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Spectroscopy Problems
47
Copyright © 2017 Pearson Education, Inc.
12.
The molecular formula indicates that the compound is a hydrocarbon with one degree of unsaturation. The
IR spectrum can tell us whether the degree of unsaturation is due to a cyclic system or a double bond. The
absorption of moderate intensity near
1660 cm
-
1
indicates a
C
“
C
stretch. The absorption at
3100 cm
-
1
,
due to
C
¬
H
stretches of hydrogens attached to
sp
2
carbons, reinforces the presence of the double bond.
The two relatively high-frequency singlets (4.7 ppm) is given by vinylic protons. Because the signal
integrates to 2 protons, we know that the compound has two vinylic protons. Because the signals are not
split, the vinyl protons must not be on adjacent carbons. Thus, they must be on the same carbon.
The singlet (1.8 ppm) that integrates to 3 protons must be a methyl group. Because it is a singlet, the
methyl group must be bonded to a carbon that is not attached to any protons.
The doublet (1.1 ppm) that integrates to 6 protons and the septet (2.2 ppm) that integrates to 1 proton is
characteristic of an isopropyl group.
C
H
H
H C C H
H
HH
isopropyl group
Because the compound has a methyl group, an isopropyl group, and two vinylic hydrogens attached to the
same carbon, we know that the compound must be
2,3-dimethyl-1-butene
.
13.
The signals in the
1
H NMR
spectrum between 6.7 and 6.9 ppm indicate the presence of a benzene ring.
Because the signals integrate to 3 protons, it must be a trisubstituted benzene ring.
The triplet (6.7 ppm) that integrates to 1 proton and the doublet (6.9 ppm) that integrates to 2 protons tell
us that the three substituents are adjacent to one another. (The
H
d
protons are split into a doublet by the
H
c
proton, and the
H
c
proton is split into a triplet by the two
H
d
protons.)
Y
H
d
H
d
H
c
X
Z
Subtracting the trisubstituted benzene
(C
6
H
3
)
from the molecular formula leaves
C
2
H
7
O
unaccounted
for. The singlet (2.2 ppm) that integrates to 6 protons indicates that two methyl groups are in identical
environments. Now only OH is left from the molecular formula. The singlet at 4.6 ppm is due to the proton
of the OH group. The compound is
2,6-dimethylphenol
.
OH
CH
3
CH
3