirror image, but only chiral molecules have enantiomers. A is not chiral, and therefore, does not have an enantiomer.

a. OH

B has one stereogenic center, and therefore, has an enantiomer. Only compounds with two or more stereogenic centers have diastereomers.

b. Cl

C is chiral and has two stereogenic centers, and therefore, has both an enantiomer and a diastereomer.

c. Cl OH

HO

rotate

OH

D has two stereogenic centers, but is a meso compound. Therefore, it has a diastereomer, but no enantiomer since it is achiral. plane of symmetry

d. OH

e. HO

E has two stereogenic centers, but is a meso compound. Therefore, it has a diastereomer, but no enantiomer since it is achiral.

OH

plane of symmetry

5.59 C2 C3

OHC

OH H

C R

H HO

C R

OHC

a. CH2OH

D-erythrose

HO

H

HOCH2

CH2OH C S

C R

b. HO

H OH

2S,3R diastereomer

2R,3R

H

C R

CHO C R

OHC

c. H

C S

HO H

OH

2R,3R identical

OHC

H OH C S

d. H HO

CH2OH

2S,3S enantiomer

H OH C R

C S

CH2OH

2R,3S diastereomer

5.60 Re-draw each Newman projection and determine the R,S configuration. Then determine how the molecules are related. CHO

CHO

H

CH2OH

H

OH

HO

H

OH A

OHC

re-draw

OH

OH H

H HO

CH2OH

R,R

a. A and B are identical. b. A and C are enantiomers.

CHO OH

HO

H

H

CH2OH

HO

H

CH2OH

B re-draw

OHC CH2OH

R,R

OH

H

CH2OH

H

H HO

CHO H

OH

C re-draw

OHC

H OH

HO H

D re-draw

OHC

H CH2OH

HO

CH2OH

S,S

c. A and D are diastereomers. d. C and D are diastereomers.

H

OH

S,R

139

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Stereochemistry 5–23

5.61 R

trans

R

NH2

A

R NH2

NH2

S

cis

trans

trans

B

C

D

NH2

NH2

E NH2 group is in a different location.

A (trans, R) and B (cis, R) are diastereomers. A (trans, R) and C (trans, R) are identical molecules. A (trans, R) and D (trans, S) are enantiomers. A (trans, R) and E are constitutional isomers.

5.62 Cl

and

a.

f.

C I

enantiomers

Br H

C

Br I

H

enantiomers CH3

CH3

b.

Cl

and

g.

and

CH3

and

C

H Br

same molecular formula different connectivity constitutional isomers

CH3

CH3

C

H

2S,3S

H Br

C

C

H Br

Br

identical

CH3

2S,3S OH

H CH3

c.

CHO

C

and

C

H HO

OHC

H

OH

CH3 C

h.

and

H HO

OH

HO

C

H

H

H OH

one different configuration diastereomers

1,4-cis

1,4-trans

2R,3R

2R,3S

diastereomers i. and

H

CH3 CH3

d.

diastereomers

different molecular formulas not isomers

Br on end HO CH3

Cl

H

1,3-trans

1,3-cis

and

e.

H

HO

Cl

j.

and mirror images not superimposable enantiomers

5.63 a. A and B are constitutional isomers. A and C are constitutional isomers. B and C are diastereomers (cis and trans). C and D are enantiomers.

and

C H

CH2Br

CH3

CH2OH C Br H

Br in middle different connectivity constitutional isomers

140

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 5–24

plane of symmetry b.

A A has two planes of symmetry. achiral

B

C

D

achiral

chiral

chiral mirror images and not superimposable enantiomers

c. Alone, C and D would be optically active. d. A and B have a plane of symmetry. e. A and B have different boiling points. B and C have different boiling points. C and D have the same boiling point. f. B is a meso compound. g. An equal mixture of C and D is optically inactive because it is a racemic mixture. An equal mixture of B and C would be optically active. 5.64 H HO H

N

ee =

CH3O

[α] mixture

x 100%

[α] pure enantiomer quinine = A quinine's enantiomer = B

N

quinine

a. −50 x 100% = 30% ee −165

b. 30% ee = 30% excess one compound (A) remaining 70% = mixture of 2 compounds (35% each A and B) Amount of A = 30 + 35 = 65% Amount of B = 35%

−83 x 100% = 50% ee −165

50% ee = 50% excess one compound (A) remaining 50% = mixture of 2 compounds (25% each A and B) Amount of A = 50 + 25 = 75% Amount of B = 25%

73% ee = 73% excess of one compound (A) remaining 27% = mixture of 2 compounds (13.5% each A and B) Amount of A = 73 + 13.5 = 86.5% Amount of B = 13.5% [α] mixture x 100% e. 60% = –165

−120 x 100% = 73% ee −165 c. [α] = +165 d. 80% − 20% = 60% ee

[α] mixture = −99

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

141

Stereochemistry 5–25

5.65 OH HO

OH

O

HO

O

O OH

O

HO

HCl, H2O

OH

amygdalin

COOH

only one of the products formed

CN

mandelic acid

OH

a. The 11 stereogenic centers are circled. Maximum number of stereoisomers = 2 11 = 2048 b. Enantiomers of mandelic acid: HO H

H OH

HOOC

COOH

R c. 60% − 40% = 20% ee 20% = [α] mixture/−154 x 100% [α] mixture = −31

+50

d. ee =

+154

S

x 100% = 32% ee

[α] for (S)-mandelic acid = +154

32% excess of the S enantiomer 68% of racemic R and S = 34% S and 34% R S enantiomer: 32% + 34% = 66% R enantiomer = 34%

5.66 sp2

a. Each stereogenic center is circled. b. The stereogenic centers in mefloquine are labeled. c. Artemisinin has seven stereogenic centers. 2n = 27 = 128 possible stereoisomers d. One N atom in mefloquine is sp2 and one is sp3. e. Two molecules of artemisinin cannot intermolecularly H-bond because there are no O–H or N–H bonds.

CF3

H

N

CF3

O O

O H

H

O

R HO

O

H H N

sp3

H S

H

artemisinin

mefloquine

CF3

CF3 N

f.

CF3

N

CF3

H Cl H H N

HO H

H HO H

HH N

+

Cl

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 5–26

5.67 c. diastereomer

a. Each stereogenic center is circled.

H S N

N

O N S H

O

HS

R

OH CONH2 O

saquinavir Trade name: Invirase

S

N

O

H N

N

S

H N H

O

N

OH CONH2 O

H NH

H NH

(CH3)3C

(CH3)3C

b. enantiomer

d. constitutional isomer

H N

N O

O

H N H

N

N

OH CONH2 O

H N H

O H

O H

NH (CH3)3C

new amine

NH

O

H N

N OH O

H NH

new aldehyde

(CH3)3C

5.68 Allenes contain an sp hybridized carbon atom doubly bonded to two other carbons. This makes the double bonds of an allene perpendicular to each other. When each end of the allene has two like substituents, the allene contains two planes of symmetry and it is achiral. When each end of the allene has two different groups, the allene has no plane of symmetry and it becomes chiral.

CH3

H C C C

CH3

H

CH3 H

CH3 C C C

B

H

HC C C C CH C CH CH CH CH CHCH2CO2H

allene

no plane of symmetry A

mycomycin re-draw

chiral

These two substituents are at 90o to these two substituents. Allene A contains two planes of symmetry, making it achiral.

CH CH CH CHCH2CO2H

HC C C C C C C H

H

The substituents on each end of the allene in mycomycin are different. Therefore, mycomycin is chiral.

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143

Stereochemistry 5–27

5.69 NH2 N

S

N

S

CH3

N

O

R H

HOOC OH

S

N

+

H NH2 S

OH

R

S

six stereogenic centers

5.70 HO O

O

OH

O

OH

NH2 O

discodermolide

OH

a. The 13 tetrahedral stereogenic centers are circled. b. Because there is restricted rotation around a C–C double bond, groups on the end of the double bond cannot interconvert. Whenever the substituents on each end of the double bond are different from each other, the double bond is a stereogenic site. Thus, the following two double bonds are isomers: R

R

R

C C H

H C C

H

H

R

These compounds are isomers.

There are three stereogenic double bonds in discodermolide, labeled with arrows. c. The maximum number of stereoisomers for discodermolide must include the 13 tetrahedral stereogenic centers and the three double bonds. Maximum number of stereoisomers = 216 = 65,536.

5.71 When the spiro compound has a plane of symmetry, it is achiral.

a.

O

achiral

b.

c. chiral

d. achiral

chiral

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Chapter 5–28

5.72 racemic mixture of 2-phenylpropanoic acid

salts formed by proton transfer COO

COOH C

S

H NH2 H CH3

C

+ S

(R)-sec-butylamine

COO

COOH H NH2

C

+ R

R

diastereomers

enantiomers

H CH3

H CH3

+

H NH3

(R)-sec-butylamine

H CH3

C

+

H NH3

R

R These salts are diastereomers, and they are now separable by physical methods since they have different physical properties.

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Understanding Organic Reactions 6–1

Chapter 6 Understanding Organic Reactions Chapter Review Writing organic reactions (6.1) •

Use curved arrows to show the movement of electrons. Full-headed arrows are used for electron pairs and half-headed arrows are used for single electrons. +

C

C Z

C Z

Z

+

Z

full-headed arrow

half-headed arrows

•

C

Reagents can be drawn either on the left side of an equation or over an arrow. Catalysts are drawn over or under an arrow.

Types of reactions (6.2) [1] Substitution

C Z

+ Y

C Y

+ Z

Z = H or a heteroatom

Y replaces Z

[2] Elimination

C

C

X

Y

+ reagent

Two σ bonds are broken.

[3] Addition

C C

+

+

C C

X Y

π bond

X Y

C C X Y

This π bond is broken.

Two σ bonds are formed.

Important trends Values compared Bond dissociation energy and bond strength

Trend The higher the bond dissociation energy, the stronger the bond (6.4). Increasing size of the halogen CH3 F

ΔHo = 456 kJ/mol

CH3 Cl

CH3 Br

351 kJ/mol

293 kJ/mol

Increasing bond strength

CH3

I

234 kJ/mol

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Chapter 6–2

Ea and reaction rate

The larger the energy of activation, the slower the reaction (6.9A).

Energy

Ea

larger Ea → slower reaction slower reaction

Ea

faster reaction

Reaction coordinate

Ea and rate constant

The higher the energy of activation, the smaller the rate constant (6.9B).

Go products ΔGo > 0 Go reactants

Keq < 1

more stable reactants

Free energy

Free energy

Equilibrium always favors the species lower in energy.

Go reactants ΔGo < 0 Go products

Equilibrium favors the starting materials.

Keq > 1

more stable products

Equilibrium favors the products.

Reactive intermediates (6.3) • • •

Breaking bonds generates reactive intermediates. Homolysis generates radicals with unpaired electrons. Heterolysis generates ions. Reactive intermediate

General structure

Reactive feature

Reactivity

radical

C

unpaired electron

electrophilic

carbocation

C

carbanion

C

positive charge; only six electrons around C net negative charge; lone electron pair on C

electrophilic nucleophilic

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147

Understanding Organic Reactions 6–3

Energy diagrams (6.7, 6.8)

Energy

transition state

Ea reactants

Ea determines the rate. ΔHo is the difference in bonding energy between the reactants and products.

ΔHo products

Reaction coordinate

Conditions favoring product formation (6.5, 6.6) Variable Keq

Value Keq > 1

Meaning More product than starting material is present at equilibrium.

ΔGo

ΔGo < 0

The energy of the products is lower than the energy of the reactants.

ΔHo

ΔHo < 0

Bonds in the products are stronger than bonds in the reactants.

ΔS

ΔS > 0

The product is more disordered than the reactant.

o

o

Equations (6.5, 6.6) ΔGo = −2.303RT log Keq

ΔGo

Keq depends on the energy difference between reactants and products. R = 8.314 J/(K•mol), the gas constant T = Kelvin temperature (K)

free energy change

=

ΔHo

TΔSo

change in bonding energy

T = Kelvin temperature (K)

Factors affecting reaction rate (6.9) Factor energy of activation concentration temperature

−

Effect higher Ea → slower reaction higher concentration → faster reaction higher temperature → faster reaction

change in disorder

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Chapter 6–4

Practice Test on Chapter Review 1. Label each statement as TRUE (T) or FALSE (F) for a reaction with Keq = 0.5 and Ea = 18 kJ/mol. Ignore entropy considerations. a. b. c. d. e.

The reaction is faster than a reaction with Keq = 8 and Ea = 18 kJ/mol. The reaction is faster than a reaction with Keq = 0.5 and Ea = 12 kJ/mol. ΔG° for the reaction is a positive value. The starting materials are lower in energy than the products of the reaction. The reaction is exothermic.

2. a. Which of the following statements is true about an endothermic reaction, ignoring entropy considerations? 1. 2. 3. 4. 5.

The bonds in the products are stronger than the bonds in the starting materials. Keq < 1. A catalyst speeds up the rate of the reaction and gives a larger amount of product. Statements (1) and (2) are both true. Statements (1), (2), and (3) are all true.

b. Which of the following statements is true about a reaction with Keq = 103 and Ea = 2.5 kJ/mol? Ignore entropy considerations. 1. 2. 3. 4. 5.

The reaction is faster than a reaction with Ea = 4 kJ/mol. The starting materials are higher in energy than the products of the reaction. ΔG° is positive. Statements (1) and (2) are both true. Statements (1), (2), and (3) are all true.

3. a. Draw the transition state for the following reaction. CH3 C

CH3 CH3

H 2O

CH3

C

CH2

H 3O +

CH3

b. Draw the transition state for the following one-step elimination reaction. H H CH3CH2 C C H H

Br

OCH3

CH3CH2CH CH2

CH3OH

Br

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149

Understanding Organic Reactions 6–5

Answers to Practice Test 1. a. F b. F c. T d. T e. F

2. a. 2 b. 4

3. a.

b.

CH3 + δ C CH2 CH3

H

δ− H

δ+ O

H

OCH3 H

CH3CH2 C

C

H

H

H

Br

δ−

Answers to Problems 6.1

[1] In a substitution reaction, one group replaces another. [2] In an elimination reaction, elements of the starting material are lost and a π bond is formed. [3] In an addition reaction, elements are added to the starting material. OH

O

Br

c.

a.

C

CH3

Br replaces OH = substitution reaction

b.

CH3

CH3

C

CH2Cl

Cl replaces H = substitution reaction

H

O

O

d.

CH3CH2CH(OH)CH3

CH3CH CHCH3

OH

π bond formed elements lost (H + OH) elimination reaction

addition of 2 H's addition reaction

6.2 Heterolysis means one atom gets both of the electrons when a bond is broken. A carbocation is a C with a positive charge, and a carbanion is a C with a negative charge. heterolysis

heterolysis

CH3

a.

CH3 C OH

b.

heterolysis

c. CH3CH2 Li

Br

CH3

Electrons go to the more electronegative atom, O.

Electrons go to the more electronegative atom, Br.

Electrons go to the more electronegative atom, C.

CH3 CH3 C

OH

Br

CH3

carbocation

carbocation

CH3CH2

carbanion

Li

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Chapter 6–6

6.3 Use full-headed arrows to show the movement of electron pairs, and half-headed arrows to show the movement of single electrons. CH3

a. (CH3)3C

N N

(CH3)3C

+

c.

N N

CH3

+

CH3 C

Br

CH3 C Br

CH3

b.

CH3

+

CH3

CH3 CH3

d.

CH3

2 HO

HO OH

6.4 Increasing number of electrons between atoms = increasing bond strength = increasing bond dissociation energy = decreasing bond length. Increasing size of an atom = increasing bond length = decreasing bond strength.

a. H Cl

or H Br Br is larger than Cl. longer, weaker bond higher bond dissociation energy

b. CH3 OH

or

CH3 SH

c. (CH3)2C O

S is larger than O. longer, higher bond weaker bond dissociation energy

or

higher bond dissociation energy

CH3 OCH3

single bond fewer electrons

6.5 To determine ΔHo for a reaction: [1] Add the bond dissociation energies for all bonds broken in the equation (+ values). [2] Add the bond dissociation energies for all of the bonds formed in the equation (− values). [3] Add the energies together to get the ΔHo for the reaction. A positive ΔHo means the reaction is endothermic. A negative ΔHo means the reaction is exothermic. a. CH3CH2 Br + H2O [1] Bonds broken

CH3CH2 OH

+ HBr

[2] Bonds formed

ΔHo (kJ/mol)

ΔHo (kJ/mol)

CH3CH2 Br

+ 285

CH3CH2 OH

− 393

H OH

+ 498

H Br

− 368

+ 783 kJ/mol

Total

− 761 kJ/mol

Total

[3] Overall ΔHo = sum in Step [1] + sum in Step [2] + 783 kJ/mol − 761

kJ/mol

ANSWER: + 22 kJ/mol endothermic

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Understanding Organic Reactions 6–7

b. CH4 + Cl2

CH3Cl + HCl

[1] Bonds broken

[3] Overall ΔHo =

[2] Bonds formed

ΔHo (kJ/mol)

ΔHo (kJ/mol)

CH3 H

+ 435

CH3 Cl

− 351

Cl Cl

+ 242

H Cl

− 431

Total

+ 677 kJ/mol

sum in Step [1] + sum in Step [2] + 677 kJ/mol

− 782 kJ/mol

Total

− 782 kJ/mol ANSWER: − 105 kJ/mol exothermic

6.6 Use the directions from Answer 6.5. In determining the number of bonds broken or formed, you must take into account the coefficients needed to balance an equation. a. CH4 + 2 O2

CO2 + 2 H2O

ΔHo (kJ/mol) CH3 H O O

ΔHo (kJ/mol)

sum in Step [1] + sum in Step [2]

OC O − 535 x 2 = − 1070

+ 435 x 4 = + 1740 + 497 x 2 = + 994

HO H − 498 x 4 = − 1992

+ 2734 kJ/mol

Total

[3] Overall ΔHo =

[2] Bonds formed

[1] Bonds broken

+ 2734 kJ/mol

− 3062 kJ/mol

Total

− 3062 kJ/mol ANSWER:

b. 2 CH3CH3 + 7 O2

[3] Overall ΔHo =

[2] Bonds formed

[1] Bonds broken ΔHo (kJ/mol) CH3CH2 H + 410 x 12 = + 4920 O O

+ 497 x 7 = + 3479

C C

+ 368 x 2 =

Total

− 328 kJ/mol

4 CO2 + 6 H2O

+736

ΔHo (kJ/mol) OC O

sum in Step [1] + sum in Step [2]

− 535 x 8 = − 4280

HO H − 498 x 12 = − 5976 Total

+ 9135 kJ/mol

− 10256 kJ/mol

− 10256 kJ/mol

+ 9135 kJ/mol

ANSWER: − 1121 kJ/mol

6.7 Use the following relationships to answer the questions: If Keq = 1, then G° = 0; if Keq > 1, then G° < 0; if Keq < 1, then G° > 0. a. A negative value of G° means the equilibrium favors the product and Keq is > 1. Therefore, Keq = 1000 is the answer. b. A lower value of G° means a larger value of Keq, and the products are more favored. Keq = 10–2 is larger than Keq = 10–5, so G° is lower.

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Chapter 6–8

6.8 Use the relationships from Answer 6.7. a. Keq = 5.5. Keq > 1 means that the equilibrium favors the product. b. G° = 40 kJ/mol. A positive G° means the equilibrium favors the starting material. 6.9 When the product is lower in energy than the starting material, the equilibrium favors the product. When the starting material is lower in energy than the product, the equilibrium favors the starting material. a. G° is positive, so the equilibrium favors the starting material. Therefore the starting material is lower in energy than the product. b. Keq is > 1, so the equilibrium favors the product. Therefore the product is lower in energy than the starting material. c. G° is negative, so the equilibrium favors the product. Therefore the product is lower in energy than the starting material. d. Keq is < 1, so the equilibrium favors the starting material. Therefore the starting material is lower in energy than the product. 6.10 H H

OCH3

Keq = 2.7

OCH3

a. The Keq is > 1, so the product (the conformation on the right) is favored at equilibrium. b. The G° for this process must be negative, because the product is favored. c. G° is somewhere between 0 and −6 kJ/mol. 6.11 A positive H° favors the starting material. A negative H° favors the product. a. H° is positive (80 kJ/mol). The starting material is favored. b. H° is negative (–40 kJ/mol). The product is favored. 6.12 a. b. c. d. e.

False. True. False. True. False.

The reaction is endothermic. This assumes that G° is approximately equal to H°. Keq < 1.

a. b. c. d. e.

True. False. G° for the reaction is negative. True. False. The bonds in the product are stronger than the bonds in the starting material. True.

The starting material is favored at equilibrium.

6.13

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Understanding Organic Reactions 6–9

6.14 transition state

Energy

Ea product ² H° starting material Reaction coordinate

6.15 A transition state is drawn with dashed lines to indicate the partially broken and partially formed bonds. Any atom that gains or loses a charge contains a partial charge in the transition state. CH3

CH3

a. CH3 C OH2

H2O

CH3 C

CH3

b. CH3O H

CH3 CH3

+

CH3O

δ−

transition state:

δ+

transition state: CH3 δ C

OH

CH3O

H2O

δ− H

OH

OH2

CH3

6.16 transition state 2: E

Energy

transition state 1: D

Ea(A–C)

Ea(A–B)

B H°A–B = endothermic H°A–C = exothermic

A

a. Reaction A–C is exothermic. Reaction A–B is endothermic. b. Reaction A–C is faster. c. Reaction A–C generates a lowerenergy product. d. See labels. e. See labels. f. See labels.

C Reaction coordinate

6.17 reactive intermediate

Energy

transition state 1

Ea1

a. b. c. d. e.

transition state 2 Ea2 H°1

Reaction coordinate

H°2 H°overall

Two steps, because there are two energy barriers. See labels. See labels. One reactive intermediate is formed (see label). The first step is rate-determining, because its transition state is at higher energy. f. The overall reaction is endothermic, because the energy of the products is higher than the energy of the reactants.

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Chapter 6–10

6.18

Energy

Ea(B–C) B

Ea(A–B)

relative energies: C < A < B B C is rate-determining.

A C Reaction coordinate

6.19 Ea, concentration, and temperature affect reaction rate. H°, G°, and Keq do not affect reaction rate. a. Ea = 4 kJ/mol corresponds to a faster reaction rate. b. A temperature of 25 °C will have a faster reaction rate, because a higher temperature corresponds to a faster reaction. c. No change: Keq does not affect reaction rate. d. No change: H° does not affect reaction rate. 6.20 a. b. c. d. e.

False. The reaction occurs at the same rate as a reaction with Keq = 8 and Ea = 80 kJ/mol. False. The reaction is slower than a reaction with Keq = 0.8 and Ea = 40 kJ/mol. True. True. False. The reaction is endothermic.

6.21 All reactants in the rate equation determine the rate of the reaction. [1] rate = k[CH3CH2Br][–OH]

[2] rate = k[(CH3)3COH]

a. Tripling the concentration of CH3CH2Br only → The rate is tripled. b. Tripling the concentration of –OH only → The rate is tripled. c. Tripling the concentration of both CH3CH2CH2Br and –OH → The rate increases by a factor of 9 (3 × 3 = 9).

a. Doubling the concentration of (CH3)3COH → The rate is doubled. b. Increasing the concentration of (CH3)3COH by a factor of 10 → The rate increases by a factor of 10.

6.22 The rate equation is determined by the rate-determining step. a.

CH3CH2 Br

b.

(CH3)3C

+

CH2 CH2

OH

(CH3)3C+

Br slow

+

Br

OH fast

+ H2O + Br

(CH3)2C

one step rate = k[CH3CH2Br][–OH]

CH2 + H2O

two steps The slow step determines the rate equation. rate = k[(CH3)3CBr]

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Understanding Organic Reactions 6–11

6.23 A catalyst is not used up or changed in the reaction. It only speeds up the reaction rate. OH and H are added to the starting material. H2O

CH2 CH2

a.

H2 adds to the starting material.

I– not used up = catalyst. b.

CH3CH2OH

I–

CH3Cl

CH3OH

–

H2SO4

O

Pt

OH

−OH

H2SO4 is not used up = catalyst.

substitutes for Cl−.

OH

H2

c.

Pt not used up = catalyst.

6.24 H

H

H

a.

H

radical H

b.

H

CH3 C OH

OH

CH3 C

H

H

carbocation

6.25 propane

propene

CH3 CH2CH3

CH3

ΔHo = 356 kJ/mol This bond is formed from two sp3 hybridized C's.

CH2CH3

CH3 CH CH2

CH3

CH CH2

ΔHo = 385 kJ/mol This bond is formed from one sp2 and one sp3 hybridized C. The higher percent scharacter in one C makes a stronger bond; thus, the bond dissociation energy is higher.

6.26 Use the directions from Answer 6.1. O

HO OH

O

H OH

c.

a. π bond formed elements lost (H + OH) elimination reaction

b.

H H

Cl replaces H = substitution reaction

H Cl

addition of 2 H's addition reaction O

d.

C

O Cl

H replaces Cl = substitution reaction

C

H

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Chapter 6–12

6.27 Use the rules in Answer 6.3 to draw the arrows. Br

a.

+

Br O

b.

CH3

Cl

+

CH3

d.

+ Cl

e.

+

+ Br

Br Br

O

CH3 C CH3

c.

Br

Cl

C

CH3

f.

CH3 Cl

CH3CH2 Br

+

CH3CH2OH +

OH

H

CH3 C CH3

CH3

C H +

H + H2O

C C

OH CH3

H

Br

H

6.28 a.

+

I

OH +

OH

O

I

OH

O

b. CH3 C CH2CH2CH3

CH3

OCH2CH3

C

CH2CH2CH3

H H C

c.

d.

H

H

C H

H

Br

H C

H

H + H2O + Br

C C H

+

H H C

Cl

H

H + HCl

OCH2CH3

6.29 Draw the curved arrows to identify the product X.

O

+

H Br

[1]

O H

+

Br

Br

[2]

OH

A

B

X

6.30 Follow the curved arrows to identify the intermediate Y. CO2H

CO2H [1]

[2]

O

COOH

O

O

O

O

O

D

C

Y

6.31 Use the rules from Answer 6.4. a.

I

CCl3

Br CCl3

largest halogen intermediate weakest bond bond strength

Cl CCl3

smallest halogen strongest bond

b. H2N NH2 single bond weakest bond

HN NH

double bond intermediate bond strength

N N

triple bond strongest bond

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Understanding Organic Reactions 6–13

6.32 Use the directions from Answer 6.5. a. CH3CH2 H

+ Br2

CH3CH2 Br

[1] Bonds broken

+ HBr

[3] Overall ΔHo =

[2] Bonds formed

ΔHo (kJ/mol)

ΔHo (kJ/mol)

CH3CH2 H

+ 410

CH3CH2 Br

− 285

+ 602 kJ/mol

Br Br

+ 192

H Br

− 368

− 653 kJ/mol

+ 602 kJ/mol

Total

− 653 kJ/mol

Total

b. OH + CH4

CH3 + H2O

[1] Bonds broken

CH3 H

[3] Overall ΔHo =

[2] Bonds formed

ΔHo (kJ/mol)

c.

ANSWER: − 51 kJ/mol

− 498 kJ/mol

H OH

+ 435 kJ/mol

CH3 OH + HBr

+ 435 kJ/mol

ΔHo (kJ/mol)

CH3 Br

ANSWER: − 63 kJ/mol

+ H2O

[3] Overall ΔHo =

[2] Bonds formed

[1] Bonds broken

− 498 kJ/mol

ΔHo (kJ/mol)

ΔHo (kJ/mol) CH3 OH

+ 389

CH3 Br

− 293

+ 757 kJ/mol

H Br

+ 368

H OH

− 498

− 791 kJ/mol

Total

+ 757 kJ/mol

d. Br + CH4

ANSWER: − 34 kJ/mol

H + CH3Br

[1] Bonds broken

[3] Overall ΔHo =

[2] Bonds formed

ΔHo (kJ/mol) CH3 H

− 791 kJ/mol

Total

+ 435 kJ/mol

ΔHo (kJ/mol)

+ 435 kJ/mol

CH3 Br

−293 kJ/mol

− 293 kJ/mol ANSWER: + 142 kJ/mol

6.33 H CH2 CH C H H

H

H CH2 CH C H

CH2 CH

C H

hybrid:

δ

H

CH2 CH C H

δ

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Chapter 6–14

6.34 The more stable radical is formed by a reaction with a smaller ΔH°. H CH3 CH2

C H

CH3 CH2

H

C H

ΔHo = 410 kJ/mol = less stable radical

H

This C–H bond is stronger.

A

H CH3 C

CH3

CH3 C

H

CH3

ΔHo = 397 kJ/mol = more stable radical

H

This C–H bond is weaker.

B

Because the bond dissociation for cleavage of the C–H bond to form radical A is higher, more energy must be added to form it. This makes A higher in energy and therefore less stable than B. 6.35 Use the rules from Answer 6.9. a. Keq = 0.5. Keq is less than one, so the starting material is favored. b. ΔGo = −100 kJ/mol. ΔGo is less than 0, so the product is favored. c. ΔHo = 8.0 kJ/mol. ΔHo is positive, so the starting material is favored. d. Keq = 16. Keq is greater than one, so the product is favored. e. ΔGo = 2.0 kJ/mol. ΔGo is greater than zero, so the starting material is favored. f. ΔHo = 200 kJ/mol. ΔHo is positive, so the starting material is favored. g. ΔSo = 8 J/(K•mol). ΔSo is greater than zero, so the product is more disordered and favored. h. ΔSo = −8 J/(K•mol). ΔSo is less than zero, so the starting material is more disordered and favored. 6.36 a. A negative G° must have Keq > 1. Keq = 102. b. Keq = [products]/[reactants] = [1]/[5] = 0.2 = Keq. G° is positive. c. A negative G° has Keq > 1, and a positive G° has Keq < 1. G° = −8 kJ/mol will have a larger Keq. 6.37 R

axial equatorial H

R H

R –CH3 –CH2CH3 –CH(CH3)2 –C(CH3)3

Keq 18 23 38 4000

a. The equatorial conformation is always present in the larger amount at equilibrium, because the Keq for all R groups is greater than 1. b. The cyclohexane with the –C(CH3)3 group will have the greatest amount of equatorial conformation at equilibrium, because this group has the highest Keq. c. The cyclohexane with the –CH3 group will have the greatest amount of axial conformation at equilibrium, because this group has the lowest Keq. d. The cyclohexane with the –C(CH3)3 group will have the most negative G°, because it has the largest Keq. e. The larger the R group, the more favored the equatorial conformation.

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f. The Keq for tert-butylcyclohexane is much higher because the tert-butyl group is bulkier than the other groups. With a tert-butyl group, a CH3 group is always oriented over the ring when the group is axial, creating severe 1,3-diaxial interactions. With all other substituents, the larger CH3 groups can be oriented away from the ring, placing a H over the ring, making the 1,3-diaxial interactions less severe. Compare: tert-butylcyclohexane

isopropylcyclohexane

CH3 CH3 H C CH3 H

H

H

C

H

severe 1,3-diaxial interactions with the CH3 group and the axial H's

CH3 CH3

less severe 1,3-diaxial interactions

6.38 Calculate Keq, and then find the percentage of axial and equatorial conformations present at equilibrium. F (axial)

a.

H

F (equatorial) H

fluorocyclohexane 1 part

1.5 parts

G° = –5.9log Keq G° = –1.0 kJ/mol –1.0 kJ/mol = –5.9log Keq Keq = 1.5

b. Keq = [products]/[reactants] 1.5 = [products]/[reactants] 1.5[reactants] = [products] [reactants] = 0.4 = 40% axial [products] = 0.6 = 60% equatorial

6.39 Reactions resulting in an increase in entropy are favored. When a single molecule forms two molecules, there is an increase in entropy. increased number of molecules ΔS° is positive. products favored decreased number of molecules ΔS° is negative. starting material favored increased number of molecules H2O ΔS° is positive. products favored

+

a. b.

CH3

c.

(CH3)2C(OH)2

d.

+

CH3COOCH3

CH3CH3

CH3

(CH3)2C=O +

H2 O

+

CH3COOH

+

CH3OH

no change in the number of molecules neither favored

6.40 Use the directions in Answer 6.15 to draw the transition state. Nonbonded electron pairs are drawn in at reacting sites. Br

+

a. transition state:

δ− δ+ Br

F

+

Br

b.

BF3

+

transition state:

Cl

F B Cl F F

F B F

Cl

δ− δ−

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Chapter 6–16

OH

c.

O

+

NH2

δ−

+

δ−

NH3

C

C H

CH3

+ H2O

transition state:

H

CH3 H

CH3 δ+ C CH3

+

C C

H

O H NH2

transition state:

CH3

H

CH3

d.

H3O

H

δ+

C H OH2 H

6.41

A

H°

B

B

EaA–B

A

Reaction coordinate • one step A B • endothermic since B higher than A • high Ea (large energy barrier)

Energy

Ea A

B

Reaction coordinate • one step A B • exothermic since B lower than A • low Ea (small energy barrier)

Energy

Ea H°

d.

c.

b.

Energy

Energy

a.

EaB–C

Ea = 16 kJ/mol A

C

Reaction coordinate H°overall • two steps • A lowest energy • B highest energy • Ea(A-B) is ratedetermining, since the transition state for Step [1] is higher in energy.

H° = −80 kJ/mol

B

Reaction coordinate • one step A B • exothermic since B lower than A

6.42 a.

CH3 H

b.

Cl + CH4

CH3 + HCl

Cl

CH3 + HCl

[1] Bonds broken ΔHo (kJ/mol)

+ 435 kJ/mol

Energy

c.

Ea = 16 kJ/mol H° = 4 kJ/mol

Reaction coordinate

+ 435 kJ/mol

ΔHo (kJ/mol) H Cl

− 431 kJ/mol

− 431 kJ/mol

ANSWER:

+ 4 kJ/mol

d. The Ea for the reverse reaction is the difference in energy between the products and the transition state, 12 kJ/mol. 16 kJ/mol

Energy

CH3 H

[3] Overall ΔHo =

[2] Bonds formed

12 kJ/mol 4 kJ/mol

Reaction coordinate

+

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161

Understanding Organic Reactions 6–17

6.43 D

Energy

B

a. b. c. d.

B, D, and F are transition states. C and E are reactive intermediates. The overall reaction has three steps. A–C is endothermic. C–E is exothermic. E–G is exothermic. e. The overall reaction is exothermic.

F C

E

A

G Reaction coordinate

6.44 O CH3

C

O

CH3 CH3 C O

OH

CH3

CH3

C

CH3 CH3 C OH

O

CH3

Since pKa (CH3CO2H) = 4.8 and pKa [(CH3)3COH] = 18, the weaker acid is formed as product, and equilibrium favors the products. Thus, ΔH° is negative, and the products are lower in energy than the starting materials. transition state

O CH3

− C δ O

Energy

transition state:

δ− CH3

H

O C CH3 CH3

Ea starting materials

² H° products

Reaction coordinate

6.45 H H

H H

H Cl

[1]

H

+H

H

+ Cl [2]

Cl H

a. Step [1] breaks one π bond and the H−Cl bond, and one C−H bond is formed. The H° for this step should be positive, because more bonds are broken than formed. b. Step [2] forms one bond. The H° for this step should be negative, because one bond is formed and none is broken. c. Step [1] is rate-determining, because it is more difficult. d. Transition state for Step [1]: Transition state for Step [2]: δ−

H H

δ+ H

Cl

δ+

H H H

− Cl δ

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Chapter 6–18

e.

Energy

Ea2 H°1 is positive.

Ea1

H°2 is negative. H°overall is negative.

Reaction coordinate

6.46 B

C Ea

Energy

162

Ea (CH3)3C +

A

+ H2O + I

Ea (CH3)3C

+ HI

OH (CH3)3C

+

H°3 H°2

H°1 = 0

OH2

H°overall

+ I

(CH3)3C

I

Reaction coordinate

a. The reaction has three steps, because there are three energy barriers. b. See above. c. Transition state A (see graph for location): Transition state B: Transition state C: δ+ (CH3)3C

OH H

δ− I

(CH3)3C

δ+

OH2

δ+

− δ+ δ I

(CH3)3C

d. Step [2] is rate-determining, because this step has the highest energy transition state. 6.47 Ea, concentration, catalysts, rate constant, and temperature affect reaction rate so (c), (d), (e), (g), and (h) affect rate. 6.48 a. b. c. d.

rate = k[CH3Br][NaCN] Double [CH3Br] = rate doubles. Halve [NaCN] = rate halved. Increase both [CH3Br] and [NaCN] by factor of 5 = [5][5] = rate increases by a factor of 25.

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163

Understanding Organic Reactions 6–19

6.49 O

acetyl chloride

CH3

C

[1] Cl

slow

O CH3O

O

[2]

CH3 C Cl fast

CH3

C

+ Cl– OCH3

methyl acetate

CH3O

a. Only the slow step is included in the rate equation: Rate = k[CH3O–][CH3COCl] b. CH3O– is in the rate equation. Increasing its concentration by 10 times would increase the rate by 10 times. c. When both reactant concentrations are increased by 10 times, the rate increases by 100 times (10 × 10 = 100). d. This is a substitution reaction (OCH3 substitutes for Cl). 6.50 a. b. c. d. e. f.

True: Increasing temperature increases reaction rate. True: If a reaction is fast, then it has a large rate constant. False: Corrected—There is no relationship between G° and reaction rate. False: Corrected—When the Ea is large, the rate constant is small. False: Corrected—There is no relationship between Keq and reaction rate. False: Corrected—Increasing the concentration of a reactant increases the rate of a reaction only if the reactant appears in the rate equation.

6.51 a. The first mechanism has one step: Rate = k[(CH3)3CI][–OH] b. The second mechanism has two steps, but only the first step would be in the rate equation, because it is slow and therefore rate-determining: Rate = k[(CH3)3CI] c. Possibility [1] is second order; possibility [2] is first order. d. These rate equations can be used to show which mechanism is plausible by changing the concentration of –OH. If this affects the rate, then possibility [1] is reasonable. If it does not affect the rate, then possibility [2] is reasonable. CH3

e.

CH3 C

CH2

Energy

δ– I

H

δ– OH

Ea H°

B

A

Reaction coordinate

A = (CH3)3CI + –OH B = (CH3)2C=CH2 + I– + H2O

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Chapter 6–20

transition state [1]

f.

transition state [2]

B

Energy

164

Ea[1] H°[1]

(CH3)3C+ + I– + –OH

CH3

[1] CH3 C +CH3 δ I δ–

Ea[2]

H°[2] H°overall

(CH3)3CI

(CH3)2C=CH2 + H2O

CH3

[2] CH3 C CH2 δ+ H

OH

δ–

Reaction coordinate

6.52 The difference in both the acidity and the bond dissociation energy of CH3CH3 versus HC≡CH is due to the same factor: percent s-character. The difference results because one process is based on homolysis and one is based on heterolysis. Bond dissociation energy: CH3CH2 H

sp3 hybridized 25% s-character

HC C H

sp hybridized 50% s-character Higher percent s-character makes this bond shorter and stronger.

Acidity: To compare acidity, we must compare the stability of the conjugate bases: CH3CH2

sp3 hybridized 25% s-character

HC C

sp hybridized 50% s-character Now a higher percent s-character stabilizes the conjugate base making the starting acid more acidic.

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Understanding Organic Reactions 6–21

6.53 a. Re-draw A to see more clearly how cyclization occurs. σ bonds formed +

re-draw +

+

σ bond formed

A

O

B

O

b.

6.54 Ha Hb

a.

Hb

Hb

C C CH3

C C CH3

C C CH3

H H

H H

H H

H H

Hb

b.

Hb

C C CH3

Hb

C C CH3

C C CH3

H H

H H

Ha Hb

Ha

Ha

C C CH3

C C CH3

C C CH3

H H

H H

H H

c. C–Ha is weaker than C–Hb since the carbon radical formed when the C–Ha bond is broken is highly resonance stabilized. This means the bond dissociation energy for C–Ha is lower. 6.55 In Reaction [1], the number of molecules of reactants and products stays the same, so entropy is not a factor. In Reaction [2], a single molecule of starting material forms two molecules of products, so entropy increases. This makes G° more favorable, thus increasing Keq. 6.56 O CH3

C

O OH

+ CH3CH2OH

CH3

C

OCH2CH3

+ H 2O

Keq = 4

ethyl acetate

To increase the yield of ethyl acetate, H2O can be removed from the reaction mixture, or there can be a large excess of one of the starting materials.

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Chapter 6–22

6.57 a.

O H

O

O

CH3CH2 O H

ethanol phenol

O

resonance stabilized less energy for homolysis O O H

Csp2–O higher % s-character shorter bond

no resonance stabilization

Less energy is required for cleavage of C6H5O–H because homolysis forms the more stable radical. O

b.

CH3CH2 O

CH3CH2 O H

Csp3–O lower % s-character longer bond

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167

Alkyl Halides and Nucleophilic Substitution 7–1

Chapter 7 Alkyl Halides and Nucleophilic Substitution Chapter Review General facts about alkyl halides • • • •

Alkyl halides contain a halogen atom X bonded to an sp3 hybridized carbon (7.1). Alkyl halides are named as halo alkanes, with the halogen as a substituent (7.2). Alkyl halides have a polar C–X bond, so they exhibit dipole–dipole interactions but are incapable of intermolecular hydrogen bonding (7.3). The polar C–X bond containing an electrophilic carbon makes alkyl halides reactive towards nucleophiles and bases (7.5).

The central theme (7.6) •

Nucleophilic substitution is one of the two main reactions of alkyl halides. A nucleophile replaces a leaving group on an sp3 hybridized carbon. R X

+

Nu

R Nu

nucleophile

+

X

leaving group

The electron pair in the C−Nu bond comes from the nucleophile.

• •

One σ bond is broken and one σ bond is formed. There are two possible mechanisms: SN1 and SN2.

SN1 and SN2 mechanisms compared SN2 mechanism One step (7.11B)

•

[2] Alkyl halide

•

Order of reactivity: CH3X > RCH2X • > R2CHX > R3CX (7.11D)

[3] Rate equation

• •

rate = k[RX][:Nu–] second-order kinetics (7.11A)

• •

rate = k[RX] first-order kinetics (7.13A)

[4] Stereochemistry

•

backside attack of the nucleophile (7.11C) inversion of configuration at a stereogenic center favored by stronger nucleophiles (7.17B) better leaving group → faster reaction (7.17C)

•

trigonal planar carbocation intermediate (7.13C) racemization at a stereogenic center favored by weaker nucleophiles (7.17B) better leaving group → faster reaction (7.17C)

favored by polar aprotic solvents (7.17D)

•

• [5] Nucleophile

•

[6] Leaving group

•

[7] Solvent

•

•

SN1 mechanism Two steps (7.13B)

[1] Mechanism

• • •

Order of reactivity: R3CX > R2CHX > RCH2X > CH3X (7.13D)

favored by polar protic solvents (7.17D)

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Chapter 7–2

Increasing rate of an SN1 reaction H

H

R

H C Br

H

R C Br

R C Br

R C Br

H

H

R

R

methyl

1

o

3o

both SN1 and SN2

SN1

o

2

SN2

Increasing rate of an SN2 reaction

Important trends •

The best leaving group is the weakest base. Leaving group ability increases left to right across a row and down a column of the periodic table (7.7). Increasing basicity NH3

Increasing basicity

H2O

F

Increasing leaving group ability

•

Cl

Br

I

Increasing leaving group ability

Nucleophilicity decreases across a row of the periodic table (7.8A). For 2nd row elements with the same charge:

CH3

NH2

OH

F

Increasing basicity Increasing nucleophilicity

•

Nucleophilicity decreases down a column of the periodic table in polar aprotic solvents (7.8C). Down a column of the periodic table

F

Cl

Br

I

Increasing nucleophilicity in polar aprotic solvents

•

Nucleophilicity increases down a column of the periodic table in polar protic solvents (7.8C). Down a column of the periodic table

F

Cl

Br

I

Increasing nucleophilicity in polar protic solvents

•

The stability of a carbocation increases as the number of R groups bonded to the positively charged carbon increases (7.14). + CH3

+ RCH2

+ R2CH

+ R3C

methyl

1o

2o

3o

Increasing carbocation stability

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169

Alkyl Halides and Nucleophilic Substitution 7–3

Important principles

•

Principle Electron-donating groups (such as R groups) stabilize a positive charge (7.14A).

•

Example 3 arbocations (R3C+) are more stable than 2o arbocations (R2CH+), which are more stable than 1o carbocations (RCH2+). o

•

Steric hindrance decreases nucleophilicity but not basicity (7.8B).

•

(CH3)3CO– is a stronger base but a weaker nucleophile than CH3CH2O–.

•

Hammond postulate: In an endothermic reaction, the more stable product is formed faster. In an exothermic reaction, this fact is not necessarily true (7.15).

•

SN1 reactions are faster when more stable (more substituted) carbocations are formed, because the rate-determining step is endothermic.

•

Planar, sp2 hybridized atoms react with reagents from both sides of the plane (7.13C).

•

A trigonal planar carbocation reacts with nucleophiles from both sides of the plane.

Practice Test on Chapter Review 1. Give the IUPAC name for the following compound, including the appropriate R,S prefix. Cl

2. a. Which of the following carbocations is the most stable?

1.

2.

3.

4.

5.

b. Which of the following anions is the best leaving group? 1. CH3–

2. –OH

3. H–

4. –NH2

5. Cl–

c. Which species is the strongest nucleophile in polar protic solvents? 1. F–

2. –OH

3. Cl–

4. H2O

5. –SH

d. Which of the following statements is true about the given reaction? Br

OCH2CH3 Na+ –OCH2CH3

Br

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Chapter 7–4

1. 2. 3. 4. 5.

The reaction follows second-order kinetics. The rate of the reaction increases when the solvent is changed from CH3CH2OH to DMSO. The rate of the reaction increases when the leaving group is changes from Br to F. Statements (1) and (2) are both true. Statements (1), (2), and (3) are all true.

3. Rank the following compounds in order of increasing reactivity in an SN1 reaction. Rank the least reactive compound as 1, the most reactive compound as 4, and the compounds of intermediate reactivity as 2 and 3. Cl

Cl

OH

A

B

Cl

C

D

4. Consider the following two nucleophilic substitution reactions, labeled Reaction [1] and Reaction [2]. (Only the starting materials are drawn.) Then answer True (T) or False (F) to each of the following statements. Reaction [1]

(CH3CH2)3CBr

CH3OH

Reaction [2]

CH3CH2CH2Br

OCH3

The rate equation for Reaction [1] is rate = k[(CH3CH2)3CBr][CH3OH]. Changing the leaving group from Br– to Cl– decreases the rate of both reactions. Changing the solvent from CH3OH to (CH3)2S=O increases the rate of Reaction [2]. Doubling the concentration of both CH3CH2CH2Br and –OCH3 in Reaction [2] doubles the rate of the reaction. e. If entropy is ignored and Keq for Reaction [1] is < 1, then the reaction is exothermic. f. If entropy is ignored and ΔH° is negative for Reaction [1], then the bonds in the product are stronger than the bonds in the starting materials. g. The energy diagram for Reaction [2] exhibits only one energy barrier. a. b. c. d.

5. Draw the organic products formed in the following reactions. Use wedges and dashes to show stereochemistry in compounds with stereogenic centers. a.

H

C

C

H

Br D CH3OH

b.

(Consider substitution only.) Cl NaOH

c.

Br

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171

Alkyl Halides and Nucleophilic Substitution 7–5

d.

Br

CH3COO

Answers to Practice Test 1. (7S)-7-chloro-3-ethyldecane

3. A–4 B–1 C–3 D–2 4. a. F b. T c. T d. F e. F f. T g. T

2. a. 4 b. 5 c. 5 d. 4

5. a.

c. H D C

OH

CH

b.

d. O2CCH3

OCH3

CH3O

Answers to Problems 7.1 Classify the alkyl halide as 1°, 2°, or 3° by counting the number of carbons bonded directly to the carbon bonded to the halogen. C bonded to 2 C's 2° alkyl halide

C bonded to 1 C 1° alkyl halide

I

CH3

a.

F

b.

CH3CH2CH2CH2CH2 Br

c. CH3 C

CHCH3

d.

CH3 Cl

C bonded to 3 C's 3° alkyl halide

C bonded to 3 C's 3° alkyl halide

7.2 Use the directions from Answer 7.1. 3°, neither 2°, neither

2°, neither

a, b.

Cl

1°, neither

Br Cl

Cl

3°, allylic

Cl Br Cl

3°, neither telfairine

Br

vinyl

2°, neither halomon

Cl

vinyl

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Chapter 7–6

7.3 Draw a compound of molecular formula C6H13Br to fit each description. Br

a.

b.

Br

c. Br

1° alkyl halide one stereogenic center

2° alkyl halide two stereogenic centers

3° alkyl halide no stereogenic centers

7.4 To name a compound with the IUPAC system: [1] Name the parent chain by finding the longest carbon chain. [2] Number the chain so the first substituent gets the lower number. Then name and number all substituents, giving like substituents a prefix (di, tri, etc.). To name the halogen substituent, change the -ine ending to -o. [3] Combine all parts, alphabetizing substituents, and ignoring all prefixes except iso. a.

(CH3)2CHCH(Cl)CH2CH3

re-draw [1]

2-methyl [2]

Cl

5 carbon alkane = pentane b.

[1]

[3] 3-chloro-2-methylpentane 2 Cl

Br

3-chloro

2-bromo [2]

Br

[3] 2-bromo-5,5-dimethylheptane

2 7 carbon alkane = heptane

5,5-dimethyl

c.

2-methyl

[2]

[1] Br

Br

[3] 1-bromo-2-methylcyclohexane

1-bromo

1 6 carbon cycloalkane = cyclohexane

2-fluoro

d. [1]

F

[2]

F

[3] 2-fluoro-5,5-dimethylheptane 5,5-dimethyl

1 2 3 4 7 carbon alkane = heptane

7

7.5 To work backwards from a name to a structure: [1] Find the parent name and draw that number of carbons. Use the suffix to identify the functional group (-ane = alkane). [2] Arbitrarily number the carbons in the chain. Add the substituents to the appropriate carbon.

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173

Alkyl Halides and Nucleophilic Substitution 7–7

a. 3-chloro-2-methylhexane [1]

6 carbon alkane

[2]

methyl at C2

1 2 3 4 5 6

chloro at C3

Cl

b. 4-ethyl-5-iodo-2,2-dimethyloctane [1]

8 carbon alkane

[2] ethyl at C4

1 2 3 4 5 6 7 8

I

2 methyls at C2

iodo at C5

c. cis-1,3-dichlorocyclopentane [1]

5 carbon cycloalkane

[2]

chloro groups at C1 and C3, both on the same side Cl

Cl

C3

C1 d. 1,1,3-tribromocyclohexane [1]

6 carbon cycloalkane

3 Br groups

[2]

Br

C3

Br Br

C1

e. propyl chloride [1] 3 carbon alkyl group CH3CH2CH2

[2] chloride on end CH3CH2CH2 Cl

f. sec-butyl bromide [1] 4 carbon alkyl group CH3 CHCH2CH3

[2] bromide CH3 CHCH2CH3 Br

7.6 a. Because an sp2 hybridized C has a higher percent s-character than an sp3 hybridized C, it holds electron density closer to C. This pulls a little more electron density towards C, away from Cl, and thus a Csp2–Cl bond is less polar than a Csp3–Cl bond. b.

Cl

lowest boiling point

Cl

sp3 C–Cl bond intermediate boiling point

Br

larger halogen, sp3 C–Br bond highest boiling point

7.7 a. Since chondrocole A has 10 C’s and only one functional group capable of hydrogen bonding to water (an ether), it is insoluble in H2O. Because it is organic, it is soluble in CH2Cl2.

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Chapter 7–8

R b.

H

Br

O

*

S

*

Three stereogenic centers are labeled with (*).

* Cl

S Br

c.

Br

O

O

Cl

Cl

enantiomer

constitutional isomer

7.8 To draw the products of a nucleophilic substitution reaction: [1] Find the sp3 hybridized electrophilic carbon with a leaving group. [2] Find the nucleophile with lone pairs or electrons in π bonds. [3] Substitute the nucleophile for the leaving group on the electrophilic carbon. +

a.

+ Br

OCH2CH3

Br

OCH2CH3

nonbonded e− pairs nucleophile leaving group Cl

OH

b.

+

nonbonded e− pairs nucleophile

leaving group

+

I

c.

+ Na+Cl–

Na+ −OH

N3

+ I

N3

nonbonded e− pairs nucleophile

leaving group Br

CN

d.

+ Na+Br–

Na+ −CN

+

nonbonded e− pairs nucleophile

leaving group

7.9 Use the steps from Answer 7.8 and then draw the proton transfer reaction.

a.

+

Br

N(CH2CH3)3

substitution

+

N(CH2CH3)3 + Br

nucleophile leaving group

b. (CH3)3C Cl

leaving group

+

H2O

substitution

nucleophile

(CH3)3C

O H H

+ Cl

proton (CH3)3C

transfer

O H

+ HCl

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Alkyl Halides and Nucleophilic Substitution 7–9

7.10 Draw the structure of CPC using the steps from Answer 7.8. +

N

Cl

nucleophile

substitution

leaving group

+ N

+

Cl

CPC

7.11 Compare the leaving groups based on these trends: • Better leaving groups are weaker bases. • A neutral leaving group is always better than its conjugate base. a. Cl–, I–

b. NH3, NH2–

farther down a column of the periodic table less basic better leaving group

c. H2O, H2S

neutral compound less basic better leaving group

farther down a column of the periodic table less basic better leaving group

7.12 Good leaving groups include Cl–, Br–, I–, and H2O. a.

CH3CH2CH2 Br

c. CH3CH2CH2 OH2

b. CH3CH2CH2OH

Br− is a good leaving group.

No good leaving group. is too strong a base.

d. CH3CH3

No good leaving group. H− is too strong a base.

H2O is a good leaving group.

−OH

7.13 To decide whether the equilibrium favors the starting material or the products, compare the nucleophile and the leaving group. The reaction proceeds towards the weaker base. a.

CH3CH2 NH2

+

Br

nucleophile better leaving group weaker base pKa (HBr) = –9 b.

I

+

+

CH3CH2 Br

CN

nucleophile pKa (HCN) = 9.1

NH2

leaving group

Reaction favors starting material.

pKa (NH3) = 38

CN

+

I

leaving group better leaving group weaker base pKa (HI) = –10

Reaction favors product.

7.14 It is not possible to convert CH3CH2CH2OH to CH3CH2CH2Cl by nucleophilic substitution with NaCl because –OH is a stronger base and poorer leaving group than Cl–. The equilibrium favors the reactants, not the products. CH3CH2CH2OH

Na Cl

weaker base

CH3CH2CH2Cl

Na

OH

stronger base

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Chapter 7–10

7.15 Use these three rules to find the stronger nucleophile in each pair: [1] Comparing two nucleophiles having the same attacking atom the stronger base is a stronger nucleophile. [2] Negatively charged nucleophiles are always stronger than their conjugate acids. [3] Across a row of the periodic table, nucleophilicity decreases when comparing species of similar charge. O

a. NH3, NH2

b. CH3NH2, CH3OH

A negatively charged nucleophile is stronger than its conjugate acid. stronger nucleophile

c.

C CH3

Across a row of the periodic table, nucleophilicity decreases with species of the same charge. stronger nucleophile

O

CH3CH2O

same attacking atom (O) stronger base stronger nucleophile

7.16 Polar protic solvents are capable of hydrogen bonding, and therefore must contain a H bonded to an electronegative O or N. Polar aprotic solvents are incapable of hydrogen bonding, and therefore do not contain any O–H or N–H bonds. a. HOCH2CH2OH contains 2 O–H bonds polar protic

b. CH3CH2OCH2CH3 no O–H bonds polar aprotic

c. CH3COOCH2CH3 no O–H bonds polar aprotic

7.17 • In polar protic solvents, the trend in nucleophilicity is opposite to the trend in basicity down a column of the periodic table so that nucleophilicity increases. • In polar aprotic solvents, the trend is identical to basicity so that nucleophilicity decreases down a column. a. Br− and Cl− in polar protic solvent

farther down the column more nucleophilic in protic solvent b. −OH and Cl− in polar aprotic solvent

farther up the column and to the left in the row more basic more nucleophilic

In polar aprotic solvents: nucleophilicity increases O F nucleophilicity Cl increases

c. HS− and F− in polar protic solvent In polar protic solvents: nucleophilicity increases farther down the column O F nucleophilicity and left in the row S increases more nucleophilic in protic solvent

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7.18 The stronger base is the stronger nucleophile except in polar protic solvents when nucleophilicity increases down a column. For other rules, see Answers 7.15 and 7.17. a.

H2O

OH

no charge weakest nucleophile

NH2

negatively charged intermediate nucleophile

negatively charged farther left in periodic table strongest nucleophile

b. Br F Basicity decreases down a Basicity decreases column in polar aprotic solvents. across a row. weakest nucleophile intermediate nucleophile c.

H2O

OH

strongest nucleophile

CH3COO

OH

weaker base than −OH intermediate nucleophile

weakest nucleophile

strongest nucleophile

7.19 To determine what nucleophile is needed to carry out each reaction, look at the product to see what has replaced the leaving group. a. (CH3)2CHCH2CH2 Br

(CH3)2CHCH2CH2

c. (CH3)2CHCH2CH2 Br

SH

SH replaces Br. HS− is needed.

b. (CH3)2CHCH2CH2 Br

(CH3)2CHCH2CH2

(CH3)2CHCH2CH2

OCOCH3

OCOCH3 replaces Br. CH3COO− is needed.

OCH2CH3

d. (CH3)2CHCH2CH2 Br

OCH2CH3 replaces Br. CH3CH2O− is needed.

(CH3)2CHCH2CH2

C CH

C≡CH replaces Br. HC≡C− is needed.

7.20 The general rate equation for an SN2 reaction is rate = k[RX][:Nu–]. a. [RX] is tripled, and [:Nu–] stays the same: rate triples. b. Both [RX] and [:Nu–] are tripled: rate increases by a factor of 9 (3 × 3 = 9). c. [RX] is halved, and [:Nu–] stays the same: rate halved. d. [RX] is halved, and [:Nu–] is doubled: rate stays the same (1/2 × 2 = 1). 7.21 The transition state in an SN2 reaction has dashed bonds to both the leaving group and the nucleophile, and must contain partial charges. a.

CH3CH2CH2 Cl

+

OCH3

CH3CH2CH2 OCH3

+

Cl

CH3CH2CH2 CH3O δ−

b.

Br

+

SH

SH

+

Br

δ−

SH δ− Br

Cl δ−

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Chapter 7–12

7.22 All SN2 reactions have one step. Cl δ−

CH3CH2CH2

Energy

CH3O δ−

Ea CH3CH2CH2 Cl

ΔH°

+ OCH3

CH3CH2CH2 OCH3 + Cl

Reaction coordinate

7.23 To draw the products of SN2 reactions, replace the leaving group by the nucleophile, and then draw the stereochemistry with inversion at the stereogenic center. CH3CH2 D

a.

C

D CH CH 2 3 Br +

CH3CH2O

OCH2CH3

b.

C

H

+

I

CN

CN

H

7.24 Increasing the number of R groups increases crowding of the transition state and decreases the rate of an SN2 reaction. Cl

or

a. 2° alkyl halide

b.

Cl

Br

or

2° alkyl halide faster reaction

1° alkyl halide faster reaction

Br

3° alkyl halide

7.25 +

H N

N H

CH3 SR2 H N

A

+ SR2

N H

loss of

CH3

a proton

H N

N CH3

nicotine

7.26 In a first-order reaction, the rate changes with any change in [RX]. The rate is independent of any change in [:Nu–]. a. [RX] is tripled, and [:Nu–] stays the same: rate triples. b. Both [RX] and [:Nu–] are tripled: rate triples. c. [RX] is halved, and [:Nu–] stays the same: rate halved. d. [RX] is halved, and [:Nu–] is doubled: rate halved.

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7.27 In SN1 reactions, racemization always occurs at a stereogenic center. Draw two products, with the two possible configurations at the stereogenic center. leaving group

nucleophile

CH3

a.

C (CH3)2CH

CH3

H2O Br

C (CH3)2CH

CH2CH3

CH3

H

CH3COO

CH3

CH3CH2

Cl

(CH3)2CH

CH2CH3

+

HBr

OH

enantiomers

nucleophile

b.

C

+

OH CH2CH3

H

CH3

CH3CH2

H

+

OOCCH3

CH3CH2

OOCCH3 CH3

diastereomers leaving group

7.28 Carbocations are classified by the number of R groups bonded to the carbon: 0 R groups = methyl, 1 R group = 1°, 2 R groups = 2°, and 3 R groups = 3°. a.

+ b. (CH3)3CCH2

+

1 R group 1° carbocation

2 R groups 2° carbocation

+

c.

d. + 2 R groups 2° carbocation

3 R groups 3° carbocation

7.29 For carbocations: Increasing number of R groups = Increasing stability. +

+

+

CH3CH2CH2CH2

CH3CHCH2CH3

CH3 C CH3

1° carbocation least stable

2° carbocation intermediate stability

3° carbocation most stable

CH3

7.30 For carbocations: Increasing number of R groups = Increasing stability. + CH2

+ +

1° carbocation least stable

3° carbocation most stable

2° carbocation intermediate stability

7.31 The rate of an SN1 reaction increases with increasing alkyl substitution. CH3

a.

(CH3)3CBr

or

(CH3)3CCH2Br

1° alkyl halide 3° alkyl halide faster SN1 reaction slower SN1 reaction

b.

Br

Br

or

3° alkyl halide faster SN1 reaction

2° alkyl halide slower SN1 reaction

+

Cl–

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Chapter 7–14

7.32 • For methyl and 1° alkyl halides, only SN2 will occur. • For 2° alkyl halides, SN1 and SN2 will occur. • For 3° alkyl halides, only SN1 will occur. CH3 H

a. CH3 C

C

Br

b.

Br

CH3 CH3

Br

d.

2° alkyl halide SN1 and SN2

1° alkyl halide S N2

2° alkyl halide SN1 and SN2

Br

c.

3° alkyl halide SN1

7.33 • Draw the product of nucleophilic substitution for each reaction. • For methyl and 1° alkyl halides, only SN2 will occur. • For 2° alkyl halides, SN1 and SN2 will occur and other factors determine which mechanism operates. • For 3° alkyl halides, only SN1 will occur. Strong nucleophile favors SN2. I

CH3OH Cl

a.

OCH3

c.

+ HCl

3° alkyl halide only SN1

Br

b.

OCH2CH3

CH3CH3O

+

I

+

HBr

2° alkyl halide Both SN1 and SN2 are possible. Weak nucleophile favors SN1. SH

SH

+ Br

CH3OH

d. Br

1° alkyl halide only SN2

OCH3

2° alkyl halide Both SN1 and SN2 are possible.

7.34 First decide whether the reaction will proceed via an SN1 or SN2 mechanism. Then draw the products with stereochemistry. +

a. H Br

Weak nucleophile favors SN1.

2° alkyl halide SN1 and SN2 Cl

b. H D

1° alkyl halide SN2 only

H 2O

+

C C H

+ HBr

+ H OH

HO H

SN1 = racemization at the stereogenic C

enantiomers

HC C D H

+

Cl

SN2 = inversion at the stereogenic C

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7.35 Compounds with better leaving groups react faster. Weaker bases are better leaving groups. a. CH3CH2CH2Cl

c. (CH3)3C OH or

CH3CH2CH2I

or

weaker base better leaving group b. (CH3)3CBr

+

OH2

weaker base better leaving group

(CH3)3CI

or

(CH3)3C

d. CH3CH2CH2OH

weaker base better leaving group

or

CH3CH2CH2 OCOCH3

weaker base better leaving group

7.36 • Polar protic solvents favor the SN1 mechanism by solvating the intermediate carbocation and halide. • Polar aprotic solvents favor the SN2 mechanism by making the nucleophile stronger. a. CH3CH2OH polar protic solvent contains an O–H bond favors SN1

b. CH3CN polar aprotic solvent no O–H or N–H bond favors SN2

c. CH3COOH polar protic solvent contains an O–H bond favors SN1

d. CH3CH2OCH2CH3 polar aprotic solvent no O–H or N–H bond favors SN2

7.37 Compare the solvents in the reactions below. For the solvent to increase the reaction rate of an SN1 reaction, the solvent must be polar protic. + CH3OH

a. Cl

CH3OH

+

or

HCl

3o RX – SN1 reaction Br

b.

OH

+

1o RX – SN2 reaction

c.

+

CH3O

H Cl

H2O

OH

+

Br

or

DMF [HCON(CH3)2]

Polar aprotic solvent increases the rate of an SN2 reaction.

DMF

CH3OH

+

or HMPA

CH3OH

Polar protic solvent increases the rate of an SN1 reaction.

OCH3

DMSO

Cl

H OCH3

2o RX strong nucleophile SN2 reaction

HMPA [(CH3)2N]3P=O

Polar aprotic solvent increases the rate of an SN2 reaction.

7.38 To predict whether the reaction follows an SN1 or SN2 mechanism: [1] Classify RX as a methyl, 1°, 2°, or 3° halide. (Methyl, 1° = SN2; 3° = SN1; 2° = either.) [2] Classify the nucleophile as strong or weak. (Strong favors SN2; weak favors SN1.) [3] Classify the solvent as polar protic or polar aprotic. (Polar protic favors SN1; polar aprotic favors SN2.) a.

CH2Br

+

CH3CH2O

CH2OCH2CH3

+

Br

SN2 reaction

1° alkyl halide SN2

b.

Br

2° alkyl halide SN1 or SN2

+

N3

Strong nucleophile favors SN2.

N3

+

Br

SN2 reaction = inversion at the stereogenic center The leaving group was "up." The nucleophile attacks from below.

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Chapter 7–16

I

c.

OCH3

CH3OH

+

SN1 reaction

HI

+

Weak nucleophile favors SN1.

3° alkyl halide SN1

d.

+

Cl

3° alkyl halide SN1

+

H2O

Weak nucleophile favors SN1.

+ HCl

OH

SN1 reaction forms two enantiomers.

HO

7.39 Vinyl carbocations are even less stable than 1° carbocations. CH3CH2CH2CH2CH=CH

CH3CH2CH2CH2CH2CH2

CH3CH2CH2CH2CHCH3

vinyl carbocation least stable

1° carbocation intermediate stability

2° carbocation most stable

7.40 Convert each ball-and-stick model to a skeletal or condensed structure and draw the reactants. Na+ −CN

a.

CN

carbon framework

Cl

CN

nucleophile

b. (CH3)3CCH2CH2

SH

(CH3)3CCH2CH2

Cl

Na+ −SH

(CH3)3CCH2CH2

SH

nucleophile

carbon framework OH

Cl

c.

OH

Na+ −OH

nucleophile carbon framework Na+ −C

d.

CH3CH2 C C H

carbon framework

CH3CH2 Cl

CH CH3CH2 C C H

nucleophile

7.41 CH3O

Cl CH2CH3

CH3OCH2CH3

CH3CH2O

Cl CH3

CH3OCH2CH3

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Alkyl Halides and Nucleophilic Substitution 7–17

7.42 Use the directions from Answer 7.4 to name the compounds. 2

1 7 C chain = heptane bromo at C2 ethyl at C5 one stereogenic center–R (2R)-2-bromo-5-ethylheptane

a. H

Br

5

R

3

5 C ring = cyclopentane chloro at C1 isopropyl at C3 trans isomer–one substituent up, one down (1R, 3R)-trans-1-chloro-3-isopropylcyclopentane

b. 2

1

R

Cl

R

7.43 H

CN (axial) Br

a. (CH3)3C

CN

(eq) acetone

H

(CH3)3C

H

inversion (equatorial to axial)

H

polar aprotic solvent SN2 reaction Large tert-butyl group is in more roomy equatorial position. Br (axial) H

b. (CH3)3C

H CN

acetone

inversion (axial to equatorial) CN (eq)

(CH3)3C

H

H

polar aprotic solvent SN2 reaction

7.44 Use the directions from Answer 7.4 to name the compounds. CH3 [1] CH3 C CH2CH2 F a.

[2]

[3] 1-fluoro-3,3-dimethylbutane

1

3 CH3

CH3

1-fluoro 3,3-dimethyl

4 carbon alkane = butane

b. [1]

CH3

CH3 C CH2CH2 F

I

[2]

3-ethyl 3 2-methyl

6 carbon alkane = hexane

2

1 I 1-iodo

[3]

3-ethyl-1-iodo-2-methylhexane

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Chapter 7–18

c. [1] (CH3)3CCH2Br

CH3

[2]

[3] 1-bromo-2,2-dimethylpropane

CH3 C CH2 Br

CH3

2 CH3 1

CH3 C CH2 Br

1-bromo 2,2-dimethyl

CH3

3 carbon alkane = propane d. [1]

[2] Br

6

2

Br

Cl

8 carbon alkane = octane

6-methyl

Br

[2]

2

I

1-bromo [3] cis-1-bromo-3-iodocyclopentane

I

5 carbon cycloalkane = cyclopentane Cl

[1]

f.

Cl 6-bromo 2-chloro

Br

e. [1]

[3] 6-bromo-2-chloro-6-methyloctane

1

3-iodo

1 Cl

[2]

[3] trans-1,2-dichlorocyclohexane

trans-1,2-dichloro

Cl

Cl

6 carbon cycloalkane = cyclohexane

2

g. [1] (CH3)3CCH2CH(Cl)CH2Cl CH3

CH3

[2]

H

CH3

CH3 C CH2 C CH2 Cl CH3

H

[3] 1,2-dichloro-4,4-dimethylpentane

CH3 C CH2 C CH2 Cl

Cl

Cl

4,4-dimethyl

1,2-dichloro

5 carbon alkane = pentane h.

[1]

[2]

4

I H

I H

6 carbon alkane = hexane (Indicate the R,S designation also)

[3] (2R)-2-iodo-4,4-dimethylhexane

2 1

4,4-dimethyl

2

(2R)-2-iodo

3 I H4

Clockwise R

1

7.45 To work backwards to a structure, use the directions in Answer 7.5. c. 1,1-dichloro-2-methylcyclohexane

a. isopropyl bromide Br CH3 CHCH3

1 Cl

Bromine on middle C makes it an isopropyl group.

2

b. 3-bromo-4-ethylheptane

3 Br

4-ethyl

1,1-dichloro

Cl

2-methyl

d. trans-1-chloro-3-iodocyclobutane 1-chloro Cl 1 3

4

3-bromo

I

3-iodo

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Alkyl Halides and Nucleophilic Substitution 7–19

g. (1R,2R)-trans-1-bromo-2-chlorocyclohexane 1R 1-bromo Br 1

e. 1-bromo-4-ethyl-3-fluorooctane 4-ethyl Br

3

1 1-bromo

4

F

Cl 2-chloro 2R h. (5R)-4,4,5-trichloro-3,3-dimethyldecane 5R 3,3-dimethyl Cl H 1 4 5 3

3-fluoro

f. (3S)-3-iodo-2-methylnonane 2-methyl 4 3 1 I H 3S 3-iodo

Cl Cl

4,4,5-trichloro

7.46 H H

CH3

a.

CH3 C CH2CH2F CH3

g.

d. Br

1° halide

h. I H

e.

2° halide

1° halide 2° halide

I

c.

1° halide

2° halide

2° halide

2° halide

Br I

C C Cl Cl H

Cl

3° halide

b.

(CH3)2CCH2

(CH3)3CCH2Br Cl

1° halide

f. Cl

Both are 2° halides.

7.47 1-chloro 1

Cl 1

3 3-chloropentane

Cl

1-chloropentane 1-chloro

Cl

Cl

1 1-chloro-2,2-dimethylpropane

2 2-chloro-2-methylbutane

Two stereoisomers

2-chloro

2 * Cl

2-chloropentane [* denotes stereogenic center]

2

3

Cl H 4 1 Clockwise "4" in back = R

2

3

1-chloro 1-chloro-3-methylbutane

2-chloro 2-methyl

2

3-methyl

Cl

3-chloro

3

4 1 Cl H Clockwise "4" in front = S

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Chapter 7–20

Two stereoisomers 3-methyl 3

2 3 *

4H

2-chloro Cl 2-chloro-3-methylbutane [* denotes stereogenic center]

2

2

3 4H

Cl

1 Counterclockwise "4" in front = R

1Cl Counterclockwise "4" in back = S

Two stereoisomers 1-chloro 2-methyl

*

3

4

3

H

4

H

Cl

Cl

2

1-chloro-2-methylbutane [* denotes stereogenic center]

Cl

1

Clockwise "4" in front = S

2 1 Clockwise "4" in back = R

7.48 Br a. (CH3)3CBr

or

CH3CH2CH2CH2Br

b.

larger surface area = stronger intermolecular forces = higher boiling point

I

or

Br

or

c.

more polar = nonpolar larger halide = more polarizable = only VDW forces higher boiling point higher boiling point

7.49 a.

CH3CH2CH2CH2 Br

b.

CH3CH2CH2CH2 Br

SH

c.

CH3CH2CH2CH2 Br

CN

d.

CH3CH2CH2CH2 Br

OCH(CH3)2

e.

CH3CH2CH2CH2 Br

C CH

f.

CH3CH2CH2CH2 Br

g.

CH3CH2CH2CH2 Br

NH3

h.

CH3CH2CH2CH2 Br

Na+ I−

i.

CH3CH2CH2CH2 Br

Na+ N3−

CH3CH2CH2CH2OH + Br

OH

H2O

CH3CH2CH2CH2SH + Br CH3CH2CH2CH2CN + Br− CH3CH2CH2CH2OCH(CH3)2 + Br− CH3CH2CH2CH2C CH + Br− CH3CH2CH2CH2OH2 + Br− CH3CH2CH2CH2NH3 + Br− CH3CH2CH2CH2I + Na+ Br− CH3CH2CH2CH2N3 + Na+ Br−

CH3CH2CH2CH2OH + HBr CH3CH2CH2CH2NH2 + HBr

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7.50 Use the steps from Answer 7.8 and then draw the proton transfer reaction, when necessary. O

a.

+ Cl

CH3

C

CH3

nucleophile I

leaving group

Cl

OH + HI

H2O

nucleophile

leaving group

d.

+ CH3CH2OH

OCH2CH3

Br

OCH3

+ Na+ –OCH3

f.

leaving group

+ NaBr

nucleophile

leaving group Cl

+ HCl

nucleophile

leaving group

e.

+ NaI

nucleophile

+

I

c.

CN

O

Na+ –CN

+

Cl

O

leaving group

b.

+ O

CH3

+

SCH3 + Cl

CH3SCH3

nucleophile

7.51 A good leaving group is a weak base. OH

c.

a. bad leaving group OH is a strong base.

b.

CH3CH2CH2CH2 Cl

Cl good leaving group weak base

e.

This has only C–C and C–H bonds. No good leaving group. OH2

d.

good leaving group H2O is a weak base.

CH3CH2NH2

bad leaving group NH2 is a strong base.

f.

CH3CH2CH2

I

I good leaving group weak base

7.52 Use the rules from Answer 7.11. a. increasing leaving group ability: −NH2 < −OH < F− most basic worst leaving group

least basic best leaving group

b. increasing leaving group ability: −NH2 < −OH < H2O most basic least basic worst leaving best leaving group group

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Chapter 7–22

c. increasing leaving group ability: Cl− < Br− < I−

d. increasing leaving group ability: NH3 < H2O < H2S most basic worst leaving group

least basic most basic worst leaving best leaving group group

least basic best leaving group

7.53 Compare the nucleophile and the leaving group in each reaction. The reaction will occur if it proceeds towards the weaker base. Remember that the stronger the acid (lower pKa), the weaker the conjugate base. I

NH2

+

+ I

a.

weaker base pKa (HI) = –10 b.

CH3CH2I

+ CH3O

stronger base pKa (NH3) = 38

CN

+ I

CH3CH2OCH3

stronger base pKa (CH3OH) = 15.5

c.

Reaction will not occur.

NH2

+ I

weaker base pKa (HI) = –10

Reaction will occur.

weaker base pKa (HI) = –10 I

+

CN

Reaction will not occur.

stronger base pKa (HCN) = 9.1

7.54 Use the directions in Answer 7.15. a. Across a row of the periodic table nucleophilicity decreases. −OH < −NH < 2

d.

CH3−

b. • In a polar protic solvent (CH3OH), nucleophilicity increases down a column of the periodic table, so: −SH is more nucleophilic than −OH. • Negatively charged species are more nucleophilic than neutral species so −OH is more nucleophilic than H2O.

Compare the nucleophilicity of N, S, and O. In a polar aprotic solvent (acetone), nucleophilicity parallels basicity. CH3SH < CH3OH < CH3NH2

e.

In a polar aprotic solvent (acetone), nucleophilicity parallels basicity. Across a row and down a column of the periodic table nucleophilicity decreases. Cl− < F− < −OH

H2O < −OH < −SH

c. • In a polar protic solvent (CH3OH), nucleophilicity increases down a column of the periodic table, so: CH3CH2S− is more nucleophilic than CH3CH2O−. • For two species with the same attacking atom, the more basic is the more nucleophilic so CH3CH2O− is more nucleophilic than CH3COO−. CH3COO− < CH3CH2O− < CH3CH2S−

f.

Nucleophilicity decreases across a row so −SH is more nucleophilic than Cl−. In a polar protic solvent (CH3OH), nucleophilicity increases down a column so Cl− is more nucleophilic than F−. F− < Cl− < −SH

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7.55 Polar protic solvents are capable of hydrogen bonding, so they must contain a H bonded to an electronegative O or N. Polar aprotic solvents are incapable of hydrogen bonding, so they do not contain any O–H or N–H bonds. a.

c.

(CH3)2CHOH

e.

no O–H or N–H bond aprotic

contains O–H bond protic d.

CH3NO2

b.

CH2Cl2

no O–H or N–H bond aprotic

N(CH3)3

no O–H or N–H bond aprotic f. HCONH2 contains an N–H bond protic

NH3 contains N–H bond protic

7.56 O CH3

C

O N

N

H

H

CH3

CH3

The amine N is more nucleophilic since the electron pair is localized on the N.

C

O N

N

H

H

CH3

CH3

C

N

N

H

H

CH3

The amide N is less nucleophilic since the electron pair is delocalized by resonance.

7.57 1° alkyl halide SN2 reaction Br

CN

CN

+

+ Br

acetone

c. Transition state:

Ea

Br

y

b. Energy diagram:

+

Energy

a. Mechanism:

CN

² H° CN

Br δ− + Br

δ−CN

Reaction coordinate

d. Rate equation: one-step reaction with both nucleophile and alkyl halide in the only step: rate = k[R–Br][–CN] e. [1] The leaving group is changed from Br− to I−: Leaving group becomes less basic → a better leaving group → faster reaction. [2] The solvent is changed from acetone to CH3CH2OH: Solvent changed to polar protic → decreases reaction rate. [3] The alkyl halide is changed from CH3(CH2)4Br to CH3CH2CH2CH(Br)CH3: Changed from 1° to 2° alkyl halide → the alkyl halide gets more crowded and the reaction rate decreases. [4] The concentration of −CN is increased by a factor of 5. Reaction rate will increase by a factor of 5. [5] The concentration of both the alkyl halide and −CN are increased by a factor of 5: Reaction rate will increase by a factor of 25 (5 x 5 = 25).

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7.58 Use the directions from Answer 7.24. Br

Br

a.

Br

2° alkyl halide intermediate reactivity

3° alkyl halide least reactive

b.

1° alkyl halide most reactive

Br

Br Br

vinyl halide least reactive

2° alkyl halide intermediate reactivity

1° alkyl halide most reactive

7.59 better leaving group a. CH3CH2Br CH3CH2Cl

+

OH

+

OH

faster reaction

stronger nucleophile b.

Cl

+

NaOH

Cl

+

NaOCOCH3

I

+

OCH3

+

OCH3

c.

I

faster reaction

CH3OH

faster reaction

DMSO

polar aprotic solvent less steric hindrance d.

Br

+

OCH2CH3

Br

+

OCH2CH3

faster reaction

7.60 All SN2 reactions proceed with backside attack of the nucleophile. When nucleophilic attack occurs at a stereogenic center, inversion of configuration occurs. CH3 a.

C H

b.

CH3 Cl

* D *

H

+

OCH3

Cl

+

OCH2CH3

C H

+ Cl

D

* OCH3

inversion of configuration OCH2CH3

* H

+ Cl

No bond to the stereogenic center is broken, since the leaving group is not bonded to the stereogenic center.

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191

Alkyl Halides and Nucleophilic Substitution 7–25

c.

*

+

Br

*

CN

CN

* *

+ Br

inversion of configuration

[* denotes a stereogenic center]

7.61 Follow the definitions from Answer 7.28. CH2CH3 a.

c.

CH3CH2CHCH2CH3

2° carbocation

e. CH2

3° carbocation

1° carbocation

d.

b.

3° carbocation

2° carbocation

7.62 For carbocations: Increasing number of R groups = Increasing stability.

a.

CH2

b. 1° carbocation least stable

2° carbocation intermediate stablity

1° carbocation least stable

3° carbocation most stable

2° carbocation intermediate stablity

3° carbocation most stable

7.63 Both A and B are resonance stabilized, but the N atom in B is more basic and therefore more willing to donate its electron pair. A

B

O CH2

O CH2

N CH2

N H

H

more basic N atom

CH2

This resonance form stabilizes the carbocation more than the equivalent resonance structure for A.

Thus, B is more stable then A.

7.64 Step [1]

CH3

a. Mechanism: SN1 only

CH3 C CH2CH3 I

A

+ H2O

CH3 CH3

Step [2] C CH2CH3 + H2O

B

+ I

CH3 CH3 C CH2CH3 OH2 C

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 7–26

b. Energy diagram:

Energy

192

Ea1

B ² H°1

Ea2

² H°2 ² H°overall = 0

C CH3

A CH3

CH3 C CH2CH3

CH3 C CH2CH3

OH2

I

Reaction coordinate c. Transition states:

CH3 δ C CH2CH3 CH3 I δ– +

CH3 δ+ C CH2CH3 CH3 OH2 δ+

d. rate equation: rate = k[(CH3)2CICH2CH3] e. [1] Leaving group changed from I− to Cl−: rate decreases since I− is a better leaving group. [2] Solvent changed from H2O (polar protic) to DMF (polar aprotic): rate decreases since polar protic solvent favors SN1. [3] Alkyl halide changed from 3° to 2°: rate decreases since 2° carbocations are less stable. [4] [H2O] increased by factor of five: no change in rate since H2O is not in rate equation. [5] [R–X] and [H2O] increased by factor of five: rate increases by a factor of five. (Only the concentration of R–X affects the rate.)

7.65 The rate of an SN1 reaction increases with increasing alkyl substitution. Br

Br

Br

1° alkyl halide least reactive

a.

3° alkyl halide most reactive

2° alkyl halide intermediate reactivity Br

Br

Br

b. aryl halide least reactive

2° alkyl halide intermediate reactivity

3° alkyl halide most reactive

7.66 The rate of an SN1 reaction increases with increasing alkyl substitution, polar protic solvents, and better leaving groups. a.

(CH3)3CCl

+ H2O

(CH3)3CI

+ H2O

b.

better leaving group faster reaction

Br

CH3OH

3° halide faster SN1 reaction

Br + CH OH 3

1° halide slower SN1 reaction

+

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193

Alkyl Halides and Nucleophilic Substitution 7–27

Cl c.

aryl halide slower reaction

+ H O 2

Cl

I

d.

I

2° halide faster SN1 reaction

+ H2O

+ CH3CH2OH

CH3CH2OH

+ CH3CH2OH

DMSO

polar protic solvent faster reaction polar aprotic solvent slower reaction

7.67 Br

a.

OCH2CH3

+

+ CH3CH2OH Br

CH3CH2O

+ HBr

OH

b.

+

H2O

OH + HBr

+

7.68 The 1o alkyl halide is also allylic, so it forms a resonance-stabilized carbocation. Increasing the stability of the carbocation by resonance, increases the rate of the SN1 reaction. CH3CH2CH2CH CHCH2

CH3CH2CH2CH CH CH2

Br

CH3CH2CH2CH CH CH2

Br

resonance-stabilized carbocation

Use each resonance structure individually to continue the mechanism: CH3CH2CH2CH CH CH2

CH3OH

CH3CH2CH2CH CHCH2 H OCH3

CH3CH2CH2CH CHCH2OCH3

HBr

Br CH3CH2CH2CH CH CH2

CH3CH2CH2CH CH CH2

CH3CH2CH2CHCH CH2

H OCH3

HBr

OCH3

Br

CH3OH

7.69 H

H Br

a.

+

acetone

1° alkyl halide SN2 only b.

CN + Br

CN

+

OCH3

DMSO

+

Br H

strong nucleophile polar aprotic solvent Both favor SN2.

2° alkyl halide SN1 and SN2 c.

Br

H OCH3

+ CH3OH

CH2CH2CH3

3° alkyl halide SN1 only

OCH3

+ HBr CH2CH2CH3

Br

reaction at a stereogenic center inversion of configuration

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Chapter 7–28

CH2CH3 d.

C H

CH2CH3 C

+ CH3COOH

I CH3

H

CH2CH3 C CH3COO H CH3

OOCCH3 CH3

HI

2° alkyl halide Weak nucleophile favors SN1. SN1 and SN2 e.

+

OCH2CH3

Br

2° alkyl halide SN1 and SN2

OCH2CH3

+ CH3CH2OH 2° alkyl halide SN1 and SN2

reaction at a stereogenic center inversion of configuration

+ Br

strong nucleophile polar aprotic solvent Both favor SN2.

Cl

f.

OCH2CH3

DMF

reaction at a stereogenic center racemization of product

OCH2CH3

+

+ HCl

two products – diastereomers Nucleophile attacks from above and below.

Weak nucleophile favors SN1.

7.70 Br [2] [1] (CH3)2NCH2CH2O H

Na+H

C OCH2CH2N(CH3)2

H (CH3)2NCH2CH2O

Na+

C

+ NaBr

+ H2 H diphenhydramine

7.71 First decide whether the reaction will proceed via an SN1 or SN2 mechanism (Answer 7.38), and then draw the mechanism.

H

Br

OCH2CH3 Br

3° alkyl halide SN1 only

CH3CH2OH

can attack from above or below

OCH2CH3 H

+

Br OCH2CH3 OCH2CH3

+

+ HBr

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Alkyl Halides and Nucleophilic Substitution 7–29

7.72 nucleophile O

O

C O H

O

C O

C

OH

+ Br O

+ H2O Br

Br

C7H10O2 leaving group

7.73 Br

H

Br

H H N CH3

N

CO32–

Br– N

CH3

N H

CO32–

CH3

N

N

Br

–

+ HCO3–

NaHCO3

+ NaBr

N

+

CH3 N

nicotine

7.74 a. Two diastereomers (C and D) are formed as products from the two enantiomers of A. H CO2CH2CH3

R Br

O

H2N

+

OC(CH3)3

A

(CH3CH2)3N

nucleophile

leaving group

C

B

CO2CH2CH3 H

S Br

CO2CH2CH3 H CH3 O S N H S OC(CH3)3

H

+

B

(CH3CH2)3N

A

b.

CO2CH2CH3 H CH3 O R N H S OC(CH3)3

H

H CH3

D

CH3CH2O H

S

Both have the S configuration quinapril

O

H CH3

N S H

O N

CO2H

Since both stereogenic centers in D have the S configuration and the corresponding stereogenic centers in quinapril are also S, D is needed to synthesize the drug.

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Chapter 7–30

7.75 CH3NH2 N

Cl

Cl

CH3NH2

+

N

Cl

+

N

H N H CH3

Cl

N CH3

+

N

Cl

N

H

+

CH3NH3

CH3

N

CH3NH3

N

H

CH3NH2

+

Cl

CH3

7.76 In the first reaction, substitution occurs at the stereogenic center. Since an achiral, planar carbocation is formed, the nucleophile can attack from either side, thus generating a racemic mixture. 3° alkyl halide CH3OH

OCH3

two steps

SN1

Br

(6R)-6-bromo-2,6-dimethylnonane

Br OCH3

achiral, planar carbocation

racemic mixture optically inactive

In the second reaction, the starting material contains a stereogenic center, but the nucleophile does not attack at that carbon. Since a bond to the stereogenic center is not broken, the configuration is retained and a chiral product is formed. 3° alkyl halide Br

two steps

CH3OH

SN1 (5R)-2-bromo-2,5-dimethylnonane

OCH3

configuration retained

Br

optically active

Reaction does not occur at the stereogenic center.

7.77 The nucleophile has replaced the leaving group. Missing reagent:

a.

O

I

O Cl

b.

C CH

The nucleophile has replaced the leaving group. Missing reagent: C CH

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197

Alkyl Halides and Nucleophilic Substitution 7–31

N3

c.

N3

SH SH

d.

The nucleophile has replaced the halide. Starting material: Cl

The nucleophile has replaced the halide. Starting material: Cl The leaving group must have the opposite orientation to the position of the nucleophile in the product.

7.78 To devise a synthesis, look for the carbon framework and the functional group in the product. The carbon framework is from the alkyl halide and the functional group is from the nucleophile. SH

a.

SH

SH

functional group

carbon framework

Na O

b.

Na

Cl

O

Cl

O

functional carbon group framework Na

c.

CH3CH2CN

carbon framework

CH3CH2Cl

CN

CH3CH2CN

functional group

O

Cl

d. functional group carbon framework

Na

2o halide O Na

or

O

O

O

O

Cl

1o halide carbon functional framework group e.

CH3CH2 OCOCH3

carbon framework

functional group

CH3CH2Cl

Na

OCOCH3

CH3CH2OCOCH3

This path is preferred. The strong nucleophile favors an SN2 reaction so an unhindered 1o alkyl halide reacts faster.

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Chapter 7–32

7.79 Work backwards to determine the alkyl chloride needed to prepare benzalkonium chloride A. CH3

a.

CH3(CH2)17Cl

CH2N(CH3)2

CH2

N

(CH2)17CH3

Cl–

CH3

B

A

b.

CH3

CH3(CH2)17N(CH3)2

CH2Cl

CH2

C

N

(CH2)17CH3

Cl–

CH3

A

7.80 OCH3

Na+ OCH3

I

C

B very crowded 3° halide

E

CH3I

O Na

preferred method The strong nucleophile favors SN2 reaction so the alkyl halide should be unhindered for a faster reaction.

OCH3

A

D

E unhindered methyl halide

7.81 H C C H

CH3(CH2)7CH2Br

NaH

H C C

CH3(CH2)7CH2C CH

A

B

NaH

CH3(CH2)7CH2C C

+ H2

C

+ H2

CH3(CH2)11CH2Br H

H

addition of H2

C C CH3(CH2)7CH2

CH3(CH2)7CH2C CCH2(CH2)11CH3

(1 equiv)

CH2(CH2)11CH3

D

muscalure

7.82 O

a.

H

[1] Na+H

[2] CH3–Br O

O

CH3

+ H2

(Chapter 9)

b.

CH3CH2CH2 C C H

[1] Na+ NH2

[2] CH3CH2–Br CH3CH2CH2 C C

CH3CH2CH2 C C CH2CH3

(Chapter 11)

c. H CH(CO2CH2CH3)2 (Chapter 23)

+ Na+ Br–

[1] Na+ –OCH2CH3

CH(CO2CH2CH3)2

[2] C6H5CH2–Br

+ Na+ Br– + NH3

C6H5CH2 CH(CO2CH2CH3)2

+ Na+ Br– + CH3CH2OH

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Alkyl Halides and Nucleophilic Substitution 7–33

7.83 quinuclidine The three alkyl groups are "tied back" in a ring, making the electron pair more available.

triethylamine

CH3CH2

N

This electron pair is more hindered by the three CH2CH3 groups. N These bulky groups around the N CH2CH3 cause steric hindrance and this CH2CH3 decreases nucleophilicity.

This electron pair on quinuclidine is much more available than the one on triethylamine. less steric hindrance more nucleophilic

7.84 O

O H

O

[1]

H

+

H2

O

O CH3 Br

CH3

[2]

minor product O

+ NaBr

O CH3 Br

CH3

[2]

major product

7.85 Br

Br

a. OH

base

(CH3)3C

b.

O (CH3)3C

OH

O

intramolecular SN2

(CH3)3C

O

O

base (CH3)3C

c. (CH3)3C

(CH3)3C

Br

Br

Br

(CH3)3C

Br

base OH

O (CH3)3C

O

3° alkyl halide harder reaction OH

d. Br (CH3)3C

intramolecular SN2

intramolecular S N2

(CH3)3C

O O

base

Br (CH3)3C

3° alkyl halide harder reaction

intramolecular S N2

(CH3)3C

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Chapter 7–34

7.86 Cl bonded to sp2 C cannot undergo SN1.

Cl O

Cl

Cl O

Cl

CH3OH

Cl bonded to sp3 C no resonance stabilization possible for the carbocation formed here

Cl

Cl Cl bonded to sp3 C Resonance-stabilized carbocation forms. best for SN1

Cl

Cl

Cl

Cl

+

(1 equiv)

O

J

Cl

O

re-draw Cl

Cl

Cl

Cl

+

CH3OH

O

O

O CH3

Cl

H

Cl

Cl

+ O

HCl

OCH3

K

7.87 a.

H OH

H OH

ee =

[α] mixture [α] pure enantiomer

x 100%

+ 5.0 x 100% = 10.% excess of R isomer +48 90% racemic mixture = 45% R and 45% S Total R isomer = 45 + 10 = 55% R isomer =

(S)-1-phenyl-1-propanol (R)-1-phenyl-1-propanol [α] = –48 [α] = +48

b. The R product is the product of inversion and it predominates. H Br

H OH

H2O

H OH

+

S retention

R inversion

c. The weak nucleophile favors an SN1 reaction, which occurs by way of an intermediate carbocation. Perhaps there is more inversion than retention because H2O attacks the intermediate carbocation while the Br– leaving group is still in the vicinity of the carbocation. The Br– would then shield one side of the carbocation and backside attack would be slightly favored.

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201

Alkyl Halides and Elimination Reactions 8–1

Chapter 8 Alkyl Halides and Elimination Reactions Chapter Review A comparison between nucleophilic substitution and β-elimination Nucleophilic substitution⎯A nucleophile attacks a carbon atom (7.6). Nu

H

H Nu +

C C

C C X

substitution product

X

good leaving group

β-Elimination⎯A base attacks a proton (8.1). B

H C C

C C

+ H B+ +

elimination product

• •

X

good leaving group

X

Similarities In both reactions RX acts as an • electrophile, reacting with an electron-rich reagent. • Both reactions require a good leaving group X:– willing to accept the electron density in the C–X bond.

Differences In substitution, a nucleophile attacks a single carbon atom. In elimination, a Brønsted–Lowry base removes a proton to form a π bond, and two carbons are involved in the reaction.

The importance of the base in E2 and E1 reactions (8.9) The strength of the base determines the mechanism of elimination. • Strong bases favor E2 reactions. • Weak bases favor E1 reactions. strong base OH CH3

E2

CH3 C CH2

CH3 C CH3 Br H2O

weak base

E1

+

H2O

+ Br

CH3 CH3 C CH2 CH3

same product different mechanism

+

H3O

+

+ Br

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Chapter 8–2

E1 and E2 mechanisms compared E2 mechanism one step (8.4B)

•

E1 mechanism two steps (8.6B)

[1] Mechanism

•

[2] Alkyl halide

•

rate: R3CX > R2CHX > RCH2X (8.4C)

•

rate: R3CX > R2CHX > RCH2X (8.6C)

[3] Rate equation [4] Stereochemistry

• • •

• • •

[5] Base

•

rate = k[RX][B:] second-order kinetics (8.4A) anti periplanar arrangement of H and X (8.8) favored by strong bases (8.4B)

rate = k[RX] first-order kinetics (8.6A) trigonal planar carbocation intermediate (8.6B) favored by weak bases (8.6C)

[6] Leaving group

•

•

[7] Solvent

•

[8] Product

•

better leaving group → faster reaction (8.4B) favored by polar aprotic solvents (8.4B) more substituted alkene favored (Zaitsev rule, 8.5)

•

• •

better leaving group → faster reaction (Table 8.4) favored by polar protic solvents (Table 8.4) more substituted alkene favored (Zaitsev rule, 8.6C)

Summary chart on the four mechanisms: SN1, SN2, E1, and E2 Alkyl halide type 1o RCH2X 2o R2CHX 3o R3CX

Conditions strong nucleophile strong bulky base strong base and nucleophile strong bulky base weak base and nucleophile weak base and nucleophile strong base

Mechanism SN2 E2 SN2 + E2 E2 SN1 + E1 SN1 + E1 E2

Zaitsev rule • •

β-Elimination affords the more stable product having the more substituted double bond. Zaitsev products predominate in E2 reactions except when a cyclohexane ring prevents trans diaxial arrangement.

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203

Alkyl Halides and Elimination Reactions 8–3

Practice Test on Chapter Review 1. Which of the following is true about an E1 reaction? 1. The reaction is faster with better leaving groups. 2. The reaction is fastest with 3° alkyl halides. 3. The reaction is faster with stronger bases. 4. Statements (1) and (2) are true. 5. Statements (1), (2), and (3) are all true. 2. Consider the SN2 and E1 reaction mechanisms. What effect on the rate of the reaction is observed when each of the following changes is made? Fill in each box of the table with one of the following phrases: increases, decreases, or remains the same. Change a. The alkyl halide is changed from (CH3)3CBr to CH3CH2CH2CH2Br. b. The solvent is changed from (CH3)2CO to CH3CH2OH. c. The nucleophile/base is changed from –OH to H2O. d. The alkyl halide is changed from CH3CH2Cl to CH3CH2I. e. The concentration of the base/nucleophile is increased by a factor of five.

SN2 Mechanism

E1 Mechanism

3. Rank the following compounds in order of increasing reactivity in an E2 elimination reaction. Rank the most reactive compound as 3, the least reactive compound as 1, and the compound of intermediate reactivity as 2.

Br

A

Br

B

Br

C

4. Draw the organic products formed in the following reactions. Cl

Br KOH

a.

K+ –OC(CH3)3

c. CH3

CH3

Br

b.

NaNH2 Br

(excess)

d.

Br

NaOCH3

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Chapter 8–4

5. a. Fill in the appropriate alkyl halide needed to synthesize the following compound as a single product using the given reagents. K+ –OC(CH3)3

C

CH3 CH3

b. What starting material is needed for the following reaction? The starting material must yield product cleanly, in one step without any other organic side products. K+ –OC(CH3)3

D

CH3CH2CH

CHCH3

(cis and trans mixture)

6. Draw all products formed in the following reaction. CH3OH

Cl

CH3CH 2

CH 3

Answers to Practice Test 1. 4

5.

4.

6. Cl

2. SN2 a. increases b. decreases c. decreases d. increases e. increases 3. A–2 B–1 C–3

E1 decreases increases same increases same

a.

b.

CH3CH2

a.

CH 3 C

CH3

C

H CH3 CH3CH2

CH3

C CH3CH2

b. Br

c. CH3

d.

CH3 OCH3

CH 3

OCH3

(or Cl or I) D

OCH3 CH3

CH3CH2

CH3

CH3CH2

CH2

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205

Alkyl Halides and Elimination Reactions 8–5

Answers to Problems 8.1 • The carbon bonded to the leaving group is the α carbon. Any carbon bonded to it is a β carbon. • To draw the products of an elimination reaction: Remove the leaving group from the α carbon and a H from the β carbon and form a π bond. a.

b.

β α CH3CH2CH2CH2CH 2 Cl

β1

α

β2

K+ −OC(CH3)3

β1

CH3CH2CH2CH CH 2

(CH3CH2)2C

CH2

CH3CH=C(CH3)CH2CH3

Cl β1

c.

K + −OC(CH3)3

α

β2 Br

K+ −OC(CH3)3

β1

8.2 Alkenes are classified by the number of carbon atoms bonded to the double bond. A monosubstituted alkene has one carbon atom bonded to the double bond, a disubstituted alkene has two carbon atoms bonded to the double bond, etc. 4 C's bonded to C=C tetrasubstituted

2 C's bonded to each C=C disubstituted

OH

a.

b. vitamin D3

3 C's bonded to each C=C trisubstituted vitamin A

H CH2

3 C's bonded to each C=C trisubstituted 2 C's bonded to the C=C disubstituted

HO

8.3 To have stereoisomers at a C=C, the two groups on each end of the double bond must be different from each other. two different groups (CH3CH2 and H) a. two CH3 groups no stereoisomers possible

two different groups (H and CH3)

b. CH3CH2CH CHCH3 stereoisomers possible

two different groups two different groups (cyclohexyl and H) (cyclohexyl and H) c.

CH CH

stereoisomers possible

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Chapter 8–6

8.4 a.

2 CH3's on one end

2 H's on one end

Only this C=C exhibits stereoisomerism.

b. A diastereomer has a different 3-D arrangement of groups but the carbon skeleton and the double bonds must stay in the original positions.

diastereomer

different arrangement of groups around this double bond

8.5 Two definitions: • Constitutional isomers differ in the connectivity of the atoms. • Stereoisomers differ only in the 3-D arrangement of the atoms in space. a.

and

c.

different connectivity of atoms constitutional isomers and

b.

and cis trans different arrangement of atoms in space stereoisomers

d.

trans

and

trans identical

different connectivity of atoms constitutional isomers

8.6 Two rules to predict the relative stability of alkenes: [1] Trans alkenes are generally more stable than cis alkenes. [2] The stability of an alkene increases as the number of R groups on the C=C increases. a.

or monosubstituted

b.

CH2CH3

CH3CH2 C C

cis

CH3CH2

H

H

CH2CH3

trans more stable

CH3

or

c.

C C

or H

H

CH3

disubstituted more stable

trisubstituted more stable

disubstituted

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207

Alkyl Halides and Elimination Reactions 8–7

8.7

A

B

Alkene A is more stable than alkene B because the double bond in A is in a six-membered ring. The double bond in B is in a four-membered ring, which has considerable angle strain due to the small ring size. 8.8 In an E2 mechanism, four bonds are involved in the single step. Use curved arrows to show these simultaneous actions: [1] The base attacks a hydrogen on a β carbon. [2] A π bond forms. [3] The leaving group comes off. CH3CH2 H CH3CH2 C Br

OCH2CH3

CHCH3

(CH3CH2)2C=CHCH3 +

transition state: δ− HOCH2CH3

+

CH3CH2 C

new π bond

β carbon

CH3CH2 H

Br

OCH2CH3

CHCH3

δ− Br

8.9 In both cases, the rate of elimination decreases. stronger base faster reaction a. CH CH Br 3 2

+

OC(CH3)3

CH3CH2 Br

+

OH

better leaving group faster reaction b. CH3CH2 Br

+

OC(CH3)3

CH3CH2 Cl

+

OC(CH3)3

8.10 As the number of R groups on the carbon with the leaving group increases, the rate of an E2 reaction increases. a. (CH3)2CHCH2CH2CH2Br

(CH3)2CHCH2CH(Br)CH3

1° alkyl halide least reactive

3° alkyl halide most reactive CH3

Cl

Cl

b. Cl

1° alkyl halide least reactive

(CH3)2C(Br)CH2CH2CH3

2° alkyl halide intermediate reactivity

CH3

2° alkyl halide intermediate reactivity

3° alkyl halide most reactive

8.11 Use the following characteristics of an E2 reaction to answer the questions: [1] E2 reactions are second order and one step. [2] More substituted halides react faster. [3] Reactions with strong bases or better leaving groups are faster. [4] Reactions with polar aprotic solvents are faster.

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Chapter 8–8

Rate equation: rate = k[RX][Base] a. tripling the concentration of the alkyl halide = rate triples b. halving the concentration of the base = rate halved c. changing the solvent from CH3OH to DMSO = rate increases (Polar aprotic solvent is better for E2.) d. changing the leaving group from I– to Br– = rate decreases (I– is a better leaving group.) e. changing the base from –OH to H2O = rate decreases (weaker base) f. changing the alkyl halide from CH3CH2Br to (CH3)2CHBr = rate increases (More substituted halide reacts faster.) 8.12 The Zaitsev rule states: In a β-elimination reaction, the major product has the more substituted double bond. CH3 H

a.

α

CH3 C

C

H

Br

CH2CH3

(CH3)2C

loss of H and Br

CH3

b.

α

CHCH2CH3

+

trisubstituted major product

Br

(CH3)2CHCH

CHCH3

disubstituted minor product

CH3

CH3

CH2

CH3

CH3

CH3

loss of H and Br CH3

trisubstituted minor product

tetrasubstituted major product

disubstituted minor product

Cl

CH3

monosubstituted minor product

CH3

disubstituted major product

CH3

α Cl

d.

CH3CH2CH2CH2CH=CHCH3

loss of H and Cl

α

c.

trisubstituted CH3 ONLY product

loss of H and Cl

CH3

CH3

8.13 An E1 mechanism has two steps: [1] The leaving group comes off, creating a carbocation. [2] A base pulls off a proton from a β carbon, and a π bond forms. CH3 CH3 C

CH2CH3

[1]

+ CH3OH

CH3 CH3 C CH CH3 CH3OH + Cl

[2]

+

(CH3)2C=CHCH3 + CH3OH2 + Cl

H

Cl

transition state [1]:

transition state [2]:

CH3 CH3 +C

δ

CH2CH3

Cl δ−

CH3 CH3 C

δ+

CH CH3 H

OCH3

δ+

H

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8.14 The Zaitsev rule states: In a β-elimination reaction, the major product has the more substituted double bond. β2

CH3

CH3

β1

α

C CHCH3 CH3CH2

CH3CH2 C CH2CH3 + H2O

a.

Cl

CH3CH2

β1

trisubstituted major product CH2CH2CH3

β1

CH3

α

b. β3

I

β2

CH2CH2CH3 +

CH3

CH3OH

+

C CH2 CH3CH2

disubstituted

CH2CH2CH3 +

tetrasubstituted major product

CH2

CH2CH2CH3 CH3

+

trisubstituted

disubstituted

8.15 Use the following characteristics of an E1 reaction to answer the questions: [1] E1 reactions are first order and two steps. [2] More substituted halides react faster. [3] Weaker bases are preferred. [4] Reactions with better leaving groups are faster. [5] Reactions in polar protic solvents are faster. Rate equation: rate = k[RX]. The base doesn’t affect rate. a. doubling the concentration of the alkyl halide = rate doubles b. doubling the concentration of the base = no change (Base is not in the rate equation.) c. changing the alkyl halide from (CH3)3CBr to CH3CH2CH2Br = rate decreases (More substituted halides react faster.) d. changing the leaving group from Cl– to Br– = rate increases (better leaving group) e. changing the solvent from DMSO to CH3OH = rate increases (Polar protic solvent favors E1.) 8.16 Both SN1 and E1 reactions occur by forming a carbocation. To draw the products: [1] For the SN1 reaction, substitute the nucleophile for the leaving group. [2] For the E1 reaction, remove a proton from a β carbon and create a new π bond. CH3

CH3

Br

a.

+ H2O leaving group nucleophile and base

SN1 product

CH3 Cl

nucleophile and base

CH3

E1 products CH3

b. CH3 C CH2CH2CH3 + CH3CH2OH

leaving group

CH2

OH

CH3 C CH2CH2CH3 OCH2CH3

SN1 product

CH3

CH3 C CH2 CH3CH2CH2

C CHCH2CH3 CH3

E1 products

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Chapter 8–10

8.17 E2 reactions occur with anti periplanar geometry. The anti periplanar arrangement uses a staggered conformation and has the H and X on opposite sides of the C–C bond. H

HO

C CH3 CH3

H C

CH3

H

Br

C

base

H C

CH3

H

H and Br are on opposite sides = anti periplanar

8.18 The E2 elimination reactions will occur in the anti periplanar orientation as drawn. To draw the product of elimination, maintain the orientation of the remaining groups around the C=C. CH3CH2O H

CH3

a.

C6H5

C

CH3

C6H5 C H Br

C6H5 C C

C6H5

H

The two benzene rings remain on opposite sides of the newly formed C=C. This makes them trans.

The two benzene rings are anti in this conformation (one wedge, one dash). diastereomers CH3CH2O H

b.

C6H5

C6H5 C6H5 C C H CH3 Br

C6H5 C C

CH3

H

The two benzene rings are gauche in this conformation (both drawn on dashes, behind the plane).

The two benzene rings remain on the same side of the newly formed C=C. This makes them cis.

8.19 Note: The Zaitsev products predominate in E2 elimination except when substituents on a cyclohexane ring prevent a trans diaxial arrangement of H and X. axial H's H

CH(CH3)2

two conformations

a. CH3

Cl

CH3

H CH3

H H H

CH(CH3)2

Cl

A Use this conformation. It has Cl axial and two axial H's.

H

H

H H

B

H Cl CH(CH3)2

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Alkyl Halides and Elimination Reactions 8–11

CH3

H H H

H

β1

H H

H

CH(CH3)2

−OH

CH3

Cl

β2

CH(CH3)2 H H

CH3

H H

H

[loss of H(β2) + Cl]

A

re-draw

CH(CH3)2

CH(CH3)2

CH3

CH3

trisubstituted major product

disubstituted H CH(CH3)2 CH3

CH3

H

two conformations

b.

CH3

Cl

Cl H H

CH(CH3)2

H

H

β1

β2

H

H

B

CH(CH3)2

CH3 −OH

H H

H H

H

Use this conformation. It has Cl axial and one axial H.

Cl

H

Cl

H

A

CH3

H

[loss of H(β1) + Cl]

re-draw

two different axial H's

CH(CH3)2

CH(CH3)2

B only one axial H on a β carbon

CH(CH3)2

H

H H

H

CH(CH3)2

[loss of H(β1) + Cl]

= CH3

disubstituted only product

8.20 Draw the chair conformations of cis-1-chloro-2-methylcyclohexane and its trans isomer. For E2 elimination reactions to occur, there must be a H and X trans diaxial to each other. Two conformations of the cis isomer: Cl CH3

H

H

H H

A reacting conformation (axial Cl)

This reacting conformation has only one group axial, making it more stable and present in a higher concentration than B. This makes a faster elimination reaction with the cis isomer.

H

CH3 H

H

H

Two conformations of the trans isomer: Cl

H H

H Cl

H CH3 H

Cl H

H H

CH3

B reacting conformation (axial Cl)

This conformation is less stable than A, since both CH3 and Cl are axial. This slows the rate of elimination from the trans isomer.

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Chapter 8–12

8.21 E2 reactions are favored by strong negatively charged bases and occur with 1°, 2°, and 3° halides, with 3° being the most reactive. E1 reactions are favored by weaker neutral bases and do not occur with 1° halides since they would have to form highly unstable carbocations. CH3

a.

C(CH3)3

CH3 C CH3

+

Cl

+

strong negatively charged base E2

I

b.

Cl

c.

OCH3

+

CH3OH

weak neutral base E1

H2O

d. CH3CH2Br

weak neutral base E1

+

OC(CH3)3

strong negatively charged base E2

8.22 Draw the alkynes that result from removal of two equivalents of HX. Br

Cl Cl

a.

NH2

C C CH2CH3

C C CH2CH3

c. CH3 C CH2CH3

NH2 CH3C CCH3

Br

H H

+ HC CCH2CH3 Br

KOC(CH3)3

b. CH3CH2CH2CHCl2

CH3CH2C

DMSO

NH2

d.

CH

C

C

Br

8.23 1° halide SN2 or E2 H

b.

K+ −OC(CH3)3

Cl

a.

CH3 C CH2CH3 Cl

2° halide any mechanism

strong bulky base E2

OH

strong base SN2 and E2

CH2CH3 I

c.

+

weak base SN1 and E1

CH3 CH CHCH3

+

SN2 product

disubstituted major E2 product

OCH2CH3

SN1 product

+

+ E1 product

CH3CH2O Cl

3° halide no SN2

CH3CH2OH

E2

major E2 product

monosubstituted minor E2 product CH2CH3

CHCH3

strong base

d.

CH2 CH CH2CH3

OH

CH2CH3 CH3CH2OH

3° halide no SN2

H CH3 C CH2CH3

minor E2 product

E1 product

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Alkyl Halides and Elimination Reactions 8–13

8.24 3° halide no SN2

weak base SN1 and E1

CH3

CH3

Br

CH3OH

CH3

overall reaction

SN1

HBr

CH3

CH3 CH3OH

CH3

or E1

+

+

CH3

The steps: CH3

CH3

OCH3

O CH3 H CH3 Br

+ Br

CH3 H CH3

Br

8.25 More substituted alkenes are more stable. Trans alkenes are generally more stable than cis alkenes. Order of stability: < B least stable

< C

A most stable

8.26 The trans isomer reacts faster. During elimination, Br must be axial to give trans diaxial elimination. In the trans isomer, the more stable conformation has the bulky tert-butyl group in the more roomy equatorial position. In the cis isomer, elimination can occur only when both the tert-butyl and Br groups are axial, a conformation that is not energetically favorable. Br Br

=

slow

This conformation must react, cis but it contains two axial groups. D cis-1-bromo-3-tert-butylcyclohexane Br

= Br

trans preferred conformation E trans-1-bromo-3-tert-butylcyclohexane

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Chapter 8–14

8.27 Translate each model to a structure and arrange H and Br to be anti periplanar.

a.

CH3

H

C6H5 C

C

CH3

Br H

CH3CH2

Br

CH3CH2

rotate

C

C6H5

C

CH3CH2

E2

H

H

C6H5

C

C

CH3

H

H and Br 180° away from each other Br

CH3CH2

b.

CH3 H

C

H

H

H

rotate

C

C

CH3CH2 CH3

C6H5

C

C6H5

CH3CH2

E2

H

C

Br

C

CH3

C6H5

H and Br 180° away from each other

8.28 a. CH3CH2CH2CH2CH2CH2Br

CH3CH2CH2CH2CH CH2

Br

b.

CH3CH2CH2CH2CH CHCH2CH3 +

CH3CH2CH2CH2CH2CH CHCH3

CH3

c. CH3CH2CHCHCH3

CH3CH2CH C(CH3)2

+

CH3CH CHCH(CH3)2

Cl CH2

I

d.

8.29 To give only one product in an elimination reaction, the starting alkyl halide must have only one type of β carbon with H’s. a.

CH2 CHCH2CH2CH3

β

β

α

CH2

CH2CH2CH2CH3

Cl

β b. (CH3)2CHCH CH2 CH2

c.

(CH3)2CHCH2

CH3

CH3

α

β

Two β carbons are identical.

CH2Cl

α

β

α Cl

d.

CH2Cl

e.

C(CH3)3

β Cl α β

C(CH3)3

Two β carbons are identical.

8.30 To have stereoisomers, the two groups on each end of the double bond must be different from each other. 2 H's on one end a. 2 CH3's on one end two different groups at each end can have stereoisomers

OH

b.

2 H's on one end

two different groups at each end can have stereoisomers

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Alkyl Halides and Elimination Reactions 8–15

8.31 Use the definitions in Answer 8.5. CH3

a.

CH2

and

and

c. trans trans

trans trans

different connectivity constitutional isomers

identical H

CH3CH2

CH3

CH3CH2

and

C C

b.

CH3

CH2CH3

CH2CH3

C C CH3

CH3

C

CH3 CH3

d.

C

and

CH3

stereoisomers

H

CH3

stereoisomers

8.32 Double bond can be cis or trans. OH

a. five sp3 stereogenic centers (four circled, one labeled) b. Two double bonds can both be cis or trans. c. 27 = 128 stereoisomers possible

CH2CH CH(CH2)3COOH

PGF2α HO

CH CHCH(OH)(CH2)4CH3

sp3 stereogenic center Double bond can be cis or trans.

8.33 Use the rules from Answer 8.6 to rank the alkenes. CH2 CHCH(CH3)2

CH2 C(CH3)CH2CH3

monosubstituted least stable

disubstituted intermediate stability

(CH3)2C

CHCH3

trisubstituted most stable

8.34 A larger negative value for H° means the reaction is more exothermic. Since both 1-butene and cis-2-butene form the same product (butane), these data show that 1-butene was higher in energy to begin with, since more energy is released in the hydrogenation reaction. 1-butene

+ H2

ΔHo = −127 kJ/mol

1-butene CH3

CH3

+ H2

C C H

H

cis-2-butene

cis-2-butene

CH3CH2CH2CH3

CH3CH2CH2CH3

ΔHo = −120 kJ/mol

Energy

CH2 CHCH2CH3

larger ² H° for 1-butene higher in energy

butane smaller ² H° for cis-2-butene lower in energy, more stable

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Chapter 8–16

8.35 Cl

a. β1

α

(CH3)3CO

CH3CH=CHCH2CH2CH(CH3)2

β2

(loss of β2 H) major product disubstituted

(loss of β1 H) monosubstituted

DBU

b. O

only product O

O

β3

I

CH3

β1

c.

β α

O

Cl

α

OH CH3CH2C(CH3)=C(CH3)CH2CH2CH3

β2

CH3CH2CH(CH3)C(CH3) CHCH2CH3

(loss of β1 H) major product tetrasubstituted

(loss of β2 H) trisubstituted

CH2

(loss of β3 H) disubstituted β d.

Cl

α

OC(CH3)3

only product

β2 e.

β1

β2

f.

α

OH

α I

Br

CH3CH2CH2CH2CH=CHCH3

(loss of β1 H) major product disubstituted

(loss of β2 H) monosubstituted

OH

β1 (loss of β2 H) major product trisubstituted

(loss of β1 H) disubstituted

8.36 To give only one alkene as the product of elimination, the alkyl halide must have either: • only one β carbon with a hydrogen atom • all identical β carbons, so the resulting elimination products are identical CH3 H

a. CH3 C H

C H Cl

CH3 C CH3

H

CH3 H CH3 C

C H

Cl

C H H Cl

b.

CH=CH2 Cl

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Alkyl Halides and Elimination Reactions 8–17

Cl Cl

c.

8.37 Draw the products of the E2 reaction and compare the number of C’s bonded to the C=C.

β1 A

α

β2

β2 β1 α major product trisubstituted

Br Br

β1

α

(CH3)2CHCH=CHCH3

α disubstituted

α β2

B

α

(CH3)2CHCH=CHCH3

β1 major product disubstituted

β2

monosubstituted

A yields a trisubstituted alkene as the major product and a disubstituted alkene as minor product. B yields a disubstituted alkene as the major product and a monosubstituted alkene as minor product. Since the major and minor products formed from A have more alkyl groups on the C=C (making them more stable) than those formed from B, A reacts faster in an elimination reaction. 8.38 a. Mechanism: by-products Br H

OC(CH3)3

+ HOC(CH3)3 + Br−

(CH3)3COH

b. Rate = k[R–Br][–OC(CH3)3] [1] Solvent changed to DMF (polar aprotic) = rate increases [2] [–OC(CH3)3] decreased = rate decreases [3] Base changed to –OH = rate decreases (weaker base) [4] Halide changed to 2° = rate increases (More substituted RX reacts faster.) [5] Leaving group changed to I– = rate increases (better leaving group) 8.39 K+ –OC(CH3)3

+

Cl

A 1-chloro-1-methylcyclopropane

B

The dehydrohalogenation of an alkyl halide usually forms the more stable alkene. In this case, A is more stable than B even though A contains a disubstituted C=C whereas B contains a trisubstituted C=C. The double bond in B is part of a three-membered ring, and is less stable than A because of severe angle strain around both C’s of the double bond.

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Chapter 8–18

8.40 H

a.

KOH

CH3CH2CH2 C

NaOCH2CH3

b. Cl

Br

trans isomer more stable major product

trans isomer more stable major product

8.41 CH3

a.

CH3

CH3

Cl CH3

CH3

tetrasubstituted major product

CH2

CH3

CH3

trisubstituted

disubstituted

Br

b. trisubstituted This isomer is more stable— large groups farther away. major product

disubstituted trisubstituted

Cl

c.

disubstituted

trisubstituted major product

8.42 Use the rules from Answer 8.21. a.

OCH3 Br

2° halide

CH3CH CHCH3

strong base E2 CH3OH

b. Br

2° halide

weak base E1

CH2CH2CH3

H2O

Cl CH3

weak base 3° halide E1 OH Cl

e. 2° halide

strong base E2 OH

f. 2° halide Cl

CH3CH2CH CH2

(cis and trans)

strong base E2

1° halide d.

CH3CH CHCH3

OC(CH3)3

I

c.

CH3CH2CH CH2

(cis and trans)

strong base E2

CHCH2CH3 CH3

CH2CH2CH3 CH3

CH2CH2CH3 CH3

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Alkyl Halides and Elimination Reactions 8–19

8.43 The order of reactivity is the same for both E2 and E1: 1° < 2° < 3°. CH3 Br

CH3

Cl

CH3

Cl

3° halide

2° halide

3° halide + better leaving group

Increasing reactivity in E1 and E2

8.44 CH3 Cl

a.

Cl OH

3° halide—faster reaction CH3

Cl

H2O

b. CH3 Cl

c. (CH3)3CCl

OH H2O

strong base—E2

H2O (CH3)3CCl

OH DMSO

OH

3° halide—faster reaction polar aprotic solvent faster reaction

8.45 In a ten-membered ring, the cis isomer is more stable and, therefore, the preferred elimination product. The trans isomer is less stable because strain is introduced when two ends of the double bond are connected in a trans arrangement in this medium-sized ring.

Br

cis-cyclodecene

bromocyclodecane

8.46 With the strong base –OCH2CH3, the mechanism is E2, whereas with dilute base, the mechanism is E1. E2 elimination proceeds with anti periplanar arrangement of H and X. In the E1 mechanism there is no requirement for elimination to proceed with anti periplanar geometry. In this case the major product is always the most stable, more substituted alkene. Thus, C is the major product under E1 conditions. (In Chapter 9, we will learn that additional elimination products may form in the E1 reaction due to carbocation rearrangement.) Cl OCH2CH3

A

strong base E2

B

Since this is an E2 mechanism, dehydrohalogenation needs an anti periplanar H to form the double bond. There is only one H trans to Cl, so the disubstituted alkene B must form.

Cl CH3OH

A

weak base E1

B

C

disubstituted alkene

trisubstituted alkene more stable

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Chapter 8–20

8.47 Br + CH3CH=CHCH3

β1

β2

2-butene

1-butene from loss of β1 H

from loss of β2 H

Na+ –OCH2CH3

19%

81%

K+ –OC(CH3)3

33%

67%

The H’s on the CH2 group of the β2 carbon are more sterically hindered than the H’s on the CH3 group of the β1 carbon. Since K+ –OC(CH3)3 is a much bulkier base than Na+ –OCH2CH3, it is easier to remove the more accessible H on β1, giving it a higher percentage of 1-butene. 8.48 H and Br must be anti during the E2 elimination. Rotate if necessary to make them anti; then eliminate. CH3

a.

C6H5

C6H5

Br

H

CH2CH3

CH3

CH3

E2

CH2CH3

CH3 Br

Br

H

CH3

C6H5

CH3

b.

CH3

rotate

CH2CH3

C6H5

E2

CH3

CH2CH3

CH3

CH3

CH2CH3

C6H5

CH2CH3

H

C6H5

c.

C6H5

C6H5

CH3

H

H

CH3

Br

rotate

CH2CH3

CH3

CH2CH3

E2

Br

CH3

CH3

CH3

8.49 Cl

a.

two chair conformations H

CH3

Cl

H

H H

H H CH3

H

B

H

H CH3

Choose this conformation. axial Cl

H

H

Cl

B

(CH3)2CH

CH3 (CH3)2CH A H

CH(CH3)2

(CH3)2CH

H axial Cl

H H

H

one axial H

(CH3)2CH

=

CH3

H

only product CH3 CH(CH3)2

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Alkyl Halides and Elimination Reactions 8–21

Cl

b.

H

two chair conformations

H (CH3)2CH

CH(CH3)2 H

H

H

H

c.

β1

re-draw

CH3

CH3

CH(CH3)2

CH(CH3)2

D

D

= D

D

H

(loss of β2 H)

(loss of β2 H)

D

Cl H Cl

D H H

(CH3)2CH CH3

(loss of β1 H) major product trisubstituted re-draw

D

H

H

H (CH3)2CH CH3

H β1 H two axial H's B

B

Choose this conformation. axial Cl

H

H

A

H

Cl

Cl (CH3)2CH CH3

β2

Cl

axial

H H Cl

H

CH3

(CH3)2CH CH3

CH3

H

β2

D H

H

=

D

H H

D

enantiomers

−OH

D

This conformation reacts. axial Cl

D

D = H

D

H

(loss of β1 H) (loss of β2 D)

D Cl

d. =

D

H

Cl

H

H

β1

H H D

D

β2

This conformation reacts. axial Cl

H

D Cl H

D

=

D

H

D

H

enantiomers

−OH

D H

H H

D

(loss of β1 D)

=

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Chapter 8–22

8.50 enantiomers

a. H CH3

H

CH3 C C CH2CH3

CH3CH2 CH3

C3

C2

H

H H CH3 CH3

C C

Cl H

enantiomers

C C CH CH 2 3 CH3 Cl

Cl

–HCl

H and Cl are arranged anti in each stereoisomer, for anti periplanar elimination.

CH3CH2

Cl

–HCl

CH2CH3

H

CH3

CH3

–HCl CH3

CH3 C C

C C

C

CH3

Cl

D

C

H

CH3

CH3

H CH3

CH3 C C

CH3 CH2CH3 CH3CH2

–HCl

H C

C C

B

A

2-chloro-3-methylpentane

H

H

H CH3

H

C C CH2CH3 CH3CH2

identical

CH3

identical

b. Two different alkenes are formed as products. c. The products are diastereomers: Two enantiomers (A and B) give identical products. A and B are diastereomers of C and D. Each pair of enantiomers gives a single alkene. Thus, diastereomers give diastereomeric products. 8.51 CH2CHCl2

a.

C CH

NaNH2 (2 equiv)

CH 3

b.

CH3CH 2 C

NaNH2

CHCH 2Br

(2 equiv)

CH 3 Br

Cl

c.

CH3 C CH2CH3 Cl

NaNH2

NaNH2

C C

CH3 C C CH3

C C

(2 equiv)

Cl Cl

C CH

CH3

HC C CH2CH3

(excess)

H H

d.

CH3 CH 3CH2 C

8.52 H H

Br

a. CH3C CCH3

CH3 C CH2CH3

or

CH3 C C CH3 Br Br

Br CH3

b. CH3 C C CH CH3

CH3 Br CH3 C

C CH3

CH3

CH3 Br

or

CH3 C

CH3 Br

CH CH2Br

CH3

CH3 Br H

c.

C C

or

CH3 C

C C Br H

Br Br

or

C C H H

CH2CHBr2

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Alkyl Halides and Elimination Reactions 8–23

8.53 H H CH3 C C CH3

CH3 C C CH3

CH3 CH=C=CH2

CH2=CH CH=CH2

C

Br Br

sp sp A

2,3-dibromobutane

sp B

8.54 Use the “Summary chart on the four mechanisms: SN1, SN2, E1, or E2” on p. 8–2 to answer the questions. a. b. c. d. e. f. g. h. i. j.

Both SN1 and E1 involve carbocation intermediates. Both SN1 and E1 have two steps. SN1, SN2, E1, and E2 all have increased reaction rates with better leaving groups. Both SN2 and E2 have increased rates when changing from CH3OH (a protic solvent) to (CH3)2SO (an aprotic solvent). In SN1 and E1 reactions, the rate depends on only the alkyl halide concentration. Both SN2 and E2 are concerted reactions. CH3CH2Br and NaOH react by an SN2 mechanism. Racemization occurs in SN1 reactions. In SN1, E1, and E2 mechanisms, 3° alkyl halides react faster than 1° or 2° halides. E2 and SN2 reactions follow second-order rate equations.

8.55 Br

a.

1° halide SN2 or E2

OC(CH3)3

OCH2CH3

I

b.

E2

sterically hindered base

OCH2CH3

SN2

strong nucleophile

1° halide SN2 or E2 Cl

NH2

c. CH3 C CH3

HC C CH3

(2 equiv)

Cl

strong base

dihalide Br

d. 1° halide SN2 or E2

DBU

sterically hindered base

CH2CH3

e. Br

2° halide SN1, SN2, E1, E2

E2

OC(CH3)3

sterically hindered base

CH2CH3

CH2CH3

E2 major product

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Chapter 8–24

OCH2CH3

Br

f.

CH3CH2OH

CH2CH3

weak base

3° halide no SN2 g.

(CH3)2CH

SN1 product

CHCH2Br

CHCH3

CH2CH3

CH2CH3

2 NaNH2

(CH3)2CH

E1 products

C CH

dihalide Br Cl Cl

h.

KOC(CH3)3 (2 equiv)

CH3 CH3 C C CH CH3

DMSO

dihalide

OCH2CH3

I

CH3CH2OH

i. 2° halide SN1, SN2, E1, E2

weak base

CH3CH CHCH3

SN1 product

E1 product

H2O

j.

Cl

CH3CH2C(CH3) CHCH3 OH

weak base

3° halide no SN2

(cis and trans) E1 product

(cis and trans) E1 product

SN1 product

E1 product

8.56 [1] NaOCOCH3 is a good nucleophile and weak base, and substitution is favored. [3] KOC(CH3)3 is a strong, bulky base that reacts by E2 elimination when there is a β hydrogen in the alkyl halide. a.

CH3Cl

[1] NaOCOCH3

[2] NaOCH3

c.

CH3OCOCH3

[1] NaOCOCH3 Cl

OCOCH3 [2] NaOCH3

CH3OCH3

OCH3

SN2 [3] KOC(CH3)3

Cl

b.

[3] KOC(CH3)3

CH3OC(CH3)3

[1] NaOCOCH3

[2] NaOCH3

[3] KOC(CH3)3

OCOCH3

d. OCH3

Cl

[1] NaOCOCH3

[2] NaOCH3

[3] KOC(CH3)3

OCOCH3

+ E2

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Alkyl Halides and Elimination Reactions 8–25

8.57 a. two enantiomers:

(CH3)3C

(CH3)3C

A

B

b. The bulky tert-butyl group anchors the cyclohexane ring and occupies the more roomy equatorial position. The cis isomer has the Br atom axial, while the trans isomer has the Br atom equatorial. For dehydrohalogenation to occur on a halo cyclohexane, the halogen must be axial to afford trans diaxial elimination of H and X. The cis isomer readily reacts since the Br atom is axial. The only way for the trans isomer to react is for the six-membered ring to flip into a highly unstable conformation having both (CH3)3C and Br axial. Thus, the trans isomer reacts much more slowly. Br

trans diaxial

Br

(CH3)3C

(CH3)3C H

H

cis-1-bromo-4-tert-butylcyclohexane

trans-1-bromo-4-tert-butylcyclohexane OCH3

c. two products: OCH3

(CH3)3C

C=

D = (CH3)3C

d. cis-1-Bromo-4-tert-butylcyclohexane reacts faster. With the strong nucleophile –OCH3, backside attack occurs by an SN2 reaction, and with the cis isomer, the nucleophile can approach from the equatorial direction, avoiding 1,3-diaxial interactions. 1,3-diaxial interactions Br

OCH3

H H

OCH3

(CH3)3C

equatorial approach preferred cis-1-bromo-4-tert-butylcyclohexane

(CH3)3C

Br

axial approach

trans-1-bromo-4-tert-butylcyclohexane

e. The bulky base –OC(CH3)3 favors elimination by an E2 mechanism, affording a mixture of two enantiomers, A and B. The strong nucleophile –OCH3 favors nucleophilic substitution by an SN2 mechanism. Inversion of configuration results from backside attack of the nucleophile. 8.58 Cl

a.

H

H

strong base 2° halide SN2 and E2 SN1, SN2, E1, E2 Cl

H

b. 2° halide SN1, SN2, E1, E2

OH

OH

H2O

weak base SN1 and E1

SN2 product major E2 product inversion at stereogenic center HO

H

H

minor E2 product

OH

SN1 products

major E1 product

minor E1 product

minor E1 product

minor E2 product

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Chapter 8–26

CH3 CH3

c.

CH3 CH3

CH3OH

Cl C6H5

OCH3 C6H5

weak base SN1 and E1

3° halide no SN2

CH3 CH3

CH3 CH3

C6H5 OCH3

C6H5

SN1 products

E1 product

NaOH

d.

strong base SN2 and E2

Cl

OH

2° halide SN1, SN2, E1, E2 CH3 Br

e.

Br

achiral SN1 product

OH

KOH

f. D

2° halide SN1, SN2, E1, E2

strong base SN2 and E2

minor E2 product

CH3 OOCCH3

CH3COO

weak base good nucleophile SN1

3° halide no SN2

major E2 product

SN2 product

(trans diaxial elimination of D, Br) D

SN2 product inversion at stereogenic center

E2 product

8.59 CH3OH

a.

Cl

weak base SN1 and E1

OCH3 OCH3

S N1

SN1

E1

E1

E1

KOH

b.

Cl

strong base E2

8.60 CH3

a.

Br

3° halide

CH3 OC(CH3)3

strong bulky base E2

OC(CH3)3 CH3

CH2

major product more substituted alkene

No substitution occurs with a strong bulky base and a 3o RX. The C with the leaving group is too crowded for an SN2 substitution to occur. Elimination occurs instead by an E2 mechanism.

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Alkyl Halides and Elimination Reactions 8–27

Br

b. 1° halide

OCH3

strong nucleophile SN2

OCH3

All elimination reactions are slow with 1° halides. The strong nucleophile reacts by an SN2 mechanism instead. CH3 Cl

c.

minor product only

OH

CH3

strong base E2

3° halide

More substituted alkene is favored.

I

d. Cl

minor product only

good nucleophile, weak base SN2 favored

2° halide

I

major product

The 2o halide can react by an E2 or SN2 reaction with a negatively charged nucleophile or base. Since I– is a weak base, substitution by an SN2 mechanism is favored.

8.61 3° halide, weak base: SN1 and E1 CH3CH2OH

a.

overall reaction

Cl

+

+

+ HCl

OCH2CH3

The steps:

+ HCl

SN1 OCH2CH3

CH3CH2OH

or E1 H Cl

Cl

H

+ HCl

or

E1

+ HCl H

Cl

Any base (such as CH3CH2OH or Cl–) can be used to remove a proton to form an alkene. If Cl– is used, HCl is formed as a reaction by-product. If CH3CH2OH is used, (CH3CH2OH2)+ is formed instead.

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Chapter 8–28

CH3

3° halide strong base E2

CH3

OH

Cl

b.

CH2 + H2O + Cl

+

overall reaction CH3

CH3

Cl

Each product:

H

one step OH

or

+ H2O + Cl OH

CH2 H

CH2

Cl

one step

8.62 Draw the products of each reaction with the 1° alkyl halide. Cl

a.

NaOCH2CH3

OCH2CH3

H

H

H

CN

strong nucleophile SN2

H

DBU

sterically hindered base E2

H

KCN

Cl

b.

Cl

c.

strong nucleophile SN2

H

8.63 H H

H Br

H

H

H2O

H

O H

OH

H2O Br

above H

H

+

HBr

+

HBr

H

Br– H H

H

O H

re-draw

Br

below H

H

H

H

+ H

OH

H2O

H

H

Br

HBr

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229

Alkyl Halides and Elimination Reactions 8–29

8.64 good nucleophile O CH3

C

CH3 CHCH3

O

O

C

CH3COO− is a good nucleophile and a weak base and so it favors substitution by SN2.

CH3

O CH3 CHCH3

(only)

Br CH3 CHCH3 + CH3CH CH2 The strong base gives both SN2 and E2 products, but since the 2° RX is OCH2CH3

CH3CH2O

strong base

20%

80%

somewhat hindered to substitution, the E2 product is favored.

8.65 Cl

OCH3 CH3OH

+

+

+

+

OCH3

3° halide weak base SN1 and E1

Cl CH3OH

Cl

H OCH3

OCH3

Cl–

HCl

Cl H

or

HCl

Cl–

CH3OH

HCl OCH3

Cl

or

H

HCl

Cl H

OCH3

HCl

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Chapter 8–30

8.66 E2 elimination needs a leaving group and a hydrogen in the trans diaxial position. Cl Cl

Two different conformations:

Cl

Cl Cl Cl

Cl Cl

Cl Cl

Cl

Cl

This conformation has Cl's axial, but no H's axial.

This conformation has no Cl's axial.

For elimination to occur, a cyclohexane must have a H and Cl in the trans diaxial arrangement. Neither conformation of this isomer has both atoms—H and Cl—axial; thus, this isomer only slowly loses HCl by elimination. 8.67 H Br CH 3

H and Br are anti periplanar. Elimination can occur.

CH3O−

O O H H

H (in the ring) and Br are NOT anti periplanar. Elimination cannot occur using this H. Instead elimination must occur with the H on the CH3 group.

O O

–HBr

CH3 Br

H major product

Elimination can occur here. H CH3O−

O O

O O

–HBr

H

H

Elimination cannot occur in the ring because the required anti periplanar geometry is not present.

8.68 leaving group H N

C6H5O O

S

N

C6H5O

DBN

N O

B

H

H H

H H N

O

overall reaction

S O

O

H C CH2Cl O

SN2

E2

H

A sequence of two reactions forms the final product: E2 elimination opens the five-membered ring. Then the sulfur nucleophile displaces the Cl– leaving group to form the six-membered ring.

H H

N

C6H5O

Cl

S

O

N

CH2 C

O

O

8.69 a.

H D

Δ

SeOC6H5

SeOC6H5 and H are on the same side of the ring. syn elimination

CH3

D

b.

C H Br

CH3

Br

rotate

C H Br

H C

CH3

H

C

CH3

Zn

CH3

Br

Both Br atoms are on the opposite sides of the C–C bond. anti elimination

H

C C H

CH3

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Alkyl Halides and Elimination Reactions 8–31

8.70 One equivalent of NaNH2 removes one mole of HBr in an anti periplanar fashion from each dibromide. Two modes of elimination are possible for each compound. C1

Br

H Br CH3

a.

H Br

CH3 H

C2

A

Rotate to make CH3 H and Br anti periplanar.

Br H

C1

CH3 Br

loss of H from C1 loss of Br from C2 C

H and Br are anti periplanar on C1 and C2 as drawn. H Br

H

NaNH2

loss of H from C2 loss of Br from C1 D

H Br

Br H Br

NaNH2

H

C2

B Two different rotations are needed.

Rotate to make H and Br anti periplanar. C1

CH3

CH3

loss of H from C1 loss of Br from C2 F

C2 H Br

H Br

NaNH2

b. C and F are diastereomers. D and E are diastereomers. C and D are constitutional isomers. E and F are constitutional isomers.

Br

H CH3 CH3

loss of H from C2 loss of Br from C1 E

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233

Alcohols, Ethers, and Epoxides 9–1

Chapter 9 Alcohols, Ethers, and Epoxides Chapter Review General facts about ROH, ROR, and epoxides • All three compounds contain an O atom that is sp3 hybridized and tetrahedral (9.2). CH3

O

CH3

H

109o an alcohol

•

O

60o

O

CH3 H H

111o an ether

H H

an epoxide

All three compounds have polar C–O bonds, but only alcohols have an O–H bond for intermolecular hydrogen bonding (9.4). H C H

H

=

O

H

=

H

O C

H H

H

hydrogen bond

•

Alcohols and ethers do not contain a good leaving group. Nucleophilic substitution can occur only after the OH (or OR) group is converted to a better leaving group (9.7A). +

R OH

+

R OH2

H Cl

+ Cl

strong acid weak base good leaving group

•

Epoxides have a leaving group located in a strained three-membered ring, making them reactive to strong nucleophiles and acids HZ that contain a nucleophilic atom Z (9.15). leaving group With strong nucleophiles, Nu

O

O C C

OH

H OH

C C

[1]

C C

[2]

Nu

+ OH

Nu

Nu

A new reaction of carbocations (9.9) • Less stable carbocations rearrange to more stable carbocations by shift of a hydrogen atom or an alkyl group. Besides rearrangement, carbocations also react with nucleophiles (7.13) and bases (8.6). 1,2-shift

C C

C C

R (or H)

R (or H)

Preparation of alcohols, ethers, and epoxides (9.6) [1] Preparation of alcohols R X

+

OH

R OH

+ X

• •

The mechanism is SN2. The reaction works best for CH3X and 1o RX.

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Chapter 9–2

[2] Preparation of alkoxides (a Brønsted–Lowry acid–base reaction) R O H + Na+ H

R O

Na+ + H2

alkoxide

[3] Preparation of ethers (Williamson ether synthesis) R X

+

OR'

R OR'

+

X

• •

The mechanism is SN2. The reaction works best for CH3X and 1o RX.

[4] Preparation of epoxides (intramolecular SN2 reaction) •

B H O

[1]

C C

O C

X

+

halohydrin

O C C

[2]

C

+X

X

H B+

A two-step reaction sequence: [1] Removal of a proton with base forms an alkoxide. [2] Intramolecular SN2 reaction forms the epoxide.

Reactions of alcohols [1] Dehydration to form alkenes [a] Using strong acid (9.8, 9.9)

C C H OH

H2SO4 or TsOH

C C

• +

H2O

• •

Order of reactivity: R3COH > R2CHOH > RCH2OH. The mechanism for 2o and 3o ROH is E1; carbocations are intermediates and rearrangements occur. The mechanism for 1o ROH is E2. The Zaitsev rule is followed.

• •

The mechanism is E2. No carbocation rearrangements occur.

•

Order of reactivity: R3COH > R2CHOH > RCH2OH. The mechanism for 2o and 3o ROH is SN1; carbocations are intermediates and rearrangements occur. The mechanism for CH3OH and 1o ROH is SN2.

•

[b] Using POCl3 and pyridine (9.10) C C H OH

POCl3

C C

pyridine

+ H2O

[2] Reaction with HX to form RX (9.11) R OH + H X

R X

+ H2O

• •

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235

Alcohols, Ethers, and Epoxides 9–3

[3] Reaction with other reagents to form RX (9.12) R OH R OH

+ SOCl2 +

R Cl

• •

pyridine

PBr3

R Br

Reactions occur with CH3OH and 1o and 2o ROH. The reactions follow an SN2 mechanism.

[4] Reaction with tosyl chloride to form alkyl tosylates (9.13A) O R OH + Cl S

CH3

O

pyridine

R O S

O

CH3

O R OTs

• The C–O bond is not broken so the configuration at a stereogenic center is retained.

Reactions of alkyl tosylates Alkyl tosylates undergo either substitution or elimination depending on the reagent (9.13B). Nu

C C

+

OTs

•

Substitution is carried out with strong :Nu–, so the mechanism is SN2.

•

Elimination is carried out with strong bases, so the mechanism is E2.

H Nu C C H OTs

B

+ HB+

C C

–OTs

+

Reactions of ethers Only one reaction is useful: Cleavage with strong acids (9.14) R O R'

+

H X

R X

+

R' X

+

H2O

(2 equiv) (X = Br or I)

• •

With 2o and 3o R groups, the mechanism is SN1. With CH3 and 1o R groups the mechanism is SN2.

Reactions of epoxides Epoxide rings are opened with nucleophiles :Nu– and acids HZ (9.15). • OH

O C

C

[1] Nu

[2] H2O

or HZ

C

C

•

Nu

(Z)

•

The reaction occurs with backside attack, resulting in trans or anti products. With :Nu–, the mechanism is SN2, and nucleophilic attack occurs at the less substituted C. With HZ, the mechanism is between SN1 and SN2, and attack of Z– occurs at the more substituted C.

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Chapter 9–4

Practice Test on Chapter Review 1. Give the IUPAC name for each of the following compounds.

a.

b. O HO

2. Draw the organic products formed in each reaction. Draw all stereogenic centers using wedges and dashes. O

HI

a.

(2 equiv)

[1] NaH b.

H

D

[1] TsCl, pyr

c.

d.

e.

[2] CH3CH2Br

OH

OH

OH

O

[2] NaOCH3

HBr

[1] NaCN [2] H2O

H2SO4

f. OH

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Alcohols, Ethers, and Epoxides 9–5

3. What starting material is needed for the following reaction? [1] TsCl, pyridine

OCH3

[2] NaOCH3 CH3CH2CH2

4. What alkoxide and alkyl halide are needed to make the following ether? O

CH2CH2CH(CH3)2

Answers to Practice Test 1. a. 3-ethoxy-2methylhexane

2. a.

Br

d.

3.

Br

I

CH2I

OH CH3CH2CH2

b. 4-ethyl-7-methyl3-octanol

4.

OH

e.

O

b.

CN OCH2CH3 XCH2CH2CH(CH3)2 D

H

c.

f.

OCH3

Answers to Problems 9.1 • Alcohols are classified as 1°, 2°, or 3°, depending on the number of carbon atoms bonded to the carbon with the OH group. • Symmetrical ethers have two identical R groups, and unsymmetrical ethers have R groups that are different. OH OH

1° alcohol

O

2° alcohol

OH OH

3° alcohol

1° alcohol

symmetrical ether

CH3

O

unsymmetrical ether

CH3

O

unsymmetrical ether

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Chapter 9–6

9.2 Use the definitions in Answer 9.1. O HO

2°

CH2OH 1° OH 3°

CH3

CH3

H

CH3

F

H

O

9.3 To name an alcohol: [1] Find the longest chain that has the OH group as a substituent. Name the molecule as a derivative of that number of carbons by changing the -e ending of the alkane to the suffix -ol. [2] Number the carbon chain to give the OH group the lower number. When the OH group is bonded to a ring, the ring is numbered beginning with the OH group, and the “1” is usually omitted. [3] Apply the other rules of nomenclature to complete the name. 1 OH

a. [1]

OH

[2]

[3] 3,3-dimethyl-1-pentanol

5 carbons = pentanol b. [1]

CH3

CH3

[2]

OH

2-methyl

[3] cis-2-methylcyclohexanol

OH

1

6 carbon ring = cyclohexanol

6-methyl c.

OH

[1]

OH

[2]

[3] 5-ethyl-6-methyl-3-nonanol

3 9 carbons = nonanol

5-ethyl

9.4 To work backwards from a name to a structure: [1] Find the parent name and draw its structure. [2] Add the substituents to the long chain. OH

7

3

2

c. 2-tert-butyl-3-methylcyclohexanol

a. 7,7-dimethyl-4-octanol 4

OH

1 5 b. 5-methyl-4-propyl-3-heptanol

d. trans-1,2-cyclohexanediol 3

OH

OH

OH

or OH

OH

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239

Alcohols, Ethers, and Epoxides 9–7

9.5 To name simple ethers: [1] Name both alkyl groups bonded to the oxygen. [2] Arrange these names alphabetically and add the word ether. For symmetrical ethers, name the alkyl group and add the prefix di. To name ethers using the IUPAC system: [1] Find the two alkyl groups bonded to the ether oxygen. The smaller chain becomes the substituent, named as an alkoxy group. [2] Number the chain to give the lower number to the first substituent. IUPAC name:

a. common name: CH3 O CH2CH2CH2CH3

CH3 O CH2CH2CH2CH3

substituent: methoxy

butyl

methyl

larger group – 4 C's butane

butyl methyl ether

1-methoxybutane IUPAC name:

b. common name: OCH3

OCH3

substituent – methoxy

methyl larger group – 6 C's cyclohexane

cyclohexyl cyclohexyl methyl ether

methoxycyclohexane IUPAC name:

c. common name: CH3CH2CH2 O CH2CH2CH3

propyl

CH3CH2CH2 O CH2CH2CH3

propyl

propoxy

dipropyl ether

propane

1-propoxypropane

9.6 Name the ether using the rules from Answer 9.5. 1 CH3 CH3 C

O CH3

CH3

2-methyl

2-methoxy

propane 2-methoxy-2-methylpropane

9.7 Three ways to name epoxides: [1] Epoxides are named as derivatives of oxirane, the simplest epoxide. [2] Epoxides can be named by considering the oxygen as a substituent called an epoxy group, bonded to a hydrocarbon chain or ring. Use two numbers to designate which two atoms the oxygen is bonded to. [3] Epoxides can be named as alkene oxides by mentally replacing the epoxide oxygen by a double bond. Name the alkene (Chapter 10) and add the word oxide.

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Chapter 9–8

H

Three possibilities: [1] methyloxirane [2] 1,2-epoxypropane [3] propene oxide

O

a. CH3

Three possibilities: [1] cis-2-methyl-3-propyloxirane [2] cis-2,3-epoxyhexane [3] cis-2-hexene oxide

O

c.

H

1

b.

CH3

1-methyl

O

epoxy group

2

Two possibilities: [1] 6 carbons = cyclohexane 1,2-epoxy-1-methylcyclohexane [2] 1-methylcyclohexene oxide

9.8 Two rules for boiling point: [1] The stronger the forces the higher the bp. [2] Bp increases as the extent of the hydrogen bonding increases. For alcohols with the same number of carbon atoms: hydrogen bonding and bp’s increase: 3° ROH < 2° ROH < 1° ROH. CH3

OH

CH3

a.

b.

O

OH

OH OH

VDW lowest bp

VDW DD intermediate bp

VDW DD hydrogen bonding highest bp

3° ROH lowest bp

2° ROH intermediate bp

1° ROH highest bp

9.9 Strong nucleophiles (like –CN) favor SN2 reactions. The use of crown ethers in nonpolar solvents increases the nucleophilicity of the anion, and this increases the rate of the SN2 reaction. The nucleophile does not appear in the rate equation for the SN1 reaction. Nonpolar solvents cannot solvate carbocations, so this disfavors SN1 reactions as well. 9.10 Draw the products of substitution in the following reactions by substituting OH or OR for X in the starting material. a. CH3CH2CH2CH2 Br

+ OH

b.

Cl

+

OCH3

c.

CH2CH2–I

+

OCH(CH3)2

CH2CH2–OCH(CH3)2

d.

Br +

OCH2CH3

OCH2CH3

alcohol

CH3CH2CH2CH2 OH + Br

OCH3

+

Cl

+

unsymmetrical ether

+

Br

I

unsymmetrical ether

unsymmetrical ether

9.11 Two possible routes to X are shown. Path [2] with a 1° alkyl halide is preferred. Path [1] cannot occur because the leaving group would be bonded to an sp2 hybridized C, making it an unreactive aryl halide.

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Alcohols, Ethers, and Epoxides 9–9

F O O

[1]

+

X

Br

O

aryl halide

N CH3 F

O O

[2]

+ Cl

X

preferred

O

N CH3

1° alkyl halide

9.12 NaH and NaNH2 are strong bases that will remove a proton from an alcohol, creating a nucleophile. a. CH3CH2CH2 O H + Na+ H CH3

b.

CH3CH2CH2 O Na+ + H2 CH3

+ + Na NH2

C O H

C O

H

Na+ + NH3

+

H

c.

O

H

Na+ H

CH3CH2CH2–Br + Na+ + H2

O

O

O

d.

H

CH2CH2CH3 + Br–

O

Na+ H

+ Na+ + H2

Br

O

+ Br–

Br C6H10O

9.13 Dehydration follows the Zaitsev rule, so the more stable, more substituted alkene is the major product. H

a.

CH3 C CH3

TsOH

CH2 CH CH3

+ H2O

OH CH3CH2

b.

TsOH OH

+ H2O

C CHCH3 CH3

trisubstituted major product

CH2

disubstituted minor product

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Chapter 9–10

CH3 OH

c.

CH2

CH3

TsOH

+

+ H2O

trisubstituted disubstituted major product minor product

9.14 The rate of dehydration increases as the number of R groups increases. CH3

OH

OH CH3

HO

1° alcohol slowest reaction

3° alcohol fastest reaction

2° alcohol intermediate reactivity

9.15 transition state [1]:

transition state [2]:

H

H

CH3CH2 C H

δ+ OH

CH3 C

δ−

H

H

OSO3H

δ−

OSO3H CH2 OH2

δ+

9.16 H H

HSO4

H

+

This alkene is also formed in addition to Y from the rearranged carbocation.

β

+

rearranged 3° carbocation

H2SO4

The initially formed 2° carbocation gives two alkenes: H

HSO4

H

or

H

H + CH 2

+

H

HSO4

9.17 CH3 H

a. CH3 C H

C CH2CH3

+

rearrangement

CH3 H CH3 C

+

1,2-H shift

2° carbocation

H

b.

+

2° carbocation

C CH2CH3

+ 3° carbocation more stable

+ rearrangement 1,2-methyl shift

H

3° carbocation more stable rearrangement 1,2-H shift

c.

2° carbocation

+

3° carbocation more stable

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Alcohols, Ethers, and Epoxides 9–11

9.18 CH3 H CH3 C H

CH3 H

H2O

C CH3

CH3 C

overall reaction

Cl

CH3 H

C CH3

H

CH3 C

OH

C CH3

HCl

OH H

The steps:

CH3 H CH3 C Cl

CH3 H

C CH3

H

CH3 C

H2O

and

O H H

CH3 H CH3 C

C CH3

H

Cl

CH3 H

C CH3

CH3 C

CH3 H

C CH3

H

CH3 C

H H2O

2° carbocation

H O

3° carbocation

C CH3 H

H

Rearrangement of H forms a more stable carbocation.

Cl

9.19 CH3

CH3

HCl

a. CH3 C CH2CH3

CH3 C CH2CH3

OH

c.

HBr

Br

+ H 2O

Cl

OH HI

b.

+ H2 O

I

OH

+ H2 O

9.20 • CH3OH and 1° alcohols follow an SN2 mechanism, which results in inversion of configuration. • Secondary (2°) and 3° alcohols follow an SN1 mechanism, which results in racemization at a stereogenic center. H OH

a.

C

I H

HI

D

CH3CH2CH2 CH3

b.

HBr

Br CH3

OH

HO

D

1° alcohol so inversion of configuration

CH3CH2CH2

achiral starting material

c.

C

HCl

3° alcohol, so Br− attacks from above and below. The product is achiral.

achiral product Cl

Cl

3° alcohol = racemization

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Chapter 9–12

9.21 OH

a.

HCl

Cl

HCl

c.

Cl

OH

(product formed after a 1,2-H shift) Cl HCl

b. OH

(product formed after a 1,2-CH3 shift)

9.22 Substitution reactions of alcohols using SOCl2 proceed by an SN2 mechanism. Therefore, there is inversion of configuration at a stereogenic center. H OH

SOCl2

Reactions using SOCl2 proceed by an SN2 mechanism = inversion of configuration.

Cl H

pyridine S

R

9.23 Substitution reactions of alcohols using PBr3 proceed by an SN2 mechanism. Therefore, there is inversion of configuration at a stereogenic center. PBr3

H OH

Reactions using PBr3 proceed by an SN2 mechanism = inversion of configuration.

Br H

S

R

9.24 Stereochemistry for conversion of ROH to RX by reagent: [1] HX—with 1°, SN2, so inversion of configuration; with 2° and 3°, SN1, so racemization. [2] SOCl2—SN2, so inversion of configuration. [3] PBr3—SN2, so inversion of configuration. OH

a.

SOCl2

c.

Cl

OH

PBr3

Br

pyridine OH

HI

I

I

S N2 = inversion

b. 3° alcohol, SN1 = racemization

9.25 To do a two-step synthesis with this starting material: [1] Convert the OH group into a good leaving group (by using either PBr3 or SOCl2). [2] Add the nucleophile for the SN2 reaction.

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Alcohols, Ethers, and Epoxides 9–13

H H

H

PBr3

CH3 C OH

N3

CH3

CH3 C Br

CH3

CH3 C N3

CH3CH2O

CH3

H CH3 C OCH2CH3 CH3

bad leaving group

good leaving group

9.26 O

+ CH3

a. CH3CH2CH2CH2 OH H OH

b. CH3CH2CH2

C

CH3CH2CH2CH2 O S

SO2Cl

pyridine

CH3 + Cl

O

H OTs TsCl

CH3

pyridine

CH3CH2CH2

C

+ Cl

CH3

9.27 OTs +

a.

1° tosylate

b.

CN

SN2 CN

+

CH3CH2CH2 OTs

1° tosylate

K+ −OC(CH3)3

E2

CH3CH CH2 + K+ −OTs + HOC(CH3)3

strong bulky base

H OTs

c.

CH3

HS H +

C

CH2CH2CH3

2° tosylate

+ OTs

strong nucleophile

SH

SN2 strong nucleophile

CH3

C

SN2 product (inversion of configuration)

CH2CH2CH3

(Substitution is favored over elimination.)

9.28 HO H

S

TsCl

TsO H

NaOH

H OH

SN 2

pyridine

retention S enantiomers

One inversion from starting material to product.

inversion R

9.29 These reagents can be classified as: [1] SOCl2, PBr3, HCl, and HBr replace OH with X by a substitution reaction. [2] Tosyl chloride (TsCl) makes OH a better leaving group by converting it to OTs. [3] Strong acids (H2SO4) and POCl3 (pyridine) result in elimination by dehydration.

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Chapter 9–14

H

a.

H

SOCl2

CH3 C OH

pyridine

CH3 H

b.

H

CH3 C Cl CH3

CH3

H TsCl

CH3 C OH

pyridine

CH3

H

CH3 C OH

CH3

H CH3 C Br CH3

H

e.

CH3 C OTs

CH3

H

c.

HBr

CH3 C OH

d.

[1] PBr3 [2] NaCN

CH3 C CN CH3

H H2SO4

CH3 C OH

CH2 CHCH3

CH3 C OH

f.

CH3

CH3

POCl3

pyridine

CH2 CHCH3

9.30 a. CH3CH2 O CH2CH3 CH3

b.

CH3 C O CH2CH3

HBr

HBr

H

2 CH3CH2 Br

c.

+ H2O

O CH3

HBr

CH3

Br + CH3Br

CH3 C Br

+

+ H2O

CH3CH2Br + H2O

H

9.31 Ether cleavage can occur by either an SN1 or SN2 mechanism, but neither mechanism can occur when the ether O atom is bonded to an aromatic ring. An SN1 reaction would require formation of a highly unstable carbocation on a benzene ring, a process that does not occur. An SN2 reaction would require backside attack through the plane of the aromatic ring, which is also not possible. Thus, cleavage of the Ph–OCH3 bond does not occur. OCH3

anisole

HBr

OH

+

CH3Br

Br

bromobenzene

phenol

NOT formed SN1:

SN2: O CH3 H

CH3

CH3OH

highly unstable carbocation

O

H

Br

9.32 Two rules for reaction of an epoxide: [1] Nucleophiles attack from the back side of the epoxide. [2] Negatively charged nucleophiles attack at the less substituted carbon. CH3

a.

O

[1] CH3CH2O [2] H2O

Attack here: less substituted C backside attack

CH3 OH OCH2CH3

b. CH3 H

O C C

[1] H

H

C C

H [2] H O 2

Attack here: less substituted C backside attack

HO CH3 H

H H

C C C CH

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Alcohols, Ethers, and Epoxides 9–15

9.33 In both isomers, –OH attacks from the back side at either C–O bond. cis-2,3-dimethyloxirane [1] –OH

O H CH3

C

C

H CH3

CH3

OH

H C

[2] H2O

C

HO

+ H CH3

HO

H C

CH3

C

H CH3

OH

[1] –OH [2] H2O

O H CH3

C

C

H CH3

enantiomers HO

HO O H CH3

C

[1] –OH C

CH3 H

CH3

OH

H C

[2] H2O

HO

HO

+

C H

CH3

CH3 H C

C

H CH3

OH

[1]

O

–OH

H CH3

[2] H2O

C

C

CH3 H

identical Rotate around the C–C bond to see the plane of symmetry.

HO

trans-2,3-dimethyloxirane

HO

OH C

CH3

H

HO

C H

CH3

meso compound

9.34 Remember the difference between negatively charged nucleophiles and neutral nucleophiles: • Negatively charged nucleophiles attack first, followed by protonation, and the nucleophile attacks at the less substituted carbon. • Neutral nucleophiles have protonation first, followed by nucleophilic attack at the more substituted carbon. But, trans or anti products are always formed regardless of the nucleophile. CH3

a.

O

HBr

Br CH3 OH

CH3CH2 CH3CH2

H H

CH3

b.

[1] CN [2] H2O

negatively charged nucleophile: attack at less substituted C

CH3CH2OH H2SO4

neutral nucleophile: attack at more substituted C

neutral nucleophile: attack at more substituted C O

CH3CH2 CH3CH2

O

c.

CH3 OH CN

O

d.

CH3CH2 CH3CH2

OH C

C

CH3CH2O

H

HO H H

[1] CH3O [2] CH3OH

H C

CH3CH2 CH3CH2

negatively charged nucleophile: attack at less substituted C

H

H

C OCH3

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Chapter 9–16

9.35 2 3

HO

cyclohexanol 6 C ring 2 CH3's at C3 3,3-dimethylcyclohexanol

1

a.

b. 3

2

1 c.

O

2

1,2-epoxy-1-ethylcyclopentane or 1-ethylcyclopentene oxide

ethyl isobutyl ether or 1-ethoxy-2-methylpropane

O

1

9.36 a. (1R,2R)-2-isobutylcyclopentanol

f.

b. 2° alcohol

[1]

NaH HO

O

A

H2SO4

HO

A

[2] HO

c. stereoisomer

POCl3

[3]

pyridine HO

HO

Cl

HCl

(1R,2S)-2-isobutylcyclopentanol

[4] HO

SOCl2

d. constitutional isomer [5]

pyridine

HO

Cl

OH

TsCl (1S,3S)-3-isobutylcyclopentanol

[6]

pyridine

HO

TsO

e. constitutional isomer with an ether O

butoxycyclopentane

9.37 HO

H

HBr

a.

Br

S HO

H

b.

H

PBr3

HO

H

HCl

Cl

S

Br

2° alcohol SN1 = racemization

R Br

R

c.

H

H

H

PBr3 follows SN2 = inversion.

H

Cl

R

2° alcohol SN1 = racemization

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249

Alcohols, Ethers, and Epoxides 9–17

HO

H

H

SOCl2

d.

pyridine

SOCl2 follows SN2 = inversion.

Cl

R

9.38 Draw the structure of each alcohol, using the definitions in Answer 9.1. OH

OH

a.

OH

1°

b.

OH

2°

3°

enol

9.39 Use the directions from Answer 9.3. a.

(CH3)2CHCH2CH2CH2OH

[1]

[2]

H CH3 C CH2CH2CH2OH

H

CH3

CH3

5 carbons = pentanol

b.

[3]

1

4-methyl-1-pentanol

CH3 C CH2CH2CH2OH

4-methyl

(CH3)2CHCH2CH(CH2CH3)CH(OH)CH2CH3

[1]

H

[2]

H OH

CH3 C CH2 C CH CH3

CH2CH3

H

CH2CH3

CH3

7 carbons = heptanol

c. [1]

[3]

4-ethyl-6-methyl-3-heptanol

CH2CH3

CH2CH3

6-methyl

4-ethyl

5-methyl [2]

[3]

OH OH

8 carbons = octanol

d. [1]

3

H OH

CH3 C CH2 C CH

HO

OH

HO

OH

HO H

HO H

[2]

3

2 HO H

[3]

(2R,3R)-2,3-butanediol

2R,3R

4 carbons = butanediol [2]

OH

4

[1] OH

g. [1]

[3]

5-methyl-2,3,4-heptanetriol

2

OH

5-methyl [2]

CH(CH3)2

OH

OH

7 carbons = heptanetriol

HO

cis-1,4-cyclohexanediol

HO H

f.

OH

[3]

cis

cyclohexanediol

e. [1]

4-ethyl

3 1

4 [2]

4-ethyl-5-methyl-3-octanol

[3]

1 HO

CH(CH3)2

5 carbons = cyclopentanol 3-isopropyl

trans-3-isopropylcyclopentanol

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Chapter 9–18

9.40 Use the rules from Answers 9.5 and 9.7. a.

c.

O

O

1,2-epoxy-2-methylhexane or 2-butyl-2-methyloxirane or 2-methylhexene oxide

dicyclohexyl ether 4,4-dimethyl

CH3

b.

d. OCH2CH2CH3

CH3

longest chain = heptane

substituent = 3-propoxy 4,4-dimethyl-3-propoxyheptane

CH3

CH3 C O C CH3 CH3

tert-butyl tert-butyl di-tert-butyl ether

9.41 Use the directions from Answer 9.4. a. 4-ethyl-3-heptanol

3

f. diisobutyl ether

4

O

1

g. 1,2-epoxy-1,3,3-trimethylcyclohexane

OH

1 OH

b. trans-2-methylcyclohexanol

O

OH

or

3

CH3 HO

CH3

3

h. 1-ethoxy-3-ethylheptane O 1

c. 2,3,3-trimethyl-2-butanol

3

2 i. (2R,3S)-3-isopropyl-2-hexanol

d. 6-sec-butyl-7,7-diethyl-4-decanol

OH

3

6

2 OH

4 1

e. 3-chloro-1,2-propanediol HO

2 OH

7

3

j. (2S)-2-ethoxy-1,1-dimethylcyclopentane 1

Cl

2

O

1

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251

Alcohols, Ethers, and Epoxides 9–19

9.42 Eight constitutional isomers of molecular formula C5H12O containing an OH group: OH 1-pentanol OH

2-pentanol 2° alcohol

OH

1 OH 3-methyl-1-butanol 1° alcohol

3

1° alcohol

OH 2 1

2,2-dimethyl-1-propanol 1° alcohol

OH

3-pentanol 2° alcohol

1 OH 2-methyl-1-butanol 1° alcohol 2 OH

3-methyl-2-butanol 2° alcohol

2-methyl-2-butanol 3° alcohol

2

9.43 Use the boiling point rules from Answer 9.8. a.

CH3CH2OCH3

(CH3)2CHOH

ether no hydrogen bonding lowest bp b.

2° alcohol hydrogen bonding intermediate bp

CH3CH2CH2OH

1° alcohol hydrogen bonding highest bp

CH3CH2CH2CH2CH2CH3

CH3CH2CH2CH2CH2CH2OH

no OH group lowest water solubility

one OH group intermediate water solubility

HOCH2CH2CH2CH2CH2CH2OH

two OH groups highest water solubility

9.44 Melting points depend on intermolecular forces and symmetry. (CH3)2CHCH2OH has a lower melting point than CH3CH2CH2CH2OH because branching decreases surface area and makes (CH3)2CHCH2OH less symmetrical so it packs less well. Although (CH3)3COH has the most branching and least surface area, it is the most symmetrical, so it packs best in a crystalline lattice, giving it the highest melting point. OH

OH

–108 °C lowest melting point

OH

–90 °C intermediate melting point

26 °C highest melting point

9.45 Stronger intermolecular forces increase boiling point. All of the compounds can hydrogen bond, but both diols have more opportunity for hydrogen bonding since they have two OH groups, making their bp’s higher than the bp of 1-butanol. 1,2-Propanediol can also intramolecularly hydrogen bond. Intramolecular hydrogen bonding decreases the amount of intermolecular hydrogen bonding, so the bp of 1,2-propanediol is somewhat lower. Increasing boiling point OH

HO

HO

OH

OH

1-butanol 118 °C

1,2-propanediol 187 °C

1,3-propanediol 215 °C

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Chapter 9–20

9.46 H2SO4

a. CH3CH2CH2OH

NaH

b. CH3CH2CH2OH

CH3CH2CH2Cl

ZnCl2 HBr

d. CH3CH2CH2OH

PBr3

f. CH3CH2CH2OH

[1] NaH [1] TsCl

CH3CH2CH2OH

CH3CH2CH2Cl

CH3CH2CH2OTs

pyridine

h. CH3CH2CH2OH

+ H 2O

CH3CH2CH2Br

TsCl

g. CH3CH2CH2OH

+ H2

CH3CH2CH2Br + H2O

SOCl2 pyridine

e. CH3CH2CH2OH

+ H 2O

CH3CH2CH2O Na+

HCl

c. CH3CH2CH2OH

i.

CH3CH CH2

[2] CH3CH2Br

CH3CH2CH2O Na+

[2] NaSH

CH3CH2CH2OTs

CH3CH2CH2OCH2CH3

CH3CH2CH2SH

9.47 a.

OH

NaH

b.

OH

NaCl

c.

OH

HBr

d.

OH

HCl

e.

OH

H2SO4

f.

OH

NaHCO3

Br

g.

OH

[1] NaH

Cl

h.

OH

POCl3 pyridine

O

N.R.

Na+ + H2

+ H2O

N.R.

O

[2] CH3CH2Br

O CH2CH3

9.48 Dehydration follows the Zaitsev rule, so the more stable, more substituted alkene is the major product. TsOH

a. OH

tetrasubstituted major product

b.

CH2CH3 OH

c.

OH

disubstituted CH2CH3

TsOH

TsOH

trisubstituted disubstituted major product

CHCH3

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253

Alcohols, Ethers, and Epoxides 9–21

d.

TsOH

CH 3CH 2CH2CH2OH

CH3CH 2CH CH2

OH TsOH

e. tetrasubstituted major product

disubstituted

Two products formed by carbocation rearrangement

9.49 The most stable alkene is the major product.

OH H2SO4

monosubstituted

trans and disubstituted major product

cis and disubstituted

9.50 OTs is a good leaving group and will easily be replaced by a nucleophile. Draw the products by substituting the nucleophile in the reagent for OTs in the starting material. a.

CH3CH2CH2CH2 OTs

b.

CH3CH2CH2CH2 OTs

c.

CH3CH2CH2CH2 OTs

d.

CH3CH2CH2CH2 OTs

CH3SH

CH3CH2CH2CH2 SCH3 + HOTs

S N2 NaOCH2CH3

CH3CH2CH2CH2 OCH2CH3

S N2 NaOH

CH3CH2CH2CH2 OH

SN2 K+

–

OC(CH3)3

+ Na+

+ Na+

–OTs

–OTs

CH3CH2CH CH2 + (CH3)3COH + K+ –OTs

E2

9.51 HBr

a.

b.

+

HO H OH

H D

Br H

HCl ZnCl2

Cl

HO H

H D

pyridine

H Cl

TsCl pyridine

d. HO H

2° Alcohol will undergo SN1. racemization

1° Alcohol will undergo SN2. inversion

SOCl2

c.

H Br

SOCl2 always implies SN2. inversion KI

TsO H

Configuration is maintained. C–O bond is not broken.

SN2 inversion

H

I

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Chapter 9–22

9.52 NaH

CH3I

A=

(a)

B=

O H

(b)

TsCl

C=

pyridine

HO H

TsO H

PBr3

(c)

CH3O H CH3O

D= H OCH3

CH3O

F=

E=

(2R)-2-hexanol

H Br

CH3O H

Routes (a) and (c) given identical products, labeled B and F.

9.53 Acid-catalyzed dehydration follows an E1 mechanism for 2o and 3o ROH with an added step to make a good leaving group. The three steps are: [1] Protonate the oxygen to make a good leaving group. [2] Break the C−O bond to form a carbocation. [3] Remove a β hydrogen to form the π bond. OH CH3 H OSO3H

+

overall reaction

The steps:

CH3

CH3

CH2 +

+ H2O

H O H CH3

H H

+ HSO4

CH3 H

CH3

+ H2O

+ H2SO4

HSO4

2° carbocation and CH3

H

1,2-H shift

+ HSO4

CH2

CH2

H

2° carbocation

+ H2SO4

3° carbocation and H

+ HSO4 CH3

CH3

+ H2SO4

9.54 With POCl3 (pyridine), elimination occurs by an E2 mechanism. Since only one carbon has a β hydrogen, only one product is formed. With H2SO4, the mechanism of elimination is E1. A 2° carbocation rearranges to a 3° carbocation, which has three pathways for elimination.

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Alcohols, Ethers, and Epoxides 9–23

O OH

Cl

P Cl

H

POCl2

O

Cl

N

POCl2

N

O H

+ N H

V

+

W

Cl

+ –OPOCl2

+ N H

H OSO3H

OH

+ OH2

+ + H 2O

V

+

1,2–CH3 shift

2° carbocation

+ HSO4–

3° carbocation

HSO4–

H

H

+

+

+ HSO4–

3° carbocation

H

3° carbocation

X

3° carbocation

Y

+ H2SO4

HSO4–

+

Z +

H2SO4

H2SO4

9.55 To draw the mechanism: [1] Protonate the oxygen to make a good leaving group. [2] Break the C−O bond to form a carbocation. [3] Look for possible rearrangements to make a more stable carbocation. [4] Remove a β hydrogen to form the π bond. Dark and light circles are meant to show where the carbons in the starting material appear in the product.

OH H OSO3H

O H

+ H2O

H + HSO4–

2o carbocation

H + HSO4–

3o carbocation

+ H2SO4

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Chapter 9–24

9.56 OH

[1] HBr

3-methyl-2-butanol

1,2-H shift

H

H

Br

Br

[2] –H2O

2° carbocation 2-methyl-1-propanol

HBr

HO

3° carbocation

H2 O

Br Br

H Br

H2O

SN2 no carbocation

The 2° alcohol reacts by an SN1 mechanism to form a carbocation that rearranges.

The 1° alcohol reacts with HBr by an SN2 mechanism. no carbocation intermediate = no rearrangement possible

9.57 H Cl

A

Cl

OH

H2O

OH2

two resonance structures for the carbocation B

Cl

Cl Cl Cl

C

9.58 H2SO4, NaBr

CH3CH2CH2CH2OH

overall reaction

CH3CH2CH2CH2Br

CH3CH2CH CH2

CH3CH2CH2CH2OCH2CH2CH2CH3

Step [1] for all products: Formation of a good leaving group CH3CH2CH2CH2

OH

H OSO3H

CH3CH2CH2CH2

O H

+ HSO4

H

Formation of CH3CH2CH2CH2Br: CH3CH2CH2CH2

O H

SN2

H

CH3CH2CH2CH2Br + H2O

Br

Formation of CH3CH2CH=CH2: CH3CH2CH CH2 O H H HSO4–

H

E2

CH3CH2CH CH2

+ H2 O

H2SO4

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Alcohols, Ethers, and Epoxides 9–25

Ether formation (from the protonated alcohol): CH3CH2CH2CH2

CH3CH2CH2CH2

OH2

CH3CH2CH2CH2

S N2

O CH2CH2CH2CH3 + H2O H

HSO4

O H

CH3CH2CH2CH2 O

CH2CH2CH2CH3

H2SO4

9.59 H OSO3H

OH O

OH2

+

O

+ HSO4–

O

+ HSO4–

H

1,2-shift O +

O + H2SO4

+ H 2O

H O

+

9.60 HSO4– H OSO3H

OH

OH2

OH

O

OH + HSO4

–

H

O

OH

+ H 2O

+ H2SO4

9.61 a.

OCH2CH2CH3

b. O

OCH2CH2CH3

O

Br O

Br

2° halide

Br

O

1° halide less hindered RX preferred path

O

OCH2CH2CH3

+ BrCH2CH2CH3

1° halide less hindered RX preferred path

2° halide

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Chapter 9–26

c. CH3CH2OCH2CH2CH3

CH3CH2OCH2CH2CH3

CH3CH2O + BrCH2CH2CH3

CH3CH2Br + OCH2CH2CH3

1° halide

1° halide

Neither path preferred.

9.62 A tertiary halide is too hindered and an aryl halide too unreactive to undergo a Williamson ether synthesis. Two possible sets of starting materials:

CH3

Br

O C CH3 CH3

CH3

CH3

O C CH3

Br C CH3

CH3

aryl halide unreactive in SN2

O

CH3

3° alkyl halide too sterically hindered for SN2

9.63 a.

(CH3)3COCH2CH2CH3

b.

HBr

HBr

O

(2 equiv)

2

c.

H2 O

(CH3)3CBr + BrCH2CH2CH3

(2 equiv)

OCH3

HBr

Br

(2 equiv)

CH3Br H2O

H2 O

Br

9.64 overall reaction

CH3

a. O

I

CH3 H

I

+ H2O

I

The steps: I CH3 O

CH3

+ I

H

+

CH3

O

H

H

CH3

O

I

CH3

CH3 H

b.

Cl

O

H

Na+ H Cl

O

+ H2 + Na+

I

H

O H

I CH3 I CH3

+ H2 + NaCl O

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Alcohols, Ethers, and Epoxides 9–27

9.65 OC(CH3)3

O C(CH3)3

H–OCOCF3

CH3

OH

H

CH2

+ CF3CO2–

C

OH

CH3

+

CH3

C CH2

H

CH3

CF3CO2–

+ CF3CO2H

9.66 O C

a. H

O C

H

HBr

H H

BrCH2CH2OH

C

d. H

b. H

C

H

C

H

HOCH2CH2OH

e. H C

H2SO4

H H

HOCH2CH2OH

H H

[2] H2O

O CH3CH2OCH2CH2OH

[2] H2O

f. H C

CH2 CH2OH

[1] −OH C

H

[1] CH3CH2O−

HC C

[2] H2O

O

H2O

H H

O

c. H C

H H

H

O C

[1] HC C–

C

C

[1] CH3S−

H H

H

CH3SCH2CH2OH

[2] H2O

9.67 O

a. CH 3

H

CH3

CH3

b.

CH3 C

H2SO4

H

H

Br

O

CH3 OH

[2] H2O

OH HBr

c.

C H

CH3CH2O

[1] CH3CH2O− Na+

O

O

CH3 OH

CH3CH2OH

OH CN

[1] NaCN

d.

[2] H2O

OCH2CH3

9.68 CH3

H

Cl C

a.

CH3

H O

Cl

H

C

C H

CH3

C

O

H

CH3

Cl

CH3 C

b. H O

H

Cl

CH3 H C

C CH3

C

O

H

CH3

Na+ H OH

Cl C

c. CH3

H

C H

CH3

CH3 H

Cl

rotate

C CH3

H

C

The 2 CH3 groups are anti in the starting material, making them trans in the product.

CH3 H

O C4H8O

Na+ H

H

H CH3 C

CH3 H C

C

O C4H8O

C

Na+ H

CH3

The 2 CH3 groups are gauche in the starting material, making them cis in the product.

CH3 H

Cl

C O H

CH3 H

H

C O

backside attack

CH3 H C

C

O C4H8O

H CH3

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Chapter 9–28

9.69 First, use the names to draw the structures of the starting material and both products. Because the product has two OH groups, one OH must come from the epoxide oxygen, and one must come from the nucleophile, either –OH or H2O. With –OH, the nucleophile attacks at the less substituted end of the epoxide to form the R isomer of the product.

OH

[1] Na+ –OH

O

HO

[2] H2O

R

from the nucleophile

(2R)-2-ethyl-2-methyloxirane

(2R)-2-methyl-1,2-butanediol

With H2O and H2SO4, H2O O H2O attacks at the HO stereogenic center, from from the nucleophile OH H2SO4 the back side, and the S (2R)-2-ethyl-2-methyloxirane (2S)-2-methyl-1,2-butanediol isomer is formed.

9.70 O

O

O

CH3CH2O

CH3CH2O

+

Cl

CH3CH2OCH2

Cl

Cl–

CH3

9.71 KOC(CH3)3

OTs

a.

b.

OH

*

Bulky base favors E2.

HBr

H CH3

c.

H CH3

O CH3CH2 C C H CH2CH3 H

[1] −CN

CH3CH2 H

[2] H2O

Keep the stereochemistry at the stereogenic center [*] the same here since no bond is broken to it.

+

C

H CH3CH2 CH2CH3 H

NC

H

OH

e.

H C

C

CH2CH3 CN

H KCN

(CH3)3C

HO

OH C

OTs

d.

SN2 inversion

PBr3

CH3

f.

Br

*

CN

(CH3)3C

Br

SN2 inversion

CH3 OH H D

TsCl pyridine

OTs H D

CH 3CO2

CH3CO2 H D

identical

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Alcohols, Ethers, and Epoxides 9–29

g.

h.

OH

Br

Br

OH

HBr

O

O

[2] H2O

OH

OCH3

OH

[1] NaOCH3

OH

OCH3 O

NaH

OCH2CH3

CH3CH2I

i. CH2CH3 CH3

j.

CH2CH3

CH2CH3

CH3CH2 C O CH3 CH3

HI

(2 equiv)

CH3 CH3CH2 C

I

+

I

+ H2O

CH3

CH3

9.72 a.

HBr or PBr3

OH

b.

OH

c.

OH

Br

HCl, ZnCl2 or SOCl2, pyridine

Cl

[1] Na+ H−

[2] CH3CH2Cl

O

O

[1] TsCl, pyridine

d.

OH

OTs

[2] N3− N3

Make OH a good leaving group (use TsCl); then add N3−.

9.73 a.

OH

HCl or SOCl2, pyridine

b.

OH

H2SO4

c.

OH [1] Na+ H−

d.

OH [1] TsCl, pyridine

Cl

O

[2] CH3Cl

OTs [2] –CN

OCH3

CN

Make OH a good leaving group (use TsCl); then add −CN.

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Chapter 9–30

9.74 Br

(a) = HBr or PBr3

or other strong base

OH OTs

(c) = (e) =

(d) = KOC(CH3)3 or other strong base

TsCl, pyridine

H2SO4 or TsOH

(b) = KOC(CH3)3

Br (f) = KOC(CH3)3

NBS

or other strong base HOCl OH

(g) = NaH

OH

(h) =

+ enantiomer

O −OH,

Cl

H2O

OH

9.75 OH

O

O

Cl

O

O O

NaH

(CH3)2CHNH2

N H H

O

proton transfer O OH

N H

propranolol

9.76 With the cis isomer, –OH acts as a nucleophile to displace Br– from the back side, forming a trans diol (A). With the trans isomer, the two functional groups are arranged in a manner that allows an intramolecular SN2. –OH removes a proton to form an alkoxide, which can then displace Br– by intramolecular backside attack to afford an ether (B). Such a reaction is not possible with the cis isomer because the nucleophile and leaving group are on the same side.

HO

OH

Br

HO

OH

SN2 cis-4-bromocyclohexanol

A O

OH

HO

Br

trans-4-bromocyclohexanol

proton transfer

O

Br

+ H2O

SN2 B

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Alcohols, Ethers, and Epoxides 9–31

9.77 If the base is not bulky, it can react as a nucleophile and open the epoxide ring. The bulky base cannot act as a nucleophile, and will only remove the proton. –

N(CH2CH3)2

H

Li+ O

+ H OH

O

+

HN(CH2CH3)2

LiOH

OH

9.78 First form the 2° carbocation. Then lose a proton to form each product. H

H CH3CH2 C CH2 OH

CH3CH2 C CH2

H OH2

H

H

1o alcohol

no 1o carbocation at this step

H

H

1,2-shift

OH2

CH3CH2 C CH2 H

+ H2O

2o carbocation

H

CH3CH2 C CH2 H H

CH3CH2 C CH2

H2O

+ CH3 H

H

CH3CH2CH=CH2

H3 O

H

H

+

C C

CH3 CH C CH3

=

CH3

H

+

C C CH3

CH3

H3O

H2O

9.79 HO

H OSO3H

H2O + H2O

2o carbocation + HSO4

H2O

1,2-shift

H

+ H3O+

3o and resonance-stabilized allylic carbocation

+

H2O

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Chapter 9–32

9.80 H OSO3H OH

a.

+ H2O

OH2

X +

HSO4– 1,2-shift

1,2-shift H

Y

+ H3O+ H2O b. Other elimination products can form from carbocations X and Y.

+ H3O+ H X H2O

H

X

+ H3O+

H2O

+ H3O+ H Y H2O

9.81 OH OH

a. CH3 C

C CH3

CH3 CH3

pinacol

H OSO3H

OH

OH2 OH CH3 C

CH3 C

C CH3

C CH3

shift

CH3 CH3

CH3 CH3

1,2-CH3

CH3 CH3 C CH3

OH C CH3

+ H2O

+ HSO4

CH3 H2SO4 +

CH3 C CH3

O C CH3

pinacolone

CH3 O H CH3 C CH3

C CH3

HSO4

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Alcohols, Ethers, and Epoxides 9–33

b. Two different carbocations can form. The carbocation with the (+) charge adjacent to the benzene rings (A) is more stable, so it is preferred.

HO

HO

H2SO4

or

(two steps) OH

OH D

A

B

resonance-stabilized preferred 1,2-CH3 shift

O

O

+ H3O+

H

H2O

9.82 O O H C I

O

H H O

Na+ OH

C I

O

O

O

O

O

H H O

C

I

O

H H O

O H

C

C

H I

O

H H

O

I

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Chapter 9–34

9.83 The conversion of X to Y requires two operations. X contains both a nucleophile (NH2) and a leaving group (OSO2CH3), so an intramolecular SN2 reaction forms an aziridine. Since the aziridine is strained, the amine nucleophile (CH2=CHCH2NH2) opens the ring by backside attack, resulting in the trans stereochemistry of the two N’s on the six-membered ring.

O

CO2CH2CH3

O

backside attack

H2N OSO2CH3

H

CO2CH2CH3

New C–N bond on opposite side to C–OSO2CH3 bond

N H

X

OSO2CH3

O

CO2CH2CH3

H2N

CO2CH2CH3

HN

Y

aziridine H2N

proton transfer

H2N

CO2CH2CH3

backside attack HN

HN

O

O

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267

Alkenes 10–1

Chapter 10 Alkenes Chapter Review General facts about alkenes • Alkenes contain a carbon–carbon double bond consisting of a stronger σ bond and a weaker π bond. Each carbon is sp2 hybridized and trigonal planar (10.1). • Alkenes are named using the suffix -ene (10.3). • Alkenes with different groups on each end of the double bond exist as a pair of diastereomers, identified by the prefixes E and Z (10.3B). Two higher priority groups on the same side

Two higher priority groups on opposite sides CH3

CH3

CH3

C C H

CH2CH3

H

E isomer (2E)-3-methyl-2-pentene

•

CH3

Z isomer (2Z)-3-methyl-2-pentene

Alkenes have weak intermolecular forces, giving them low mp’s and bp’s, and making them water insoluble. A cis alkene is more polar than a trans alkene, giving it a slightly higher boiling point (10.4). cis-2-butene more polar isomer

trans-2-butene

CH3

CH3

CH3

C C

H

less polar isomer

C C H

H

H

CH3

no net dipole

a small net dipole higher bp

•

CH2CH3

C C

lower bp

Since a π bond is electron rich and much weaker than a σ bond, alkenes undergo addition reactions with electrophiles (10.8).

Stereochemistry of alkene addition reactions (10.8) A reagent XY adds to a double bond in one of three different ways: • Syn addition—X and Y add from the same side. C C

•

H

BH 2 C C

•

Syn addition occurs in hydroboration.

Anti addition—X and Y add from opposite sides. C C

•

H BH2

X2 or X2, H2O

X

•

C C X(OH)

Anti addition occurs in halogenation and halohydrin formation.

Both syn and anti addition occur when carbocations are intermediates. H X(OH) H • Syn and anti addition occur in H X C C C C C C and hydrohalogenation and or X(OH) H2O, H+ hydration.

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Chapter 10–2

Addition reactions of alkenes [1] Hydrohalogenation—Addition of HX (X = Cl, Br, I) (10.9–10.11) • The mechanism has two steps. • Carbocations are formed as intermediates. RCH CH2 + H X R CH CH2 • Carbocation rearrangements are possible. X H alkyl halide • Markovnikov’s rule is followed. H bonds to the less substituted C to form the more stable carbocation. • Syn and anti addition occur. [2] Hydration and related reactions—Addition of H2O or ROH (10.12) For both reactions: • The mechanism has three steps. + H OH RCH CH2 R CH CH2 H2SO4 • Carbocations are formed as intermediates. OH H • Carbocation rearrangements are possible. alcohol • Markovnikov’s rule is followed. H bonds to the less substituted C to form the more RCH CH2 + H OR R CH CH2 stable carbocation. H2SO4 OR H • Syn and anti addition occur. ether

[3] Halogenation—Addition of X2 (X = Cl or Br) (10.13–10.14) • The mechanism has two steps. • Bridged halonium ions are formed as RCH CH2 + X X R CH CH2 intermediates. X X • No rearrangements occur. vicinal dihalide • Anti addition occurs. [4] Halohydrin formation—Addition of OH and X (X = Cl, Br) (10.15) • The mechanism has three steps. + • Bridged halonium ions are formed as X X RCH CH2 R CH CH2 H2O intermediates. OH X • No rearrangements occur. halohydrin • X bonds to the less substituted C. • Anti addition occurs. • NBS in DMSO and H2O adds Br and OH in the same fashion. [5] Hydroboration–oxidation—Addition of H2O (10.16) • Hydroboration has a one-step mechanism. [1] BH3 or 9-BBN • No rearrangements occur. RCH CH2 R CH CH2 [2] H2O2, HO • OH bonds to the less substituted C. H OH • Syn addition of H2O results. alcohol

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Alkenes 10–3

Practice Test on Chapter Review 1. Give the IUPAC name for the following compounds.

a.

b.

2. Draw the organic products formed in the following reactions. Draw all stereogenic centers using wedges and dashes. CH2

a.

b.

HCl

[1] BH3 [2] H2O2, –OH

Br2 c.

d.

H2O H2SO4

3. Fill in the table with the stereochemistry observed in the reaction of an alkene with each reagent. Choose from syn, anti, or both syn and anti. Reagent a. [1] 9-BBN; [2] H2O2, –OH b. H2O, H2SO4 c. Cl2, H2O d. HI e. Br2

Stereochemistry

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Chapter 10–4

4. a. In which of the following reactions are carbocation rearrangements are observed? 1. hydrohalogenation 2. halohydrin formation 3. hydroboration–oxidation 4. Carbocation rearrangements are observed in reactions (1) and (2). 5. Carbocation rearrangements are observed in reactions (1), (2), and (3). b. Which of the following products are formed when HCl is added to 3-methyl-1-pentene? 1. 2-chloro-2-methylpentane 2. 3-chloro-3-methylpentane 3. 1-chloro-3-methylpentane 4. Both (1) and (2) are formed. 5. Products (1), (2), and (3) are all formed. Answers to Practice Test 1. a. (3E)-4-ethyl-2,5dimethyl-3-nonene b. (3E)-4-isopropyl-2methyl-3-octene

2. Cl

Cl

a.

b. HO

HO

c. Br

Br

Br

Br

d.

3. a. syn b. both c. anti d. both e. anti

OH

Answers to Problems 10.1 Six alkenes of molecular formula C5H10:

trans cis diastereomers

10.2 To determine the number of degrees of unsaturation: [1] Calculate the maximum number of H’s (2n + 2). [2] Subtract the actual number of H’s from the maximum number. [3] Divide by two.

4. a. 1 b. 2

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Alkenes 10–5

a. C2H2 [1] maximum number of H's = 2n + 2 = 2(2) + 2 = 6 [2] subtract actual from maximum = 6 – 2 = 4 [3] divide by two = 4/2 = 2 degrees of unsaturation b. C6H6 [1] maximum number of H's = 2n + 2 = 2(6) + 2 = 14 [2] subtract actual from maximum = 14 – 6 = 8 [3] divide by two = 8/2 = 4 degrees of unsaturation c. C8H18 [1] maximum number of H's = 2n + 2 = 2(8) + 2 = 18 [2] subtract actual from maximum = 18 – 18 = 0 [3] divide by two = 0/2 = 0 degrees of unsaturation

d. C7H8O Ignore the O. [1] maximum number of H's = 2n + 2 = 2(7) + 2 = 16 [2] subtract actual from maximum = 16 – 8 = 8 [3] divide by two = 8/2 = 4 degrees of unsaturation e. C7H11Br Because of Br, add one more H (11 + 1 H = 12 H's). [1] maximum number of H's = 2n + 2 = 2(7) + 2 = 16 [2] subtract actual from maximum = 16 – 12 = 4 [3] divide by two = 4/2 = 2 degrees of unsaturation f. C5H9N Because of N, subtract one H (9 – 1 H = 8 H's). [1] maximum number of H's = 2n + 2 = 2(5) + 2 = 12 [2] subtract actual from maximum = 12 – 8 = 4 [3] divide by two = 4/2 = 2 degrees of unsaturation

10.3 One possibility for C6H10: a. a compound that has 2 π bonds

c. a compound with 2 rings

b. a compound that has 1 ring and 1 π bond

d. a compound with 1 triple bond

10.4 To name an alkene: [1] Find the longest chain that contains the double bond. Change the ending from -ane to -ene. [2] Number the chain to give the double bond the lower number. The alkene is named by the first number. [3] Apply all other rules of nomenclature. To name a cycloalkene: [1] When a double bond is located in a ring, it is always located between C1 and C2. Omit the “1” in the name. Change the ending from -ane to -ene. [2] Number the ring clockwise or counterclockwise to give the first substituent the lower number. [3] Apply all other rules of nomenclature. 3-methyl 1 CH3

a. [1] CH2 CHCH(CH3)CH2CH3

[2] CH2 CHCHCH2CH3

5 C chain with double bond pentene

[3] 3-methyl-1-pentene

1-pentene

3 CH3CH2

b. [1] (CH3CH2)2C CHCH2CH2CH3 7 C chain with double bond heptene

C CHCH2CH2CH3

[2] CH3CH2

3-heptene 3-ethyl

[3] 3-ethyl-3-heptene

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Chapter 10–6

2-ethyl

c. [1]

[2]

[3] 2-ethyl-4-methyl-1-pentene

4-methyl 1 1-pentene

5 C chain with double bond pentene

1

[2]

d. [1]

4

[3] 3,4-dimethylcyclopentene

3

3,4-dimethyl

5 C ring with a double bond cyclopentene

1-methyl

e. [1]

[2]

[3] 5-tert-butyl-1-methylcyclohexene 1

5-tert-butyl 6 C ring with a double bond cyclohexene

10.5 Use the rules from Answer 10.4 to name the compounds. Enols are named to give the OH the lower number. Compounds with two C=C’s are named with the suffix -adiene.

a.

1

3

[2]

[1]

[3] 4-ethyl-3-hexen-1-ol

OH

OH

4-ethyl 6 C chain with double bond hexene [1]

OH

1

4

[2]

[3] 5-ethyl-6-methyl-7-octen-4-ol

OH

b. 6-methyl

5-ethyl 8 C chain with double bond octene [1]

5 [2]

2

c.

[3] 2,6-dimethyl-2,5-heptadiene 2-methyl

6-methyl 7 C chain with two double bonds heptadiene

10.6 To label an alkene as E or Z: [1] Assign priorities to the two substituents on each end using the rules for R,S nomenclature. [2] Assign E or Z depending on the location of the two higher priority groups. • The E prefix is used when the two higher priority groups are on opposite sides. • The Z prefix is used when the two higher priority groups are on the same side of the double bond.

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273

Alkenes 10–7

higher priority

Cl

CH3

a.

higher priority

C C

higher priority Two higher priority groups are on opposite sides: E isomer.

OCH3

Br

H

higher priority

H

H

c.

higher priority

C6H5

O H

higher priority

b.

higher priority

CH2CH3

CH3CH2

kavain

C C

In both double bonds, the two higher priority groups are on opposite sides: E isomers.

CH3

H

higher priority

Two higher priority groups are on the same side: Z isomer.

10.7 E

E

Z E

Z

skeletal structure of 11-cis-retinal

CHO

10.8 To work backwards from a name to a structure: [1] Find the parent name and functional group and draw, remembering that the double bond is between C1 and C2 for cycloalkenes. [2] Add the substituents to the appropriate carbons. a. (3Z)-4-ethyl-3-heptene

The higher priority groups are on the same side = Z.

c. (1Z)-2-bromo-1-iodo-1-hexene 6 carbons

7 carbons

The double bond is between C1 and C2. I Br

4-ethyl The double bond is between C3 and C4. b. (2E)-3,5,6-trimethyl-2-octene

The higher priority groups are on the same side = Z.

The double bond is between C2 and C3.

8 carbons The higher priority groups are on opposite sides = E.

3,5,6-trimethyl

10.9 Draw all of the stereoisomers and then use the rules from Answer 10.6 to name each diene.

E

E

(2E,4E)-2,4-hexadiene

E

Z

(2E,4Z)-2,4-hexadiene

Z

Z

(2Z,4Z)-2,4-hexadiene

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Chapter 10–8

10.10 To rank the isomers by increasing boiling point: Look for polarity differences: small net dipoles make an alkene more polar, giving it a higher boiling point than an alkene with no net dipole. Cis isomers have a higher boiling point than their trans isomers. CH3

CH3 CH3

H

CH3CH2

C C

CH3CH2

C C CH3

H

All dipoles cancel. smallest surface area no net dipole lowest bp

CH2CH3

C C CH2CH3

H

Two dipoles cancel. no net dipole trans isomer intermediate bp

H

Two dipoles reinforce. net dipole cis isomer highest bp

10.11 Increasing number of double bonds = decreasing melting point. O O

OH

stearic acid no double bonds highest melting point

OH

stearidonic acid

O

4 double bonds lowest melting point

OH

linolenic acid 3 double bonds intermediate melting point

10.12 Br

a.

H2SO4

NaOCH2CH3

b.

OH

CH2=CHCH2CH2CH2CH3

+ CH3CH=CHCH2CH2CH3

10.13 To draw the products of an addition reaction: [1] Locate the two bonds that will be broken in the reaction. Always break the π bond. [2] Draw the product by forming two new σ bonds. HCl

H

two new σ bonds

a.

CH3

c.

Cl

b.

CH3CH2CH2CH CHCH2CH2CH3

CH3

HCl

H HCl

CH3 CH3 Cl

H H CH3CH2CH2 C C CH2CH2CH3 H Cl

two new σ bonds

10.14 Addition reactions of HX occur in two steps: [1] The double bond attacks the H atom of HX to form a carbocation. [2] X– attacks the carbocation to form a C−X bond.

two new σ bonds

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Alkenes 10–9

transition state step [1]: H +

H

H

H

[1]

H Cl

transition state step [2]:

H

H

[2] Cl

H

Cl δ−

δ+

δ+ H

Cl

Cl δ−

10.15 Addition to alkenes follows Markovnikov’s rule: When HX adds to an unsymmetrical alkene, the H bonds to the C that has more H’s to begin with. 2 H's H adds here.

no H's Cl adds here. CH3

CH3 Cl

HCl

a.

c.

H H

H

CH2

HCl

Cl CH3

no H's Cl adds here.

one H H adds here.

H

no H's Cl adds here CH3

b.

CH3

HCl

C CH2

Cl C

CH2

CH3 H

CH3

2 H's H adds here.

10.16 To determine which alkene will react faster, draw the carbocation that forms in the ratedetermining step. The more stable, more substituted the carbocation, the lower the Ea to form it and the faster the reaction. CH3

CH3

H

CH3

CH3

C

C C

CH2 H

CH3

H H X

H

C C H

CH3CHCH2 H

H

2° carbocation slower reaction

H X

3° carbocation faster reaction

10.17 Look for rearrangements of a carbocation intermediate to explain these results.

H Cl

CH3

Cl

H

H

H H

Cl CH3

1-chloro-3methylcyclohexane

H CH3

H

H

2° carbocation

CH3 H

+ Cl–

2° carbocation

Rearrangement would not further stabilize this carbocation.

H

1,2-H shift Cl CH3

3° carbocation

CH3 Cl

1-chloro-1methylcyclohexane

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Chapter 10–10

10.18 Addition of HX to alkenes involves the formation of carbocation intermediates. Rearrangement of the carbocation will occur if it forms a more stable carbocation. H

a.

H CH3

CH3 C C

H

H CH3

H C C CH2CH3 CH2CH3

H C C CH2CH3

H

H Br

3° carbocation no rearrangement b.

CH3

H H

H

C C H

CH2CH3

CH3

H

H H

CH3 C C CH2CH3

CH3 C C CH2CH3

H Br

H

Br H

H H

H H

H

C C H

H H

CH3 C C CH2CH3

2° carbocation 2° carbocation Rearrangement would not further stabilize either carbocation. no rearrangement

(cis or trans)

c.

H H

CH3 C C CH2CH3

CH(CH3)2

CH3 C C

H

H

(cis or trans)

H H

1,2-H shift

CH3 C C CH(CH3)2

CH3 C C CH(CH3)2

2° carbocation

C CH3

H H

2° carbocation rearrangement

Rearrangement would not further stabilize the carbocation.

CH3

3° carbocation more stable

H H

H H

CH3 C C CH(CH3)2

CH3 C C

Br H

H H

CH3 C CH3 Br

10.19 To draw the products, remember that addition of HX proceeds via a carbocation intermediate. Addition of H+ (from HBr) from above and below gives an achiral, trigonal planar carbocation. a.

H H

CH3 CH3

CH3 CH3

Addition of Br– from above and below.

CH3

CH3

CH3

Br CH3

CH3 CH3

Br CH3

enantiomers

Addition of H+ (from HCl) from above and below by Markovnikov's rule forms an achiral 3° carbocation. b. CH3

CH3 CH3

CH3

Cl– attacks from above and below. CH3

H

CH3

CH3

CH3

Cl

H

achiral, trigonal planar 3° carbocation

diastereomers

CH3 Cl

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Alkenes 10–11

10.20 The product of syn addition will have H and Cl both up or down (both on wedges or both dashes), while the product of anti addition will have one up and one down (one wedge, one dash). CH3 H

CH3 H

CH3 H

CH3 H

Cl CH3

Cl CH3

Cl CH3

Cl CH3

A syn addition

B anti addition

C anti addition

D syn addition

10.21 CH3

CH2CH2CH3

CH3

or

CH2 C

a. CH3 C CH2CH2CH3 OH

CH3

H+ would add here to form a 3° carbocation. CH3 OH

b.

CH3

H+ would add here to form a 3° carbocation. or

c. OH

H+ would add here to form a 3° carbocation.

CH2

or

H+ would add here to form a 3° carbocation. H+

C CHCH2CH3 CH3

CH3CH CHCH3

would add here to form a 2° carbocation.

(cis or trans)

10.22 H

H2O H2SO4

1-pentene

OH

HO

H

enantiomers

10.23 Halogenation of an alkene adds two elements of X in an anti fashion. Br

Br2

a.

Br

Cl2

CH3 Cl

CH3 Cl

Cl

Cl

b. Br

Br

10.24 To draw the products of halogenation of an alkene, remember that the halogen adds to both ends of the double bond but only anti addition occurs.

a.

Cl

Cl

Cl CH3

Cl CH3

Cl2

enantiomers

H

b. C H

Br2 Br H

C C

C

H Br

achiral meso compound

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Chapter 10–12

c.

Br2

CH3

CH3

CH3 Br

Br

Br

Br

diastereomers

10.25 The two steps in the mechanism for the halogenation of an alkene are: [1] Addition of X+ to the alkene to form a bridged halonium ion [2] Nucleophilic attack by X– CH3

trans-2-butene

C C

H

Addition of Br+ can occur from above or below:

Br Br

CH3

overall

CH3 H

H

Br

H C

H

enantiomers

C

H

Attack of Br– can occur from the left or right:

CH3

below

Br C

CH3

C C

Br Br

above

CH3

H

Br

H

C CH3

Br

CH3

right

left

left

right + Br

Br

CH3

C

H

H

C

CH3

CH3

H

Br C

C

+ Br CH3 H C Br

CH3

CH3

C

H

H

C

CH3

CH3

Br

C

H

C Br

H CH3

+ Br

Br C

H

+ Br

Br

H C

H CH3

CH3 H

C

Br CH3

H C

C

CH3 H

Br

CH3 Br

CH3 H C Br

Br C

H CH3

All four compounds are identical—an achiral meso compound.

10.26 Halohydrin formation adds the elements of X and OH across the double bond in an anti fashion. The reaction is regioselective so X ends up on the carbon that had more H’s to begin with. a.

Br

NBS DMSO, H2O

HO

CH3

b. HO

Cl2 H2O

Br

CH3 OH

CH3 OH

Cl

Cl

Cl bonds to the carbon with more H's to begin with.

10.27 H

a. (CH3)2S

H

H CH2CH3

CH3

H B S

b. (CH3CH2)3N CH3

H B N CH2CH3 H CH2CH3

H CH2CH2CH2CH3

c. (CH3CH2CH2CH2)3P

H B P CH2CH2CH2CH3 H CH2CH2CH2CH3

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Alkenes 10–13

10.28 In hydroboration the boron atom is the electrophile and becomes bonded to the carbon atom that had more H’s to begin with. CH3 C CH2

a.

CH3

BH3

CH3 C CH2

CH3

C with more H's. B will add here.

CH2BH2

C with more H's. B will add here.

BH3

b.

BH3

CH2

c.

H BH2

BH2 H

C with more H's. B will add here.

10.29 The hydroboration–oxidation reaction occurs in two steps: [1] Syn addition of BH3, with the boron on the less substituted carbon atom [2] OH replaces the BH2 with retention of configuration. a. CH3CH2CH CH2

BH3

CH3CH2CH CH2 H

CH2CH3

b.

c. CH3

CH3

H2O2, –OH

CH3CH2CH CH2

BH2

H

OH

CH2CH3 H

CH2CH3 H2O2, –OH H

CH2CH3 H

CH2CH3 H

BH2

BH2

OH

OH

BH3

CH3

CH3

H

CH3 H

CH3

BH2

BH2 H2O2, –OH

CH3 H

CH3

CH3

CH3

H

OH

OH

10.30 Remember that hydroboration results in addition of OH on the less substituted C. OH HO

a.

c.

b.

(E or Z isomer can be used.)

OH

10.31 a.

CH2

CH2

H2O H2SO4 [1] BH3 [2] H2O2, HO

OH CH3 CH2OH

Hydration places the OH on the more substituted carbon. Hydroboration–oxidation places the OH on the less substituted carbon.

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Chapter 10–14

OH

Hydration places the OH on the more substituted carbon.

H2O H2SO4

b.

OH

[1] BH3 [2] H2O2, HO OH

H2 O H2SO4

c.

Hydroboration–oxidation places the OH on the less substituted carbon.

[1] BH3 [2] H2O2, HO

Hydration places the OH on the more substituted carbon.

Hydroboration–oxidation places the OH on the less substituted carbon.

HO

10.32 There are always two steps in this kind of question: [1] Identify the functional group and decide what types of reactions it undergoes (e.g., substitution, elimination, or addition). [2] Look at the reagent and determine if it is an electrophile, nucleophile, acid, or base. acid: catalyzes loss of H2O

acid

CH2

a.

HBr

CH3

CH2CH2CH3

c.

Br

alkene: addition reactions

H2SO4

CHCH2CH3

OH

CH2CH2CH3

alcohol: substitution and elimination

nucleophile and base Cl

OCH3 NaOCH3

b.

CH3CH2CH=CHCH3

2° alkyl halide: substitution and elimination

(cis and trans)

10.33 To devise a synthesis: [1] Look at the starting material and decide what reactions it can undergo. [2] Look at the product and decide what reactions could make it. CH3

? Br

a.

Cl

alkyl halide Can undergo substitution and elimination. Br

K+ –OC(CH3)3

halohydrin: Can form from an alkene with Cl2 and H2O. Cl2 H2 O

Cl OH

OH is added to the more substituted C.

CH3

?

OH

b.

OH

OH

alcohol Can undergo substitution and elimination. CH3 OH

H2SO4

alcohol Can be formed by substitution and addition. CH3 [1] BH 3 [2] H2O2, HO–

(major product)

CH3 OH

OH is added to the less substituted C.

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281

Alkenes 10–15

10.34 Convert each ball-and-stick model to a skeletal structure and then name the molecule. 2

3

6 C chain with a C=C hexene C=C at C2 2-hexene 2 CH3's at C3 and C5 two higher priority groups on opposite sides Answer: (2E)-3,5-dimethyl-2-hexene

5

a.

b.

E isomer

5 C ring with a C=C cyclopentene sec-butyl at C1 methyl at C2 Answer: 1-sec-butyl-2-methylcyclopentene

2 1

10.35 H

a.

The two higher priority groups are on the same side of the C=C, making it a Z alkene.

higher priority H

higher priority

A b. Add H2O in a Markovnikov fashion to form two products. OH

OH

H

H

H2 O H2SO4

H

10.36 C1 8-methyl-1-nonene C8 a.

b.

Br2

Br2

Br Br

OH Br

H2O c.

Br2 CH3OH

OCH3 Br

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Chapter 10–16

10.37 Use the directions from Answer 10.2 to calculate degrees of unsaturation. a. C3H4 [1] maximum number of H's = 2n + 2 = 2(3) + 2 = 8 [2] subtract actual from maximum = 8 – 4 = 4 [3] divide by 2 = 4/2 = 2 degrees of unsaturation

f. C8H9Br Because of Br, add one H (9 + 1 = 10 H's). [1] maximum number of H's = 2n + 2 = 2(8) + 2 = 18 [2] subtract actual from maximum = 18 – 10 = 8 [3] divide by 2 = 8/2 = 4 degrees of unsaturation

b. C6H8 [1] maximum number of H's = 2n + 2 = 2(6) + 2 = 14 [2] subtract actual from maximum = 14 – 8 = 6 [3] divide by 2 = 6/2 = 3 degrees of unsaturation

g. C8H9ClO Ignore the O; count Cl as one more H (9 + 1 = 10 H's). [1] maximum number of H's = 2n + 2 = 2(8) + 2 = 18 [2] subtract actual from maximum = 18 – 10 = 8 [3] divide by 2 = 8/2 = 4 degrees of unsaturation

c. C40H56 [1] maximum number of H's = 2n + 2 = 2(40) + 2 = 82 [2] subtract actual from maximum = 82 – 56 = 26 [3] divide by 2 = 26/2 = 13 degrees of unsaturation d. C8H8O Ignore the O. [1] maximum number of H's = 2n + 2 = 2(8) + 2 = 18 [2] subtract actual from maximum = 18 – 8 = 10 [3] divide by 2 = 10/2 = 5 degrees of unsaturation e. C10H16O2 Ignore both O's. [1] maximum number of H's = 2n + 2 = 2(10) + 2 = 22 [2] subtract actual from maximum = 22 – 16 = 6 [3] divide by 2 = 6/2 = 3 degrees of unsaturation

h. C7H9Br Because of Br, add one H (9 + 1 = 10 H's). [1] maximum number of H's = 2n + 2 = 2(7) + 2 = 16 [2] subtract actual from maximum = 16 –10 = 6 [3] divide by 2 = 6/2 = 3 degrees of unsaturation i. C7H11N Because of N, subtract one H (11 – 1 = 10 H's). [1] maximum number of H's = 2n + 2 = 2(7) + 2 = 16 [2] subtract actual from maximum = 16 – 10 = 6 [3] divide by 2 = 6/2 = 3 degrees of unsaturation j. C4H8BrN Because of Br, add one H, but subtract one for N (8 + 1 – 1 = 8 H's). [1] maximum number of H's = 2n + 2 = 2(4) + 2 = 10 [2] subtract actual from maximum = 10 – 8 = 2 [3] divide by 2 = 2/2 = 1 degree of unsaturation

10.38 First determine the number of degrees of unsaturation in the compound. Then decide which combinations of rings and π bonds could exist. C10H14 [1] maximum number of H's = 2n + 2 = 2(10) + 2 = 22 [2] subtract actual from maximum = 22 – 14 = 8 [3] divide by two = 8/2 = 4 degrees of unsaturation

possibilities: 4 π bonds 3 π bonds + 1 ring 2 π bonds + 2 rings 1 π bond + 3 rings 4 rings

10.39 The statement is incorrect because when naming isomers with more than two groups on a double bond, one must use an E,Z label, rather than a cis, trans label. higher priority

Cl

Cl

higher priority higher priority

higher priority O

enclomiphene E isomer

N(CH2CH3)2

O

zuclomiphene Z isomer

N(CH2CH3)2

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283

Alkenes 10–17

10.40 Name the alkenes using the rules in Answers 10.4 and 10.6. CH3

a.

CH2=CHCH2CH(CH3)CH2CH3

CH2 CHCH2CHCH2CH3

6 C chain with a double bond = hexene

1-hexene 4-methyl

4-methyl-1-hexene

5-ethyl

2-methyl b.

5-ethyl-2-methyl-2-octene 2-octene

8 C chain with a double bond = octene 2-isopropyl 2-isopropyl-4-methyl-1-pentene

c. 1-pentene 5 C chain with a double bond = pentene

4-methyl

d. 1-ethyl 5-isopropyl

6 C ring with a double bond = cyclohexene

E double bond (higher priority groups on opposite sides with bold bonds)

4-isopropyl

e.

1-ethyl-5-isopropylcyclohexene

3-ol

OH

(4E)-4-isopropyl-4-hepten-3-ol OH

7 C chain with a double bond = heptene

4-heptene

2-cyclohexene 5-sec-butyl-2-cyclohexenol 1-ol

f. OH

6 C ring with a double bond = cyclohexene

OH

5-sec-butyl

10.41 Use the directions from Answer 10.8.

a. (3E)-4-ethyl-3-heptene 7 carbons

4-ethyl 3 Higher priority groups on opposite sides = E.

b. 3,3-dimethylcyclopentene 5 carbon ring

3,3-dimethyl 1

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Chapter 10–18

c. cis-4-octene 8 carbons

1

f. cis-3,4-dimethylcyclopentene 4 Higher priority groups on the same side = cis. 2

d. 4-vinylcyclopentene 5 carbon ring

5 carbon ring

or 3

4

3,4-dimethyl g. trans-2-heptene

1

4

2 Higher priority groups on opposite sides = trans.

7 carbons 2

h. 1-isopropyl-4-propylcyclohexene

e. (2Z)-3-isopropyl-2-heptene 3-isopropyl 7 carbons

6 carbon ring 1-isopropyl

Higher priority groups on the same side = Z.

4-propyl

10.42 a.

(2E,4S)-4-methyl-2-nonene (2E,4R)-4-methyl-2-nonene (2Z,4S)-4-methyl-2-nonene A B C b. A and B are enantiomers. C and D are enantiomers. c. Pairs of diastereomers: A and C, A and D, B and C, B and D.

(2Z,4R)-4-methyl-2-nonene D

10.43 CH3

CH3

a.

CH3

H

c.

(1Z,4S)-1,4-dimethylcyclodecene diastereomer

(1E,4R)-1,4-dimethylcyclodecene CH3

CH3 CH3

b.

H

(1E,4S)-1,4-dimethylcyclodecene enantiomer

(1Z,4R)-1,4-dimethylcyclodecene diastereomer

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285

Alkenes 10–19

10.44 Name the alkene from which the epoxide can be derived and add the word oxide. 1-ethyl H

derived

a.

O

H

O

derived

c.

from

from

H

6 carbon ring cyclohexene 1-ethylcyclohexene 1-ethylcyclohexene oxide

(3E)- 3-heptene oxide or trans-3-heptene oxide

2-methyl O

b.

derived

derived

d.

O

C(CH3)3

C(CH3)3

from

from

2-methyl-2-hexene oxide

H

7 carbon chain heptene (3E)- 3-heptene or trans-3-heptene

5 carbon ring cyclopentene 4-tert-butylcyclopentene oxide 4-tert-butylcyclopentene

6 carbon chain hexene 2-methyl-2-hexene

10.45

2-sec-butyl

a. 2-butyl-3-methyl-1-pentene As written, this is the parent chain, but there is another longer chain containing the double bond.

new name: 2-sec-butyl-1-hexene

b. (Z)-2-methyl-2-hexene new name: 2-methyl-2-hexene Two groups on one end of the C=C are the same (2 CH3's), so no E and Z isomers are possible. 1 c. (E)-1-isopropyl-1-butene new name: (3E)-2-methyl-3-hexene

As written, this is the parent chain, but there is another longer chain containing the double bond.

1

d. 5-methylcyclohexene

4 As written the methyl is at C5. Re-number to put it at C4. e. 4-isobutyl-2-methylcylohexene

new name: 4-methylcyclohexene

2

1 5 As written this methyl is at C2. Re-number to put it at C1.

1

new name: 5-isobutyl-1-methylcyclohexene

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Chapter 10–20

f.

1-sec-butyl-2-cyclopentene

3 1

1 3

2

new name: 3-sec-butylcyclopentene

This has the double bond between C2 and C3. Cycloalkenes must have the double bond between C1 and C2. Re-number. g.

OH

1-cyclohexen-4-ol

OH

3

1 The numbering is incorrect. When a compound contains both a double bond and an OH group, number the C skeleton to give the OH group the lower number.

3-cyclohexenol (The "1" can be omitted.)

OH

h.

OH

6

3-ethyl-3-octen-5-ol The numbering is incorrect. When a compound contains both a double bond and an OH group, number the C skeleton to give the OH group the lower number.

4 5 6-ethyl-5-octen-4-ol

10.46 a, b. E,Z, and R,S designations are shown. CH3O

E

E

R E

S

S

E

O

O

N H

O

Z

H N

CHO

OH

OCH3

E

S

E S

S

c. Nine double bonds that can be E or Z Six tetrahedral stereogenic centers Maximum possible number of stereoisomers = 215 10.47 O

stearic acid

OH

hIghest melting point no double bonds

OH

intermediate melting point one E double bond

O

elaidic acid O

oleic acid

OH

lowest melting point one Z double bond

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Alkenes 10–21

10.48 O

a.

O

OH

all trans double bonds higher melting point

OH

O

b.

eleostearic acid

OH

all cis double bonds lower melting point

10.49 The more negative the H°, the larger the Keq assuming entropy changes are comparable. Calculate the H° for each reaction and compare. CH2 CH2 + HI

CH3CH2

[1] Bonds broken

I

[2] Bonds formed

ΔHo (kJ/mol) C–C π bond

+ 267

I

+ 297

H

Total

ΔHo (kJ/mol) CH2ICH2 H

− 410

I

− 222

C

+ 564 kJ/mol

CH2 CH2 + HCl

− 632 kJ/mol

Total

[3] Overall ΔHo = sum in Step [1] + sum in Step [2] + 564 kJ/mol − 632

kJ/mol

– 68

kJ/mol

CH3CH2 Cl

[1] Bonds broken

[2] Bonds formed

[3] Overall ΔHo =

C–C π bond

+ 267

CH2ClCH2 H

− 410

sum in Step [1] + sum in Step [2]

H Cl

+ 431

C Cl

− 339

+ 698 kJ/mol

+ 698 kJ/mol

Total

− 749 kJ/mol

ΔHo (kJ/mol)

Total

ΔHo (kJ/mol)

− 749

kJ/mol

– 51 kJ/mol Compare the ΔH°: Addition of HI: –68 kJ/mol = more negative ΔH°, larger Keq Addition of HCl: –51 kJ/mol

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Chapter 10–22

10.50 H

HBr

a.

Br

Br2, H2O

f. Br H

HI

b.

NBS

g.

H

H2O H2SO4

H

H2SO4

Br OH

OH H

[1] 9-BBN [2] H2O2, HO–

OCH2CH3

Cl2

e.

[2] H2O2, HO–

i.

Br

H

OH

CH3CH2OH

d.

OH

OH

[1] BH3

h.

c.

OH

+

aqueous DMSO

I

Br +

OH

Cl +

Cl

Cl

Cl

10.51 CH3

a.

HBr

C CH2

HI

C CH2

c.

C CH2

C CH2

H2SO4 CH3CH2OH

C CH2

h.

OH

CH3

NBS

CH3 C CH2Br

aqueous DMSO

OH CH3

[2] H2O2, HO–

CH3 C CH2OH H

CH3

CH3 C CH3

C CH2

i.

OCH2CH3

CH3

CH3

Cl2

OH

[1] BH3

C CH2 CH3

CH3

CH3 CH3

CH3

CH3 C CH3

H2SO4

CH3

e.

C CH2 CH3

CH3 CH3

d.

g.

CH3

H2O

CH3 C CH2Br

CH3

CH3 C CH3 I

CH3

Br2, H2O

CH3

CH3

CH3

CH3

C CH2

f.

Br

CH3 CH3

b.

CH3

CH3 CH3 C CH3

CH3

[1] 9-BBN [2] H2O2, HO–

CH3 C CH2OH H

CH3 C CH2Cl Cl

10.52 Br

a.

Br

d.

CH3CH2CH2CH=CHCH3 Br

Cl

b.

e.

CH3

c.

C CH3 Br

CH3

CH2 or

Cl

Cl CH3 C CH2

CH3CH2

f.

(CH3CH2)3CBr

C CHCH3 CH3CH2

CH3

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Alkenes 10–23

10.53 Hydroboration–oxidation results in addition of an OH group on the less substituted carbon, whereas acid-catalyzed addition of H2O results in the addition of an OH group on the more substituted carbon. a.

OH

c.

hydroboration–oxidation and acid-catalyzed addition

e.

OH

acid-catalyzed addition

OH

or

OH

Both methods would give product mixtures.

OH

b.

d. hydroboration–oxidation

hydroboration–oxidation

10.54 a.

Cl (CH3CH2)2C=CHCH2CH3 CH3CH2

H

HCl

C C CH3CH2

CH3CH2 H CH3CH2 C Cl

CH2CH3

Cl

C H CH2CH3

e.

Br2

CH2Br OH

Br adds here to less substituted C.

(CH3CH2)2C=CH2 CH3CH2 CH3CH2

[1] 9-BBN

CH3CH2

H2O

C CH2

H2SO4

f.

CH3CH2 C CH3 OH

[1] BH3

(CH3)2C=CHCH3

CH2

CH2OH

[2] H2O2, HO−

OH adds here to less substituted C.

H adds here to less substituted C. c.

CH2

H2O

H adds here to less substituted C. b.

Cl2

d.

H (CH3)2C

DMSO, H2O

CHCH3

H (CH3)2C

OH Br

Br adds here to less substituted C.

[2] H2O2, HO−

BH3 adds here to less substituted C.

NBS

g.

BH2

OH

h.

CHCH3

Br2 Br Br

10.55 CH3CH2

H

CH3CH2CH2

CH2CH2CH3

or

C C H

or

CH3CH C CH3

CH3 C CHCH2CH3 CH3CH2

Cl CH3CH2 C CH2CH2CH3 CH3

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Chapter 10–24

10.56 H

a.

CH3

CH3

H

C C

C C

=

H

C

H

CH3

b.

CH3

CH3CH2

H

CH3CH2

CH3

=

CH3

CH3 C C

CH3

H

H2O

H

CH3CH2

NBS

HO

CH3 H

CH3 H C

CH3CH2 CH3

CH2

H 2O

CH 3

(CH3)3C

H2SO 4

I H

CH3

c.

CH3

CH3

CH3

Cl

[1] BH3 CH3

OH

Br CH2CH3

CH3 H CH3

only syn addition

OH

Br CH2CH3

HBr

e.

Cl

Cl2

OH

only anti addition

+

H2O OH NBS

h.

only anti addition

CH3

CH3 H

[2] H2O2, HO−

CH3

I

Cl CH3

Cl

d.

H

Cl Cl2

CH3

g.

OH

HI

b.

f.

CH3

(CH 3)3C

OH

DMSO, H2O

CH3CH=CHCH2CH3

H2O H2SO4

Cl Br

Br HO CH3 H OH

HO CH3 HO H OH

CH3 H

HO

10.57 a. (CH3)3C

C

HO

Br C

Cl

CH3 CH3CH2 C

Cl

CH3CH2 CH3 C

C Br

C

CH3 CH3CH2

C

CH3

CH3 H C

DMSO, H2O

H

H

HO

Cl2

C C

CH3 CH3CH2

C C

c.

CH3

=

C C

H

C

Br

CH3 H

Br

Br

H

Br2

C Br

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Alkenes 10–25

10.58 Draw each reaction. (a) The cis isomer of 4-octene gives two enantiomers on addition of Br2. (b) The trans isomer gives a meso compound. H

H

H

Br

Br2

a.

H

Br

H

H Br

cis-4-octene

Br

(4R,5R)-4,5-dibromooctane

(4S,5S)-4,5-dibromooctane

enantiomers

H

H

Br

Br2

b.

H

H

Br

trans-4-octene

(4R,5S)-4,5-dibromooctane meso compound

10.59 CH3CH2

CH3CH2

CH2CH3

C C H

H

H

H Cl CH3CH2

CH3CH2

CH2CH3

H

C Cl

H

H

Cl H

By protonation of the alkene, the cis and trans isomers produce identical carbocation intermediates.

CH2CH3 Cl–

CH3CH2

CH2CH2CH3

H C C H

Cl– CH3CH2

CH2CH3 H Cl

C C H H

H

C C

C

CH3CH2 C

CH2CH2CH3

H

Cl

CH3CH2 C

CH2CH2CH3

CH2CH2CH3

Cl H

Both cis- and trans-3-hexene give the same racemic mixture of products, so the reaction is not stereospecific.

10.60 a.

Cl H

H

H Cl

H

Cl

H

H

H H

and O

H H H

HO

C

Cl

O C H H

CH3 H

O

O

O C CH3

H H H

CH3 + HCl

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Chapter 10–26

H

H Br

b.

H

Br

H

1,2-H shift

Br

C H CH3

C CH3

2o carbocation

3o carbocation

+ Br–

10.61 a.

H CH3

CH3

CH3

1,2-shift

H2O

CH3

CH3

CH3

O H – CH3 HSO4

OH CH3

H2SO4 HSO4–

H OSO3H

H2O

2o carbocation

3o carbocation

+ H2SO4

H OSO3H

b.

OH

OH

+ H2SO4

H2SO4

O HSO4–

CH3

O

CH3

H –OSO

3H

10.62 H

H OH2

H + H3O+ O H

H2O

OH H2O

H

H O H

1,2-shift

H2O

H2O

H

H

OH + H3O+

H2O

1,2-shift

O H

H2O

OH

H

H + H3O+

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Alkenes 10–27

10.63 The isomerization reaction occurs by protonation and deprotonation. CH3

H

CH2

OSO3H

CH3 C C H

CH3

CH3

C

CH3

CH3 C H CH3

CH3

CH3

C C

HSO4–

CH3

2,3-dimethyl-1-butene

CH3

2,3-dimethyl-2-butene

10.64 H

H Br

H C C

H

H

H

CH3 C

C C H

H

CH3CH CHCH2

C C H

H

Br

CH3CH=CH–CH2Br

H

Br

Since two resonance structures can be drawn for the intermediate carbocation, two different products result from attack by Br–.

Br CH3CHCH CH2

10.65 Br HBr

A

OCH3

OCH3

B

H Br

COOCH3

D

H Br

Br

O OCH3

OCH3

H

Br

HBr

COOCH3

C

H

This carbocation is resonance stabilized by the O atom, and therefore preferentially forms and results in B.

C

H O

CH3

O C

CH3

δ+ O H This carbocation is destabilized by the δ+ on the adjacent C, so it does not form. This carbocation is formed preferentially and results in product D. It is not destabilized by an adjacent electron-withdrawing COOCH3 group.

10.66 Br Br

H OH

Br

OH

O

O

Br

+ HBr + Br

+ Br

Br

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Chapter 10–28

10.67 OH PBr3

OH

K+ –OC(CH3)3

Br

H2O + H2SO4

a. OH adds to more substituted C.

POCl3, pyridine Br

b.

CH3 C CH3

K+ –OC(CH3)3

Br

Br2

CH3 CH CH2

CH3 C CH2Br

H

H

OH

H2SO4

c.

[1] BH3 [2] H2O2,

d.

CH3 CH CH2I

NaH

OH

HO–

+ H2 CH3

K+ –OC(CH3)3

(CH3)2C

CH2

HCl

CH3 C Cl

CH3

CH3

Cl

e.

Br

Br2

K+ –OC(CH3)3

H2O

f.

CH3I

O–

CH3CH CH2

Br2

Br

OH

Br

CH3CH CH2

NaNH2 (2 equiv)

CH3C CH

10.68 Cl2

a.

H2 O

HBr

b. Br

Cl

O O

OH

+ H2 Br

CN NaCN

OH

NaH

NaOH

c.

Cl

Na+ H–

(from b.) [1] –SH

d.

O

[2] H2O

OH

+ enantiomer SH

(from a.) e.

O

(from a.)

[1] –C≡CH

OH

[2] H2O

C

+ enantiomer CH

O

O CH3CH2CH2I

OCH3

295

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Alkenes 10–29

10.69 Br

Br K+ –OC(CH3)3

a.

d.

Br

NaNH2

C CH

(2 equiv) Br Br2

b.

Br

(from b.) OH

OH

Br2

c.

O

NaH

Br

e.

(from a.) Br

(from c.)

H2O

(from a.)

10.70 OH

Br OH POCl3

Br2

pyridine

A

O NaH

H2O

B

major product

H2SO4 OH Br2

NaH O

H2O

Br

C

10.71 Having two rings joined together as in A and B creates a very rigid ring system and constrains bond angles. Evidently the bond angles around the C=C in A are close enough to the trigonal planar bond angle of 120° so that A is stable. With B, however, the C=C is located at a carbon shared by both rings and the bond angles around the C=C deviate greatly from the desired angle, so that B is not a stable compound.

296

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Chapter 10–30

10.72 a. Br2 adds in an anti fashion to form a meso dibromide. Rotate around the C–C bond to place H and Br anti periplanar in the second step. HBr can be eliminated in two ways, but both give the same product. C6H5

H C C C6H5

H

Br Br2

C

H C6H5

anti

C

Br

C6H5 H

rotate

C

C

H C6H5

Br

Br C6H5

H C

or

C

C6H5 Br

H

C6H5 H Br

H and Br must be anti.

meso compound

KOH (–HBr) Br

H C C

C6H5

C6H5

Two higher priority groups are on opposite sides = E.

b. Br2 addition forms two enantiomers. Anti periplanar elimination of H and Br gives the same alkene from both compounds. Br

H

H

Br2

C C C6H5

H C6H5

C6H5 anti

H C

C6H5

C

or

C H C6H5

Br

Br

H C6H5

C

Br

C6H5

two enantiomers rotate

Br

H

C6H5 Br

H C

C

C

Br

rotate H

H

C6H5 Br

C6H5

C

C

C6H5 Br

KOH (–HBr)

H

Br C H C6H5

Br

C6H5

H

C C C6H5

C6H5 Br C

KOH (–HBr) C6H5

H

or

C

Br

H

H C6H5

C C

identical

Z isomer

Br

C6H5

Z isomer

c. The products in (a) and (b) are diastereomers. 10.73 TsO H

H

TsO–

+ TsOH

A

+ TsO–

isocomene

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Alkenes 10–31

10.74 Na+ HCO3– O

H O

O

O

O

O

O

O I

I

I

OCH3 HO

B

HO

HO Na+

I

OCH3

OCH3

OCH3 HO

I

C

H2CO3

10.75

OH

OH2 H OSO3H

H

2° carbocation

+ HSO4–

H2O

1,2-H shift H

HSO4–

H

3° carbocation

+ H2SO4

10.76

TsO H OH

H2O

OH2

nerol OTs

+ TsOH H2O O H H H

OH

OH H OSO2Cl

OTs

OSO2Cl

OH

OH

α-cyclogeraniol

nerol OSO2Cl

OH

α-terpineol

HSO3Cl

298

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Chapter 10–32

10.77 O

H OSO3H

O H OH

+ HSO4

–

H2 O

H O H

HSO4–

OH

HO

OH

+ H2SO4

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299

Alkynes 11–1

Chapter 11 Alkynes Chapter Review General facts about alkynes • Alkynes contain a carbon–carbon triple bond consisting of a strong σ bond and two weak π bonds. Each carbon is sp hybridized and linear (11.1). 180o H C C H

=

acetylene

sp hybridized

• • •

Alkynes are named using the suffix -yne (11.2). Alkynes have weak intermolecular forces, giving them low mp’s and low bp’s, and making them water insoluble (11.3). Since its weaker π bonds make an alkyne electron rich, alkynes undergo addition reactions with electrophiles (11.6).

Addition reactions of alkynes [1] Hydrohalogenation—Addition of HX (X = Cl, Br, or I) (11.7) •

X H

H X

R C C H

R C C H

(2 equiv)

X H

Markovnikov’s rule is followed. H bonds to the less substituted C in order to form the more stable carbocation.

geminal dihalide

[2] Halogenation—Addition of X2 (X = Cl or Br) (11.8) •

X X

X X

R C C H

R C C H

(2 equiv)

•

X X

Bridged halonium ions are formed as intermediates. Anti addition of X2 occurs.

tetrahalide

[3] Hydration—Addition of H2O (11.9)

R C C H

H2O H2SO4 HgSO4

H

R C C

H

HO

enol

•

O R

C

CH3

ketone

•

Markovnikov’s rule is followed. H bonds to the less substituted C in order to form the more stable carbocation. The unstable enol that is first formed rearranges to a carbonyl group.

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Chapter 11–2

[4] Hydroboration–oxidation—Addition of H2O (11.10)

R C C H

R

C C

[2] H2O2, HO

H

•

O

H

R

[1] R2BH

OH

C

C

H

H H

enol

The unstable enol, first formed after oxidation, rearranges to a carbonyl group.

aldehyde

Reactions involving acetylide anions [1] Formation of acetylide anions from terminal alkynes (11.6B) +

R C C H

B

R C C

+

+

HB

•

Typical bases used for the reaction are NaNH2 and NaH.

[2] Reaction of acetylide anions with alkyl halides (11.11A) H C C

+

R X

H C C R + X

• •

The reaction follows an SN2 mechanism. The reaction works best with CH3X and RCH2X.

[3] Reaction of acetylide anions with epoxides (11.11B) [1] O H C C CH2CH2OH

H C C

[2] H2O

• •

The reaction follows an SN2 mechanism. Ring opening occurs from the back side at the less substituted end of the epoxide.

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

301

Alkynes 11–3

Practice Test on Chapter Review 1. Draw the structure for the compound with the following IUPAC name: 5-tert-butyl-6,6-dimethyl3-nonyne. 2. a. Which of the following compounds is an enol tautomer of compound A? O

OH

OH

1.

2.

OH

3.

A

4. A can be a tautomer of both compounds (1) and (2). 5. A can be a tautomer of compounds (1), (2), and (3). b. Which of the following bases is strong enough to deprotonate CH3C≡CH (propyne, pKa = 25)? The pKa’s of the conjugate acids of the bases are given in parentheses. 1. 2. 3. 4. 5.

CH3Li (pKa = 50) NaOCH3 (pKa = 15.5) NaOCOCH3 (pKa = 4.8) The bases in (1) and (2) are both strong enough. The bases in (1), (2), and (3) are all strong enough.

3. Draw the organic products formed in the following reactions. [1] R2BH

a.

[2] H2O2, –OH

b.

[1] NaH C

C

H

[2] O [3] H2O HO H

c.

d.

D

C

C

H

[1] TsCl, pyridine [2]

C

CH

H2O H2SO4 HgSO4

e.

C

C

H

2 HBr

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Chapter 11–4

4. Draw two different enol tautomers for the following compound. O

OH

5. What acetylide anion and alkyl halide are needed to make the following alkyne? CH3 CH3 C C

C

CH2CH3

CH3

Answers to Practice Test 1. H CH3CH2 C C C (CH3)3C

2. a. 2 b. 1

CH3 C

4.

3.

HO

a.

CH2CH2CH3

H

CH3

O

b.

OH C

C

(E + Z)

CH2CH2OH

C D

HC

c.

HO

H

OH

(E + Z) CH3

d.

5.

C

CH3

O Br H

e.

C

C

Br

H

CH3X H

C C

C CH3

CH2CH3

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

303

Alkynes 11–5

Answers to Problems 11.1 • An internal alkyne has the triple bond somewhere in the middle of the carbon chain. • A terminal alkyne has the triple bond at the end of the carbon chain. HC C CH2CH 2CH 3

CH3 C C CH2CH 3

HC C CH CH3 CH3

terminal alkyne

internal alkyne

terminal alkyne

11.2 Csp–Csp3 (c)

Csp2–Csp (b)

OH

(a) Csp3–Csp2

11.3

O

santalbic acid

To name an alkyne: [1] Find the longest chain that contains both atoms of the triple bond, change the -ane ending of the parent name to -yne and number the chain to give the first carbon of the triple bond the lower number. [2] Name all substituents following the other rules of nomenclature. CH2CH2CH3

a.

Increasing bond strength: a < c < b

H C C CH2CCH2CH2CH3

CH2=CHCH2CHC

CH2CH2CH3

CCCH2CH2CH3 number to the first site of

unsaturation.)

CH3

4,4-dipropyl-1-heptyne b. CH3C CCClCH2CH3

(Number to give the lower

CH2CH3 CH3

c.

4-ethyl-7,7-dimethyl-1-decen-5-yne d.

1 5

CH3

3

(The longest chain must contain both functional groups.)

4-chloro-4-methyl-2-hexyne 3-isopropyl-1,5-octadiyne

11.4

To work backwards from a name to a structure: [1] Find the parent name and the functional group. [2] Add the substituents to the appropriate carbon. a. trans-2-ethynylcyclopentanol 5 C ring with OH at C1

b. 4-tert-butyl-5-decyne

OH

OH

OH on C1 C CH

ethynyl at C2

tert-butyl at C4

10 C chain with a triple bond triple bond at C5

or

C CH

304

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 11–6

c. 3-methylcyclononyne 9 C ring with a triple bond at C1

3-methyl

triple bond at C1

11.5

Two factors cause the boiling point increase. The linear sp hybridized C’s of the alkyne allow for more van der Waals attraction between alkyne molecules. Also, since a triple bond is more polarizable than a double bond, this increases the van der Waals forces between two molecules as well.

11.6

To convert an alkene to an alkyne: [1] Make a vicinal dihalide from the alkene by addition of X2. [2] Add base to remove two equivalents of HX and form the alkyne. a. Br2CH(CH2)4CH3

Na+ –NH2

Na+ –NH2

BrCH CHCH2CH2CH2CH3

HC CCH2CH2CH2CH3

not isolated b. CH2=CCl(CH2)3CH3

Na+ –NH2 Cl2

c. CH2=CH(CH2)3CH3

HC CCH2CH2CH2CH3 CH2CHCH2CH2CH2CH3 Cl Cl

11.7

HC CCH2CH2CH2CH3

(2 equiv)

Acetylene has a pKa of 25, so bases having a conjugate acid with a pKa above 25 will be able to deprotonate it. a. CH3NH– [pKa (CH3NH2) = 40] pKa > 25 = Can deprotonate acetylene. b. CO32– [pKa (HCO3–) = 10.2] pKa < 25 = Cannot deprotonate acetylene.

11.8

Na+ –NH2

c. CH2=CH– [pKa (CH2=CH2) = 44] pKa > 25 = Can deprotonate acetylene. d. (CH3)3CO– {pKa [(CH3)3COH] = 18} pKa < 25 = Cannot deprotonate acetylene.

To draw the products of reactions with HX: • Add two moles of HX to the triple bond, following Markovnikov’s rule. • Both X’s end up on the more substituted C. a. CH3CH2CH2CH2 C C H

Br

2 HBr

CH3CH2CH2CH2 C CH3 Br

b. CH3 C C CH2CH3

2 HBr

Br CH3 CH2 C CH2CH3 Br Br

c.

C CH

2 HBr

C CH3 Br

Br CH3 C CH2 CH2CH3 Br

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Alkynes 11–7

11.9 a.

+

Cl

c.

Cl

+

+

CH3

b. CH3 O CH2

O

NH

NH

CH2

11.10 Addition of one equivalent of X2 to alkynes forms trans dihalides. Addition of two equivalents of X2 to alkynes forms tetrahalides. 2 Br2

CH3CH2 C C CH2CH3

Br Br CH3CH2 C C CH2CH3 Br Br

CH3CH2 C C CH2CH3

Cl

Cl2

CH2CH3 C C

CH3CH2

trans dihalide Cl

11.11 Cl2

CH3 C C CH3

Cl

CH3

The two Cl atoms are electron withdrawing, making the π bond less electron rich and therefore less reactive with an electrophile.

C C CH3

Cl

11.12 To draw the keto form of each enol: [1] Change the C–OH to a C=O at one end of the double bond. [2] At the other end of the double bond, add a proton. H H C H

CH2

a.

OH

c.

O

OH H

H

new C–H bond OH

O H

new C–H bond

O

b. H

new C–H bond

H H

11.13 The treatment of alkynes with H2O, H2SO4, and HgSO4 yields ketones. H2O

CH3CH=C(OH)CH2CH3

H2SO4, HgSO4

CH3C(OH)=CHCH2CH3

CH3 C C CH2CH3

O

O

Two enols form.

two ketones after tautomerization

11.14 OH

OH

a.

OH

OH

b.

enol tautomers

constitutional isomers, but not tautomers

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Chapter 11–8

11.15 Reaction with H2O, H2SO4, and HgSO4 adds the oxygen to the more substituted carbon. Reaction with [1] R2BH, [2] H2O2, –OH adds the oxygen to the less substituted carbon. a. (CH3)2CHCH2 C C H

(CH3)2CHCH2

b.

C C H

CH3

H2O H2SO4, HgSO4

C CH

O CH2 C CH3

H

Forms a ketone. H2O is added with the O atom on the more substituted carbon.

CH3

[1] R2BH [2] H2O2, HO

C CH

CH3 C

CH3 C

O CH2 CH2 C H

H

Forms an aldehyde. H2O is added with the O atom on the less substituted carbon.

O

H2O

C

H2SO4, HgSO4

CH3

Forms a ketone. H2O is added with the O atom on the more substituted carbon.

H H Forms an aldehyde. H2O is added with the C O atom on the less substituted carbon. C O H

[1] R2BH [2] H2O2, HO

11.16 [1] NaH

a. H C C H

[1] NaH

H C C

+ H2

C C

+ H2

[2] (CH3)2CHCH2–Cl

(CH3)2CHCH2 C C H

[2] CH3CH2–Br

C C CH2CH3 + NaBr

b.

C CH [1] NaNH2

C C

+ NH3

[2] (CH3)3CCl

CH3 C CH2

+ NaCl

+ NaCl

1° alkyl halide substitution product 3° alkyl halide elimination product

CH3

+

C CH

11.17 a.

(CH3)2CHCH2C CH3 CH3 C CH2 H

CH

C C H

CH3 CH3 C CH2Cl H

C C H

terminal alkyne only one possibility

1° RX b. CH3C CCH2CH2CH2CH3 CH3

C C

[1]

CH2CH2CH2CH3

[2]

[1]

CH3Cl

[2] CH3 C C

C C CH2CH2CH2CH3 Cl CH2CH2CH2CH3

1° RX

internal alkyne two possibilities

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307

Alkynes 11–9

c. (CH3)3CC CCH2CH3 CH3 CH3 C C C CH3

CH3 CH3 C C C CH3

CH2CH3

internal alkyne only one possibility

Cl CH2CH3

1° RX

CH3 CH3 C Cl CH3

The 3° alkyl halide would undergo elimination.

C CCH2CH3

3° RX too crowded for SN2 reaction

11.18 HC C H

Na+ H

Na+ H H C C CH2CH3

CH3CH2–Br

HC C

C C CH2CH3 + H2 H

+ H2

+ Na+Br–

+ Na+

(CH3)2CHCH2C

Br

H (CH3)2CHCH2CH2

– C C CH2CH3 + Br

11.19 CH3

CH3

CH3

CH3 C C C C CH3 CH3

CH3

+

CH3 C C C

CH3

X C CH3

CH3

CH3

The 3° alkyl halide is too crowded for nucleophilic substitution. Instead, it would undergo elimination with the acetylide anion.

2,2,5,5-tetramethyl-3-hexyne

11.20 CH3

a.

[1]

CH3 OH

C C H

O

[2] H2O

Epoxide is drawn up, so the acetylide anion attacks from below at less substituted C.

C CH

O

b.

[1]

CH

C

OH

C C H

+

[2] H2O

C

OH

Backside attack of the nucleophile (–C≡CH) at either C since both ends are equally substituted

CH

enantiomers

11.21 a. CH3CH2CH2Br

CH3CH2C C–Na+

b. (CH3)2CHCH2CH2Cl

c. (CH3CH2)3CCl

CH3CH2CH2C CCH2CH3

CH3CH2C C–Na+

CH3CH2C C–Na+

d. BrCH2CH2CH2CH2OH

(CH3)2CHCH2CH2C

(CH3CH2)2C

CH3CH2C C–Na+

CHCH3

CH3CH2C CH

CCH2CH3

CH3CH2C CH

BrCH2CH2CH2CH2O–Na+

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Chapter 11–10

e. f.

O

CH3CH2C C–Na+

O

CH3CH2C C–Na+

H 2O

CH3CH2C CCH2CH2OH

H 2O

CH3CH2C CCH2CHCH3 OH

11.22 To use a retrosynthetic analysis: [1] Count the number of carbon atoms in the starting material and product. [2] Look at the functional groups in the starting material and product. • Determine what types of reactions can form the product. • Determine what types of reactions the starting material can undergo. [3] Work backwards from the product to make the starting material. [4] Write out the synthesis in the synthetic direction. ?

CH3CH2C CCH2CH3

HC CH

6 C's CH3CH2C CCH2CH3

HC C H

Na+ H

CH3CH2C C

CH3CH2–Br

HC C

2 C's + CH3CH2Br

CH3CH2C C H

Na+ H

HC C

CH3CH2C C

+ CH3CH2Br

CH3CH2–Br

CH3CH2C CCH2CH3

11.23 O HC CH

CH3CH2CH2 C H

product: 4 carbons, aldehyde functional group (can be made by hydroboration–oxidation of a terminal alkyne)

starting material: 2 carbons, C≡C functional group (can form an acetylide anion by reaction with NaH)

O

Retrosynthetic analysis:

CH3CH2CH2 C

CH3CH2C C H

H C C H

H

Forward direction: H C C H

Na+ H C C H

CH3CH2–Br

CH3CH2C C H

O

[1] R2BH [2] H2O2,

HO–

11.24 a.

4

6

3 2

5

1

6 C alkyne hexyne C≡C at C1 1-hexyne CH3 and CH2CH3 at C3 3-ethyl-3-methyl-1-hexyne

1-hexyne 3 b.

5 11

1

12 C chain with 2 C C's dodecadiyne C≡C's at C3 and C5 3,5-dodecadiyne CH3 at C11 11-methyl-3,5-dodecadiyne

CH3CH2CH2 C H

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Alkynes 11–11

11.25 O OH

b.

a. O

OH

keto form

enol form

H

enol form

keto form

11.26 a, b.

N

O

O O

N

O

c. d. * = sp hybridized C e. 3 < 1 < 2

H

(1)

(2)

*

* *

O

*

C*

HN

(3)

HO

erlotinib

most acidic C–H proton

phomallenic acid C most acidic proton

shortest C–C single bond Csp–Csp2

11.27 Use the rules from Answer 11.3 to name the alkynes.

3-hexyne

1-hexyne

2-hexyne

3-methyl-1-pentyne

3,3-dimethyl-1-butyne

4-methyl-2-pentyne

4-methyl-1-pentyne

11.28 Use the rules from Answer 11.3 to name the alkynes. 5

3

a. CH3CH2CH(CH3)C CCH2CH3 5-methyl

5-methyl-3-heptyne

3-heptyne e.

CH3CHC CCHCH3 CH3

HC C CH(CH2CH3)CH2CH2CH3

1-hexyne

3-hexyne b.

d.

3-ethyl

CH3CH2C CCH2C

5

2,5-dimethyl-3-hexyne 2

CH3

3-ethyl-1-hexyne

CCH3

2,5-octadiyne

2 5-ethyl

f. 2,5-dimethyl 4-nonyne c.

6 4-ethyl CH3

7-methyl

CH3CH2CHC CCHCHCH2CH3 CH3CH2

CH2CH3

3,6-diethyl

3,6-diethyl-7-methyl-4-nonyne

6-methyl g.

(2E)-4,5-diethyl-2-decen-6-yne

1 ethynyl 1-ethynyl-6-methylcyclohexene

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Chapter 11–12

11.29 Use the directions from Answer 11.4 to draw each structure. a. 5,6-dimethyl-2-heptyne

d. cis-1-ethynyl-2-methylcyclopentane 1-ethynyl

2-heptyne

or

C CH

C CH

2-methyl

b. 5-tert-butyl-6,6-dimethyl-3-nonyne 6,6-dimethyl

e. 3,4-dimethyl-1,5-octadiyne 3,4-dimethyl

3-nonyne 5-tert-butyl

diyne

c. (4S)-4-chloro-2-pentyne Cl H

diyne

f. (6Z)-6-methyl-6-octen-1-yne

4-chloro S configuration

6-methyl Z

2-pentyne 1

11.30 Keto–enol tautomers are constitutional isomers in equilibrium that differ in the location of a double bond and a hydrogen. The OH in an enol must be bonded to a C=C. O

a.

CH3

C

OH CH3

and

CH2

C

• C=O • C=C • one more CH bond • OH on C=C keto–enol tautomers

H

and

O • C=O • one more CH bond

• C=C • OH on C=C

keto–enol tautomers O

OH

O

b.

OH

c.

CH3

OH

and

d.

and OH is not bonded to the C=C. NOT keto–enol tautomers

OH is not bonded to the C=C.

NOT keto–enol tautomers

11.31 To draw the enol form of each keto form: [1] Change the C=O to a C–OH. [2] Change one single C–C bond to a double bond, making sure the OH group is bonded to the C=C. Use the directions from Answer 11.12 to draw each keto form. a.

O H O

b.

OH

OH

CH3CH2CHO = H OH

E/Z isomers possible

H

OH

d.

+

CH2CH3

O

c.

CH2CH3

CH2CH3

OH

O

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Alkynes 11–13

11.32 Tautomers are constitutional isomers that are in equilibrium and differ in the location of a double bond and a hydrogen atom. O

O

OH

O

O

a.

A

OH

OH

tautomer

O

O

c.

b. constitutional isomer

OH

d.

constitutional isomer

neither

11.33 O

OH

OH

(E and Z) 2-butanone

C=C has one C bonded to it.

C=C has two C's bonded to it. The more substituted double bond is more stable.

11.34 O H

O

HO

O

H OH

HO

O

H OH

11.35 NHCH3

NHCH3

X H O H enamine 2

+ H 2O

H

NCH3

NCH3

+ H 3O

H + H2 O

Y imine

11.36 HC CCH2CH2CH2CH3

a.

HCl (2 equiv)

b.

HBr

O

Cl CH3 C CH2CH2CH2CH3

e.

Cl Br CH3 CCH2CH2CH2CH3

(2 equiv)

f.

Cl2 (2 equiv)

d.

HC CCH2CH2CH2CH3

g.

Cl Cl O

H2O H2SO4, HgSO4

[2] H2O2, HO–

NaH

H

C

CH2CH2CH2CH2CH3

Na+ C CCH2CH2CH2CH3

Br Cl Cl

c.

[1] R2BH

CH3

C

h. CH2CH2CH2CH3

[1] –NH2

CH3CH2C CCH2CH2CH2CH3

[2] CH3CH2Br [1] –NH2 [2] O [3] H2O

HOCH2CH2 C CCH2CH2CH2CH3

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Chapter 11–14

11.37 Br Br

HBr

a.

H2O

c.

(2 equiv)

b.

H2SO4 O

Br Br

Br2

O

[1] R2BH

d.

(2 equiv)

[2] H2O2, HO–

Br Br

11.38 O

a.

(CH3CH2)3C

H2O

C CH

(CH3CH2)3C

H2SO4, HgSO4

C

CH3

[1] R2BH

b.

(CH3CH2)3C

C CH

c.

(CH3CH2)3C

C CH

d.

(CH3CH2)3C

C CH

(CH3CH2)3C

[2] H2O2, HO– HCl

CH2CHO

(CH3CH2)3CCCl2CH3

(2 equiv) [1] NaH

(CH3CH2)3C

[2] CH3CH2Br

C CCH2CH3

11.39 Reaction rate (which is determined by Ea) and enthalpy (H°) are not related. More exothermic reactions are not necessarily faster. Since the addition of HX to an alkene forms a more stable carbocation in an endothermic, rate-determining step, this carbocation is formed faster by the Hammond postulate. H R C C R'

HX

H X

R C C R'

H

R C C R'

R C C R'

HX

H H

sp hybridized carbocation less stable slower reaction

R C C R'

H H

H H

sp2 hybridized carbocation more stable faster reaction

11.40 O

OH C

a.

CH3

C

CH3 O

C

c.

CH3

CH3

C

CH3 C CH

CH2

OH C

d.

C

CH2

CH3CH2 O

CH

OH

O

b.

OH

CH

C C

H X

R C C R'

CH2CH3

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

313

Alkynes 11–15

11.41 To determine what two alkynes could yield the given ketone, work backwards by drawing the enols and then the alkynes. H HC C CH2CH3

OH C C

H

HO

O C

CH3

CH2CH3

H CH3 C C CH3

C C

CH2CH3

CH3

2-butanone

CH3

11.42 CH2CHO

a.

C CH

b.

(CH3)2CHC

CCH(CH3)2

O

11.43

Cl

2 Cl2

b.

(CH3)3CC

Br Br

2 HBr

C CH

a.

(CH3)3CC

CH

CHCl2

Cl

c.

CH

Cl Cl

[1] Cl2

CH

[2] NaNH2

C C

C C

(2 equiv)

H H

e.

[1] R2BH

C CH

d.

C

[2] H2O2, HO HC C + D2O

HC CD

O +

DO– O

O

f.

g.

C C CH3

CH2

CH3CH2C C + CH3CH2CH2–OTs CH3

h.

H2O H2SO4

H

C

C CH3

CH2CH3

CH3CH2C CCH2CH2CH3 CH3

[1] HC C

O

[2] HO–H

+ OTs CH3

O

OH

C CH

C CH CH2–I

i.

CH3CH2C C–H

[1] Na+ –NH2

C C–H [1] Na+ H–

j.

CH3CH2C C

C C

[2] O

[2]

CH3CH2C C

CH2

C CCH2CH2O– [3] HO–H

C C CH2CH2OH

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Chapter 11–16

11.44 CH2CH2Br

KOC(CH3)3

H H

Br2

CH=CH2

C C H

KOC(CH3)3

A

B

C CH

DMSO (2 equiv)

Br Br C

NaNH2

D CH3I

C CCH3

C C E

11.45 most acidic H [1] NaNH2 [2] CH3I H C C CH2CH2CH2OH

CH3 C C CH2CH2CH2OH

NaNH2 will remove the proton from the OH since it is more acidic.

A

H C C CH2CH2CH2OCH3 = B

H C C CH2CH2CH2O CH3

I

11.46 stereogenic center at the site of reaction

identical

Cl HC C

a.

C CH D H

H D

stereogenic center NOT at the site of reaction

CH3 H

O H CH3

[1] HC C H

CH3 [2] H2O

C H CH3

CH3

C C

OH

H C

C

C CH

inversion

H

CH3

HC

Configuration is retained. O

Cl HC C

b.

c.

CH3 H

HO

C CH CH3 H

d.

H CH3

[1] HC C H CH3

[2] H2O

HO

H C

H CH3

CH3

CH3

OH

H

C

C C

C CH

HC

enantiomers

C H CH3

315

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Alkynes 11–17

11.47 H D

H D

TsCl

OH pyridine

D H

HC C

OTs

SN2 B

A

retention H D

D H

PBr3

OH

inversion

Br

SN2

H D

HC C

SN2

C

A

inversion

inversion

11.48 H–C CCH2OR'

new C–C bond

CH3 Li+ OH

a.

Br

PBr3

C CCH2OR'

C CCH2OR' B

OCOR

OCOR

OCOR

A

C OH

O

[1] D = HC C OR

b.

[2] H2O

OH OR

H C C

OH

H C C

E = TsCl, pyridine

These 2 C's are added. OCH2CH3

OCH2CH3 [1] NaH CH3 C C

[2] CH3I

OH

OH

[1] NaH

H C C

H C C

H C C [2] F = CH3CH2Br

G

OTs

new C–C bond

11.49 Br H CH3CH2 C C

CH3CH2 H

Br H H CH3–C H

H

C C Br

NH2

Br C CH3

H

CH3CH2 C C H

1-butyne

H2N H

CH3

CH3 C C CH3

C C CH3

Br

NH2

Br

(E and Z isomers possible)

2-butyne

NH2

Reaction by-products: H

Br H CH3CH2 C C

major product more substituted alkyne

H

Br H NH2

NH2 H

C H

CH3 C C Br H

CH3CH C CH2

1,2-butadiene

2 NH3 + 2 Br–

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Chapter 11–18

11.50 A carbanion is more stable when its lone pair is in an orbital with a higher percentage of the smaller s orbital. A carbocation is more stable when its positive charge is due to a vacant orbital with a lower percentage of the smaller s orbital. In HC≡C+, the positively charged C uses two p orbitals to form two π bonds. If the σ bond is formed using an sp hybrid orbital, the second hybrid orbital would have to remain vacant, a highly unstable situation. HC C

CH2 CH

sp hybridized higher % s-character more stable

HC C

sp2 hybridized lower % s-character less stable

CH2 CH

sp hybridized sp hybridized Vacant orbital has 50% s-character. Vacant orbital is a p orbital. less stable more stable

11.51 CH3 Cl CH3C CCH3

H Cl

CH3

H

+

C

Cl

C C CH3

CH3

C CH3

CH3

Cl– attack on the same side as the H yields the E isomer.

C C

H Cl

Cl– attack on the opposite side to the H yields the Z isomer.

H

Cl

11.52 a. C

C

C

C

C

C

C

C

H

H

C O

H

H CH3CH2 Li+

OH H OH

+ CH3CH3

+

C H O

Li+

H OSO3H OH

HSO4

OH2

b. CH3 C C CH

CH3 C C CH

H

+ OH

CH3 C C CH

H2O C C CH

CH3

H

H

H

H O H

CH3 C C C H H

resonance structures

HSO4

HSO4 CH3 CH CH

O H C O

H2SO4

H

CH3 CH C C H

OH CH3 CH C C

H

H

O H CH3 CH C C H

H H OSO3H

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

317

Alkynes 11–19

11.53 H C C OCH3 CH3CH2CH2 H OH2

OH C C OCH3 CH3CH2CH2

H CH3CH2CH2C C OCH3

OH C C OCH3 CH3CH2CH2

H

C C OCH3 CH3CH2CH2

H OH2

H2O

H

H

H2O

O CH3CH2CH2CH2 C OCH3

O H H2O CH3CH2CH2CH2 C OCH3

OH CH3CH2CH2CH2 C OCH3 H2O

H OH2

11.54 a. C6H5CH2CHBr2

KOC(CH3)3 (2 equiv) DMSO

KOC(CH3)3

b. C6H5CHBrCH3

H2SO4

c. C6H5CH2CH2OH

C6H5C CH

C6H5CH=CH2

C6H5CH=CH2

Br2

Br2

NaNH2

C6H5CHBrCH2Br

excess NaNH2

C6H5CHBrCH2Br

excess

C6H5C CH

C6H5C CH

11.55 The alkyl halides must be methyl or 1°. a. HC C

CH2CH2CH(CH3)2

HC C

Cl CH2CH2CH(CH3)2

CH3

b. CH3

CH3

C C C CH2CH3

CH3 Cl

C C C CH2CH3

CH3

c.

C C

1° RX

CH3 CH2CH2CH3

C C

Cl CH2CH2CH3 1° RX

11.56 a.

HC C–H

b.

HC C–H

Na+ H–

Na+ H–

HC C

HC C

(CH3)2CHCH2–Cl

CH3CH2CH2–Cl

(CH3)2CHCH2C

CH

NaH CH3CH2CH2C CH

CH3CH2CH2C C CH3CH2CH2–Cl CH3CH2CH2C CCH2CH2CH3

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Chapter 11–20

c.

[1] R2BH

CH3CH2CH2C CH

[2] H2O2, HO–

(from b.) d.

O

H2O

CH3CH2CH2C CH

H2SO4 HgSO4 2 HCl

(from b.)

e.

CH3CH2CH2CH2CHO

CH3CH2CH2C CH

CH3CH2CH2

C

CH3

CH3CH2CH2CCl2CH3

(from b.) f.

O

H 2O

CH3CH2CH2C CCH2CH2CH3

H2SO4, HgSO4

CH3CH2CH2

C

CH2CH2CH2CH3

(from b.)

11.57 HBr

b. CH3CH2C CH (from a.)

(2 equiv) Cl2

c. CH3CH2C CH (from a.)

(2 equiv) Br2

d. CH3CH2CH CH2 e.

Cl

Cl2

a. CH3CH2CH CH2

NaH

CH3CH2C CH

Cl

2 –NH2

CH3CH2CH CH2

CH3CH2C CH

CH3CH2CBr2CH3

CH3CH2CCl2CHCl2

CH3CH2CHBrCH2Br

CH3CH2C C

CH3CH2CH2 Br

CH3CH2C C CH2CH2CH3

(from a.) [1]

f.

CH3CH2C C

O CH3CH2C C CH2CH2OH

[2] H2O

(from e.) O

g. CH3CH2C C (from e.)

OH

[1] [2] H2O

C CCH2CH3

(+ enantiomer)

11.58 NaH

a.

HC CH

b.

CH3CH2CH2CH2CH2CH2C CH

HC C

CH3CH2CH2CH2CH2CH2Br

NaH

CH3CH2CH2CH2CH2CH2C CH

CH3CH2CH2CH2CH2CH2C C

CH3CH2Br

CH3CH2CH2CH2CH2CH2C CCH2CH3

(from a.)

c.

CH3CH2CH2CH2CH2CH2C C

(from b.) d.

[1]

O

[2] H2O

CH3CH2CH2CH2CH2CH2C CCH2CH2OH

(from c.)

CH3CH2CH2CH2CH2CH2C CCH2CH2OH

[1] NaH [2] CH3CH2Br

CH3CH2CH2CH2CH2CH2C CCH2CH2OCH2CH3

319

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Alkynes 11–21

11.59 K+ –OC(CH3)3

Br

Br2

CH2 CH2

Br

NaNH2

NaH

HC CH

HC C

Br (2 equiv) Br H2 O

Br

C C

NaH

C C

HC C

H2SO4, HgSO4

O

11.60 H2SO4

a.

2 –NH2

Br2

OH

NaH

CH3 C CH

Br

CH3 C C

Br

CH3 C CCH2CH2CH3 SOCl2 OH

Cl2

b. (from a.)

[1] CH3 C C

NaH Cl

H2O

(from a.) [2] H2O

O

OH

Cl

OH CH3C CCH2CHCH3

11.61 a.

H2SO4

OH

CH2 CH2

Br

Br2

NaNH2 (2 equiv)

Br

NaH

HC CH

HC C Br

PBr3

OH

[1] NaH C C HO

HC C

[2] O [3] H2O

OH

b.

H2SO4

C C

CH2 CH2

Br

Br2 H2O

O

OH

C C

NaH

HO

NaH

O

(from a.)

C C

CH3CH2Br

(from a.)

CH3CH2 O

11.62 Two resonance structures can be drawn for an enol. OH

OH

negative charge on C that is part of the enol C=C

Since the second resonance structure places an electron pair (and therefore a negative charge) on an enol carbon, this makes the C=C more nucleophilic than the C=C of an alkene for which no additional resonance forms can be drawn. Thus, the OH group donates electron density to the C=C by a resonance effect.

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Chapter 11–22

11.63 O H

H

O

H

O

H

O

H2O

H

O

OH2

+ H3O H

H

H

H

H

H

H

11.64 Br Br

Br

CH3 C C H

CH3

CH3

Br C C

CH3 C C H

H

H O

H2O

C C H H

HO

H

H O

H

Br–

Br

CH3

Br C C

+ H2O

H OH2

H2O

CH3

C

Br

CH3

O

C C H H O

CH2Br

H3O+

H

H2O

11.65 Only this carbocation forms because it is resonance stabilized. The positive charge is delocalized on oxygen. H TsOH O TsO H

H

O

O

re-draw

O

+ TsO–

H

not resonance stabilized

(not formed)

H

OCH3

NOT O

O CH3OH

O

OCH3 H

CH3OH

OCH3

X

O

Y

+ CH3OH2

11.66 A more stable internal alkyne can be isomerized to a less stable terminal alkyne under these reaction conditions because when CH3CH2C≡CH is first formed, it contains an sp hybridized C–H bond, which is more acidic than any proton in CH3–C≡C–CH3. Under the reaction conditions, this proton is removed with base. Formation of the resulting acetylide anion drives the equilibrium to favor its formation. Protonation of this acetylide anion gives the less stable terminal alkyne.

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Alkynes 11–23

H

H CH3 C C CH2

NH2

CH3 C C CH2

CH3 C C CH2

H

H

2-butyne

NH3

NH2 H

H2O

NH2

NH2 NH3

CH3CH2 C C

CH3 C C CH

CH3CH2 C C H

acetylide anion

1-butyne

H

A

C C C(CH3)2

(CH3)2CH

NH2

NH3

CH3 C C CH

H

H (CH3)2CH

NH2

CH3 C C CH

B

C C C(CH3)2

(CH3)2CH

C C C(CH3)2 H

2,5-dimethyl-3-hexyne

NH3

NH2

(CH3)2CH

NH2

C C C(CH3)2 H

2,5-dimethyl-2,3-hexadiene

In this case the reaction stops with formation of 2,5-dimethyl-2,3-hexadiene because a terminal alkyne (with an acidic sp hybridized C–H bond) is not formed. Removal of the circled H in the diene re-forms the anion shown in resonance structures A and B. 11.67 CH3 C C

O H

C

O

H CH3

CH3 OH

OH2 C C

C C

+ H

C

CH3 H2O C

CH3

O

C C

O H2O

CH3

O

C OH H

O

C

O CH3

C O

CH3

C O H

H

C

O

CH3

C OH

CH3

H

HCOOH

resonance structures

C OH H

enol

O O

C

H

H

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Chapter 11–24

11.68 In the presence of acid, (R)-α-methylbutyrophenone enolizes to form an achiral enol. H

O

OH2

OH

H

OH

H H2 O

(R)-α-methylbutyrophenone

(+ 1 resonance structure)

(E and Z isomers) achiral enol

The achiral enol can then be protonated from above or below the plane to form a racemic mixture that is optically inactive.

O H

H2O

O H3O+

below

OH

H O H

H H2O

O

above H

OH2

H3O+ H

H

racemic mixture

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Oxidation and Reduction 12–1

Chapter 12 Oxidation and Reduction Chapter Review Summary: Terms that describe reaction selectivity • A regioselective reaction forms predominately or exclusively one constitutional isomer (Section 8.5). CH3

CH3

I

OH

major product trisubstituted alkene

•

minor product disubstituted alkene

A stereoselective reaction forms predominately or exclusively one stereoisomer (Section 8.5).

H Br

H

Na+ –OCH2CH3

C C

+

C C

H H

C C

H

H

trans alkene major product

•

CH2

+

H

cis alkene minor product

An enantioselective reaction forms predominately or exclusively one enantiomer (Section 12.15). C C CH2OH

allylic alcohol

O C C

Sharpless reagent

or

CH2OH

C C O

CH2OH

One enantiomer is favored.

Definitions of oxidation and reduction Oxidation reactions result in: • an increase in the number of C–Z bonds, or • a decrease in the number of C–H bonds.

Reduction reactions result in: • a decrease in the number of C–Z bonds, or • an increase in the number of C–H bonds. [Z = an element more electronegative than C]

Reduction reactions [1] Reduction of alkenes—Catalytic hydrogenation (12.3)

R

CH CH R

H2 Pd, Pt, or Ni

H H R C C R H H

alkane

• •

Syn addition of H2 occurs. Increasing alkyl substitution on the C=C decreases the rate of reaction.

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Chapter 12–2

[2] Reduction of alkynes

[a]

H H

2 H2

R C C R

•

Two equivalents of H2 are added and four new C–H bonds are formed (12.5A).

R

•

H

•

Syn addition of H2 occurs, forming a cis alkene (12.5B). The Lindlar catalyst is deactivated so that reaction stops after one equivalent of H2 has been added.

R C C R

Pd-C

H H

alkane

[b]

R

H2 R C C R

C C

Lindlar catalyst

H

cis alkene

[c]

R

Na

R C C R

H

•

C C

NH3 H

R

Anti addition of H2 occurs, forming a trans alkene (12.5C).

trans alkene

[3] Reduction of alkyl halides (12.6) R X

[1] LiAlH4 [2] H 2O

R H

alkane

• •

The reaction follows an SN2 mechanism. CH3X and RCH2X react faster than more substituted RX.

• •

The reaction follows an SN2 mechanism. In unsymmetrical epoxides, H– (from LiAlH4) attacks at the less substituted carbon.

• • •

The mechanism has one step. Syn addition of an O atom occurs. The reaction is stereospecific.

•

Ring opening of an epoxide intermediate with –OH or H2O forms a 1,2-diol with two OH groups added in an anti fashion.

[4] Reduction of epoxides (12.6) O C C

OH

[1] LiAlH4 [2] H2O

C C H

alcohol

Oxidation reactions [1] Oxidation of alkenes [a] Epoxidation (12.8) C C

+

O C C

RCO3H

epoxide

[b] Anti dihydroxylation (12.9A)

[1] RCO3H C C

HO C

C

[2] H2O ( H+ or HO–)

OH

1,2-diol

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Oxidation and Reduction 12–3

[c] Syn dihydroxylation (12.9B) HO

[1] OsO4; [2] NaHSO3, H2O C C

OH C

or [1] OsO4, NMO; [2] NaHSO3, H2O or KMnO4, H2O, HO–

C

•

Each reagent adds two new C–O bonds to the C=C in a syn fashion.

•

Both the σ and π bonds of the alkene are cleaved to form two carbonyl groups.

1,2-diol

[d] Oxidative cleavage (12.10) R'

R C C R

H

[1] O3

R

[2] Zn, H2O or CH3SCH3

R

R' C O

O C

+

H

ketone

aldehyde

[2] Oxidative cleavage of alkynes (12.11) [a]

[1] O3

R C C R'

internal alkyne

[2] H2O

R

R' C O

+

•

O C

HO

The σ bond and both π bonds of the alkyne are cleaved.

OH

carboxylic acids

[b]

R C C H

terminal alkyne

R

[1] O3 [2] H2O

+

C O

CO2

HO

[3] Oxidation of alcohols (12.12, 12.13) H

[a]

[b]

PCC or HCrO4– H A-26 1o alcohol Amberlyst resin H

Oxidation of a 1o alcohol with PCC or HCrO4– (Amberlyst A-26 resin) stops at the aldehyde stage. Only one C–H bond is replaced by a C–O bond.

•

Oxidation of a 1o alcohol under harsher reaction conditions—CrO3 (or Na2Cr2O7 or K2Cr2O7) + H2O + H2SO4—affords a RCOOH. Two C–H bonds are replaced by two C–O bonds. Since a 2o alcohol has only one C–H bond on the carbon bearing the OH group, all Cr6+ reagents—PCC, CrO3, Na2Cr2O7, K2Cr2O7, or HCrO4– (Amberlyst A-26 resin)—oxidize a 2o alcohol to a ketone.

H

aldehyde

R C O

H2SO4, H2O

1o alcohol

• C O

CrO3

R C OH H

HO

carboxylic acid

H

[c]

R

R C OH

PCC or CrO3 or R HCrO4– o 2 alcohol Amberlyst A-26 resin

R C OH

•

R C O R

ketone

[4] Asymmetric epoxidation of allylic alcohols (12.15) H

CH2 OH C C

R

H

(CH3)3C

OOH

Ti[OCH(CH3)2]4

O H C C R

CH2OH H

with (–)-DET

or

H CH2OH R C C H O

with (+)-DET

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Chapter 12–4

Practice Test on Chapter Review 1.a. Compound X has a molecular formula of C9H12 and contains no triple bonds. X is hydrogenated to a compound of molecular formula C9H14 with excess H2 and a palladium catalyst. What can be said about X? 1. X has four rings. 2. X has three rings and one double bond. 3. X has two rings and two double bonds.

4. X has one ring and three double bonds. 5. X has four double bonds.

b. Syn addition to an alkene occurs exclusively with which reagents? 1. 2. 3. 4. 5.

OsO4 KMnO4, H2O, –OH mCPBA, then H2O, –OH Both reagents (1) and (2) give syn addition exclusively. Reagents (1), (2), and (3) give syn addition exclusively.

c. Which of the following reagents adds to an alkene exclusively in an anti fashion? 1. Br2 2. H2, Pd-C 3. BH3, then H2O2, –OH

4. Reagents (1) and (2) both add in an anti fashion. 5. Reagents (1), (2), and (3) all add in an anti fashion.

2. Label each statement as True (T) or False (F). a. b. c. d. e. f. g. h.

PCC oxidizes 1° alcohols to aldehydes. CrO3 oxidizes 2° alcohols to ketones. Treatment of 2-hexyne with Na in NH3 forms cis-2-hexene. Reduction of propene oxide with LiAlH4 forms 1-propanol. Ozonolysis of 2-methyl-2-octene forms one ketone and one aldehyde. mCPBA is an oxidizing agent that converts alkenes to trans diols. 1-Octen-5-yne reacts with H2 and Pd-C, but does not react with H2 and Lindlar catalyst. Treatment of cyclohexene with OsO4 affords an optically inactive product mixture that contains two enantiomers.

3. Label each reagent as an oxidizing agent, reducing agent, or neither. a. b. c. d. e. f.

O3 LiAlH4 mCPBA H2O, H2SO4 PCC Na, NH3

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327

Oxidation and Reduction 12–5

4. Draw the organic products formed in each reaction and indicate stereochemistry when necessary. Na a.

NH3

H2

b.

Pd–C

OH

c.

PCC

HO

d.

OH

[1] SOCl2 [2] LiAlH4

5. a. Fill in the appropriate starting material (including any needed stereochemistry) in the following reaction. [1] OsO4 [2] H2O, NaHSO3

HO

H C

H C6H5

C

C6H5

+

OH

b. Fill in the appropriate reagent in the following reaction. O H

OH

c. What starting material is needed for the following reaction? [1] O3

CHO

[2] (CH3)2S O

OH

enantiomer

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Chapter 12–6

Answers to Practice Test 1.a. 2 b. 4 c. 1

2.a. T 3.a. oxidizing b. T b. reducing c. F c. oxidizing d. F d. neither e. T e. oxidizing f. F f. reducing g. F h. F

4.

5. H

H

a.

a. H

H

b.

b. Sharpless reagent (+)-DET

c.

CHO

c.

HO

d.

Chapter 12: Answers to Problems 12.1 Oxidation results in an increase in the number of C–Z bonds (usually C–O bonds) or a decrease in the number of C–H bonds. Reduction results in a decrease in the number of C–Z bonds (usually C–O bonds) or an increase in the number of C–H bonds. O

O

oxidation

a.

c.

reduction

b.

d.

CH3

C

O CH3

CH2 CH2

CH3

C

OCH3

CH3CH2Cl

oxidation

neither

1 new C–H bond and 1 new C–Cl bond

12.2 Hydrogenation is the addition of hydrogen. When alkenes are hydrogenated, they are reduced by the addition of H2 to the π bond. To draw the alkane product, add a H to each C of the double bond. CH3 CH3

b.

CH2CH(CH3)2 C C

a.

H

CH3 CH2CH(CH3)2 CH3 C H

C H H

c.

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Oxidation and Reduction 12–7

12.3 Draw the alkenes that form each alkane when hydrogenated. a.

or

H2 Pd-C

or

c.

or or

or H2 Pd-C

or or

H2 Pd-C

or

b.

12.4 Cis alkenes are less stable than trans alkenes, so they have larger heats of hydrogenation. Increasing alkyl substitution increases the stability of a C=C, thus decreasing the heat of hydrogenation.

cis alkane less stable larger heat of hydrogenation

or

b.

or

a.

trans alkane

trisubstituted

disubstituted less stable larger heat of hydrogenation

12.5 Hydrogenation products must be identical to use hydrogenation data to evaluate the relative stability of the starting materials. Different products are formed. Hydrogenation data can't be used to determine the relative stability of the starting materials.

2-methyl-2-pentene

3-methyl-1-pentene

12.6 Two enantiomers are formed in equal amounts:

new stereogenic center H

a.

b.

CH3 C CH2CH2CH3

C C H

CH3

CH2CH3

CH2CH3

CH2CH3 CH2CH2CH3

CH2

H

CH3

CH3 H

=

C

CH3

+ CH3

CH2CH3

+ CH2CH2CH3 H H CH3

diastereomers c.

C(CH3)3

C(CH3)3

+

diastereomers

C(CH3)3

CH3CH2CH2 C CH 3 H

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Chapter 12–8

12.7

A

Molecular formula before hydrogenation C10H12

Molecular formula after hydrogenation C10H16

Number of rings 3

Number of π bonds 2

B C

C4H8 C6H8

C4H10 C6H12

0 1

1 2

Compound

12.8 H2, Pd-C CH2OCO(CH2)16CH3 CHOCO(CH2)6(CH2CH=CH)2(CH2)4CH3 CH2OCO(CH2)16CH3

CHOCO(CH2)6(CH2CH2CH2)2(CH2)4CH3 B

excess

CH2OCO(CH2)16CH3

H2, Pd-C

CH2OCO(CH2)16CH3

CH2OCO(CH2)16CH3

CHOCO(CH2)6CH2CH=CH(CH2)7CH3 CHOCO(CH2)9CH2CH=CH(CH2)4CH3

1 equiv

A

CH2OCO(CH2)16CH3

CH2OCO(CH2)16CH3

CH2OCO(CH2)16CH3

or

C A has 2 double bonds. lowest melting point

C has 1 double bond. intermediate melting point

C

B has 0 double bonds. highest melting point

12.9 Hydrogenation of HC≡CCH2CH2CH3 and CH3C≡CCH2CH3 yields the same compound. The heat of hydrogenation is larger for HC≡CCH2CH2CH3 than for CH3C≡CCH2CH3 because internal alkynes are more stable (lower in energy) than terminal alkynes. 12.10 O CH2 C

O

C CH2CH3

H2

CH2CH3 C C

Lindlar catalyst

CH3

H H C

CH3

A

H

cis-jasmone (perfume component isolated from jasmine flowers)

H

12.11 In the presence of Pd-C, H2 adds to alkenes and alkynes to form alkanes. In the presence of the Lindlar catalyst, only alkynes react with H2 to form cis alkenes.

H2

a. C6H10

Pd-C

H2

b. Lindlar catalyst

no reaction

C6H10

Pd-C

Lindlar catalyst

12.12 Use the directions from Answer 12.11. a. CH3OCH2CH2C CCH2CH(CH3)2

H2 (excess) Pd-C

CH3OCH 2CH2CH2CH2CH2CH(CH3)2

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Oxidation and Reduction 12–9

b. CH3OCH2CH2C CCH2CH(CH3)2

CH3OCH2CH2

H2 (1 equiv)

CH2CH(CH3)2

C C

Lindlar catalyst H

c.

CH3OCH2CH2C CCH2CH(CH3)2

H

CH3OCH2CH2

H2 (excess)

CH2CH(CH3)2

C C

Lindlar catalyst

H

H

CH3OCH2CH2 Na, NH3

d. CH3OCH2CH2C CCH2CH(CH3)2

H

C C H

CH2CH(CH3)2

12.13 H2 Lindlar catalyst

A

H

R B

H

12.14 LiAlH4 reduces alkyl halides to alkanes and epoxides to alcohols. CH3 [1] LiAlH4

Cl

a.

H

b.

[2] H2O

O

[1] LiAlH4

H

[2] H2O

H replaces Cl.

CH3 OH H H

12.15 To draw the product, add an O atom across the π bond of the C=C. O

mCPBA

a.

(CH3)2C

CH2

b.

(CH3)2C

C(CH3)2

(CH3)2C

c.

CH2

CH2

mCPBA

O mCPBA

(CH3)2C

C(CH3)2

12.16 For epoxidation reactions: • There are two possible products: O adds from above and below the double bond. • Substituents on the C=C retain their original configuration in the products. H

CH3

mCPBA

C C

a.

H

H CH3CH2

b.

CH2CH3 mCPBA

C C H

O CH3 C C H H H

H

CH3 H H C C H enantiomers O

CH3CH2 CH2CH3 O H C C H CH3CH2 C C CH2CH3 O H H

identical

CH3

c.

mCPBA

CH3

CH3

O

O

H

H

enantiomers

O CH2

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Chapter 12–10

12.17 Treatment of an alkene with a peroxyacid followed by H2O, HO– adds two hydroxy groups in an anti fashion. cis-2-Butene and trans-2-butene yield different products of dihydroxylation. cis-2-Butene gives a mixture of two enantiomers and trans-2-butene gives a meso compound. The reaction is stereospecific because two stereoisomeric starting materials give different products that are also stereoisomers of each other. CH3 C C H

HO

[1] RCO3H

CH3 H

[2] H2O, HO−

C CH3

cis-2-butene

CH3 C C H

C

CH3 H

CH3 H

OH

H

OH C

C

HO

H

CH3

enantiomers

H

[1] RCO3H

CH3

[2] H2O, HO−

HO

H C

CH3

CH3 H

CH3

C

OH

H

trans-2-butene

OH C

C H CH3

HO

identical meso compound

12.18 Treatment of an alkene with OsO4 adds two hydroxy groups in a syn fashion. cis-2-Butene and trans-2-butene yield different stereoisomers in this dihydroxylation, so the reaction is stereospecific. HO CH3 C C H

CH3

[1] OsO4

H

[2] NaHSO3, H2O

OH C

CH3

H

H

[1] OsO4

HO

H CH3

[2] NaHSO3, H2O

trans-2-butene

C

HO

OH C

CH3

CH3

C

CH3 H

C H CH3

H

HO

CH3 H OH

identical meso compound

cis-2-butene CH3 C C H

CH3 H

C

H C

CH3

C OH

enantiomers

12.19 To draw the oxidative cleavage products: • Locate all the π bonds in the molecule. • Replace all C=C’s with two C=O’s. Replace this π bond with two C=O's. [1] O3

a. (CH3)2C

CHCH2CH2CH2CH3

[2] Zn, H2O

[1] O3 [2] Zn, H2O

+

O CHCH2CH2CH2CH3

aldehyde

+

O

One ketone and one aldehyde are formed.

Two aldehydes are formed.

O

[2] Zn, H2O

c.

O

ketone H

[1] O3

b.

(CH3)2C

H

O O H

A dicarbonyl compound is formed.

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Oxidation and Reduction 12–11

12.20 To find the alkene that yields the oxidative cleavage products: • Find the two carbonyl groups in the products. • Join the two carbonyl carbons together with a double bond. This is the double bond that was broken during ozonolysis. CH3

a. (CH3)2C O

+ (CH3CH2)2C

O

(CH3)2C

c.

C(CH2CH3)2

CH3 O only

C

C

CH3

CH3

O

CHCH3 +

CH3

With only one product, the alkene must be symmetrical around the double bond. Join this C to the same C in another identical molecule.

Join these two C's.

b.

CH3 C

CH3CHO

Join these two C's.

12.21 O

a.

[1] O3

O

[2] CH3SCH3

H O

[2] CH3SCH3 H

c.

[2] CH3SCH3

O O

H

O

O

[1] O3

H

[1] O3

b.

O

O

O

H

H

H O

O H

H

12.22 To draw the products of oxidative cleavage of alkynes: • Locate the triple bond. • For internal alkynes, convert the sp hybridized C to COOH. • For terminal alkynes, the sp hybridized C–H becomes CO2. a.

[2] H2O

O

O

[1] O3

CH3CH2 C C CH2CH2CH3

CH3CH2

C

OH

+

HO

C

CH2CH2CH3

internal alkyne O [1] O3

b.

C

C C [2] H2O

internal alkyne terminal alkyne c.

O

+ OH

HO

C

identical compounds

internal alkyne

H C C CH2 CH2 C C CH3

[1] O3 [2] H2O

O CO2 +

HO

C O

C

O OH

+ HO

C

CH3

H

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Chapter 12–12

12.23 a.

c.

CH3CH2CH(CH3)CO2H

b.

CO2 + CH3(CH2)8CO2H

CH3CH2CO2H, HO2CCH2CO2H, CH3CO2H

CH3CH2CH C C CHCH2CH3

CH3(CH2)8 C CH

CH3

CH3

CH3CH2 C C CH2 C C CH3

d.

HO2C(CH2)14CO2H

12.24 For the oxidation of alcohols, remember: • 1° Alcohols are oxidized to aldehydes with PCC. • 1° Alcohols are oxidized to carboxylic acids with oxidizing agents like CrO3 or Na2Cr2O7. • 2° Alcohols are oxidized to ketones with all Cr6+ reagents. O OH

a.

PCC

CrO3

OH

H

c.

OH

H2SO4, H2O

O OH OH

O

O PCC

b.

CrO3

d.

H2SO4, H2O

12.25 Upon treatment with HCrO4––Amberlyst A-26 resin: • 1° Alcohols are oxidized to aldehydes. • 2° Alcohols are oxidized to ketones. H OH

HCrO4

a.

O

–

c.

Amberlyst A-26 resin

b. HO

OH

HCrO4–

O

OH OH

H

HCrO4

–

Amberlyst A-26 resin

O O

O

Amberlyst A-26 resin

12.26 NaOCl

a. OH

b.

O

+ NaCl + H2O

The by-products of the reaction with sodium hypochlorite are water and table salt (NaCl), as opposed to the by-products with HCrO4–– Amberlyst A-26 resin, which contain carcinogenic Cr3+ metal.

Oxidation with NaOCl has at least two advantages over oxidation with CrO3, H2SO4 and H2O. Since no Cr6+ is used as oxidant, there are no Cr by-products that must be disposed of. Also, CrO3 oxidation is carried out in corrosive inorganic acids (H2SO4) and oxidation with NaOCl avoids this.

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Oxidation and Reduction 12–13

12.27 To draw the products of a Sharpless epoxidation: • With the C=C horizontal, draw the allylic alcohol with the OH on the top right of the alkene. • Add the new oxygen above the plane if (−)-DET is used and below the plane if (+)-DET is used. OH (CH3)3C

a.

OH

OOH

Ti[OCH(CH3)2]4 (+)-DET

O H

(+)-DET adds O below the plane.

OH

b.

OH

re-draw

(CH3)3C

H O

OOH

OH

Ti[OCH(CH3)2]4 (–)-DET

(−)-DET adds O above the plane.

12.28 Sharpless epoxidation needs an allylic alcohol as the starting material. Alkenes with no allylic OH group will not undergo reaction with the Sharpless reagent. This alkene is part of an allylic alcohol and will be epoxidized. OH

geraniol

This alkene is not part of an allylic alcohol and will not be epoxidized.

12.29 OH

a.

OH

H2 Pd-C

A OH

b.

mCPBA

O

O OH

A OH

c.

PCC

CHO

A OH

d. A

CrO3 H2SO4 H2O

COOH

OH

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Chapter 12–14

e.

OH

O

Sharpless reagent (+)-DET

A

OH

12.30 O

[1] O3

CH3

[2] CH3SCH3

H

O

H

H

H O

O

12.31 [1] NaH HC CH

[2] CH3CH2Br

CH3CH2C CH

[1] NaH [2]

OH

Na

O

OH

NH3

[3] H2O

12.32 Use the rules from Answer 12.1. OH

a.

b.

CH3CH2Br

c.

CH2CH3 reduction

C CH

CH2 CH2 neither 1 C–H and 1 C–Br bond are removed.

O

reduction

d. HO

OH

O

O oxidation

12.33 Use the principles from Answer 12.2 and draw the products of syn addition of H2 from above and below the C=C. H2 Pd-C

a. CH3

Pd-C

CH3 CH2CH3

c. CH3

C CH2 (CH3)2CH

+ CH3

CH3 CH2CH3

H2

H2 Pd-C

CH2CH3

+

Pd-C

CH3CH2

d.

CH3

CH3

H2

b.

CH3

CH3 CH2CH3 H C (CH3)2CH

CH2CH3

+ CH3

C

(CH3)2CH H

CH3

12.34 Increasing alkyl substitution increases alkene stability, thus decreasing the heat of hydrogenation.

2-methyl-2-butene trisubstituted smallest ΔHo = –112 kJ/mol

2-methyl-1-butene disubstituted intermediate ΔHo = –119 kJ/mol

3-methyl-1-butene monosubstituted largest ΔHo = –127 kJ/mol

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Oxidation and Reduction 12–15

12.35 A possible structure: a. Compound A: molecular formula C5H8: hydrogenated to C5H10. 2 degrees of unsaturation, 1 is hydrogenated. 1 ring and 1 π bond b. Compound B: molecular formula C10H16: hydrogenated to C10H18. 3 degrees of unsaturation, 1 is hydrogenated. 2 rings and 1 π bond c. Compound C: molecular formula C8H8: hydrogenated to C8H16. 5 degrees of unsaturation, 4 are hydrogenated. 1 ring and 4 π bonds

12.36 c.

a. monosubstituted largest heat of hydrogenation b. fastest reaction rate

O

[1] O3

H

[2] Zn, H2O

A

c.

a. tetrasubstituted smallest heat of hydrogenation b. slowest reaction rate

+

C

H

[1] O3 O [2] Zn, H2O

B

+ O

identical a. trisubstituted intermediate heat of hydrogenation b. intermediate reaction rate

c. O

[1] O3 [2] Zn, H2O

O

+

H

C

12.37 O

O OH

a.

H2 (excess)

OH

Pd-C

stearidonic acid

stearic acid O

b.

H

O

H2 (1 equiv)

O OH

OH

Pd-C O

O OH

O OH

c. trans one possibility

OH

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Chapter 12–16

O

d.

O

O

OH

OH


5. Common bases are listed in Table 19.3. Factors that affect acidity Resonance effects. A carboxylic acid is more acidic than an alcohol or phenol because its conjugate base is more effectively stabilized by resonance (19.9). OH ROH

pKa = 16–18

O R C OH

pKa = 10

pKa | 5

Increasing acidity

Inductive effects. Acidity increases with the presence of electron-withdrawing groups (like the electronegative halogens) and decreases with the presence of electron-donating groups (like polarizable alkyl groups) (19.10).

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Chapter 19–2

Substituted benzoic acids. x Electron-donor groups (D) make a substituted benzoic acid less acidic than benzoic acid. x Electron-withdrawing groups (W) make a substituted benzoic acid more acidic than benzoic acid. COOH

COOH

COOH

D

W

less acidic higher pKa

more acidic lower pKa pKa < 4.2

pKa = 4.2

pKa > 4.2 Increasing acidity

Other facts x Extraction is a useful technique for separating compounds having different solubility properties. Carboxylic acids can be separated from other organic compounds by extraction, because aqueous base converts a carboxylic acid into a water-soluble carboxylate anion (19.12). x A sulfonic acid (RSO3H) is a strong acid because it forms a weak, resonance-stabilized conjugate base on deprotonation (19.13). x Amino acids have an amino group on the D carbon to the carboxy group [RCH(NH2)COOH]. Amino acids exist as zwitterions at pH | 6. Adding acid forms a species with a net (+1) charge [RCH(NH3)COOH]+. Adding base forms a species with a net (–1) charge [RCH(NH2)COO]– (19.14). Practice Test on Chapter Review 1. a. Give the IUPAC name for the following compound. CO2H

b. Draw the structure corresponding to the following name: sodium m-bromobenzoate. 2. a. Which of the labeled atoms is least acidic? Ha

O

O

O

O Hd

He

1. Ha

3. Hc

2. Hb

5. He

4. Hd

Hc Hb

O

b. Which of the following carboxylic acids has the lowest pKa? COOH

1.

COOH

2. CH3

COOH

3. Cl

COOH

COOH

4.

5.

CH3O

O2 N

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Carboxylic Acids and the Acidity of the O–H Bond 19–3

c. Which compound(s) can be converted to A by an oxidation reaction? OH

1.

CO2H

4. Both [1] and [2] can be converted to A. O

O

5. Compounds [1], [2], and [3] can all be converted to A.

OH

A

2. OH

3. OH

3. Rank the following compounds in order of increasing basicity. Label the least basic compound as 1, the most basic compound as 3, and the compound of intermediate basicity as 2. COO–

COO–

O2N

A

O–

B

C

4. Draw the organic products formed in each of the following reactions. +

NH3

a. CH3 H

OH

NaOH

C COOH

(1 equiv)

N

HO

[2] CH3I

NaH

COOH

b.

[1] NaH

c.

(1 equiv)

Answers to Practice Test 1.a. cis-2-methylcyclopentanecarboxylic acid

2.a. 1 b. 5 c. 5

3. A–2 B–1 C–3

4.a.

OCH3

NH3 CH3 H

b.

c. +

C

N

COO

b. COO–Na+

COO HO

Br

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 19–4

Answers to Problems To name a carboxylic acid: [1] Find the longest chain containing the COOH group and change the -e ending to -oic acid. [2] Number the chain to put the COOH carbon at C1, but omit the number from the name. [3] Follow all other rules of nomenclature.

19.1

3 a.

2

4 H

CH3

c.

CH3CH2CH2 C CH2COOH CH3 2

H

CH3CH2 C CH2

1

CH3CH2

1

C COOH CH2CH3

Number the chain to put COOH at C1. 6 carbon chain = hexanoic acid 2,4-diethylhexanoic acid

Number the chain to put COOH at C1. 6 carbon chain = hexanoic acid 3,3-dimethylhexanoic acid

O

4 b.

6

H

CH3 C CH2CH2COOH

d.

1

Cl

4

1 OH

8 9

Number the chain to put COOH at C1. 5 carbon chain = pentanoic acid 4-chloropentanoic acid

Number the chain to put COOH at C1. 9 carbon chain = nonanoic acid 4-isopropyl-6,8-dimethylnonanoic acid

19.2 c. 3,3,4-trimethylheptanoic acid

a. 2-bromobutanoic acid O

e. 3,4-diethylcyclohexanecarboxylic acid

O

O HO

OH HO

Br

b. 2,3-dimethylpentanoic acid

d. 2-sec-butyl-4,4-diethylnonanoic acid

O

O

HO

f. 1-isopropylcyclobutanecarboxylic acid COOH

OH

19.3 O

D

aD-methoxyvaleric acid

c. DE-dimethylcaproic acid

OH

E

OCH3

O O

b. E-phenylpropionic acid

OH

D

E

d. D-chloro-E-methylbutyric acid OH

Cl

E

OH

D O

19.4 O

O

a.

O

Li+

lithium benzoate

b.

Na+

O

O

O H

sodium formate or sodium methanoate

c.

O

K+

d.

O

Na+

Br

potassium 2-methylpropanoate

sodium 4-bromo-6-ethyloctanoate

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Carboxylic Acids and the Acidity of the O–H Bond 19–5

19.5 COO–Na+

COOH

C2 2-propylpentanoic acid

19.6

19.7

sodium 2-propylpentanoate

More polar molecules have a higher boiling point and are more water soluble. COOCH3

CH2CH2CH2OH

least polar lowest boiling point least H2O soluble

intermediate polarity intermediate boiling point

CH2COOH

most polar highest boiling point most H2O soluble

Look for functional group differences to distinguish the compounds by IR. Besides sp3 hybridized C–H bonds at 3000–2850 cm–1 (which all three compounds have), the following functional group absorptions are seen: O CH3CH2CH2CH2

C

OH

O OH

CH3CH2CH2CH2

C

O

OCH3

carboxylic acid 2 strong absorptions ~1710 (C=O) ~2500–3500 (OH) cm–1

ester 1 strong absorption ~1700 (C=O) cm–1

Molecular formula: C4H8O2 one degree of unsaturation

1

alcohol 1 strong absorption ~3600–3200 (OH) cm–1

19.8 H NMR data (ppm): 0.95 (triplet, 3 H) 1.65 (multiplet, 2 H) 2.30 (triplet, 2 H) 11.8 (singlet, 1 H)

O HO

19.9 HO

HO COOH

HO

COOH

HO

OH

PGF2D a prostaglandin

OH

enantiomer

There are five tetrahedral stereogenic centers. Both double bonds can exhibit cis–trans isomerism. Therefore, there are 27 = 128 stereoisomers.

19.10 1° Alcohols are converted to carboxylic acids by oxidation reactions. O

a.

OH

OH

O

b.

(CH3)2CH

C

OH

(CH3)2CH

CH2 OH

c.

COOH

CH2OH

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Chapter 19–6

19.11 a.

Na2Cr2O7

CH2OH

COOH

H2SO4, H2O

A

b.

c. O2N

KMnO4

CH3

O2 N

COOH

C (Any R group with benzylic H's can be present para to NO2.) OH

[1] O3

CH3C CCH3

[2] H2O

B

2° OH

d.

CH3COOH (2 equiv)

O

CrO3

OH

CO2H

H2SO4, H2O

D

1° OH

19.12 a.

b.

COOH

CH3

CH3

NaOH

OH

COO

NaOCH3

CH3

Na+ + H2 O

O

CH3

NaH

c. CH3 C OH

CH3

CH3

Na+

d.

C O Na+ + H2 CH3

COOH

NaHCO3

+ HOCH3

COO

Na+

+ H2CO3

19.13 CH3COOH has a pKa of 4.8. Any base having a conjugate acid with a pKa higher than 4.8 can deprotonate it. a. F– pKa (HF) = 3.2 not strong enough b. (CH3)3CO– pKa [(CH3)3COH]= 18 strong enough c. CH3– pKa (CH4) = 50 strong enough

d. –NH2 pKa (NH3) = 38 strong enough e. Cl– pKa (HCl) = –7.0 not strong enough

19.14 Ha

H OHb

Increasing acidity: Ha < Hb < Hc OHc

O

H OHb – Ha

OHc

negative charge on C unstable conjugate base

O

mandelic acid Ha

H O

– Hb

OHc O Ha

negative charge on O more stable conjugate base Ha

H OHb

– Hc

H OHb

O O

O O

negative charge on O, resonance stabilized most stable conjugate base

19.15 Electron-withdrawing groups make an acid more acidic, lowering its pKa. CH3CH2 COOH

least acidic pKa = 4.9

ICH2

COOH

one electron-withdrawing group intermediate acidity pKa = 3.2

CF3 COOH

three electron-withdrawing F's most acidic pKa = 0.2

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

523

Carboxylic Acids and the Acidity of the O–H Bond 19–7

19.16 a. CH3COOH

least acidic

HSCH2COOH

HOCH 2COOH

intermediate acidity

most acidic

b. ICH2CH2COOH

least acidic

ICH2COOH

I2CHCOOH

intermediate acidity

most acidic

19.17 a.

CH3

COOH

least acidic

COOH

Cl

intermediate acidity

COOH

most acidic O

b. CH3O

COOH

CH3

C

COOH

COOH

CH3

least acidic

intermediate acidity

most acidic

19.18 OH

Phenol A has a higher pKa than phenol because of its substituents. Both the OH and CH3 are electron-donating groups, which make the conjugate base less stable. Therefore, the acid is less acidic.

HO

A

19.19 To separate compounds by an extraction procedure, they must have different solubility properties. a. CH3(CH2)6COOH and CH3CH2CH2CH2CH=CH2: YES. The acid can be extracted into aqueous base, while the alkene will remain in an organic layer. b. CH3CH2CH2CH2CH=CH2 and (CH3CH2CH2)2O: NO. Both compounds are soluble in organic solvents and insoluble in water. Neither is acidic enough to be extracted into aqueous base. c. CH3(CH2)6COOH and NaCl: one carboxylic acid, one salt: YES. The carboxylic acid is soluble in an organic solvent while the salt is soluble in water. d. NaCl and KCl: two salts: NO. 19.20 weaker conjugate base better leaving group

stronger conjugate base worse leaving group

CF3SO3–

CF3SO3H

CH3SO3–

CH3SO3H

CF3 is electron withdrawing. stronger acid lower pKa

CH3 is electron donating. weaker acid higher pKa

19.21 phenylalanine COOH C H2N

H

methionine

COOH H

COOH

C

C NH2

H2N

H CH2CH2SCH3

R

R

S

COOH H CH3SCH2CH2

C NH2

S

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Chapter 19–8

19.22 Since amino acids exist as zwitterions (i.e., salts), they are too polar to be soluble in organic solvents like diethyl ether. Thus, they are soluble in water. 19.23 COOH H 3N C H

COO

COO

H3 N C H

H2N C H

H

H

H

pH = 1

glycine

pH = 11

neutral form

19.24 COO

pKa(COOH) + pKa(NH3+)

pI =

=

(2.58) + (9.24)

= 5.91

H3N C H

2

2

CH2

19.25 +

+

H3N CH COO–

H3N CH COOH H

H

electron-withdrawing group

The nearby (+) stabilizes the conjugate base by an electron-withdrawing inductive effect, thus making the starting acid more acidic.

19.26

6

5 4

2

1

3

OH

A a. 2,5-dimethylhexanoic acid

2

B a. 3-ethyl-3-methylcyclohexanecarboxylic acid

NaOH

NaOH

O

b.

COOH

1

O 3

COO Na+

b. O Na+

c. sodium 2,5-dimethylhexanoate

c. sodium 3-ethyl-3-methylcyclohexanecarboxylate

d. An alcohol or ether would have a much higher pKa than a carboxylic acid.

d.

OH

OH OH

O H

19.27 COOH

COOH

CH3

least acidic

COOH CF3

intermediate acidity

most acidic

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and the Acidity of the O–H Bond 19–9

19.28 Use the directions from Answer 19.1 to name the compounds. a.

(CH3)2CHCH2CH2CO2H

b.

BrCH2COOH

COOH

4-methylpentanoic acid

g.

o-bromobenzoic acid

2-bromoacetic acid or 2-bromoethanoic acid

Br

O

h.

c.

CH3CH2

COOH p-ethylbenzoic acid

OH

4,4,5,5-tetramethyloctanoic acid d.

–

+

CH3CH2CH2COO Li

O

lithium butanoate

O– Na+

i.

sodium 2-methylhexanoate 1-ethylcyclopentanecarboxylic acid

e.

COOH

COOH

10

j.

3 7

2,4-dimethylcyclohexanecarboxylic acid

f.

5

7-ethyl-5-isopropyl-3-methyldecanoic acid

COOH

19.29 OH

f. o-chlorobenzoic acid

a. 3,3-dimethylpentanoic acid

COOH

O Cl OH

b. 4-chloro-3-phenylheptanoic acid

CH3

O

c. (2R)-2-chloropropanoic acid

Cl

Cl

Cl

e. m-hydroxybenzoic acid

C

O

K+

O

h. sodium D-bromobutyrate

OH

O

Na+

Br

O

d. EE-dichloropropionic acid

Cl

g. potassium acetate O

i. 2,2-dichloropentanedioic acid O

O

O

HO

HOOC

OH Cl Cl

OH

j. 4-isopropyl-2-methyloctanedioic acid OH

O HO

OH O

19.30 O

O OH

pentanoic acid

OH

3-methylbutanoic acid

O– Na+

sodium pentanoate

O– Na+

sodium 3-methylbutanoate

OH

OH

2-methylbutanoic acid

2,2-dimethylpropanoic acid O

O

O

O

O

O

O– Na+

sodium 2-methylbutanoate

O– Na+

sodium 2,2-dimethylpropanoate

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Chapter 19–10

19.31 C5 or G carbon OH

a.

OH

C2 or D carbon

b.

CO2H

CO2H

HO

C3 or E carbon IUPAC: 2-hydroxypropanoic acid common:D-hydroxypropionic acid

IUPAC: 3,5-dihydroxy-3-methylpentanoic acid common: EG-dihydroxy-E-methylvaleric acid

19.32 O

a.

OH

OH O

lowest boiling point

intermediate boiling point

O

highest boiling point O

OH

b. HO

lowest boiling point

intermediate boiling point

highest boiling point

19.33 O OH

a.

CrO3

OH

[1] O3

c.

CH3

CO2

[2] H2O

H2SO4, H2O

b. (CH3)2CH

COOH +

C C H

KMnO4

COOH d. CH3(CH2)6CH2OH

HOOC

Na2Cr2O7

CH3(CH2)6COOH

H2SO4, H2O

19.34 [1] BH3

a.

CH2

[2] H2O2, OH

HC CH [2] CH3I

B

[1] NaNH2 HC CCH3

C

[2] CH3CH2I

CH(CH3)2

(CH3)2CHCl

c.

COOH

H2SO4, H2O

A

[1] NaNH2

b.

CrO3

CH2OH

–

CH3CH2C CCH3

D

[1] O3 [2] H2O

CH3CH2COOH

CH3COOH

E + F

COOH

KMnO4

AlCl3

G

H

19.35 Bases: [1] –OH pKa (H2O) = 15.7; [2] CH3CH2– pKa (CH3CH3) = 50; [3] –NH2 pKa (NH3) = 38; [4] NH3 pKa (NH4+) = 9.4; [5] HC{C– pKa (HC{CH) = 25.

a.

CH3

COOH

pKa = 4.3 All of the bases can deprotonate this.

b.

–OH,

Cl

OH

pKa = 9.4 CH3CH2–, –NH2, and HC{C– can deprotonate this.

c. (CH3)3COH pKa = 18 CH3CH2–, –NH2, and HC{C– can deprotonate this.

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Carboxylic Acids and the Acidity of the O–H Bond 19–11

19.36 a.

K+

COOH

OC(CH

COO K+

3)3

+ HOC(CH3)3 pKa = 18

Reaction favors products.

pKa = 4.2 OH

b.

NH4+

O

NH3

pKa | 16 +

OH

c.

Na+ NH2

+ NH3 + Na+

O

CH3–Li+

COOH

COO Li+

CH3

Na+H

OH

Na+

O

Reaction favors products.

H2

Reaction favors products.

pKa = 35

pKa | 16 CH3

f.

CH4

pKa = 50

CH3

pKa | 4 e.

Reaction favors products.

pKa = 38

pKa = 10 d.

Reaction favors reactants.

pKa = 9.4

OH

Na2CO3

O– Na+

CH3

With the same pKa for the starting acid and the conjugate acid, an equal amount of starting materials and products is present.

Na+ HCO3

pKa = 10.2 pKa = 10.2

19.37 The stronger acid has a lower pKa and a weaker conjugate base. COOH

a.

CH2OH

or

carboxylic acid stronger acid lower pKa weaker conjugate base

c.

or

COOH

or ClCH2COOH FCH2COOH F is more electronegative. weaker acid stronger acid higher pKa lower pKa stronger conjugate base weaker conjugate base

d.

Cl

or

NCCH2COOH

CN is electron withdrawing. stronger acid lower pKa weaker conjugate base

CH3COOH

weaker acid higher pKa stronger conjugate base

19.38 Br

a.

COOH

least acidic

Br is electronegative intermediate acidity

OH

b.

CH3

least acidic

Cl COOH

COOH

Cl more electronegative most acidic

OH Cl

intermediate acidity

COOH

Cl is electron withdrawing. stronger acid lower pKa weaker conjugate base

CH3 is electron donating. weaker acid higher pKa stronger conjugate base

alcohol weaker acid higher pKa stronger conjugate base

b.

CH3

OH O2N

most acidic

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 19–12

19.39 O

a. BrCH2COO–

BrCH2CH2COO–

weakest base

(CH3)3CCOO–

intermediate basicity

O2N

strongest base

weakest base b.

O

NH

weakest base

O

O

c. intermediate basicity

strongest base

CH2

intermediate basicity

strongest base

19.40 Increasing acidity ICH2COOH

pKa values

BrCH2COOH

least acidic 3.12

FCH2COOH

2.86

2.66

F2CHCOOH

F3CCOOH

most acidic 0.28

1.24

19.41 The OH of the phenol group in morphine is more acidic than the OH of the alcohol (pKa  10 versus pKa  16). KOH is basic enough to remove the phenolic OH, the most acidic proton. most acidic proton HO The OH is part of a phenol. Methylation occurs here.

O

O

an alcohol

H

HO

N

H

CH3

CH3O

[2] CH3I H

HO

H

CH3

HO

CH3

O

H

H

HO

N CH3

O

O

O

N

H

N

Many resonance structures stabilize the conjugate base.

O

O

O

HO

morphine

H

O

[1] KOH

H

O

H

N

H

CH3

HO

O H

N CH3

H HO

H

N CH3

codeine

19.42 The closer the electron-withdrawing CH3CO– group is to the carboxylic acid, the more it will stabilize the conjugate base, making the acid stronger. O C

CH3 G+ C O

O OH

pyruvic acid stronger acid

O

C CH3 G+ CH2 OH C

farther away acetoacetic acid weaker acid

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and the Acidity of the O–H Bond 19–13

19.43 a. The negative charge on the conjugate base of p-nitrophenol is delocalized on the NO2 group, stabilizing the conjugate base, and making p-nitrophenol more acidic than phenol (where the negative charge is delocalized only around the benzene ring). OH

O

O

O + N

O + N

O + N

O

O

O

p-nitrophenol pKa = 7.2

OH

O

phenol pKa = 10

two of the possible resonance structures for the conjugate base [See part (b) for all the possible resonance structures.]

b. In the para isomer, the negative charge of the conjugate base is delocalized over both the benzene ring and onto the NO2 group, whereas in the meta isomer it cannot be delocalized onto the NO2 group. This makes the conjugate base from the para isomer more highly resonance stabilized, and the para substituted phenol more acidic than its meta isomer. OH O2N

O O2N

O

O O

O2N

pKa = 7.2 p-nitrophenol

O

N O

N

O O2N

O

negative charge on two O atoms very good resonance structure more stable conjugate base O2N stronger acid

OH

NO2

O

O

O

NO2

O

NO2

O

NO2

NO2

O

O

NO2

pKa = 8.3 m-nitrophenol

19.44 A CH3O group has an electron-withdrawing inductive effect and an electron-donating resonance effect. In 2-methoxyacetic acid, the OCH3 group is bonded to an sp3 hybridized C, so there is no way to donate electron density by resonance. The CH3O group withdraws electron density because of the electronegative O atom, stabilizing the conjugate base, and making CH3OCH2COOH a stronger acid than CH3COOH. – H+

O CH3O

OH

more acidic acid

O CH3O

O

more stable conjugate base

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 19–14

In p-methoxybenzoic acid, the CH3O group is bonded to an sp2 hybridized C, so it can donate electron density by a resonance effect. This destabilizes the conjugate base, making the starting material less acidic than C6H5COOH. O

– H+

CH3O

O

O CH3O

CH3O

OH

O

O

less acidic acid

like charges nearby less stable conjugate base

19.45 Phenol has a pKa of 10, making p-methylthiophenol (pKa = 9.53) the stronger acid. A substituent that increases the acidity of a phenol must withdraw electron density to stabilize the negative charge of the conjugate base. An electron-withdrawing substituent deactivates a benzene ring towards electrophilic aromatic substitution, making p-methylthiophenol less reactive than phenol. OH

OH CH3S

stronger acid less reactive in electrophilic substitution

19.46 The O in A is more electronegative than the N in C so there is a stronger electron-withdrawing inductive effect. This stabilizes the conjugate base of A, making A more acidic than C. CO2H

O

A pKa = 3.2

CO2H

O

B pKa = 3.9

N H

CO2H

C pKa = 4.4

Since the O in A is closer to the COOH group than the O atom in B, there is a stronger electron-withdrawing inductive effect. This makes A more acidic than B.

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Carboxylic Acids and the Acidity of the O–H Bond 19–15

19.47 CO2H

CO2H O2N

O2N

D

CO2H

C

E – H+

– H+

O

– H+

O O

O

O O2N

O2N

O

Since the benzene ring is bonded to the D carbon (not the carbonyl carbon), this compound is not much different than any alkylsubstituted carboxylic acid. least acidic

The electron-withdrawing Since the NO2 group is bonded to a inductive effect of the NO2 benzene ring that is bonded directly to the – group helps stabilize the COO carbonyl group, inductive effects and group. resonance effects stabilize the conjugate intermediate acidity base. For example, a resonance structure can be drawn that places a (+) charge close to the COO– group. Two of the resonance structures for the conjugate base of C: most acidic O

O

O

N

O N

O

O

O

O

unlike charges nearby stabilizing

19.48 O CH3

C * OH

NaOH CH3

O

O

C * O

C * O

H3O

labeled O atom

CH3 +

H3O+ OH

O CH3

The resonance-stabilized carboxylate anion can now be protonated on either O atom, the one with the label and the one without the label.

C * OH

CH3

C * O

The label is now in two different locations.

19.49 O

a.

Ha

O

Hc

Hb

loss of Hb:

O

Hb

Hc

O

one Lewis structure least stable conjugate base

O Hc O

2 resonance structures intermediate stability

O Hc

Hb

The most acidic proton forms the most stable conjugate base.

O

1,3-cyclohexanedione increasing acidity: Hb < Ha < Hc loss of Ha:

Ha

O

O

loss of Hc: H a Hb

O

Ha O

Hb

Ha O

Hb

3 resonance structures most stable conjugate base

O

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 19–16

Hb N

N

b.

Hc O

Ha

N

Hc

loss of Hb: O

Ha

O

Ha

acetanilide N Ha

Hc

N

Hc O

Hb N

Hc O

Ha

7 resonance structures most stable conjugate base

increasing acidity: Ha < Hc < Hb

loss of Ha:

N

Hc

O

Ha

Hb

Hb

N

N

N

Hc Ha

N

loss of Hc:

O

O

Ha

one Lewis structure least stable conjugate base

O

Ha

Hc O

Ha

Hc O

2 resonance structures that delocalize the negative charge intermediate stability

19.50 O

O

O

H

H

H

H

H H

HO

O

O H

H

H

OH

HO

O

O

O

H

H

H

H

The conjugate base is resonance stabilized. Two of the structures place a negative charge on an O atom. weaker conjugate base stronger acid

The conjugate base has only one Lewis structure. stronger conjugate base weaker acid

O

19.51 The negatively charged C is more nucleophilic than the negatively charged O atom.

CH3COOH

CH2COO–

strong base (2 equiv)

CH3CH2CH2CH2 Br

X

CH3CH2CH2CH2CH2COO– H3O+

Two equivalents of strong base remove both the O–H and C–H protons.

CH3CH2CH2CH2CH2COOH

hexanoic acid

19.52 CH3

O

O

C

C

NH2

acetamide

CH3

CH3

N

CH3

C

N

O

C

C

CH3

O O

H

O is more electronegative than N, making the conjugate base of CH3COOH more stable than the conjugate base of acetamide. Therefore, acetamide is less acidic.

somewhat less stable with the (–) charge on N

O OH

O H

CH3

C

O

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

533

Carboxylic Acids and the Acidity of the O–H Bond 19–17

19.53 COOH

A

x x

B

x

Dissolve both compounds in CH2Cl2. Add 10% NaHCO3 solution. This makes a carboxylate anion (C10H7COO–) from B, which dissolves in the aqueous layer. The other compound (A) remains in the CH2Cl2. Separate the layers.

19.54 OH

OH

and

x x

x

Dissolve both compounds in CH2Cl2. Add 10% NaOH solution. This converts C6H5OH into a phenoxide anion, C6H5O–, which dissolves in the aqueous solution. The alcohol remains in the organic layer (neutral) since it is not acidic enough to be deprotonated to any significant extent by NaOH. Separate the layers.

19.55 To separate two compounds in an aqueous extraction, one must be water soluble (or be able to be converted into a water-soluble ionic compound by an acid–base reaction), and the other insoluble. 1-Octanol has greater than 5 C’s, making it insoluble in water. Octane is an alkane, also insoluble in water. Neither compound is acidic enough to be deprotonated by a base in aqueous solution. Since their solubility properties are similar, they cannot be separated by an extraction procedure. 19.56 O C one double bond or ring a. Molecular formula: C3H5ClO2 ClCH2CH2 OH C=O and O–H IR: 3500–2500 cm–1, 1714 cm–1 NMR data: 2.87 (triplet, 2 H), 3.76 (triplet, 2 H), and 11.8 (singlet, 1 H) ppm

b. Molecular formula: C8H8O3 5 double bonds or rings CH3O COOH IR: 3500–2500 cm–1, 1688 cm–1 C=O and O–H NMR data: 3.8 (singlet, 3 H), 7.0 (doublet, 2 H), 7.9 (doublet, 2 H), and 12.7 (singlet, 1 H) ppm para disubstituted benzene ring 5 double bonds or rings c. Molecular formula: C8H8O3 OCH2COOH IR: 3500–2500 cm–1, 1710 cm–1 C=O and O–H NMR data: 4.7 (singlet, 2 H), 6.9–7.3 (multiplet, 5 H), and 11.3 (singlet, 1 H) ppm monosubstituted benzene ring

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 19–18

19.57 Compound A: Molecular formula C4H8O2 (one degree of unsaturation) IR absorptions at 3600–3200 (O–H), 3000–2800 (C–H), and 1700 (C=O) cm–1 1 H NMR data: Absorption singlet singlet triplet triplet

ppm 2.2 2.55 2.7 3.9

# of H’s 3 1 2 2

Explanation a CH3 group 1 H adjacent to none or OH 2 H’s adjacent to 2 H’s 2 H’s adjacent to 2 H’s

Structure: O CH3

C

A

CH2CH2OH

Compound B: Molecular formula C4H8O2 (one degree of unsaturation) IR absorptions at 3500–2500 (O–H) and 1700 (C=O) cm–1 1 H NMR data: Absorption doublet septet singlet (very broad)

ppm 1.6 2.3 10.7

# of H’s 6 1 1

Explanation 6 H’s adjacent to 1 H 1 H adjacent to 6 H’s OH of RCOOH

Structure: CH3 CH3 C COOH H

B

19.58 Compound C: Molecular formula C4H8O3 (one degree of unsaturation) IR absorptions at 3600–2500 (O–H) and 1734 (C=O) cm–1 1 H NMR data: Absorption triplet quartet singlet singlet

ppm 1.2 3.6 4.1 11.3

# of H’s 3 2 2 1

Explanation a CH3 group adjacent to 2 H’s 2 H’s adjacent to 3 H’s 2 H’s OH of COOH

Structure: O O

OH

C

19.59 Compound D: Molecular formula C9H9ClO2 (five degrees of unsaturation) 13 C NMR data: 30, 36, 128, 130, 133, 139, 179 = 7 different types of C’s 1 H NMR data: Absorption triplet triplet two signals singlet

ppm 2.7 2.9 7.2 11.7

# of H’s 2 2 4 1

Explanation 2 H’s adjacent to 2 H’s 2 H’s adjacent to 2 H’s on benzene ring OH of COOH

Structure: COOH Cl

D

19.60 Molecular formula: C8H6O4: 6 degrees of unsaturation IR 1692 cm–1 (C=O) 1H NMR 8.2 and 10.0 ppm (singlets) aromatic H

COOH

O HO

O OH

535

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and the Acidity of the O–H Bond 19–19

19.61 A

COOH

COOH

B

O

C

OH

3 different C's Spectrum [2]: peaks at 27, 39, 186 ppm 5 different C's Spectrum [1]: peaks at 14, 22, 27, 34, 181 ppm 4 different C's Spectrum [3]: peaks at 22, 26, 43, 180 ppm

19.62 GBL: Molecular formula C4H6O2 (two degrees of unsaturation) IR absorption at 1770 (C=O) cm–1 1 H NMR data: Absorption multiplet triplet triplet

ppm 2.28 2.48 4.35

# of H’s 2 2 2

Explanation 2 H’s adjacent to several H’s 2 H’s adjacent to 2 H’s 2 H’s adjacent to 2 H’s

Structure: O O

GBL

19.63 HOOC

H C

threonine

OH

HOOC

H

C

H H2N

C H2N

CH3

OH

HOOC

C CH3

H

HOOC

C

H H2N

2S,3S

2R,3S

OH H C

C CH3

H2N

H

C

OH H CH3

2S,3R naturally occurring

2R,3R

19.64 NH

NH

COOH

NH2

COOH

proline

enantiomer

O

O

COO

zwitterion

19.65 a. methionine O H3N CH C OH

b. serine

O

O

H2N CH C O

CH2

CH2

CH2

CH2

CH2

CH2

CH2SCH3

CH2SCH3

CH2SCH3

OH

OH

OH

pH = 1

pH = 6 form at isoelectric point

pH = 11

H3N CH C OH

pH = 1

H3N CH C O

O

H3N CH C O

pH = 6 form at isoelectric point

19.66 a. cysteine pI = b. methionine pI =

pKa(COOH) + pKa(NH3+)

= (2.05) + (10.25) / 2 = 6.15

2 pKa(COOH) + pKa(NH3+) 2

H2N CH C O

= (2.28) + (9.21) / 2 = 5.75

pH = 11

536

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 19–20

19.67 O

O H2N

C

C OH N H

H2N H

lysine This lone pair is localized on the N atom, making it a base.

OH

H2N H

tryptophan This lone pair is delocalized in the S system to give 10S electrons, making it aromatic. This is similar to pyrrole (Chapter 17). Since these electrons are delocalized in the aromatic system, this N atom in tryptophan is not basic.

19.68 The first equivalent of NH3 acts as a base to remove a proton from the carboxylic acid. A second equivalent then acts as a nucleophile to displace X to form the ammonium salt of the amino acid. O R

OH X

O

O

NH3

R

acid–base reaction

NH3

O– NH4+

SN2 reaction

X

R

O– NH4+ NH3

19.69 a. At pH = 1, the net charge is (+1). NH3

b. increasing pH: As base is added, the most acidic proton is removed first, then the next most acidic proton, and so forth. NH3

C

HOOCCH2CH2

COOH H

NH3 C

OOCCH2CH2

COO– Na+

H

C

HOOCCH2CH2

c. monosodium glutamate

COOH H

base (1 equiv) NH3 C

HOOCCH2CH2

COO–

H

base (2nd equiv) NH3 C

OOCCH2CH2

COO–

H

base (3rd equiv) NH2 C

OOCCH2CH2

COO–

H

19.70 The first equivalent of NaH removes the most acidic proton—that is, the OH proton on the phenol. The resulting phenoxide can then act as a nucleophile to displace I to form a substitution product. With two equivalents, both OH protons are removed. In this case the more nucleophilic O atom is the stronger base—that is, the alkoxide derived from the alcohol (not the phenoxide), so this negatively charged O atom reacts first in a nucleophilic substitution reaction.

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

537

Carboxylic Acids and the Acidity of the O–H Bond 19–21

nucleophile most acidic proton

NaH

–

O

(CH2)4OH

–O

(CH2)4O–

(1 equiv)

[1] CH3I

CH3O

(CH2)4OH

[2] H2O

(CH2)4OH

HO

NaH (2 equiv)

[1] CH3I

HO

(CH2)4OCH3

[2] H2O

nucleophile

19.71 O HO

COOH

HO

C O

p-hydroxybenzoic acid less acidic than benzoic acid

+

HO

O C O

like charges on nearby atoms destabilizing The OH group donates electron density by its resonance effect and this destabilizes the conjugate base, making the acid less acidic than benzoic acid. O H

OH

O C

COOH

O

o-hydroxybenzoic acid more acidic than benzoic acid

Intramolecular hydrogen bonding stabilizes the conjugate base, making the acid more acidic than benzoic acid.

19.72

O

Hd OHe

HaO HbO Hc O

2-hydroxybutanedioic acid increasing acidity: Hd < Hc < Hb < He < Ha

Ha and He must be the two most acidic protons since they are part of carboxylic acids. Loss of a proton forms a resonance-stabilized carboxylate anion that has the negative charge delocalized on two O atoms. Ha is more acidic than He because the nearby OH group on the D carbon increases acidity by an electron-withdrawing inductive effect. Hb is the next most acidic proton because the conjugate base places a negative charge on the electronegative O atom, but it is not resonance stabilized. The least acidic H’s are Hc and Hd since these H’s are bonded to C atoms. The electronegative O atom further acidifies Hc by an electron-withdrawing inductive effect.

538

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 19–22

19.73 most acidic H OH

O

O

O

O

O

base O

O

O

O

O

A

O

B

O

O

O

O

C

The conjugate base has three resonance structures, two of which place a negative charge on the oxygens. In this way the conjugate base resembles a carboxylate anion. In addition, the C=C’s in A and C are conjugated.

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

539

Introduction to Carbonyl Chemistry 20–1

Chapter 20 Introduction to Carbonyl Chemistry Chapter Review Reduction reactions [1] Reduction of aldehydes and ketones to 1o and 2o alcohols (20.4) O R

C

OH

NaBH4, CH3OH H(R')

R C H(R')

or

H

[1] LiAlH4 [2] H2O

1o or 2o alcohol

or

H2, Pd-C

[2] Reduction of α,β-unsaturated aldehydes and ketones (20.4C) OH NaBH4 CH3OH

• reduction of the C=O only R

O

O H2 (1 equiv)

R

Pd-C

• reduction of the C=C only

R

OH H2 (excess) Pd-C

• reduction of both π bonds

R

[3] Enantioselective ketone reduction (20.6) [1] (S)- or (R)CBS reagent

O C

R

HO H C

[2] H2O

H OH

R

(R) 2o alcohol

C

or

R

(S) 2o alcohol

•

A single enantiomer is formed.

[4] Reduction of acid chlorides (20.7A) [1] LiAlH4 [2] H2O O R

C

RCH2OH

1o alcohol

•

LiAlH4, a strong reducing agent, reduces an acid chloride to a 1o alcohol.

•

With LiAlH[OC(CH3)3]3, a milder reducing agent, reduction stops at the aldehyde stage.

Cl O

[1] LiAlH[OC(CH3)3]3 [2] H2O

R

C

H

aldehyde

540

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–2

[5] Reduction of esters (20.7A) [1] LiAlH4

RCH2OH

[2] H2O

1o alcohol

O R

C

•

LiAlH4, a strong reducing agent, reduces an ester to a 1o alcohol.

•

With DIBAL-H, a milder reducing agent, reduction stops at the aldehyde stage.

OR' O

[1] DIBAL-H

C

R

[2] H2O

H

aldehyde

[6] Reduction of carboxylic acids to 1o alcohols (20.7B) O R

C

[1] LiAlH4 OH

[2] H2O

RCH2OH

1o alcohol

[7] Reduction of amides to amines (20.7B) O R

C

[1] LiAlH4 [2] H2O

N

RCH2 N

amine

Oxidation reactions Oxidation of aldehydes to carboxylic acids (20.8) • O R

C

O

CrO3, Na2Cr2O7, K2Cr2O7, KMnO4 H

or Ag2O, NH4OH

C

R

•

OH

carboxylic acid

Preparation of organometallic reagents (20.9) [1] Organolithium reagents: R X [2] Grignard reagents:

All Cr6+ reagents except PCC oxidize RCHO to RCOOH. Tollens reagent (Ag2O + NH4OH) oxidizes RCHO only. Primary (1°) and secondary (2°) alcohols do not react with Tollens reagent.

R X

+

2 Li

R

Li

+ Mg

(CH3CH2)2O

[3] Organocuprate reagents:

R X 2 R Li

+

2 Li

+ CuI

+

LiX

R Mg X

R Li + LiX R2Cu Li+

+

LiI

541

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Introduction to Carbonyl Chemistry 20–3

[4] Lithium and sodium acetylides:

Na+ –NH2

R C C H

R C C

Na+

+

NH3

a sodium acetylide R Li

R C C H

R C C Li

+ R H

a lithium acetylide

Reactions with organometallic reagents [1] Reaction as a base (20.9C) R M

+

H O R

+

R H

M+

• •

O R

RM = RLi, RMgX, R2CuLi This acid–base reaction occurs with H2O, ROH, RNH2, R2NH, RSH, RCOOH, RCONH2, and RCONHR.

[2] Reaction with aldehydes and ketones to form 1o, 2o, and 3o alcohols (20.10) O R

C

OH

[1] R"MgX or R"Li

R C H (R')

[2] H2O

H (R')

R"

1o, 2o, or 3o alcohol

[3] Reaction with esters to form 3o alcohols (20.13A) [1] R"Li or R"MgX (2 equiv)

O R

C

OR'

[2] H2O

OH R C R" R"

3o alcohol

[4] Reaction with acid chlorides (20.13) [1] R"Li or R"MgX (2 equiv) [2] H2O O R

C

OH

•

More reactive organometallic reagents—R"Li and R"MgX—add two equivalents of R" to an acid chloride to form a 3o alcohol with two identical R" groups.

•

Less reactive organometallic reagents— R'2CuLi—add only one equivalent of R' to an acid chloride to form a ketone.

R C R" R"

3o alcohol Cl O [1] R'2CuLi [2] H2O

R

C

R'

ketone

542

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–4

[5] Reaction with carbon dioxide—Carboxylation (20.14A) O

[1] CO2

R MgX

R C

[2] H3O+

OH

carboxylic acid

[6] Reaction with epoxides (20.14B) OH

O C

[1] RLi, RMgX, or R2CuLi

C

C

[2] H2O

C

R

alcohol

[7] Reaction with α,β-unsaturated aldehydes and ketones (20.15B)

[1] R'Li or R'MgX

OH R

[2] H2O R

C

C

C

•

More reactive organometallic reagents— R'Li and R'MgX—react with α,βunsaturated carbonyls by 1,2-addition.

•

Less reactive organometallic reagents— R'2CuLi— react with α,β-unsaturated carbonyls by 1,4-addition.

R'

O C

C

allylic alcohol

C

O H C C R

[1] R'2CuLi [2] H2O

C R'

ketone

Protecting groups (20.12) [1] Protecting an alcohol as a tert-butyldimethylsilyl ether CH3 R O H

+ Cl

CH3

Si C(CH3)3 CH3

R O Si C(CH3)3 N

CH3

NH

R O TBDMS

Cl TBDMS

tert-butyldimethylsilyl ether

[2] Deprotecting a tert-butyldimethylsilyl ether to re-form an alcohol CH3 R O Si C(CH3)3

CH3

(CH3CH2CH2CH2)4N+ F– R O H

+

F Si C(CH3)3

CH3 R O TBDMS

CH3 F

TBDMS

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

543

Introduction to Carbonyl Chemistry 20–5

Practice Test on Chapter Review 1. Which compounds undergo nucleophilic addition and which undergo substitution? O

a. CH3

C

O

b. CH3

CH3CH2CH2

C

O

O

c. Cl

CH3

C

C

d. OCH3

H

2. What product is formed when CH3CH2CH2Li reacts with each compound, followed by quenching with water and acid? a. CH3CH2CHO b. (CH3)2CO c. CH3CH2CO2CH3 d. CH3CH2COCl

e. CO2 f. CH2=CHCOCH3 g. ethylene oxide h. CH3COOH

3. What product is formed when HO(CH2)4CHO is treated with each reagent? a. NaBH4, CH3OH b. PCC

c. Ag2O, NH4OH d. Na2Cr2O7, H2SO4, H2O

4. What reagent is needed to convert (CH3CH2)2CHCOCl into each compound? a. (CH3CH2)2CHCOCH2CH3 b. (CH3CH2)2CHCHO

c. (CH3CH2)2CHC(OH)(CH2CH3)2 d. (CH3CH2)2CHCH2OH

5. Draw the organic products formed in the following reactions. OH N

a.

[1] LiAlH4

[2] CH3Li [3] H2O [Indicate stereochemistry.]

c. O

[2] H2O

[1] TBDMS–Cl, imidazole

O O

[2]

b.

[1] Mg

OCH3

d.

Br

[3] H3O+

O

O

NaBH4 CH3OH

544

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–6

6. What starting materials are needed to synthesize each compound using the indicated reagent or functional group? OH OH

c. Synthesize:

a. Synthesize:

using a Grignard reagent

from an ester O

b. Synthesize:

using an organocuprate reagent

Answers to Practice Test 1. a. addition 4.a. (CH3CH2)2CuLi b. substitution b. DIBAL-H c. substitution c. CH3CH2MgBr d. addition d. LiAlH4

5.

6. CO2R

a.

N

a.

+ CH3MgBr O

2. a. CH3CH2CH(OH)CH2CH2CH3 b. (CH3)2C(OH)CH2CH2CH3 c. (CH3CH2)3COH d. (CH3CH2)3COH e. CH3CH2CH2CO2H f. CH2=CHC(OH)(CH3)CH2CH2CH3 g. CH3(CH2)4OH h. CH3CH2CH3 + CH3CO2H

OH

b.

b.

CuLi

2

OTBDMS

c.

HO

3. a. HO(CH2)5OH b. OHC(CH2)3CHO c. HO(CH2)4CO2H d. HO2C(CH2)3CO2H

c. OTBDMS

MgBr + CH3CHO

HO

OCH3

d. OH

O

Answers to Problems 20.1 [1] Csp2–Csp2

a.

H

[3] Csp2–Csp2

[2] σ: C –O π: Cp–Op sp2

α-sinensal

O

sp2

b. The O is sp2 hybridized. Both lone pairs occupy sp2 hybrid orbitals.

545

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Introduction to Carbonyl Chemistry 20–7

20.2

A carbonyl compound with a reasonable leaving group (NR2 or OR bonded to the C=O) undergoes substitution reactions. Those without good leaving groups undergo addition. O

no good leaving group addition reactions

O O

O

O

OH N

O

H

All other C=O's have leaving groups. substitution reactions

OH HO O

O

H O

O O

20.3

Aldehydes are more reactive than ketones. In carbonyl compounds with leaving groups, the better the leaving group, the more reactive the carbonyl compound. O

a.

CH3CH2CH2

O

C

H

or

CH3CH2CH2

C

O

c.

CH3

O CH3CH2

C

O

or CH3

CH3CH(CH3)

C

CH3

C

OCH3

CH2CH3

CH3

C

O

or OCH3

CH3

C

NHCH3

better leaving group more reactive

NaBH4 reduces aldehydes to 1° alcohols and ketones to 2° alcohols. O

a.

Cl

O

d.

less hindered carbonyl more reactive

20.4

O

or

better leaving group more reactive

less hindered carbonyl more reactive b.

CH3CH2

C

CH3CH2CH2

C

OH

NaBH4 H

CH3OH

CH3CH2CH2 C H

NaBH4

c.

H

CH3OH

O

OH

NaBH4 b.

20.5

O

OH

CH3OH

1° Alcohols are prepared from aldehydes and 2° alcohols are from ketones. H OH

O

a.

b.

OH

O

OH

c.

20.6 OH

1-methylcyclohexanol

3° Alcohols cannot be made by reduction of a carbonyl group, because they do not contain a H on the C with the OH.

O

546

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–8

20.7 O

a.

H2 (excess)

d.

[2] H2O O

OH

O

OH

[1] LiAlH4

Pd-C

OH

NaBH4

O

NaBH4 (excess)

e.

b.

CH3OH

CH3OH O

O

O

H2 (1 equiv)

c.

OH

D OH

NaBD4

f.

CH3OH

Pd-C

20.8 O

OH

CH3OH NaBH4

CHO

b.

CH3OH

c. (CH3)3C

20.9

OH

NaBH4

a.

OH NaBH4

O

(CH3)3C

CH3OH

OH

(CH3)3C

OH

The 2° alcohol comes from a carbonyl group. Since hydride was delivered from the back side, the (R)-CBS reagent must be used. O

O O

H OH

H drawn back

F

N

N

F

2° OH O

O

F

X

+ (R)-CBS

F

20.10 Part [1]: Nucleophilic substitution of H for Cl O CH3

C

O Cl

[1]

H3Al H

CH3 C Cl H

O [2]

CH3

C

+

H

aldehyde

+ AlH3

Cl–

H replaces Cl.

Part [2]: Nucleophilic addition of H– to form an alcohol O

O CH3

C

H3Al H

H

CH3 C H [3]

H + AlH3

H OH [4]

OH CH3 C H

+ –OH

H

1o alcohol

20.11 Acid chlorides and esters can be reduced to 1° alcohols. Keep the carbon skeleton the same in drawing an ester and acid chloride precursor.

547

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Introduction to Carbonyl Chemistry 20–9

O O CH2OH

a.

C

C

or Cl

OH

CH2OH

OCH3

Cl

b.

C

O

or O CH3O

or

O

CH3O

c.

OCH3

Cl

C

O

OCH3

CH3O

20.12 O

a.

O

[1] LiAlH4 OH

C

c.

OH

[2] H2O

N(CH3)2

[1] LiAlH4

CH2N(CH3)2

[2] H2O O

O

b.

[1] LiAlH4 NH2

NH2

[2] H2O

NH

d.

[1] LiAlH4

NH

[2] H2O

20.13 O CH2NH2

C

a.

O NH2

c.

O N

b.

CH2CH3

C

CH2CH3

N H

O N

CH2CH3

CH2CH3

N H

N

or

CH2CH3

20.14 O COOCH3

a.

[1] LiAlH4

O

b. CH3O

OH

NaBH4

COOCH3

CH3OH

[1] LiAlH4 OH [2] H2O NaBH4 CH3OH

HO

OH

+ HOCH3

Neither functional group reduced

20.15 CO2CH2CH3

a. O

b.

CO2CH2CH3 O

H2 (1 equiv) Pd-C

CO2CH2CH3 O

H2 (2 equiv) Pd-C

[1] LiAlH4

CO2CH2CH3 OH

CH3O

OH

CH3O

OH

[2] H2O

+ HOCH3

CH3OH O

O

c.

OH

[2] H2O NaBH4

CH3O

OH

548

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–10

CO2CH2CH3

c.

[1] LiAlH4 + CH3CH2OH

OH

[2] H2O

O

OH

d.

NaBH4

CO2CH2CH3

CO2CH2CH3

CH3OH O

OH

20.16 Tollens reagent reacts only with aldehydes. Ag2O, NH4OH

a.

C O

COOH

Na2Cr2O7 H2SO4, H2O

O

CHO

b.

CH2OH

OH

Ag2O, NH4OH

OH

No reaction

OH

O

Na2Cr2O7 H2SO4, H2O

C

OH

20.17 O OH CHO

c. B

CHO

PCC O

HO

B

H

OH

OH a. B

CH2OH

NaBH4

d. B

O

Ag2O, NH4OH

C HO

CH3OH HO

O

O

OH b. B

[1] LiAlH4 [2] H2O

OH

CH2OH

e. B

C

CrO3 H2SO4, H2O

HO

OH

O OH

20.18 a. CH3CH2Br + 2 Li

c. CH3CH2Br + 2 Li

CH3CH2Li + LiBr

b. CH3CH2Br + Mg

CH3CH2MgBr

CH3CH2Li

2 CH3CH 2Li + CuI

+ LiBr

LiCu(CH2CH3)2 + LiI

20.19 HC CCH2CH2CH2CH2CH2CH3 + NaH

Na+ C CCH2CH2CH2CH2CH2CH3 + H2

hydrogen gas

HC CCH2CH2CH2CH2CH2CH3

BrMgC CCH2CH2CH2CH2CH2CH3 + CH4

methane gas

+ CH3MgBr

20.20 a.

b.

Li + H2O CH3 CH3 C MgBr + H2O CH3

+ LiOH

c.

MgBr

CH3 CH3 C H CH3

+ HOMgBr

d. CH3CH2C C Li

+ H2O

+ H2O

+ HOMgBr

CH3CH2C CH + LiOH

549

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Introduction to Carbonyl Chemistry 20–11

20.21 To draw the products, add the alkyl or phenyl group to the carbonyl carbon and protonate the oxygen. [1] CH3CH2CH2Li

a.

[2] H2O

O

b.

H

C

Li

[1] H

O [2] H2O

HO CH2CH2CH3

OH

O

OH

[1] C6H5Li

c.

d.

[1] CH2=O

Na+

C C

C C CH2OH

[2] H2O

H C H

[2] H2O

20.22 Addition of RM always occurs from above and below the plane of the molecule. H OH

O

a.

CH3

C

H

CH3

b.

H OH

[1] CH3CH2MgBr

+

[2] H2O O

[1] CH3CH2Li

CH2CH3

CH3

+

OH

CH3

OH

[2] H2O

CH2CH3

20.23 OH OH

a.

O

CH3 C CH3

CH3MgBr +

H

b.

c.

CH2OH

H

MgBr

C

CH3

+

C

+ CH3CH2MgBr

H

+ H

C

O

OH

O

MgBr

O

OH

O

C CH2CH3

MgBr

+

d.

H

OH

O

+ CH3MgBr

CH2CH3

H OH

or O

C CH2CH3

+ CH3CH2MgBr H

H

20.24 OH

O

a.

+ CH3Li

linalool (three methods)

OH

Li

O Li

lavandulol O

+ CH2=O Li

550

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–12

b.

c. Linalool is a 3° ROH. Therefore, it has no H on the carbon with the OH group, and cannot be prepared by reduction of a carbonyl compound.

CHO

OH NaBH4 CH3OH

20.25 N(CH3)2

N(CH3)2

OH

O

BrMg

OCH3

OCH3

venlafaxine

20.26 CH3 O

CH3 O TBDMS–Cl

CH3OH C CH [1] Li C CH

imidazole [2] H2O HO

TBDMSO

TBDMSO

estrone

(CH3CH2CH2CH2)4NF CH3OH C CH

ethynylestradiol HO

20.27 O

a.

CH3CH2

C

Cl

b.

CH3CH2 C CH2CH2CH2CH3 CH2CH2CH2CH3

[2] H2O

O C

OH

[1] CH3CH2CH2CH2MgBr (2 equiv)

OCH3

[1] CH3CH2CH2CH2MgBr (2 equiv)

OH CH3CH2CH2CH2 C CH2CH2CH2CH3

[2] H2O O

c.

OCH2CH3

[1] CH3CH2CH2CH2MgBr (2 equiv)

OH CH3CH2CH2CH2 C CH2CH2CH2CH3

[2] H2O

CH2CH2CH2CH2CH3

20.28 O

OH

a.

C CH3

CH3O

C

+

MgBr

CH3 (2 equiv)

O

b. (CH3CH2CH2)3COH

CH3O

C

CH2CH2CH3 + CH3CH2CH2MgBr (2 equiv)

551

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Introduction to Carbonyl Chemistry 20–13

O

OH

c. CH3 C CH2CH(CH3)2

C

CH3O

CH2CH(CH3)2

CH3

+ CH3MgBr (2 equiv)

20.29 The R group of the organocuprate has replaced the Cl on the acid chloride. O

a. CH3CH2

Cl

C

CH3CH2

[2] H2O

O

[1] (Ph)2CuLi

c. CH3CH2 C Cl

CH3

[2] H2O

Ph

O

[1] [CH3CH2CH(CH3)]2CuLi

O

b.

O

O

[1] (CH3)2CuLi

C

[2] H2O

Cl

20.30 O

a. (CH3)2CHCH2

Cl [2] H O 2

O

O

[1] LiAlH[OC(CH3)3]3

C

(CH3)2CHCH2

C

H

c. (CH3)2CHCH2

O

b.

(CH3)2CHCH2

C

[2] H2O O

C

Cl

[1] CH3CH2Li (2 equiv)

HO

d.

(CH3)2CHCH2

O

[1] (CH3CH2)2CuLi Cl

C

[1] LiAlH4 Cl

[2] H2O

(CH3)2CHCH2CH2OH

[2] H2O

20.31 O C

a.

O 2CuLi

CH3

Cl

C

CH3

Cl + [(CH3)3C]2CuLi

b. O

or O C

O

or O C

CH3

(CH3)2CuLi Cl

Cl O

+ [(CH3)2CH]2CuLi O

20.32 O [1] Mg

Br

MgBr

[2] CO2

a.

b. c. CH3O

Cl

[1] Mg

CH2Br

MgCl [1] Mg

CH3O

C

O O

[2] CO2

[3] H3O+

COO–

CH2MgBr

[2] CO2

C

OH

[3] H3O+

CH3O

COOH

CH2COO– [3] H3O+

CH3O

CH2COOH

552

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–14

20.33 OH

a.

O

c.

O

O

OH

b.

OH

OH

O

d.

(+ enantiomer)

20.34 The characteristic reaction of α,β-unsaturated carbonyl compounds is nucleophilic addition. Grignard and organolithium reagents react by 1,2-addition and organocuprate reagents react by 1,4-addition. O

O CH3

O

a.

[1] (CH3)2CuLi

c.

[2] H2O

CH3

O

CH3

[1] (CH3)2CuLi

[2] H2O

CH3 CH3 HO C CH CH3

[1] H C C Li [2] H2O

HO C CH [1] H C C Li [2] H2O

CH3

[1] (CH3)2CuLi

b.

[2] H2O

O

O

CH3

[1] H C C Li [2] H2O HO C CH

20.35 a. CH3CH2OH

HBr or PBr3 PCC

OH

Mg

CH3CH2Br

O

CH3CH2MgBr

[1] CH3CH2MgBr

OH

[2] H2O

b.

HBr

OH

Br

(from a.) c.

OH

H2SO4

H2

+

Pd-C

(from a.) d.

OH

HBr or PBr3

Br

O MgBr

MgBr H2O

CH3CH2OH

e.

Mg

PCC

CH3CHO

OH

H2O

(from d.) CH3CH2OH

H2SO4

CH2=CH2

mCPBA

O

PCC OH

O

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

553

Introduction to Carbonyl Chemistry 20–15

20.36 O

OH a. NaBH4 CH3OH

A

OH

b. [1] LiAlH4 [2] H2O

OH

c. [1] CH3MgBr (excess) [2] H2O

HO C6H5

d. [1] C6H5Li (excess) [2] H2O e. Na2Cr2O7

No reaction

H2SO4 H2O O CH3O OCH3

a. NaBH4

No reaction

CH3OH

B

CH3O

b. [1] LiAlH4

OH

[2] H2O CH3O

c. [1] CH3MgBr (excess)

OH

[2] H2O d. [1] C6H5Li (excess)

C6H5 C6H5

CH3O

OH

[2] H2O e. Na2Cr2O7 H2SO4 H2O

No reaction

20.37 PBr3

a.

OH

Mg Br O

PCC H

[1] BrMg CH2CH3 [2] H2O

OH

CH3OH PBr3 CH3Br Mg O

b.

[1] CH3MgBr H

(from a.)

[2] H2O

OH

PCC

O

[1] CH3MgBr [2] H2O

OH

554

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–16

OH

Br

PBr3

[1] Mg

c. (from b.) CH3OH

d.

OH

OH

[2] H2C=O [3] H2O

H2SO4

PCC

CH2 CH2

O

mCPBA

[1] BrMg CH2CH3 OH

(from a.) [2] H2O

20.38 O

O

NaBH4

a.

H

CH3OH

OH

O H

[1] LiAlH4

OH

h.

H [2] H2O O

H2

H

OH

[1] (CH3)2CuLi

i. H

Pd-C

No reaction [2] H2O

O

d.

O

PCC

H

No reaction

O H

OH

H2SO4, H2O

[2] H2O

OH

NH4OH

OH

[1] CH3C≡CLi H

[2] H2O

O

Ag2O

H

H

k.

OH

[1] HC≡CNa

O

O

f.

j.

O

Na2Cr2O7

e.

OH

[1] C6H5Li

[2] H2O

O

c.

H [2] H2O

O

b.

OH

[1] CH3MgBr

g.

TBDMSCl

l.

OH

O–TBDMS

imidazole

20.39 Li (2 equiv)

a.

Br

b.

Br

c.

Mg

Br

+

Li

LiBr

MgBr

[1] Li (2 equiv) [2] CuI (0.5 equiv)

(CH3CH2CH2CH2)2CuLi

H2O

d.

Li

e.

MgBr

f.

+ LiOH

D2 O

D

+ DOMgBr

CH3C≡CH

+

Li

LiC≡CCH3

20.40 a.

MgBr

CH2 O

H2 O OH

d. b.

MgBr

O

MgBr CH3CH2COOCH3

OH H2 O

HO

H2O

e.

MgBr CH3COOH

f.

MgBr

+ CH3COO

OH

c.

MgBr

CH3CH2COCl

H2O

HC CH

HC C

555

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Introduction to Carbonyl Chemistry 20–17

g.

MgBr

h.

MgBr

O

H3O+

CO2

OH

i.

MgBr

j.

MgBr

D2O

O

OD

O

OH

H2O

D +

H2O

OH

20.41 O C

O Cl

a.

C

(CH3CH2CH2CH2)2CuLi

CH2CH2CH2CH3

O C

OCH3

b.

(CH3CH2CH2CH2)2CuLi

No reaction

O

O CH3

c.

CH3

[1] (CH3CH2CH2CH2)2CuLi [2] H2O

O

d.

CH3

[1] (CH3CH2CH2CH2)2CuLi

CH3

[2] H2O

CH2CH2CH2CH3 HO

20.42 Arrange the larger group [(CH3)3C–] on the left side of the carbonyl. HO H

H OH

a.

NaBH4, CH3OH

S

O

R

HO H

b.

[1] (S)-CBS reagent; [2] H2O

c.

[1] (R)-CBS reagent; [2] H2O

R

H OH S

20.43 HO

OH CH3

a. NaBH4, CH3OH

C

d. [1] CH3Li; [2] H2O

O

O

CH3 C

CH3 HO

CH2CH3 C

CH3 b. H2 (1 equiv), Pd-C

CH3

e. [1] CH3CH2MgBr; [2] H2O

OH

O

A c. H2 (excess), Pd-C

CH3

CH3

C

f.

[1] (CH2=CH)2CuLi; [2] H2O

CH3

CH=CH2

556

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–18

20.44 O

a. (CH3)2CHCH2CH2

C

Cl

C

c. (CH3)2CHCH2CH2

C

H

O [1] (CH2=CH)2CuLi Cl

(CH3)2CHCH2CH2

[2] H2O [1] C6H5MgBr (2 equiv)

O C

(CH3)2CHCH2CH2

[2] H2O

O

b. (CH3)2CHCH2CH2

O

[1] LiAlH[OC(CH3)3]3

Cl

O

CH=CH2

OH (CH3)2CHCH2CH2

C C6H5

[2] H2O

C6H5 OH

[1] LiAlH4

d. (CH ) CHCH CH C Cl 3 2 2 2

C

(CH3)2CHCH2CH2

[2] H2O

C H H

20.45 O

a.

CH3CH2

C

[1] LiAlH4 OCH2CH2CH3

O

b.

CH3CH2

C

[1] CH3CH2CH2MgCl (2 equiv) OCH2CH2CH3

O

c.

CH3CH2

C

CH3CH2CH2OH

[2] H2O OH CH3CH2 C CH2CH2CH3

[2] H2O

CH2CH2CH3 O

[1] DIBAL-H OCH2CH2CH3

CH3CH2

[2] H2O

C

H

20.46 a.

NaBH4

O COOCH3

OH

O

[1] LiAlH4

c.

CH3OH

COOCH3

OH

(CH3)2N

[2] H2O

OH

(CH3)2N

O [1] LiAlH4

b. O

OH

d.

COOCH3 [2] H2O

CH2OH

[1] LiAlH[OC(CH3)3]3

O OH

Cl

[2] H2O

O OH

H O

O

20.47 CH3

a. CH3

CH3

MgBr

[1] CO2

CH3

COCl

COOH

c.

C

CH3

O [1] CH3CH2MgBr

b.

OH

[2] H2O

[2] H3O+ CH3

[1] C6H5MgBr (excess)

[2] H2O

OH

d.

COOCH2CH3

[1] CH3MgCl (excess)

CH3

[2] H2O

OH

C CH3

557

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Introduction to Carbonyl Chemistry 20–19

e.

OH

[1] CH2=O

MgBr

O

CH2OH

g.

[2] H2O O

f.

CH3

[2] H2O OH

O

[1] (CH3)2CuLi

[1] CH3MgBr

O

h.

[2] H2O

[1] (CH3)2CuLi

C6H5

[2] H2O

20.48 [1] CH3Li

a. (CH3)3C

O [2] H2O

b.

(CH3)3C

[1] CH3CH2MgBr

O

OH

OH

OH

CH3

CH3 CH=CH2

OH

c.

O

OH

CH2CH3

+

CH2CH3

[2] H2O

+ (CH3)3C

[1] (CH2=CH)2CuLi

+

[2] H2O

OH

CH=CH2

[1] Mg

d.

Br

COOH [2] CO2

H

H

[3] H3O+

O

H OH

[1] (S)-CBS reagent

e. [2] H2O

O

f.

OCH2CH3 H

[1] LiAlH4

CH2OH + HOCH2CH3

[2] H2O

H

20.49 Three carbons bear a δ+ because they are bonded to electronegative O atoms. C3 O

C2

O

O

C1

C1 is an sp3 hybridized C bonded to an O so it bears a δ+. There are no additional resonance structures that affect C1. C2 is part of a carbonyl that has three resonance structures, only one of which places a (+) charge on C. The O atom of the ester donates its electron pair, making the carbonyl C less electrophilic than C3. C3 is most electrophilic. Its carbonyl is stabilized by two resonance structures, one of which places a (+) charge on carbon.

558

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–20

O

O

O

C2 O

O

O

O

O

O

O donates electron density. C3 O O

O

O

O

full (+) charge most electrophilic site

O

20.50 The organolithium reagent is a nucleophile and a base. As a base it can remove the most acidic proton (between the benzene ring and C=O) to form an enolate that is protonated by D3O+. OCH3 H

OCH3

OCH3 D

O

O

O

D OD2

+ D2O

HC C Li OCH3

OCH3

OCH3

HC CH

B C12H13DO3

Li

A OCH3

O

OCH3

20.51 Both ketones are chiral molecules with carbonyl groups that have one side more sterically hindered than the other. In both reductions, hydride approaches from the less hindered side. The CH3 groups on the bridgehead carbon make the top more hindered. H– attacks from below to afford an exo OH group. Attack comes from below.

2 H's less hindered Attack comes from above. H

H CH3

CH3

[1] LiAlH4 H

CH3

H

[2] H2O O

[1] LiAlH4

from above

OH

endo OH group

The concave shape of the six-membered ring makes the bottom face of the C=O more sterically hindered. Addition of the H– occurs from above to place the new C–H bond exo, making the OH endo.

O

OH

[2] H2O H

from

exo OH group below

559

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Introduction to Carbonyl Chemistry 20–21

20.52 Since a Grignard reagent contains a carbon atom with a partial negative charge, it acts as a base and reacts with the OH of the starting halide, BrCH2CH2CH2CH2OH. This acid–base reaction destroys the Grignard reagent so that addition cannot occur. To get around this problem, the OH group can be protected as a tert-butyldimethylsilyl ether, from which a Grignard reagent can be made. basic site Br

OH

Mg

acidic site

δ−

OH

BrMg

proton transfer

CH3

O

+ HOMgBr

These will react.

INSTEAD: Use a protecting group. Br

OH

TBDMS–Cl imidazole

OTBDMS

Br

Mg ether

BrMg

OTBDMS O [1]

protected OH

[2] H2O OH

(CH3CH2CH2CH2)4NF OH

OH OTBDMS

A

20.53 Compounds F, G, and K are all alcohols with aromatic rings so there will be many similarities in their proton NMR spectra. These compounds will, however, show differences in absorptions due to the CH protons on the carbon bearing the OH group. F has a CH2OH group, which will give a singlet in the 3–4 ppm region of the spectrum. G is a 3o alcohol that has no protons on the C bonded to the OH group so it will have no peak in the 3–4 ppm region of the spectrum. K is a 2o alcohol that will give a doublet in the 3–4 ppm region of the spectrum for the CH proton on the carbon with the OH group.

560

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–22

[1] O3 [2] CH3SCH3

C

O

CHO O

OH

[1] Mg H2SO4

[1] C6H5MgBr

HCl

Cl

[2] CH2=O

D

[2] H2O

CH2OH F

[3] H2O

B

A

mCPBA

[1] LiAlH4

[1] (CH3)2CuLi

O

E

[2] H2O

OH

Br PBr3

G OH CH3

[2] H2O

OH

MgBr [1] C6H5CHO

Mg

[2] H2O

I

H

J

K

20.54 O Cl

O

R Si R

H OH

[1] (R)-CBS reagent

R Si

[2] H2O

R

O

H OH Cl

O

NaI

R Si

A

O

I

O

R

B

O

(CH3CH2)3SiCl imidazole H OH HO

H N

KF

H OSi(CH2CH3)3 NHCH(CH3)2

O

R Si R

HO

O

O

R Si

(CH3)2CHNH2

D

H OSi(CH2CH3)3 I

R

C

O

20.55 H OH

O

a.

O O

CH3 MgBr

O

+ + MgBr

O

O O

O

H OH

+

+ MgBr

CH3 MgBr

OH HOMgBr

+ OH

561

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Introduction to Carbonyl Chemistry 20–23

H OH

H OH O

O

b.

O

O

O

O

CH2CH2CH2CH2

MgBr

O

BrMg

O

+

+ + MgBr

BrMg CH2CH2CH2CH2 MgBr

+ MgBr OH

HOMgBr

OH

+

20.56 O MgBr

O

+ CH O C OCH 3 3

O

CH3O C OCH3

CH3O

C

+ CH3O

MgBr MgBr HO H

O O

O

C

C

CH3O C

MgBr

MgBr

+ CH3O

MgBr

H2O OH C

+ CH3OH + HOMgBr

20.57 O O

O

O H

H AlH3

O

O

O

O H AlH3

H

O

O

H

O O

O

O

OH

HO OH

H AlH3

O H2O

Protonate four alkoxides.

O

O

O

O O O

+ AlH3

O

AlH3

OH

H O

+ AlH3

+ AlH3

O O

H

O O H

O H AlH3

562

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–24

20.58 MgBr

CH2OH

a.

O H

C

BrMg

d.

H

HO

O

or b.

H

MgBr OH

or H

O MgBr

O

or

MgBr

CH3MgBr

O OH

c.

(C6H5)2C=O

(C6H5)3COH

O

BrMg

O MgBr

e.

20.59 (a)

(a)

N

N

(c)

(b)

MgBr

O

OH

(b) BrMg

N

O

(c) N

MgBr

O

20.60 O

OH

a.

C

CH3O

O

C

MgBr

(2 equiv)

CH3

CH3O

C

CH3O

C

CH3

BrMg

(2 equiv)

O

OH

b. CH3 C CH2CH2CH(CH3)2

c. (CH3CH2CH2CH2)2C(OH)CH3

CH2CH2CH(CH3)2

+ CH3–MgBr

(2 equiv)

563

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Introduction to Carbonyl Chemistry 20–25

20.61 OH Li

O

OH

+

a.

H

(two ways)

C CH2CH3

O

or

(three ways) H

+

Li

O

OH

b.

c.

O

C +

or

Li

+

O

or (three ways)

O Li CH2CH3 +

Li

+

Li CH2CH3

O

or

or +

Li

CH3CH2

O C

Li OCH3

+

(2 equiv)

20.62 OH

a.

+

MgBr

O

HO

H

H

c.

O

C6H5

b.

O

H

+ BrMg C6H5

H

+

OH

MgBr

20.63 C6H5

C6H5

C6H5

Br

Mg

Br

KOC(CH3)3

Br

C6H5

H2O

MgBr

C6H5 H2

C6H5

C6H5

Pd-C

LiAlH4 C6H5

20.64 Br

Mg

O

MgBr

OH

H2O

H CHO

MgBr Mg Br

564

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–26

20.65 OH

Br

PBr3

a.

MgBr

Mg

CH2OH

[1] CH2=O [2] H2O

MgBr

COOH

[1] CO2

b.

[2] H3O+

(from a.) OH

c.

O

CrO3

[1] CH3MgBr

H2SO4, H2O

H2SO4

OH

[2] H2O

mCPBA

O

major product OH

d.

mCPBA

H2SO4

O

OH

[1] C6H5MgBr [2] H2O

O [1] CH3CH2CH2MgBr

e.

Pd-C

+

(from c.)

O

OH

[1] CH3MgBr

H2SO4

[1] O3

CH3

[2] H2O

O CHO

[2] (CH3)2S

(from c.)

C6H5

H2

[2] H2O

f.

C6H5

H2SO4

OH

O

PCC

major product OH

[1]

O

MgBr

OH

O

(from a.) H

PCC

g.

PCC [2] H2O

(from a.) [1]

MgBr

(from a.)

O

OH H2SO4

h. [2] H2O

major product

(from c.)

20.66 a.

OH

Br

PBr3

MgBr

[1] O OH

b. [2] H2O

(from a.)

Mg OH

[1] CH3CHO [2] H2O

MgBr

OH

c.

PCC

O

[1] [2] H2O

MgBr OH

565

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Introduction to Carbonyl Chemistry 20–27

MgBr

[1]

O

d. (from a.)

OH

[1] 2 Li

e.

[(CH3)2CH]2CuLi

Br [2] CuI (0.5 equiv)

[2] H2O

(from a.) [1] O [2] H2O O

20.67 OH H

O

H

O

H

H

PCC H

[1] NaH

H

H

HO

H

H

[2] CH3I

HO

H

CH3O

estradiol [1] HC CLi [2] H2O OH C CH H H

H

CH3O

mestranol

20.68 Br

a.

CH3CH2Cl

Br2

AlCl3

hν

Br2

Br

K+ –OC(CH3)3

OH

[1] BH3 [2] H2O2, –OH

MgBr

Mg

FeBr3

O

OH

[1] [2] H2O

CH3Cl

Br2

AlCl3

hν

OH Br Mg

MgBr [1] H2C=O [2] H2O

566

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–28

O

OH MgBr [1] CH3CH=O

b.

PCC

[2] H2O

(from a.)

O

O

CH3

Cl

CH3Br

AlCl3

O

Br NaOH

OH PCC

O

OH

Mg [1] CH3MgBr

H

PCC

[2] H2O

(from a.)

20.69 Br

Br2

a.

MgBr

Mg

[2] H3O+

FeBr3

MgBr

b.

COOH

[1] CO2

CH3OH PCC

OH CH2OH

[1] CH2=O

CHO [1] PhMgBr

PCC

(from a.)

[2] H2O

O PCC

[2] H2O

(from a.) PBr3

c. CH3CH2CH2OH

Mg CH3CH2CH2Br

PCC

CH3CH2CH2OH

CH3CH2CH2MgBr [1] CH3CH2CH2MgBr

CH3CH2CHO HO

[1] PhMgBr

PCC

[2] H2O

OH

O

(from a.) [2] H2O

O

d.

PCC

OH

H

OH

O

O

H2O

O Br2

PCC

Br

FeBr3

MgBr

(from a.)

20.70 PCC

a. HO

[1] CH3CH2MgBr

H O

OH

b.

PCC

OH

Mg

PBr3

CH3CH2OH

PCC

[2] H2O

CH3CH2Br

O

CH3CH2MgBr

[1] CH3CH2CH2MgBr

OH

[2] H2O

CH3CH2CH2OH

PBr3

Mg CH3CH2CH2Br

CH3CH2CH2MgBr

O

567

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Introduction to Carbonyl Chemistry 20–29

PCC

OH

c.

[1] CH3CH2CH2CH2MgBr

H

PCC

[2] H2O O PBr3

CH3CH2CH2CH2OH

CH3CH2CH2CH2Br

O

CH3CH2CH2CH2MgBr

[1] (CH3)2CHCH2MgBr

H

d.

OH Mg

PBr3

Mg

[2] H2O O

OH

Br

MgBr

(from c.) PBr3

(CH3)2CHCH2OH

[1] CO2

Mg

(CH3)2CHCH2Br

[2] H3O+

(CH3)2CHCH2MgBr

CO2H

e.

HBr

OH

Mg

Br

MgBr

H2O

K+ –OC(CH3)3

PBr3

+ OH

H

(from c.)

Br

O

20.71 a.

O

[1] (CH3)2CuLi

[1] CH3MgBr [2] H2O O

[2] H2O

OH

H2SO4

H2, Pd-C

+

b.

PCC

CHO

OH

OH

[1] A

OTs TsCl

[2] H2O

pyridine

PBr3

CN

–CN

Mg Br

MgBr

A O

c.

OCH3

[1] LiAlH4

OH

Br2 FeBr3

[2] H2O

Br

Br

(+ ortho isomer) HBr

d.

ROOR

[1] Mg

Br

[2] O [3] H2O

O CH3CH2I

OH NaH

OH

Br

O

568

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–30

HO

[1] NaH

e. HC CH

TBDMS–Cl

CH3CHC CH [2] CH3CH=O [3] H2O

imidazole

OTBDMS

NaH

CH3CHC CH

OTBDMS CH3CHC C

[1] O [2] H2O

OTBDMS

OH

CH3CHC C OH

(CH3CH2CH2CH2)4NF

OH

20.72 Hb Hb

O

CH3 Ha

H

CH3

Hb

cm–1 (C=O)

IR peak: 1716 1H NMR: 2 signals (ppm) CH3 doublet 1.2 (Hb) H CH3 Ha septet 2.7 (Ha)

Hb

Hd

Hd H

NaBH4

OHa CH3

CH3

CH3OH

H Hc

H CH3 CH3 Hd

Hd

Hb

Hc

C7H16O

C7H14O

IR peak: 3600–3200 cm–1 (OH) 1H NMR: 4 signals (ppm) doublet 0.9 (Hd) singlet 1.5 (Ha) multiplet 1.7 (Hc) triplet 3.0 (Hb)

B

A

20.73

Hb

H

Hb

H

O

CH3 CH3

C4H8O

Ha Ha

Hc

Hc

H

H

[2] H2O

C 1H NMR: 2 signals (ppm) singlet (6 H) 1.3 (Ha) singlet (2 H) 2.4 (Hb)

IR peak 3600–3200 cm–1 (OH) NMR: 4 signals (ppm) singlet (6 H) 1.2 (Ha) singlet (1 H) 1.6 (Hb) singlet (2 H) 2.7 (Hc) multiplet (5 H) 7.2 (benzene ring)

OHb

[1] C6H5MgBr

1H

CH3 CH3 Ha Ha C10H14O

D

20.74 O CH3

C

Hc OCH2CH3

Ha Hb C4H8O2

E

IR peak 1743 cm–1 (C=O) 1H NMR: 2 signals (ppm) triplet (3 H) 1.2 (Hc) singlet (3 H) 2.0 (Ha) quartet (2 H) 4.1 (Hb)

[1] CH3CH2MgBr (excess) [2] H2O

Hb OH Hb d CH3CH2 C CH2CH3 Ha

CH3 Hc C6H14O

F

Ha

IR peak 3600–3200 cm–1 (OH) 1H NMR: 4 signals (ppm) triplet (6 H) 0.9 (Ha) singlet (3 H) 1.1 (Hc) quartet (4 H) 1.5 (Hb) singlet (1 H) 1.55 (Hd)

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

569

Introduction to Carbonyl Chemistry 20–31

20.75 Molecular ion at m/z = 86: C5H10O (possible molecular formula). Determine the number of integration units per H: Total number of integration units: 25 + 17 + 24 + 17 = 83 83 units/10 H's = 8.3 units per H Divide each integration value by 8.3 to determine the number of H's per signal: 25 units/ 8.3 = 3 H's 24 units/ 8.3 = 3 H's 17 units/ 8.3 = 2 H's H Hb

O

[1] CH3MgBr

CH3CH2CH2C N

b

H H

CH3

+

[2] H3O

CH3 H H

Hc

Ha Hd Hd

G

IR peak 1721 cm–1 (C=O) 1H NMR: 4 signals (ppm) triplet (3 H) 0.9 (Ha) sextet (2 H) 1.6 (Hb) singlet (3 H) 2.1 (Hc) triplet (2 H) 2.4 (Hd)

20.76 Molecular ion at m/z = 86: C5H10O (possible molecular formula). Hb Hd Hd CH3 H H

[1] (CH3)3CLi

H

[2] CH2=O [3] H2O

OH

He H Hf

H H

Ha

Hc Hc

H

IR peaks: 3600–3200 cm–1 (OH) 1651 cm–1 (C=C) 1H NMR: 6 signals (ppm) singlet (1 H) 1.7 (Ha) singlet (3 H) 1.8 (Hb) triplet (2 H) 2.2 (Hc) triplet (2 H) 3.8 (Hd) two signals at 4.8 and 4.9 due to 2 H's: He and Hf

20.77 Hb O

He

[1] CH3MgBr [2] H2O

O

IR peak 3600–3200 cm–1 (OH) 1H NMR: 5 signals (ppm) triplet (3 H) 0.94 (Ha) multiplet (2 H) 1.39 (Hb) multiplet (2 H) 1.53 (Hc) singlet (1 H) 2.24 (Hd) triplet (2 H) 3.63 (He)

OH Ha

Hc

Hd

CH3 MgBr O H OH

OH

HOMgBr

570

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–32

20.78 O

OH [1] (R)-CBS reagent

Br

[2] H2O

HO

OTBDMS Br imidazole (2 equiv)

HO

COOCH3

COOCH3

[1] NaH

HO

Br TBDMSCl

C6H5

COOCH3

Br

[2] Br

A

TBDMS O

C6H5

O

Br NH3 H2N

C6H5

O

B

A

+

OTBDMS H N

B

C6H5

O

TBDMS O

[1] LiAlH4

COOCH3

[2] H2O

OTBDMS H N

C6H5

O

TBDMS O

(CH3CH2CH2CH2)4NF OH

(R)-salmeterol

20.79 larger group equatorial axial

OH [1] L-selectride (CH3)3C

O [2] H2O

(CH3)3C

H

= (CH3)3C

OH

equatorial

cis-4-tert-butylcyclohexanol major product

4-tert-butylcyclohexanone

L-Selectride adds H– to a C=O group. There are two possible reduction products—cis and trans isomers— but the cis isomer is favored. The key element is that the three sec-butyl groups make L-selectride a large, bulky reducing agent that attacks the carbonyl group from the less hindered direction. O– O (CH3)3C

(CH3)3C R3B H

OH axial H

equatorial

H2O (CH3)3C

H equatorial

cis product

When H– adds from the equatorial direction, the product has an axial OH and a new equatorial H. Since the equatorial direction is less hindered, this mode of attack is favored with large bulky reducing agents like L-selectride. In this case, the product is cis.

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

571

Introduction to Carbonyl Chemistry 20–33

R3B H

Axial H's hinder H axial attack. H

H O

(CH3)3C

(CH3)3C

H axial

axial O–

H2O

OH equatorial

(CH3)3C

trans product The axial H's hinder H– attack from the axial direction. As a result, this mode of attack is more difficult with larger reducing agents. In this case the product is trans. This product is not formed to any appreciable extent.

20.80 The β carbon of an α,β-unsaturated carbonyl compound absorbs farther downfield in the 13C NMR spectrum than the α carbon, because the β carbon is deshielded and bears a partial positive charge as a result of resonance. Since three resonance structures can be drawn for an α,β-unsaturated carbonyl compound, one of which places a positive charge on the β carbon, the decrease of electron density at this carbon deshields it, shifting the 13C absorption downfield. This is not the case for the α carbon. 122.5 ppm O

150.5 ppm

O

O

Oδ−

hybrid:

β α mesityl oxide

δ+

δ+

20.81 CH2CN

OH

CN

OH

[1] Li+ –N[CH(CH3)2]2 [2]

N(CH3)2

CH2NH2

OH

[1] LiAlH4

O

OCH3

[2] H2O

OCH3

OCH3

OCH3

W

H CN H C

[3] H2O

–

N[CH(CH3)2]2

X C15H19NO2

H C CN

O

H OH

O

OCH3

venlafaxine

Y

OCH3

HN[CH(CH3)2]2

CN

OH

CN

OCH3

OCH3 OH

572

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 20–34

20.82 H MgCl

O

O

H

OH

H OH

P

MgCl

OH

MgCl MgCl

H OH

H O

H

H

HB

O

O

H OH

B

H2O

Any base (such as the alkoxide) can deprotonate the intermediate.

CH3 OH

N OH

20.83 O

NH2OH

O

O

H2NOH

H2NOH

proton transfer H O

OH

H

N

OH

OH

N

N

= proton transfer

H NOH O

O

O

(+ 1 resonance structure)

20.84 O N

H A O CH2CH3

CH3CH2 MgBr N

CH3

CH3

(any proton source)

OH CH2CH3 N

+ A

CH3

+ MgBr+ CH2CH3

MgBr+

CH2CH3

+ N

CH3

CH2CH3

CH3CH2 MgBr N

+ CH3

OH

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

573

Aldehydes and Ketones—Nucleophilic Addition 21–1

Chapter 21 Aldehydes and Ketones—Nucleophilic Addition Chapter Review General facts • Aldehydes and ketones contain a carbonyl group bonded to only H atoms or R groups. The carbonyl carbon is sp2 hybridized and trigonal planar (21.1). • Aldehydes are identified by the suffix -al, while ketones are identified by the suffix -one (21.2). • Aldehydes and ketones are polar compounds that exhibit dipole–dipole interactions (21.3). Summary of spectroscopic absorptions of RCHO and R2CO (21.4) C=O ~1715 cm–1 for ketones IR absorptions • increasing frequency with decreasing ring size ~1730 cm–1 for aldehydes • For both RCHO and R2CO, the frequency decreases with conjugation. Csp2–H of CHO CHO C–H α to C=O C=O

1

H NMR absorptions 13

C NMR absorption

~2700–2830 cm–1 (one or two peaks) 9–10 ppm (highly deshielded proton) 2–2.5 ppm (somewhat deshielded Csp3–H) 190–215 ppm

Nucleophilic addition reactions [1] Addition of hydride (H–) (21.8) O R

C

OH

NaBH4, CH3OH

R C H(R')

or

H(R')

H

[1] LiAlH4 [2] H2O

• •

The mechanism has two steps. H:– adds to the planar C=O from both sides.

• •

The mechanism has two steps. R:– adds to the planar C=O from both sides.

1o or 2o alcohol

[2] Addition of organometallic reagents (R–) (21.8) O R

C

OH

[1] R"MgX or R"Li

R C H(R')

[2] H2O

H(R')

R"

1o, 2o, or 3o alcohol

[3] Addition of cyanide (–CN) (21.9) O R

C

NaCN H(R')

HCl

OH R C H(R') CN

cyanohydrin

• •

The mechanism has two steps. – CN adds to the planar C=O from both sides.

574

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 21–2

[4] Wittig reaction (21.10) R

R C O

+ Ph3P

C C

•

alkene

•

C

(R')H

(R')H

Wittig reagent

The reaction forms a new C–C σ bond and a new C–C π bond. Ph3P=O is formed as by-product.

[5] Addition of 1o amines (21.11) R

R

R"NH2

C O

C N

mild acid

(R')H

(R')H

R"

• •

The reaction is fastest at pH 4–5. The intermediate carbinolamine is unstable, and loses H2O to form the C=N.

• •

The reaction is fastest at pH 4–5. The intermediate carbinolamine is unstable, and loses H2O to form the C=C.

•

The reaction is reversible. Equilibrium favors the product only with less stable carbonyl compounds (e.g., H2CO and Cl3CCHO). The reaction is catalyzed with either H+ or –OH.

imine

[6] Addition of 2o amines (21.12) O

NR2 R2NH

H

(R')H

(R')H

mild acid

enamine

[7] Addition of H2O—Hydration (21.13) H2O H+ or –OH

O R

C

H(R')

OH R C H(R') OH

•

gem-diol

[8] Addition of alcohols (21.14) O R

C

H+ H(R')

OR" R C H(R')

+ R"OH (2 equiv)

+

OR''

acetal

H2O

• • •

The reaction is reversible. The reaction is catalyzed with acid. Removal of H2O drives the equilibrium to favor the products.

Other reactions [1] Synthesis of Wittig reagents (21.10A) RCH2X

[1] Ph3P [2] Bu

Li

Ph3P CHR

• •

Step [1] is best with CH3X and RCH2X since the reaction follows an SN2 mechanism. A strong base is needed for proton removal in Step [2].

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

575

Aldehydes and Ketones—Nucleophilic Addition 21–3

[2] Conversion of cyanohydrins to aldehydes and ketones (21.9) OH

O

–OH

R C H(R') CN

C

H(R')

+ H2 O

aldehyde or ketone

+ –CN

R

•

This reaction is the reverse of cyanohydrin formation.

[3] Hydrolysis of nitriles (21.9) OH R C H(R') CN

OH

H2 O H+ or –OH Δ

R C H(R') COOH

α-hydroxy carboxylic acid

[4] Hydrolysis of imines and enamines (21.12) NR (R')H

O

NR2 H

or

H2O, H+

(R')H

imine

(R')H

H

+ RNH2 or R2NH

aldehyde or ketone

enamine

[5] Hydrolysis of acetals (21.14) OR" R C H(R') + H2O OR"

•

O

H+ R

C

H(R')

aldehyde or ketone

+ R"OH (2 equiv)

•

The reaction is acid catalyzed and is the reverse of acetal synthesis. A large excess of H2O drives the equilibrium to favor the products.

576

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 21–4

Practice Test on Chapter Review 1. Give the IUPAC name for the following compounds. O O

a.

b.

H

c.

O

2. (a) Considering compounds A–D, which compound forms the smallest amount of hydrate? (b) Which compound forms the largest amount of hydrate? CH3O

CHO

O2N

Cl

CHO

CH3O O

O

A

B

C

D

3. (a) Considering compounds A–D, which compound absorbs at the lowest wavenumber in its IR spectrum? (b) Which compound absorbs at the highest wavenumber in its IR spectrum? O

O

O

A

B

O

C

D

4. Fill in the lettered reagents (A–G) in the following reaction scheme. HO

CHO

B O

HO

COOH

C

HO C CH A

O

O

HO D

TBDMSO E

TBDMSO

OH

F

G HO

OH

577

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Aldehydes and Ketones—Nucleophilic Addition 21–5

5. Draw the organic products formed in the following reactions. a. CH3CH2Br

NH

[1] Ph3P

d.

[2] BuLi [3]

b.

mild acid

O O NH2

CHO

+

O

e.

HO

OH

TsOH

mild acid O

c.

O

[1] NaCN, HCl

f.

HO

OH

CH3CH2OH H+

[2] H2O, H+, Δ

[Indicate stereochemistry.]

Answers to Practice Test 1.a. 54. isopropyl-2,4- A = [1] LiC≡CH; [2] H2O dimethylB = [1] R2BH; [2] H2O2, –OH cyclohexanone C = Ag2O, NH4OH b. 3,3D = H2O, H2SO4, HgSO4 dimethyl-5E = TBDMS–Cl, pyridine phenyl-2F = [1] CH3Li; [2] H2O pentanone G = (CH3CH2CH2CH2)4N+F– c. 4-ethyl-2methylcyclohexanecarbaldehyde 2. a. D b. C 3. a. B b. C

5. O

a.

e. O

b.

N

O

f.

O

+

OH

c.

O

CO2H

d. N

O

(E + Z)

Answers to Problems 21.1

As the number of R groups bonded to the carbonyl C increases, reactivity towards nucleophilic attack decreases. Steric hindrance decreases reactivity as well. O

O

Increasing reactivity decreasing steric hindrance

OH

O

OH

578

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 21–6

21.2

More stable aldehydes are less reactive towards nucleophilic attack. CHO

CHO

benzaldehyde Several resonance structures delocalize the partial positive charge on the carbonyl carbon, making it more stable and less reactive towards nucleophilic attack.

21.3

cyclohexanecarbaldehyde This aldehyde has no added resonance stabilization.

O

O

O

C

C

C

H

H

O

O

C

C

H

O C

H

O C

H

(CH3)3CC(CH3)2CH2CHO

3 2 1 H

5

O

5 C chain = pentanal

O

Cl Cl

CHO

Cl 3 4 Cl

3,3-dichlorocyclobutane4 C ring = carbaldehyde cyclobutanecarbaldehyde

3,3,4,4-tetramethylpentanal 8

1

7

CHO

b.

6 8 C chain = octanal

2 1

c.

4

H

21.4

H

• To name an aldehyde with a chain of atoms: [1] Find the longest chain with the CHO group and change the -e ending to -al. [2] Number the carbon chain to put the CHO at C1, but omit this number from the name. Apply all other nomenclature rules. • To name an aldehyde with the CHO bonded to a ring: [1] Name the ring and add the suffix-carbaldehyde. [2] Number the ring to put the CHO group at C1, but omit this number from the name. Apply all other nomenclature rules. CHO

a.

H

CHO

5

4

3

2

2,5,6-trimethyloctanal

Work backwards from the name to the structure, referring to the nomenclature rules in Answer 21.3. a. 2-isobutyl-3-isopropylhexanal 6 C chain

O

b. trans-3-methylcyclopentanecarbaldehyde 5 carbon ring

CHO

CHO

or

H CH3

CH3

579

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Aldehydes and Ketones—Nucleophilic Addition 21–7

d. 3,6-diethylnonanal

c. 1-methylcyclopropanecarbaldehyde

9 C chain

CHO

3 carbon ring

H

CH3 O

21.5 • To name an acyclic ketone: [1] Find the longest chain with the carbonyl group and change the -e ending to -one. [2] Number the carbon chain to give the carbonyl C the lower number. Apply all other nomenclature rules. • To name a cyclic ketone: [1] Name the ring and change the -e ending to -one. [2] Number the C’s to put the carbonyl C at C1 and give the next substituent the lower number. Apply all other nomenclature rules. a.

1

3

O

4

c.

5

O

8 C chain = octanone

1

8 O

5-ethyl-4-methyl-3-octanone

CH3

(CH3)3C

(CH3)3CCOC(CH3)3

b.

3

O

4

5

2,2,4,4-tetramethyl-3-pentanone

2 1

5 C ring = cyclopentanone

3 O

5 C chain = pentanone

CH3

(CH3)3C

2

O

3-tert-butyl-2-methylcyclopentanone

Most common names are formed by naming both alkyl groups on the carbonyl C, arranging them alphabetically, and adding the word ketone.

21.6

a. sec-butyl ethyl ketone

d. 3-benzoyl-2-benzylcyclopentanone benzyl group:

benzoyl group:

e. 6,6-dimethyl-2-cyclohexenone 6 C ketone

5 C ketone

O

O

O

CH2

b. methyl vinyl ketone

C

f. 3-ethyl-5-hexenal O

O

CHO

c. p-ethylacetophenone

O

CH2

O

21.7

Compounds with both a C–C double bond and an aldehyde are named as enals. a. (2Z)-3,7-dimethyl-2,6-octadienal 3

7 6

2 1 CHO

neral

b. (2E,6Z)-2,6-nonadienal 3

7 6

1 CHO

2 cucumber aldehyde

580

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 21–8

Even though both compounds have polar C–O bonds, the electron pairs around the sp3 hybridized O atom of diethyl ether are more crowded and less able to interact with electron-deficient sites in other diethyl ether molecules. The O atom of the carbonyl group of 2-butanone extends out from the carbon chain, making it less crowded. The lone pairs of electrons on the O atom can more readily interact with the electron-deficient sites in the other molecules, resulting in stronger forces and a higher boiling point.

21.8

O O

diethyl ether

2-butanone

For cyclic ketones, the carbonyl absorption shifts to higher wavenumber as the size of the ring decreases and the ring strain increases. Conjugation of the carbonyl group with a C=C or a benzene ring shifts the absorption to lower wavenumber.

21.9

CHO

a.

CHO

or

conjugated C=O lower wavenumber

b.

or

O

higher wavenumber

O

smaller ring higher wavenumber

21.10 The number of lines in their 13C NMR spectra can distinguish the constitutional isomers. O

O

2-pentanone 5 lines

O

3-pentanone 3 lines

3-methyl-2-butanone 4 lines

21.11 [1] DIBAL-H

a. CH3CH2CH2COOCH3

PCC

b. CH3CH2CH2CH2OH

CH3CH2CH2CHO

[2] H2O

CH3CH2CH2CHO

[1] R2BH

c.

HC CCH2CH3

[2] H2O2, HO–

d. CH3CH2CH2CH CHCH2CH2CH3

CH3CH2CH2CHO [1] O3 [2] Zn, H2O

CH3CH2CH2CHO

21.12 O Cl

a.

O CH3

c.

AlCl3

Cl

[1] (CH3)2CuLi [2] H2O

C CH

H2O H2SO4 HgSO4

O

O

b.

O

581

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Aldehydes and Ketones—Nucleophilic Addition 21–9

21.13 CH3O

OCH3

CH3O OCH3 [1] O3

OHC

CHO

[2] (CH3)2S

Cleave this C=C with O3.

21.14 Addition of hydride or R–M occurs at a planar carbonyl C, so two different configurations at a new stereogenic center are possible. O

a.

Add stereochemistry:

H OH

NaBH4 CH3OH

new stereogenic center O

Add

CH=CH2

CH=CH2 OH

OH

[2] H2O

(CH3)3C

H OH

OH CH=CH2 stereochemistry:

[1] CH2 CHMgBr

b.

H OH

(CH3)3C

(CH3)3C

(CH3)3C

21.15 Treatment of an aldehyde or ketone with NaCN, HCl adds HCN across the double bond. Cyano groups are hydrolyzed by H3O+ to replace the three C–N bonds with three C–O bonds. OH CHO

CHCN

NaCN

a.

OH

b.

HCl

OH

H3 O+, Δ

CN

COOH

21.16 HO HO HO

O O OH HO HO

O O OH

CN

HO

enzyme

C

O

CN

enzyme H

C

21.17 CH3

CH3 C O

+

CH3

b.

Ph3P=CH2

C CH2 CH3

O +

+

HCN

toxic by-product

amygdalin

a.

H

Ph3P=CHCH2CH2CH2CH3

CHCH2CH2CH2CH3

582

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 21–10

21.18 BuLi

a. Ph3P + Br−CH2CH3

Ph3P−CH2CH3 Br–

b. Ph3P + Br−CH(CH3)2

Ph3P−CH(CH3)2

Ph3P=CHCH3 BuLi Ph3P=C(CH3)2

Br–

c. Ph3P + Br−CH2C6H5

Ph3P−CH2C6H5

BuLi Ph3P=CHC6H5

Br–

21.19 CHO

a.

CH2CH3 Ph3P CHCH2CH3

+

CH2CH3

CHO Ph3P CHC6H5

+

b. CHO +

c.

COOCH3 Ph3P CHCOOCH3

COOCH3

21.20 Ph3P=CHCO2CH2CH3

N

N

CHO

A

CO2CH2CH3

(+ cis isomer) B H2, Pd–C

N

CO2CH2CH3

C

21.21 To draw the starting materials of the Wittig reactions, find the C=C and cleave it. Replace it with a C=O in one half of the molecule and a C=PPh3 in the other half. The preferred pathway uses a Wittig reagent derived from a less hindered alkyl halide. a.

CH3 CH2CH3 C C CH3 H

CH3 C PPh3 CH3

CH2CH3

+

O C

or H

2° halide precursor (CH3)2CHX

b.

CH2CH3

CH3CH2 H

CH2CH3 C

H

(cis or trans)

H

PPh3

CH2CH3 C H

1° halide precursor XCH2CH2CH3 preferred pathway

CH3CH2

C C

CH3 C O + Ph3P CH3

+

(only one route possible)

O C H

583

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Aldehydes and Ketones—Nucleophilic Addition 21–11

C6H5

c.

CH3 C

C6H5

C

H

H

C6H5

CH3 C

PPh3

+

O

or

C

H

H

(cis or trans)

CH3 C O

+

Ph3P

H

C H

(both routes possible)

1° halide precursor C6H5CH2X

1° halide precursor XCH2CH3

21.22 a. Two-step sequence: O

[1] CH3MgBr

HO CH3

H2SO4

[2] H2O

One-step sequence: O

minor product

Ph3P=CH2

tetrasubstituted major product

(E and Z isomers)

only product

b. Two-step sequence: O

[1] C6H5CH2MgBr

OH

[2] H2O

CH2C6H5

H2SO4

CHC6H5

trisubstituted conjugated C=C

One-step sequence: O

Ph3P=CHC6H5

CH2C6H5

trisubstituted

CHC6H5 only product

21.23 When a 1o amine reacts with an aldehyde or ketone, the C=O is replaced by C=NR. a.

CHO

O

CH3CH2CH2CH2NH2

NCH2CH2CH2CH3

NCH2CH2CH2CH3

CH3CH2CH2CH2NH2

b.

CH

c.

O

CH3CH2CH2CH2NH2

NCH2CH2CH2CH3

21.24 Remember that the C=NR is formed from a C=O and an NH2 group of a 1° amine. CH3

a.

CH3 C

C NCH2CH2CH3 H

b.

CH3

O + NH2CH2CH2CH3

H N

CH3

O

NH2

21.25 O

N(CH3)2

N(CH3)2

CH3 N CH3

H

+

584

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 21–12

21.26 • Imines are hydrolyzed to 1° amines and a carbonyl compound. • Enamines are hydrolyzed to 2° amines and a carbonyl compound. O

H2O CH N

a.

+

C H+

H2N

H

imine

1° amine H2O

b.

CH2 N

(CH3)2NCH

O

CH3

enamine c.

+

CH2 N H

H+

CH3

2° amine O

H2O

C(CH3)2

H+

enamine

+

(CH3)2NH

C CH(CH3)2 H

2° amine

21.27 • A substituent that donates electron density to the carbonyl C stabilizes it, decreasing the percentage of hydrate at equilibrium. • A substituent that withdraws electron density from the carbonyl C destabilizes it, increasing the percentage of hydrate at equilibrium. a. CH3CH2CH2CHO

or

b.

CH3CH2COCH3

one R group on C=O higher percentage of hydrate

2 R groups on C=O

CH3CF2CHO

or

CH3CH2CHO

F atoms are electron withdrawing. higher percentage of hydrate

21.28 H O H

H OH2

O H

O H

O H

O H +

O

+ H2O

+ H3O

(+ 1 resonance structure)

H2O

21.29 Treatment of an aldehyde or ketone with two equivalents of alcohol results in the formation of an acetal (a C bonded to two OR groups). O

a.

O

+

2 CH3OH

TsOH

OCH3

+ HO

b.

OH

TsOH

O

O

OCH3

21.30 OCH3

OCH3

a.

b.

OCH3

O

OCH3

2 OR groups on different C's 2 ethers

O

c.

2 OR groups on same C acetal

CH3 CH3

2 OR groups on same C acetal

d.

OCH3 OH

1 OR group and 1 OH group on same C hemiacetal

585

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Aldehydes and Ketones—Nucleophilic Addition 21–13

21.31 The mechanism has two parts: [1] nucleophilic addition of ROH to form a hemiacetal; [2] conversion of the hemiacetal to an acetal. O O

O

TsO−H

+ HOCH CH OH 2 2

+ H2O

overall reaction TsO

H O H

O H

HO O CH2CH2OH

HO O CH2CH2OH

+ TsO−H

+ TsO

hemiacetal

HOCH2CH2OH H HO O CH2CH2OH

H O

O CH2CH2OH

O CH2CH2OH

O CH2CH2OH

TsO−H

+ H2O

+ TsO–

O CH2CH2OH

O

TsO

O H

O

O + TsO−H

carbocation re-drawn

acetal

21.32 CH3O OCH3

a.

+

O

H2SO4

H2O

CH3

C

O CH3

+

b.

+ 2 CH3OH

H 2O

CH3

O

OH

O

H2SO4

C

+ CH3

OH

21.33 HO

O

O H2 O O

H2SO4

safrole

+ H

H

HO

formaldehyde

21.34 Use an acetal protecting group to carry out the reaction. O

O

O

[1] CH3Li (2 equiv)

HOCH2CH2OH

O

O H3

O+

[2] H2O

TsOH COOCH2CH3

O

COOCH2CH3 OH

OH

586

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 21–14

21.35 OH

O

C4

C1

OH O

a. HO

O

H

b.

C1 O

H

C4

OH

C5

C1

C5

C1

21.36 acetal

O

HO H O

O

H

OH

O O

H OCH3

O

acetal

H

HO

O

H O

HO

COOH HO

acetal

O

H

O

HO

O

H O

OH

H

OH

acetal

OH

monensin

O

digoxin

hemiacetal Ether O atoms are indicated in bold.

21.37 The hemiacetal OH is replaced by an OR group to form an acetal. OH

OCH2CH3

O

a.

H+

+ CH3CH2OH

O

OH HO

O

b.

OCH2CH3 + CH3CH2OH

H+

HO

HO

O

HO

21.38 HO OH

a.

HO OH O

*

*

*

HO

*

OH * OH

5 stereogenic centers (labeled with *)

HO

HO OH

b.

OH

d.

HO OH O

HO

hemiacetal C

OH OH

HO OH O OH

HO OH

β-D-galactose

e.

HO OH O OCH3

HO OH

α-D-galactose

c.

CHO OH

+

O HO OH OCH3

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

587

Aldehydes and Ketones—Nucleophilic Addition 21–15

21.39 2

O

a.

3 4

O 1

1

cis-5-isopropyl-2methylcyclohexanone B

3,3-dimethylbutanal A

b.

[1] A

NaBH4 OH

CH3OH

[2] A

[3] A

[4] A

[5] A

[1] B

CH3MgBr

OH Ph3P=CHOCH3

+

OH

OCH3 (+ cis isomer)

CH3CH2CH2NH2 NCH2CH3

mild H+ O

HOCH2CH2CH2OH H+

O

NaBH4 HO

HO

H2O

CH3MgBr

HO

[3] B

H

H

H2O

CH3OH

[2] B

4 5

CH 2

3

HO

Ph3P=CHOCH3 CH3O OCH3

[4] B

[5] B

CH3CH2CH2NH2 mild H+

HOCH2CH2CH2OH H+

CH3CH2CH2N

O O

21.40 The least hindered carbonyl group is the most reactive. O

O H O

least reactive

intermediate reactivity

most reactive

588

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 21–16

21.41 O

CHO HO

O

a.

O

OH

O

OH

b. O

OH

21.42 Use the rules from Answers 21.3 and 21.5 to name the aldehydes and ketones. O

a. Ph

O

8 C = octanone 8-phenyl-3-octanone

e.

CH3

o-nitroacetophenone

NO2

b. CHO

5 C ring 2-methylcyclopentanecarbaldehyde

f.

6 C ring = cyclohexanone 5-ethyl-2-methylcyclohexanone

g.

CHO

6 C = hexanal 3,4-diethylhexanal

O

c.

O

E 8 C = octenone (5E)-2,5-dimethyl-5-octen-4-one

CHO

d.

trans-2-benzylcyclohexanecarbaldehyde CH2Ph

h.

CHO

6 C = hexenal 3,4-diethyl-2-methyl-3-hexenal

21.43 f. 2-formylcyclopentanone

a. 2-methyl-3-phenylbutanal

O

O

CHO

H O

O

b. dipropyl ketone CHO

g. (3R)-3-methyl-2-heptanone

c. 3,3-dimethylcyclohexanecarbaldehyde h. m-acetylbenzaldehyde CHO

O

d. α-methoxypropionaldehyde

O H

i. 2-sec-butyl-3-cyclopentenone

O

OCH3 O

j. 5,6-dimethyl-1-cyclohexenecarbaldehyde

e. 3-benzoylcyclopentanone

CHO O

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

589

Aldehydes and Ketones—Nucleophilic Addition 21–17

21.44 O

O

H

O

H

H

2-ethylbutanal

hexanal

O H

2,2-dimethylbutanal

O

O

H

(2S)-2,3-dimethylbutanal

O

3,3-dimethylbutanal

O

H

(2R)-2,3-dimethylbutanal

H

4-methylpentanal

O

H

O

(2R)-2-methylpentanal

H

(2S)-2-methylpentanal

O

H

H

(3R)-3-methylpentanal

(3S)-3-methylpentanal

21.45 H

= C6H5CH2CHO

phenylacetaldehyde

O

a. NaBH4, CH3OH

g.

C6H5CH2CH2OH

b. [1] LiAlH4; [2] H2O

C6H5CH2CH2OH

h.

C6H5CH2CH(OH)CH3 NaCN, HCl

OCH2CH3 H NH

C6H5CH2CH=CHCH3

i.

(E and Z isomers) f.

OCH2CH3

CH3CH2OH (excess), H+

C6H5CH2CH(OH)CN

e. Ph P CHCH 3 3

(E and Z isomers)

N(CH2CH3)2

c. [1] CH3MgBr; [2] H2O d.

H

(CH3CH2)2NH, mild acid

mild acid

H

(CH3)2CHNH2 mild acid

(E and Z isomers)

N

NCH(CH3)2

j.

CH3CH=C(CH3)2

c.

O

OH

HO

O

H+

21.46 a. CH3CH2Cl

[1] Ph3P

[1] Ph3P CH2Cl

CH=CHCH2CH2CH3 [2] BuLi [3] CH3CH2CH2CHO

[2] BuLi [3] (CH3)2C=O

(E and Z isomers)

[1] Ph3P

b.

CH2Br

[2] BuLi [3] C6H5CH2CH2CHO

CH=CHCH2CH2C6H5

(E and Z isomers)

21.47 a.

Ph3P CHCH2CH2CH3

b.

Ph3P C(CH2CH2CH3)2

BrCH2CH2CH2CH3 BrCH(CH2CH2CH3)2

c. Ph3P CHCH=CH2

BrCH2CH=CH2

590

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 21–18

21.48 Ph3P

CHO

a.

H

b. CH3CH2CHO + c.

mild

H2N

O

CH3CH2CH N

acid

HOCH2CH2OH

O

H+

O

H3O+

d.

H2N

N

O

O

e.

mild

+

C6H5

N H

N

acid

(E and Z isomers)

C6H5 HO

H3O+, Δ

CN C

f. C6H5

HO

COOH

C6H5

C6H5

C6H5

CH3CH2OH OH

g.

i.

OCH3

CH3O

O H3O+

N

h.

OCH2CH3

H+

O

O

H3O+

OCH3 O

j.

+

O + HOCH3

CH3O

Ph3P CH(CH2)5COOCH3

HN

CH(CH2)5COOCH3

21.49 CH3CH2O

O

OCH2CH3

+ HOCH2CH3

a. CH3O

O

OCH3 OCH3

OCH3

c.

b.

HO

O OCH2CH3

OCH3

OCH3 + HOCH3

H O + HOCH2CH3

591

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Aldehydes and Ketones—Nucleophilic Addition 21–19

21.50 Consider para product only, when an ortho, para mixture can result. Br Br2

O

CH3COCl

FeBr3

O

HOCH2CH2OH Br

AlCl3

O

Br

B

A

O

Mg

O

H+

MgBr D

C

[1] CH3CHO [2] H2O O

O

O

H2 O

O

PCC

O

H+ G

O

OH

O

F

E

21.51 Ph3P=CHCH2CH2CH3

a. CH3CH2CH2CHO

CH3CH2CH2

CH2CH2CH3

H O

HO CN

NaCN

b.

+

+

C C H

CH3CH2CH2 H C C H CH2CH2CH3

NC OH

HCl NaBH4

O

c. CH3CH2

d. HO

+

OH

OH

CH3OH CH3CH2 CH3OH

O

HO

CH3CH2

+

O

O

HO

HCl OCH3

OH

OCH3

21.52 new stereogenic center O

OH

O

OH

O

OH

CHO

HO

A achiral S

new stereogenic center H

CHO

S HO H B

chiral

An equal mixture of enantiomers results, so the product is optically inactive.

H O

H O

OH

O OH

A mixture of diastereomers results. Both compounds are chiral and OH they are not enantiomers, so the mixture is optically active.

592

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 21–20

21.53 OH H

HO

a.

acetal O

acetal O O

O

H

H O

acetal

etoposide

O O H

O

CH3O

OCH3 OH

b. Lines of cleavage are drawn in. OH H

HO O

H

O

O O

O O

CH3O

HO

H+

O H

OH H

H

O

H2O

O

+

HO

OCH3

H

CH3O

O

HO

OH H

O CH HO H

OH OH

+ CH3CHO + CH2=O

OCH3

OH

OH

21.54 Use the rules from Answer 21.1. O

O

O

Increasing reactivity decreasing steric hindrance

21.55 The less stable carbonyl compound forms the higher percentage of hydrate. O H

δ+ O

H O

The aldehyde is destabilized since there is a δ+ on its adjacent carbon, which is part of a C=O. Thus, PhCOCHO has the higher concentration of hydrate. 21.56 Electron-donating groups decrease the amount of hydrate at equilibrium by stabilizing the carbonyl starting material. Electron-withdrawing groups increase the amount of hydrate at equilibrium by destabilizing the carbonyl starting material. Electron-donating groups make the IR absorption of the C=O shift to lower wavenumber because they stabilize the chargeseparated resonance form, giving the C=O more single bond character.

593

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Aldehydes and Ketones—Nucleophilic Addition 21–21

O2N

COCH3

CH3O

COCH3

a.

p-nitroacetophenone NO2 withdrawing group less stable

b.

higher percentage of hydrate

p-methoxyacetophenone CH3O donating group more stable lower percentage of hydrate

c.

higher wavenumber

lower wavenumber

21.57 Use the principles from Answer 21.21. CH3CH2

a.

CH3CH2

CH2CH2CH3

C C CH3CH2

H

CH3CH2

CH2CH2CH3

C O

Ph3P C

CH3CH2

or H

CH2CH3

CH3

C C

b. H

H

CH3

or

Ph3P C

H

O C

H

H

(Both routes possible.)

1° alkyl halide precursor (XCH2CH3)

O

PPh3

c.

Ph3P CH2

methyl halide precursor (CH3X) preferred pathway C6H5

C6H5 O

CH2CH3

C PPh3

H

1° alkyl halide precursor (XCH2CH2CH3)

d.

H

2° alkyl halide precursor [(CH3CH2)2CHX]

CH2CH3

C O

O C

CH3CH2

1° alkyl halide precursor (XCH2CH2CH2CH3) preferred pathway CH3

CH2CH2CH3

C PPh3

or

Ph3P

or

CH2 O

2° alkyl halide precursor (CH3CH2CH2CHXCH3) C6H5

Ο

PPh3

H

1° alkyl halide precursor (C6H5CH2X) preferred pathway

2° alkyl halide precursor X

21.58 a.

N

O

O

O

H

+ HOCH2CH2OH

c.

O

+ NH2

b.

N

O

+

HN

d. CH3O

OCH3 OCH3

CH3O + HOCH3

H O

594

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 21–22

21.59 PCC

a. C6H5–CH2OH b. C6H5–COCl

C6H5–CHO

[1] LiAlH[OC(CH3)3]3 [2] H2O

c. C6H5–COOCH3

[1] DIBAL-H

f. C6H5–CH=CH2

[1] O3

C6H5–CHO

g. C6H5–CH=NCH2CH2CH3

[1] LiAlH4

e. C6H5–CH3

KMnO4

PCC

C6H5–CH2OH

[2] H2O

C6H5–COOH

C6H5–CHO

H+

C6H5–CHO

H+

C6H5–CHO

[1] LiAlH4 C6H5–CH2OH

[2] H2O

H2O

H2O

h. C6H5–CH(OCH2CH3)2

C6H5–CHO

[2] H2O

d. C6H5–COOH

C6H5–CHO

[2] Zn, H2O

PCC

C6H5–CHO

21.60 PCC

a. OH Cl

b.

c. CH3COCl (CH3CH2)2CuLi

e. CH3C CCH3

O

O

(CH3)2CuLi

O

H2 O

d. CH3CH2C CH

O

H2O H2SO4 HgSO4

H2SO4 HgSO4

O

O

21.61 O

a. One-step sequence:

preferred route only one product formed

Ph3P

or O

Two-step sequence:

OH H2SO4

[1] CH3CH2CH2MgBr [2] H2O

b. One-step sequence:

Ph3P

CHO

preferred route only one product formed

or

OH CHO [1] (CH ) CHMgBr 3 2

H2SO4

Two-step sequence: [2] H2O

+ other alkenes that result from carbocation rearrangement

21.62 a.

CH3CH2CH2CH CHCH3 PCC

One possibility:

CHO

OH PBr3 OH

Br

[1] Ph3P [2] BuLi

CHO Ph3P CHCH3

CH3CH2CH2CH CHCH3

(E and Z)

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

595

Aldehydes and Ketones—Nucleophilic Addition 21–23

b.

C6H5CH CHCH2CH2CH3 CH3OH

One possibility:

SOCl2 CH3Cl

Br2

AlCl3

hν

CHO

Br [1] Ph3P [2] BuLi

C6H5CH PPh3

(from a.)

C6H5CH CHCH2CH2CH3

(E and Z)

c. PCC

OH OH

O Br

PBr3

O

[1] Ph3P

Ph3P CHCH(CH3)2

[2] BuLi

21.63 H2 O

a.

PCC OH

H2SO4

O

Ph3P CHCH2CH3

PBr3

CH3CH2CH2OH O

b.

H2NCH2CH2CH3

CHCH2CH3

CH3CH2CH2Br

[1] Ph3P [2] BuLi

Ph3P CHCH2CH3

NCH2CH2CH3

mild acid NH3 (excess)

(from a.)

BrCH2CH2CH3 PBr3 HOCH2CH2CH3

OH

c.

Br

HBr

MgBr

Mg

CH3OH PCC [1] H2C=O

OH

O H

PCC

MgBr

[2] H2O

[1]

(from a.)

C(OCH2CH3)2

O

CH3CH2OH (2 equiv)

PCC

TsOH 2

d.

[1] OsO4 [2] NaHSO3, H2O

OH OH

O TsOH PCC OH

OH

O O

[2] H2O

596

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 21–24

21.64 CH3OH SOCl2 CH3Cl

a.

KMnO4 CH3

AlCl3

OH

COOH

Br

PBr3

OH

[1] LiAlH4

PCC CHO

[2] H2O

PPh3

[1] PPh3 [2] BuLi

PPh3 CH CH

CHO

(E and Z isomers) OH

b.

CH3Cl (from a.) AlCl3

OH

OCH3

[1] NaH [2] CH3Br

CH3

OCH3

P(C6H5)3

hν

CH3

PBr3

(+ ortho isomer)

OCH3

Br2

Br

P(C6H5)3

CH3OH

Br

BuLi OCH3

CH3O

CH CH

(CH3)3COH

C(CH3)3 P(C6H5)3

(E and Z isomers)

HCl

CH3Cl (from a.)

C(CH3)3

(CH3)3CCl AlCl3

AlCl3

C(CH3)3

C(CH3)3

KMnO4

CH3

C(CH3)3

[1] LiAlH4 [2] H2O [3] PCC OHC

HOOC

(+ ortho isomer)

21.65 O

a.

OH

HOCH2CH2OH

O

O OH

H+

PBr3

O

[1] Mg

O

PCC OH

O

O

Br [2] (CH3)2CO [3] H2O

OH H2O, H+

O [1] LiAlH4

OH

O

[2] H2O OH

OH CH3CH2CH2OH PCC

b.

O

[1] Ph3P

O Br

(from a.)

[2] BuLi

O

O

PPh3

CH3CH2CHO O

H3O+

O CH=CHCH2CH3

(E/Z mixture)

O CH=CHCH2CH3

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Aldehydes and Ketones—Nucleophilic Addition 21–25

21.66 O

a. CH3CH2OH

PCC CH3

C

H

PBr3 CH3CH2Br

Mg

CH3CH2MgBr

H2SO4

b. CH3CH2OH

O [1] C CH3 H [2] H2O

CH2=CH2

O

H OH PCC C CH3 CH2CH3

Br2

Br

Br

CH3

2 NaNH2

C

CH3CH2OH CH2CH3

TsOH

NaH

HC CH

[1] CH3

O HO

O

O

OH

O C

[2] H2O H

(from a.)

OH

PCC

CH2CH3

HC C Na+

[1] OsO4 [2] NaHSO3, H2O

OCH2CH3 CH3 C OCH2CH3

TsOH

21.67 O

O O

O

mCPBA O

(CH3)3CNH2

OH NHC(CH3)3

O

NHC(CH3)3

O

H2O

O

O

O H3O+

X

OH NHC(CH3)3

HO

+ (CH3)2C=O

HO

albuterol

21.68 [1]

HOCH2CH2OH Br

O

Br

CHO TsOH

Mg

O

O

O

O

OH

O

BrMg

O

[2] H2O

H2O H+ OH A

21.69 a.

O H

Na+H–

O

Cl

+ H2 + Na+

b.

O O

acetal

O

CH3

O

+ Cl– O

methoxy methyl ether

CHO

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Chapter 21–26

H O

O

c.

O

H OH2

H H2O

O

O

+ H2O

O

O

O

O

H2O H

+ HOCH3

(The three organic products are boxed in.)

O CH2

H

+ H3O H O

O

H

H OH2

O

O

OH

H

CH2 OH

H2O CH2 O H

CH2 O

H2O

+ H3O

21.70 H

H

N

N

H2O

H2O

H

OH

H

HO

N

H N

HO

H OH2

H H OH2

H2O

(+ 1 resonance structure) O

O H2O

H H N

NH3

O NH2 H OH2

H

H2O NH2

(+ 1 resonance structure)

599

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Aldehydes and Ketones—Nucleophilic Addition 21–27

21.71 O

O

O CF3

Ar

O

O

CF3 Ar'NHNH2

Ar

proton transfer

H Cl

OH

O

CF3 Ar'NHNH

Ar

OH2

CF3 Ar'NHNH

+ Cl–

Ar NHNH2 CF3 H2NSO2

O

N

O

Ar Ar'

CF3

Ar

N

Ar

CF3 H2O

H

N

Ar'NH

H

+ H3O+

proton transfer

H Cl HO

O

N

N Ar Ar' H

Ar'

CF3

H2O

CF3

N

N

H2O

N

N

N

Ar

Ar'

CF3

H Ar

Ar'

+

N

N Ar'

+ H3O+

H2NSO2

Cl–

CF3 N

celecoxib

21.72 The OH groups react with the C=O in an intramolecular reaction, first to form a hemiacetal, and then to form an acetal. O

O OH

HO

OH

O O

OH

OH adds here to form a hemiacetal. Then, the acetal is formed by a second intramolecular reaction.

hemiacetal

acetal C9H16O2

21.73 a.

H H OH O

OCH2CH3

O

OCH2CH3

O

H

H OH2

+

HO

O

O

OH

H

H2O

H OH

H OH2

+ H OCH2CH3 H2O

O H3O+

O

H

HO

H O H

O H

OH + H2O

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Chapter 21–28

O

b.

H OH2 H

OH

OH

+

H

H

O

O

OH

H

H OH2

O

OH

H

OH

O H OH

O

H OH

OH2

O

H H OH

H O

H2O

H

O

O

O

O

H OH

+ H3O+

O

21.74 HSO4– H–OSO3H

O

H

OH

O

O

HO

O

HO

O H–OSO3H

+ HSO4–

HSO4– H

H2SO4 +

O

O +

H2O

O

+ HSO4–

H2O

21.75 O

H Cl

O

H O

OH

OH

O

O CH2

O

CH2

+ Cl– NH2 S

N S

O

OH

N S

H

O

OH

H H

H Cl

Cl

Cl H

S

N

O

H

H

OH

N S

H HO

NH

CH2

OH

S

NH S H Cl

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Aldehydes and Ketones—Nucleophilic Addition 21–29

21.76 H O HO

O

NH2

+ CH3

C

HO

H CH3

H

HO

HO

H OH2

H OH

proton

N C H

HO

N C H CH3

transfer

HO

dopamine

H OH2

HO

HO

HO NH

HO

H

CH3 H3 O

N H

HO

CH3

CH3

H2O

salsolinol

N C H CH3

NH

HO

HO

(+ 3 more resonance structures)

C

HO H2O

H

H2 O

21.77 O O

O (CH3)2S

CH2 H

Bu Li

(CH3)2S

CH2

+ (CH3)2S

S(CH3)2

sulfur ylide

X

sulfonium salt

X

+ Bu H + LiX

21.78 Hemiacetal A is in equilibrium with its acyclic hydroxy aldehyde. The aldehyde can undergo hydride reduction to form 1,4-butanediol and a Wittig reaction to form an alkene. OH

O

a.

OH O

OH

C

A

OH H

This can now be reduced with NaBH 4. OH O

O

b.

OH

A

C

Ph3P=CHCH2CH(CH3)2 H

reacts with the Wittig reagent

1,4-butanediol

(CH3)2CHCH2CH=CHCH2CH2CH2OH

(E and Z isomers)

602

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 21–30

+

c.

PPh3

Y

NaOCH2CH3

HO

O

X

PPh3

O +

H

O

O HO

aldehyde for Wittig reaction

HO

O

New C=C could be cis or trans. The more stable trans C=C is drawn. isotretinoin

21.79 O

H OH2

CH3–MgI

O

HO

OCH3

HO

OCH3

OCH3

OCH3

OCH3

OCH3

H2O

5,5-dimethoxy-2-pentanone

OCH3 OCH3 H

H OH2

HO

H

H OH2

HO

OCH3

HO

HO

OCH3

OH

OCH3

OH

H

O

OCH3

(+ 1 resonance structure)

H

H2O

H2O

H2O

HO

H O CH3 O OH

(+ 1 resonance structure) H O CH3

OH

O

H H2O

Y

OH

H3O

H2O

603

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Aldehydes and Ketones—Nucleophilic Addition 21–31

21.80 cyclopropenone (1640 cm–1) O

O

2-cyclohexenone (1685 cm–1) O

O

O

These three resonance structures include an aromatic ring; 4n + 2 = 2 π electrons. Although they are charge separated, the stabilized aromatic ring makes these three structures contribute to the hybrid more than usual. Since these three resonance contributors have a C–O single bond, the absorption is shifted to a lower wavenumber.

O

O

There are three resonance structures for 2-cyclohexenone, but the charge-separated resonance structures are not aromatic so they contribute less to the resonance hybrid. The C=O absorbs in the usual region for a conjugated carbonyl.

21.81 A. Molecular formula C5H10O IR absorptions at 1728, 2791, 2700 cm–1 NMR data: singlet at 1.08 (9 H) singlet at 9.48 (1 H) ppm B. Molecular formula C5H10O IR absorption at 1718 cm–1 NMR data: doublet at 1.10 (6 H) singlet at 2.14 (3 H) septet at 2.58 (1 H) ppm C. Molecular formula C10H12O IR absorption at 1686 cm–1 NMR data: triplet at 1.21 (3 H) singlet at 2.39 (3 H) quartet at 2.95 (2 H) doublet at 7.24 (2 H) doublet at 7.85 (2 H) ppm D. Molecular formula C10H12O IR absorption at 1719 cm–1 NMR data: triplet at 1.02 (3 H) quartet at 2.45 (2 H) singlet at 3.67 (2 H) multiplet at 7.06–7.48 (5 H) ppm

1 degree of unsaturation C=O, CHO 3 CH3 groups CHO

CHO CH3 C CH3 CH3

1 degree of unsaturation C=O 2 CH3's adjacent to H CH3 CH adjacent to 2 CH3's

O

5 degrees of unsaturation (4 due to a benzene ring) C=O CH3 adjacent to 2 H's CH3 CH2 adjacent to 3 H's 2 H's on benzene ring 2 H's on benzene ring

O

CH3

5 degrees of unsaturation (4 due to a benzene ring) C=O CH3 adjacent 2 H's O 2 H's adjacent to 3 H's CH2 a monosubstituted benzene ring

21.82 C7H16O2: 0 degrees of unsaturation IR: 3000 cm–1: C–H bonds NMR data (ppm): Ha: quartet at 3.5 (4 H), split by 3 H's Hb: singlet at 1.4 (6 H) Hc: triplet at 1.2 (6 H), split by 2 H's

CH3

Hb

CH3CH2 O C O CH2CH3

Hc Ha

CH3

Hb

Ha Hc

604

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 21–32

21.83 B. Molecular formula C9H10O A. Molecular formula C9H10O 5 degrees of unsaturation 5 degrees of unsaturation IR absorption at 1720 cm–1 → C=O IR absorption at 1700 cm–1 → C=O IR absorption at ~2700 cm–1 → CH of RCHO IR absorption at ~2700 cm–1 → CH of RCHO NMR data (ppm): NMR data (ppm): 2 triplets at 2.85 and 2.95 (suggests –CH2CH2–) triplet at 1.2 (2 H's adjacent) multiplet at 7.2 (benzene H's) quartet at 2.7 (3 H's adjacent) signal at 9.8 (CHO) doublet at 7.3 (2 H's on benzene) doublet at 7.7 (2 H's on benzene) O singlet at 9.9 (CHO) CH2 CH2 C

H

O C

CH2 CH3

H

21.84 C. Molecular formula C6H12O3 1 degree of unsaturation IR absorption at 1718 cm–1 → C=O To determine the number of H's that give rise to each signal, first find the number of integration units per H by dividing the total number of integration units (7 + 40 + 14 + 21 = 82) by the number of H's (12); 82/12 = 6.8. Then divide each integration unit by this number (6.8). O OCH3 NMR data (ppm): singlet at 2.2 (3 H's) OCH3 doublet at 2.7 (2 H's) singlet at 3.2 ( 6 H's – 2 OCH3 groups) triplet at 4.8 (1 H)

21.85 D. Molecular ion at m/z = 150: C9H10O2 (possible molecular formula) 5 degrees of unsaturation IR absorption at 1692 cm–1 → C=O O NMR data (ppm): triplet at 1.5 (3 H's – CH3CH2) quartet at 4.1 (2 H's – CH3CH2) doublet at 7.0 (2 H's – on benzene ring) doublet at 7.8 (2 H's – on benzene ring) singlet at 9.9 (1 H – on aldehyde)

21.86 OH OH OH

a. HO

CHO OH

b.

HO HO HO

OH CHO OH

CHO

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Aldehydes and Ketones—Nucleophilic Addition 21–33

21.87 OH

OH O

HO HO

O

HO HO

H Cl

OH

OH

OH

β-D-glucose

OH

above

H

OH

Cl

OCH3

CH3

OH

+

HCl

acetal

OH O

HO HO

below

O

HO HO

O OH

CH3OH

OH

OH

Cl

OH

O

O

H2O

O

HO HO

CH3OH

HO HO

OH

Cl OH

HO HO

O

HO HO

OH2

OH

OH

O

HO HO

OH O CH3 H

+ acetal

OH

HCl

OCH3

The carbocation is trigonal planar, so CH3OH attacks from two different directions, and two different acetals are formed.

21.88 H+

H O

H O

OH

O

OH

O OH

O

O

O

OH

OH

O H

O

O

H2O H3O

21.89 O

a.

O

brevicomin O

HO

b.

brevicomin

Br

H3O+

OH

O

O

O

[1] Ph3P

O

O

Br [2] BuLi

H+

OH

PPh3 CH3CH2CHO

H3O+ O

O

OH

[1] OsO4 [2] NaHSO3, H2O

OH

OH

O

O

PCC CH3CH2CH2OH CH=CHCH2CH3

(E and Z isomers)

606

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 21–34

21.90 acetal carbon

OH

a.

OCH3

O

HO HO

c.

OH HO O HO

CH3O CH3O

O

OCH3

O CH3O

HO CH3O

OH

O CH3O

OH

OH

(OH can be up or down in both products.)

HO

hemiacetal carbon OH + b. [1] H3O HO

O

HO

HO

(OH can be up or down.)

OH OH

[2] CH3OH, HCl

O

HO HO

OH HO O HO

O OCH3

HO

(OCH3 can be up or down.)

OCH3 [3] NaH (excess)

O

CH3O CH3O

CH3O O CH3O

CH3I (excess)

OCH3 O OCH3 OCH3

21.91 OH

OH

[1] O3

O

O CHO

[2] (CH3)2S CH3O

CH3O H OSO2R

RSO3H

H

H OH

CH3O

OH

O

R

S C6H10O3

OH

OH

O OH

O

S:

OH O

O

CH3OH

CH3O

Mechanism of R

OH

OH

NaBH4

O

O

CH3O H OSO2R

OH

OH

H

H

H

H O

H CH3O H

OH O O

S OSO2R

RSO3H

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

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Aldehydes and Ketones—Nucleophilic Addition 21–35

21.92 The mechanism involves SN2 displacement of Br, followed by intramolecular enamine formation.

H2N

+ H

H2N

N

N

+

SN2

H

H2N

+

HCO3–

CO32–

O

Br

H

H

N

H2N

N

O

CO32–

N R

N

+ H N

H CO32–

H2O

+

(any acid)

N

H A

H H

HO

N HN

+ HCO3– + H2O O NH

conivaptan

proton transfer

N R

N

R

Br–

N

R

O

N R

N

H N

+ HCO3–

O

N+

N

O N R

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

609

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–1

Chapter 22 Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution Chapter Review Summary of spectroscopic absorptions of RCOZ (22.5) IR absorptions • All RCOZ compounds have a C=O absorption in the region 1600–1850 cm–1. • RCOCl: 1800 cm–1 • (RCO)2O: 1820 and 1760 cm–1 (two peaks) • RCOOR': 1735–1745 cm–1 • RCONR'2: 1630–1680 cm–1 • Additional amide absorptions occur at 3200–3400 cm–1 (N–H stretch) and 1640 cm–1 (N–H bending). • Decreasing the ring size of a cyclic lactone, lactam, or anhydride increases the frequency of the C=O absorption. • Conjugation shifts the C=O to lower wavenumber. 1 H NMR absorptions • C–H α to the C=O absorbs at 2–2.5 ppm. • N–H of an amide absorbs at 7.5–8.5 ppm. •

13

C NMR absorption

C=O absorbs at 160–180 ppm.

Summary of spectroscopic absorptions of RCN (22.5) IR absorption • C≡N absorption at 2250 cm–1 13 C NMR absorption • C≡N absorbs at 115–120 ppm. Summary: The relationship between the basicity of Z– and the properties of RCOZ • Increasing basicity of the leaving group (22.2) • Increasing resonance stabilization (22.2)

R

O

O

O

C

C

C

Cl

R

acid chloride

O

anhydride

O R

R

O OH

carboxylic acid

~

R

C

O OR'

ester

R

C

NR'2

amide

• Increasing leaving group ability (22.7B) • Increasing reactivity (22.7B) • Increasing frequency of the C=O absorption in the IR (22.5)

General features of nucleophilic acyl substitution • The characteristic reaction of compounds having the general structure RCOZ is nucleophilic acyl substitution (22.1). • The mechanism consists of two steps (22.7A): [1] Addition of a nucleophile to form a tetrahedral intermediate [2] Elimination of a leaving group • More reactive acyl compounds can be used to prepare less reactive acyl compounds. The reverse is not necessarily true (22.7B).

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Chapter 22–2

Nucleophilic acyl substitution reactions [1] Reaction that synthesizes acid chlorides (RCOCl) From RCOOH (22.10A):

O R

C

O +

SOCl2

OH

R

+

C

+

SO2

Cl

HCl

[2] Reactions that synthesize anhydrides [(RCO)2O] [a] From RCOCl (22.8):

O R

C

+–

Cl

O

O

O

C

C

R'

R

O

C

O

Cl–

+

R'

O Δ

OH

[b] From dicarboxylic acids (22.10B):

O

O

+

OH

H2O

O

O

cyclic anhydride

[3] Reactions that synthesize carboxylic acids (RCOOH) O

O

[a] From RCOCl (22.8): [b] From (RCO)2O (22.9):

R

R

C

+

Cl

O

O

C

C

O

H2O

pyridine

R

R

C

+

OH

R

H 2O

2

R

C

Cl–

OH

O +

OR'

H2 O (H+ or –OH)

R

C

O OH

(with acid)

or R

R

C

C

O–

+

(with base)

O

H2O, H+

[d] From RCONR'2 (R' = H or alkyl, 22.13):

+

N H

O +

O

[c] From RCOOR' (22.11):

C

+

+

R'2NH2

+

R'2NH

+

+ N H

OH

O R

C

NR'2

R' = H or alkyl

O

H2O, –OH R

C

O–

[4] Reactions that synthesize esters (RCOOR') O

[a] From RCOCl (22.8):

R

C

O +

Cl

R'OH

pyridine

R

C

OR'

Cl–

R'OH

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Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–3

[b] From (RCO)2O (22.9):

R

O

O

C

C

O

O +

R'OH

R

O

[c] From RCOOH (22.10C):

R

C

OH

+

+

RCOOH

OR'

O

H2SO4

R'OH

C

R

R

C

+ H 2O

OR'

[5] Reactions that synthesize amides (RCONH2) [The reactions are written with NH3 as the nucleophile to form RCONH2. Similar reactions occur with R'NH2 to form RCONHR', and with R'2NH to form RCONR'2.] O

[a] From RCOCl (22.8):

R

C

O +

NH3

Cl

R

(2 equiv)

[b] From (RCO)2O (22.9):

R

O

O

C

C

O

R

C

R

NH3

R

(2 equiv)

C

OH

R

[2] Δ

C

NH2

R

C

C

NH2

+

RCOO– NH4+

+

H2O

O OH

R'NH2

+

C

R

DCC

O

[d] From RCOOR' (22.11):

NH4+Cl–

O

[1] NH3

O R

+ NH2

O +

O

[c] From RCOOH (22.10D):

C

+ H2O

NHR'

O +

NH3

OR'

R

C

NH2

+

R'OH

Nitrile synthesis (22.18) Nitriles are prepared by SN2 substitution using unhindered alkyl halides as starting materials. R X R = CH3,

+

–

CN

R C N

X–

+

SN 2

1o

Reactions of nitriles [1] Hydrolysis (22.18A) O

H 2O

R C N (H+

or

–OH)

R

C

O OH

(with acid)

or

R

C

O–

(with base)

612

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–4

[2] Reduction (22.18B) [1] LiAlH4

R CH2NH2

[2] H2O

1o amine

R C N O

[1] DIBAL-H [2] H2O

R

C

H

aldehyde

[3] Reaction with organometallic reagents (22.18C) R C N

O

[1] R'MgX or R'Li [2] H2O

R

C

R'

ketone

Practice Test on Chapter Review 1. Give the IUPAC name for each of the following compounds. CN

a.

O

O

b.

c.

O

N CH3

2. (a) Which compound absorbs at the lowest wavenumber in the IR? (b) Which compound absorbs at the highest wavenumber? O O

H N

O NH

O

O

O

A

B

C

D

3. a. Which of the following reaction conditions can be used to synthesize an ester? 1. 2. 3. 4. 5.

RCOCl + R'OH + pyridine RCOOH + R'OH + H2SO4 RCOOH + R'OH + NaOH Both methods [1] and [2] can be used to synthesize an ester. Methods [1], [2], and [3] can all be used to synthesize an ester.

b. Which of the following compounds is most reactive in nucleophilic acyl substitution? 1. CH3COCl 2. CH3COOCH3

3. CH3CON(CH3)2 4. (CH3CO)2O

5. CH3COOH

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

613

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–5

4. What reagent is needed to convert (CH3CH2)2CHCOOH into each compound? a. (CH3CH2)2CHCOO–Na+ b. (CH3CH2)2CHCOCl c. (CH3CH2)2CHCON(CH3)2 d. (CH3CH2)2CHCO2CH2CH3 e. [(CH3CH2)2CHCO]2O 5. What reagent is needed to convert CH3CH2CH2CN to each compound? a. CH3CH2CH2COOH b. CH3CH2CH2CH2NH2 c. CH3CH2CH2COCH2CH3 d. CH3CH2CH2CHO 6. Draw the organic products formed in the following reactions. H D

a.

Br

CO2H

[1] NaCN

[1] SOCl2

d.

[2] LiAlH4

[2] (CH3CH2)2NH

[3] H2O [Indicate stereochemistry.]

b.

O

O

O

O CH2CH3 O

OH

c.

OH

O

NaOH, H2O

+

O O

e.

+ CH3CH2NH2 (excess)

O

[1] NaCN

f. Br

O

[2] CH3CH2MgBr [3] H2O

O

Answers to Practice Test 1. a. 5-ethyl-2-methylheptanenitrile b. cyclohexyl 2methylbutanoate c. N-cyclohexyl-Nmethylbenzamide 2. a. C b. A 3. a. 4 b. 1 4. a. NaOH b. SOCl2 c. HN(CH3)2, DCC d. CH3CH2OH, H2SO4 e. heat

5. a. H3O+ b. [1] LiAlH4; [2] H2O c. [1] CH3CH2Li; [2] H2O d. [1] DIBAL-H; [2] H2O 6. D H

a.

b.

O

CH2NH2

O

O

O

O

N(CH2CH3)2

d. O

NHCH2CH3

e. O

O– H3NCH2CH3

HOCH2CH3 OCOCH3

c.

OH O

O

f.

+ OH

O

614

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–6

Answers to Problems 22.1

The number of C–N bonds determines the classification as a 1o, 2o, or 3o amide. NH2 1° amide O

O

O

1° amide H2N O

NH

oxytocin All seven others are 2° amides.

NH O

N H

O

O

O

HN

N

H N

O N H H2N

NH2 S

H N

O

1° amide

S

3° amide

O HO

22.2

As the basicity of Z increases, the stability of RCOZ increases because of added resonance stabilization. R

O

O

C

C

O

R + Br

Br

R

C

+

Br

The basicity of Z determines how much this structure contributes to the hybrid. Br– is less basic than –OH, so RCOBr is less stable than RCOOH.

22.3 O CH3 Cl

CH3

C

O Cl

CH3

H

C

Cl

O

O CH3 NH2

C

NH2

H

C

NH2

This resonance structure contributes little to the hybrid since Cl– is a weak base. Thus, the C–Cl bond has little double bond character, making it similar in length to the C–Cl bond in CH3Cl. This resonance structure contributes more to the hybrid since –NH2 is more basic. Thus, the C–N bond in HCONH2 has more double bond character, making it shorter than the C–N bond in CH3NH2.

22.4 a. (CH3CH2)2CH

COCl

b.

C6H5COOCH3

re-draw

re-draw

O

O Cl 2-ethylbutanoyl chloride

2-ethyl

OCH3

methyl benzoate

alkyl group = methyl acyl group = benzoate

615

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–7

O

e.

c. CH3CH2CON(CH3)CH2CH3

CH3CH2

O

C

C

O

benzoic propanoic anhydride

re-draw O N

acyl group = propanoic

N-ethyl-N-methyl

acyl group = benzoic

N-ethyl-N-methylpropanamide

acyl group = propanamide

1

O

d. H

CN

f.

C

OCH2CH3

3

6 carbon chain = hexanenitrile

alkyl group = ethyl acyl group = formate

3-ethylhexanenitrile

ethyl formate

22.5 d. N-isobutyl-N-methylbutanamide

a. 5-methylheptanoyl chloride

g. sec-butyl 2-methylhexanoate

O

O

O N

Cl

b. isopropyl propanoate

O

h. N-ethylhexanamide

e. 3-methylpentanenitrile

O O

CN O

f. o-cyanobenzoic acid

c. acetic formic anhydride O

O

O

H

O

N H

OH

CH3

CN

22.6

CH3CONH2 has two H’s bonded to N that can hydrogen bond. CH3CON(CH3)2 does not have any H’s capable of hydrogen bonding. This means CH3CONH2 has much stronger intermolecular forces, which leads to a higher boiling point.

22.7 a. CH3

O

O

O

C

C

C

OCH2CH3

and CH3

N(CH2CH3)2

c. CH3CH2CH2

b.

O

O O

and

O

d.

O O

smaller ring: C=O at a higher wavenumber

NHCH3

and

CH3CH2CH2

2° amide: 1 N–H absorption at 3200–3400 cm–1

amide: C=O at lower wavenumber

O

O

anhydride: 2 C=O peaks

NH2

1° amide: 2 N–H absorptions O

and

C

Cl

616

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–8

22.8 Hb

O

Hc signal from the 5 H's on the aromatic ring

Ha

O

H H

CH3

Hb

Ha

A

Molecular formula C9H10O2 5 degrees of unsaturation IR: 1743 cm–1 from ester C=O 3091–2895 cm–1 from sp2 and sp3 C–H 1 H NMR: Ha = 2.06 ppm (singlet, 3 H) – CH3 Hb = 5.08 ppm (singlet, 2 H) – CH2 Hc = 7.33 ppm (broad singlet, 5 H)

22.9

H H

O

O

Hb signal from the 10 H's on the two B aromatic rings Molecular formula C14H12O2 9 degrees of unsaturation IR: 1718 cm–1 from conjugated ester C=O 3091–2953 cm–1 from sp2 and sp3 C–H 1H NMR: Ha = 5.35 ppm (singlet, 2 H) Hb = 7.26–8.15 ppm (multiplets, 10 H)

More reactive acyl compounds can be converted to less reactive acyl compounds. a.

CH3COCl

CH3COOH

more reactive

YES

b. CH3CONHCH3 less reactive

NO

c.

CH3COOCH3

less reactive CH3COOCH3

d.

more reactive

(CH3CO)2O

more reactive

CH3COCl

NO

more reactive

YES

less reactive

less reactive

CH3CONH2

22.10 The better the leaving group is, the more reactive the carboxylic acid derivative. The weakest base is the best leaving group. O

O

C

C

a.

NH2

−NH 2

CH3CH2

C

OCH3

−OCH 3

strongest base least reactive

b.

O

intermediate

−Cl

weakest base most reactive

O

O

O

C

C

C

CH3CH2

NHCH3

−NHCH 3

OH

−OH

strongest base least reactive

intermediate

Cl

CH3CH2

O O

C

CH2CH3

−OOCCH

2CH3 weakest base most reactive

22.11 CH3

O

O

O

O

C

C

C

C

O

CH3

Cl3C

O

CCl3

acetic anhydride trichloroacetic anhydride

The Cl atoms are electron withdrawing, which makes the conjugate base (the leaving group, CCl3COO–) weaker and more stable.

617

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–9

22.12 O

O H2O

a.

O

OH

+

N – H Cl

pyridine Cl

O

NH2 + NH4+Cl–

NH3 excess O

O

CH3COO−

b.

c.

CH3 +

O

Cl–

(CH3)2NH

N(CH3)2

d. excess

+ (CH3)2NH2 Cl–

22.13 The mechanism has three steps: [1] nucleophilic attack by O; [2] proton transfer; [3] elimination of the Cl– leaving group to form the product. OCH3

OCH3

O OH

+

OCH3

O Cl O

Cl

O Cl O H N

H OCH3

OCH3

OCH3

N

OCH3

O Cl

O

OCH3

A

22.14 O

O O

O

O OH

H2O

O NH3

HO

a.

c. O

b.

CH3OH

O

NH2

O

NH4

excess

O CH3

O

O (CH3)2NH

HO

d.

excess

O N(CH3)2

O (CH3)2NH2

22.15 Reaction of a carboxylic acid with thionyl chloride converts it to an acid chloride. O

a. CH3CH2

C

O SOCl2 OH

CH3CH2

O

b.

C

C

Cl

O OH

[1] SOCl2

C

O [2] (CH3CH2)2NH (excess) Cl

C

N(CH2CH3)2

(CH3CH2)2NH2 Cl–

618

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–10

22.16 H2SO4

COOH + CH3CH2OH

a.

OCH2CH3 + H2O

C O O

COOH

b.

+

C

H2SO4

OH

+ H2O

O

O COOH

c.

+

C

O

Na+

+ CH3OH

O

H2SO4

d. HO

O

NaOCH3

+ H2O

OH

O

22.17 O C

O CH318OH

OH

C

18

OCH3 + H2O

22.18 O C

HO

OH O

H

H−OSO3H

C

HO

O

HO OH H

O

HO OH H

+ HSO4–

O –OSO

O

O H O

+ H2SO4

3H

HSO4–

HO OH2 O

O

H2O +

H−OSO3H

+ HSO4–

22.19 O

a. O

CH3NH2

+

O− H3NCH3

OH

b.

O

O

[1] CH3NH2 OH

[2] Δ

NHCH3 + H2O

c.

O

O CH3NH2 OH DCC

NHCH3

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

619

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–11

22.20 O [1] H3O+

OH

+

HO

O

+

HO

O

a.

CH3O

O

CH3O

O

[2] H2O, –OH

octinoxate CH3O

O [1] H3O+

O

b.

OH

O

+

HO

+

HO

OH O

OH [2] H2O, –OH

octyl salicylate

O OH

22.21 Bonds broken during hydrolysis are indicated. O HO HO

H C(CH3)3

O O

O CH3 OH

COOH OH

HO HO

O H3O+

HO HOOC

O

H C(CH3)3

CH3 OH HO H HOOC

O H O

ginkgolide B

22.22 HOCH2 O

OH HOCH2

HO

RCOOH, H2SO4

O O HO

RCOOCH2

OH

OH

RCOO

CH2OH

a long-chain fatty acid

sucrose

RCOO OOCR OOCR CH2 O O CH2OOCR OOCR

O

RCOO

olestra

22.23 CH2OCO(CH2)15CH3 CHOCO(CH2)15CH3 CH2OCO(CH2)7CH=CH(CH2)7CH3 cis

CH2 OH

Na+ −OOC(CH2)15CH3

CH OH

Na+ −OOC(CH2)15CH3

CH2 OH

Na+ −OOC(CH2)7CH=CH(CH2)7CH3

hydrolysis

glycerol

soap

cis

620

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–12

22.24 O

O NH

OH

H2O, OH

proton transfer

O

N H H

O

O O

NH

NH2

22.25 Aspirin has an ester, a more reactive acyl group, whereas acetaminophen has an amide, a less reactive acyl group. a. The ester makes aspirin more easily hydrolyzed with water from the air than acetaminophen. Therefore, Tylenol can be kept for many years, whereas aspirin decomposes. b. Similarly, aspirin will be hydrolyzed and decompose in the aqueous medium of a liquid medication, but acetaminophen is stable due to the less reactive amide group, allowing it to remain unchanged while dissolved in H2O. ester O

C

amide H N

CH3

O COOH

CH3

O

HO

acetylsalicylic acid

C

acetaminophen

22.26 "Regular" amide is not hydrolyzed. H H H N S

R O

H H N

R

H3O+

O

N O COOH

O

H S HN OH COOH

22.27 O

H N

O

H N

N H

N H

O

O

nylon 6,10

O Cl

Cl O

+

H2N

NH2

621

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–13

22.28 OH

+

HOOC

COOH

In the polyester Kodel, most of the bonds in the polymer backbone are part of a ring, so there are fewer degrees of freedom. Fabrics made from Kodel are stiff and crease resistant, due to these less flexible polyester fibers.

HO

1,4-dihydroxymethylcyclohexane

terephthalic acid

O

O

O

O

O

O O

O

Kodel

22.29 O O

O

OH +

OH

OH

O O

O

OH

PLA polyl(lactic acid)

O

O

O

O

Reaction occurs here (–H2O).

22.30 Acetyl CoA acetylates the NH2 group of glucosamine, since the NH2 group is the most nucleophilic site. HO

HO O

O HO

+

OH

CH3

C

O HO

NH2

HO

OH

SCoA HO

glucosamine

HN

NAG

O

22.31 a. CH3CH2CH2 Br

NaCN

CH3CH2CH2 CN

H2O, –OH

c. CN

CN

H2O, H+

b.

COO

COOH

CN

COOH

22.32 a.

CH3

O

OH

NH

C

C

C

NH2 O

b.

C

CH3

c.

NH OH

NHCH3

C

NCH3

CH3CH2

OH

NH2 CH3CH2

C

O

622

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–14

22.33 [1] NaCN

a. CH3CH2 Br

b. CH3CH2CH2 CN

CH3CH2 CH2NH2

[2] LiAlH4 [3] H2O

[1] DIBAL-H

O CH3CH2CH2 C

[2] H2O

H

22.34 OCH3 [1] CH3CH2MgCl

a.

CN

[2] H2O

OCH3 O C CH2CH3

O

CN

C

[1] C6H5Li

b.

[2] H2O

22.35 O

O CH2 CN

a.

[1] CH3MgBr

CH2 CN

CH2 C CH3

[2] H2O

c.

[1] DIBAL-H

CH2 C H

[2] H2O O

CH2 CN

b.

[1] (CH3)3CMgBr

O

CH2 C

CH2 CN C(CH3)3

CH2 C

d.

H 3 O+

OH

[2] H2O

22.36 CH3 C N

O

O

[1] CH3CH2MgBr

[1] CH3MgBr

CH3CH2 C N

[2] H2O

[2] H2O

22.37 The better the leaving group, the more reactive the acyl compound. O

O OH

O SH

worst leaving group least reactive

intermediate reactivity

22.38 O

a.

3

2

1

1 CN

O

A isobutyl 2,2-dimethylpropanoate

6

5

4

3

2

B 2-ethylhexanenitrile

Cl

best leaving group most reactive

623

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–15

b.

A

A

[1] H3O+

COOH

[2] –OH

COO

HO

HO

H2O OH [3] CH3CH2CH2MgBr

A

HO

COO

[2] –OH

B

OH

COOH

[1] H3O+

B

HO

H2O

[4] LiAlH4

A

H2O

H2O O [3] CH3CH2CH2MgBr

B

H2O NH2

H2O

[4] LiAlH4

B

22.39 Better leaving groups make acyl compounds more reactive. C has an electron-withdrawing NO2 group, which stabilizes the negative charge of the leaving group, whereas D has an electron-donating OCH3 group, which destabilizes the leaving group. O CH3

C

O O

NO2

C

CH3

C

O

OCH3

D

an electron-withdrawing substituent an electron-donating substituent O O

N

O O

O

leaving group from C

N

O

OCH3

O

OCH3

O

one possible resonance structure

Delocalizing the negative charge on the NO2 stabilizes the leaving group, making C more reactive than D.

one possible resonance structure leaving group from D Adjacent negative charges destabilize the leaving group.

624

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–16

22.40 Cl

a.

f.

2,2-dimethylpropanoyl chloride

m-chlorobenzonitrile Cl

O

O

O

b.

CN

cyclohexyl pentanoate

g.

Cl

O O

c.

3-phenylpropanoyl chloride O O

h.

O

cis-2-bromocyclohexanecarbonyl chloride

C Cl Br

cyclohexanecarboxylic anhydride

O

i. O

d.

N

phenyl phenylacetate

N,N-diethylcyclohexanecarboxamide

O O

O

e.

N

N-cyclohexylbenzamide

H

j.

cyclopentyl cyclohexanecarboxylate

O

22.41 e. isopropyl formate

a. propanoic anhydride O

i. benzoic propanoic anhydride O

O

O H

O

b. α-chlorobutyryl chloride

O O

O

f. N-cyclopentylpentanamide

O

j. 3-methylhexanoyl chloride

NH

O

Cl Cl

O

Cl

c. cyclohexyl propanoate

g. 4-methylheptanenitrile

k. octyl butanoate

O

O CN

O

d. cyclohexanecarboxamide

h. vinyl acetate

O C

O

l. N,N-dibenzylformamide O

O NH2

O

H

CH3

N

22.42 Rank the compounds using the rules from Answer 22.10. O

a.

O NH2

−NH

2 strongest base least reactive

O OCH2CH2CH3

−OCH CH CH 2 2 3

intermediate

Cl

Cl– weakest base most reactive

625

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–17

O

b.

O

O

O

O

O

F3C

ester least reactive

anhydride intermediate

O O

CF3

anhydride with electronwithdrawing F's most reactive

22.43 resonance structures for the leaving group O

O C N

N

Nu

O N

R C N

R

R

Nu

C

N

N

Nu

imidazolide

N

N

N

N

N

N

N

N

The leaving group is both resonance stabilized and aromatic (6 π electrons), making it a much better leaving group than exists in a regular amide.

22.44 Reaction as an acid: O CH3

C

O

B NH2

CH3

C

O NH

CH3

C

CH3CH2 NH2 NH

B

CH3CH2 NH

no resonance stabilization of the conjugate base

These two resonance structures make the conjugate base more stable, and therefore CH3CONH2 a stronger acid. Reaction as a base: O CH3

C

O NH2

CH3

C

This electron pair is localized on N.

CH3CH2 NH2 NH2

This electron pair is delocalized by resonance, making it less available for electron donation. Thus, CH3CONH2 is a much weaker base.

22.45 CH3CH2CH2CH2COCl

a.

pyridine

b.

NH3

H2O

CH3CH2OH

d.

CH3CH2CH2CH2COOH N Cl– H

c.

O

CH3COO− CH3CH2CH2CH2

O

excess

N Cl– H

O CH3

NH4 Cl–

(CH3CH2)2NH

e.

CH3CH2CH2CH2COOCH2CH3

pyridine

CH3CH2CH2CH2CONH2

excess

Cl–

CH3CH2CH2CH2CON(CH2CH3)2 (CH3CH2)2NH2 Cl–

C6H5NH2

f.

excess

CH3CH2CH2CH2CONHC6H5 C6H5NH3 Cl–

626

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–18

22.46 O

O O

a.

b.

SOCl2

d.

no reaction

O

e.

2

no reaction

(CH3CH2)2NH

O

H2O

NaCl

OH

excess

N(CH2CH3)2 O O H2N(CH2CH3)2

O CH3OH

O OCH3

c.

f.

O

CH3CH2NH2

NHCH2CH3 O

excess OH

O

H3NCH2CH3

22.47 C6H5CH2COOH

a.

O

NaHCO3

+ H2CO3 g. O Na

b.

NaOH

h.

O

+ H2O

i.

O

SOCl2

OCH3 O

CH3OH –OH

O Na

c.

O

CH3OH H2SO4

O

[1] NaOH

O

Cl

d.

NaCl

no reaction

j.

O

CH3NH2 DCC

e.

NH3

O

k.

(1 equiv) O NH4

f.

O

[1] NH3 [2] Δ

l.

NH2

O

O

[2] CH3COCl

NHCH3

[1] SOCl2

O

[2] CH3CH2CH2NH2

NHCH2CH2CH3 O

[1] SOCl2 [2] (CH3)2CHOH

OCH(CH3)2

22.48 c.

CH3CH2CH2CO2CH2CH3

H2O, –OH

O O

HO

O

a.

SOCl2

no reaction

d.

NH3

O

b.

NH2

HO

O

H3O+ OH

HO

e. CH3CH2NH2

NHCH2CH3

HO

627

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–19

22.49 NH2

H3 O +

a.

CH2COOH

CH2COO

H2O, –OH

b.

O

22.50 H3O+

a.

COOH

d.

[1] CH3CH2Li O

[2] H2O H2O, –OH

b.

C6H5CH2CN

COO

C

[2] H2O

O

[1] LiAlH4

CH2NH2

CH3

c. [1] CH3MgBr

f.

[2] H2O

H

e. [1] DIBAL-H

O

[2] H2O

22.51 O

O COOH

C

SOCl2

a. OH

O

f.

Cl

O H3O+

OH

OH O

O [1] NaCN

O

b. C6H5COCl

+

+ C6H5

N H (excess)

c. C6H5CN

OH

N

N

Cl–

CH3CH2CH2CH2Br

O

[2] H2O, –OH

C6H5

[2] H2O

h. C6H5CH2COOH

C

[1] SOCl2

O

C6H5(CH2)2NH(CH2)3CH3

[2] CH3CH2CH2CH2NH2 [3] LiAlH4 [4] H2O

CH2CH2CH3

O

O

H2SO4 (CH3)2CH

CH3CH2CHOH

C

OCH(CH3)CH2CH3 i.

Δ HOOC

O

COOH

CH3

e.

CH3CH2CH2CH2

H H O

[1] CH3CH2CH2MgBr

d. (CH3)2CHCOOH +

g.

O

NHCOCH3

H2 O –OH

O

j. +

(CH3CO)2O +

NH2

NHCOCH3

NH2

O

(excess) + CH3COO– H3N

22.52 Both lactones and acetals are hydrolyzed with aqueous acid, but only lactones react with aqueous base. O

a.

O O

O

X

H3O+

O

OH O OH

COOH OH

b.

NaOH

O

H 2O

O

O O

O

X

CO2 Na+ OH

628

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–20

22.53 Br NaCN

CN

H3O+

COOH SOCl2

A

COCl [1] (CH3)2CuLi

[1] LiAlH4

CH3

O

[2] H2O

D

B

C E

CH3OH, H+

[2] H2O

PCC

CH2NH2

CO2CH3 [1] DIBAL-H

CHO [1] CH3Li

[2] H2O C

G

F

OH

[2] H2O

H

[1] LiAlH4

(CH3CO)2O

[2] H2O

CH2OTs NaCN

CH2OH

CH2NHCOCH3

CH2CN

TsCl, pyridine I

K

J

L [1] CH3MgBr [2] H2O O

M

22.54 O

H D

C CH3

a.

CH3COCl

OH

c.

O

pyridine CH3

CH3 NaCN

Br

b.

CH3

H+

COOH

O CN

H D

C

H D CH3CH2OH

D H

d.

CH3

C

+ Cl C H 6 5

CH3

C

CH3

CH3

C

C

H NH2 (2 equiv)

C6H 5

COOCH2CH3 CH3

C H + H C6H5 NH NH3 + C Cl– CH3 O

22.55 Hydrolyze the amide and ester bonds in both starting materials to draw the products.

a.

O

O

CO2CH2CH3

H2 O

O

CO2H

O HO

N H

OH

H 2N NH2

oseltamivir

NH2

629

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–21

NH2 HOOC

b.

H N

NH2

H2 O

O CH3O

HOOC

H2 N

+ CH3OH

OH COOH

O

O

aspartame

phenylalanine

22.56 F F

F (–H+)

CH3SO2Cl

O

(CH3CH2)3N

O

O

OH

CH3O

OSO2CH3

CH3O

N H

CH3O

PhCH2NH2

intramolecular amide formation F

O

N Ph

F

C18H18FNO

22.57 O C

a.

Cl

O

C Cl

C Cl

H

b.

HO O

O

O

proton

O O

OH

NHNH2 + Cl

+ HN

N O

C

NH2NH

NH2NH

NH2NH2

O

O

OH

transfer HO

O O

Ph

630

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–22

22.58 H+

O CH3

C

OH

OH

CH3

C

OH

OH

OH

CH3 C

18

O H

H

18

OH

CH3 C OH2

18

18

O H

+ H2O

OH

H+

CH3 C OH

CH3 C OH

18

O H

O H

A

+ H3O+

+ H2O

+ H2O

Two possibilities for A:

OH

O H

CH3 C

CH3

18

O H

C

O

H2O

18

C

CH3

OH

+ H3O+

18

OH

A OH

OH

CH3 C

CH3

18

O H

OH

C 18 O H

C

CH3

H2O

+ H3O+

18

O

A

22.59 H2O

H

H2O

O H O O

OH

OH

OH

H−OH2

O

O

OH

O

OH

O

H2O

OH

H

γ-butyrolactone

H−OH2 O H3O

O

HO

HO

OH

H

H2O

OH

4-hydroxybutanoic acid GHB

22.60 enzyme CH2OH

CH2 O H

O O CO2H

H A

aspirin

O

O

OH CO2H

OH CO2H

CH2 A

H A

O O OH CO2H

A

CH2 H O O HA

CH2O C CH3

inactive enzyme

A

O H CH2O C

OH

O

CO2H

OH CO2H

+ CH3

salicylic acid

A

631

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–23

22.61 O

Ar Ar F

O RO

O

Ar'

N

CH3CH3

Ar

RO

N

O

MgBr+

Ar'

N OR

Ar'

H CH3CH2 MgBr O

N

RO–

O F

22.62 CH3O

CH3O O

C

O

O

OH

OH O

O

proton

O

OH

O O

OCH3

transfer

HO

OCH3

CO2H HO

OCH3

D

H OH2

632

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–24

22.63 H OCH2CH3 O

O CH2CH3

OH

O

O

H–Cl

A

Cl

H

OCH2CH3

OH

O

OH

O

Cl

H–Cl

OCH2CH3 OH2

OCH2CH3

O

OH

OCH2CH3

O

OH

O

OCH2CH3 OH

O

H

Cl

H

H–Cl Cl Cl CO2CH2CH3 OCH2CH3

OCH2CH3

O

Cl

+ Cl

Cl

=

O

H2O

B

22.64 The mechanism is composed of two parts: hydrolysis of the acetal and intramolecular Fischer esterification of the hydroxy carboxylic acid. H2O

H O

O OH

H OH2

O

O

H2O O

HO

O

O

OH O

O

H2O

H

O

O

OH

OH

O

+ H2O

HO

H HO HO

H OH2

O

HO

HO

HO HO

O

OH

O

OH

HO

OH O

O

+ H3O+

+ H2O

H

H OH2 H2O HO

H O

OH OH

OH

HO

HO

O

OH

H OH2 OH OH

HO

+ H 2O O

O

O O

HO

O H

HO

HO

+ H3O+

H2O

OH2 OH

+ H 2O

633

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–25

22.65 O H3O+ HO

Ha

CN

A C6H10O2: 2 degrees of unsaturation O

IR: 1770 cm–1 from ester C=O in a five-membered ring NMR: Ha = 1.27 ppm (singlet, 6 H) – 2 CH3 groups Hb = 2.12 ppm (triplet, 2 H) – CH2 bonded to CH2 Hc = 4.26 ppm (triplet, 2 H) – CH2 bonded to CH2

1H

Hb Hc A

proton C N

HO

HO

C N OH2

HO

transfer

C NH

H OH2

OH

imidic acid H2O

C NH2

HO

C NH2

HO

OH

C NH2

HO

O H OH2

O H

amide

H2O

H2O

H2O HO

NH2

HO

O H

NH2

H O

H OH2

O

H O

NH3 O

O

NH3

O

H3O

H2O

H2O

O

22.66 Fischer esterification is the treatment of a carboxylic acid with an alcohol in the presence of an acid catalyst to form an ester. a. (CH3)3CCO2CH2CH3

(CH3)3CCOOH

O

c.

+ HOCH2CH3

b.

O

OH

HO

O

OH

O

O

O

d.

HO

OH

HO

O

O

O

22.67 a.

NH3 Br excess

NH2 H3O+

NaCN Br

H N

OH NH2

CN O

DCC

O

634

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–26

[1]

b.

(from a.)

CN

MgBr

Mg

[2] H2O

Br

MgBr

H2SO4

O

O O CrO3

NaOH

c.

OH

Br

OH

H2SO4, H2O

OH

O

(from b.) [1] O

d.

MgBr

OH

(2 equiv) [2] H2O

O

(from c.)

22.68 a.

Br

b.

COOH

(from a.)

c.

–CN

CN

H2O, H+

HOCH2CH3

(from a.)

C

[2] H2O

CN

(from a.)

COCl

CH3

O

[1] LiAlH4

d.

SOCl2

CO2CH2CH3

H2SO4

[1] CH3MgBr CN

COOH

CH2NH2

[2] H2O

CH3COCl

CH2NHCOCH3

22.69 a. CH3Cl

+ NaCN

CH3 CN

H3 O +

CH3 COOH

[1] CO2 CH3 Cl + Mg

Br + NaCN

b.

sp2

CH3 MgCl

MgBr

+ Mg

(CH3)3C

Cl + Mg

d. HOCH2CH2CH2CH2Br

CH3 COOH

This method can't be used because an SN2 reaction can't be done on an sp2 hybridized C.

Br

c. (CH3)3CCl + NaCN

[2] H3O+

COOH

[1] CO2 [2] H3

O+

This method can't be used because an SN2 reaction can't be done on a 3° C. (CH3)3C + NaCN

HOCH2CH2CH2CH2 Br + Mg

MgCl

[1] CO2 [2] H3O+

(CH3)3C

HOCH2CH2CH2CH2 CN

COOH H3O+

HOCH2CH2CH2CH2 COOH

This method can't be used because you can't make a Grignard reagent with an acidic OH group.

635

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–27

22.70 CH3OH O

SOCl2

CH3

CH3Cl

CH3

HNO3 H2SO4

AlCl3

COOH

KMnO4

O2N

C

CH3CH2OH H2SO4

O 2N

OCH2CH3

O2N H2 Pd-C

(+ ortho isomer)

O C

OCH2CH3

H2N

22.71 O NH2

HO

HO

CH3COCl N H

N H

NaH

O

N H

SOCl2

serotonin

O N H N H CH3Cl

CH3COOH

SOCl2

CH3OH

O

CrO3 H2SO4, H2O

N H

CH3O

CH3CH2OH

N H

melatonin

22.72 O

O CH3CH2Cl

a.

AlCl3

OH

COH

KMnO4

NO2

H2, Pd-C

HO

HO

(+ ortho isomer) c. HO

H N

C

CH3

O

acetaminophen (from b.)

H N

NaH O

OH

salicylamide NH2

H N

CH3COCl

H2SO4

OH

CNH2

NH3

OH

OH

(+ para isomer)

b.

Cl

(2 equiv)

OH

HNO3

SOCl2

O

C

CH3

acetaminophen H N

CH3CH2Br

O CH3CH2O

CH3

O

HO

(More nucleophilic NH2 reacts first.)

C

C

CH3

O

p-acetophenetidin

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–28

22.73 HO [1] CrO3, H2SO4, H2O [2] SOCl2 O

a.

O

O Cl

Cl2

Cl AlCl3

FeCl3

CH3OH O

SOCl2

b.

CH3Cl

CH3

AlCl3

CH3

HNO3 H2SO4

COOH

KMnO4

O2N

C

CH3OH H+

O2N

O2N

(+ ortho isomer)

H2, Pd-C O

Br

FeBr3

H2SO4

Cl [1] CrO3, H2SO4, H2O [2] SOCl2

Br

HNO3

C

OCH3

N H

Br2

O

O

C

O

c.

OCH3

O2N (+ ortho isomer)

H2N

CH3CH2CH2OH O Br Cl

H2, Pd-C

Br

[1] CrO3, H2SO4, H2O [2] SOCl2

H2 N

OCH3

H

N O

CH3CH2OH (CH3)3COH

CH3OH

HCl

d.

SOCl2

(CH3)3CCl

CH3Cl

AlCl3

KMnO4

SOCl2 HO

AlCl3 (+ ortho isomer)

Br2 FeBr3

Br

CH3Cl

Br

Cl O

O Br

KMnO4 HO

AlCl3 (+ ortho isomer)

Br

NaOH O

O

O O

O O

(CH3)3C

Br

637

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–29

22.74 CH3Cl

a.

CH3OH

Br2

CH3

AlCl3 SOCl2

NaCN

CH2Br

hν

CH2CN

H2O H+

CH2COOH HOCH2CH3 H2SO4

CH3Cl

CH2COOCH2CH3

ethyl phenylacetate NO2

(from a.) CH3

CH3Cl

b.

NO2 CH3

HNO3

NH2 COOH

KMnO4

H2SO4

AlCl3

NH2 COOH

H2

HOCH3

Pd-C

H2SO4

COOCH3

methyl anthranilate

(+ para isomer)

(from a.)

O CH3 KMnO 4

CH3Cl

c.

COOH

AlCl3 CH3CH2OH

OH CH3COOH H2SO4

[1] LiAlH4 [2] H2O

CrO3

O

benzyl acetate

CH3COOH

H2SO4, H2O

22.75 O C

a. CH3

OH

c. CH3CH2Br

d. CH3 CH2OH

13

OCH2CH3

O

O

CrO3

CH3CH2OH

13

H2SO4, H2O H218O

CH3

C

–

13

CH3 CH2Br

OCH2CH3

O

CH3COCl CH3

C

18

OCH2CH3 18

18

OH

13C

CH3

H2SO4

OH

CH3CH218OH

+ base (H18O–) PBr3

C

CH3

H2SO4

b. CH313CH2OH

13

O

CH313CH2OH

CH3

13

CH218OH

CrO3 H2SO4, H218O

18

O

13 C 18 OH CH3

CH3CH2OH H2SO4

O

13 C

CH3

+

OCH2CH3

H218O

22.76

a. HO

OH

and

HO

C

C

O

O

O

O

O

OH

O

O O

O O

O NH

b. ClOC

COCl

and

H2N

NH2

O NH HN

HN O

O

638

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–30

22.77 O O

a.

O O

O

O

O

O O O

O O

b.

O

OH

HO O

HO

OH

O O O

O

O

O

HO

OH

HO

OH

22.78 a. Docetaxel has fewer C’s and one more OH group than taxol. This makes docetaxel more water soluble than taxol. OH O

b.

O

O

O

OH O

N

O

H

carbamate

=

OH

docetaxel

HO O

O

RO

NHR

carbamate

H O

O O

O RO

O NHR

RO

1 most stable

O NHR

O

RO

4 least stable

NHR

RO

3

Increasing stability: 4 < 3 < 2 < 1

O

c.

H OH2

OH O

O

NHR

O

NHR H2O

NHR

2 More basic N atom allows N to donate electron density more than O, so this structure contributes more than 3 to the hybrid.

H

O

C

H

H2O

NHR

H2O

H OH2

O C

+ NH2R

H OH2

O H3O H3NR

H2O

639

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–31

OH O

d.

O

O

OH O

N

O

H

H3O+

OH HO O

docetaxel

O

COOH

O

O

O

CO2

H

OH

H3N OH

OH

O

O OH CH3CO2H

HO HO

H HO HO

O

22.79 O

a. CH3

C

O

and OCH3

CH3CH2

C

c.

OH

contains a broad, strong OH absorption at 3500–2500 cm–1 O

N(CH3)2

and

NH2

2 NH absorptions at 3200–3400 cm–1 C=O at < 1700 cm due to the stabilized amide C=O absorption at higher wavenumber –1

O

O

and

b.

O

O

d. O

Cl

OH

O

and CH3

Acid chloride CO absorbs at much higher wavenumber.

Cl

OH absorption at 3500–3200 cm–1 + C=O

ketone

only C=O

22.80 O

a. C6H5COOCH2CH3

CH3CH2COOCH2CH3

O

b.

most resonance stabilized CH3CONH2

Increasing wavenumber

least resonance stabilized CH3COOCH3

CH3COCl

Increasing wavenumber

22.81 a.

C6H12O2 → one degree of unsaturation IR: 1738 cm–1 → C=O NMR: 1.12 (triplet, 3 H), 1.23 (doublet, 6 H), 2.28 (quartet, 2 H), 5.00 (septet, 1 H) ppm

O

O O

b.

c. C8H9NO IR: 3328 (NH), 1639 (conjugated amide C=O) cm–1 NMR: 2.95 (singlet, 3 H), 6.95 (singlet, 1 H), 7.3–7.7 (multiplet, 5 H) ppm

C4H7N IR: 2250 cm–1 → triple bond NMR: 1.08 (triplet, 3 H), 1.70 (multiplet, 2 H), 2.34 (triplet, 2 H) ppm CH3CH2CH2C N

N H

CH3

d. C4H7ClO → one degree of unsaturation IR: 1802 cm–1 → C=O (high wavenumber, RCOCl) NMR: 0.95 (triplet, 3 H), 1.07 (multiplet, 2 H), 2.90 (triplet, 2 H) ppm O Cl

640

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–32

e. C5H10O2 → one degree of unsaturation IR: 1750 cm–1 → C=O NMR: 1.20 (doublet, 6 H), 2.00 (singlet, 3 H), 4.95 (septet, 1 H) ppm

f. C10H12O2 → five degrees of unsaturation IR: 1740 cm–1 → C=O NMR: 1.2 (triplet, 3 H), 2.4 (quartet, 2 H), 5.1 (singlet, 2 H), 7.1–7.5 (multiplet, 5 H) ppm

O

O O

O

g. C8H14O3 → two degrees of unsaturation IR: 1810, 1770 cm–1 → 2 absorptions due to C=O (anhydride) NMR: 1.25 (doublet, 12 H), 2.65 (septet, 2 H) ppm O

O O

22.82 A. Molecular formula C10H12O2 → five degrees of unsaturation IR absorption at 1718 cm–1 → C=O NMR data (ppm): triplet at 1.4 (CH3 adjacent to 2 H's) singlet at 2.4 (CH3) quartet at 4.4 (CH2 adjacent to CH3) doublet at 7.2 (2 H's on benzene ring) doublet at 7.9 (2 H's on benzene ring)

O O

B. IR absorption at 1740 cm–1 → C=O NMR data (ppm): singlet at 2.0 (CH3) triplet at 2.9 (CH2 adjacent to CH2) triplet at 4.4 (CH2 adjacent to CH2) multiplet at 7.3 (5 H's, monosubstituted benzene)

O O

22.83 Molecular formula C10H13NO2 → five degrees of unsaturation IR absorptions at 3300 (NH) and 1680 (C=O, amide or conjugated) cm–1 NMR data (ppm): triplet at 1.4 (CH3 adjacent to CH2) singlet at 2.2 (CH3C=O) quartet at 3.9 (CH2 adjacent to CH3) CH3CH2O doublet at 6.8 (2 H's on benzene ring) singlet at 7.2 (NH) doublet at 7.4 (2 H's on benzene ring)

H N

CH3 O

phenacetin

22.84 Molecular formula C11H15NO2 → five degrees of unsaturation IR absorption 1699 (C=O, amide or conjugated) cm–1 NMR data (ppm): triplet at 1.3 (3 H) (CH3 adjacent to CH2) singlet at 3.0 (6 H) (2 CH3 groups on N) quartet at 4.3 (2 H) (CH2 adjacent to CH3) doublet at 6.6 (2 H) (2 H's on benzene ring) doublet at 7.9 (2 H) (2 H's on benzene ring)

CH3

O N

CH3

OCH2CH3

C

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–33

22.85 a. Molecular formula C6H12O2 → one degree of unsaturation IR absorption at 1743 cm–1 → C=O 1 O H NMR data (ppm): triplet at 0.9 (3 H) – CH 3 adjacent to CH2 O multiplet at 1.35 (2 H) – CH2 D multiplet at 1.60 (2 H) – CH2 singlet at 2.1 (3 H – from CH3 bonded to C=O) triplet at 4.1 (2 H) – CH 2 adjacent to the electronegative O atom and another CH 2 b. Molecular formula C6H12O2 → one degree of unsaturation IR absorption at 1746 cm–1 → C=O O 1 H NMR data (ppm): doublet at 0.9 (6 H) – 2 CH3's adjacent to CH O E multiplet at 1.9 (1 H) singlet at 2.1 (3 H) – CH3 bonded to C=O doublet at 3.85 (2 H) – CH2 bonded to electronegative O and CH

22.86 The extent of resonance stabilization affects the position of the C=O absorption in the IR of an amide. N

A

N O

O

NH

NH

O

O

This resonance structure does not contribute significantly to the hybrid because it places a double bond at the bridgehead N, an impossible geometry in small rings. The C=O has more double bond character, so the absorption is at a higher wavenumber. This resonance structure contributes significantly to the hybrid, so the C=O has more single bond character. The absorption shifts to a lower wavenumber.

B

22.87 O

O

ClCH2 C

ClCH2 C NH2

N H

4.02 ppm

H

There is restricted rotation around the amide C–N bond. The 2 H's are in different environments (one is cis to an O atom, and one is cis to CH2Cl), so they give different NMR signals.

different environments

7.35 and 7.60 ppm

This resonance structure makes a significant contribution to the resonance hybrid.

22.88 18 18

O C6H5

C

–OH

OCH2CH3

ethyl benzoate

18

O C6H5 C OCH2CH3 OH

H2O

OH

C6H5 C OCH2CH3

18 –OH

OH

OH

O

C6H5 C OCH2CH3 O

C6H5

C

OCH2CH3 18

–

Two OH groups are now equivalent and either can lose H2O to form labeled or unlabeled ethyl benzoate.

OH

Unlabeled starting material was recovered.

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–34

22.89 NH2

abbreviate as: R

N H

CH3O

NH2

R H

NH2

C

RCH2CH2NH2 H C O

O CH3OOC

CH3OOC CH3OOC

OCOCH3

CH3OOC

OCOCH3 OCH3

OCH3

proton transfer H RCH2CH2N

H2O

RCH2CH2NH OH2 H C

CH3OOC

any proton CH3OOC source

CH3OOC

CH3OOC

OCOCH3

CH3OOC

OCH3

RCH2CH2NH OH H C

OCOCH3

CH3OOC

OCH3

OCOCH3 OCH3

R

RCH2CH2N

H BH3

RCH2CH2N CH3O

CH3OOC CH3OOC

OCOCH3 OCH3 H3O

CH3O

CH3O

N

N O H

O C O OCOCH3

CH3OOC OCH3

CH3OOC

H H

OCOCH3 OCH3

N

CH3OOC

OCOCH3 OCH3

BH3 CH3O

22.90 Both acetals are hydrolyzed by the usual mechanism for acetal hydrolysis (Steps [1 –[6]), forming four new OH groups. Intramolecular esterification forms a lactone (Steps [10]–[15]), followed by conversion of a carbonyl tautomer to an enol (Steps [16]–[17]). Both acetals are hydrolyzed at once in the given mechanism.

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carboxylic Acids and Their Derivatives—Nucleophilic Acyl Substitution 22–35

H O H

H2O H Cl

O OO O

CO2H

O

O H OO

[1]

O

OH OO

CO2H

O

[2]

H

H Cl

CO2H

O HO

Cl

OH OO

[3]

CO2H

O HO O H

+ 2 Cl–

H2O

H Cl

[4]

H

OH OH

O

O

[7] CO2H

HO HO

+

Cl–

Cl

O O

[5]

CO2H

O HO

H Cl

O H

OH

O H

+ 2 HCl

Cl

+ H

O H H Cl

CO2H

O HO

H

O

O

H

[6]

CO2H

HO HO

[8]

OH

O H OH OH O

OH OH

H Cl

(2 equiv)

OH HO

CO2H OH O

+

+ 2 HCl

[9]

HO

(2 equiv) OH

H Cl

O O

OH HO

OH

C +

HO

[10]

OH

OH

OH

OH

C

HO

HO

O

[12]

HO HO

O H

O

Cl

HO HO

HO

+OH2

O

[14]

HO

HO

H Cl OH

O

+ H2O

O

+ Cl–

[15] HO HO HO

HO

HO

O

O O

HO

O

[17] H

H Cl

O H

Cl

+ Cl–

OH O

O

[16]

OH

O

[13]

OH

O

HO

OH

O

+ Cl–

HO

Cl

OH

[11]

OH

OH

HO

H O

HO

O OH

vitamin C

+ HCl

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 22–36

22.91 O O

N

N

O Ph

O

O

Ar N

Ph

O

Ar N

O

Ph O

O

H F

R Li

O

Ar

O Ar N

Ar N

O O

O

O

H

+ RH +

Li+

O

O

O H

O Ph

O

O O

O

O O

O

Ph

O

O Ar N

O

O O

O H OH

Ph

Ph O

O Ar N

Ar N

O

+ –OH OH

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

645

Substitution Reactions of Carbonyl Compounds at the α Carbon 23–1

Chapter 23 Substitution Reactions of Carbonyl Compounds at the α Carbon Chapter Review Kinetic versus thermodynamic enolates (23.4) O R

kinetic enolate

Kinetic enolate • The less substituted enolate • Favored by strong base, polar aprotic solvent, low temperature: LDA, THF, –78 oC

O R

thermodynamic enolate

Thermodynamic enolate • The more substituted enolate • Favored by strong base, protic solvent, higher temperature: NaOCH2CH3, CH3CH2OH, room temperature

Halogenation at the α carbon [1] Halogenation in acid (23.7A) O R

C

O C

H

X2 R

CH3COOH

C

X

C

• •

The reaction occurs via enol intermediates. Monosubstitution of X for H occurs on the α carbon.

•

The reaction occurs via enolate intermediates. Polysubstitution of X for H occurs on the α carbon.

α-halo aldehyde or ketone

X2 = Cl2, Br2, or I2

[2] Halogenation in base (23.7B) O R

C

C

R

O

X2 (excess) –OH

R

C

C

R

•

X X

H H X2 = Cl2, Br2, or I2

[3] Halogenation of methyl ketones in base—The haloform reaction (23.7B) O R

C

CH3

–OH

X2 = Cl2, Br2, or I2

•

O

X2 (excess) R

C

O–

+ HCX3

haloform

The reaction occurs with methyl ketones and results in cleavage of a carbon–carbon σ bond.

646

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 23–2

Reactions of α-halo carbonyl compounds (23.7C) [1] Elimination to form α,β-unsaturated carbonyl compounds O R

α

Li2CO3

Br

LiBr DMF

O R

•

Elimination of the elements of Br and H forms a new π bond, giving an α,β-unsaturated carbonyl compound.

•

The reaction follows an SN2 mechanism, generating an α-substituted carbonyl compound.

β

α

[2] Nucleophilic substitution O

O R

Nu

α Br

α Nu

R

Alkylation reactions at the α carbon [1] Direct alkylation at the α carbon (23.8) O

•

O

C α H C

[1] Base [2] RX

C α R C

X–

+

• •

The reaction forms a new C–C bond to the α carbon. LDA is a common base used to form an intermediate enolate. The alkylation in Step [2] follows an SN2 mechanism.

[2] Malonic ester synthesis (23.9) [1] NaOEt

α

[2] RX [3] H3O+, Δ

H

•

R CH2COOH

•

H Cα COOEt COOEt

diethyl malonate

[1] NaOEt

[1] NaOEt

[2] RX

[2] R'X [3] H3O+, Δ

α

The reaction is used to prepare carboxylic acids with one or two alkyl groups on the α carbon. The alkylation in Step [2] follows an SN2 mechanism.

R CHCOOH R'

[3] Acetoacetic ester synthesis (23.10)

O CH3

C α H C H

[1] NaOEt [2] RX [3] H3O+, Δ

•

O CH3

C

CH2 R

•

COOEt

ethyl acetoacetate

O [1] NaOEt [2] RX

[1] NaOEt [2] R'X [3] H3O+, Δ

CH3

C

CH R R'

The reaction is used to prepare ketones with one or two alkyl groups on the α carbon. The alkylation in Step [2] follows an SN2 mechanism.

647

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Substitution Reactions of Carbonyl Compounds at the α Carbon 23–3

Practice Test on Chapter Review 1. a. Which of the following compounds can be prepared using either the acetoacetic ester or malonic ester syntheses? COOH O

1.

O

3.

2.

4. Both [1] and [2] can be prepared by one of these routes. 5. Compounds [1], [2], and [3] can all be prepared by these routes.

b. Which of the following compounds is not an enol form of dicarbonyl compound A? O

O

OH

O

OH

1.

O

O

OH

3.

2.

A

5. Compounds [1][4] are all enols of A. 2. a. Which proton in compound A has the lowest pKa? b. Which proton in compound A has the highest pKa? Hb

Hc Hd O

Ha

A

c. What proton in compound B is the least acidic? d. What proton in compound B is the most acidic? Hb

O

O

Hd H c

O Ha

B

3. What reagents are needed to convert 4-heptanone to each compound? a. 3-bromo-4-heptanone b. 3-methyl-4-heptanone c. 3,5-dimethyl-4-heptanone d. 2-hepten-4-one e. 2-methyl-4-heptanone

O

4.

OH

648

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 23–4

4. Draw the organic products formed in each of the following reactions. O I2 (excess)

a.

O

d.

O

[1] NaOEt OEt

–OH

[2] CH3CH2CH2CH2Br [3] H3O+, Δ

O CO2Et

O

[2] CH3CH2Br [3] H3O+, Δ

b.

c.

[1] LDA

CH2(CO2Et)2

[1] LDA

e.

[2] CH3CH2Br

[1] NaOEt

[1] NaOEt

[2] CH3CH2Br

[2] C6H5CH2Cl [3] H3O+, Δ

5. a. What starting materials are needed to synthesize carboxylic acid A by a malonic ester synthesis? O OH

A OCH3

b. What starting materials are needed to synthesize ketone B by the acetoacetic ester synthesis? O

B

Answers to Practice Test 1. a. 1 b. 4 2. a. Hb b. Hd c. Hb d. Hc 3. a. Br2, CH3CO2H b. [1] LDA; [2] CH3I c. []1 LDA; [2] CH3I; [3] LDA; [4] CH3I d. [1] Br2, CH3CO2H; [2] Li2CO3, LiBr, DMF e. [1] Br2, CH3CO2H; [2] Li2CO3, LiBr, DMF; [3] (CH3)2CuLi; [4] H2O

O

4. CO2

a.

5. a.

d.

+ HCI3

CH3O

O

CH2(CO2Et)2

O

e.

b.

Br Br

Br

b. O

c.

Br

OH O C6H5

CO2Et

649

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Substitution Reactions of Carbonyl Compounds at the α Carbon 23–5

Answers to Problems 23.1 • To convert a ketone to its enol tautomer, change the C=O to C–OH, make a new double bond to an α carbon, and remove a proton at the other end of the C=C. • To convert an enol to its keto form, find the C=C bonded to the OH. Change the C–OH to a C=O, add a proton to the other end of the C=C, and delete the double bond. [In cases where E and Z isomers are possible, only one stereoisomer is drawn.] HO

a. O

OH O

b.

O

d.

C6H5

OH C6H5

H

C6H5

H

c.

C6H5

O O

O

OH

O

e. O

O O

OH

OH

C6H5

f.

C6H5 OH C6H5

Draw mono enol tautomers only.

O

OH

OH

O

(Conjugated enols are preferred.)

23.2

The mechanism has two steps: protonation followed by deprotonation. OH

OH

O H

H2O

H H

H H OH2

O H H

H H

H 3O +

23.3 Cl H

H H

O

O D D Cl

D Cl H O D

Cl

D H

D

D Cl H

H O D

O D H Cl

H Cl

D D

D D

H

H D Cl

O

O D Cl

650

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 23–6

23.4 a. CH3CH2O

O

O

O

O

O

C

C

C

C

C

C

OCH2CH3

CH3CH2O

C

H

b.

CH3

OCH2CH3

CH3CH2O

H

O

O

O

O

C

C

C

C

C

OCH2CH3

CH3

C

c.

CH3

O

C

C

CH3

CHC N

C

OCH2CH3

OCH2CH3

CH3

O C C

H O

C H

O CH

O

OCH2CH3

H O CH3

CHC N

C

C C N H

23.5

The indicated H’s are α to a C=O or C≡N group, making them more acidic because their removal forms conjugate bases that are resonance stabilized. O

a.

CH3

C

O

b.

CH2CH2CH3

c.

CH3CH2CH2 CN

CH3CH2CH2

C

d. O

OCH2CH3

O CH3

23.6 O

O

O

–H+ O

O

–H+ O

O

–H+

O O

O

no resonance stabilization least acidic

23.7

O

O

O

O

O

O

O

O

Two resonance structures stabilize the conjugate base. intermediate acidity

Three resonance structures stabilize the conjugate base. most acidic

In each of the reactions, the LDA pulls off the most acidic proton. O

O O LDA

a.

c.

THF

CHO

b.

LDA THF

O LDA

CH3 C OCH2CH3

CHO

CN

d.

THF

LDA THF

CH2 C OCH2CH3

CN

651

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Substitution Reactions of Carbonyl Compounds at the α Carbon 23–7

23.8 the gas O

O

O

O

O CH4

OCH2CH3

H3

OCH2CH3

O

O+

OCH2CH3

+ MgBr+

H CH3–MgBr

The CH2 between the two C=O’s contains acidic H’s, so CH3MgBr reacts as a base to remove a proton. Thus, proton transfer (not nucleophilic addition) occurs. 23.9

In addition to being strong bases, organolithiums are good nucleophiles that can add to a carbonyl group instead of pulling off a proton to generate an enolate.

23.10 • LDA, THF forms the kinetic enolate by removing a proton from the less substituted C. • Treatment with NaOCH3, CH3OH forms the thermodynamic enolate by removing a proton from the more substituted C. O

O

LDA, THF O

a.

O

O

NaOCH3

LDA, THF O

c.

CH3OH

NaOCH3

O

CH3OH

LDA, THF

O

b.

O

NaOCH3 CH3OH

23.11 a. This acidic H is removed with base to form an achiral enolate. O

O H CH3

NaOH

(2R)-2-methylcyclohexanone

b.

O α

O H2O

NaOH H CH3

CH3

H

H

achiral Protonation of the planar achiral enolate occurs with equal probability from two sides, so a racemic mixture is formed. The racemic mixture is optically inactive. O

α

O CH3

O H CH3

or

O

H CH3

H2O

α

α H CH3

(3R)-3-methylcyclohexanone This stereogenic center is not located at the α carbon, so it is not deprotonated with base. Its configuration is retained in the product, and the product remains optically active.

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 23–8

23.12 O

Cl

a.

O

H2O, HCl O

b.

CH3CH2CH2

C

Br

Br2, CH3CO2H

Cl2

O

O

c. O

Br2 H

CH3CO2H

CH3CH2CH

C

H

Br

23.13 O

O

O Br Br

–

Br2, OH

a.

O I2, –OH

b.

I I

O

I I

O I2, –OH

c.

O

+ HCI3

23.14 O

O

Br

O

O

Li2CO3 LiBr DMF

a.

CH3SH

c. Br

O

SCH3

O CH3CH2NH2

b. Br

NHCH2CH3

23.15 Bromination takes place on the α carbon to the carbonyl, followed by an SN2 reaction with the nitrogen nucleophile. CH3

Br O

O

O

N O

O

O

O

NHCH3

Br2

LSD

CH3CO2H N

N

COC6H5

(Section 18.5)

N

COC6H5

M

COC6H5

23.16 O

a. CH3CH2

C

CH2CH3

[2] CH3CH2I

O

O

[1] LDA, THF CH3CH2

C

CHCH3

c.

CH2CH3 O

b.

O [1] LDA, THF O

O [2] CH3CH2I

O [1] LDA, THF

[1] LDA, THF

d. [2] CH3CH2I

CH2CH2CN

[2] CH3CH2I

CH2CHCN CH2CH3

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Substitution Reactions of Carbonyl Compounds at the α Carbon 23–9

23.17 O

O [1] LDA, THF

a.

[2] CH3I

O

O

O +

[1] LDA, THF

b.

[2] CH3I

COOCH3

c.

[1] LDA, THF

C

[2] CH3I

O

CH3 +

C

O

O

CH3

O

23.18 Three steps are needed: [1] formation of an enolate; [2] alkylation; [3] hydrolysis of the ester.

CO2CH2CH3

CO2CH2CH3

[1] LDA

CH3O

CO2CH2CH3

[2] CH3I

CH3O

CH3O

A [3] H3O+

The product is racemic because the new stereogenic center is formed by alkylation of a planar enolate with equal probability from above and below. CH3O

CO2H

naproxen

23.19 LDA

O

a.

O

O

CH3CH2Br

THF

O

KOC(CH3)3

b.

O

O

CH3Br

(CH3)3COH

O

(CH3)3COH

O

O CH3I

LDA

c. (from a.) O

d.

THF LDA THF

(from b.)

KOC(CH3)3

O

O CH3Br

O

CH3Br

O

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 23–10

23.20 LDA THF

O O

CH3I

O

CH2

Br Br2

O O

O

A

Li2CO3 LiBr DMF

O

CH3CO2H

O

C

B

O O

α-methyleneγ-butyrolactone

23.21 Decarboxylation occurs only when a carboxy group is bonded to the α C of another carbonyl group. O COOH

a.

COOH

α β

b.

COOH

O

α

c.

COOH

α β

d.

COOH

COOH

α

YES

NO

YES

NO

23.22 H3O+

[1] NaOEt

a. CH2(CO2Et)2

[2] Br

CH2 CH2COOH

Δ

b. CH2(CO2Et)2 [1] NaOEt

[1] NaOEt

H3O+

[2] CH3Br

[2] CH3Br

Δ

CH3 CH3 C COOH H

23.23 COOH

a.

Cl

b.

Cl

Br

O

O

Br

COOH

23.24 Locate the α C to the COOH group, and identify all of the alkyl groups bonded to it. These groups are from alkyl halides, and the remainder of the molecule is from diethyl malonate. a. (CH3)2CHCH2CH2CH2CH2 CH2COOH α

H3O+

[1] NaOEt CH2(CO2Et)2 [2] (CH3)2CHCH2CH2CH2CH2Br

b.

(CH3)2CHCH2CH2CH2CH2CH2COOH

Δ

COOH

α H3O+

[1] NaOEt

[1] NaOEt

[2] CH3CH2CH2Br

[2] CH3CH2CH(CH3)CH2CH2Br

CH2(CO2Et)2

c.

Δ

(CH3CH2CH2CH2)2 CHCOOH

α [1] NaOEt

[1] NaOEt

[2] CH3CH2CH2CH2Br

[2] CH3CH2CH2CH2Br

CH2(CO2Et)2

H3O+ Δ

(CH3CH2CH2CH2)2CHCOOH

COOH

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Substitution Reactions of Carbonyl Compounds at the α Carbon 23–11

23.25 The reaction works best when the alkyl halide is 1° or CH3X, since this is an SN2 reaction. (CH3)3C–CH2COOH

a.

α

COOH

b.

(CH3)3CX 3° alkyl halide (too crowded)

c.

X

aryl halide (leaving group on an sp2 hybridized C)

Aryl halides are unreactive in SN2 reactions.

(CH3)3C–COOH

This compound has 3 CH3 groups on the α carbon to the COOH. The malonic ester synthesis can be used to prepare mono- and disubstituted carboxylic acids only: RCH2COOH and R2CHCOOH, but not R3CCOOH.

23.26 O

O

a.

CH3

C

[1] NaOEt CH2CO2Et

CH3

[2] CH3I [3] H3O+, Δ

C

O

O

b.

CH2CH3

CH3

C

[1] NaOEt CH2CO2Et

CH3

[2] CH3CH2CH2Br [3] NaOEt [4] C6H5CH2I [5] H3O+, Δ

C

CH CH2 CH2CH2CH3

23.27 Locate the α C. All alkyl groups on the α C come from alkyl halides, and the remainder of the molecule comes from ethyl acetoacetate. O

a.

C

CH3

O CH2 CH2CH3

CH3

C

CH2

COOEt

O

H3O+

[1] NaOEt

Δ

[2] CH3CH2Br

CH3

C

CH2CH2CH3

α O

O

b. CH3

C

CH3

CH(CH2CH3)2 α

α

CH2

COOEt

[1] NaOEt

[2] CH3CH2Br

[2] CH3CH2Br

O CH3

C

CH2

[1] NaOEt

[1] NaOEt

[2] CH3CH2Br

[2] CH3(CH2)3Br

COOEt

23.28 O CH3

C

CH2CO2Et

O

H3O+ Δ

CH3

C

CH(CH2CH3)2

O

O

c.

C

[1] NaOEt

+

Br

Br

O

NaOEt (2 equiv)

CH3

X

CO2Et

H3O+ Δ

656

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 23–12

23.29 O

O O

a.

CH3

C

H3

[1] NaOEt CH2CO2Et

[2]

Δ

CO2Et

CH3O Br

O+ CH3O

nabumetone

CH3O O

b. CH3

C

O

[1] LDA, THF [2]

CH3

Br CH3O

CH3O

nabumetone

23.30 Use the directions from Answer 23.1 to draw the enol tautomer(s). In cases where E and Z isomers can form, only one isomer is drawn. O

a.

OH H

O

O

OH

O

OH

O

b.

H

OEt

OEt

OEt

unconjugated enol (less stable) O

OH

OH

OEt

OH OEt

23.31 a.

O (CH3)3C

NaOH H2O

b.

O (CH3)3C

A

(CH3)3C

B COCH3 H

(CH3)3C

one axial and one equatorial group, less stable

C O

H COCH3

(CH3)3C

Both groups are equatorial. more stable

This isomerization will occur since it makes a more stable compound.

(CH3)3C

COCH3

Both groups are equatorial = cis. Compound C will not isomerize since it already has the more stable arrangement of substituents.

23.32 Use the directions from Answer 23.1 to draw the enol tautomer(s). In cases where E and Z isomers can form, only one isomer is drawn. a.

O O

OH

conjugated enol (more stable) b.

O

HO

O

(mono enol form) OH

O OH

O

O

HO

O

c.

O

OH

d. OH

OH

conjugated enol (more stable)

657

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Substitution Reactions of Carbonyl Compounds at the α Carbon 23–13

23.33 O

O

O OCH2CH3

O OCH2CH3

ethyl acetoacetate

The ester C=O is resonance stabilized, and is therefore less available for tautomerization. Since the carbonyl form of the ester group is stabilized by electron delocalization, less enol is present at equilibrium.

23.34 a. CH3CH2CH2CO2CH(CH3)2

c.

O C CH2CH3

e. NC

O OH

O

b.

d. CH3O

O

f.

CH2CN

O

HOOC

23.35 CH2

a.

C

H

Hb O

CH3

O

O

O

b.

c.

CH3

Ha H

Hc

H

HO

Ha

Hc

H

H

Hb

H

c Hc is bonded to an sp2 hybridized C = least acidic. Ha Hb Ha is bonded to an α C = intermediate acidity. Hb is bonded to an sp3 hybridized Hc is bonded to an α C = Hb is bonded to an α C, and is adjacent C = least acidic. least acidic. to a benzene ring = most acidic. Hc is bonded to an α C = Ha is bonded to an α C, and is intermediate acidity. adjacent to a benzene ring = Ha is bonded to O = most acidic. intermediate acidity. Hb is bonded to an α C between two C=O groups = most acidic.

23.36 CN

LDA

a.

d.

THF

O

CN

LDA THF

O O

OCH3

b. O

O

OCH3

THF

O

O

LDA

e.

THF

O LDA

c.

LDA

THF

f.

LDA O

O

THF

O

O

23.37 Enol tautomers have OH groups that give a broad OH absorption at 3600–3200 cm–1, which could be detected readily in the IR.

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 23–14

23.38 O

O Hb

Ha Ha

remove Ha

O Hb

Ha

Hb O

Hb

Ha

Hb

Hb

O Hb

Ha Ha

remove Hb

O

Ha Ha

Hb

Removal of Ha gives two resonance structures. The negative charge is never on O.

O

Ha Ha

Hb

Hb

Ha Ha

Hb

Removal of Hb gives three resonance structures. The negative charge is on O in one resonance structure, making the conjugate base more stable and Hb more acidic (lower pKa).

23.39 O

O

HO

O

5,5-Dimethyl-1,3-cyclohexanedione exists predominantly in its enol form because the C=C of the enol is conjugated with the other C=O of the dicarbonyl compound. Conjugation stabilizes this enol.

5,5-dimethyl-1,3cyclohexanedione O

O

HO

O

2,2-dimethyl-1,3cyclohexanedione

The enol of 2,2-dimethyl-1,3-cyclohexanedione is not conjugated with the other carbonyl group. In this way it resembles the enol of any other carbonyl compound, and thus it is present in low concentration.

23.40 a. O N

HN H2N

OH

N

N

N

N

OH H2N

OH N

N

O

O

keto form acyclovir

enol form

In the enol form, the bicyclic ring system has four π bonds (eight π electrons) and a lone pair on N, for a total of 10 π electrons. This makes it aromatic by Hückel’s rule. b. The keto form of acyclovir can also be drawn in a resonance form that gives it 10 π electrons, making it aromatic as well. O HN H2N

O N OH

N

N

H2N O

N

HN

OH N

N O

aromatic completely conjugated 10 π electrons

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

659

Substitution Reactions of Carbonyl Compounds at the α Carbon 23–15

23.41 O CH3

C

O CH3

O

CH2

C

O O

CH3

CH2

C

O O

CH3

CH2

C

O

CH3

ester The O atom of the ester OR group donates electron density by a resonance effect. The resulting resonance structure keeps a negative charge on the less electronegative C end of the enolate. This destabilizes the resonance hybrid of the conjugate base, and makes the α H's of the ester less acidic.

CH3

O

O

C

C

CH3

CH2

O CH3

CH2

C

CH3

no additional resonance structures

ketone

This structure, which places a negative charge on the O atom, is the major contributor to the hybrid, stabilizing it, and making the α H's of the ketone more acidic.

23.42 O

O

2,4-pentanedione base (1 equiv) O

O

[1] CH3I

O

O

base

O

O

(2nd equiv)

[2] H2O A

[1] CH3I

O

O

[2] H2O B

more nucleophilic site

With a second equivalent of base a dianion is formed. Since the second enolate is less resonance stabilized, it is more nucleophilic and reacts first in an alkylation with CH3I, forming B after protonation with H2O.

One equivalent of base removes the most acidic proton between the two C=O's, to form A on alkylation with CH3I.

23.43 The mechanism of acid-catalyzed halogenation consists of two parts: tautomerization of the carbonyl compound to the enol form and reaction of the enol with halogen. A higher percentage of the more stable enol is present. O

CH3COOH

OH

OH

Br2

O

O

Br

2-pentanone

C=C has 1 bond to C

C=C has 2 bonds to C more stable (E and Z isomers)

A

Br

B major product formed from the more stable enol

660

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 23–16

23.44 • The mechanism of acid-catalyzed halogenation [Part (a)] consists of two parts: tautomerization of the carbonyl compound to the enol form and reaction of the enol with halogen. • In the haloform reaction [Part (b)], the three H’s of the CH3 group are successively replaced by X, to form an intermediate that is oxidatively cleaved with base. O

O O

H

O H

O

H H

a.

O O H

O

H H

O H

O

O Br

O

H Br

O C

b.

H

C

H H OH

H

I

H

Br

[2]

I

C I

C

H

+

H

+ I

O

O O

O

H

Repeat [1] and [2] two times.

O

+

Br

H

+ H2O

O

O

O H

H

–

H

O C

H

Br Br

Br

O

[1]

H

Br

O H H

O H

–CI

CI3

CI3

OH

3

CHI3

–OH

23.45 Use the directions from Answer 23.24. a.

α

CH3OCH2Br

CH3OCH2 CH2COOH α

b.

C6H5

COOH

Br

COOH

c.

and Br

Br

C6H5

α

23.46 a. CH3CH2CH2CH2CH2 CH2COOH

[2] CH3CH2CH2CH2CH2Br

α b.

H3O+

[1] NaOEt CH2(CO2Et)2

COOH

[1] NaOEt

[1] NaOEt

[2] CH3CH2CH2CH2CH2Br

[2] CH3Br

CH2(CO2Et)2

Δ

CH3CH2CH2CH2CH2CH2COOH

H3O+ Δ

COOH

α c.

COOH

α

[1] NaOEt

[1] NaOEt

[2] (CH3)2CHCH2CH2CH2CH2Br

[2] CH3Br

CH2(CO2Et)2

H3O+ Δ

COOH

661

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Substitution Reactions of Carbonyl Compounds at the α Carbon 23–17

23.47 O

O

O

O

[1] NaOEt O

[2] CH3CH2CH2Br

O

O

[1] NaOEt

O

O

O

O

[2] CH3CH2CH2Br

O

H3 O + Δ O

H

O

valproic acid

23.48 H 3O +

[1] NaOEt

a.

CH2(CO2Et)2 [2] BrCH2CH2CH2CH2CH2Br [3] NaOEt

b.

COOH

CH2OH

[2] H2O

(from a.)

c.

[1] LiAlH4

COOH

Δ

COOH

CH3OH

CH3

[1] CH3MgBr (2 equiv)

CO2CH3

H2SO4

C OH

[2] H2O

CH3

(from a.)

COOH

d.

CH3CH2OH H2SO4

COOCH2CH3

CH3

[1] LDA

COOCH2CH3

[2] CH3I

(from a.)

23.49 a.

O CH3

[1] Na+ –CH(COOEt)2 HO [2] H2O CH3CHCH2 CH(COOEt)2

O

c. CH3

C

O [1] Na+ –CH(COOEt)2 C [2] H2O CH3 CH(COOEt)2

Cl

nucleophilic attack here b.

CH2 O

[1] Na+ –CH(COOEt)2 [2] H2O

HOCH2CH(COOEt)2

d.

CH3

O

O

C

C

O

CH3

[1] Na+ –CH(COOEt)2 [2] H2O CH

O C 3

CH(COOEt)2 + CH3COOH

23.50 Use the directions from Answer 23.27. O

α

a.

CH3 O

b.

O

α

C

[1] NaOEt CH2

COOEt

O CH3

C

CH2

[2] Br

H3O+ Δ

[1] NaOEt

[1] NaOEt

[2] CH3CH2Br

[2] Br

COOEt

O

H3O+ Δ

O

662

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 23–18

α

O

O

c.

CH3 O

H3O+

[1] NaOEt

C

CH2

COOEt

O

Δ

[2] Br

α O

d.

C

CH3

[1] NaOEt CH2

COOEt

O

H3O+

[2]

Δ

Br

Br [3] NaOEt

23.51 O

a.

CH3

[1] NaOEt

C

COOEt

CH2

b.

CH3

C

Δ

CH3

[1] NaOEt

[1] NaOEt

H3O+

[2] CH3Br

[2] CH3Br

[2] CH3CH2Br

O CH2

COOEt

O

c.

CH3

C

O

H3O+

THF

CH(CH3)2

C

CH2

CH2CH2CH3 O

Δ

O

LDA

C

CH3

C

CH(CH3)2 O

CH3I CH3CH2

CH(CH3)2

C

CH(CH3)2

(from b.) O CH3

d.

C

CH(CH3)2

O

KOC(CH3)3 (CH3)3COH

C

CH3

O

CH3I C(CH3)2

CH3

C

C(CH3)3

(from b.)

23.52 O

O

O

Li2CO3

a.

LiBr DMF

Br O

f.

[2] Li2CO3, LiBr, DMF

O COOH

b.

(E and Z)

O [1] Br2, CH3CO2H

O

Δ

O I2 (excess)

g.

O

+ CHI3

–OH

COOH

c. CH3CH2CH2CO2Et

[1] LDA

COOH CH3CH2 CH3CH2CHCO2Et

h.

Cl

NaH

CN

C N

[2] CH3CH2I O

O Br

d. O

O NHCH(CH3)2

(CH3)2CHNH2

O Br2 (excess)

i.

–OH

O

O

Br Br O

[1] LDA

e.

[2] CH3CH2I

NaI

j. Cl

I

663

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Substitution Reactions of Carbonyl Compounds at the α Carbon 23–19

23.53 O

O

O

O

O

[1] LDA

[1] LDA

a.

c.

Cl

[2]

[2] CH3I

H

H

O D H

O

[1] LDA

C

b.

CH3

H D [2]

C

CH3

I

23.54 CHO

a.

OH

NaBH4 CH3OH

[1] PBr3

CN

[2] CH3I

A

p-isobutylbenzaldehyde

CN

[1] LDA

[2] NaCN B

C H3O+ Δ

COOH

ibuprofen

COOH

b.

COO

[1] LDA

removal of the most acidic H

D

[2] CH3I

COOCH3

substitution reaction

+ I–

Removal of the most acidic proton with LDA forms a carboxylate anion that reacts as a nucleophile with CH3I to form an ester as substitution product. 23.55 CO2CH3

CO2CH3

Cl

a.

NH

+

sp2

Cl on hybridized C does not react. Cl on sp3 hybridized C reacts. H COOH

H2SO4

Cl

A

pyridine

Cl

B

Cl

clopidogrel (racemic)

H COOCH3

TsCl

HO

S

The N atom acts as a nucleophile to displace Cl–.

H COOCH3

CH3OH

b. HO

K2CO3

S

Cl

N

NH

S isomer

S

TsO Cl

C

inversion of configuration SN2

CO2CH3 N

S

Cl

clopidogrel (single enantiomer)

664

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 23–20

23.56 O

O

O

O

LDA

KOC(CH3)3

THF

Br

Br

–78 oC

A

B

(CH3)3COH

room temperature

A

C

23.57 O

O O

α a.

Δ

β

+

CO2

c.

O [1] LDA

H

COOH

[2] CH3CH2I

In order for decarboxylation to occur readily, the COOH group must be bonded to the α C of another carbonyl group. In this case, it is bonded to the β carbon. [1] NaOEt (CH3CH2)3CCH(CO2Et)2

b. CH2(CO2Et)2 [2] (CH3CH2)3CBr

LDA removes a H from the less substituted C, forming the kinetic enolate. This product is from the thermodynamic enolate, which gives substitution on the more substituted α C.

The 3° alkyl halide is too crowded to react with the strong nucleophile by an SN2 mechanism.

23.58 Protonation in Step [3] can occur from below (to re-form the R isomer) or from above to form the S isomer as shown. H A

A H

H O

H O

[1]

H

O

O

H

OH

R isomer inactive enantiomer

O

[2]

OH

H

OH

achiral enol

H A A

[3] H

H O

HA

H

+

H O

[4]

H

O

O

S isomer active enantiomer

O

H

OH

H A

(+ one more resonance structure)

23.59 O

O

CH3

LDA

O

O

O

I

+

I–

O CH3

I

+

I–

H

665

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Substitution Reactions of Carbonyl Compounds at the α Carbon 23–21

23.60 O O

O

O

O

O

O

EtO O

O

OEt EtO

OEt

EtO

OEt

H H

H

EtO

OEt

O

O

+ HOEt

–

OEt –

OEt

O

O

CO2Et H3O+

O

O

H

O

OEt

O

O

OEt O

+ HOEt

23.61 Protons on the γ carbon of an α,β-unsaturated carbonyl compound are acidic because of resonance. There is no H on this C, so a planar enolate cannot form and this stereogenic center cannot change. γ O

α

β

γ

X

Remove the H on this γ C.

O

O H

OH H OH

Removal of this proton forms a resonance-stabilized anion. One resonance structure places a negative charge on O. O

O

Protonation of the O planar enolate can occur from below (to re-form starting material X), or from above to form Y.

H

Y

666

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 23–22

23.62 LDA = B O

C6H5 Br

B

+

Br

C6H5

C6H5

O

Br

+ HB+

O Br

+

+ HB+

C6H5 H

Br

C6H5

B B

O

O

O

H

O

H

O

Br

O

Br

C6H5

Br

C6H5

C6H5

O

C6H5

+ HB+ C 6H 5

O Br

Br

H C6H5 3)3

+ Br

O

C6H5

–OC(CH

This reaction occurs with both bases [LDA and KOC(CH3)3]. + Br

+ HOC(CH3)3

23.63 1st new C–C bond O H H

OCH2CH3 LDA THF –78 oC

α

Remove proton here.

O H

[1]

A

O

OCH2CH3

[2] H

OCH2CH3

[3]

Cl Cl

B

O OCH2CH3 LDA THF –78 oC Cl

Cl

[4] O

Cl

OCH2CH3

C 2nd new C–C bond Cl

23.64 a. Since there are two C’s bonded to the α carbon, there are two possible intramolecular alkylation reactions. Path [1] Br

Br

O

O

Path [1] O

A

Path [2] Path [2] Br

O

Br

O

two different halo ketones possible

667

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Substitution Reactions of Carbonyl Compounds at the α Carbon 23–23

b. Form both bonds to the α carbon during acetoacetic ester synthesis.

CO2Et

O O

[1] NaOEt

Br

CO2Et

H3O+

NaOEt

O

Br

[2]

EtO2C

O

Δ

O

Br

23.65 O COCH3

a.

O

Cl2

Cl

NHC(CH3)3

H2NC(CH3)3

H2O, HCl COO–

COCH3 –OH, I 2 excess

b.

COOH

H3O+

+ CHI3 O COCH3

OH [1] LiAlH4 [2] H2O

[1] LDA, THF

c.

[2] CH3CH2CH2Br O Br2

[1] LDA, THF

d.

Li2CO3

CH3COOH

[2] CH3Br

Br

23.66 O

O

O Br

Br2

a.

OCH3

NaOCH3

CH3COOH

O

OH

O [1] LDA, THF

b.

[1] LiAlH4

[2] CH2=CHCH2Br

c.

[1] LDA, THF

[1] LDA, THF [2] CH3CH2Br

O Br

(from a.)

O

[2] CH3Br

O

d.

[2] H2O O

O

Li2CO3 LiBr DMF

O

O

COCH3

CH3CH2Br [1] Li (2 equiv) [2] CuI (0.5 equiv) [1] (CH3CH2)2CuLi [2] H2O

O

LiBr DMF

668

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 23–24

O

O

O

e.

[1] LDA, THF

[1] LDA, THF

[2] CH3CH2Br

[2] CH3CH2Br

O

Li2CO3

CH3COOH

O

Br

LiBr DMF

O

[1] KOC(CH3)3

f.

O

O Br2

O

Br

Br2

Li2CO3 LiBr DMF

CH3COOH

[2] CH3CH2Br

(from e.)

23.67 O

O

O

Cl AlCl3

O

Cl2

Br2

FeCl3

CH3CO2H

Br

Cl

Cl H2NC(CH3)3 O NHC(CH3)3

Cl

bupropion

23.68 NaCN Br

Br

[1] LDA [2] CH3I (4 equiv each)

NC

(2 equiv)

CN

N HN

NC

CN

Br2 hν

N

NaH

N

CH2Br

N

N

N N

N NC

NC

CN

CN

anastrozole

23.69 O

a.

OH

PBr3

Br

O

O COOEt

[1] NaOEt

COOEt

H3O+ Δ

[2] Br

mCPBA O O

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

669

Substitution Reactions of Carbonyl Compounds at the α Carbon 23–25

O

O

b.

O

[1] NaOEt

COOEt

[2] HC CCH2Br

H 3 O+

HOCH2CH2OH

Δ

TsOH

O

COOEt

[1] NaH [2] CH3I

O

H3 O +

O

H2

O

O

O CH3

Lindlar catalyst

23.70 O

O

Br2

a.

O

CH3CH2NH2

CH3COOH

NH3 (excess)

Br

CH3CH2Br

NHCH2CH3

PBr3 CH3CH2OH

O

b.

O

Li2CO3

OH

NaBH4

LiBr DMF

Br

CH3OH

(from a.) Br2

c.

Br

FeBr3

O

(from b.)

[1] Li (2 equiv) [2] CuI (0.5 equiv)

O

[1] (C6H5)2CuLi [2] H2O

C6H5

SOCl2

d.

CH2Br

O

(from b.)

Br2 hν

CH3

CH3OH

CH3Cl AlCl3

[1] Li (2 equiv) [2] CuI (0.5 equiv) [1] (C6H5CH2)2CuLi [2] H2O

O

O C6H5

NaBH4 CH3OH

OH C6H5

670

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 23–26

23.71 O

O

O

CH3CH2I

LDA THF HO

–78 oC

A

O

CH3CH2O

most acidic H To synthesize the desired product, a protecting group is needed: O

O

O LDA

TBDMS–Cl imidazole HO

THF TBDMSO

A

TBDMSO

CH3CH2I

O O (CH3CH2CH2CH2)4N+F– HO

B

TBDMSO

23.72 CH2(COOEt)2 NaOEt OH

PBr3

Br

–

CH(COOEt)2

CH(COOEt)2 H3O+, Δ

Cl

Y

SOCl2 COOH

O

23.73 O

a.

O

Y =

O

[1] CH3Li W

[2] CH3I

(CH3)2CH

Ha O

Y

Hb

C

Hc Hd

CH2CH2CH3

He

C7H14O → one degree of unsaturation IR peak at 1713 cm–1 → C=O 1 H NMR signals at (ppm) He: triplet at 0.8 (3 H) Ha: doublet at 0.9 (6 H) Hd: sextet at 1.4 (2 H) Hc: triplet at 1.9 (2 H) Hb: septet at 2.1 (1 H)

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

671

Substitution Reactions of Carbonyl Compounds at the α Carbon 23–27

O

O

b. CH3 Li

O

O

O

O

O

O

+ I–

Y

CH3

I

23.74 Removal of Ha with base does not generate an anion that can delocalize onto the carbonyl O atom, whereas removal of Hb generates an enolate that is delocalized on O. CO2CH3

CO2CH3

CO2CH3

Delocalization of this sort can't occur by removal of Ha, making Ha less acidic. H

H

H

B

O

H O

O

Ha Hb

B H CO2CH3

CO2CH3

H

CO2CH3

H

O

CO2CH3

H

O

H

O

O

Removal of Hb gives an anion that is resonance stabilized so Hb is more acidic.

Mechanism: CO2CH3

H

OCH3

CO2CH3

CO2CH3

O

O

CO2CH3

Br O

O

+ CH3OH CO2CH3

O

CO2CH3

CO2CH3 HO H

+ O

B

+ Br–

CO2CH3

O

–OH

H

O

+ HB+

672

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 23–28

23.75 O

OH

OH

OH

OH

H OSO3H

1,2-shift

OH H

H2O

HSO4

HSO4

23.76 CH3 Li O

OCH2CH3

H OH2 O CH3

OCH2CH3

OH H 3 O+

OH2

OCH2CH3

CH3

+ Li+

+ H 2O H

OH OCH2CH3

H OH2 OH OCH2CH3

CH3

CH3

OCH2CH3

+ H2O

X CH3

OCH2CH3

CH3

CH3

+ H2O

+ H2O

OH OCH2CH3

CH3

OCH2CH3

H + H2 O O H

+ H 2O O

+ H2O

H3O+

HOCH2CH3

23.77 O

O

O

O

O

O R X

e

H

e

H NH2

or O

H NH2

OH

H

proton transfer OH O R

+ X– + NH2

e

673

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Substitution Reactions of Carbonyl Compounds at the α Carbon 23–29

23.78 a. In the presence of base an achiral enolate that can be protonated from both sides is formed. CH3

OH

N

H

front H

OH

OH

O

OH

H O

RO

conjugated H2O

RO

OH O

H

O

back

achiral A

(–)-hyoscyamine

RO O

racemic mixture optically inactive

b. The enolate A formed from (–)-hyoscyamine is conjugated with the benzene ring, making it easier to form. The enolate B formed from (–)-littorine is not conjugated, so it is less readily formed. CH3

CH3

N

H H OH O

N

OH

H OH O

O

(–)-littorine

O

NOT conjugated B

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

675

Carbonyl Condensation Reactions 24–1

Chapter 24 Carbonyl Condensation Reactions Chapter Review The four major carbonyl condensation reactions Reaction type

Reaction

[1] Aldol reaction (24.1)

2

O RCH2

OH

–OH

C

H

H2O

[2] Claisen reaction (24.5)

RCH2

C

OR'

[1] NaOR'

O

[2] H3O+

C

RCH2

C

or

R

H

H3O+

CHO C R

(E and Z) α,β-unsaturated carbonyl compound

β-hydroxy carbonyl compound

O

2

CHCHO

H

aldehyde (or ketone)

RCH2

–OH

RCH2 C

O C

CH

ester

OR'

R

β-keto ester O

O

[3] Michael reaction (24.8)

–OR'

+

R

–

α,β-unsaturated carbonyl compound

[4] Robinson annulation (24.9)

R

1,5-dicarbonyl compound

carbonyl compound

–OH

H2O

O

O

α,β-unsaturated carbonyl carbonyl compound compound

2-cyclohexenone

Useful variations [1] Directed aldol reaction (24.3) O

O R'CH2

C

[1] LDA R"

R" = H or alkyl

[2] RCHO [3] H2O

HO

O

R C CH C H R'

O

H2O

OH

+

O

O or

R"

β-hydroxy carbonyl compound

–

OH

or H3O+

C

R C H

R"

C R'

(E and Z) α,β-unsaturated carbonyl compound

676

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 24–2

[2] Intramolecular aldol reaction (24.4) [a] With 1,4-dicarbonyl compounds:

O

O NaOEt O

[b] With 1,5-dicarbonyl compounds:

EtOH

O

O NaOEt O

EtOH

[3] Dieckmann reaction (24.7) OEt

[a] With 1,6-diesters:

O [1] NaOEt

O O

O C

[2] H3O+

OEt

OEt

[b] With 1,7-diesters:

OEt O

O O

[1] NaOEt

O C

OEt

[2] H3O+

OEt

Practice Test on Chapter Review 1. a. Which compounds are possible Michael acceptors? 1.

3. O O

2.

O

4. Both compounds [1] and [2] are Michael acceptors. OEt

5. Compounds [1], [2], and [3] are all Michael acceptors.

b. Which of the following compounds can be formed by an aldol reaction? O

1.

CHO

HO

4. Both [1] and [2] can be formed by aldol reaction.

O

2.

3. HO

OH

5. Compounds [1], [2], and [3] can all be formed by aldol reaction.

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

677

Carbonyl Condensation Reactions 24–3

c. Which compounds can be formed in a Robinson annulation? O

O

1.

3.

CO2Et

2.

4. Compounds [1] and [2] can be formed by Robinson annulation. O

5. Compounds [1], [2], and [3] can be formed by Robinson annulation.

d. What compounds can be used to form A by a condensation reaction? O

O

O

O

CHO

CO2Et

1.

3.

and

A

O

and

O

2.

OEt

4. Compounds [1] and [2] can be used to form A. 5. Compounds [1], [2], and [3] can be used to form A.

2. Give the reagents required for each step. O

O

O

(a)

O

(b) H

HO

(c) O

O

(e) O

O

(d)

O

EtO

O

3. Draw the organic products formed in the following reactions. O

O O

a.

[1] LDA

c.

NaOEt +

[2] CH3CH2CHO [3] H2O

O

–

CHO

O

b.

+

HCO2Et

[1] NaOEt, EtOH [2] H3

O+

d.

EtOH O

+

OH, H2O

678

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 24–4

4. a. What organic starting materials are needed to synthesize D by a Robinson annulation reaction? O

CO2CH3

D

b. What organic starting materials are needed to synthesize β-keto ester B by a Dieckmann reaction? O CO2Et

B

c. What starting materials are needed to synthesize A by an aldol reaction? O

A

Answers to Practice Test 1.a. 4 b. 2 c. 1 d. 4 2.a. [1] O3; [2] (CH3)2S b. CrO3, H2SO4, H2O c. CH3CH2OH, H2SO4 d. [1] NaOEt, EtOH; [2] H3O+ e. [1] LDA; [2] CH3I

4.

3. O

OH

a.

O CO2CH3

a.

O

OEt

b.

b.

O

CHO

CO2Et

O O

c. O

c. O

d.

O

+

O

679

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbonyl Condensation Reactions 24–5

Chapter 24: Answers to Problems 24.1 O

CH2CHO

H

a. OH

OH

c. CH3 C CH3

O

CH3 C CH2 C CH3 CH3

O

O

(CH3)3CCH2CHO

b.

HO

C(CH3)3

(CH3)3CCH2C

C CHO

H

O

HO

d.

H

24.2 O O

O

CHO

a.

c.

b.

(CH3)3C

αH

no α H no aldol reaction

d.

C

(CH3)3C

H

no α H no aldol reaction

yes

C

CHO

e.

αH

CH3

αH

yes

yes

24.3 base

O

a.

O

O

OH

O

base

c. HO HO

CHO

CHO

base

b.

(E and Z isomers)

24.4 H OSO3H O C C C H H H

HO H CH3

OH2 H CH3 C H

H

O

C C H

CH3 C H

H

+

+ HSO4

24.5

O

C C H

O

HSO4

CH3CH CH C H

H +

H2O

H2SO4

Locate the α and β C’s to the carbonyl group, and break the molecule into two halves at this bond. The α C and all of the atoms bonded to it belong to one carbonyl component. The β C and all of the atoms bonded to it belong to the other carbonyl component. β

β OH α

CHO

a.

b. C6H5

c.

C6H5

OH O

CHO

H C6H5 O

β

O

α

H

CHO CHO

α O

C6H5 O

680

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 24–6

24.6 –OH

O

H

H H C

H OH

O

H

C

O

C

H

H C H

C H

O

O

+ H2O OH

OH

H C

+

H C

–OH

C H

CHO

H

O

+ H2 O

C H O

–OH

24.7 O

a.

CHO b. C6H5CHO

CHO

CH3CH2CH2CHO and CH2=O

or

and

OH O

(E and Z isomers)

24.8 CO2Et H

a.

CO2Et

CH2(CO2Et)2

O

H

c.

O

H CH3COCH2CN NC

COCH3 H

b.

COCH3

(E and Z isomers)

COCH3

CH2(COCH3)2 O

24.9 2

–

CHO

OH

CHO

a.

H2O

NaBH4

OH

CH3OH

OH

OH CHO

b.

CHO [1] (CH3)2CuLi

–OH

(E and Z mixture)

[2] H2O

CHO NaBH4

c. (from b.)

CH3OH

CH3CH2CH2CH C(CH2CH3)CH2OH

CHO [1] CH3MgBr

d. (from b.)

[2] H2O

CH3CH2CH2CH C(CH2CH3)CH(CH3)OH

681

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbonyl Condensation Reactions 24–7

24.10 Find the α and β C’s to the carbonyl group and break the bond between them. O

OH

O

a.

b. O

OH

O

c.

O

O

H

H

O

O H

O

24.11 O

O H N

CH3O CH3O O

O H2

N

CH3O

Pd-C

X

CH3O

N

CH3O

donepezil

CH3O

24.12 All enolates have a second resonance structure with a negative charge on O. O

O

O

EtO

H

O

O

O

H

EtOH

H

EtOH

H H

H

EtO O

O

+ OH

OH

H

H OH

O

O

O

EtO H EtOH

24.13 O

O –

a. CHO

O O

OH

O

b.

H2O

[2] (CH3)2S

A

CHO O

OH

H2O

24.14 [1] O3

–

–OH

O

H 2O

B

682

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 24–8

24.15 Join the α C of one ester to the carbonyl C of the other ester to form the β-keto ester. O

a.

α

OCH3

O

O

O

b.

O

α

OCH2CH3

OCH3

O

α

α

OCH2CH3

24.16 In a crossed Claisen reaction between an ester and a ketone, the enolate is formed from the ketone, and the product is a β-dicarbonyl compound. O

a. CH3CH2CO2Et and HCO2Et

H

O OEt

Only this compound can form an enolate. b.

O

and

CO2Et

H

HCO2Et

C

OEt

O

Only this compound can form an enolate. O

c.

CH3

C

O

and CH3

C

CH3

O

O

OEt

The ketone forms the enolate. d.

O

O

O

C

C

and

OEt

O

The ketone forms the enolate.

24.17 A β-dicarbonyl compound like avobenzone is prepared by a crossed Claisen reaction between a ketone and an ester. O

O O

O

O CH3O

CH3O

O

C(CH3)3

Break the molecule into two components at either dashed line. avobenzone

C(CH3)3

or O O

CH3O

C(CH3)3

683

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbonyl Condensation Reactions 24–9

24.18 O

O

O

[1] NaOEt

a.

O

OEt

b.

[2] (EtO)2C=O

C6H5CH2

O

[1] NaOEt

C

C

[2] ClCO2Et

OEt

EtO

OEt

O

24.19 O

O

CO2Et

OEt

OEt CH3 O

[1] NaOEt

[1] NaOEt

[2] CH3I

O

[2] (EtO)2C=O A

EtO

B

EtO

H3O+ Δ CO2H

ibuprofen

24.20 O

O

O

O

O

OEt

EtO

EtO

OEt O

24.21 A Michael acceptor is an α,β-unsaturated carbonyl compound. O

O

O

α,β-unsaturated yes—Michael acceptor

O

d.

c.

b.

a.

O

not α,β-unsaturated

CH3O

not α,β-unsaturated

α,β-unsaturated yes—Michael acceptor

24.22 O

a.

CH2 CHCO2Et

+

CH3

C

[1] NaOEt CH2CO2Et

[2] H2O

CH3

O

O

C

C COOEt

+ CH2(CO2Et)2

b.

[1] NaOEt [2] H2O

O

O EtO2C

O

O

c.

+ CH2

CH3

C

[1] NaOEt CH2CN [2] H2O

O

CO2Et CN

O

OEt

684

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 24–10

24.23 O

O

O

O EtO2C

a. CO2Et

b.

O

O

O

O

24.24 The Robinson annulation forms a six-membered ring and three new carbon–carbon bonds: two σ bonds and one π bond. new C–C bond O

O O

re-draw

+

a.

O O–

CH3CH2 CH3CH2OH

O

O

O

O

new σ and π bonds

new C–C bond O

O

COOEt

COOEt +

b.

re-draw O

CH3CH2O– CH3CH2OH

O

COOEt O

new σ and π bonds new C–C bond

O

c.

O CH3CH2O– CH3CH2OH

re-draw

+

O

O

O

new σ and π bonds new C–C bond EtO2C

O

d. O

COOEt

EtOOC

re-draw

+

O

CH3CH2O– CH3CH2OH

O

O

new σ and π bonds

24.25 a.

O

O

O

b.

c.

O

O

O

O

O

O

685

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbonyl Condensation Reactions 24–11

24.26 O

O

a. (CH CH ) C=O + CH =O 3 2 2 2

–OH

O + C6H5CHO

b.

H2O

or

–OH

H2O

(E and Z isomers)

O

OH

24.27 CHO [1] O3

CHO

[2] (CH3)2S

CHO

NaOEt EtOH

A

B

24.28 The product of an aldol reaction is a β-hydroxy carbonyl compound or an α,β-unsaturated carbonyl compound. The α,β-unsaturated carbonyl compound is drawn as product unless elimination of H2O cannot form a conjugated system. OH

a. (CH3)2CHCHO only

–OH

(CH3)2CHCHC(CH3)2

H2O

CHO

–OH

c. C6H5CHO + CH3CH2CH2CHO

H2O

CHO

(E and Z isomers) b. (CH3)2CHCHO + CH2=O

O

CH3

–OH

CH3 C CHO

H2O

d.

(CH3CH2)2C=O only

CH2OH

–OH

H 2O

24.29 HO

CHO

O H

OH

CHO

H O

OH

OH

CHO

CHO

24.30 O

O

a. CH3

C

OH

[1] LDA CH3 [2] CH3CH2CH2CHO

O

CH3CH2CH2 C CH2 C CH3

H

[3] H2O

b. CH CH C OEt 3 2

OH

[1] LDA [2] O

O

CHO

O

C

CH

H

CH3

C OEt

O O

[3] H2O

24.31 O O CHO

a.

O

c.

O

b. OHC

CHO CHO

O

686

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 24–12

24.32 Locate the α and β C’s to the carbonyl group, and break the molecule into two halves at this bond. The α C and all of the atoms bonded to it belong to one carbonyl component. The β C and all the atoms bonded to it belong to the other carbonyl component. O O

a.

OH

β

b.

β

O

β

d.

c.

C6H5

O

O

α C6H5

O

H

CH CHCN

β

α O

e.

C6H5

α

α

CH3

O

O

O

C6H5

CH2=O O

C6H5

α

β CH3

CH3CN

CHO

C6H5

24.33 O CHO (CH3)2C=O NaOEt EtOH

CH3O

CH3O

X

24.34 Base removes the most acidic proton between the two C=O’s in B. This enolate reacts with the aldehyde in A to form a product that loses H2O. Form enolate here. O F

+

CO2Et

H

A

base

H OH F

F

CO2Et

O

electrophilic C

– H2O

B

CO2Et

O

O

(E and Z isomers can form.)

24.35 O

O

O

O

c.

b.

a.

d. HO

O

O

O H

H

CH3

CH3 O

O

O O H

O

687

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbonyl Condensation Reactions 24–13

24.36 Ozonolysis cleaves the C=C, and base catalyzes an intramolecular aldol reaction. O

O

NaOH

[1] O3

H2O

[2] (CH3)2S O C

D

C10H14O

24.37 Aldol reactions proceed via resonance-stabilized enolates. K can form an enolate that allows for delocalization of the negative charge on O. Delocalization is not possible in J, because a double bond would be placed at a bridgehead carbon, which is geometrically impossible. O

O O

O

Bond angles don't allow this.

from J

from K

24.38 O a. C6H5CH2CH2CH2CO2Et

O

C6H5CH2CH2CH2

OEt CH2CH2C6H5 O

b. (CH3)2CHCH2CH2CH2CO2Et

O

(CH3)2CHCH2CH2CH2

OEt CH2CH2CH(CH3)2

CH3O

c. CH3O

O

O

CH2COOEt

OEt

OCH3

24.39 O CH3CH2CH2CH2CO2Et + CH3CH2CO2Et

O

O

O

OEt O

OEt O

O OEt

O OEt

688

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 24–14

24.40 O

O

O a. CH3CH2CH2CO2Et + C6H5CO2Et

d. CH3CH2CO2Et + (EtO)2C=O

OEt CH2CH3 O

EtO

O

O

e.

O + HCO2Et

O

O f.

OEt

c. EtO2CC(CH3)2CH2CH2CH2CO2Et

O

H

Cl

C

O

O

O +

OEt

O

O

b. CH3CH2CH2CO2Et + (CH3)2C=O O

O

EtO

OEt

24.41 O

O

O

CH3O

CH3O

CHO

c.

a. O

OEt

or

H

CH3

O CH3O

O

O

(EtO)2C=O

OEt

or

O

EtO

OEt OEt

C6H5

b. O

OEt

O

C6H5

O

O

CH3CHO

d. C6H5CH(COOEt)2

O

O

O C6H5

OEt

EtO

OEt

24.42 Only esters with two H’s or three H’s on the α carbon form enolates that undergo Claisen reaction to form resonance-stabilized enolates of the product β-keto ester. Thus, the enolate forms on the CH2 α to one ester carbonyl, and cyclization yields a five-membered ring. This is the only α carbon with 2 H's. O

O OCH3

CH3O CH3O

O

NaOCH3

OCH3 [1] nucleophilic attack

CH3O

CH3OH

O

CH3O

O

O

CO2CH3

CH3O2C

CH3O–

[2] loss of [3] deprotonation

O

B O CO2CH3

CH3O2C

H3

O+

CH3O2C

OCH3

CO2CH3

CH3O2C

H O

acidic H between 2 C=O's

O

highly resonance-stabilized enolate Formation of this enolate drives the reaction to completion.

O

689

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbonyl Condensation Reactions 24–15

24.43 O

O

a.

+

C 6H 5

C6 H5

–OEt,

O

O

EtOH

C6 H5 C 6H 5

O

O –

b.

OEt, EtOH

+ CH2(CN)2

CN CN

24.44 O

O

O

a.

c.

b. O O

CO2Et

O EtO2C

O

d. CN

CO2Et

O

O

O

O

O

O

CO2Et

O

C6H5

EtO2C

CN

CO2Et

C6H5

24.45 O

O H

a.

Michael reaction

CH3 O

A

O

(E or Z isomer can be used.)

O O

b.

H OH O

O H

OH

O

O H2O

OH O

O

OH

H2O

H OH

OH O

690

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 24–16

24.46 O

O −OH

a.

+

H2O

O

O

O O

O +

b.

−OH

re-draw C6H5

+

O

O

H2O

C6H5

O

C6H5

O O re-draw

c.

−OH

+

+

H2O

O

O

O

O O

−OH

re-draw

d.

+

O

H2O

O

O

24.47 O

O

c.

a.

O

O

O

O

O

O

O

O

d. O

b. O

O

O

24.48 CH3CH2CH2CHO

a.

–OH

CH3CH2CH2

H2O

b.

H

–OH

CH2

CH2=O, H2O

CHO

(E and Z) CH2CH3 CHO C CH2CH3

g. h.

CHO

c.

i.

[1] LDA [2] CH3CHO; [3] H2O CH3CH2CH2

d.

e.

j. EtO2C OH

[1] CH3Li

NaBH4 CH3OH

HOCH2CH2OH TsOH CH3NH2 mild acid

O

(CH3)2NH mild acid

k.

CrO3

H C

CH3CH2CH2CH2OH

CH3 N

H CH3CH2CH=CHN(CH3)2

(E and Z) CH3CH2CH2COOH

H2SO4

l.

O

CH3CH2CH2

CO2Et

[2] H2O

f.

CH3CH2CH2CH2OH

OH H

CH2(CO2Et)2 NaOEt, EtOH

H2 Pd-C

O

Br2

H

CH3COOH Br

691

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbonyl Condensation Reactions 24–17

m.

CH3CH2CH2

Ph3P=CH2

O

OH

n.

C CH2

NaCN, HCl

o.

CH3CH2CH2 C H

H

[1] LDA; [2] CH3I H

CN

24.49 O

O

O

–OH

a.

e.

H2O O NaOEt, EtOH

f.

(CH3)2C=O

+

CHO

CH3

[2] CH3CH2CHO

O C

C6H5 NaOCH3

C6H5 +

CH3OH

O

O O

CO2Et

OH

H2O

O

C6H5

+

–OH

C6H5

H2O

g.

–

CH3

CHO

CN

O

d.

O

O

NaOEt, EtOH

c. NCCH2CO2Et

CH3

O

b. O

OH

[1] LDA

[3] H2O

O

O

CH3

C

CH2CO2Et

[1] NaOEt, EtOH

CH2CO2Et

[2] H3O+

(E and Z)

h.

O

(E and Z) O

O

O

OEt

24.50 O

A [1] NaOEt

O CO2Et

[2](EtO)2CO [3] H3O+ G

[1] LDA [2] CH3CH2CHO [3] H2O, –OH H H2, Pd-C O

(1 equiv)

O

B [1] NaOEt [2] CH3CH2I

C H3O+ Δ CO2Et

O

HO D [1] CH3CH2MgBr [2] H2O E [1] LDA [2] CH3I

O

I [1] LDA [2] HCO2Et [3] H3O+

O

O

F H2SO4

O

H

major product J [1] O3 O O K –OH

H2O O

[2] (CH3)2S

692

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 24–18

24.51 O

[1] O3

O

[2] (CH3)2S

O

NaOH

O O

EtOH

H

O

B

O H

O

O OH

Form the enolate here to generate a five-membered ring in the product.

A

new C–C bond

The RCHO has the more accessible carbonyl.

24.52 O H

C

C

O

O

[1]

H

H

H H

C

C

+ H C H

H

O

C

C

C

H

HO

C

C

C

H H

H OCO2

O

O

HH

[3]

H H

H + HCO3–

CO32–

O

HH

[2]

HOCH2 H

C

C

H

HOCH2 CH2OH

+ CO32–

Repeat steps [1]–[3] with these two H's and CO32–.

24.53 Enolate A is more substituted (and more stable) than either of the other two possible enolates and attacks an aldehyde carbonyl group, which is sterically less hindered than a ketone carbonyl. The resulting ring size (five-membered) is also quite stable. That is why 1-acetylcyclopentene is the major product. O

O H

H OH

O

H

O H H –

OH H O

–OH

OH O

A O most stable enolate less hindered carbonyl

O

H2 O

H

OH

O

H2 O

–

OH

1-acetylcyclopentene major product O

H H

–OH

O

–

H

O

H O

H O

B The more hindered ketone carbonyl makes nucleophilic attack more difficult. H2O

O H

H OH

OH H O

OH

OH O

H

H H2O CHO –OH

693

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbonyl Condensation Reactions 24–19

O H

H

H H –

H

O

H OH

O

O

–OH

OH

H H

OH H2O

H O

O

O

O

OH

These two reacting functional groups are farther away than the reacting groups in the first two reactions, making it harder for them to find each other. Also, the product contains a less stable seven-membered ring.

C less stable enolate H2O

O

–OH

24.54 Na+ –OCH3 H H

H

CH3O

CH3O O O

O

O

O

O

O

O

COOCH3

H OH

+ CH3OH –OH

O O COOCH3

24.55 Removal of a proton from CH3NO2 forms an anion for which three resonance structures can be drawn. O

O

H CH2 N

O

CH2 N O

CH2 N O

O CH2 N

O

O

–OH

O C6H5

C

O H

CH2 N O

O

H OH

C6H5 C CH2 H

NO2

OH

OH

C6H5 C CH NO2 H H

C6H5 C CH NO2 H

–OH

+ H2O

C6H5CH CHNO2

+

–OH

694

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 24–20

24.56 All enolates have a second resonance structure with a negative charge on O. O O

O

O

H H

O

+ H2O

OH

H OH H OH

O

O

O

O H H OH H OH

HO

OH

O

O

HO OH

H

O

O

O H OH

24.57 Et3N reacts with phenylacetic acid to form a carboxylate anion that acts as a nucleophile to displace Br, forming Y. Then an intramolecular crossed-Claisen reaction yields rofecoxib. O CO2H

CO2

Et3N

O

SN2

CH3SO2

H

Y

+ Et3NH

phenylacetic acid

O

X

Et3N O

O

O

O

Et3N

H

O

O

proton source

OH CH3SO2

CH3SO2

O

CH3SO2

H A

+ Et3NH

O

O

O

O

+ –OH

OH CH3SO2

+ Et3NH

CH3SO2

rofecoxib

O

695

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbonyl Condensation Reactions 24–21

24.58 Polymerization occurs by repeated Michael reactions. B H

B

H [1]

B

[2]

O

O

O

O

O

O

O O

O

O

O

O

Repeat step [2].

new C–C bonds B O

O

O

O

O

O

polytulipalin

24.59 O

O H

O H

O H

CH3

C

O

O O

C

O H

CH3

CH3

O

O

CH3COO–

+

C

H O

O O

O

CH3COO

CH2 H

O H

CH3COO–

CH3COO– OH

OH

O

H OCOCH3

H

+ CH3COOH O

O

O

O

H

O

O

O

O

CH2 O

+ CH3COOH O

O

+ –OH

coumarin

24.60 C6 CO2Et

a.

CO2Et

+ EtOH

H

ethyl 2,4-hexadienoate CO2Et

OEt

O

EtO O

OEt

CO2Et

O EtO

O

OEt

diethyl oxalate

O CO2Et

EtO2C O OEt

CO2Et

+

OEt

696

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 24–22

b. The protons on C6 are more acidic than other sp3 hybridized C–H bonds because a highly resonance-stabilized carbanion is formed when a proton is removed. One resonance structure places a negative charge on the carbonyl O atom. This makes the protons on C6 similar in acidity to the α H’s to a carbonyl. c. This is a crossed Claisen because it involves the enolate of a conjugated ester reacting with the carbonyl group of a second ester. 24.61 O

O –OH,

a.

H2 O C6 H5

C6H5CHO O

b.

–

O

OH, H2O

O

O

PCC

CH2OH

CHO

H2C=O

–OEt,

O

O –OH,

c.

HCO2Et CH3CH2OH

O

H 2O O

O

O

OH [1] LiAlH4

[1] LDA, THF

d.

[2] CH3CH2CH2CH2Br

[2] H2O

or

O

[1] LDA, THF

H2 (excess)

[2] CH3CH2CH2CHO

Pd-C

[3] H2O, –OH

(E and Z isomers)

O

e.

O

OH

–OH

H2 (excess)

H2 O

Pd-C

24.62 O

O

O

[1] Br2, CH3COOH

a.

O

[1] NaOEt

[2] Li2CO3, LiBr, DMF

O

H3O+

O

Δ

O OEt

[2] H2O

COOEt

O –OH,

b. O

[1] (CH3)2CuLi

H2O O

[2] H2O

O

O

697

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbonyl Condensation Reactions 24–23

24.63 O

OH

O [1] LDA

a. [2] CH3CH2CHO [3] H2O PCC OH O

O

O

O

O

[1] LDA

[1] NaOEt

[2] CH3CO2Et

[2] CH3Br

b. OH H2SO4 O

CrO3 OH

PBr3

H2SO4 H2O

CH3OH OH O

OH

O [1] LDA

c.

CH3OH

C6H5

[2] C6H5CHO

SOCl2 [1] Br2, hν

O

HO

OH

C6H5

O

OH H2SO4

[1] CH3MgBr

O

[1] O3

[2] H2O

–

CHO

OH, H2O

[2] (CH3)2S Mg

PBr3

H2 (excess) Pd-C

unstable

[2] –OH

AlCl3

CH3OH

C6H5

– H2O

PCC

CH3Cl

d.

O

major product

CH3Br

24.64 OEt

a.

[2]

O

[1] NaOEt

[1] NaOEt CO2Et

H3O+

[2]

O

CO2Et

Br PBr3

O

OH C6H5

[1] NaOEt

b. CO2Et O

[2]

C6H5

CO2Et

Br

O

(from a.) CH3OH + SOCl2 [1] CH3Cl, AlCl3 [2] Br2, hν

Br

C6H5

[1] LiAlH4

OH

[2] H2O OH

698

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 24–24

C6H5

OEt

c.

O

O

[1] NaOEt

C6H5

[2] H3O+

O

O

AlCl3 O Cl SOCl2 O

CrO3 OH OH H2SO4 H2O

24.65 O [1] O3

CHO

CrO3

[2] (CH3)2S

CHO

H2SO4

COOH COOH

OH H2SO4

CO2Et

[1] NaOEt

CO2Et

[2] H3O+

CO2Et

H2O

[1] NaOEt O

O H3

O+

CO2Et

Δ

24.66 O

O CHO

O

a. CH3O

CH3O

octinoxate

+

[2] CH3I

O

699

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbonyl Condensation Reactions 24–25

[1] NaH

b.

[2] CH3Cl

HO

Br

Br2 CH3O

[2] H2C=O

CH3O

SOCl2

CH2OH

[1] Mg CH3O

PCC

(+ ortho isomer)

PCC

CH3OH

CH3OH O O

H2SO4

HO

HO

CHO O CH3O

CrO3 H2SO4, H2O

H2O

HO

CH2=O [1] PBr3

BrMg

NaOR, ROH

HO

octinoxate

[2] Mg H2O O

Mg

BrMg

Br

PCC

PBr3

HO

HO

24.67 O O

O

base

d.

a. EtO

O

O

O

O

O EtO

b.

O

O

[1] NaOEt

c.

base

CH3 O

O O

Two products are possible.

O

O

O

O

O

O OH

O

e.

O

O

OH

or

[2] CH3I conjugated tautomers favored

24.68 O

a.

O

[1] HCO2Et NaOEt, EtOH

[2] H3

O+

CHO

700

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 24–26

O

O CHO

CHO

O

O

b.

NaOEt, EtOH

O O

O O

O

H

OEt

c.

OEt H

O

H OEt O

O

+

O

O

+

O

HCO2Et

OEt

+

O

H OH2

d.

O

=

O

O

+ H O

+OH

OH

H2O

H2O

+ OH

H OH2

OH

no α H

OH

O

OH

OH

+ H OH2

H2O

+O H

O

β-hydroxy ketone + OH2

H

+

H2O

H2O

H2 O H3O+

O

O

O

H2O

This reaction is an acid-catalyzed aldol that proceeds by way of enols not enolates. The β-hydroxy ketone initially formed cannot dehydrate to form an α,β-unsaturated carbonyl because there is no H on the α carbon. Thus, dehydration occurs, but the resulting C=C is not conjugated with the C=O. O

e.

H CHO

This H is now more acidic because it is located between two carbonyl groups. As a result, it is the most readily removed proton for the Michael reaction in the next step.

701

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbonyl Condensation Reactions 24–27

24.69 O

O

O

O

O Ar'

CH3O

OCH3

CH3O

OCH3

CH3O

H

N

O

F

OCH3

R2N

Ar

N OCH2Ph

Ar

Ar' OCH2Ph O

O Ar'

CH3O

CH3O N

+ CH3O–

O

O

N OCH3 Ar

F

24.70 Rearrangement generates a highly resonance-stabilized enolate between two carbonyl groups. OEt O

O

O OEt O

O

OEt

O

O

OEt

OEt

O

H OCH2CH3

OCH2CH3

OCH2CH3

OCH2CH3 CH3CH2O

H OCH2CH3 OCH2CH3 O

O CO2CH2CH3

(+ 2 resonance structures) This product is a highly resonance stabilized enolate. This drives the reaction. H3O+ O

H

CO2CH2CH3

H

O OEt

O OEt CO2CH2CH3

CO2CH2CH3

O OCH2CH3

H OCH2CH3

702

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 24–28

24.71 All enolates have a second resonance structure with a negative charge on O. O O

O

O

B

H OH

[2]

[1] +

O

O H OH

[3]

B

O

[4]

OH

[5]

+ OH

HB+ + HB+

+ –OH

O OH

O

O HO

+

Repeat steps [1]–[5] by deprotonating the indicated CH3.

O

O

H

isophorone

B

H

+ HB+

(+ 2 resonance structures)

24.72 All enolates have a second resonance structure with a negative charge on O. B O

O

O

COOEt

H H

O CO2CH3

CO2CH3 CO2CH3

new bond

new bond H2O

O

+ –OH

CO2CH3

703

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbonyl Condensation Reactions 24–29

24.73 a.

–

H CH2

OH

N

CH2

CH2

N

(+ other resonance structures)

one possible resonance structure The negative charge is delocalized on the electronegative N atom. This factor is what makes the CH3 group bonded to the pyridine ring more H OH acidic, and allows the condensation to occur.

O C6H5

N

O H

CH2

OH

C6H5 C CH2

N

N

C6H5 C CH

H

OH N

C6H5 C CH

H H

N

H

H2O

OH

CH CH

N

A OH

b. The condensation reaction can occur only if the CH3 group bonded to the pyridine ring has acidic hydrogens that can be removed with –OH. OH

N CH2

N

H

CH2

OH

H CH2

3-methylpyridine

Since the negative charge is delocalized on the N, the CH3 contains acidic H's and reaction will occur.

CH2

(+ other resonance structures)

2-methylpyridine N

N

N CH2

N CH2

N CH2

N CH2

No resonance structure places the negative charge on the N, so the CH3 is not acidic and condensation does not occur.

N CH2

704

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 24–30

24.74 Since the reaction takes place in acid, enols are involved. After the initial condensation reaction, the NH2 and C=O groups form an imine by an intramolecular reaction. H OH2 O

OH CO2Et

OH CO2Et

CO2Et

enol nucleophile

H H2O

Ar OH CO2Et

H OH2

NH2 O

NH2 OH

NH2 OH

Ar

–H+

F

Ar H OH2

H2O Ar

H

Ar OH

Ar OH2 CO2Et

NH2 O

CO2Et

CO2Et

NH2 O

NH2 O

Ar

Ar

Ar CO2Et

NH2 O

N H H

H3O

CO2Et

CO2Et

proton transfer

O

N H

OH H OH2

Ar

Ar

Ar

N

N

H H3O

CO2Et

CO2Et

CO2Et

H2O

N H

OH2

705

Carbonyl Condensation Reactions 24–31

24.75 B

B

H

O CO2CH3

O O

OCH3

O

O

O

OCH3 O

O

O

O O

O

+ HB+

H

O

O O

O

O O

O

+ CH3O–

+ HB+

O

The enolate opens the epoxide ring.

O O

O

O

HO

O H+

O

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

707

Amines 25–1

Chapter 25 Amines Chapter Review General facts • Amines are organic nitrogen compounds having the general structure RNH2, R2NH, or R3N, with a lone pair of electrons on N (25.1). • Amines are named using the suffix -amine (25.3). • All amines have polar C–N bonds. Primary (1°) and 2° amines have polar N–H bonds and are capable of intermolecular hydrogen bonding (25.4). • The lone pair on N makes amines strong organic bases and nucleophiles (25.8). Summary of spectroscopic absorptions (25.5) Molecular ion Amines with an odd number of N atoms give an odd molecular ion. Mass spectra IR absorptions N–H

3300–3500 cm–1 (two peaks for RNH2, one peak for R2NH)

1

H NMR absorptions

NH CH–N

0.5–5 ppm (no splitting with adjacent protons) 2.3–3.0 ppm (deshielded Csp3–H)

13

C–N

30–50 ppm

C NMR absorption

Comparing the basicity of amines and other compounds (25.10) • Alkylamines (RNH2, R2NH, and R3N) are more basic than NH3 because of the electron-donating R groups (25.10A). • Alkylamines (RNH2) are more basic than arylamines (C6H5NH2), which have a delocalized lone pair from the N atom (25.10B). • Arylamines with electron-donor groups are more basic than arylamines with electronwithdrawing groups (25.10B). • Alkylamines (RNH2) are more basic than amides (RCONH2), which have a delocalized lone pair from the N atom (25.10C). • Aromatic heterocycles with a localized electron pair on N are more basic than those with a delocalized lone pair from the N atom (25.10D). • Alkylamines with a lone pair in an sp3 hybrid orbital are more basic than those with a lone pair in an sp2 hybrid orbital (25.10E). Preparation of amines (25.7) [1] Direct nucleophilic substitution with NH3 and amines (25.7A) R X

+

NH3

excess

R NH2

+ NH4+ X–

1o amine

• •

R' R X + R' N R' R'

+

R N R' X– R'

ammonium salt

•

The mechanism is SN2. The reaction works best for CH3X or RCH2X. The reaction works best to prepare 1o amines and ammonium salts.

708

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–2

[2] Gabriel synthesis (25.7A) • •

O R X

+

CO2–

–OH

N

R NH2

H 2O

+

CO2–

1o amine

O

The mechanism is SN2. The reaction works best for CH3X or RCH2X. Only 1o amines can be prepared.

•

[3] Reduction methods (25.7B) [a] From nitro compounds

H2, Pd-C or

R NO2

[b] From nitriles

R NH2

Fe, HCl or Sn, HCl

1o amine

[1] LiAlH4 [2] H2O

R C N

R CH2NH2

1o amine O

[c] From amides

C

R

NR'2

[1] LiAlH4 [2] H2O

RCH2 N R' R'

R' = H or alkyl 1o, 2o, and 3o amines

[4] Reductive amination (25.7C) R C O

•

R

NaBH3CN

R2"NH

+

R' C N R"

R'

H R"

•

R', R" = H or alkyl 1o, 2o, and 3o amines

Reductive amination adds one alkyl group (from an aldehyde or ketone) to a nitrogen nucleophile. Primary (1°), 2°, and 3° amines can be prepared.

Reactions of amines [1] Reaction as a base (25.9) R NH2

+

+

H A

R NH3

+

A

[2] Nucleophilic addition to aldehydes and ketones (25.11) With 1o amines:

With 2o amines:

O R

C

O

NR' C

H

R = H or alkyl

R'NH2

R

C

C

imine

H

R

C

C

H

R = H or alkyl

NR'2

R'2NH R

C

C

enamine

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

709

Amines 25–3

[3] Nucleophilic substitution with acid chlorides and anhydrides (25.11) O R

C

O R'2NH (2 equiv)

+

Z

R

Z = Cl or OCOR R' = H or alkyl

C

NR'2

1o, 2o, and 3o amides

[4] Hofmann elimination (25.12) C C H NH2

[1] CH3I (excess)

•

C C

[2] Ag2O [3] Δ

The less substituted alkene is the major product.

alkene

[5] Reaction with nitrous acid (25.13) With 1o amines: R NH2

With 2o amines:

NaNO2

+

Cl–

R N N

NaNO2

R N H

HCl

HCl

R

alkyl diazonium salt

R N N O R

N-nitrosamine

Reactions of diazonium salts [1] Substitution reactions (25.14) N2

Cl–

F

X

OH

phenol

With HBF4:

With CuX:

With H2O: +

aryl chloride or aryl bromide

aryl fluoride

X = Cl or Br With NaI or KI:

With CuCN:

I

aryl iodide

With H3PO2:

CN

H

benzonitrile

benzene

[2] Coupling to form azo compounds (25.15) N2+ Cl– +

Y

Y = NH2, NHR, NR2, OH (a strong electrondonor group)

N N

azo compound

Y

+

HCl

710

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–4

Practice Test on Chapter Review 1. Give a systematic name for each of the following compounds. CH3

a.

b. NHCH2CH3

NHCH2CH3

2. (a) Which compound is the weakest base? (b) Which compound is the strongest base? O

H N

N H

CH3

N H

NH2

CH3 CH3CH2

A

B

C

D

3. (a) Which compound is the weakest base? (b) Which compound is the strongest base? O NH2

CH3

NH2

NH2

Cl

NH2

H 2N

A

B

C

D

4. Draw the organic products formed in each of the following reactions. a. Cl

NH2

[1] NaNO2, HCl

CHO d.

NaBH3CN

+

NH

[2] CuCN O NH2

[1] KOH b.

NH O NH2

c.

[2] PhCH2CH2Br [3] –OH, H2O

e.

[1] CH3I (excess) [2] Ag2O [3] heat

[1] NaNO2, HCl [2] NaI

Cl

5. Draw the products formed when the given amine is treated with [1] CH3I (excess); [2] Ag2O; [3] , and indicate the major product. You need not consider any stereoisomers formed in the reaction.

N H

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

711

Amines 25–5

6. What organic starting materials are needed to synthesize B by reductive amination?

N C(CH3)3

B

Answers to Practice Test 1.a. N-ethyl-2,4-dimethyl-3heptanamine b. N-ethyl-3methylcyclohexanamine

4.

5. a. Cl

CN N CO2

b.

2.a. B b. C

+

Ph

CO2

3.a. C b. B

NH2

I

c.

N

N

Cl

major N

d.

6. +

e.

O

H

N C(CH3)3

Answers to Problems 25.1

Amines are classified as 1o, 2o, or 3o by the number of alkyl groups bonded to the nitrogen atom. 1o amine

2o amine a. H N 2

N H

C6H5

H N

2o

NH2

b. CH3CH2O

amine

O

1o amine

25.2 3o amine a.

CH3

b.

HO C CH2NH2 CH3

3o alcohol

CH3 N CH2CH2OH CH3

1o amine

N CH3

1o alcohol

3o amine

712

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–6

The N atom of a quaternary ammonium salt is a stereogenic center when the N is surrounded by four different groups. All stereogenic centers are circled.

25.3

OH +

CH3

HO

CH3 +

a. CH3 N CH2CH2 N CH2CH3 CH3

H N

b. HO

H

N has 3 similar groups.

25.4 NHCH2CH3

a.

c.

CH3CH2CH(NH2)CH3

2-butanamine or sec-butylamine

e.

N(CH3)2

N,N-dimethylcyclohexanamine

N-ethyl-3-hexanamine CH3

b.

(CH3CH2CH2CH2)2NH

d.

f.

dibutylamine

2-methyl-5-nonanamine

25.5

NHCH2CH2CH3

NH2

2-methyl-N-propylcyclopentanamine

An NH2 group named as a substituent is called an amino group.

a. 2,4-dimethyl-3-hexanamine

c. N-isopropyl-p-nitroaniline

e. N,N-dimethylethylamine

NHCH(CH3)2

g. N-methylaniline NHCH3

N O2N

NH2

b. N-methylpentylamine

d. N-methylpiperidine

f. 2-aminocyclohexanone O

NHCH3

h. m-ethylaniline NH2

NH2 N CH2CH3

25.6

Primary (1°) and 2o amines have higher bp’s than similar compounds (like ethers) incapable of hydrogen bonding, but lower bp’s than alcohols that have stronger intermolecular hydrogen bonds. Tertiary amines (3°) have lower boiling points than 1o and 2o amines of comparable molecular weight because they have no N–H bonds. CH3

alkane lowest boiling point

O

CH3

ether intermediate boiling point

NH2

amine N–H can hydrogen bond. highest boiling point

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

713

Amines 25–7

25.7

The NH signal occurs between 0.5 and 5.0 ppm. The protons on the carbon bonded to the amine nitrogen are deshielded and typically absorb at 2.3–3.0 ppm. The NH protons are not split. molecular formula C6H15N 1H NMR absorptions (ppm): 0.9 (singlet, 1 H) 1.10 (triplet, 3 H) 1.15 (singlet, 9 H) 2.6 (quartet, 2 H)

25.8

NH CH3 adjacent to CH2 (CH3)3C CH2 adjacent to CH3

N H

The atoms of 2-phenylethylamine are in bold. O

a.

N

(CH3CH2)2N

CH3

b.

H

CH3O

O H

N H LSD lysergic acid diethyl amide

25.9

H

HO

N CH3

codeine

SN2 reaction of an alkyl halide with NH3 or an amine forms an amine or an ammonium salt. Cl

a.

NH3

NH2

excess

NH2

b.

CH3CH2Br

N(CH2CH3)3 Br–

excess

25.10 The Gabriel synthesis converts an alkyl halide into a 1o amine by a two-step process: nucleophilic substitution followed by hydrolysis. NH 2

a.

b.

Br

(CH3)2CHCH 2CH2NH2

c. CH3O

NH2

CH3O

Br

(CH3)2CHCH 2CH2Br

25.11 The Gabriel synthesis prepares 1° amines from alkyl halides. Since the reaction proceeds by an SN2 mechanism, the halide must be CH3 or 1°, and X can’t be bonded to an sp2 hybridized C. NH2

a. aromatic An SN2 does not occur on an aryl halide. cannot be made by Gabriel synthesis

b.

NH2

can be made by Gabriel synthesis

c.

N H

2° amine cannot be made by Gabriel synthesis

NH2

d. N on 3° C An SN2 does not occur on a 3° RX. cannot be made by Gabriel synthesis

714

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–8

25.12 Nitriles are reduced to 1o amines with LiAlH4. Nitro groups are reduced to 1o amines using a variety of reducing agents. Primary (1°), 2o, and 3o amides are reduced to 1o, 2o, and 3o amines respectively, using LiAlH4. O

a.

CH3CHCH2NH2

CH3CHCH2NO2

CH3

CH3CHC N

CH3CHCNH2

CH3

CH3

CH3

O

b.

CH2NH2

CH2NO2

C N

C NH2

c.

NH2

C

NO2

O

N

NH2

25.13 Primary (1°), 2o, and 3o amides are reduced to 1o, 2o, and 3o amines respectively, using LiAlH4. O CONH2

CH2NH2

a.

NHCH3

c.

NHCH3

O

b.

N

N

25.14 Only amines with a CH2 or CH3 bonded to the N can be made by reduction of an amide. NH2

a.

N bonded to benzene cannot be made by reduction of an amide

NH2

NH2

b.

c.

N bonded to a 3° C cannot be made by reduction of an amide

N bonded to CH2 can be made by reduction of an amide

N H

d.

N on 2° C on both sides cannot be made by reduction of an amide

25.15 Reductive amination is a two-step method that converts aldehydes and ketones into 1o, 2o, and 3o amines. Reductive amination replaces a C=O by a C–H and C–N bond. a.

CHO

NHCH3

CH3NH2

c.

NaBH3CN

b.

O

O

(CH3CH2)2NH

N(CH2CH3)2

NaBH3CN NH2

NH3 NaBH3CN

d.

O

+

NH2

NHCH(CH3)2

NaBH3CN

25.16 Reductive amination occurs using the ketone in E and the amine in D. CO2CH2CH3 O OCH2CH3 O

E

O

H2N

+

N H

O N OH

D

O O N OH

enalapril

715

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Amines 25–9

25.17 NH2

a.

O NH3

OH

OH

b.

NHCH3

O NH2CH3

or

OH NH2

CH2=O

25.18 a. NH2

Only amines that have a C bonded to a H and N atom can be made by reductive amination; that is, an amine must have the following structural feature: H C N

phentermine In phentermine, the C bonded to N is not bonded to a H, so it cannot be made by reductive amination. b. systematic name: 2-methyl-1-phenyl-2-propanamine

25.19 The pKa of many protonated amines is 10–11, so the pKa of the starting acid must be less than 10 for equilibrium to favor the products. Amines are thus readily protonated by strong inorganic acids (e.g., HCl and H2SO4) and by carboxylic acids. a. CH3CH2CH2CH2 NH2

CH3CH2CH2CH2 NH3 + Cl–

+ HCl

pKa ≈ 10 weaker acid products favored

pKa = –7

b. C6H5COOH

+

c.

pKa = 10.7 weaker acid products favored

pKa = 4.2

N H H

pKa ≈ 10 pKa = 15.7 weaker acid reactants favored

(CH3)2NH2 + C6H5COO–

(CH3)2NH

+ HO–

+ H2O

N H

25.20 An amine can be separated from other organic compounds by converting it to a water-soluble ammonium salt by an acid–base reaction. In each case, the extraction procedure would employ the following steps: • Dissolve the amine and either X or Y in CH2Cl2. • Add a solution of 10% HCl. The amine will be protonated and dissolve in the aqueous layer, while X or Y will remain in the organic layer as a neutral compound. • Separate the layers. a.

NH2

and

CH3

X

H Cl

+

NH3 Cl–

• soluble in H2O • insoluble in CH2Cl2

CH3

X • insoluble in H2O • soluble in CH2Cl2

716

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–10

b.

(CH3CH2CH2CH2)3N

and

+

H Cl

(CH3CH2CH2CH2)2O

(CH3CH2CH2CH2)3NH Cl–

Y

(CH3CH2CH2CH2)2O

Y • insoluble in H2O • soluble in CH2Cl2

• soluble in H2O • insoluble in CH2Cl2

25.21 Primary (1°), 2o, and 3o alkylamines are more basic than NH3 because of the electrondonating inductive effect of the R groups. a. (CH3)2NH

and

NH3

2° alkylamine CH3 groups are electron donating. stronger base

b. CH3CH2NH2 1° alkylamine stronger base

and

ClCH2CH2NH2

1° alkylamine Cl is electron withdrawing. weaker base

25.22 Arylamines are less basic than alkylamines because the electron pair on N is delocalized. Electron-donor groups add electron density to the benzene ring making the arylamine more basic than aniline. Electron-withdrawing groups remove electron density from the benzene ring, making the arylamine less basic than aniline. NH2

NH2

NH2

NH2

a.

NH2

NH2

b.

CH3OOC

O2N

CH3O

electronwithdrawing group least basic

arylamine intermediate basicity

electrondonating group most basic

electronwithdrawing group least basic

arylamine intermediate basicity

alkylamine most basic

25.23 Amides are much less basic than amines because the electron pair on N is highly delocalized. CONH2

amide least basic

NH2

NH2

arylamine intermediate basicity

alkylamine most basic

25.24 sp2 hybridized more basic

This N is also sp2 hybridized but the electron pair CH3 occupies a p orpital, so it N N can delocalize onto the a. Electron pair on N CH3 aromatic ring. Delocalization occupies an sp2 hybrid orbital. makes this N less basic. DMAP 4-(N,N-dimethylamino)pyridine

b.

H N

N CH3

sp3 hybridized N stronger base

nicotine sp2 hybridized

N higher percent s-character weaker base

717

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Amines 25–11

25.25 This electron pair is delocalized, making it a weaker base.

Br H

N

N

stronger base sp3 hybridized N 25% s-character

HO H

NH2 CH3O

a.

b.

N

sp2 hybridized N 33% s-character N

N(CH3)2

c.

stronger base sp3 hybridized N 25% s-character

sp2 hybridized N 33% s-character

stronger base This compound is similar to DMAP in Problem 25.24a.

25.26 Amines attack carbonyl groups to form products of nucleophilic addition or substitution. CH3CH2CH2NH2

O

a. O

NCH2CH2CH3

O

b. CH3 C O C CH3 COCl

c.

CH3CH2CH2NH2 CH3

O

O

O

C

C

C

NHCH2CH2CH3

CH3

O

CONHCH2CH2CH3

CH3CH2CH2NH2

(CH3CH2)2NH

O

N(CH2CH3)2 O

(CH3CH2)2NH CH3

COCl

C

CH3

N(CH2CH3)2 CON(CH2CH3)2

(CH3CH2)2NH

25.27 [1] Convert the amine (aniline) into an amide (acetanilide). [2] Carry out the Friedel–Crafts reaction. [3] Hydrolyze the amide to generate the free amino group. O

a.

NH2

CH3

C

O C CH3 (CH3)3CCl (CH ) C NH 3 3 AlCl3

Cl

NH2

CH3

C

C CH3 H3O+ NH (CH3)3C

(+ ortho isomer)

O

b.

O

H N

Cl

Cl C

CH3

O

H N O AlCl3

C

CH3

H3O+

O CCH2CH3

NH2

O

O

(+ para isomer)

25.28 transition state

δ+

N(CH3)3 H

OH

δ− (no 3-D geometry shown here)

Ea

Energy

transition state:

starting materials

H°

Reaction coordinate

products

NH2

718

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–12

25.29 [1] CH3I (excess)

a. CH3CH2CH2CH2 NH2

b.

[2] Ag2O [3] Δ

[1] CH3I (excess)

(CH3)2CHNH2

[1] CH3I (excess)

NH2

c.

CH3CH2CH=CH2

[2] Ag2O [3] Δ

CH3CH=CH2

[2] Ag2O [3] Δ

25.30 In a Hofmann elimination, the base removes a proton from the less substituted, more accessible β carbon atom, because of the bulky leaving group on the nearby α carbon. [1] CH3I (excess)

a.

CH2CHCH3

CH=CHCH3

[2] Ag2O [3] Δ

NH2

major product

[1] CH3I (excess)

b. H2N

major product +

β [1] CH3I (excess)

β

c. β

[2] Ag2O [3] Δ

CH2CH=CH2

N H

[2] Ag2O

[3] Δ (3 β C's) least substituted β carbon

CH3CH=CH(CH2)3N(CH3)2 + CH2=CH(CH2)2CHN(CH3)2 + CH3 CH2=CH(CH2)4N(CH3)2

major product, formed by removal of a H from the least substituted β C

25.31 K+ –OC(CH3)3

a.

Cl

Br [1] CH3I (excess)

b.

K+ –OC(CH3)3

c.

[2] Ag2O [3] Δ

NH2

[1] CH3I (excess)

d. (E and Z)

[2] Ag2O [3] Δ

NH2

25.32 N2+ Cl–

NH2 NaNO2

a. CH3

b.

H CH3CH2 N CH3

HCl

HCl

HCl

N H

CH3

NaNO2

NaNO2

c.

CH3CH2 N CH3 NO

N NO

NaNO2

d.

HCl NH2

N2 Cl–

719

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Amines 25–13

25.33 a.

NH2

b.

NH2

[1] NaNO2, HCl

c. CH3O

Br

[2] CuBr [1] NaNO2, HCl

d.

OH O2 N

CH3O

[2] HBF4 [1] CuCN

N2+ Cl–

[2] H2O O2N

[1] NaNO2, HCl

NH2

CH2NH2

[2] LiAlH4 [3] H2O

Cl

F

Cl

25.34 NH2

NO2

a.

H2

H2SO4

Pd-C

NO2

b.

F [1] NaNO2, HCl

HNO3

[2] HBF4

NO2

OH

NH2

HNO3

H2

[1] NaNO2, HCl

H2SO4

Pd-C

[2] H2O

(from a.)

NO2

OH

NH2 CH3

[1] NaNO2, HCl

c. NH2

CH3Cl

[2] NaI

(+ para isomer)

AlCl3

I

I

(from a.) NH2

NH2 Cl

Cl

Cl2 (excess)

d.

Cl

Cl

[1] NaNO2, HCl [2] H3PO2

FeCl3

(from a.)

Cl

Cl

25.35 NH2

a.

OH N N

NH2

c.

HO

OH

N N HO

b.

HO

N N

OH

25.36 To determine what starting materials are needed to synthesize a particular azo compound, always divide the molecule into two components: one has a benzene ring with a diazonium ion, and one has a benzene ring with a very strong electron-donor group. O2N

a. H2N

Cl

b. HO

N N

N N

O2N H2N

Cl– N2

CH3 Cl

HO

Cl– N2

CH3

720

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–14

25.37 a. N N OH

b. O2N

NO2

N N

OH

alizarine yellow R

para red

COOH

+

+

N N Cl–

O2N

NO2

N N Cl–

OH COOH

OH

25.38 O

CH3

O

O

O O

O

–

O3S

N N

O

Dacron

methyl orange

O

To bind to fabric, methyl orange (an anion) needs to interact with positively charged sites. Since Dacron is a neutral compound with no cationic sites on the chain, it does not bind methyl orange well.

25.39 a.

4,6-dimethyl-1-heptanamine

N(CH2CH3)2

b.

NH2

N,N-diethylcycloheptanamine

25.40 or N H

A weaker base (delocalized electron pair on N)

NH

B stronger base

25.41 N

N

a, b.

NH

NH2 + Cl–

HCl

N

N

varenicline

most basic only sp3 hybridized N

N CH3

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

721

Amines 25–15

25.42 a. CH3NHCH2CH2CH2CH3 N-methyl-1-butanamine (N-methylbutylamine)

e. (C6H5)2NH diphenylamine

h.

CH2CH3

N H

2-ethylpyrrolidine b.

N C(CH3)3

f.

NH2

1-octanamine (octylamine)

CH3CH2CH2CH(NH2)CH(CH3)2

i.

CH2CH3

2-methyl-3-hexanamine

N-tert-butyl-N-ethylaniline

CH3 N

c.

g. O

CH2CH2CH3

N-methyl-N-propylcyclohexanamine d.

NH2

j.

NH2

3-ethyl-2-methylcyclohexanamine

4-aminocyclohexanone

(CH3CH2CH2)3N

tripropylamine

25.43 e. N-methylpyrrole

a. cyclobutylamine

h. 3-methyl-2-hexanamine NH2

NH2 N

b. N-isobutylcyclopentylamine

CH3

f. N-methylcyclopentylamine N

i. 2-sec-butylpiperidine

NHCH3

H

c. tri-tert-butylamine

N H

g. cis-2-aminocyclohexanol

N[C(CH3)3]3

NH2

d. N,N-diisopropylaniline

NH2

or OH

N[CH(CH3)2]2

j. (2S)-2-heptanamine H NH2

OH

25.44 NH2

1-butanamine H N

N-methyl-1-propanamine

NH2

2-butanamine

N H

diethylamine

NH2

2-methyl-1-propanamine

N H

N-methyl-2-propanamine

NH2

2-methyl-2-propanamine

N

N,N-dimethylethanamine

722

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–16

25.45 [* denotes a stereogenic center.] a.

*

N

CH3

b.

CH2CH3 * * CH3CH2CHCH 2CH2CH2 N CH2CH2CH2CH3

CH2CH3

CH3

1 stereogenic center 2 stereoisomers

N H

H CH3

CH3 CH3CH2

CH2CH2CH2 H CH3

H

CH3 CH3CH2

H CH3

N Cl

CH2CH2CH2

N Cl

CH3 CH2CH3

C CH3CH2

CH2CH2CH2CH3

CH2CH2CH2 H CH3

CH3 CH2CH3

C

CH2CH3

Cl

CH3 CH2CH3

C

CH2CH3

N

CH3

2 stereogenic centers 4 stereoisomers

CH3CH2

CH2CH2CH2CH3

CH3 CH2CH3

C CH2CH2CH2CH3

N Cl

CH2CH2CH2

N Cl

CH2CH2CH2CH3

25.46 a.

(CH3CH2)2NH

b. (CH3CH2)2NH

or

sp3 hybridized N stronger base

N

or

2° alkylamine Cl is electron withdrawing. weaker base

2° alkylamine stronger base

sp2 hybridized N weaker base

(ClCH2CH2)2NH

25.47 a.

NH2

NH2

NH3

arylamine intermediate least basic basicity

alkylamine most basic

O2N

b.

d.

N H

N

delocalized electron pair on N least basic

sp2 hybridized N intermediate basicity

NH2

c.

N H

sp3 hybridized N most basic

NH2 Cl

CH3

electronwithdrawing group least basic

intermediate basicity

(C6H5)2NH

C6H5NH2

diarylamine least basic

NH2

arylamine intermediate basicity

electrondonating group most basic NH2

alkylamine most basic

25.48 The electron-withdrawing inductive effect of the phenyl group stabilizes benzylamine, making its conjugate acid more acidic than the conjugate acid of cyclohexanamine. The conjugate acid of aniline is more acidic than the conjugate acid of benzylamine, since loss of a proton generates a resonance-stabilized amine, C6H5NH2. NH3

pKa = 10.7

NH2

alkylamine cyclohexanamine

CH2NH3

pKa intermediate

CH2NH2

electron-withdrawing inductive effect of the sp2 hybridized C's benzylamine

NH3

pKa = 4.6

NH2

resonance-stabilized aromatic amine aniline

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

723

Amines 25–17

25.49 The most basic N atom is protonated on treatment with acid. O

O

O

O NH

a. N

O CH2COOH

+ CH3CO2–

NH2

CH3CO2H N

more basic

O CH2COOH

benazepril

a 3° alkylamine with an sp3 hybridized N most basic O

O Cl

b. N

Cl

N

NH

CH3CO2H

N NH

aripiprazole

O

Cl

O

Cl

N

H CH3CO2–

25.50 Nc

O C

a.

NHNH2

N

Nb < Na < Nc

Nb Na

Na

Nc

N

NH2

b. N Nb H

Nb < Na < Nc

Order of basicity: Nb < Na < Nc Nb – The electron pair on this N atom is delocalized on the O atom; least basic. Na – The electron pair on this N atom is not delocalized, but is on an sp2 hybridized atom. Nc – The electron pair on this N atom is on an sp3 hybridized N; most basic.

Order of basicity: Nb < Na < Nc Nb – The electron pair on this N atom is delocalized on the aromatic five-membered ring; least basic. Na – The electron pair on this N atom is not delocalized, but is on an sp2 hybridized atom. Nc – The electron pair on this N atom is on an sp3 hybridized N; most basic.

25.51 The para isomer is the weaker base because the electron pair on its NH2 group can be delocalized onto the NO2 group. In the meta isomer, no resonance structure places the electron pair on the NO2 group, and fewer resonance structures can be drawn:

724

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–18

O2N

NH2

O2N

NH2

O2N

NH2

O2N

NH2

O2N

NH2

meta NH2

NH2

O2 N

NH2 O

O2N

N

para

NH2

NH2

O2N

O2N

O NH2 O

NH2 O

N O

N O

25.52 N

N

A

This two-carbon bridge makes it difficult for the lone pair on N to delocalize on the aromatic ring.

B

pKa of the conjugate acid = 5.2 pKa of the conjugate acid = 7.29 Resonance structures that place a double bond between the N atom and the benzene ring are stronger conjugate acid weaker conjugate acid destabilized. Since the electron pair is more weaker base stronger base localized on N, compound B is more basic. The electron pair of this arylamine is delocalized on the benzene ring, decreasing its basicity. N

N

B

Geometry makes it difficult to have a double bond here.

25.53 NH

B

N

pyrrole pKa = 23 stronger acid

N

N

N

N

weaker conjugate base The electron pair is delocalized, decreasing the basicity. The N atom is sp2 hybridized.

NH

pyrrolidine pKa = 44 weaker acid

B

N

stronger conjugate base The electron pair is not delocalized on the ring. The N atom is sp3 hybridized.

25.54 a. C6H5CH2CH2CH2Br

b. C6H5CH2CH2Br

NH3

excess

NaCN

c. C6H5CH2CH2CH2NO2

C6H5CH2CH2CH2NH2 [1] LiAlH4

Pd-C

[1] LiAlH4 [2] H2O

C6H5CH2CH2CN H2

d. C6H5CH2CH2CONH2

[2] H2O

C6H5CH2CH2CH2NH2

C6H5CH2CH2CH2NH2

C6H5CH2CH2CH2NH2 e. C6H5CH2CH2CHO

NH3 NaBH3CN

C6H5CH2CH2CH2NH2

725

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Amines 25–19

25.55 a. (CH3CH2)2NH

NH2

b.

N(CH3)2

c.

H N

d.

O C

CH3

H N

N(CH3)2

NH2

NHCH2CH3 O

or

O

O CH3 N

C

H

O

25.56 In reductive amination, one alkyl group on N comes from the carbonyl compound. The remainder of the molecule comes from NH3 or an amine. NH2

a.

H

+ NH3

O H N

b.

H

C6H5

or

NH2

C6H5 O (CH3CH2CH2)2NH

or

H

H

N H

C

O

CH3CH2CH2

CH2CH3

O

d.

C6H5

O

O

c. (CH3CH2CH2)2N(CH2)2CH(CH3)2

H2N

H

N H

NH2

H

25.57 a.

NH2

C6H5 O

b.

O

NaBH3CN

C6H5

c. NH

O

CH2NH2 NH3 excess

CN

b.

[1] LiAlH4

CH2NH2

[2] H2O CONH2

CH2NH2 [1] LiAlH4

c.

[2] H2O CHO

d.

NH3 NaBH3CN

C6H5

CH2NH2

d.

N(CH3)2

25.58 Br

CHO

NH2

(CH3)2NH NaBH3CN

a.

C6H5

NH3 NaBH3CN

CH2NH2

NaBH3CN

NH

726

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–20

CH3

CH2Br

e.

CH2NH2

Br2

NH3

hν

excess

COOH

CONH2

[1] SOCl2

f.

[2] NH3 NH2

g.

[2] H2O CN

[1] NaNO2, HCl

[2] H2O NO2

HNO3

CH2NH2

[1] LiAlH4

[2] CuCN

h.

CH2NH2

[1] LiAlH4

NH2

H2

Then as in (g).

Pd-C

H2SO4

25.59 Use the directions from Answer 25.20. Separation can be achieved because benzoic acid reacts with aqueous base and aniline reacts with aqueous acid according to the following equations: COO–Na+ + H2O

COOH NaOH

benzoic acid • soluble in CH2Cl2 • insoluble in H2O

(10% aqueous) • soluble in H2O • insoluble in CH2Cl2 +

NH2

NH3 Cl– H Cl

aniline • soluble in CH2Cl2 • insoluble in H2O

(10% aqueous) • soluble in H2O • insoluble in CH2Cl2

Toluene (C6H5CH3), on the other hand, is not protonated or deprotonated in aqueous solution, so it is always soluble in CH2Cl2 and insoluble in H2O. The following flow chart illustrates the process. CH3

COOH

NH2

CH2Cl2 10% NaOH CH2Cl2

H2O COO–Na+

CH3

NH2

10% HCl H2O

CH2Cl2 +

NH3 Cl–

CH3

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

727

Amines 25–21

25.60 H N

a.

b.

N-ethylaniline f.

H H N

HCl

Δ

+ CH2=CH2

Cl–

O CH3CH2COCl

CH3COOH

N

g.

H H N CH3COO–

O HNO3

h. The product in (g) c.

N(CH3)2

Ag2O

CH3I (excess)

N

N

H2SO4

N

(CH3)2C=O

O

O2N

NO2

CH3 N

d.

e.

i. The product in (g)

CH2O

[1] LiAlH4

N

[2] H2O

NaBH3CN CH3I

CH3 CH3 N

(excess)

I–

O H2

j. The product in (h)

O N

N

Pd-C NH2 H2N

25.61 NH2 p-methylaniline

CH3

f. HCl

a.

CH3

CH3COCl

CH3

(CH3CO)2O

CH3

g.

h.

CH3I excess

e.

CH3COOH

NH3 CH3COO–

CH3

NaNO2

N2+Cl–

CH3

HCl

C CH 3 NH

i. d.

NH2

C CH3 NH O

c.

AlCl3

AlCl3 CH3

NH3 Cl– O

b.

CH3COCl

O

Step (b), then

CH3COCl

C CH3 NH

CH3

AlCl3 CH3

(CH3)2C=O

CH3

N(CH3)3 I–

N C(CH3)2

C O CH3

j.

CH3CHO NaBH3CN

25.62 a. CH3(CH2)6NH2

[1] CH3I (excess) [2] Ag2O [3] Δ

CH3(CH2)4CH=CH2

CH3

NHCH2CH3

728

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–22

[1] CH3I (excess)

b. NH2

[2] Ag2O [3] Δ

N H

β1

β2 N H

[2] Ag2O [3] Δ

β3

N(CH3)2

+

β2

major product + (CH3)2CHN(CH3)2

+

CH3

(E + Z)

(CH3)2N

β3

(E + Z)

CH2=CHCH3 + CH3CH2N(CH3)2

d.

major product

(CH3)2N

CH2=CH2

[2] Ag2O [3] Δ

β1

[1] CH3I (excess)

major product

[1] CH3I (excess)

c.

e.

(E + Z)

[1] CH3I (excess) CH3

[2] Ag2O [3] Δ

NH2 CH3

CH3

CH2

major product CH3

CH3

CH3

25.63 N(CH3)2

O HN(CH3)2

a.

mild acid O NH2CH2CH2CH3

b.

NCH2CH2CH3

mild acid NH2

O NH3

c.

NaBH3CN

NH2

O

d.

[1] NaBH4, CH3OH

[1] CH3I (excess)

[2] H2SO4

[2] Ag2O [3] Δ

O NH3 NaBH3CN

OH

O CH3NH2

mCPBA

NHCH3

e.

(from d.) Br2

f.

O

O

O

Br

CH3COOH

25.64 CH3 N

a. one stereogenic center benzphetamine

NH2CH2CH2CH2CH3

NHCH2CH2CH2CH3

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

729

Amines 25–23

b. Amides that can be reduced to benzphetamine: CHO N

N

and O

c. Amines + carbonyl compounds that form benzphetamine by reductive amination: NHCH3

O

or

H O

or

β1

H N

NH CH2=O

CH3

α N

[1] CH3I

β2

[2] Ag2O [3] Δ

d.

(CH3)2NCH2

major product elimination across α, β1 (elimination across α, β2)

25.65 a.

CH2CH2Cl

NH3

g.

CH2CH2NH2

NH

NaNO2

N N O

HCl

excess O CO2

[1] KOH

b.

NH O

c. Br

NH2 +

[2] (CH3)2CHCH2Cl [3] –OH, H2O

NO2

Sn

NH + C6H5CHO

NaBH3CN

Br

NH2

O

+

i.

N

N H

[1] LiAlH4

d. CN

[2] H2O CONHCH2CH3

N CH2C6H5

CO2

HCl

e.

h.

j. CH3CH2CH2 N CH(CH3)2

CH2NH2 [1] LiAlH4

H CH2NHCH2CH3

[2] H2O

f. C6H5CH2CH2NH2 + (C6H5CO)2O C6H5CH2CH2NHCOC6H5 + C6H5CH2CH2NH3+ C6H5COO–

[1] CH3I (excess) [2] Ag2O [3] Δ

CH3CH=CH2

(CH3)2NCH(CH3)2

CH3CH2CH2N(CH3)2

730

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–24

25.66 NH2 and H must be anti for the Hofmann elimination. Rotate around the C–C bond so the NH2 and H are anti. a.

C6H5

NH2

H

C6H5

NH2

CH3

b.

H

CH3

C6H5

CH2CH3

c.

CH3 (CH3)3C

C(CH3)3

rotate 120° counterclockwise

three steps (CH3)3C

NH2 CH2CH3

CH2CH3

C(CH3)3

H

rotate 120° counterclockwise C6H5

NH2 CH3

CH2CH3

H

C6H5

C(CH3)3

(CH3)3C

CH2CH3 NH2 CH3

H

three steps

three steps

(CH3)3C

(CH3)3C

25.67 O F

KOH

F

NO2 F

DMSO

F F

NO2 OH

Cl

F

KI, K2CO3

F

H2 NO2

Ni catalyst

F F

NH2

O

O

O

O

A B

C not isolated

731

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Amines 25–25

25.68 N2+ Cl– Cl HO

Cl

A

Br

a. H2O

Cl

h. C6H5NH2

d. CuBr

b. H3PO2

H

Cl

NC

Cl

F

Cl

N N

OH

e. CuCN

Cl

Cl

NH2

Cl

i. C6H5OH c. CuCl

N N

Cl

f. HBF4 I

I

Cl

Cl

j. KI

g. NaI

25.69 R

CH3 C

CH3CH2CH2

[1] CH3I (excess)

H NH2

A

H

[2] Ag2O [3] Δ

H

B

O

[1] O3

H C C

H

[2] CH3SCH3

CH2CH2CH3

C

O H

H

C

CH2CH2CH3

25.70 a.

Br

Br

H H Na+ N

Br

OH

H N

Br

Br

Na+

+ CH3CH2NH2

Br

N H CH2CH3

+ H2O

Na+

OH

N

+ H2O + Na+

CH2CH3

H A

H O

O

H N

b. NH2

O

H

N

proton transfer

OH2

H

N

H3B–H H

H N + H2O

proton source + A

H N + BH3

732

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–26

25.71 OH

O

OH HN

H N

H2C=O

H A

OH

OH N

proton transfer A

OH

OH

OH H

OH2

OH H2C

N

N

N

N

A

H2O

H A OH H

OH H

OH H

N

N

N

25.72 NaNO2

Overall reaction:

HCl, H2O

NH2

The steps:

OH

+

+

OH

NaNO2 + HCl

NH2

N N O

Cl

H

+

+ HCl

+

N O

N N O

N N O

H

H

N N O H

H

+

H Cl

+

H Cl

– N N OH2 + Cl

N N O H

H Cl

+

O H

OH

H

+ HCl

H +

B

A

O H H

H OH Cl

H

B

+ HCl

Cl

+

+N

A H OH

Cl

+

1,2-H shift

N N +

OH

+ HCl

N

+ H2O

733

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Amines 25–27

25.73 A nitrosonium ion (+NO) is a weak electrophile so electrophilic aromatic substitution occurs only with a strong electron-donor group that stabilizes the intermediate carbocation. O N O

H Cl

NaCl

HO N O

H Cl

H O N O H

Na N(CH3)2

O N H

N(CH3)2

O N H

Cl

N(CH3)2

O N H

O N H O N H

N(CH3)2

O N

N(CH3)2

especially good resonance structure All atoms have octets

resonance-stabilized carbocation

N O

H2O

N O

N(CH3)2

N(CH3)2 H2O

OH

25.74 O

NH2

a.

NH3

[1] CH3I

NaBH3CN

[2] Ag2O [3] Δ

O

(Hofmann elimination, less substituted C=C favored)

OH NaBH4

b.

H2SO4

(more substituted C=C favored)

CH3OH

(E + Z isomers formed)

25.75 HNO3 H2SO4

a.

NO2

CH3Cl AlCl3

b.

CH3

Br2 FeBr3

ClCOCH3

H2 Br

HNO3 O 2N H2SO4

NO2

CH3

Pd-C

H2 Pd-C

Br

NH 2

H2 N

CH3

Br [1] NaNO2, HCl [2] NaI

NHCOCH3

I

CH3

(+ ortho isomer)

O2N

H2N

H2

c.

[1] NaNO2, HCl

NC

Br2

[2] CuCN

Pd-C

NC

Br

FeBr3

(from a.)

d.

Br2 FeBr3

e. O2N

CH3

(from b.)

NO2

HNO3 Br KMnO4

O 2N

H2SO4

Br

H2 Pd-C

(+ para isomer) COOH

H2 Pd-C

H 2N

NH2 Br COOH

[1] NaNO2, HCl [2] H2O

[1] NaNO2, HCl HO [2] H2O

OH Br

COOH

734

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–28

f.

NH2

[1] NaNO2, HCl

OH

C6H5N2+Cl–

N N

OH

[2] H2O

(from c.)

(from c.)

25.76 NH2

CN

[1] NaNO 2, HCl

a.

COOH

H 3 O+

[2] NaCN

[2] NH2CH3 Br

Br

Br2

CH3COCl

b.

NH2

CONHCH3

[1] SOCl2

Br –

CH3Cl

NHCOCH3 FeBr3

OH NHCOCH3 H2O

NHCOCH3 AlCl3

(+ ortho isomer)

NH2 [1] NaNO2, HCl [2] H3PO2 CH3

Br

Br

c.

[1] LiAlH4

COOH

CH2OH

[2] H2O

Br2

Br

FeBr3

CH2OH

(3x)

(from a.)

Br

HO H2N

HO

HO [1] NaNO2, HCl

d.

C6H5N2+Cl–

CH3Cl AlCl3

[2] H2O

CH3

N N

(from a., Step [1])

CH3

(+ ortho isomer)

25.77 CH3Cl

[1]

Br2

–OH

Br

[2]

Br

hν

AlCl3

NH3

OH

PCC

H2C=O

NH2

(excess) CHO CH NH 3 2 NaBH3CN

(from [1])

N H

[1] LiAlH4 [2] H2O KMnO4

O

COOH [1] SOCl 2 [2] CH3NH2

[3]

NCH3

[1] LiAlH4

H

[2] H2O

(from [1]) [1] HNO3, H2SO4

[4]

N H

NaBH3CN

[2] H2, Pd-C [3] NaNO2, HCl [4] CuCN

CN [1] DIBAL-H [2] H2O [1] LiAlH4 [2] H2O NH2

Then route [1].

CHO

Then route [2].

N H

735

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Amines 25–29

Br

Br2

[5]

COOH

MgBr [1] CO 2

Mg

[2]

FeBr3

Then route [3].

H3O+

[1] H2C=O [2] H2O OH

Then route [2].

25.78 O

O

mCPBA

O

O

O

[1] LiAlH4

O

[2] H2O

O

PBr3 OH

O Br

O

PCC O

O

CH3NH2 NHCH3

O

CH3NH2

NaBH3CN

O O

O

NHCH3

O

MDMA

MDMA

[part (b)]

[part (a)]

25.79 CH3O

CH3O

Br

CN [1] LiAlH 4

NaCN

a. CH3O

[2] H2O

CH3O CH3O

CH3O

Br

CH3O HBr

b. CH3O

ROOR

CH3O

Mg

CH3O

NH2

[1]

CH3O

CH3O CH3O

O

[2] H2O

CH3O CH3O

excess

MgBr

c. CH3O

CH3O

CH3O Br

mescaline

NH3

CH3O CH3O

NH2

CH3O

CH3O

CH3O

CH3O

CH3O

OH

CH3O

PBr3

CH3O

Br

CH3O CH3O

CH3O NH3 excess CH3O

NH2

CH3O CH3O

736

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–30

25.80 NO2

HNO3

a.

H2SO4

NH2

H2

CN

[1] NaNO2, HCl

COOH

H3 O +

[2] CuCN

Pd-C

COOH

Cl2 FeCl3 Cl

NH2

NHCOCH3 CH3COCl

b.

NHCOCH3

HNO3

Cl2

H2SO4

FeCl3 (2X)

(from a.)

NHCOCH3 Cl

Cl

NH2

–OH

Cl

Cl Cl

H2O

NO2

Cl

[1] NaNO2, HCl [2] H3PO2 NO2

NO2

NO2

Cl

H2 Pd-C Cl

Cl

Cl [1] NaNO2, HCl [2] H2O

NH2

OH

CH3Cl

c.

HNO3

CH3

AlCl3

CH3

H2SO4

H2

NO2

CH3

NH2

Pd-C

[1] NaNO2, HCl [2] NaCN

(+ ortho isomer) CH3 CH3CH2Cl

d.

HNO3

CH3CH2

AlCl3

H2SO4

CH2NH2

[1] LiAlH4

CH3

[2] H2O H2 CH3CH2

NO2

CN

CH3CH2

Pd-C

NH2 [1] NaNO2, HCl [2] CuCN

(+ ortho isomer) CH3CH2

e. O2N

CH3

Br2

O2N

CH3

(from c.)

CH3

Br

H3O+

CH3CH2

[1] NaNO2, HCl

CN HO

[2] H2O

CH3

Br

Br

Cl NH2

Cl

H2 N

Pd-C

FeBr3

Cl

f.

H2

COOH

[1] NaNO2, HCl

N N

NH2

[2] NH2

(from b.)

Cl

(from a.)

25.81 O HNO3

a.

H2SO4

NO2

NO2 H 2

Cl2 FeCl3

NH2

O

H N

O

O

Pd-C Cl

Cl2 FeCl3 H N

Cl

Cl

O

Cl Cl

(+ isomer)

737

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Amines 25–31

H2

b.

Pd-C H2N

O2N

NO2

[1] NaNO2, HCl

HNO3

[2] H2O

H2SO4 HO

HO

NH2

H2 Pd-C HO O

(+ ortho isomer)

(from a.)

O O H N O

HO O

O

ClCOCH2CH3

c.

Cl2

AlCl3

O

OH NHCH3 NaBH4

Cl NH2CH3

NHCH3

CH3OH

H2O, HCl

25.82 CH3OH OH

NH2

a.

[1] HNO3, H2SO4

[1] NaNO2, HCl

[2] H2, Pd-C

[2] H2O

OH

SOCl2

OH

CH3Cl

Br2

AlCl3

FeBr3 (2X)

OCH3

Br

Br

[1] NaH

Br

[2] CH3Cl

CH3

CH3

(+ ortho isomer) CH3CH2OH

Br

CH3 [1] Br2, hν

Br

[2] NH3 (excess)

NH2 CH3O

CrO3, H2SO4, H2O Br

CH3COOH NH2

SOCl2

H N

CH3COCl

b.

O

H N

Cl O

H2O

O NH2

O

AlCl3

(from a.)

–OH,

O

(+ ortho isomer)

CH3OH SOCl2

c.

Cl2 FeCl3

[1] CH3Cl, AlCl3 Cl

[2] KMnO4

HOOC

Cl

[1] LiAlH4 [2] H2O

HO Cl

PCC

O Cl H

(+ ortho isomer) NH2

(from a.) mild acid N CH

Cl

CH3OH SOCl2

d.

CH3Cl AlCl3

CH3

CH3

HNO3 H2SO4

O2N

COOCH3

[1] KMnO4 [2] CH3OH, H+

O2N

COOCH3

H2 Pd-C

H2N PCC

(+ ortho isomer)

O (CH3)2CHNH

CO2CH3 NaBH3CN

OH

738

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–32

NH2

e. O2N

CH3

(from d.)

[1] H2, Pd-C

F

[1] KMnO4

CH3

[2] NaNO2, HCl [3] HBF4

O

O F

F

C

[2] SOCl2

Cl

HN

(from a.)

Probably a strong enough activator that the Friedel–Crafts reaction will still occur. O NHCOCH3

O

Make two parts:

AlCl3

Cl

CH3

C

NH

O

f. I

I

CN CN NHCOCH3

NHCOCH3

NHCOCH3

[1] HNO3, H2SO4

[1] NaNO2, HCl

[2] H2, Pd-C

[2] CuCN

(from b.)

NH2

CN

(+ ortho isomer) CH3 O2N

CH3

[1] H2, Pd-C [2] NaNO2, HCl

(from d.)

O [1] KMnO4

Cl

[2] SOCl2

I

I

[3] NaI

25.83 molecular weight = 87 C5H13N two IR peaks = 1° amine

H

H

H

H

H

H

N

N

N

N

H

H

H

H

H

H

H

N

N

N

N

H

H

H

25.84 H2N NHCH2CH3

CH2CH3

CH2NHCH3

Compound A: C8H11N Compound B: C8H11N IR absorption at 3400 cm–1→ 2° amine IR absorption at 3310 cm–1 → 2° amine 1H NMR signals at (ppm): 1H NMR signals at (ppm): 1.3 (triplet, 3 H) CH3 adjacent to 2 H's 1.4 (singlet, 1 H) amine H 3.1 (quartet, 2 H) CH2 adjacent to 3 H's 2.4 (singlet, 3 H) CH3 3.8 (singlet, 2 H) CH2 3.6 (singlet, 1 H) amine H 6.8–7.2 (multiplet, 5 H) benzene ring 7.2 (multiplet, 5 H) benzene ring

Compound C: C8H11N IR absorption at 3430 and 3350 cm–1 → 1° amine 1H NMR signals at (ppm): 1.3 (triplet, 3 H) CH3 near CH2 2.5 (quartet, 2 H) CH2 near CH3 3.6 (singlet, 2 H) amine H's 6.7 (doublet, 2 H) para disubstituted 7.0 (doublet, 2 H) benzene ring

739

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Amines 25–33

25.85 C N

HO

HO

Compound D: Molecular ion at m/z = 71: C3H5NO (possible formula) IR absorption at 3600–3200 cm–1→ OH 2263 cm–1→ CN Use integration values and the molecular formula to determine the number of H's that give rise to each signal. 1 H NMR signals at (ppm): 2.6 (triplet, 2 H) CH2 adjacent to 2 H's 3.2 (singlet, 1 H) OH 3.9 (triplet, 2 H) CH2 adjacent to 2 H's

NH2

Compound E: Molecular ion at m/z = 75: C3H9NO (possible formula) IR absorption at 3600–3200 cm–1→ OH 3636 cm–1 → N–H of amine 1H NMR signals at (ppm): 1.6 (quintet, 2 H) CH2 split by 2 CH2's 2.5 (singlet, 3 H) NH2 and OH 2.8 (triplet, 2 H) CH2 split by CH2 3.7 (triplet, 2 H) CH2 split by CH2

25.86 Guanidine is a strong base because its conjugate acid is stabilized by resonance. This resonance delocalization makes guanidine easily donate its electron pair; thus it’s a strong base. NH H2 N

C

HA NH2

H2N

NH2

NH2

NH2

C

C

C

NH2

H2 N

NH2

H2 N

NH2

NH2

H2 N

C

NH2

pKa = 13.6

guanidine

25.87 The compound with the most available electron pair or the compound with the highest electron density on an atom (N in this case) is the strongest base. Pyrrole is the weakest base because its lone pair is delocalized on the five-membered ring to make it aromatic. Both imidazole and thiazole contain sp2 hybridized N atoms with electron pairs that are localized on N. Imidazole is a stronger base than thiazole, because its second N atom is more basic than thiazole’s S atom, so it places more electron density on N by a resonance effect. N

N

NH

imidazole

N

NH

N

pyrrole least basic

N

S

thiazole This form contributes less to the hybrid than the equivalent resonance structure in imidazole.

This N atom is the strongest base. NH

S

N

S

thiazole intermediate basicity

NH

imidazole most basic

25.88 CH3

N

[1] CH3I (excess)

CH3

N

CH3 I

[2] Ag2O

[2] Ag2O

[1] CH3I (excess)

[3] Δ

[3] Δ N(CH3)2

I

N(CH3)3

C8H10 Y

740

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 25–34

25.89 One possibility: O

O

O

CH3COCl

a. O

O

Br2 O

AlCl3

H2NC(CH3)3

CH3CO2H

O

O

Br

O

O

(+ isomer)

NaBH4 CH3OH HO

O H3O+

HO

O

HN(CH2CH3)2

+

A [1] NaNO2, HCl

H2 Pd-C

H2SO4 O N 2

N H

OH

N(CH2CH3)2

HO

HNO3

O

N H

OH

albuterol

b.

N H

O

[2] H2O

H2N

[1] NaH HO

O

Cl

[2]

Br2 O2N

COOH

HNO3

O

Mg O

O

COCl

O2N

O N(CH2CH3)2

O

H2

H2 N

N(CH2CH3)2

O

Pd-C

A

O

H3O+ CO2

O

SOCl2 O2N

Br

COOH

H2SO4

O

O

25.90 CH2=O reacts with the amine to form an intermediate imine, which undergoes an intramolecular Diels–Alder reaction. O

Ph

O

Ph

O

Ph

H

CH2=O H2 N

H2O

X

OH

OH

H

H

one step CH2

N

N

Y C17H23NO

N

lupinine

+

N

epilupinine

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

741

Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–1

Chapter 26 Carbon–Carbon Bond-Forming Reactions in Organic Synthesis Chapter Review Coupling reactions [1] Coupling reactions of organocuprate reagents (26.1) • R'X can be CH3X, RCH2X, 2o cyclic + RCu R' X + R2CuLi R' R halides, vinyl halides, and aryl halides. + LiX X = Cl, Br, I • X may be Cl, Br, or I. • With vinyl halides, coupling is stereospecific. [2] Suzuki reaction (26.2) Y R' X

+

Pd(PPh3)4

R B

NaOH

Y

X = Br, I

R' R

+

HO BY2

+

NaX

• •

R'X is most often a vinyl halide or aryl halide. With vinyl halides, coupling is stereospecific.

[3] Heck reaction (26.3) Pd(OAc)2

R' X

Z

•

R'

Z

P(o-tolyl)3

X = Br or I

• •

(CH3CH2)3N (CH3CH2)3NH X–

•

R'X is a vinyl halide or aryl halide. Z = H, Ph, COOR, or CN With vinyl halides, coupling is stereospecific. The reaction forms trans alkenes.

Cyclopropane synthesis [1] Addition of dihalocarbenes to alkenes (26.4) C

X X C

CHX3

C

KOC(CH3)3

C

• •

The reaction occurs with syn addition. The position of substituents in the alkene is retained in the cyclopropane.

• •

The reaction occurs with syn addition. The position of substituents in the alkene is retained in the cyclopropane.

•

Metathesis works best when CH2=CH2, a gas that escapes from the reaction mixture, is formed as one product.

C

[2] Simmons–Smith reaction (26.5) CH2I2

C C

Zn(Cu)

H H C C C

+ ZnI2

Metathesis (26.6) 2 RCH CH2

Grubbs RCH CHR

catalyst Grubbs catalyst diene

+ CH2 CH2

+ CH2 CH2

742

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 26–2

Practice Test on Chapter Review 1. a. Which functional groups react with lithium dialkyl cuprates? 1. epoxides 2. vinyl halides 3. acid chlorides 4. Compounds [1] and [2] both react with R2CuLi. 5. Compounds [1], [2], and [3] all react with R2CuLi. b. Which of the following statements is (are) true for the Suzuki reaction? 1. Arylboranes can serve as one reactant. 2. The reaction is stereospecific. 3. The reaction occurs between a vinyl or aryl halide and an alkene in the presence of a palladium catalyst. 4. Statements [1] and [2] are both true. 5. Statements [1], [2], and [3] are all true. c. Which of the following compounds can react with CH2=CHCN in a Heck reaction? 1.

4. Both [1] and [2] can react.

Br I

2.

5. Compounds [1], [2], and [3] can all react.

3. CH2 CH2

d. Which of the following compounds yields a pair of enantiomers on reaction with the Simmons– Smith reagent? 1.

3.

4. Compounds [1] and [2] both yield a pair of enantiomers. 2.

5. Compounds [1], [2], and [3] all yield a pair of enantiomers.

e. Which of the following compounds can be made by a ring-closing metathesis reaction? O

1.

3. OH

4. Compounds [1] and [2] can both be prepared.

2.

5. Compounds [1], [2], and [3] can all be prepared.

HO O

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

743

Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–3

f. Which of the following compounds can be prepared from CH3–C≡C–H by a Suzuki reaction? You may use other organic compounds or inorganic reagents. 1.

3.

4. Compounds [1] and [2] can both be prepared.

2.

5. Compounds [1], [2], and [3] can all be prepared. 2. Draw the product formed in each reaction. Indicate the stereochemistry around double bonds and stereogenic centers when necessary. a. HC CCH2CH2CH3

[1] catecholborane [2]

e.

I , Pd(PPh3)4, NaOH

[1] 2 Li Br

b.

[2] 0.5 CuI [3] Ph

f.

Grubbs cataylst

Br

[1] 2 Li c.

CHBr3 KOC(CH3)3

g.

Br

[2] B(OCH3)3 [3]

Grubbs cataylst

Br, Pd(PPh3)4, NaOH OCH3

d.

CH3O

Pd(OAc)2 CO2CH3

I

P(o-tolyl)3 (CH3CH2)3N

3. What starting material is needed to synthesize each compound by ring-closure metathesis? O

a.

b.

c. HO OH

OH

O

O

744

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 26–4

Answers to Practice Test 1.a. 5 2. b. 4 a. c. 4 d. 2 b. e. 3 f. 4

3. Br Br

e.

a.

Ph

O

b.

f.

c.

HO OH OCH3

d.

OH

g.

CH3O

O

c.

O

CO2CH3

Answers to Problems 26.1 A new C–C bond is formed in each coupling reaction. a.

Cl

(CH2=CH)2CuLi

new C–C bond b.

(CH3)2CuLi

Br

CH3

new C–C bond c.

CH3O

CH3O

I (CH3CH2CH2CH2)2CuLi

CH3O

new C–C bond

CH3O

Br

d.

2

new C–C bond

CuLi

26.2 I OH

A=

(CH3CH2)2CuLi

OH B= several steps

I OH

(CH3)2CuLi

several steps OH

O

CO2CH3 C18 juvenile hormone

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

745

Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–5

26.3 Br [1] Li

a. (CH3)2CHCH2CH2I

[(CH3)2CHCH2CH2]2CuLi

CH2CH2CH(CH3)2

[2] CuI

or [1] Li

Br

CH2CH2CH(CH3)2

2

[1] Li

Cl

b.

(CH3)2CHCH2CH2I

CuLi

[2] CuI

Br

CuLi

[2] CuI 2

Cl Cl

c.

[1] Li

CuLi

[2] CuI

2

26.4 I +

a.

b.

O

Pd(PPh3)4

O

NaOH

B

B(OCH3)2

Pd(PPh3)4

+

Br NaOH

c.

[1] Li

Br

B(OCH3)2 + LiOCH3

[2] B(OCH3)3

O

O

B

C C H + H B

d.

O

O

26.5 The Suzuki reaction forms a new carbon–carbon bond between a vinyl halide and an arylborane. O CH3S

B(OH)2

+

O

O

O

Br

B

A CH3S

26.6 O O

H B

a. CH3CH2CH2 C C H

b.

c. CH3O

I

Li

O

Li

Br2

I

B

H

O

Br

CH3O

(+ ortho isomer)

B(OCH3)3

Li CH3O

B(OCH3)2 Li

B(OCH3)3 CH3O

Pd(PPh3)4, NaOH

I

Pd(PPh3)4 NaOH

B(OCH3)2

Br Pd(PPh3)4 NaOH

CH3O

746

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 26–6

26.7 Br

Heck

a.

+

b.

I

reaction OCH3

+

OCH3

Heck reaction

Br

c.

+

CN

CN

Br

OCH3

d.

Heck reaction

Heck

+

O

OCH3

reaction

O

26.8 Locate the double bond with the aryl, COOR, or CN substituent, and break the molecule into two components at the end of the C=C not bonded to one of these substituents. O

Br

O

OCH3

a.

OCH3

Br

b.

O

c.

O CO2CH3

CO2CH3

Br

26.9 Add the carbene carbon from either side of the alkene. H

CH3

a.

C C H

H

Cl Cl C

CHCl3 KOC(CH3)3

CH3 C H

C

H

+

CH3 H

C

H enantiomers

H H

C C

Cl Cl CH3CH2

b.

CH2CH3 C C H

H

Cl Cl C

CHCl3

C KOC(CH3)3 CH3CH2 H

C

+

CH2CH3 H

CH3CH2 H

CH3

CHCl3

KOC(CH3)3

CH3 CCl2

CH3

+

H

CCl2 H

enantiomers

C C

identical c.

C

Cl Cl

CH2CH3 H

747

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–7

26.10 CHCl3

a.

LiCu(CH3)2

c.

Cl Cl

KOC(CH3)3

Br Br

(from b.)

CHBr3

b.

Br Br

KOC(CH3)3

26.11 CH2I2

a.

ZnI2

+

CH2I2

c.

+

Zn(Cu)

Zn(Cu)

ZnI2

H CH2I2

b.

ZnI2

+

Zn(Cu) H

26.12 The relative position of substituents in the reactant is retained in the product.

CH3CH2 C C H

H H C

H

CH2I2

CH2CH3

Zn(Cu)

CH3CH2 C H

trans-3-hexene

C

CH3CH2 H

C

C

H CH2CH3

C H H

H CH2CH3

two enantiomers of trans-1,2-diethylcyclopropane

26.13 Grubbs catalyst

a.

Grubbs catalyst

c. (E and Z) OCH3

Grubbs catalyst

b.

(CH2=CH2 is also formed in each reaction.) (E and Z)

OCH3

OCH3

26.14 + cis-2-pentene

+

There are four products formed in this reaction including stereoisomers, and therefore, it is not a practical method to synthesize 1,2-disubstituted alkenes.

26.15 O

a.

Ph

+

O

O

Ph

CO2CH3

b. CO2CH3

O

CO2CH3 CO2CH3

748

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 26–8

26.16 O

O

RCM

O

H

O

OR

O

O

new C–C bond here

H

V

OR

26.17 Cleave the C=C bond in the product, and then bond each carbon of the original alkene to a CH2 group using a double bond.

a.

O

O

c.

CO2CH3

CO2CH3

CHO

CHO

O

O

b. CH3O

CH3O OH

OH

26.18 Inversion of configuration occurs with the substitution of the methyl group for the tosylate.

O

SO2

CH3

(CH3)2CuLi

a. CH3

(equatorial) (CH3)2CuLi

b.

O SO2

CH3 CH3

(axial)

26.19 a.

O

O

RCM

b. O

O

EtO2C CO2Et

EtO2C CO2Et

RCM

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

749

Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–9

26.20 Cl

CH2CH2CH3

+

a.

(CH3CH2CH2)2CuLi

N

B(OCH3)2

Pd(PPh3)4

Br

+

b.

N

NaOH O O

Br

+

c. N

d.

Cl

Pd(OAc)2 P(o-tolyl)3 (CH3CH2)3N [1] Li [2] CuI [3]

Br

O

e.

Br

Pd(PPh3)4

B

+

O

f.

g.

NaOH

Br +

CH3O

Br

N

CO2CH3

Pd(OAc)2

CH3O

P(o-tolyl)3 (CH3CH2)3N

CO2CH3

[1] Li [2] B(OCH3)3 Br + Pd catalyst [3] O [1] H B

h. (CH3)3C C C H

O [2] C6H5Br, Pd(PPh3)4, NaOH

26.21 Br

a.

c. CO2CH3

Br

CO2CH3

Br

d. b.

Br

CH3O

CH3O

750

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 26–10

26.22 Each coupling reaction uses Pd(PPh3)4 and NaOH to form the conjugated diene.

I

A I O

H

H B

O

C C H

B

B

O

O

H

ethynylcyclohexane

I

C It is not possible to synthesize diene D using a Suzuki reaction with ethynylcyclohexane as starting material. Hydroboration of ethynylcyclohexane adds the elements of H and B in a syn fashion, affording a trans vinylborane. Since the Suzuki reaction is stereospecific, one of the double bonds in the product must be trans.

D

26.23 Locate the styrene part of the molecule, and break the molecule into two components. The second component in each reaction is styrene, C6H5CH=CH2.

a.

Br

b.

c.

Br

Br

styrene part

styrene part

styrene part

26.24 HBr

1-butene

Br

ROOR

+

CuLi 2

octane

[1] Li [2] CuI CuLi 2

26.25 Add the carbene carbon from either side of the alkene. H

a.

Cl

CHCl3

c.

Cl

KOC(CH3)3

CHBr3

b.

Br

KOC(CH3)3

H H

CH3 Br

CH3 Br

+

Br H

H H

CH2I2

H

d.

Zn(Cu)

CH2I2

H

+

Zn(Cu) H

H

H

H

751

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–11

Cl Cl C

CHCl3

e. Cl Cl C C

H

CHCl3

f.

KOC(CH3)3

KOC(CH3)3

H

C

C

H

H

+

C

C

C

H

C

H

Cl Cl

26.26 Since the new three-membered ring has a stereogenic center on the C bonded to the phenyl group, the phenyl group can be oriented in two different ways to afford two stereoisomers. These products are diastereomers of each other. CHI2 H

H H

+ Zn(Cu)

H H

H

26.27 High dilution conditions favor intramolecular metathesis. H

OH

H

Grubbs catalyst

N

a.

OH

N

OH

Grubbs catalyst

c. CO2CH3

OH

O

O

O

Grubbs catalyst

O

b.

CO2CH3

26.28 Retrosynthetically break the double bond in the cyclic compound and add a new =CH2 at each end to find the starting material. O

a.

O

O O O

b.

O

CO2CH3 O

CO2CH3

c. O

26.29 Alkene metathesis with two different alkenes is synthetically useful only when both alkenes are symmetrically substituted; that is, the two groups on each end of the double bond are identical to the two groups on the other end of the double bond.

752

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 26–12

a.

CH3CH2CH CH2

CH3CH2CH2CH CHCH2CH3 + CH3CH2CH CHCH2CH3 + CH3CH2CH2CH CHCH2CH2CH3

+

+ CH2 CH2 (Z + E)

CH3CH2CH2CH CH2

(Z + E)

b.

(Z + E)

This reaction is synthetically useful since it yields only one product.

+ c.

+

+

+

+

+

+ CH3CH CHCH3 + CH3CH2CH CHCH2CH3 (Z + E) (Z + E)

+

26.30 O O

O

M

O

O

26.31 All double bonds can have either the E or Z configuration. a.

c.

b.

26.32 Br Br CHBr3

a.

KOC(CH3)3 Br

COOH

+

b.

CO2CH3

COOH

CH3O2C

Pd(OAc)2 P(o-tolyl)3 (CH3CH2)3N

Br

+

c.

NHCOCH3

Pd(OAc)2 P(o-tolyl)3 (CH3CH2)3N

NHCOCH3

CH3

CH3

d.

B(OCH3)2

O

e.

B

Br

+ O

CH3O

+

Br

Pd(PPh3)4 NaOH Pd(PPh3)4 NaOH

CH3O

O

753

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–13

O

Grubbs

f.

O

catalyst

O

O CH2I2

g.

Zn(Cu) Br Br

h. (CH3)2CuLi

Product in (a) [1] Li

Cl

i.

[2] CuI Br

[3]

O

[1]

j.

H B

H

O CH3CH2 C C H

[2]

Br

H

Pd(PPh3)4 NaOH

26.33 O

Cl

Δ

Cl3C C

Cl

O Na+

C Na+ +

O C O

Cl

Cl

C

+

CO2

+

NaCl

Cl

26.34 This reaction follows the Simmons–Smith reaction mechanism illustrated in Mechanism 26.5. Br CHBr2

Zn(Cu)

[1]

CH Zn

Br

+

[2]

ZnBr2

26.35 Ph2S

O OCH3 Ph2S

a sulfur ylide

1,4-addition of the sulfur ylide to the β carbon

O OCH3

OCH3

a resonance-stabilized enolate

O

methyl trans-chrysanthemate

Ph2S

754

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 26–14

26.36 + N2

A N

N

CO2CH3

N

N

CO2CH3

CO2CH3

diazo compound

CO2CH3

B

26.37 I

PAr3

[1]

a.

I

Ar3P Pd

Pd(PAr3)2 oxidative addition

PAr3

[2]

+

Ar3P Pd

I

syn addition

PAr3 I

Ar3P Pd

PAr3

[3]

H syn elimination

H

H

+

[4] I

reductive elimination

H

E

This H is trans to Pd.

The H cis to Pd is eliminated.

Ar3P Pd

Pd(PAr3)2 + HI

b. This suggests that the stereochemistry in Step [3] must occur with syn elimination of H and Pd to form E. Product F cannot form because the only H on the C bonded to the benzene ring is trans to the Pd species, so it cannot be removed if elimination occurs in a syn fashion. 26.38 O H B

Br NaNH2

O

C CH

Br

O

(– HBr)

B O

(Z)-2-bromostyrene

Pd(PPh3)4 NaOH

A

26.39

Br2 FeBr3 O

O

H

H B O

B O

Br Pd(PPh3)4, NaOH

755

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–15

26.40 HO

SOCl2

OH

HO

Cl TBDMS–Cl imidazole

NaH

HC CH

HC C

TBDMSO

Cl

O B

OH

O

O

Bu4NF

BH O

O HC C

B

OTBDMS

OTBDMS

O

26.41 a. CH3O

CH3O HNO3

NO2

H2SO4

H2

NH2

Pd-C Br2

[1] NaH OH

NaNO2

N2+Cl–

HCl

CH3O

CH3O

[2] CH3Br PBr3

Synthesize these two components, and then use a Heck reaction to synthesize the product.

Br

H2O

OH

Br

CH3OH CH3CH2OH SOCl2 CH3CH2Cl

CH2CH3

AlCl3

CHCH3

hν

KOC(CH3)3

Br

Pd(OAc)2

Br

CH3O

Br2

CH3O

P(o-tolyl)3 (CH3CH2)3N Br

CH3CH2OH SOCl2

Br

CH3CH2Cl

b. CH3O

AlCl3

(from a.) Br CO2Et

CH3O

CH3O CO2Et

Pd(OAc)2 P(o-tolyl)3 (CH3CH2)3N

CO2Et H2 Pd-C

CH3O

Br2

Br

CH3O MgBr

Mg

[1] (2 equiv) [2] H2O

FeBr3 OH

CH3O

756

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 26–16

26.42 Br Br

a.

OH

OH Br Br

NBS

Br

NaOH

CHBr3

OH

hν

OH

KOC(CH3)3

Br

b.

Br

CH2I2

(CH3CH2CH2CH2)2CuLi

Zn(Cu)

(from a.)

Br

c.

CuLi

CH2I2

2

Br

Zn(Cu)

(from a.)

26.43 Br

Br2

a.

FeBr3

CH2I2

CH2=CH2 Pd(OAc)2 P(o-tolyl)3 (CH3CH2)3N

Zn(Cu)

styrene

CHCl3

b. styrene (from a.)

Cl

KOC(CH3)3

Br

Cl

CO2CH3

CO2CH3

c. Pd(OAc)2 P(o-tolyl)3 (CH3CH2)3N

(from a.) Br

d.

[1] Li

Br CuLi

[2] CuI

(from a.)

2

For an alternate synthesis of styrene, see Problem 26.41.

757

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–17

26.44 [1] NaH

a.

C CH

HC CH Cl

[2]

[1] NaH

H2 Lindlar catalyst

[2] Cl

CH2I2 Zn(Cu)

+ enantiomer

[1] NaH

b.

[1] NaH

C CH

HC CH [2]

Na Cl

[2]

NH3

Cl

CH2I2 Zn(Cu)

+ enantiomer

c.

O

CH2I2

CH2

Ph3P=CH2

Zn(Cu)

Br

d.

O

Br2 CH3CO2H

O

LiBr

O

Li2CO3, DMF

[1] (CH3)2CuLi

O

[2] H2O Ph3P=CH2

[1] (CH3)2CuLi

CHCl3

[2] H2O Cl

Cl

KOC(CH3)3

758

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 26–18

26.45 Either compound can be used to

a. CH3

OCH3

CH3

Br

Br

OCH3 synthesize the organoborane, so

two routes are possible. Possibility [1]: CH3Cl AlCl3 Br2

Br2 CH3

[1] Li CH3

FeBr3 HNO3

Br

FeBr3

Br [2] B(OCH3)3

Br

H2

NO2

Br

CH3

B(OCH3)2

NH2

Pd-C

H2SO4

[1] NaNO2, HCl

Br

OH

[2] H2O [1] NaH [2] CH3I Br

Pd(PPh3)4 CH3

B(OCH3)2

Br

OCH3

NaOH

CH3

OCH3

OCH3

Possibility [2]: Br

OCH3

Pd(PPh3)4

[1] Li

(CH3O)2B

OCH3

[2] B(OCH3)3

CH3

Br

NaOH

(from Possibility [1])

(from Possibility [1])

CH3

HO

OCH3

HO

b.

Br

Br

The acidic OH makes it impossible to prepare an organolithium reagent from this aryl halide, so this compound must be used as the aryl halide that couples with the organoborane from bromobenzene. O2N

O2N

HNO3

Br2

H2SO4

FeBr3

Br2 FeBr3

Br

H2N Br

[1] Li

(CH3O)2B

[2] B(OCH3)3 HO Pd(PPh3)4 Br NaOH

[1] NaNO2, HCl Br

Pd-C

HO (CH3O)2B

HO

H2

Br [2] H2O

759

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–19

Br

O

c.

O

Br

O

O

This can't be converted to an organoborane reagent via an organolithium reagent.

three B(OCH3)2

(as in 26.45b)

steps Br2 FeBr3

HNO3 Br

[1] NaNO2, HCl

H2 NO2

Br

H2SO4

Pd-C

Br

NH2

[2] H2O

Br

OH CH3CO Cl pyridine

B(OCH3)2

Br

Pd(PPh3)4

O

O

Br

O

NaOH

O O

O

26.46 EtO2C CO2Et EtO2C CO2Et

Grubbs catalyst

a.

Synthesis of starting material: [1] NaOEt

CH2(CO2Et)2

[2]

CH(COEt)2

EtO2C CO2Et

[1] NaOEt [2]

Cl

Cl SOCl2 OH

SOCl2 OH

OTBDMS

b.

OTBDMS

Grubbs catalyst

Synthesis of starting material: OH

PCC

H

OH

O OH

PBr3

H2O Br

Mg

MgBr

TBDMS–Cl imidazole OTBDMS

760

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 26–20

26.47 O

O

Grubbs catalyst

a.

Synthesis of starting material: O

SOCl2

CrO3 OH

H2SO4 H2O

OH

O

O Cl AlCl3

OH

O

NaBH4

[1] NaOEt

CH3OH

[2]

Cl SOCl2 OH

b.

O O

Grubbs catalyst

O O

Synthesis of starting material:

Br2

Br

MgBr

Mg

OH

FeBr3

OH

H2SO4

O CH2 CH2

PCC

CHO

H2O

mCPBA

OH TsOH O O

761

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–21

26.48 Cl

CH3

CH3O

N

Cl

CH3

CH3O

CO2CH3

N

[1] (CH3CH2CH2CH2)4N+ F—

CO2CH3 [2] PCC

Cl

CH3

CH3O

N

CO2CH3

OTBDMS

LiCu

OTBDMS

I

CHO

2

directed aldol [1] CH3CHO + LDA → −CH2CHO [2] H3O+

O N CH3O

O

O

Cl

O

Cl

O

N

CH3

CH3O

N

CO2CH3

several steps O

CH3O

OH

N H

O

CHO

A

maytansine

26.49 NO2 Br

NH2 Br H2

HNO3

a.

OH Br [1] NaNO2, HCl

Pd-C

H2SO4

OCH3 Br [1] NaH

[2] H2O

OCH3 CuLi

Br [1] Li

[2] CH3I

[2] CuI

2

Br OCH3

OCH3

OH H2O

(2 enantiomers)

H2SO4

OH

b.

[1] OsO4

OTBDMS TBDMS–Cl

HO

Br

Br

[2] NaHSO3, H2O

TBDMSO

imidazole

[1] Li [2] CuI

Br

OTBDMS

Br

OTBDMS

TBDMSO TBDMSO

CuLi 2

(CH3CH2CH2CH2)4N+F– OH HO

(2 enantiomers) O H

HB

c.

O

O B O

Br Pd(PPh3)4, NaOH

762

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 26–22

Cl

Br

d.

Br

AlCl3

CO2CH3

CO2CH3 Pd(OAc)2 P(o-tolyl)3 (CH3CH2)3N

CO2CH3 Br

e.

CO2CH3

Pd(OAc)2 P(o-tolyl)3 (CH3CH2)3N

Br

CO2CH3 H

(+ enantiomer)

[1] Li

f.

H

Zn(Cu) CH2I2

O

mCPBA CuLi

[2] CuI

Br

2

(2 enantiomers)

26.50 O

a.

O

O

O

b. O O

O

O

Δ

Δ

2

[1] LiAlH4 CO2Et

EtO2C

CO2Et

OH

[2] H2O

CO2Et

OH [1] NaH (2 equiv) Cl (2 equiv) [2]

O O

26.51 There is more than one way to form Z by metathesis reactions. One possibility involves ring opening of the bicyclic alkene followed by successive ring closures to generate the five- and seven-membered rings. O

O O

RCM

H

O

O

H

O

RCM

O

H

O Ru

H

Z ring open

763

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbon–Carbon Bond-Forming Reactions in Organic Synthesis 26–23

26.52 I

O

O

+ A

Pd(PPh3)4

B O

N H

OCH2CH3

N

NaOH

B

C

OCH2CH3

OCH2CH3 H3O+

H

D C11H11NO

N O

O

OCH2CH3

H OH2

H OH2 NH

OCH2CH3

NH

O

OCH2CH3

N H + H2O

O

N + H2O

O

O

C H

H

H2O N

N

D

OCH2CH3

O H3O

N O

O

+ H2O

resonance stabilized

+

H OCH2CH3

26.53 O CH3O

O PPh3

PPh3 CH3O

H O

X

CH3O H O

Y

+ Ph3P=O

O Ph3P CH3O

CH3O

O Ph3P

O

+ PPh3

26.54 a. Reaction of a terminal alkene with the catalyst forms a metal–carbene that undergoes an intramolecular reaction with the triple bond, generating a new metal–carbene. A second intramolecular reaction forms the bicyclic product. OSiEt3

OSiEt3

OSiEt3

OSiEt3

M M

A

cyclization

two new C=C's

cyclization B

764

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 26–24

b. Two products are possible because the cascade of reactions can begin at two different double bonds. OSiEt3

OSiEt3 OSiEt3

M

OSiEt3

M

C

Begin here. OSiEt3

OSiEt3

OSiEt3 M M

C Begin here.

OSiEt3

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

765

Pericyclic Reactions 27–1

Chapter 27 Pericyclic Reactions Chapter Review Electrocyclic reactions (27.3) Woodward–Hoffmann rules for electrocyclic reactions Number of π bonds Even Odd

Thermal reaction Conrotatory Disrotatory

Photochemical reaction Disrotatory Conrotatory

Examples The stereochemistry of a thermal electrocyclic reaction is opposite to that of a photochemical electrocyclic reaction. CH3 • A thermal electrocyclic reaction with an even Δ number of π bonds occurs in a conrotatory CH3 fashion. CH3 + enantiomer

CH3

CH3

hν

(2E,4E)-2,4-hexadiene 2 π bonds

CH3

• A photochemical electrocyclic reaction with an even number of π bonds occurs in a disrotatory fashion.

Cycloaddition reactions (27.4) Woodward–Hoffmann rules for cycloaddition reactions Number of π bonds Even Odd

Thermal reaction Antarafacial Suprafacial

Photochemical reaction Suprafacial Antarafacial

Examples [1] A thermal [4 + 2] cycloaddition takes place in a suprafacial fashion with an odd number of π bonds. An antarafacial photochemical [4 + 2] cycloaddition to form a six-membered ring cannot occur, because of the geometrical constraints of forming a six-membered ring. CH3

H

H

+ CH2

CH2

CH3

Δ suprafacial

CH3 H

CH3 H

cis product only

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 27–2

[2] A photochemical [2 + 2] cycloaddition takes place in a suprafacial fashion with an even number of π bonds. An antarafacial thermal [2 + 2] cycloaddition to form a four-membered ring cannot occur, because of the geometrical constraints of forming a four-membered ring. C6H5

C6H5

C H CH2

C

+

C6H5

hν

C6H5

suprafacial

H

cis product only

CH2

Sigmatropic rearrangements (27.5) Woodward–Hoffmann rules for sigmatropic rearrangements Number of electron pairs Even Odd

Thermal reaction Antarafacial Suprafacial

Photochemical reaction Suprafacial Antarafacial

Examples [1] A Cope rearrangement is a thermal [3,3] sigmatropic rearrangement that converts a 1,5-diene into an isomeric 1,5-diene. Δ

isomeric 1,5-diene

1,5-diene

[2] An oxy-Cope rearrangement is a thermal [3,3] sigmatropic rearrangement that converts a 1,5-dien-3-ol into a δ, ε-unsaturated carbonyl compound, after tautomerization of an intermediate enol. O

HO

Δ [3,3]

1,5-dien-3-ol

δ,ε-unsaturated carbonyl compound

[3] A Claisen rearrangement is a thermal [3,3] sigmatropic rearrangement that converts an unsaturated ether into a γ,δ-unsaturated carbonyl compound. Δ O

unsaturated ether

O

γ,δ-unsaturated carbonyl compound

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

767

Pericyclic Reactions 27–3

Practice Test on Chapter Review 1. a. Which of the following pericyclic reactions is symmetry allowed and readily occurs? 1. a photochemical conrotatory electrocyclic ring closure of a conjugated triene 2. a disrotatory thermal electrocyclic ring opening of a substituted cyclohexadiene 3. a thermal [2 + 2] cycloaddition 4. Reactions [1] and [2] will both occur. 5. Reactions [1], [2], and [3] will all occur. b. Which of the following reactions requires suprafacial stereochemistry to be symmetry allowed? 1. a photochemical [1,5] sigmatropic rearrangement 2. a thermal [8 + 2] cycloaddition 3. a photochemical [4 + 2] cycloaddition 4. Reactions [1] and [2] are both suprafacial. 5. Reactions [1], [2], and [3] are all suprafacial. c. What product(s) are formed from the photochemical [2 + 2] cycloaddition of (3E)-3-hexene?

A

1. 2. 3. 4. 5.

B

C

A only B only C only A and B A, B, and C

d. What product(s) are formed from the photochemical electrocyclic ring opening of cis-3,4-dimethylcyclobutene? 1. 2. 3. 4. 5.

(2E,4E)-2,4-hexadiene (2E,4Z)-2,4-hexadiene (3Z)-1,3,5-hexatriene Compounds [1] and [2] are both formed. Compounds [1], [2], and [3] are all formed.

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 27–4

e. What product(s) are formed by the [3,3] sigmatropic rearrangement of 1,5-cyclodecadien-3-ol? OH

A

1. 2. 3. 4. 5.

CHO

CHO

B

C

A only B only C only A and B A, B, and C

2. Consider the p orbitals of the terminal carbons of a conjugated polyene with like phases on the same side of the molecule (as in A) or opposite sides of the molecule (as in B), and answer each question.

A

B

a. Which drawing is consistent with the ground state HOMO of a conjugated triene? b. Which drawing is consistent with the excited state LUMO of a conjugated diene? c. Which drawing is consistent with the ground state LUMO for a conjugated tetraene? 3. What type of sigmatropic rearrangement is depicted in each reaction? a.

b.

Answers to Practice Test 1. a. 4 b. 2 c. 4 d. 1 e. 3

2. a. A b. B c. A

3. a. [3,3] b. [1,3]

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

769

Pericyclic Reactions 27–5

Answers to Problems 27.1

Use the following definitions: • An electrocyclic ring closure is an intramolecular reaction that forms a cyclic product containing one more σ bond and one fewer π bond than the reactant. An electrocyclic ring opening is a reaction in which a σ bond of a cyclic reactant is cleaved to form a conjugated product with one more π bond. • A cycloaddition is a reaction between two compounds with π bonds that forms a cyclic product with two new σ bonds. • A sigmatropic rearrangement is a reaction in which a σ bond is broken in the reactant, the π bonds rearrange, and a σ bond is formed in the product. electrocyclic reaction

a. 2 π bonds

σ bond broken

3 π bonds

O

O

σ bond formed

+

b.

H2C CH2

cycloaddition σ bond formed σ bond formed

c. 4 π bonds

3 π bonds

d.

σ bond formed 1 π bond σ bond broken

27.2

electrocyclic reaction

sigmatropic rearrangement

1 π bond

a. For a bonding molecular orbital, the number of bonding interactions is greater than the number of nodes. b. For an antibonding molecular orbital, the number of bonding interactions is less than the number of nodes. For 1,3-butadiene: ψ1 ψ2 ψ 3* ψ 4*

Bonding 3 2 1 0

Nodes 0 1 2 3

Type of MO bonding MO bonding MO antibonding MO antibonding MO

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 27–6

The molecular orbitals of all conjugated dienes look similar.

27.3

a.

c. Excited state

b. Ground state ψ4∗

excited state LUMO

ψ3∗

excited state HOMO

ground state LUMO

Energy

770

ψ2 ground state HOMO

ψ1

a. There are 10 molecular orbitals from the 10 p orbitals of the five π bonds. b. Five molecular orbitals are bonding and five molecular orbitals are antibonding. c. The lowest energy molecular orbital (ψ1) has zero nodes.

27.4

ψ1

d. The highest energy molecular orbital (ψ10*) has nine nodes. ψ10∗

To draw the product of an electrocyclic reaction, use curved arrows and begin at a π bond. Move the π electrons to an adjacent carbon–carbon bond, and continue in a cyclic fashion.

27.5

a.

b.

Δ

c.

Δ

re-draw

Δ

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

771

Pericyclic Reactions 27–7

27.6

Thermal electrocyclic reactions occur in a disrotatory fashion for a conjugated polyene with an odd number of π bonds, and in a conrotatory fashion for a conjugated polyene with an even number of π bonds. C6H5

C6H5

Δ

a.

disrotatory

C6H5

Δ

b.

conrotatory

C6H5

2 π bonds

3 π bonds

27.7

For an even number of π bonds, thermal electrocyclic reactions occur in a conrotatory fashion. re-draw

Δ

re-draw

Δ

+ enantiomer

a.

b.

27.8

Photochemical electrocyclic reactions occur in a conrotatory fashion for a conjugated polyene with an odd number of π bonds, and in a disrotatory fashion for a conjugated polyene with an even number of π bonds. C6H5

C6H5

hν

a.

conrotatory

C6H5

hν

b.

disrotatory

C6H5

+ enantiomer

27.9

[The (E,E) diene is favored over the (Z,Z) diene.]

The photochemical electrocyclic reaction cleaves a six-membered ring to form a hexatriene.

hν H

H

H

HO

HO provitamin D3

7-dehydrocholesterol

27.10 Use the rules for electrocyclic reactions found in Answers 27.6 and 27.8. Δ a.

re-draw

disrotatory 3 π bonds

hν conrotatory

+ enantiomer

772

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 27–8

Δ

re-draw b.

+ enantiomer

disrotatory 3 π bonds

hν conrotatory

27.11 Use the rules for electrocyclic reactions found in Answer 27.6. A reaction with three π bonds and a disrotatory cyclization is thermal. H

H

27.12 Count the number of π electrons in each reactant to classify the cycloaddition. O

O

[2 + 2] cycloaddition

CH2

+

a.

CH2

2π electrons

2π electrons

(one possibility)

O

O

b.

[4 + 2] cycloaddition

CH2 CH2

4π 2π electrons electrons O

c.

O

[6 + 2] cycloaddition

CH2 CH2

6π 2π electrons electrons

27.13 A thermal suprafacial addition is symmetry allowed in a [4 + 2] cycloaddition because like phases interact. LUMO of the diene

Like phases interact.

HOMO of the alkene

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

773

Pericyclic Reactions 27–9

27.14 A thermal [4 + 2] cycloaddition is suprafacial.

a.

CH2

re-draw

+ CH2 CH2

Δ

CH2

(E,E) diene

CN

b.

NC

re-draw

+

CN

Δ

CN CN

+ enantiomer

CN

(E,Z) diene

27.15 CO2CH3 H O CO2CH3

a.

O

Δ

H

endo addition

H

O

X

O

CO2CH3 HH O

CO2CH3

b.

diene HOMO

H O

H H

O

dienophile LUMO

X

H

O

The dienophile is under the diene, by the rule of endo addition (Section 16.13). The H's at the ring fusion are cis to each other, but trans to the CO2CH3 group.

27.16 A photochemical [2 + 2] cycloaddition is suprafacial. C6 H5

a.

hν

+

C6 H5

C 6H 5

+ enantiomer C 6H 5

O

O H

H

b.

+ CN

CH2 CH2

+ enantiomer CN

27.17 a. The photochemical [6 + 4] cycloaddition involves five π bonds (the total number of π electrons divided by two) and is antarafacial. b. A thermal [8 + 2] cycloaddition involves five π bonds and is suprafacial.

774

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 27–10

27.18 A photochemical [4 + 2] cycloaddition like the Diels–Alder reaction must proceed by an antarafacial pathway. This would require either the 1,3-diene or the alkene component to twist 180° in order for the like phases of the p orbitals to overlap. Such a rotation is not possible in the formation of a six-membered ring. excited state HOMO of the diene

180° twist required

LUMO of the alkene

27.19 Locate the σ bonds broken and formed, and count the number of atoms that connects them. D 1

3

a.

2 1

D1 D

2

3

1

D

σ bond formed

σ bond broken [1,3] sigmatropic rearrangement

1

2

3

1

3

1

2

3

b.

σ bond formed 1

2

2

3

σ bond broken [3,3] sigmatropic rearrangement

27.20 D 2

a.

CD3

re-draw

1

3 4

D D1

3

1

4

7 5

D

2

[1,7]

6

D 7

5

D1

6

b, c. The reaction involves four electron pairs (three π bonds and one σ bond), so it proceeds by an antarafacial pathway under thermal conditions, and by a suprafacial pathway under photochemical conditions. 27.21 Draw the products of each reaction. CO2CH3

a.

re-draw

CH3O2C

[3,3]

CH3O2C

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

775

Pericyclic Reactions 27–11

[3,3]

b.

O

OH

[3,3]

tautomerize

c.

OH

27.22 Draw the product after protonation. HO

O

[3,3]

base

protonation

O

O

27.23 Draw the product of Claisen rearrangement. CHO

[3,3]

a.

[3,3] c. O

OHC

O

[3,3]

b.

O CHO

27.24 O

a.

O O

O

RO

O

Z

O

O

O

O

O

RO

O

b.

O

O

O

O O

O

O

O

O

+ metathesis

RO

RO

CH2 CH2

776

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 27–12

27.25 Predict the stereochemistry of each reaction using Table 27.4. a. A [6 + 4] thermal cycloaddition involves five electron pairs, making the reaction suprafacial. b. A photochemical electrocyclic ring closure of 1,3,5,7,9-decapentaene involves five electron pairs, making the reaction conrotatory. c. A [4 + 4] photochemical cycloaddition involves four electron pairs, making the reaction suprafacial. d. A thermal [5,5] sigmatropic rearrangement involves five electron pairs, making the reaction suprafacial. 27.26 Use the rules found in Answers 27.6 and 27.8.

A

Δ

Δ

(a) disrotatory

(a) disrotatory

3 π bonds

B

hν

hν

(b) conrotatory

(b) conrotatory

3 π bonds

27.27 Draw the product of [3,3] sigmatropic rearrangement of each compound. OH

O

a. [3,3]

tautomerization

OH

O

O

OH

b. [3,3]

tautomerization

27.28 An electrocyclic reaction forms a product with one more or one fewer π bond than the starting material. A cycloaddition forms a ring with two new σ bonds. A sigmatropic rearrangement forms a product with the same number of π bonds, but the π bonds are rearranged. Use Table 27.4 to determine the stereochemistry. H

thermal electrocyclic ring closure • 3 π bonds • disrotatory

Δ

a.

H

3 π bonds

2 π bonds

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

777

Pericyclic Reactions 27–13

Δ

H

b.

hν

2 π bonds

2 π bonds

thermal [1,5] sigmatropic rearrangement • 3 electron pairs (2 π + 1 σ) • suprafacial

c.

3 π bonds

photochemical electrocyclic ring opening • 2 π bonds in diene formed • disrotatory

2 new σ bonds

Δ

N(CH2CH3)2

N(CH2CH3)2

thermal [6 + 4] cycloaddition • 5 π bonds • suprafacial

27.29 Use the rules for thermal electrocyclic reactions found in Answer 27.6. E Δ

a.

disrotatory

Z

Δ

b. re-draw

Z

conrotatory

2 π bonds

3 π bonds

27.30 Use the rules for photochemical electrocyclic reactions found in Answer 27.8. E a.

hν conrotatory

Z E 3 π bonds

Although conrotatory ring opening could also form, at least in theory, an all-(Z) triene, steric hindrance during ring opening would cause the terminal CH3’s to crash into one another, making this process unlikely.

hν

b. re-draw

disrotatory + enantiomer

778

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 27–14

27.31 Use the rules found in Answer 27.8. hν a.

re-draw

disrotatory

4 π bonds

hν

b. re-draw

4 π bonds

+ enantiomer

disrotatory

27.32 Use the rules found in Answers 27.6 and 27.8. 2E a, b.

Δ 4Z

+ enantiomer

disrotatory 6Z 3 π bonds

hν

+ enantiomer

conrotatory

E

Δ

c.

disrotatory

E 3 π bonds

(one enantiomer)

E

hν conrotatory

d.

E 3 π bonds

(one enantiomer)

27.33 The trans product is indicative of a disrotatory ring closure from the cyclic triene with the given stereochemistry at the double bonds. A disrotatory ring closure with a polyene having three π bonds must occur under thermal conditions. E

Z

H

H

trans

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

779

Pericyclic Reactions 27–15

27.34 A disrotatory cyclization of a reactant with an even number of π bonds must occur under photochemical conditions. hν H H

cis

M 2 π bonds

27.35 Use the rules found in Answers 27.6 and 27.8. H

H

Δ a, c.

hν

b, c.

disrotatory

conrotatory

H

N 3 π bonds

H

N 3 π bonds

cis

trans + enantiomer

27.36 Use the rules found in Answers 27.6 and 27.8. H

E

Δ conrotatory H

Z Q 2 π bonds

P hν disrotatory

Z (A 10-membered ring cannot contain two E double bonds.) Z

R

27.37 Since the reaction involves three π bonds in one reactant and two π bonds in the second reactant, the reaction is a [6 + 4] cycloaddition. A suprafacial cycloaddition with five π bonds must proceed under thermal conditions. two new σ bonds formed on the same side

[1] + O

O O

O C6H5 C6H5

[2] +

C6H5 C6H5

780

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 27–16

27.38 The Diels–Alder reaction is a thermal, suprafacial [4 + 2] cycloaddition. O

a.

+

H

Δ

O

O

O

O H

O

+

b.

H

Δ

O

O

O H

O

O

O

+ enantiomer

27.39 A photochemical [2 + 2] cycloaddition is suprafacial. hν

a. 2

b.

hν

2

27.40 A thermal [4 + 2] cycloaddition is suprafacial.

CO2CH3

a.

CO2CH3

Δ

CO2CH3

C

c.

C

CO2CH3

CO2CH3

CO2CH3

CO2CH3

b. CO2CH3

Δ

CO2CH3

Δ CH3O2C

CO2CH3 CO2CH3

27.41 1,3-Butadiene can react with itself in a symmetry-allowed thermal [4 + 2] cycloaddition to form 4-vinylcyclohexene. [4 + 2] Δ 4-vinylcyclohexene

1,5-Cyclooctadiene would have to be formed from 1,3-butadiene by a [4 + 4] cycloaddition, which is not allowed under thermal conditions. Δ [4 + 4] 1,5-cyclooctadiene

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

781

Pericyclic Reactions 27–17

27.42 A series of three [2 + 2] cycloadditions with E alkenes forms X.

CO2CH2CH3

CO2CH2CH3

[2 + 2]

CO2CH2CH3

X

CO2CH2CH3

27.43 Re-draw the reactant and product to more clearly show the relative location of the bonds broken and formed. 2

1

3

2

1

3

a. re-draw

1

[3,3]

3

2

1

D

5

5 4

b. D re-draw

3

1

2

[1,5]

H

3

2

O

SC6H5

re-draw

1

2

1

3

c.

CD2H

4

D

D

3

2

2 3

1

O S2 1

[2,3]

2

1

S

O

1

C6H5

C6H5

27.44 Draw the products of each reaction.

H

[3,3]

a.

[3,3]

c.

O H

O

[3,3]

b.

O

tautomerize OH

OH

27.45 a. Two [1,5] sigmatropic rearrangements occur. H1

[1,5]

1

2 3

3

2 1

5 4

5-methyl1,3-cyclopentadiene

H

[1,5]

H1

4 5

1-methyl1,3-cyclopentadiene

H H

2-methyl1,3-cyclopentadiene

782

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 27–18

b. H 1 2 3

[1,3]

H H

5-methyl isomer

2-methyl isomer

A [1,3] sigmatropic rearrangement requires photochemical conditions not thermal conditions, so 5-methyl-1,3-cyclopentadiene cannot rearrange directly to its 2-methyl isomer by a [1,3] shift. 27.46 3 4

3

1 2

2

O

4

4

1

3

5

4

2

[5,5]

5

1

O

OH

H 5

3

1 2

tautomerization

5

27.47 O OH

+

O Cl

O

LDA

O

pyridine

O

O

O

B

A

re-draw OH

The conversion of B to C is a [3,3] sigmatropic rearrangement of an intermediate enolate.

O

H3O+

CO2H

re-draw

O

O

O

[3,3] C

27.48 Use the definitions found in Answer 27.1. Δ

an electrocyclic ring opening

[1] 2 π bonds 3 4

3 1

5

3 π bonds

2

7

Δ H [2] 1

4

6

6

3 π bonds

2 1

5

7

3 π bonds Δ

an electrocyclic ring closure

[3] 3 π bonds

a [1,7] sigmatropic rearrangement

H

2 π bonds

O

Δ

783

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Pericyclic Reactions 27–19

27.49 CO2CH3

a.

CH3O2C

+

CO2CH3

D

hν

CH2 CH2

disrotatory

D

D

H

[4 + 2]

c.

Δ

D

H

D

hν

d.

CO2CH3

D

Δ

b.

+ enantiomer

[2 + 2]

D

conrotatory

D

27.50 O

O

Δ

a.

[3,3] Claisen

HO

O

[1] KH

b.

[2] H3O+

anionic oxy-Cope O

18-crown-6

O

1 2

c.

d.

O 1

Δ

2 3

3

Claisen

hν [2 + 2] C7H8

O

OH

tautomerization

784

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 27–20

27.51 The mechanism consists of sequential [3,3] sigmatropic rearrangements, followed by tautomerization. O

O

O

OH

H+ source [3,3]

[3,3] H

H

OH OH

H B

27.52 O

H+

B O

H

HO

HB+ O

O O

O

O

[3,3] O

H+

O

O

O

allyl vinyl ether H

O H B

+

HB

OH

H

27.53 Z

H

Δ H

conrotatory electrocyclic ring opening forming 4 π bonds

H

Δ E

disrotatory ring closure involving only 3 of the 4 π bonds

H

785

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Pericyclic Reactions 27–21

27.54 SCH3

SCH3

Δ

OCH3

OCH3

electrocyclic ring opening forming two π bonds

O

CH3S

H

Δ

CH3S

[4 + 2] cycloaddition suprafacial

CH3O

O CH3O

O

O O

H

O

27.55 The mechanism consists of a [4 + 2] cycloaddition, followed by intramolecular imine formation. H H NH2

O

NH2

O

H

N

O

N

H HO

[4 + 2]

O

O

H

O

proton transfer

H

O

H

H H2O

H

N

O

H

O

H

O

H3O

27.56 This bridged bicyclic system is cleaved. HO H

=

H

H

OH OH

H H

[3,3]

This bond forms a new bridged ring system.

N

H2O

N

H

H

786

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Chapter 27–22

27.57 O

O

O

B O H

O

OCH2CH3

O

OCH2CH3 O

+ HB+ H OH O

O

O

O O

O

O O

O

CH3CH2O

H B O

H

CH3CH2OH

B O O O

H

H2O

O

O

OH H OH

CO2

O

H

HO–

787

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Pericyclic Reactions 27–23

27.58 Conrotatory cyclization of Y using four π bonds forms X. Disrotatory ring closure of X can occur in two ways—on the top face or bottom face of the eight-membered ring to form diastereomers. Δ C6H5

C6H5

4 π bonds conrotatory

CO2CH3

CO2CH3

+ enantiomer

X

Y

disrotatory

H

C6H5

diene

=

H

H

H

H

C6H5 CO2CH3

C6H5

CO2CH3

CO2CH3

endiandric acid E methyl ester

dienophile

endiandric acid D methyl ester

Only this stereoisomer has the side chain close to the sixmembered ring for Diels–Alder. [4 + 2] C6H5 H H

H H

H

Endiandric acid A has a free COOH group instead of the CO2CH3 group.

methyl ester of endiandric acid A

H CO2CH3

27.59 O

O

a.

O

O

O

O

C6H5

dienophile

O

diene dienophile O

b.

or O O

diene

C6H5

or C6H5

diene

O

C6H5

dienophile

O O

dienophile O

diene

O

O

788

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 27–24

O2CCH3 O2CCH3 O2CCH3

O C4H9

O

O

[4 + 2]

c.

C4H9

O

[3,3]

cycloaddition O2CCH3 C4H9

O2CCH3

sigmatropic rearrangement

O2CCH3

O C4H9

O

C4H9 C4H9

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

789

Carbohydrates 28–1

Chapter 28 Carbohydrates Chapter Review Important terms A monosaccharide containing an aldehyde (28.2) • Aldose A monosaccharide containing a ketone (28.2) • Ketose A monosaccharide with the O bonded to the stereogenic center farthest from the • D-Sugar carbonyl group drawn on the right in the Fischer projection (28.2C) Two diastereomers that differ in configuration around one stereogenic center • Epimers only (28.3) Monosaccharides that differ in configuration at only the hemiacetal OH • Anomers group (28.6) • Glycoside An acetal derived from a monosaccharide hemiacetal (28.7) Acyclic, Haworth, and 3-D representations for D-glucose (28.6) CHO CH2OH O H H OH

H

Haworth projection

H H

HO

OH

HO H

OH

CH2OH O H H OH

OH H

H

H

OH

H

OH

H

HO H

CH2OH

α anomer OH H

HO HO

3-D representation

H

HO HO

H C OH

OH OH

OH H

HO C H H

H

H

H C OH O

OH

β anomer

acyclic form CHO

H

OH

OH H

H C OH

O

H

OH

H

CH2OH

Reactions of monosaccharides involving the hemiacetal [1] Glycoside formation (28.7A) OH

OH O

HO HO

ROH HCl

OH

α-D-glucose

OH

OH

•

OH

O

HO HO

O

+ HO HO

OR OH

OR

•

β glycoside

α glycoside

Only the hemiacetal OH reacts. A mixture of α and β glycosides forms.

[2] Glycoside hydrolysis (28.7B) OH

OH HO HO

O OH

H 3O +

HO HO

OH O

OH OR

HO + HO

OH OH

OH

β anomer

α anomer

+

O

ROH

•

A mixture of α and β anomers forms.

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–2

Reactions of monosaccharides at the OH groups [1] Ether formation (28.8) OR

OH O

HO HO

OH OH

RX

• •

O

RO RO

Ag2O

All OH groups react. The stereochemistry at all stereogenic centers is retained.

OR OR

[2] Ester formation (28.8) OH HO HO

O

Ac2O or OH

AcCl

OH

OAc O

AcO AcO

• •

OAc

All OH groups react. The stereochemistry at all stereogenic centers is retained.

AcO

pyridine

Reactions of monosaccharides at the carbonyl group [1] Oxidation of aldoses (28.9B) CHO

COOH

COOH

•

[O]

or

• CH2OH

CH2OH

aldose

aldonic acid

Aldonic acids are formed using: • Ag2O, NH4OH • Cu2+ • Br2, H2O Aldaric acids are formed with HNO3, H2O.

COOH

aldaric acid

[2] Reduction of aldoses to alditols (28.9A) CHO

CH2OH NaBH4 CH3OH

CH2OH

CH2OH

aldose

alditol

[3] Wohl degradation (28.10A) This C–C bond is cleaved.

•

CHO H

OH

CHO

•

[1] NH2OH [2] Ac2O, NaOAc [3] NaOCH3 CH2OH

• CH2OH

The C1–C2 bond is cleaved to shorten an aldose chain by one carbon. The stereochemistry at all other stereogenic centers is retained. Two epimers at C2 form the same product.

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

791

Carbohydrates 28–3

[4] Kiliani–Fischer synthesis (28.10B) CHO CHO

H

CHO

OH

HO

[1] NaCN, HCl

H

•

+

[2] H2, Pd-BaSO4

•

[3] H3O+ CH2OH

CH2OH

One carbon is added to the aldehyde end of an aldose. Two epimers at C2 are formed.

CH2OH

Other reactions [1] Hydrolysis of disaccharides (28.12) O

H3O+

This bond is cleaved.

O

O

+

OH

O

O

OH

OH

A mixture of anomers is formed.

[2] Formation of N-glycosides (28.14B) CH2OH O H H OH

H

RNH2

H

mild

OH OH

H+

CH2OH O H H OH

H +

H

NHR OH

CH2OH O H H OH

NHR H H OH

•

Two anomers are formed.

Practice Test on Chapter Review 1. a. How are the following two representations related to each other?

HO HO

CH2OH O

H OH

HO

A

H

CH2OH O H OH H

OH

H OH

H

OH

B

1. A and B are anomers of each other. 2. A and B are epimers of each other. 3. A and B are diastereomers of each other. 4. Statements [1] and [2] are both true. 5. Statements [1], [2], and [3] are all true.

b. Which of the following statements is (are) true about monosaccharide C? 1. C is a D-sugar. OH 2. The β anomer is drawn. HOO HO 3. C is an aldohexose. OH 4. Statements [1] and [2] are both true. HO C 5. Statements [1], [2], and [3] are all true.

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–4

c. Which of the following are different representations for monosaccharide D? CHO H

OH

H

OH

HO

HO

H

OH

O

1.

H

CH2OH

OH

2.

HO HO OH

OH

O H H

HO

OH OH

H

H

OH O

3.

H

OH

CH2OH

HO

OH

OH

4. Both [1] and [2] are representations for D.

D

5. Compounds [1], [2], and [3] all represent D. d. Which aldoses give an optically active compound upon reaction with NaBH4 in CH3OH? CHO H

1. HO HO

CHO

OH

H

H

2. HO

H

H

CH2OH

CHO

OH H

3.

OH

H

OH

H

OH

H

CH2OH

4. Both [1] and [2] give an optically active product.

OH CH2OH

5. Compounds [1], [2], and [3] all give optically active products.

2. Answer each question about monosaccharide D as True (T) or False (F). OH HO O

HO HO

OH

D

D is a D-sugar. D is drawn as an α anomer. D is an aldohexose. Reduction of D with NaBH4 in CH3OH forms an optically inactive alditol. e. Oxidation of D with Br2, H2O forms an optically active aldonic acid.

f. Oxidation of D with HNO3 forms an optically active aldaric acid. g. C2 has the R configuration. h. Treatment of D with CH3OH, HCl forms two products. i. Treatment of D with Ag2O, and CH3I (excess) forms two products. j. An epimer of D at C3 has an axial OH group.

a. b. c. d.

3. Answer the following questions about the three monosaccharides (A–C) drawn below. CHO H

OH

HO HO H

a. b. c. d.

CHO HO

H

H

H

OH

H

H

OH

OH

H

OH

CH2OH

CH2OH

A

B

CH2OH HO

O

H H

OH

H H

H OH

OH

C

Draw the α anomer of A in a Haworth projection. Draw the β anomer of B in a three-dimensional representation using a chair conformation. Convert C into the acyclic form of the monosaccharide using a Fischer projection. What two aldoses yield A in a Wohl degradation?

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbohydrates 28–5

4. Draw the product of each reaction with the starting material D-xylose. a. b. c. d. e.

CHO H HO H

OH H OH CH2OH

CH3OH, HCl NaBH4, CH3OH Br2, H2O [1] NaCN, HCl; [2] H2, Pd-BaSO4; [3] H3O+ Ac2O, pyridine

D-xylose

Answers to Practice Test 1. a. 5 b. 5 c. 4 d. 1

3.

4. CH2OH HO

a.

CH2OH O

H OH

OH H

b. HO

c.

OH

HO H

OCH3 OH

CH2OH

b.

HO H

HO CHO H

H

H OH

OH

H

+ β anomer

OH OH HO O

H

c. HO

OH

H

CHO

CHO

OH

HO

H

OH

H

HO

H

H

OH CH2OH

OH H OH CH2OH

H

H

H OH COOH

H

H

HO

OH

CH2OH

CH2OH

d.

H

O

OH H

H

H

2. a. T b. F c. T d. F e. T f. T g. F h. T i. F j. T

a.

H

+

HO

OH

HO

H

HO

H

H

CHO

H

d.

H HO H

OH

CHO

H

H

OH

OH

H

OH

H

+ HO

OH

H

CH2OH

CH2OH

CH2OAc O OAc

e.

H H

H OH CH2OH

H H

+ β anomer

OAc OAc

Answers to Problems 28.1

A ketose is a monosaccharide containing a ketone. An aldose is a monosaccharide containing an aldehyde. A monosaccharide is called: a triose if it has three C’s; a tetrose if it has four C’s; a pentose if it has five C’s; a hexose if it has six C’s, and so forth.

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–6

CHO

CH2OH C O

a. a ketotetrose

b. an aldopentose

H C OH

H C OH

H C OH H C OH CH2OH

CH2OH

Rotate and re-draw each molecule to place the horizontal bonds in front of the plane and the vertical bonds behind the plane. Then use a cross to represent the stereogenic center in a Fischer projection formula.

28.2

COOH

COOH

a. CH3 C OH

=

CH3

CHO HO CH3

OHC

OH

HO H

C

CH3

CHO

CH2CH3

re-draw

H

C

HOCH2 H

CHO

re-draw

C

C

c.

CH2CH2OH

CH2CH2OH

CHO

= HO

CH3

CHO

re-draw

d. H C CHO

HO

C

CH3

CH2OH CH2CH3

CH2CH3

OH

H

H

CHO

CH2OH = H

CHO

= HO

CH3

CH3 H

H

For each molecule: [1] Convert the Fischer projection formula to a representation with wedges and dashes. [2] Assign priorities (Section 5.6). [3] Determine R or S in the usual manner. Reverse the answer if priority group [4] is oriented forward (on a wedge).

28.3

CH2NH2

a. Cl

[1]

H

[1]

Cl C H

CH2NH2

[2]

H

CH2NH2

CH2NH2

[3]

1 Cl C CH2Br 2

1

H

H

2

[2]

CHO

CHO

[3]

1 Cl C H 4

1

[1]

3

[2]

Cl C H

CH2OH

d. Cl

CH2Br H

[1]

COOH Cl C CH2Br H

[2]

2 1 Cl C H

CH2OH

CH2OH

3

3

COOH

1 Cl C CH2Br 2 H

4

S configuration

CHO

[3]

3

COOH

H forward

3

1 Cl C H 4

CH2OH

Cl C H CH2NH2

CH2NH2

CHO

S configuration

2

2 CHO

Cl C CH2Br 2

4

CH2NH2

CHO Cl

CHO

3

3

H

CHO

b. Cl

CH2NH2 Cl C CH2Br

CH2Br H

c.

c. an aldotetrose

H C OH

CH2OH

b.

CHO

H C OH

H forward S configuration

3 COOH

[3] 1

Cl C CH2Br 2 H

S configuration

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbohydrates 28–7

28.4 OHC HO H H OH HO

HO

rotate

CHO

H OH

H H HO

HO H HO H

OHC

H OH

Convert staggered to eclipsed.

HO

OH H OH H OH H H OH

CHO

re-draw

CHO

HO

H

HO

H

HO

H

HO

H

HO

H

H

=

OH

HO H

CH2OH

H OH CH2OH

OH

28.5 R

CHO

S

H C OH

R

HO C H

R

H C OH H C OH CH2OH D-glucose

A D sugar has the OH group on the stereogenic center farthest from the carbonyl on the right. An L sugar has the OH group on the stereogenic center farthest from the carbonyl on the left.

28.6

CHO

a.

CHO

H

OH

HO

H

OH

H

HO

H

HO

CH2OH

CHO

H

HO

OH

HO

H

C

OH group on the left: L sugar

b. A and B are diastereomers. A and C are enantiomers. B and C are diastereomers.

28.7

There are 32 aldoheptoses; 16 are D sugars. C2 R

CHO

CHO

CHO

CHO

C3 R H H

OH

H

OH

H

OH

H

OH

OH

H

OH

H

OH

H

OH

H

OH

HO

H

HO

H

H

OH

H

OH

HO

H

H

OH

HO

H

OH

H

OH

H

OH

H

CH2OH

CH2OH

CH2OH

OH CH2OH

B

A

H

H

CH2OH

OH group on the left: L sugar

H

H OH CH2OH

OH group on the right: D sugar

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–8

28.8

Epimers are two diastereomers that differ in the configuration around only one stereogenic center. epimers CHO

28.9

a. b. c. d. e. f.

CHO

H

OH

HO

H

OH

H

CHO

H

and

OH

CH2OH

CH2OH

D-erythrose

D-threose

H HO

OH H CH2OH

L-threose

D-allose and L-allose: enantiomers D-altrose and D-gulose: diastereomers but not epimers D-galactose and D-talose: epimers D-mannose and D-fructose: constitutional isomers D-fructose and D-sorbose: diastereomers but not epimers L-sorbose and L-tagatose: epimers

28.10 a.

HO H H

CH2OH

CH2OH

CH2OH

C O

C O

C O

H

H

OH

HO

OH

HO

CH2OH

b.

OH

HO

H

H

HO

H

H

H

CH2OH

D-fructose

CH2OH C O

c.

HO

OH

OH

HO

CH2OH

H CH2OH

D-tagatose

L-fructose

H

H

L-sorbose

enantiomers

28.11 CH2OH

S

CH2OH

S

C O

C O

H

HO

H

OH

HO

H

OH

H

HO

H H OH CH2OH

CH2OH

D-tagatose

D-fructose

28.12 Step [1]: Place the O atom in the upper right corner of a hexagon, and add the CH2OH group on the first carbon counterclockwise from the O atom. Step [2]: Place the anomeric carbon on the first carbon clockwise from the O atom. Step [3]: Add the substituents at the three remaining stereogenic centers, clockwise around the ring. a. Draw the α anomer of: CHO

[1]

CH2OH O H

OH

H

OH

D sugar, CH2OH

H

OH

is drawn up.

OH CH2OH

H OH

H

H

[2]

CH2OH O H

farthest away C, OH on right = D sugar

α anomer OH is down for a D sugar.

[3]

H

CH2OH O H H H

HO OH

H OH

OH

First three substituents are on the right so they are drawn down.

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbohydrates 28–9

b. Draw the α anomer of: CHO HO

H

HO

H

H

[2]

O CH2OH

is drawn down.

H

farthest away C, OH on left = L sugar CH2OH O H

c. Draw the β anomer of: [1]

CHO

[2]

CH2OH O H

OH

H

OH

H

[3]

HO

OH

HO

OH CH2OH

β anomer OH is up for a D sugar.

D sugar, CH2OH

is drawn up.

farthest away C, OH on right = D sugar

H

[3]

H

H

H

O OH CH2OH OH OH H H H

H

The α anomer The first two substituents are has the OH and on the left so they are CH2OH trans. In drawn up. The third is on an L sugar, the the right, drawn down. OH must be drawn up.

L sugar, CH2OH

CH2OH

HO

OH H

OH

HO

O CH2OH

[1]

H

H

H

CH2OH O OH H OH H H OH H

The first substituent is on the left so it is drawn up. The other two are on the right, drawn down.

28.13 To convert each Haworth projection into its acyclic form: [1] Draw the C skeleton with the CHO on the top and the CH2OH on the bottom. [2] Draw in the OH group farthest from the C=O. A CH2OH group drawn up means a D sugar; a CH2OH group drawn down means an L sugar. [3] Add the three other stereogenic centers, counterclockwise around the ring. “Up” groups go on the left, and “down” groups go on the right. "up" group on left

CH2OH

CH2OH is up = D sugar O

HO

H H

a.

OH

[1]

CHO

CHO

[2]

H H

H OH

OH CH2OH

"down" groups on right

H

b. HO

"down" groups on right

OH

H

OH

OH

H

CH2OH

H OH CH2OH

OH on right = D sugar

CH2OH is down = L sugar

O H CH2OH OH H OH OH

H HO H

H

CHO

[3]

[1]

CHO

CHO

[2]

HO

H

"up" group on left

CHO

[3]

HO CH2OH

H CH2OH

OH on left = L sugar

H

H

OH

H

OH

HO

H CH2OH

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–10

28.14 To convert a Haworth projection into a 3-D representation with a chair cyclohexane: [1] Draw the pyranose ring as a chair with the O as an “up” atom. [2] Add the substituents around the ring. O is an "up" atom. CH2OH

HO O

HO

H H

a.

OH

O

[1]

OH H

[2]

H

H

OH

H

H OH

HO

OH

H

O is an "up" atom. O

H

b.

H OH

H

H

[1]

[2] HO

O

H

H

OH

HO OH

H HO

H

H

OH O

H

CH2OH OH H

O

H

With so many axial groups, this is not the more stable conformation of this sugar.

OH

OH

28.15 Cyclization always forms a new stereogenic center at the anomeric carbon, so two different anomers are possible. Two anomers of D-erythrose: CHO H

H

OH

H

OH

H

OH

OH OH

H

CH2OH

H

O H

H

OH

O H

H

OH

OH

H

D-erythrose

H

28.16 HO

HO CH3CH2OH HO OH HO HCl

HO O

a. HO HO

H

OH

OH CH3CH2OH H

OH

OH OH

CH2OH O OH CH3CH2OH HO H HCl H CH2OH H OH

β-D-fructose

OH

H

OH O

OH O

+

H

HCl

α-D-gulose

c.

OCH2CH3

HO

OCH2CH3

O

b.

HO O

+ HO

H

β-D-mannose OH

HO

HO O

OH

HO

OCH2CH3

OCH2CH3 OH HO H

CH2OH O OCH2CH3 + HO H H CH2OH H OH

CH2OH O CH2OH HO H H OCH2CH3 H OH

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbohydrates 28–11

28.17 H OH2

H

+

HOCH2

O

H

H

OH

H OH

H

+

HOCH2

OCH2CH3

OCH2CH3

O

H

H

OH

H OH

H

HOCH2 O H H OH

+ H2O

H

+

H

+ CH3CH2OH

OH

resonance-stabilized carbocation +

HOCH2

O

H

H

H

OH

OH

H H H2O

above HOCH2

O

+

H

H

OH

OH

+

HOCH2

O H

O

H

H

OH

H OH

H

H

H2O

HOCH2

OH

O

+ H3O+

H

H

OH

H OH

H

H

planar carbocation

HOCH2

H2O

below

H

H HO

OH

HOCH2

H

O

H

H

O

H +

H

O H H

+ H3O+

H

OH

HO

OH

H2O

28.18 a. All circled O atoms are part of a glycoside.

OH

HO

OH

OH

b. Hydrolysis of rebaudioside A breaks each bond indicated with a dashed line and forms four molecules of glucose and the aglycon drawn. OH

OH O

O O

HO

O OH

OH

O

O HO

OH O

HO

O

OH

+

OH

HO

OH HO

+

OH

HO

OH

OH

OH HO

H

OH H O

O HO

O

rebaudioside A Trade name: Truvia

HO

OH

+

OH

O

H

+

H O

OH HO

OH

HO

HO

OH O OH

aglycon

HO

Both anomers of glucose are formed, but only the β anomer is drawn.

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–12

28.19 OH

CH3O

OH Ag2O

O

a.

OH

HO

NaH OH

HO

c.

C6H5CH2O

OH

C6H5CH2O

O

H

O

pyridine

OAc

AcO AcO

OH

H C6H5COO

O + OH

C6 H5

C

Cl

pyridine

OOCC6H5 O

C6H5COO

OH H

OOCC6H5 C6H5COO

O

product in (c)

+ α anomer + HOCH2C6H5

OAc

Ac2O OH

O

f.

OH

C6H5CH2O

OAc

HO

H OCH2C6H5

H3O+

H

OH H

e.

OCH2C6H5 C6H5CH2O

OH

HO

OH

C6H5CH2O

C6H5CH2O

O

d.

O

C6H5CH2Cl

OH H OCH2C6H5 O OCH2C6H5

C6H5CH2O

H OCH2C6H5

C6H5CH2O

OH

C6H5CH2O

OCH3 CH3O

O

b.

O CH3O

CH3I

OH H OH

OCH3

+

C6H5

C

C6H5CH2O Cl

H

OCH2C6H5 O

pyridine

C6H5CH2O

+ α anomer OCOC6H5

C6H5CH2O H

28.20 CH2OH

CH2OH

CH2OH

O

H

OH

HO

H H

HO

H

NaBH4

HO

H

HO

HO

H

CH3OH

HO

H

HO

H

OH

H

CH2OH

OH CH2OH

CH2OH

D-tagatose

H

H

OH

D-galactitol

D-talitol

28.21 Carbohydrates containing a hemiacetal are in equilibrium with an acyclic aldehyde, making them reducing sugars. Glycosides are acetals, so they are not in equilibrium with any acyclic aldehyde, making them nonreducing sugars. acetal

hemiacetal HO

a.

CH2OH O H H OH

OH

b. H

H H

OH

reducing sugar

CH2OH O H H OH

hemiacetal HO

HO

H H OCH2CH3 OH

nonreducing sugar

O

c.

O

HO

O HO

OH

OH HO

lactose reducing sugar

OH

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbohydrates 28–13

28.22 CHO

a. HO

COOH

H

H

OH

H

OH

HO

Ag2O NH4OH

H

OH

H

OH

H

OH

CH2OH

HO

Br2 H2O

OH

H

HO

HNO3

H

H

H2O

OH

H

OH

CH2OH

COOH

COOH

H

H

COOH

H

OH

CHO

b.

c. HO

H

CH2OH

HO

CHO

H

H

H

OH

OH

H

CH2OH

OH CH2OH

28.23 Molecules with a plane of symmetry are optically inactive. CHO CHO

a.

COOH

H

OH

H

OH

H

OH

H

CH2OH D-erythrose

HO

HO

H

HO

OH

H

H

HO

OH

HO

H

HO

COOH

H

H

CH2OH

OH

OH H H

H

CH2OH

OH COOH

optically inactive

D-galactose

COOH

H

COOH

OH

HO

CHO

H

c.

optically inactive

HO

b.

H

H H OH COOH

D-lyxose

optically active

28.24 CHO HO H HO

H

OH

OH

H

OH

H

H

CHO

H

OH

or

HO H

CH2OH D-idose

CHO H

H

HO

OH

H

OH H OH

CH2OH

CH2OH

D-gulose

D-xylose

28.25 CHO HO

a.

H

H OH CH2OH

D-threose

CHO HO

H

H

HO

H

HO

H

OH CH2OH

CHO

CHO

H

OH H OH CH2OH

b.

H

OH

H

OH

H

OH CH2OH

D-ribose

CHO HO

CHO

H

H

OH

H

OH

H

OH

H

OH

H

OH

H

OH

H

CH2OH

OH CH2OH

802

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–14

CHO

CHO H

c.

HO

OH

HO

H

HO

H

H

H

OH

H

OH

OH

H

OH

HO

H

HO

HO

H

HO

H

CH2OH

CHO

H

OH

H

H

CH2OH

D-galactose

H OH CH2OH

28.26 Possible optically inactive D-aldaric acids: CHO

COOH

H

OH

H

OH

H

OH

H

OH

H

OH

H

OH

CH2OH

COOH H

plane of symmetry

HO

H

H

H

COOH

A'

CHO

OH

HO

OH

H

COOH

OH H OH CH2OH

A"

This OH is on the right for a D sugar.

There are two possible structures for the D-aldopentose (A' and A''), and the Wohl degradation determines which structure corresponds to A. H

OH

H

OH

H

OH CH2OH

This is A.

Product of Wohl degradation:

CHO

CHO CHO

Wohl

COOH

H

OH

H

OH

[O]

CH2OH

COOH

H

OH

HO

H

OH

H

COOH

A'

H

CHO

[O]

HO

OH

H

COOH

Wohl

OH CH2OH

optically active

optically inactive

H

H

OH

HO

H

H

OH CH2OH

A"

B

Since this compound has no plane of symmetry, its precursor is B, and thus A'' = A.

28.27 CHO HO H HO H

H OH H OH CH2OH

CH2OH HO H HO H

CHO

H

HO

OH

rotate

H

H

180°

HO

OH CHO

identical

H

H OH H OH CH2OH

D-idose

803

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbohydrates 28–15

28.28 Optically inactive alditols formed from NaBH4 reduction of a D-aldohexose. CHO

CH2OH

CH2OH

H

OH

H

OH

H

H

OH

H

OH

HO

H

H

OH

H

OH

HO

H

H

OH

H

OH

H

NaBH4

CH2OH

CH2OH

H NaBH4

OH

OH

H

OH

H H

H

HO

H

H

OH CH2OH

optically inactive

A''

This is A.

This is B.

CHO H

OH

HO

CH2OH

optically inactive

A'

CHO

OH

CHO K–F

COOH

H

OH

OH

H

OH

H

CH2OH

H

OH

HO

H

OH

H

OH

HO

H

OH

H

OH

H

[O]

CH2OH

A'

COOH

COOH

[O]

OH

HO

H

HO

H

H

COOH

optically inactive

B'

CHO

CHO

optically active

H K–F

OH

OH

HO

H

HO H

H OH

CH2OH

CH2OH

B''

A''

Two D-aldohexoses (A' and A'') give optically inactive alditols on reduction. A'' is formed from B'' by Kiliani–Fischer synthesis. Since B'' affords an optically active aldaric acid on oxidation, B'' is B and A'' is A. The alternate possibility (A') is formed from an aldopentose B' that gives an optically inactive aldaric acid on oxidation. 28.29 OH HO HO

O OH O HO

OH

OH

H3O+

OH

O

HO HO

O

+

OH OH

HO OH

α-D-glucose

HO HO

O OH

The same products are formed on hydrolysis of the α and β anomers of maltose.

HO

β-D-glucose

α anomer

28.30 β glycoside bond OH O HO

HO

4 1 O

OH O

HO

OH

cellobiose

OH

OH

Two possible anomers here. β OH is drawn.

804

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–16

28.31 a. 4

OH HO

OH HO

4

O

O HO

4

O

O HO

OH HO

O O

O HO

b.

O HO

1

1

dextran

1

HO HO

O

1

6

O HO

HO

1

HO O 6

O HO

HO HO

28.32 O

HO

OHO NHAc

O HO NHAc

HO

NHAc

OH O

OH O

OH O

OH O

O

NHAc

chitin—a polysaccharide composed of NAG units OH O

OH O HO

O

NH2

HO

OH O

O HO

NH2

OH O OHO

NH2

chitosan

O

NH2

28.33 OH

a.

CH2OH O H H OH

H OH

H H

OH

CH3NH2 mild H+

CH2OH O H H OH H

OH

b.

C6H5NH2 OH

OH

+

OH

CH2OH O H H OH

OH

H

mild

OH OH

NHC6H5 +

H+ OH

+ H2O

H H

O

HO

NHCH3

H

OH O

HO

OH

NHCH3

H

OH

H

OH

H

O

HO

H OH

OH

+ H2O

NHC6H5

805

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbohydrates 28–17

28.34 OH

OH

OH O

HO HO

O

HO HO

OH OH

+

OH2

OH

OH

H OH2

OH O

HO HO

HO HO

+

O+

+

H2 O

OH

+ H2O

CH3CH2NH2

above

OH

OH

H

H2O NHCH2CH3

O

HO HO

O

HO HO

+

NHCH2CH3 + OH

HO

OH

H3O+

O

HO HO

+

OH

OH

OH HO HO

below

HO

CH3CH2NH2

O

HO HO

O

+

OH

+

NHCH2CH3 H

H3O+

NHCH2CH3

H2O

28.35 O

O NH

a. N

HO CH2

O

N HO CH2

O

N

N

NH2

O H

H OH

OH

NH

b.

OH

28.36 a. Two purine bases (A and G) are both bicyclic bases. Therefore they are too big to hydrogen bond to each other on the inside of the DNA double helix. b. Hydrogen bonding between guanine and cytosine has three hydrogen bonds, whereas between guanine and thymine there are only two. This makes hydrogen bonding between guanine and cytosine more favorable. H H N N H

H O N H

guanine G

H

cytosine C

H

O

NH

N N

N

O

N

O

N N H

N N

H N H

guanine G

O H

CH3

N N H

thymine T

806

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–18

28.37 OHC HO H H OH

a. HO

H

rotate

CHO

H OH

HO HO H

H OH H OH

CHO

OHC

H OH

H OH OH H H OH H OH

Convert staggered to eclipsed.

HO

H HO

re-draw

CHO

OH

H

H

H

=

OH

H

OH

HO

H

H

OH

OH

H

CH2OH

OH CH2OH

OH

OHC HO H H OH

b. HO

H

rotate

CHO

OH H

HO HO H

HO H H OH

CHO

OHC

H OH

H OH OH H OH H H OH

Convert staggered to eclipsed.

HO

H

re-draw

CHO

OH

HO

H

HO

H

H

H

OH

HO

=

H

HO

OH

H

H

CH2OH

OH CH2OH

OH

28.38 CHO HO

CHO

OH

O

= H OH

OH

H

OH

H

OH

H

OH

OH O

HO

OH OH OH

CH2OH

A β anomer

=

H

OH

H

OH

H

OH

H

B α anomer

D-ribose

OH CH2OH

D-allose

28.39 Label the compounds with R or S and then classify. CHO H

H

OH CH2CH3

c.

R identical

A R

d.

C CH3CH2

H OH

CHO

CHO

OH

H

CHO

b.

a. CH3CH2 C OH

R identical

C HO

CHO

S enantiomer

CH2CH3 H

S enantiomer

28.40 Use the directions from Answer 28.2 to draw each Fischer projection. Cl

COOH

a.

CH3 C Br

COOH

=

CH3 S

Br

b. CH3 C Cl

H

H CH3O OCH2CH3 CH3

CH3CH2

re-draw CH3

CH2CH3

S C Br = H Cl OCH3 C

H

re-draw

CH3CH2

H Br

CH3 Br

CH2CH3 = CH3

OCH2CH3

=

Cl C H

Cl

S

CH2CH3

OCH3 CH2CH3 f. H OCH2CH3

R

C Br Cl

Cl = CH3CH2 S

S

Cl

C H Br

S

Br

S Cl

H

H

CH2CH3 Cl

S re-draw Br C H Cl C

R

H

H Cl Br

CH3

H C Br

e.

C Br

CH3

CH3

re-draw

C

Br

C

H

H

c.

d.

Br

Cl

=

Br Cl

S R

H

H Br

807

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbohydrates 28–19

CHO CH3

H

C

g. H

CH3 Br

H C Br = H S Br C H Br S

re-draw

C CH3

Br

CH3 Br

h. HO

CHO

H

HO

H C OH

HO H

CH3

CH3

CHO

S = S HO C H HO R

HO H H OH re-draw HO C H

H H

H

OH

CH2OH

CH2OH

28.41 Epimers are two diastereomers that differ in the configuration around only one stereogenic center. CHO H

CHO

OH

HO H

H

H

HO

OH

HO

OH H H

CH2OH

C4

CH2OH

D-xylose

L-arabinose

28.42 CHO HO

a.

CHO

H

H H

H

OH

HO

OH

HO

CH2OH

CHO

OH

HO

b. HO

H H

H

C3

H

H

c. HO

OH

CH2OH

D-arabinose

CHO

H

enantiomer

O

d. HO

H

H

CH2OH

OH

OH

OH

OH

OH

CH2OH

epimer

diastereomer (but not epimer)

constitutional isomer

28.43 CHO HO

CHO

H

HO

CHO

H

H

CH2OH C O

OH

CH2OH

H

OH

H

OH

HO

H

H

OH

H

OH

H

OH

HO

H

H

OH

H

OH

HO

OH

H

OH

H

H

CH2OH

CH2OH

CH2OH

A

B

C

a. A and B epimers b. A and C diastereomers

CH2OH

O

H H H

OH

OH H

OH OH

D

HO

OH HO

F

H

E

c. B and C enantiomers d. A and D constitutional isomers

e. E and F diastereomers

28.44

OH H

H

OH

OH

H

a. anomers, epimers, diastereomers, reducing sugars b. CHO H

OH OH

B

H HO

OH H

A

H

H

OH OH

O

O

H

H

O OH

HO

OH H OH CH2OH

D-xylose

This is the acyclic form of both A and B.

808

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–20

28.45 Use the directions from Answer 28.12. a. β-D-talopyranose

CH2OH O H

[1]

CHO HO

H

HO

H

HO

H

H

OH CH2OH

CH2OH O H

[2]

OH

[3]

OH H

H

β anomer OH is up for a D sugar.

D sugar, CH2OH

is drawn up.

farthest away C, OH on right = D sugar

CH2OH O OH H OH OH H H H

D-talose

b. β-D-mannopyranose

CH2OH O H

[1]

CHO HO

CH2OH O H

[2]

OH H

H

HO

H

H

OH CH2OH

β anomer OH is up for a D sugar.

D sugar, CH2OH

is drawn up.

OH

H

CH2OH O OH H OH OH H OH H H H

[3]

farthest away C, OH on right = D sugar

D-mannose

c. α-D-galactopyranose

CH2OH O H

[1]

CHO H

CH2OH O H

[2]

H OH

OH

HO

H

HO

H

H

OH CH2OH

α anomer OH is down.

D sugar, CH2OH

is drawn up.

CH2OH O H

[3] OH H OH H H

H OH

OH

farthest away C, OH on right = D sugar

D-galactose

d. α-D-ribofuranose

[1]

[2]

H

CHO H

OH

H

OH

H

CH2OH O

OH CH2OH

CH2OH O

H

H

OH

[3]

CH2OH H O H OH

α anomer OH is down.

D sugar, CH2OH

is drawn up.

H H OH OH

farthest away C, OH on right = D sugar

D-ribose

e. α-D-tagatofuranose CH2OH C O HO

H

HO

H

H

OH CH2OH

D-tagatose

[1]

CH2OH O H

[2]

CH2OH O

CH2OH

H

OH

[3]

CH2OH OH O H

OH H

D sugar, CH2OH

is drawn up.

farthest away C, OH on right = D sugar

α anomer OH is down.

CH2OH OH

H

809

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbohydrates 28–21

28.46 CHO

a. HO

H

HO

H

HO

H

HO

OH

α anomer OH

H

D sugar HOOH OH

D sugar

HOOH HO

O

O HO

OH

β anomer

CH2OH

farthest away C, OH on right = D sugar CHO

b.

H

OH

H

OH

HO

O OH

OH OHOH

OH OH

α anomer

OH

D sugar

OH HO

O

H

H

D sugar

OH HO

β anomer

CH2OH

farthest away C, OH on right = D sugar c.

CHO HO

D sugar HOOH OH

H

H

O

OH

HO

H

H

OH

OH

D sugar

HOOH HO O

OH

OH

OH

α anomer

β anomer

CH2OH

farthest away C, OH on right = D sugar

28.47 Use the directions from Answer 28.13. "up" group on left a.

OH

CH2OH

CH2OH is up = D sugar O

H OH

OH

[3]

H

OH

HO H

"down" group on right

CH2OH

OH

H

CH2OH

[1]

CHO

CHO

[2]

H

"down" group on right

HO HO

H

"up" group on left

H OH

CHO

[3] H

OH OH

H

CH2OH

HO

CH2OH OH H

H

OH

OH on right = D sugar

CH2OH is down = L sugar O

CHO H HO

"up" group on left

b.

CHO

[2]

H H

OH

CHO

H

H

"up" group on left

[1]

CH2OH

H CH2OH

OH on left = L sugar

HO

H OH H H CH2OH

810

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–22

OH

c.

H

HO O

HO

CH2OH is up = D sugar O [1] OH

CH2OH H H

OH =

CHO

[2]

CHO

[3] HO

OH H

OH

OH

CHO

OH

H

H CH2OH

OH

H

H

OH

H

OH

H

CH2OH

OH CH2OH

OH on right = D sugar CHO

d.

OH

H

OH

O

H

OH

HO

OH OH

H

OH

CHO

OH

H

OH

H

O

f. HO

=

HO HO

CH2OH

O OH

e. HOCH2

OH

O

H

H

H

OH H

H

OH

H

OH

H

OH CH2OH

28.48 CHO HO H H

a.

H OH OH

H

CH2OH

b.

CH2OH OH CH2OH H H O HO H O HO H OH

H

β anomer

D-arabinose

H

H

OH OH

H

O OH OH

OH

H

OH

H

α anomer

H

H H

O H OH

OH

H

OH

OH

two anomers in the pyranose form

28.49 Two anomers of D-idose, as well as two conformations of each anomer:

α anomer

axial O

HO OH

OH OH OH O

OH OH

HO

OH

4 axial substituents

β anomer

OH

equatorial CH2OH group

OH OH OH O

4 equatorial OH groups More stable conformation for the α anomer—the CH2OH is axial, but all other groups are equatorial.

equatorial CH2OH group OH

OH

3 axial substituents

OH O

HO

axial

OH

OH H OH CH2OH

OH

C O

CH2OH OH

OH

H

= HO

OH

OH

CH2OH

D sugar

H

axial

OH

HO

3 equatorial OH groups The more stable conformation for the β anomer—the CH2OH is axial, as is the anomeric OH, but three other OH groups are equatorial.

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

811

Carbohydrates 28–23

28.50 CH3O

HO

CH3O HO

OH

f. C6H5COCl, pyridine

OH

g. The product in (a), then H3O+

O

OH

D-gulose

HO C6H5CH2O

c. C6H5CH2Cl, Ag2O

C6H5COO

OCH3

OOCC6H5 O O OOCC H 6 5 COC6H5 OCH3 OCH 3

O

+ β anomer

+ β anomer

HO

HO OH

CH3O

b. CH3OH, HCl

O

C6H5COO

OCH3 O

a. CH3I, Ag2O

OCH3 OCH2C6H5 O

CH3O

CH3O

h. The product in (b), then Ac2O, pyridine

OH

OAc

OAc O

+ β anomer

O

OCH2C6H5 CH2C6H5

C6H5CH2O HO

OAc

O

CH3O

HO OH

OCH3 O

+ β anomer

d. C6H5CH2OH, HCl

OAc

OCH2C6H5 CH3O

OAc

CH3O

+ β anomer

OCH2C6H5

j. The product in (d), then CH3I, Ag2O

O

e. Ac2O, pyridine

OAc OCH3

i. The product in (g), then C6H5CH2Cl, Ag2O

OH

CH3O

OCH3 O

OAc OAc OAc CH3O

CH3O

+ β anomer OCH2C6H5

28.51 OH OH O

a. CH3OH, HCl HO CHO HO

OH

COOH

d. Br2, H2O

+ β anomer OCH3

H

H

OH

H

OH

H

OH

b. (CH3)2CHOH, HCl HO

OH OH O OH

c. NaBH4, CH3OH

HO

OH

H

OH OH CH2OH

+ β anomer

OCH(CH3)2

COOH

e. HNO3, H2O

HO

H

H

OH

CH2OH

H

OH

H

H

H

OH

H

OH

H

H

H H

CH2OH D-altrose

HO

OH CH2OH

OH COOH

812

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–24

CHO

f. [1] NH2OH [2] (CH3CO)2O, NaOCOCH3 [3] NaOCH3

H

OH

H

OH

H

OH

OCH3 OCH3 O

h. CH3I, Ag2O CH3O

OCH3 OCH3

CH2OH CHO HO

H

H

OH

H H

H

HO

+

AcO

OH H OH

OH

H

OH

OH

H

OH

+ β anomer

OAc

OAc

H

CH2OH

OAc OAc O

i. Ac2O, pyridine

CHO

g. [1] NaCN, HCl [2] H2, Pd-BaSO4 HO H [3] H3O+

+ β anomer

j. C6H5CH2NH2, mild H+ OH

OH O

HO

+ β anomer

CH2OH

NHCH2C6H5

OH

28.52 OH

H

H

OH

H HO HO

O H H

OH

H3O+

H

O

HO

H

H

N H

O

H

H

HO

H3O+

OH

O

solanine HO

OH

H H HO

HO

OH

OH

O

HO

+ HO

OH

HO

HO H H

H

monosaccharide (both anomers)

monosaccharide (both anomers)

CHO HO

OH

H

OH

H

CH2OH D-glucose

OH OH

CHO OH

O

OH

28.53 H

H

aglycon

monosaccharide (both anomers)

OH

N

H

+

O

HO

O O

H

OH H

HO

aglycon

H

OH

O OH

HO

+

OH

OH

monosaccharide (both anomers) H

HO

OH O

HO HO H

salicin

OH

OH H

O

CHO

HO

H

HO

OH

H

OH

H

CH2OH D-arabinose

H H OH OH CH2OH

D-mannose

813

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbohydrates 28–25

28.54 CHO CHO

HO

a. H

OH

H

OH

H

H

H

OH

H

OH

H

OH

CH2OH

CH2OH

CHO

b.

HO

H

H

HO

H

HO

H

HO

HO

H

HO

H

HO

OH

H

OH

CH2OH

HO

H

H

HO

H

HO

H

HO

H

HO

H

HO

H

HO

H

HO

H

HO

H

HO

H

OH

CHO

CHO

H

c.

H

CH2OH

OH

OH

OH

H

H

CH2OH

OH CH2OH

H H

H

CH2OH

CHO

CHO

OH

H

OH

CH2OH

CHO

CHO

OH CH2OH

28.55 OH

OH CH3OH

O

a. HO

OH

HO

HO HO

HCl

OH

+ α anomer H

H HO

H

OH CH3OH

H

OH

H

OH

H

OH

CH2OH

H H H

H

H

Ag2O

OCH3

H

OCH3 CH2OCH3

COOH

OH

HO

OCH3

CH3O

CH2OH

CHO

c.

H

H

Br2

HO

OH

H2O

H

OH

H

CH2OH

OCH3 OCH2CH3

+ α anomer

H CH3I

H

O

CH3CH2O CH3CH2O

CH2OCH3

OH

NaBH4

H

Ag2O

OH

OCH2CH3

+ α anomer

OH

HO

OCH3

CH2OH

CHO

b.

CH3CH2I

O

COOH

OH

H Ac2O

H OH

OAc

AcO

pyridine

H

H

OH

OAc

H

OAc

CH2OH

CH2OAc

28.56 Molecules with a plane of symmetry are optically inactive. CHO H

OH

H

NaBH4

OH

H

CHO

CH2OH

CH3OH

OH

H H H

CH2OH

OH

H

OH

HO

OH

H

CH2OH

OH

NaBH4

H OH

CH3OH

H HO H

CH2OH

CH2OH

OH H OH CH2OH

D-xylose

D-ribose

28.57 OH

OH O

a. HO

OCH3

H3

O+

HO

OH

OH OH

OH O

O

HO

OH OH

OH OH

+ CH3OH OH

814

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–26

b.

OCH3 HO O

HO

H3O+

HOCH2

HO

OCH2CH3

OH

c.

OCH3 HO O

OH

HOCH2

H3O+

OH

HO OH

OH

NHCH2CH3

O

OCH3 HO O

OH

+ CH3CH2OH

OH HOCH2

OH

O

OH

OH

O

+ NH2CH2CH3 OH OH

OH

28.58 OH

OH O

HO HO

HO HO

OH OH

OH O

+

HO HO

OH2 OH

OH

OH H OH

H OH2 OH

OH O

HO HO

O

HO HO

resonance-stabilized carbocation

H OH

+

OH O

HO HO

OH

OH O

O

HO HO

H OH2

OH OH

OH + O H H

H OH

H OH

28.59 H

O C6H5

C

H

OH C6H5

C

H

C6H5 CH2OH HO HO

O

H OH

B (any base)

OH H O HO HO

C6H5

O OCH3

OCH3

O HO HO

OH

O

+ HB+ OCH3

OH

OH OH2 C6H5

O H O HO

OH

B

C6H5

O OCH3

O O HO

+ H2O

O OH

+ HB+ OCH3

C6H5

O OH

(any base) C6H5

O HO HO

OCH3

O HO HO

O OCH3 OH

815

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbohydrates 28–27

28.60 OH

OH O

HO HO

OH OH

OH HO H O HO

O

O HO

OH

O

HO HO OCH3

HO HO

O

+

OH

OCH3

OH

OH

OH

O

A

OH +

+ H2O

H OH2

O+

HO HO

+

OH O

HO HO

OCH3

B HO OH

OH

O

HO HO

HO HO

OCH3

OH

O

HO

HO

B

HO HO

OCH3 H

OH

H OH2

+

O+

HO HO

OH

A

+ H2O

+

OH

O

+ CH3OH OH

H2O

above

OH

H

O

HO HO

H2O

+

OH HO

OH

O

HO HO

OH HO

O

HO HO

+ H3O+

+

A OH

OH

below

OH

O

HO HO

H2O

HO

O

HO HO

+

HO

OH H

OH

H2O

28.61 COOH H

OH

H

OH

H

OH

H

H

=

CH2OH OH O H H H

OH OH

OH

H A

H

OH

OH

OH

CH2OH + OH OH H H H OH OH

HO

H

O OH

HO OH

A

=

H H

H

OH

OH

OH

H

O

H O

H

OH OH +

H A

OH

H H

H

OH

OH

OH

OH

OH O

OH OH

OH

H H

H A

OH

A

OH

O

A

O+

OH

OH +

CH2OH

H

OH H

OH O

H H

H

H O H

A

+

O

H H

OH2

H

OH

OH

OH

OH

+ H2O

OH

OH +

A

816

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–28

28.62 OH

CHO H

OH

HO

H

H

H

C O

C O

C OH

C OH

HO

OH

H

H

H

H

H2O

OH

H

CH2OH

HO

OH

H

OH

H

CH2OH

Protonation of this enolate can occur from two directions.

D-glucose

H C OH

H OH

C OH

H

HO

OH

H

OH

H

H OH

+

OH

OH

CH2OH

CH2OH

enediol A

Protonation on O forms an enediol.

H2O

two protonation products

CHO HO

H

HO

H

H

OH

H

OH CH2OH H OH

H

H

C OH

C OH

C O H HO

OH

H H

HO

OH

H

OH

H

H

CH2OH

enediol A Deprotonation of the OH at C2 of the enediol forms a new enolate that goes on to form the ketohexose.

H C OH

C O HO

OH

H

OH

H

CH2OH

H

C OH

C O

H

H

C O

H

HO

OH

H

OH

H

CH2OH

H OH

+

OH CH2OH

+ H2O

28.63 Two D-aldopentoses (A' and A'') yield optically active aldaric acids when oxidized. Optically active D-aldaric acids: CHO HO

H

H

OH

H

OH CH2OH

A'

COOH

COOH

[O]

H

HO

H

H

OH

HO

H

H

OH

H

HO

COOH

optically active

OH COOH

optically active

CHO

[O]

HO HO H

H H OH CH2OH

A" D-lyxose

OH

817

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbohydrates 28–29

CHO HO

CHO

H

H

OH

H

OH

Wohl

H

OH

H

OH

[O]

H

OH

HO

H

OH

H

CH2OH

CH2OH

COOH

COOH

A'

[O] HO

H OH

H

HO

Wohl

H

H

HO

OH

COOH

COOH

optically inactive

CHO

CHO

H

H

CH2OH

OH CH2OH

no plane of symmetry optically active

A"

Only A" undergoes Wohl degradation to an aldotetrose that is oxidized to an optically active aldaric acid, so A'' is the structure of the D-aldopentose in question. 28.64 CHO HO

CH2OH

H

HO

CH2OH HO

H

HO

H

HO

H

OH

H

OH

HO

H

OH

H

OH

H

CH2OH

CHO

H

CH2OH

OH

H

H

OH

CH2OH

CH2OH

identical (by rotating 180°)

D-arabinose

H

D-lyxose

28.65 Only two D-aldopentoses (A' and A'') yield optically inactive aldaric acids (B' and B''). CHO

COOH

H

OH

H

OH

H

OH

[O]

COOH

H

OH

H

H

OH

HO

H

OH

H

CHO

OH H

H

[O]

OH

CH2OH

COOH

A'

B'

B''

optically inactive

optically inactive

HO H

COOH

OH H OH CH2OH

A"

Product of Kiliani–Fischer synthesis: CHO

CHO H

OH

H

OH

H

OH

K–F

CH2OH

CHO

OH

HO

H

OH

H

OH

H

H

OH

H

OH

HO

OH

H

OH

H

H

CH2OH

A'

D' [O] COOH

plane of symmetry

H

CHO

H

HO

CHO

H

H

OH

OH

H

OH

H OH

HO H

H OH

CH2OH

CH2OH

CH2OH

C'

C"

D"

[O]

[O]

[O]

COOH

COOH

COOH

H

OH

HO

H

OH

H

OH

H

H

H

OH

H

OH

HO

H

OH

H

OH

H

HO

H

H

OH

OH

H

OH

H OH

HO H

H OH

COOH

COOH

COOH

COOH

optically inactive

optically active

optically active

optically active

CHO H

K–F HO H

OH H OH CH2OH

A"

818

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–30

Only A' fits the criteria. Kiliani–Fischer synthesis of A' forms C' and D', which are oxidized to one optically active and one optically inactive aldaric acid. A similar procedure with A'' forms two optically active aldaric acids. Thus, the structures of A–D correspond to the structures of A'–D'. 28.66 Only two D-aldopentoses (A' and A'') are reduced to optically active alditols. CHO HO

CH2OH

H

H

OH

H

OH

HO

[H]

CH2OH

H

HO

H

H

OH

HO

H

H

OH

H

CH2OH

[H]

H

H

CH2OH

optically active

H

HO

OH

CH2OH

A'

CHO HO

OH CH2OH

A"

optically active

Product of Kiliani–Fischer synthesis: CHO

CHO HO

H

H

H

OH K–F

H

OH

HO H H

CH2OH

CHO H

HO

H

H

H

HO

H

HO

H

HO

H

OH

HO

H

HO

H

OH

H

OH

H

OH

H

CH2OH

C'

C"

[O]

[O]

COOH HO

OH

CH2OH

B'

H

CHO

HO

CH2OH

A'

CHO

OH

H

COOH HO

H

H

H

HO

H

HO

H

HO

H

HO

OH

HO

H

OH

H

OH

H

COOH

COOH

optically active

OH

OH

CH2OH

A"

OH H H

H

COOH

optically active

OH

COOH

H

H

H

H

[O]

HO

OH

H

K–F HO

B''

OH

H

HO

CH2OH

[O]

COOH

CHO

OH

OH COOH

optically active

optically inactive

Only A'' fits the criteria. Kiliani–Fischer synthesis of A'' forms B'' and C'', which are oxidized to one optically inactive and one optically active diacid. A similar procedure with A' forms two optically active diacids. Thus, the structures of A–C correspond to A''–C''. 28.67 D-gulose

CHO

CH2OH

CHO H

OH

H

OH

H

H

OH

H

OH

HO

HO

H

H

OH CH2OH

A

HO

H

H

OH CH2OH

B

OH H

H

OH CH2OH

C

CHO

CH2OH H

OH

HO

H

H

OH CH2OH

D

HO

H

H

OH CH2OH

E

COOH HO

CH2OH

H

H

OH

OH

H

OH

COOH

HO

H

F

H

H

OH CHO

G

819

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbohydrates 28–31

28.68 A disaccharide formed from two galactose units in a 1→4-β-glycosidic linkage: β glycoside bond OH O

HO HO

OH O

1 O

OH 4 HO

OH

OH

28.69 A disaccharide formed from two mannose units in a 1→4-α-glycosidic linkage: α glycoside bond HO

HO O

HO HO

HO

1

4

O

OH O

OH

HO

28.70 OH

a.

OH O

OH O

HO

OH

HO OH

OH

OCH3 OCH 3 O

CH3I

O

Ag2O

CH3O

OCH3 OCH3

(Both anomers of B and C are formed, but only one is drawn.)

O CH3O

OCH3

OH O

CH3I

b. HO

OH O HO HO

HO

H3O+ OCH3 O

CH3O

OCH3 +

O OH

D

CH3O OCH3 OH E

OCH3 O

OH OCH3 + CH3OH

C

OCH3 OCH 3 O

H3O+

OCH3 O

HO CH3O

OH

OCH3 O

CH3O OCH3 O CH3O CH3O

Ag2O

OCH3 OCH3

A

B OH

OCH3 O

HO CH3O CH3O +

OCH3 O OH F + CH3OH

(Both anomers of E and F are formed, but only one is drawn.)

OCH3

28.71 OH

a.

β glycoside bond

OH O

1

HO OH

b. c.

4

O

OH

OH

OH O

O OH

HO OH

1 4-β-glycoside bond reducing sugar (hemiacetal)

HO

HO

1

O HO HO

reducing sugar (hemiacetal)

α glycoside bond 6

OH O

1 6-α−glycoside bond OH

820

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–32

28.72 a, b.

OH

OH

CH2OH O

HO O CH2 OH

1 6-α-glycoside bond

HO

O

OH OH

stachyose

1 2-α-glycoside bond

OH HOCH2 O

OH O

c.

OH CH2OH OH O

CH2OH O

OH

O OH

O OH OH

HO

OH

OH HOCH2

O

HO OH

OH

OH

O

O

CH2OH

OH

OH

OH

HO OH

OH

OH

O

2

HO

CH2OH O

O

=

O CH2

OH

O

HO

O

OH OH

H3O+

OH

1 6-α-glycoside bond

OH

OH

CH2OH

OH

OH

OH

identical Two anomers of each monosaccharide are formed, but only one anomer is drawn.

d. Stachyose is not a reducing sugar since it contains no hemiacetal. e.

CH3I Ag2O

OCH3 CH3O

OCH3 O

CH3O O CH3O

O

CH3O

O OCH3 CH3O

CH3O

f.

product in (e)

H3O+

OCH3

O

O

O O OCH3 OCH3 CH3

CH2OCH3 O

CH3O OH

OCH3 OCH3

OCH3 OCH3 CH2OH O

CH2OH O OH

OCH3 OCH3

OH

OCH3 CH3O

OCH3

Two anomers of each monosaccharide are formed.

CH3OCH2 CH3O OH O CH3O

CH2OCH3

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

821

Carbohydrates 28–33

28.73 OH O

HO HO

Isomaltose must be composed of two glucose units in an α-glycosidic linkage. Since it is a reducing sugar it contains a hemiacetal. The free OH groups in the hydrolysis products show where the two monosaccharides are joined.

isomaltose + α anomer

OH O O

HO HO

OH

the hemiacetal

OH [1] CH3I, Ag2O [2] H3O+ OCH3 O

CH3O CH3O

OH

6 1

CH3O

O

CH3O CH3O

OH

CH3O

(Both anomers are present.) OH

28.74 OH O

HO HO

CH3O CH3O

CH3I

trehalose

OH O

Ag2O

OH

OCH3 O CH3O

H3O+ O

O

OCH3

O

HO

OH

CH3O CH3O

CH3O

CH3O

(both anomers)

OCH3

OH

OCH3 O

OCH3

Trehalose must be composed of D-glucose units only, joined in an α-glycosidic linkage. Since trehalose is nonreducing it contains no hemiacetal. Since there is only one product formed after methylation and hydrolysis, the two anomeric C’s must be joined. 28.75 HO a. HO

O H2N

CH2OH

1

O HO HO

6

α

c. O

H

O HO

H H

H

OH

NH2 O

NHCH2C6H5

O CH3

b. HO HO

OH OH O

4 1

d.

OH

O HO

OH

mannose

glucose

N HO CH2

O

NH

OH OH

O H OH

O

OH

822

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–34

28.76 OHC H OH HO H

HO

rotate

a. OHC

OH H H OH H OH OH H

H OH

H H HO

H OH HO H

CHO

OHC

OH H

HO

re-draw

H

H

OH

H

OH

HO

H CH3

b. OH on left in Fischer L-monosaccharide

OH O

Ring

OH

OH OH

flip.

more stable chair

OH O

HO

OH

OH

The α anomer has the CH3 on C5 and the anomeric OH trans.

c. Fucose is unusual because it is an L-monosaccharide and it contains a CH3 group rather than a CH2OH group on its terminal carbon. 28.77 CH2OH OH

a.

O HO

H

OH

OH

O

OH O H

OH

OH OH

HO

OH

=

HO

H

H

OH

H

H

OH CH2OH

D-fructose

H HO

O

b. HO OH

CHO

OH

H

OH

OH

OH H CHO OH

=

H

OH

H

OH

H

OH CH2OH

HO

D-ribose CHO

HO

H

c.

HO H

OH O

OH

HO H

OH H OH CHO = H H

OH

OH

HO H H

OH H OH OH CH2OH D-glucose

CHO OH HO d. HO

O OH

OH

HO

H H

HO HO H

OH

OH CHO = OH H

H HO HO

H OH H H CH2OH

L-glucose

823

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Carbohydrates 28–35

28.78 Ignoring stereochemistry along the way:

A HO

OH

OH H

H A O

O

[1]

OH

A

OH A O O

OH

H

HO

HO

OH

HO

C

[5]

O

CH3

OH

O

CH3 O

OH

C

CH3

[8]

OH

[4] OH

OH HO H A

[7]

HO

HO

OH

O

[6]

OH2

HO

OH

HO

[3]

HO OH

OH

OH

OH

O

OH

[2]

OH

OH

OH

H

OH

OH

HO

H A

HO

OH + HA

CH3

– H2O

O

O O

HO HO

[9]

OH

O O

O H

OH

HO

A

[10]

OH

OH O

O HO

OH

C

OH

CH3

OH

+ H2O

[11] O

O O

O

[14]

OH

O

O

– H2O HO

O

[13]

OH

O O

O

H2O HO

O

OH

O

O

OH

HO HO

O

HO

O

H A

A

[15]

[12]

HO

O H A

A

O

H

O

O

O HO

O

O

[16]

O O

O

O HO

O + HA

=

O

O

H O

HO

H

O

CH3

824

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 28–36

28.79 The hydrolysis data suggest that the trisaccharide has D-galactose on one end and D-fructose on the other. D-Galactose must be joined to its adjacent sugar by a β-glycosidic linkage. D-Fructose must be joined to its adjacent sugar by an α-glycosidic linkage. HO

CH3O

OH O

HO OH

galactose

β O HO HO

HO

O HO

α O

CH3I Ag2O

CH3O

OCH3 O O CH3O CH3O CH3O

O CH3O

O OH OH

glucose

HO

CH3O

O OCH3

(Hydrolysis cleaves all acetals, indicated with arrows.)

OCH3 O CH3O

O

CH3O CH3O

H3O+

fructose 2 anomers CH3O

CH3O

CH3O

OH OH

CH3O CH3O

CH3O O

O CH3O

OH

OCH3 OH

OH

2,3,4,6-tetra-O-methyl2,3,4-tri-O-methyl-D-glucose 1,3,6-tri-O-methyl-D-fructose D-galactose (Both anomers of each compound are formed.)

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

825

Amino Acids and Proteins 29–1

Chapter 29 Amino Acids and Proteins Chapter Review Synthesis of amino acids (29.2) [1] From α-halo carboxylic acids by SN2 reaction NH3

R CHCOOH

R CHCOO–NH4+

(large excess) SN2

Br

+

NH4+ Br–

•

Alkylation works best with unhindered alkyl halides—that is, with CH3X and RCH2X.

NH2

[2] By alkylation of diethyl acetamidomalonate O

H H C N C COOEt

CH3

COOEt

R

[1] NaOEt

H2N C COOH

[2] RX

H

[3] H3O+, Δ

[3] Strecker synthesis O R

C

NH4Cl H

NaCN

NH2

NH2

H3O+

R C CN

R C COOH

H

H

α-amino nitrile

Preparation of optically active amino acids [1] Resolution of enantiomers by forming diastereomers (29.3A) • Convert a racemic mixture of amino acids into a racemic mixture of N-acetyl amino acids [(S)- and (R)-CH3CONHCH(R)COOH]. • React the enantiomers with a chiral amine to form a mixture of diastereomers. • Separate the diastereomers. • Regenerate the amino acids by protonation of the carboxylate salt and hydrolysis of the N-acetyl group. [2] Kinetic resolution using enzymes (29.3B) H2N

C

COOH

(CH3CO)2O

AcNH

H R

C

COOH

acylase

H R

H2N

C

COOH

H R

(S)-isomer

(S)-isomer separate

H2N

C

COOH

R H

(CH3CO)2O

AcNH

C

COOH

R H

acylase

AcNH

C

COOH

R H

(R)-isomer enantiomers

enantiomers

NO REACTION

826

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–2

[3] By enantioselective hydrogenation (29.4) R

NHAc

H

AcNH

H2

C C

Rh*

COOH

C

H2O, –OH

COOH

H2 N

H CH2R

C

COOH

H CH2R

S enantiomer

S amino acid

Rh* = chiral Rh hydrogenation catalyst

Summary of methods used for peptide sequencing (29.6) • Complete hydrolysis of all amide bonds in a peptide gives the identity and amount of the individual amino acids. • Edman degradation identifies the N-terminal amino acid. Repeated Edman degradations can be used to sequence a peptide from the N-terminal end. • Cleavage with carboxypeptidase identifies the C-terminal amino acid. • Partial hydrolysis of a peptide forms smaller fragments that can be sequenced. Amino acid sequences common to smaller fragments can be used to determine the sequence of the complete peptide. • Selective cleavage of a peptide occurs with trypsin and chymotrypsin to identify the location of specific amino acids (Table 29.2). Adding and removing protecting groups for amino acids (29.7) [1] Protection of an amino group as a Boc derivative R H2N

H C

R

[(CH3)3COCO]2O (CH3CH2)3N

CO2H

Boc N H

H C

CO2H

[2] Deprotection of a Boc-protected amino acid R

H C

Boc N H

CO 2H

CF3CO2H or HCl or HBr

R H2N

H C

CO2H

[3] Protection of an amino group as an Fmoc derivative

H2N

C

C

R H

O

R H OH

+

CH2O

C

O Fmoc Cl

Na2CO3 Cl

H2O

Fmoc N H

C

C O

OH

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

827

Amino Acids and Proteins 29–3

[4] Deprotection of an Fmoc-protected amino acid R H C

Fmoc N H

R H C

OH H2N

C

O

C

OH

O

N H

[5] Protection of a carboxy group as an ester O

O H2N

C

C

OH

CH3OH, H+ H2N

C

C

OCH3

H R

H R

O

O H2N

C

C

C6H5CH2OH,

OH

H+

H2N

C

C

OCH2C6H5

H R

H R

methyl ester

benzyl ester

[6] Deprotection of an ester group O H2N

C

C

O

O –OH

OCH3

H2N

C

H2O

H R

C

OH

H R

H2N

C

C

O H2O, –OH

OCH2C6H5

H2N

or H2, Pd-C

H R

C H R

benzyl ester

methyl ester

Synthesis of dipeptides (29.7) [1] Amide formation with DCC R H Boc N H

C

OH + C O

O H2N

C H R

C

R H OCH2C6H5

DCC Boc N H

C

C O

H N

O C

C

OCH2C6H5

H R

[2] Four steps are needed to synthesize a dipeptide: a. Protect the amino group of one amino acid using a Boc or Fmoc group. b. Protect the carboxy group of the second amino acid using an ester. c. Form the amide bond with DCC. d. Remove both protecting groups in one or two reactions. Summary of the Merrifield method of peptide synthesis (29.8) [1] Attach an Fmoc-protected amino acid to a polymer derived from polystyrene. [2] Remove the Fmoc protecting group. [3] Form the amide bond with a second Fmoc-protected amino acid using DCC. [4] Repeat steps [2] and [3]. [5] Remove the protecting group and detach the peptide from the polymer.

C

OH

828

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–4

Practice Test on Chapter Review 1. a Which statement is true about the peptide Ala–Gly–Tyr–Phe? 1. The N-terminal amino acid is Ala. 2. The N-terminal amino acid is Phe. 3. The peptide contains four peptide bonds. 4. Statements [1] and [3] are true. 5. Statements [2] and [3] are true. b. Which of the following peptides is hydrolyzed by trypsin? 1. Glu–Ser–Gly–Arg 2. Arg–Gln–Trp–Asp 3. Glu–Val–Leu–Lys 4. Peptides [1] and [2] are hydrolyzed. 5. Peptides [1], [2], and [3] are all hydrolyzed. c. In which types of protein structure is hydrogen bonding observed? 1. α-helix 2. β-pleated sheet 3. 3° structure 4. Hydrogen bonding is present in [1] and [2]. 5. Hydrogen bonding is present in [1], [2], and [3]. 2. Answer the following questions about peptides. O

O H 2N H

C

C

CH3

Ala

H 2N OH

C

C

O OH

H 2N

H CH(CH3)2

Val

C

C

OH

H CH2OH

Ser

a. Draw the structure of the following tripeptide: Val–Ser–Ala. b. Give the three-letter abbreviation for the N-terminal amino acid. c. Give the three-letter abbreviation for the C-terminal amino acid. 3. Answer the following questions about the amino acid leucine (2-amino-4-methylpentanoic acid), which has pKa’s of 2.33 and 9.74 for its ionizable functional groups. a. Draw a Fischer projection for L-leucine and label the stereogenic center as R or S. b. What is the pI of leucine? c. Draw the structure of the predominant form of leucine at its isoelectric point. d. Draw the structure of the predominant form of leucine at pH 10. e. Is leucine an acidic, basic, or neutral amino acid? 4. What product is formed when the amino acid phenylalanine is treated with each reagent? a. PhCH2OH, H+ b. Ac2O, pyridine c. PhCOCl, pyridine d. (Boc)2O e. C6H5N=C=S

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

829

Amino Acids and Proteins 29–5

5. Draw the amino acids and peptide fragments formed when the octapeptide Tyr–Gly–Ala–Lys–Val–Ser–Phe–Met is treated with each reagent or enzyme: a. chymotrypsin b. trypsin c. carboxypeptidase d. C6H5N=C=S Answers to Practice Test 1. a. 1 b. 2 c. 5 2. a

4. O

Val

OCH2Ph

a.

H CH2OH O H H2N C C C N N C C OH C H O H CH(CH3)2 H CH3 O

H2N H O OH

b.

HN H

Ala

Ser

5.a. Tyr, Gly–Ala–Lys–Val–Ser–Phe, Met b. Tyr–Gly–Ala–Lys, Val–Ser–Phe– Met c. Tyr–Gly–Ala–Lys–Val–Ser–Phe, Met d. Tyr, Gly–Ala–Lys–Val–Ser–Phe–Met

O

b. Val c. Ala

O

3. a. H2N S

OH

c.

COOH

HN H

H

O Ph

CH2CH(CH3)2

O

b. 6.04

O

c.

OH

d.

O

HN H

NH3

Boc O

O C6H5

d.

O

e.

NH2

CH2 S

N H

e. neutral

Answers to Problems 29.1 H CH3 O

S

S C

CH3 OH

H2N H

L-isoleucine

R

O

H CH3 O

S C

R C

H

H2N H

OH

S

H NH2

CH3 OH

H O

R

R C

H NH2

OH

830

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–6

29.2 a.

b.

NH3+

(CH3)2CH

C

COO–

c.

NH3+

COO–

+

HOOCCH2CH2 C COO–

N

H

H

NH3+

d.

C COO–

(CH3)2CHCH2

H

H H

29.3

In an amino acid, the electron-withdrawing carboxy group destabilizes the ammonium ion (–NH3+), making it more readily donate a proton; that is, it makes it a stronger acid. Also, the electron-withdrawing carboxy group removes electron density from the amino group (–NH2) of the conjugate base, making it a weaker base than a 1o amine, which has no electron-withdrawing group.

29.4

The most direct way to synthesize an α-amino acid is by SN2 reaction of an α-halo carboxylic acid with a large excess of NH3.

a.

NH3

Br CH COOH H

b. Br CH COOH CH CH3

c. Br CH COOH

H2N CH COO– NH4+

large excess

H

NH3

CH2

glycine

NH3

large excess

H2N CH COO– NH4+ CH2

H2N CH COO– NH4+

large excess

CH CH3

CH2

phenylalanine

CH2

CH3

CH3

isoleucine

29.5 a.

CH3

CH3I

H2N C COOH

alanine

c.

CH3CH2CH(CH3)Br

H

b.

CH(CH3)CH2CH3 H2N C COOH

isoleucine

H CH2CH(CH3)2

(CH3)2CHCH2Cl

H2N C COOH

leucine

H

29.6 O

H H C N C COOEt

CH3

COOEt

[1] NaOEt [2] CH2=O [3] H3O+, Δ

CH2OH H2N C COOH H

serine

29.7 O

a. H2N CHCOOH CH CH3 CH3

valine

(CH3)2CH

C

O

b. H2N CHCOOH H

CH2

(CH3)2CHCH2

C

O

c. H2N CHCOOH H

CH2

CH CH3 CH3

leucine phenylalanine

C6H5CH2

C

H

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

831

Amino Acids and Proteins 29–7

29.8 a. BrCH2COOH

NH3 large excess

H

b. CH3CONH C COOEt COOEt

NH2CH2COO– NH4+

H

[1] NaOEt [2] (CH3)2CHCl

CH(CH3)CH2CH3 H

[1] NaOEt

d. CH3CONH C COOEt

CH(CH3)2

H2N CHCOOH

[2] H3O+

H

H2N C COOH

[3] H3O+, Δ

[1] NH4Cl, NaCN

c. CH3CH2CH(CH3)CHO

H2N C COOH

[2] BrCH2CO2Et

COOEt

CH2CO2H

[3] H3O+, Δ

A chiral amine must be used to resolve a racemic mixture of amino acids.

29.9

N CH3 H

a. C6H5CH2CH2NH2 achiral

c.

b.

d.

C

N

CH3CH2

achiral

H

NH2

N

chiral (can be used)

H

O

H HO

chiral (can be used)

29.10 NH2

+

Ac2O AcNH

COOH

C

C

R H2 N

proton transfer

C6H5

C

(R isomer only)

CH3 H

+

COO–

H3N

C

C6H5

CH3 H

H CH2CH(CH3)2

enantiomers

(CH3)2CHCH2 H

S

AcNH

COOH

C

+

H CH2CH(CH3)2

Step [1]: React both enantiomers with the R isomer of the chiral amine.

enantiomers

R

S

AcNH

COOH

C

(CH3)2CHCH2 H

H CH2CH(CH3)2

To begin: Convert the amino acids into N-acetyl amino acids (two enantiomers).

NH2

COOH

C

S

R

AcNH

COO–

C

(CH3)2CHCH2 H

+

H3 N

C

C6H5

CH3 H

diastereomers

R

R

These salts have the same configuration around one stereogenic center, but the opposite configuration about the other stereogenic center.

separate

Step [2]: Separate the diastereomers.

AcNH

C

COO–

H CH2CH(CH3)2

S

+

H3N

C

CH3 H

R

C6H5

AcNH

C

(CH3)2CHCH2 H

R

COO–

+ H3N

C

CH3 H

R

C6H5

832

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–8

H2O, –OH

Step [3]: Regenerate the amino acid by hydrolysis of the amide.

NH2

C

H2O, –OH

COOH

NH2

H CH2CH(CH3)2

COOH

C

H2N

+

(CH3)2CHCH2 H

(S)-leucine

C

C6H5

CH3 H

(R)-leucine

The chiral amine is also regenerated.

The amino acids are now separated.

29.11 COOH

H2N

[1] (CH3CO)2O

H2N C H

[2] acylase

CH2CH(CH3)2

C

H COOH N C C O (CH3)2CHCH2 H

COOH

CH3

+

H CH2CH(CH3)2

(S)-leucine

(mixture of enantiomers)

N-acetyl-(R)-leucine

29.12 H

a. H2N CHCOOH CH3

(CH3)2CH

NHAc

b. H2N CHCOOH C C H COOH CH2

H2NCOCH2

NHAc C C

H

c. H2N CHCOOH CH2

COOH

CH CH3

CH2

CH3

CONH2

NHAc

C C COOH

H

29.13 Draw the peptide by joining adjacent COOH and NH2 groups in amide bonds. amide O

a.

H2N CH C OH

(CH3)2CH H

O H2N CH C OH

CH CH3

CH2

CH3

CH2

Val

COOH

H2N

C

N-terminal

O

O

H2N CH C OH H2N CH C OH

H

CH2

Gly

C

C

C-terminal OH

H CH2CH2COOH

amide

O H2N CH C OH

O

O

Val–Glu

Glu

b.

C

H N

NH N

CH2

Leu His

H2N

CH CH3 CH3

(CH3)2CHCH2 H O

H H C

C O

N-terminal

H N

C

C

H CH2

N H

amide

NH

Gly–His–Leu N

C

C O

OH

C-terminal

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

833

Amino Acids and Proteins 29–9

O

O

O

O

c. H2N CH C OH H2N CH C OH H2N CH C OH H2N CH C OH CH2 CH3 CH OH CH OH CH2SCH3

M

A

CH3

CH3

T

T

N-terminal

CH2SCH3 amide

CH3

CH2 H

O

CHOH H

C

C

C

H2N

H N

C O

C

CH3

H

N H

amide

C

O

H N

O

C

H

OH

C-terminal

CHOH CH3

M–A–T–T

29.14 N HN

CONH2

a.

H2N

C

O

CH2 H

C

C

N H

C O

H CH2

O

H N

C

C

H2N

b.

OH

H CH(CH3)2

C

Arg–Asn–Val R–N–V

NH

C

N H

C

O

H N

O

C

C

OH

H CH2

CH2

CH2

CH2

CONH2

CH2

C HN

CH2 H

C

H CH2

CH2 CH2

O

Lys–His–Gln K–H–Q

NH2

NH2

29.15 There are six different tripeptides that can be formed from three amino acids (A, B, C): A–B–C, A–C–B, B–A–C, B–C–A, C–A–B, and C–B–A. 29.16 O H2 N

N H

H N O

O N H

O

H N

OH

O

leu-enkephalin

HO

29.17 O HO HS O

a. H2N

H N

N COOH

H

O

glutathione

O

N O

H

H O

H2N

NH2 O

S OH

COOH

N

S

H N

N COOH

H

O

O OH

834

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–10

b.

O

HS

H2N

α COOH

H

O

N

N COOH

H2N

H

The peptide bond beween glutamic acid and its adjacent OH amino acid (cysteine) is formed from the COOH in the R

group of glutamic acid, not the α COOH.

O

This comes from the amino acid glutamic acid. O

α

OH

This carboxy group is used to form the amide bond in the peptide, not the α COOH, as is usual. That's what makes glutathione's structure unusual.

COOH

glutamic acid

29.18 O C6H5

a. S

O C6H5

N

CH3 N H

b. S

from Ala

N N H

from Val

29.19 Determine the sequence of the octapeptide as in Sample Problem 29.2. Look for overlapping sequences in the fragments. common amino acids Answer: Ala–Leu–Tyr Tyr–Leu–Val–Cys

Ala–Leu–Tyr–Leu–Val–Cys–Gly–Glu

Val–Cys–Gly–Glu

29.20 Trypsin cleaves peptides at amide bonds with a carbonyl group from Arg and Lys. Chymotrypsin cleaves at amide bonds with a carbonyl group from Phe, Tyr, and Trp. a. [1] Gly–Ala–Phe–Leu–Lys + Ala [2] Phe–Tyr–Gly–Cys–Arg + Ser [3] Thr–Pro–Lys + Glu–His–Gly–Phe–Cys–Trp–Val–Val–Phe b. [1] Gly–Ala–Phe + Leu–Lys–Ala [2] Phe + Tyr + Gly–Cys–Arg–Ser [3] Thr–Pro–Lys–Glu–His–Gly–Phe + Cys–Trp + Val–Val–Phe

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

835

Amino Acids and Proteins 29–11

29.21 Edman degradation gives N-terminal amino acid: Carboxypeptidase identifies the C-terminal amino acid:

Leu–___–___–___–___–___–___ Leu–___–___–___–___–___–Glu

Partial hydrolysis common amino acids Ala–Ser–Arg Gly–Ala–Ser Leu–Gly–Ala–Ser–Arg–___–Glu or

Gly–Ala–Ser–Arg

Leu–___–Gly–Ala–Ser–Arg–Glu Cleavage by trypsin is after Arg and yields a dipeptide; therefore, this must be the peptide:

Leu–Gly–Ala–Ser–Arg–Phe–Glu

29.22 a.

O

(CH3)2CHCH2 H H2 N

C

Leu

OH

C O

(CH3)2CHCH2 H Boc N H

C

(CH3)2CHCH2 H [(CH3)3COCO]2O C OH Boc N C (CH3CH2)3N H O

O

C OH + H2N OCH2C6H5 C C O

H CH(CH3)2

H2N

C

O

C

C6H5CH2OH, H+

OH

H CH(CH3)2

H2N

C

(CH3)2CHCH2 H Boc N H

C

C O

O

H N

C

C

OCH2C6H5

H CH(CH3)2 HBr, CH3COOH

(CH3)2CHCH2 H H2N

C

C O

H N

OCH2C6H5

H CH(CH3)2

Val new amide bond

DCC

C

O C

C

OH

H CH(CH3)2

Leu–Val

836

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–12

O CH3 H

b.

C

H2N

Ala

C

CH3 H [(CH3)3COCO]2O

OH

Boc N H

(CH3CH2)3N

O

C

H2N

C

OH

C6H5CH2OH, H+

H2N

H CH(CH3)CH2CH3

O

A

C

OH

O

C

C

OCH2C6H5

CH(CH3)CH2CH3

H

Ile

C

B

new amide bond A

+

CH3 H

DCC

B

Boc N H

C

C O

O H2N

C

C H H

H N

C

O

H2

C

Pd-C

OCH2C6H5

CH3 H Boc N H

C

OH

H2N

C

C

C

C

OH

H CH(CH3)CH2CH3

C

O C6H5CH2OH, H+

C O

H CH(CH3)CH2CH3

O

H N

OCH2C6H5

H H

Gly

new amide bond O

C

+ H2N

C H H

C

CH3 H OCH2C6H5

DCC Boc N H

C

C O

O

H N

C

H

C

N H

C

OCH2C6H5

HBr CH3COOH

O CH(CH3)CH2CH3

CH3 H

Ala–Ile–Gly

H2N

C

C O

H N H

O C

C

N H

C

OH

O CH(CH3)CH2CH3

837

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Amino Acids and Proteins 29–13

CH3 H

c.

H2N

C

CH3 H

C

OH [(CH3)3COCO]2O (CH3CH2)3N

Ala O

C

Boc N H

O OH

C

H2N

O

C

C H H

A

O C6H5CH2OH, H+

OH

+

A

DCC

B

Boc N H

CH3 H H2N

C

C O

C

C

H2 OCH2C6H5

Boc N H

Pd-C

C

C

H H

C

OH

H2N

+

H2N

C

C O

C

OH

H H

C

C

C

OCH2C6H5

new amide bond OCH2C6H5

DCC

CH3 H

BOC N H

C

C O

H N

O C

C

CH3 H N H

H H

C

C

H2

OCH2C6H5

Pd-C

O CH3 H Boc N H

C

B

DCC

CH3 H CH3 H O O H H C N C N C C C C C C N OCH2C6H5 Boc N H H H H O O H H

O

H N

C

CH3 H

C

C

N H

C

OH

O

H H

D HBr CH3COOH

CH3 H H2N

C O

new amide bond +

C

O CH3 H

D

B O

H N

O

OCH2C6H5

CH3 H C6H5CH2OH, H+

Ala O

C

C

CH3 H

O

H N

C

C H H

Gly

new amide bond CH3 H

H2N

C

C O

H N

O C

C

H H

CH3 H N H

C

C O

Ala–Gly–Ala–Gly

H N

O C H H

C

OH

838

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–14

29.23 The dipeptide depicted in the 3-D model has alanine as the N-terminal amino acid and cysteine as the C-terminal amino acid. O H2N

H Boc N

[(CH3)2COCO]2O OH

O

OH

H2N

OH

(CH3CH2)3N

H CH3

HSCH2 H

O

A O

H Boc N

OCH2C6H5

H2N

H CH3

H Boc N

DCC

HSCH2 H O

H CH3

O

A

O

B

Cys

HSCH2 H OH

OCH2C6H5

H2 N

H+

H CH3

Ala

HSCH2 H

C6H5CH2OH

OCH2C6H5

N H

O

B HBr, CH3COOH HSCH2 H O H2N H CH3

N H

Ala–Cys

29.24 All Fmoc-protected amino acids are made by the following general reaction: R H2N

H C

C O

R OH

Fmoc–Cl Na2CO3, H2O

Fmoc N H

H C

C O

OH

OH O

839

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Amino Acids and Proteins 29–15

H H

The steps:

Fmoc N H

C

OH

C

[1] base [2] Cl CH2 POLYMER

O

H H C

Fmoc N H

C

O CH2 POLYMER

O [1]

[2] DCC

N H H Fmoc N

O C

C

OH

H CH(CH3)CH2CH3 (CH3)2CHCH2 H C

Fmoc N H

C

H N

O C C

O H

H

H C

N H

[1]

C

O CH2 POLYMER

O CH(CH3)CH2CH3

N H [2] DCC

H Fmoc N

Fmoc N H

N H

CH3 H [2] DCC C OH Fmoc N C H O

C

C

C

C N H

C

O CH2 POLYMER

O H CH(CH )CH CH 3 2 3

(CH3)2CHCH2 H

[1]

H H

O

OH

C O

[1] (CH3)2CHCH2 O H H O H H H2N C C C C OH N C N N C C C H H O O H CH3 H CH(CH )CH CH

N (CH3)2CHCH2 H H H O O H H [2] HF H Fmoc N C C C C O CH2 POLYMER N C N N C C C H H O O H CH3 H CH(CH3)CH2CH3

3

2

3

Ala–Leu–Ile–Gly

+ F CH2 POLYMER

29.25 In a parallel β-pleated sheet, the strands run in the same direction from the N- to C-terminal amino acid. In an antiparallel β-pleated sheet, the strands run in the opposite direction.

H

H

O CH3 H

N

C

C

N

H CH3 H

H

C

C O

H N

O CH3 H C

C

N

H CH3 H

C

C

OH

O

H CH3 O H H CH3 O H HO C C N C C N C N C C N C H O H CH H O H CH H 3

H

H

N

N

O CH3 H H O CH3 H C C N C C OH C N C C N C H H O H CH3 H CH3 O

parallel

H

C

O C

H CH3

3

CH3 H

H

C

N

N H

C O

O CH3 H C

C

H CH3

antiparallel

N H

C

C O

OH

840

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–16

29.26 a. Ser and Tyr

b. Val and Leu

H2N CH COOH

H2N CH COOH

CH2

CH2

H2N CH COOH

OH

side chains with OH groups

c. 2 Phe residues

H2N CH COOH

CH CH3

CH2

CH3

CH CH3

H2N CH COOH

H2N CH COOH

CH2

CH2

CH3

side chains with only C–C and C–H bonds

OH

van der Waals forces

van der Waals forces

hydrogen bonding

29.27 a. The R group for glycine is a hydrogen. The R groups must be small to allow the β-pleated sheets to stack on top of each other. With large R groups, steric hindrance prevents stacking. b. Silk fibers are water insoluble because most of the polar functional groups are in the interior of the stacked sheets. The β-pleated sheets are stacked one on top of another, so few polar functional groups are available for hydrogen bonding to water. 29.28 H CH2C6H5 H2N

C

c.

H CH2C6H5

a. CH3OH, H+ H2N

C

C

H3N COOCH3 H CH2C6H5

O

b. CH3COCl, pyridine

CH3

C

N H

COOH

H CH2C6H5

d. NaOH (1 equiv)

COOH

phenylalanine

H CH2C6H5

HCl (1 equiv)

C

C O

H2 N

C

COO

OH

O C6H5

e. C6H5N=C=S

N

CH2C6H5

N H

S

29.29 a. N-terminal amino acid: alanine C-terminal amino acid: serine b. A–Q–C–S

c. Amide bonds are bold. C-terminal HSCH2 O

CH3 H H2N

C

C O

H N H

C

C

H C

N H

C

O H N

C C

OH

O H CH2OH CH2CH2CONH2

N-terminal Ala–Gln–Cys–Ser A–Q–C–S

841

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Amino Acids and Proteins 29–17

29.30 The dipeptide is composed of phenylalanine and leucine. O

C6H5CH2 H H2N

C

C

C6H5CH2 H

[(CH3)3COCO]2O OH

(CH3CH2)3N

Boc N H

O

Phe

C

C

OH

C

+ B

DCC

Boc N H

C

H N

C O

O

C

C6H5CH2OH, H+

OH

O

C

C

OCH2C6H5

B C6H5CH2 H

HBr OCH2C6H5

C

H CH2CH(CH3)2

Leu

O C

H2N

H CH2CH(CH3)2

A C6H5CH2 H

A

H2N

CH3COOH

H2N

H CH2CH(CH3)2

C

C O

H N

O C

C

OH

H CH2CH(CH3)2

Phe–Leu

29.31 R COOH

a.

S

H2 N C H

COOH

CH3 C SH

COOH

b.

H C NH2

H2 N C H

CH3 C SH

CH3

CH3 C S

CH3

(R)-penicillamine

CH3

(S)-penicillamine

COOH H2 N C H S C CH3 CH3

29.32 Amino acids are insoluble in diethyl ether because amino acids are highly polar; they exist as salts in their neutral form. Diethyl ether is weakly polar, so amino acids are not soluble in it. N-Acetyl amino acids are soluble because they are polar but not salts. NH3+ C

R H

NHCOCH3 R H

COO–

C COOH

N-acetyl amino acid ether soluble

amino acid, a salt H2O soluble and ether insoluble

29.33 The electron pair on the N atom not part of a double bond is delocalized on the fivemembered ring, making it less basic. H2N CH COOH CH2

HA

H2N CH COOH CH2

+

NH N

N

When this N is protonated... H2N CH COOH CH2 NH N

sp3 hybridized N atom

NH2

...the ring is no longer aromatic. H2N CH COOH

HA

preferred path

When this N is protonated...

H2N CH COOH

CH2

CH2

+

NH

NH

+N H

N H

...the ring is still aromatic.

6 π electrons

842

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–18

29.34 H2N CH COOH

H2N CH COOH

CH2

CH2

The ring structure on tryptophan is aromatic since each atom contains a p orbital. Protonation of the N atom would disrupt the aromaticity, making this a less favorable reaction.

H2N

HN

+

no p orbital on N This electron pair is delocalized on the bicyclic ring system (giving it 10 π electrons), making it less available for donation, and thus less basic.

29.35 At its isoelectric point, each amino acid is neutral. a. +

COO–

H3N C H

b.

COO–

+

H3N C H

CH3

alanine

CH2CH2SCH3

methionine

c.

+

COO–

H3N C H CH2COOH

aspartic acid

d.

COO– H2N C H

+

CH2CH2CH2CH2NH3

lysine

29.36 a. [1] glutamic acid: use the pKa’s 2.10 + 4.07 [2] lysine: use the pKa’s 8.95 + 10.53 [3] arginine: use the pKa’s 9.04 + 12.48 b. In general, the pI of an acidic amino acid is lower than that of a neutral amino acid. c. In general, the pI of a basic amino acid is higher than that of a neutral amino acid. 29.37 a. threonine pI = 5.06 (+1) charge at pH = 1 H3N CH COOH

b. methionine pI = 5.74 (+1) charge at pH = 1 H3N CH COOH

c. aspartic acid pI = 2.98 (+1) charge at pH = 1 H3N CH COOH

d. arginine pI = 5.41 (+2) charge at pH = 1 H3N CH COOH

CH OH

CH2

CH2

CH2

CH3

CH2

COOH

CH2

S

CH2

CH3

NH C NH2 NH2

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

843

Amino Acids and Proteins 29–19

29.38 a. valine pI = 6.00 (–1) charge at pH = 11

b. proline pI = 6.30 (–1) charge at pH = 11

c. glutamic acid pI = 3.08 (–2) charge at pH = 11

COO–

H2N CH COO–

H2N CH COO–

CH2

CH2

CH2

CH2

COO–

CH2

H2N CH COO– CH CH3 HN

CH3

d. lysine pI = 9.74 (–1) charge at pH = 11

CH2 NH2

29.39 The terminal NH2 and COOH groups are ionizable functional groups, so they can gain or lose protons in aqueous solution. O

a.

CH3 H

H2N CH C OH CH3

H2N

Ala b. At pH = 1

C

C O

CH3 H H3N

C

C O

H N

O CH3 C

C

H CH3

N H

H N

O CH3 C

C

H CH3

N H

H C

C

OH

A–A–A

O

H C

C

OH

O

c. The pKa of the COOH of the tripeptide is higher than the pKa of the COOH group of alanine, making it less acidic. This occurs because the COOH group in the tripeptide is farther away from the –NH3+ group. The positively charged –NH3+ group stabilizes the negatively charged carboxylate anion of alanine more than the carboxylate anion of the tripeptide because it is so much closer in alanine. The opposite effect is observed with the ionization of the –NH3+ group. In alanine, the –NH3+ is closer to the COO– group, so it is more difficult to lose a proton, resulting in a higher pKa. In the tripeptide, the –NH3+ is farther away from the COO–, so it is less affected by its presence.

844

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–20

29.40

H2N

H CH2CH(CH3)2 H2N

C

C

g. C6H5COCl, pyridine

C6H5

COOCH3

b. CH3COCl, pyridine

C

CH3

N H

C

C

C

H2N

N H

C

C O

Boc N H

C

COOH

O

i. The product in (d), then NH2CH2COOCH3 + DCC

H CH2CH(CH3)2 H C N COOCH3 AcNH C C

COOCH2C6H5

O H CH2CH(CH3)2

d. Ac2O, pyridine

C

AcNH

e.

C

H3N

H CH2CH(CH3)2 H C N COOCH Boc N C C 3 H O H H

H2N

C

H CH2CH(CH3)2

k. Fmoc–Cl, Na2CO3, H2O Fmoc N H O

COOH

H CH2CH(CH3)2

f. NaOH (1 equiv)

C6H5

l. C6H5N=C=S

COO

N

S

Br

b. CH3CONHCH(COOEt)2

NH3

(CH3)2CHCH2CHCOO– NH4+ + NH4+Br–

excess

NH3 NH2

[1] NaOEt [2] O C O CH3

HO

CH2 CHCOOH

CH2Br

[3] H3O+, Δ O

NH2

O

c.

[1] NH4Cl, NaCN H

N H

[2] H3O+

HOOC

O

NH2

O [1] NH4Cl, NaCN

d. CH3O

CHO

e. CH3CONHCH(COOEt)2

[2] H3O+

COOH

HO

[1] NaOEt [2] ClCH2CH2CH2CH2NHAc [3] H3O+, Δ

NH2 CH2CH2CH2CH2NH2 H2N C COOH H

C

COOH

CH2CH(CH3)2 N H

29.41 a. (CH3)2CHCH2CHCOOH

H H

j. The product in (h), then NH2CH2COOCH3 + DCC

COOH

H CH2CH(CH3)2

HCl (1 equiv)

OH

H CH2CH(CH3)2

h. [(CH3)3COCO]2O (CH3CH2)3N

OH

H CH2CH(CH3)2

c. C6H5CH2OH, H+

C

H CH2CH(CH3)2

O

COOH

leucine

H CH2CH(CH3)2

O

H CH2CH(CH3)2

+ a. CH3OH, H

845

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Amino Acids and Proteins 29–21

29.42 a. Asn

b. His

O

H2N CH C OH

c. Trp

O

H2N CH C OH CH2Br

CH2 C O

C O

NH2

NH2

CH2Br

CH2

O CH2Br

H2N CH C OH CH2

NH

NH

HN

N

N

HN

29.43 OH O

H

H C N C COOEt

CH3

CHCH3

[1] NaOEt

H2N C COOH

[2] CH3CHO [3] H3O+, Δ

COOEt

H

threonine

29.44 Br2

a. CH3CHO

CH3

C

+

NH3

H3NCH2COO–

excess

glycine

NH2

[1] NH4Cl, NaCN H

BrCH2COOH

H2SO4, H2O

CH3COOH

O

b.

CrO3

BrCH2CHO

CH3 C COOH

[2] H3O+

H

alanine

29.45 O N–K+ CH2(COOEt)2

Br2 Br CH(COOEt)2 CH3COOH

O

O

N

CH(COOEt)2

A B

O [1] NaOEt [2] ClCH2CH2SCH3 O

H H2N C COOH CH2CH2SCH3 D

[1] NaOH, H2O

COOEt N

[2] H3O+, Δ

C COOEt CH2CH2SCH3

O

C

846

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–22

29.46 O

OEt H H C N C COOEt

CH3

COOEt

[1]

O

NaOEt

CH2 CHCOOEt H EtOH C N C COOEt +

CH3

O

[2]

COOEt H C N C COOEt

CH3

COOEt

CH2–CHCOOEt H OH2 [3]

H2N CHCOOH

H3O+

CH2

Δ

CH2COOH

O

COOEt H C N C COOEt

CH3

CH2CH2COOEt

glutamic acid + H O 2

29.47 CN N

CHO

[1] DIBAL-H [2] H2O

CO2Et

Ph3P=CHCO2Et

N

N

CO2Et

H2 Pd-C

(+ cis isomer)

N

B

A

C KOH H2O

O H2N

O NH N

OCH3

O

E

OCH3 Boc NH H

CO2H

Boc NH H

DCC

N

D

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

847

Amino Acids and Proteins 29–23

29.48 CH3O

O

H

CH3O

Cl

O

H

N

Cl

enantiomers

N

S

S

R isomer

S isomer

proton transfer O

HO3S CH3O

O

H

CH3O

Cl

O

H

NH

Cl

NH

S

diastereomers

S

O

O3S

O

O3S

separate

CH3O

O

H

CH3O

Cl

O

H

NH

Cl

NH

S

S

O

O3S

O

O3S

–OH

–OH

CH3O

O

H

CH3O

Cl

N S

O

H N S

R isomer

S isomer

Cl HO3S

O

The chiral sulfonic acid is regenerated.

848

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–24

29.49 H CH3 C

NH2

S

COOH

To begin: Convert the amino acids into amino acid esters (two enantiomers).

H C

H2N

enantiomers

COOH

R

CH3OH, H+ H CH3 C

H2N

S

COOCH3

CH3 H2N

H C

enantiomers

COOCH3

R

H OH

Step [1]: React both enantiomers with the R isomer of mandelic acid.

proton transfer

C

COO–

H3N

C

S

COOCH3

COOH

CH3

H OH

H CH3

R

(R)-mandelic acid

C

C6H5

H OH C6H5

CH3

C

C6H5

R

COO– H3N

H C

R

COOCH3 diastereomers

These salts have the same configuration around one stereogenic center, but the opposite configuration about the other stereogenic center.

separate

Step [2]: Separate the diastereomers.

H OH C6H5

Step [3]: Regenerate the amino acids by hydrolysis of the esters.

C

R

H CH3 –

COO

H3N

C

S

COOCH3

C6H5

C

COO–

H2N

C

S

COOH

H3N

R

H2O, –OH

H CH3

CH3

H OH

H C

R

COOCH3

H2O, –OH

CH3 H2N

H C

R

COOH

+

H OH C6H5

C

COOH

The chiral acid is regenerated.

849

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Amino Acids and Proteins 29–25

29.50 NH2

COOH

C

NH2

+

R

S

AcNH

enantiomers

C6H5CH2 H

H CH2C6H5

To begin: Convert the amino acids into N-acetyl amino acids (two enantiomers).

COOH

C

Ac2O COOH

C

AcNH

+

COOH

C

enantiomers

C6H5CH2 H

H CH2C6H5

R

S

N

proton transfer

Step [1]: React both enantiomers with the R isomer of the chiral amine.

CH3O CH3O

N

brucine H

O

O

H

H

N AcNH

C

COO–

H CH2C6H5

S

AcNH

CH3O CH3O

N

COO–

C

C6H5CH2 H

R

H

O

N CH3O CH3O

N

H

O

O

O

diastereomers separate

Step [2]: Separate the diastereomers. H AcNH

C

N

COO– CH3O

H CH2C6H5

S

H AcNH

COO–

C

N CH3O

C6H5CH2 H CH3O

N O

Step [3]: Regenerate the amino acid by hydrolysis of the amide.

R

H

CH3O

H2O, –OH

C

COOH

H CH2C6H5

(S)-phenylalanine

H

O

O

NH2

N

O

H2O, –OH NH2

C

N

COOH

C6H5CH2 H

(R)-phenylalanine

The amino acids are now separated.

+

CH3O CH3O

N O

H O

The chiral amine is also regenerated.

850

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–26

29.51 NH2

Ac2O

a. (CH3)2CH CH COOH racemic mixture

H2N

acylase

C

AcNH

COOH

H CH(CH3)2

H CH(CH3)2

CH3CONH NHCOCH3

b.

COOH

H2

–OH

chiral Rh catalyst

H2O

H2N

COOH

C

+

COOH

C

CH2CH2CH2CH2NH2

H

(S)-isomer

COOH

c.

NHAc

OH

chiral Rh catalyst

N H

H2N

–

H2

H2O

C

COOH

(S)-isomer

CH2

H

N H

29.52 CH3O

COOH

O CH3

N

H O

H2

CH3O

chiral Rh catalyst

CH3

COOH CH3

H N H

O

HO

strong acid

COOH H NH2

HO

O

O

CH3

A

L-dopa

O

29.53 a.

C6H5CH2 H H2N

C

C

O

H N

O

C

C

H

c.

H2NCH2CH2CH2CH2 H

OH

H2N

CH3

H H H2N

C

C O

H N

C

O C

C

OH

H H

Lys–Gly d.

O C

H N

O

Phe–Ala b.

C

C

NH2C(NH)NHCH2CH2CH2 H

OH

H2N

C

C O

H CH2

H N

O C

C

OH

CH2

H

R–H

CH2CONH2

NH

Gly–Gln

N

29.54 Amide bonds are bold lines (not wedges). C-terminal (CH3)2CH O H O H H C N C OH C C N C C C N C H2N H H CH2C6H4OH O O H CH2CH2CH2NHC(NH)NH2

HO2CCH2 H

N-terminal Asp–Arg–Val–Tyr D–R–V–Y

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

851

Amino Acids and Proteins 29–27

29.55 Name a peptide from the N-terminal to the C-terminal end. C-terminal

CH2COOH H H

a.

H2N

C

C

H N

O

C

O

CH2 H

C

C

N H CH H 2 COOH

C

b.

OH

HOOC

H N

C

H CH2

O

H H C O

C

O N H

C

N-terminal C

NH2

H CH3

CH2 CH2

Gly–Asp–Glu G–D–E

NH

Ala–Gly–Arg A–G–R

C HN

NH2

29.56 A peptide C–N bond is stronger than an ester C–O bond because the C–N bond has more double bond character due to resonance. Since N is more basic than O, an amide C–N bond is more stabilized by delocalization of the lone pair on N. O R

C

A

O R

NH2

C

B

+

NH2

Structure B contributes greatly to the resonance hybrid and this shortens and strengthens the C–N bond.

29.57 O CH3 H H2 N

C

C

N

C

H CH3 H

C

O OH

O

s-trans

H 2N

C

C

N

H

H CH3 C OH C s-cis HCH3 O

29.58 a. b. c. d.

A–P–F + L–K–W + S–G–R–G A–P–F–L–K + W–S–G–R + G A–P–F–L–K–W–S–G–R + G A + P–F–L–K–W–S–G–R–G

29.59 common amino acids

a.

Answer:

Gly–Ala Ala–His

Gly–Ala–His–Tyr

His–Tyr common amino acids

b.

Answer:

Lys–His His–Gly–Glu Gly–Glu–Phe

Lys–His–Gly–Glu–Phe

852

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–28

29.60 common amino acids

Arg–Arg–Val

Answer:

Val–Tyr Tyr–Ile–His

Arg–Arg–Val–Tyr–Ile–His–Pro–Phe

Ile–His–Pro–Phe

29.61 Gly is the N-terminal amino acid (from Edman degradation), and Leu is the C-terminal amino acid (from treatment with carboxypeptidase). Partial hydrolysis gives the rest of the sequence. common amino acids Answer:

Gly–Ala–Phe–His

Gly–Gly–Ala–Phe–His–Ile–His–Leu

Phe–His–Ile Ile–His–Leu

29.62 Edman degradation data give the N-terminal amino acid for the octapeptide and all smaller peptides. A

B A: Glu–Arg–Val–Tyr B: Ile–Leu–His–Phe C: Glu–Arg D: Val–Tyr

C D Glu–Arg–Val–Tyr–Ile–Leu–His–Phe

octapeptide

cleavage with trypsin

cleavage with chymotrypsin

carboxypeptidase

Glu–Arg–Val–Tyr–Ile–Leu–His + Phe

29.63 A and B can react to form an amide, or two molecules of B can form an amide. H H Boc N H

C

H

O OH + H2N

C O

C

C

H

DCC OH

C

Boc N H

O

H CH(CH3)2

A

C

H N

C

29.64 O C

C

O CH3OH, H+

OH

H2 N

H CH(CH3)2 O

b.

H2 N

C

C

C

C

OCH3

H CH(CH3)2 O C6H5CH2OH, H+

OH

H2 N

H CH2CH(CH3)2

c. NH2CH2COOH

[(CH3)3COCO]2O (CH3CH2)3N

H Boc N

C

C

OCH2C6H5

H CH2CH(CH3)2 O C H H

C

OH

+ OH

H CH(CH3)2

B

a. H2N

(CH3)2CH

O

H2 N

H C

C O

H N

O C

OH

H CH(CH3)2

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

853

Amino Acids and Proteins 29–29

d. product in (b) + product in (c)

H

O

H Boc N

DCC

C

C

N H

CH2CH(CH3)2 C

C O

H H

e. (CH3)3CO

C

H

O

N

C

O

C

H H2

OCH2C6H5

(CH3)3CO

Pd-C

H CH(CH3)2

O

N

C

OCH2C6H5

O

C

C

OH +

C6H5CH3

H CH(CH3)2

O HBr

f. starting material in (e)

H2N

CH3COOH

C

C

OH

H CH(CH3)2 O

CF3COOH

g. product in (e)

H2N

C

C

OH

H CH(CH3)2 O C

h.

O

Na2CO3

+ Fmoc–Cl

OH

C

H2O

H2N H

OH

HN H Fmoc

29.65 O H H

a. H2N

C

Gly

C

H H

[(CH3)3COCO]2O OH

(CH3CH2)3N

Boc N H

O

DCC Boc N H

C

C O

H N

C

OH

O

A H H

A + B

C

H2N H

C

C

O OH

C

H CH3

H2N

CH3

B H H

OCH2C6H5

HBr CH3COOH

C

C

H CH3

Ala

O C

C6H5CH2OH, H+

H2N

C

C O

H N

O C

C

H CH3

Gly–Ala

OH

OCH2C6H5

854

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–30

CH3 CH3CH2CH

b.

C

H2N

CH3 H

Ile

OH

C

(CH3CH2)3N

O

C6H5CH2 H2N

C

C

C

C

OH

O OH

C6H5CH2OH, H+

H CH3

O

C

Phe

C O

C6H5CH2OH, H

C

OCH2C6H5

B

Ala

H N

C O

H

O C

C

CH3CH2CH H

H2 Pd-C

OCH2C6H5

C

Boc N H

CH3

C6H5CH2 H

OH

C

H CH3

C

C

+

H2N

C

C O

DCC

Boc N H

C

H N

O

C

C

C

OH

H CH3

C6H5CH2 H O

CH3CH2CH H C

OCH2C6H5

O

H N

O

CH3 H

H2N

CH3

DCC Boc N H

H2N

A

CH3CH2CH H

+ B

C

Boc N H

CH3

A

O

CH3CH2CH H

[(CH3)3COCO]2O

C

C

N H

C

C

OCH2C6H5

O

H CH3 HBr CH3COOH

CH3 C6H5CH2 H O

CH3CH2CH H H2N

C

C O

H N

C

C

H CH3

Ile–Ala–Phe

N H

C

C O

OH

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

855

Amino Acids and Proteins 29–31

29.66 Make all the Fmoc derivatives as described in Problem 29.24. C6H5CH2 H

a. Fmoc N H

C

C

OH

C6H5CH2 H

[1] base [2] Cl CH2 POLYMER

C

Fmoc N H

O

C

O CH2 POLYMER

O

[1]

(Fmoc-Phe)

N H

H Fmoc N

[2] DCC

O C

C

OH

H CH2C6H5 C6H5CH2 H O

(CH3)2CHCH2 H Fmoc N H

C

C O

H N

N H C O CH2 POLYMER N C [2] DCC + H (CH3)2CHCH2 H H CH2C6H5 O C

C6H5CH2 H O

[1]

C

Fmoc N H

[1] N H

C

C

H Fmoc N

(Fmoc-Phe)

C

C O CH2 POLYMER N C H H CH2C6H5 O C

OH

O

(Fmoc-Leu)

[2] DCC CH3 H

+

Fmoc N H

C

C O

OH

(Fmoc-Ala)

(CH3)2CHCH2 C6H5CH2 O H H O H H Fmoc N C C N C C O CH2 POLYMER C N C C N C H H H CH3 O H CH2C6H5 O

[1] N H [2] HF

(CH3)2CHCH2 C6H5CH2 H O H O H H2N C C N C C OH C N C C N C H H H CH3 O H CH2C6H5 O

Ala–Leu–Phe–Phe

+

F CH2 POLYMER

856

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 29–32

CH3

CH3

CH3CH2CH H

b. Fmoc N H

C

C

OH

CH3CH2CH H

[1] base [2] Cl CH2 POLYMER

C

Fmoc N H

O

C O

(Fmoc-Ile)

O CH2 POLYMER

[1] N H H Fmoc N

[2] DCC

CH3CH2CH H O H C N C C O CH2 POLYMER Fmoc N C C N C H H O O H CH3 H H

[1] N H

H Fmoc N

C

N H

C

C

N H

C

H CH3

H H Fmoc N H

[1]

C

C

OH

(Fmoc-Ala)

CH3CH2CH H O

+

[2] DCC

C

H CH3

CH3

CH3

O

C

O CH2 POLYMER

O

OH

O

(Fmoc-Gly)

C6H5CH2 H

[2] DCC C OH + Fmoc N C H (Fmoc-Phe) O CH3 CH3CH2CH H H O H O H H Fmoc N C C N C C O CH2 POLYMER C N C C N C H H H O H O CH2C6H5 CH3

CH3 [1]

H H

O N H

CH3CH2CH H O

H C N C C OH N C C N C H H O H CH2C6H5 O H CH3

H2N

[2] HF

C

C

Phe–Gly–Ala–Ile

+

F CH2 POLYMER

29.67 a. A p-nitrophenyl ester activates the carboxy group of the first amino acid to amide formation by converting the OH group into a good leaving group, the p-nitrophenoxide group, which is highly resonance stabilized. In this case the electron-withdrawing NO2 group further stabilizes the leaving group. O

NO2

O

NO2

O

NO2

O

NO2

O

O N+ O

p-nitrophenoxide

O

O N+ O

O O

N

The negative charge is delocalized on the O atom of the NO2 group.

b. The p-methoxyphenyl ester contains an electron-donating OCH3 group, making CH3OC6H4O– a poorer leaving group than NO2C6H4O–, so this ester does not activate the amino acid to amide formation as much.

O

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Amino Acids and Proteins 29–33

29.68 R H

a. H2N

C C

= RNH 2

OH

O O

O CH2O

C

O

O

O O

N

CH2O

O

C

O N

CH2O

HNR O

RNH2

CO3

O

O H OCO2–

O

O +

HO N

CH2O

O N

O

H

2–

C HNR

C

O CH2O

NHR

C

O N

HNR H O

N-hydroxysuccinimide

O

Fmoc-protected amino acid + CO32–

N H H

b.

CH2 O

O

R H

C

C

N H

C

OH

O

CH2 O

O

R H

C

C

N H

R H C

OH

+ CO2 +

DMF

HN

O

C

C

OH

O +

+

Fmoc-protected amino acid

N H2

proton transfer R H H2N

C

C O

OH

+

N H

29.69 Reaction of the OH groups of the Wang resin with the COOH group of the Fmoc-protected amino acids would form esters by Fischer esterification. After the peptide has been synthesized, the esters can be hydrolyzed with aqueous acid or base, but the conditions cannot be too harsh to break the amide bond or cause epimerization.

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Chapter 29–34

H Fmoc N CHCOOH

H2SO4

R O

O

O

O

Fischer esterification

Wang resin

OH

OH

O

new ester

C O

new ester

O C O

CH R

CHR

NHFmoc

NHFmoc

29.70 Amino acids commonly found in the interior of a globular protein have nonpolar or weakly polar side chains: isoleucine and phenylalanine. Amino acids commonly found on the surface have COOH, NH2, and other groups that can hydrogen bond to water: aspartic acid, lysine, arginine, and glutamic acid. 29.71 The proline residues on collagen are hydroxylated to increase hydrogen bonding interactions. O

O [O]

N

N

The new OH group allows more hydrogen bonding interactions between the chains of the triple helix, thus stabilizing it.

OH

29.72 O C

(CH3)2CHCH2

O C

(CH3)2CHCH2

Br

Br2 H

CH3COOH

(CH3)2CHCHCHO

[1] NH4Cl, NaCN H [2] H3

O +,

Δ

Br

CrO3 H2SO4, H2O

(CH3)2CHCHCOOH

NH2 (CH3)2CHCH2

NH3+

NH3

excess

(CH3)2CHCHCOO–

valine

(Racemic valine and leucine are formed as products, but the synthesis of the tripeptide is drawn with one enantiomer only.)

C COOH H

leucine O H CH(CH3)2 H2N

C

C

[(CH3)3COCO]2O

OH

(CH3CH2)3N

O

valine

H CH(CH3)2 C

C O

Boc N H

CH(CH3)2 C

A C6H5CH2OH, H+

H2N

H

OCH2C6H5

C

C O

H2N

OH H

C

C

O OH

C6H5CH2OH, H+

CH2CH(CH3)2

leucine

H2N

C

C

OCH2C6H5

H CH2CH(CH3)2

B

859

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Amino Acids and Proteins 29–35

DCC

A + B

H CH(CH3)2 O H C N C C C Boc N OCH2C6H5 H O CH CH(CH H 2 3)2

H CH(CH3)2 O H C N C C C Boc N OH H O H CH2CH(CH3)2

DCC

C

H CH(CH3)2 H C N C C Boc–NH

H2 Pd-C

O

O C

OH

H CH2CH(CH3)2

H CH(CH3)2 O H CH(CH3)2 H C N C C C C Boc N N COOCH2C6H5 H H O H CH2CH(CH3)2 HBr CH3COOH H CH(CH3)2 H C N C C H2N O

O

H CH(CH3)2 C N COOH H CH CH(CH H 2 3)2 C

Val–Leu–Val

29.73 Perhaps using a chiral amine R*NH2 (or related chiral nitrogen-containing compound) to make a chiral imine, will now favor formation of one of the amino nitriles in the Strecker synthesis. Hydrolysis of the CN group and removal of R* would then form the amino acid. O R

NR*

NH2R* H

R

H

chiral imine chiral amine

CN

NR* NR* R C H Hydrolyze R C H CN

nitrile

amino nitrile Perhaps a large excess of one stereoisomer will be formed.

COOH

NH2

Remove R*

R C H COOH

amino acid enantiomerically enriched (?)

29.74 This reaction is similar to the reaction of penicillin with the glycopeptide transpeptidase enzyme discussed in Section 22.14. Serine has a nucleophilic OH, which can open the strained β-lactone to form a covalently bound, inactive enzyme.

860

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Chapter 29–36

HO

Enzyme

NHCHO

from serine residue

O O

O

O

orlistat Three operations occur: nucleophilic addition; loss of the alkoxide leaving group; proton transfer.

NHCHO O

O OH

O

O

Enzyme

29.75

H OH2

S C6H5

OH

S

R C6H5

N H

N

H

H2O

O

N

H O OH R

N H

C6H5

N

HO OH

proton

S

R N H

S

transfer C6H5

R N H

N H

thiazolinone H2O OH2 OH N R N S H

O H C6H5 H2O

N

C6H5

R N H

S

O C6H5 S

N

R N H

N-phenylthiohydantoin

H3O

H proton C6H5 N

transfer S

OH OH R N H

HO

OH

S C6H5

N H

N H

R

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

861

Lipids 30–1

Chapter 30 Lipids Chapter Review Hydrolyzable lipids [1] Waxes (30.2)—Esters formed from a long-chain alcohol and a long-chain carboxylic acid O R

R, R' = long chains of C's

C

OR'

[2] Triacylglycerols (30.3)—Triesters of glycerol with three fatty acids O O

R O

R, R', R" = alkyl groups with 11–19 C's

O R' O

R" O

[3] Phospholipids (30.4) [a] Phosphatidylethanolamine [b] Phosphatidylcholine (cephalin) (lecithin) O O

[c] Sphingomyelin

O R

O

(CH2)12CH3 R

O

O NH

O

O O

HO

O R' +

O P O CH2CH2NH3 O–

O

O R' +

O P O CH2CH2N(CH3)3 O–

R, R' = long carbon chain

R, R' = long carbon chain

O

R +

P O CH2CH2NR'3 –

O

R = long carbon chain R' = H or CH3

Nonhydrolyzable lipids [1] Fat-soluble vitamins (30.5)—Vitamins A, D, E, and K [2] Eicosanoids (30.6)—Compounds containing 20 carbons derived from arachidonic acid. There are four types: prostaglandins, thromboxanes, prostacyclins, and leukotrienes. [3] Terpenes (30.7)—Lipids composed of repeating five-carbon units called isoprene units Isoprene unit C

Types of terpenes [1] monoterpene

10 C’s

[4] sesterterpene

25 C’s

[2] sesquiterpene [3] diterpene

15 C’s 20 C’s

[5] triterpene [6] tetraterpene

30 C’s 40 C’s

C C C

C

862

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Chapter 30–2

[4] Steroids (30.8)—Tetracyclic lipids composed of three six-membered and one five-membered ring 18

1 2

A 3

CH3

12

19

17

11

CH3 10

D

9 8

B

14

16

15

7

5 4

13

C

6

Practice Test on Chapter Review 1. a. Which of the following compounds contains an ester? 1. a wax 2. a cephalin 3. a sphingomyelin 4. Both [1] and [2] contain esters. 5. Compounds [1], [2], and [3] all contain esters. b. Which of the following compounds is an eicosanoid? 1. a leukotriene 2. a thromboxane 3. a prostaglandin 4. Both [1] and [2] are eicosanoids. 5. Compounds [1], [2], and [3] are all eicosanoids. c. Which of the following statements is (are) true about A? (CH2)12CH3 HO NHCO(CH2)14CH3 O O

+

P O CH2CH2N(CH3)3 O–

A

1. 2. 3. 4. 5.

A is a lecithin. A is a phosphoacylglycerol. A has two tetrahedral stereogenic centers. Both [1] and [2] are true. Statements [1], [2], and [3] are all true.

2. Label each compound as a (1) hydrolyzable or (2) nonhydrolyzable lipid. a. a triacylglycerol e. oleic acid b. vitamin A f. PGE1 c. cholesterol g. a wax d. a lecithin h. a phosphoacylglycerol

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Lipids 30–3

3. For each compound: How many isoprene units does the compound contain? Classify the compound as a monoterpene, sesquiterpenoid, etc.

OH

O O O O

A

B

C

4. Answer True (T) or False (F) for each statement. a. Eicosanoids are biosynthesized from dimethylallyl diphosphate and isopentenyl diphosphate. b. Prostaglandins are biosynthesized from arachidonic acid. c. Fats have lower melting points than oils, and are generally solids at room temperature. d. A cephalin is one type of phosphoacylglycerol. e. Sphingomyelins possess a polar head and two nonpolar tails. f. A sesterterpene contains six isoprene units. g. The typical steroid skeleton has four six-membered rings joined with trans ring fusions. h. Waxes are high molecular weight lipids formed from a fatty acid and a long-chain amine. 5. Which terms describe each compound: (a) hydrolyzable lipid; (b) triacylglycerol; (c) phosphoacylglycerol; (d) phospholipid; (e) nonhydrolyzable lipid; (f) prostaglandin; (g) terpenoid? More than one term may apply to a compound. O O

O

(CH2)12CH3 O

COOH O

OH HO

O

B

OH

(CH2)12CH3

+

O P O CH2CH2NH3

A

O–

C

Answers to Practice Test 1. a. 4 b. 5 c. 3

2. a. 1 b. 2 c. 2 d. 1 e. 2 f. 2 g. 1 h. 1

3. A: 2, monoterpenoid B: 4, diterpenoid C: 3, sesquiterpenoid

4. a. F b. T c. F d. T e. T f. F g. F h. F

5. A: e, f B: e, g C: a, c, d

864

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Chapter 30–4

Answers to Problems Join the carboxylic acid and alcohol together to form an ester.

30.1

O O

Eicosapentaenoic acid has 20 C’s and 5 C=C’s. Since an increasing number of double bonds decreases the melting point, eicosapentaenoic acid should have a melting point lower than arachidonic acid; that is, < –49 °C.

30.2

30.3 O O

HO

O

a.

O

O

OH H2O

O

HO OH

H+

O

HO

OH

O

A

O

b. O

H2 (excess)

c.

O

Pd-C

H2 (1 equiv)

O O

Pd-C O

O

B

two possible products: O

O

O

O O

O

O

O

or

O

O O

O

C

A 2 double bonds lowest melting point


Z configuration.

31.14 RO

2 RO

OR

+

RO

RO

RO

RO

A

The resonance-stabilized radical can react at two carbons.

RO

B

Cl E

OH

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Chapter 31–8

31.15 O HOOC

+

a.

O

b.

O

COOH

H

O

O

HO

NH

H2 N

c.

+

O

NH2

O

NH2

N

N

H2N(CH2)6NH2

H2O

H

H

H

N

N

OH

HO

N

N

H

H O

O

O

31.16 H OH2

O C

+

OH

[1]

C

OCH3

CH3O2C

CH3O2C

H

OCH3

O

OH

[2] CH3O2C

C OCH3 +

O

+ H O 2

H

H2O

HO

OH

[3] OH CH3O2C

H OH2

C OCH3 OCH2CH2OH

OH CH3O2C

C OCH3

[4]

OCH2CH2OH

H H O 2 O+

OH H

H OH2

+

CH3O2C

C OCH3 OCH2CH2OH

CH3O2C

[5]

O CH2CH2OH + CH3OH [6]

O

O

O

O

Repeat for all ester bonds. CH3O2C O CH2CH2OH

O +

H OH2

31.17 O OH

HO O

HO

O

OH

O

O O

This compound is less suitable than either nylon 6,6 or PET for use in consumer products because esters are more easily hydrolyzed than amides, so this polyester is less stable than the polyamide nylon. This polyester has more flexible chains than PET, and this translates into a less strong fiber.

885

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Synthetic Polymers 31–9

31.18 O C6H5O

C

OH

H A OC6H5

C6H5O

C

OH

OC6H5

OH C6H5O C OC6H5 H A

C6H5O C OC6H5 OH

O

OH

O

H + A A HO

OH A O H

O C6H5O

C

OH H

C6H5O C

C6H5O C OC6H5 OH

O O

OH

O

OH

+ HA

+ C6H5OH

+ A

Repeat this process for all other CO bonds. O O

O

C

O O

O

+

2 C6H5OH

31.19 O HO

OH

1,4-dihydroxybenzene

+

O

O

O

O

Cl O

epichlorohydrin (excess)

O OH

n

A H2N

NH2

OH

OH O

NH

O O

O OH

HN

n

OH O

O

NH

HN O

O OH

B

n

OH

886

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Chapter 31–10

31.20 HO AlCl3

O

H

O

OH

OH

OH

OH

OH

H

OH

H

AlCl3

+ 3 resonance structures

resonance-stabilized carbocation

AlCl3

OH

HO AlCl3

OH + AlCl3 + H2O

31.21 HO OH OH

OH

R H2C=O

=

H+

R

R CH2

cardinol

OH

R

31.22 Chemical recycling of HDPE and LDPE is not easily done because these polymers are both long chains of CH2 groups joined together in a linear fashion. Since there are only C–C bonds and no functional groups in the polymer chain, there are no easy methods to convert the polymers to their monomers. This process is readily accomplished only when the polymer backbone contains hydrolyzable functional groups. 31.23 a. b. c. d.

Combustion of polyethylene forms CO2 + H2O. Combustion of polyethylene terephthalate forms CO2 + H2O. These reactions are exothermic. HDPE and PET must be separated from poly(vinyl chloride) prior to incineration because combustion of hydrocarbons (like HDPE) and oxygen-containing organics (like PET) releases only CO2 + H2O into the atmosphere. Poly(vinyl chloride) also contains Cl atoms bonded to a hydrocarbon chain. On combustion this forms HCl, which cannot be released directly into the atmosphere, making incineration of halogen-containing polymers more laborious and more expensive.

31.24 OH

a.

O

O OH

O

H

O O

b.

OH

H2N O

N

N H

O

O

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Synthetic Polymers 31–11

31.25 a.

b.

O

HO

NC NC NC CO2CH3 CO2CH3 CO2CH3

OH

O n

31.26 CN

a. CN

and

CN O

b.

O

O

O HO

O

or

OH

O

O

31.27 O

O

H N

O

a.

b. HO

OH

and Cl

O

Cl

O

N H

O

31.28 Draw the polymer formed by chain-growth polymerization as in Answer 31.2. COCH3 COCH3 COCH3 F

F

F F

F F

F

O

a.

c.

F

OCH3

CH3O

b.

C CH3CH2O2C

NH

O

d.

C CH2

NH

N H

OCH3 CH2

NH

O

O

CO2CH2CH3 CO2CH2CH3

31.29 Draw the copolymers. a.

CN

and

d. CN

and

n

n

Cl

b.

Cl

and Cl Cl

c.

e. Ph

n

n

CN and CN

and

n

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Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 31–12

31.30 a.

CO2Et CO2Et CO2Et

b.

CO2Et O

H N

N H

O H2N

OH

O C(CH3)3 O

c.

O

O

C(CH3)3

C(CH3)3 O

d.

O

O O

O

O O

Cl

and

Cl

HO

OH

31.31 =

a. CO2Et CO2Et

b.

CO2Et

O

H N

N H

n

CO2Et

O

H N

=

O

n

C(CH3)3 O

c.

O

C(CH3)3 O

=

n

C(CH3)3

O

d.

O

O O

O

O

=

O

O

O

n

31.32 An isotactic polymer has all Z groups on the same side of the carbon backbone. A syndiotactic polymer has the Z groups alternating from one side of the carbon chain to the other. An atactic polymer has the Z groups oriented randomly along the polymer chain. a.

c.

b. Cl

Cl

Cl

CN

CN

CN

CN

Ph

Ph

Ph

Ph

Ph

889

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Synthetic Polymers 31–13

31.33 from ethylene oxide

OH

CCl3

n

31.34 O H2N

NH2

a.

and

HO2C

H N

CO2H

H N O H N

O

b. O C N

and

N C O

HO

OH

O

O

O

N H O O O

COCl

c.

and

HO

OH

COCl O O

d.

HO

COOH O

31.35 CN

ABS

Ph

31.36 H2N

NH2

O

a. N H

+

N H

O

HO

O

Quiana OH

O ClOC

b.

COCl

H2N

+

NH2

O

O N H

Nomex

N H

890

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 31–14

31.37 O H N

O

O

O

H N

N

H

H

N O

O

O

N

H

H

N

N H

Kevlar

O

N H

31.38 O O O

O

O

O

H N

O

O

O

n

polyester A

O

n

nylon 6,6

PET

T g = < 0 oC Tm = 50 oC

Tg = 70 oC Tm = 265 oC

N H n

Tg = 53 oC Tm = 265 oC

a. Polyester A has a lower Tg and Tm than PET because its polymer chain is more flexible. There are no rigid benzene rings, so the polymer is less ordered. b. Polyester A has a lower Tg and Tm than nylon 6,6 because the N–H bonds of nylon 6,6 allow chains to hydrogen bond to each other, which makes the polymer more ordered. c. The Tm for Kevlar would be higher than that of nylon 6,6, because in addition to extensive hydrogen bonding between chains, each chain contains rigid benzene rings. This results in a more ordered polymer. 31.39 O

O O

O

O

O

A

dibutyl phthalate

O

O

Diester A is often used as a plasticizer in place of dibutyl phthalate because it has a higher molecular weight, giving it a higher boiling point. A should therefore be less volatile than dibutyl phthalate, so it should evaporate from a polymer less readily.

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

891

Synthetic Polymers 31–15

31.40 Initiation: (CH3)3CO

OC(CH3)3

[1]

[2]

2 (CH3)3CO

(CH3)3CO

+

Propagation: (CH3)3CO

[3]

(CH3)3CO

(CH3)3CO

new C–C bond Repeat Step [3] over and over to form gutta-percha.

Termination:

[4]

31.41 H

F

[1]

O

F 3B

F H

H

F

[2]

Ph

+

B O

Ph

+ F3B OH

H

[3]

1,2-H shift

H

a highly resonance-stabilized carbocation Ph Ph

[4]

Ph

+

Repeat Steps [3] and [4].

A

Ph

new C–C bond Ph

Ph

Ph

A is the major product formed due to the 1,2-H shift (Step [3]) that occurs to form a resonance-stabilized carbocation. B is the product that would form without this shift.

Ph

A major product

B

31.42 H F3B O

CN

C N

δ+

H H F3B O

CN

C

This carbocation is unstable because it is located next to an electron-withdrawing CN group that bears a δ+ on its C atom. This carbocation is difficult to form, so CH2=CHCN is only slowly polymerized under cationic conditions.

N

H H

This 2° carbocation is more stable because it is not directly bonded to the electron-withdrawing CN group. As a result, it is more readily formed. Thus, cationic polymerization can occur more readily.

892

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 31–16

31.43 Initiation:

[1]

+

Bu Li

Li+

Bu

Ph

Ph

Repeat Step [2]

[2]

Propagation: Bu

Bu Ph

Ph

Ph

over and over.

Ph

O [3]

Termination:

O

O C O Ph

Ph

Ph

Ph

31.44 The substituent on styrene determines whether cationic or anionic polymerization is preferred. When the substituent stabilizes a carbocation, cationic polymerization will occur. When the substituent stabilizes a carbanion, anionic polymerization will occur. OCH3

a.

NO2

b.

cationic polymerization

CF3

c.

anionic polymerization

CH2CH3

d.

anionic polymerization

cationic polymerization

31.45 The rate of anionic polymerization depends on the ability of the substituents on the alkene to stabilize an intermediate carbanion: the better a substituent stabilizes a carbanion, the faster anionic polymerization occurs. O

CN OCH3

O

OCH3

O

O

least

most

increasing ability to undergo anionic polymerization

31.46 The reason for this selectivity is explained in Figure 9.9. In the ring opening of an unsymmetrical epoxide under acidic conditions, nucleophilic attack occurs at the carbon atom that is more able to accept a δ+ in the transition state; that is, nucleophilic attack occurs at the more substituted carbon. The transition state having a δ+ on a C with an electrondonating CH3 group is more stabilized (lower in energy), permitting a faster reaction. [1]

O

O

H OH2

H O

+

HO

H

H

OH

+

[4]

[2]

H2O

HO

HO

O H

H2O

Repeat Steps [4] and [5] over and over.

O

OH

[3] H

+ H2O

[5]

HO

HO

O

OH

OH +

+ H3O+

H3O+

893

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Synthetic Polymers 31–17

31.47 AlCl3

H ClCH2

C Cl

H ClCH2

H

C Cl AlCl3

H

ClCH2

+ HO

C

H

H

1,4-cyclohexanediol

resonance-stabilized carbocation

A

OH

+ AlCl4–

Cl AlCl3 H H

ClCH2

OH + AlCl3 + HCl

CH2 O

ClCH2

C O

OH

H

Repeat. CH2

CH2 O

O n

B

31.48 O

H

O

+ H OSO3H

+

+

OH

H

HSO4

H

+ resonance-stabilized cation

HSO4

H

O OH

OH

(+ 3 additional resonance structures)

H OSO3H

H HO

H

OH

bisphenol A H2SO4

HO

+

HSO4

O H

OH

HO

OH

O

HSO4

(+ 5 additional resonance structures)

(+ 3 additional resonance structures)

OH

H2O

31.49 H N N C O

+ CH3 OH

C

O

H CH3

N

O

C

O

CH3 OH CH3

O H CH3 O H H N

CH3 OH

+

OCH3 O

a urethane

N

C O

O

CH3

894

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 31–18

31.50 (CH3)3CO–OC(CH3)3

a. OCH3

OCH3 OCH3 OCH3 Ziegler–Natta

b.

catalyst

CH3 CH2 C

BuLi (initiator)

c. O

CH3 CH2 C

CH3 CH2

C

CO

CO

CO

CH2CH3

CH2CH3

CH2CH3

BF3 + H2O

d.

O

e.

–OH

O

O

OH

f. OH O

OH

O

O

+

O

Cl

O

O

O

(excess)

n

O

O

–

g.

HO

OH

OH

H2 O

O O

H N

h.

O

O

O

O

O O

O –OH

NH2

O

H2O

or

NH

H N

i.

OCN

NCO + HO

OH

O

O

N H

O

O

O

j.

HO

Cl2C=O OH

O

O

31.51 Polyethylene bottles are resistant to NaOH because they are hydrocarbons with no reactive sites. Polyester shirts and nylon stockings both contain functional groups. Nylon contains amides and polyester contains esters, two functional groups that are susceptible to hydrolysis with aqueous NaOH. Thus, the polymers are converted to their monomer starting materials, creating a hole in the garment.

895

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Synthetic Polymers 31–19

31.52 O HO

CH2

OH

Cl

+

OH

O

O

CH2

CH2

O

O

O

n

prepolymer NH2

H2N

OH

OH

OH O

CH2

O

O

O

CH2

O

n

HN

NH

NH

HN O

CH2

O

OH

O

CH2

O

OH

OH

n

31.53 a.

Poly(vinyl alcohol) cannot be prepared from vinyl alcohol because vinyl alcohol is not a stable monomer. It is the enol of acetaldehyde (CH3CHO), and thus it can't be converted to poly(vinyl alcohol).

OH OH

vinyl alcohol

OH

poly(vinyl alcohol) b.

OAc

vinyl acetate

ROOR radical polymerization

–OH

OAc OAc

poly(vinyl acetate)

H2O hydrolysis

+ CH3CO2– OH

OH

poly(vinyl alcohol)

O C

c. OH

H

H+

OH

O

O

poly(vinyl alcohol)

an acetal poly(vinyl butyral)

31.54 NBS hν

Br

–

OH

OH

[1] BH3 HO [2] H2O2, –OH

OH

1,3-propanediol

896

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 31–20

31.55 CH3 CH3Cl

CH3Cl

AlCl3

AlCl3

KMnO4 HOOC

CH3

CH3

[1] OsO4 CH2 CH2

HO

[2] NaHSO3, H2O

COOH

terephthalic acid

OH

ethylene glycol or mCPBA

O

H2O, –OH

OH

HO

31.56 OH

OH

OH OH

+ CH2 O

OH

phenol OH

Since phenol has no substituents at any ortho or para position, an extensive network of covalent bonds can join the benzene rings together at all ortho and para positions to the OH groups.

Bakelite OH

CH3

OH

OH

OH

OH

+ CH2 O

p-cresol

Since p-cresol has a CH3 group at the para position to the OH group, new bonds can be formed only at two ortho positions, so that a less extensive three-dimensional network can form.

31.57 O

a.

O O

ε-caprolactone

O O

polycaprolactone

O

b.

O O

p-dioxanone

O

O

polydioxanone

897

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Synthetic Polymers 31–21

31.58 O

O

H N

O O

O

O

H N

N H

O

H N

O

O

O

poly(ester amide) A

O

H2N

O

H2N

OH

NH2

OH

HO

O

OH

HO

O

O

the benzyl ester of lysine

leucine

31.59 CO2CH2Ph

O

O

OH

O

O

O

n

salicylic acid (2 equiv) +

O

+

O

OH

O

O

benzyl salicylate (2 equiv)

CO2H

PolyAspirin

OH

HO

Cl

Cl

O

sebacic acid

O

sebacoyl chloride

31.60 H H N

O H2N

NH2

N

H2N H

N

N

H

N

N

[1]

NH2

N

O

H2N

proton transfer

N

N

OH H A

[2]

NH2

H N

N

NH2

[3]

melamine A H2N

H N

N N

N

N

N NH2

H2N

H N

N

N

N

N NH2

N NH2

NH2

N

NH2 H2N

N

[5]

NH2

[6] H N

N

NH2

N

N

NH2 H2N

H H N

H A

H N

N N

CH2

N NH2

[4]

H2N N

H2O

H N

N N NH2

OH2 A

898

Student Study Guide/Solutions Manual to accompany Organic Chemistry, Fourth Edition

Chapter 31–22

31.61 [1]

a. H

re-draw [2]

intramolecular H abstraction

[3] Repeat Step [2].

b. Abstraction of the H is more facile than abstraction of the other H's because the H atom that is removed is six atoms from the radical. The transition state for this intramolecular reaction is cyclic, and resembles a six-membered ring, the most stable ring size. Other H's are too far away or the transition state would resemble a smaller, less stable ring.

n

butyl substituent

31.62 O O H2N

O

O NH2

H

H2N

H

O N H

OH

H2N

H2N

O

NH2 H2N

O N H

N H

formaldehyde

urea

O

N

N

N

H2N

O

N O

O N

H N

N O

N

O NH2

repeat

H2N

O N

N

N

NH N

NH2

repeat O

N

N

N H

CH2

N HN

O

O

N H NH

NH2

O

NH2

NH2