Give the structure of 'A' in the following reaction sequence. 4-methyl-2-nitroaniline (an aniline bearing a –C

NCERT Exemplar · Q32

Q. Give the structure of 'A' in the following reaction sequence. 4-methyl-2-nitroaniline (an aniline bearing a -CH3 group para to the -NH2 and an -NO2 group ortho to the -NH2) is treated with

(i) NaNO2 + HCl at 273–278 K, then with

(ii) H3PO2 and H2O, giving A.

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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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The amino group is turned into a diazonium salt and then replaced by hydrogen. Removing –NH2 from 4-methyl-2-nitroaniline leaves –CH3 and –NO2 meta to each other, so A is m-nitrotoluene.

Step 1 – Diazotisation

4-Methyl-2-nitroaniline + NaNO2 + HCl at 273–278 K converts the –NH2 into a diazonium group:

Ar–NH2 → Ar–N2+ Cl− (4-methyl-2-nitrobenzenediazonium chloride).

Step 2 – Deamination with H3PO2

Hypophosphorous acid (H3PO2) in water reduces the diazonium group, replacing –N2+ by –H and releasing N2:

Ar–N2+ + H3PO2 + H2O → Ar–H + N2 + H3PO3 + HCl.

Result …

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

KCET 2026Set D31 markMCQ
$Q. The compound from which chlorobenzene cannot be prepared easily is (A) Aniline (B) Benzene (C) Phenol (D) Benzene diazonium chloride$
›Reveal solutionSolution

The C–OH bond in phenol is strengthened by resonance with the aromatic ring, so phenol resists conversion to chlorobenzene, unlike benzene, aniline, or benzene diazonium chloride.

Step 1 — Routes that DO give chlorobenzene easily

Benzene undergoes direct electrophilic aromatic chlorination with $Cl_2$ in the presence of a Lewis acid catalyst such as anhydrous $FeCl_3$ , giving chlorobenzene in one step. Aniline can first be converted to benzene diazonium chloride by diazotisation ( $NaNO_2$ / $HCl$ , 0–5°C), and this diazonium salt then undergoes the Sandmeyer reaction with $CuCl$ / $HCl$ to directly replace $-N_2^+Cl^-$ with $-Cl$ , giving chlorobenzene. So benzene, aniline, and benzene diazonium chloride are all genuine, well-established starting materials for chlorobenzene.

Step 2 — Why phenol is the exception …
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KCET 2023Set D-21 markMCQ
$\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\ P\ } \text{C}_6\text{H}_5\overset{+}{\text{N}}_2\text{Cl}^- \xrightarrow{\ Q\ } \text{C}_6\text{H}_5\overset{+}{\text{N}}_2\overset{-}{\text{B}}\text{F}_4 \xrightarrow{\ R\ } \text{C}_6\text{H}_5\text{NO}_2$
›Reveal solutionSolution

Aniline → diazonium chloride (NaNO₂/dil. HCl, 273–278 K) → diazonium fluoroborate (HBF₄) → nitrobenzene (Cu + NaNO₂) — the standard Balz–Schiemann-type sequence.

Step 1 — Identify P (diazotisation).

The conversion $\mathrm{C_6H_5NH_2 \rightarrow C_6H_5\overset{+}{N}_2Cl^-}$ is diazotisation. A primary aromatic amine reacts with nitrous acid, which is unstable and must be generated in situ:
$\mathrm{NaNO_2 + HCl \longrightarrow HNO_2 + NaCl}$ $\mathrm{C_6H_5NH_2 + HNO_2 + HCl \xrightarrow{273-278\ K} C_6H_5\overset{+}{N}_2Cl^- + 2H_2O}$
The reagent is therefore $\mathrm{NaNO_2}$ + dilute HCl at 0–5 °C (ice-cold; the diazonium salt decomposes above ~278 K).

Why not $\mathrm{NaNO_3}$ ? Nitrate gives no nitrous acid — it cannot diazotise. That eliminates option (D). (Dil. HCl is the textbook condition, which is what (A) and (C) carry.)

Step 2 — Identify Q (the counter-ion swap).

The chloride $\mathrm{Cl^-}$ must be replaced by the tetrafluoroborate ion $\mathrm{BF_4^-}$ . This is done with fluoroboric acid, $\mathrm{HBF_4}$ :
$\mathrm{C_6H_5\overset{+}{N}_2Cl^- + HBF_4 \longrightarrow C_6H_5\overset{+}{N}_2\overset{-}{B}F_4 \downarrow + HCl}$
The fluoroborate salt is insoluble and precipitates out — that is the point of this step (it is stable enough to be isolated and dried).

Why not $\mathrm{F_2}$ or $\mathrm{BF_3}$ ? Elemental $\mathrm{F_2}$ would destroy the substrate and supplies no $\mathrm{BF_4^-}$ (eliminates (B) and (D)); $\mathrm{BF_3}$ alone is only the Lewis acid, not the $\mathrm{HBF_4}$ source used in the standard preparation (eliminates (C)).

Step 3 — Identify R (nitro-de-diazoniation). …
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COMEDK 2022Set 20221 markMCQ
$Q. Which among the following will not liberate nitrogen on reaction with nitrous acid? (A) dimethylamine (B) 2-aminopropane (C) ethylamine (D) methylamine$
›Reveal solutionSolution
Checking: (A) Dimethylamine (CH3)2NH - SECONDARY amine -> N-nitrosodimethylamine, NO N2 liberated. (B) 2-aminopropane (CH3)2CH-NH2 - primary -> N2 evolved. (C) Ethylamine C2H5NH2 - primary -> N2 evolved. (D) Methylamine CH3NH2 - primary -> N2 evolved.

Concept: Reaction of amines with nitrous acid (HNO2, from NaNO2 + HCl).

PRIMARY aliphatic amines give an unstable diazonium salt that decomposes at once, evolving N2 gas (brisk effervescence) and giving an alcohol.

SECONDARY amines give a yellow oily N-nitrosamine; NO nitrogen gas is evolved.

TERTIARY amines give soluble salts, no N2.

Checking: …
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