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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper II covers organic chemistry and spectroscopy.

UPSC Civil Services (Mains), Chemistry optional · 2018 · Paper II

Q8(d). UV and Woodward–Fieser rules: β-ionone and its ring analogue

  • 10 marks
  • Answered
Question

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.

IncrementX (β-ionone)Y
Base value (acyclic enone)215215
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 nm281 nm
  1. 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.
  2. 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.
  3. 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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Model answer to Q8(d), 2018 Paper II: solution page 22
Solution page 22 of 24