See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the target molecule: The product is 1-bromo-3-nitrobenzene (meta-bromonitrobenzene), where NO2 and Br are in a meta relationship on the benzene ring. Step 2 - Determine the correct order of substitution: To achieve meta orientation of Br relative to NO2, we need to exploit directing effects. The nitro group (NO2) is a meta director (deactivating), while bromine is an ortho/para director (weakly activating via lone pair donation). If we introduce NO2 first, it will direct the next incoming electrophile (Br+) to the meta position, giving us the desired meta product. Step 3 - Evaluate option D: (1) HNO3, H2SO4 (nitration of benzene to give nitrobenzene), then (2) Br2, FeBr3 (electrophilic aromatic bromination). The nitro group on nitrobenzene is a meta director, so Br2/FeBr3 will substitute Br predominantly at the meta position, yielding 1-bromo-3-nitrobenzene. This is the correct two-step sequence. Step 4 - Why other options fail: - Option (a): Uses HNO3, H2SO2 (not standard; H2SO4 is required) and then Br2 without a Lewis acid catalyst — Br2 alone cannot efficiently brominate an aromatic ring without FeBr3 or similar Lewis acid, and the reagents are non-standard. - Option (b): Bromination first gives bromobenzene (Br is ortho/para director), then nitration would place NO2 ortho or para to Br, giving ortho- or para-bromonitrobenzene, NOT the meta product. - Option (c): Friedel-Crafts alkylation with CH3Br/AlBr3 introduces a methyl group (ortho/para director), which is irrelevant to making bromonitrobenzene and uses H2SO3 which is not standard for nitration. - Option (d) correctly uses HNO3/H2SO4 first (nitration), then Br2/FeBr3 (bromination directed meta by the nitro group). Therefore, the correct answer is D.