See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the starting material. The structure shown is maleic anhydride: a five-membered cyclic anhydride containing a cis C=C double bond (but-2-enedioic anhydride). Step 2 – Reaction with H3O+ (acid hydrolysis) gives product (A). Opening the anhydride with water yields maleic acid: HOOC-CH=CH-COOH (cis-butenedioic acid). Step 3 – Reaction of maleic acid with NaOH gives product (B). Neutralisation of both carboxylic acid groups produces the disodium salt: NaOOC-CH=CH-COONa (sodium maleate / disodium maleate). Step 4 – Kolbe electrolysis of (B). In Kolbe electrolysis, carboxylate anions are oxidised at the anode: each –COO⁻ loses one electron to form a carboxyl radical, which then loses CO2 to give a carbon radical. The two carbon radicals couple. For NaOOC-CH=CH-COONa: - Each –COO⁻ group is on a vinyl carbon (=CH–). - Oxidation at the anode: –CH=CH–(COO⁻)2 → two •CH=CH– fragments after loss of 2 CO2? Actually, each end loses –COO⁻ as a radical → –CO2, leaving a vinyl radical on each carbon. Because the two radical centres are on adjacent carbons connected by a double bond, the result after radical coupling is: •CH=CH• → the diradical recombines internally to give HC≡CH (acetylene/ethyne). More precisely: the two carboxylate radicals each expel CO2, generating =CH• radicals at each end. Since both radicals are on the same two-carbon unit connected by a double bond, the C=C with two radical termini collapses to a triple bond, giving HC≡CH. Step 5 – Why other options fail: (a) H2C=CH2 would require a saturated dicarboxylate (succinate), not maleate. (b) CH3-C≡C-CH3 would require a four-carbon dicarboxylate with internal radical coupling of two separate CH3-C• units. (d) CH3-CH=CH-CH3 would result from Kolbe coupling of two propionate radicals. Therefore, the correct answer is C.