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246 Chapter 7
Copyright © 2017 Pearson Education, Inc.
b.
CH
3
CH CCH
2
CH
2
CH
3
OH
CHCH
2
CH
3
CH
3
CH
2
C
OH
and
C C
CH
2
CH
2
CH
3
OH
H
Z
C C
OH
CH
2
CH
2
CH
3
H
E
C C
CH
2
CH
3
H
CH
3
CH
2
HO
Z
C C
H
CH
2
CH
3
CH
3
CH
2
HO
E
E
and
Z
isomers are possible for each of these enols.
CH
3
CH
3
c.
C CH
2
OH
CH
3
C
OH
and
Because each enol has identical
groups bonded to one of its
sp
2
carbons,
E
and
Z
isomers are not possible
for either enol.
17.
a.
CH
3
CH
2
CCH
3
O
CH
3
CH
2
CH
2
CH
O
(1)
(2)
b.
CH
3
CH
2
CCH
3
O
CH
3
CH
2
CCH
3
O
(1)
(2)
c.
CH
3
CH
2
CH
2
CCH
3
O
and CH
3
CH
2
CCH
2
CH
3
O
CH
3
CH
2
CH
2
CCH
3
O
and CH
3
CH
2
CCH
2
CH
3
O
(1)
(2)
18.
Ethyne (acetylene)
An alkyne can form an aldehyde only if the OH group adds to a terminal
sp
carbon. In the acid-catalyzed
addition of water to a terminal alkyne, the proton adds to the terminal
sp
carbon. Therefore, the only way
the OH group can add to a terminal
sp
carbon under these conditions is if there are two terminal
sp
carbons
in the alkyne. In other words, the alkyne must be ethyne.
19.
a.
CH
3
CH
2
CH
2
C CH CH
3
CH
2
C CCH
3
CH
3
CH
2
CH
2
CH
2
CH
3
1-pentyne
2-pentyne
H
2
Pd/C
or
b.
CH
3
C CCH
3
C C
CH
3
H
H
3
C
H
H
2
Lindlar
2-butyne
catalyst
c.
CH
3
CH
2
CH
3
CH
2
C CCH
3
Na
NH
3
(liq)
–
78 °C
C C
H
H
CH
3
2-pentyne
d.
CH
3
CH
2
CH
2
CH
2
C CH
H
2
Lindlar
1-hexyne
CH
3
CH
2
CH
2
CH
2
CH CH
2
catalyst
Na
>
NH
3
1
liq
2
cannot be used to reduce terminal alkynes because Na removes the hydrogen that is attached
to the
sp
carbon of the terminal alkyne.
2 RC CH 2 Na
2 Na
+
2 RC C
–
H
2