Question

Does $$LiAl{H}_4$$ reduce Amides?

Answer 1

Hint: $$LiAl{H}_4$$ 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 $$1^o,2^o\text{ or }{3}^o$$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.