WORKED PROBLEMS
661
(b) This is a standard Strecker synthesis of an amino acid; it is not important that you remember the name in
‘named’ reactions. An analysis of the transformation should lend some clues. You are informed it is a twostage process; acid hydrolysis is a necessary part, and you can relate this to hydrolysis of a nitrile, supplied by
the cyanide in the first part. The initial reaction is between the aldehyde and the ammonia to form an imine.
This then acts as carbonyl analogue for attack of the cyanide. Alternative thoughts starting with aldehyde
plus cyanide giving a cyanohydrin are unproductive, in that you then wonder how to use the ammonia.
H
O
NH 3
NH 3
O
H
NH 2
OH
NH
C N
NH
C
N
NH 2
C
N
NH 2
CO 2 H
nucleophilic attack
of ammonia onto
carbonyl
− H + , + H
+
nucleophilic
attack of
cyanide onto
imine
+ H
+
H
+
H 2 O
NH
OH 2
− H 2 O
H
acid-catalysed
hydrolysis of
nitrile to acid
The initial part of the reaction is the same as in part (a), and leads to the imine intermediate. Nucleophilic
attack by the cyanide onto this carbonyl analogue provides the next intermediate, an aminonitrile. The last
part of the synthesis is hydrolysis of the nitrile function. It involves another intermediate, an amide, and this
also has to be subjected to acid hydrolysis. As a result, it all adds up to a lengthy mechanism that can become
rather tedious, because it contains so many repetitive steps. It is shown in general terms below.
C N H
C N
C
NH
OH 2
C
NH 2
O
amide
nucleophilic attack
onto conjugate acid
H
+
protonation of N
C
NH 2
O H
− H
+
C
NH 2
O
H
+
C
NH 2
OH
H 2 O
C
NH 2
OH
H 2 O
C
NH 3
OH
HO
C
OH
OH
C
OH
O
− H
+
, + H
+
amide
− H
+
, + H
+
− H
+
H
OH 2
(c) This is another ‘named’ reaction, the Mannich reaction. Three reagents join together in a two-stage process.
CH 3
O
HN
H +
O
N
O
H
H
We have an aldehyde, an amine, and a ketone. As in part (b), the amine reacts first to give an imine, and
this behaves as a carbonyl analogue, which in the Mannich reaction is then the electrophile for an enolate
anion equivalent. How can we remember the sequence of events? The most common mistake is to react
the aldehyde and ketone via an aldol reaction, but this then leads to an alcohol and one is faced with a
substitution reaction to incorporate the amine. It is the mild acidic conditions that help us to avoid wrong
661
(b) This is a standard Strecker synthesis of an amino acid; it is not important that you remember the name in
‘named’ reactions. An analysis of the transformation should lend some clues. You are informed it is a twostage process; acid hydrolysis is a necessary part, and you can relate this to hydrolysis of a nitrile, supplied by
the cyanide in the first part. The initial reaction is between the aldehyde and the ammonia to form an imine.
This then acts as carbonyl analogue for attack of the cyanide. Alternative thoughts starting with aldehyde
plus cyanide giving a cyanohydrin are unproductive, in that you then wonder how to use the ammonia.
H
O
NH 3
NH 3
O
H
NH 2
OH
NH
C N
NH
C
N
NH 2
C
N
NH 2
CO 2 H
nucleophilic attack
of ammonia onto
carbonyl
− H + , + H
+
nucleophilic
attack of
cyanide onto
imine
+ H
+
H
+
H 2 O
NH
OH 2
− H 2 O
H
acid-catalysed
hydrolysis of
nitrile to acid
The initial part of the reaction is the same as in part (a), and leads to the imine intermediate. Nucleophilic
attack by the cyanide onto this carbonyl analogue provides the next intermediate, an aminonitrile. The last
part of the synthesis is hydrolysis of the nitrile function. It involves another intermediate, an amide, and this
also has to be subjected to acid hydrolysis. As a result, it all adds up to a lengthy mechanism that can become
rather tedious, because it contains so many repetitive steps. It is shown in general terms below.
C N H
C N
C
NH
OH 2
C
NH 2
O
amide
nucleophilic attack
onto conjugate acid
H
+
protonation of N
C
NH 2
O H
− H
+
C
NH 2
O
H
+
C
NH 2
OH
H 2 O
C
NH 2
OH
H 2 O
C
NH 3
OH
HO
C
OH
OH
C
OH
O
− H
+
, + H
+
amide
− H
+
, + H
+
− H
+
H
OH 2
(c) This is another ‘named’ reaction, the Mannich reaction. Three reagents join together in a two-stage process.
CH 3
O
HN
H +
O
N
O
H
H
We have an aldehyde, an amine, and a ketone. As in part (b), the amine reacts first to give an imine, and
this behaves as a carbonyl analogue, which in the Mannich reaction is then the electrophile for an enolate
anion equivalent. How can we remember the sequence of events? The most common mistake is to react
the aldehyde and ketone via an aldol reaction, but this then leads to an alcohol and one is faced with a
substitution reaction to incorporate the amine. It is the mild acidic conditions that help us to avoid wrong