Transcript Slide 1

Derivatives of
Carboxylic Acid
:O:
R
: O:
R
C
..
Cl
.. :
carboxylate
acid chloride
: O:
C
R
:O:
R
C
..
O
..
C
.. _
O
..:
R
..
OH
..
N:
C
nitrile
: O:
C
acid anhydride
R
:O:
R
C
: O:
..
O
..
ester
R
R
C
amide
..
H
N
H
1
Nomenclature of Acid Halides
• IUPAC: alkanoic acid  alkanoyl halide
• Common: alkanic acid  alkanyl halide
NH2
O
NO2
CH2 CH2 C Cl
Cl C
CH2
4
C Cl
C
O
I: 4-nitropentanoyl chloride
c: g-nitrovaleryl chloride
c: b-aminopropionyl chloride
O
R
..
Cl
.. :
H3C CH CH2 CH2 C Cl
I: 3-aminopropanoyl chloride
O
: O:
I: hexanedioyl chloride
c: adipoyl chloride
Rings: (IUPAC only): ringcarbonyl halide
C Cl
O
I: 3-cylcopentenecarbonyl chloride
O
C Br
I: benzenecarbonyl bromide
c: benzoyl bromide
2
:O:
Nomenclature of Acid Anhydrides
•
R
C
..
O
..
: O:
R
C
Acid anhydrides are prepared by dehydrating carboxylic acids
CH3

O
O
C OH
+
O
H O C CH3
CH3
- H2O
acetic acid
O
O
C O C
HO C
I: benzenecarboxylic anhydride
c: benzoic andhydride
C O C CH3
acetic anhydride
ethanoic anhydride
ethanoic acid
O
O

O
C OH
I: butanedioic acid
c: succinic acid
O C
O
C O
- H2O
I: butanedioic anhydride
c: succinic anhydride
Some unsymmetrical anhydrides
I: cis-butenedioic anhydride
c: maleic anhydride
O C
O
C O
O
O
O
C O C H
I: benzoic methanoic anhydride
c: benzoic formic anhydride
CH3
C
O
O
C
H
I: ethanoic methanoic anhydride
c: acetic formic anhydride
3
Nomenclature of Esters
•
: O:
Esters occur when carboxylic
acids react with alcohols
I: alkanoate
c: alkanate
O
alkyl
CH3 O C CH3
I: methyl ethanoate
c: methyl acetate
O
C O CH(CH3)2
R
C
..
OH
..
+
H
carboxylic
acid
..
O
..
R
alcohol
-
:O:
H2O
H+
R
C
..
O
..
R
ester
O
O
C O C(CH3)3
O C H
I: phenyl methanoate
c: phenyl formate
I: t-butyl benzenecarboxylate
c: t-butyl benzoate
O
O C CH(CH3)2
I: isobutyl cyclobutanecarboxylate
c: none
I: cyclobutyl 2-methylpropanoate
c: cyclobutyl a-methylpropionate
O O
CH3 O C C O CH3
I: dimethyl ethanedioate
c: dimethyl oxalate
4
Nomenclature of Cyclic Esters, “Lactones”
O
C
OH
H
CH2
H+
CH2
O
C
C
CH2
O
CH2
O
- H2O
I: 4-hydroxybutanoic acid
c: g-hydroxybutyric acid
CH2
O
CH2
O
Cyclic esters, “lactones”, form
when an open chain hydroxyacid
reacts intramolecularly. 5 to 7membered rings are most stable.
I: 4-hydroxybutanoic acid lactone
c: g-butyrolactone
 ‘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.
O
I: 5-hydroxypentanoic acid lactone
O
O c: d-valerolactone
I: 4-hydroxypentanoic acid lactone
O c: g-valerolactone
CH3
I: 3-hydroxypentanoic acid lactone
c: b-valerolactone
O
O
O
H3C
I: 6-hydroxy-3-methylhexanoic acid lactone
c: b-methyl-e-caprolactone
O
5
Nomenclature of Amides
1° amide
O
R C N
O
O
3° amide
R
R
C
N
N,N-disubstituted amide
R
H
2° amide
R C N
R
N-substituted amide
H
H
 1° amides: ‘alkanoic acid’ + amide  ‘’alkanamide’
 a ring is named ‘ringcarboxamide’
O
CH3CH2CH2 C N
Cl
H
C N
H
I: butanamide
c: butyramide
O
O
NO2
H
C N
H
I: 3-chlorocyclopentanecarboxamide
c: none
H
H
I: p-nitrobenzenecarboxamide
c: p-nitrobenzamide
 2° and 3° amides are N-substituted amides
I: N-phenylethanamide
c: N-phenylacetamide
c: acetanilide
I: N,2-dimethylpropanamide
c: N,a-dimethylpropionamide
O
CH3 O
CH3CH
C N
CH3
H
C N
CH3
CH2CH3
I: N-ethyl-N-methylcyclobutanecarboxamide
c: none
O
H3C C N
H
6
Nomenclature of Cyclic Amides, “Lactams”
O
O
C
OH
H
CH2
N
CH2
H
CH2

- H2O
I: 4-aminobutanoic acid
c: g-aminobutyric acid
O
C
CH2
CH2
NH
NH
CH2
Cyclic amides, “lactams”, form
when an open chain aminoacid
reacts intramolecularly. 5 to 7membered rings are most stable.
I: 4-aminobutanoic acid lactam
c: g-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.
O
NH
CH3
O
I: 3-amino-2-bromopropanoic acid lactam
c: a-bromo-b-propionolactam
I: 5-aminohexanoic acid lactam
c: d-caprolactam
Br
NH
O
H3C
NH
I: 4-amino-3-methylbutanoic acid lactam
c: b-methyl-g-butyrolactam
7
Nomenclature of Nitriles
O
R C N
•
•
•
H
H
POCl3
R C N
- H2O
Nitriles are produced when 1° amides are
dehydrated with reagents like POCl3
IUPAC: alkane + nitrile  ‘alkanenitrile’
IUPAC rings: ‘ringcarbonitrile’
Common: alkanic acid + ‘onitrile’  ‘alkanonitrile’
N C CH2CH2CH2
CH3 C N
I: 4-iodobutanenitrile
c: g-iodobutyronitrile
I: ethanenitrile
c: acetonitrile
CH3O
CN
I
HS
CN
I: p-thiobenzenecarbonitrile
c: p-mercaptobenzonitrile
COOH
CN
I: 3-methoxycyclohexanecarbonitrile
c: none
I: 2-cyanocyclopentanecarboxylic acid
c: none
8
Nomenclature Practice Exercise
O
CN
I: cyclobutanecarbonitrile
c: none
O
O
O
O
CH3 C O CH2Br
CH3 C O- Na+
I: bromomethyl ethanoate
c: bromomethyl acetate
I: sodium ethanoate
c: sodium acetate
O
O
I: 3-bromo-N-methylpentanamide
c: b-bromo-N-methylvaleramide
CH3 C CH2 C Cl
I: pentanedioic anhydride
c: glutaric anhydride
NHCH3
I: 3-oxobutanoyl chloride
c: b-oxobutyryl chloride
O
Br
O
O
O
NH
Cl
I: 2-ethyl-5-hydroxypentanoic acid lactone
c: a-ethyl-d-valerolactone
I: 6-amino-6-chlorohexanoic acid lactam
c: e-chloro-e-caprolactam
9
Relative Reactivity of Carbonyl Carbons
 Nucleophiles (electron donors), like OH-, bond with the sp2 hybridized carbonyl carbon.
 The order of reactivity is shown.
:O:
R
C
..
O
..
acid chloride
: O:
C
R
C
: O:
R
C
..
OH
..
C N:
C
most
reactive
: O:
R
R
..
Cl
.. :
acid anhydride
C
H
ketone
ester
:O:
R
carboxylic acid
amide
R
R
aldehyde
: O:
R
: O:
: O:
R
C
:O:
nitrile
carboxylate
C
R
C
..
O
..
..
R
H
N
H
.. _
O
..:
least
reactive
10
Nucleophilic Addition to Aldehydes and Ketones
•
•
•
•
Recall that electron donors (Nu: -’s) add to the electrophilic carbonyl C in aldehydes
and ketones. The C=O p 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.
-
O
O
R C H
R C H
CH3 MgBr
tetrahedral alkoxide with
sp3 carbon.
CH3
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’.
O
-
R C H
CH3
O H
H
+
H
O+H
R C H
+
H
O H
CH3
11
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 p bond breaks and the pair of electrons are stabilized on the
electronegative O atom. A tetrahedral alkoxide is temporarily formed.
O
R C Cl
R C Cl
CH3 MgBr
sp2 carbonyl C
•
•
-
O
O
R C CH3 +
CH3
Cl
Chlorine is a fair
leaving group.
sp2 carbonyl reforms
alkoxide C js sp3
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.
O
O
-
R C CH3
R C CH3
CH3 MgBr
CH3
O H
H
H
O+H
R C CH3 + H O H
CH3
Note that because the C=O group reforms, the nucleophile can react a second time.
12
Nucleophilic Acyl Substitution in Acid Derivatives
•
In carboxylic acid derivatives, the acyl group (RCO) is bonded to a leaving group (-Y).
O
-
O
R C
R C Y
Nu:-
acyl group
•
•
O
O
R C Y
R C Nu +
Y:-
Draw the
mechanism.
Nu
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
Cl
+21
non basic
+9
weak base
O R
-2
strong base
NH2
-21
v. strong base
O
R C Cl
O
O
R C O C R
O
R C O R
O
R C NH2
-
O
O C R
-
-
13
Nucleophilic Acyl Substitution in Acid Derivatives
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)
O
R C Y +
O
H OH
H OR
alcoholysis
R C Y +
O
R C Y +
O
R C Y +
H NH2
R MgX
LiAlH3 H
R C O H
O
R C O R
R C Y +
O
O
hydrolysis
O
aminolysis
R C NH2
Grignard
reduction
hydride
reduction
O H
R C R
R
O H
R C H
H
14
Nucleophilic Acyl Substitution of Carboxylic Acids
O
O
R C O H
•
R C Y
Nucleophilic acyl substitution converts carboxylic acids into carboxylic acid
derivatives, i.e., acid chlorides, anhydrides, esters and amides.
: O:
R
C
..
Cl
.. :
SOCl2
R
acid chloride
:O:
R
C
..
O
..
: O:

-H2O
: O:
C
acid anhydride
R
C
..
OH
..
: O:
NH3, , -H2O
R
C
amide
ROH
H+
:O:
R
C
..
O
..
..
H
N
H
R
ester
15
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.
1
:O :
R
C
..
OH
..
3
•
•
+
:O :
S
Cl
Cl
3
2
:O :
1
2
thionyl chloride
R
C
- HCl
..
O
..
:O :
S
:O:
R
Cl
C
Cl
+
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.
SOCl2
O
H3C
C OH
O
H3C C O H
O
HCl + SO2 + H3C
SOCl2
C Cl
O
HCl
+ SO2 + H3C C Cl
I: p-methylbenzenecarbonyl chloride
c: p-methylbenzoyl chloride
I: ethanoyl chloride
c: acetyl chloride
16
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

O
O
C OH
H O C CH3
+
O
CH3
- H2O
acetic acid
C O C CH3
acetic anhydride
ethanoic anhydride
ethanoic acid
•
O
Cyclic anhydrides with 5 or 6-membered rings are prepared by dehydration of diacids.
O
O
C
2HC
OH

2HC
2HC
OH
- H2O
2HC
C
O
C
O
O
•
C
I: butanedioic acid
I: butanedioic anhydride
c: succinic acid
c: succinic anhydride
Draw a reaction showing the preparation of cyclohexanecarboxylic anhydride.
O
O
O

2
C O C
- H2O
C O H
17
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.
: O:
CH3CH2
C
+
: O:
-
Na OH
..
O
..
H
_
propanoic acid
C
CH3CH2
_
..
:
O
..
Na
: O:
I
CH3
CH3CH2
+
C
..
O
..
CH3
+ NaI
SN2
H2O
acid/base
: O:
CH3CH2 C
CH3
_
.. :
O
..
Na+
+
H3C
O H
CH2Br
NaOH
- H2O
I: 5-bromopentanoic acid
c: d-bromovaleric acid
E2
CH3CH2C OH
I
O Na+
O CH2 Br
O
CH2 C
CH3
isobutylene
O
CH2
I: sodium 5-bromopentanoate
c: sodium d-bromovalerate
+
O
O
- NaBr
NaI
CH3
CH3
sodium propionoate
O
C
+
O
I: 5-hydroxypentanoic acid lactone
c: d-valerolactone
18
Conversion of Carboxylic Acids to Esters
-
-
-
alkyl
halide
(substrate)
good Nu
strong base
e.g., bromide
e.g., ethoxide
strong bulky base
e.g., t-butoxide
Br
C2H5O
(CH3)3CO
Me
SN2
SN2
SN2
no reaction
1°
SN2
SN2
E2 (SN2)
no reaction
2
SN2
E2
3
SN1
E2
alkyl
halide
(substrate)
-
-
good Nu
-
good Nu
nonbasic
-
-
e.g., cyanide
e.g., alkyl sulfide
e.g., carboxylate
CN
RS , also HS
RCOO
9
pkb = ………
4.7
pkb = ………
-
-
6.0 / 7.0
pkb = ………
CH3COOH
SN1, E1
fair Nu
………..
weak base
………..
-
e.g., acetic acid
SN1, E1
E2
E2
v. gd. Nu………..
moderate
……….. base
v. gd. Nu………..
moderate
………. .base
very poor Nu
nonbasic
-
alcohol
(substrate)
HI
HBr
HCl
Me
SN2
SN2
SN2
Me
SN2
1
SN2
SN2
SN2
1
SN2
2
SN2
SN2
E2
2
SN1
3
E2
E2
E2
3
SN1
19
Conversion of Carboxylic Acids to Esters
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.
+
: O:
R
..
C
H
: O+
-
H HSO4
..
OH
R
C
: O:
..
OH
..
R
R'
'
•
..
OH
..
C
+
H
H
H
..
OH
..
: O:
R
C
+
R' ..
O
..
OH
..
H
proton transfer
: O:
R
R'
C
O:
..
: O:
-
HSO4
..
+OH
H
R
..
C OR'
..
+
H2O
+
H2 SO4
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.
20
Conversion of Carboxylic Acids to Esters
•
Draw and name the products of the following reactions.
O
O
HO CCH2C OH
2 CH3CH2 OH
H+
I: propanedioic acid
c: malonic acid
O
C OH
•
CH2Br
CH3CH2O CCH2C OCH2CH3
I: diethyl propanedioate
c: diethyl malonate
C OCH2
cyclopentylmethyl benzoate
Draw the reagents that will react to produce the following ester.
O
H3C
CH3
HC C OCH
H3C
CH3
I: isopropyl 2-methylpropanoate
c: isopropyl isobutyrate
•
O
O
1. NaOH
2.
O
H+
CH3
HOCH
CH3
+
O
H3C
HC C OH
H3C
Why will an SN2 reaction of a
carboxylate and an alkyl
halide not work here?
Isopropyl 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.
21
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+)
: O:
CH3
C
..
O
..
: O:
..
H
+
CH3NH2
H3C C
..
:
O
..
_
+
NH3CH3
: O:

-
H2O
CH3
C
..
NH2
• 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.
• Explain why methylamine is a Bronsted base.
Proton (H+) acceptor
• Explain why methylamine is a Lewis base.
Electron pair donor
• Explain why methylamine is not an Arrhenius base
Has no OH- group
22
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.
?
H3C
CH3

KMnO4
Cl
O
O
O
C
C Cl
O
HO C
SOCl2
C OH
MgBr
O
1. CO2
2.
C OH

- H2O
H3O+
Mg
ether
Br
?
KMnO4

O
C O CH3
C O CH3
O
O
?
O
C O C
NaCN
CN
H2O 
O

- H2O
C OH
H+
O
C OH
C OH
2 CH3OH
 H+
O
23
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).
: O:
R
C
: O:
R
C
..
OH
O
(or PCl3)
+
SOCl2
R
C
+
Cl
SO2
+
HCl
O
..
OH
+
R
PBr3
C
PBr2OH
+
Br
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
R
C
R
Cl ...... AlCl3
+ ..
O
C ..
R
R
+
C O
+ AlCl4-
acyl cation = acylium ion
O
+
+
O
C ..
C
H
+
O
C
R
-
Cl
R
+
HCl
24
Chemistry of Acid Halides
• Draw a reaction showing how propylbenzene can be produced by a Friedel Crafts
acylation reaction. O
O
Cl
CCH2CH3
AlCl3
H2/Pt
CCH2CH3
CH2CH2CH3
I: 1-phenyl-1-propanone
c: ethyl phenyl ketone
• 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.
: O:
R
C
..
H2O
OH
O
R
C
acid
: O:
R
C
ester
..
O
.. R
R
Cl
R
..
C NH2
amide
C
R' ketone
R':- MgBr+
: O:
R
: O:
C
[H]
NH3
ROH
O
R':- MgBr+
H aldehyde
OH
R C
[H]
RCH2OH
1º alcohol
R'
R'
3º alcohol
25
Hydrolysis: Conversion of Acid Halides into Acids
• Acid chlorides react via nucleophilic attack by H2O producing carboxylic acids and HCl.
_
..
: O:
: O:
R
C
Cl
R
..
..
H2O
C
O
H
_
..
-
Cl
: O:
-
Cl
Cl
R
: O:
+ H
O:
C
+
R
+
H
H
C
..
OH
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.
.. _
.. _
: O:
: O:
R
C
Cl
+
: O:
..
R
R'3N
C
: O:
Cl
R
C
+
NR'3
R'3N +
..
R'3N
pyridine
..
H2O
..
+
R'3NH Cl +
R
C
+
NR'3
C
+O
H
..
H
+ HCl
: O:
N:
R
..
: O:
..
OH
R
C
..
O
+
H
R'3N
H
• Draw the mechanism of the reaction of cyclopentanecarbonyl chloride with water.
26
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.
O
H3C C Cl
CH3
+
HO CH
CH3
I: ethanoyl chloride
c: acetyl chloride
O
CH3
H3C C O CH
+ HCl
CH3
I: isopropyl ethanoate
c: isopropyl acetate
• Draw the mechanism of the reaction of benzoyl chloride and ethanol.
• 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).
NaOH, H2O
O
HCl
R C Cl
CH2Cl2 ROH
27
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.
O
H2SO4
C O CH2
O
CH2OH
C OH
+
O
C O-Na+
CH2Br +
N
O
CH2OH
C Cl
+
• Answer the same question as above but for t-butyl butanoate
O
CH3
CH3CH2CH2C O C CH3
N
O
CH3
CH3CH2CH2C Cl + HO C CH3
CH3
CH3
This is the only
method that will
work.
• Explain why the other methods will fail.
28
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.
: O:
C Cl
O
..
+
2 NH(CH3)2
C
H
+
O
..
NH(CH3)2
Cl-
CH3
C
N CH3
N
CH3
+
+
NH2(CH3)2 Cl-
CH3
benzoyl chloride
demethyl amine
I: N,N-dimethylbenzenecarboxamide
c: N,N-dimethylbenzamide
• Write equations showing how the following products can be made from an acid chloride.
N-methylacetamide
O
N
CH3 C NHCH3
propanamide
O
CH3CH2C NH2
O
CH3 C Cl
N
+
NH2CH3
O
CH3CH2C Cl
+
NH3
29
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.
Cl
O
R C Cl
1. LiAlH4
-
O
O
R C H
R C H
LiAlH4
H
2. H3O+
O H
R C H + H2O
H
• Draw the reaction and name the product when 2,2-dimethylpropanoyl chloride is reduced
with LiAlH4
CH3 O
H3C C
CH3
excess
1. LiAlH4
CH3
H3C C
C Cl
2.
H3O+
CH2OH
I: 2,2-dimethyl-1-propanol
c: neopentyl alcohol
CH3
30
Reduction of Acid Chlorides to Aldehydes with Hydride
• 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.
CH3
H
CH3CHCH2
Al
CH3
CH3
CH2CHCH3
CH3CHCH2
+
Al
CH3
CH2CHCH3
+
H:
_
Al
diisobutyl aluminum hydride (DIBAH)
H


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.
O
NO2
- 78°C
1 equiv.
1. DIBAH
C Cl
+
2. H3O
H
H
aluminum
hydride
O
NO2
C H
p-nitrobenzaldehyde
31
Reduction of Acid Chlorides to Alcohols with Grignards
• 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.
O
O
C Cl
O
CH3 MgBr
C CH3
CH3 MgBr
H3O+
O H
C CH3
C CH3
CH3
CH3
I: 2-phenyl-2-propanol
• 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 0C)
1 equiv.
O
1. Li(CH3)2Cu
H3C
- 78°C
CH C Cl
H3C
2. H3O+
H3C
O
CH C CH3
I: 3-methyl-2-butanone
c: isopropyl methyl ketone
H3C
32
Practice Questions for Acid Chloride Reductions
• 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.
I: ethanoyl chloride
excess
OH
1.
O
I: 1,1-dicyclopentylethanol
MgBr
CH C
3
H3C C Cl
2.
O
C CH2CH3
H3C C
O
1.
C H
2.
C Cl
O
1 equiv. - 78°C
CH3CH2 C Cl
+
H3O
I: 1-phenyl-1-propanone
CH3 O
H3C C
H3O+
DIBAH
1 equiv. - 78°C
H3O+
or
2.
excess
1. LiAlH4
2.
1.
O
H3O+
CH2
CH3
2CuLi
CH3CH2 2CuLi
1.
1 equiv. - 78°C
2.
CH3 OH
H3O+
c: ethyl phenyl ketone
C Cl
I: 2,2-dimethylpropanoyl chloride
I: 2,2-dimethyl-1-propanol
CH3
O
C Cl
I: cyclohexanecarbonyl chloride
I: cyclohexanecarbaldehyde
33
:O:
Preparations of Acid Anhydrides
R
C
..
O
..
: O:
C
R
Preparation of Acid Anhydrides:
Dehydration of carboxylic acids as previously discussed is difficult and therefore limited to
a few cases.
O
O
O
O

CH3 C OH + H O C CH3
CH3 C O C CH3
- H2O
acetic anhydride
acetic acid
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.
: O:
: O:
..
_
+
:
O
.. Na
H C
sodium formate
+
Cl
C
: O:
ether 25ºC
CH3
SN2
H C
..
O
..
: O:
C
CH3
+
NaCl
acetic formic anhydride
acetyl chloride
• Draw all sets of reactants that will produce the anhydride shown with an acid chloride.
O
O
O
CH3 C O C
O
CH3 C O- Na+ +
O
O
CH3 C Cl
Cl C
+
Na+ -O C
34
:O:
Reactions of Acid Anhydrides
R
..
O
..
C
: O:
C
R
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.
O
O
: O:
2 R
C
acid
H2O
R
..
C
O
C
R
R
OH
R
C
ester
acid
C
[H]
: O:
+
O
R':- MgBr+
..
HO
..
O
.. R'
: O:
R
:O:
C
2 R
R
..
C
NH2
amide
C
ketone
R':- MgBr+
: O:
NH3
R'OH
R'
H aldehyde
OH
R C
[H]
2 RCH2OH
1º alcohol
R'
R'
3º alcohol
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).
35
Practice Questions for Acid Anhydrides
• Show the product of methanol reacting with phthalic anhydride
O
O
C
C OCH3
2-(methoxycarbonyl)benzoic acid
O + CH3OH
C
C OH
O
O
• Draw acetominophen; formed when p-hydroxyaniline reacts with acetic anhydride
O
HO
O
O
NH2 + CH3 C O C CH3
HO
O
N C CH3 + CH3 C OH
H
N-(4-hydroxyphenyl)acetamide
36
Chemistry of Esters
• 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
Flavor
Name
pineapple
methyl butanoate
Structure
O
CH3CH2CH2C OCH3
CH3
O
bananas
isopentyl acetate
CH3C
OCH2CH2CHCH3
O
apple
isopentyl pentanoate
rum
isobutyl propanoate
oil of wintergreen
methyl salicylate
[methyl 2-hydroxybenzoate)
nail polish remover
ethyl acetate
new car smell
(plasticizer for PVC)
dibutyl phthalate
CH3
CH3(CH2)3COCH2CH2CHCH3
O
CH3
CH2CH2C OCH2CHCH3
O
C O CH3
OH
O
O
CH3C
OCH2CH3
C O (CH2)3CH3
C O (CH2)3CH3
O
37
Preparation of Esters
1. SN2 reaction of a carboxylate anion with a methyl or 1 alkyl halide
: O:
R
C
.. _
:
O
..
: O:
Na+
R'
+
R
Br
SN2
..
O
.. R'
C
2. Fischer esterification of a carboxylic acid + alcohol + acid catalyst
: O:
R
C
: O:
+
..
H
OH
+
R'
R
OH
C
..
O
.. R'
3. Acid chlorides react with alcohols in basic media
: O:
O
R
C
Cl
+
R'
OH
R
C
..
O
.. R'
+
HCl
38
Reactions of Esters
• 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 1alcohols, and react
with Grignards to 3 alcohols.
: O:
R
C
H2O
: O:
R
C
acid
NH3
+
..
..
H
OH
: O:
R
R
C
NH2
amide
C
[H]
C
R'
ketone
R':- MgBr+
: O:
R
..
O
R':- MgBr+
O
.. R'
H aldehyde
OH
R C
[H]
RCH2OH
+
R'OH
R'
R'
3º alcohol
1º alcohols
39
Base Hydrolysis of Esters
• 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.
O
bar soap
I: sodium dodecanoate
C
c: sodium laurate
O- Na+
• 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.
O
-
-
c: potassium laurate
OCH3
CH3(CH2)10C O CH3
O
CH3(CH2)10C O H
O
CH3(CH2)10C O
liquid soap
-
K+
+
CH3OH
K+OH-
• 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.
40
Acid Hydrolysis of Esters
• 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.
..
O
H+
O H
O
H
H2O
C OCH3
+
H
O
H
+
+
H2O
O H
C OCH3
C OCH3
C OCH3
..
O H
H
O
C 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.
41
Alcoholysis of Esters
:O:
R
C
..
O
..
R
+
H
ester
..
O
..
:O:
R'
H+
alcohol
R
C
..
O
..
R'
..
O
..
H
+
ester
R
alcohol
• 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
O
O C
H2SO4
O
dicyclobutyl 1,4-benzenedicarboxylate
C O
dicyclobutyl terephthalate
2 CH3CH2OH
O
diethyl 1,4-benzenedicarboxylate
CH3CH2O C
O
C OCH2CH3 + 2
OH
diethyl terephthalate
cyclobutanol
42
Aminolysis of Esters
• 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.
H3C
-
O
CH C OCH3
H3C
-
OCH3
NH3

H3C
O
H
CH C N H
+
H3C
H
H3C
O
CH C N H
H3C
H
+
CH3OH
I: 2-methylpropanamide
c: a-methylpropionamide
• Write an equation showing how the following amide can be prepared from an ester.
O
(CH3)2N C
O
C N(CH3)2
O
CH3O C
O
C OCH3 +
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°.
43
Hydride Reduction of Esters
• 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.
O
LiAlH4
R C O R'
O
R C H
-
-
O
LiAlH4
I: 4-hydroxybutanoic acid lactone
R C H
H
H
excess
1. LiAlH4
O
c: g-butyrolactone
2.
O H
H3O+
R C H
OR'
• Draw and name the products. O
-
H3O
CH2OH
OH
+ R'OH
I: 1,4-butanediol
c: none
+
• 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).
O
O
I: 4-hydroxypentanoic acid lactone
c: g-valerolactone
O
1.
CH3 2.
DIBAH
1 equiv. - 78°C
H3O+
I: 4-hydroxypentanal
H c: g-hydroxyvaleraldehyde
OH
CH3
44
Grignard Reduction of Esters
• 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.
MgBr
O
C OCH3
methyl benzoate
MgBr
O
triphenylmethoxide
O
C
-
-
C
H
OCH3
H3O+
O
benzophenone
C
+ CH3OH
triphenylmethanol
O
O
O
CH3 MgBr
O
I: 4-hydroxybutanoic acid lactone
CH3 MgBr
CH3
O
-
-
CH3
CH3
O
-
OH
H3O+
CH3
CH3
OH
4-methyl-1,4-pentanediol
c: g-butyrolactone
45
Practice with Esters
• What ester and Grignards will combine to produce the following
2-phenyl-2-propanol
CH3
C O +
OR
C OH
CH3
1. 2 CH3MgBr
2. H3O+
1,1-diphenylethanol
OH
C CH3
O
RO C CH3 +
1. 2 CH3MgBr
2. H3O+
46
Chemistry of Amides
• Amides are usually prepared by reaction of an acid chloride with an amine. Ammonia,
monosubstituted and disubstituted amines (but not trisubstituted amines) all react.
O
..
NH3
O
R
C
R
NH2
C
NR3
Cl
NH2R
1º amine
1º amide
3º amine
NHR2
2º amine
O
O
R
C
NHR
2º amide
no reaction
R
C
NR2
3º amide
• 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)
O
- - NH2
C NH2
aq Na+ OH-
O
O
C O H
C O
-
Na+
+ NH3
I: sodium cyclohexanecarboxamide
47
Hydrolysis of Amides
• 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).
O
H3O+
CH3
C N
H
H2O
O H CH
3
C N
+
H
O H CH
3
+
C N H
O H CH
3
C N
+
H
O
H
O H
H
N-methylcyclohexanecarboxamide
H
H3O+
+
H2O
NH3CH3
O
NH2CH3
+
C O H
cyclohexanecarboxylic acid

NaOH
O
O
-
NH2
O
NH
H2O
H3O+

O
OH
+
NH3
5-aminopentanoic acid lactam
d-valerolactam
48
Alcoholysis of Amides (to Esters)
• 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.
O
C N(CH3)2
OH
CH3CHCH2CH3
H2SO4
N,N-dimethylcyclopentanecarboxamide
O
C
+
+
NH2(CH3)2
O
CH3CHCH2CH3
sec-butyl cyclopentanecarboxylate
•Write a mechanism for this reaction. Refer to acid hydrolysis mechanism if necessary.
49
Hydride Reduction of Amides
• 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.
H - H
Al
O
C N CH3
CH3
-
H AlH3
H
AlH3
-
Li+
O
O
C N CH3
C N CH3
H
H
CH3
CH3
AlH2O-
N,N-dimethylcyclopentanecarboxamide
H
+
C N CH3
H
O
CH3
O
?
C Cl
C NHCH3
• Write equations showing how the above transformation can be carried out.
O
NH2CH3
C Cl
benzoyl chloride
O
C NHCH3
N
1. LiAlH4
2.
H3O+
H
C NHCH3
H
N-methylbenzamide
50
Grignard Reduction of Amides
• 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.
: O:
R
C
.. _
: O:
+
..
:CH3 MgBr
H
N
R
_ CH
4
R
2º amide
C
..
N
R
:CH3 +MgBr
imide anion
• Write equations showing how the following transformation can be carried out.
Br
Mg, ether
O
1. CO2
C OH
MgBr
?
CH2 N(CH3)2
O
SOCl2
2. H3O+
1. LiAlH4
NH(CH3)2
O
C N(CH3)2
C Cl
2. H3O+
O H
C N CH2CH3
H2O 
H+
or
OH-
?
CH2OH
1. LiAlH4
O
C OH
2. H3O+
51
dN:
+
Chemistry of Nitriles
d
C
R
• The carbon atom in the nitrile group is electrophilic because it is bonded to an electronegative
N atom and a p bond in the nitrile is easily broken, i.e., as if it were providing a leaving group.
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)
bromoethane
ethyl bromide
SN2
CH3CH2 Br
propanenitrile
CH3CH2 C N
Na+ CN-
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.
O
C
NH2
SOCl2 , benzene
C
N
+
SO2
+
2 HCl
80ºC
SOCl2
Recall:
R
R
Cl
OH
POCl3 in pyridine
OH
PBr3
R
_
H2O
Br
52
+
Reactions of Nitriles
d
C
R
dN:
• Like carbonyl groups, the nitrile group is strongly polarized and the nitrile C is
electrophilic. Nucleophiles thus attack yielding an sp2 hybridized imine anion.
_
..
: O:
: O:
C
sp 2
products
C
E
R
sp
Nu
Nu:-
+
sp 3
d+
C
+
E
dN:
..
C
R
Nu:-
_
N:
products
Nu
sp 2 imine anion
• 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
H2O
N:
C
O
R
C
1º amide
NH2
O
RMgX
R
R
ketone
[H]
_HO
2
C
[H]
POCl3
O
R
CH2
1º amine
NH2
R
C
H
aldehyde
53
dN:
+
Hydrolysis of Nitriles into Carboxylic Acids
d
C
R
• Nitriles are hydrolyzed in either acidic or basic aqueous solution to yield carboxylic
acids plus ammonia or an amine.
H2O 
O
R C N
H+
or
OH-
+ NH3
R C O H
• 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.
H3O+
R
C
N:
R
C
+
N
..
..
R
H
C
+
N
:O
..
R
..
C OH
acid
: O:
H3O+
R
C
amide
..
NH2
N
+O
H
..
H
H
..
..
H
H
: O:
C
R
H
H2O
..
R
N
C
H
: O:
hydroxyimine
H
54
+
Hydrolysis of Nitriles into Carboxylate Salts
R
d
C
dN:
• 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.
:O
..
H
: O:
-
OH
R
C
R
N:
C
H
H
: O:
H
.. _
N:
_
OH-
H
N
R
C
hydroxy imine
: O:
NH3 +
R
C
: O:
..
.. _ +
:
O
.. Na
R
C
..
NH2
amide

OHH2O
• Show how the following transformation can be carried out without using a Grignard.
Br
?
O
C OH

H3O +
Na+CNSN2
CN

H3O+
O
C NH2
55
+
Reduction of Nitriles
R
d
C
dN:
Alcoholysis of Nitriles doesn’t work. Alcohols are weak nucleophiles and nitriles are
weak electrophiles
Aminolysis 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 p bonds forming a dianion.
H
H
1. H:R
C
R
N:
C
.. _
N:
R
C
-
H:
.. _ 2
N:
H
2.
R
..
H
dianion
..
H2O
H2O
C
NH2
H
+ 2 OH-
• If less powerful DIBAH is used, only 1 equivalent of hydride can add. Subsequent
addition of HOH yields the aldehyde.
1. LiAlH4
C
CH2NH2
N
2.
CH3
o-methylbenzonitrile
H2O
1 equiv.
1. DIBAH , toluene , -78ºC
2.
CH3
O
C H
2-methylbenzaldehyde
H2O or H3O+
CH3
56
+
Reduction of Nitriles with Grignards
R
d
C
dN:
• 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.
1.
R
C
R
R:- MgBr+
R
N:
C
.. _
N:
2.
R
R
H3O+
.. -
..
R
H
N
C
imine anion
:O
..
NH4 +
NH3
+
R
H
H
R
: O:
H3O +
C
N
..
+O
H
R
..
H
H
..
C
R
R
C
N
: O:
H
H
H
C
N
1. CH3CH3 MgBr
2.
benzonitrile
+
H3O
O
C CH2CH3
1-phenyl-1-propanone
ethyl phenyl ketone
57
Multistep Synthesis Problems
• Write equations to show how the following transformations can be carried out.
Br
CH2NH2
?
CN
Na+ CN-
1. LiAlH4
2. H3O+
O
Br
C H
?
1.
2.
DIBAH
1 equiv. - 78°C
H3O+
O
Br
C
?
CN
1.
2.
from above
MgBr
ether
H3O+
Br
?
C
1.
2.
O
H2O 
MgBr
ether
H3O+
C OH
H+
from above
2 equiv.
HO
CN
SOCl2
O
C Cl
58