Unit II MCQs: Phenols, Aromatic Amines, Aryl Diazonium Salts and Aromatic Acids
Strengthen your understanding of phenols, aromatic amines, aryl diazonium salts and aromatic acids through topic-wise, examination-oriented MCQs. Attempt each question before opening the answer and use the explanations to revise important concepts, reactions and substituent effects.
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Practice questions on the classification, nomenclature, preparation, physical properties and important chemical reactions of phenols.
Practice questions on the acidic character of phenols, phenoxide-ion stability, resonance effects and the influence of electron-donating and electron-withdrawing substituents.
Practice questions on the basic character of aromatic amines, aniline basicity, resonance and inductive effects, and the influence of electron-donating and electron-withdrawing substituents.
Practice questions on diazotization, preparation and stability of aryl diazonium salts, replacement reactions, azo coupling and their important synthetic applications.
Practice questions on benzoic-acid acidity, resonance stabilization, substituent effects, the ortho effect, and important reactions of benzoic acid.
Topic 1: Phenols—Preparation and Important Reactions MCQs
Attempt each question before opening the answer. Each explanation highlights the concept commonly tested in university examinations.
1. Which industrial process produces phenol together with acetone?
A. Dow process
B. Cumene process
C. Raschig process
D. Wurtz reaction
Correct answer: B. Cumene process
Explanation: Cumene first undergoes air oxidation to form cumene hydroperoxide. Acidic cleavage then produces phenol and acetone.
Reaction conditions: Air oxidation at approximately 100–130°C (373–403 K), followed by acid-catalyzed cleavage.
Reaction:
C₆H₅CH(CH₃)₂ + O₂ → C₆H₅C(CH₃)₂OOH
C₆H₅C(CH₃)₂OOH → C₆H₅OH + (CH₃)₂CO
Exam point: The cumene process simultaneously produces phenol and acetone.
2. In the Dow process, chlorobenzene is converted into phenol by heating it with:
A. Dilute NaOH at room temperature
B. Aqueous NaOH at high temperature and pressure
C. Alcoholic KOH under reflux
D. Concentrated H₂SO₄ at 0°C
Correct answer: B. Aqueous NaOH at high temperature and pressure
Explanation: Chlorobenzene reacts with aqueous sodium hydroxide at approximately 623 K and high pressure to form sodium phenoxide. Acidification of sodium phenoxide produces phenol.
Reaction conditions: Aqueous NaOH at approximately 350°C (623 K) and about 300 atm pressure, followed by acidification.
Reaction:
C₆H₅Cl + 2NaOH → C₆H₅ONa + NaCl + H₂O
C₆H₅ONa + HCl → C₆H₅OH + NaCl
Exam point: Chlorobenzene requires severe conditions because resonance gives the C–Cl bond partial double-bond character.
3. Fusion of sodium benzene sulphonate with sodium hydroxide initially produces:
A. Benzene
B. Sodium phenoxide
C. Chlorobenzene
D. Benzoic acid
Correct answer: B. Sodium phenoxide
Explanation: Sodium benzene sulphonate undergoes alkaline fusion with sodium hydroxide to form sodium phenoxide. Acidification then produces phenol.
Reaction conditions: Fused NaOH at approximately 300°C (573 K), followed by acidification.
Reaction:
C₆H₅SO₃Na + 2NaOH → C₆H₅ONa + Na₂SO₃ + H₂O
C₆H₅ONa + HCl → C₆H₅OH + NaCl
4. Hydrolysis of benzenediazonium chloride on warming produces:
A. Aniline
B. Chlorobenzene
C. Phenol
D. Benzonitrile
Correct answer: C. Phenol
Explanation: On warming with water, the diazonium group is replaced by –OH. Nitrogen gas evolves and phenol forms.
Reaction conditions: Gentle warming with water, generally above room temperature.
Reaction: C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂↑ + HCl
Exam point: Evolution of highly stable N₂ gas drives the reaction forward.
5. Phenol reacts with bromine water to form:
A. Bromobenzene
B. 2-Bromophenol only
C. 4-Bromophenol only
D. 2,4,6-Tribromophenol
Correct answer: D. 2,4,6-Tribromophenol
Explanation: The –OH group strongly activates the aromatic ring and directs substitution toward both ortho positions and the para position.
Reaction conditions: Bromine water at room temperature, approximately 25°C (298 K).
Reaction: C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr
Observation: Bromine water becomes colorless, and a white precipitate of 2,4,6-tribromophenol forms.
6. Treatment of phenol with dilute nitric acid at room temperature produces:
A. Nitrobenzene only
B. 2-Nitrophenol and 4-nitrophenol
C. 2,4-Dinitrophenol only
D. 2,4,6-Trinitrophenol only
Correct answer: B. 2-Nitrophenol and 4-nitrophenol
Explanation: The activating and ortho/para-directing –OH group promotes nitration at the ortho and para positions.
Reaction conditions: Dilute HNO₃ at approximately 25°C (298 K).
Reaction: C₆H₅OH + HNO₃ → o-HOC₆H₄NO₂ + p-HOC₆H₄NO₂ + H₂O
Exam point: Dilute HNO₃ produces a mixture of mononitrophenols.
7. Phenol reacts with concentrated nitric acid to produce:
A. 2-Nitrophenol
B. 4-Nitrophenol
C. 2,4-Dinitrophenol
D. 2,4,6-Trinitrophenol
Correct answer: D. 2,4,6-Trinitrophenol
Explanation: Strong nitration introduces nitro groups at both ortho positions and the para position. The resulting compound is picric acid.
Reaction conditions: Concentrated HNO₃ with controlled heating.
Reaction: C₆H₅OH + 3HNO₃ → C₆H₂(NO₂)₃OH + 3H₂O
Exam point: 2,4,6-Trinitrophenol is commonly called picric acid.
8. Which reagent commonly gives a violet color with phenol?
A. Bromine water
B. Neutral ferric chloride solution
C. Tollens’ reagent
D. Fehling’s solution
Correct answer: B. Neutral ferric chloride solution
Explanation: Phenol reacts with neutral ferric chloride to form a colored ferric–phenolate complex.
Test conditions: Neutral FeCl₃ solution at room temperature, approximately 25°C (298 K).
Test representation: Phenol + Fe³⁺ → Violet ferric–phenolate complex
Observation: Phenol generally produces a violet coloration.
9. The Kolbe–Schmitt reaction of sodium phenoxide mainly produces:
A. Benzoic acid
B. Salicylic acid
C. Phenyl acetate
D. Benzaldehyde
Correct answer: B. Salicylic acid
Explanation: Sodium phenoxide reacts with carbon dioxide under pressure to produce sodium salicylate. Acidification gives salicylic acid.
Reaction conditions: CO₂ under pressure at approximately 125–130°C (398–403 K), followed by acidification.
Reaction:
C₆H₅ONa + CO₂ → o-HOC₆H₄COONa
o-HOC₆H₄COONa + HCl → o-HOC₆H₄COOH + NaCl
Exam point: The –COOH group enters mainly at the ortho position to –OH.
10. Phenol gives salicylaldehyde when treated with:
A. CHCl₃ and aqueous NaOH
B. CCl₄ and alcoholic KOH
C. CH₃Cl and AlCl₃
D. CO₂ under pressure
Correct answer: A. CHCl₃ and aqueous NaOH
Explanation: This reaction is the Reimer–Tiemann reaction. It introduces a formyl group mainly at the ortho position of phenol.
Reaction conditions: CHCl₃ and aqueous NaOH with heating at approximately 60–70°C (333–343 K), followed by acidification.
Reaction: C₆H₅OH + CHCl₃ + 3NaOH → o-HOC₆H₄CHO + 3NaCl + 2H₂O
Major product: Salicylaldehyde or 2-hydroxybenzaldehyde.
11. Phenol reacts with sodium hydroxide to form:
A. Sodium benzoate
B. Sodium phenoxide
C. Sodium benzene sulphonate
D. Sodium benzylate
Correct answer: B. Sodium phenoxide
Explanation: Phenol possesses sufficient acidic character to react with sodium hydroxide and form sodium phenoxide.
Reaction conditions: Aqueous NaOH at room temperature, approximately 25°C (298 K).
Reaction:
C₆H₅OH + NaOH → C₆H₅ONa + H₂O
Exam point: Phenol reacts with NaOH, whereas ordinary alcohols generally do not.
12. Which gas evolves when phenol reacts with metallic sodium?
A. Oxygen
B. Nitrogen
C. Hydrogen
D. Carbon dioxide
Correct answer: C. Hydrogen
Explanation: Sodium replaces the hydrogen atom of the phenolic –OH group, producing sodium phenoxide and hydrogen gas.
Reaction conditions: Metallic sodium under dry conditions at room temperature, approximately 25°C (298 K).
Reaction: 2C₆H₅OH + 2Na → 2C₆H₅ONa + H₂↑
13. Distillation of phenol with zinc dust produces:
A. Cyclohexane
B. Benzene
C. Toluene
D. Chlorobenzene
Correct answer: B. Benzene
Explanation: Zinc dust removes oxygen from phenol, converting it into benzene. Zinc oxide forms as the second product.
Reaction conditions: Heating and distillation with zinc dust.
Reaction: C₆H₅OH + Zn → C₆H₆ + ZnO
Exam point: Zinc-dust distillation demonstrates the presence of the benzene nucleus in phenol.
14. In the phthalein test, phenol reacts with phthalic anhydride to form:
A. Fluorescein
B. Phenolphthalein
C. Picric acid
D. Salicylic acid
Correct answer: B. Phenolphthalein
Explanation: Heating phenol with phthalic anhydride in the presence of concentrated sulphuric acid produces phenolphthalein. The product becomes pink in alkaline solution.
Reaction conditions: Concentrated H₂SO₄ and heating at approximately 180°C (453 K).
Reaction:
2C₆H₅OH + phthalic anhydride → Phenolphthalein + H₂O
Observation: Addition of aqueous NaOH produces a pink color.
15. Azo coupling between phenol and benzenediazonium chloride occurs most readily in:
A. Strongly acidic medium
B. Alkaline medium
C. Neutral water only
D. Anhydrous ether
Correct answer: B. Alkaline medium
Explanation: In alkaline medium, phenol converts into the more reactive phenoxide ion. Coupling occurs mainly at the para position to produce p-hydroxyazobenzene.
Reaction conditions: Mildly alkaline medium maintained at 0–5°C (273–278 K).
Reaction: C₆H₅N₂⁺Cl⁻ + C₆H₅OH + NaOH → p-HOC₆H₄–N=N–C₆H₅ + NaCl + H₂O
Product: p-Hydroxyazobenzene, an intensely colored azo compound.
Topic 2: Acidity of Phenols and Effect of Substituents MCQs
Attempt each question before opening the answer. Each explanation highlights the concept commonly tested in university examinations.
1. Why is phenol more acidic than ethanol?
A. Phenol has a higher molecular mass
B. Phenol contains an aromatic ring
C. The phenoxide ion is stabilized by resonance
D. Ethanol contains more hydrogen atoms
Correct answer: C. The phenoxide ion is stabilized by resonance
Explanation: Loss of H⁺ from phenol produces the phenoxide ion. Its negative charge becomes delocalized over the oxygen atom and the ortho and para carbon atoms. The ethoxide ion does not receive comparable resonance stabilization.
Ionization:
C₆H₅OH ⇌ C₆H₅O⁻ + H⁺
Exam point: Greater conjugate-base stability produces greater acidity.
2. The approximate pKₐ value of phenol is:
A. 4.8
B. 10
C. 16
D. 25
Correct answer: B. 10
Explanation: Phenol has a pKₐ of approximately 10. Therefore, it is more acidic than common alcohols, whose pKₐ values are generally around 16–18, but less acidic than carboxylic acids.
3. Which compound is the most acidic?
A. Phenol
B. p-Cresol
C. p-Methoxyphenol
D. p-Nitrophenol
Correct answer: D. p-Nitrophenol
Explanation: The para nitro group withdraws electron density through both negative inductive (–I) and negative resonance (–R) effects. It stabilizes the p-nitrophenoxide ion and increases acidity.
4. Electron-withdrawing groups generally affect phenolic acidity by:
A. Decreasing acidity
B. Increasing acidity
C. Completely preventing ionization
D. Converting phenol into an alcohol
Correct answer: B. Increasing acidity
Explanation: Electron-withdrawing groups stabilize the negative charge of the phenoxide ion through inductive or resonance effects. This favors the loss of H⁺ and increases acidity.
5. Electron-donating groups generally decrease the acidity of phenol because they:
A. Stabilize the phenoxide ion
B. Destabilize the phenoxide ion
C. Remove the oxygen atom
D. Increase O–H bond polarity
Correct answer: B. Destabilize the phenoxide ion
Explanation: Electron-donating groups increase electron density in the negatively charged phenoxide ion. This destabilizes the conjugate base and makes proton loss less favorable.
6. Which order correctly represents the acidity of nitrophenol isomers?
A. m-Nitrophenol > o-Nitrophenol > p-Nitrophenol
B. p-Nitrophenol > o-Nitrophenol > m-Nitrophenol
C. m-Nitrophenol > p-Nitrophenol > o-Nitrophenol
D. o-Nitrophenol > m-Nitrophenol > p-Nitrophenol
Correct answer: B. p-Nitrophenol > o-Nitrophenol > m-Nitrophenol
Explanation: At the ortho and para positions, –NO₂ stabilizes the phenoxide ion through both –I and –R effects. At the meta position, it primarily operates through the –I effect. The para isomer generally shows slightly greater acidity than the ortho isomer because intramolecular hydrogen bonding stabilizes neutral o-nitrophenol.
Exam point:
p-Nitrophenol > o-Nitrophenol > m-Nitrophenol > phenol
7. Which compound is commonly known as picric acid?
A. 2-Nitrophenol
B. 4-Nitrophenol
C. 2,4-Dinitrophenol
D. 2,4,6-Trinitrophenol
Correct answer: D. 2,4,6-Trinitrophenol
Explanation: Three strongly electron-withdrawing –NO₂ groups greatly stabilize the conjugate base. Therefore, 2,4,6-trinitrophenol is much more acidic than phenol.
Formula: C₆H₂(NO₂)₃OH
8. Why is p-cresol less acidic than phenol?
A. The methyl group shows a –I effect
B. The methyl group shows a +I effect
C. The methyl group stabilizes the phenoxide ion
D. The methyl group removes the phenolic hydrogen
Correct answer: B. The methyl group shows a +I effect
Explanation: The –CH₃ group releases electron density through its positive inductive effect. This additional electron density destabilizes the p-cresoxide ion and decreases acidity.
Acidity order:
Phenol > p-Cresol
9. p-Chlorophenol is more acidic than phenol mainly because chlorine shows:
A. A strong +I effect
B. A negative inductive effect
C. No electronic effect
D. Only hyperconjugation
Correct answer: B. A negative inductive effect
Explanation: Chlorine withdraws electron density through its –I effect and stabilizes the chlorophenoxide ion. Although chlorine can donate through resonance, its electron-withdrawing inductive effect produces a net increase in acidity.
Acidity order:
p-Chlorophenol > Phenol
10. Phenol reacts with sodium hydroxide to produce:
A. Sodium benzoate
B. Sodium phenoxide
C. Sodium ethoxide
D. Sodium benzene sulphonate
Correct answer: B. Sodium phenoxide
Explanation: Phenol transfers its acidic proton to the hydroxide ion, producing sodium phenoxide and water.
Reaction conditions: Aqueous NaOH at room temperature, approximately 25°C (298 K).
Reaction:
C₆H₅OH + NaOH → C₆H₅ONa + H₂O
11. Phenol generally does not react with sodium bicarbonate because phenol is:
A. Stronger than carbonic acid
B. Weaker than carbonic acid
C. A strong inorganic acid
D. Completely nonacidic
Correct answer: B. Weaker than carbonic acid
Explanation: Sodium bicarbonate reacts readily with acids stronger than carbonic acid. Phenol is weaker than carbonic acid and cannot release CO₂ from NaHCO₃ under ordinary conditions.
Comparison:
Carboxylic acids react with NaHCO₃, whereas phenol generally does not.
12. Which order of acidity is correct?
A. Ethanol > Phenol > Benzoic acid
B. Phenol > Benzoic acid > Ethanol
C. Benzoic acid > Phenol > Ethanol
D. Phenol > Ethanol > Benzoic acid
Correct answer: C. Benzoic acid > Phenol > Ethanol
Explanation: The benzoate ion receives strong resonance stabilization over two equivalent oxygen atoms. The phenoxide ion receives resonance stabilization, but its negative charge extends partly onto carbon atoms. The ethoxide ion lacks resonance stabilization.
Approximate pKₐ values:
Benzoic acid ≈ 4.2
Phenol ≈ 10
Ethanol ≈ 16
13. In the resonance structures of the phenoxide ion, negative charge appears mainly at:
A. The meta positions only
B. The ortho and para positions
C. Every ring position equally
D. The hydrogen atom
Correct answer: B. The ortho and para positions
Explanation: Conjugation between the oxygen lone pair and the aromatic π system delocalizes the negative charge onto oxygen and the two ortho and one para carbon atoms.
Exam point: This delocalization explains the enhanced stability of the phenoxide ion.
14. o-Nitrophenol is more steam volatile than p-nitrophenol because o-nitrophenol forms:
A. Strong intermolecular hydrogen bonding
B. Intramolecular hydrogen bonding
C. An ionic lattice
D. A covalent polymer
Correct answer: B. Intramolecular hydrogen bonding
Explanation: In o-nitrophenol, the –OH and –NO₂ groups lie close enough to form intramolecular hydrogen bonding. This reduces intermolecular association and increases steam volatility. Conversely, p-nitrophenol forms intermolecular hydrogen bonds.
15. Which compound is the most acidic?
Topic 3: Basicity of Aromatic Amines and Effect of Substituents MCQs
Attempt each question before opening the answer. Each explanation highlights the concept commonly tested in university examinations.
1. Aromatic amines act as bases because the nitrogen atom contains:
A. A positive charge
B. A vacant d-orbital
C. A lone pair of electrons
D. An acidic hydrogen only
Correct answer: C. A lone pair of electrons
Explanation: The nitrogen atom can donate its lone pair to a proton. Therefore, aromatic amines behave as Lewis bases and Brønsted–Lowry bases.
Reaction conditions: Dilute HCl at room temperature, approximately 25°C (298 K).
Reaction:
C₆H₅NH₂ + HCl → C₆H₅NH₃⁺Cl⁻
Product: Anilinium chloride
2. Why is aniline less basic than methylamine?
A. Aniline has no lone pair
B. The nitrogen lone pair in aniline is delocalized into the benzene ring
C. Methylamine contains an aromatic ring
D. Aniline cannot form salts
Correct answer: B. The nitrogen lone pair in aniline is delocalized into the benzene ring
Explanation: Resonance delocalizes the nitrogen lone pair into the benzene ring. Consequently, the lone pair becomes less available for protonation.
Exam point: Resonance decreases the basicity of aniline.
3. The conjugate acid of aniline is:
A. Phenoxide ion
B. Anilinium ion
C. Benzenediazonium ion
D. Amide ion
Correct answer: B. Anilinium ion
Explanation: Aniline accepts H⁺ through its nitrogen lone pair and forms the anilinium ion.
Reaction:
C₆H₅NH₂ + H⁺ ⇌ C₆H₅NH₃⁺
Formula of anilinium ion: C₆H₅NH₃⁺
4. Which substituent generally increases the basicity of aniline?
A. –NO₂
B. –CN
C. –CH₃
D. –COOH
Correct answer: C. –CH₃
Explanation: The methyl group releases electron density through its positive inductive effect. This increases the availability of the nitrogen lone pair for protonation.
Basicity comparison:
p-Toluidine > Aniline
5. Which substituent most strongly decreases the basicity of aniline?
A. –CH₃
B. –OCH₃
C. –NH₂
D. –NO₂
Correct answer: D. –NO₂
Explanation: The nitro group strongly withdraws electron density through both negative inductive and negative resonance effects. This decreases the availability of the nitrogen lone pair.
Basicity comparison:
Aniline > p-Nitroaniline
6. Which order of basicity is generally correct?
A. p-Nitroaniline > p-Chloroaniline > Aniline > p-Toluidine
B. p-Toluidine > Aniline > p-Chloroaniline > p-Nitroaniline
C. Aniline > p-Toluidine > p-Nitroaniline > p-Chloroaniline
D. p-Chloroaniline > p-Toluidine > Aniline > p-Nitroaniline
Correct answer: B. p-Toluidine > Aniline > p-Chloroaniline > p-Nitroaniline
Explanation: The electron-donating –CH₃ group increases basicity. Chlorine decreases basicity mainly through its –I effect, whereas the strongly electron-withdrawing –NO₂ group produces the greatest reduction.
7. p-Nitroaniline is less basic than m-nitroaniline because the para nitro group operates through:
A. Only a +I effect
B. Only hyperconjugation
C. Both –I and –R effects
D. No electronic effect
Correct answer: C. Both –I and –R effects
Explanation: At the para position, the nitro group withdraws electron density through both inductive and resonance effects. At the meta position, its influence occurs mainly through the inductive effect.
Basicity order:
m-Nitroaniline > p-Nitroaniline
8. The reduced basicity of many ortho-substituted anilines is commonly explained by:
A. Aromaticity loss
B. Steric inhibition of solvation
C. Complete removal of the nitrogen lone pair
D. Formation of a carbon–carbon triple bond
Correct answer: B. Steric inhibition of solvation
Explanation: An ortho substituent can hinder effective solvation of the protonated anilinium ion. Poor stabilization of the conjugate acid decreases the observed basicity in aqueous solution.
Exam point: This behavior contributes to the ortho effect.
9. Which order of basicity in aqueous solution is correct?
A. Aniline > Ammonia > Methylamine
B. Ammonia > Aniline > Methylamine
C. Methylamine > Ammonia > Aniline
D. Aniline > Methylamine > Ammonia
Correct answer: C. Methylamine > Ammonia > Aniline
Explanation: The methyl group increases electron density on nitrogen through the +I effect. Aniline is least basic because resonance delocalizes its nitrogen lone pair.
10. A lower pK₍b₎ value indicates:
A. A weaker base
B. A stronger base
C. A stronger acid only
D. A neutral compound
Correct answer: B. A stronger base
Explanation: A stronger base has a larger base-ionization constant, K₍b₎. Because pK₍b₎ = −log K₍b₎, a larger K₍b₎ corresponds to a lower pK₍b₎.
Exam point: Basic strength increases as pK₍b₎ decreases.
11. Acetylation of aniline produces:
A. Acetanilide
B. Phenol
C. Benzamide
D. Benzoic acid
Correct answer: A. Acetanilide
Explanation: The amino group reacts with an acetylating agent to form acetanilide. In the amide product, the nitrogen lone pair becomes strongly delocalized toward the carbonyl group.
Reaction conditions: Acetyl chloride or acetic anhydride at approximately 20–25°C (293–298 K), usually in the presence of a base.
Reaction:
C₆H₅NH₂ + CH₃COCl → C₆H₅NHCOCH₃ + HCl
Product: Acetanilide
12. Why is acetanilide less basic than aniline?
A. Acetanilide contains no nitrogen atom
B. The nitrogen lone pair is delocalized toward the carbonyl group
C. The methyl group removes the carbonyl oxygen
D. Acetanilide is an aliphatic amine
Correct answer: B. The nitrogen lone pair is delocalized toward the carbonyl group
Explanation: In acetanilide, the nitrogen lone pair participates in resonance with the carbonyl group. It is therefore less available for protonation.
13. Aniline reacts with nitrous acid at 0–5°C to produce:
A. Phenol directly
B. Benzenediazonium chloride
C. Nitrobenzene
D. Benzyl alcohol
Correct answer: B. Benzenediazonium chloride
Explanation: Nitrous acid, produced in situ from sodium nitrite and hydrochloric acid, converts aniline into benzenediazonium chloride. This reaction is called diazotization.
Reaction conditions: NaNO₂ and HCl at 0–5°C (273–278 K).
Reaction:
C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
Exam point: The temperature must remain below approximately 5°C (278 K) because the diazonium salt becomes unstable at higher temperatures.
14. Which factor directly controls the basic strength of an aromatic amine?
A. Availability of the nitrogen lone pair
B. Number of hydrogen atoms in the benzene ring
C. Molecular color
D. Melting point alone
Correct answer: A. Availability of the nitrogen lone pair
Explanation: Any factor that increases lone-pair availability generally increases basicity. Conversely, resonance delocalization and electron-withdrawing groups decrease lone-pair availability and basicity.
15. Which compound is the strongest base?
Topic 4: Aryl Diazonium Salts—Preparation, Reactions and Synthetic Uses MCQs
Attempt each question before opening the answer. Each explanation highlights the concept commonly tested in university examinations.
1. The general formula of an aryl diazonium salt is:
A. ArNH₂
B. ArN₂⁺X⁻
C. ArNO₂
D. ArNCO
Correct answer: B. ArN₂⁺X⁻
Explanation: Aryl diazonium salts contain an aromatic group attached to the diazonium group, –N₂⁺. An anion such as Cl⁻, Br⁻ or BF₄⁻ balances the positive charge.
Example: Benzenediazonium chloride, C₆H₅N₂⁺Cl⁻
2. Benzenediazonium chloride is prepared from aniline using:
A. NaNO₃ and H₂SO₄
B. NaNO₂ and HCl
C. NH₄Cl and NaOH
D. HNO₃ and HCl
Correct answer: B. NaNO₂ and HCl
Explanation: Sodium nitrite reacts with hydrochloric acid to generate nitrous acid in situ. Nitrous acid converts aniline into benzenediazonium chloride through diazotization.
Reaction conditions: NaNO₂ and HCl at 0–5°C (273–278 K).
Reaction:
C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
3. Why is diazotization maintained at 0–5°C?
A. Aniline freezes above 5°C
B. Sodium nitrite becomes inactive above 5°C
C. Benzenediazonium chloride becomes unstable at higher temperatures
D. Hydrochloric acid boils at 5°C
Correct answer: C. Benzenediazonium chloride becomes unstable at higher temperatures
Explanation: Benzenediazonium chloride remains sufficiently stable in cold aqueous solution. At higher temperatures, it decomposes readily and may undergo hydrolysis to phenol.
Required temperature: 0–5°C (273–278 K)
4. Warming an aqueous solution of benzenediazonium chloride produces:
A. Aniline
B. Phenol
C. Nitrobenzene
D. Benzonitrile
Correct answer: B. Phenol
Explanation: Water replaces the diazonium group during hydrolysis. Stable nitrogen gas evolves and phenol forms.
Reaction conditions: Gentle warming with water at approximately 40–50°C (313–323 K).
Reaction:
C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂↑ + HCl
5. Benzenediazonium chloride produces chlorobenzene when treated with:
A. CuCl and HCl
B. KI
C. HBF₄
D. H₃PO₂
Correct answer: A. CuCl and HCl
Explanation: Cuprous chloride replaces the diazonium group with chlorine. This transformation is a Sandmeyer reaction.
Reaction conditions: CuCl in hydrochloric acid, generally with gentle warming.
Reaction:
C₆H₅N₂⁺Cl⁻ → C₆H₅Cl + N₂↑
Product: Chlorobenzene
6. Conversion of a benzenediazonium salt into bromobenzene requires:
A. CuBr and HBr
B. NaOH and water
C. KI only
D. H₃PO₂
Correct answer: A. CuBr and HBr
Explanation: Cuprous bromide replaces the diazonium group with bromine through the Sandmeyer reaction.
Reaction conditions: CuBr in hydrobromic acid, generally with gentle warming.
Reaction:
C₆H₅N₂⁺Br⁻ → C₆H₅Br + N₂↑
Product: Bromobenzene
7. Benzenediazonium chloride gives benzonitrile when treated with:
A. CuCN
B. CuCl
C. KI
D. NaOH
Correct answer: A. CuCN
Explanation: Cuprous cyanide replaces the diazonium group with –CN. This Sandmeyer reaction introduces one additional carbon atom into the functional group.
Reaction conditions: CuCN with gentle warming.
Reaction:
C₆H₅N₂⁺Cl⁻ + CuCN → C₆H₅CN + N₂↑ + CuCl
Product: Benzonitrile
8. In the Gattermann reaction, the diazonium group is replaced by chlorine or bromine using:
A. Copper powder and the corresponding hydrogen halide
B. Zinc dust and water
C. Sodium metal and alcohol
D. Concentrated nitric acid
Correct answer: A. Copper powder and the corresponding hydrogen halide
Explanation: Copper powder and HCl or HBr replace the diazonium group with Cl or Br. Unlike the Sandmeyer reaction, the Gattermann reaction uses copper powder rather than a cuprous salt.
Reaction conditions: Copper powder with HCl or HBr; gentle warming.
General reaction:
ArN₂⁺X⁻ → ArCl or ArBr + N₂↑
9. Benzenediazonium chloride produces iodobenzene when treated with:
A. CuI and HCl
B. Potassium iodide
C. HBF₄
D. Sodium fluoride
Correct answer: B. Potassium iodide
Explanation: Iodide ion directly replaces the diazonium group. A copper catalyst is not normally required.
Reaction conditions: Aqueous KI, generally at room temperature, approximately 25°C (298 K).
Reaction:
C₆H₅N₂⁺Cl⁻ + KI → C₆H₅I + KCl + N₂↑
Product: Iodobenzene
10. Fluorobenzene is prepared from benzenediazonium chloride through:
A. Sandmeyer reaction
B. Balz–Schiemann reaction
C. Reimer–Tiemann reaction
D. Kolbe–Schmitt reaction
Correct answer: B. Balz–Schiemann reaction
Explanation: Benzenediazonium chloride first reacts with fluoroboric acid to form solid benzenediazonium tetrafluoroborate. Heating this salt produces fluorobenzene.
Reaction conditions: HBF₄ at 0–5°C (273–278 K), followed by isolation and heating of the dry diazonium tetrafluoroborate.
Reaction:
C₆H₅N₂⁺Cl⁻ + HBF₄ → C₆H₅N₂⁺BF₄⁻ + HCl
C₆H₅N₂⁺BF₄⁻ → C₆H₅F + BF₃ + N₂↑
Product: Fluorobenzene
11. Replacement of the diazonium group by hydrogen is achieved using:
A. H₃PO₂
B. Concentrated HNO₃
C. CuCN
D. Bromine water
Correct answer: A. H₃PO₂
Explanation: Hypophosphorous acid reduces the diazonium group and replaces it with hydrogen. This reaction is useful for removing an amino group after it has served as a temporary directing group.
Reaction conditions: Hypophosphorous acid in aqueous medium, generally with gentle warming.
Reaction:
C₆H₅N₂⁺Cl⁻ + H₃PO₂ + H₂O → C₆H₆ + N₂↑ + H₃PO₃ + HCl
Product: Benzene
12. Azo coupling of benzenediazonium chloride with phenol occurs most readily in:
A. Strongly acidic medium
B. Mildly alkaline medium
C. Concentrated hydrochloric acid
D. Anhydrous ether
Correct answer: B. Mildly alkaline medium
Explanation: Alkaline medium converts phenol into the more reactive phenoxide ion. Coupling occurs mainly at the para position.
Reaction conditions: Mildly alkaline medium at 0–5°C (273–278 K).
Reaction:
C₆H₅N₂⁺Cl⁻ + C₆H₅OH + NaOH → p-HOC₆H₄–N=N–C₆H₅ + NaCl + H₂O
Product: p-Hydroxyazobenzene
13. Azo coupling of benzenediazonium chloride with aniline mainly produces:
A. Azobenzene
B. p-Aminoazobenzene
C. Phenylhydrazine
D. Nitrobenzene
Correct answer: B. p-Aminoazobenzene
Explanation: The activating –NH₂ group directs azo coupling mainly toward the para position. A weakly acidic or buffered medium prevents excessive protonation of aniline.
Reaction conditions: Weakly acidic or buffered medium at 0–5°C (273–278 K).
Reaction:
C₆H₅N₂⁺Cl⁻ + C₆H₅NH₂ → p-H₂NC₆H₄–N=N–C₆H₅ + HCl
Product: p-Aminoazobenzene
14. Azo compounds are generally strongly colored because they contain:
A. An isolated C–C single bond
B. An extended conjugated system containing –N=N–
C. Only saturated carbon atoms
D. An ionic metal lattice
Correct answer: B. An extended conjugated system containing –N=N–
Explanation: The azo group connects aromatic rings and creates an extended conjugated π-electron system. Absorption of visible light gives many azo compounds yellow, orange, red or related colors.
Functional group: –N=N–
15. Why are aryl diazonium salts important in organic synthesis?
A. They produce only one type of compound
B. The diazonium group can be replaced by several different functional groups
C. They cannot release nitrogen gas
D. They react only with alkanes
Correct answer: B. The diazonium group can be replaced by several different functional groups
Explanation: The diazonium group can be replaced by –OH, –Cl, –Br, –I, –F, –CN or –H. Diazonium salts also undergo azo coupling to produce colored compounds.
Synthetic conversion summary:
ArN₂⁺ → ArOH, ArCl, ArBr, ArI, ArF, ArCN or ArH
Exam point: Aryl diazonium salts provide routes to substituted aromatic compounds that may be difficult to prepare through direct substitution.
COACH VNA ACADEMY · BP301T · UNIT II
Aromatic Acids — MCQ Practice
Acidity, substituent effects, and important reactions of benzoic acid
15 questions · 1 mark each · No negative marking. Select your answers, then check your score. Unanswered questions score zero. Answers are revealed after submission.
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