See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Nucleophilic Aromatic Substitution (SNAr). In SNAr, a nucleophile displaces a leaving group on an aromatic ring. The reaction is facilitated by electron-withdrawing groups (EWGs) ortho/para to the leaving group, which stabilize the Meisenheimer complex intermediate. Step 1: Identify the electrophilic positions on the ring. The starting material is 2-bromo-1-fluoro-4-nitrobenzene. The ring bears F at C1, Br at C2, and NO2 at C4. Step 2: Assess which halogen is the better leaving group in SNAr. In SNAr, fluorine is the best leaving group (despite being poorest in SN2) because the rate-determining step is formation of the Meisenheimer complex, not C-X bond breaking. Fluorine's strong electron-withdrawing inductive effect stabilizes the negative charge in the Meisenheimer complex better than Br. Step 3: Assess activation. For displacement of F at C1: the NO2 group at C4 is para to F (C1), providing strong activation of C1 toward nucleophilic attack. The Br at C2 is ortho to F and also provides some electron withdrawal. This makes C1 highly activated for SNAr. For displacement of Br at C2: NO2 at C4 is meta to Br (C2), which does NOT effectively stabilize the Meisenheimer complex at C2. Therefore, Br at C2 is not well-activated for SNAr. Step 4: The nucleophile CH3S- (from NaSCH3) attacks C1 (bearing F), which is activated para to NO2. F is displaced, giving a product where F is replaced by SCH3, while Br and NO2 remain intact. Step 5: The product is 2-bromo-1-(methylthio)-4-nitrobenzene: ring with SCH3 at C1, Br at C2, NO2 at C4. This corresponds to answer choice (b). Why other options fail: - (a) shows NO2 replaced by SCH3 at C4, but NO2 is not a leaving group in SNAr. - (c) shows SCH3 added at a new position without displacing any group, which is not the reaction mechanism. - (d) shows Br replaced by SCH3, but Br at C2 is meta to NO2 (C4) and is not activated for SNAr; also F is a better leaving group than Br in SNAr. Therefore, the correct answer is B.