Exhaustive Guide: Amines - Properties & Reactions | Class 12 Chemistry | ChemCA

Amines: Properties & Chemical Reactions

Module 2 | CBSE Class 12 Chemistry | Organic Chemistry

1. Physical Properties

The lower aliphatic amines are gases with a fishy odour. Primary amines with three or more carbon atoms are liquid and still higher ones are solid. Aniline and other arylamines are usually colourless but get coloured on storage due to atmospheric oxidation.

Boiling Points & Hydrogen Bonding:
Primary and secondary amines engage in intermolecular hydrogen bonding because they have hydrogen atoms directly attached to the electronegative Nitrogen atom. Tertiary amines do not have intermolecular hydrogen bonding as they lack hydrogen atoms on Nitrogen.
Order of Boiling Point for isomeric amines: Primary > Secondary > Tertiary.

Comparison with Alcohols: Alcohols have higher boiling points than amines of comparable molecular masses because oxygen is more electronegative than nitrogen, resulting in stronger hydrogen bonds in alcohols.

2. Basic Character of Amines (Highly Tested)

Amines, being derivatives of ammonia, have an unshared pair of electrons (lone pair) on the nitrogen atom. This makes them behave as Lewis bases. They react with acids to form ammonium salts.

Basicity is expressed in terms of K_b (Basicity constant) or pK_b. Larger the K_b, or smaller the pK_b, stronger is the base.

2.1 Basicity of Aliphatic Amines in Gas Phase

In the gas phase (or non-aqueous solvents), basicity depends purely on the +I effect (inductive effect) of the alkyl groups. Alkyl groups are electron-donating; they push electron density towards the nitrogen atom, making the lone pair more available for protonation. They also stabilize the resulting ammonium ion.

Gas Phase Basicity Order: Tertiary (3°) > Secondary (2°) > Primary (1°) > NH₃

2.2 Basicity in Aqueous Phase (The Anomaly)

In aqueous solution, the basicity order changes dramatically. It is no longer governed solely by the +I effect. The stability of the substituted ammonium cation depends on a delicate balance of three factors:

+I Effect: Favors 3° > 2° > 1°.
Solvation Effect (Hydration): The protonated amine is stabilized by hydrogen bonding with water molecules. Greater the number of N-H bonds, greater the hydration. Favors 1° > 2° > 3°.
Steric Hindrance: Bulky alkyl groups hinder the approach of water molecules for solvation and the approach of the proton. Favors 1° > 2° > 3°.

Crucial Board Memory Tool: The combined interplay of these three opposing factors leads to specific, anomalous orders in aqueous solutions that must be memorized:

For Methyl group (-CH₃): Secondary > Primary > Tertiary > NH₃
(2° > 1° > 3° > NH₃) i.e., (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃

For Ethyl group (-C₂H₅): Secondary > Tertiary > Primary > NH₃
(2° > 3° > 1° > NH₃) i.e., (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃

2.3 Basicity of Arylamines (Aniline) vs. Ammonia

Aniline is a much weaker base than ammonia and aliphatic amines. (Its pK_b value is quite high).

Why is Aniline less basic?
In aniline, the -NH₂ group is attached directly to the benzene ring. The unshared electron pair on the nitrogen atom is in conjugation with the pi-electron system of the benzene ring (+R effect). Because the lone pair is delocalized over the ring, it is less available for protonation.
Furthermore, the anilinium ion (formed after protonation) is stabilized by only 2 resonating structures, whereas aniline is stabilized by 5. Thus, the equilibrium heavily favors the unprotonated aniline.

3. Distinguishing Tests for Amines

3.1 Carbylamine Reaction (Isocyanide Test)

This is a highly specific and extremely important test used to detect the presence of Primary (1°) Amines (both aliphatic and aromatic).

When a primary amine is heated with chloroform (CHCl₃) and ethanolic potassium hydroxide (KOH), it forms an isocyanide (carbylamine), which possesses an extremely foul, offensive, and unbearable odour.

R-NH₂ + CHCl₃ + 3KOH(alc.) →(Δ) R-NC (Foul smelling Isocyanide) + 3KCl + 3H₂O

Secondary and tertiary amines do NOT give this test.

3.2 Hinsberg's Test

Used to distinguish between 1°, 2°, and 3° amines. The reagent used is Benzenesulphonyl chloride (C₆H₅SO₂Cl), known as Hinsberg's reagent.

Primary (1°) Amines: React to form N-alkylbenzenesulphonamide. This product has a highly acidic hydrogen attached to Nitrogen. Hence, it is soluble in alkali (aq. NaOH).
Secondary (2°) Amines: React to form N,N-dialkylbenzenesulphonamide. This product has NO acidic hydrogen attached to Nitrogen. Hence, it is insoluble in alkali.
Tertiary (3°) Amines: Do not react with Hinsberg's reagent at all because they lack a replaceable hydrogen atom on the nitrogen.

4. Electrophilic Substitution of Aromatic Amines (Aniline)

The -NH₂ group is strongly activating and ortho, para-directing. It directs the incoming electrophile to ortho and para positions and dramatically increases the reactivity of the ring.

4.1 Bromination & Protection of Amino Group

Because aniline is so highly activated, reaction with bromine water yields a white precipitate of 2,4,6-tribromoaniline almost instantaneously.

How do we get a mono-substituted product (e.g., p-bromoaniline)?
To stop the reaction at mono-substitution, the immense activating power of the -NH₂ group must be controlled. This is done by protecting the amino group by acetylation.
1. Aniline is reacted with acetic anhydride to form acetanilide.
2. In acetanilide, the lone pair of nitrogen is delocalized towards the carbonyl group of the acetyl part, making it less available to the benzene ring. This moderates the activating effect.
3. Bromination now yields majorly p-bromoacetanilide.
4. Hydrolysis (with H⁺ or OH^-) regenerates the -NH₂ group, yielding p-bromoaniline.

4.2 Nitration (The Meta-Product Anomaly)

Direct nitration of aniline with a nitrating mixture (conc. HNO₃ + conc. H₂SO₄) yields a complex mixture, including a surprisingly high amount of the meta-derivative.

Board Question: Why is a significant amount (47%) of m-nitroaniline formed during the direct nitration of aniline, even though -NH₂ is o/p directing?
In the strongly acidic nitrating medium, the basic aniline is protonated to form the anilinium ion (C₆H₅-NH₃⁺). The -NH₃⁺ group is strongly deactivating and meta-directing. Thus, alongside the o/p products from unprotonated aniline, a massive amount of the meta product is formed from the anilinium ion.

To get only p-nitroaniline: We must again protect the -NH₂ group by acetylation, perform nitration, and then hydrolyze.

4.3 Sulphonation (Zwitterion Formation)

Aniline reacts with concentrated H₂SO₄ to form anilinium hydrogensulphate, which on heating at 453-473 K produces p-aminobenzenesulphonic acid, commonly known as Sulphanilic acid, as the major product.

Zwitterion: Sulphanilic acid contains both an acidic group (-SO₃H) and a basic group (-NH₂). The acidic group transfers its proton to the basic group, creating an internal salt called a Zwitterion: ⁺H₃N-C₆H₄-SO₃^-. It is electrically neutral but contains positive and negative charges.

4.4 Why Aniline fails Friedel-Crafts Reaction

Aniline does NOT undergo Friedel-Crafts alkylation or acylation.

Reason: The catalyst used in Friedel-Crafts reactions is anhydrous Aluminium Chloride (AlCl₃), which is a Lewis acid. Aniline is a Lewis base due to the lone pair on nitrogen. They immediately react to form a stable complex (salt). This creates a strong positive charge on the nitrogen atom, turning it into a powerful deactivating group that entirely halts any electrophilic substitution on the ring.

5. NCERT Solved Examples (Step-by-Step)

NCERT Example 13.4: Arrange the following in decreasing order of their basic strength:
C₆H₅NH₂, C₂H₅NH₂, (C₂H₅)₂NH, NH₃

Solution:
1. Aniline (C₆H₅NH₂): Due to resonance, the lone pair is delocalized over the benzene ring, making it the weakest base.
2. Ammonia (NH₃): Reference point.
3. Ethylamine (C₂H₅NH₂) vs Diethylamine ((C₂H₅)₂NH): Both are stronger bases than ammonia due to the +I effect of ethyl groups. In aqueous solution, for ethyl substituted amines, the secondary amine is stronger than the primary amine.
Decreasing order: (C₂H₅)₂NH > C₂H₅NH₂ > NH₃ > C₆H₅NH₂

NCERT Example 13.5: How will you convert:
Ethanoic acid into methanamine?

Solution:
This requires a "step-down" conversion (losing one carbon atom). We use the Hoffmann bromamide reaction.
Step 1: Convert acid to amide by heating with ammonia.
CH₃COOH + NH₃ →(Δ) CH₃CONH₂ (Ethanamide) + H₂O
Step 2: Treat the amide with Bromine and NaOH.
CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ (Methanamine) + Na₂CO₃ + 2NaBr + 2H₂O

6. Previous Year Questions (PYQs) & Exhaustive Question Bank

Part A: Distinguishing Tests (2-3 Marks)

[CBSE 2017, 2020]

Q1. Give a simple chemical test to distinguish between the following pairs of compounds:
(a) Aniline and N-methylaniline
(b) Ethylamine and Dimethylamine

Answer:
(a) Carbylamine Test: Aniline is a primary (1°) amine and gives a foul-smelling isocyanide when heated with chloroform and alcoholic KOH. N-methylaniline is a secondary (2°) amine and does not give this test.
(b) Carbylamine Test: Ethylamine (1°) gives the foul smell of isocyanide. Dimethylamine (2°) does not.
(Alternatively, Hinsberg test can be used for both).

Part B: Conceptual & Reasoning (1-2 Marks)

[CBSE 2018, 2021]

Q2. Account for the following: pK_b of aniline is more than that of methylamine.

Answer: A higher pK_b means a weaker base. Aniline is a weaker base than methylamine because the lone pair of electrons on the nitrogen atom in aniline is in conjugation with the benzene ring and gets delocalized due to resonance. Thus, it is less available for protonation. In methylamine, the +I effect of the methyl group increases electron density on nitrogen, making it a stronger base.

[CBSE 2016, 2019]

Q3. Why does acetylation of -NH₂ group of aniline reduce its activating effect?

Answer: In acetanilide (formed after acetylation of aniline), the lone pair of electrons on the nitrogen atom is involved in resonance not only with the benzene ring but also heavily with the carbonyl group (>C=O) of the acetyl moiety. Due to this cross-conjugation, the availability of the lone pair to activate the benzene ring is significantly reduced.

Part C: Assertion-Reason Type (1 Mark)

[CBSE Sample Paper 2023]

Q4. Assertion (A): Aniline does not undergo Friedel-Crafts reaction.
Reason (R): Friedel-Crafts reaction is an electrophilic substitution reaction.

Answer: Both Assertion and Reason are correct, but Reason is NOT the correct explanation for Assertion. The correct reason is that Aniline (a Lewis base) forms an insoluble complex (salt) with the catalyst AlCl₃ (a Lewis acid). This creates a positive charge on the nitrogen atom, strongly deactivating the ring for electrophilic substitution.

Part D: Synthesis & Reactions (3 Marks)

[CBSE 2015, 2022]

Q5. Write the structures of the main products when aniline reacts with the following reagents:
(i) Br₂ water
(ii) CHCl₃ + alc. KOH
(iii) Conc. H₂SO₄ (heated at 453-473 K)

Answer:
(i) Aniline is highly activated. Reaction with Br₂ water yields a white precipitate of 2,4,6-Tribromoaniline.

(ii) This is the Carbylamine reaction. Aniline (a 1° amine) forms a foul-smelling isocyanide.
Product: Phenyl isocyanide (C₆H₅-NC).

(iii) This is sulphonation. It yields Sulphanilic acid, which exists predominantly as a zwitterion.
Product: p-Aminobenzenesulphonic acid (or its Zwitterion: ⁺H₃N-C₆H₄-SO₃^-).

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Exhaustive Guide: Amines - Properties & Reactions | Class 12 Chemistry