Question
Question: Complete the following reaction:  are generally reduced to the primary and secondary alcohol.
However, the LAH reduces the amides group to the amine. The lone pair of the nitrogen assists the mechanism. The lone pair pushes or knocks out the oxygen. The oxygen is better leaving the group than nitrogen.
The reaction mechanism involves the following steps:
Step 1) The LAH attacks on the carbonyl carbon, such that the LAH transfers its proton to the carbonyl atom, and the oxygen acquires the lone pair of electrons.
Step 2) The lone pair on the electrons shift to the adjacent carbon atom followed by the removal of the oxygen atom. The nitrogen acquires a positive charge. The resonating structures are drawn to show the removal of the oxygen atom.
Step 3) In the next step, the next molecule of LAH attacks on the carbon followed by nitrogen accommodating its lone pair.
Thereby, the amide is converted into an amine.
Hence, the obtained product is N-methyl pyrrolidine.
Additional Information:
Among the complex hydrides, Lithium Aluminium Hydride is the strongest reducing agent because it reduces four molecules at a time.
Complex hydrides in general cannot reduce the non-polar carbon-carbon pi-bonds. That is, these reducing agents are not used in the reduction or hydrogenation of alkenes and alkynes.
Since Lithium Aluminium Hydride is a strong reducing agent, it can reduce carboxylic acids, esters, and amides.
Sodium borohydride cannot reduce an ester, an amide, or carboxylic acids. It can be used to selectively reduce an aldehyde or a ketone group in a compound.
Note: At a glance, it looks like the carbonyl group is reduced to the hydroxyl group. However, that cannot be true. The amide contains the lone pair of electrons on the nitrogen atom which pushes out the oxygen. But, in aldehyde or ketone , no atom contains the lone pair, thus the reduction ends at the hydroxyl group. Note that acid workup is used for the protonation, but it can be assumed as there is a tiresome step.