A tertiary alkyl halide undergoes substitution in a polar protic solvent. Why does it follow a two-step mechanism instead of a single-step pathway?
Q1. A tertiary alkyl halide undergoes substitution in a polar protic solvent. Why does it follow a two-step mechanism instead of a single-step pathway?
Answer: Carbocation stability
Explanation: Tertiary carbocations are stabilized by three alkyl groups via hyperconjugation; Walden inversion is tempting but is a characteristic of SN2 reactions.