Q.(a) An organic compound (A) on treatment with Lucas reagent gives compound (B). Oxidation of (A) by gives compound (E) which is resistant to oxidation by mild oxidising agents. When (B) is heated with alcoholic KOH it gives compound (C) which can decolourise bromine water. Compound (C) undergoes hydroboration-oxidation to produce an alcohol (D) . Predict the structure of compounds (A), (B), (C), (D) and (E). OR
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Start your 14-day free trial to unlock the full solution →This question offers a choice between deducing five related C3 alcohol/haloalkane/alkene structures from their reactions (primary), or the classic phenol-to-aspirin synthesis sequence (the alternative). Both are answered below.
(Primary, a) — Deducing A–E ( system):
Clue: (A) + Lucas reagent → (B); (A) + → (E), resistant to further mild oxidation. A secondary alcohol gives a ketone on oxidation (which resists further easy oxidation, unlike a primary alcohol's aldehyde, which oxidises on to a carboxylic acid). This points to A = propan-2-ol (isopropanol), :
Clue: (B) + alcoholic KOH → (C), decolourises bromine water. Elimination (dehydrohalogenation) of the secondary halide gives an alkene:
Clue: (C) undergoes hydroboration-oxidation to give (D), . Hydroboration-oxidation adds and across the double bond with anti-Markovnikov regiochemistry, putting on the less substituted (terminal) carbon:
OR (b) — Phenol to aspirin (classic named sequence):
Step 1 (, i.e. phenol, + benzenediazonium chloride): Phenol couples with the diazonium salt to give an orange azo dye (p-hydroxyazobenzene) — this simply confirms A is phenol (a classic diazo-coupling test for phenols/anilines), and is not itself part of the synthetic chain toward aspirin.
Step 2 (A + Na metal): Phenol reacts with sodium metal, releasing hydrogen gas and forming sodium phenoxide, B:
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