Q8(d). UV and Woodward–Fieser rules: β-ionone and its ring analogue
How do the UV spectral data λ_(max) 296 nm, ε_(max) 10700 and λ_(max) 281 nm, ε_(max) 20800 help in deciding between structures X and Y?
[X] [Y]
Answer in brief
Woodward–Fieser gives X: 215+30+3(18) = 299 nm and Y: 215+30+2(18) = 281 nm. Hence 296 nm (ε 10700, lowered by steric twisting) is X (β-ionone) and 281 nm (ε 20800, planar) is Y.
Key points
- The calculated values (299 and 281 nm) match the observed values 296 nm for X and 281 nm for Y. The extra δ-methyl of X shifts its band to longer wavelength.
- Y has ε = 20800, normal for a planar s-trans dienone. In X the ring methyl on C-2' and the gem-dimethyl group on C-6' crowd the side chain; the ring C=C is twisted out of the plane of the side-chain C=C–C=O. Overlap is…
- λ_(max) 296 nm, ε 10700 belongs to X; λ_(max) 281 nm, ε 20800 to Y.
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. Both are dienones (C=C–C=C–C=O, extended conjugation); their π→π* λ_(max) is estimated by the Woodward–Fieser rules for enones (ethanol): base value for an acyclic or six-ring α,β-unsaturated ketone 215 nm; extended conjugation +30; each alkyl group or ring residue at the γ or δ carbon +18. The intensity ε tells how well the two C=C lie in one plane.
| Increment | X (β-ionone) | Y |
| Base value (acyclic enone) | 215 | 215 |
| Extended conjugation (one more C=C) | +30 | +30 |
| γ ring residue (C-6 of the ring on the γ carbon) | +18 | +18 |
| δ ring residue | +18 | +18 |
| δ alkyl substituent (ring CH₃) | +18 | – |
| Calculated λ_(max) | 299 nm | 281 nm |
- The calculated values (299 and 281 nm) match the observed values 296 nm for X and 281 nm for Y. The extra δ-methyl of X shifts its band to longer wavelength.
- Intensity: Y has ε = 20800, normal for a planar s-trans dienone. In X the ring methyl on C-2' and the gem-dimethyl group on C-6' crowd the side chain; the ring C=C is twisted out of the plane of the side-chain C=C–C=O. Overlap is reduced, so ε falls to 10700 and λ_(max) is a little below the calculated 299 nm.
- Both criteria agree: λ_(max) 296 nm, ε 10700 belongs to X; λ_(max) 281 nm, ε 20800 to Y.
Conclusion. Woodward–Fieser gives X: 215+30+3(18) = 299 nm and Y: 215+30+2(18) = 281 nm. Hence 296 nm (ε 10700, lowered by steric twisting) is X (β-ionone) and 281 nm (ε 20800, planar) is Y.
Application. β-Ionone is the perfumery and vitamin A starting material; UV of such dienones is used to follow its purity, and the steric loss of intensity is a general test for twisted conjugated systems.
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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