Important Questions Class 12 Chemistry Chapter 9 Amines With Answers

Amines are organic compounds formed when hydrogen atoms of ammonia get replaced by alkyl or aryl groups. Methylamine, aniline, benzene diazonium chloride, and azo dyes are common examples from Class 12 Chemistry.

Organic chemistry becomes easier when students connect structure, basicity, reagents, and products. Important Questions Class 12 Chemistry Chapter 9 help Class 12 students revise Amines for CBSE 2026-27 board exams, school tests, and pre-board papers. NCERT Chapter 9 includes classification, nomenclature, preparation methods, reactions of amines, diazonium salts, and aromatic conversions. The chapter gives many 1-mark, 2-mark, 3-mark, and conversion-based questions in Indian school exam patterns.

Key Takeaways

• Amines: Amines are ammonia derivatives where one or more hydrogen atoms are replaced by alkyl or aryl groups.
• Basicity Trend: Alkylamines are generally stronger bases than ammonia, while aniline is weaker due to resonance.
• Diazonium Salts: Benzene diazonium chloride helps prepare chlorobenzene, bromobenzene, phenol, iodobenzene, and azo dyes.
• CBSE 2026-27 Focus: School exams often ask named reactions, basicity order, distinguishing tests, and organic conversions from Amines.

Important Questions Class 12 Chemistry Chapter 9 Structure 2026-27

Important Questions Class 12 Chemistry Chapter 9 for CBSE 2026-27

NCERT Chapter 9 gives direct reactions, reasoning-based questions, and conversion chains. CBSE and state boards following NCERT often test the same reaction logic with changed starting compounds.

1. What are amines in Class 12 Chemistry?

Amines are derivatives of ammonia in which one or more hydrogen atoms are replaced by alkyl or aryl groups. They contain nitrogen with one lone pair.

1. Primary amine: R-NH2
2. Secondary amine: R-NH-R'
3. Tertiary amine: R-NR'R''

Example:
CH3NH2 is methylamine or methanamine.

2. Why do amines show basic character?

Amines show basic character because nitrogen has a lone pair of electrons. This lone pair accepts a proton from acids.

Reaction:
R-NH2 + HCl → R-NH3+Cl-

1. Nitrogen donates its lone pair to H+.
2. The amine forms a substituted ammonium salt.
3. This proves the Lewis basic nature of amines.

Final product:
Alkylammonium chloride

3. Why do amines have pyramidal geometry?

Amines have pyramidal geometry because nitrogen is sp3 hybridised and has one lone pair. The lone pair compresses the bond angle.

1. Nitrogen forms three sigma bonds.
2. One sp3 orbital contains the lone pair.
3. Lone pair-bond pair repulsion reduces the angle below 109.5°.

Example:
Trimethylamine has a bond angle close to 108°.

Amines Class 12 Important Questions on Classification and Nomenclature

The number of carbon groups attached to nitrogen decides the class of an amine. This part helps students answer direct NCERT-style questions in CBSE 2026-27 school exams.

4. How are amines classified as primary, secondary, and tertiary?

Amines are classified by counting how many alkyl or aryl groups are attached to nitrogen. The number also affects their reactions.

1. Primary amine: one alkyl or aryl group attached to nitrogen.
   Example: CH3NH2
2. Secondary amine: two alkyl or aryl groups attached to nitrogen.
   Example: (CH3)2NH
3. Tertiary amine: three alkyl or aryl groups attached to nitrogen.
   Example: (CH3)3N

Final fact:
Primary amines give carbylamine reaction, but secondary and tertiary amines do not.

5. Write the IUPAC name of CH3NHCH2CH3.

The IUPAC name of CH3NHCH2CH3 is N-methylethanamine. The ethyl group forms the parent amine.

1. The longest carbon chain attached to nitrogen is ethyl.
2. The parent name becomes ethanamine.
3. The methyl group is attached to nitrogen.
4. The prefix N-methyl shows substitution on nitrogen.

Final answer:
N-methylethanamine

6. Write the IUPAC name and class of (CH3CH2)2NCH3.

The IUPAC name is N-ethyl-N-methylethanamine, and it is a tertiary amine. Nitrogen is attached to three alkyl groups.

1. Two ethyl groups and one methyl group attach to nitrogen.
2. No hydrogen remains attached to nitrogen.
3. This makes it a tertiary amine.

Final answer:
N-ethyl-N-methylethanamine, tertiary amine

Class 12 Chemistry Chapter 9 Important Questions on Basicity of Amines

Basicity depends on lone pair availability, inductive effect, resonance, solvation, and steric hindrance. CBSE board-style questions often ask students to justify the order of basic strength.

7. Why are aliphatic amines stronger bases than ammonia?

Aliphatic amines are stronger bases than ammonia due to the +I effect of alkyl groups. Alkyl groups increase electron density on nitrogen.

1. Alkyl groups push electrons towards nitrogen.
2. The nitrogen lone pair becomes more available.
3. The protonated ammonium ion gains stability.

Example:
C2H5NH2 is more basic than NH3.

8. Why is aniline less basic than methylamine?

Aniline is less basic than methylamine because nitrogen’s lone pair delocalises into the benzene ring. This reduces proton acceptance.

1. In aniline, -NH2 is directly attached to benzene.
2. The nitrogen lone pair participates in resonance.
3. The lone pair becomes less available for H+.
4. Methylamine has no resonance loss.

Final order:
CH3NH2 > NH3 > C6H5NH2

9. Arrange C6H5NH2, NH3, C2H5NH2, and (C2H5)2NH in increasing basic strength.

The increasing order is C6H5NH2 < NH3 < C2H5NH2 < (C2H5)2NH. Aniline is the weakest base.

1. C6H5NH2 has resonance delocalisation.
2. NH3 has no alkyl electron donation.
3. C2H5NH2 has one electron-releasing ethyl group.
4. (C2H5)2NH has stronger +I effect and good solvation.

Final answer:
C6H5NH2 < NH3 < C2H5NH2 < (C2H5)2NH

10. Why is ethylamine soluble in water but aniline is not?

Ethylamine is soluble in water because it forms hydrogen bonds with water molecules. Aniline has a large hydrophobic benzene ring.

1. Ethylamine has a small alkyl group.
2. It forms N-H···O hydrogen bonds with water.
3. Aniline contains a bulky phenyl group.
4. The phenyl group reduces solubility in water.

Final fact:
Water solubility decreases as the hydrocarbon part increases.

Important Questions of Amines Class 12 on Preparation Methods

Amines form from nitro compounds, alkyl halides, nitriles, amides, and phthalimide derivatives. Indian school exams often ask reagent-based and named-reaction questions from this section.

11. How is nitrobenzene converted into aniline?

Nitrobenzene converts into aniline by reduction. The common reagents are Fe/HCl or H2 with Ni, Pd, or Pt.

Reaction:
C6H5NO2 + 6[H] → C6H5NH2 + 2H2O

Steps:

1. Take nitrobenzene as the starting compound.
2. Reduce the -NO2 group to -NH2.
3. Obtain aniline as the final product.

Final product:
Aniline or benzenamine

12. What is Gabriel Phthalimide Synthesis Class 12?

Gabriel phthalimide synthesis prepares primary aliphatic amines from alkyl halides. It does not prepare aromatic primary amines.

Reaction sequence:

1. Phthalimide + KOH → Potassium phthalimide
2. Potassium phthalimide + R-X → N-alkyl phthalimide
3. N-alkyl phthalimide + hydrolysis → R-NH2

Reason for limitation:
Aryl halides do not undergo nucleophilic substitution with phthalimide ion.

13. What is Hoffmann Bromamide Reaction Class 12?

Hoffmann bromamide reaction converts an amide into a primary amine with one carbon atom less. It uses Br2 and NaOH.

General reaction:
RCONH2 + Br2 + 4NaOH → RNH2 + Na2CO3 + 2NaBr + 2H2O

Example:
CH3CONH2 + Br2 + 4NaOH → CH3NH2 + Na2CO3 + 2NaBr + 2H2O

Final fact:
The amine product has one carbon atom less than the starting amide.

14. How is ethanamine prepared from chloroethane?

Ethanamine forms when chloroethane reacts with excess alcoholic ammonia. This reaction is called ammonolysis.

Reaction:
C2H5Cl + 2NH3 → C2H5NH2 + NH4Cl

Steps:

1. Treat chloroethane with alcoholic ammonia.
2. Heat the mixture in a sealed tube.
3. Use excess ammonia to favour primary amine formation.

Final product:
Ethanamine

Amines Class 12 Questions With Answers on Chemical Reactions

Amines react as bases and nucleophiles due to nitrogen’s lone pair. Reaction-based questions from Amines NCERT Class 12 Questions usually test tests, acylation, bromination, and aniline behaviour.

15. What is Carbylamine Reaction Class 12?

Carbylamine reaction is a test for primary amines. Primary amines form foul-smelling isocyanides with chloroform and alcoholic KOH.

General reaction:
R-NH2 + CHCl3 + 3KOH → R-NC + 3KCl + 3H2O

Example with aniline:
C6H5NH2 + CHCl3 + 3KOH → C6H5NC + 3KCl + 3H2O

Final fact:
Secondary and tertiary amines do not give this reaction.

16. How does Hinsberg Test Class 12 distinguish primary, secondary, and tertiary amines?

Hinsberg test distinguishes amines using benzenesulphonyl chloride or p-toluenesulphonyl chloride. Each class shows different solubility behaviour.

1. Primary amine forms sulphonamide soluble in alkali.
2. Secondary amine forms sulphonamide insoluble in alkali.
3. Tertiary amine does not react with Hinsberg reagent.

Final use:
Hinsberg test separates primary, secondary, and tertiary amines.

17. Why does aniline give 2,4,6-tribromoaniline with bromine water?

Aniline gives 2,4,6-tribromoaniline because -NH2 strongly activates the benzene ring. It directs substitution at ortho and para positions.

Reaction:
C6H5NH2 + 3Br2 → C6H2Br3NH2 + 3HBr

Steps:

1. The -NH2 group increases electron density on benzene.
2. Ortho and para positions become highly reactive.
3. Bromination occurs at three activated positions.

Observation:
A white precipitate of 2,4,6-tribromoaniline forms.

18. Why does aniline not undergo Friedel-Crafts reaction?

Aniline does not undergo Friedel-Crafts reaction because it forms a salt with AlCl3. The benzene ring becomes strongly deactivated.

1. AlCl3 acts as a Lewis acid.
2. Nitrogen donates its lone pair to AlCl3.
3. The nitrogen gains positive character.
4. The ring becomes unsuitable for electrophilic substitution.

Final fact:
Aniline fails in Friedel-Crafts alkylation and acylation.

19. Why is acetylation used before bromination of aniline?

Acetylation controls the strong activating effect of the -NH2 group in aniline. It helps form mainly p-bromoaniline after hydrolysis.

Steps:

1. Aniline reacts with acetic anhydride to form acetanilide.
2. Acetanilide undergoes controlled bromination.
3. Hydrolysis gives p-bromoaniline.

Reaction sequence:
C6H5NH2 → C6H5NHCOCH3 → p-bromoacetanilide → p-bromoaniline

Final product:
p-bromoaniline

Amines Class 12 Board Questions on Diazonium Salts

Aromatic diazonium salts help convert aniline into several substituted benzene compounds. CBSE 2026-27 papers can ask diazotisation, Sandmeyer reaction, coupling reaction, and hydrolysis.

20. What is Diazotisation Class 12 Chemistry?

Diazotisation is the conversion of a primary aromatic amine into a diazonium salt at 273-278 K. Aniline forms benzene diazonium chloride.

Reaction:
C6H5NH2 + NaNO2 + 2HCl → C6H5N2+Cl- + NaCl + 2H2O

Conditions:

1. Temperature: 273-278 K
2. Reagents: NaNO2 and HCl
3. Product: Benzene diazonium chloride

Final fact:
The diazonium salt is used immediately because it decomposes easily.

21. Why are aromatic diazonium salts more stable than aliphatic diazonium salts?

Aromatic diazonium salts are more stable because the diazonium ion gets resonance stabilisation with the benzene ring. Aliphatic diazonium salts lack this support.

1. Arenediazonium ions show resonance.
2. Alkyldiazonium ions decompose easily.
3. Aromatic diazonium salts remain stable briefly in cold solution.

Final fact:
Benzene diazonium chloride is stable only at low temperature.

22. What are Diazonium Salts Class 12 Questions based on Sandmeyer reaction?

Sandmeyer reaction replaces the diazonium group with Cl, Br, or CN using cuprous salts. It prepares substituted aromatic compounds.

Examples:
C6H5N2+Cl- + CuCl → C6H5Cl + N2

C6H5N2+Cl- + CuBr → C6H5Br + N2

C6H5N2+Cl- + CuCN → C6H5CN + N2

Final use:
Sandmeyer reaction prepares aryl halides and benzonitrile from diazonium salts.

23. What is coupling reaction of diazonium salts?

Coupling reaction joins a diazonium salt with phenol or aniline to form an azo compound. These azo compounds are often coloured.

Example with phenol:
C6H5N2+Cl- + C6H5OH → p-hydroxyazobenzene + HCl

Example with aniline:
C6H5N2+Cl- + C6H5NH2 → p-aminoazobenzene + HCl

Final group:
The azo linkage is -N=N-.

24. Why are diazonium salts important in organic synthesis?

Diazonium salts are important because they introduce groups that direct substitution may not introduce easily. They help prepare aryl halides, phenols, nitriles, and azo dyes.

Groups introduced through diazonium salts:

1. -Cl
2. -Br
3. -I
4. -F
5. -CN
6. -OH
7. -NO2

Final fact:
Diazonium salts connect aniline chemistry with aromatic synthesis.

Class 12 Chemistry Amines Conversion Questions With Answers

Conversion questions need a correct sequence of reagents and products. Indian pre-board papers often test aniline, benzamide, nitrobenzene, diazonium salts, and protected substitution reactions.

25. Convert benzene into aniline.

Benzene converts into aniline through nitration followed by reduction.

Step 1: Nitration
C6H6 + HNO3 → C6H5NO2 + H2O
Reagent: Conc. HNO3 / Conc. H2SO4

Step 2: Reduction
C6H5NO2 + 6[H] → C6H5NH2 + 2H2O
Reagent: Fe/HCl or H2/Ni

Final product:
Aniline

26. Convert aniline into p-bromoaniline.

Aniline converts into p-bromoaniline by protection, bromination, and hydrolysis.

Step 1: Acetylation
C6H5NH2 + (CH3CO)2O → C6H5NHCOCH3 + CH3COOH

Step 2: Bromination
C6H5NHCOCH3 + Br2 → p-bromoacetanilide + HBr

Step 3: Hydrolysis
p-bromoacetanilide + H3O+ → p-bromoaniline + CH3COOH

Final product:
p-bromoaniline

27. Convert aniline into phenol.

Aniline converts into phenol through diazotisation followed by hydrolysis.

Step 1: Diazotisation
C6H5NH2 + NaNO2 + 2HCl → C6H5N2+Cl- + NaCl + 2H2O

Step 2: Hydrolysis
C6H5N2+Cl- + H2O → C6H5OH + N2 + HCl

Final product:
Phenol

28. Convert benzamide into aniline.

Benzamide converts into aniline through Hoffmann bromamide reaction.

Reaction:
C6H5CONH2 + Br2 + 4NaOH → C6H5NH2 + Na2CO3 + 2NaBr + 2H2O

Steps:

1. Treat benzamide with Br2 and NaOH.
2. Remove the carbonyl carbon as carbonate.
3. Obtain aniline with one carbon less.

Final product:
Aniline

29. Convert aniline into chlorobenzene.

Aniline converts into chlorobenzene through diazotisation and Sandmeyer reaction.

Step 1: Diazotisation
C6H5NH2 + NaNO2 + 2HCl → C6H5N2+Cl- + NaCl + 2H2O

Step 2: Sandmeyer reaction
C6H5N2+Cl- + CuCl → C6H5Cl + N2

Final product:
Chlorobenzene

30. Convert aniline into benzene.

Aniline converts into benzene through diazotisation followed by reduction of the diazonium salt.

Step 1: Diazotisation
C6H5NH2 + NaNO2 + 2HCl → C6H5N2+Cl- + NaCl + 2H2O

Step 2: Reduction
C6H5N2+Cl- + H3PO2 + H2O → C6H6 + N2 + H3PO3 + HCl

Final product:
Benzene

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