Chapter 13: Organic Compounds Containing Nitrogen Class 12 Chemistry NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 13

Amines – Exercise Questions with Answers

Question 13.1: Classify the following amines as primary, secondary, or tertiary:

Answer:
(i) C₂H₅NH₂ – Primary amine (1°)
(ii) (C₂H₅)₃N – Tertiary amine (3°)
(iii) CH₃NHCH₃ – Secondary amine (2°)
(iv) (CH₃)₂CHNH₂ – Primary amine (1°)

Question 13.2: Write the structures of different isomeric amines corresponding to the molecular formula C₄H₁₁N. (i) Write the IUPAC names. (ii) What type of isomerism is exhibited?

Answer:
IUPAC Names:
– Butan-1-amine
– Butan-2-amine
– 2-Methylpropan-1-amine
– 2-Methylpropan-2-amine
– N-Methylpropan-1-amine
– N-Methylpropan-2-amine
– N-Ethylethanamine
– N,N-Dimethylethanamine

Types of Isomerism:
– Chain isomerism
– Position isomerism
– Metamerism
– Functional isomerism (Primary, secondary, and tertiary amines are functional isomers)

Question 13.3: How will you convert the following?

Answer:
(i) Benzene → Nitrobenzene (nitration) → Aniline (reduction with Sn/HCl)
(ii) Benzene → Aniline → N,N-Dimethylaniline (methylation with excess CH₃I and Na₂CO₃)
(iii) Cl-(CH₂)₄-Cl → Hexanedinitrile (reaction with KCN) → Hexane-1,6-diamine (reduction with H₂/Ni)

Question 13.4: Arrange the following in increasing order of their basic strength:

Answer:
(i) C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH
(ii) C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH
(iii) C₆H₅NH₂ < CH₃NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃N < (CH₃)₂NH

Explanation: Basic strength depends on electron density on nitrogen and solvation effects. Greater +I effect and solvation → stronger base.

Question 13.5: Complete the following acid-base reactions and name the products:

Answer:
(i) CH₃CH₂CH₂NH₂ + HCl → CH₃CH₂CH₂NH₃⁺Cl⁻ (Propan-1-aminium chloride)
(ii) (C₂H₅)₃N + HCl → [(C₂H₅)₃NH]⁺Cl⁻ (Triethylaminium chloride)

Question 13.6: Write the reaction of aniline with excess methyl iodide in the presence of sodium carbonate solution.

Answer:
C₆H₅NH₂ + 2CH₃I + Na₂CO₃ → C₆H₅N(CH₃)₂ + 2NaI + CO₂ + H₂O

Product: N,N-Dimethylaniline

Question 13.7: Write the chemical reaction of aniline with benzoyl chloride. What is the name of the product formed?

Answer:
C₆H₅NH₂ + C₆H₅COCl → C₆H₅NHCOC₆H₅ + HCl (in presence of NaOH)

Product: Benzanilide

Question 13.8: Write structures of isomers of C₃H₉N. Identify those which release N₂ gas upon treatment with nitrous acid.

Answer:
Isomers:
– Propan-1-amine
– Propan-2-amine
– N-Methylethanamine
– N,N-Dimethylmethanamine

Primary amines (Propan-1-amine and Propan-2-amine) evolve N₂ with nitrous acid.

Question 13.9: Convert:

Answer:
(i) 3-Methylaniline → 3-Nitrotoluene
(Protect the amino group → Nitrate → Deprotect to form 3-nitrotoluene)

(ii) Aniline → 1,3,5-Tribromobenzene
(Direct bromination of aniline with excess Br₂ produces 1,3,5-tribromobenzene)

NCERT Solutions for Class 12 Chemistry Chapter 13

Amines – Exercise Questions with Answers

Question 13.1: Write the IUPAC names of the following compounds and classify them into primary, secondary, and tertiary amines:

Answer:
(i) (CH₃)₂CHNH₂ – Propan-2-amine – Primary (1°) amine
(ii) CH₃(CH₂)₂NH₂ – Propan-1-amine – Primary (1°) amine
(iii) CH₃NHCH(CH₃)₂ – N-Methylpropan-2-amine – Secondary (2°) amine
(iv) (CH₃)₃CNH₂ – 2-Methylpropan-2-amine – Primary (1°) amine
(v) C₆H₅NHCH₃ – N-Methylaniline – Secondary (2°) amine
(vi) (CH₃CH₂)₂NCH₃ – N-Ethyl-N-methylethanamine – Tertiary (3°) amine
(vii) m-BrC₆H₄NH₂ – 3-Bromoaniline – Primary (1°) amine

Question 13.2: Give one chemical test to distinguish between the following pairs of compounds:

Answer:
(i) Methylamine vs Dimethylamine: Carbylamine Test – Methylamine gives foul smell of isocyanide, dimethylamine does not.
(ii) Secondary vs Tertiary Amines: Liebermann’s Nitroso Test – Secondary amines give a green/blue color, tertiary do not.
(iii) Ethylamine vs Aniline: Dye Test – Aniline forms azo dye; ethylamine does not.
(iv) Aniline vs Benzylamine: Oxidation Test – Benzylamine gives benzaldehyde on oxidation, aniline does not.
(v) Aniline vs N-Methylaniline: Acetylation Test – Forms different acetanilides with different melting points.

Question 13.3: Give reasons for the following:

Answer:
(i) Aniline has higher pKb than methylamine due to lone pair delocalization over benzene ring.
(ii) Ethylamine is more soluble than aniline due to better hydrogen bonding with water.
(iii) Methylamine increases hydroxide ion concentration causing Fe³⁺ to precipitate as hydrated ferric oxide.
(iv) Aniline forms substantial meta-nitroaniline because anilinium ion is meta-directing.
(v) Aniline forms a complex with AlCl₃ preventing Friedel-Crafts reaction.
(vi) Aromatic diazonium salts are resonance-stabilized; aliphatic ones are not.
(vii) Gabriel synthesis avoids multiple alkylation and provides pure primary amines.

Question 13.4: Arrange the following as indicated:

Answer:
(i) Decreasing pKb: C₆H₅NH₂ < C₆H₅NHCH₃ < C₂H₅NH₂ < (C₂H₅)₂NH
(ii) Increasing basic strength: C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₂NH
(iii) (a) p-Nitroaniline < Aniline < p-Toluidine
(b) C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅CH₂NH₂
(iv) Decreasing gas phase basicity: (C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃
(v) Increasing boiling point: (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH
(vi) Increasing solubility: C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂

Question 13.5: How will you bring about the following conversions?

Answer:
(i) Ethanoic acid → Ethanamide → Methanamine (Hofmann degradation)
(ii) Hexanenitrile → Hexylamine → 1-Aminopentane
(iii) Methanol → Methanal → Ethanoic acid
(iv) Ethanamine → Ethanenitrile → Methanamine
(v) Ethanoic acid → Ethyl bromide → Propanenitrile → Propanoic acid
(vi) Methanamine → Acylation → Hofmann degradation → Ethanamine
(vii) Nitromethane → Methylamine → Dimethylamine
(viii) Propanoic acid → Decarboxylation → Ethane → Oxidation → Ethanoic acid

Question 13.6: Describe the method for the identification of primary, secondary, and tertiary amines.

Answer:
Hinsberg’s Test:
– Primary amine: Forms soluble sulphonamide; precipitates on acidification.
– Secondary amine: Forms insoluble sulphonamide.
– Tertiary amine: No reaction with Hinsberg’s reagent; soluble in dilute acid.

Reaction with Nitrous Acid:
– Primary aliphatic: Evolution of N₂ gas.
– Primary aromatic: Formation of diazonium salts.
– Secondary: Formation of N-nitrosoamines.
– Tertiary: Formation of nitroso derivatives.

Question 13.7: Write short notes on the following:

Answer:
(i) Carbylamine Reaction: Primary amines react with CHCl₃ + alcoholic KOH to form isocyanides.
(ii) Diazotisation: Primary aromatic amine reacts with NaNO₂ + HCl at 273-278 K forming diazonium salt.
(iii) Hoffmann’s Bromamide Reaction: Amide reacts with Br₂ + NaOH to form primary amine with one C less.
(iv) Coupling Reaction: Diazonium salt reacts with phenols/aromatic amines to form azo dyes.
(v) Ammonolysis: Alkyl halides react with NH₃ to form amines.
(vi) Acetylation: Amines react with acetyl chloride to form acetamides.
(vii) Gabriel Phthalimide Synthesis: Produces primary amines by reaction of potassium phthalimide with alkyl halides.

Question 13.8: Accomplish the following conversions:

Answer:
(i) Nitrobenzene → Aniline → Phenol → Oxidation → Benzoic acid
(ii) Benzene → Nitrobenzene → m-Bromonitrobenzene → m-Bromoaniline → Hydrolysis → m-Bromophenol
(iii) Benzoic acid → Benzoyl chloride → Benzamide → Aniline
(iv) Aniline → 2,4,6-Tribromofluorobenzene via bromination, diazotisation, and fluorination.
(v) Benzyl chloride → Benzyl cyanide → 2-Phenylethanamine (reduction)
(vi) Chlorobenzene → p-Nitrochlorobenzene → p-Nitroaniline → p-Chloroaniline
(vii) Aniline → p-Bromoaniline
(viii) Benzamide → Aniline → Benzylamine → Toluene
(ix) Aniline → Benzene → Benzyl chloride → Benzyl alcohol

Question 13.9: Give the structures of A, B, and C in the following reaction.

Answer:
(As per NCERT figure, A → B → C are specific transformations involving amines, diazonium salts, etc.)

Question 13.10: An aromatic compound A reacts with aqueous ammonia to form B. B reacts with Br₂ and KOH to form C (C₆H₇N). Identify A, B, and C.

Answer:
A: Benzoic acid (C₆H₅COOH)
B: Benzamide (C₆H₅CONH₂)
C: Aniline (C₆H₅NH₂)

Question 13.11: Complete the following reactions:

Answer:
Provide complete balanced chemical equations based on amine reactions, diazotisation, bromination, acylation, etc. (Refer NCERT.)

Question 13.12: Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Answer:
Aromatic halides like chlorobenzene are resistant to nucleophilic substitution due to resonance stabilization, making Gabriel synthesis ineffective for aromatic primary amines.

Question 13.13: Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

Answer:
(i) Aromatic amines: Form diazonium salts (e.g., aniline forms C₆H₅N₂⁺Cl⁻)
(ii) Aliphatic amines: Form unstable diazonium salts which decompose into alcohols and nitrogen gas.

Question 13.14: Give plausible explanations for the following:

Answer:
(i) Amines are less acidic than alcohols due to lower electronegativity of nitrogen compared to oxygen.
(ii) Primary amines have higher boiling points than tertiary amines due to stronger hydrogen bonding.
(iii) Aliphatic amines are stronger bases than aromatic amines due to lone pair delocalization in aromatic amines.