I've got a $\ce{^{13}C}$ NMR spectrum for bromobenzene. The ipso carbon is significantly more shielded (appearing upfield) than the two ortho carbons. Why is this? Wouldn't the electron-withdrawing effects of the halogen be enough to deshield the ipso carbon?

Does it have to do with the fact the $\ce{Br}$ is a deactivating group? I see that on an NMR database site, the ipso carbon for the $\ce{^{13}C}$ spectrum of fluorobenzene is deshielded. Does it have to do with the size difference between fluorine and bromine?

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    – airhuff
    Apr 30, 2017 at 19:49

1 Answer 1


The principle influence on chemical shift for 13C is the paramagnetic term, which essentially describes the electron density in the p orbitals, and this largely mirrors the influence of electronegativity that is seen in 1H chemical shifts.

However, there is something called the 'heavy atom effect' where the effect of substitution runs counter to electronegativity. This is the case for Bromine, Iodine, Tellurium etc. This comes from an increased level of diamagnetic shielding introduced by the electrons from the halogen atom.

To understand why this is the case, we can consider that the overall shielding on the ipso carbon relative to a methyl susbituent. Even though the bromine is more electronegative than the methyl group, it is so large that it still contributes significant electron density around the ipso carbon nucleus, and hence provides greater overall shielding than a methyl group.

While trends in electronegativity are useful tools to justify observations on chemical shift, it is always good to relate the effects of electronegativity, and other properties, on what is happening to the electron density around the nucleus on question. Ultimately, chemical shift is a measurement of electron shielding around the nucleus.


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