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528 CHAPTER 15 
 
15.11. Among the integration values provided, the 
lowest number is 33.2, so we divide all integration 
values by 33.2, giving the following ratio: 
1 : 1.5 : 1 : 1.5 
Since there is no such thing as a half of a proton, these 
numbers must represent 2, 3, 2, and 3 protons, 
respectively. This is confirmed by the molecular 
formula, which indicates that the compound has ten 
hydrogen atoms. 
Therefore, 
The signal at 4.0 ppm represents two protons. 
The signal at 2.0 ppm represents three protons. 
The signal at 1.6 ppm represents two protons. 
The signal at 0.9 ppm represents three protons. 
 
15.12. Among the integration values provided, the 
lowest number is 17.1, so we divide all integration 
values by 17.1, giving the following ratio: 
1 : 5 : 1 : 3 
The molecular formula indicates that the compound has 
ten hydrogen atoms, so the numbers above are not only 
relative values, but they are also exact values. 
Therefore, 
The signal at 9.6 ppm represents one proton. 
The signal at 7.5 ppm represents five protons. 
The signal at 7.3 ppm represents one proton. 
The signal at 2.1 ppm represents three protons. 
 
15.13. Among the integration values provided, the 
lowest number is 18.92, so we divide all integration 
values by 18.92, giving the following ratio: 
1 : 1 : 1 
The molecular formula indicates that the compound has 
six hydrogen atoms (not just three), so the numbers 
above are only relative values. Each signal must actually 
represent two protons. 
 
15.14. We expect a total of eight signals in the 1H 
NMR spectrum of phenylalanine; note the symmetry in 
the benzene ring and the nonequivalent diastereotopic 
protons next to the chiral center bearing the NH2 group. 
 
 
 
Cinnamic acid has a simpler 1H NMR spectrum with 
only six signals. 
 
 
15.15. 
(a) This compound has four different kinds of protons, 
highlighted here. In each case, we apply the n+1 rule, 
giving the multiplicities shown: 
 
 
 
(b) This compound has four different kinds of protons, 
highlighted here. In each case, we apply the n+1 rule, 
giving the multiplicities shown: 
 
 
 
(c) This compound has four different kinds of protons, 
highlighted here. In each case, we apply the n+1 rule, 
giving the multiplicities shown: 
 
 
(d) This compound has six different kinds of protons, 
highlighted here. In each case, we apply the n+1 rule, 
giving the multiplicities shown: 
 
 
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