Question

How reduction of the following takes place:
i.Reduction of acetaldehyde
ii.Reduction of ester

Answer 1

Hint: When a loss in oxygen atom i.e., an electronegative element or a gain in hydrogen atom i.e., electropositive element is observed within a compound during an organic reaction, then it is said to be a reduction reaction. The reduction of a compound leads to net decrease in the electronegativity of the compound.

Complete answer: The reduction of the given compounds takes place as follows:
(a) Reduction of acetaldehyde:
Step-1: Reaction of acetaldehyde with $NaB{H}_4$ to form the respective sodium derivative along with the removal of $B{H}_3$. The reaction takes place as follows:

Step-2: Protonation of the given compound occurs in the presence of water molecules to form primary alcohol. The reaction proceeds as follows:

Hence, formation of ethanol takes place on the reduction of acetaldehyde.
(b) Reduction of ester:
Step-1: The hydrogen ion from $LiAl{H}_4$ attacks the carbonyl centre of the ester and formation of respective aldehyde takes place along with the removal of $OR$ group. The reaction proceeds as follows:

Step-2: The hydrogen ion of $LiAl{H}_4$ again attacks the carbonyl centre of aldehyde and the compound is reduced to its respective alcohol. The reaction proceeds as follows:

Hence, on reduction of ester the formation of primary alcohol takes place.

Note:
It is important to note that $NaB{H}_4$ is a mild reducing agent and can only be used in the reduction reaction of aldehydes and ketones. It cannot be used for the reduction of carboxylic acids and esters. Only strong reducing agents like $LiAl{H}_4$ can be used for the reduction of esters and acids.