See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: In nucleophilic aromatic substitution (SNAr), the reaction proceeds via a Meisenheimer complex intermediate. The rate-determining step is the addition of the nucleophile to the aromatic ring to form this anionic sigma complex, NOT the departure of the leaving group. Step 1: Identify the mechanism. SNAr proceeds by addition-elimination: the nucleophile adds to the ipso carbon bearing the halogen, forming a negatively charged Meisenheimer complex, followed by expulsion of the halide as the leaving group. Step 2: Determine what controls reactivity. Since the rate-determining step is the formation of the Meisenheimer complex (addition step), the identity of the leaving group has minimal influence on the rate of this step. Instead, the ability of the halogen to stabilize the partial negative charge at the ipso carbon in the transition state (through its inductive/field electron-withdrawing effect) is what matters. Step 3: Compare electronegativity and inductive withdrawal of halogens. Fluorine is the most electronegative halogen (electronegativity = 3.98). It exerts the strongest inductive electron-withdrawing effect on the ring, thereby best stabilizing the developing negative charge in the Meisenheimer complex transition state. This lowers the activation energy for the addition step. Step 4: Consider C-X bond strength and leaving group ability. Although F- is normally a poor leaving group in aliphatic substitution (strong C-F bond), in SNAr the leaving group departs AFTER the rate-determining step. Therefore, fluorine's poor leaving group ability is irrelevant to the overall reaction rate; what matters is its superior ability to stabilize the Meisenheimer intermediate via inductive withdrawal. Step 5: Reactivity order. The reactivity order in SNAr for 2-halonitrobenzenes is: F >> Cl > Br > I, which is the reverse of the order seen in SN2 reactions. Why other options fail: - (b) 2-chloronitrobenzene: Cl is less electronegative than F, provides less inductive stabilization of the Meisenheimer complex, so it is less reactive than F. - (c) 2-bromonitrobenzene: Br is even less electronegative, further reducing reactivity. - (d) 2-iodonitrobenzene: I is the least electronegative halogen, providing the least stabilization; least reactive in SNAr. Therefore, the correct answer is A.