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. ![](https://i.stack.imgur.com/yQ3Uw.png)




All of the protons in $\ce{^{12}C\_6{}^{1}H\_6}$ benzene are equivalent by symmetry, therefore they are **chemical shift equivalent**.


You wrote, "I understand that for two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other **nuclei** in the molecule."


I think the correct statement would be, for two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other **non-chemical shift equivalent nuclei** in the molecule. If there are no other non-chemical shift equivalent nuclei in the molecule, then the chemical shift equivalent nuclei are also magnetically 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.


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