The mechanism is a specialized, highly conditional elimination pathway. Which of the following speci — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

The $E1cB$ pathway specifically involves the fast, highly reversible removal of a proton to form a transient carbanion, followed by the slow, reluctant ejection of the leaving group. To properly form a stable carbanion, the proton must be unusually acidic, which absolutely requires the presence of a strong electron-*withdrawing* group (EWG), not an electron-donating group, on or near the $\beta$ -carbon (Option A is false; Option C is true). Furthermore, if the leaving group were excellent (like an iodide or tosylate), it would leave instantly during proton abstraction, forcing the reaction into a standard concerted E2 mechanism. The $E1cB$ mechanism therefore inherently requires a stubborn, poor leaving group (like fluoride) that severely delays its departure long enough for the discrete carbanion intermediate to fully localize (Option B is true).