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I'm fairly new to NMR especially $\ce{^{19}F}$ and $\ce{^{31}P}$ NMR, we've not been formally taught NMR yet but we're expected to know $\ce{^{19}F}/\ce{^{31}P}$ NMR for our exam for basic molecules.

I know $\ce{PPh2F3}$ will have a trigonal bipyramidal geometry, I know in $\ce{PPh2F3}$ the phenyl groups will be equatorial due to the torsional angle and sterics. This leaves two fluorines axial and one fluorine equatorial, but what I don't understand is why this gives a doublet of triplets signal (for $\ce{^{31}P}$). Any explanation to why you get this pattern would be helpful, taking into account my fundamental understanding of $\ce{^{31}P}$ NMR is not so great.

also what would the $\ce{^{19}F}$ NMR be for the same molecule?

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    $\begingroup$ 1) Coupling to 1 equatorial fluorine gives 1+1 = 2 peaks 2) Coupling to 2 axial fluorines means that both of those two peaks is split into 2+1 = 3 peaks. Hence two triplets. $\endgroup$ – orthocresol Apr 30 '16 at 18:14
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    $\begingroup$ Technically it should be called a "doublet of triplets", not two triplets, because two triplets would imply that there are two different phosphorus environments (obviously not true since there's only one phosphorus). As for the 19F NMR, it's a bit more fiddly, but you just need to work through it slowly. There is F-P coupling across one bond as well as F-F coupling across two bonds, but the two axial fluorines, which are equivalent, do not couple to each other. $\endgroup$ – orthocresol Apr 30 '16 at 18:21
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    $\begingroup$ This earlier answer for an analogous compound may be helpful. $\endgroup$ – ron Apr 30 '16 at 18:30
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    $\begingroup$ The thing is that the ax. and eq. fluorines are not equivalent, so each of them will have their own peak, and both of these peaks will be split accordingly. $\endgroup$ – orthocresol Apr 30 '16 at 18:36
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    $\begingroup$ There's a significant typo in the first sentence 19F -> 19P. But SE is being an arse, as ever. Could someone with enough kudos to make a one character change fix it? $\endgroup$ – Dan Sheppard Apr 30 '16 at 18:57
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Well, these are the fundamentals of NMR that are important:

  • Signals will be split if they couple to neighbouring atoms with a spin $\ne 0$ (though coupling to spin ½ nuclei is most common especially in organic NMR).

  • Every different environment of a coupling nucleus will produce a distinct coupling.

  • Depending on the number of nuclei in a given environment, the number of signals (coupling pattern) is given by the equation $2nI+ 1$, where $n$ is the number of nuclei and $I$ is that nucleus’ spin.

In your example you have correctly identified two different fluorine environments with one and two fluorines in them respectively.

The single equatorial fluorine will give a doublet because $2nI +1 = 2\ (n = 1, I = \frac{1}{2})$.

For the same reason, the axial fluorines will give a triplet: $2nI + 1 = 3\ (n =2, I = \frac{1}{2})$.

If one coupling gives a triplet and the other a doublet, the final result is either a doublet of triplets or a triplet of doublets. (The larger coupling constant i.e. the greater splitting is given first.)


The fluorine NMR would be more complex: They can couple to both phosphorus and the other fluorine. It is important to know that the axial fluorines are homotopic, i.e. the molecule can be rotated exactly exchanging their places — this means that the two give an identical signal. The equatorial one cannot be brought to overlap with the axial fluorines by any mean of symmetry; the two environments are heterotopic, i.e. they will produce different signals (and typically couple).

  • The axial fluorines will couple to the equatorial fluorine and phosphorus.

  • The equatorial fluorine will couple to the axial ones and phosphorus.

I’ll leave you to work out the coupling patterns. (Hint: The less bonds between two coupling nuclei, the larger the coupling constant usually is.)


A final note: It is due to electronic reasons that the phenyl groups occupy the equatorial positions rather than going axial. A four-electron three-centre bond is more stable for electronegative elements than electropositive ones.

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  • $\begingroup$ This is very helpful. Okay so if the axial fluorines couple to the equatorial fluorines you will get a triplet (2+1) and then couple to the phosphorus will split this into two triplets? The equatorial fluorine will couple to the axial fluorines and give a doublet (1+1) and which will then subsequently couple with phosphorus and split into two doublets. Overall giving a doublet of triplets and doublet of doublets? $\endgroup$ – Petrichorr Apr 30 '16 at 18:50
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    $\begingroup$ @Petrichorr Yes, a doublet of doublets (the axial fluorines) and a doublet of triplets (the equatorial one). If the ppm scale works like it does for carbon and hydrogen, then the doublet of doublets should be shifted downfield when compared to the doublet of triplets. $\endgroup$ – Jan Apr 30 '16 at 18:52
  • $\begingroup$ What do you mean by shifted downfield? That the peaks will be of greater intensity? $\endgroup$ – Petrichorr Apr 30 '16 at 18:53
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    $\begingroup$ No, have a lower ppm value @Petri ;) $\endgroup$ – Jan Apr 30 '16 at 19:13

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