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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 · 2017 · Paper II

Q5(b). Cyclohexanone to cyclohexanecarbaldehyde in two steps

  • 10 marks
  • Answered
Question

How will you convert the following reactant into product by a two-step process? Also give the mechanism.
[structure] → [structure]

Answer in brief

(1) Ph₃P=CHOCH₃ (Wittig) gives methoxymethylenecyclohexane; (2) H₃O⁺ hydrolyses it to cyclohexanecarbaldehyde. (A Darzens glycidic ester with ClCH₂CO₂Et/NaOEt, then hydrolysis and decarboxylation, is an alternative two-stage route.)

Key points

  • Ylide. Ph₃P+CH₂OCH₃ Cl⁻ (from Ph₃P and ClCH₂OCH₃) is deprotonated by a strong base (PhLi or KOtBu) to Ph₃P=CH-OCH₃.
  • Step 1, Wittig. The ylide carbon adds to the C=O; the betaine/oxaphosphetane collapses (driving force: the strong P=O bond) to Ph₃P=O and the enol ether C₆H₁₀=CH-OCH₃.
  • Step 2, hydrolysis. H₃O⁺ protonates the enol ether at carbon (the ring carbon), giving an oxocarbenium ion C₆H₁₁-CH=O+CH₃; water adds, the hemiacetal loses methanol, and the aldehyde C₆H₁₁CHO results.

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 conversion adds one carbon to the carbonyl carbon (a one-carbon homologation). It is done by a Wittig reaction with (methoxymethylene)triphenylphosphorane, which gives an enol ether, followed by acid hydrolysis of the enol ether.

Scheme: Step 1: Wittig olefination; step 2: hydrolysis of the enol ether.

  1. Ylide. Ph₃P+CH₂OCH₃ Cl⁻ (from Ph₃P and ClCH₂OCH₃) is deprotonated by a strong base (PhLi or KOtBu) to Ph₃P=CH-OCH₃.
  2. Step 1, Wittig. The ylide carbon adds to the C=O; the betaine/oxaphosphetane collapses (driving force: the strong P=O bond) to Ph₃P=O and the enol ether C₆H₁₀=CH-OCH₃.
  3. Step 2, hydrolysis. H₃O⁺ protonates the enol ether at carbon (the ring carbon), giving an oxocarbenium ion C₆H₁₁-CH=O+CH₃; water adds, the hemiacetal loses methanol, and the aldehyde C₆H₁₁CHO results.

Conclusion. (1) Ph₃P=CHOCH₃ (Wittig) gives methoxymethylenecyclohexane; (2) H₃O⁺ hydrolyses it to cyclohexanecarbaldehyde. (A Darzens glycidic ester with ClCH₂CO₂Et/NaOEt, then hydrolysis and decarboxylation, is an alternative two-stage route.)

Application. Methoxymethylene Wittig homologation is used in steroid and prostaglandin syntheses to convert a ketone into the next higher aldehyde.

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Model answer to Q5(b), 2017 Paper II: solution page 15
Solution page 15 of 23
Read the whole 2017 solution2017 question paper

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