Because of the effects of solvation, halide ions have differing nucleophilicity in protic solvents vs aprotic solvents, iodide being the best nucleophile in protic solvents and fluoride being best in aprotic solvents. For the sake of simplicity, fluoride will be ignored due to its high electronegativity and unusual reactive properties, and chloride will be used instead.

So would that mean that an alkyl chloride will react with HI in a protic solvent to form the alkyl iodide? And the alkyl chloride will not react with HI in an aprotic solvent? And by the same token, would an alkyl iodide react with HCl to form alkyl chloride in an aprotic solvent but not in a protic solvent?

I also realize that iodide has a much greater polarizability than chloride and that must be taken into account as well. Iodide is more a thus a more stable anion than chloride, making iodide much better leaving group than chloride.

But which factor will have a greater overall effect? Do aprotic/protic solvents affect the leaving group capability (ie although chloride is less polarizable and more unstable relative to iodide, after leaving in a protic solvent the H-bonding "shell" will stabilize the anion and prevent it from reacting with the alkyl iodide formed. Conversely in aprotic solvent, the chloride ion would not be stabilized after leaving and thus could reattack the alkyl iodide to reform the alkyl chloride)?

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    $\begingroup$ en.wikipedia.org/wiki/Finkelstein_reaction $\endgroup$
    – DHMO
    Commented Oct 4, 2016 at 15:53
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    $\begingroup$ @DHMO Exploiting solubility of products would be the more efficient route, but my question is more conceptually oriented, though it does make implications that support my statements. Acetone is polar aprotic and thus chlorides and bromides would be more nucleophilic than iodide, causing the reaction to favor the alkyl chloride or bromide side over the iodide. However the reaction is "driven forward" by exploiting the solubility of sodium bromide and chloride salts in acetone. There wasnt an example of an iodide salt displacing an alkyl chloride or bromide in a protic solvent though. $\endgroup$
    – KeatonB
    Commented Oct 4, 2016 at 16:04


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