1.The following question came up on a recent practice quiz: In which of the following solvents will a bromoalkane with NaF SN2 reaction proceed faster, acetone or hexane? 2. No equations. 3. I reasoned that in order for the reaction to proceed the salt must first dissociate. Although acetone will do a poor job of dissociating NaF, hexane-being nonpolar-will dissociate far less still. As such, there will be much more free F- in the acetone solution, allowing the reaction to proceed faster. However, the TA giving the quiz claimed that hexane will be the faster solvent as, although acetone is generally thought of as being aprotic polar, enough of it will tautomerize into propen-2-ol to allow it to behave as a polar protic solvent, thereby solvating the F-. I am somewhat skeptical about his entire logic, as the amount of propen-2-ol present at any given moment will be extremely low. However, even if acetone/propen-2-ol could solvate an anion as well as a regular alcohol or water (clearly not the case), I would still mantain my position as it seems to me that while a solvated F- may have a low reactivity, a F- in an ionic bond with Na+ will have essentially no reactivity. I would appreciate any input anyone can give on the issue, however I would be especially grateful if anyone is aware of any experimental data to indicate either way.