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:



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> 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.
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(Quoted from [Hans Reich’s NMR course online syllabus](http://www.chem.wisc.edu/areas/reich/nmr/05-hmr-08-symmetry.htm#05-hmr-08-symmetry-magnequiv); emphasis present in the original.)


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


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

**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 $\ce{H^4}$.




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‡: 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.


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