See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Alcoholic KOH (alc. KOH) is a reagent that promotes dehydrohalogenation (E2 elimination), removing HX from alkyl halides to form alkenes. However, when a gamma-halo carbonyl compound (or a suitable 1,3-dihalo compound) is treated with base, intramolecular cyclization via an SN2 pathway can compete with elimination, especially when a three-membered ring can be formed. Step 1: Identify the starting material. The compound is 1,7-dichloroheptan-4-one: Cl-CH2CH2CH2-C(=O)-CH2CH2CH2-Cl. The carbonyl is at C4, with three-carbon chains bearing Cl at each end (C1 and C7). Step 2: Consider the role of alc. KOH (2 moles). Alcoholic KOH can act as a base for elimination OR as a nucleophilic base that deprotonates alpha to the carbonyl. The alpha carbons to the ketone are C3 and C5. Deprotonation at C3 generates an enolate which can displace the Cl at C1 intramolecularly (C1 is 3 atoms away from C3, forming a 3-membered ring: C1-C2-C3 cyclizes to give a cyclopropane ring on one side of the ketone). Similarly, deprotonation at C5 generates an enolate which displaces Cl at C7 intramolecularly to form another cyclopropane ring on the other side. Step 3: First equivalent of KOH deprotonates C3 (alpha to ketone), the enolate carbon (C3) attacks C1 (bearing Cl) in an intramolecular SN2, displacing Cl- and forming a cyclopropane ring fused to the ketone on the left side, giving cyclopropyl-C(=O)-CH2CH2CH2Cl. Step 4: Second equivalent of KOH deprotonates C5 (alpha to ketone in the intermediate), the enolate carbon attacks C7 intramolecularly, displacing the second Cl- and forming the second cyclopropane ring. Step 5: The product is dicyclopropyl ketone: cyclopropyl-C(=O)-cyclopropyl, which corresponds to option (b). Why other options fail: - Option (a): Simple double dehydrohalogenation would give divinyl ketone only if elimination occurred, but with alc. KOH and a gamma-chloro ketone, intramolecular cyclization (forming cyclopropane) is strongly favored over intermolecular elimination because the enolate is an excellent intramolecular nucleophile at the gamma position. - Option (c): This would require a different elimination pattern inconsistent with the structure. - Option (d): Aqueous KOH (nucleophilic substitution with OH-) would give the diol, not alcoholic KOH which favors elimination/cyclization. Therefore, the correct answer is B.