Chemistry · Amines · NEET
With LiAlH4 (strong reagent) an amide is fully reduced to an amine (R-CH2-NH2), NOT an aldehyde. The C=O becomes a CH2. Only a milder, controlled reagent stops at the aldehyde; the standard NEET reagent LiAlH4 goes all the way to the amine.
In reduction, no carbon is added or removed. The carbonyl carbon (C=O) simply becomes CH2. So CH3-CO-NH2 (2 carbons) gives CH3-CH2-NH2 (2 carbons). This is different from nitrile reduction, where the carbon count also stays same because the nitrile carbon becomes CH2 too. But it is the OPPOSITE of adding a carbon.
The nitrogen stays in the molecule and ends up as the -NH2 (or -NHR) of the amine. LiAlH4 only reduces the C=O part. So a primary amide (R-CO-NH2) gives a primary amine, and an N-substituted amide (R-CO-NHR') gives a secondary amine.
Amides are very hard to reduce by simple catalytic hydrogenation (H2/Ni) under normal NEET conditions, so LiAlH4 is the standard reagent for this in Class 12. For NEET, remember LiAlH4 as the reagent that turns amides into amines.
An N-substituted amide gives a higher amine. R-CO-NHCH3 on LiAlH4 reduction gives R-CH2-NHCH3, a SECONDARY amine, because the CH3 on nitrogen stays. The C=O becomes CH2 and the N keeps whatever groups it already had.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Ethanamine, CH3-CH2-NH2 (a primary amine with 2 carbons).
LiAlH4 is a strong reducing agent. It supplies hydride (H-) to reduce the C=O of the amide to CH2.
No. A primary amide gives a primary amine, an N-monosubstituted amide gives a secondary amine, and an N,N-disubstituted amide gives a tertiary amine. The nitrogen keeps its existing groups.
It is one clean way to make amines while keeping the same carbon skeleton, and NEET often tests the reagent (LiAlH4) and the same-carbon-count point in one-liner questions.
