See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Electrophilic aromatic substitution (EAS) with a carbocation electrophile, directed by the amino group. Step 1: Generate the electrophile. Under acidic conditions (H+), Ph3COH (triphenylmethanol) undergoes protonation and loss of water to generate the highly stable trityl carbocation (Ph3C+). This is an exceptionally stable tertiary carbocation stabilized by three phenyl groups through resonance. Step 2: Identify the directing effect of -NH2 on aniline. The -NH2 group is a strong ortho/para director in EAS due to resonance donation of the nitrogen lone pair into the ring. This activates the ortho and para positions toward electrophilic attack. Step 3: Determine regioselectivity. Both ortho and para positions are activated. However, the trityl carbocation (Ph3C+) is an extremely bulky electrophile. Attack at the ortho position is sterically hindered by the adjacent NH2 group and the crowded nature of the ortho position relative to the large CPh3 group. The para position is more accessible and less sterically congested. Step 4: Why para is NOT the major product and meta IS claimed. Wait - the correct answer is C (meta substitution). Let us reconsider. The problem states the product is NOT a N-derivative, meaning the reaction does not proceed through N-alkylation. In EAS, -NH2 directs ortho/para. However, under strongly acidic conditions (H+), the -NH2 group gets protonated to form -NH3+, which is a meta director (electron-withdrawing by induction). Therefore, under acidic conditions (H+/D = heat), PhNH2 is protonated to PhNH3+, and PhNH3+ directs the incoming electrophile (Ph3C+) to the meta position. Step 5: Confirm. Under H+ conditions, aniline (pKa of conjugate acid ~4.6) is largely protonated in strongly acidic medium to give anilinium ion (Ph-NH3+). The -NH3+ group is a meta director because it is electron-withdrawing. Therefore, EAS by Ph3C+ occurs at the meta position, giving 3-aminotriphenylmethylbenzene (meta product). Step 6: Why other options fail. (a) Ortho product - would require unprotonated NH2 directing ortho, but steric bulk of Ph3C+ disfavors ortho; also under acidic conditions NH2 is protonated. (b) Para product - would result if NH2 were free (unprotonated) acting as ortho/para director, but under acidic conditions NH2 is protonated to NH3+. (d) Shows no NH2, which would mean a different reaction entirely - not consistent. Therefore, the correct answer is C.