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 $\ce{NaI}?$

  • 6
    $\begingroup$ Search for the solubility of NaF in acetone. $\endgroup$ – Waylander May 19 at 20:34

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.

| improve this answer | |
  • $\begingroup$ Thanks a ton... $\endgroup$ – TheChemist May 29 at 18:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.