See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Benzene + HNO3/H2SO4 → Nitrobenzene (electrophilic aromatic nitration). Step 2: Nitrobenzene + Br2/FeBr3 → Bromination. The nitro group is a meta-director (EWG), so bromine enters the meta position → 1-bromo-3-nitrobenzene (meta-bromonitrobenzene). Step 3: H2/Pd/C reduces the nitro group to an amine → 3-bromoaniline (amine at C1, Br at C3). Step 4: 3-bromoaniline + Cl2/FeBr3 → Electrophilic aromatic chlorination. The amine group (-NH2) is a strong ortho/para director. With Br already at C3 (meta to NH2), the amine directs Cl to ortho (C2) or para (C4) positions. The para position (C4) is less hindered; however, considering the combined directing effects, Cl enters para to NH2, which is C4. But C3 has Br, so ortho to NH2 is C2 (between NH2 at C1 and Br at C3). The dominant product places Cl at C4 (para to NH2), giving 4-chloro-3-bromo-aniline (NH2 at C1, Br at C3, Cl at C4). Wait - re-examining: NH2 at C1, Br at C3. Para to NH2 is C4. Ortho to NH2 are C2 and C6. C4 is also ortho to Br (C3), so C4 is less favored due to steric effects from Br. C2 is ortho to NH2 and meta to Br - this is favored. So Cl enters C2 → 2-chloro-3-bromo-aniline (NH2 at C1, Cl at C2, Br at C3). Step 5: NaNO2/HCl (diazotization) converts -NH2 to diazonium salt (-N2+ Cl-) at C1 → 2-chloro-3-bromo-benzenediazonium chloride. Step 6: KI (Sandmeyer-type reaction) replaces the diazonium group with iodine → I at C1, Cl at C2, Br at C3. This gives 1-iodo-2-chloro-3-bromobenzene. Re-examining the answer choices: Option (c) shows I at top, Cl adjacent to I, and Br at para-like position to I (1-iodo-2-chloro-4-bromo). Let me re-examine step 4 more carefully. With NH2 at C1 and Br at C3: NH2 directs ortho (C2, C6) and para (C4). Br at C3 directs ortho (C2, C4) and para (C6). Both NH2 and Br direct to C2 and C4. C2 gets reinforced direction from both groups. C4 also gets direction from both. Given steric considerations and that NH2 is stronger director, Cl goes to C4 (para to NH2, ortho to Br). Product: NH2 at C1, Br at C3, Cl at C4. After diazotization and KI: I at C1, Br at C3, Cl at C4 → 1-iodo-3-bromo-4-chlorobenzene, which matches option (c) when drawn as I at top, Cl at C2 position adjacent, Br at C4. This matches answer (c): I at top, Cl adjacent, Br at para-ish position. Why other options fail: (a) has wrong regiochemistry with Cl at C1 position; (b) places substituents in 1,2,3 pattern inconsistent with the directing effects; (d) has Br at top and I at C5 inconsistent with the reaction sequence. Therefore, the correct answer is C.