Consider the following complex alkyl bromides carefully subjected to pure, highly polarizing solvoly — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Under extreme E1 conditions, the leaving group eventually departs to form a transient carbocation. Neopentyl bromide ($1^\circ$) forms an impossibly unstable primary carbocation that instantly and violently undergoes a 1,2-methanide shift to form a highly stable, completely rearranged tertiary carbocation: $CH_3-C^+(CH_3)-CH_2-CH_3$ (tert-pentyl cation). The subsequent elimination of a proton yields 2-methyl-2-butene, a heavily substituted alkene. While Option B already forms a tertiary carbocation and yields the same exact alkene, it does so *without* any skeletal rearrangement. The prompt explicitly asks for the "most highly substituted, heavily rearranged" alkene, specifically highlighting the extensive, mandatory structural rearrangement unique to neopentyl bromide.