See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: NaCN is a nucleophile that undergoes SN2 substitution reactions with alkyl halides. The selectivity between an aryl halide and a benzylic/alkyl halide is key here. Step 1: Identify the two C-halogen bonds in the starting material. The starting material is 4-bromobenzyl chloride: it has (1) an aryl C-Br bond (bromobenzene-type, sp2 carbon) and (2) a benzylic C-Cl bond (sp3 carbon, CH2Cl). Step 2: Determine reactivity toward NaCN (nucleophilic substitution). Aryl halides (C-Br on the benzene ring) are generally unreactive toward SN2 substitution because the sp2 carbon is not accessible for backside attack, and there is no leaving group ability under mild conditions. Benzylic C-Cl bonds, however, are highly reactive toward SN2 substitution due to the sp3 carbon being accessible and the good leaving group ability of Cl-. Step 3: Reaction outcome. CN- (from NaCN in ethanol) selectively attacks the benzylic CH2Cl via SN2, displacing Cl- and forming CH2CN. The aryl C-Br bond remains intact because aryl halides do not undergo SN2 under these conditions. Step 4: Product is 4-bromobenzyl cyanide: Br-[para-benzene ring]-CH2CN, which corresponds to option (a). Why other options fail: - (b): No reaction occurred - this is just the starting material, incorrect. - (c): This would require both the aryl C-Br and benzylic C-Cl to be substituted by CN, which does not happen under mild SN2 conditions with aryl halides. - (d): This shows an ortho-substituted product with CN added to the ring, which is not the mechanism here. Therefore, the correct answer is A.