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In Finkelstein reaction we know when a primary alkyl chloride or bromide is treated with NaI in acetone, we get the corresponding alkyl iodide Now acetone is a polar aprotic solvent and order of nucleophilicity in acetone is F->Cl->Br->I-. I have read some articles on chemistry.stackexchange which explain why the driving force behind Finkelstein reaction is the precipitation of NaCl or NaBr. But one thing I am still not able to get why the Carbon chlorine or Carbon Bromine bond breaks in the first place by a relatively weaker nucleophile ie I-.

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First, $\ce{NaI}$ is soluble in acetone while $\ce{NaBr}$ and $\ce{NaCl}$ are not. Here, the reaction is $\ce{S_N2}$ so the nucleophile, $\ce{I-}$, will displace the other halide. The other leaving group halide will enter the solution as $\ce{NaBr}$ or $\ce{NaCl}$ which will precipitate. It doesn't matter if $\ce{Cl-}$ or $\ce{I-}$ is a stronger nucleophile in acetone since they are the leaving group and won't be present in the solution once they are precipitated.

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