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JEE Advanced 2022 Paper 2 · Chemistry · Carbonyl Compounds, Carboxylic Acids and Amines · hard

Cyclohexanone is treated successively with (i) CH3MgBr in dry ether followed by aqueous NH4Cl; (ii) conc. H2SO4, heat; (iii) O3 at 195 K followed by…

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Cyclohexanone is treated successively with (i) CH3MgBr\mathrm{CH_3MgBr} in dry ether followed by aqueous NH4Cl\mathrm{NH_4Cl}; (ii) conc. H2SO4\mathrm{H_2SO_4}, heat; (iii) O3\mathrm{O_3} at 195 K195\ \mathrm{K} followed by Zn/H2O\mathrm{Zn/H_2O}; (iv) dilute aqueous NaOH\mathrm{NaOH}, heat (an intramolecular reaction leading to a five-membered ring). The final major organic product is
  1. A. 1-acetylcyclopent-1-ene
  2. B. 2-methylcyclohept-2-en-1-one
  3. C. cyclohex-1-enecarbaldehyde
  4. D. 3-methylcyclohex-2-en-1-one
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Answer: (A) 1-acetylcyclopent-1-ene

Step (i): Grignard addition gives 1-methylcyclohexanol. Step (ii): acid-catalysed dehydration follows Saytzeff's rule via the tertiary carbocation and gives 1-methylcyclohexene (endocyclic alkene). Step (iii): reductive ozonolysis cleaves the ring C=C and gives the open-chain keto-aldehyde 6-oxoheptanal, OHC(CH2)4COCH3\mathrm{OHC(CH_2)_4COCH_3}, C7H12O2\mathrm{C_7H_{12}O_2}. Step (iv): in base, the ketone forms an enolate. Attack through the ketone's ring-side α\alpha-carbon (C-5) on the aldehyde C-1 closes a five-membered ring (C1 to C5), favoured over the seven-membered ring that the CH3\mathrm{CH_3} enolate would form. Dehydration of the aldol gives the conjugated enone 1-acetylcyclopentene, C7H10O\mathrm{C_7H_{10}O}. The other options are not possible products: 2-methylcyclohept-2-en-1-one is C8H12O\mathrm{C_8H_{12}O} (one carbon too many; the seven-membered closure through the CH3\mathrm{CH_3} enolate would give plain cyclohept-2-en-1-one), and cyclohex-1-enecarbaldehyde and 3-methylcyclohex-2-en-1-one contain six-membered rings, which no enolate of 6-oxoheptanal can form (the available closures give five- or seven-membered rings only).
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