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In a prior question, I asked about the (a)symmetry of the potential energy surface of the methyl rotation of propene. In that context, the kinetic energy of the nuclear motions is of course assumed to be negligible, and thus nuclear inertial effects are irrelevant.

Here, I want to ask about the (a)symmetry of this same internal rotation, but instead in the context of molecular dynamics (MD), where the Born-Oppenheimer approximation still applies but the kinetic energy of the nuclei is non-negligible compared to the variations in the potential energy over the course of the motion of the system. For the sake of argument, assume that the propene is in a sufficiently high rovibrational energy state for the methyl rotation that it behaves as a hindered internal rotor, and thus the nuclear motion is not "trapped" in one of the wells of the potential energy surface.


Depending on how one imagines the methyl rotation of propene to occur, it seems like it could behave as either a symmetric or an asymmetric internal rotor in an MD context. If the $\ce{H2C=CH\! -}$ portion were held fixed, the rotating methyl would be inertially symmetric due to the local $\mathrm C_3$ axis:

propene with rotating methyl

On the other hand, if the methyl group were held fixed, the rotating $\ce{C=C}$ portion would be highly inertially asymmetric:

propene with rotating C=C moiety

Obviously this rotation can't simultaneously be both inertially symmetric and inertially asymmetric. So which is it, and why?

My inclination is to assume that the internal rotor is asymmetrical, since the overall system does not have any axes of three-fold or higher symmetry. However, I don't know a good mathematical or group theoretic 'language' in which to define such internal rotor (a)symmetry, and so I'm unable to prove this to my own satisfaction.


The best qualitative argument I've thought of in favor of it being an asymmetric internal rotor is that the axis of the rotation (green bar) does not pass through the center of mass of the system (purple sphere / orange bar):

Stick representation of propene with c.o.m. and axes indicated

FWIW, I calculate the skew angle between the two colored bars in the above geometry to be about $18^\circ$. Anyways, as a consequence of this skewed orientiation, I think the methyl hydrogens would oscillate parallel to the axis of internal rotation while they rotate around the methyl carbon. Simultaneously, I would expect a sort of corresponding "waggle" of the ethenyl fragment.

Thus, directly related to the "why" question above: If this qualitative argument is correct, how can I express it quantitatively?

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  • $\begingroup$ I still think that you are convolving whole body motion with that of internal motion. The two can be separated just as we separate out various normal modes from one another and from whole body translations. Sure, the centre of mass of the whole thing does not have to be placed at an atom or bond and can be 'in space' as you draw. $\endgroup$ – porphyrin Jun 14 '17 at 11:20
  • $\begingroup$ @porphyrin I agree, they can be fully separated -- that's what I mean by "internal rotation" -- the "true" internal methyl rotation occurs in the absence of net overall rotation of the system. It is this very constraint (zero net total system rotation) that makes me think that there is some sort of 'asymmetric' character to the methyl internal rotation here, intuitively since the nominal axis of internal rotation doesn't include the center of mass. I may be using the wrong language to frame the question, though. $\endgroup$ – hBy2Py Jun 14 '17 at 11:26
  • $\begingroup$ As I wrote in reply to your other question the internal rotation occurs in a 3-fold potential; at low energy (cold molecules) motion is hindered and may be restricted to be within the wells. At high temp it is more or less 'free' as there is enough energy to overcome the barrier. I would not call the 3-fold potential 'asymmetric' though as it has some symmetry. $\endgroup$ – porphyrin Jun 14 '17 at 14:05
  • $\begingroup$ @porphyrin <nod>, and I'm interested here in the inertial dynamics of the "hot" molecule (ignoring electronic excitation effects), where the potential energy barrier is more or less irrelevant to the character of the motion. $\endgroup$ – hBy2Py Jun 28 '17 at 19:31
  • $\begingroup$ Then have a look at the J Chem. Educ. article in the comments your other question and put the potential to zero or so small that it becomes irrelevant. $\endgroup$ – porphyrin Jun 29 '17 at 8:13

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