Q.Predict the missing reagents (labelled 1, 3 and 5) and the missing products (labelled 2 and 4) in the following reaction sequence. p-nitrotoluene (a benzene ring with –CH3 and –NO2 para to each other) is treated with reagent 1 to give p-toluidine (–CH3 and –NH2 para). p-Toluidine is treated with (CH3CO)2O / pyridine to give p-methylacetanilide (–CH3 and –NHCOCH3 para). p-Methylacetanilide is treated with HNO3/H2SO4 to give product 2. Product 2 is treated with reagent 3 to give 4-methyl-2-nitroaniline (–CH3 para to –NH2, with –NO2 ortho to the –NH2). 4-methyl-2-nitroaniline is treated with NaNO2/HCl to give compound 4. Compound 4 is treated with reagent 5 to give m-nitrotoluene (a benzene ring with –CH3 and –NO2 meta to each other).
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🔒 Start your 14-day free trial to unlock the full solution →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.
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:
Ar–NH2+NaNO2+2HCl0−5∘CAr–N2+Cl−+NaCl+2H2O
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 |
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:
C6H5–N2+Cl−+C6H5–OHNaOH, 0–5∘CC6H5–N=N–C6H4–OH (p-hydroxyazobenzene, orange dye)
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? …
Reagent 1 reduces –NO2 to –NH2; the acetamido group directs nitration ortho, giving product 2; reagent 3 hydrolyses off the acetyl group; NaNO2/HCl makes the diazonium salt (compound 4); reagent 5 removes it by deamination. …
The sequence reduces the nitro group to an amine, protects it as the acetanilide, nitrates ortho to nitrogen, hydrolyses the protecting group, diazotises the amine, and finally removes it by deamination — shifting the substitution pattern to give m-nitrotoluene.
Reagent 1 – reduction
p-Nitrotoluene → p-toluidine requires reduction of –NO2 to –NH2. Reagent 1 = Sn/HCl (equivalently Fe/HCl or H2/catalyst).
Product 2 – nitration of p-methylacetanilide
p-Methylacetanilide has –CH3 and –NHCOCH3 para. On nitration (HNO3/H2SO4), the strongly directing acetamido group sends –NO2 to the position ortho to itself. Product 2 = 4-methyl-2-nitroacetanilide (–CH3, –NHCOCH3 para, –NO2 ortho to –NHCOCH3).
Reagent 3 – hydrolysis
Product 2 → 4-methyl-2-nitroaniline means the acetyl protecting group is removed. Reagent 3 = H3O+ (dilute acid hydrolysis; aqueous base then acidification also works).
Compound 4 – diazotisation
4-Methyl-2-nitroaniline + NaNO2/HCl (273–278 K) → the diazonium salt. Compound 4 = 4-methyl-2-nitrobenzenediazonium chloride. …
Method: Reverse-Engineering a Multistep Reagent/Product Sequence
Core Concept
Each blank in a reaction sequence can be filled by comparing the structures immediately before and after it, identifying which named reaction type explains that specific change, and then recalling the standard reagent/conditions (or product) for that reaction.
Steps
- Look at each adjacent pair of given/missing structures in turn and note exactly which functional group changed.
- Match that change to a known reaction type: -NO2 to -NH2 is reduction; -NH2 to -NHCOCH3 is acetylation/protection; a new -NO2 appearing on a protected amine is nitration; -NHCOCH3 to -NH2 is hydrolysis; -NH2 to -N2+ is diazotisation; -N2+ to -H is deamination.
- Recall the standard reagents/conditions for that reaction type.
- Fill each blank (reagent or product) in sequence, checking that the resulting structure is a valid input for the next step. …
Showing the 12 most recent of 14 on this concept.
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.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
…
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.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. …
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.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 …
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.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.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.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.
…
- GUJCET 2024Set 131 markMCQQ.Which diazonium salt is water insoluble and stable at room temperature? (A) C6H5⋅N2+BF4− (B) C6H5N2+Br− (C) C6H5N2+HSO4− (D) C6H5N2+Cl−
›Reveal solutionSolution
Benzenediazonium tetrafluoroborate C6H5N2+BF4− 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 …
- GUJCET 2023Set 091 markMCQQ.Benzene diazonium chloride reacts with phenol in basic medium to give product. How many σ (sigma) and π (pi) bonds are present in that product? (A) 16 - σ and 7 - π (B) 16 - σ and 6 - π (C) 26 - σ and 7 - π (D) 26 - σ and 6 - π
›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: C6H5-N=N-C6H4-OH. Each C=C of an aromatic ring is one σ + one π; every single bond is a σ bond.
Solution
Product structure: Ring A (C6H5) — N=N — Ring B (C6H4) with −OH para.
π bonds:
- Ring A benzene: 3
- Ring B benzene: 3
- N=N: 1
- Total π=7
σ bonds: …
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.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. …
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.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: …
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.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 …
- GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQQ.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 …
- GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQQ.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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