Derivatives (1)

1

Derivatives of Carboxylic Acid

R C

O

Cl

: :

:....

R RC

O

O C

O: : : :....

R RC

O

O....

: :

R C

O

OH

: :

..

..

R N

H

HC

O: :..

R

_

C

O

O:

::....

R C N:acid chloride

acid anhydride

esteramide

nitrile

carboxylate

2

Nomenclature of Acid Halides

• IUPAC: alkanoic acid alkanoyl halide• Common: alkanic acid alkanyl halide

C

O

Cl

CH2

NH2

CH2 C

O

Cl

I: 3-aminopropanoyl chloride

c: -aminopropionyl chloride

I: benzenecarbonyl bromide

c: benzoyl bromide

Rings: (IUPAC only): ringcarbonyl halide

I: 3-cylcopentenecarbonyl chloride

CH2 C

O

ClC

O

Cl 4

I: hexanedioyl chloride

c: adipoyl chloride

C

O

Br

CH

NO2

CH2 CH2H3C C

O

Cl

I: 4-nitropentanoyl chloride

c: -nitrovaleryl chloride

R C

O

Cl

: :

:....

3

Nomenclature of Acid Anhydrides

• Acid anhydrides are prepared by dehydrating carboxylic acids

CH3 C

O

OH H O C

O

CH3+

CH3 C

O

O C

O

CH3- H2O

acetic acid acetic anhydrideethanoic anhydrideethanoic acid

R RC

O

O C

O: : : :....

CH3 C

O

O C

O

HC

O

O C

O

H

C

O

O C

O

C CO OO

I: benzenecarboxylic anhydride

c: benzoic andhydride

I: butanedioic acid

c: succinic acid

I: benzoic methanoic anhydride

c: benzoic formic anhydride

C C

O

OH

O

HOC C

O OO

- H2O

I: butanedioic anhydride

c: succinic anhydride

I: cis-butenedioic anhydride

c: maleic anhydride

Some unsymmetrical anhydrides

I: ethanoic methanoic anhydride

c: acetic formic anhydride

4

Nomenclature of Esters

CH3 O C

O

CH3

alkyl

I: alkanoatec: alkanate

R C

O

OH

: :

..

..R RC

O

O....

: :

+H2O-

carboxylicacid

alcohol esterH+

H O R....

O C

O

H

• Esters occur when carboxylic acids react with alcohols

I: methyl ethanoate

c: methyl acetate

C

O

O CH(CH3)2 O C CH(CH3)2

O

C O C(CH3)3

O

CH3 O C

O

C

O

O CH3

I: phenyl methanoate

c: phenyl formateI: t-butyl benzenecarboxylate

c: t-butyl benzoate

I: isobutyl cyclobutanecarboxylate

c: noneI: cyclobutyl 2-methylpropanoate

c: cyclobutyl -methylpropionate

I: dimethyl ethanedioate

c: dimethyl oxalate

5

CH2

CH2 CH2

OH

C

O

OHH+

H2O-

CH2

CH2 CH2

O

C

O

OC

O

OO

I: 4-hydroxybutanoic acid

c: -hydroxybutyric acid

I: 4-hydroxybutanoic acid lactone

c: -butyrolactone

O

O

O

H3C

O

O

CH3

O

Nomenclature of Cyclic Esters, “Lactones”

Cyclic esters, “lactones”, form when an open chain hydroxyacid reacts intramolecularly. 5 to 7-membered rings are most stable.

‘lactone’ is added to the end of the IUPAC acid name. ‘olactone’ replaces the ‘ic acid’ of the common name and ‘hydroxy’ is dropped but its

locant must be included.

I: 4-hydroxypentanoic acid lactone

c: -valerolactone

I: 6-hydroxy-3-methylhexanoic acid lactone

c: -methyl--caprolactone


c: -valerolactone


c: -valerolactone

6

R C

O

NH

H

R C

O

NH

R R C

O

NR

R

CH3CH2CH2 C

O

NH

HC

O

NH

H

Cl

CH3CH

CH3

C

O

NCH3

HC

O

NH

H3CC

O

NCH3

CH2CH3

Nomenclature of Amides

1° amide2° amideN-substituted amide

3° amideN,N-disubstituted amide

1° amides: ‘alkanoic acid’ + amide ‘’alkanamide’ a ring is named ‘ringcarboxamide’

I: butanamide

c: butyramideI: 3-chlorocyclopentanecarboxamide

c: none

I: p-nitrobenzenecarboxamide

c: p-nitrobenzamide

C

O

NH

HNO2

2° and 3° amides are N-substituted amides

I: N,2-dimethylpropanamide

c: N,-dimethylpropionamide

I: N-phenylethanamide

c: N-phenylacetamide

I: N-ethyl-N-methylcyclobutanecarboxamide

c: none

c: acetanilide

7

Nomenclature of Cyclic Amides, “Lactams”

CH2

CH2 CH2

NH

C

O

OH

H H2O-

CH2

CH2 CH2

NH

C

O

NH

O

NH

OBr

Cyclic amides, “lactams”, form when an open chain aminoacid reacts intramolecularly. 5 to 7-membered rings are most stable.

I: 4-aminobutanoic acid

c: -aminobutyric acid

I: 4-aminobutanoic acid lactam

c: -butyrolactam

‘lactam’ is added to the end of the IUPAC acid name. ‘olactam’ replaces the ‘ic acid’ of the common name and ‘amino’ is dropped but its

locant must be included.

I: 5-aminohexanoic acid lactam

c: -caprolactam

I: 3-amino-2-bromopropanoic acid lactam

c: -bromo--propionolactam

I: 4-amino-3-methylbutanoic acid lactam

c: -methyl--butyrolactam

NH

O

H3C

NH

O

CH3

8

CH3 C N CN CH2CH2CH2 I

CH3O CN

CNHS

R C

O

NH

H H2O-

POCl3R C N

Nomenclature of Nitriles

Nitriles are produced when 1° amides are dehydrated with reagents like POCl3

• IUPAC: alkane + nitrile ‘alkanenitrile’• IUPAC rings: ‘ringcarbonitrile’• Common: alkanic acid + ‘onitrile’ ‘alkanonitrile’

I: ethanenitrilec: acetonitrile

I: 4-iodobutanenitrile

c: -iodobutyronitrile

I: 3-methoxycyclohexanecarbonitrile

c: none

I: p-thiobenzenecarbonitrile

c: p-mercaptobenzonitrile

COOH

CN

I: 2-cyanocyclopentanecarboxylic acid

c: none

9

CN

NH

O

Cl

NHCH3

Br O

O

O

CH3 C

O

O CH2Br CH3 C

O

O- Na+

CH3 C

O

CH2 C

O

Cl

Nomenclature Practice Exercise

I: cyclobutanecarbonitrilec: none

I: bromomethyl ethanoatec: bromomethyl acetate

I: sodium ethanoatec: sodium acetate

I: pentanedioic anhydridec: glutaric anhydride

O OO

I: 3-oxobutanoyl chloridec: -oxobutyryl chloride

I: 3-bromo-N-methylpentanamidec: -bromo-N-methylvaleramide

I: 2-ethyl-5-hydroxypentanoic acid lactonec: -ethyl--valerolactone

I: 6-amino-6-chlorohexanoic acid lactamc: -chloro--caprolactam

10

most

reactive

acid chloride

acid anhydride

aldehyde

ketone

ester

carboxylic acid

amide

nitrile

carboxylate

least

reactive

R C

O

Cl

: :

:....

R RC

O

O C

O: : : :....

R C

O

H

: :

R RC

O: :

R RC

O

O....

: :

R C

O

OH

: :

..

..

R N

H

HC

O: :..

R

_

C

O

O:

::....

R C N:

Relative Reactivity of Carbonyl Carbons

Nucleophiles (electron donors), like OH-, bond with the sp2 hybridized carbonyl carbon. The order of reactivity is shown.

11

• Recall that electron donors (Nu: -’s) add to the electrophilic carbonyl C in aldehydes and ketones. The C=O bond breaks and the pair of electrons are stabilized on the electronegative O atom.

• R (alkyl groups) and hydrogens (H) bonded to the C=O carbon remain in place. R- and H- are too reactive (pKb of – 40 and -21). R and H are not leaving groups, so the carbonyl group becomes an alkoxide as the sp2 C becomes a tetrahedral sp3 C.

• A second addition of a nucleophile cannot occur since alkoxides are not nucleophilic. The reaction is usually completed by protonation of the alkoxide with H3O+ forming an alcohol. This later reaction is simply an acid/base reaction.

• The characteristic reaction of aldehydes and ketones is thus ‘nucleophilic addition’.

R C

O

HCH3 MgBr

R C

O

H

CH3

-tetrahedral alkoxide with sp3 carbon.

R C

O

H

CH3

-

OH

HH ++ R C

O

H

CH3

H

Nucleophilic Addition to Aldehydes and Ketones

OH H+

12

Nucleophilic Acyl Substitution in Acid Derivatives

• Carboxylic acid derivatives commonly undergo nucleophilic substitution at the carbonyl carbon rather than addition. The first step of the mechanism is the same.

• The C=O bond breaks and the pair of electrons are stabilized on the electronegative O atom. A tetrahedral alkoxide is temporarily formed.

• In carboxylic acid derivates, one of the groups that was bonded to the carbonyl C is a leaving group. When this group leaves, the sp3 tetrahedral alkoxide reverts back to an sp2 C=O group. Thus substitution occurs instead of addition.

• In many cases, the substitution product contains a carbonyl that can react again.

R C

O

ClCH3 MgBr

R C

O

Cl

CH3

-

R C

O

CH3 + Cl- Chlorine is a fair

leaving group.

sp2 carbonyl reforms

sp2 carbonyl C alkoxide C js sp3

R C

O

CH3CH3 MgBr

R C

O

CH3

CH3

-

OH

HH +

OH H+R C

O

CH3

CH3

H

Note that because the C=O group reforms, the nucleophile can react a second time.

13

• In carboxylic acid derivatives, the acyl group (RCO) is bonded to a leaving group (-Y).


R C

O

Y Nu:- R C

O

Y

Nu

-

R C

O

Nu + Y:-

acyl group

R C

O

• The leaving group (-Y) becomes a base (Y:-) . The acid derivative is reactive If the base formed is weak (unreactive). Weak bases are formed from good leaving groups.

• For the carboxylic acid derivatives shown, circle the leaving group. Then draw the structure of the base formed, give its pKb, and describe it as a strong or weak base.

acid derivative leaving group pKb strength as base

R C

O

Cl

R C

O

O C

O

R

R C

O

O R

R C

O

NH2

Cl-

-O C

O

R

O R-

NH2-

non basic

weak base

strong base

v. strong base

+21

+9

-2

-21

Draw the mechanism.

14

We will study the reaction of only a few nucleophiles with various carboxylic acid derivatives and we will see that the same kinds of reactions occur repeatedly.


•Hydrolysis: Reaction with water to produce a carboxylic acid•Alcoholysis: Reaction with an alcohol to produce an ester•Aminolysis: Reaction with ammonia or an amine to produce an amide•Grignard Reaction: Reaction with an organometallic to produce a ketone or alcohol•Reduction: Reaction with a hydride reducing agent to produce an aldehyde or alcohol

Draw the structures of the expected products of these nucleophilic substitution reactions, then circle the group that has replaced the leaving group (-Y)

hydrolysis

alcoholysis

aminolysis

Grignard reduction

hydride reduction

R C

O

Y +

R C

O

Y +

R C

O

Y +

R C

O

Y +

R C

O

Y +

R MgX

LiAlH3 H

R C

O

O H

R C

O

O R

R C

O

NH2

R C

O

R

R

H

R C

O

H

H

H

H OH

H OR

H NH2

15

• Nucleophilic acyl substitution converts carboxylic acids into carboxylic acid derivatives, i.e., acid chlorides, anhydrides, esters and amides.

Nucleophilic Acyl Substitution of Carboxylic Acids

R C

O

Cl

: :

:....

R RC

O

O C

O: : : :.... R RC

O

O....

: :

R C

O

OH

: :

..

..R N

H

HC

O: :..

acid chloride

acid anhydride ester

amide

SOCl2

-H2O

ROH H+

NH3,, -H2O

R C

O

O H R C

O

Y

16

Conversion of Carboxylic Acids to Acid Halides

• The S atom in SOCl2 is a very strong electrophile. S is electron deficient because it is bonded to 3 electronegative atoms (Cl and O). Cl is a leaving group.

• The hydroxyl O atom in a carboxylic acid has non bonded pairs of electrons, making it a nucleophile. This O atom bonds with S (replacing a Cl) and forming a chlorosulfite intermediate. The chlorosulfite group is a very good leaving group. It is easily displaced by a Cl- ion via an SN2 mechanism yielding an acid chloride.

• Use curved arrows to draw the initial steps of the mechanism shown below.

C

O

OHR S

O

Cl Cl

thionyl chloride

+ C

O

OR S

O

Cl- HCl

C

O

ClR + SO2 + HCl

H+

chlorosulfite intermediate

..

..

: :..

..

: : : : : : : :

• PBr3 will substitute Br for OH converting a carboxylic acid to an acid bromide• Draw and name the products of the following reactions.

C

O

OHH3C

SOCl2

C

O

O HH3C

SOCl2

C

O

ClH3C+HCl + SO2

I: p-methylbenzenecarbonyl chloride

c: p-methylbenzoyl chlorideC

O

ClH3C+HCl + SO2

I: ethanoyl chloride

c: acetyl chloride

1

12

23 3

17

Conversion of Carboxylic Acids to Acid Anhydrides

• High temperature dehydration of carboxylic acids results in two molecules of the acid combining and eliminating one molecule of water.

CH3 C

O

OH H O C

O

CH3+

CH3 C

O

O C

O

CH3- H2O

acetic acid acetic anhydrideethanoic anhydrideethanoic acid

• Cyclic anhydrides with 5 or 6-membered rings are prepared by dehydration of diacids.

2HC

2HCC

OH

OHC

O

O

2HC

2HCC

O

C

O

O

H2O-

I: butanedioic acid

c: succinic acid

I: butanedioic anhydride

c: succinic anhydride

• Draw a reaction showing the preparation of cyclohexanecarboxylic anhydride.

C

O

OC

O

H2O-C

O

O H2

18

H2O-CH2

O

O

Br

Na+

-

+CH3CH2 C

O

O H....

: :Na+ OH-

CH3CH2 C

O

O....

: :

:_ CH3 I

CH3CH2 C

O

O....

: :

CH3 NaI

propanoic acidH2O

_Na+

SN2

acid/base

CH3CH2 C

O

O....

: :

:_

+Na+ C

CH3

H3C

CH3

I

Conversion of Carboxylic Acids to Esters

Two methods are used: SN2 reaction of a carboxylate and Fischer Esterification

1. SN2 reaction of a carboxylate with a methyl halide or 1 alkyl halide is straightforward. 2 and 3 alkyl halides are not used because carboxylate is only a fair nucleophile and is basic enough (pKb = 9) that elimination of HX from the alkyl halide will compete with substitution. The carboxylate will be protonated and the alkyl halide eliminates HX becoming an alkene.

CH3CH2C

O

OH + CH2 CCH3

CH3

+ NaIE2

isobutylene

CH2Br

O

O

H NaOH

CH2

O

O

O

O

NaBr-

I: 5-hydroxypentanoic acid lactonec: -valerolactone

I: 5-bromopentanoic acid

c: -bromovaleric acid

sodium propionoate

I: sodium 5-bromopentanoate

c: sodium -bromovalerate

19

alkyl halide

(substrate)

good Nu-

nonbasic e.g., bromide

Br-

good Nu-

strong base e.g., ethoxide

C2H5O

-

good Nu-

strong bulky base e.g., t-butoxide

(CH3)3CO

-

very poor Nu-

nonbasic e.g., acetic acid

CH3COOH

Me

1°

2

3

SN1, E1

SN2

E2

SN2

SN2

SN1

SN2

SN2

E2

E2

SN2

E2 (SN2)

E2

no reaction

no reaction

SN1, E1

alkyl halide

(substrate)

……….. Nu- ………. .base

e.g., cyanide CN-

pkb = ………

……….. Nu- ……….. base

e.g., alkyl sulfide RS-, also HS-

pkb = ………

……….. Nu-

……….. base e.g., carboxylate

RCOO-

pkb = ………

alcohol (substrate)

HI

HBr

HCl

Me Me

1 1

2 2

3 3

4.7

v. gd.

moderate

SN2

SN2

SN2

E2

6.0 / 7.0

moderate

v. gd.

SN2

SN2

E2

SN2

9

weak

fair

SN2

SN2

E2

E2

SN2

SN2

SN1

SN1


20

2. Fischer Esterification: (RCOOH RCOOR) Esters are produced from carboxylic acids by nucleophilic acyl substitution by a methyl or 1º alcohol. Heating the acid and alcohol in the presence of a small quantity of acid catalyst (H2SO4 or HCl (g)) causes ester formation (esterification) along with dehydration. The equilibrium constant is not large (Keq ~ 1) but high yields can be obtained by adding a large excess of one of the reactants and removing the H2O formed. The reaction is reversible. A large excess of H2O favors the reverse reaction. Bulky (sterically hindered) reagents reduce yields.

Since alcohols are weak nucleophiles, acid catalyst is used to protonate the carbonyl oxygen which makes the carbonyl C a better electrophile for nucleophilic attack by ROH. Proton transfer from the alcohol to the hydroxyl creates a better leaving group (HOH). Learn the mechanism since it is common to other reactions.


R C

O

OH

: :..H+ HSO4

-

R C

O

OH

:..

H+

R' OH....

R C

O

OH

: :..

H

+

'

.. .. ..R C

O

OH

: :..

H

R' O.. H+

..

HSO4-

R C

O

OH

H

: :..

H

R' O.. :

+

R C

O

OR'+

H2O+

H2 SO4

: :....

proton transfer

• The net effect of Fischer esterification is substitution of the –OH group of a carboxylic acid with the –OR group of a methyl or 1° alcohol.

21

CCH2C

O

OH

O

HOCH3CH2 OH2H+ CCH2C

O

OCH2CH3

O

CH3CH2O

I: propanedioic acid

c: malonic acid

I: diethyl propanedioate

c: diethyl malonate


• Draw and name the products of the following reactions.

C

O

OHNaOH1.

2.CH2Br

C

O

OCH2

cyclopentylmethyl benzoate

• Draw the reagents that will react to produce the following ester.

C

O

OCHHCH3C

H3C

CH3

CH3

C

O

OHHCH3C

H3C

HOCHCH3

CH3H+

+Why will an SN2 reaction of a carboxylate and an alkyl halide not work here?

I: isopropyl 2-methylpropanoate

c: isopropyl isobutyrateIsopropyl bromide is a 2° alkyl halide and would undergo an E2 rather than SN2 reaction.

• Draw the complete mechanism for Fischer esterification of benzoic acid with methanol.

22

Conversion of Carboxylic Acids to Amides

• Amides are difficult to prepare by direct reaction of carboxylic acids with amines (RNH2) because amines are bases that convert carboxylic acids to non electrophilic carboxylate anions and themselves are protonated to non nucleophilic amine cations, (RNH3

+)

• High temperatures are required to dehydrate these quaternary amine salts and form amides. This is a useful industrial method but poor laboratory method.

• In the lab amides are often prepared from acid chloride after converting the carboxylic acid to the acid chloride.

- H2OH3C C

O

O

: :.... :

_

NH3CH3

+CH3NH2

..+CH3 C

O

O

: :

H.... CH3 C

O

NH2

: :..

• Explain why methylamine is a Bronsted base.

• Explain why methylamine is a Lewis base.

• Explain why methylamine is not an Arrhenius base

Proton (H+) acceptor

Electron pair donor

Has no OH- group

23

Synthesis Problems Involving Carboxylic Acids

• Write equations showing how the following transformations can be carried out. Form a carboxylic acid at some point in each question.

?CC

O

Cl

O

ClCH3H3C

CC

O

OH

O

HOSOCl2

KMnO4

?Br C

O

O C

O

C

O

OH

-

H2OH2O

H+

CN

NaCN

C

O

OH

-

H2O

MgBr

2. H3O+

CO21.

Mgether

CC

O

OH

O

OH H+CH3OH2

KMnO4

?

CC

O

O CH3

O

O CH3

24

Chemistry of Acid Halides

• In the same way that acid chlorides are produced by reacting a carboxylic acid with thionyl chloride (SOCl2), acid bromides are produced by reacting a carboxylic acid with phosphorus tribromide (PBr3).

R C

O

OH

: :..

+ SOCl2 R C

O

Cl + SO2 + HCl

R C

O

OH

: :..

+ PBr3R C

O

Br + PBr2OH

(or PCl3)

Reactions of Acid Halides:Acid halides are among the most reactive of the carboxylic acid derivatives and are readily

converted to other compounds. Recall that acid chlorides add to aromatic rings via electrophilic aromatic substitution (EAS) reactions called Friedel-Crafts Acylation with the aid of Friedel-Crafts catalysts.

O

ClR C

AlCl3

R C O+

R C O+

+ AlCl4-

acyl cation = acylium ion......

..

.. ..

R C O+

+

O

RH

+Cl-

C C

O

R+ HCl

25

Chemistry of Acid Halides

• Draw a reaction showing how propylbenzene can be produced by a Friedel Crafts acylation reaction.

CH2CH2CH3CCH2CH3

O H2/Pt

I: 1-phenyl-1-propanone

c: ethyl phenyl ketone

Cl CCH2CH3

O

AlCl3

R C

O

ClR C

O

OH

: :..

H2O

R C

O

O

: :..

R..

ROH

R C

O

NH2

: :..

NH3

R C

O

H

: :

[H]

RCH2OH[H]

R':- MgBr+

R C

O

R'

R C

OH

R'

R'

R':- MgBr+

3º alcohol1º alcoholamide

ester

acid

ketone

aldehyde

• Most acid halide reactions occur by a nucleophilic acyl substitution mechanism. The halogen can be replaced by -OH to produce an acid, -OR to produce an ester, -NH2 to produce an amide. Hydride reduction produces a 1 alcohol, and Grignard reaction produces a 3 alcohol.

26

+R C

O

Cl

: :

R'3N..

R C

O

Cl

: :.. _

R'3N +

R C

O

NR'3

: :+

H2O....

R C

O

NR'3

: :.. _

+

O H

H..+

R C

O

O

: :H

H+

..R'3N

..

R C

O

OH

: :..

+R'3NH+ Cl-

R'3N..

+ HCl

N:

pyridine

R C

O

Cl

: :

H2O....

R C

O

O

Cl

:

H H..

:.. _

+

R C

O: :

OH

H

:+

_Cl-

Cl-

R C

O

OH

: :..

+ HCl

Hydrolysis: Conversion of Acid Halides into Acids

• Acid chlorides react via nucleophilic attack by H2O producing carboxylic acids and HCl.

• Tertiary amines, such as pyridine, are sometimes used to scavenge the HCl byproduct and drive the reaction forward. 3º amines will not compete with water as a nucleophile because their reaction with acid halide stops at the intermediate stage (there is no leaving group). Eventually, water will displace the amine from the tetrahedral intermediate, regenerating the 3º amine and forming the carboxylic acid.

• Draw the mechanism of the reaction of cyclopentanecarbonyl chloride with water.

27

Alcoholysis: Conversion of Acid Halides into Esters

• Acid chlorides react with alcohols producing esters and byproduct HCl by the same mechanism as hydrolysis above.

• Draw and name the products of the following reaction.

• Once again, 3º amines such as pyridine may be used to scavenge the HCl byproduct or for water insoluble acid halides, aqueous NaOH can be used to scavenge HCl since it will not enter the organic layer and attack the electrophile (thus it cannot compete with the alcohol as the nucleophile).

C

O

ClH3C HO CHCH3

CH3

+


c: acetyl chloride

C

O

H3C O CHCH3

CH3

+ HCl

I: isopropyl ethanoate

c: isopropyl acetate

NaOH, H2O

R C

O

Cl CH2Cl2 ROH

HCl

• Draw the mechanism of the reaction of benzoyl chloride and ethanol.

28

Practice on Synthesis of Esters

• Write equations showing all the ways that benzyl benzoate can be produced. Consider Fischer esterification, SN2 reaction of a carboxylate with an alkyl halide, and alcoholysis of an acid chloride.

C

O

O CH2C

O

OH+CH2OHH2SO4

C

O

O-Na+CH2Br +

C

O

Cl+CH2OH

N

• Answer the same question as above but for t-butyl butanoate

CH3CH2CH2C

O

O C

CH3

CH3

CH3

CH3CH2CH2C

O

Cl + C

CH3

CH3

CH3

HO

N This is the only method that will work.

• Explain why the other methods will fail.

29

C

O

N

H

CH3

CH3

+ Cl-

Aminolysis: Conversion of Acid Halides into Amides

• Acid chlorides react rapidly with ammonia or 1 or 2 but not 3 amines producing amides. Since HCl is formed during the reaction, 2 equivalents of the amine are used. 1 equivalent is used for formation of the amide and a second equivalent to react with the liberated HCl, forming an ammonium chloride salt. Alternately, the second equivalent of amine can be replaced by a 3º amine or an inexpensive base such as NaOH (provided it is not soluble in the organic layer). Using NaOH in an aminolysis reaction is referred to as the Schotten-Baumann reaction.

C

O

Cl + NH(CH3)2..

2

benzoyl chloride demethyl amine

::

NH(CH3)2..

C

O

NCH3

CH3+

+NH2(CH3)2 Cl-

I: N,N-dimethylbenzenecarboxamide

c: N,N-dimethylbenzamide

• Write equations showing how the following products can be made from an acid chloride.

N-methylacetamide

propanamide

CH3 C

O

NHCH3 CH3 C

O

Cl + NH2CH3

N

CH3CH2C

O

NH2

N

CH3CH2C

O

Cl + NH3

30

Reduction of Acid Chlorides to Alcohols with Hydride

• Acid chlorides are reduced by LiAlH4 to produce 1 alcohols. The alcohols can of course be produced by reduction of the carboxylic acid directly.

• The mechanism is typical nucleophilic acyl substitution in which a hydride (H: -) attacks the carbonyl C, yielding a tetrahedral intermediate, which expels Cl-. The result is substitution of -Cl by -H to yield an aldehyde, which is then immediately reduced by LiAlH4 in a second step to yield a 1 alcohol.

R C

O

Cl 1. LiAlH4R C

O

H

Cl-

LiAlH4 R C

O

H

H

-H3O

+R C

O

H

H

H+ H2O

2.

• Draw the reaction and name the product when 2,2-dimethylpropanoyl chloride is reduced with LiAlH4

C

O

ClC

CH3

CH3

H3C1. LiAlH4

2. H3O+

excess

CH2OHC

CH3

CH3

H3C

I: 2,2-dimethyl-1-propanolc: neopentyl alcohol

31

• The aldehyde cannot be isolated if LiAlH4 (and NaBH4) are used. Both are too strongly nucleophilic.

• However, the reaction will stop at the aldehyde if exactly 1 equivalent of a weaker hydride is used, i.e., diisobutylaluminum hydride (DIBAH) at a low temperature (-78°C).

• Under these conditions, even nitro groups are not reduced.

Reduction of Acid Chlorides to Aldehydes with Hydride

CH3CHCH2

CH3

CH2CHCH3

CH3

Al

H

CH3CHCH2

CH3

CH2CHCH3

CH3

Al+ H:

_

diisobutyl aluminum hydride (DIBAH)

+

DIBAH is weaker than LiAlH4. DIBAH is neutral; LiAlH4 is ionic.

DIBAH is similar to AlH3 but is hindered by its bulky isobutyl groups.

Only one mole of H:- is released per mole of DIBAH.

Al

H

H H

aluminumhydride

1. DIBAH

2. H3O+

1 equiv.78°C-

NO2 C

O

Cl NO2 C

O

H p-nitrobenzaldehyde

32

• Grignard reagents react with acid chlorides producing 3 alcohols in which 2 alkyl group substituents are the same. The mechanism is the same as with LiAlH4 reduction. The 1st equivalent of Grignard reagent adds to the acid chloride, loss of Cl- from the tetrahedral intermediate yields a ketone, and a 2nd equivalent of Grignard immediately adds to the ketone to produce an alcohol.

Reduction of Acid Chlorides to Alcohols with Grignards

C

O

Cl CH3 MgBr C

O

CH3 CH3 MgBrC

O

CH3

CH3

-H3O

+

C

O

CH3

CH3

H

• The ketone intermediate can’t be isolated with Grignard reaction but can be with Gilman reagent (diorganocopper), R2CuLi. Only 1 equivalent of Gilman is used at -78°C to prevent reaction with the ketone product. Recall the preparation of ketones (Ch. 19). This reagent does not react other carbonyl compounds (although it does replace halogens in alkyl halides near 0C)

CH C

O

ClH3C

H3C

Li(CH3)2Cu1.

2. H3O+

78°C-

1 equiv.

CH C

O

CH3

H3C

H3C

I: 3-methyl-2-butanonec: isopropyl methyl ketone

I: 2-phenyl-2-propanol

33

• Draw the reagents that can be used to prepare the following products from an acid chloride by reduction with hydrides, Grignards and Gilman reagent. Draw all possible combinations.

Practice Questions for Acid Chloride Reductions

CH3 C

OH1.

2. H3O+

MgBr

excess

C

O

ClH3C

C

O

CH2CH3 1 equiv.

1.

2. H3O+

78°C-C

O

Cl

CH3CH2 2CuLi 1 equiv.

1.

2. H3O+

78°C-CH3CH2 C

O

Cl

2CuLi

or

C

CH3

H3C

CH3

CH2

OH 1.

2. H3O+

excessLiAlH4

C

CH3

H3C

CH3

C

O

Cl

C

O

H 1 equiv.1.

2. H3O+

78°C-

DIBAH

C

O

Cl

I: 1,1-dicyclopentylethanol


I: 2,2-dimethyl-1-propanolI: 2,2-dimethylpropanoyl chloride

I: cyclohexanecarbaldehyde

I: cyclohexanecarbonyl chloride

c: ethyl phenyl ketone

I: 1-phenyl-1-propanone

34

+H C

O

O

: :.... :

_

Na+ C

O

Cl

::

CH3

sodium formate acetyl chloride

H C

O

O

: :.... C

O::

CH3

ether

SN2

25ºC

NaCl+

acetic formic anhydride

CH3 C

O

OH H O C

O

CH3+

CH3 C

O

O C

O

CH3- H2O

acetic acid

Preparation of Acid Anhydrides:Dehydration of carboxylic acids as previously discussed is difficult and therefore limited to

a few cases.

acetic anhydride

A more versatile method is by nucleophilic acyl substitution of an acid chloride with a carboxylate anion. Both symmetrical and unsymmetrical anhydrides can be prepared this way.

Preparations of Acid AnhydridesR RC

O

O C

O: : : :....

• Draw all sets of reactants that will produce the anhydride shown with an acid chloride.

CH3 C

O

O C

OCH3 C

O

O- Na+ C

O

Cl+

CH3 C

O

Cl + C

O

Na+ -O

35

The chemistry of acid anhydrides is similar to that of acid chlorides except that anhydrides react more slowly. Acid anhydrides react with HOH to form acids, with ROH to form esters, with amines to form amides, with LiAlH4 to form 1 alcohols and with Grignards to form 3 alcohols. Note that ½ of the anhydride is wasted so that acid chlorides are more often used to acylate compounds. Acetic anhydride is one exception in that it is a very common acetylating agent.

Reactions of Acid AnhydridesR RC

O

O C

O: : : :....

R C

O

OH

: :..

H2O

R C

O

O

: :..

R'..

R'OH

R C

O

NH2

: :..

NH3

R C

O

H

: :

[H]

RCH2OH[H]

R':- MgBr+

R C

O

R'

R C

OH

R'

R'

R':- MgBr+

3º alcohol1º alcoholamide

ester

acid

ketone

aldehyde

R C

O

O C

O

R

2

2

RC

O

HO

::..

+

acid

2

Write the mechanism for the following reactions and name all products:•aniline with acetic anhydride (2 moles aniline are needed or use 1 mole + aq. NaOH)•cyclopentanol with acetic formic anhydride (the formic carbonyl is more reactive).•methyl magnesium bromide with acetic propanoic anhydride (Grignards are not nucleophilic enough to react with carboxylate by products)•lithium aluminum hydride with acetic formic anhydride (LiAlH4 is so powerful a nucleophile that it will reduce even carboxylates).

36

• Show the product of methanol reacting with phthalic anhydride

Practice Questions for Acid Anhydrides

C

O

C

O

O CH3OH+

C

O

C

O

OCH3

OH

2-(methoxycarbonyl)benzoic acid

• Draw acetominophen; formed when p-hydroxyaniline reacts with acetic anhydride

NH2HO CH3 C

O

O C

O

CH3+ NHO C

O

CH3

H

CH3 C

O

OH+

N-(4-hydroxyphenyl)acetamide

37

• Esters are among the most widespread of all naturally occurring compounds. Most have pleasant odors and are responsible for the fragrance of fruits and flowers. Write chemical formulas for the following esters

Chemistry of Esters

Flavor NameStructure

pineapple methyl butanoate

bananas isopentyl acetate

apple isopentyl pentanoate

rum isobutyl propanoate

oil of wintergreenmethyl salicylate

[methyl 2-hydroxybenzoate)

nail polish remover ethyl acetate

new car smell(plasticizer for PVC)

dibutyl phthalate

CH3(CH2)3COCH2CH2CHCH3

CH3O

C

OH

O

O CH3

C

O

O (CH2)3CH3

O

O (CH2)3CH3C

CH3CH2CH2C

O

OCH3

CH2CH2C

O

OCH2CHCH3

CH3

CH3C

O

OCH2CH3

CH3C

O

OCH2CH2CHCH3

CH3

38

1. SN2 reaction of a carboxylate anion with a methyl or 1 alkyl halide

2. Fischer esterification of a carboxylic acid + alcohol + acid catalyst

3. Acid chlorides react with alcohols in basic media

Preparation of Esters

R C

O

O

: :.... :

_

Na+ + R' Br R C

O

O

: :..

R'..SN2

R C

O

O

: :..

R'..+ R' OHR C

O

OH

: :.. H+

R C

O

O

: :..

R'..+ R' OHR C

O

Cl+ HCl

39

• Esters react like acid halides and anhydrides but are less reactive toward nucleophiles because the carbonyl C is less electrophilic. Both acyclic esters and cyclic esters (lactones) react similarly. Esters are hydrolyzed by HOH to carboxylic acids, react with amines to amides, are reduced by hydrides to aldehydes, then to 1alcohols, and react with Grignards to 3 alcohols.

Reactions of Esters

R C

O

OH

: :..

H2O

R C

O

NH2

: :..

NH3

R C

O

H

: :

[H]

RCH2OH[H]

R':- MgBr+

R C

O

R'

R C

OH

R'

R'

R':- MgBr+

3º alcohol

1º alcohols

amide

acid

ketone

aldehyde

R C

O

O

: :..

R'..

+R'OH

H+

40

• Esters are hydrolyzed (broken down by water) to carboxylic acids or carboxylates by heating in acidic or basic media, respectively.

• Base-promoted ester hydrolysis is called saponification (Latin ‘soap-making’). Boiling animal fat (which contains ester groups) in an aqueous solution of a strong base (NaOH, KOH, etc.) makes soap. A soap is long hydrocarbon chain with an ionic end group.

Base Hydrolysis of Esters

• The mechanism of base hydrolysis is nucleophilic acyl substitution in which OH - adds to the ester carbonyl group producing a tetrahedral intermediate. The carbonyl group reforms as the alkoxide ion leaves, yielding a carboxylate.

CH3(CH2)10C

O

O CH3

K+OH-

CH3(CH2)10C

O

O H

OCH3-

-

c: potassium laurate

CH3(CH2)10C

O

O - K+ + CH3OH

liquid soap

• The leaving group, methoxide (OCH3-), like all alkoxides, is a strong base (pKb = -2).

It will deprotonate the carboxylic acid intermediate converting it to a carboxylate. The alkoxide, when neutralized, becomes an alcohol.

c: sodium laurate

I: sodium dodecanoateC

O

O- Na+

bar soap

41

C

O

OCH3

H

OHH

+

• Acidic hydrolysis of an ester yields a carboxylic acid (and an alcohol). The mechanism of acidic ester hydrolysis is the reverse of Fischer esterification. The ester is protonated by acid then attacked by the nucleophile HOH. Transfer of a proton and elimination of ROH yields the carboxylic acid. The reaction is not favorable. It requires at least 30 minutes of refluxing.

• Draw the complete mechanism of acid hydrolysis of methyl cyclopentanecarboxylate.

Acid Hydrolysis of Esters

C

O

OCH3

H+

H2O

H2O

C

O

OCH3

H

+

..

C

O

OCH3

H

OH

H+

..

C

O

OH + +H3O+ CH3OH

• Acid hydrolysis of an ester can be reversed by adding excess alcohol. The reverse reaction is called Fischer Esterification. Explain why base hydrolysis of an ester is not reversible.

42

+H+

R RC

O

O....

: :

ester alcohol

H O R'....

R R'C

O

O....

: :

ester

+

alcohol

H O R....

• Nucleophilic acyl substitution of an ester with an alcohol produces a different ester. The mechanism is the same as acid hydrolysis of esters except that that the nucleophile is an alcohol rather than water. A dry acid catalyst must be used, e.g., HCl(g) or H2SO4. If water is present, it will compete with the alcohol as the nucleophile producing some carboxylic acid in place of the ester product.

• The process is also called Ester Exchange or Transesterification

Alcoholysis of Esters

CH3CH2OH2H2SO4

C

O

OC

O

O

C

O

OCH2CH3C

O

CH3CH2O + OH2

dicyclobutyl terephthalate

dicyclobutyl 1,4-benzenedicarboxylate

diethyl terephthalate

diethyl 1,4-benzenedicarboxylate

cyclobutanol

43

CH C

O

OCH3

H3C

H3CNH3

• Amines can react with esters via nucleophilic acyl substitution yielding amides but the reaction is difficult, requiring a long reflux period. Aminolysis of acid chlorides is preferred.

• Draw the mechanism aminolysis of methyl isobutyroxide with ammonia.

Aminolysis of Esters

CH C

O

NH3C

H3CH

H

H- OCH3-

+

CH C

O

NH3C

H3CH

H+ CH3OH

c: -methylpropionamide

I: 2-methylpropanamide

• Write an equation showing how the following amide can be prepared from an ester.

C

O

N(CH3)2C

O

(CH3)2N C

O

OCH3C

O

CH3O + 2 NH(CH3)2

• Note that the amide intermediate must deprotonate to form a stable, neutral amide. Thus the amine must have at least one H. NH3, 1° and 2° amines will work but not 3°.

44

• Esters are easily reduced with LiAlH4 to yield 1 alcohols. The mechanism is similar to that of acid chloride reduction. A hydride ion first adds to the carbonyl carbon temporarily forming a tetrahedral alkoxide intermediate. Loss of the –OR group reforms the carbonyl creating an aldehyde and an OR - ion. Further addition of H: - to aldehyde gives the 1 alcohol. Draw the mechanism and show all products.

Hydride Reduction of Esters

• The hydride intermediate can be isolated if DIBAH is used as a reducing agent instead of LiAlH4. 1 equivalent of DIBAH is used at very low temp. (-78 C).

1.

2. H3O+

excessLiAlH4

O

O• Draw and name the products.

1 equiv.1.

2. H3O+

78°C-

DIBAHO

O

CH3

O

CH3

H

OH

CH2OHOH

c: -butyrolactone

I: 4-hydroxybutanoic acid lactonec: none

I: 1,4-butanediol

c: -valerolactone

I: 4-hydroxypentanoic acid lactoneI: 4-hydroxypentanal

c: -hydroxyvaleraldehyde

R C

O

O R'LiAlH4

R C

O

H- OR'-

LiAlH4R C

O

H

H

-

R C

O

H

H

H

+ R'OHH3O+

45

• Esters and lactones react with 2 equivalents of Grignard reagent to yield 3 alcohols in which the 2 substituents are identical. The reaction occurs by the usual nucleophilic substitution mechanism to give an intermediate ketone, which reacts further with the Grignard to yield a 3 alcohol.

Grignard Reduction of Esters

C

O

OCH3

MgBr

C

O

- OCH3-

MgBr

C

O-

C

O

H

+ CH3OH

H3O+

O -

O

CH3

CH3 MgBr

O -CH3

CH3O -

H3O+

OH

CH3

CH3OH

I: 4-hydroxybutanoic acid lactone

methyl benzoate benzophenone

triphenylmethanol

triphenylmethoxide

c: -butyrolactone

4-methyl-1,4-pentanediol

CH3 MgBr

O

O

46

• What ester and Grignards will combine to produce the following

Practice with Esters

C

CH3

CH3

OH +C O

OR

CH3MgBr

2. H3O+

21.

C CH3

OH

+ CH3MgBr

2. H3O+

21.CRO

O

CH3

2-phenyl-2-propanol

1,1-diphenylethanol

47

• Amides are usually prepared by reaction of an acid chloride with an amine. Ammonia, monosubstituted and disubstituted amines (but not trisubstituted amines) all react.

Chemistry of Amides

• Amides are much less reactive than acid chlorides, acid anhydrides, or esters. Amides undergo hydrolysis to yield a carboxylic acids plus an amine on heating in either aqueous acid or aqueous base.

• Basic hydrolysis occurs by nucleophilic addition of OH- to the amide carbonyl, followed by elimination of the amide ion, NH2‑,(a very reactive base – a difficult step requiring reflux)

R C

O

ClNH3

..

R C

O

NH2

1º amideNH2R

1º amine

R C

O

NHR

NHR22º amine

R C

O

NR2

3º amide2º amide

NR3

3º amineno reaction

C

O

NH2

aq Na+ OH-C O

O

H

- NH2-

C O

O-

Na+ + NH3

I: sodium cyclohexanecarboxamide

48

• Acidic hydrolysis occurs by nucleophilic addition of HOH to the protonated amide, followed by loss of a neutral amine (after a proton transfer to nitrogen).

Hydrolysis of Amides

C

O

NH

CH3 H3O+

C

O

NH

CH3H

+

H2O

C

O

NH

CH3H

OHH

+

C

O

NCH3

H

O

H

H

H

+ H2O

C

O

O H+NH2CH3

N-methylcyclohexanecarboxamideH3O

+

NH3CH3

+

NH

O

H3O+

H2O

NaOH O

OH

NH3

+

O

O-

NH2

cyclohexanecarboxylic acid

5-aminopentanoic acid lactam

-valerolactam

49

• Alcoholysis of amides occurs by the same acid catalyzed mechanism as acid hydrolysis except that the amido group of the amide is replaced with by an alcohol rather than water. Dry acid, e.g., HCl(g) or H2SO4 must be used otherwise water would compete with the alcohol as the nucleophile producing some carboxylic acid product in place of an ester.

• The reaction will require a long reflux period because amides are weak electrophiles and alcohols are weak nucleophiles.

Alcoholysis of Amides (to Esters)

C

O

N(CH3)2

CH3CHCH2CH3

OH

H2SO4

C

O

OCH3CHCH2CH3

+ NH2(CH3)2+

N,N-dimethylcyclopentanecarboxamide sec-butyl cyclopentanecarboxylate

•Write a mechanism for this reaction. Refer to acid hydrolysis mechanism if necessary.

50

C

O

N CH3

CH3H

AlH H

H

-

C

O

N CH3

CH3H

- AlH3

• Amides are reduced by LiAlH4. The product is an amine rather than an alcohol. The amide carbonyl group is converted to a methylene group (-C=O -CH2). This is unusual. It occurs only with amides and nitriles. Initial hydride attack on the amide carbonyl eliminates the oxygen. A second hydride ion is added to yield the amine. The reaction works with lactams as well as acyclic amides.

Hydride Reduction of Amides

C

O

N CH3

CH3

AlH3H-

Li+

C N CH3

CH3H

H

+AlH2O-

•Write equations showing how the above transformation can be carried out.

C

O

Cl?

C

O

NHCH3

C

O

Cl

N

NH2CH3

C

O

NHCH3

LiAlH41.

2. H3O+C NHCH3

H

H

N,N-dimethylcyclopentanecarboxamide

benzoyl chloride N-methylbenzamide

51

• Grignards deprotonate 1º and 2º amides and are not reactive enough to add to the imide ion product. N-H protons are acidic enough (pKa = 17) to be abstracted by Grignards.

Grignard Reduction of Amides

:CH3 +MgBr

R C

O

N

: :.. H

R

R C

O

N

: :..

R

.. _

imide anion2º amide:CH3

+MgBrCH4_

• Write equations showing how the following transformation can be carried out.

Br?

CH2 N(CH3)2

MgBr2. H3O

+

CO21.C

O

OH

2. H3O+LiAlH41.Mg, ether

C

O

N

H

CH2CH3?

CH2OH

C

O

OH

H2O

H+ or OH-

2. H3O+

LiAlH41.

C

O

ClSOCl2 C

O

N(CH3)2

NH(CH3)2

52

• The carbon atom in the nitrile group is electrophilic because it is bonded to an electronegative N atom and a bond in the nitrile is easily broken, i.e., as if it were providing a leaving group.

Chemistry of Nitriles R C N : +

Preparation of Nitrile:

1. Nitriles are easily prepared by SN2 reaction of cyanide ion (CN-) with methyl halides or a 1 alkyl halide. 2º alkyl halides also work but some E2 product also forms. 3º alkyl halides will result in mostly an alkene (E2) product instead of a nitrile. (pKb of CN - = 4.7)

CH3CH2 BrNa+ CN-

SN2CH3CH2 C N propanenitrilebromoethane

ethyl bromide

2. Another method of preparing nitriles is by dehydration of a 1 amide using any suitable dehydrating agent such as SOCl2, POCl3, P2O5, or acetic anhydride. Initially, SOCl2 reacts with the amide oxygen atom and elimination follows. This method is not limited by steric hindrance.

SOCl2 , benzene

80ºC

C

O

NH2 C N + +SO2 HCl2

SOCl2

R OH

PBr3 R Br

R Cl

Recall:OH

POCl3 in pyridine

H2O_

53

• Like carbonyl groups, the nitrile group is strongly polarized and the nitrile C is electrophilic. Nucleophiles thus attack yielding an sp2 hybridized imine anion.

Reactions of Nitriles R C N : +

Nu:-E+

sp2

sp3

: :

.. _

C

OC

O: :

Nu

products

R C N : +

sp Nu:-

E+

R C N

Nu

:.. _

sp2 imine anion

products

• Nitriles are hydrolyzed by HOH to amides and subsequently to carboxylic acids, reduced by hydrides to amines or aldehydes, and by Grignards to ketones.

R C

O

NH2

1º amide

R C N :

POCl3

H2O

H2O_

R CH2 NH2

[H][H]

R C

O

H

RMgXR C

O

R

1º aminealdehyde

ketone

54

R C N :H3O

+

R C N H+

R C N H..

+

O H

H

:..

H2O....

R C N H

O H

H..

+

R C H

R C N H

O

R C H

..

: :

H

..R C

O

NH2

: :..

hydroxyimine

amide

H3O+

R C

O

OH

: :..

acid

Hydrolysis of Nitriles into Carboxylic Acids R C N : +

• Nitriles are hydrolyzed in either acidic or basic aqueous solution to yield carboxylic acids plus ammonia or an amine.

• In acid media, protonation of N produces a cation that reacts with water to give an imidic acid (an enol of an amide). Keto-enol isomerization of the imidic acid gives an amide. The amide is then hydrolyzed to a carboxylic acid and ammonium ion. It is possible to stop the reaction at the amide stage by using only 1 mole of HOH per mole of nitrile. Excess HOH forces carboxylic acid formation.

R C NH2O

H+ or OH-

R C

O

O H + NH3

55

R C N :OH- ..

:_

O H

H

:..

R C N

OH

: :

R C N

OH

: :..

H R C

O

NH2

: :..

hydroxy imine amideOH-_

OH-

H2O

R C

O

O

: :.... :

_

Na++NH3

Hydrolysis of Nitriles into Carboxylate Salts R C N : +

• In basic media, hydrolysis of a nitrile to a carboxylic acid is driven to completion by the reaction of the carboxylic acid with base. The mechanism involves nucleophilic attack by hydroxide ion on the electrophilic C producing a hydroxy imine, which rapidly isomerizes to an amide. Further hydrolysis yields the carboxylate salt.

Br ?C

O

OH

• Show how the following transformation can be carried out without using a Grignard.

CN

H3O+

C

O

NH2

Na+CN-

SN2

H3O+

56

_

R C N :H:- ..

:R C N

H

H:-

2. H2O1.

..

_ 2..:R C N

H

H

..R C NH2

H

H

+ OH-2dianion

H2O

Reduction of Nitriles R C N : +

Alcoholysis of Nitriles doesn’t work. Alcohols are weak nucleophiles and nitriles are weak electrophilesAminolysis of Nitriles doesn’t work. Amines are weak nucleophiles and nitriles are weak electrophiles. Reduction with Hydrides:Reduction of nitriles with 2 equivalents of LiAlH4 gives 1 amines. LiAlH4 is a very good nucleophile and can break 2 bonds forming a dianion.

• If less powerful DIBAH is used, only 1 equivalent of hydride can add. Subsequent addition of HOH yields the aldehyde.

C

CH3

N

o-methylbenzonitrile

LiAlH41.

2. H2O

1.

2.

DIBAH , toluene , -78ºC

H2O or H3O+

1 equiv.

CH3

CH2NH2

CH3

C

O

H

2-methylbenzaldehyde

57

_

R C N :1. ..

:R:- MgBr+

imine anion

+NH3

2. H3O+

R C N

R

O H

H

:..

R C N

R

H..

R C N

R

H..

O H

H..+

R C N

R

H..

OH

: : H

R C

O

R

: :

..

-

Reduction of Nitriles with Grignards R C N : +

• Grignards add to nitriles giving intermediate imine anions which when hydrolyzed yield ketones. The mechanism is similar to hydride reduction except that the attacking nucleophile is a carbanion (R-). Grignards are not as strongly nucleophilic as LiAlH4 and so can only add once – a dianion is not formed.

C NCH3CH3 MgBr1.

2. H3O+

benzonitrile

H3O+

NH4+

C

O

CH2CH3 1-phenyl-1-propanone

ethyl phenyl ketone

58

Multistep Synthesis Problems• Write equations to show how the following transformations can be carried out.

Br?

CH2NH2

Na+ CN- CN

2. H3O+1. LiAlH4

Br?

C

O

HCN

from above2. H3O+

1 equiv.1.

78°C-

DIBAH

Br?

C

O CN

from above2. H3O+

1. MgBrether

Br?

CHO

H2O

H+

C

O

OH

SOCl2

C

O

Cl

2. H3O+

1. MgBrether

2 equiv.

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