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I have been learning about magnetic inequivalence today, and I find it quite hard to undersand. See, for example:

para-nitrophenol

In this case, according to the book I am using, $\ce{H_\mathrm{a}}$ and $\ce{H_\mathrm{b}}$ may be chemical equivalents, but they are not magnetic equivalents, due to the fact that $J_\mathrm{{\ce{a-c}}}$ and $J_\mathrm{{\ce{b-c}}}$ are different. Which I understand, but should we not consider that there is a $J_\mathrm{{\ce{b-d}}}$ which is equal to $J_\mathrm{{\ce{a-c}}}$? I do not understand why they are magnetically different, but both have to chemical equivalent protons, one in ortho and one in para position.

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If, as you say, you understand that Ja-c is different to Jb-c, then you should be able to rationalise that A is different to B. Therefore A will see B as being different to itself and there will be observable coupling between the two. That is the crux of magnetic non-equivalence; not so much as having different coupling to other nuclei around them, but being different to each other and therefore manifesting coupling between themselves.

As always, I'd recommend reading Reich's online NMR resource (symmetry section) to learn more.

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The problem is that from the perspective of $\ce{H_{d}}$ and $\ce{H_{c}}$, $\ce{H_{a}}$ and $\ce{H_{b}}$ are different (you can tell which is which by their coupling constants). This difference means that there is a magnetic inequivalence between the two hydrogens $\ce{H_{a}}$ and $\ce{H_{b}}$, so there will be coupling between them.

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