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CHAPTER 19 755 
 
The spectrum also exhibits the characteristic pattern of 
an ethyl group (a quartet with an integration of 2 and a 
triplet with an integration of 3): 
 
 
 
If we inspect the two fragments that we have determined 
thus far (the monosubstituted aromatic ring and the ethyl 
group), we will find that these two fragments account for 
nearly all of the atoms in the molecular formula 
(C9H10O). We only need to account for one more carbon 
atom and one oxygen atom. And let’s not forget that our 
structure still needs one more degree of unsaturation, 
suggesting a carbonyl group: 
 
 
 
There is only one way to connect these three fragments. 
 
 
 
This structure is consistent with the 13C NMR spectrum. 
The signal near 200 ppm corresponds with the carbon 
atom of the carbonyl group. A monosubstituted aromatic 
ring gives four signals between 100 and 150 ppm, and 
there are two signals between 0 and 50 ppm, 
corresponding to the carbon atoms of the ethyl group. 
The signal in the IR spectrum (at 1687 cm-1) is consistent 
with a conjugated carbonyl group. 
 
19.80. The molecular formula (C13H10O) indicates nine 
degrees of unsaturation (see Section 14.16), which is 
highly suggestive of two aromatic rings, in addition to 
either one double bond or one ring. The 13C NMR 
spectrum exhibits only five signals, which must account 
for all thirteen carbon atoms in the compound. 
Therefore, many of the carbon atoms are identical, as a 
result of symmetry. There are four signals between 100 
and 150 ppm, indicating a monosubstituted aromatic 
ring. To account for so many degrees of unsaturation, as 
well as the symmetry that must be present, we propose 
two monosubstituted aromatic rings, rather than just one: 
 
 
If we inspect these two fragments, we will find that they 
account for nearly all of the atoms in the molecular 
formula (C13H10O). We only need to account for one 
more carbon atom and one oxygen atom. And let’s not 
forget that our structure still needs one more degree of 
unsaturation, suggesting a carbonyl group: 
 
 
 
There is only one way to connect these fragments, as 
shown: 
 
 
The carbonyl group in this compound (benzophenone) is 
conjugated to each of the rings, which explains why it 
produces a signal at a relatively low wavenumber (1660 
cm-1) for a carbonyl group of a ketone. 
 
 
 
19.81. The molecular formula (C9H18O) indicates one 
degree of unsaturation (see Section 14.16). The problem 
statement indicates that the compound is a ketone, which 
accounts for the one degree of unsaturation: 
 
 
 
With only one signal in the 1H NMR spectrum, the 
structure must have a high degree of symmetry, such that 
all eighteen protons are equivalent. This can be achieved 
with two tert-butyl groups: 
 
 
 
This compound is a ketone with a parent chain of five 
carbon atoms. The carbonyl group is located at C3, and 
there are four methyl groups (two at C2 and two at C4): 
 
 
 
 
 
19.82. 
(a) There are certainly many acceptable solutions to this 
problem. One such solution derives from the following 
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