Maximum (Z) to (E) ratio in the photoisomerization of azobenzene?

Question

Considering the photoswitching of azobenzene, I was told the following argument: Considering that the photoexcited state of azobenzene has a about 50% to 50% chance of yielding either the (E)- or (Z)-isomer, the maximum (Z)/(E) ratio achievable should be 1:1.

However, I have seen UV-VIS spectra where the (Z)-isomer seems to be in excess. So I figured that the above argument must be wrong. I would analyze the situation in the following way:

In this picture, the fact that the photoexcited state has about equal probability to yield (E) or (Z), i.e. $k_{-1} = k_2$.

So wouldn't the overall (Z) to (E) ratio be given by the expression $\frac{k_1\cdot k_2}{k_{-1}\cdot k_{-2}}$?

I would then argue that in many cases, the cis isomer exhibits lower absorption at the wavelength of excitation, meaning that $k_{-2} << k_{1}$, which would explain that in many cases, an excess of the cis isomer is obtained. By this reasoning, the initial argument would be completely wrong, as in principle an infinite excess of the cis isomer could be obtained if $k_{-2}$ is sufficiently smaller than $k_{1}$.

Is this analysis correct or am I overlooking something?

Answer 1

Your question is not completely clear to me, so I am answering based on thinking that you are asking about why, after photoisomerization, the system can relax with roughly equal probability to either the E or Z minimum.

Strangely I know something about the photochemistry of azobenzene. One overarching theme is that azobenzene molecules undergo very complicated isomerization processes which are sensitive both to how you substitute them and to the solvent in which the experiment is done. Additionally, the photoisomerization and thermal isomerization processes don't necessarily share anything in common. There is a pretty in-depth interesting literature on different types of azobenzene molecules discussing the relative roles of inversion and rotation relaxation pathways, solvent effects, and so on... I just bring this up in case you are interested to read more.

The case you describe, I think has a relatively straightforward explanation actually. In ref. [1], a computational study is done on azobenzene to find which pathway, rotation or inversion, is dominanant upon photoisomerization of azobenzene. To clarify, rotation means rotation about the CNNC dihedral, and inversion means moving one of the rings along the CNC angle.

The paper shows that there is a conical intersection along the rotation pathway which is only 2 kcal/mol above the $\ce{S1}$ excited state minimum. Thus, the molecule will follow this conical intersection upon relaxation. Because the molecule is already above both states, and the molecule can essentially end up going toward either the E or Z minimum from this conical intersection, to first order, the two should end up equally populated on relaxation. Of course, from there, one would expect to move back to a Boltzmann distribution of the two states thermally.

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[1]: Cembran, A., Bernardi, F., Garavelli, M., Gagliardi, L., & Orlandi, G. (2004). On the mechanism of the cis− trans isomerization in the lowest electronic states of azobenzene: S0, S1, and T1. Journal of the American Chemical Society, 126(10), 3234-3243.