Why is nucleophilicity is solvent dependent?

1 Answer
Jul 11, 2015

Here's an example.

Let's hypothetically react #Li^((+)) [(CH_2)_3CH_3]^((-))# (commonly #BuLi#) with acetone. Normally, #BuLi# is a fantastic nucleophile due to lithium's lewis acid characteristics.

http://www.umich.edu/

If you solvate #BuLi# in the optimal amount of ethanol (commonly #EtOH#), you have now in solution, before anything happens, #BuLi#, #EtOH#, and acetone.

Acetone:
https://upload.wikimedia.org/

What would most likely happen is that since #BuLi# has such a high nucleophilicity, instead of reacting with acetone all the time, there is a good chance it would also steal a proton from #EtOH#.

At that point, #BuLi# would become butane, which is clearly nonreactive as a poor nucleophile. Then, #EtO^(-)# forms and it becomes a potential nucleophile to attack acetone (but less often, as it's a worse nucleophile).

At this point, you may realize that you now have a situation where:

  1. #BuLi# grabs a proton and loses its reactivity, allowing #EtO^(-)# to be an additional nucleophile (there's still some #BuLi# leftover)
  2. #BuLi# attacks acetone and the reaction proceeds to #EtOH# protonating the tetrahedral intermediate to form a tertiary alcohol.

The result then is a mixture of the butane, #EtOH#, acetone, the tertiary alcohol, and the product of the mechanism where #EtO^(-)# attacks acetone. Ideally you don't want a mixture that you'd have to separate and purify later. If you got a pure product, that's what you should want.

So naturally, it's a good idea, for example, to not use a protic solvent when using an anionic nucleophile, because it may actually deactivate the nucleophile.