Why is nucleophilicity is solvent dependent?

1 Answer
Jul 11, 2015

Here's an example.

Let's hypothetically react Li^((+)) [(CH_2)_3CH_3]^((-))Li(+)[(CH2)3CH3]() (commonly BuLiBuLi) with acetone. Normally, BuLiBuLi is a fantastic nucleophile due to lithium's lewis acid characteristics.

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

Acetone:
![https://upload.wikimedia.org/](useruploads.socratic.org)

What would most likely happen is that since BuLiBuLi 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 EtOHEtOH.

At that point, BuLiBuLi would become butane, which is clearly nonreactive as a poor nucleophile. Then, EtO^(-)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. BuLiBuLi grabs a proton and loses its reactivity, allowing EtO^(-)EtO to be an additional nucleophile (there's still some BuLiBuLi leftover)
  2. BuLiBuLi attacks acetone and the reaction proceeds to EtOHEtOH protonating the tetrahedral intermediate to form a tertiary alcohol.

The result then is a mixture of the butane, EtOHEtOH, acetone, the tertiary alcohol, and the product of the mechanism where EtO^(-)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.