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Spectroscopy Problems
Copyright © 2017 Pearson Education, Inc.
54.
A.
The broad singlet at
∼
2.0 ppm indicates an N
¬
H bond. The compound has 23 hydrogens but only
4 additional signals, suggesting that it is a symmetrical compound. The doublet at 0.9 ppm indicates
an isopropyl group. The triplet at
∼
2.6 ppm indicates a group split only by an adjacent CH
2
group.
Therefore, this compound is
H
N
B.
This spectrum is similar to the spectrum in part A. However, the signal indicating a hydrogen bonded
to a nitrogen is missing, and the triplet that indicates a group split only by an adjacent CH
2
group is at a
higher frequency (
∼
3.4 ppm), indicating that it is adjacent to the oxygen. Therefore, this compound is
O
55.
D.J. made the mistake of thinking that the H
a
and H
b
protons are equivalent. This made him conclude
that there would be only 2 signals in the spectrum. However, the H
a
and H
b
protons are not in the same
environment—H
a
is trans to Br, and H
b
is cis to Br. Therefore, there are 3 signals and each is a multiplet.
Br
Br
H
a
H
c
H
a
H
b
H
c
H
b
56.
Although D.J. is now more experienced, he made the same mistake, still thinking that the H
a
and H
b
protons
are equivalent. The H
a
and H
b
protons are not in the same environment —H
a
is trans to OH, and H
b
is cis to
OH. Therefore, there are 3 signals (all multiplets) in addition to the signal for the H that is attached to the
oxygen.
H
b
H
b
H
c
H
a
H
a
H
d
O
57.
A.
The highest-frequency signal is for the H that is attached to the same carbon that Br is attached to.
The signal at
∼
0.9 ppm that integrates to 3 hydrogens is a methyl group that is attached to a CH
2
group. The signal at
∼
2.8 ppm that integrates to 3 hydrogens is a methyl group that is close to
an electron-withdrawing group. Integration shows that the compound has two CH
2
groups. Thus,
compound
A
is
Br