# Polar Protic, Polar Aprotic and Non-Polar Solvents

## Key Questions

Non-polar solvents are non-polar molecules that can be used as solvent.

#### Explanation:

Non-polar solvents are any non-polar molecules that can be used as a solvent.

Example:
Hexane, pentane, heptane, etc.
Carbon tetrachloride $\text{C} C {l}_{4}$.

• Here's an example.

Let's hypothetically react $L {i}^{\left(+\right)} {\left[{\left(C {H}_{2}\right)}_{3} C {H}_{3}\right]}^{\left(-\right)}$ (commonly $B u L i$) with acetone. Normally, $B u L i$ is a fantastic nucleophile due to lithium's lewis acid characteristics.

If you solvate $B u L i$ in the optimal amount of ethanol (commonly $E t O H$), you have now in solution, before anything happens, $B u L i$, $E t O H$, and acetone.

Acetone:

What would most likely happen is that since $B u L i$ 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 $E t O H$.

At that point, $B u L i$ would become butane, which is clearly nonreactive as a poor nucleophile. Then, $E t {O}^{-}$ 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. $B u L i$ grabs a proton and loses its reactivity, allowing $E t {O}^{-}$ to be an additional nucleophile (there's still some $B u L i$ leftover)
2. $B u L i$ attacks acetone and the reaction proceeds to $E t O H$ protonating the tetrahedral intermediate to form a tertiary alcohol.

The result then is a mixture of the butane, $E t O H$, acetone, the tertiary alcohol, and the product of the mechanism where $E t {O}^{-}$ 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.