Amines — Previous Year Questions (Class 12 Chemistry)
Amines are organic compounds derived from ammonia. Study their classification, properties, preparation, and reactions.
TL;DR: Amines are organic compounds derived from ammonia. Study their classification, properties, preparation, and reactions.
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Amines are organic compounds derived from ammonia. Study their classification, properties, preparation, and reactions.
Amines — Previous Year Questions with Solutions
Q (2023, 3 marks): Classify amines based on their structure and give examples of each class.
Answer: Amines are classified based on the number of alkyl/aryl groups attached to nitrogen:
Primary amines (1° amines): One alkyl or aryl group attached to nitrogen.
General formula: RNH₂
Examples: Methylamine (CH₃NH₂), Aniline (C₆H₅NH₂), Ethylamine (C₂H₅NH₂)
Secondary amines (2° amines): Two alkyl or aryl groups attached to nitrogen.
General formula: R₂NH or RR'NH
Examples: Dimethylamine ((CH₃)₂NH), N-methylaniline (C₆H₅NHCH₃), Diethylamine ((C₂H₅)₂NH)
Tertiary amines (3° amines): Three alkyl or aryl groups attached to nitrogen.
General formula: R₃N or RR'R''N
Examples: Trimethylamine ((CH₃)₃N), Triethylamine ((C₂H₅)₃N), N,N-dimethylaniline ((CH₃)₂NC₆H₅)
Quaternary ammonium compounds: Four alkyl groups attached to nitrogen with a positive charge.
General formula: R₄N⁺ X⁻ (where X⁻ is an anion)
Example: Tetramethylammonium chloride (CH₃)₄N⁺Cl⁻
Final Answer: Three classes (primary, secondary, tertiary) based on number of alkyl/aryl groups; examples provided
Q (2022, 3 marks): Explain the basic character of amines in terms of the lone pair on nitrogen.
Answer: Amines contain a lone pair of electrons on nitrogen, which makes them basic (proton acceptors).
Basic nature mechanism:
The lone pair on nitrogen can accept a proton (H⁺) from acids:
R₃N + H⁺ ⇌ R₃NH⁺
Or in aqueous solution:
R₃N + H₂O ⇌ R₃NH⁺ + OH⁻
Amines are weak bases with Kb values typically in the range 10⁻³ to 10⁻¹⁰.
Factors affecting basicity:
1. Alkyl groups: Increase electron density on N, increasing basicity.
Order: 3° > 2° > 1° > NH₃ (in solution, but order differs in gas phase)
2. Aromaticity: Aniline (C₆H₅NH₂) is less basic than aliphatic amines because the lone pair is involved in resonance with the benzene ring, reducing its availability.
3. Steric hindrance: Bulky alkyl groups can decrease basicity due to steric effects.
4. Electron-withdrawing groups: Decrease basicity by reducing electron density.
PKa values (for conjugate acids):
Aniline: pKa ≈ 4.6 (weak base, Kb ≈ 4 × 10⁻¹⁰)
Methylamine: pKa ≈ 10.6 (stronger base, Kb ≈ 4 × 10⁻⁴)
Trimethylamine: pKa ≈ 9.8
Final Answer: Lone pair on N makes amines basic; can accept protons; aromaticity and alkyl groups affect basicity
Q (2023, 3 marks): How is aniline prepared in the laboratory from nitrobenzene?
Answer: Preparation of aniline from nitrobenzene:
Reaction:
C₆H₅NO₂ + 6H⁺ + 6e⁻ → C₆H₅NH₂ + 2H₂O
Laboratory method (reduction of nitrobenzene):
Method 1: Using tin and hydrochloric acid
C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O
[H] from Sn/HCl
Procedure:
1. Add granulated tin to nitrobenzene
2. Add conc. HCl carefully with cooling
3. Nitrobenzene gets reduced to aniline salt (anilinium chloride: C₆H₅NH₃⁺Cl⁻)
4. Heat the mixture
5. Cool and add NaOH solution to neutralize
6. Aniline separates as oily layer (insoluble in water)
7. Extract with ether or distill
Method 2: Using iron and acetic acid
C₆H₅NO₂ + 3Fe + 4H⁺ → C₆H₅NH₂ + 3Fe²⁺ + 2H₂O
Method 3: Using catalytic hydrogenation
C₆H₅NO₂ + 3H₂ → C₆H₅NH₂ + 2H₂O (with Ni/Pt catalyst at high temperature and pressure)
Obtained aniline is a colorless oily liquid with characteristic pungent smell.
Final Answer: Reduce nitrobenzene with Sn/HCl, then neutralize with NaOH to obtain aniline as oily layer
Q (2021, 3 marks): Write the equation for the Hofmann elimination reaction of an alkylammonium salt.
Answer: Hofmann elimination: Conversion of a quaternary ammonium hydroxide into an alkene and tertiary amine upon heating.
General reaction:
R₄N⁺ OH⁻ → R₃N + R-OH (if R is primary)
Or for secondary/tertiary R groups:
R₄N⁺ OH⁻ --Δ--> R₃N + RH + (R₃N: R-OH via E2 mechanism)
Most commonly, the more substituted alkene forms (Zaitsev's rule applies in reverse here).
Example 1: Trimethylethylammonium hydroxide
(CH₃)₃N⁺ C₂H₅ OH⁻ --Δ--> (CH₃)₃N + C₂H₄ + H₂O
Example 2: Sequential Hofmann elimination (exhaustive methylation)
Starting with primary amine RCH₂NH₂:
1. CH₃I (methylation) → RCH₂NHCH₃ (2° amine)
2. CH₃I (methylation) → RCH₂N(CH₃)₂ (3° amine)
3. CH₃I (methylation) → RCH₂N⁺(CH₃)₃ I⁻ (quaternary ammonium salt)
4. Ag₂O/H₂O (hydroxide formation) → RCH₂N⁺(CH₃)₃ OH⁻
5. Heat (elimination) → CH₂=CHR + N(CH₃)₃ + H₂O
Mechanism: E2 (bimolecular elimination)
Final Answer: R₄N⁺OH⁻ --Δ--> Alkene + R₃N + H₂O via E2 mechanism
Q (2022, 3 marks): Explain why aniline does not undergo Friedel-Crafts alkylation but phenol does.
Answer: Aniline does NOT undergo Friedel-Crafts alkylation, while phenol does. The reason:
In Friedel-Crafts alkylation, AlCl₃ (or other Lewis acid) coordinates with the benzene ring to activate it.
Aniline (C₆H₅NH₂):
1. The lone pair on nitrogen forms a strong complex with AlCl₃: C₆H₅NH₂·AlCl₃
2. This removes the lone pair from the nitrogen, preventing its resonance donation to the ring.
3. The amino group can no longer activate the ring by resonance.
4. The complex becomes electron-withdrawing, deactivating the ring.
5. Also, the nitrogen atom in the complex becomes positively charged and can be alkylated instead of the benzene ring.
6. Result: No alkylation occurs on the benzene ring.
Phenol (C₆H₅OH):
1. The hydroxyl group is less basic than the amino group.
2. While oxygen can weakly coordinate with AlCl₃, the interaction is not as strong.
3. The OH group retains sufficient electron-donating ability through resonance.
4. The benzene ring remains activated and undergoes substitution.
5. Phenol undergoes Friedel-Crafts alkylation (though it's slower than benzene derivatives with alkyl groups).
Conclusion: Aniline's basic nitrogen prevents Friedel-Crafts alkylation by complexing with the acid catalyst. Phenol's less basic oxygen allows alkylation.
Final Answer: Aniline's lone pair complexes with AlCl₃, deactivating the ring; phenol's OH is less basic, allowing alkylation
Q (2023, 3 marks): Describe the structure of aniline and explain its properties as a weak base.
Answer: Structure of aniline (C₆H₅NH₂):
Aniline is a primary amine with a benzene ring directly attached to the NH₂ group.
Structure: [Benzene ring]-NH₂
The nitrogen is sp² hybridized with the lone pair in a p orbital.
Its weak basicity (Kb ≈ 4 × 10⁻¹⁰, pKa ≈ 4.6) compared to aliphatic amines is due to:
1. Resonance stabilization of the free base:
The lone pair on nitrogen participates in resonance with the π system of the benzene ring, delocalizing electron density. This makes the lone pair less available for proton acceptance.
Resonance structures show the lone pair is shared with the aromatic ring.
2. In the conjugate acid (anilinium ion C₆H₅NH₃⁺), this resonance stabilization is lost, making the protonated form less stable.
3. Comparison of basicities:
Aliphatic amines (e.g., methylamine CH₃NH₂): Kb ≈ 4 × 10⁻⁴ (stronger base)
Aniline (C₆H₅NH₂): Kb ≈ 4 × 10⁻¹⁰ (weaker base)
Amonia (NH₃): Kb ≈ 1.8 × 10⁻⁵
4. pKa values confirm this:
Anilinium ion: pKa ≈ 4.6 (weak acid, weak base)
Methylammonium ion: pKa ≈ 10.6 (stronger conjugate base)
Conclusion: Aniline is weakly basic because the aromatic NH₂ group's lone pair is delocalized through resonance.
Final Answer: Aniline's NH₂ lone pair is delocalized by resonance with benzene ring, reducing basicity to Kb ≈ 4 × 10⁻¹⁰
Frequently Asked Questions
What is the difference between amines and amides?
Amines (R₃N) are compounds where nitrogen is bonded to alkyl/aryl groups and have lone pair. Amides (RCONH₂, RCONHR, RCONR₂) have nitrogen bonded to a carbonyl group. Amines are basic due to lone pair; amides are neutral or weakly acidic. Amines are derived from ammonia by replacing H with alkyl/aryl groups. Amides are formed from carboxylic acids or acyl chlorides. Amides have significantly different chemical properties from amines.
Why is aniline less basic than methylamine?
Aniline (C₆H₅NH₂, Kb ≈ 4 × 10⁻¹⁰) is much weaker than methylamine (CH₃NH₂, Kb ≈ 4 × 10⁻⁴). The aromatic benzene ring withdraws electron density from nitrogen through resonance, reducing the lone pair's availability for proton acceptance. Methylamine lacks this resonance delocalization, so its lone pair remains on nitrogen, making it a stronger base. The resonance in aniline is a major factor reducing basicity.
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