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Spectroscopy Problems
53
Copyright © 2017 Pearson Education, Inc.
In the second spectrum, the triplet
( 1.0 ppm)
that integrates to 3 protons is a methyl group that is attached
to a methylene group. The triplet
( 3.4 ppm)
that integrates to 2 protons is a methylene group that is
also attached to a methylene group; the highly deshielded nature of the signal indicates that the carbon is
attached to an electronegative group. Thus, the compound is
1-bromopropane
.
C Br
H
H
C
H
H HH
H
C
The structure is confirmed by the multiplet
( 1.8)
that integrates to 2 protons; the signal is split by both the
adjacent methyl and methylene groups.
Notice that the pattern of a triplet that integrates to 3 protons, a multiplet that integrates to 2 protons, and a
triplet that integrates to 2 protons is characteristic of a propyl group.
24.
A strong and sharp absorption in the IR spectrum at
1730 cm
-
1
indicates a carbonyl
(C
“
O)
group. The
two absorptions at 2710 and
2810 cm
-
1
tell us that the product of ozonolysis is an aldehyde. The aldehydic
proton is also visible in the NMR as a singlet (9.0 ppm) that integrates to 1 proton.
The NMR spectrum has two additional signals. One is a doublet (1.1 ppm) that integrates to 6 protons,
and the other is a septet (2.4 ppm) that integrates to 1 proton. This is characteristic of an isopropyl group.
Therefore, we know that the product of ozonolysis is
2-methylpropanal
.
H
O
Because only one product is formed, we know that the alkene that formed the aldehyde must be symmetrical.
The identification of the aldehyde also agrees with the molecular formula of the alkene that underwent
ozonolysis—that is, an eight-carbon symmetrical alkene will form a four-carbon carbonyl compound.
Two symmetrical alkenes will form 2-methylpentanal—
trans
-2,5-dimethyl-3-hexene
and
cis
-2,5-
dimethyl-3-hexene
. We are not given any information that distinguishes between the two stereoisomers.
Therefore, the unknown alkene can be either of the two stereoisomers.
or
25.
The strong and sharp absorption in the IR spectrum at
1720 cm
-
1
indicates the presence of a carbonyl
group. The broad absorption centered at
3000 cm
-
1
tells us that the carbonyl-containing compound is a
carboxylic acid. The broad singlet (12.0 ppm) in the NMR spectrum (shown as offset by 0.2 ppm from
where it is placed on the spectrum) confirms the presence of a carboxylic acid group.
The only other signal in the NMR spectrum is a singlet (2.0 ppm) that integrates to six protons, indicating
two methyl groups in the same environment. Because the signal is a singlet, the methyl groups must be