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540 CHAPTER 15 
 
(a) This compound has four different kinds of carbon 
atoms (highlighted below), giving rise to four signals. 
 
These four carbon atoms are all
equivalent (giving one signal)
These six methyl groups are all
equivalent (giving one signal)
These two carbon atoms are
equivalent (giving one signal)
These two carbon atoms are
equivalent (giving one signal)
 
 
 
(b) Each of the six carbon atoms is in a unique 
environment (because of its unique proximity to the two 
substituents). Therefore, we expect six signals. 
 
(c) Each of the six carbon atoms is in a unique 
environment (because of its unique proximity to the two 
substituents). Therefore, we expect six signals. 
 
(d) This compound has four different kinds of carbon 
atoms (highlighted below), giving rise to four signals. 
 
Cl
Br
Cl
Br
Cl
Br
This carbon atom
gives one signal
Cl
Br
These two carbon atoms are
equivalent (giving one signal)
This carbon atom
gives one signal
These two carbon atoms are
equivalent (giving one signal) 
 
 
(e) The carbon atom of the methine (CH) group gives 
one signal, and the two methyl groups give one signal, 
for a total of two signals. 
 
(f) Each of the four carbon atoms is in a unique 
environment, because of its proximity to the substituent. 
Therefore, we expect four signals. 
 
15.37. The first compound exhibits symmetry that 
causes some of the carbon atoms to be equivalent 
(similar to the symmetry present in 15.36d). As such, 
the first compound will have five signals in its 13C NMR 
spectrum. In contrast, the second compound lacks this 
symmetry. Each carbon atom occupies a unique 
environment, so the second compound is expected to 
produce seven signals in its 13C NMR spectrum. 
 
15.38. This compound has six different kinds of protons, 
highlighted here. In each case, we apply the n+1 rule, 
giving the multiplicities shown: 
 
O
CH3
CH3
H3C
H3C
H3C
H H
H H H H
H
triplet
singlet triplet
doublet
doublet
septet of triplets
or triplet of septets
(multiplet)
 
 
15.39. 
(a) The first compound has a very high degree of 
symmetry, and will produce only four only signals in its 
13C NMR spectrum, while the second compound will 
produce twelve signals. 
Also, the first compound will produce only two signals 
in its 1H NMR spectrum, while the second compound 
will produce eight signals. 
 
(b) The first compound is a meso compound. Two of the 
protons are enantiotopic (the protons that are alpha to the 
chlorine atoms) and are therefore chemically equivalent. 
As such, the first compound will only have two signals 
in its 1H NMR spectrum, while the second compound 
will have three signals. For a similar reason, the first 
compound will only have two signals in its 13C NMR 
spectrum, while the second compound will have three 
signals. 
 
(c) The 13C NMR spectrum of the second compound will 
have one more signal than the 13C NMR spectrum of the 
first compound. The 1H NMR spectra will differ in the 
following way: the OH group in the first compound will 
produce a singlet somewhere between 2 and 5 ppm with 
an integration of 1, while the methoxy group in the 
second compound will produce a singlet at 
approximately 3.4 ppm with an integration of 3. 
 
(d) The first compound has symmetry that is not present 
in the second compound. As such, the first compound 
will have three signals in its 13C NMR spectrum, while 
the second compound will have five signals. 
For similar reasons, the first compound will have two 
signals in its 1H NMR spectrum, while the second 
compound will have four signals. 
 
15.40. The molecular formula (C8H18) indicates no 
degrees of unsaturation (see Section 14.16), which 
means that the compound does not have a  bond or a 
ring. With only one signal in its 1H NMR spectrum, the 
structure must have a high degree of symmetry, such that 
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