Question

The major product of the following reaction is:
$\xrightarrow{{{\text{LiAl}}{{\text{H}}_4}}}$
A. ${\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{CHO}}$
B.
C.
D. ${\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{OH}}$

Answer 1

${\text{LiAl}}{{\text{H}}_4}$ is a very strong reducing agent. It reduces carbonyl, carboxylic acid and ester. It also reduces nitrile, amide and aryl nitro group to amine. Reduction is the exchange of electrons in association with hydrogen and oxygen or some other atom.

Complete step by step answer:
Esters are the carboxylic acid derivatives in which the hydroxyl group is replaced by an alkoxy group. They are the chemical organic compounds which are formed when an alcohol reacts with a carboxylic acid. It has the molecular formula ${\text{RCOOR}}$.
Lithium aluminium hydride is a very strong hydride $\left( {{{\text{H}}^ - }} \right)$. It can reduce carboxylic acids, esters, aldehydes and ketones.
Both reduction and hydrolysis of esters produces alcohol. But reduction of esters using ${\text{LiAl}}{{\text{H}}_4}$ produces two alcohols while hydrolysis of esters produces an alcohol and an acid.
Generally primary alcohols are produced by reducing esters. The mechanism involves the nucleophilic attack by the hydride, leaving group removal, nucleophilic attack by hydride anion and the protonation of alkoxide. Thus alcohol is formed.
When the given compound is reacted with ${\text{LiAl}}{{\text{H}}_4}$, it produces two alcohols. One alcohol is produced from the alcohol portion of ester and the other alcohol is produced from the reduction of carboxylate portion.
$\xrightarrow{{{\text{LiAl}}{{\text{H}}_4}}}$

Thus, the correct option is B.

Note:
This type of reaction occurs in two steps- nucleophilic acyl substitution reaction and nucleophilic addition reaction. ${\text{LiAl}}{{\text{H}}_4}$ is more reactive than ${\text{NaB}}{{\text{H}}_4}$. Therefore ${\text{NaB}}{{\text{H}}_4}$ cannot be used for reducing the acids or esters. ${\text{NaB}}{{\text{H}}_4}$ can reduce aldehydes, ketones and acid halides to alcohols. Reduction of esters using ${\text{NaB}}{{\text{H}}_4}$ is a slow process. ${\text{LiAl}}{{\text{H}}_4}$ can reduce ketones to secondary alcohols.