Four of the 13C NMR signals are due to the carbon atoms within the Me, CH, CO, and OCH groups, these are shown in black in the diagram below. The two phenyl rings are diastereotopic, since replacement of each of them in turn by a different group (X) creates a pair of diastereomers as shown below. Diastereotopic groups have different NMR spectra, so each phenyl ring gives rise to four signals in the 13C NMR spectrum, one signal for the ipso-carbon, one for the two homotopic ortho-carbons, one for the two homotopic meta-carbons and one for the para-carbon. Thus, in total the compound shows 12 lines in the 13C NMR spectrum.
In this case, the two phenyl groups are enantiotopic as illustrated
below, and so are not distinguished by NMR. Thus, the compound will show
just five lines in its 13C NMR spectrum, one for the OCH, one
for the aromatic ipso-carbon, one for the ortho-carbons, one for the meta-carbonds
and one for the para-carbons.
This compound contains only three carbon atoms, all of which are obviously
different (a CH3 group, a CH, and a C=O), and so will show three
lines in its 13C NMR spectrum.
The only difference between compounds A and D is the absence of the
methyl group in compound D. However, this change also removes the stereocentre
that was present in A, as a result of which the phenyl rings are no longer
diastereotopic, rather they are enantiotopic as illustrated below. Enantiotopic
groups are not distinguished by NMR, so the 13C NMR spectrum
will consists of just seven lines: one for the CH2, one for
the C=O, one for the OCH, one for the two ipso-aromatic carbons, one for
the four ortho-carbons, one for the four meta-carbons, and one for the
two para-carbons.
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