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I was studying the Finkelstein reaction and all sources through which I have studied this reaction indicate that the reaction uses $\ce{NaI},$ where $\ce{I-}$ acts as the nucleophile which will then replace $\ce{X}$ from the alkyl halide $\ce{R-X}.$

This is an $\mathrm{S_N2}$ reaction, hence we use a polar aprotic solvent (PAS, usually acetone) as PAS stabilizes the $\ce{Na+}$ cation. We know that while using a PAS the order of nucleophilicity is

$$\ce{F-} > \ce{Cl-} > \ce{Br-} > \ce{I-}.$$

The part I am struggling with is that even though $\ce{F-}$ is a better nucleophile than $\ce{I-}$ in a PAS, why do we still use $\ce{NaF}$ rather than $\ce{NaI}?$

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    $\begingroup$ Search for the solubility of NaF in acetone. $\endgroup$
    – Waylander
    Commented May 19, 2020 at 20:34

1 Answer 1

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Further to my comment - the solubility of Sodium Fluoride in Acetone is 2.4 microgrammes per 100g of solvent i.e. essentially insoluble source. There will be no F- in solution so the reaction will not occur.

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