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Chapter 14 479
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c.
The IR absorption band at
3400 cm
-
1
indicates that the compound is an alcohol. The two doublets
in the
1
H
NMR spectrum at 7.3 and 8.1 ppm that integrate to 4 protons indicate a 1,4-disubstituted
benzene ring with a strongly electron-withdrawing substituent. The two triplets that each integrate to
2 protons and the two multiplets that each integrate to 2 protons indicate that the four-carbon substituent
is not branched. The broad signal at 2.1 ppm is due to the OH proton.
OH
O
2
N
d.
The IR absorption bands at
1700
and
2700 cm
-
1
indicate that the compound is an aldehyde. The two
doublets at
7.0
and 7.8 ppm in the
1
H
NMR spectrum indicate a 1,4-disubstituted benzene ring. That
none of the remaining NMR signals is a doublet suggests that the aldehyde group is attached directly to
the benzene ring. The two triplets and two multiplets indicate an unbranched substituent. The triplet at
4.0
ppm indicates that the group giving this signal is next to an electron-withdrawing group.
H
O
O
78.
a. 1.
7
2.
3
3.
4
4.
3
5.
4
6.
4
b. 1.
5
2.
4
3.
3
4.
3
5.
3
6.
3
79.
a.
The IR absorption band at
1730 cm
-
1
and the absence of an aldehyde C—H stretch suggest that the
compound is a ketone. The signals in the
1
H
NMR spectrum are consistent with a
CH
3
CH
2
CH
2
1
C
“
O
2
group. That there are no other signals in the
1
H
NMR spectrum suggests that it is a symmetrical ketone.
The mass spectrum shows a molecular ion with
m
>
z
=
114;
this is consistent with the ketone shown
below. Also, the large fragments at
m
>
z
=
71
1
M
-
propyl
2
and at
m
>
z
=
43
(a propyl cation) are
consistent with the expected
a
-cleavage fragmentation. The fact that there are only three signals in the
NMR spectrum suggests that it is a symmetrical ketone. The splitting pattern confirms the structure.
O
b.
The
M
+
2
peak at
m
>
z
=
156
in the mass spectrum indicates that the compound contains chlorine;
the IR spectrum indicates that it is a ketone; the NMR spectrum indicates that it has a monosubstituted
benzene ring. The singlet at
4.7
ppm indicates that the
CH
2
group giving this signal is in a strongly
electron-withdrawing environment. The major fragment ions at
m
>
z
=
105
1
M
-
CH
2
Cl
2
and at
m
>
z
=
77
1
C
6
H
5
+
2
are consistent with the structure shown below.
Cl
O