This question sparked from a long discussion in chat about the nature of $\ce{H2O2}$ and whether that molecule can be considered to rotate around the $\ce{O-O}$ axis (and hence display axial chirality) or not.

Considering two rather clear cases:
Ethane is considered to rotate freely around the $\ce{C-C}$ bond. The activation energy for rotation (equivalent to the energy difference between the staggered and eclipsed conformations) is given as $12.5\,\mathrm{\frac{kJ}{mol}}$.

Ethene on the other hand is considered to not freely rotate around the $\ce{C=C}$ bond. The energy difference between the planar and perpendicular conformations (the rotation barrier) is given at $250\,\mathrm{\frac{kJ}{mol}}$. (I am unsure of this value which I found on Yahoo answers. If I understand correctly, this should correspond with the excitation from $\unicode[Times]{x3C0}$ to $\unicode[Times]{x3C0}^*$ which I found elsewhere as $743\,\mathrm{\frac{kJ}{mol}}$.)

Somewhere between these two extreme cases must lie some kind of barrier around which rotation is severely hindered to entirely inhibited.

Attempting to calculate this myself, I remembered the Boltzmann distribution: If two states differ in energy by $\Delta E$, their relative populations $F_1$ and $F_2$ can be calculated as follows:

$$\frac{F_2}{F_1} = \mathrm{e}^{-\frac{\Delta E}{k_{\mathrm{B}} T}}$$

($F_2$ being the higher energy state and $k_\mathrm{B}$ being the Boltzmann constant.)

If I calculate relative populations from the energy differences $12.5, 19, 25$ and $250\,\mathrm{\frac{kJ}{mol}}$ for ethane, butane, $\ce{H2O2}$ and ethene, respectively (which correspond to $0.13, 0.20, 0.26$ and $2.6\,\mathrm{\frac{eV}{particle}}$) I get the following results at $300\,\mathrm{K}$:

  • Ethane: $6.55 \times 10^{-3}$
  • Butane: $4.79 \times 10^{-4}$
  • $\ce{H2O2}$: $4.29 \times 10^{-5}$
  • Ethene: $2.09 \times 10^{-44}$

Or logarithmic (ln) values of:

  • Ethane: $5.03$
  • Butane: $7.64$
  • $\ce{H2O2}$: $10.1$
  • Ethene: $101$

Again, somewhere between butane and ethene must be a threshold-ish value, above which free rotation at $300\,\mathrm{K}$ cannot be considered, but what range are we looking at; similar to what energy difference does this correspond to?

Asked the other way round, if I have an energy barrier of $25\,\mathrm{\frac{kJ}{mol}}$, what would I need to put on into the equation to calculate the temperature above which free rotation can be assumed?

  • 1
    $\begingroup$ The undergraduate organic chemistry text I teach form (Klein, Wiley) lists the barrier in ethene as 63 kcal/mol =264 kJ/mol, but gives no references. Propagations of values like this one in undergrad texts are probably the source of the 250 kJ/mol value you found. $\endgroup$
    – Ben Norris
    Commented Sep 24, 2015 at 11:44
  • $\begingroup$ @BenNorris Do you have a different value with a reputable source? $\endgroup$
    – Jan
    Commented Sep 24, 2015 at 11:54
  • $\begingroup$ Alas, I do not. I wish I did. $\endgroup$
    – Ben Norris
    Commented Sep 24, 2015 at 21:36
  • $\begingroup$ M. H. Wood, Chem. Phys. Lett. 1974, 24 (2), 239-242 calculates 63.2 kcal/mol and references the exp. value as 65 kcal/mol. They also reference a more accurate calculation, i.e. CI, as 63.7 kcal/mol. This is just the first one I opened, there might be more accurate measurements and calculations by now, however, I doubt revolutionary change. (cc. @Ben) $\endgroup$ Commented Sep 29, 2015 at 3:07
  • $\begingroup$ Please be aware that with your Boltzmann approach, you calculated the population of the transition state based on your assumed ground state. This is probably fine when you deal with ethene/ethane, but it might be problematic for butene/butane. $\endgroup$ Commented Sep 29, 2015 at 3:13

1 Answer 1


I'm not a transition-state physical chemist, but I think a good approach to this problem is transition-state theory, specifically the Eyring equation:

$$k = \kappa \frac{k_b T}{h} e^{\frac{-\Delta G^{\ddagger}}{RT}}$$

This equation tries to predict the rate constant $k$ from an assumed pseudo-equilibrium between the transition-state and the starting material, which are assumed to differ in energy by $\Delta G^{\ddagger}$. $k_b$ is Boltzmann's constant, $h$ is Planck's constant, $T$ is temperature, and $\kappa$ is the "fudge factor" or transmission coefficient, the fraction of transition states that react productively instead of converting back to starting material.

In the conformational reactions you are talking about, I think a good starting place would be to assume the eclipsed conformations are the transition state, although I don't know if this has been shown to be 100% true experimentally or computationally. As a first approximation, let's assume there isn't a huge entropy term in $\Delta G^{\ddagger}$, so that it is well-approximated by the energy level differences you found. If we also assume that $\kappa$ is 1.0, then we can just multiply the $e^{\frac{-\Delta E}{RT}}$ you've already calculated by $\frac{k_b T}{h} = 6.2\times10^{12}\mathrm{~s^{-1}}$ at 298 K to get (very) approximate rate constants.

  • Ethane: $6.55 \times 10^{-3} \rightarrow 4.1 \times 10^{10}\mathrm{~s^{-1}}$
  • Butane: $4.79 \times 10^{-4} \rightarrow 3.1 \times 10^{9}\mathrm{~s^{-1}}$
  • $\ce{H2O2}$: $4.29 \times 10^{-5} \rightarrow 2.0 \times 10^{8}\mathrm{~s^{-1}}$
  • Ethene: $2.09 \times 10^{-44} \rightarrow 1.3 \times 10^{-31}\mathrm{~s^{-1}}$

That shows that rotational equilibrium of everything except ethene should be very rapid, but that the half-life for ethene equilibration much greater than the age of the universe ($4 \times 10^{17} \mathrm{~s}$).

Suppose we want to find an energy difference that should lead to a rate constant that is neither very fast nor very slow, say we'd like $k\approx\mathrm{1~hr^{-1}}$. Inverting our very very crude approximation gives:

$$\Delta E = - RT \ln{\left(\frac{k h}{\kappa k_b T}\right)}$$

Plugging in $k=\mathrm{1~hr^{-1}=2.7\times10^{-4} s^{-1}}$ gives (if I did my math right) 93 kJ per mol. Some well-known atropoisomers do have estimated energy differences in that range.

  • 2
    $\begingroup$ If $\kappa = 1$, wouldn't that mean that all transition states react productively? If so, wouldn't a fudge factor of $\kappa = 0.5$ be a much better model? Actually, no it probably wouldn't, because the molecule is rotating towards the transition state, so it's more likely it'll rotate on. Maybe $\kappa = 0.75$? ;) Also, I'll take a large slice of fudge cake, please. $\endgroup$
    – Jan
    Commented Sep 23, 2015 at 17:14
  • 1
    $\begingroup$ @Jan: In any case, if we're just comparing similar molecules that presumably have similar $\kappa$ values, the actual value doesn't really matter much (since it just scales all the rates proportionally). $\endgroup$ Commented Sep 23, 2015 at 17:41
  • $\begingroup$ Wow, thanks for the reverse calculation! Assuming it to be correct that actually is right what that article said (and what I skimmed across before but chose not to mention)! I would +1 a second time ;) $\endgroup$
    – Jan
    Commented Sep 23, 2015 at 18:44
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    $\begingroup$ TST suffers from many more problems, than the "correct" value (choice) for $\kappa$. I think for small energy barriers as they are expected for a free rotation it is quite fair to assume, that all molecules entering in the TS ensemble also convert to the other rotamer. This has probably by far a lesser effect, than ignoring one degree of freedom in the TS. $\endgroup$ Commented Sep 25, 2015 at 6:53
  • 1
    $\begingroup$ @Curt I think wikipedia gives a brief summary of the general problems. I thought we have a post here, but the closest I could find was this. You can certainly ask about that, if the above does not help you. $\endgroup$ Commented Sep 26, 2015 at 20:53

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