Chemistry 29b

Practice NMR Spectral Problem A

Spectrum A:

Four DBEs.  At first glance the pair of doublets at 7-8 ppm suggests a para-disubstituted aromatic ring (although the extra H in the “aromatic” region seems confusing), but this is wrong!  Count the no. of sp2 carbons in the C-13 spectrum, we would expect four for a typical para with 2 different substituents, but we have only three!  An ortho disubstituted ring with two identical substituents fits the bill.

But what about this peak at 7.2 ppm?  Again the C-13 helps, it indicates we have an sp3 carbon (68 ppm) with electronegative substituents on it.  Since we have 4 Cls to account for, this is reasonable.  Cl is similar to O in electronegativity, so one Cl takes the chemical shift to around 3.5.  If we have a second Cl and an aromatic ring attached to this CH, it is reasonable that the H is moved way downfield to around 7 ppm.  The lesson- don’t be too tied to simple tables of shifts, you have to use your brain to think about the effects of substituents sometimes, and predict where things should come!

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