769 / 912
Chapter 23 761
Copyright © 2017 Pearson Education, Inc.
10.
Notice that the only difference in this reaction and the one in Problem 9 is that the species to which the
two-carbon fragment is transferred has an ethyl group in place of the methyl group.
−
H B
CH
3
C
HO
CH
3
C
HO
H B
+
CO
2
CH
3
C
O
C
OH
CH
2
CH
3
H B
_
+
−
_
C
O
O
C CH
3
C
O
O
O
_
C
CH
3
CH
2
C
O
O
O
_
C
O
O
_
S
C
RN
+
S
C
RN
+
S
C
RN
S
C
RN
+
S
C
RN
+
_
HO CH
2
CH
3
C CH
3
C
O
O
C
O
-
aceto- -hydroxybutyrate
11.
Solved in the text.
12.
a.
CH
3
C
O
b.
CH
3
C
O
c.
HOCH
2
C
O
13.
The compound on the right is more easily decarboxylated because the electrons left behind when
CO
2
is eliminated are delocalized onto the positively charged nitrogen of the pyridine ring. The electrons left
behind if the other compound is decarboxylated cannot be delocalized.
N
H
+
N
H
+
_
O
CH
2
CH
2
C
O
_
O CH
2
C
O