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
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.
KCN displaces the chloride (SN2) to give a nitrile, A. H2/Pd then reduces the –C≡N to –CH2NH2, giving a primary amine B (which, being a δ-amino ketone, can cyclise intramolecularly). …
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.
Method: Chain Extension by Cyanide Substitution, then Selective Nitrile Reduction
Core Concept
A primary alkyl halide reacts with CN- by SN2 substitution to give a nitrile with one extra carbon; catalytic hydrogenation (H2/Pd) then reduces that nitrile to a primary amine while leaving an existing ketone carbonyl untouched.
Steps
Identify the leaving group and its position: here the -Cl sits on a primary, unhindered -CH2- carbon at the end of the side chain - ideal for SN2.
React with KCN: CN- (a good nucleophile) displaces Cl-, converting -CH2CH2Cl to -CH2CH2CN. This adds one carbon to the chain and installs the nitrile group - this is product A.
Reduce the nitrile with H2/Pd (catalytic hydrogenation): the nitrile carbon plus 2 H2 gives -CH2NH2, a primary amine. Under these mild catalytic conditions the ring ketone (C=O) is NOT reduced - this is product B. …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
COMEDK 2026Set 2026-A1 markMCQ
Q.A Nitrogen containing organic compound with molecular formula C3H5N undergoes the following reactions. C3H5N+Na/Hg in C2H5OH→[A][A]+CH3COCl/NaOH(aq)→[B][A]+CHCl3/KOH on heating →[C] (foul smelling compound) Identify compounds [B] and [C]
(A) [B]: (CH3)3−C−NH−COCl [C]: CH3−CH2−CN
(B) [B]:CH3−CH2−NH−COCH2−CH3 [C]: CH3−(CH2)2−CN
(C) [B]:CH3−(CH2)2−NH−COCH3 [C]: CH3−(CH2)2−NC
(D) [B]:(CH3)2−CH−NH−COCH3 [C]: CH3−CH2−NC
›Reveal solutionSolution
The compound C₃H₅N is propionitrile (CH₃CH₂CN); reduction gives propylamine (A), which with acetyl chloride yields N‑propylacetamide (B) and with chloroform/KOH gives propyl isocyanide (C) — the foul‑smelling product. The correct option is (C).
Concept & Intuition
The molecular formula C₃H₅N has three carbons, five hydrogens, and one nitrogen. The degree of unsaturation is 22C+2+N−H=26+2+1−5=2. Two degrees of unsaturation suggest either a nitrile (–C≡N) or an isocyanide (–N⁺≡C⁻). Nitriles are far more common and stable; isocyanides have a foul smell (which appears later in the problem). The first reaction — reduction with Na/Hg in ethanol — is a classic reduction of a nitrile to a primary amine. So A is a primary amine. Then A reacts with acetyl chloride (Schotten–Baumann) to give an amide (B), and with chloroform/KOH (carbylamine reaction) to give an isocyanide (C) — which is indeed foul‑smelling. This matches the pattern exactly.
Step‑by‑step reasoning
Identify the starting compound
C₃H₅N with two degrees of unsaturation is most likely propionitrile: CH₃CH₂C≡N. (The alternative, an isocyanide like CH₃CH₂N⁺≡C⁻, is less stable and would not be the usual starting material.)
First reaction — reduction to amine [A]
Na/Hg in ethanol reduces a nitrile to a primary amine:
CH3CH2C≡N+4[H]Na/Hg,C2H5OHCH3CH2CH2NH2
So [A] is propylamine (n‑propylamine).
Second reaction — acetylation to amide [B]
Primary amines react with acetyl chloride (CH₃COCl) in aqueous NaOH to give N‑substituted acetamides:
Q.An organic compound A(C5H9N) upon reaction with Na/Hg/C2H5OH gives compound B . B reacts with NaNO2/HCl at 274 K to form C with quantitative liberation of N2 gas. B also reacts with Hinsberg's reagent to form a compound which is soluble in alkali. Identify compound B.
(A) [(CH3)3−N−CH2CH3]+
(B) CH3−(CH2)3−NH−CH3
(C) CH3−(CH2)4−NH2
(D) CH3−(CH2)2−N−(CH3)2
›Reveal solutionSolution
The key is that B must be a primary aliphatic amine (R–NH₂) because it liberates N₂ with HNO₂ and gives an alkali-soluble Hinsberg product. The molecular formula of A (C₅H₉N) and its reduction to B point to B being n‑pentylamine, which is option (C).
Concept & Intuition
We start with an organic compound A (C₅H₉N). The reaction with Na/Hg/C₂H₅OH is a reduction (often used for nitriles, amides, or nitro compounds). The product B then reacts with NaNO₂/HCl at 274 K (cold nitrous acid) with quantitative liberation of N₂ gas — that is a classic test for a primary aliphatic amine (R–NH₂). Secondary and tertiary amines do not liberate N₂ under these conditions; they form N‑nitrosoamines or simply dissolve.
Additionally, B reacts with Hinsberg’s reagent (benzenesulfonyl chloride) to give a compound soluble in alkali. That is characteristic of a primary amine: the sulfonamide formed has an acidic N–H hydrogen and dissolves in NaOH. Secondary amines give insoluble sulfonamides; tertiary amines do not react.
Thus B must be a primary aliphatic amine with formula derived from A (C₅H₉N) after reduction.
Step‑by‑step reasoning
Determine the reduction product from A
A has formula C₅H₉N. The reduction with Na/Hg/C₂H₅OH typically adds hydrogen across multiple bonds (e.g., –C≡N → –CH₂NH₂). If A were a nitrile (R–C≡N), reduction gives a primary amine R–CH₂NH₂.
For C₅H₉N, a saturated primary amine would be C₅H₁₁NH₂ (i.e., C₅H₁₃N). But B is not yet saturated — we need to check the degree of unsaturation.
A: C₅H₉N → unsaturation = (2×5 + 2 + 1 – 9)/2 = (10+2+1–9)/2 = 4/2 = 2. So A has two degrees of unsaturation (e.g., a triple bond or two double bonds). A common structure is a nitrile (one triple bond = 2 unsaturations). So A is likely a pentanenitrile: CH₃(CH₂)₃C≡N.
Reduction of a nitrile with Na/Hg/EtOH gives the primary amine:
CH3(CH2)3C≡NNa/Hg/C2H5OHCH3(CH2)4NH2
That is n‑pentylamine (C₅H₁₁NH₂), formula C₅H₁₃N. But wait — B’s formula must be consistent with the reactions. Let’s verify.
Confirm B’s identity from the nitrous acid test
Primary aliphatic amines react with HNO₂ to give an unstable diazonium salt that decomposes to a carbocation and N₂:
RNH2+HNO2→[R−N2+]→R++N2
The quantitative liberation of N₂ means every mole of B gives one mole of N₂ — exactly what a primary amine does. Secondary amines give yellow oils (N‑nitrosoamines), no N₂. Tertiary amines give salts, no N₂. So B is primary.
Confirm B’s identity from the Hinsberg test
With benzenesulfonyl chloride (Hinsberg’s reagent), a primary amine forms a sulfonamide with an N–H bond:
RNH2+C6H5SO2Cl→C6H5SO2NHR+HCl …