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542 CHAPTER 15 
 
(i) The replacement test gives the same compound, so the 
protons are homotopic: 
 
 
 
(j) The replacement test gives the same compound, so 
the protons are homotopic: 
 
 
 
 
(k) The replacement test gives the same compound, so 
the protons are homotopic: 
 
 
 
 
(l) The replacement test gives diastereomers, so the 
protons are diastereotopic: 
 
 
 
 
(m) The replacement test gives enantiomers, so the 
protons are enantiotopic: 
 
 
 
(n) The replacement test gives diastereomers, so the 
protons are diastereotopic: 
 
 
 
(o) The replacement test gives the same compound, so 
the protons are homotopic: 
 
 
 
15.42. This compound has four aromatic protons. 
Because of their relationship with the respect to the ring 
(symmetry), we expect two types of protons, giving rise 
to a pair of doublets between 7 and 8 ppm: 
 
 
 
The structure also has an ethyl group, so we expect the 
characteristic pattern of signals for an ethyl group. 
Specifically, we expect a triplet with an integration of 3 
(corresponding to the CH3 of the ethyl group) and a 
quartet with an integration of 2 (corresponding to the 
CH2 of the ethyl group). 
 
 
 
The signal for the CH2 of the ethyl group is expected to 
appear at 1.2 + 1 = 2.2 ppm, while the signal for the CH3 
of the ethyl group is expected to appear at 0.9 + 0.2 = 1.1 
ppm. 
Finally, there are two neighboring methylene groups, 
shown here: 
 
 
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