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546 CHAPTER 15 
 
Now let’s consider the remaining four protons. The two 
protons on wedges (highlighted below) are 
interchangeable via reflectional symmetry, so they are 
enantiotopic and therefore chemically equivalent. 
 
 
 
Similarly, the two protons on dashes (highlighted below) 
are also interchangeable via reflectional symmetry, so 
they too are enantiotopic and therefore chemically 
equivalent. 
 
 
In summary, we expect this compound to produce three 
signals in its 1H NMR spectrum. 
 
15.51. Among these three compounds, the first one 
(benzene) has aromatic protons, which are expected to 
produce a signal the farthest downfield (between 6.5 and 
8 ppm). Acetylenic protons give signals that are 
relatively upfield (near 2.5 ppm) while vinylic protons 
are expected to produce a signal in the range of 4.5 – 6.5 
ppm. 
 
 
 
15.52. In Section 15.5, the term “chemical shift” was 
defined in the following way: 
 
 
 
The problem statement indicates that the chemical shift 
of the proton is 1.2 ppm and the operating frequency of 
the spectrometer is 300-MHz. We then plug these values 
into the equation above, as shown: 
 
 
 
which gives the following observed shift from TMS (in 
Hz): 
 
 
 
15.53. The molecular formula (C13H28) indicates no 
degrees of unsaturation (see Section 14.16), which 
means that the compound does not have a  bond or a 
ring. The 1H NMR spectrum exhibits the characteristic 
pattern of an isopropyl group (a septet with an 
integration of 1 and a doublet with an integration of 6): 
 
 
 
However, there are no other signals in this spectrum, and 
the molecular formula indicates that there are 28 protons 
(not just 7 protons, as we would expect for an isopropyl 
group). So the compound must be highly symmetrical, 
with four equivalent isopropyl groups (to account for all 
28 protons). This also accounts for 12 of the 13 carbon 
atoms in this compound. The remaining carbon atom 
must be at the center, connected to all four isopropyl 
groups: 
 
 
15.54. The molecular formula (C8H10) indicates four 
degrees of unsaturation (see Section 14.16), which is 
highly suggestive of an aromatic ring. This accounts for 
six of the eight carbon atoms in the structure. The other 
two carbon atoms must be connected to the ring, either 
as an ethyl group or as two methyl groups. Ethylbenzene 
would give an 1H NMR spectrum with four signals and a 
13C NMR spectrum with six signals. The problem 
statement indicates fewer signals in each of these 
spectra, which means that the compound must have more 
symmetry than ethylbenzene. If we explore the three 
possible ways to connect two methyl groups to a ring 
(1,2 or 1,3 or 1,4), we will find that only 1,4-
dimethylbenzene has the necessary symmetry to give 
only two signals in the 1H NMR spectrum and three 
signals in the 13C NMR spectrum. 
 
 
 
15.55. The molecular formula (C3H8O) indicates no 
degrees of unsaturation (see Section 14.16), which 
means that the compound does not have a  bond or a 
ring. The broad signal between 3200 and 3600 cm-1 
indicates the presence of an OH group. The molecular 
formula indicates that the structure has three carbon 
atoms, yet the 13C NMR spectrum exhibits only two 
signals (not three), indicating the presence of symmetry. 
This is only true for 2-propanol (not for 1-propanol): 
 
 
 
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