See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: In Nucleophilic Aromatic Substitution (SNAr), the mechanism proceeds via an addition-elimination pathway. The rate-determining step is the addition of the nucleophile to the aromatic ring to form a Meisenheimer complex (a negatively charged cyclohexadienyl anion intermediate). The leaving group departs in the second (fast) step. Reasoning: Unlike aliphatic nucleophilic substitution (SN2), where a better leaving group (weaker conjugate base, e.g., I- > Br- > Cl- > F-) leads to faster reaction, in SNAr the rate-determining step is the ADDITION of the nucleophile, not the departure of the leaving group. Therefore, the ability of the leaving group to stabilize negative charge in the transition state/Meisenheimer complex matters more than its leaving group ability in the classical sense. Fluorine is the most electronegative halogen. Its strong inductive electron-withdrawing effect (through the sigma framework) stabilizes the negative charge developing on the ring carbon bearing X in the Meisenheimer complex transition state. This lowers the activation energy for the addition step, making fluorine the best 'leaving group' in SNAr despite being the worst leaving group in SN2 reactions. Additionally, the C-F bond has partial ionic character and the high electronegativity of F stabilizes the anionic Meisenheimer intermediate. Why other options fail: - Cl, Br, and I are progressively less electronegative than F, so they provide less stabilization of the Meisenheimer complex through inductive effects, making their SNAr reactions slower despite being better leaving groups in SN2 chemistry. - The order of reactivity in SNAr is: F > Cl > Br > I (opposite to SN2 order). Therefore, the correct answer is A.