Match the compounds given in Column I with the items given in Column II. Column I: (i) Benzene sulphonyl chlor

NCERT Exemplar · Q70

Q. Match the compounds given in Column I with the items given in Column II. Column I:

(i) Benzene sulphonyl chloride

(ii) Sulphanilic acid

(iii) Alkyl diazonium salts

(iv) Aryl diazonium salts
Column II:

(a) Zwitter ion

(b) Hinsberg reagent

(d) Conversion to alcohols

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Concept understanding — Diazonium Salt Reactions

Diazonium Salt Reactions – A First Look

Imagine you have a benzene ring, and you want to attach a new group — say a chlorine, a bromine, a cyano group, or even a hydroxyl — directly onto the ring. The benzene ring is stubborn; it doesn't easily let go of its hydrogen atoms for simple substitution. But there is a clever trick: first convert the ring into a diazonium salt, a highly reactive intermediate that will let you swap in almost any group you want.

That is the core idea. A diazonium salt is a temporary, energetic handle on the benzene ring that you can then replace with a wide variety of substituents. It is one of the most powerful tools in aromatic synthesis.

What is a Diazonium Salt?

A diazonium salt has the general formula Ar–N₂⁺ X⁻, where Ar is an aryl group (like phenyl, C₆H₅–), N₂⁺ is a diazonium cation (two nitrogen atoms triple-bonded, with a positive charge on the terminal nitrogen), and X⁻ is a counterion like chloride, bromide, or hydrogensulfate.

The key structural feature: the –N₂⁺ group is attached directly to the benzene ring. This group is unstable — it wants to leave as N₂ gas. That instability is exactly what makes it useful: when the N₂ leaves, the ring is left with a highly reactive carbocation-like intermediate that can be attacked by a nucleophile.

Watch out

Diazonium salts are thermally unstable and can explode if dried. They are almost always prepared and used in cold solution (0–5 °C) without isolation.

How Do You Make One? (Diazotization)

You start with a primary aromatic amine (Ar–NH₂). Treat it with nitrous acid (HNO₂) at low temperature (0–5 °C). The reaction is:

\text{Ar–NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0-5^\circ\text{C}} \text{Ar–N}_2^+ \text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}

The nitrous acid is generated in situ from sodium nitrite and a mineral acid. The amine gets converted into the diazonium salt almost instantly. You must keep the solution cold; if it warms up, the diazonium salt decomposes and you get phenol and nitrogen gas.

Two Major Classes of Reactions

Once you have the diazonium salt in solution, you can do two fundamentally different things with it:

1. Substitution Reactions (N₂ leaves)

Here the –N₂⁺ group is replaced by another group. The nitrogen gas bubbles away, and the ring gets a new substituent. This is called dediazoniation. The leaving group is N₂, which is extremely stable, so the reaction is thermodynamically driven.

The most common substitutions:

Reagent/Condition	Product	Name
CuCl / HCl, heat	Ar–Cl	Sandmeyer reaction
CuBr / HBr, heat	Ar–Br	Sandmeyer reaction
CuCN / KCN, heat	Ar–CN	Sandmeyer reaction
KI, heat	Ar–I	Direct substitution
H₂O, heat	Ar–OH	Hydrolysis
H₃PO₂ (hypophosphorous acid)	Ar–H	Reduction (replaces N₂ with H)
Cu₂O, Cu(NO₃)₂, H₂O	Ar–NO₂	Replacement with nitro group

Tip

The Sandmeyer reaction uses copper(I) halide or cyanide as a catalyst. The copper helps transfer the halide or cyanide to the ring. Without copper, the reaction is much slower or gives different products.

The mechanism for Sandmeyer: the diazonium salt accepts an electron from Cu⁺, forming an aryl radical, which then abstracts a halogen from CuX₂. The N₂ leaves as a gas.

2. Coupling Reactions (N₂ stays)

Here the diazonium salt keeps its N₂ group and attacks another aromatic ring (usually an activated one like phenol or aniline). The result is an azo compound with the general structure Ar–N=N–Ar'. These compounds are intensely coloured — many are used as dyes.

The reaction is an electrophilic aromatic substitution. The diazonium cation is a weak electrophile, so it only attacks rings that are strongly activated (with –OH, –NH₂, –NHR, –NR₂ groups). The coupling occurs at the para position if available; otherwise ortho.

Example: coupling with phenol in alkaline medium:

\text{C}_6\text{H}_5\text{–N}_2^+ \text{Cl}^- + \text{C}_6\text{H}_5\text{–OH} \xrightarrow{\text{NaOH, 0–5}^\circ\text{C}} \text{C}_6\text{H}_5\text{–N=N–C}_6\text{H}_4\text{–OH} \text{ (p-hydroxyazobenzene, orange dye)}

Important

Coupling requires the coupling component (phenol or aniline) to be in its reactive form: phenol is used in alkaline solution (phenoxide ion is more activating), aniline is used in slightly acidic or neutral solution (to avoid protonation of the amino group).

Why Are Diazonium Salts So Versatile? …

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This question tests your ability to match organic reagents and compounds with their characteristic properties or uses. The key is to recall the specific role of each compound: Benzene sulphonyl chloride is the Hinsberg reagent for distinguishing amines, Sulphanilic acid exists as a zwitter ion, Alkyl diazonium salts are unstable and convert to alcohols, and Aryl diazonium salts are used in dye formation. The correct matches are (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c).

Let’s break down each compound and its matching item by understanding the chemistry behind it.

1. Benzene sulphonyl chloride (i) → Hinsberg reagent (b)

Benzene sulphonyl chloride, $\ce{C6H5SO2Cl}$ , is famously known as the Hinsberg reagent. It is used to distinguish between primary, secondary, and tertiary amines. The reaction works because:

Primary amines react to form a sulphonamide that is soluble in alkali (due to an acidic N–H bond).
Secondary amines form a sulphonamide that is insoluble in alkali (no N–H bond).
Tertiary amines do not react at all.

Tip

The name "Hinsberg reagent" is a direct giveaway. If you see benzene sulphonyl chloride in a matching question, it almost always pairs with this test for amines.

2. Sulphanilic acid (ii) → Zwitter ion (a)

Sulphanilic acid has the structure $\ce{H2N-C6H4-SO3H}$ . Notice it contains both a basic amino group ( $\ce{-NH2}$ ) and an acidic sulphonic acid group ( $\ce{-SO3H}$ ). In the solid state or in neutral solution, the acidic group donates a proton to the basic group, forming an internal salt called a zwitter ion:

\ce{^{+}H3N-C6H4-SO3^{-}}

This is why sulphanilic acid is insoluble in organic solvents but soluble in water — the ionic character dominates.

Watch out

A common mistake is to think that all amino acids form zwitter ions. While true, sulphanilic acid is not an amino acid — it’s an aromatic compound with both acidic and basic groups. The principle is the same: any molecule with both a strong acid and a strong base group can form a zwitter ion.

3. Alkyl diazonium salts (iii) → Conversion to alcohols (d)

Alkyl diazonium salts, $\ce{R-N2^{+}X^{-}}$ , are extremely unstable. Even at low temperatures (0–5°C), they decompose readily. When treated with water, they undergo hydrolysis to give alcohols:

$\ce{R-N2^{+} + H2O -> R-OH + N2 + H+}$ …

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Concept: Reactions of Amines and Diazonium Salts

This question tests your understanding of functional group transformations and characteristic reactions of nitrogen-containing organic compounds.

Method: Reagent–Function Matching

Step 1: Identify the key property or use of each compound in Column I.

(i) Benzene sulphonyl chloride — This is the Hinsberg reagent. It reacts with primary and secondary amines to form sulphonamides, distinguishing them from tertiary amines.
(ii) Sulphanilic acid — Contains both an amino group ( $-NH_2$ ) and a sulphonic acid group ( $-SO_3H$ ). In solution, it exists as a zwitter ion (internal salt).
(iii) Alkyl diazonium salts — Unstable; decompose readily to give alcohols (via $S_N1$ or $S_N2$ with water).
(iv) Aryl diazonium salts — Stable at low temperatures; used in coupling reactions to form azo dyes.

Step 2: Match each with the correct item in Column II.

Column I	Column II	Reason
(i) Benzene sulphonyl chloride	(b) Hinsberg reagent	Classic test for amines
(ii) Sulphanilic acid	(a) Zwitter ion	Has both acidic and basic groups

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Here are the common mistakes students make when matching these compounds, along with the correct reasoning to avoid them.

Mistake 1: Confusing Alkyl and Aryl Diazonium Salts

The Mistake: Students often match Alkyl diazonium salts (iii) with (d) Conversion to alcohols and Aryl diazonium salts (iv) with (c) Dyes. While the second match is correct, the first is dangerously wrong.
Why it happens: Students remember that diazonium salts react with water to give alcohols. They forget that alkyl diazonium salts are extremely unstable and cannot be isolated or used in practical organic synthesis.
How to Avoid:
- Remember the stability rule: Aryl diazonium salts are stable at 0-5°C. Alkyl diazonium salts decompose instantly into carbocations and nitrogen gas ( $N_2$ ).
- Focus on the reaction: The conversion to alcohols (via hydrolysis) is a classic reaction of aryl diazonium salts (e.g., $C_6H_5N_2^+Cl^- + H_2O \rightarrow C_6H_5OH + N_2 + HCl$ ).
- Correct Match: (iv) Aryl diazonium salts → (d) Conversion to alcohols.

Mistake 2: Forgetting the Zwitterion Structure of Sulphanilic Acid

The Mistake: Students match Sulphanilic acid (ii) with (b) Hinsberg reagent or (c) Dyes, missing the zwitterion property.
Why it happens: Students see "acid" in the name and think it behaves like a normal acid. They don't visualize the internal salt formation.
How to Avoid:
- Draw the structure: Sulphanilic acid has both an acidic $-SO_3H$ group and a basic $-NH_2$ group. In the solid state and in solution, a proton transfers from the acid to the amine.
- Learn the definition: A zwitterion is a molecule with equal positive and negative charges but a net zero charge. The structure is $^+H_3N-C_6H_4-SO_3^-$ .
- Correct Match: (ii) Sulphanilic acid → (a) Zwitter ion.

Mistake 3: Misidentifying the Hinsberg Reagent

The Mistake: Students match Benzene sulphonyl chloride (i) with (c) Dyes or (d) Conversion to alcohols, failing to recognize its specific test-tube function.
Why it happens: The name "benzene sulphonyl chloride" sounds like a complex compound, and students don't connect it to the familiar "Hinsberg reagent" from the amines chapter.
How to Avoid:
- Memorize the common name: The Hinsberg reagent is benzenesulphonyl chloride ( $C_6H_5SO_2Cl$ ). It is used to distinguish between primary, secondary, and tertiary amines.
- Link the function: It reacts with amines to form sulphonamides, which are either soluble or insoluble in alkali, allowing identification.
- Correct Match: (i) Benzene sulphonyl chloride → (b) Hinsberg reagent.

Mistake 4: Overlooking the Dye-Making Ability of Aryl Diazonium Salts

The Mistake: Students match Aryl diazonium salts (iv) with only (d) Conversion to alcohols and forget the coupling reaction that makes dyes. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 14 on this concept.

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. Benzene diazonium chloride is water ____ and ____ at room temperature.$
(a) Insoluble, Unstable
(b) Soluble, Stable
(c) Insoluble, Stable
(d) Soluble, Unstable
›Reveal solutionSolution

Benzenediazonium chloride is water-soluble (an ionic salt) but thermally unstable at room temperature, decomposing to phenol and nitrogen gas.

Diazonium salts like C6H5N2+Cl⁻ are ionic compounds and dissolve readily in water. However, the N≡N+ group is only stable at LOW temperatures (0–5°C, which is why diazotisation is always carried out in an ice bath). At room temperature (and above), the diazonium salt rapidly decomposes:

C6H5N2+Cl⁻ + H2O → C6H5OH + N2↑ + HCl

…

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GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. Which reagent can be used to distinguish between aniline and benzylamine?$
(a) CHCl3 / KOH
(b) NaNO2 + HCl
(c) C6H5SO2Cl
(d) CH3COCl / Base
›Reveal solutionSolution
Nitrous acid (from NaNO2 + HCl) distinguishes aromatic from aliphatic primary amines by how their diazonium salts behave: aromatic diazonium salts are stable in the cold, while aliphatic ones decompose immediately with brisk N2 evolution.

Both aniline and benzylamine are primary amines, so simple tests for amine class (like the carbylamine test, which both would give positively) cannot distinguish them. But their behaviour with nitrous acid (NaNO2/HCl, 0–5°C) differs sharply:

Aniline (aromatic 1° amine, –NH2 directly on the ring) forms a stable diazonium salt, C6H5N2+Cl⁻, that persists in the cold reaction mixture without effervescence. …
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GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. Identify compounds A and B, and the reducing agent respectively used in the reaction. [benzene ring]-N2+ Cl- --CH3CH2OH--> A + B$
(a) A = C6H6, B = CH3COOH, CH3COOH
(b) A = C6H6, B = CH3CHO, CH3CH2OH
(c) A = C6H6, B = CH3CHO, C6H6
(d) A = C6H5OH, B = CH3, CHO, C6H5OH
›Reveal solutionSolution
Ethanol acts as a reducing agent on a diazonium salt, replacing the -N2+ group with -H (deamination) while ethanol itself is oxidised to acetaldehyde.

When a diazonium salt is treated with a reducing agent such as ethanol (or hypophosphorous acid, H3PO2), the diazonium group is reductively replaced by hydrogen:

C6H5N2+Cl⁻ + CH3CH2OH → C6H6 + N2↑ + CH3CHO + HCl

A = C6H6 (benzene) — from reductive replacement of –N2+ by –H …
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GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. Among the following which one is Gattermann - Reagent.$
(a) CuCl / HCl
(b) Cu / NaNO2
(c) CO + HCl
(d) Cu (powder) / HCl
›Reveal solutionSolution

The Gattermann reaction is a variant of the Sandmeyer reaction that uses copper powder together with the corresponding halogen acid (HCl or HBr) directly on the diazonium salt, instead of a preformed cuprous halide.

Both the Sandmeyer and Gattermann reactions convert an aryldiazonium salt into the corresponding aryl halide.

Sandmeyer reaction: uses a preformed cuprous halide (CuCl/HCl for chloro, CuBr/HBr for bromo) as catalyst.

…

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GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. Benzene diazonium Fluoroborate is water _____ and _____ at room temperature.$
(a) Soluble, Stable
(b) Insoluble, Stable
(c) Insoluble, Unstable
(d) Soluble, Unstable
›Reveal solutionSolution

Unlike most diazonium salts, benzenediazonium fluoroborate is a stable solid that is only sparingly soluble in cold water, which is exactly why it can be isolated pure and used in the Balz-Schiemann reaction.

Most diazonium salts (chloride, bromide, sulphate) are water-soluble and decompose readily even at low temperature, so they are typically used in solution immediately after preparation, without isolation.

…

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GUJCET 2024Set 131 markMCQ
$C_6H_5\cdot N_2^+BF_4^-$
›Reveal solutionSolution

Benzenediazonium tetrafluoroborate $C_6H_5N_2^+BF_4^-$ is water-insoluble and stable at room temperature.

Concept. Most diazonium salts (chloride, bromide, hydrogen sulphate) are water-soluble and decompose readily. The tetrafluoroborate salt precipitates as a water-insoluble solid and is stable enough t …
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GUJCET 2023Set 091 markMCQ
$\sigma$
›Reveal solutionSolution
[!TLDR]

The coupling product p-hydroxyazobenzene has 26 sigma bonds and 7 pi bonds.

Concept

In a coupling reaction, benzene diazonium chloride reacts with phenol in mildly basic medium to give the azo dye p-hydroxyazobenzene: $C_6H_5\text{-}N{=}N\text{-}C_6H_4\text{-}OH$ . Each C=C of an aromatic ring is one $\sigma$ + one $\pi$ ; every single bond is a $\sigma$ bond.

Solution

Product structure: Ring A ( $C_6H_5$ ) — $N{=}N$ — Ring B ( $C_6H_4$ ) with $-OH$ para.

$\pi$ bonds:

Ring A benzene: 3

Ring B benzene: 3

$N{=}N$ : 1

Total $\pi = 7$

$\sigma$ bonds: …
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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. Which salt is insoluble in water?$
(a) C6H5N2+ HSO4^-
(b) C6H5N2+ BF4^-
(c) C6H5N2+ Br^-
(d) C6H5N2+ Cl^-
›Reveal solutionSolution

Benzenediazonium tetrafluoroborate (C6H5N2+ BF4-) is water-insoluble and precipitates out.

Most benzenediazonium salts (chloride, bromide, hydrogen sulphate) are soluble in water. However, benzenediazonium tetrafluoroborate, C6H5N2+ BF4-, is only sparingly soluble; it separates out of solution as a stable solid. …

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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. Which gas is evolved during the reaction of methyl amine with HNO2?$
(a) NO2
(b) NH3
(c) H2
(d) N2
›Reveal solutionSolution

Primary aliphatic amine + HNO2 -> alcohol + N2 gas; the gas evolved is N2.

Primary aliphatic amines react with nitrous acid (generated in situ from NaNO2 + HCl) to form very unstable diazonium salts that immediately decompose, giving an alcohol and liberating nitrogen gas: …

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GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$Q. Which reagent is used in the Gattermann reaction?$
(a) Cu/HX
(b) CHCl3 + NaOH
(c) Cu2X2/HX
(d) Zn-Hg/HCl
›Reveal solutionSolution

The Gattermann reaction is a variant of the Sandmeyer reaction that uses copper POWDER directly with the halogen acid, instead of a cuprous halide salt.

Both reactions convert an aryl diazonium salt to an aryl halide: ArN2+ + Cu/HX (Gattermann) or ArN2+ + Cu2X2/HX (Sandmeyer) -> ArX + N2. Option (b), CHCl3+NaOH, is the carbylamine test …

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GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
$Q. Which of the following is structural formula of orange dye?$
(a) C6H5-N=N-C6H4-OH (hydroxy group para to the azo linkage)
(b) C6H5-N=N-C6H4-NH2 (amino group drawn off-set from the azo linkage, i.e. not directly para)
(c) C6H5-N=N-C6H4-NH2 (amino group para to the azo linkage)
(d) HO-C6H4-N=N-C6H4-OH (hydroxy groups on both rings, each para to the azo linkage)
›Reveal solutionSolution

Diazonium salts couple with phenols in weakly alkaline medium at the position para to the -OH group, giving a coloured azo dye - p-hydroxyazobenzene.

Coupling reaction: C6H5N2+Cl- + C6H5OH --(weakly alkaline, pH 9-10)--> C6H5-N=N-C6H4-OH (p-hydroxyazobenzene) + HCl. The strongly activating -OH group of phenol directs the electrophilic diazonium ion to attack predominantly at the para position (with respect to -OH), giving the e …

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GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$Q. p-Toluenediazonium chloride, on reaction with SnCl2 + HCl, what will be the product of the reaction?$
(a) C6H5-NH-NH2 (phenylhydrazine, unsubstituted ring)
(b) H3C-C6H4-CH3 (p-xylene, -CH3 on both sides of ring)
(c) H3C-C6H4-NH-NH2 (p-tolylhydrazine, -CH3 on one side, -NH-NH2 on the other)
(d) C6H5-CH3 (toluene, ring with only -CH3)
›Reveal solutionSolution

Reduction of a diazonium salt with SnCl2/HCl converts the -N2+ group into -NH-NH2 (a hydrazine), without touching the rest of the ring.

p-Toluenediazonium chloride is CH3-C6H4-N2+Cl- (the diazonium group at the position para to the methyl group). Treating a diazonium salt with SnCl2 (a reducing agent) in HCl reduces the terminal nitrogen (-N2+) to a hydrazine group (-NH-NH2), while the ring and its existing -CH3 substituent are left unchanged:

CH3-C6H4-N2+Cl- + 4[H] (from SnCl2/HCl) -> CH3-C6H4-NH-NH2 + NH4Cl (as HCl salt) …

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Q.Match the compounds given in Column I with the items given in Column II. Column I:

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Q.Match the compounds given in Column I with the items given in Column II.
Column I:

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