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562 Chapter 17
Copyright © 2017 Pearson Education, Inc.
14.
a.
1. LDA
2. ICH
2
CH CH
2
b.
CH
2
Br
2.
1. LDA/THF
O
O
O
O
15.
The compound formed in
14a
has no stereoisomers.
The compound formed in
14b
has a new asymmetric center, so both the R and S stereoisomers are obtained.
16.
a.
1. LDA/THF
O
O
2.
Br
b.
1. LDA/THF
O
O
2.
Br
c.
1. LDA/THF
O
O
2.
Br
17.
a.
O
N
H
+
trace
acid
N
N
CH
2
CH
2
CH
3
+
Br
−
HCl H
2
O
O
CH
2
CH
2
CH
3
+
N
H H
+
Br
−
CH
3
CH
2
CH
2
Br