Question

Which type of reaction in the reduction of carbonyl compound with $LAH$ and $NaB{H}_4$ occurs, and which nucleophile takes part in the reaction?

• A. Nucleophilic addition and $Al{H}_4^{\ominus }$ or $B{H}_4^{\ominus }$
• B. Nucleophilic addition and $H^{\ominus }$
• C. Nucleophilic substitution and $Al{H}_4^{\ominus }$ or $B{H}_4^{\ominus }$
• D. Nucleophilic substitution and $H^{\ominus }$

Answer 1

The correct option is B Nucleophilic addition and $H^{\ominus }$

Reaction of carbonyl compounds with $LiAl{H}_4$ and $NaB{H}_4$ is written as:

$R\left(CO\right)R^{\prime }+LiAl{H}_4$ or $NaB{H}_4\to RCH\left(OH\right)R^{\prime }$

The reaction is nucleophilic addition.

Aldehydes and ketones are most readily reduced with hydride reagents.

The reducing agents $LiAl{H}_4$ and $NaB{H}_4$ act as a source of $4H^{-}$ (hydride ion). Overall, $2H$ atoms are added across the $C=O$ to give $-HC-O-H$.

Option B is correct.