See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: This reaction is the Gattermann-Koch aldehyde synthesis variant using HN=CH-Cl (formimidoyl chloride, also written as chloromethyleneamine or the reagent derived from HCN + HCl) with AlCl3, which is essentially the Gattermann synthesis of aldehydes. The reagent HN=CH-Cl with AlCl3 generates an electrophilic iminium species (or equivalent to a formyl cation equivalent [CHO]+) that undergoes electrophilic aromatic substitution (EAS) on the aromatic ring, followed by hydrolysis. Step 1 - Reagent activation: HN=CH-Cl reacts with AlCl3 (Lewis acid) to generate an electrophilic species equivalent to [CH=NH2]+ (or a formyl cation equivalent), which acts as the electrophile in EAS. Step 2 - Electrophilic aromatic substitution on toluene: The methyl group on toluene is an ortho/para director. EAS therefore occurs preferentially at the para position (para-substitution is favored over ortho due to steric reasons). The electrophile attacks para to the methyl group, giving intermediate (A): 4-methyl-N-(benzylidene)amine or imine intermediate at the para position, i.e., a Schiff base/aldimine: 4-CH3-C6H4-CH=NH. Step 3 - Hydrolysis with H3O+: The imine intermediate (A) undergoes acid hydrolysis (H3O+) to give the aldehyde (B): 4-methylbenzaldehyde (para-tolualdehyde), with CH3 at position 1 and CHO at position 4. Step 4 - Why other options fail: - Option (a): ortho-tolualdehyde would be the minor product due to steric hindrance at ortho position; the major product is para. - Option (c): No-reaction is incorrect because toluene readily undergoes EAS with this activated electrophile. - Option (d): Acetophenone would result from Friedel-Crafts acylation with CH3COCl, not with HN=CH-Cl/AlCl3/H3O+. The para product (4-methylbenzaldehyde) is the major product (B). Therefore, the correct answer is B.