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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

Q5(a). Favorskii rearrangement: one product from two chloroketones

  • 10 marks
  • Answered
Question

Each of PhCH₂COCH₂Cl and PhCHClCOCH₃ in reaction with NaOMe/MeOH furnishes the same product. What mechanistic implication can be drawn from the above fact?

Answer in brief

Both chloroketones give methyl 3-phenylpropanoate. The identical product shows that the Favorskii rearrangement passes through a common cyclopropanone intermediate (2-phenylcyclopropanone), which is opened by MeO⁻ to the more stable benzylic carbanion.

Key points

  • MeO⁻ removes an α'-H on the side away from Cl: from PhCH₂ in the first ketone, from CH₃ in the second. An enolate forms.
  • The enolate carbon displaces Cl⁻ on the other α-carbon (internal SN2, or via the oxyallyl zwitterion): 2-phenylcyclopropanone from both substrates.
  • MeO⁻ adds to the strained cyclopropanone C=O: a tetrahedral alkoxide.
  • The ring opens by breaking the C(O)–C bond that gives the more stable carbanion: the bond to the CHPh carbon breaks, giving a benzylic carbanion (stabilised by the ring), not the primary CH₂⁻.

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. α-Halo ketones with an α'-H rearrange with alkoxide to esters (Favorskii rearrangement). Both 1-chloro-3-phenylpropan-2-one and 1-chloro-1-phenylpropan-2-one give the same ester, methyl 3-phenylpropanoate, PhCH₂CH₂CO₂Me. The two chlorides can lead to one product only through a common intermediate in which the two α-carbons have become equivalent in connectivity: a cyclopropanone.

Scheme: Both chloroketones close the same three-membered ring.

Mechanism

  1. MeO⁻ removes an α'-H on the side away from Cl: from PhCH₂ in the first ketone, from CH₃ in the second. An enolate forms.
  2. The enolate carbon displaces Cl⁻ on the other α-carbon (internal SN2, or via the oxyallyl zwitterion): 2-phenylcyclopropanone from both substrates.
  3. MeO⁻ adds to the strained cyclopropanone C=O: a tetrahedral alkoxide.
  4. The ring opens by breaking the C(O)–C bond that gives the more stable carbanion: the bond to the CHPh carbon breaks, giving a benzylic carbanion (stabilised by the ring), not the primary CH₂⁻.
  5. Protonation by MeOH gives methyl 3-phenylpropanoate.

Scheme: The cyclopropanone always opens towards the benzylic carbanion.

Mechanistic implications

  • A direct substitution or a one-step 1,2-shift (semibenzilic route) would give different esters: PhCH₂CH₂CO₂Me from one ketone and PhCH(CH₃)CO₂Me from the other. Only one is found, so the reaction goes through a symmetrical-type intermediate, the cyclopropanone (Loftfield; also shown by 14C labelling of 2-chlorocyclohexanone, where the label ends equally on both α-carbons).
  • The direction of ring opening is governed by carbanion stability, not by where Cl was.
  • The rearrangement needs an α'-H; without one (e.g. α-chloro ketones with a quaternary α' carbon) the slower semibenzilic path takes over.

Conclusion. Both chloroketones give methyl 3-phenylpropanoate. The identical product shows that the Favorskii rearrangement passes through a common cyclopropanone intermediate (2-phenylcyclopropanone), which is opened by MeO⁻ to the more stable benzylic carbanion.

Application. The Favorskii rearrangement is used for ring contraction (2-chlorocyclohexanone gives methyl cyclopentanecarboxylate) and in the synthesis of cubane.

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Model answer to Q5(a), 2018 Paper II: solution page 15
Solution page 15 of 24