# Suggest possible structures for the compound PF3(Ph)(NMe2) using the NMR data

use these data to suggest the possible structure for the compound PF3(Ph)(NMe2):

19F Spectrum, two signals:
a doublet of doublets, J=56 and 819Hz, -39.5ppm, relative intensity 2
a doublet of triplets J=959 and 56Hz, -68.3ppm, relative intensity 1

31P Spectrum, one signal:
a doublet of triplets, J=959 and 819Hz, -53.0ppm

part b,
suggest an experiment or measurement that might be helpful to discern possible isomers of the compound

There are two types of structure that a Phosphorus with five bonds could be, trigonal bipyramidal and square based pyramid. I think that because there are two equivalent fluorines that this must be the trigonal bipyramidal structure as it had the two axial positions that would be equivalent.

• What causes the 31P spectrum to give a double of triplets? I thought only adjacent phosphorus atoms with a different chemical shift would cause splitting. Also, the 19F suggests that you have 3 different environments for the fluorine atoms since you have a doublet of doublets instead of a single doublet. – LDC3 Nov 2 '14 at 15:41
• I understood it as there is coupling between the P and F. So if you look at the coupling constants each peak for F is coupling with each other and the Phosphorous. – user5181 Nov 2 '14 at 15:50
• That's correct. – ron Nov 2 '14 at 18:48

I agree that the trigonal bipyramid structure is what we are dealing with, so let me focus on that structure.

The nmr data shows that 2 of the fluorines are equivalent and one is different. This results in 3 possible isomers as shown below.

Both $\ce{^31P}$ and $\ce{^19F}$ are spin 1/2 nuclei, so each of these isomers would produce phosphorous and fluorine nmr spectra with splitting patterns and integration ratios consistent with the reported data. That is, the phosphorous will split each different fluorine into a doublet, the two equivalent fluorines will split the singular fluorine's doublet lines further into triplets. The one non-equivalent fluorine will split the other two equivalent fluorine doublet lines into doublets. Similar analysis of the phosphorous signal leads to the expectation of a doublet (from the singular fluorine) of triplets (from the two equivalent fluorines).

2. Because the orbitals involved in axial and equatorial bonding in pentacoordinate phosphorous compounds are different, the axial and equatorial coupling constants will be different. Generally, $\ce{J_{PF(eq)}}$ > $\ce{J_{PF(ax)}}$ because there is increased s-orbital density in the equatorial bond (see here, for example)
The larger reported P-F coupling constant (959 Hz) is associated with the intensity = 1 fluorine signal. This suggests that structure I, with one equatorial fluorine (with the larger $\ce{P-F}$ coupling constant) and 2 axial fluorines (with the smaller coupling constant and as preferred electronically - point 1 above) is the structure being observed.