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During my PhD, I prepared the two compounds below (there was actually a slightly longer chain but it's not inherently relevant).

When looking at the 1H NMR spectra of these compounds, something slightly weird occurred.

In one of them, the CH2 group of the benzyl appeared as a singlet (s, 2H). In the other, the CH2 group of the benzyl was observed as a multiplet (an AB quartet like structure, to be specific).

Is there any rationale for why the two cases would be different, given the distal nature of the closest stereocentre (i.e. why are the two CH2 protons diastereotopic to begin with given how distal they are)?

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    $\begingroup$ Is the left compound the singlet and the right compound the multiplet? I drew out some chair conformations, and if I'm on the right track, then hopefully, I've correctly identified which is which. $\endgroup$
    – Zhe
    Commented Aug 2, 2018 at 1:40
  • $\begingroup$ Yes. The 1,2-anti 1,4-anti is a singlet and the 1,2-syn 1,4-syn is a multiplet $\endgroup$
    – NotEvans.
    Commented Aug 2, 2018 at 7:02
  • $\begingroup$ They should both be multiplets due to coupling with the unequivalent aryl protons and between each other. Were both of them run on the same instrument? Was the singlet a bit too broad? The CH2 protons are diastereotopic since there is no plane of symmetry or Cn rotation that can relate them due to the asymmetry of the right part of the molecule. $\endgroup$
    – Outlander
    Commented Aug 2, 2018 at 8:31
  • $\begingroup$ @AMM I think he's saying that the chiral environment that results in those protons being diastereotopic is very far away. $\endgroup$
    – Zhe
    Commented Aug 2, 2018 at 12:27
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    $\begingroup$ Well, it has to be an apparent singlet. It's diastereotopic like you said... But the fact that's under consideration is that it is an apparent singlet rather than not. $\endgroup$
    – Zhe
    Commented Aug 2, 2018 at 14:31

1 Answer 1

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This is a bit speculative. I'm still thinking about it, and maybe this shouldn't even been an answer. I drew out the left structure and the enantiomer of the right structure because I want the configuration at the carbon with the OMOM group to be the same.

structures

I think you're calling these the 1,2-anti-1,4-anti and 1,2-syn-1,4-syn, respectively. The main focus here was on the conformation about the 1,2-carbon-carbon axis. I put the two hydrogens anti to each other. In the anti-anti case, the alkyl chain extends off away from the benzyl moiety while in the syn-syn case, the alkyl chain is pointed more towrads the benzyl moiety.

I'm not in love with this explanation. It's a bit speculative, but I used the tools I have and at least, it's consistent with your observation about splitting.

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  • $\begingroup$ Nice rationalisation! I set some calculations going to see what the low energy conformations in chloroform look like, will update to let you know if they match your predictions $\endgroup$
    – NotEvans.
    Commented Aug 3, 2018 at 7:54
  • $\begingroup$ I dont understand this one guys, in both structures you will have rotation around the CH-CH bond where the grey protons are drawn (and all other bonds as well). Therefore in both cases the alkyl chain will spend the same amount of time close to the aryl for both structures drawn. $\endgroup$
    – Outlander
    Commented Aug 3, 2018 at 8:09
  • $\begingroup$ @AMM That's only true if all of the conformations are equally likely. The issue is that there is enough substitution on the 1- and 2- carbons that you might have a strong conformational bias. Again, this is speculative at best, so the real answer may be something else. It's also likely solvent dependent, so if the spectra were taken in a different solvent, I suspect you might not get the same results. $\endgroup$
    – Zhe
    Commented Aug 3, 2018 at 13:34

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