Organic Chemistry 2 Summary Sheets

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"Master Organic Chemistry"

Introduction to Alcohols and Ethers Ethers from alcohols and alkyl halides The Williamson Ether synthesis: alcohol, base, alkyl halide (or tosylate) base (e.g. NaOH) OH

O Br

+ H2O

+ NaBr

This is an SN2 reaction; it works best for primary alkyl halides (and alkyl tosylates)

Conversion of alcohols to good leaving groups

Due to ring strain, epoxides are highly reactive towards nucleophiles. They will react with nucleophiles under both acidic and basic conditions. However the patterns are different.

The hydroxide group (HO–) of alcohols is a strong base and a poor leaving group. Converting to a halogen or "sulfonate" (e.g. tosylate or mesylate) greatly facilitates substitution reactions. Alcohols to alkyl chlorides

Under basic conditions

OH

SOCl2

Cl

Under basic conditions, nucleophiles will attack epoxides at the least sterically hindered position (primary [fastest] > secondary > tertiary [slowest]) The reaction is essentially an SN2 reaction!

+ HCl + SO2

OH

PCl3

Cl

+ HOPCl2

+ HBr + SO2

"Nu"

Alcohols to alkyl bromides

OH Nu

Example: reaction of epoxides with Grignard reagents

H2SO4

R

O

ROH

O

The reaction is similar to the hydration of alkenes with aqueous acid. Acid leads to the formation of a carbocation, which is then trapped by the alcohol as solvent. Carbocation rearrangements (hydride and alkyl shifts) can occur in certain cases.

OH

1) CH3MgBr

O

2) NaBH4

+ HOAc + Hg (s) + NaOAc + BH3

Ethers from alcohols through dehydration OH

H2SO4

O

heat

+ H2O

OH

O H

H + δ O preferred site of nucleophilic attack: the "most substituted" carbonmore substituted positive charge of the epoxide more stabilized

Br

PBr3

inversion!

+ HOPBr2

Alcohols to tosylates and mesylates ("sulfonate esters") Methanesulfonyl chloride (mesyl chloride, MsCl) O Cl S CH3 O OH OMs + HCl

O H + δ

p-toluenesulfonyl chloride (Tosyl chloride, TsCl) O Cl S O OH OTs

less substituted positive charge less stabilized

OH

Oxidation of primary alcohols to aldehydes OH

PCC

O

O

m-CPBA (meta-chloroperoxybenzoic acid, a peroxyacid) converts alkenes to epoxides, a cyclic ether. Other peroxyacids can be used (e.g. CH3CO3H)

Alternative reagents for oxidation of primary alcohols to aldehydes and secondary alcohols to ketones (not seen in all courses): Dess-Martin Periodinane (DMP)

O H Cl Cr O N H O PCC = pyrdinium chlorochromate

AcO

Oxidation of secondary alcohols to ketones OH

Epoxides from halohydrins

Swern oxidation

O

OH

Can also use KMnO4 or H2CrO4 (or DMP or Swern, see right)

OH O

Br Formation of the halohydrin from the alkene is stereospecific for the trans product. Deprotonation of the alcohol by base results in SN2 (with inversion at carbon bearing the leaving group) to give the epoxide.

Oxidation of primary alcohols to carboxylic acids OH

KMnO4 or H2CrO4

OAc I OAc O

+ HCl

O OH

Common source of confusion: Another way of writing H2CrO4 is K2Cr2O7 / H2SO4 or Na2Cr2O7/H2SO4

•oxidizes secondary alcohols to ketones

COCl2, DMSO

O

O

+ HCl

retention!

2) H2O

OH

Reduction of carboxylic acids by lithium aluminum hydride (LiAlH4) 1) LiAlH4 OH

2) acid

OH

Reduction of aldehydes by sodium borohydride (NaBH4) LiAlH4 can also do this reaction. O NaBH4 H

NEt3 (often just written, "Swern") (COCl)2 is oxalyl chloride DMSO is dimethyl sulfoxide and NEt3 is triethylamine (base)

OTs

Reduction of esters by lithium aluminum hydride (LiAlH4) O 1) LiAlH4

O •oxidizes primary alcohols to aldehydes •oxidizes secondary alcohols to ketones

TsCl

Alcohols through reduction

•oxidizes primary alcohols to aldehydes

O

PCC

base

+ HOPBr2

OH

Oxidation of alcohols

Epoxides from alkenes

Br2, H2O

Br

One thing to note - these reactions do not change the stereochemistry of the alcohol.

Strong acid (and heat) leads to protonation of the alcohol, followed by nucleophilic attack of a second molecule of alcohol to give the ether. Only practical for the synthesis of symmetrical ethers.

m-CPBA

OH

HCl, HBr, or HI R Cl R Br R I R OH These reactions can proceed through an SN1 or SN2 pathway depending on the structure of the alcohol

H2SO4) The nucleophile will attack the carbon best able to stabilize positive charge - which is the more substituted carbon Just like Markovnikov's rule!

The reaction is similar to the hydration of alkenes with aqueous acid. The key difference is that it does not proceed through a carbocation, so no rearrangements can occur.

OH

Aqueous acid (e.g. H2O

Br

PBr3

Alcohols to alkyl halides by using acids

Under acidic conditions

O

SOBr2

OH

Under acidic conditions, the epoxide oxygen is protonated:

R

OH

One thing to note - these reactions occur with inversion of configuration. For example:

2) acid workup (e.g. H+, H3O+, H2O)

Ethers from alkenes through oxymercuration 1) Hg(OAc)2, ROH

Note - this sheet is not meant to be comprehensive. Your course may provide additional material, or may not cover some of the reactions shown here. Your course instructor is the final authority.

Opening of epoxides

O Ethers from alkenes

masterorganicchemistry.com 2015 Version

OH

Reduction of ketones by sodium borohydride (NaBH4) LiAlH4 can also do this reaction. O

NaBH4

OH Omissions, Mistakes, Suggestions? [email protected] This sheet copyright 2015, James A. Ashenhurst masterorganicchemistry.com

"Master Organic Chemistry"

Introduction to the Diels Alder Reaction

Note - this sheet is not meant to be comprehensive. Your course may provide additional material, or may not cover some of the reactions shown here. Your course instructor is the final authority.

masterorganicchemistry.com

1. The Basics The Diels-Alder reaction is a concerted reaction between a compound with two adjacent double bonds (a "diene") and an alkene (the "dienophile"), usually attached to an electron withdrawing group. The transition state is "pericyclic", meaning it has cyclic geometry and is concerted The Diels-Alder reaction always forms a six-membered ring with an alkene. A Very Simple Diels Alder

1

1 2

2

6

3

6 5

3

5

4

4

Form

Break

C1–C6

C1–C2 (π)

C2–C3 (π)

C3–C4 (π)

C4–C5

C5–C6 (π)

Rule #3: The two "outside" groups (red)and the two "inside" groups (blue) always end up on the same side of the new cyclohexene

What this means: 1 2

the C2-C3 bond can rotate freely

When the diene is "locked" in an s-trans configuration, no Diels-Alder reaction is possible (the two ends are too far apart)

1 2

3

3 4

4

note how the two alkenes are on the same side of the C2–C3 sigma bond

note how the two alkenes are on opposite sides of the C2–C3 sigma bond

"s-cis"

The end carbons (C1 and C4) are too far apart no Diels-Alder reaction possible

6 5

3

top view

3

7

2

6

7 8

5

3

4

4

4

1

2

6

7

3

5

8

6 8

5

side view

1

1

top view

5

side view

Form

Break

C1–C6

C1–C2 (π)

C2–C3 (π)

C3–C4 (π)

C4–C5

C5–C6 (π)

note how the pattern of bonds forming and bonds breaking is exactly the same as for the simple case, above.

Rule #2: Stereochemistry in the dienophile is always preserved cis relationship is preserved in product O O

O

A CH3 O

B

C

D

CH3 O

CH3 O

CH3 O

CH3

CH3

+ CH3

CH3

+ CH3

enantiomers Let's just pick one set of diastereomers. A and C. C A

Exo/endo examples: O

O

O

these are the same molecule (it's meso) trans relationship is preserved in product O O

CH3

CH3

CH3 Endo

H H

CH3 Exo

How to tell if a product is exo or endo One way to do it: if the outside groups (red) end up on the same side of the ring as the electron withdrawing group, the product is endo. If the outside groups end up on the opposite side of the ring as the electron withdrawing group, the product is exo.

O

CH3

O

CH3 O endo

CH3 exo

O

O

CH3

Note how in A, the two methyl groups are on the same side of the ring as the electron withdrawing group (ketone), whereas in C, they are on the opposite side of the ring

Rule #4: Under normal conditions, endo products dominate

O

enantiomers

CH3 O

CH3 O

A is referred to as the endo product. C is referred to as the exo product (note: B is also endo, and D is also exo)

or O cis dienophile

H3C H

enantiomers

+

How the stereochemistry of the dienophile works:

O

H3C H

Note how the two "outside" groups (CH3 and H) end up on the same side of the ring, as do the two "inside" groups (CH3 and H)

Note how the two "outside" groups (CH3 and CH3) end up on the same side of the ring, as do the two "inside" groups (H and H)

3. Stereochemistry: The Dienophile

Example:

H CH3 and

CH3

H3C H

6

4

4

4

2

H3C H

CH3

1

2

7 3

5

CH3

H CH3 CH3 H

=

Up to 4 products are possible!

7

1 2

7 3

H H

Example:

Diels-Alder of cyclic dienes 6

H

H3C H

What if there's a substituted diene and a substituted dienophile?

These look a little weird, but they're no different than a normal Diels-Alder

1

H3C H

5. The Endo Rule

2. Special Cases Diels-Alder Reactions of Cyclic Dienes:

2

CH3

these are the same molecule These dienes cannot participate in the Diels-Alder

"s-trans"

The end carbons (C1 and C4) are close together

Example 2:

Example 1:

The diene must always be in the "s-cis" conformation

Rule #1

4. Stereochemistry: The Diene How the stereochemistry of the diene works:

O H 2

6

1

4

H

2

1

H

5

CH2 3

OCH3

7

3

5

1

7

3

CO2Me

1

2

6 4

5

4

H exo (top view)

EWG

O

endo (side view)

OCH3 5

6 5

OCH3

6 7

3

4

4

H endo (top view) H O 2

1

2

OCH3

7 3

7

O 6

exo (side view)

EWG

+

Omissions, Mistakes, Suggestions? O trans dienophile

O

these are enantiomers

O

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"Master Organic Chemistry"

Introduction to Aromaticity 1. An introduction to resonance energy and aromaticity

3. How to tell if a molecule is aromatic? Rule 4. It must be flat

Hydrogenation of alkenes liberates 119 kJ/mol of energy ΔH = –119 kJ/mol (28.4 kcal/mol)

Pd/C H2

Rule 1. It must be a ring. No acyclic molecule is aromatic. Ever. Most molecules that obey the first 3 rules are also flat. One exception is [10]-annulene, which is bent due to repulsion of the hydrogens.

We would expect hydrogenation of benzene to liberate 3 × 119 = 357 kJ/mol. Pd/C H2

ΔH = –207 kJ/mol (49 kcal/mol)

1,3,5 hexatriene not aromatic

benzene aromatic

N H pyrrole aromatic

butadiene not aromatic

H H not aromatic

Instead, 207 kJ/mol is liberated (150 kJ mol less than we expect!) So it is 150 kJ/mol (36 kcal/mol) more stable.

4. Antiaromaticity Rule 2. The molecule must be conjugated

The extra stability of benzene is called the "resonance energy". Benzene has a particularly large resonance energy, which leads us to classify it as "aromatic". Resonance energy of some compounds:

Meaning: there must be a continuous line of p orbitals around the ring. p orbitals can come from 1) π-bonds 2) lone pairs 3) carbocations

aromatic

H Conjugated

2. Two major ways in which reactions of aromatic compounds differ from alkenes

H2

easy

H

H

H

Reaction with electrophiles (such as bromine): aromatics give substitution, not addition Br2 Br Addition reaction

Alkene

Br

Br2 AlBr3 benzene

Substitution reaction difficult

easy

not aromatic (antiaromatic)

4

2

4

aromatic

O 6 [NOT 8] aromatic

6

not aromatic (antiaromatic)

aromatic

N 6 [NOT 8] aromatic

6 [NOT 8] aromatic

N H 6 aromatic

H 6 aromatic

10

H N

8 not aromatic

N

O

6 [NOT 8]

6 [NOT 8]

aromatic

aromatic

5. Some physical evidence for aromaticity: H

1. Shows a different reactivity profile than for alkenes (see section on reactivity at left)

Rule 3. There must be 4n + 2 π electrons

6 aromatic

2 aromatic

All bonds in benzene are the same length (1.4 Angstrom) i.e. 2, 6, 10, 14.... π electrons

π electrons can come from double bonds or lone pairs. Note: A carbocation indicates the absence of π electrons. One tricky part: for a given atom, you can only count electrons from a lone pair if the atom is not part of a π bond. And in that case you can only count a maximum of one lone pair. This is due to the fact that each atom can only share one p orbital with the π system of the molecule. the electrons from the lone pair on N are at 90° to the π system, and do not contribute to aromaticity.

N pyridine 6 π electrons (3 π bonds, 2 electrons per π bond)

Similar example: lone pair on this anion does not contribute towards aromaticity.

ignore lone pair for this calculation

furan 6 π electrons (2 π bonds, 2 electrons per π bond) + 2 electrons from lone pair only count one lone pair for this calculation

Compare this to cyclobutadiene, which has short double bonds and long single bonds - like a rectangle. 3. Ring currents in NMR Resonances for aromatic protons in NMR typically show up in the region 6.8–8.0 ppm, whereas those for "normal" alkenes show up in the region from 5.0–6.5 ppm.

6. "Frost circles" - a trick for obtaining the molecular orbital structures of aromatic rings. General idea: Inscribe a polygon of n sides in a circle. Make sure one of the apices is pointing down. Then, each apex will represent a level in the molecular orbital energy diagram. Frost, J. Chem. Phys. 1953, 21, 572 Examples:

only one of the lone pairs can align itself with the pi system at any given time, so only one lone pair is counted towards aromaticity.

O

aromatic

extremely unstable

2. All π bonds are of the same length & do not alternate

no reaction without a catalyst such as AlCl3

π electrons - some examples

•cyclic •conjugated •4π electrons

cyclooctatetraene has 8π electrons, but can adopt a "tub" shape, "escaping" antiaromaticity.

H

Not Conjugated

no reaction at normal temperature and pressure

Br

Conjugated

difficult

benzene

Alkene

Conjugated

O

antiaromatic

N

Pd-C H2

H3C CH3

Conjugated H

Hydrogenation: more difficult with aromatic compounds Pd-C

N H

O H Conjugated

H2C CH2

Molecules that obey rules 1,2 and 4 but have (4n) π electrons instead of (4n +2) π electrons have special instabiity. This special instability is called "antiaromaticity".

A good test. Can you push electrons all the way around the ring through resonance? If not, it's not conjugated.

N O N H 121 kJ/mol 67 kJ/mol 92 kJ/mol 12.2 kJ/mol (29 kcal/mol) (16 kcal/mol) (22 kcal/mol) (3 kcal/mol) aromatic aromatic aromatic not aromatic

150 kJ/mol (36 kcal/mol)

Note - this sheet is not meant to be comprehensive. Your course may provide additional material, or may not cover some of the reactions shown here. Your course instructor is the final authority.

masterorganicchemistry.com

Similar examples: only count one lone pair

S

N

For more examples, see section on left

3

4

6

7

this shows the arrangement of molecular orbitals for benzene

5

8

Omissions, Mistakes, Suggestions? [email protected] This sheet copyright 2015, James A. Ashenhurst masterorganicchemistry.com

"Master Organic Chemistry"

Reactions of Aromatic Compounds Six reactions for electrophilic aromatic substitution Cl2

Chlorination

Electrophilic aromatic substitution: Mechanism E

Cl

X

H E

FeCl3

Bromination

Step 1: attack of the double bond at the electrophile (E )

FeBr3 NO2

HNO3

Nitration

SO3H

SO3

R–Cl AlCl3 (or FeCl3) O R C Cl Friedel-Crafts Acylation

X

H

E

X

X

X

E

X

This resonance form is available for the ortho and para adducts, but NOT the meta.

meta

E

E

X

E

H

X

H

X

–NH2 –NHR –NR2 –OR –OH –F –Cl –Br –I

B

X

O

E

Oxidation of side chain

CH3

Halogenation of side chain

–SH –SR

para H

Cl 1,2 (ortho) 1,3 (meta) 1,4 (para) "ortho - meta - para" only applies to di-substituted benzenes

E

H

E

H

E

O

Alkyl groups such as CH3, CH2CH3, etc. are also ortho-para directors. Why? Notice the partial charge:

X

δ The partial negative charge on the carbon H δ− δ+ will stabilize an adjacent positive charge. C H δ+ H

Activating vs. Deactivating : What does it mean? An activating group is a group that causes electrophilic aromatic substitution to be faster, relative to H

conditions Fe / HCl Sn / HCl Zn / HCl

Similarly, if there is a partial positive charge on the atom adjacent to the ring, this will destabilize resonance forms A and B

A deactivating group is a group that causes electrophilic aromatic substitution to be slower, relative to H either of these conditions will work

CH3

Cl

δ+

CO2Me

X

bad!

CH2Br

Benzylic halogenation

light (hν) O C

Reduction of ketone to alkane

relative rate for nitration:

1

NH2NH2/ KOH (Wolff-Kishner) Zn(Hg) / HCl

O

O

all of these conditions will convert a ketone to an alkane

(Clemmensen)

Pd/C / H2 (5 atm) R

m-CPBA

O

KI

strongly activating o/p director

O

Cl Br

CuBr

N N

CuCl

O

Cl CuCN CN

Diazonium salt

NaOH, heat OH

–CH3

–NH2

–NHR

–OH

–OR

–R

H E para

meta

–NR2

NOT OBSERVED Key concept: more activating donor wins

–CN O C H

Not as unstable

–CN

O C OH

O C OR

O C R

O S OH O

O C OR O S OH O

O C H

O C R

–NR3

Deactivating o-p directors –F

–Cl

–Br

–I

O

N H

Moderately deactivating meta directors

Cl

H

O

O

–SH

NO2 Cl

deactivating o/p director

Moderately activating o-p

–NH

NO2

HNO3 H2SO4

Cl

O

–CF3

O C OH

ortho (Minor)

–O

I

–NO2

Key concept: para is generally favored due to steric hindrance

Strongly activating o-p

HNO2 acid (e.g. HCl)

H3PO2

NO2

NO2 para (Major)

Question 2. When two groups "direct" to different carbons What product dominates? H

Xδ

This is why the following groups are meta directors. (Although "ortho-para avoider" is more appropriate)

H2SO4

NH2

Formation and reactions of diazonium salts

+

X

Destabilized! adjacent positive charges

O

HNO3

R

Baeyer-Villiger reaction

.004

Cl and CO2Me are deactivating

O Conversion of ketone to ester

.03

Question 1: Ortho or para: what product dominates?

R

O

24

NBS = N-bromosuccinimide

H C

conditions

δ+

E

ortho

CH3 is activating

conditions H

R

bad!

H E

COOH will convert any carbon with a C–H bond to a carboxylic acid Benzylic oxidation

NBS

O

N H

Cl

NH2

KMnO4

E

This is why the following groups are ortho-para "directors"

Pd-C, H2 CH3

H

E H

E

X

H

E

ortho

Transformations of aromatic functional groups NO2 conditions

A

X

H

+

R

AlCl3 (or FeCl3)

Reduction of nitro groups

X

H E

Ortho-meta-para : What does it mean? Cl Cl Cl Cl

O C

X

E

Step 2: Removal of a proton from the carbon regenerates the aromatic group

Substituents that stabilize this carbocation (relative to H) are called activating groups.

R

Resonance forms A and B are KEY If X has an electron pair, these resonance forms will be stabilized through formation of a double bond:

The group X on the ring will affect the stability of the carbocation

+ H–X

Certain substituents will stabilize this carbocation Stabilize the carbocation, speed up the reaction

H2SO4

Friedel-Crafts Alkylation

E

This forms a carbocation. [Breaks C–C, forms C–E] [Breaks C–H, forms C–C (and X–H)] rate determining step

H2SO4

Sulfonation

What if there's already a group on the ring?

X

Br

Br2

X

Note - this sheet is not meant to be comprehensive. Your course may provide additional material, or may not cover some of the reactions shown here. Your course instructor is the final authority.

masterorganicchemistry.com

–SR

Strongly deactivating meta directors –CF3

–NO2

–NR3

Omissions, Mistakes, Suggestions? [email protected] This sheet copyright 2015 James A. Ashenhurst masterorganicchemistry.com

"Master Organic Chemistry"

Key Reactions of Aldehydes and Ketones A Simple "Formula" for Seven Key Reactions of Aldehydes & Ketones

H

R H Nu

2. Protonation

Aldehyde ( but also applies to ketones) Reaction

C–Nu O–H

Product OH

Nucleophile R MgX

Grignard Reaction

C–O ( π )

R

H

R

OH

Addition of Organolithiums

R Li

R R

H

OH Reduction by sodium borohydride (NaBH4)

Na

R

H–BH3

H

H

OH

Reduction by lithium aluminum hydride (LiAlH4)

H–AlH3

Li

Addition of cyanide ion to form cyanohydrins

R H

R H NC OH

R H HO OH

Addition of alkoxide ions to form hemiacetals

OR

R H RO

Each of these reactions follows a "two-step" pattern: 1) addition 2) protonation How to draw the mechanism for this pattern: Step 1 is donation of a lone pair of electrons from the nucleophile to the carbonyl carbon, which bears a partial positive charge [oxygen is more electronegative than carbon]. A bond forms between the nucleophile and carbon, and the carbon-oxygen double bond [π bond] breaks. The oxygen then bears a negative charge. A Step 1: Addition of nucleophile δ– O O δ+ R R H H Nu Nu

Bonds formed

Bonds broken

C–Nu

C–O (π)

P Step 2: Protonation of oxygen In the second step of this reaction, acid is added the oxygen is protonated to give a neutral hydroxyl [O-H] group: O

OH

H

R H Nu

R H Nu

Specific example: Grignard addition Step 1: Addition of nucleophile

Step 2: Protonation

MgBr O R

O H BrMg CH3

draw the arrow as coming from the Mg–C bond

1. Electronic effects In aldehydes and ketones, the C=O bond is polarized; due to the greater electronegativity of oxygen relative to carbon [3.4 vs. 2.5]; the carbon bears a partial positive charge and the oxygen bears a partial negative charge δ– O δ+ R H

R H H3C

H

X

OH R H H3C

MgBrX

Oxymercuration of alkynes

R The oxygen is electron-rich (nucleophilic) The carbon is electron-poor (electrophilic)

The more electron-poor the aldehyde or ketone, the more reactive it will be with nucleophiles – – Oδ Oδ δ+ δ+ H H CF3 H3C more electrophilic less electrophilic this aldehyde is more electron-poor than the aldehyde on the right due to the electron-withdrawing fluorines

1

2

H

1) O3

O

2) Zn

2

O

O




forms readily!

O OR >

O CH3>

O CF3

>

RO

O OR

increasing reactivity toward nucleophilic attack increasing stability of enolate This means we can predict the course of the reaction by pKa!

Key Reaction: Enolate Formation Enolate = deprotonated enol i-Pr

O H3C

OEt

i-Pr N Li (LDA)

O Li

O Li H

(or other strong base) Carbonyl compound

OEt

H ester enolate O-bound form

H

OEt H

C-bound form

Important: the Enolate is a NUCLEOPHILE Amphiphilic =nucleophilic at both O and C; note: though an ester enolate here we focus on the reactions at C. is shown here, the reaction of any enolate with an aldehyde is generally called an "Aldol". Two key examples: O

O M

pKa = 13

OEt

The rate of keto/enol interconversion isgreatly enhanced by acid: Acid makes carbonyl more electrophilic, increasing acidity of αprotons, facilitating formation of enol: this increases K1

4. Conjugation is stabilizing.

H

O OH

Me

O

O

NEt2

Enolate (oxy-anion resonance form)

Me Me

does not form

R

OH H

O

R

OEt

Aldol reaction

Me

3. Strongly hydrogen bonding solvents can disrupt this, however. The above equilibrium is 81:19 using water as solvent.

O

O

CH3 NEt2

H

2

O

> OH > OR, NHAc > CH3, R > Cl, Br, F, I > C(O)OR CF3, etc.

O

t-Bu LiNEt

H

least acidic

Me

X

NEt2

The reactivity of the alkene toward nucleophilic attack is directly related to the stability of the enolate that forms -

The aromatic electrophilic substitution chart is a good proxy for the ability of a functional group to donate to a carbonyl:

Exclusive

H

CH2

enolate: stabilized!

OMe CH3

most acidic

NR2 , O

M

LiNEt2

Li O

Answer: The more electron-poor the carbonyl, the greater will be its ability to stabilize negative charge. Conversely, the greater the donating ability of a substitutent on the carbonyl, the less it will be able to stabilize negative charge.

2. Hydrogen bonding stabilizes the enol form. H-bond H O O O O Me

H

Question: How do you explain the relative acidity of the following series? O O O O > > > NR2 H3C H3C CF3 H3C CH3 H3C OR

Five factors that influence the relative proportion of keto/enol: 1. Aromaticity O OH H H Not observed

H

H

Substitution of the α-carbon by a second carbonyl derivative makes the α-proton even more acidic:

25

H

highly unstable

Why the huge difference in acidity? The lone pair is stabilized by donation into the carbonyl π system.

enol form

H

O

CH3

M Conjugate base (enolate) ~1025 more acidic just by replacing H with a carbonyl!

R H

Acetone in D2O will slowly incorporate deuterium at the α− Note 1 carbon. The enol form is responsible for this behavior. The rate of keto/enol tautomerism is greatly increased by acid (see below right)

O

O anhydride

N Me

NH(R) H C 2

OH

R

R

C H2

H2C CH3

OMe O

CH3

H Methyl propionate pKa = ~25

The enol tautomer is most significant for ketones and aldehydes. (You may also encounter it with acid chlorides in the mechanism of the Hell-Vollhard-Zolinsky reaction). Esters and amides are less acidic and exist almost exclusively as the keto form (e.g. >106 : 1 keto: enol for ethyl acetate)

O

O acid chloride

30

NMe2

H2C

O carboxylic R acid

Conjugate base

Tautomerism: a form of isomerism where a keto converts to an enol through the movement of a proton and shifting of bonding electrons

O NR2

M

Ethane pKa = ~50

H

OH

keto form

NH2 = 1° amide NHR = 2 ° amide NR1R2 = 3 ° amide O lactam (cyclic amide)

O

)n

R

H3C CH3

OMe

Key Concept: Tautomerism

25

Effect on reactivity of alkenes: Likewise, the presence of a carbonyl group activates alkenes toward nucleophilic attack:

The carbonyl is an electron withdrawing π system with low-lying π* orbitals. It stabilizes adjacent negative charge.

O

O

O amide

20

Me

H3C

O lactone (cyclic ester)

δ+

O

O ester

Oδ –

H

H3C

Effects on acidity of alkyl groups

•Carbon, oxygen: sp2 hybridized •O–C–C bond angle ~120 ° •C=O bond strongly polarized toward oxygen. •Carbonyl carbon is partially positive therefore electrophilic! •Lone pairs render oxygen weakly nucleophilic (will react with strong acid)

Me Preferred enol form

5. As with alkenes, increasing substitution increases thermodynamic stability (assuming equal steric factors)

R

R

H X R

H H Keto

O

K1 R

H H

K2

H O R

R

H X

H Enol

X:

Enol is nucleophilic at α-carbon, acid is excellent electrophile: this increases K2

Net result: Addition of acid speeds proton exchange between the keto and enol forms.

O

O M OEt

R

OH OR

R

O

Claisen Condensation OEt

2015 Version copyright James Ashenhurst, 2015 suggestions/questions/comments? [email protected]

1

2

Copyright 2014 James Ashenhurst masterorganicchemistry.com R–MgX [email protected]

O

A

OH

R

H

R

OH

R

2° alcohol

Aldehyde

OH

Aldol Reaction

OH

NR2

R

O

OH

R

R

Cl

Acyl chloride

O

C

R

R'

O

R'

Ketone

R

R'

NR2

O

O R

R

O

R

O

O

R

OR

OH

R

R

3° alcohol

O

NR2

O R

OH R

OR

R'

Aldol Reaction

R

R

OH

O R

R

R'

R

R

O

O

NR2

R

R

Aldol Reaction

N R

R

O R α, β unsaturated ketone (enone)

F

Varies with conditions: 1,2 adduct is kinetic pdt.

Ester

O OR

Michael Reaction

OH OR

O

O

R

NR2

O R

R

R

O

R

R

R

3° alcohol

O

O OR

β−keto ester: Claisen Condensation

R

Deprotonation

Deprotonation

Acid nitrile

NH2

Deprotonation

1° and 2° amides: deprotonation 3°amides: NR

Acetal

R

CN

Ester

Acid nitrile

R

Ester

Borderline

R

OR

HO

OH

R'

R

R

R'

Acetal Requires acid catalysis

1,3 diketone: Claisen Condensation

Deprotonation

NR

Deprotonation

R

1° and 2° amides: deprotonation 3°amides: NR

R

R

O R

X

Mix of addition /deprotonation

R

R

R

R

R

O

H H 1° alcohol

R

H H

OH R'

H

2° alcohol

OH R

R'

H

2° alcohol

Varies with conditions: 1,2 adduct is kinetic product, 1,4 adduct is thermodynamic.

O OR'

R

OH OH

NR

OR'

Borderline reaction: requires strong acid, alcohol as solvent, heat

OH

O R

H H 1° alcohol R

H OH

H H

1° alcohol

NR

–––

R

NR

R

H NH2

Amine

O R

RO

OH R

1° alcohol

Amide Hydrolysis Requires strong conc. acid, heat

NR2

O OR

Borderline

Fischer esterification (requires acid, heat)

NR

1° alcohol

Note: Easily reversible

NR

NHR

H

RO

Amide

I

H 1° alcohol

Transesterification Saponification Can be done under basic (basic conditions) or acidic conditions. Can also hydrolyze with aqueous acid O

NR

NR

NR

R

O

Amide

H

R

H

R

O R

RO

NR

1° alcohol

R

Hydrate see above: even less favored than with aldehydes due to sterics

O R

NHR

H

Carboxylic Acid

Note: in both cases, very prone to the reverse reaction (elimination)

Borderline

OH

H

OH OH

R

O R

OH

O R

RO

NC

LiAlH4 Lithium aluminum hydride

OH OH

Carboxylic acid

OR

O

NaBH4

O R

OR

11

Sodium borohydride

R

Hydrate (usu. thermodynaically disfavored, except for electron poor aldehydes) If aldehyde is enolizable, hydroxide can form enolate.

O

R

Usually requires dehydration agent (e.g. DCC)

O R

H

OH

RHN

Carboxylic acid

H

CN

O OH

OH

R

R CN R' Cyanohydrin

R

O

G

HO

H

O

R

R

O

R

OR

R

Requires acid catalysis to form

O

R

Michael Reaction

O

R

OH

RO

O

O R

O R

RO

R

Imine (ketimine)

Note: best when ketones are identical or when only one can enolize (to avoid scrambling)

E

Cyanohydrin

10

H2O/ M–OH water/ hydroxide

O NHR

Amide

OR

CN

H

O R

O

RO

9

R–OH/ R–OM alcohol/ alkoxide

O NHR

Amide (Schotten-Bauman reaction)

R

O

R

R

R

RO

H

OH R

O

R O

8

Cyanide

R

Imine (aldimine)

OH

RO

O

OH

R

O

R

R R R 3° alcohol

O

D

O R

OR

M–CN

N R

Knoevenagel Condensation

β−keto ester

OH

Anhydride

O

R

R 3° alcohol

O O

O

R

R–NH2 amine (primary amine)

R

Aldol Reaction

7

O

RO

heating under basic conditions will lead to elimination of OH - Aldol condensation also note that reaction can be reversible under basic conditions : Retro-aldol reaction

B

R

O

R

6 M

β−keto ester enolate

Enamine

R

O

RO

R

O

R

OR

5 O

NR2

R Ketone enolate

O

R

M

O

OR Ester enolate

Grignard

4

3 M

O

R

NHR

R

CN

R

OR

R

OH

NR

R

H

R Alkyl halide

Enolate Alkylation

Enolate Alkylation

Amine

Stork enamine reaction note: capable of alkylating a second time

caution! product is a good nucleophile; multiple alkylations usually result

Williamson Ether Synthesis requires basic conditions

requires basic conditions

Alkane

from "Master Organic Chemistry" masterorganicchemistry.com 2015 Version, Copyright 2015 James A. Ashenhurst [email protected]

Identifying the Patterns in Carbonyl Reaction Mechanisms

O Grignard Reaction

R–MgX

Cyanohydrin formation

NaCN

R

R O

R

NaBH4 or LiAlH4

R

O CH2R H

R

R

HO

CN

R

R

HO

H

R

R

R

O

Base-catalyzed aldol reaction

R

R O

Ketone or aldehyde reduction

R

HO

O

base R

A

P

A

P

Reactions of anionic nucleophiles with alkyl halides

R HO

RO

NR2 R

X

R

X

R

X

O R

R

OR

OH

R

CN

SN2

O

base R

D

R

SN2

R O

R

D

R

R

CH2R

OH

R

SN2

R

CN

Enolate alkylation (ketones or esters)

P

A

NR2

Reactions of neutral nucleophiles with alkyl halides

D

A

P

Reduction of esters

LiAlH4

A

E

D

Grignard Reaction + acid chloride, ester, or anhydride

RMgX

E

Imine formation

RNH2

*note 1

Fischer esterification

ROH

Amide hydrolysis

H2O

R

HO

H

R

H

HO

R

R

R

OR

A

E

A

P

A

E

A

P

P

A

PT

E

D

P

A

PT

E

D

P

A

PT

E

D

P

A

PT

E

A

D

P

E

A

PT

E

D

P

T

A

PT

T

14E

R Addition of neutral nucleophiles to acid halides or anhydrides (e.g.amines, alcohols, water)

O

HO

R

Cl

O Claisen condensation

R

Cl

O

Addition of anionic nucleophiles to acid halides or anhydrides (e.g.RO , H2N , HO )

CH2R R

NHR

O

RO

O R

O

R

OH

O

base OR

O

R

H2N

O

R

O

RNH2

R

R

OR

O A

NH2

R

A

14A

P

E

N

R

O D

R

R

acid

R

O

R

Cl

R O

acid

R

R

OH

OR

R O 1,4 addition of Gilman reagent

O

O

R2CuLi R

R

O

acid

R

NR2

R

OH

RO

OR

R

R

R O Michael reaction

O

R

O

base R

CH2R

O

O

R

D

R

P

14A

Formation of acetals

ROH

Hydrolysis of acetals

H2O

R

acid R

R O RNH2

1,4 addition of neutral nucleophiles

O R

14A

R

PT

RHN

RO

OR

R

R

O Acid catalyzed aldol

R

R

O CH2R

H

O

acid

R O

acid R

R

R

D

R *note 1 - There is actually a fourth step; base removes a proton from the acidic alpha-carbon, rendering the reaction irreversible. Acidic workup gives the final product.

A Cα

X

O

Nu

R

X

E

O

Nu

R

X

SN2 R

Nu

O R

Nu

OH R

Nu

14E

R X β

Deprotonation O

O

O

R H X

R

R

R

H

[1,4] elimination

O

Nu

O

Nu

[1,4] addition 14A

SN2

Elimination [1,2-elimination]

Addition to carbonyl [1,2-addition] O

The Nine Mechanistic Components (with examples)

D

O

O

T R

H3C

H2C

R

OH R

H

R H

P

H2C

O

acid R

Intramolecular acid-base reaction example:

R

Protonation O

PT

H HO OR

Removal of a proton from substrate

Keto-Enol Tautomerization H

base R

H3C

Proton Transfer O

H3C

Addition of a proton to substrate

R

R

H2O OR R

R

This can either be drawn as an intramolecular reaction (one step) or as a deprotonation (D)followed by a protonation (P)

from "Master Organic Chemistry" masterorganicchemistry.wordpress.com 2015 Version copyright 2015 James A. Ashenhurst [email protected]

Nine Key Mechanisms In Carbonyl Chemistry Description

Mechanism Addition [sometimes "[1,2] addition"] O Cα

Nu

O X

Attack of a nucleophile at the carbonyl carbon, breaking the C–O π bond.

Cα

Nu

X

Elimination Lone pair on carbonyl oxygen comes down to carbonyl carbon, forming new π-bond and displacing leaving group X.

[sometimes "[1,2] elimination"] O

Nu

O Cα

X

Cα

Nu

Promoted by

Hindered by

Anything that makes the carbonyl carbon a better electrophile (more electron-poor)

Examples

1) Anything that makes the carbonyl carbon a poorer electrophile (more electron-rich) 2) Sterically bulky substituents next to the carbonyl X-groups that are strong π-donors (e.g. amino, hydroxy, alkoxy)

Electron withdrawing groups on α carbon Electron-withdrawing X groups that are poor π-donors (e.g. Cl, Br, I, etc.) Addition of acid (protonates carbonyl oxygen, making carbonyl carbon more electrophilic. Note: acid must be compatible with nucleophile;alcohols are OK, strongly basic nucleophiles (e.g. Grignards) are not.

Grignard reaction Imine formation Fischer esterification Aldol reaction Acetal formation Claisen condensation

Sterics: X=H (fastest) > 1° alkyl > 2° alkyl > 3° alkyl (most hindered, slowest) X=Cl (poorest π-donor, fastest addition) > OAc > OR > NH2/NHR/NR2 (best π donor, slowest rate)

The better the leaving group X, the faster the reaction will be. The rate follows pKa very well. Acid can turn poor leaving groups (NR2, OH) into good leaving groups (HNR2, H2O)

Fischer esterification Formation of amides by treatment of acid halides with amines. Claisen condensation

X groups that are strong bases are poor leaving groups. Alkyl groups and hydrogens never leave. Amines and hydroxy are poor leaving groups under basic conditions, but are much better leaving groups under acidic conditions.

I > Br > Cl > H2O > OAc > SR > OR >> NR2, O2– H alkyl >40 –8 –7 –2 4 12 17 35

–9 Nucleophile attacks alkene polarized by electron withdrawing group, leading to formation of enolate.

[1,4] addition O

O R

Nu

R

[1,4] elimination OH

O

R

Lone pair on oxygen comes down to form carbonyl, enol double bond displaces leaving group on the β-carbon

H R

X β SN2 R H

Backside attack of nucleophile onto electrophile (alkyl halide or equivalent)

O

O

R

R

R

H

Keto-Enol Tautomerization O H3C

MIchael reaction Addition of Gilman reagents (organocuprates)

[1,2]-addition can compete in the example of Grignard reagents. The more electron rich the carbonyl, the slower will be the rate of reaction (less able to stabilize negative charge). So addition to α,β-ketones > α,β−esters > α,β-amides.

Nu

Nu

X

So-called "soft" nucleophiles such as Gilman reagents (organocuprates) will add [1,4], as will amines, enolates etc. The more stable the conjugate base (enolate) of the carbonyl, the faster the reaction. Extra electron withdrawing groups on the α-carbon will promote the reaction.

OH R

H

Internal oxygen ↔ proton transfer with change in hybridization of oxygen and carbon.

Facilitated if X is a good leaving group (just like [1,2]-elimination) In the aldol condensation, addition of acid helps OH group leave as H2O.

Aldol condensation Knovenagel condensation

As with [1,2]elimination, X groups that are strong bases are poor leaving groups. Addition of acid will promote elimination of groups such as NR2 and OH/OR.

Note that in the Aldol reaction run under basic conditions, the enolate is a stronger base than OH(–), so in the base-promoted Aldol reaction, the [1,4]-elimination is favorable.

Facilitated by good leaving group on electrophile (alkyl halide or tosylate). Polar aprotic solvent is ideal. Enolate α-carbon is excellent nucleophile for SN2 The higher the pKa of the carbonyl compound, the more reactive the conjugate base will be in the SN2.

Enolate alkylation Carboxylate alkylation

Rate of reaction will go primary alkyl halide > secondary alkyl halide Tertiary alkyl halides unreactive in SN2.

Facilitated by acid The enol form is stabilized by internal hydrogen bonding if there is a carbonyl present at the β position.

Tautomerism under acidic conditions only significant for ketones, aldehydes, and acid halides (the latter under the conditions of the Hell-Vollhard-Zolinski reaction).

Acid-catalyzed aldol Acid-catalyzed bromination of ketones

R H

Deprotonation

Acid Base Reactions

Protonation

Proton Transfer

The conjugate base is always a better nucleophile than the conjugate acid. Deprotonation increases nucleophilicity. E.g. enolate > enol, alkoxide > alcohol, NH2(–) > NH3 Conjugate base can perform reactions the conjugate acid cannot. Deprotonation is also the last step in acid-catalyzed reactions, in order to generate the final (neutral) product

1) catalyzes [1,2] addition to carbonyls 2) promotes [1,2] elimination 3) to promote tautomerization. 4) quench (e.g. enolate from 1,4.

An internal acid-base reaction. Not mechanistically distinct from the above, but often drawn as one step. Can proceed either intramolecularly or intermolecularly (both pathways operate) hence distinct arrow pushing steps often not drawn, and we just say "proton transfer"

R

OH

Base R

H RO

O

O R

H

faster [1,2] addition

H R

R

R

OH2

R

R

deprotonation at end of acidcatalyzed acetal formation provides neutral product

OH

H

O

H

H2O OR R

R

R

OH

H

R

H faster enolization

R

OR

R

CH3

H HO OR

O OH

OH2

faster [1,2] elimination

O

RO

–[H]

R

O

HO X

H

OR

R

forms alkoxide (more nucleophilic)

R

reaction quench

R

R

"Master Organic Chemistry"

Introduction to Carbohydrates

Carbohydrates have the molecular formula Cn(H2O)n

What's a Carbohydrate? For example:

Glucose C6H12O6

= C6(H2O)6

a hexose

Fructose C6H12O6

= C6(H2O)6

a hexose

Sucrose C12H24O12

= C12(H2O)12

a disaccharide

= C3(H2O)3

a triose

Glyceraldehyde C3H6O3

The Fischer Projection O H

O

H

H

OH CH2OH

H OH

What's D and L?

H

H CH2OH L-glyceraldehyde

OH CH2OH D-glyceraldehyde

HO

Examples H

O

O

H

H H H

OH OH OH CH2OH

D-Ribose

HO HO HO

H H H CH2OH

L-Ribose

H HO H H

In the Fischer projection, assign D and L by placing the most oxidized carbon of the sugar (i.e. the aldehyde) at the top, and then if the stereocenter furthest from it is on the right, it's D. If it's on the left, it's L.

CHO OH H OH OH CH2OH

D-Glucose

HO H H HO

CHO H OH OH H CH2OH

CHO OH H OH OH CH2OH CH2OH D-glyceraldehyde D-erythrose a tetrose a triose H H

O

3

H OH

H H H

OH OH OH CH2OH

D-ribose a pentose

O

H

HO HO

H H

OH H

OH

D-Glucose

CHO OH H H OH CH2OH

HO

HO

HO HO

H

OH OH

H

β−D-Glucose

6

4

HO HO

5

1

OH

H2 OH H H β−anomer C-1 OH is up Mnemonic: "β" stands for bird, which flies up in the sky 3

H

H

OH OH OH

HOH

H

OH

OH H

H H

α−D-Fructose

OH

2

1

OH

2

3

Step 4 OH

O H H

HO H

O H

H

HOH

O H H OH

OH H Maltose

OH

O

H

Ag, NH3

OH CH2OH

H2O

H

OH OH CH2OH

Aldonic Acid synthesis: Br2, H2O; oxidizes aldehyde to carboxylic acid O H

O

H

Br2

OH CH2OH

H

H2O

OH OH CH2OH

Aldaric acid synthesis: HNO3, H2O; oxidizes both ends to acids O

O

H

HNO3 H2O

OH CH2OH

OH

H

OH

O

OH

Ruff degradation: Br2, H2O, then FeCl3, H2O2; shortens sugar by one carbon O H 1 O 1) Br2 H2O 2 + CO2 H 2 OH H 1 2) H2O CH2OH CH2OH 3

CH2OH

HO

H HO

4

Tollens Test: Ag+, NH3, H2O : oxidizes aldehyde to carboxylic acid

CH2OH

O

α−D-Ribose

CH2OH O H HO

CH2OH O OH

5 1

Step 3 OH

3

D-galactose a hexose

Aldoses contain an aldehyde; ketoses contain a ketone CH2OH H OH O HO CH2OH O HO H HO H HO HO OH HO H OH H H OH OH H H H OH OH H It's trickier to see this in CH2OH Ring closed form the ring-closed form. D-Fructose (a ketose) D-glucose (an aldose)

4

1

Step 2

H

CH2OH

HO OH

HO

6

Reactions (not a complete list)

HO

H OH

OH

β−D-Galactose

CH2OH H O O H H H HO HOH H O CH2OH HO H HO OH H Sucrose not a reducing sugar - no hemiacetal!

Kiliani-Fischer Synthesis: Convert aldehyde to cyanohydrin, then reduce and hydrolyze. Extends sugar chain by one carbon. O H O H 1) HCN H HO 2) H2, Pd H OH H OH then H O 3 CH2OH CH2OH Glycosides: Acetal formation

Aldoses and Ketoses O C H H OH H HO H OH H OH CH2OH

1

Step 1

H OH

OH H OH O H CH2OH

O H H H

6

CH2OH O OH

5

H

Examples of reducing sugars (the hemiacetal is highlighted in red) H OH

CH2OH O

5

OH 3

A cyclic sugar with a hemiacetal is in equilibrium with the open chain form. And since the open chain form can be reduced to an alcohol with NaBH4, it is termed a "reducing sugar".

HO

H

H HO HO H

4

"Reducing Sugars"

L-Galactose

H

H HO H H

HO 1

Triose, Tetrose, Pentose, Hexose

H

3

Mnemonic: "α" looks like a fish, Open chain form which swims down in the sea The two anomers (α and β) are in equilibrium with each other. A pure sample of either α-D-glucose (optical rotation +112°) or β-D-glucose (optical rotation +19°) will change over time to +52.7° which is an equilibrium mixture of 64% alpha and 36% beta. This change in optical rotation is known as mutarotation.

Each of these terms simply denotes the number of carbons in the sugar O

5

H 2 OH H OH α−anomer C-1 OH is down

He guessed right!

H

6

4

HO HO

D and L are arbitrary terms assigned by Emil Fischer in 1891 to denote the enantiomers of glucose. The absolute configuration wasn't determined until 1951. O

OH H OH 5 O H 6 CH2OH

Note that a stereocenter is formed on C1 ! This leads to two isomers of the cyclic sugar, called "anomers" (designated alpha and beta) O

H

A sugar with a five-membered ring is called a furanose

1. A bond is forming between the oxygen on C-5 and C-1 . So draw a six membered ring with oxygen at the top right. 2. Since the sugar is D, on C-5, place the CH2OH pointing "up" on the ring (if it was L, CH2OH would point "down") 3. For C-2, C-3, and C-4, every group on the right of the Fischer will be down on the ring, and every group on the left of the Fischer will be up on the ring. 4. The configuration of the anomeric carbon (C-1) will be a mixture of alpha (α) and beta (β)

H

2

The alcohol on C5 attacks the aldehyde on C1, forming a cyclic hemiacetal (after proton transfer)

Fischer projection

O

H HO H H

A sugar with a six-membered ring is called a pyranose

OH H OH O CH2OH

1

Ring form

Open chain form

Mnemonic: "the arms come out to hug you" (or strangle if you prefer)

OH CH2OH

H H HO H H

OH H OH O H CH2OH

O

new stereocenter

OH

H

O

A 3-D molecule is "projected" as a flat molecule. H

Converting a Fischer to a Ring Form

Chain and Ring Forms of Sugars Sugars can exist as mixtures of their open chain and ring forms.

H HO H H

Note - this sheet is not meant to be comprehensive. Your course may provide additional material, or may not cover some of the reactions shown here. Your course instructor is the final authority.

masterorganicchemistry.com

H OH

Haworth Projections 6

CH2OH O H H OH

5

H 4

OH

3

H

2

OH

α−D-Glucose (Haworth)

H OH H 1

OH

The Haworth projection makes it more clear which groups are "up" and "down" in a cyclic sugar. The ring isn't "actually" flat; it's really in a chair conformation.

4

HO HO

6

5

HO 2

3

H

H OH

1

H H

OH OH

H

ROH, H

HO

HO HO

H H

OR OH

H

H

OH

α−D-Glucose (Chair)

H OH HO

HO HO

Omissions, Mistakes, Suggestions? [email protected] This sheet copyright 2015, James A. Ashenhurst masterorganicchemistry.com

"Master Organic Chemistry"

Introduction to Amines Important nitrogen-containing functional groups

Basicity of amines (cont'd)

Primary amine

N H

N

Secondary amine

Tertiary Amine

(1° amine)

(2° amine)

R

O NH2

N

N H A secondary amide

sp hybridized least basic

R

R

Acidity of amine derivatives (ammonium salts) The weaker the base, the stronger the conjugate acid. NH3

N

O

N H2 less acidic

The nitrogen of amides is considerably less basic than that of amines due to the contribution of the right-hand resonance form

R

NH2

less acidic

more acidic

N H more acidic

R

R

N

H

R

imine "Aldimine"

NR2

N

R

OH

R

imine "Ketimine"

enamine

N

R

R

oxime

R

hydrazone

"exotic" variations of imines, less frequently seen

1. Resonance NH2

NH2

CH3–I

resonance stabilization less basic

N

NH2

N

more basic

NH2

key resonance form

N

N

less basic

more basic

N

NO2 less basic

N

An azide "R–N3 "

R

H

Note: under acidic conditions, the conjugate acid of the imine is formed ("the iminium ion") and this is what is actually reduced here

R

H

R'

N R

R'

O + H2O

H

R

Imine

H2N R

R'

N R

H

Ketone

N H H

N

R

+ H2O

Imine

Enamines are good nucleophiles, due to the importance of this resonance form: N

O

R'

+ H2O

N

R

R

R enamine

This carbon is nucleophilic, and will react with electrophiles such as alkyl halides, etc.

N

NH2NH2

NH2

The Hoffmann elimination of ammonium salts

(hydrazine)

Ag2O

N

The less substituted alkene is formed here.

heat

Non-Zaitsev product The Hofmann rearrangement:

NH2

Pd/C H2

Nitro groups can be reduced to amines with Pd-C/H2, or Zn/HCl, or Sn/HCl

O R

NaOH Br2

NH2

R

NH2

+ CO2 These reactions both proceed through an isocyanate intermediate

The Curtius Rearrangement Nitriles can also be reduced NH2 with Pd-C or Pt-C and H2

LiAlH4

Both nitriles and azides can be prepared through SN2 reactions with alkyl halides

O

O R

heat N3

R

R–OH

N H

R O

R + N2

NH2

can also use LiAlH4

C

O

isocyanate

Formation of diazonium salts

N Pd-C H2

N

NH2

NaNO2 H

N

N

For reactions of diazonium salts, see the summary sheet on aromatic chemistry

diazonium salt

Reduction of amides

Electron withdrawing groups will make the nitrogen less electron-rich, and therefore more stable (less basic)

>

N

R'

O

(phthalimide)

Reduction of azides

Other examples of electron donors - OH, OR, CH3, NR2

N

H

Reduction of nitriles N

H

H

HN

NaCNBH3 R

primary amine O Alkylation of phthalimide with alkyl halides stops after one alkylation. The phthalimide is then removed with hydrazine (NH2NH2) to give the free amine.

C

>

H2N

R

O

2. Electron donating and withdrawing groups Electron donating groups will make the nitrogen more electron rich, and therefore more unstable (more basic)

N

NaCNBH3 is sodium cyanoborohydride, a reducing agent. NaBH(OAc)3 is also sometimes used

R'

Formation of enamines When ketones (or aldehydes) are treated with a secondary amine and acid, an enamine is formed. This is also a condensation reaction.

N

O

NO2

R

Imine

Aldehyde

Gabriel synthesis - a way of making primary amines

NH2

no resonance stabilization more basic

R

doesn't stop after just one reaction. Goes on to form ammonium salt.

Br

H

R

H

Reduction of nitro groups

O NH2

resonance stabilization less basic

R'

R'

HN

NaCNBH3

Formation of imines

Alkylation of amines with alkyl halides is not a good way of making amines; it will lead to ammonium salts

Basicity of amines

no resonance stabilization more basic

R

O

N

Factors that affect the basicity of amines The more stable the lone pair, the less basic it will be. Examples of factors that stabilize the lone pair are 1) resonance 2) electron donating/withdrawing groups, and 3) orbitals

NH2

R

R'

When ketones or aldehydes are treated with a primary amine, imines are formed. Mild acid can catalyze this reaction. The byproduct is one equivalent of water, so this is a condensation reaction. Imines can be hydrolyzed back to the aldehyde/ketone by adding water.

Synthesis of amines

NR2

N

Imine

Ketone

Imines are the nitrogen-containing analogues of ketones and aldehydes. Enamines are the nitrogen-containing analogues of enols R

R'

H

H2N

R

Imines and Enamines

N

H

O

NH3

A tertiary amide

O NH2

H2N

Aldehyde

Important to note: Amides have a significant resonance form

R

O

sp2 hybridized less basic

O

R

A primary amide

A very versatile method for amine synthesis involves making the imine, and then reducing it to the amine. Primary or secondary amines can be used; often, imine formation and reduction is done under slightly acidic conditions.

HC N N H sp3 hybridized more basic

Quaternary ammonium technically not an amine not a base (no lone pair)

Amides An amide has a carbonyl group adjacent to an amine nitrogen O

Reductive amination

The more s-character the orbital, the more stable the nitrogen lone pair will be (and therefore less basic) N

(3° amine)

Synthesis of amines (cont'd)

3. Orbitals

An amine is classifed as primary, secondary, or tertiary depending on how many carbons the nitrogen is attached to NH2

Note - this sheet is not meant to be comprehensive. Your course may provide additional material, or may not cover some of the reactions shown here. Your course instructor is the final authority.

←

Amines

masterorganicchemistry.com 2015 Version

O N O

O

LiAlH4 NH2

NH2

Depending on the type of amide (primary, secondary, tertiary) the corresponding amine will be formed. Omissions, Mistakes, Suggestions? [email protected] This sheet copyright 2015, James A. Ashenhurst masterorganicchemistry.com