Why don't trigonal S and P compounds undergo inversion at room temperature?

Most molecules containing nitrogen atoms in trigonal pyramid configuration undergo a relatively fast process of inversion at room temperature. On the other hand, the free energy barrier for phosphines, sulfoniums and sulfoxides are high enough that they are optically stable: the rate of racemization is slow at room temperature.

I wonder what effect is responsible for this difference in behaviour (the larger energy barrier). I expect that it has to do with the size of the central atom (N being one row higher than P and S), but is that all there is to it? And how does size impact the free energy barrier for the inversion: energetically or entropically (or both, of course)?

• IIRC, quantum tunnelling is significant in trigonal nitrogen inversion. – Richard Terrett May 20 '12 at 14:38
• – orthocresol Sep 7 '16 at 17:15

Azane all by itself has an inversion barrier of about $6\ \mathrm{kcal/mol}$, which is low compared to phosphine's approximately $30\ \mathrm{kcal/mol}$ barrier. Pyramidal nitrogen inversion requires a planar transition state. One can use a classical theory to model the thermal rate and we could write $$k \propto \mathrm e^{-E_\mathrm a/(RT)}$$ and using models like these we would predict that ammonia at $300\ \mathrm K$ would have a rate on the order of $10^8\ \mathrm{s^{-1}}$, instead what is observed is something on the order of $10^{10}\ \mathrm{s^{-1}}$

Assuming the potential energy surface has two minima equal in energy and a higher energy TS, two routes are possible: a thermal process going over the maxima or quantum-mechanical tunneling. Determining a tunneling frequency requires writing down vibrational wave functions, finding their linear combination and overlap. If we were to do all this for a particular vibration, we'd see that the tunneling frequency decreases exponentially with a dependence on increasing $\mu$, thickness of the barrier/shape, and the barrier height. As an example, $\ce{ND3}$ has approximately an order of magnitude slower rate of inversion than $\ce{NH3}$ at a sufficiently low temperature – this has been attributed to a decrease in its tunneling frequency.

But why the difference in the energy barriers and corresponding rates depending on the atomic position in the periodic table? Perhaps the easiest/quickest explanation rests on considering what the TS requires for interconversion. Ideally the classical conversion requires passing through a trigonal planar structure with bond angles on the order of 120 degrees. If you consider the series of azane, phosphane, arsane then you'll see bond angles of 108, 94, 92 degrees respectively. In the absence of other effects, the barrier goes from smaller to larger. Indeed, asane-like molecules were the first of this series resolved, followed by phosphane-like. To the best of my knowledge the fastest have yet to be resolved, because rates on the order of $10^{-5}\ \mathrm{s^{-1}}$ are required at RT for a decent shot at isolating them. Naturally, this isn't the whole story, as I've previously alluded to above. Let me know if you would like to know more.

Neither of the two enantiomers is actually an energy eigenstate of the molecule. If the right-handed and left-handed enantiomers are states $|R\rangle$ and $|L\rangle$ respectively, then the ground and first-excited energy eigenstates are going to be something like $|g\rangle=\left( |R\rangle +|L\rangle \right)/\sqrt{2}$ and $|e\rangle=\left( |R\rangle -|L\rangle \right)/\sqrt{2}$.

Because each enantiomer is in a superposition of energy eigenstates, a molecule initially in one chiral state will oscillate back and forth from one to the other, with a frequency determined by the difference in energy between $|g\rangle$ and $|e\rangle$.

In general, the difference in energy between $|g\rangle$ and $|e\rangle$ is determined by the barrier between the $|R\rangle$ and $|L\rangle$ states: the higher and wider the energy barrier, the slower the oscillation. Since P and S are much larger than N, the energy barrier between the $|R\rangle$ and $|L\rangle$ states is larger, and the frequency of oscillation between $|R\rangle$ and $|L\rangle$ is much lower.

• Sorry, my question is: why is the barrier higher for larger atoms? – F'x May 20 '12 at 20:25
• I don't know why (or even if) larger atoms would have a higher barrier, but they would have a wider barrier by virtue of just being larger. Part of the point was that a wider barrier is enough to slow down the oscillation. – Dan May 20 '12 at 20:28
• Indeed, that's a good point (not the answer to my question, but something I hadn't considered). As you said, the wider the barrier, the slower the oscillation… but between N and P, the energy barrier actually is much higher for phosphine (132 kJ/mol) than ammonia (24 kJ/mol). I would expect that contribution to be more important than the barrier widening. – F'x May 20 '12 at 20:31