CBSE Important Questions Class 12 Chemistry Chapter 9 – Amines

Class 12 Chemistry Chapter 9 Amines is part of the CBSE Class 12 Chemistry syllabus. Amines are compounds in which one or more hydrogen atoms of ammonia are replaced by alkyl or aryl groups. Students learn their classification, nomenclature and structure, methods of preparation, physical properties, basic strength and the factors affecting it, chemical reactions, and the preparation and importance of diazonium salts.

These CBSE important questions for Class 12 Chemistry Chapter 9 cover the order of basic strength of amines in the gas phase and in water, why aniline is a weaker base than ammonia, the Hinsberg test for distinguishing primary, secondary and tertiary amines, the carbylamine reaction, Gabriel phthalimide synthesis, Hoffmann bromamide degradation, and the coupling reactions of diazonium salts. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Arrange the following in increasing order of basic strength in aqueous solution: NH₃, CH₃NH₂, (CH₃)₂NH, (CH₃)₃N.

Answer: In aqueous solution the order is
NH₃ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH
Three factors work together and pull in different directions:
1. Inductive effect: Alkyl groups release electrons, making the nitrogen lone pair more available. By this alone the order would be 3° > 2° > 1° > NH₃.
2. Solvation (hydration) of the ion formed: The more hydrogen atoms left on nitrogen, the more hydrogen bonds the resulting ammonium ion can form with water, and the more stable it is. By this alone the order would be 1° > 2° > 3°.
3. Steric hindrance: Three bulky methyl groups make it hard for the proton to reach the nitrogen.
The net result of all three is that the secondary amine is the strongest base and the tertiary amine falls back below the primary one.
(In the gas phase, where there is no solvent, only the inductive effect matters, so the order is simply 3° > 2° > 1° > NH₃.)

Q2. Why is aniline a weaker base than ammonia?

Answer: In aniline, the lone pair of electrons on the nitrogen is in conjugation with the benzene ring. Through resonance, this lone pair is pulled into the ring and spread over the carbon atoms.
Since the lone pair is delocalised and no longer freely available, it cannot easily accept a proton.
In ammonia, the lone pair sits entirely on the nitrogen and is fully available to take up a proton.
There is a second reason too: in aniline the nitrogen is attached to an sp² hybridised carbon, which is more electronegative than an sp³ carbon and pulls electron density away from the nitrogen.
Both effects make the lone pair less available, so aniline is a much weaker base than ammonia.

Q3. What is the Hinsberg test? How does it distinguish primary, secondary and tertiary amines?

Answer: In the Hinsberg test, the amine is shaken with benzenesulphonyl chloride (C₆H₅SO₂Cl) and then treated with alkali.
• Primary amine: forms a sulphonamide which still has one hydrogen on nitrogen. This hydrogen is acidic because of the two strongly electron-withdrawing sulphonyl oxygens, so the product dissolves in alkali.
• Secondary amine: forms a sulphonamide with no hydrogen left on the nitrogen. So it is insoluble in alkali and stays as a solid.
• Tertiary amine: has no hydrogen on nitrogen at all, so it does not react with the reagent.
In short: soluble product = primary, insoluble product = secondary, no reaction = tertiary.

Q4. What is the carbylamine reaction? Which amines give this test?

Answer: In the carbylamine reaction, an amine is heated with chloroform and alcoholic KOH. An isocyanide (carbylamine) is formed, which has a strong and extremely unpleasant smell.
R-NH₂ + CHCl₃ + 3KOH → R-NC + 3KCl + 3H₂O
Which amines: Only primary amines, both aliphatic and aromatic, give this reaction. Secondary and tertiary amines do not.
Because of this, the carbylamine reaction is used as a test for primary amines. The foul smell makes it unmistakable.

Q5. What is the Gabriel phthalimide synthesis? Why can it not be used to prepare aromatic primary amines?

Answer: Gabriel phthalimide synthesis is a method of preparing pure primary amines.
Phthalimide is treated with alcoholic KOH to give potassium phthalimide. This is then heated with an alkyl halide, and the product is finally hydrolysed with alkali to give the primary amine.
Advantage: It gives only the primary amine, with no mixture of secondary and tertiary amines.
Why it fails for aromatic amines: The step with the alkyl halide is a nucleophilic substitution. Aryl halides such as chlorobenzene do not undergo nucleophilic substitution easily, because the C-X bond has partial double bond character due to resonance.
So the phthalimide ion cannot displace the halogen from an aryl halide, and aniline cannot be made this way.

Q6. What is Hoffmann bromamide degradation? Write the equation.

Answer: In Hoffmann bromamide degradation, an amide is treated with bromine and aqueous or alcoholic NaOH to give a primary amine with one carbon atom less.
R-CONH₂ + Br₂ + 4NaOH → R-NH₂ + Na₂CO₃ + 2NaBr + 2H₂O
Example: CH₃CONH₂ (acetamide) gives CH₃NH₂ (methylamine).
Note carefully: the amine produced has one carbon fewer than the starting amide, because the carbonyl carbon is lost as carbonate. This is what makes it a "degradation" and why it is a useful step for shortening a carbon chain.

Q7. Why do primary amines have higher boiling points than tertiary amines of the same molecular mass?

Answer: The difference is again hydrogen bonding.
• A primary amine has two hydrogen atoms on nitrogen, so it can form several intermolecular hydrogen bonds with neighbouring molecules.
• A secondary amine has only one such hydrogen, so it forms fewer.
• A tertiary amine has no hydrogen on nitrogen at all, so it cannot form intermolecular hydrogen bonds with other amine molecules.
More hydrogen bonds mean stronger attraction between molecules and more energy needed to separate them.
So the order of boiling points is primary > secondary > tertiary.

Q8. How is benzene diazonium chloride prepared? Why is the reaction carried out at 273-278 K?

Answer: Preparation (diazotisation): Aniline is treated with nitrous acid, made in the flask itself from sodium nitrite and hydrochloric acid.
C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
Why 273-278 K (0-5°C): Benzene diazonium chloride is very unstable at ordinary temperature. Above about 278 K it decomposes, reacting with water to give phenol and releasing nitrogen gas.
C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂ + HCl
Keeping the mixture in an ice bath prevents this decomposition and keeps the diazonium salt available for further reactions.

Q9. What is a coupling reaction? Write the reaction of benzene diazonium chloride with phenol.

Answer: In a coupling reaction, a diazonium salt joins with an electron-rich aromatic compound such as phenol or aniline to give a brightly coloured azo compound, which contains the −N=N− group.
With phenol: In a mildly alkaline medium, benzene diazonium chloride couples at the para position of phenol:
C₆H₅N₂⁺Cl⁻ + C₆H₅OH → p-hydroxyazobenzene (an orange dye)
With aniline: In a mildly acidic medium, the product is p-aminoazobenzene, a yellow dye.
Importance: These reactions are the basis of the whole azo dye industry, which gives us a huge range of coloured dyes for cloth.

Q10. Why cannot aniline be nitrated directly? How is nitration carried out?

Answer: The problem: Nitration needs a strongly acidic mixture of concentrated HNO₃ and H₂SO₄. In this acidic medium aniline is protonated to give the anilinium ion, C₆H₅NH₃⁺.
The −NH₃⁺ group is strongly electron withdrawing and meta directing, exactly the opposite of the −NH₂ group. So instead of the expected ortho and para products, a large amount of the meta product is obtained.
The solution — protect the group:
1. Acetylate the aniline first with acetic anhydride to give acetanilide. The −NHCOCH₃ group is still ortho-para directing but much less basic, so it is not protonated.
2. Nitrate the acetanilide, which gives mainly the para product.
3. Hydrolyse the amide back with acid or alkali to recover the free −NH₂ group.
This gives p-nitroaniline in good yield.

Q11. How will you convert nitrobenzene into aniline? Write the equation.

Answer: Nitrobenzene is reduced to aniline. There are two common methods.
Method 1: Catalytic hydrogenation
C₆H₅NO₂ + 3H₂ → C₆H₅NH₂ + 2H₂O (with Ni, Pd or Pt as catalyst)
Method 2: Metal and acid
C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O (using Sn/HCl or Fe/HCl)
The second method is cheaper and is often used in the laboratory. Note that if a C=C double bond is also present in the molecule, catalytic hydrogenation would reduce that too, so the metal-acid method is preferred in such cases.

Q12. How do you distinguish between ethylamine and aniline by a simple chemical test?

Answer: Two simple tests work well.
1. Azo dye test: Both are cooled with NaNO₂/HCl at 273-278 K and then added to an alkaline solution of 2-naphthol.
• Aniline gives a bright orange or red dye, because it forms a stable diazonium salt which couples.
• Ethylamine gives no dye, only brisk effervescence of nitrogen gas, because aliphatic diazonium salts are unstable even at that temperature and break down at once.
2. Solubility test: Ethylamine is soluble in water and turns red litmus blue strongly, while aniline is almost insoluble in water and is only a very weak base.

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