# NMR magnetically equivalent protons for a 1,4-disubstituted benzene ring

Below I have drawn a molecule with random substituents $$\ce{X}$$ and $$\ce{Y}$$:

If the distances from $$\ce{H_b}$$ and $$\ce{H_a}$$ to $$\ce{H_d}$$ are considered, they are different - $$\ce{H_b}$$ is 5 bonds away and $$\ce{H_a}$$ is 3 bonds away, so $$\ce{H_b}$$ sees $$\ce{H_d}$$ differently compared to $$\ce{H_c}$$. The same argument can be made vice versa and also between the same pair of protons towards $$\ce{H_c}$$. This was what the lecturer told me regarding this compound below.

This led me to believe then that all the protons in the molecule are magnetically inequivalent and using up to $$\ce{^4J}$$ coupling, all protons show a doublet of doublet.

However, when I was in school, I was told that if symmetry was a property of a molecule, as above, $$\ce{H_a}$$ and $$\ce{H_b}$$ were equivalent, as are $$\ce{H_c}$$ and $$\ce{H_d}$$, and each pair of protons gives a doublet. Researching this matter on Google still held true in my undergraduate studies.

What am I getting wrong here?

I read a quote on another question highlighted here

For two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other non-chemical shift equivalent nuclei in the molecule.

So as $$\ce{H_a}$$ and $$\ce{H_b}$$ are chemically equivalent but $$\ce{H_b}$$ couples differently to $$\ce{H_c}$$ than $$\ce{H_a}$$ does, that therefore means that surely $$\ce{H_a}$$ and $$\ce{H_b}$$ are magnetically inequivalent?

## 1 Answer

when I was in school, I was told that [...] each pair of [equivalent] protons gives a doublet

This is only strictly true for magnetically equivalent protons. If there is magnetic inequivalence, such as in the p-disubstituted benzene ring, then it is no longer true.

It's not as simple as a dd either; when you have magnetically inequivalent protons that see each other, the first-order rules (the so-called $$n+1$$ rule) doesn't work so well anymore. You can see some real-life examples of how these molecules look like here: https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-15-AABB.htm For the p-disubstituted benzene it sort of looks like a doublet if you zoom out far enough, but if you look closely, there are additional peaks flanking the major peaks.

I say "see each other" because if you think about it, the two protons labelled HA above are also magnetically inequivalent; but since they are so far apart, the two benzene rings are essentially independent of each other, and give rise to the same NMR spectrum.

• Ah right that makes sense then. So if I considered $\ce{^4J}$ coupling as I mentioned under first-order rules (because this is being examined soon) I would be correct given the information I know of at this moment in time in my studies? – vik1245 Jan 1 at 2:41
• I'm not sure what answer to give you: on one hand, I could tell you that technically, given what you know at this stage, you could claim that it's a dd. It is certainly a good guess, because you can't realistically be expected to produce the second-order spectrum shown above. But that's not in line with reality, and I don't think it makes sense to pretend that it's right and then unlearn it later... – orthocresol Jan 1 at 2:43
• I understand - I think I am covering the disadvantages of first-order interpretations later on so I will tick your answer anyway but thanks so much for the help! I have read your content but I will wait for the lecturer to offer his clarification before proceeding onwards! – vik1245 Jan 1 at 2:45
• No problem. Don't be too overwhelmed by the second-order spectra when you get to them; there isn't all that much rhyme or reason to them. It suffices to know that they are, well, weird. (There are equations that describe them, and the website I linked explains it very well, but in practice it's... not very useful.) – orthocresol Jan 1 at 3:05