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Spectroscopy Problems
55
Copyright © 2017 Pearson Education, Inc.
28.
The short signal at
185
ppm in the
13
C NMR
spectrum suggests the presence of the carbonyl group of a
carboxylic acid.
The broad singlet (12.2 ppm) in the
1
H NMR
spectrum that integrates to 1 proton confirms that the
compound contains a carboxylic acid group.
The doublet (1.2 ppm) that integrates to 6 protons and the septet (2.6 ppm) that integrates to 1 proton are
characteristic of an isopropyl group.
C H
H
H
H
C
H
H
C
H
Therefore, the compound is
2-methylpropanoic acid
.
OH
CH
3
O
CH
3
C CH
29.
The breadth of the singlet (11.8 ppm) that integrates to 1 proton indicates a hydrogen that is attached to an
oxygen. The chemical shift of the signal indicates that it is due to the OH group of a carboxylic acid.
C
O
OH
The triplet
( 0.9 ppm)
that integrates to 3 protons is a methyl group that is attached to a methylene group.
The triplet
( 2.3 ppm)
that integrates to 2 protons indicates a methylene group that is also attached to a
methylene group; the chemical shift of this signal indicates that the protons that give this signal are closest
to the electron-withdrawing carboxylic acid group. The multiplet at 1.7 ppm that integrates to 2 protons is
given by the two protons that split the other two signals into triplets.
We can conclude that the compound responsible for the spectrum is
butanoic acid
.
CH
3
CH
2
CH
2
C
O
OH
30.
The singlet
( 9.7 ppm)
that integrates to 1 proton and the molecular formula that contains one oxygen
suggest that an aldehyde is present.
C
O
H
The signals at 7.7 and 6.7 ppm are due to benzene-ring protons. The fact that they are both doublets that
integrate to 2 protons tells us that substituents are on the 1- and 4-positions of the benzene ring.