Determining the coupling on a substituted benzene ring

Question

I can't seem to find any clear explanation of how benzene ring protons couple with each other when mono and disubstituted with the same constituent.

My professor claims that if a benzene ring is para disubstituted with X, the remaining 4 protons exhibit a doublet interference pattern. This seems strange to me because they are equivalent, and my last two organic chemistry professors said the information isn't useful diagnostically, but never explained why.

Also, he claims that if a benzene ring is monosubstituted the proton in the para position will display a triplet interference pattern. Is this right? Symmetry seems to say that the proton would be a doublet.

Do benzene rings have their own rules for coupling?

Answer 1

Benzene rings follows the same general principles as other spin systems for coupling.

A symmetrically para-substituted benzene, such as 1,4-dichlorobenzene, will have a 1H spectrum which will contain just a singlet. All four proton environments are identical. An XY para-disubstituted benzene will have the aromatic peaks that appear somewhat like two sets of doublets. The appearance is actually incredibly useful diagnostically,as it gives a useful indication of the type of ring substitution you have. They are, however, not true doublets, and consist of second order coupling arising from coupling between magnetically non-equivalent sites. That is, the splitting of the apparent doublets is not a true reflection of the coupling constant.

The para proton on a mono-substituted benzene ring does appear as a triplet. The ortho coupling giving rise to the ~7Hz splitting comes from coupling from two chemically equivalent protons at the meta position. Coupling to 2 equivalent protons will give rise to a 3 line splitting (from your 2nI+1 rule). Strictly speaking, a mono-substituted benzene ring is just a very special subset of a para-disubstituted benzene ring system, but does not typically show the classic splitting expected in a para system, because the 5J coupling is essentially zero.