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Spectroscopy Problems
51
Copyright © 2017 Pearson Education, Inc.
19.
The molecular formula shows that the compound has one degree of unsaturation, indicating a cyclic
compound, an alkene, or a carbonyl group.
A cyclic system containing an oxygen (a cyclic ether) would have the most deshielded signal at
3.5
ppm,
which would be due to the hydrogens attached to the carbon adjacent to the oxygen. Therefore, a cyclic
ether would not give a signal at 6.4 ppm, so it can be ruled out.
Protons attached to a carbon adjacent to a carbonyl group show a signal at
2.1
ppm. Because there is no
signal in that region, a carbonyl group can also be ruled out.
Vinylic protons would account for the signals in the 3.9–4.2 ppm range that integrate to 2 protons, so we
can conclude that the compound is an alkene. Because a highly deshielding oxygen is also present, the
high-frequency signal (6.4 ppm) is not unexpected.
The triplet (1.3 ppm) that integrates to 3 protons and the quartet (3.8 ppm) that integrates to 2 protons
indicate the presence of an ethyl group. The fact that the quartet is deshielded suggests that the ethyl’s
methylene group is attached to the oxygen.
The highly deshielded doublet of doublets (6.4 ppm) that integrates to 1 proton suggests that the proton
that gives this signal is attached to an
sp
2
carbon that is attached to the oxygen. The fact that the signal is
a doublet of doublets indicates that it is split by each of two nonidentical protons on the adjacent carbon.
Thus, the compound is
ethyl vinyl ether
.
OCH
2
CH
3
H
H H
CC
The identification is confirmed by the two doublets (
4.0
and 4.2 ppm) that each integrate to 1 proton.
When those signals are magnified, we can see that each is actually a doublet of doublets. The doublets of
doublets are not well defined because of the small coupling constant (
J
value) for geminal coupling on
sp
2
carbons.
20.
The doublet (1.2 ppm) that integrates to 6 protons and the septet (5.0 ppm) that integrates to 1 proton are
characteristic of an isopropyl group. (The two methyl groups are split by a single proton, and the single
proton is split by six protons.)
The remaining signal (a singlet at a tiny bit more than 2.0 ppm) that integrates to 3 protons indicates an
unsplit methyl group.
When the isopropyl and methyl groups are subtracted from the molecular formula, one carbon and two
oxygens are left over. Thus, the compound has the following fragments:
C H
H
H
H
C
H
H
C
H
C H H
H
O
C
O