Why do fish smell fishy, why does adrenaline surge through your veins when you are startled, and how do industrial dyers color vibrant blue jeans with synthetic azo dyes? Amines are the nitrogenous bases of biology and color.
Why This Chapter Matters
In Class 12 Chemistry, "Amines" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.
Explain preparations: Gabriel Phthalimide Synthesis (selective for $1^\circ$ aliphatic amines) and Hoffmann Bromamide Degradation (reduces 1 carbon!).
Analyze Basicity of Amines: Gas phase vs aqueous phase ($2^\circ > 1^\circ > 3^\circ > \text{NH}_3$ for methyl).
Distinguish $1^\circ, 2^\circ, 3^\circ$ amines using the Hinsberg Test (benzenesulphonyl chloride).
Explain Diazotisation of Aniline ($\text{NaNO}_2 + \text{HCl}$ at $0-5^\circ\text{C}$) and Azo Dye coupling reactions.
Chapter Roadmap & Progression
11. Hoffmann Bromamide & Gabriel Pht...
22. Basicity Order in Aqueous Soluti...
33. The Hinsberg Test ($1^\circ, 2^\...
Complete Concept Guide (100% Curriculum Coverage)
1. Hoffmann Bromamide & Gabriel Phthalimide
Hoffmann Bromamide Degradation: Amide treated with $\text{Br}_2 + 4\text{KOH}$ degrades into a primary amine with one carbon less: $$\mathbf{\text{R-CONH}_2 + \text{Br}_2 + 4\text{KOH} \to \text{R-NH}_2 + \text{K}_2\text{CO}_3 + 2\text{KBr} + 2\text{H}_2\text{O}}$$
Gabriel Phthalimide Synthesis: Pure primary aliphatic amines only! (Aryl halides cannot react due to partial double bond character).
2. Basicity Order in Aqueous Solution
Basicity depends on balance of Inductive effect ($+I$), Steric hindrance, and Solvation energy: • Methyl group: $\mathbf{(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3}$ ($2^\circ > 1^\circ > 3^\circ > \text{NH}_3$). • Ethyl group: $\mathbf{(C_2H_5)_2NH > (C_2H_5)_3N > C_2H_5NH_2 > NH_3}$ ($2^\circ > 3^\circ > 1^\circ > \text{NH}_3$). • Aniline: Far weaker base because nitrogen lone pair delocalizes into aromatic ring!
3. The Hinsberg Test ($1^\circ, 2^\circ, 3^\circ$)
Reagent: Benzenesulphonyl chloride ($\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$): • $1^\circ$ Amine: Forms N-alkylbenzenesulphonamide, which has an acidic $N-H$ and dissolves in $\text{KOH}$. • $2^\circ$ Amine: Forms product with NO acidic hydrogen; insoluble in $\text{KOH}$. • $3^\circ$ Amine: Does not react at all!
Amines - Key Molecular Architecture & Reaction Mechanism Model
Chapter Summary & 10 Key Takeaways
Takeaway 1
Hoffmann Degradation: Carbon-shortening conversion of amides to primary amines.
Takeaway 2
Gabriel Synthesis: Chemoselective alkylation producing unadulterated primary aliphatic amines.
Takeaway 3
Aqueous Basicity Balance: Complex interplay of inductive $+I$, steric crowding, and water hydration.
Check Your Understanding (Diagnostic Practice Questions)
Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.
1
Describe the Hoffmann Bromamide Degradation reaction with a balanced chemical equation.
Reveal Answer & Explanation
Answer: When an acid amide is treated with bromine in an aqueous or ethanolic solution of sodium/potassium hydroxide, it is degraded to a primary amine containing one carbon atom less than the starting amide: $\text{CH}_3\text{CONH}_2 + \text{Br}_2 + 4\text{KOH} \to \text{CH}_3\text{NH}_2 + \text{K}_2\text{CO}_3 + 2\text{KBr} + 2\text{H}_2\text{O}$. Amide + Br2 + 4KOH gives 1° amine with 1 fewer carbon.
2
Why is Gabriel Phthalimide Synthesis preferred for the preparation of primary aliphatic amines, and why can it not prepare aniline?
Reveal Answer & Explanation
Answer: It yields exclusively pure primary aliphatic amines free from secondary and tertiary amine impurities. It cannot prepare aniline because aryl halides do not undergo nucleophilic substitution with the phthalimide anion due to resonance partial double bond character. Pure 1° amine yield; aryl halides resist nucleophilic substitution.
3
How will you distinguish between Ethylamine and Diethylamine using the Hinsberg Reagent?
Reveal Answer & Explanation
Answer: Treat both with benzenesulphonyl chloride (Hinsberg reagent): Ethylamine ($1^\circ$) forms N-ethylbenzenesulphonamide, which dissolves in aqueous $\text{KOH}$; Diethylamine ($2^\circ$) forms N,N-diethylbenzenesulphonamide, which has no acidic hydrogen and remains insoluble in $\text{KOH}$. 1° product dissolves in KOH; 2° product is insoluble in KOH.
4
Why is Aniline a much weaker base than Ethylamine?
Reveal Answer & Explanation
Answer: In aniline, the lone pair of electrons on the nitrogen atom is delocalized over the benzene ring through resonance across 5 structures, making it less available for protonation. In ethylamine, the electron-donating $+I$ ethyl group increases electron density on nitrogen, making it a stronger base. Nitrogen lone pair delocalized into benzene ring via resonance.
5
What is the Carbylamine Test? Write the chemical reaction.
Reveal Answer & Explanation
Answer: An exclusive test for primary ($1^\circ$) aliphatic and aromatic amines. When heated with chloroform and alcoholic potassium hydroxide, primary amines form extremely foul-smelling isocyanides (carbylamines): $\text{R-NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{R-NC} + 3\text{KCl} + 3\text{H}_2\text{O}$. (Secondary and tertiary amines do NOT react). 1° amine + CHCl3 + KOH gives foul-smelling isocyanide.
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