Identify A and B in the following reaction sequence: 2-(2-chloroethyl)cyclohexan-1-one (a cyclohexanone ring c

NCERT Exemplar · Q48

Q. Identify A and B in the following reaction sequence: 2-(2-chloroethyl)cyclohexan-1-one (a cyclohexanone ring carrying a -CH2CH2Cl side chain on the carbon next to the C=O) reacts with KCN to give A; A is then treated with H2/Pd to give B.

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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 $-\mathrm{C}\equiv\mathrm{N}$ 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 ( $\mathrm{H}^-$ ).

When you treat a nitrile with a strong reducing agent like lithium aluminium hydride ( $\mathrm{LiAlH}_4$ ) or hydrogen gas with a metal catalyst ( $\mathrm{H}_2$ /Ni, Pd, or Pt), you are essentially adding hydrogen atoms across that triple bond. The reaction does not stop at an imine (a $\mathrm{C}=\mathrm{N}$ intermediate) because the conditions are strongly reducing — it pushes all the way to a saturated $\mathrm{C}-\mathrm{N}$ single bond.

The key structural change: the nitrile carbon becomes a methylene ( $-\mathrm{CH}_2-$ ) group, and the nitrogen becomes an amino ( $-\mathrm{NH}_2$ ) group. So a nitrile $\mathrm{R}-\mathrm{C}\equiv\mathrm{N}$ becomes a primary amine $\mathrm{R}-\mathrm{CH}_2-\mathrm{NH}_2$ .

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.

\mathrm{R-C\equiv N \xrightarrow[\text{or } \mathrm{H_2}/\text{catalyst}]{\mathrm{LiAlH_4}} R-CH_2-NH_2}

The mechanism in brief (for $\mathrm{LiAlH}_4$ )

A hydride ion ( $\mathrm{H}^-$ ) attacks the electrophilic nitrile carbon, forming an imine anion intermediate: $\mathrm{R-C}^-=\mathrm{N}^-$ .
A second hydride adds to the imine carbon, giving a dianion: $\mathrm{R-CH}_2-\mathrm{N}^{2-}$ .
Aqueous work-up (adding water or dilute acid) protonates the nitrogen, yielding the free amine $\mathrm{R-CH}_2-\mathrm{NH}_2$ .

Watch out

A common mistake is to think the product is $\mathrm{R-NH}_2$ (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, $\mathrm{CH}_3\mathrm{C}\equiv\mathrm{N}$ (acetonitrile) gives $\mathrm{CH}_3\mathrm{CH}_2\mathrm{NH}_2$ (ethylamine), not methylamine.

Catalytic hydrogenation …

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Cyanide substitutes the chloride to make a nitrile (A); catalytic hydrogenation reduces that nitrile to a primary amine (B). The ketone is retained, giving a δ-amino ketone that can cyclise.

Step 1 – Nucleophilic substitution (A)

The cyanide ion (from KCN) is a good nucleophile and displaces chloride from the primary –CH2–Cl end of the side chain by SN2:

–CH2CH2Cl + KCN → –CH2CH2CN + KCl.

So A = 2-(2-cyanoethyl)cyclohexan-1-one (the cyclohexanone now carries a –CH2CH2CN side chain). One carbon is added to the chain.

Step 2 – Reduction of the nitrile (B)

H2 over Pd reduces the nitrile group to a primary amine:

–CH2CH2C≡N + 2 H2 → –CH2CH2CH2NH2.

The ketone C=O is not reduced under H2/Pd. So B = 2-(3-aminopropyl)cyclohexan-1-one.

Note (enrichment) …

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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 $\ce{C≡N}$ triple bond, converting it to a primary amine with the same carbon count. Ethanenitrile ( $\ce{CH3CN}$ ) becomes ethylamine ( $\ce{CH3CH2NH2}$ ).

Nitriles contain a carbon-nitrogen triple bond ( $\ce{C≡N}$ ), 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 ( $\ce{LiAlH4}$ ), you're forcing hydride ions ( $\ce{H-}$ ) onto that electron-deficient carbon.

The key insight: $\ce{LiAlH4}$ is a strong enough reductant to break the $\ce{C≡N}$ triple bond completely and add hydrogen across it, but it stops at the amine stage—it doesn't break the $\ce{C–N}$ single bond. The carbon skeleton stays intact; you simply convert $\ce{–C≡N}$ into $\ce{–CH2–NH2}$ .

Let's trace what happens to ethanenitrile:

Start with the structure of ethanenitrile

Ethanenitrile is $\ce{CH3–C≡N}$ . It has two carbons: one methyl group attached to the nitrile carbon.

The reduction mechanism (simplified)

$\ce{LiAlH4}$ delivers hydride ions in stages. The nitrile carbon is electrophilic, so $\ce{H-}$ attacks it. After the first hydride addition and protonation, you get an imine intermediate ( $\ce{CH3–CH=NH}$ ). 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.

Identify the product

$\ce{CH3–C≡N} \xrightarrow{\ce{LiAlH4}} \ce{CH3–CH2–NH2}$

This is ethylamine, a two-carbon primary amine. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which one of the following reactions will not form primary amine?$
(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. …
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CBSE 2024Set ANNUAL1 mark
$C_6H_5CN \xrightarrow{LiAlH_4} ?$
›Reveal solutionSolution

Lithium aluminium hydride reduces the $C\equiv N$ triple bond of a nitrile completely, adding four hydrogens to give a primary amine with one extra $CH_2$ carbon relative to the nitrile carbon.

Benzonitrile, $C_6H_5{-}C\equiv N$ , is reduced by the powerful hydride donor $LiAlH_4$ . The nitrile carbon accepts hydride and (after aqueous work-up) protons across both $\pi$ bonds of the $C\equiv N$ triple bond, converting it into a $-CH_2NH_2$ group:

$C_6H_5{-}C\equiv N \xrightarrow[\text{(ii) }H_2O]{\text{(i) }LiAlH_4} C_6H_5{-}CH_2{-}NH_2$

…

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CBSE 2021Set A1 markMCQ
$Q. CH3CN on reduction gives$
(a) CH4
(b) CH3COOH
(c) C2H5NH2
(d) C2H5COOH
›Reveal solutionSolution

Nitrile + [H] (e.g. LiAlH4 or Na/alcohol) -> primary amine; CH3CN gives C2H5NH2.

Reduction of a nitrile R-C≡N adds four hydrogen atoms across the C≡N triple bond to give a primary amine R-CH2-NH2. For methyl cyanide (acetonitrile):

CH3-C≡N + 4[H] -> CH3-CH2-NH2 (ethylamine, C2H5NH2)

…

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Q.Identify A and B in the following reaction sequence: 2-(2-chloroethyl)cyclohexan-1-one (a cyclohexanone ring carrying a –CH2CH2Cl side chain on the carbon next to the C=O) reacts with KCN to give A; A is then treated with H2/Pd to give B.

Concept understanding — Reduction of Nitriles to Amines

Reduction of Nitriles to Amines

The mechanism in brief (for $\mathrm{LiAlH}_4$ )

Catalytic hydrogenation …

Step 1 – Nucleophilic substitution (A)

Step 2 – Reduction of the nitrile (B)

Note (enrichment) …

Method: Chain Extension by Cyanide Substitution, then Selective Nitrile Reduction

Core Concept

Steps

Q.Identify A and B in the following reaction sequence: 2-(2-chloroethyl)cyclohexan-1-one (a cyclohexanone ring carrying a –CH2CH2Cl side chain on the carbon next to the C=O) reacts with KCN to give A; A is then treated with H2/Pd to give B.

Step 1 – Nucleophilic substitution (A)

Step 2 – Reduction of the nitrile (B)

Note (enrichment) …

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