Carbon-13 NMR for chloroform

Question

I am slightly confused by what the spectrum would show for carbon-13 NMR of $\ce{CHCl3}$.

My initial guess would be that the peak would be split by coupling to both the proton and the 3 chlorines, as both nuclei have a net spin.

If the peak were split by the chlorine only, then as there are three chlorine atoms we would get a quartet peak. However this cannot be correct because the spectrum for CDCl3 shown here has only three peaks:

I also don't understand here why the areas of the three peaks are the same. Should they not be in some other ratio (for the four peaks I expected it would be $1:3:3:1$ are ratio)?

Then I would expect that the presence of the proton would split every one of the four peaks from $\ce{C-Cl}$ coupling further into a doublet, giving a quartet of doublets.

I didn't find any spectra showing this, only showing very tiny peaks for $\ce{CHCl3}$ compared with $\ce{CDCl3}$ which I don't understand:

To make it more confusing, my lecture notes say that for chloroform, $\ce{CHCl3}$,

each $\ce{^13C}$ is attached to a spin $1/2$ proton so unless we applied broadband proton decoupling, coupling to the proton would mean that the $\ce{^13C}$ signal would appear as a doublet.

And it makes no mention of the effect of the chlorine atoms on the spectrum.

I would be very grateful is someone could expalin what the spectrum for carbon-13 NMR of chloroform, $\ce{CHCl3}$, would actually look like!

Answer 1

You assumed that coupling to the three chlorides would yield some type of quartet. This is correct in principle. However, chlorine is one of the many quadrupole nuclei that are basically unobservable by NMR due to their rapid relaxation. I hope another answer is around explaining that as I am not good at it.

I can, however, help you interpret the $1:1:1$ triplet of your spectrum. It is not, as you have assumed, the signal for $\ce{CHCl3}$ but that for $\ce{CDCl3}$. The difference between the two in NMR terms is that chloroform contains the spin ½ nucleus protium while deuterochloroform contains the spin 1 nucleus deuterium — and the latter does not relax rapidly as the chlorine isotopes do. Hence, deuterium is well-observable by NMR.

Now your first thought should be something like the following:

But I am only coupling to one nucleus. That should give me a $1:1$ doublet and not a $1:1:1$ triplet, shouldn’t it?

That simplification is only correct for spin ½ nuclei. The actual formula for calculating the multiplicity of a peak is $2\,n\,I + 1$, where $n$ is the number of chemically equivalent nuclei your observed nucleus is coupling to and $I$ is the spin of the coupling nuclei. Since deuterium is spin 1 ($I = 1$), the formula gives us:

$$2\,n\,I + 1 = 2 \times 1 \times 1 + 1 = 3$$

Or an expected triplet. Since this triplet does not derive from two equivalent couplings to spin ½ nuclei, it is not a $1:2:1$ triplet but a $1:1:1$ one.

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Your second spectrum shows a $\ce{^13C}$ NMR of deuterochloroform which still contains residues of chloroform — at least, that is what it is supposed to show. For some reason unknown to me, the protochloroform signal has not been decoupled, i.e. the spectrum is in fact $\ce{^13C}$ and not $\ce{^13C\{^1H\}}$. Thus, you are observing a strong coupling to the single hydrogen nucleus giving a $1:1$ doublet as expected if chlorine is ignored.