56 / 912
48
Spectroscopy Problems
Copyright © 2017 Pearson Education, Inc.
14.
A major clue to the compound’s structure comes from the IR spectrum. The strong absorption at
1740 cm
-
1
indicates the presence of a carbonyl
(C
“
O)
group. Because the compound has only one
oxygen, the compound must be an aldehyde or a ketone. The absence of absorptions at 2820 and
2720 cm
-
1
tells us that the compound is not an aldehyde.
The NMR spectrum shows a singlet (2.3 ppm) that integrates to 3 protons, indicating that it is due to a
methyl group. The chemical shift of the signal (hydrogens attached to carbons adjacent to carbonyl carbons
typically have shifts between 2.1 and 2.3 ppm) and the fact that the signal is a singlet suggest that the
methyl group is attached directly to the carbonyl group.
C
O
CH
3
The two remaining signals are split, indicating that the protons that give these signals are attached to
adjacent carbons. Because the signal at 4.3 ppm is a quartet, we know that the proton that gives this signal
is bonded to a carbon that is attached to a methyl group. The other signal (1.6 ppm) is a doublet, so the
proton that gives this signal is bonded to a carbon that is attached to one hydrogen.
C
H
c
H
a
C
H
a
H
a
When these two fragments are subtracted from the molecular formula, only a Cl remains.
The only possible arrangement has the alkyl group bonded directly to the other side of the carbonyl group
and the chlorine on the last available bond. The relatively high-frequency chemical shift of the quartet
(4.3 ppm) reinforces this assignment because it must be attached to an electronegative atom. Thus, the
compound is
3-chloro-2-butanone
.
O
Cl
15.
Given the simplicity of the
1
H NMR
spectrum, the product must be highly symmetrical.
The singlet (7.4 ppm) that integrates to 4 protons is due to benzene-ring protons. Because there are four
aromatic protons, we know that the benzene ring is disubstituted. Because the signal is a singlet, we know
that the four protons are chemically equivalent. Therefore, the two substituents must be the same and they
must be on the 1- and 4-positions of the benzene ring.
Y
Y