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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 · 2026 · Paper II

Q1(c). Friedel–Crafts, Wittig, dihydroxylation and pinacol rearrangement

  • 10 marks
  • Answered
Question

Identify the major products V, W, X, Y and Z in the following reaction sequences:
[structure] + [structure] → (AlCl₃, excess benzene) V
[structure] → ((i) PPh₃, (ii) BuLi) W → (V) X → (OsO₄ (cat.), NMO, ^tBuOH, H₂O) Y → (H₂SO₄) Z

Answer in brief

V = benzophenone; W = C₅H₈=PPh₃; X = (diphenylmethylidene)cyclopentane; Y = the 1,2-diol; Z = 2,2-diphenylcyclohexan-1-one, formed by a pinacol rearrangement with ring expansion.

Key points

  • Introduction. The sequence links five standard steps: Friedel–Crafts acylation, ylide formation, Wittig olefination, syn-dihydroxylation and an acid-catalysed pinacol rearrangement, which here enlarges the five-membered ring.
  • V. AlCl₃ forms the benzoyl (acylium) ion PhC#O⁺, which substitutes benzene: V = benzophenone, Ph₂C=O. The ketone complexes AlCl₃, so no second acylation occurs.

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 sequence links five standard steps: Friedel–Crafts acylation, ylide formation, Wittig olefination, syn-dihydroxylation and an acid-catalysed pinacol rearrangement, which here enlarges the five-membered ring.

Scheme: V = benzophenone; W = cyclopentylidenetriphenylphosphorane; X = (diphenylmethylidene)cyclopentane

V. AlCl₃ forms the benzoyl (acylium) ion PhC#O⁺, which substitutes benzene: V = benzophenone, Ph₂C=O. The ketone complexes AlCl₃, so no second acylation occurs.

W. PPh₃ displaces chloride (SN2) to give cyclopentyltriphenylphosphonium chloride; BuLi removes the α-H: W = the ylide C₅H₈=PPh₃, cyclopentylidenetriphenylphosphorane.

X. The ylide adds to the C=O of V to form an oxaphosphetane, which collapses to Ph₃P=O and the alkene: X = (diphenylmethylidene)cyclopentane, C₅H₈=CPh₂.

Y. OsO₄/NMO (Upjohn conditions) adds two OH groups syn through a cyclic osmate ester; NMO reoxidises osmium, so only a catalytic amount is needed: Y = 1-[hydroxy(diphenyl)methyl]cyclopentan-1-ol, a pinacol (1,2-diol).

Z. Acid protonates the OH that leaves the more stable cation, the one on the diphenyl carbon (tertiary and doubly benzylic). A ring C–C bond then migrates to this cation (1,2-alkyl shift, ring expansion, the step that relieves strain), and the oxocarbenium ion loses H⁺.

Scheme: The ring bond migrates to the cation: the five-membered ring becomes a cyclohexanone.

Conclusion. V = benzophenone; W = C₅H₈=PPh₃; X = (diphenylmethylidene)cyclopentane; Y = the 1,2-diol; Z = 2,2-diphenylcyclohexan-1-one, formed by a pinacol rearrangement with ring expansion.

Application. Pinacol ring expansion of cyclic 1,2-diols is a laboratory route to spiro and cyclohexanone frameworks; the Wittig reaction is used industrially in the synthesis of vitamin A.

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Model answer to Q1(c), 2026 Paper II: solution page 4
Solution page 4 of 26