Mechanism for Ketone and aldehyde reduction by NaBH4 to alcohol

Sodium borohydride is a salt which is made up of a sodium cation (Na+) and a borohydride anion (BH4-) The sodium ion plays no important role in the reaction, so we will ignore it (ions like sodium and potassium are seldom directly involved in reactions. They are present merely to maintain charge balance so that stable compounds can be added to reaction mixtures. They are often called spectator ions.) The borohydride ion is the important player in this process, and it is the B-H bond that we want to examine.

From periodic it is understood that both boron and aluminum are metals with relatively low electronegativities. Each is less electronegative than carbon, and since hydrogen has about the same electronegativity as carbon we can conclude that the B-H bond is polarized with the boron positive and the hydrogen negative. So B-H bond effectively serves as a source of hydride ion (:H-). This is given below.

 Mechanism of aldehyde to alcohol conversion

 Below attack by a nucleophile at the carbonyl carbon, followed by protonation of the carbonyl oxygen.

The same mechanism applies to the reduction of ketones by sodium borohydride and to reactions of carbonyl groups with lithium aluminum hydride. In case of lithium aluminum hydride, the reagent itself highly reactive with water, so the water is added after the lithium aluminum hydride has reacted.

Mechanism (LAH reduction nitrile to amine)

Below is the mechanism of the reduction of an nitrile to amine using LiAlH4 (LAH).  I have kept it simple by not going into much details.

 Transfer of a hydride ion from lithium aluminum hydride to the carbon atom of the nitrile generates an imine intermediate, which is in equilibrium with the corresponding imino alanate.

 The imino alanate converts to the amino alanate after adddition of an additional hydride ion.

Hydrolysis yields the primary amine.

Mechanism (LAH reduction ester to alcohol)

Below is the mechanism of the reduction of an ester to alcohol using LiAlH₄ (LAH).  I have kept it simple by not going into details.

Step 1: The nucleophilic H from the hydride reagent adds to the electrophilic C in the polar carbonyl group of the ester. Electrons from the C=O move to the electronegative O creating the tetrahedral intermediate a metal alkoxide complex.

Step 2: The tetrahedral intermediate collapses and displaces the alcohol portion of the ester as a leaving group, in the form of the alkoxide, RO^-. This produces an aldehyde as an intermediate.

Step 3: Now we are reducing an aldehyde.  The nucleophilic H from the hydride reagent adds to the electrophilic C in the polar carbonyl group of the aldehyde. Electrons from the C=O move to the electronegative O creating an intermediate metal alkoxide complex.

Step 4: This is the work-up step, a simple acid/base reaction. Protonation of the alkoxide oxygen creates the primary alcohol product from the intermediate complex.