Question

Does $LiAl{H_4}$ reduce Amides?

Answer 1

$LiAl{H_4}$ is a very strong reducing agent. It reduces aldehydes, amides, ketones, esters, carboxylic acid and even carboxylate salts to alcohols.
Amide is a functional group that contains both nitrogen and carboxyl groups in it. It is usually represented as below:

$LiAl{H_4}$ will reduce Amides to Amines and this means that the oxygen that was present in the amide is removed from the molecule after the reaction completes.

Complete answer:
Let’s see how the reaction occurs when amide is treated with $LiAl{H_4}$
First there is a nucleophilic attack by one of the hydrides of $LiAl{H_4}$ . Because the carbonyl group is highly polar the oxygen becomes negative and thus $Al$ having negatively charge can now attack the positively charged carbonyl carbon and thus the following step occurs

Now since there is an oxygen that can act as a leaving group we will get the following reaction which leads to the formation of an iminium ion

Now there is a nucleophilic attack by the hydride of $LiAl{H_4}$ causing for the formation of the final product, amine
That is given by the following reaction:

Thus we get Amine as the product when amides are treated with $LiAl{H_4}$ which is a strong reducing agent.
Thus we can write the general form of the reaction as:

Note:
$NaB{H_4}$ is also a reducing agent but it is very less reactive than $LiAl{H_4}$. $NaB{H_4}$ can reduce aldehydes, ketones and acid chlorides but cannot reduce amides or esters.
Amides can be converted to ${1^o},{\text{ }}{2^o}{\text{ or }}{3^o}$amine just by using $LiAl{H_4}$. For doing this we have to select the suitable amides to begin the reaction with.
The purpose of using ${H_2}O$ at the end is for a bit of workup. It neutralizes strongly basic reagents at the end of the reaction thus helping in the rate of the reaction.