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Spectroscopy Problems
49
Copyright © 2017 Pearson Education, Inc.
Subtracting the disubstituted benzene ring from the molecular formula, only
C
2
H
4
Br
2
remains. Thus, each
substituent must contain 1 carbon, 2 hydrogens, and 1 bromine. The compound that gives the spectrum,
therefore, is the one shown here.
CH
2
Br
BrCH
2
16.
The molecular formula indicates that the compound has two degrees of unsaturation. The weak absorption
at
2120 cm
-
1
is due to a carbon–carbon triple bond, which accounts for the two degrees of unsaturation.
The intense and sharp absorption at
3300 cm
-
1
is due to the
C
¬
H
stretch of a hydrogen attached to an
sp
carbon. The intensity and shape of this absorption distinguishes it from an alcohol (intense and broad) and
an amine (weaker and broad). Thus, we know that the compound is a terminal alkyne.
The absorptions between 2800 and
3000 cm
-
1
are due to the
C
¬
H
stretch of hydrogens attached to
sp
3
carbons.
All three signals in the
1
H NMR
spectrum are singlets, indicating that none of the protons that give these
signals have neighboring protons. The singlet (2.4 ppm) that integrates to 1 proton is the proton of the
terminal alkyne.
C C H
The two remaining signals (3.4 and 4.1 ppm) that integrate to 3 protons and 2 protons, respectively, can
be attributed to a methyl group and a methylene group. When the alkyne fragment and the methyl and
methylene groups are subtracted from the molecular formula, only an oxygen remains.
C
H
H
O
C H
H
H
The arrangement of these groups can be determined by the splitting and the chemical shift of the signals.
Because each signal is a singlet, the methyl and methylene groups cannot be adjacent or they would split
each other’s signal. Because the terminal alkyne and the methyl group must be on the ends of the molecule,
the only possible arrangement is shown below. Thus, the compound is
3-methoxy-1-propyne
.
CH
3
OCH
2
C CH
Notice that both the methyl and methylene groups show strong deshielding because of their direct
attachment to the oxygen. The methylene hydrogens are also deshielded by the neighboring alkyne.
17.
The signals in the
1
H NMR
spectrum between 6.5 and 7.2 ppm indicate the presence of a benzene ring.
Because the signals integrate to 3 protons, it must be a trisubstituted benzene ring.
The singlet (2.1 ppm) that integrates to three protons must be a methyl group; 2.1 ppm is characteristic of
protons bonded to a benzylic carbon.
When the trisubstituted benzene ring
(C
6
H
3
)
and the methyl group
(CH
3
)
are subtracted from the molecular
formula,
NH
2
Br
is all that remains. Thus, the three substituents must be a methyl group, bromine, and
an amino group
(NH
2
).
The amino group gives the broad singlet (3.6 ppm) that integrates to 2 protons.
Hydrogens attached to nitrogens and oxygens typically give broad signals.