COMEDK 2025 Morning Solved Paper

Concept:2‑Bromo‑2‑methylpropane is a tertiary alkyl halide. In aqueous KOH (a polar protic solvent) it undergoes an $\mathrm{S_N1}$ mechanism, which is fast because the tertiary carbocation intermediate is highly stabilised.Explanation:2‑Bromo‑2‑methylpropane $(\mathrm{(CH_3)_3CBr})$ is tertiary.Aqueous KOH provides a polar protic medium.The reaction follows an $\mathrm{S_N1}$ pathway.The rate‑determining step is ionisation of the C–Br bond to form a tertiary carbocation $(\mathrm{(CH_3)_3C^+})$ and a bromide ion.Tertiary carbocations are stabilised by inductive and hyperconjugation effects, so this step is fast.Polar protic solvents further stabilise both the carbocation and the leaving bromide ion, speeding up the reaction.Therefore the overall reaction occurs at a fast rate.Option A is incorrect because in $\mathrm{S_N1}$ the rate depends only on the concentration of the haloalkane, not on the nucleophile $\mathrm{OH^-}$.Option C is false because polar protic solvents actually favour $\mathrm{S_N1}$.Option D is false because tertiary halides cannot undergo $\mathrm{S_N2}$ (the mechanism would be very slow).Answer:Option B – the reaction occurs at a fast rate since the substrate is a tertiary alkyl halide and follows $\mathrm{S_N1}$ mechanism.