Diazonium Salts

8 MCQs9-step worked example
Source: NCERT Polymers and Chemistry in Everyday LifePYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The diazonium salt question on NEET is a reagent-recall trap. You see a benzenediazonium chloride (C₆H₅N₂⁺Cl⁻) reacting with a copper salt or water, and the answer hinges on whether you remember which copper reagent gives which product. Confuse Sandmeyer with Gattermann and you lose four marks — the wrong option is designed to look right.

What is a diazonium salt? When a primary aromatic amine (like aniline) reacts with nitrous acid (NaNO₂ + HCl) at 273–278 K, the amino group is replaced by the diazonium group (–N₂⁺). The product, benzenediazonium chloride, is the gateway to a wide range of aromatic substitutions that are otherwise difficult or impossible by direct methods (NCERT Class 12 Chemistry Chapter 9, page 16).

Why is this synthetically important? The –N₂⁺ group can be replaced by –Cl, –Br, –CN, –I, –OH, –H, –F, and –NO₂ through specific named reactions. This makes diazonium chemistry the single most versatile route to substituted benzene rings in NCERT organic chemistry (NCERT Class 12 Chemistry Chapter 9, page 18).

The high-frequency trap — Sandmeyer vs. Gattermann reagent confusion. Both reactions replace –N₂⁺ with a halide, but the reagents differ:

  • Sandmeyer reaction: ArN₂⁺ + CuCl → ArCl (or CuBr → ArBr, CuCN → ArCN). Reagent is cuprous salt in solution.
  • Gattermann reaction: ArN₂⁺ + Cu powder + HCl → ArCl (or HBr → ArBr). Reagent is copper metal powder with the hydrogen halide.

Other key conversions: hydrolysis with warm water gives ArOH (phenol). Reaction with HBF₄ followed by heating gives ArF (Balz–Schiemann). Reaction with H₃PO₂ gives ArH (deamination). Azo coupling with phenol or aniline in alkaline medium gives azo dyes.

Watch out: NEET distractors routinely swap the Sandmeyer reagent (CuX) with the Gattermann reagent (Cu + HX). Anchor the distinction: Sandmeyer = cuprous salt; Gattermann = copper powder + acid.


Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

Benzenediazonium chloride is obtained by treating aniline with NaNO₂ and HCl at 273–278 K. This reaction is called:

MCQ 2Easy RecallPractice

Which of the following is the correct reagent and product for the Sandmeyer reaction starting from benzenediazonium chloride?

MCQ 3Easy RecallPractice

Which reagent is used in the Gattermann reaction to convert benzenediazonium chloride to chlorobenzene?

MCQ 4Direct ApplicationPractice

Benzenediazonium chloride on treatment with CuCN gives product X. What is X?

MCQ 5Direct ApplicationPractice

Benzenediazonium chloride reacts with warm water to form:

MCQ 6Direct ApplicationPractice

Which of the following reactions of benzenediazonium chloride follows the Balz–Schiemann pathway?

MCQ 7Concept TrapPractice

Benzenediazonium chloride couples with phenol in alkaline medium. The site of coupling on phenol is preferentially at the:

MCQ 8CalculationPractice

A student needs to convert aniline to iodobenzene. The correct sequence is: (i) NaNO₂ + HCl at 273–278 K (ii) Reagent X Identify Reagent X.

Worked Example

  1. 1

    Given

    - Starting material: Aniline (C₆H₅NH₂) - Reagent 1: NaNO₂ + HCl, 273–278 K - Reagent 2: CuBr

  2. 2

    Required

    Final organic product and reaction name.

  3. 3

    Concept

    Primary aromatic amines undergo diazotisation with NaNO₂ + HCl at low temperature to form diazonium salts. The diazonium group can then be replaced by various nucleophiles. Cuprous salts (CuX) drive the Sandmeyer reaction.

  4. 4

    Formula / Reaction scheme

    - Step 1: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O (at 273–278 K) - Step 2: C₆H₅N₂⁺Cl⁻ + CuBr → C₆H₅Br + N₂ + CuCl

  5. 5

    Substitution

    Aniline → benzenediazonium chloride → bromobenzene. The cuprous bromide replaces –N₂⁺ with –Br.

  6. 6

    Calculation

    No numerical calculation. This is a product-identification problem. The key reasoning step is recognising CuBr as a Sandmeyer reagent (cuprous salt) rather than Gattermann (which would require Cu powder + HBr).

  7. 7

    Final answer

    The product is **bromobenzene (C₆H₅Br)**. The second step is the **Sandmeyer reaction**.

  8. 8

    Common trap

    Students confuse the Sandmeyer reagent (CuBr) with the Gattermann reagent (Cu + HBr). Both give bromobenzene, but the question asks you to identify the reaction by the reagent. If the reagent is a cuprous salt → Sandmeyer. If the reagent is copper powder + hydrogen halide → Gattermann. Picking "Gattermann" when CuBr is given loses you 4 marks.

  9. 9

    Similar NEET-style question

    "Aniline is diazotised and then treated with CuCN. Name the reaction and the product." (Answer: Sandmeyer reaction; product = cyanobenzene / benzonitrile.) ---

Before solving, remember these

Definition

Diazonium salts

Aryl diazonium chloride: ArN₂⁺Cl⁻. Prepared by diazotisation: ArNH₂ + HCl + NaNO₂ at 0-5°C → ArN₂⁺Cl⁻. Highly versatile in synthesis.

-- NCERT, p. 16

Sandmeyer: ArN₂⁺Cl⁻ + Cu/HX → ArX (X = Cl, Br, CN). Gattermann (Cu powder), Hofmann (KI for ArI). Hydrolysis: ArN₂⁺ + H₂O → ArOH. Coupling with phenols/anilines → azo dyes (orange/red).

-- NCERT, p. 18

Formulas

pKb of common amines (aqueous)

Aqueous basicity order: 2° ≈ 1° > 3° > NH3. Aromatic amines much weaker due to resonance delocalisation of lone pair.

SymbolQuantitySI Unit
pKb-log Kb-

Valid when

  • Aqueous solution; gas-phase order is 3°>2°>1°

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Category: Inorganic Exception

Student treats aniline as stronger base than methylamine. Aniline weaker due to lone-pair delocalization into ring.

When it triggers

Basicity ordering across aromatic and aliphatic amines.

How to avoid

Aliphatic amines (CH3NH2) > NH3 > aromatic amines (C6H5NH2). Aniline lone pair delocalised into ring → less available for protonation.

Category: Organic Reaction Conditions

Quaternary ammonium hydroxide (Hofmann) elimination favours LESS substituted alkene (anti-Saytzeff). Halide elimination (E2) follows Saytzeff (more substituted).

When it triggers

E2 elimination question with quaternary ammonium hydroxide vs alkyl halide.

How to avoid

Hofmann: bulky leaving group + base → least substituted alkene (Hofmann product). Saytzeff: alkyl halide + base → most substituted alkene (Zaitsev product).

Past Year Questions

11 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2023

The correct order of energies of molecular orbitals of N molecule, is 2

11s < *1s < 2s < *2s < 2p < (2p = 2p ) < (*2p = *2p ) < *2p z x y x y z
21s < *1s < 2s < *2s < 2p < *2p < (2p = 2p ) < (*2p = *2p ) z z x y x y
31s < *1s < 2s < *2s < (2p = 2p ) < (*2p = *2p ) < 2p < *2p x y x y z z
41s < *1s < 2s < *2s < (2p = 2p ) < 2p < (*2p = *2p ) < *2p x y z x y z
NTA Answer: Option 4(final)
NEET 2022

Match List-I with List-II. List – I List – II (Products formed) (Reaction of carbonyl compound with) (a) Cyanohydrin (i) NH OH 2 (b) Acetal (ii) RNH 2 (c) Schiff's base (iii) alcohol (d) Oxime (iv) HCN Choose the correct answer from the options given below

1(a) – (iv), (b) – (iii), (c) – (ii), (d) – (i)
2(a) – (iii), (b) – (iv), (c) – (ii), (d) – (i)
3(a) – (ii), (b) – (iii), (c) – (iv), (d) – (i)
4(a) – (i), (b) – (iii), (c) – (ii), (d) – (iv)
NTA Answer: Option 1(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Sources

NCERT refs: Class 12 Chemistry Chapter 9, p.16 | Class 12 Chemistry Chapter 9, p.18

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