Q5(a). Favorskii rearrangement: one product from two chloroketones
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
- 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⁻.
- 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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All parts of Question 5
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