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Spectroscopy Problems
Copyright © 2017 Pearson Education, Inc.
50.
Each of the doublets (7.3 and 7.7 ppm) that integrates to 4 protons is given by benzene-ring protons.
Because a benzene ring does not have 8 protons, there must be two benzene rings in the compound. The
doublets indicate that the benzene rings have substituents at the 1- and 4-positions and, because each
doublet integrates to 4 protons, the two substituents on each of the benzene rings must be the same.
The singlet (3.8 ppm) that integrates to 6 protons suggests the compound has two methyl groups in an
identical environment. The chemical shift of the singlet indicates that each is attached to an electronegative
atom. The molecular formula indicates that the electronegative atom is an oxygen.
2
¬
OCH
3
When the two disubstituted benzene rings and the two
CH
3
O
groups are subtracted from the molecular
formula, all that remains is CO. Therefore, a carbonyl group must connect the two benzene rings.
OCH
3
CH
3
O
C
O
51.
A.
The broad singlet at
∼
5.2 ppm that integrates to 2 protons indicates an NH
2
group. The broad singlet
at
∼
3.7 ppm that integrates to 1 proton indicates an OH group. Subtracting NH
2
and OH from the
molecular formula leaves C
3
H
6
. The two triplets and the quintet that each integrate to 2 hydrogens
suggests a
¬
CH
2
CH
2
CH
2
¬
unit with a group on either end that does not cause splitting. Therefore,
this isomer is
HO
NH
2
B.
The broad singlet at
∼
3.7 ppm that integrates to 1 proton indicates an NH group. The broad singlet at
∼
2.0 ppm that integrates to 1 proton indicates an OH group. The 3H singlet at
∼
3.3 ppm indicates
a methyl group attached to an electron-withdrawing group. Subtracting NH, OH, and CH
3
from
the molecular formula leaves C
2
H
4
. The two triplets that each integrate to 2 hydrogens suggest
a
¬
CH
2
CH
2
¬
unit with a group on either end that does not cause splitting. Therefore, this isomer is
HO
N
H
C.
The broad singlet at
∼
5.1 ppm that integrates to 2 protons indicates an NH
2
group. The singlet at
∼
3.3 that integrates to 3 protons indicates a methyl group attached to an electron-withdrawing group.
The electron-withdrawing group must be the oxygen because the nitrogen has two protons attached
to it, so it can be attached to only one other group. Subtracting NH
2
and OCH
3
from the molecular
formula leaves C
2
H
4
. The two triplets that each integrate to 2 hydrogens suggest a
¬
CH
2
CH
2
¬
unit
with a group on either end that does not cause splitting. Therefore, this isomer is
NH
2
O