See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Friedel-Crafts alkylation (FC alkylation) using CH3Cl/AlCl3 generates a methyl carbocation (electrophile: CH3+) that attacks an aromatic ring. The key question is which aromatic ring gets alkylated and at which position. Step 1: Identify the substrate. The starting material is benzanilide, PhC(=O)NHPh, which contains two aromatic rings: - Ring A: the benzoyl ring (bearing the C=O group, which is electron-withdrawing, meta-director) - Ring B: the anilide ring (bearing the NHCOPh group, where N is an electron-donating group via resonance despite the adjacent carbonyl, making it an ortho/para-director) Step 2: Determine which ring is more reactive. The NHCOPh group (amide nitrogen) on Ring B still donates electron density into the ring through the nitrogen lone pair, making Ring B more electron-rich and more reactive toward electrophilic aromatic substitution than Ring A (which bears the deactivating C=O group). Step 3: Determine regioselectivity on Ring B. The -NHCOPh substituent is an ortho/para-director. Therefore, electrophilic attack occurs preferentially at the para position relative to -NHCOPh (para is favored over ortho due to steric reasons), giving a methyl group at the para position of Ring B. Step 4: Product identification. Methylation at the para position of the aniline ring gives N-(4-methylphenyl)benzamide, which corresponds to option (c): a benzene ring bearing -NHCOPh at C1 and -CH3 at C4. Why other options fail: - Options (a) and (b): These show methylation on the benzoyl ring (Ring A), which is deactivated by the electron-withdrawing carbonyl group and is less reactive. Ring A would not be preferentially alkylated. - Option (d): This shows meta-methylation on Ring B, but -NHCOPh directs ortho/para, not meta. Meta product would be the minor product. Therefore, the correct answer is C.