See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material and reagents. Starting material: 1-butanol, CH3-(CH2)3-OH (n-butanol). Step 2: First reaction — treatment with methanesulfonyl chloride (CH3-S(=O)2-Cl, mesyl chloride) and triethylamine (ET3N). This converts the primary alcohol into a mesylate (A): CH3-(CH2)3-OMs. The hydroxyl is converted to a good leaving group. ET3N acts as a base to neutralize HCl produced. So (A) = CH3-(CH2)3-OMs (butyl mesylate). Step 3: Second reaction — treatment with K14CN (potassium cyanide labeled with carbon-14). The cyanide ion (14CN-) acts as a nucleophile and displaces the mesylate via SN2 reaction, inverting at the carbon bearing the leaving group (though primary carbon, so no stereochemical issue). This gives (B): CH3-(CH2)3-14CN (pentanenitrile with 14C at the nitrile carbon, i.e., 1-cyanobutane with labeled carbon). The carbon chain is extended by one carbon (the 14C-labeled carbon from CN). Step 4: Third reaction — treatment with H3O+ (aqueous acid hydrolysis). The nitrile group (14CN) is hydrolyzed under acidic aqueous conditions to give a carboxylic acid. CH3-(CH2)3-14CN + H3O+ → CH3-(CH2)3-14CO2H The labeled carbon (14C) ends up in the carboxyl group. Step 5: Identify product (C). Product (C) = CH3-(CH2)3-14CO2H, which is pentanoic acid with the carboxyl carbon labeled with 14C. This matches option (b). Why other options fail: (a) CH3-(CH2)3-CO2H — this would be correct if the CN were not labeled (14C); the structure is right but lacks the 14C label, so it is incorrect here. (c) CH3-CO2H — this is acetic acid, a 2-carbon acid, which would result from a much shorter chain; incorrect. (d) CH3-C(14)(=O)-OH — this implies the label is on a 2-carbon acid or the carbonyl carbon of acetic acid; incorrect chain length and labeling position. Therefore, the correct answer is B.