See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting compound is maleic acid or fumaric acid (butenedioic acid), represented as CH-CO2H || CH-CO2H, meaning both carbons of a double bond each bear a -CO2H group. This is specifically the structure of an alkene diacid (maleic/fumaric acid). Step 2 - Reaction with NaOH (two moles): Two moles of NaOH neutralize both carboxylic acid groups to give the disodium salt: NaOOC-CH=CH-COONa. This is compound (A), the disodium salt of the diacid. Step 3 - Electrolysis (Kolbe's electrolysis): In Kolbe's electrolysis, carboxylate anions are oxidized at the anode, losing CO2 and generating carbon radicals, which then couple. For the disodium salt of an alkene diacid (NaOOC-CH=CH-COONa), electrolysis at the anode causes each -COO⁻ group to lose an electron and release CO2. This leaves radical centers on the two carbons: •CH=CH•. These two radical centers are already bonded to each other (intramolecular), so after loss of both CO2 molecules, the result is a diradical that forms a triple bond: HC≡CH (acetylene/ethyne). Step 4 - Why other options fail: - (a) CH3-CH3: This would result from Kolbe electrolysis of sodium acetate (CH3COONa), not from a diacid with a double bond. - (b) H2C=CH2: This would result from Kolbe electrolysis of sodium succinate (NaOOC-CH2-CH2-COONa), where after loss of two CO2, two -CH2• radicals couple to give ethylene, but that requires a single bond between the two carbons in the starting salt. - (d) CH2=CH-CH=CH2 (1,3-butadiene): This would result from intermolecular coupling of two vinyl radicals, not the intramolecular product expected here. - (c) H-C≡C-H: Correct. The disodium salt of maleic/fumaric acid (with CH=CH backbone) upon Kolbe electrolysis loses two CO2, and the two remaining carbon centers, already double-bonded, form a triple bond giving acetylene. Therefore, the correct answer is C.