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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 · 2016 · Paper II

Q8(a). Woodward–Fieser rules: which structure absorbs at 229 nm

  • 15 marks
  • Answered
Question

Which of the following structures has λ_(max) 229 nm? Give the calculations of each:
tabular@c@c@c@c@c@
[structure] & [structure] & [structure] & [structure] & [structure]
(i) & (ii) & (iii) & (iv) & (v)
tabular

Answer in brief

Structure (ii), β-phellandrene: λ_(max)=214+10+5=229 nm. The others: (i) 268, (iii) 219, (iv) 259, (v) 244 nm.

Key points

  • (i) Both C=C are in the same ring: homoannular base 253. Substituents on the diene carbons: two ring residues (the ring CH₂ and the ring CH carrying isopropyl) and the methyl: 253+3×5=268 nm.
  • (ii) One ring C=C conjugated with an exocyclic =CH₂: transoid base 214; two ring residues (+10); the C=CH₂ is exocyclic to the ring (+5): 214+10+5=229 nm. This structure has λ_(max) 229 nm (observed about 232 nm).
  • (iii) Myrcene: the conjugated part is CH₂=C(R)-CH=CH₂; one alkyl residue: 214+5=219 nm (observed about 224 nm). The isolated trisubstituted C=C is not conjugated and adds nothing.
  • (iv) The C=C is α,β to C=O: α carbon carries one ring residue (+10), β carbon two ring residues (+24); the C=C is exocyclic both to the cyclohexanone ring and to the cyclohexylidene ring (2×5): 215+10+24+10=259 nm.

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. The π→π* band of a conjugated diene or enone is estimated by the Woodward–Fieser rules: a base value for the parent chromophore plus increments for each substituent on it. Dienes: base 214 nm (transoid/heteroannular or acyclic) or 253 nm (homoannular); +5 for each alkyl group or ring residue, +5 for each exocyclic C=C, +30 for each C=C extending conjugation. Six-membered ring or acyclic enones: base 215 nm; +10 (α), +12 (β), +18 (γ and beyond) for each alkyl group or ring residue; +5 for each exocyclic C=C.

ChromophoreIncrementsλ_(max)
(i)homoannular diene (both C=C in one ring): α-phellandrenebase 253; 2 ring residues +10; 1 alkyl (CH₃) +5268 nm
(ii)transoid diene, one C=C in the ring and one exocyclic (=CH₂): β-phellandrenebase 214; 2 ring residues +10; 1 exocyclic C=C +5229 nm
(iii)acyclic transoid diene: myrcenebase 214; 1 alkyl residue +5219 nm
(iv)six-ring enone: 2-cyclo-hexylidene-cyclohexanonebase 215; α ring residue +10; 2 β ring residues +24; C=C exocyclic to two rings +10259 nm
(v)cross-conjugated dienone: the longer-wavelength enone (C=O with the C=C at the ring fusion)base 215; 2 β ring residues +24; exocyclic to ring B +5244 nm

Scheme: The diene: ring C=C and the exocyclic =CH₂ on the same ring carbon chain, s-trans.

  1. (i) Both C=C are in the same ring: homoannular base 253. Substituents on the diene carbons: two ring residues (the ring CH₂ and the ring CH carrying isopropyl) and the methyl: 253+3×5=268 nm.
  2. (ii) One ring C=C conjugated with an exocyclic =CH₂: transoid base 214; two ring residues (+10); the C=CH₂ is exocyclic to the ring (+5): 214+10+5=229 nm. This structure has λ_(max) 229 nm (observed about 232 nm).
  3. (iii) Myrcene: the conjugated part is CH₂=C(R)-CH=CH₂; one alkyl residue: 214+5=219 nm (observed about 224 nm). The isolated trisubstituted C=C is not conjugated and adds nothing.
  4. (iv) The C=C is α,β to C=O: α carbon carries one ring residue (+10), β carbon two ring residues (+24); the C=C is exocyclic both to the cyclohexanone ring and to the cyclohexylidene ring (2×5): 215+10+24+10=259 nm.
  5. (v) Cross-conjugated: C=O is conjugated with two C=C bonds that are not conjugated with each other. Each enone is calculated separately and the larger value is reported: C3=O with C4=C5 (two β ring residues, exocyclic to ring B) gives 215+24+5=244 nm; C3=O with C1=C2 (one β ring residue) gives 215+12=227 nm. λ_(max)≈244 nm.

Conclusion. Structure (ii), β-phellandrene: λ_(max)=214+10+5=229 nm. The others: (i) 268, (iii) 219, (iv) 259, (v) 244 nm.

Application. Woodward–Fieser calculations were used to assign structures of terpenes and steroids (distinguishing homo- and heteroannular dienes, locating C=C in ring A of steroid enones) before NMR became routine.

Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram

Model answer to Q8(a), 2016 Paper II: book edition page 34
Book edition, page 34
Model answer to Q8(a), 2016 Paper II: book edition page 35
Book edition, page 35
Read the whole 2016 solution2016 question paper

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