MHTCET 22 Apr 2025 Shift 1 Paper

Concept:$\mathrm{S_N1}$ reactions proceed through a carbocation intermediate, and the rate depends on the stability of that carbocation.Explanation:In $\mathrm{S_N1}$, the rate-determining step is the ionization of the alkyl halide to form a carbocation.A more stable carbocation lowers the activation energy, making the reaction faster.The stability order of carbocations is: $3^\circ > 2^\circ > 1^\circ >$ methyl.n-Butyl iodide on ionization gives a $1^\circ$ carbocation, which is the least stable among the options.sec-Butyl iodide forms a $2^\circ$ carbocation, which is moderately stable.Isobutyl iodide also gives a $1^\circ$ carbocation at the carbon bearing iodine, so it is less stable than a $2^\circ$ carbocation.tert-Butyl iodide forms a $3^\circ$ carbocation, which is highly stabilized by hyperconjugation and the inductive effect of three methyl groups.Thus, tert-butyl iodide produces the most stable carbocation and shows the highest $\mathrm{S_N1}$ reactivity.Answer:Option D — tert-Butyl iodide