Amines Practice

Take timed practice tests on Amines for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.

Timer-based practiceDetailed answer reviewMobile-friendly flow

Chemistry/Amines/ Practice

Amines Practice Tests

Six 20-question timed sessions plus a 60-question chapter module. Each item is original and reframed for copyright safety.

Ready

JEE Main Session 1

3Questions

60s/QTimer

+4 / -1Scoring

Preview all 13 questions in JEE Main Session 1 (no login required)

Classification
1. Trimethylamine (CH₃)₃N is classified as:
- A. Tertiary amine (3°) (Correct)
- B. Primary amine (1°)
- C. Secondary amine (2°)
- D. Quaternary ammonium
Explanation: Amines are classified by how many alkyl/aryl groups are bonded to N: 1° = R−NH₂, 2° = R₂NH, 3° = R₃N. Trimethylamine has three methyl groups on N → tertiary.
Basicity
2. Amines are basic because:
- A. N has a lone pair that can accept a proton (H⁺) (Correct)
- B. N is a large atom
- C. Amines have high boiling points
- D. N has positive charge
Explanation: The lone pair on nitrogen makes amines Lewis bases and Brønsted-Lowry bases. They accept H⁺: R₃N + H⁺ → R₃NH⁺ (ammonium salt).
Preparation
3. The Gabriel synthesis uses phthalimide to prepare:
- A. Primary aliphatic amines (Correct)
- B. Secondary amines
- C. Tertiary amines
- D. Aromatic amines
Explanation: Gabriel synthesis: Phthalimide + KOH → potassium phthalimide → + RX (alkyl halide) → N-alkylphthalimide → hydrolysis with acid/base → primary amine (RNH₂) + phthalic acid. It gives only 1° amines free from 2° and 3° impurities.
Basicity order
4. The correct order of basicity in aqueous solution for methylamines is:
- A. (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃ (Correct)
- B. (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃
- C. NH₃ > CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N
- D. All equal
Explanation: In aqueous solution (Kb values), basicity is affected by inductive effect (more alkyl = more +I = more basic) AND solvation of ammonium ion. 2° > 1° > 3° > NH₃ due to solvation counteracting the inductive effect at 3°.
Aniline basicity
5. Aniline (C₆H₅NH₂) is a weaker base than cyclohexylamine because:
- A. Lone pair on N delocalises into the aromatic ring, reducing availability to accept H⁺ (Correct)
- B. Phenyl group is larger
- C. Aniline has fewer lone pairs
- D. Cyclohexylamine has lower pKb
Explanation: In aniline, the lone pair of N is partially delocalised into the benzene ring (+M effect), reducing its electron density and hence basicity. Cyclohexylamine has no such delocalisation. pKb: aniline (9.4) >> cyclohexylamine (3.36).
Reaction with HNO₂
6. Primary aliphatic amines react with HNO₂ (NaNO₂ + HCl) to give:
- A. Unstable diazonium salt → N₂ + alcohol (Correct)
- B. Stable diazonium salt
- C. Amide
- D. Secondary amine
Explanation: 1° aliphatic amines + HNO₂ → alkyl diazonium salt (R−N₂⁺) which is unstable and immediately decomposes: evolves N₂ gas + carbocation → alcohol + alkene (various products). Unlike aryl diazonium salts which are stable at 0°C.
Diazotisation
7. Aniline + NaNO₂ + HCl at 0–5°C gives:
- A. Benzene diazonium chloride (stable) (Correct)
- B. Phenol
- C. Nitrobenzene
- D. Azobenzene
Explanation: Diazotisation: C₆H₅NH₂ + NaNO₂ + HCl (0–5°C) → C₆H₅N₂⁺Cl⁻ (benzene diazonium chloride). Aryl diazonium salts are stable at low temperature and undergo coupling reactions.
JEE Main
8. The Hinsberg test differentiates amines using:
- A. Benzenesulfonyl chloride (C₆H₅SO₂Cl) (Correct)
- B. Acetyl chloride
- C. FeCl₃
- D. NaOH + I₂
Explanation: Hinsberg test: 1° amines react with C₆H₅SO₂Cl → product soluble in NaOH (N−H still present). 2° amines → product insoluble in NaOH. 3° amines → no reaction. Distinguishes all three classes.
Coupling reaction
9. Benzene diazonium chloride reacts with phenol in alkaline medium to give:
- A. p-hydroxyazobenzene (orange azo dye) (Correct)
- B. Benzene
- C. Chlorobenzene
- D. Phenol + N₂
Explanation: Coupling reaction: C₆H₅N₂⁺ + C₆H₅OH (alkaline) → C₆H₅−N=N−C₆H₄OH (para) + H⁺. The azo dye has −N=N− (azo) linkage and is highly coloured. Para position is favoured due to −OH being ortho/para director.
JEE Advanced
10. Carbylamine reaction (formation of isocyanide) is a test for:
- A. Primary amines only (1°) (Correct)
- B. Secondary amines
- C. Tertiary amines
- D. All amines
Explanation: Carbylamine test: 1° amine + CHCl₃ + KOH/NaOH → isocyanide (R−NC) with extremely foul smell. This is a specific test for primary amines. 2° and 3° amines do not give this reaction.
Mixed
11. The IUPAC name of CH₃−NH−CH₃ is:
- A. N-methylmethanamine (or dimethylamine) (Correct)
- B. Ethylamine
- C. Trimethylamine
- D. Dimethyl ether
Explanation: CH₃−NH−CH₃ is a secondary amine with two methyl groups. Common name: dimethylamine. IUPAC: N-methylmethanamine.
Mixed
12. Amines have lower boiling points than alcohols of similar molecular weight because:
- A. N−H···N hydrogen bonds are weaker than O−H···O hydrogen bonds (Correct)
- B. Amines are gases
- C. Amines have lower molecular weight
- D. N is less electronegative than O
Explanation: N is less electronegative than O. So N−H hydrogen bonds are weaker than O−H hydrogen bonds. Primary and secondary amines hydrogen bond, but the interactions are weaker than in alcohols → lower boiling points than alcohols of similar molecular mass.
Mixed
13. The Sandmeyer reaction replaces the diazonium group (−N₂⁺) with:
- A. −Cl, −Br, or −CN using CuCl/CuBr/CuCN as catalyst (Correct)
- B. −OH only
- C. −NO₂ only
- D. −H only
Explanation: Sandmeyer reactions: ArN₂⁺ + CuCl → ArCl + N₂; + CuBr → ArBr; + CuCN → ArCN. The Cu(I) salt acts as catalyst in the radical mechanism. Also: + CuI → ArI (Finkelstein-type); + H₂O/H⁺ → ArOH (diazonium hydrolysis); + HBF₄ → ArF (Balz-Schiemann).

Loading practice engine...

JEE Main Session 1

3Questions

60s/QTimer

+4 / -1Scoring

Preview all 13 questions in JEE Main Session 1 (no login required)

Classification
1. Trimethylamine (CH₃)₃N is classified as:
- A. Tertiary amine (3°) (Correct)
- B. Primary amine (1°)
- C. Secondary amine (2°)
- D. Quaternary ammonium
Explanation: Amines are classified by how many alkyl/aryl groups are bonded to N: 1° = R−NH₂, 2° = R₂NH, 3° = R₃N. Trimethylamine has three methyl groups on N → tertiary.
Basicity
2. Amines are basic because:
- A. N has a lone pair that can accept a proton (H⁺) (Correct)
- B. N is a large atom
- C. Amines have high boiling points
- D. N has positive charge
Explanation: The lone pair on nitrogen makes amines Lewis bases and Brønsted-Lowry bases. They accept H⁺: R₃N + H⁺ → R₃NH⁺ (ammonium salt).
Preparation
3. The Gabriel synthesis uses phthalimide to prepare:
- A. Primary aliphatic amines (Correct)
- B. Secondary amines
- C. Tertiary amines
- D. Aromatic amines
Explanation: Gabriel synthesis: Phthalimide + KOH → potassium phthalimide → + RX (alkyl halide) → N-alkylphthalimide → hydrolysis with acid/base → primary amine (RNH₂) + phthalic acid. It gives only 1° amines free from 2° and 3° impurities.
Basicity order
4. The correct order of basicity in aqueous solution for methylamines is:
- A. (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃ (Correct)
- B. (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃
- C. NH₃ > CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N
- D. All equal
Explanation: In aqueous solution (Kb values), basicity is affected by inductive effect (more alkyl = more +I = more basic) AND solvation of ammonium ion. 2° > 1° > 3° > NH₃ due to solvation counteracting the inductive effect at 3°.
Aniline basicity
5. Aniline (C₆H₅NH₂) is a weaker base than cyclohexylamine because:
- A. Lone pair on N delocalises into the aromatic ring, reducing availability to accept H⁺ (Correct)
- B. Phenyl group is larger
- C. Aniline has fewer lone pairs
- D. Cyclohexylamine has lower pKb
Explanation: In aniline, the lone pair of N is partially delocalised into the benzene ring (+M effect), reducing its electron density and hence basicity. Cyclohexylamine has no such delocalisation. pKb: aniline (9.4) >> cyclohexylamine (3.36).
Reaction with HNO₂
6. Primary aliphatic amines react with HNO₂ (NaNO₂ + HCl) to give:
- A. Unstable diazonium salt → N₂ + alcohol (Correct)
- B. Stable diazonium salt
- C. Amide
- D. Secondary amine
Explanation: 1° aliphatic amines + HNO₂ → alkyl diazonium salt (R−N₂⁺) which is unstable and immediately decomposes: evolves N₂ gas + carbocation → alcohol + alkene (various products). Unlike aryl diazonium salts which are stable at 0°C.
Diazotisation
7. Aniline + NaNO₂ + HCl at 0–5°C gives:
- A. Benzene diazonium chloride (stable) (Correct)
- B. Phenol
- C. Nitrobenzene
- D. Azobenzene
Explanation: Diazotisation: C₆H₅NH₂ + NaNO₂ + HCl (0–5°C) → C₆H₅N₂⁺Cl⁻ (benzene diazonium chloride). Aryl diazonium salts are stable at low temperature and undergo coupling reactions.
JEE Main
8. The Hinsberg test differentiates amines using:
- A. Benzenesulfonyl chloride (C₆H₅SO₂Cl) (Correct)
- B. Acetyl chloride
- C. FeCl₃
- D. NaOH + I₂
Explanation: Hinsberg test: 1° amines react with C₆H₅SO₂Cl → product soluble in NaOH (N−H still present). 2° amines → product insoluble in NaOH. 3° amines → no reaction. Distinguishes all three classes.
Coupling reaction
9. Benzene diazonium chloride reacts with phenol in alkaline medium to give:
- A. p-hydroxyazobenzene (orange azo dye) (Correct)
- B. Benzene
- C. Chlorobenzene
- D. Phenol + N₂
Explanation: Coupling reaction: C₆H₅N₂⁺ + C₆H₅OH (alkaline) → C₆H₅−N=N−C₆H₄OH (para) + H⁺. The azo dye has −N=N− (azo) linkage and is highly coloured. Para position is favoured due to −OH being ortho/para director.
JEE Advanced
10. Carbylamine reaction (formation of isocyanide) is a test for:
- A. Primary amines only (1°) (Correct)
- B. Secondary amines
- C. Tertiary amines
- D. All amines
Explanation: Carbylamine test: 1° amine + CHCl₃ + KOH/NaOH → isocyanide (R−NC) with extremely foul smell. This is a specific test for primary amines. 2° and 3° amines do not give this reaction.
Mixed
11. The IUPAC name of CH₃−NH−CH₃ is:
- A. N-methylmethanamine (or dimethylamine) (Correct)
- B. Ethylamine
- C. Trimethylamine
- D. Dimethyl ether
Explanation: CH₃−NH−CH₃ is a secondary amine with two methyl groups. Common name: dimethylamine. IUPAC: N-methylmethanamine.
Mixed
12. Amines have lower boiling points than alcohols of similar molecular weight because:
- A. N−H···N hydrogen bonds are weaker than O−H···O hydrogen bonds (Correct)
- B. Amines are gases
- C. Amines have lower molecular weight
- D. N is less electronegative than O
Explanation: N is less electronegative than O. So N−H hydrogen bonds are weaker than O−H hydrogen bonds. Primary and secondary amines hydrogen bond, but the interactions are weaker than in alcohols → lower boiling points than alcohols of similar molecular mass.
Mixed
13. The Sandmeyer reaction replaces the diazonium group (−N₂⁺) with:
- A. −Cl, −Br, or −CN using CuCl/CuBr/CuCN as catalyst (Correct)
- B. −OH only
- C. −NO₂ only
- D. −H only
Explanation: Sandmeyer reactions: ArN₂⁺ + CuCl → ArCl + N₂; + CuBr → ArBr; + CuCN → ArCN. The Cu(I) salt acts as catalyst in the radical mechanism. Also: + CuI → ArI (Finkelstein-type); + H₂O/H⁺ → ArOH (diazonium hydrolysis); + HBF₄ → ArF (Balz-Schiemann).