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388 Chapter 10
Copyright © 2017 Pearson Education, Inc.
88.
A
is the substitution product that forms when methoxide ion attacks a carbon of the three-membered ring
and eliminates the amino group, thereby opening the ring. It does not discolor
Br
2
because it does not have
a double bond to which
Br
2
can add.
C C N
CH
3
CH
3
CH
3
CH
3
O
A
C C N
CH
2
CH
3
CH
3
CH
3
B
CH
3
CH
3
CH
3
CH
3
CH
3
B
is the product of a Hofmann elimination reaction: methoxide ion removes a proton from a methyl group
bonded to a ring carbon and eliminates the amino group. The red color disappears when
Br
2
is added to
B
,
because
Br
2
adds to the double bond.
colorless
Br
2
red
C C
Br
Br
C C
+
When the aziridinium ion reacts with methanol, only
A
, the substitution product, is formed.
1.
substitution
product
OCH
3
N
N
+
CH
3
O
–
2.
eliminiation
product
colorless
N
CH
3
O
–
H
Br
Br
N
red
Br
2
N
+
+
89.
a.
The reaction of 2-chlorobutane with
HO
-
is an intermolecular reaction, so the two compounds have to
find each other in the solution.
CH
3
CHCH
2
CH
3
Cl
+
HO
−
CH
3
CHCH
2
CH
3
OH
+
Cl
−
The following reaction takes place in two steps. The first is an intramolecular
S
N
2
reaction; the reac-
tion is much faster than the above reaction because the two reactants are in the same molecule and can
find each other relatively easily.
(CH
3
CH
2 2
) N
(CH
3
CH
2 2
) N
CH
2
CHCH
2
CH
3
Cl
CH
2
CHCH
2
CH
3
N
CH
3
CH
2
CH
2
CH
3
+
HO
CHCH
2
CH
3
CH
2
OH
−