See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The haloform reaction is a named reaction in which a methyl ketone (or acetaldehyde) reacts with a halogen (Cl2, Br2, or I2) in the presence of a base (NaOH) to produce a carboxylate salt and a haloform (CHX3). Upon acidification, the carboxylate salt gives the corresponding carboxylic acid. Step 1 - Identify the substrate: Acetophenone is C6H5-CO-CH3, a methyl aryl ketone. It contains a methyl group adjacent to the carbonyl, making it susceptible to the haloform reaction. Step 2 - Reaction with I2/NaOH (iodoform reaction): The three alpha-hydrogens of the CH3 group are successively replaced by iodine in the basic medium, giving C6H5-CO-CI3 (triiodomethyl ketone intermediate). The base then cleaves the C-C bond between the carbonyl carbon and the CI3 group, producing the carboxylate C6H5-COO^- (sodium benzoate) and CHI3 (iodoform). Step 3 - Acidification: Treatment of sodium benzoate with acid gives benzoic acid (C6H5-COOH). Why other options fail: (a) NaOH alone followed by acidification: Acetophenone does not undergo simple hydrolysis under basic conditions to give benzoic acid; no reaction cleaves the ketone to the acid without a halogen. (c) Hydroxylamine followed by H2SO4: This converts a ketone to an oxime (Beckmann rearrangement with H2SO4 gives an amide/lactam), not benzoic acid. (d) m-Chloroperoxobenzoic acid (mCPBA): This is a peracid used for Baeyer-Villiger oxidation of ketones; acetophenone would give phenyl acetate (an ester), not benzoic acid. Therefore, the correct answer is B.