Imagine you have a molecule with a carbonyl group (C=O) that is directly attached to a nitrogen atom — that's an amide. The carbonyl is electron-deficient (the oxygen pulls electron density), and the nitrogen has a lone pair. Now, you want to turn this into an amine, where the carbon is fully saturated and the nitrogen is the only functional group left. How do you get there?
The key idea: you need to add hydrogen across the C=O bond and also break the C−N bond in a controlled way, so that the nitrogen stays attached to the carbon chain. That's exactly what a strong reducing agent like lithium aluminium hydride (LiAlHX4) does.
The Intuition
Think of LiAlHX4 as a source of hydride ions (HX−) — tiny, negatively charged hydrogen atoms that love to attack electron-poor carbons. In an amide, the carbonyl carbon is the most electron-poor spot. The hydride attacks there first.
But here's the twist: after the first attack, the molecule rearranges. The C=O bond becomes a C−OX− (an alkoxide), and the nitrogen's lone pair helps push electrons, eventually kicking out the oxygen as OHX− and forming an imine intermediate (RCH=NH). That imine is then attacked by another hydride, giving you the final amine.
The net result: the carbonyl carbon becomes a CHX2 group, and the nitrogen remains attached. No carbon atoms are lost — the amide carbon becomes the CHX2 next to the nitrogen.
Note
This is different from reducing an amide with other agents like HX2/Pd or NaBHX4, which either don't work or give different products. LiAlHX4 is uniquely powerful here.
The Precise Statement
Reduction of an amide with LiAlHX4 converts the −CONHX2 group into a −CHX2NHX2 group, without loss of carbon atoms.
For a primary amide:
R−CONHX2LiAlHX4R−CHX2NHX2
For a secondary amide (RCONHRX′):
R−CONHRX′LiAlHX4R−CHX2NHRX′
For a tertiary amide (RCONRX′RX′′):
R−CONRX′RX′′LiAlHX4R−CHX2NRX′RX′′
RCONHX2+4[H]LiAlHX4RCHX2NHX2+HX2O
The "4[H]" represents the net addition of four hydrogen atoms — two go to the carbon, two to the nitrogen.
Why This Matters for Exams
No carbon loss: The amide carbon becomes the CHX2 directly attached to nitrogen. This is a key difference from other carbonyl reductions (e.g., acid chlorides to alcohols lose carbon via CO). …
CH3CONH2 (acetamide) with LiAlH4/ether reduces the C=O to CH2, giving ethylamine, CH3CH2NH2.
Aniline with HNO3/H2SO4 (nitration) gives a mixture of o- and p-nitroaniline (with some oxidation/tarring, since –NH2 is protonated to the deactivating −NH3+ under the strongly acidic conditions, so the aniline is usually acetylated first in practice — but the direct/simple textbook expected answer is o-/p-nitroaniline). …
LiAlH4 reduces the amide carbonyl of acetamide all the way to a CH2 group, giving ethylamine; nitrating aniline directly with HNO3/H2SO4 gives a mixture of nitroanilines because the strongly acidic medium partly protonates –NH2 to the meta-directing −NH3+.
(i) CH3CONH2 + 4[H] (LiAlH4/ether): Amides are reduced by LiAlH4 (a strong reducing agent) at the carbonyl carbon, replacing the C=O oxygen with two hydrogens, converting the amide into the corresponding amine (one carbon fewer than nothing lost — same carbon skeleton, just C=O → CH2):
CH3CONH2+4[H]LiAlH4/etherCH3CH2NH2+H2O
The product is ethylamine (a primary amine with the same number of carbon atoms as the starting amide).
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2018Set ANNUAL1 mark
Q.How is benzamide converted into benzylamine ?
›Reveal solutionSolution
Reducing the amide carbonyl of benzamide with a strong hydride reducing agent converts −CONH2 into −CH2NH2, giving benzylamine directly, with no change in carbon count.
Benzamide, C6H5−CONH2, is the amide of benzoic acid. Amides are reduced to the corresponding primary amine (same number of carbons) by strong reducing agents such as lithium aluminium hydride (LiAlH4) or diborane (B2H6):