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.
यह अध्याय क्यों महत्वपूर्ण है
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.
अध्याय रूपरेखा एवं प्रगति
11. Hoffmann Bromamide & Gabriel Pht...
22. Basicity Order in Aqueous Soluti...
33. The Hinsberg Test ($1^\circ, 2^\...
सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन
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!
Keep an error log and revisit questions that exposed a misconception.
अवधारणात्मक हल उदाहरण एवं अनुप्रयोग (Solved Examples)
उदाहरण 1
Describe the Hoffmann Bromamide Degradation reaction with a balanced chemical equation.
विस्तृत समाधान / उत्तर:
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: $\text{CH}_3\text{CONH}_2 + \text{Br}_2 + 4\text{KOH} \to \text{CH}_3\text{NH}_2 + \text{K}_2\text{CO}_3 + 2\text{KBr} + 2\text{H}_2\text{O}$.
उदाहरण 2
Why is Gabriel Phthalimide Synthesis preferred for the preparation of primary aliphatic amines, and why can it not prepare aniline?
विस्तृत समाधान / उत्तर:
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.
उदाहरण 3
How will you distinguish between Ethylamine and Diethylamine using the Hinsberg Reagent?
विस्तृत समाधान / उत्तर:
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 $\text{KOH}$.
सामान्य गलतियाँ एवं परीक्षक के जाल (Examiner Traps)
सामान्य भ्रम / गलत उत्तर
Reciting a definition without applying it to the question or data.
सही वैज्ञानिक तथ्य
Identify the concept, show the relevant evidence or calculation, and explain the final implication.
सामान्य भ्रम / गलत उत्तर
Skipping conditions, units, domain restrictions, or adjustment effects.
सही वैज्ञानिक तथ्य
State assumptions, preserve units, check boundary cases, and verify the answer against the original problem.
सामान्य भ्रम / गलत उत्तर
Treating a correct intermediate result as proof that the whole solution is correct.
सही वैज्ञानिक तथ्य
Perform an independent reasonableness check and connect the result back to the chapter principle.
Amines - Key Molecular Architecture & Reaction Mechanism Model
अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष
मुख्य बिंदु 1
Hoffmann Degradation: Carbon-shortening conversion of amides to primary amines.
मुख्य बिंदु 2
Gabriel Synthesis: Chemoselective alkylation producing unadulterated primary aliphatic amines.
मुख्य बिंदु 3
Aqueous Basicity Balance: Complex interplay of inductive $+I$, steric crowding, and water hydration.
मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।
1
Describe the Hoffmann Bromamide Degradation reaction with a balanced chemical equation.
उत्तर एवं व्याख्या देखें
उत्तर: 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?
उत्तर एवं व्याख्या देखें
उत्तर: 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?
उत्तर एवं व्याख्या देखें
उत्तर: 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?
उत्तर एवं व्याख्या देखें
उत्तर: 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.
उत्तर एवं व्याख्या देखें
उत्तर: 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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