See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the transformation needed: Starting material is 4-chlorotoluene (Cl at C4, CH3 at C1). The product has CO2H at C1 (where CH3 was), Br at C5 (adjacent to Cl), NO2 at C2 (ortho to CO2H), and Cl at C4 (unchanged). So we need to: introduce Br onto the ring, oxidize CH3 to CO2H, and nitrate the ring. Step 2 - Determine directing effects and sequence: The CH3 group is an ortho/para director. Cl is also an ortho/para director. In 4-chlorotoluene, both groups cooperate to direct incoming electrophiles to specific positions. Bromination with Br2/FeBr3 on 4-chlorotoluene: CH3 (at C1) directs ortho/para, and Cl (at C4) directs ortho/para. The position between Cl and CH3 (i.e., C3 and C5) is activated by both groups cooperatively. Bromination occurs preferentially at C3 or equivalently C5 (ortho to Cl and meta to CH3, but since the molecule is symmetric about the Cl-CH3 axis, position ortho to Cl = C3 or C5). Given the product shows Br at C5 (ortho to Cl at C4, and meta to CO2H at C1), bromination first with Br2/FeBr3 places Br ortho to Cl (the ring position activated by both Cl ortho-para direction and CH3 ortho-para direction cooperate at C3/C5). Step 3 - After bromination, oxidize CH3 to CO2H using KMnO4 and heat. This converts the methyl group to carboxylic acid without affecting the ring substituents. Step 4 - Now the ring has CO2H (deactivating, meta director), Cl (weakly ortho/para), and Br (weakly ortho/para) at C4 and C5. Nitration with HNO3/H2SO4 introduces NO2. The CO2H group directs meta (to C4 area), but Cl and Br direct ortho/para. The combined directing effects place NO2 at C2 (ortho to CO2H is not favored by CO2H, but considering Cl at C4 and Br at C5, position C2 is ortho to CO2H, meta to Cl - the position activated by the halogen directors). In the product, NO2 is at C2 (ortho to CO2H at C1, meta to Cl at C4), which is consistent with direction by the halogen substituents. Step 5 - Why option (a) works: Br2/FeBr3 first (while CH3 is still activating and directing, ensuring correct Br placement), then KMnO4 oxidation, then nitration. This sequence is optimal. Step 6 - Why other options fail: - Option (b): KMnO4 first converts CH3 to CO2H, making the ring deactivated before bromination, making electrophilic bromination difficult and giving wrong regiochemistry for Br. - Option (c): NBS in CCl4 is a benzylic bromination reagent (free radical), which would brominate the CH3 group (benzylic position) rather than the ring, giving benzyl bromide not ring-brominated product. - Option (d): NBS gives benzylic bromination, then NaNO2/heat is not a standard nitration procedure for aromatic rings. Therefore, the correct answer is A.