At low temperature, the rotation around the N-C amide bond of dimethylformamide is slow, so the two methyl groups are in different chemical environments. The H NMR thus shows a peak for each of the methyl groups.

As the T is increased, we see that the peaks begin to coalesce, until eventually they coalesce into a single average peak. The basic explanation is that, since the methyl groups are exchanging their chemical environment at a rate faster than the NMR time scale, they are considered to be within the same average chemical environment.

I am not very comfortable with this explanation, because it could also be argued that 50% of the time the methyl will be emitting signal A, and 50% of the time it will be emitting signal B - and the FID signal should still contain two signals. Alternatively, if the rotation is continuous, then I could expect a very broad peak extending from A to B.

My guess is that there is something about the structure of the signal and the Fourrier transform that I do not understand. Could someone help me out understanding why this coalescence occurs?

Thank you!

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  • $\begingroup$ Thank you Mithoron. The answer to that question provides the basic answer, and I am hoping I can understand the why a bit deeper :-) $\endgroup$
    – Max
    Commented Mar 26, 2017 at 22:02
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    $\begingroup$ Yeah, this could be treated as sort of follow-up to DMF question. $\endgroup$
    – Mithoron
    Commented Mar 26, 2017 at 22:04
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    $\begingroup$ Do you want a mathematical answer, or would just a qualitative one suffice? The books listed in this thread (see Physical chemistry > NMR) cover the topic of chemical exchange very comprehensively. Keeler is more readable, Levitt and Gunther have the maths to back it up (QM treatment and Bloch equations respectively). $\endgroup$ Commented Mar 26, 2017 at 22:29
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    $\begingroup$ Levitt (Spin Dynamics) p 488-495 explains very clearly how when the molecule jumps between two sites the effect this has on the FID is to cause the two signals become broadened , merged & then narrowed to a single peak as the exchange rate constant is increased. There is also a mathematical treatment if you wish to follow that. $\endgroup$
    – porphyrin
    Commented Mar 27, 2017 at 9:11

1 Answer 1


The basic explanation is quite simple: a spectrum reports the rate of precession of a collection (ensemble) of nuclear spins. Precession can be pictured as orbital motion about a common axis, like earth rotating about the sun (merely for purposes of illustration). Now assume you have two groups of spins (equal number in each), spins in each group precessing at a fixed rate, that is, spins in each group precess coherently. Since the NMR signal is an average over all of the spins, and you have two populations with distinct precession rates, you see two independent peaks in the spectrum.

If on the other hand there is fast exchange (fast to be defined later) between the spins in the two populations, each spin ends up with an average precession rate, the mean of the starting values of the populations. In that case, the NMR spectrum contains only one peak. The spins have been mixed and each population is no longer pure, the original populations have lost coherence.

In the intermediate regime, when exchange rate is not so fast that the populations are thoroughly averaged, the spins will inhabit mostly one population but their rates will deviate from the pure initial rate due to sampling of the rate of the other population. This jumping between populations leads to a loss of coherence, and the NMR spectrum will show broadened peaks. These eventually coalesce into a broad central peak with a precession rate equal to the average over the two starting populations. The condition for coalescence (fast exchange) is that the exchange between populations happen rapidly relative to the precession rate about the axis, thereby blurring the precession rate of individual spins.


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