See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: The substrate shown is a tertiary alkyl bromide (tert-butyl bromide), and the solvent/nucleophile is methanol (CH3OH). Tertiary substrates undergo substitution via the SN1 mechanism because the tertiary carbocation intermediate is highly stabilized by three alkyl groups. Step 1 – Identify the substrate: The X-shaped structure with Br represents tert-butyl bromide (a tertiary carbon bearing three methyl groups and one Br). Tertiary substrates cannot undergo SN2 due to steric hindrance. Step 2 – Determine mechanism: Because the substrate is tertiary and the solvent is a weak nucleophile (methanol), the reaction proceeds by SN1. In SN1, the rate-determining step is the ionization of the C–Br bond to form a tertiary carbocation. This step involves only the alkyl bromide. Step 3 – Write the rate law: For SN1, Rate = k[alkyl bromide]. The nucleophile (CH3OH) does not appear in the rate law because it attacks after the slow step (carbocation formation). Step 4 – Evaluate options: (a) Rate depends only on alkyl bromide concentration – CORRECT for SN1. (b) Rate depends only on methanol concentration – Incorrect; methanol is not involved in the rate-determining step. (c) Rate depends on both concentrations – Incorrect; that would describe SN2. (d) Rate depends on neither – Incorrect; the substrate concentration clearly affects rate. Therefore, the correct answer is A.