# Why are protons in the same position in 1-chloro-4-nitrobenzene magnetically different?

Protons in ortho positions towards the nitro group are said to be magnetically different, even though they're chemically equivalent. Same goes for protons ortho towards the chlorine. Why is that so?

The hunch lies in the definition of magnetic equivalence:

For two protons to be magnetically equivalent they not only have to have the same chemical shift, but they must also each have the same $J$ coupling to other magnetic nuclei in the molecule.

(Quoted from Hans Reich’s NMR course online syllabus; emphasis present in the original.)

What does that mean for our specific example? Compare the molecule we have in figure 1 below.

Figure 1: 1-chloro-4-nitrobenzene with labelled protons.

Needless to say as you already figured that $\ce{H^1}$ and $\ce{H^3}$ are chemically equivalent (one is transformed into the other by $C_2$) meaning that they will display the same chemical shift; likewise for $\ce{H^2}$ and $\ce{H^4}$.

However, for magnetic equivalence we need to consider their coupling constants to each other. If $\ce{H^1}$ and $\ce{H^3}$ are magnetically equivalent, then $J_{\ce{H^1 H^2}} \overset{!}{=} J_{\ce{H^3 H^2}}$. But, $J_{\ce{H^1 H^2}}$ is an ortho coupling with a value of $J \approx 7~\mathrm{Hz}$ while $J_{\ce{H^3 H^2}}$ is a para coupling with $J \approx 0~\mathrm{Hz}$.

Thus we can easily see that $J_{\ce{H^1 H^2}} \ne J_{\ce{H^3 H^2}}$ and therefore the nuclei $\ce{H^1}$ and $\ce{H^3}$ are magnetically non-equivalent.

The same argument can be made for $\ce{H^2}$ and $\ce{H^4}$.

‡: They are also transformed into each other by the $\sigma_{\mathrm{v}1}$ plane. However, planes of symmetry create enantiotopic protons while rotations create isotopic ones. In this case — a non-chiral environment and a non-chiral solvent — enantiotopic protons cannot be distinguished and give a common signal, and thus are sometimes incorrectly called chemically equivalent. The argument is not of relevance here though, since the $C_2$ axis renders the protons isotopic thus equivalent.

• Nice answer! One small point, I agree that $\ce{H^1}$ and $\ce{H^3}$ are chemically equivalent due to the presence of the $C_2$ axis; if only the $\sigma_{\mathrm{v}1}$ plane existed wouldn't the two protons be enantiotopic instead of equivalent?
– ron
Commented Dec 8, 2015 at 16:55
• It isn't a "hunch" but a "twist" or "quirk" in the definition. You need to also explain that if the two chemically equivalent H atoms were also magnetically equivalent then there would be just two peaks in the H NMR where each peak was for two protons. Rather the observed spectra is for two sets of peaks where each set has a value of two and each set is split into multiple lines. $ce{H^1}$ and $ce{H^3}$ are each contribute half in each of the multiple peaks in one of the sets.
– MaxW
Commented Dec 8, 2015 at 17:08
• @MaxW About that first word, feel free to edit in a better one. About the rest: I disagree. It’s not like we could say one half of the peak is $\ce{H^1}$ and the other half $\ce{H^3}$. Those two signals are overlaid together into a single peak much like the three protons of a methyl group give a single peak.
– Jan
Commented Dec 8, 2015 at 17:23
• @ron Thinking about it again, you are correct. I hope I have re-edited my post to reflect that.
– Jan
Commented Dec 8, 2015 at 17:43
• @MaxW If you take a macroscopic look at that spectrum, I see a pair of doublets. Each doublet is then further (and complexly) split as all aromatic signals are. At least two minor peaks around each main peak should be $\ce{^{13}C}$ satellites, too.
– Jan
Commented Dec 8, 2015 at 17:50