See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Electrophilic aromatic substitution (EAS) directed by activating groups. Iodine monochloride (I-Cl) is an electrophilic iodinating agent. In salicylic acid (2-hydroxybenzoic acid), the OH group is a strong ortho/para director and the CO2H group is a weak meta director (deactivating). The OH group dominates directing effects. Step 1: Identify the directing effects. The OH group at position 2 strongly activates the ring toward ortho and para positions relative to itself. The CO2H group at position 1 is a meta director but is weaker than OH. Step 2: Determine available positions. With CO2H at C1 and OH at C2, the positions ortho and para to OH are: ortho to OH = C1 (blocked by CO2H) and C3; para to OH = C5. So the preferred positions for electrophilic attack are C3 (ortho to OH) and C5 (para to OH). Step 3: With 2 equivalents of I-Cl, both activated positions (C3 and C5) are iodinated, giving 3,5-diiodo-2-hydroxybenzoic acid (3,5-diiodosalicylic acid). Step 4: This matches option (c): CO2H at C1, OH at C2, I at C3 and C5 (positions ortho and para to OH, i.e., 3,5-diiodo-2-hydroxybenzoic acid). Why other options fail: - (a) shows acyl iodide and ring iodination, which would require different conditions; I-Cl does not convert -CO2H to -C(=O)I under normal EAS conditions. - (b) shows chlorination (Cl substituents) but I-Cl acts as an iodinating agent (I is the electrophile, Cl is the leaving group), so Cl does not substitute onto the ring. - (d) shows a different substitution pattern (not 3,5-diiodo) inconsistent with ortho/para direction from OH. Therefore, the correct answer is C.