Q.The reagent X in the reaction C₆H₅NH₂ (aniline) —X / 273K→ C₆H₅N₂⁺Cl⁻ (benzenediazonium chloride) is –

Showing the 12 most recent of 35 on this concept.

1. CBSE 2026Set V11 markMCQ
   Q.p-Hydroxyazobenzene is formed by the reaction of benzene diazonium chloride with phenol. It is

   (a) an electrophilic substitution reaction

   (b) a nucleophilic substitution reaction

   (c) a hydrogenation reaction

   (d) a halogenation reaction

   ›Reveal solutionSolution

   Formation of p-hydroxyazobenzene by coupling of benzene diazonium chloride with phenol is an electrophilic aromatic substitution.

   Benzene diazonium chloride, $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$, reacts with phenol in mildly alkaline medium to give the orange azo dye p-hydroxyazobenzene:

   $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{OH} \rightarrow p\text{-}\text{HO-C}_6\text{H}_4\text{-N=N-C}_6\text{H}_5 + \text{HCl}$ …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Aqueous solution of benzene diazonium chloride in presence of dil. H2SO4 is heated to produce:

   (a) Aniline

   (b) Benzene

   (c) Chlorobenzene

   (d) Phenol

   ›Reveal solutionSolution

   Heating an aqueous benzenediazonium salt solution hydrolyses it to phenol, releasing nitrogen gas.

   Benzenediazonium chloride, $C_6H_5-N_2^+Cl^-$, is thermally unstable in aqueous acidic solution. When its aqueous solution is warmed (with dilute $H_2SO_4$ or simply with water), the diazonium group is displaced by a water molecule in a nucleophilic substitution, with loss of nitrogen gas:

   $C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2\uparrow + HCl$

   …

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3. CBSE 2026Set ANNUAL1 mark
   Q.Write general formula of diazonium salt.

   ›Reveal solutionSolution

   Diazonium salts have the general formula R-N≡N+ X−, formed by diazotisation of a primary aromatic amine with NaNO2/HCl at 0-5°C.

   …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.The reagent X in the reaction C₆H₅NH₂ (aniline) —X / 273K→ C₆H₅N₂⁺Cl⁻ (benzenediazonium chloride) is –

   (a)

   (i) HNO₃

   (b)

   (ii) NaNO₂ and HCl

   (c)

   (iii) NaCl and HNO₃

   (d)

   (iv) NaNO₂ and H₂SO₄

   ›Reveal solutionSolution

   The conversion of aniline to benzenediazonium chloride at 273 K (diazotisation) uses nitrous acid generated from $\text{NaNO}_2 + \text{HCl}$. Correct option: (ii).

   Concept. Diazotisation is the reaction of a primary aromatic amine with nitrous acid ($\text{HNO}_2$) at low temperature (273–278 K) to form a diazonium salt. Nitrous acid is unstable, so it is prepared in situ by adding sodium nitrite ($\text{NaNO}_2$) to a mineral acid.

   Why HCl specifically (option ii not iv). The anion of the diazonium salt required here is chloride ($\text{C}_6\text{H}_5\text{N}_2^{+}\text{Cl}^{-}$), so HCl must be the acid supplying $\text{Cl}^{-}$; $\text{H}_2\text{SO}_4$ would give the sulphate salt.

   Reaction. …

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5. CBSE 2025Set 56/4/11 markMCQ
   Q.In the given reaction sequence, the structure of Y would be : Aniline [$C_6H_5NH_2$] $\xrightarrow{NaNO_2, HCl, 0-5^\circ C}$ X $\xrightarrow{C_2H_5OH}$ Y (A) Phenol [$C_6H_5OH$] (B) Benzene [$C_6H_6$] (C) Nitrobenzene [$C_6H_5NO_2$] (D) Benzenediazonium chloride [$C_6H_5N_2^+Cl^-$]

   ›Reveal solutionSolution

   Aniline undergoes diazotization to form a benzenediazonium salt (X), which then reacts with ethanol to undergo reductive dediazoniation, replacing the diazonium group with hydrogen to yield benzene (Y). The correct option is (B).

   This problem tests your understanding of two classic reactions in aromatic chemistry: diazotization and the replacement of a diazonium group. The key is to recognize that ethanol here acts as a reducing agent, not as a nucleophile.

   Let’s walk through the sequence step by step.

   1. Step 1: Diazotization of aniline Aniline ($C_6H_5NH_2$) is treated with sodium nitrite ($NaNO_2$) and hydrochloric acid ($HCl$) at a low temperature ($0-5^\circ C$). This is the standard condition for forming a diazonium salt. The reaction proceeds as:

   $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0-5^\circ C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$

   The product X is benzenediazonium chloride ($C_6H_5N_2^+Cl^-$). This is a highly reactive intermediate, stable only in cold solution.

   Watch out

   A common mistake is to think that the diazonium salt itself is the final product Y. But the reaction sequence continues — X is just an intermediate.

   2. Step 2: Reaction of the diazonium salt with ethanol When benzenediazonium chloride (X) is treated with ethanol ($C_2H_5OH$), a reductive dediazoniation occurs. Ethanol acts as a reducing agent, donating a hydride ion ($H^-$) or a hydrogen atom to replace the diazonium group. The overall transformation is: $C_6H_5N_2^+Cl^- + C_2H_5OH \longrightarrow C_6H_6 + N_2 + CH_3CHO + HCl$ …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.When an aqueous solution of benzenediazonium chloride is boiled with water or steam distilled the product formed is:

   (a) Benzene

   (b) Diphenyl

   (c) Chlorobenzene

   (d) Phenol.

   ›Reveal solutionSolution

   Aqueous benzenediazonium chloride is thermally unstable; warming or steam-distilling it causes hydrolysis, replacing the $-N_2^+Cl^-$ group with $-OH$ and releasing nitrogen gas.

   Benzenediazonium chloride ($C_6H_5-N_2^+Cl^-$), when its aqueous solution is warmed/boiled (or steam distilled), undergoes hydrolysis. The diazonium group is displaced by a hydroxyl group from water, with loss of nitrogen gas and hydrochloric acid:

   $C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2\uparrow + HCl$

   …

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7. CBSE 2025Set ANNUAL1 mark
   Q.Draw the structure of the major monohalo product: a benzenediazonium salt ($C_6H_5-N_2^+X^-$, drawn as a benzene ring with an $-N_2^+X^-$ substituent) $\xrightarrow{KI} ?$

   ›Reveal solutionSolution

   Treating a benzenediazonium salt with KI substitutes the diazonium group directly by iodide, giving iodobenzene and nitrogen gas — no copper catalyst is required for this particular halogen (unlike Cl/Br via Sandmeyer).

   Aromatic diazonium salts undergo nucleophilic substitution at the carbon bearing the $-N_2^+$ group. Iodide is a good enough nucleophile/reducing species that this substitution proceeds simply on warming with aqueous KI, without needing Cu(I)/Cu(III) catalysis (which is required to introduce –Cl or –Br via the Sandmeyer reaction):

   $C_6H_5-N_2^+X^- + KI \rightarrow C_6H_5-I + N_2\uparrow + KX$

   …

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8. CBSE 2025Set ANNUAL1 mark
   Q.Give one chemical test to distinguish between an aromatic primary amine from an aliphatic primary amine.

   ›Reveal solutionSolution

   Bromine water reacts instantly with an aromatic primary amine like aniline to give a white precipitate of 2,4,6-tribromoaniline, because the –NH$_2$ group strongly activates the benzene ring; an aliphatic primary amine has no such ring to brominate and gives no precipitate.

   In an aromatic primary amine such as aniline ($C_6H_5NH_2$), the lone pair on nitrogen is delocalised into the benzene ring by resonance, making the ring extremely electron-rich and strongly activating it towards electrophilic aromatic substitution — so strongly that even mild aqueous bromine (no catalyst needed) substitutes at all three free ortho/para positions at once:

   $C_6H_5NH_2 + 3Br_2(aq) \rightarrow 2,4,6\text{-tribromoaniline}\downarrow (\text{white ppt}) + 3HBr$

   …

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9. CBSE 2025Set ANNUAL1 mark
   Q.Diazonium salts of aromatic amines are more stable than those of aliphatic amines. Give reason.

   ›Reveal solutionSolution

   The aromatic ring in benzenediazonium salts can share the positive charge by resonance, giving extra stability that keeps the salt intact at low temperature; aliphatic diazonium ions have no ring to delocalise into and lose $N_2$ almost as soon as they form.

   A diazonium ion has the general structure $R-\overset{+}{N}\equiv N$ (or the resonance form $R-N=\overset{+}{N}$). In an aryl diazonium salt, $C_6H_5-N_2^+$, the ring's π system can conjugate with the $-N\equiv N$ group, so the positive charge is not confined to the terminal nitrogen but is delocalised over the ring and the nitrogen atoms through resonance structures (e.g. with positive charge shifted onto an ortho or para ring carbon).

   This delocalisation lowers the energy of the ion and makes the C–N bond to the ring stronger, so benzenediazonium chloride is reasonably stable and can be stored/handled in solution at 273–278 K (0–5 °C) and used in coupling/Sandmeyer reactions.

   …

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10. CBSE 2025Set ANNUAL1 markMCQ
    Q.Benzene diazonium chloride on hydrolysis gives

    (a) phenol

    (b) chlorobenzene

    (c) benzene

    (d) aniline

    ›Reveal solutionSolution

    Warming a diazonium salt in water lets water act as the nucleophile that displaces N2, an SN1-type substitution via the aryl cation-like transition state.

    When an aqueous solution of benzenediazonium chloride is warmed, water acts as a nucleophile, displacing nitrogen gas from the diazonium group: C₆H₅N₂⁺Cl⁻ + H₂O --(Δ)--> C₆H₅OH (phenol) + N₂↑ + HCl. This is one of the most important synthetic uses of diazonium …

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11. CBSE 2025Set ANNUAL1 markMCQ
    Q.Aniline (C6H5-NH2, shown protonated as C6H5-NH2 with a positive-charge notation) is treated with NaNO2/dil. HCl at 0°-5°C to give A, and A is then treated with C6H5OH/OH⊖ to give B (major product). The major product B of the reaction is

    (a) Diphenyl ether, C6H5-O-C6H5

    (b) C6H5-N=N-C6H4-OH (the azo-coupling product with -OH at the para position, drawn in-line)

    (c) C6H5-N=N-C6H4-OH (the azo-coupling product with -OH shown branching off the second ring)

    (d) C6H5-N=N-O-C6H5

    ›Reveal solutionSolution

    Aniline first forms a diazonium salt (A) with $NaNO_2$/HCl at 0-5°C; this then undergoes azo coupling with phenol to give p-hydroxyazobenzene (B) as the major product.

    Step 1 (formation of A): Aniline reacts with $NaNO_2$ and dilute HCl at 0-5°C (diazotisation). At this low temperature the unstable diazonium salt survives without decomposing: $C_6H_5NH_2 + NaNO_2 + 2HCl \rightarrow C_6H_5N_2^+Cl^- + NaCl + 2H_2O$. So A = benzenediazonium chloride.

    …

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12. CBSE 2025Set ANNUAL1 markMCQ
    Q.Aromatic primary amines and aliphatic primary amines can be distinguished by –

    (i) Hinsberg's test

    (ii) Tollen's test

    (iii) Azo dye test

    (iv) Litmus test

    ›Reveal solutionSolution

    Aromatic primary amines form a stable diazonium salt on treatment with HNO₂ at low temperature, which couples with a phenol (e.g. β-naphthol) to give a brightly coloured azo dye; aliphatic primary amines cannot do this because their diazonium salts are unstable and decompose instantly.

    When a primary amine is treated with nitrous acid (NaNO₂ + HCl, 0–5°C):

    • Aromatic primary amines (e.g. aniline) form a reasonably stable diazonium salt (ArN₂⁺Cl⁻) at low temperature. This diazonium salt can be coupled with an alkaline solution of β-naphthol (or a similar phenol) to give a bright orange/red azo dye. This visible colour formation is a positive azo dye test. …

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