See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The Kolbe electrolysis involves the electrolysis of carboxylate salts. At the anode, carboxylate ions (RCOO-) are oxidized, lose CO2 (decarboxylation), and the resulting carbon radicals combine to form a new C-C bond product. Reasoning: The starting material is potassium succinate (the dipotassium salt of succinic acid), which has the structure: -OOC-CH2-CH2-COO- (as K+ salts). In Kolbe electrolysis, each carboxylate group loses CO2 and an electron to form a radical. For a dicarboxylate salt like succinate, both ends undergo decarboxylation simultaneously. Each -CH2-COO- end loses CO2 to give a -CH2 radical. Since both radicals are connected intramolecularly (they are the two ends of the same molecule), the two CH2 radicals combine intramolecularly to form a ring. The -CH2-CH2- chain after losing both COO- groups from each end cyclizes to form ethylene (CH2=CH2)? Actually, let us reconsider: potassium succinate is K-OOC-CH2-CH2-COO-K. After Kolbe electrolysis, both carboxylate groups lose CO2 and electrons. The two carbon radicals remaining are -CH2• and •CH2- (connected as •CH2-CH2•). These two radicals on the same molecule combine to form a ring - cyclopropane? No, •CH2-CH2• is a diradical with only 2 carbons, which would form a C-C bond giving a 2-carbon ring, which is not possible (cycloethane does not exist). Correct interpretation: The two radicals •CH2-CH2• (1,2-diradical) form a double bond, giving CH2=CH2 (ethylene). This is the known result of Kolbe electrolysis of succinate: it produces ethylene (CH2=CH2) as the major product. Why other options fail: (a) CH3-CH3 (ethane): This would require addition of H atoms, not radical coupling of the diradical from succinate. (c) CH3-CH=CH2 (propene): This is a 3-carbon compound; succinate only provides 2 carbons after decarboxylation. (d) None of these: Incorrect since CH2=CH2 is a valid answer listed as option (b). Therefore, the correct answer is B.