Regarding leaving groups and solvent effects (SN1 and SN2 reactions), I would like to ask some questions.

As the nucleophilic character is also influenced by the solvents, there is some indication (to me at least) that the leaving group can be influenced as well. Unfortunately, there is not much information found. Let us consider the halides, the quality of the leaving group is $\ce{I-} > \ce{Br-} > \ce{Cl-} > \ce{F-}$ as the larger ions are more polarizable, more stable and less basic.

So imagining a polar protic solvent, wouldn't the solvation stabilize $\ce{F-}$ via hydrogen bonds? I totally understand, that fluorine forms tighter bonds with carbons, so it naturally is a bad leaving group.

Considering the Finkelstein reaction, the driving force of the reaction is the formation of an insoluble salt, so basically the solvent influences the leaving group ability.

So, can I generally say that the solvent can theoretically influence the leaving group ability? Is there some general thumb rule I can use? Until now, I haven't seen any reaction, where $\ce{Cl-}$ is preferred over $\ce{I-}$ as a leaving group except for the Finkelstein reaction.


1 Answer 1


The solvent generally doesn't affect the leaving group ability of an anion. But coming to your Frinkelstein case, it a special one cause here some precipitate is being formed (yes, that's correct that it is due to lower solubility product of say $\ce{NaCl}$ in acetone). As soon as some ppt. is formed, those atoms lose their ionic identity, settle at the bottom and don't interfere in the solvent as solute and so their concentration decreases in the solvent. Due to Le Chatelier's principle, the reaction goes forward to a great extent. This is also the case for other super leaving groups like gases (for eg. $\ce{N2}$, $\ce{H2}$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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