Finkelstein reaction explicitely entails on the conversion of an alkyl chloride or an alkyl bromide to an alkyl iodide by treatment with a solution of sodium iodide in acetone. $$\ce{R−X + NaI→[acetone] R−I + NaX↓ }$$ $\ce{(X= Cl, Br; R= alkyl group)}$

What I think I understand

Acetone is a polar aprotic solvent so $\ce{NaBr and NaCl}$ formed are insoluble due to being mostly ionic while $\ce{NaI}$ has much covalent character (which can be thought of by Fajans's rules) (also organic solvents can dissolve covalent compounds) so, $\ce{I-}$ will be in solution while $\ce{Br- or Cl-}$ will not be there due to being precipitated.

So, Le Chatelier's principle will make the reaction to be towards the product. Therefore I also understand how can a weaker nucleophile can substitute a stronger nucleophile.

Nucleophilicity order of halides in polar aprotic solvent: $\ce{F- > Cl- > Br- > I-}$

What I ponder

  1. Why can't we do it on alkyl fluorides?
  2. Why only acetone? Will the reaction be same in other polar aprotic solvents too?

Some specific reactions which seems to deny all this

  1. $$\ce{CH3-CH2-Cl →[NaF/DMF] CH3-CH2-F}$$
  2. $$\ce{CH3-CH2-Cl →[NaBr/DMSO] No rxn}$$
  3. $$\ce{CH3-CH2-Br' →[NaBr/DMSO] CH3-CH2-Br}$$ where $\ce{Br'}$ is isotopic bromine


  1. Here I am talking about major organic products.
  2. DMF & DMSO are polar aprotic solvents.

I would like to know the theoretical reasoning for above reactions.

  • $\begingroup$ I have searched best from my side but couldn't get any more help than the information I provided in the question. $\endgroup$
    – D13G
    May 26 at 13:32


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