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My book states that:

Finkelstein reaction is particularly useful for preparing iodoalkanes. The iodoalkanes are obtained by heating chloro or bromoalkanes with a concentrated solution of sodium iodide in dry acetone.
$$\ce{R-X~+~NaI->[\ce{acetone,~reflux}]R-I~+NaX}$$ $(\ce{X=Cl,~Br;~R=alkyl ~group})$
Sodium chloride and sodium bromide being less soluble in acetone get precipitated from the solution and can be removed by filtration. This also prevents the backward reaction to occur according to Le Chartelier's principle. The reaction gives the best results with primary halides.

My question:

Can we prepare alkyl fluoride (not aryl fluoride) by Finkelstein reaction?

My book mentions about another reaction called Swarts reaction, through which alkyl fluoride is prepared. In Swarts reaction, alkyl fluorides are prepared by heating alkyl chlorides or bromides with salts such as mercurous fluoride ($\ce{Hg2F2}$), silver fluoride ($\ce{AgF}$), cobalt fluoride ($\ce{CoF3}$) or antimony trifluoride ($\ce{SbF3}$). Why can't we prepare alkyl fluoride by treating sodium fluoride with alkyl halide (Finkelstein reaction)?

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    $\begingroup$ Fluoride ions are not good at $\mathrm{S}_{\mathrm{N}}2$ reactions since they are bad nucleophiles. $\endgroup$
    – Philipp
    Nov 15, 2013 at 15:50

2 Answers 2

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In the future I recommend consulting March's Advanced Organic Chemistry before asking such questions. It is a priceless reference work for organic synthesis methods. The following text is result of brief consultation with the book.

The equilibrium shift in Finkelstein reaction is governed by relative nucleofility of halogenide-ions. In general, in polar solvents iodine is the strongest nucleophile, because it is poorly solvated, while in nonpolar solvents chlorine is the strongest solvent as it is smaller.

Fluoride in non-polar solvents should be the strongest nucleophile of all halogenide-ions, but it so strongly solvates in protonic solvents that it almost do not migrate in organic phase even if phase-transfer catalists are present. Strong basicity of fluoride does not help either as beta-eliminations becomes significant side-reaction.

This article claims to develop practically meaningful procedure of clean fluoro-de-halogenations.

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As far as the Swarts reaction is concerned: There are indeed reports using $\ce{SbF3}$ alone.

Most of the time, however, the active component in the Swarts reaction for the fluorination through halogene exchange is not an $\ce{Sb(III)}$ fluoride, but a mixed $\ce{Sb(V)}$ halide, such as $\ce{SbF3Cl2}$. This may be prepared by reaction of $\ce{SbF3}$ with chlorine (either previously or in situ), or with $\ce{SbCl5}$ (typically in situ).

Anyway, I'd give the reaction of an alkyl iodide with sodium fluoride in dry DMF a try. Alternatively, I'd start from a tosylate.

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  • $\begingroup$ Extending Finkelstein reaction's starting materials to pseudohalides reportedly works $\endgroup$
    – Buttonwood
    Feb 9, 2014 at 12:14

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