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JEE Main 2018 · Chemistry · Hydrocarbons · hard

A hydrocarbon A (C8H14) decolourises bromine water. Reductive ozonolysis (O3, then Zn/H2O) gives a single compound B (C8H14O2), which gives a…

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A hydrocarbon A (C8H14\mathrm{C_8H_{14}}) decolourises bromine water. Reductive ozonolysis (O3\mathrm{O_3}, then Zn/H2O\mathrm{Zn/H_2O}) gives a single compound B (C8H14O2\mathrm{C_8H_{14}O_2}), which gives a positive iodoform test but no silver mirror. B on warming with dilute NaOH undergoes intramolecular aldol condensation to give C (C8H12O\mathrm{C_8H_{12}O}) with the more favourable five-membered ring. C is
  1. A. 1-acetyl-2-methylcyclopent-1-ene
  2. B. 3-methylcyclohept-2-en-1-one
  3. C. 1-acetylcyclohex-1-ene
  4. D. 2,3-dimethylcyclohex-2-en-1-one
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Answer: (A) 1-acetyl-2-methylcyclopent-1-ene

A single dicarbonyl product from a ring alkene means A is a cycloalkene (degrees of unsaturation =2=2). Both products being methyl ketones with 88 C's fits A = 1,2-dimethylcyclohexene, and B = octane-2,7-dione, CH3CO(CH2)4COCH3\mathrm{CH_3CO(CH_2)_4COCH_3} (iodoform positive, Tollens negative). Under base the enolate at C-3 attacks C-7: ring atoms C3, C4, C5, C6, C7 form a five-membered ring; dehydration gives C3=C7\mathrm{C3=C7} with the acetyl on C3 and the methyl on C7, i.e. 1-acetyl-2-methylcyclopent-1-ene. The enolate at C-1 would give a strained seven-membered ring.
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