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Prévia do material em texto

CHAPTER 15 521 
 
In the case of butane, the signal for the highlighted CH2 group is expected to be a quartet as a result of the neighboring 
methyl group. Notice that the signal is not further split by the neighboring CH2 group (C3), because the CH2 groups at 
C2 and C3 are chemically equivalent and therefore, they don’t split each other. In contrast, the signal for the 
highlighted CH2 group in pentane has a more complex multiplicity. In the case of pentane, the C2 and C3 positions are 
not identical. The C2 position is next to a methyl group, while the C3 position is not. They occupy different electronic 
environments, and they are not interchangeable by symmetry. As a result, the signal for the highlighted CH2 group 
(C2) is expected to be split into a quartet of triplets (or into a triplet of quartets, depending on which J value is larger). 
When analyzing a 1H NMR spectrum, make sure to take this into account, as students will often misinterpret a 
spectrum as a result of failing to take this into account. 
 
 
Solutions 
 
15.1. 
(a) When looking for symmetry, don’t be confused by 
the position of the double bonds in the aromatic ring. 
Recall that we can draw the following two resonance 
structures: 
 
 
 
Neither resonance structure is more correct than the 
other. For purposes of looking for symmetry, it will be 
less confusing to draw the compound like this: 
 
 
 
When drawn in this way, we can see that the two 
highlighted protons can be interchanged by rotational 
symmetry (the axis of symmetry is shown below). 
Therefore, the protons are homotopic. 
 
 
 
This conclusion can be verified by the replacement test. 
Specifically, each proton is replaced with deuterium, and 
the resulting compounds are found to be the same. 
Therefore, the compounds are homotopic: 
 
 
 
(b) These protons cannot be interchanged by rotational 
symmetry, so they are not homotopic. They can be 
interchanged by reflectional symmetry (the plane of 
symmetry is the plane of the page). Therefore, the 
protons are enantiotopic. 
 
 
 
This conclusion can be verified by the replacement test. 
Specifically, each proton is replaced with deuterium, and 
the resulting compounds are found to be enantiomers. 
Therefore, the compounds are enantiotopic: 
 
 
 
(c) These protons cannot be interchanged by rotational 
symmetry, so they are not homotopic. They also cannot 
be interchanged by reflectional symmetry so they are not 
enantiotopic either. To determine if they are 
diastereotopic, we use the replacement test. Specifically, 
each proton is replaced with deuterium, and the resulting 
compounds are found to be diastereomers. Therefore, 
the compounds are diastereotopic: 
 
 
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