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The image below is from a Nilered video How to extract Vanillin from Vanilla Sugar, where he displays the $\ce{^1H}$-$\mathrm{NMR}$ of the final product created:

1H-NMR of vanillin

The labels A, F, & E make sense to me, but I'm confused about D and C - I feel they should be swapped. To my understanding, B and D share a more similar chemical environment (though not identical), specifically they are near a $\ce{C-O}$ bond, moving their chemical shift to left. Also, the NMR peaks labelled D display a doublet, suggesting it is next to a single proton. Though B, C, and D are all next to a single proton, C is the odd one out.

Is this a case of me missing something, or is the picture mislabeled?

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    $\begingroup$ Assignment looks correct to me; don't forget the effect of the carbonyl group. Not quite sure what you're trying to say with C being the 'odd one out'. It's hard to see any multiplet structure in the screenshot, especially when two of the protons are overlapping with each other. But ideally, C should be a doublet of doublets as it is coupled to both B and D. On the other hand, B and D should both be doublets, and the D doublet should have a larger separation (coupling constant) than the B doublet. $\endgroup$
    – orthocresol
    Jul 29, 2019 at 17:49
  • $\begingroup$ Thanks for the answer. A few follow up questions then: 1) why is C coupled to both B & D (and not vice versa)? Does it have something to do with the benzene ring? You mention the carbonyl group - but what effect does that have? It appears far away from D/C/B, no? $\endgroup$
    – Cam.Davidson.Pilon
    Jul 29, 2019 at 18:22
  • $\begingroup$ Hm, thinking more about your carbonyl comment: is it because it is a meta-directing group and will change the environment of D. Likewise, -OH is meta-directing and again will change the environment of D. I still don't understand why C should be a doublet of doublets. $\endgroup$
    – Cam.Davidson.Pilon
    Jul 29, 2019 at 18:33
  • $\begingroup$ You may want to go over this with a teacher if you have one. There are several problems with your comments. (1) B and D are both coupled to C, that's exactly why they appear as doublets. (2) Whether a group is o/p- or m-directing is not of any consequence here. They affect all protons on the ring, albeit not to the same extent (it depends on the group). (3) OH is most certainly not m-directing. $\endgroup$
    – orthocresol
    Jul 29, 2019 at 18:35
  • $\begingroup$ Right, I meant ortho-para for -OH. AFAIU, coupling occurs between protons on neighbouring carbons, but B is not neighbouring to C. I'm back to square one: what effect does the carbonyl have on the NMR? $\endgroup$
    – Cam.Davidson.Pilon
    Jul 29, 2019 at 18:52

2 Answers 2

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"Normal" coupling is vicinal coupling across 3 bonds such as $\ce{H-C-C-H}$. For aromatics, ortho-coupling is of this type. But one can also get other couplings such as long range coupling, e.g., meta-coupling across 4 bonds. Long range coupling tends to be smaller than vicinal coupling (smaller $J$, the coupling constant). Thus C/D coupling is stronger than B/C coupling. And remember coupling must work both ways C to D, and D to C. The problem here is that B and C are overlapped so tough to see their multiplicities. C has higher shift than D due to position relative to the substituents.

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[...] but I'm confused about D and C - I feel they should be swapped. To my understanding, B and D share a more similar chemical environment (though not identical), specifically they are near a C-O bond, moving their chemical shift left. Also, the NMR peaks labelled D display a doublet, suggesting it is next to a single proton. Though B, C and D are all next to a single proton, C is the odd one out.

First, in this statement, your assignment of "[...] B, C and D are all next to a single proton," is not correct. Clearly, B is not next to a single proton:

NMR of Vanilin

Now, after said that, your confusion is due to your ignorance of the electron withdrawing nature of aldehyde group $(\ce{CHO})$. That group is ortho to both $\ce{H_B}$ and $\ce{H_C}$ while meta to $\ce{H_D}$. Also, it is para to the electron donating $\ce{OH}$ group. Because they are para to each other, their effect may be optimized on each other. Regardless, $\ce{H_D}$ experiences significant electron density due to the electron donating $(\ce{OH})$ group at ortho position, while $\ce{H_C}$ (and $\ce{H_B}$) experiences significantly lesser electron density. When compare electron donating ability of $\ce{OH}$ group and $\ce{OCH3}$ group, it is safe to say that $\ce{OH}$ group has been the superior (compare $\sigma_{o,p}$ values of $\ce{OH}$ and $\ce{OCH3}$ groups). Since both $\ce{H_B}$ and $\ce{H_C}$ would experience significant electron withdrawing from $(\ce{CHO})$ group ortho to them in addition to experiencing aforementioned electron density from $(\ce{OCH3})$ group. Keep in mind that since $(\ce{CHO})$ group is meta to $\ce{H_D}$, it makes less effective on deshielding. Thus, based on these facts along, one can say safely that $\ce{H_B}$ and $\ce{H_D}$ as well as $\ce{H_C}$ and $\ce{H_D}$ are significantly different chemically and magnetically. Therefore, their chemical shifts should be vastly different as assigned.

When compare the effects of substituents on $\ce{H_B}$ and $\ce{H_C}$, one can also say that they are very similar magnetically due to the effects of meta $\ce{CHO}$ group and ortho, para $\ce{OCH3}$ group (ortho to $\ce{H_B}$ and para to $\ce{H_C}$). The only difference is the $+I$ effect of $\ce{OCH3}$ group on each proton (with pata being the minimal). However, seemingly that difference is not been very significant in this case (again, compare $\sigma_{o}$ and $\sigma_{p}$ values of $\ce{OCH3}$ group for comparison). Thus, one can say that the chemical shifts of $\ce{H_B}$ and $\ce{H_C}$ should be very similar as assigned.

Note: The resonance of $\ce{H_D}$ should be a clear doublet since it does not have any meta-protons to have long-range couplings (usually $\pu{1-2 ppm}$). Yet, $\ce{H_B}$ and $\ce{H_C}$ are meta to each other and show multiplicity.

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