# Why does internal rotation of ethane give an energy spectrum asymptotically approaching a rotation based on $\nu=1$ rather than $\nu=0$?

I'm a physicist with very little knowledge of chemistry. Recently a chem colleague left a molecular model in my classroom, so I brought it back to her, and as I was walking down the hall with it, a hydroxy group started spontaneously swinging around in circles. This made me curious about whether such motions could actually be observed. After asking around and doing some web searching, my understanding is that this sort of thing is called an internal rotation, and usually there is a barrier to rotation, which is modeled by a potential $$V(\phi)=(1/2)\sum V_n(1-\cos n\phi)$$.

If I'm understanding the history correctly, then originally people thought that the spectrum would be that of a free rotor, but that produced incorrect thermodynamic predictions. Then in 1937 Pitzer published a famous paper, "Thermodynamics of Gaseous Hydrocarbons," in which he showed that you could get correct thermodynamic predictions if you put in a barrier to rotation.

The classic textbook example seems to be ethane. I found this explanation of the physics (di Lauro, "Floppy Molecules," in Rotational Structure in Molecular Infrared Spectra, 2013), which I only partly understand. He gives the following theoretical spectrum of energy states:

So at low energies we have roughly a set of vibrational energy levels with energies $$\hbar\omega (\nu+1/2)$$, which makes sense to me because I guess we're sitting in a minimum of the potential where the geometry is staggered. He says, "The high levels, on the contrary, correlate and eventually almost coincide with the free internal rotor eigenfunctions, whose energy increases parabolically as $$AK_i^2$$, where $$K_i$$ is the eigenvalue of $$J_γ$$, in ℏ units." This also sort of makes sense to me, because if you put in a lot of energy, you shouldn't see the effect of the little hills and valleys in the potential $$V(\phi)$$, so it should act more like a free rotor.

Do the units of inverse cm mean $$1/\lambda$$, or $$2\pi/\lambda$$? If the former, then $$1/\lambda= 200\ \text{cm}^{-1}$$ corresponds to have $$\hbar\omega_0/k\approx 300$$ kelvin, so it makes sense that the barrier to rotation would have significant thermodynamic effects at temperatures on the order of room temperature or below, since the rotational-vibrational modes are not available. Am I getting this right?

But why is it necessary to place this fictitious rotational band on top of the $$\nu=1$$ state, rather than $$\nu=0$$? I would think that the fictitious band should be build from the ground state, it wouldn't match the actual energies at low energies, and then at high energies they would match up.

• Wavenumber units are cycles-per-cm, which I think corresponds to your $1/\lambda$. So, if you're using $\omega_0$ to represent the wavenumber value, you would need to use $h\omega_0c\over k$ to calculate that representative temperature. By my math $200\mathrm{cyc\over cm} \rightarrow 288\ \mathrm K$. – hBy2Py May 18 at 2:21
• The rotor potential is cyclic which means that levels in each potential well interact with one another and become split. In addition there is quantised angular momentum both about the C-C axis and perpendicular to this which leads to further sets of energy levels. The unique axis is defined by quantum number $K$. Look for spectroscopy of symmetric top molecules. To me it is unclear what the levels in the figure refer to. ($k_BT \approx 208\, \mathrm{cm^{-1}}$ at 300 K) – porphyrin May 18 at 7:35
• @porphyrin: To me it is unclear what the levels in the figure refer to. (kBT≈208cm−1 at 300 K) I think that rotational band is supposed to be a fictitious one that he is saying gives the right energy levels in the limit of high spin. – Ben Crowell May 20 at 21:26