Concept understanding — Reduction of Nitriles to Amines
Reduction of Nitriles to Amines
Think of a nitrile as a molecule with a carbon atom triple-bonded to a nitrogen atom — that's the −C≡N group. That triple bond is packed with electrons and is quite polarised: the nitrogen is more electronegative, so it pulls electron density toward itself, leaving the carbon slightly positive. This makes the carbon a good target for a nucleophilic attack by a hydride ion (H−).
When you treat a nitrile with a strong reducing agent like lithium aluminium hydride (LiAlH4) or hydrogen gas with a metal catalyst (H2/Ni, Pd, or Pt), you are essentially adding hydrogen atoms across that triple bond. The reaction does not stop at an imine (a C=N intermediate) because the conditions are strongly reducing — it pushes all the way to a saturated C−N single bond.
The key structural change: the nitrile carbon becomes a methylene (−CH2−) group, and the nitrogen becomes an amino (−NH2) group. So a nitrile R−C≡N becomes a primary amine R−CH2−NH2.
Notice that the carbon chain has grown by exactly one carbon atom — the nitrile carbon is now part of the alkyl chain. This is a powerful method to extend a carbon skeleton by one unit while introducing an amine functionality.
R−C≡NLiAlH4or H2/catalystR−CH2−NH2
The mechanism in brief (for LiAlH4)
A hydride ion (H−) attacks the electrophilic nitrile carbon, forming an imine anion intermediate: R−C−=N−.
A second hydride adds to the imine carbon, giving a dianion: R−CH2−N2−.
Aqueous work-up (adding water or dilute acid) protonates the nitrogen, yielding the free amine R−CH2−NH2.
Watch out
A common mistake is to think the product is R−NH2 (an amine with the same number of carbons). It is not — the nitrile carbon is reduced and retained, so you always get one extra carbon in the chain. For example, CH3C≡N (acetonitrile) gives CH3CH2NH2 (ethylamine), not methylamine.
LiAlH4 reduces the amide's C=O down to CH2 while retaining that same carbon in the chain: CH3CONH2LiAlH4CH3CH2NH2. This is fundamentally different from the Hofmann bromamide degradation of the identical starting amide, CH3CONH2+Br2/KOH→CH3NH2 (methylamine) +CO2, which instead EXPELS the carbonyl carbon as carbon dioxide/carbonate. So the two routes, starting from the same amide, give amines that differ by exactly one carb …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2026Set 56/2/11 markMCQ
Q.Identify the compound produced by the reduction of Ethanenitrile with Lithium aluminium hydride : (A) Ethylamine (B) Ethanal (C) Propylamine (D) Methylamine
›Reveal solutionSolution
Lithium aluminium hydride reduces nitriles by adding four hydrogens across the C≡N triple bond, converting it to a primary amine with the same carbon count. Ethanenitrile (CHX3CN) becomes ethylamine (CHX3CHX2NHX2).
Nitriles contain a carbon-nitrogen triple bond (C≡N), one of the most polar and reactive functional groups in organic chemistry. When you treat a nitrile with a powerful reducing agent like lithium aluminium hydride (LiAlHX4), you're forcing hydride ions (HX−) onto that electron-deficient carbon.
The key insight: LiAlHX4 is a strong enough reductant to break the C≡N triple bond completely and add hydrogen across it, but it stops at the amine stage—it doesn't break the C−N single bond. The carbon skeleton stays intact; you simply convert −C≡N into −CHX2−NHX2.
Let's trace what happens to ethanenitrile:
Start with the structure of ethanenitrile
Ethanenitrile is CHX3−C≡N. It has two carbons: one methyl group attached to the nitrile carbon.
The reduction mechanism (simplified)
LiAlHX4 delivers hydride ions in stages. The nitrile carbon is electrophilic, so HX− attacks it. After the first hydride addition and protonation, you get an imine intermediate (CHX3−CH=NH). A second hydride attack on that imine carbon, followed by aqueous workup, gives the primary amine.
Count the carbons
Ethanenitrile has two carbons. Reduction doesn't add or remove carbons—it only converts the nitrile group to an amine. So the product must also have two carbons.
Q.Which one of the following reactions will not form primary amine?
(a) CH3CONH2 --KOH, Br2-->
(b) CH3CN --LiAlH4-->
(c) CH3NC --LiAlH4-->
(d) CH3CH2CONH2 --LiAlH4-->
›Reveal solutionSolution
Three of the four routes give a primary amine directly; only the reduction of an alkyl isocyanide (R-NC) yields a secondary amine, because the extra carbon from the isocyanide's -NC group becomes a methyl group bonded to the same nitrogen.
Checking each route:
CH3CONH2 --KOH, Br2--> is the Hofmann bromamide degradation. It converts an amide to a primary amine with loss of one carbon: CH3CONH2 -> CH3NH2 (methanamine). Forms a primary amine.
CH3CN --LiAlH4--> reduces a nitrile fully to a primary amine: CH3CN -> CH3CH2NH2 (ethanamine). Forms a primary amine.
…
Q.Complete the following reaction: C6H5CNLiAlH4?
›Reveal solutionSolution
Lithium aluminium hydride reduces the C≡N triple bond of a nitrile completely, adding four hydrogens to give a primary amine with one extra CH2 carbon relative to the nitrile carbon.
Benzonitrile, C6H5−C≡N, is reduced by the powerful hydride donor LiAlH4. The nitrile carbon accepts hydride and (after aqueous work-up) protons across both π bonds of the C≡N triple bond, converting it into a −CH2NH2 group: