# NMR: Magnetic and chemical equivalence in benzene

According to my textbook, all the protons in $\ce{^{12}C_6{}^{1}H_6}$ are magnetically equivalent, but those in $\ce{^{13}C_6{}^{1}H_6}$ are not.

I understand that for two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other nuclei in the molecule. I do not understand this with respect to benzene as I expect Hc to couple differently to Ha than Hb to Ha and thus for them to be magnetically inequivalent irrespective of the isotope.

All of the protons in $\ce{^{12}C_6{}^{1}H_6}$ benzene are equivalent by symmetry, therefore they are chemical shift equivalent.
In the case of $\ce{^{12}C_6{}^{1}H_6}$ benzene, all of the protons are equivalent by symmetry, therefore they are chemically equivalent and since there are no "different" nuclei (non-chemical shift equivalent nuclei) for them to couple with, they are also magnetically equivalent.
In the case of $\ce{^{13}C_6{}^{1}H_6}$ benzene all of the protons are equivalent by symmetry so they are all chemical shift equivalent. Since all of the carbons are $\ce{^{13}C}$ (spin=1/2), there are now "different" (the $\ce{^{13}C}$ nuclei are non-chemical shift equivalent) nuclei for the chemical shift equivalent protons to couple with; therefore the protons will be magnetically non-equivalent. The spectrum is now an AA'A'A"A"A"'XX'X'X"X"X"' (A=proton, X=carbon-13) spectrum. Since the protons are not magnetically equivalent (at least some of) the couplings will be observable.