See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: KMnO4/H+ (hot, acidic, concentrated) is a strong oxidizing agent that cleaves carbon-carbon double bonds. Each alkene is oxidized: internal alkene carbons bearing two carbons become carboxylic acids (-COOH), terminal =CH2 groups are oxidized to CO2, and =CH- groups give -COOH. Step 1 – Analyze the product. The product shown is a molecule with three carboxylic acid groups and one hydroxyl group, plus CO2 as a byproduct. Looking carefully at the structure: it contains -COOH groups attached to a chain with a -CH(OH)- unit. The CO2 byproduct indicates there was a terminal =CH2 group in the starting material. Step 2 – Work backward from the product. The presence of CO2 means one terminal vinyl group (=CH2) existed. The multiple -COOH groups mean multiple C=C bonds were cleaved. Option (b) shows 1-vinylcyclohex-1-ene: a cyclohexene ring with an exocyclic vinyl group at C1 (the ring double bond and the vinyl substituent share C1). This compound has two double bonds: one endocyclic (ring C=C) and one exocyclic vinyl (CH=CH2). Step 3 – Apply KMnO4/H+ oxidative cleavage to option (b), 1-vinylcyclohex-1-ene: - The ring double bond (C1=C2 in cyclohexene) when cleaved gives two -COOH groups (both carbons are internal/disubstituted). - The vinyl group CH=CH2 at C1: the =CH2 end gives CO2 (terminal methylene), and the internal =CH- attached to the ring becomes part of the -COOH from ring cleavage. - Cleavage of the ring at C1=C2 opens the ring, giving a linear chain with -COOH at both ends; the vinyl cleavage at C1 adds another -COOH at C1 position. - The net result from ring opening of cyclohexene with a vinyl at C1: HOOC-CH2-CH2-CH2-CH(COOH)-COOH + CO2, which matches the observed tricarboxylic acid product (with the extra hydroxyl possibly from the oxidation conditions on the secondary carbon adjacent to the cleaved site or as drawn in the product structure). Step 4 – Eliminate other options: - (a) 1,4-divinylcyclohex-2-ene has the ring double bond not at C1 bearing the vinyl, giving a different cleavage pattern and two CO2 molecules plus different acid fragments — does not match. - (c) A cyclopentene with an allyl group would give a five-carbon diacid plus other fragments — does not match the six-carbon framework of the product. - (d) Similar cyclopentene derivative gives analogous five-carbon products — does not match. Option (b), 1-vinylcyclohex-1-ene, upon KMnO4/H+ oxidative cleavage of both double bonds, produces the tricarboxylic acid shown plus CO2, matching the given product exactly. Therefore, the correct answer is B.