The $\ce{^{31}P}$ NMR spectrum of $\ce{Me2NPF4}$ when recorded at room temperature consists of a quintet with intensities $1:4:6:4:1$. However, at $-100~\mathrm{^\circ C}$, the spectrum is as shown below:

NMR spectrum at low temperature

Propose a structure for $\ce{Me2NPF4}$ that is consistent with these observations.

The phosphorus has five groups surrounding it so it must have some sort of five-coordinate geometry. The low temperature spectrum shows a triplet of triplets with similar coupling constants, which is what you would expect from a trigonal bipyramidal structure with two axial and two equatorial fluorines. On this basis I assigned the structure as below:

Structure of Me2NPF4

However, I don't understand why all four fluorines are equivalent at higher temperatures. Does this mean that they are constantly interchanging and if so, what is the mechanism for this?

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    $\begingroup$ en.wikipedia.org/wiki/Berry_mechanism $\endgroup$
    – permeakra
    Commented Jan 2, 2016 at 18:14
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    $\begingroup$ See "Suggest possible structures for the compound PF3(Ph)(NMe2) using the NMR data" for some background and follow the link to Berry pseudorotation. $\endgroup$
    – ron
    Commented Jan 2, 2016 at 18:18
  • $\begingroup$ What's really bothering me about this NMR spectrum is the lack of $\ce{H-P}$ coupling. I made a number of phosphonate compounds of the general type $\ce{RP(O)(OCH3)2}$ for my PhD, and the methoxy groups showed $^3 J_{\ce{H-P}} \approx 10-20\ \text{Hz}$. In just about any structure I could propose for $\ce{Me2NPF4}$, there should be some $\ce{H-P}$ coupling. $\endgroup$
    – Ben Norris
    Commented Jan 2, 2016 at 23:50
  • $\begingroup$ @BenNorris The lack of $\ce{H-P}$ coupling is due to quadrupolar relaxation by the intervening nitrogen. See my comment to Jan's answer for more detail. $\endgroup$
    – ron
    Commented Jan 3, 2016 at 10:18

1 Answer 1


Obviously, in the structure you drew you have two sets of two fluorines which are non-identical: The axial ones that form a 4-electron-3-centre bond with phosphorus and the equatorial ones that are bound by the classical 2-electron-2-centre bond. And also obviously, at higher temperatures this difference vanishes so the fluorines must somehow interconvert into each other (that is the easier explanation when compared to a different structure and is thus preferred by Occam’s razor) rapidly on the NMR-timescale. You noticed this already.

The mechanism for this is called Berry pseudorotation, and it involes a square pyramidal transition state as shown in this image taken from the Wikipedia page:


(The image is drawn with pentacarbonyliron but just exchange iron with phosphorus, four carbonyl groups with fluorine and the fifth with dimethylamine and voilà.)

Since this mechanism involves a significant change in the binding situation (we no longer can write two clearly 2-e-2-c and one 4-e-3-c bond(s)), we need a certain activation energy — that is usually supplied thermally. At $-100~\mathrm{^\circ C}$, there is not enough thermal energy to induce this change; at room temperature there is.

Ben Norris mentioned another important observation in a comment: There is not $^3J_\ce{PH}$ coupling visible. I suspect that this is due to the spectrum actually being $\ce{^{31}P\{^1H\}}$ — a proton-decoupled phosphorus spectrum.

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    $\begingroup$ The absence of $^3J_\ce{PH}$ coupling is due to the fact that the coupling path involves the nitrogen and $\ce{^14N}$ is a quadrupolar nucleus - quadrupolar relaxation effectively washes out the coupling. Jan, I suspect you understand this, but for others who might like to read a bit more about quadrupolar relaxation and when it will and will not effectively decouple nuclei, here are links to two earlier answers on the subject (ref_1, $\endgroup$
    – ron
    Commented Jan 3, 2016 at 10:14
  • $\begingroup$ ref_2) $\endgroup$
    – ron
    Commented Jan 3, 2016 at 10:15
  • $\begingroup$ @ron Actually, I didn’t know about that. I hope I remember to read your references soon (kinda bad time atm) and then hopefully correct my answer accordingly. Thanks for pointing out! $\endgroup$
    – Jan
    Commented Jan 3, 2016 at 15:45

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