Question

Why, in the reduction of Aldehydes and Ketones, is H- considered a base?

Answer 1

Answer:

In some reductions of aldehydes and ketones, $\rm H^{-}$ donates a pair of electrons. As a result, by the Lewis Theory of acids and bases, $\rm H^{-}$ can be considered as a base.

Explanation:

When a reducing agent such as $\rm NaBH_4$ reduces aldehydes or ketones to alcohol, the $\rm H^{-}$ ion acts as an intermediary. In other words, these ions are produced in one step but completely consumed in another.

$\rm H^{-}$ ions are electron-rich; they are attracted to the electron-deficient region of aldehyde and ketone molecules. Each $\rm H^{-}$ ion has a lone pair of electrons. In each aldehyde and ketone molecule, at least one carbon is bonded with a double bond to an oxygen atom. Oxygen is very electronegative. It would draw the electron pairs away from the carbon atom. As a result, in aldehydes and ketones, carbon atoms with a $\rm C\text{=}O$ bound are electron-deficient; they carry partial-positive $\delta^{+}$ charges. The electron-rich $\rm H^{-}$ ions would be attracted to these electron-deficient carbon atoms.

During the reaction, $\rm H^{-}$ ions shares its two electrons with the partially-positive carbon atom in the aldehyde or ketone. It does so by forming a $\rm C-H$ bond. Even though that's not a complete electron transfer, the Lewis Theory of acids and bases would consider that $\rm H^{-}$ ion as an electron-pair donor. Because of that, the $\rm H^{-}$ ion can be called a Lewis base. (Similarly, the aldehyde or ketone that "accepts" the electron-pair would be a Lewis acid.)