Consider the following complex alkyl bromides carefully subjected to pure, highly polarizing solvolysis conditions (). Which specific one will inevitably form the most highly substituted, heavily rear

Question

Consider the following complex alkyl bromides carefully subjected to pure, highly polarizing solvolysis conditions ($CH_3OH/\Delta$). Which specific one will inevitably form the most highly substituted, heavily rearranged alkene via an aggressive $E1$ mechanism?

Prashant Kotian · Accepted Answer

Correct answer: A. Under extreme E1 conditions, the leaving group eventually departs to form a transient carbocation. Neopentyl bromide () 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: (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.

💡 Solution & Explanation

Practice 22,000+ questions like this