Finkelstein Reaction

Question

Heating an alkyl chloride with sodium iodide in dry acetone steadily converts it into an alkyl iodide; what drives this exchange forward?

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The Finkelstein reaction interconverts alkyl halides using sodium iodide dissolved in dry acetone, and it is fundamentally an SN2 halide-exchange reaction. Sodium iodide is freely soluble in acetone, whereas sodium chloride and sodium bromide are only sparingly soluble. As iodide displaces chloride, the liberated chloride precipitates out as solid sodium chloride, removing a product from solution and shifting the equilibrium toward the alkyl iodide by Le Chatelier's principle. The option about higher C-I bond energy is wrong because the C-I bond is actually the weakest carbon-halogen bond, not the strongest. Greater electronegativity of iodine is incorrect since iodine is in fact the least electronegative halogen. A stable carbocation intermediate is wrong because Finkelstein proceeds by a concerted SN2 pathway, not through a carbocation. This is the standard NCERT preparation method for converting chloro/bromo compounds to iodo compounds. Sanity check: solubility-driven precipitation, not bond strength, governs the direction, consistent with an equilibrium pulled by removal of NaCl.

Finkelstein Reaction

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