Class 12 Chemistry Important Chapter 13 Amines

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Class 12 Chemistry Important Chapter 13 Amines

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Amines

Chapter: 13

PART – II

IMPORTANT QUESTION AND ANSWER

1. Explain the Hinsberg’s method for distinction of primary, secondary, and tertiary amines.

Ans: Primary amines react with benzenesulphonyl chloride to form sulphonamide soluble in alkali.

Secondary amines form sulphonamide insoluble in alkali. 

Tertiary amines do not react.

This is used for identification and separation.

2. Discuss the basicity trend among primary, secondary, and tertiary amines in aqueous solution.

Ans: In aqueous medium, the basicity order is:

Secondary > Primary > Tertiary amine > Ammonia.

This is due to a balance of inductive effect (+I), solvation, and steric hindrance.

3. Explain Gabriel Phthalimide synthesis and state its limitation.

Ans:Phthalimide reacts with KOH to form its salt, which on reacting with alkyl halide and subsequent hydrolysis gives primary amine.

Limitation: Aromatic amines cannot be prepared by this method due to low reactivity of aryl halides.

4. What is the importance of diazonium salts in organic synthesis?

Ans:Diazonium salts are versatile intermediates used to introduce functional groups like –Cl, –Br, –CN, –OH, –NO₂, and aryl groups. They are also key to azo dye synthesis via coupling reactions.

5. Explain Hinsberg’s method for distinction of primary, secondary, and tertiary amines.

Ans: Hinsberg’s method uses benzenesulphonyl chloride (C₆H₅SO₂Cl) to differentiate between primary, secondary, and tertiary amines.

(i) Primary amines react with benzenesulphonyl chloride to form N-alkylbenzenesulphonamide, which has a hydrogen attached to nitrogen. This hydrogen is acidic due to the –SO₂ group, so the compound is soluble in alkali.

(ii) Secondary amines also react with benzenesulphonyl chloride to form N,N-dialkylbenzenesulphonamide. However, no hydrogen is present on the nitrogen, so the product is insoluble in alkali.

(iii) Tertiary amines do not react with benzenesulphonyl chloride because they lack an N–H bond.

6. Discuss the basicity of amines in aqueous solution and gaseous phase.

Ans: The basicity of amines depends on the availability of the lone pair of electrons on nitrogen to accept a proton.

(i) In the gaseous phase, the order of basicity is: Tertiary > Secondary > Primary > Ammonia.

This is due to the +I effect of alkyl groups, which increases electron density on nitrogen.

(ii) In aqueous solution, solvation effects and steric hindrance also play a role. 

The order becomes:

Secondary > Primary > Tertiary > Ammonia.

(a) Secondary amines are best stabilized due to both +I effect and solvation.

(b) Tertiary amines, though more basic in gas, are poorly solvated due to bulky groups and show less basicity in water.

Thus, both electronic effects and solvation influence basic strength.

7. What is Gabriel Phthalimide Synthesis? Mention its limitations.

Ans: Gabriel Phthalimide Synthesis is used to prepare primary aliphatic amines.

Steps:

(i) Phthalimide is treated with ethanolic KOH to form potassium phthalimide.

(ii) It reacts with alkyl halide (R–X) to form N-alkylphthalimide.

(iii) On acidic or basic hydrolysis, it gives primary amine (R–NH₂) and phthalic acid.

Limitation:

(i) It cannot prepare aryl amines (like aniline), because aryl halides do not undergo nucleophilic substitution due to resonance stabilization and poor reactivity.

8. Explain the preparation and significance of diazonium salts.

Ans: Diazonium salts are formed by reacting primary aromatic amines with nitrous acid (HNO₂) at 0–5°C.

Example: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O

This reaction is called diazotisation.

Significance:

(i) Diazonium salts are highly reactive intermediates used in:

(a) Substitution reactions (to form –Cl, –Br, –I, –CN, –OH, –NO₂).

(b) Coupling reactions to make azo dyes, used in textile and pigment industries.

9. Explain the Carbylamine reaction. How is it used to distinguish amines?

Ans: Carbylamine reaction is used to detect primary amines.

Reaction: R–NH₂ + CHCl₃ + 3KOH → R–NC (isocyanide) + 3KCl + 3H₂O

Conditions: 

(a) Alcoholic KOH and chloroform.

(b) Heating is required.

Key Points:

(a) Produces foul-smelling isocyanide (carbylamine).

(b) Only primary aliphatic or aromatic amines respond.

(c) Secondary and tertiary amines do not show this reaction.

10. Write a note on electrophilic substitution reactions of aniline. How can we control the reactivity?

Ans: Aniline (C₆H₅NH₂) is highly reactive in electrophilic substitution reactions because the –NH₂ group is an electron-donating group and activates the ortho and para positions of the benzene ring.

(i) Common electrophilic substitution reactions:

(a) Bromination: Aniline reacts with bromine water at room temperature to give 2,4,6-tribro moaniline (white ppt).

(b) Nitration: Direct nitration gives meta product due to formation of protonated anilinium ion (–NH₃⁺), which is meta-directing.

(c) Sulphonation: Aniline reacts with conc. H₂SO₄ to form sulphanilic acid via anilinium hydrogensulphate.

(ii) Problem: Due to high reactivity, multiple substitutions occur.

(iii) Control method: By acetylating the NH₂ group with acetic anhydride, we get acetanilide, which is less reactive. After substitution, the acetyl group is removed by hydrolysis.