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252 Chapter 7
Copyright © 2017 Pearson Education, Inc.
43.
a. 1.
CH
3
CHC
CH
3
CH
H
2
Lindlar
catalyst
CH
3
CHCH
CH
3
CH
2
CH
3
CCHCH
3
CH
3
H
CH
3
CCH
2
CH
3
CH
3
CH
3
CCH
2
CH
3
CH
3
OH
H
2
O
H
2
SO
4
H
2
O
+
+
2.
CH
3
CHC CH
Lindlar
catalyst
H
2
CH
3
CHCH CH
2
CH
3
1. R
2
BH/THF
2. H
2
O
2
, HO
–
, H
2
O
CH
3
CHCH
2
CH
2
OH
CH
3
CH
3
b.
3-Methyl-2-butanol would be a minor product obtained from both
1
and
2
.
CH
3
CHCHCH
3
CH
3
OH
3-methyl-2-butanol
3-Methyl-2-butanol will be obtained from
1
, because occasionally water will attack the secondary car-
bocation before it has a chance to rearrange to the tertiary carbocation.
3-Methyl-2-butanol will be obtained from
2
, because in the second step of the synthesis, boron can
also add to the other
sp
2
carbon; it will be a minor product because the transition state for its formation
is less stable than the transition state leading to the major product. Because a carbocation is not formed
as an intermediate, a carbocation rearrangement cannot occur.
(The proton cannot add to the other
sp
2
carbon in the second step of part
1
because that would form a
primary carbocation. Primary carbocations are so unstable that they can never be formed.)
44.
Three of the names are correct.
a.
3-heptyne
b.
5-methyl-3-heptyne
45.
Only
c
and
e
are keto–enol tautomers. Notice that an enol tautomer has an OH group bonded to an
sp
2
carbon. The structures in
d
are not enol tautomers, because they do not have the oxygen on the same
carbon.
46.
a.
HC CH
H
2
O, H
2
SO
4
HgSO
4
CH
2
CHOH
CH
3
CH
O
enol
b.
HC CH
1. NaNH
2
2. CH
3
CH
2
Br
CH
3
CH
2
C CH
Lindlar
catalyst
CH
3
CH
2
CH CH
2
CH
3
CH
2
CHCH
2
Br
Br
H
2
Br
2
CH
2
Cl
2
c.
correct
d.
6,7-dimethyl-3-octyne
e.
correct
f.
correct