253 / 912
Chapter 7 245
Copyright © 2017 Pearson Education, Inc.
9.
alkane
=
pentane alkene
=
1
@
pentene alkyne
=
1
@
pentyne
10.
The cis isomer has a higher boiling point because it has a dipole moment, whereas the dipole moment of
the trans isomer is zero.
11.
Solved in the text.
12.
a.
CH
2
CCH
3
Br
b.
CH
3
CCH
3
Br
Br
7.
a.
(
E
)-2-hepten-4-ol
b.
1-hepten-5-yne
c.
(
E
)-4-hepten-1-yne
8.
a.
sp
2
-
sp
2
b.
sp
2
-
sp
3
c.
sp
-
sp
2
d.
sp
-
sp
3
e.
sp
-
sp
f.
sp
2
-
sp
2
g.
sp
2
-
sp
3
h.
sp
-
sp
3
i.
sp
2
-
sp
c.
CH
3
C CCH
3
Br
Br
d.
Br
HC CCH
3
Br
Br Br
e.
CH
3
CH
2
CCH
3
Br
Br
f.
CH
3
CCH
2
CH
2
CH
3
Br
Br
+
CH
3
CH
2
CCH
2
CH
3
Br
Br
13.
a.
CH
3
C CCH
3
Br
Br
Br
Br
CCH
3
+
_
Br
Br
CH
3
C
CCH
3
CH
3
C
b.
Only anti addition occurs because the intermediate is a cyclic bromonium ion.
Therefore, the product has the
E
configuration.
C C
Br
H
3
C
Br
CH
3
14.
Because the alkyne is not symmetrical, two ketones are obtained.
CH
3
CH
2
CCH
2
CH
2
CH
2
CH
3
O
CH
3
CH
2
CH
2
CCH
2
CH
2
CH
3
and
O
15.
a.
CH
3
C CH
b.
CH
3
CH
2
C CCH
2
CH
3
c.
HC C
The best answer for
b
is 3-hexyne, because it would form only the desired ketone.
2-Hexyne would form two different ketones, so only half of the product would be the desired ketone.
16.
a.
CH
2
CCH
3
OH
Because the ketone has identical substituents bonded to the
carbonyl carbon, it has only one enol tautomer.