## The Question :

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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?

*The Question Comments :*

## The Answer 1

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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.