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328 Chapter 9
Copyright © 2017 Pearson Education, Inc.
63.
CH
3
CH
2
CCH
2
CH
3
CH
3
Br
C C
H
3
C
H
CH
3
CH
2
CH
3
+
C C
H
3
C
H
CH
2
CH
3
CH
3
minor
major
CH
3
CH
2
CCH
2
CH
3
CH
3
Br
C C
H
3
C
H
CH
3
CH
2
CH
3
+
+
C C
H
3
C
H
CH
2
CH
3
CH
3
minor
HO
−
CH
3
CH
2
CCH
2
CH
3
CH
3
OH
A tertiary alkyl halide
cannot undergo a
substitution reaction
under S
N
2/E2 conditions. .
H
2
O
64.
a.
HO
Br
because it forms a six-membered ring, whereas the other compound
would form a seven-membered ring. A seven-membered ring is
more strained than a six-membered ring, so the six-membered ring
is formed more easily. (See Table 3.8 on page 124 of the text.)
b.
HO
Br
because it forms a five-membered ring, whereas the other com-
pound would form a four-membered ring. A four-membered ring is
more strained than a five-membered ring, so the five-membered ring
is formed more easily.
c.
HO
Br
because it forms a seven-membered ring, whereas the other com-
pound would form an eight-membered ring. An eight-membered
ring is more strained than a seven-membered ring, so the seven-
membered ring is formed more easily; also, the Br and OH in the
compound that leads to the eight-membered ring are less likely to
be in the proper position relative to each other for reaction because
there are more bonds around which rotation to an unfavorable
conformation can occur.
65.
a.
When hydride ion removes a proton from the OH group, the alkoxide ion cannot react in an intramo-
lecular reaction with the alkyl chloride to form an epoxide, because it cannot reach the back side of the
carbon attached to the chlorine. Therefore, the major product will result from an intermolecular reaction.
OH
+
Cl
O
−
Cl
H
2
O
NaH
Cl
OH O
b.
Hydride ion removes a proton from the OH group more rapidly than it attacks the alkyl chloride. Once
the alkoxide ion is formed, it attacks the back side of the alkyl chloride, forming an epoxide. (Removing
a proton from an oxygen is always a fast reaction.)
Cl
OH
Cl
+
O
−
O
Cl
−
NaH