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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.

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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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    $\begingroup$ In the case of 13-C benzene, the protons will couple to the spin 1/2 13-C nuclei surely ? $\endgroup$ – J. LS Mar 21 '15 at 15:23
  • $\begingroup$ Yes, that's why I mentioned the sidebands, but I was thinking natural abundance, I've edited the answer accordingly. $\endgroup$ – ron Mar 21 '15 at 16:40

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