See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: When an alkyl halide reacts with ammonia (NH3) in ethanol (EtOH), nucleophilic substitution (SN2) occurs at the sp3 carbon bearing the halide. Aryl halides (C-Br directly on the aromatic ring) are generally unreactive toward nucleophilic substitution under mild conditions (NH3/EtOH) because the sp2 carbon is much less electrophilic and the aryl C-Br bond is strengthened by resonance. Step 1: Identify the electrophilic sites in the reactant (2-bromobenzyl bromide). There are two C-Br bonds: one is a benzylic CH2-Br (sp3, highly reactive toward SN2) and one is an aryl C-Br (sp2, unreactive under these mild conditions). Step 2: NH3 acts as a nucleophile and attacks the benzylic CH2-Br via SN2. The benzylic position is activated and very reactive toward nucleophilic substitution. The aryl bromide does not react under NH3/EtOH conditions. Step 3: The reaction replaces the CH2-Br with CH2-NH2 (after deprotonation), leaving the aryl Br intact on the ring. Step 4: The major product is therefore 2-bromobenzylamine: a benzene ring with CH2-NH2 at position 1 and Br at position 2, which corresponds to option (a). Why other options fail: - Option (b): This would require selective substitution of the aryl Br while leaving the benzylic Br intact, which does not occur with NH3/EtOH; aryl halides are inert under these conditions. - Option (c): This would require both the benzylic Br and the aryl Br to be substituted by NH2, but aryl Br is unreactive with NH3/EtOH. - Option (d): This is a meta-substituted benzylamine with no second substituent, which does not correspond to the starting material or expected product. Therefore, the correct answer is A.