# Why is the residual proton resonance in deuterated benzene a singlet?

In deuterated benzene ($99\%$) the residual protio peak is a singlet at $\pu{7.16 ppm}$. This however doesn't make sense since the protons will be randomly distributed among various benzene molecules. Therefore there won't be a mixture of $99$ $\ce{C6D6}$ molecules and $1$ $\ce{C6H6}$ but $94$ $\ce{C6D6}$ and $6$ $\ce{C6D5H}$. Considering that deuterium has a spin of $1$ the residual solvent peak in D6-benzene should be a septet. Any thoughts?

To a very good approximation, $$J_\ce{HD}$$ in a deuterated molecule is related to $$J_\ce{HH}$$ in the corresponding protonated molecule by the gyromagnetic ratios:

$$\frac{J_\ce{HD}}{J_\ce{HH}} = \frac{\gamma_\ce{D}}{\gamma_\ce{H}} = \frac{\pu{41.066 * 10^6 rad s-1 T-1}}{\pu{267.522 * 10^6 rad s-1 T-1}} = 0.15$$

In benzene where $$^3J_\ce{HH} = \pu{7.6 Hz}$$ and $$^4J_\ce{HH} = \pu{1.4 Hz}$$ (Reich NMR tables), we would thus have $$^3J_\ce{HD} \approx \pu{1.2 Hz}$$ and $$^4J_\ce{HD} \approx \pu{0.2 Hz}$$. The four-bond meta coupling would be all but invisible, and the three-bond ortho coupling if resolved would give rise to a $$1:2:3:2:1$$ quintet.

On a lower-field spectrometer this relatively small coupling is usually not properly resolved. A suitable window multiplication, though, will reveal all. I don't have a d6-benzene spectrum on hand, but I do have a $$\pu{500 MHz}$$ d8-toluene spectrum here. The two spectra on the left are the aromatic residual peaks, but at the bottom-left a Gaussian window function has been used to improve resolution (at the cost of signal:noise ratio). This allows us to resolve the three-bond ortho couplings of $$\pu{1.1 Hz}$$.

At the bottom-right is the signal corresponding to residual $$\ce{C6D5CD2H}$$. Since the two-bond H–D coupling has a larger magnitude ($$^2J_\ce{HD} = \pu{2.2 Hz}$$), it is much easier to resolve, and can be seen even with the normal exponential window function (which sacrifices resolution for signal:noise ratio improvement).

Similarly, in d6-DMSO (shown below at $$\pu{500 MHz}$$), the two-bond coupling is more easily resolved.