What exchange timescales are needed to cause decoupling in NMR?

Question

I'll try to explain the source of my confusion as clear as possible:

(*) The methyl protons in toluene are decoupled because the C-C bond between the methyl group and benzene allows very fast rotation. Hence, on average, these protons will see the same environment.

It is said that the time scale for the rotation must be faster than the time scale of the instrument. For a 250 MHz apparatus this would be (1/250MHz)=4 nanoseconds.

(*) In impure solvent the methylene protons of ethanol don't couple to -OH, since fast exchange of the OH protons results in decoupling and the protons will see an average environment.

Here it is said that this occurs if the timescale of exchange must be faster than the coupling constant J for the coupling. This is typically around 5-10 Hz. So in this case, if the exchange is in the 100 millisecond timescale decoupling already occurs.

This utterly confuses me, in the first case averaging requires nanosecond timescales (related to the frequency of the instrument) and in the second case almost second timescales (related to the small difference in frequency between coupled peaks)? Why?

ALSO and maybe related to this, why do we see in the spectrum of ethanol peak broadening for the OH proton, but not for the methylene group?

Answer 1

You need to differentiate what gets averaged, a CS difference or a scalar coupling.

Methyl protons have a different chemical shift depending on the current conformation, but they rotate so fast this gets averaged out perfectly, and they have a very sharp signal. They show no coupling between themselves because it´s an $AB_2$ system with CS difference zero, so $A_3$ actually.

$J$ coupling to a methyl group (e.g. the ethyl group in ethanol) gives sharp lines and clear splitting, because the coupling varies very rapidly (with the methyl rotation) around a stable and finite value, and it only depends on the occupation of the spin states in both groups.

The coupling methylene to OH is inverted every time the proton gets exchanged for one with different spin state, so it averages to zero. The exchange happens so often that the methylene group still gives a sharp quartett from the methyl.

The OH group coalesces with the water peak. Because the exchange happens randomly and differently for each molecule, the coherence in your spin ensemble deteriorates, hence the line broadening. Note: It´s not a splitting that get´s averaged out, but a change in chemical shift.

If the proton exchange was even faster, the line would sharpen again at some point I guess. Just like the methyl protons themselves. If the exchange was even slower (milliseconds), the line would broaden so much it became invisible. If you have a really dry sample, the exchange becomes so slow it does not happen on the NMR time scale, the line becomes sharp, and you even see the $^3J$ coupling to the methylene group.