See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the functional groups in the starting material: The starting material is 2-(bromomethyl)benzonitrile, which contains (i) a nitrile group (CN) directly attached to the aromatic ring and (ii) a benzylic bromide (CH2-Br) on the ortho position. Step 2 - Reactivity of the benzylic bromide with NaOH/H2O: The CH2-Br group is a benzylic halide and undergoes SN2 hydrolysis readily with aqueous NaOH to give CH2-OH (benzyl alcohol). This is a straightforward nucleophilic substitution: CH2-Br + OH⁻ → CH2-OH + Br⁻. Step 3 - Reactivity of the nitrile group with NaOH (excess)/H2O/heat: An aryl nitrile (ArCN) undergoes base-catalyzed hydrolysis under aqueous NaOH with heat. The nitrile is first hydrolyzed to an amide, then further to a carboxylate salt, and upon workup gives a carboxylic acid (COOH). With excess NaOH and heat, the hydrolysis goes all the way to the carboxylic acid (as its sodium salt, which on acidification gives COOH). Step 4 - Combined outcome: Both reactions occur: CN → COOH and CH2-Br → CH2-OH. This gives 2-(hydroxymethyl)benzoic acid, i.e., a benzene ring bearing COOH at C1 and CH2-OH at C2. This corresponds to option (c). Step 5 - Why other options fail: - Option (a): Only the CN is hydrolyzed but CH2-Br remains unreacted — incorrect because benzylic bromides are more reactive toward SN2 with NaOH than nitriles, and under excess NaOH both would react. - Option (b): Only the CH2-Br is hydrolyzed to CH2-OH but CN remains intact — incorrect because excess NaOH with heat is sufficient to hydrolyze the nitrile to COOH. - Option (d): Shows an additional OH on the ring which is not justified by the reaction conditions; aromatic rings do not undergo hydroxylation under these conditions. Therefore, the correct answer is C.