See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: The hydrolysis of primary alkyl halides proceeds via an SN2 mechanism. The rate of an SN2 reaction depends on the nucleophilicity of the attacking species and the leaving group ability of the departing group. Step 1: Understand the role of KI. When KI is added to a reaction mixture containing a primary alkyl halide (R-X, where X = Cl or Br), the iodide ion (I⁻) acts as a nucleophile and displaces the original halide in a fast SN2 step, converting R-X into R-I. This first step is fast because I⁻ is a powerful nucleophile (large, polarizable, low solvation energy). Step 2: The alkyl iodide (R-I) formed is then hydrolyzed by water in a second SN2 step. This step is faster than the direct hydrolysis of R-Cl or R-Br because I⁻ is an excellent leaving group (weak conjugate base of HI, stable large anion, low bond energy with carbon). Step 3: The iodide ion regenerated in the second step can re-enter the catalytic cycle, effectively acting as a catalyst. This two-step sequence (halide exchange followed by hydrolysis) is faster overall than direct hydrolysis of the original alkyl halide. Why other options fail: - (a) Solubility of KI in organic solvents is not the primary reason; KI is actually sparingly soluble in many organic solvents, and solubility alone does not explain rate acceleration. - (b) Iodide ion being a poor leaving group is incorrect; I⁻ is one of the best leaving groups among halides. - (c) Iodide ion being a strong base is incorrect; I⁻ is a very weak base (conjugate base of strong acid HI), which is actually why it is a good leaving group. Therefore, the correct answer is D.