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Chapter 7 247
Copyright © 2017 Pearson Education, Inc.
d.
H
O
e.
O
O
+
f.
O
+
O
20.
a.
b.
c.
21.
A terminal alkyne has a p
K
a
=
25. A base that removes a proton from a terminal alkyne in a reaction that
favors products must have a conjugate acid that is a weaker acid than a terminal alkyne. That is, it must
have a p
K
a
> 25. (Recall that the equilibrium favors formation of the weak acid.)
CH
3
CH
2
and H
2
C
“
CH because the p
K
a
values of their conjugate acids are > 60 and 44, respectively.
The
p
K
a
values of the conjugate acids of the other choices are all < 25.
22.
The reaction of sodium amide with an alkane does not favor products because the acid that would be
formed is a stronger acid than the alkane (the reactant acid). Recall that the equilibrium favors reaction of
the strong acid (or strong base) and formation of the weak acid (or weak base); Section 2.5.
–
+
–
+
CH
3
CH
3
p
K
a
> 60
NH
2
CH
3
CH
2
NH
3
p
K
a
= 36
weaker acid weaker base
stronger base stronger acid
23.
The base used to remove a proton must be stronger than the base that is formed as a result of removing the
proton. Therefore, the base used to remove a proton from a terminal alkyne must be a stronger base than
the conjugate base of the terminal alkyne. A terminal alkyne has a p
K
a
25. In other words, any base
whose conjugate acid has a p
K
a
greater than 25 can be used.
24.
a.
−
−
−
>
>
CH
3
CH
2
CH
2
CH
2
CH
3
CH
2
CH
CH
3
CH
2
C C
CH
b.
−
−
− −
>
>
>
NH
2
CH
3
C
CH
3
CH
2
O F
C
25.
Solved in the text.
26.
a.
+
H
2
C CH
b.
+
CH
3
CH
2
A triply bonded (
sp
) carbon is more electronegative than an
sp
2
or
sp
3
carbon. Therefore, a triply bonded
carbon with a positive charge is less stable than a doubly bonded or singly bonded carbon with a positive
charge. Thus, in
a
, the vinyl cation is more stable and in
b
, the ethyl cation is more stable.
27.
Solved in the text.