Why Does the 31P NMR Spectrum of PF5 Depend on Temperature?

[Astudious]

Why is the ³¹P NMR spectrum of PF₅ temperature-dependent?

PF₅ has two different F environments (3 F, 2 F) and ¹⁹F has spin 1/2 so we'll be getting a quartet and a triplet with some overlap I guess. (Also not sure about peak intensities or if we can even predict them) So what changes, and how and why, when the temperature is altered - chemical shifts, coupling constants?

 

[Borek]

Not that I know the answer, but the obvious line of attack seems to be checking if the molecule geometry is not (somehow) temperature dependent.

 

[TeethWhitener]

Not that I know the answer, but the obvious line of attack seems to be checking if the molecule geometry is not (somehow) temperature dependent.

This.

Look up Berry pseudorotation for a good idea of what's going on here. The equatorial and axial fluorides in PF₅ can exchange positions, and the rate of exchange is temperature-dependent. The timescale that an NMR experiment probes is on the order of a microsecond or so. At low temperatures, the equatorial-axial exchange is slower (on average) than a microsecond, so the equatorial and axial fluorides are inequivalent on the NMR timescsale and can be resolved as separate peaks. At higher temperature, the exchange is faster and the equatorial and axial flourides become equivalent over the timescale of the NMR experiment, so you only get one line for the fluorides. Incidentally, this is the same reason why the protons on free methyl groups can be treated as equivalent: at room temperature, the methyls are essentially freely and rapidly rotating on the timescale of the NMR experiment. However, if for some reason there's hindered rotation for the methyl, you can get (temperature-dependent) resolved individual peaks for each proton of the methyl group.

 

[Astudious]

This.

Look up Berry pseudorotation for a good idea of what's going on here. The equatorial and axial fluorides in PF₅ can exchange positions, and the rate of exchange is temperature-dependent. The timescale that an NMR experiment probes is on the order of a microsecond or so. At low temperatures, the equatorial-axial exchange is slower (on average) than a microsecond, so the equatorial and axial fluorides are inequivalent on the NMR timescsale and can be resolved as separate peaks. At higher temperature, the exchange is faster and the equatorial and axial flourides become equivalent over the timescale of the NMR experiment, so you only get one line for the fluorides.

Thanks

Do coupling constants and shifts in general depend on temperature, or is it just relaxation time?

Incidentally, this is the same reason why the protons on free methyl groups can be treated as equivalent: at room temperature, the methyls are essentially freely and rapidly rotating on the timescale of the NMR experiment. However, if for some reason there's hindered rotation for the methyl, you can get (temperature-dependent) resolved individual peaks for each proton of the methyl group.

Wouldn't the protons have equivalent environment anyway due to symmetry?

 

[TeethWhitener]

Do coupling constants and shifts in general depend on temperature, or is it just relaxation time?

Chemical shift is dependent on the electronic environment around the nucleus, so if that changes with temperature, then the chemical shift will change as well. If there's a temperature induced structural change in a material, the chemical shifts will change. In the case of PF₅, what's going on is a little more complicated, since at high temperatures the chemical shift is an average of equatorial and axial shifts. As for J-coupling, I'm not aware of any specific situation where this changes with temperature, but I don't want to say it can never happen.

Wouldn't the protons have equivalent environment anyway due to symmetry?

They're only symmetric if there's a threefold symmetry axis in the overall molecule along the C-R bond leading out of the methyl group. As a really simple example, think about acetaldehyde. The methyl group can either have a proton aligned with the C=O (call it eclipsed) or tilted 60° from the C=O (call it staggered). (...or any angle in between, but let's keep it simple...) At room temperature, this methyl group rotates rapidly, so that, over the timescale of the experiment, the NMR can't distinguish which proton is which and gives a single line as an average of the protons' chemical environments. However, at some ridiculously low temperature where the rotation of the methyl group is much slower, the proton aligned with the C=O in the eclipsed conformation is going to have a different chemical environment from the other two protons on the methyl group; therefore its chemical shift will be different from the other two. In addition, in the staggered conformation, the antialigned proton will have a different chemical shift from the other two protons and a different chemical shift from any of the protons in the eclipsed conformation, because the geometries are different. NB--the other two protons in each of these cases will be equivalent to each other by symmetry, but of course that's an artifact of the plane of symmetry of the molecule. One can think of more complex situations where all three methyl protons are inequivalent when frozen.