Why is there a triplet at 46.7 ppm in the following spectrum of 2,2,2-tribromoethan-1-ol?

  • $\begingroup$ The spectrum is of really poor quality. We can’t even see if the triplet is a $1:2:1$ or $1:1:1$ one. From the other (poor quality) spectra I can see on SciFinder, it looks as though that should be a singlet — but I can’t see too much on $75~\mathrm{MHz}$ spectra at that resolution. Finally, it seems weird that both signals should be of similar height even though one is quarternary and the other is secondary … $\endgroup$ – Jan Jun 20 '16 at 18:50
  • $\begingroup$ I bet this is just a typo on the website. $\endgroup$ – jerepierre Jun 20 '16 at 22:39

So, there actually isn't a triplet in the spectrum shown. This is a 13C{1H}, and so proton coupling to carbons is not visible. Hence, the two peaks at 79.9 and 46.7 ppm are singlets. If you read the question carefully, on the left hand side it indicates what the multiplicity of that peak is i.e. how that peak would appear in the absence of proton decoupling. The peak at 46.7ppm, in this case, would appear as a triplet. This information helps you work out how many protons are attached to a particular carbon centre.

So for this example, one of your carbon centres will have no protons attached, and the other will have 2 protons attached, CH2. A CH2 carbon will split into a triplet, due to the normal 2nI+1 rule.

Multiplicities used to be determined by running off-resonance decoupled experiments, so that multiplicity could be seen clearly for 1J coupling. These days, multiplicity would usually be determined via a DEPT experiment, or preferably a phase edited HSQC.

Of course, that peak at ~77ppm is a 1:1:1 triplet, and that is because it is CDCl3, and a proton-decoupled carbon experiment won't remove coupling to deuterium, and so the chloroform carbon splits into 3 lines, again due to the 2nI+1 rule (2H has spin=1).

Important Edit I've just had a quick look at the solution on your page link, and it actually appears that there is an error in their assignment of the multiplicities. The -CBr3 peak will come at 46.7ppm, and would be a singlet. The -CH2OH peak wil come at 79.9ppm, and this would be a triplet. They have these two multiplicities switched on the question page.

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  • $\begingroup$ No, for some reason the website agrees that $46.7~\mathrm{ppm}$ is the $\ce{CBr3}$ group. Which would make sense if that is just a singlet. If you click on solution, it gives you the correct assignment. $\endgroup$ – Jan Jun 21 '16 at 20:43
  • $\begingroup$ @Jan - yes, this is what I mean. The error is on the question page, with the multiplicities. The solution page is quite correct. $\endgroup$ – long Jun 21 '16 at 22:15

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