See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material. The starting material is propyne (CH3-C≡CH), a terminal alkyne with the triple bond between C2 and C3. Step 2: Reaction with HgSO4/H2SO4 (acid-catalyzed hydration of alkyne). For terminal alkynes, Markovnikov's rule applies: the OH adds to the more substituted carbon (C2) and H adds to the terminal carbon (C1 of the triple bond, which is C3 of the chain). This gives an enol intermediate that tautomerizes to the more stable ketone. For propyne (CH3-C≡CH), the product A is acetone... wait, let me reconsider. Propyne is CH3-C≡CH. The triple bond carbons are C2 (internal, bearing CH3) and C3 (terminal). By Markovnikov addition, OH adds to C2 (more substituted) giving the enol CH3-C(OH)=CH2, which tautomerizes to CH3-CO-CH3 (acetone, propan-2-one). So A = acetone (a ketone). Step 3: Reaction of A (acetone) with NaBH4. NaBH4 is a mild reducing agent that reduces ketones and aldehydes to alcohols. Acetone (CH3-CO-CH3) is reduced to 2-propanol (CH3-CH(OH)-CH3). Step 4: Product B = 2-propanol. Step 5: Why other options fail: - (a) propanol is vague but if meaning 1-propanol, incorrect. - (c) 1-propanol would result if hydration followed anti-Markovnikov addition, which does not occur under HgSO4/H2SO4 conditions. - (d) propane is not formed; NaBH4 reduces the carbonyl to an alcohol, not fully to an alkane. Therefore, the correct answer is B.