CUET UG Chemistry Booster Test - 3 Diazonium Salts -

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CUET UG Chemistry Booster Test - 1 Category: Diazonium Salts

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Chemistry Booster Test - 1 Category: Diazonium Salts

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QUESTION 1 OF 20

The positive charge in the diazonium group is localized primarily on which atom?

Answer Review

1 The positive charge in the diazonium group is localized primarily on which atom?

A The terminal nitrogen atom

B The inner nitrogen atom directly attached to the aromatic ring

C The halogen anion

D The carbon atom of the aromatic ring

💡 Short Explanation

�� Diazonium group is Ar–N≡N⁺. �� Positive charge is mainly on the nitrogen attached to the ring. �� This contributes to its leaving-group ability.

✅ Correct Answer Explanation

In the diazonium ion (Ar–N₂⁺), the positive charge is primarily associated with the nitrogen atom directly bonded to the aromatic ring. This nitrogen participates in resonance stabilization with the aromatic system.

❓ Why Other Options Are Incorrect

�� Option A → Terminal nitrogen does not primarily carry the positive charge.
�� Option C → Halogen is the counter anion.
�� Option D → Aromatic carbon does not bear the positive charge.

🎯 Strategy

Used

Conceptual Analysis

Final Logic:

�� Positive charge is primarily associated with the ring-attached nitrogen.

🧠 Memory Trick

Ring N = Positive N

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Structure of Diazonium Salts

2 What is the unit of pressure commonly associated with the release of the nitrogen gas during the decomposition of diazonium salts?

A Kelvin

B Atmospheres (or Pascals)

C Moles per liter

D Grams

💡 Short Explanation

�� Pressure is measured in atm or Pa. �� Nitrogen gas evolution increases pressure. �� Kelvin measures temperature, not pressure.

✅ Correct Answer Explanation

Nitrogen gas released during diazonium decomposition is a gaseous product. Gas pressure is measured using units such as atmospheres (atm) or pascals (Pa).

❓ Why Other Options Are Incorrect

�� Option A → Temperature unit.
�� Option C → Concentration unit.
�� Option D → Mass unit.

🎯 Strategy

Used

Dimensional/Unit Analysis

Final Logic:

�� Pressure is measured in atm or Pa.

🧠 Memory Trick

Pressure → Pa / atm

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Diazonium Salt Reactions

3 Why does the stability of arenediazonium ions strictly require a low temperature (273–278 K) despite resonance stabilization?

A At higher temperatures, resonance entirely ceases.

B The thermal energy at higher temperatures is sufficient to overcome the activation energy for N₂ displacement by water.

C Water freezes above 278 K, halting the reaction.

D The diazonium salt becomes insoluble at higher temperatures.

💡 Short Explanation

�� Arenediazonium salts are only moderately stable. �� Higher temperatures promote decomposition. �� Water can replace the diazonium group.

✅ Correct Answer Explanation

Although resonance stabilizes arenediazonium ions, increasing temperature supplies enough energy for decomposition reactions such as hydrolysis. Therefore low temperatures are required during preparation and storage.

❓ Why Other Options Are Incorrect

�� Option A → Resonance does not cease.
�� Option C → Water freezes below 273 K.
�� Option D → Insolubility is not the reason.

🎯 Strategy

Used

Elimination

Final Logic:

�� Heat promotes decomposition despite resonance stabilization.

🧠 Memory Trick

Diazo + Heat = Decompose

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Stability of Arenediazonium Salts

4 Match the following: Diazonium Salt Stability

List 1 (Primary Amine Type)	List 2 (Diazonium Salt Stability)
1. Primary aliphatic amine	a. Stable for a short time at 273–278 K
2. Primary aromatic amine	b. Highly unstable, liberates nitrogen quantitatively
3. Arenediazonium salt	c. Stability enhanced by resonance delocalization
4. Alkyldiazonium salt	d. Decomposes rapidly even at low temperatures

A 1-b, 2-a, 3-c, 4-d

B 1-a, 2-b, 3-d, 4-c

C 1-d, 2-c, 3-a, 4-b

D 1-b, 2-c, 3-a, 4-d

💡 Short Explanation

Primary aliphatic amines form highly unstable diazonium salts. Primary aromatic amines form relatively stable diazonium salts at low temperature. Arenediazonium ions are stabilized by resonance. Alkyldiazonium salts decompose rapidly with evolution of nitrogen gas.

✅ Correct Answer Explanation

Primary aliphatic amines produce alkyldiazonium salts, which are extremely unstable and liberate nitrogen quantitatively. Therefore 1 → b. Primary aromatic amines form arenediazonium salts that remain stable for a short time at 273–278 K, hence 2 → a. The stability of arenediazonium salts arises from resonance delocalization of the positive charge into the aromatic ring, so 3 → c. Alkyldiazonium salts lack such resonance stabilization and therefore decompose rapidly even at low temperatures, giving 4 → d. Thus the correct matching is: 1-b, 2-a, 3-c, 4-d

❓ Why Other Options Are Incorrect

Option B → Reverses the stability behavior of aliphatic and aromatic diazonium salts.
Option C → Incorrectly assigns resonance stabilization to primary aromatic amines rather than arenediazonium salts.
Option D → Misplaces the resonance stabilization and decomposition characteristics.

🎯 Strategy

Used

Concept Mapping

Application:

Match each amine type or diazonium salt with its characteristic stability behavior based on NCERT discussion of diazotisation.

Final Logic:

Aliphatic diazonium salts are highly unstable, whereas aromatic diazonium salts are resonance-stabilized and temporarily stable at low temperatures.

🧠 Memory Trick

Alkyl = Away (decomposes quickly)

📚 Class XII – Chemistry Part II – Amines – Diazonium Salts: Stability of Alkyldiazonium and Arenediazonium Salts

5 If aniline is treated with sodium nitrite and hydrochloric acid at 298 K instead of 273 K, what will be the predominant organic product?

A Benzenediazonium chloride

B Phenol

C Chlorobenzene

D p-Aminoazobenzene

💡 Short Explanation

�� Diazonium salt forms initially. �� At higher temperature it hydrolyses. �� Phenol becomes the major product.

✅ Correct Answer Explanation

Benzenediazonium chloride is stable only at 273–278 K. At 298 K it decomposes and reacts with water to form phenol.

❓ Why Other Options Are Incorrect

�� Option A → Stable only at low temperature.
�� Option C → Requires replacement by chloride.
�� Option D → Requires coupling reaction.

🎯 Strategy

Used

Contextual/Tonal Matching

Final Logic:

�� Warm diazonium salt → phenol.

🧠 Memory Trick

Warm Diazo = Phenol

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Hydrolysis of Diazonium Salts

6 During diazotisation, what is the role of adding excess hydrochloric acid?

A To act as a solvent only.

B To react with NaNO₂ to generate nitrous acid and to maintain acidic conditions.

C To precipitate the diazonium salt immediately.

D To reduce the aniline.

💡 Short Explanation

�� HCl generates HNO₂ in situ. �� Acidic medium is essential. �� Diazotisation requires both.

✅ Correct Answer Explanation

Sodium nitrite reacts with HCl to generate nitrous acid, the actual diazotising agent. Excess HCl also maintains the required acidic medium.

❓ Why Other Options Are Incorrect

�� Option A → Not its only role.
�� Option C → Not the primary purpose.
�� Option D → HCl does not reduce aniline.

🎯 Strategy

Used

Conceptual Analysis

Final Logic:

�� HCl creates HNO₂ and maintains acidity.

🧠 Memory Trick

NaNO₂ + HCl → HNO₂

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Diazotisation

📄 Passage

Replacement by –NO₂ group: When diazonium fluoroborate is heated with aqueous sodium nitrite solution in the presence of copper, the diazonium group is replaced by –NO₂ group. Replacement by fluoride ion: When arenediazonium chloride is treated with fluoroboric acid, arene diazonium fluoroborate is precipitated which on heating decomposes to yield aryl fluoride.

Based on the passage, what is the intermediate isolated before the introduction of a nitro group or a fluorine atom?

A Benzenediazonium chloride

B Arene diazonium fluoroborate

C Sodium nitrite

D Aqueous copper solution

💡 Short Explanation

�� Fluoroborate salt is isolated. �� It serves as the key intermediate. �� Further heating gives desired products.

✅ Correct Answer Explanation

The passage specifically mentions the formation and isolation of arene diazonium fluoroborate before fluorination or nitro substitution.

❓ Why Other Options Are Incorrect

�� Option A → Starting diazonium salt.
�� Option C → Reagent.
�� Option D → Catalyst medium.

🎯 Strategy

Used

Contextual Matching

Final Logic:

�� Fluoroborate is the isolated intermediate.

🧠 Memory Trick

BF₄ = Intermediate

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Diazonium Fluoroborates

📄 Passage

Based on the passage, identify the reaction type required to convert arene diazonium fluoroborate into aryl fluoride.

A Condensation

B Thermal decomposition

C Hydrolysis

D Electrophilic addition

💡 Short Explanation

�� Heating fluoroborate is required. �� Decomposition releases N₂. �� Aryl fluoride is formed.

✅ Correct Answer Explanation

Arene diazonium fluoroborate yields aryl fluoride upon heating. This transformation occurs through thermal decomposition.

❓ Why Other Options Are Incorrect

�� Option A → No condensation involved.
�� Option C → No hydrolysis occurs.
�� Option D → Not an addition reaction.

🎯 Strategy

Used

Contextual Matching

Final Logic:

�� Heating causes decomposition to aryl fluoride.

🧠 Memory Trick

Heat BF₄ → Fluoride

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Balz–Schiemann Reaction

9 Unlike benzenediazonium chloride, benzenediazonium fluoroborate is water insoluble and stable at room temperature. This is chemically advantageous for which of the following processes?

A Carrying out the Gattermann reaction.

B Safely isolating the diazonium salt as a dry solid for fluorination or nitration.

C Performing a coupling reaction with phenol.

D Hydrolyzing it to phenol.

💡 Short Explanation

�� Fluoroborate salt is stable. �� Can be isolated safely. �� Useful for subsequent reactions.

✅ Correct Answer Explanation

The stability and insolubility of diazonium fluoroborates allow isolation and handling as solids before fluorination or nitro substitution.

❓ Why Other Options Are Incorrect

�� Option A → Not the main advantage.
�� Option C → Coupling uses diazonium solution.
�� Option D → Hydrolysis does not require fluoroborate isolation.

🎯 Strategy

Used

Elimination

Final Logic:

�� Stability enables safe isolation.

🧠 Memory Trick

BF₄ = Stable Solid

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Diazonium Fluoroborates

10 Consider the following statements:
Benzenediazonium chloride reacts with water when warmed.
Diazonium salts of aromatic amines are less stable than those of aliphatic amines.
Benzenediazonium chloride decomposes easily in the dry state.

A I and II only

B I and III only

C II and III only

D I, II, and III

💡 Short Explanation

�� Warming gives phenol. �� Dry diazonium salts decompose. �� Aromatic diazonium salts are more stable than aliphatic ones.

✅ Correct Answer Explanation

Statement I is correct because benzenediazonium chloride hydrolyses on warming. → Statement II is incorrect because aromatic diazonium salts are more stable than aliphatic diazonium salts. → Statement III is correct because benzenediazonium chloride decomposes readily in the dry state.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect Statement II.
�� Option C → Omits correct Statement I.
�� Option D → Statement II is false.

🎯 Strategy

Used

Option Grouping

Final Logic:

�� Statements I and III are correct.

🧠 Memory Trick

Warm → Phenol; Dry → Decompose

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Properties of Diazonium Salts

11 Arrange the following reaction intermediates/reagents in the sequence needed to synthesize aryl fluoride from aniline:
I. Heating the precipitate
II. Aniline + NaNO₂/HCl (273–278 K)
III. Addition of fluoroboric acid (HBF₄)

A I, II, III

B II, III, I

C III, II, I

D II, I, III

💡 Short Explanation

�� Aniline is first diazotized. �� Diazonium fluoroborate is then formed. �� Heating produces aryl fluoride.

✅ Correct Answer Explanation

The preparation of aryl fluoride from aniline follows the Balz–Schiemann route. → First, aniline is diazotized using NaNO₂/HCl at 273–278 K to form benzenediazonium chloride. → Fluoroboric acid is then added to precipitate arene diazonium fluoroborate. → Heating this fluoroborate causes thermal decomposition, yielding aryl fluoride. Sequence: Aniline + NaNO₂/HCl (273–278 K) Addition of HBF₄ Heating the precipitate

❓ Why Other Options Are Incorrect

�� Option A → Heating cannot occur before diazotization and fluoroborate formation.
�� Option C → HBF₄ cannot react before diazonium salt formation.
�� Option D → Heating before fluoroborate formation is incorrect.

🎯 Strategy

Used

Ordering

Application:

�� Follow the reaction sequence from diazotization to fluoroborate formation to decomposition.

Final Logic:

�� Diazotization → Fluoroborate formation → Heating.

🧠 Memory Trick

Diazo → BF₄ → Heat → F

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Preparation of Aryl Fluorides (Balz–Schiemann Reaction)

12 In the synthesis of iodobenzene from benzenediazonium chloride, why is the Sandmeyer reaction not required?

A Iodide is a strong enough nucleophile to easily displace the diazonium group directly upon mixing with KI.

B Copper reacts explosively with iodine.

C Iodobenzene is completely insoluble.

D The diazonium salt reduces the iodine to iodide.

💡 Short Explanation

�� KI directly reacts with diazonium salts. �� Cu(I) catalyst is unnecessary. �� Iodobenzene forms with N₂ evolution.

✅ Correct Answer Explanation

Benzenediazonium chloride reacts directly with potassium iodide. → The iodide ion replaces the diazonium group and nitrogen gas is released. → Unlike Cl and Br substitutions, no Cu(I) catalyst is required. [ C_6H_5N_2^+Cl^- + KI \rightarrow C_6H_5I + KCl + N_2 ]

❓ Why Other Options Are Incorrect

�� Option B → Copper does not react explosively with iodine under these conditions.
�� Option C → Solubility is unrelated to the reaction mechanism.
�� Option D → Diazonium salt does not reduce iodine.

🎯 Strategy

Used

Elimination

Application:

�� Identify the unique feature of iodide substitution.

Final Logic:

�� KI directly displaces the diazonium group without Cu(I).

🧠 Memory Trick

KI works directly

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Replacement by Iodide Ion

13 Which of the following represents the correct reactants for the Gattermann reaction?

A ArN₂⁺X⁻ + Cu(I)CN

B ArN₂⁺X⁻ + Cu powder / HCl

C ArN₂⁺X⁻ + KI

D ArN₂⁺X⁻ + H₃PO₂

💡 Short Explanation

�� Gattermann reaction uses copper powder. �� HCl or HBr is employed. �� Halogen replaces diazonium group.

✅ Correct Answer Explanation

In the Gattermann reaction, arenediazonium salts are treated with copper powder and hydrochloric acid or hydrobromic acid. → This introduces chlorine or bromine into the aromatic ring. → Copper powder distinguishes Gattermann reaction from Sandmeyer reaction.

❓ Why Other Options Are Incorrect

�� Option A → Represents Sandmeyer cyanation.
�� Option C → Direct iodination reaction.
�� Option D → Reduction of diazonium salt to arene.

🎯 Strategy

Used

Odd One Out

Application:

�� Identify the reagent uniquely associated with Gattermann reaction.

Final Logic:

�� Copper powder + HCl/HBr = Gattermann reaction.

🧠 Memory Trick

Gattermann = Copper Powder

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Gattermann Reaction

14 What is the IUPAC name of the compound formed when benzenediazonium chloride is subjected to the Sandmeyer reaction using Cu(I)CN / KCN?

A Phenyl isocyanide

B Cyanobenzene

C Benzylamine

D Azo cyanide

💡 Short Explanation

�� CN replaces diazonium group. �� Sandmeyer cyanation forms aryl nitrile. �� Product is cyanobenzene.

✅ Correct Answer Explanation

In Sandmeyer reaction, CuCN/KCN replaces the diazonium group with a cyano group (–CN). [ C_6H_5N_2^+Cl^- \xrightarrow{CuCN} C_6H_5CN + N_2 ] → The product C₆H₅CN is called cyanobenzene (benzonitrile).

❓ Why Other Options Are Incorrect

�� Option A → Isocyanides have –NC linkage.
�� Option C → Benzylamine contains –CH₂NH₂.
�� Option D → Not a recognized product.

🎯 Strategy

Used

Substitution

Application:

�� Replace diazonium group with CN.

Final Logic:

�� Ar–N₂⁺ → Ar–CN gives cyanobenzene.

🧠 Memory Trick

CuCN → ArCN

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Sandmeyer Cyanation

15 When reducing benzenediazonium chloride to benzene using ethanol, ethanol itself undergoes what chemical change?

A It is reduced to ethane.

B It is dehydrated to ethene.

C It is oxidised to ethanal.

D It couples to form ethylbenzene.

💡 Short Explanation

�� Ethanol acts as a reducing agent. �� Benzene is formed from diazonium salt. �� Ethanol is oxidized to ethanal.

✅ Correct Answer Explanation

Ethanol reduces benzenediazonium chloride to benzene. → During this process, ethanol itself undergoes oxidation. [ C_2H_5OH \rightarrow CH_3CHO ] → Therefore ethanol is converted into ethanal while the diazonium salt is reduced to benzene.

❓ Why Other Options Are Incorrect

�� Option A → Ethanol is oxidized, not reduced.
�� Option B → Dehydration does not occur here.
�� Option D → Ethylbenzene is not formed.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Identify the oxidation product of ethanol in diazonium reduction.

Final Logic:

�� Ethanol acts as reducing agent and becomes ethanal.

🧠 Memory Trick

Ethanol → Ethanal

📚 Class XII – Chemistry Part II – Chapter: Amines – Topic: Reduction of Diazonium Salts to Arenes

16 The replacement of the diazonium group by a hydroxyl group requires warming the solution to 283 K. What is the molecular formula of the product formed?

A C₆H₆

B C₆H₅OH

C C₆H₅NH₂

D C₆H₅CH₃

💡 Short Explanation

Warming benzenediazonium chloride with water causes hydrolysis. The diazonium group is replaced by –OH. Phenol is formed.

✅ Correct Answer Explanation

Benzenediazonium chloride undergoes hydrolysis when warmed to about 283 K. The diazonium group (–N₂⁺) is replaced by a hydroxyl group (–OH), producing phenol (C₆H₅OH). Reaction: C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂ + HCl

❓ Why Other Options Are Incorrect

A. C₆H₆ → Benzene is obtained by reduction of diazonium salts, not hydrolysis.
C. C₆H₅NH₂ → Aniline is the starting compound for diazotisation.
D. C₆H₅CH₃ → Toluene is not formed in this reaction.

🎯 Strategy

Used

Reaction identification

🧠 Memory Trick

Warm Diazonium + Water = Phenol

📚 Class XII Chemistry – Amines – Reactions of Diazonium Salts – Hydrolysis

17 The extended conjugate system in p-hydroxyazobenzene causes it to absorb light in the visible spectrum. These compounds are commonly known as:

A Azo dyes

B Fluorescent markers

C Primary pigments

D UV absorbers

💡 Short Explanation

Azo compounds contain the –N=N– linkage. Extended conjugation absorbs visible light. Hence they are brightly coloured dyes.

✅ Correct Answer Explanation

Compounds such as p-hydroxyazobenzene possess an extended conjugated π-electron system through the azo linkage (–N=N–). This allows absorption of visible light, giving intense colours. Such compounds are widely used as azo dyes.

❓ Why Other Options Are Incorrect

B. Fluorescent markers → Not the general classification.
C. Primary pigments → Not a standard chemical class.
D. UV absorbers → They mainly absorb visible light.

🎯 Strategy

Used

Concept recognition

🧠 Memory Trick

Azo Bond (–N=N–) = Azo Dye

📚 Class XII Chemistry – Amines – Coupling Reactions

18 Which of the following conditions is most accurate regarding the coupling of benzenediazonium chloride with phenol or aniline?

A It occurs via free radical mechanism.

B It is an electrophilic substitution reaction at the para position.

C It involves the elimination of nitrogen gas.

D It requires temperatures above 373 K.

💡 Short Explanation

Diazonium ion acts as an electrophile. Phenol and aniline act as activated aromatic rings. Coupling occurs mainly at the para position.

✅ Correct Answer Explanation

The coupling reaction of benzenediazonium chloride with phenol or aniline is an electrophilic aromatic substitution reaction. The diazonium ion acts as the electrophile and attacks the para position of the activated aromatic ring, producing azo compounds.

❓ Why Other Options Are Incorrect

A. Not a free-radical reaction.
C. Nitrogen is not eliminated during azo coupling.
D. High temperatures would decompose the diazonium salt.

🎯 Strategy

Used

Mechanism identification

🧠 Memory Trick

Coupling = Electrophilic Substitution

📚 Class XII Chemistry – Amines – Coupling Reactions

19 Evaluate the statement: "Diazonium salts are excellent intermediates for preparing substituted aromatic compounds that cannot be prepared by direct substitution." Which of the following compounds justifies this statement?

A Bromobenzene

B Chlorobenzene

C Aryl fluorides

D Nitrobenzene

💡 Short Explanation

Aryl fluorides are difficult to prepare directly. Diazonium salts provide an efficient route. Balz–Schiemann reaction is used.

✅ Correct Answer Explanation

Aryl fluorides cannot be conveniently prepared by direct fluorination of benzene because fluorine is extremely reactive. Diazonium salts overcome this difficulty through the Balz–Schiemann reaction, making them valuable synthetic intermediates.

❓ Why Other Options Are Incorrect

A. Bromobenzene → Easily prepared by bromination.
B. Chlorobenzene → Easily prepared by chlorination.
D. Nitrobenzene → Easily prepared by nitration.

🎯 Strategy

Used

Application of synthetic importance

🧠 Memory Trick

Diazonium → Fluorobenzene Route

📚 Class XII Chemistry – Amines – Importance of Diazonium Salts

20 Cyanobenzene cannot be easily obtained by direct nucleophilic substitution of chlorobenzene. Why is the diazonium salt route effective for this synthesis?

A The diazonium group is a very good leaving group compared to chlorine attached to a benzene ring.

B Cyanide ions destroy the benzene ring directly.

C The reaction acts via an addition-elimination mechanism only on diazonium salts.

D Copper catalysts only bind to nitrogen.

💡 Short Explanation

Diazonium group leaves as stable N₂ gas. Nitrogen gas escapes from the reaction mixture. This makes substitution highly favorable.

✅ Correct Answer Explanation

The diazonium group is one of the best leaving groups in organic chemistry because it departs as extremely stable nitrogen gas (N₂). Therefore, cyanide can readily replace the diazonium group, whereas chlorobenzene resists nucleophilic substitution due to the partial double-bond character of the C–Cl bond.

❓ Why Other Options Are Incorrect

B. Cyanide does not destroy the benzene ring.
C. Not the fundamental reason for the reaction's success.
D. Copper does not function in this manner.

🎯 Strategy

Used

Leaving-group analysis

🧠 Memory Trick

Diazonium Leaves as N₂ → Easy Substitution

📚 Class XII Chemistry – Amines – Sandmeyer Reaction and Importance of Diazonium Salts

CUET UG Chemistry Booster Test – 3 Diazonium Salts

CUET UG Chemistry Booster Test - 1 Category: Diazonium Salts

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