Draw the products for the proton transfer reaction between sodium hydride and ethanol

Sodium hydride has the formula NaH where we have a sodium ion, Na⁺ and a hydride ion, H⁻. Hydride is an incredibly powerful base. While it is capable of acting as a nucleophile, if there is an acidic proton in a molecule, the hydride will deprotonate the molecule and grab the most acidic proton.

The pka of H⁻ is 35. The pka of ethanol is 16. The species with the larger pka is the better base and is capable of deprotonating the species with the smaller pka. Therefore, the hydride will deprotonate the acidic -OH proton of the alcohol in the following reaction:

CH₃CH₂OH + NaH → CH₃CH₂O⁻Na⁺ + H₂

The result of the reaction is the hydride deprotonates the proton of the alcohol and forms the alkoxide, which is a sodium salt. This reaction also leads to the formation of H₂ gas which ensures that this reaction is not reversible as the H₂ leaves the reaction mixture upon formation.

It is called a neutralization reaction

Anions are negative charged ions