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654 Chapter 19
Copyright © 2017 Pearson Education, Inc.
Solutions to Problems
1.
a.
2,2-dimethylazacyclopropane or
2,2-dimethylaziridine
b.
4-ethylazacyclohexane
4-ethylpiperidine
c.
2-methylthiacyclopropane or
2-methylthiirane
d.
3-methylazacyclobutane or
3-methylazetidine
e.
2,3-dimethyloxacyclopentane or
2,3-dimethyltetrahydrofuran
f.
2-ethyloxacyclobutane or
2-ethyloxetane
2.
Solved in the text.
3.
The oxygen in morpholine withdraws electrons inductively, which make protonated morpholine the stron-
ger acid. Recall that inductive electron withdrawal increases acidity (Section 2.7).
N
O
+
H H
N
H
O
N
H
p
K
a
=
9.28
morpholine
p
K
a
=
11.12
piperidine
N
+
H H
+
H
+
+
H
+
4.
a.
N
O
b.
The conjugate acid of 3-quinuclidinone has a lower p
K
a
than the conjugate acid of morpholine
because the
sp
2
oxygen of 3-quinuclidinone is more electronegative than the
sp
3
oxygen of morpholine
(Section 2.6) and the oxygen is closer to the nitrogen. So we know that its p
K
a
is less than 9.
N
+
O
H
conjugate acid
p
K
a
=
7.46
p
K
a
=
9.28
of 3-quinuclidinone
N
O
+
H H
conjugate acid
of morpholine
c.
The conjugate acid of 3-chloroquinuclidine has a lower p
K
a
than the conjugate acid of
3-bromoquinuclidine because chlorine is more electronegative than bromine, so it is better at with-
drawing electrons inductively.
N
+
Cl
H
conjugate acid of
3-chloroquinuclidine
N
+
Br
H
conjugate acid of
3-bromoquinuclidine