See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the substrate types and their SN1/ionization tendencies. Solvolysis in aqueous ethanol proceeds predominantly by SN1 for substrates that can form stable carbocations. The rate depends on how easily and how stable the carbocation intermediate is. Structure (1): H2C=C(CH3)−Br is a vinylic bromide. Vinylic carbocations are extremely high in energy (sp-hybridized, no resonance stabilization, no hyperconjugation). Ionization to a vinylic carbocation is very unfavorable. SN2 on a vinylic carbon is also geometrically very difficult. Therefore, compound (1) undergoes solvolysis at an extremely slow rate. Structure (2): 1-bromo-1-methylcyclohexane is a tertiary alkyl bromide (the carbon bearing Br has three carbon substituents: two ring carbons and one methyl). Tertiary carbocations are highly stabilized by hyperconjugation and inductive effects from three alkyl groups. Additionally, the release of ring strain upon ionization can assist the process. This substrate ionizes very readily via SN1, giving a fast rate of solvolysis. Structure (3): CH3−CH(Br)CH2CH(CH3)2 is a secondary alkyl bromide (2-bromo-4-methylpentane). Secondary carbocations are moderately stable (less than tertiary). This substrate undergoes SN1 solvolysis at a moderate rate, faster than the vinylic bromide but slower than the tertiary bromide. Step 2 – Rank from fastest to slowest. Fastest: (2) tertiary bromide → most stable carbocation, fastest ionization. Middle: (3) secondary bromide → moderately stable carbocation, moderate rate. Slowest: (1) vinylic bromide → extremely unstable vinylic carbocation, negligible ionization rate. Order: 2 > 3 > 1 Step 3 – Evaluate the answer choices. (a) 2 > 1 > 3: Incorrect; places vinylic bromide (1) ahead of secondary bromide (3), which is wrong. (b) 1 > 2 > 3: Incorrect; places vinylic bromide first, which is wrong. (c) 2 > 3 > 1: Correct; tertiary > secondary > vinylic. (d) 1 > 3 > 2: Incorrect; places vinylic bromide first and tertiary last, both wrong. Therefore, the correct answer is C.