Q8(b). Photostationary state of cis- and trans-stilbene
The direct irradiation of either cis- or trans-stilbene at 313 nm results in the formation of a mixture of 93% cis- and 7% trans-olefin, no matter how long, within reason, the irradiation is continued. Give a suitable explanation for that.
Answer in brief
A photostationary state is set up: both isomers decay through a common twisted excited state to either isomer, and because trans absorbs about 13 times more strongly at 313 nm, the steady state is 93% cis: 7% trans, whichever isomer is irradiated.
Key points
- Both cis and trans absorb at 313 nm and are excited to S₁, where the C=C becomes nearly a single bond.
- Each excited isomer twists to the same perpendicular (p) geometry (about 90°), the minimum of S₁. From p the molecule drops to S₀ and falls to cis or trans in about equal amounts (Φ_(t→ c) ≈ Φ_(c→ t) ≈ 0.5).
- εₜ Φ_(t→ c)[trans] = ε_c Φ_(c→ t)[cis], so (([cis])/([trans]))_(PSS) = εₜ Φ_(t→ c)ε_c Φ_(c→ t)
- trans-Stilbene is planar and fully conjugated; it absorbs much more strongly at 313 nm than cis-stilbene, which is twisted by steric clash of the two phenyl rings (its band lies at shorter wavelength and is weaker). Light…
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Introduction. Irradiation interconverts the two isomers in both directions. When the rate of trans→cis equals the rate of cis→trans, the composition stops changing: a photostationary state (PSS). It is a steady state of two light-driven reactions, not a thermal equilibrium (at thermal equilibrium trans-stilbene, the more stable isomer, would dominate).
- Both cis and trans absorb at 313 nm and are excited to S₁, where the C=C becomes nearly a single bond.
- Each excited isomer twists to the same perpendicular (p) geometry (about 90°), the minimum of S₁. From p the molecule drops to S₀ and falls to cis or trans in about equal amounts (Φ_(t→ c) ≈ Φ_(c→ t) ≈ 0.5).
Figure: Both isomers relax in S₁ to the same twisted state p, which returns to either isomer.
- At the PSS the two rates are equal: εₜ Φ_(t→ c)[trans] = ε_c Φ_(c→ t)[cis], so (([cis])/([trans]))_(PSS) = εₜ Φ_(t→ c)ε_c Φ_(c→ t)
- trans-Stilbene is planar and fully conjugated; it absorbs much more strongly at 313 nm than cis-stilbene, which is twisted by steric clash of the two phenyl rings (its band lies at shorter wavelength and is weaker). Light therefore pumps trans away faster than cis is pumped back, and cis accumulates.
- With the observed ratio 93/7 = 13.3 and the two quantum yields about equal, εₜ/ε_c ≈ 13 at 313 nm. The composition depends only on these molecular constants and the wavelength, so it is the same from either starting isomer and does not change with time once reached.
Conclusion. A photostationary state is set up: both isomers decay through a common twisted excited state to either isomer, and because trans absorbs about 13 times more strongly at 313 nm, the steady state is 93% cis: 7% trans, whichever isomer is irradiated.
Application. Photochemical E→Z isomerisation is used to make the less stable cis (Z) isomers (e.g. in vitamin A chemistry) and underlies vision (11-cis-retinal to all-trans) and molecular photo-switches.
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All parts of Question 8
- 8(a) Fluorescence and phosphorescence10 marks
- 8(b) Photostationary state of cis- and trans-stilbene10 marks
- 8(c) Photocyclisation of cyclodecanone to 9-decalol10 marks
- 8(d) UV and Woodward–Fieser rules: β-ionone and its ring analogue10 marks
- 8(e) Charge-transfer spectra of donor–acceptor complexes10 marks