See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify compound (A): Aniline (Ph-NH2) reacts with 2 moles of CH3Cl (methylating agent) to give N,N-dimethylaniline (Ph-N(CH3)2). This is a straightforward N-alkylation: the first mole of CH3Cl methylates NH2 to NHCH3, and the second mole methylates it further to N(CH3)2. Step 2 - Identify compound (B) - Coupling reaction: N,N-dimethylaniline undergoes electrophilic aromatic substitution (azo coupling) with benzene diazonium chloride (Ph-N2Cl). The diazonium ion is a weak electrophile, so it couples preferentially with activated aromatic rings (those bearing electron-donating groups). Step 3 - Regioselectivity of azo coupling: In N,N-dimethylaniline, the -N(CH3)2 group is a strong para-director. The diazonium electrophile attacks predominantly at the para position relative to the -N(CH3)2 group. This gives 4-(dimethylamino)azobenzene, i.e., a benzene ring bearing -N(CH3)2 at one end and -N=N-Ph at the para position. Step 4 - Butter Yellow: The product is known commercially as 'Butter Yellow' (p-dimethylaminoazobenzene), which is Ph-N=N-C6H4-N(CH3)2 (para substitution). This corresponds exactly to option (c): a benzene ring with N(Me)2 at one position and N=N-Ph at the para position. Step 5 - Elimination of other options: (a) Shows ortho coupling product - not the major product; para is strongly preferred. (b) Shows an additional methyl on the ring, which is not present in the product. (d) Shows an incorrect ionic/triazene-type structure, not the azo dye product. Therefore, the correct answer is C.