Methanol and DMSO solvent peaks

Two common deuterated NMR solvents, methanol and dimethyl sulfoxide, produce very similar, characteristic peaks with an unusual splitting pattern. What is the cause of these peaks?

The figure below shows the characteristic peaks of methanol-d₄ and dimethyl sulfoxide-d₆ observed in ¹H NMR spectra. Both peaks show five lines with intensities in the ratio 1:2:3:2:1. Both peaks are from methyl groups, with no other nearby protons, so what causes these characteristic peaks?

The first thing to remember is that these signals are observed in a ¹H spectrum so must be due to ¹H. Secondly, deuterated solvents are never completely deuterated. A small percentage (0.1 or 0.2%) of the hydrogens are ¹H instead of ²H. In methanol and DMSO this means that 0.1 or 0.2% of the methyl groups contain one ¹H atom and two ²H atoms, ie -CHD₂. There is an even tinier amount (0.1x0.1=0.01% or 0.2x0.2=0.04%) of methyl groups where two of the hydrogens are ¹H and one is ²H, but this is so small that it is not detectable. The pentets then are due to -CHD₂ groups, but what causes the unusual splitting pattern?

The splitting of the peaks is due to ²H coupling. Unlike ¹H, which is a spin ½ nucleus, ²H is a spin 1 nucleus. Coupling to a spin 1 nucleus produces three states, rather than the two produced by a spin ½ nucleus. The n+1 rule for determining the splitting of a resonance is a specific case for spin ½ nuclei. The more general rule is 2nI+1, where I is the spin state. With the general rule we can calculate that coupling to two spin 1 nuclei, like the two deteurons on -CHD₂, will produce a pentet, as observed.

The relative intensities of the lines in a multiplet are typically explained by Pascal's triangle, but Pascal's triangle does not produce the 1:2:3:2:1 ratio observed for the methanol-d₄ and dimethyl sulfoxide-d₆ peaks. Using Pascal's triangle to predict multiplet intensity patterns assumes that the splitting is due to a spin ½ nucleus. In the case of a spin 1 nucleus each state must be split into three instead of two. The figure below shows how two levels of splitting into three produces the intensity ratio 1:2:3:2:1.

Thus, the characteristic pentets of methanol-d₄ and dimethyl sulfoxide-d₆ are due to residual, partially deuterated solvent. ²H coupling creates a characteristic splitting pattern that cannot be produced by spin ½ nuclei.