Class 12 : Chemistry (English) – Chapter 9: Amines
EXPLANATION & SUMMARY
✨ Introduction
🔵 Amines are derivatives of ammonia (NH₃) in which one or more hydrogen atoms are replaced by alkyl or aryl groups.
🟢 They are considered as organic bases and are also important as nucleophiles.
🟠 Found in natural products (alkaloids, hormones, vitamins) and industrially useful (dyes, drugs, polymers).
🔴 Classified into aliphatic and aromatic amines.
🌿 Classification of Amines
By type of substitution:
Primary (1°): One R group, formula R–NH₂ (e.g., CH₃NH₂).
Secondary (2°): Two R groups, formula R₂NH (e.g., (CH₃)₂NH).
Tertiary (3°): Three R groups, formula R₃N (e.g., (CH₃)₃N).
By nature of group attached:
Aliphatic amines: R = alkyl.
Aromatic amines: R = aryl (e.g., aniline).
🧪 Nomenclature
Common system: Alkyl groups + amine (e.g., ethylamine).
IUPAC system: Replace “-e” of alkane with “-amine” (e.g., ethanamine).
For aromatic amines, name derived from parent aromatic amine (e.g., aniline, N-methylaniline).
⚡ Structure of Amines
🔵 The nitrogen atom is sp³ hybridised.
🟢 Geometry: Trigonal pyramidal.
🟠 Bond angle ~ 107°, slightly less than tetrahedral angle due to lone pair repulsion.
🔬 Preparation of Amines
Reduction Methods
Nitro compounds → Amines (Sn/HCl, Fe/HCl, catalytic hydrogenation).
Nitriles (RCN) → Amines (LiAlH₄ or H₂/Ni).
Amides → Amines (LiAlH₄ or NaBH₄).
Ammonolysis of alkyl halides
R–X + NH₃ → R–NH₂ + HX (further reaction gives secondary/tertiary amines).
Gabriel phthalimide synthesis
K-phthalimide + R–X → R–NH₂ (primary amine only).
Hoffmann bromamide degradation
RCONH₂ + Br₂ + NaOH → R–NH₂ (1° amine, one carbon less).
🌍 Physical Properties
🔵 Lower aliphatic amines: gases with fishy odour.
🟢 Higher amines: liquids/solids.
🟠 Solubility: Lower amines soluble in water due to H-bonding; solubility decreases with size.
🔴 Boiling points: Primary > Secondary > Tertiary (due to H-bonding differences).
💡 Chemical Properties
Basic character
Amines act as bases due to lone pair on N.
Measured by Kb and pKb values.
Order in gas phase: 3° > 2° > 1° > NH₃.
In aqueous solution: 2° > 1° > 3° (hydration effect).
Aromatic amines less basic due to delocalisation of lone pair into ring.
Reaction with acids
R–NH₂ + HCl → R–NH₃⁺Cl⁻
Reaction with acyl halides
R–NH₂ + R′COCl → R′CONHR + HCl (acylation).
Reaction with nitrous acid
1° aliphatic amines → alcohols + N₂.
1° aromatic amines → diazonium salts.
2° amines → nitrosoamines.
3° amines → ammonium salts.
Hinsberg’s test (distinction between 1°, 2°, 3° amines).
Carbylamine reaction
1° amine + CHCl₃ + alc. KOH → isocyanide (foul smell).
🌸 Diazonium Salts
🔵 Benzene diazonium chloride (C₆H₅N₂⁺Cl⁻): obtained from aniline + NaNO₂ + HCl (0–5 °C).
🟢 Used in synthesis of azo dyes (coupling reactions).
🟠 Unstable above 5 °C → decomposes.
🔴 Reactions: Replacement by –OH, –Cl, –Br, –I, –CN, –NO₂, –H.
🌟 Uses of Amines
Aniline: precursor for dyes, drugs, rubber chemicals.
Alkylamines: corrosion inhibitors, solvents.
Diazonium salts: azo dyes, synthetic intermediates.
📝 Summary
Amines are ammonia derivatives (R–NH₂, R₂NH, R₃N). Classified as aliphatic or aromatic; primary, secondary, tertiary.
Nomenclature: Common (alkylamine, aniline derivatives), IUPAC (-amine ending).
Structure: N sp³ hybridised, trigonal pyramidal, bond angle ~107°.
Preparation: Reduction of nitro compounds, nitriles, amides; ammonolysis of halides; Gabriel synthesis; Hoffmann bromamide.
Physical properties: Fishy odour, solubility due to H-bonding, boiling point trend: 1° > 2° > 3°.
Chemical properties:
Basic nature (lone pair on N, hydration effect, aromatic vs aliphatic).
Acylation, salt formation, nitrous acid reactions.
Special tests: Hinsberg, Carbylamine.
Aromatic amines form diazonium salts at 0–5 °C. These are key intermediates for azo dyes, halogenation, substitution, cyanation.
Applications: Drugs, dyes, polymers, agrochemicals, intermediates.
🎯 Quick Recap
🟦 Amines = derivatives of NH₃.
🟩 Classified as 1°, 2°, 3°, aliphatic, aromatic.
🟨 Prepared by reduction, Gabriel synthesis, Hoffmann degradation.
🟧 Show basicity, acylation, nitrous acid, Hinsberg’s test, Carbylamine.
🟪 Aromatic amines form diazonium salts (dye synthesis).
🟫 Industrial importance: drugs, dyes, agrochemicals.
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QUESTIONS FROM TEXTBOOK
Question 9.1
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines:
(i) (CH₃)₂CHNH₂
(ii) CH₃(CH₂)₂NH₂
(iii) CH₃NHCH(CH₃)₂
(iv) (CH₃)₃CNH₂
(v) C₆H₅NHCH₃
(vi) CH₃CH₂N(CH₃)₂
(vii) m–BrC₆H₄NH₂
Answer 9.1
🟦 (i) 2-Aminopropane → Primary amine
🟩 (ii) Propan-1-amine → Primary amine
🟨 (iii) N-Methylpropan-2-amine → Secondary amine
🟧 (iv) 2-Methyl-2-aminopropane → Primary amine
🟪 (v) N-Methylaniline → Secondary aromatic amine
🟫 (vi) N,N-Dimethylethanamine → Tertiary amine
🔷 (vii) 3-Bromoaniline → Primary aromatic amine
Question 9.2
Give one chemical test to distinguish between the following pairs of compounds:
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline
Answer 9.2
🟦 (i) Methylamine vs Dimethylamine
➤ Carbylamine test: Methylamine (1°) gives foul-smelling isocyanide with CHCl₃ + alc. KOH.
➤ Dimethylamine (2°) does not respond.
🟩 (ii) Secondary vs Tertiary amines
➤ Hinsberg test: Secondary amine forms insoluble sulphonamide.
➤ Tertiary amine does not react.
🟨 (iii) Ethylamine vs Aniline
➤ Solubility: Ethylamine soluble in water, aniline insoluble.
➤ Also, aniline gives bromination test → 2,4,6-tribromoaniline.
🟧 (iv) Aniline vs Benzylamine
➤ Nitrous acid test: Aniline gives stable diazonium salt at 0–5 °C.
➤ Benzylamine gives unstable diazonium salt → immediate N₂ evolution.
🟪 (v) Aniline vs N-Methylaniline
➤ Azo coupling: Aniline couples with phenol to form azo dye.
➤ N-Methylaniline does not couple.
Question 9.3
Account for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although –NH₂ is o,p-directing, aniline on nitration gives m-nitroaniline as a major product.
(v) Aniline does not undergo Friedel–Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Answer 9.3
🟦 (i) In aniline, lone pair on N delocalises into benzene ring → less available for protonation → weaker base → higher pKb.
🟩 (ii) Ethylamine forms strong H-bonds with water → soluble. Aniline, bulky aromatic ring → insoluble.
🟨 (iii) Methylamine is alkaline → reacts with FeCl₃ forming Fe(OH)₃ precipitate (brown).
🟧 (iv) Nitration of aniline is carried out in acidic medium. –NH₂ gets protonated to –NH₃⁺ (deactivating, m-directing) → m-product predominates.
🟪 (v) In Friedel–Crafts, AlCl₃ catalyst reacts with –NH₂ to form a complex → reaction blocked.
🟫 (vi) Aromatic diazonium ions are resonance-stabilised → more stable. Aliphatic diazonium ions unstable → release N₂ immediately.
🔷 (vii) Gabriel synthesis ensures only primary amine (no secondary/tertiary), hence preferred.
Question 9.4
Arrange the following:
(i) In decreasing order of pKb values:
C₆H₅NH₂, C₆H₅N(CH₃)₂, C₆H₅CH₂NH₂, C₆H₅NHCH₃
(ii) In increasing order of basic strength:
C₆H₅NH₂, C₆H₅CH₂NH₂, (C₂H₅)₂NH, CH₃NH₂
(iii) In increasing order of basic strength:
Aniline, p-Nitroaniline, p-Toluidine
(iv) In decreasing order of basic strength in gas phase:
C₂H₅NH₂, (C₂H₅)₃N, (C₂H₅)₂NH, NH₃
(v) In increasing order of boiling point:
C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂
(vi) In increasing order of solubility in water:
C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂
Answer 9.4
🟦 (i) pKb order (higher pKb = weaker base):
C₆H₅N(CH₃)₂ < C₆H₅CH₂NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂
🟩 (ii) Basic strength order:
C₆H₅NH₂ < C₆H₅CH₂NH₂ < CH₃NH₂ < (C₂H₅)₂NH
🟨 (iii) Aniline < p-Nitroaniline < p-Toluidine
🟧 (iv) Gas phase: (C₂H₅)₃N < (C₂H₅)₂NH < C₂H₅NH₂ < NH₃
🟪 (v) Boiling point: (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH
🟫 (vi) Solubility: C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂
Question 9.5
How will you convert:
(i) Ethanoic acid into methanamine
(ii) Hexanenitrile into 1-aminopentane
(iii) Methanol to ethanoic acid
(iv) Ethanamine into methanamine
(v) Ethanoic acid into propanoic acid
(vi) Methanamine into ethanamine
(vii) Nitromethane into dimethylamine
(viii) Propanoic acid into ethanoic acid
Answer 9.5
🟦 (i) CH₃COOH → (SOCl₂) → CH₃COCl → (NH₃) → CH₃CONH₂ → (Br₂/NaOH) → CH₃NH₂
🟩 (ii) CH₃(CH₂)₄CN → (LiAlH₄) → CH₃(CH₂)₄CH₂NH₂ → 1-Aminopentane
🟨 (iii) CH₃OH → (Cu, 573 K) → HCHO → (K₂Cr₂O₇/H⁺) → HCOOH → (oxidation) → CH₃COOH
🟧 (iv) CH₃CH₂NH₂ → (HNO₂) → Ethanol → (oxidation) → CH₃CHO → (oxidation) → CH₃COOH → (Hoffmann) → CH₃NH₂
🟪 (v) CH₃COOH → (HCN/KCN) → CH₃CH(OH)CN → (hydrolysis) → CH₃CH₂COOH
🟫 (vi) CH₃NH₂ → (CH₃I + KOH) → (CH₃)₂NH → (excess CH₃I) → (CH₃)₃N → (reduction) → C₂H₅NH₂
🔷 (vii) CH₃NO₂ → (Zn/HCl) → CH₃NH₂ → (CH₃I + KOH) → (CH₃)₂NH
🔶 (viii) CH₃CH₂COOH → (oxidation) → CH₃COOH
Question 9.6
Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
Answer 9.6
🟦 Hinsberg’s Test
Reagent: Benzenesulphonyl chloride in NaOH.
➤ Primary amine (1°): Forms sulphonamide soluble in alkali.
Equation: RNH₂ + C₆H₅SO₂Cl → R–NHSO₂C₆H₅ (soluble).
➤ Secondary amine (2°): Forms sulphonamide insoluble in alkali.
Equation: R₂NH + C₆H₅SO₂Cl → R₂N–SO₂C₆H₅ (insoluble).
➤ Tertiary amine (3°): Does not react with C₆H₅SO₂Cl.
Question 9.7
Write short notes on the following:
(i) Carbylamine reaction
(ii) Diazotisation
(iii) Hofmann’s bromamide reaction
(iv) Coupling reaction
(v) Ammonolysis
(vi) Acetylation
(vii) Gabriel phthalimide synthesis
Answer 9.7
🟦 (i) Carbylamine reaction
1° amine + CHCl₃ + alc. KOH → R–NC (isocyanide, foul smell).
🟩 (ii) Diazotisation
Aromatic 1° amine + NaNO₂ + HCl (0–5 °C) → ArN₂⁺Cl⁻ (diazonium salt).
🟨 (iii) Hofmann bromamide reaction
RCONH₂ + Br₂ + NaOH → R–NH₂ (1° amine with one C less).
🟧 (iv) Coupling reaction
Diazonium salt + phenol/aniline → azo dye (Ar–N=N–Ar′).
🟪 (v) Ammonolysis
Alkyl halide + NH₃ → amine(s) + HX.
🟫 (vi) Acetylation
RNH₂ + CH₃COCl → RNHCOCH₃ + HCl.
🔷 (vii) Gabriel phthalimide synthesis
Potassium phthalimide + R–X → R–NH₂ (only primary amine).
Question 9.8
Accomplish the following conversions:
(i) Nitrobenzene to benzoic acid
(ii) Benzene to m-bromophenol
(iii) Benzoic acid to aniline
(iv) Aniline to 2,4,6-tribromoaniline
(v) Benzyl chloride to 2-phenylethanamine
(vi) Chlorobenzene to p-chloroaniline
(vii) Aniline to p-bromoaniline
(viii) Benzamide to toluene
(ix) Aniline to benzyl alcohol
Answer 9.8
🟦 (i) Nitrobenzene → (reduction, Sn/HCl) → Aniline → (oxidation, KMnO₄) → Benzoic acid
🟩 (ii) Benzene → (nitration, HNO₃/H₂SO₄) → Nitrobenzene → (reduction, Sn/HCl) → Aniline → (diazotisation, NaNO₂/HCl, 273 K) → Diazonium salt → (hydrolysis, H₂O) → Phenol → (Br₂, Fe) → m-Bromophenol
🟨 (iii) Benzoic acid → (SOCl₂) → Benzoyl chloride → (NH₃) → Benzamide → (Hoffmann bromamide) → Aniline
🟧 (iv) Aniline → (Br₂ in water) → 2,4,6-tribromoaniline (white ppt).
🟪 (v) Benzyl chloride → (KCN) → Benzyl cyanide → (reduction, LiAlH₄) → 2-Phenylethanamine
🟫 (vi) Chlorobenzene → (nitration) → p-Nitrochlorobenzene → (reduction, Sn/HCl) → p-Chloroaniline
🔷 (vii) Aniline → (acetylation with Ac₂O) → Acetanilide → (Br₂, AcOH) → p-Bromoacetanilide → (hydrolysis) → p-Bromoaniline
🔶 (viii) Benzamide → (Hoffmann bromamide, Br₂/NaOH) → Aniline → (reduction of diazonium salt) → Toluene
✨ (ix) Aniline → (diazotisation) → Benzene diazonium chloride → (reduction with H₂O/SnCl₂) → Benzyl alcohol
Question 9.9
Give the structures of A, B and C in the following reactions:
(i) CH₃CH₂I →(NaCN)→ A →(hydrolysis)→ B →(NaOH + Br₂)→ C
(ii) C₆H₅N₂Cl →(CuCN)→ A →(H₃O⁺/Δ)→ B →(NH₃)→ C
(iii) CH₃CH₂Br →(KCN)→ A →(LiAlH₄)→ B →(HNO₂, 0 °C)→ C
(iv) C₆H₅NO₂ →(Fe/HCl)→ A →(NaNO₂ + HCl, 273 K)→ B →(H₂O/H⁺/Δ)→ C
(v) CH₃COOH →(NH₃)→ A →(NaOBr)→ B →(NaNO₂/HCl)→ C
(vi) C₆H₅NO₂ →(Fe/HCl)→ A →(NaNO₂, 273 K)→ B →(C₂H₅OH)→ C
Answer 9.9
🟦 (i)
A = CH₃CH₂CN (Propanenitrile)
B = CH₃CH₂COOH (Propanoic acid)
C = CH₃CH₂NH₂ (Ethylamine)
🟩 (ii)
A = C₆H₅CN (Benzonitrile)
B = C₆H₅COOH (Benzoic acid)
C = C₆H₅CONH₂ (Benzamide)
🟨 (iii)
A = CH₃CH₂CN (Propanenitrile)
B = CH₃CH₂CH₂NH₂ (Propylamine)
C = CH₃CH₂CH₂OH (Propanol, from diazotisation–hydrolysis)
🟧 (iv)
A = C₆H₅NH₂ (Aniline)
B = C₆H₅N₂⁺Cl⁻ (Benzene diazonium chloride)
C = C₆H₅OH (Phenol)
🟪 (v)
A = CH₃CONH₂ (Acetamide)
B = CH₃NH₂ (Methylamine)
C = CH₃OH (Methanol, from diazotisation–hydrolysis)
🟫 (vi)
A = C₆H₅NH₂ (Aniline)
B = C₆H₅N₂⁺Cl⁻ (Benzene diazonium chloride)
C = C₆H₅OC₂H₅ (Phenetole)
Question 9.10
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’. On heating with Br₂ and KOH, ‘B’ forms compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.
Answer 9.10
🟦 Step 1: Compound A = C₆H₅CH₂Cl (Benzyl chloride)
➤ Reaction with aq. NH₃ → C₆H₅CH₂NH₂ (Benzylamine, B).
🟩 Step 2: B (Benzylamine) + Br₂/NaOH → Hoffmann rearrangement → C₆H₅NH₂ (Aniline, C).
✅ Final:
A = Benzyl chloride (C₆H₅CH₂Cl)
B = Benzylamine (C₆H₅CH₂NH₂)
C = Aniline (C₆H₅NH₂)
Question 9.11
Complete the following reactions:
(i) C₆H₅NH₂ + CHCl₃ + alc.KOH → ?
(ii) C₆H₅N₂Cl + H₃PO₂ + H₂O → ?
(iii) C₆H₅NH₂ + H₂SO₄ (conc.) → ?
(iv) C₆H₅N₂Cl + C₂H₅OH → ?
(v) C₆H₅NH₂ + Br₂ (aq) → ?
(vi) C₆H₅NH₂ + (CH₃CO)₂O → ?
(vii) C₆H₅N₂Cl →(i) HBF₄, (ii) NaNO₂/Cu, Δ→ ?
Answer 9.11
🟦 (i) → C₆H₅NC (Phenyl isocyanide, foul smell)
🟩 (ii) → C₆H₆ (Benzene)
🟨 (iii) → C₆H₅NH₃⁺HSO₄⁻ (Anilinium hydrogen sulfate)
🟧 (iv) → C₆H₅OC₂H₅ (Phenetole)
🟪 (v) → 2,4,6-tribromoaniline (white ppt)
🟫 (vi) → Acetanilide (C₆H₅NHCOCH₃)
🔷 (vii) → Fluorobenzene (C₆H₅F)
Question 9.12
Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Answer 9.12
🟦 Gabriel synthesis works via nucleophilic substitution (SN).
🟩 Aryl halides (Ar–X) are resistant to nucleophilic substitution due to resonance and partial double bond character.
✅ Hence, aromatic primary amines cannot be synthesised by this method.
Question 9.13
Write the reactions of:
(i) Aromatic primary amines
(ii) Aliphatic primary amines
with nitrous acid.
Answer 9.13
🟦 (i) Aromatic primary amines (ArNH₂):
➤ ArNH₂ + NaNO₂ + HCl (273 K) → ArN₂⁺Cl⁻ (stable diazonium salt) + H₂O
🟩 (ii) Aliphatic primary amines (RNH₂):
➤ RNH₂ + NaNO₂ + HCl → ROH + N₂↑ + HCl
(unstable diazonium → alcohol + N₂ gas)
Question 9.14
Give plausible explanation:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling points than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Answer 9.14
🟦 (i) O–H bond in alcohol is more polar than N–H in amine → alcohol more acidic.
🟩 (ii) 1° amines have more –NH groups for H-bonding → higher boiling point than 3° amines (no N–H bond).
🟨 (iii) In aromatic amines, lone pair on N delocalises into benzene ring → less available for protonation → weaker base.
Aliphatic amines: electron-releasing alkyl groups increase electron density on N → stronger base.
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OTHER IMPORTANT QUESTIONS FOR EXAMS
(CBSE MODEL QUESTIONS PAPER)
ESPECIALLY MADE FROM THIS LESSON ONLY
✨ Section A (Q1–Q16) – MCQs (1 mark each)
Options for Assertion–Reason questions:
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A
Both A and R are true, but R is not the correct explanation of A
A is true, but R is false
A is false, but R is true
Q1. Which of the following is the correct IUPAC name of CH₃–NH₂?
🔵 (A) Methyl amine
🟢 (B) Aminomethane
🟠 (C) Methanamine
🔴 (D) Aminomethyl
Answer: 🟠 (C) Methanamine
Q2. Among the following, which amine gives carbylamine test?
🔵 (A) Aniline
🟢 (B) Diphenylamine
🟠 (C) Triethylamine
🔴 (D) N-methyl aniline
Answer: 🔵 (A) Aniline
Q3. Primary amines can be distinguished from secondary and tertiary amines by:
🔵 (A) Tollen’s test
🟢 (B) Carbylamine test
🟠 (C) Iodoform test
🔴 (D) Fehling’s test
Answer: 🟢 (B) Carbylamine test
Q4. Which one of the following does not undergo diazotisation?
🔵 (A) Aniline
🟢 (B) p-Toluidine
🟠 (C) Benzylamine
🔴 (D) p-Aminophenol
Answer: 🟠 (C) Benzylamine
Q5. The basic strength of amines in aqueous solution decreases in the order:
🔵 (A) C₂H₅NH₂ > NH₃ > C₆H₅NH₂
🟢 (B) NH₃ > C₂H₅NH₂ > C₆H₅NH₂
🟠 (C) C₆H₅NH₂ > NH₃ > C₂H₅NH₂
🔴 (D) C₂H₅NH₂ > C₆H₅NH₂ > NH₃
Answer: 🔵 (A) C₂H₅NH₂ > NH₃ > C₆H₅NH₂
Q6. In Gabriel phthalimide synthesis, the limitation is:
🔵 (A) Primary aliphatic amines are not formed
🟢 (B) Secondary amines are not formed
🟠 (C) Aromatic primary amines are not formed
🔴 (D) Tertiary amines are not formed
Answer: 🟠 (C) Aromatic primary amines are not formed
Q7. Which reagent is used in Hoffmann bromamide reaction?
🔵 (A) Br₂/NaOH
🟢 (B) Br₂/H₂O
🟠 (C) Cl₂/NaOH
🔴 (D) NaBH₄
Answer: 🔵 (A) Br₂/NaOH
Q8. Assertion (A): Aniline undergoes bromination easily.
Reason (R): –NH₂ group is an ortho-para directing group.
Answer: 1
Q9. Assertion (A): Ethylamine is more basic than aniline.
Reason (R): In aniline, lone pair of nitrogen is delocalised over benzene ring.
Answer: 1
Q10. Which amine is soluble in water?
🔵 (A) Aniline
🟢 (B) Diphenylamine
🟠 (C) Ethylamine
🔴 (D) Triphenylamine
Answer: 🟠 (C) Ethylamine
Q11. Which compound on reduction gives aniline?
🔵 (A) Nitrobenzene
🟢 (B) Benzoic acid
🟠 (C) Benzamide
🔴 (D) Benzaldehyde
Answer: 🔵 (A) Nitrobenzene
Q12. Hinsberg’s reagent is:
🔵 (A) CHCl₃
🟢 (B) C₆H₅SO₂Cl
🟠 (C) C₂H₅Cl
🔴 (D) C₆H₅COCl
Answer: 🟢 (B) C₆H₅SO₂Cl
Q13. In diazonium salts, –N₂⁺ group is:
🔵 (A) Strongly nucleophilic
🟢 (B) Strongly electrophilic
🟠 (C) Weakly acidic
🔴 (D) Strongly basic
Answer: 🟢 (B) Strongly electrophilic
Q14. Which one is an aliphatic tertiary amine?
🔵 (A) (CH₃)₃N
🟢 (B) (C₂H₅)₂NH
🟠 (C) C₆H₅NH₂
🔴 (D) CH₃NH₂
Answer: 🔵 (A) (CH₃)₃N
Q15. Which diazonium salt is most stable?
🔵 (A) Methyl diazonium chloride
🟢 (B) Phenyl diazonium chloride
🟠 (C) Ethyl diazonium chloride
🔴 (D) n-Propyl diazonium chloride
Answer: 🟢 (B) Phenyl diazonium chloride
Q16. Which test is not given by tertiary amines?
🔵 (A) Carbylamine test
🟢 (B) Acetylation
🟠 (C) Nitrous acid test
🔴 (D) Reaction with HCl
Answer: 🔵 (A) Carbylamine test
✨ Section B (Q17–Q21) – Very Short Answer (2 marks each)
Q17. Write any two differences between primary and secondary amines.
🟦 Primary amines: one alkyl group attached to –NH₂.
🟩 Secondary amines: two alkyl groups attached to nitrogen.
📌 Primary amines give carbylamine test, secondary amines do not.
Q18. Why do tertiary amines have lower boiling points than secondary amines?
🟨 Secondary amines: hydrogen bonding present → stronger intermolecular forces.
🟪 Tertiary amines: no –NH hydrogen → no H-bonding → weaker forces → lower boiling point.
Q19. Write reaction of ethanamine with nitrous acid.
🧪 CH₃CH₂NH₂ + HNO₂ → CH₃CH₂OH + N₂↑ + H₂O
🎯 Primary aliphatic amine → alcohol + nitrogen gas.
Q20. Give IUPAC names of: (i) C₆H₅–NH₂, (ii) CH₃–NH–CH₃.
➤ (i) Aniline (benzenamine)
➤ (ii) N-methyl methanamine
Q21. Write short note on diazotisation.
🔷 Primary aromatic amine + NaNO₂ + HCl (0–5 °C) → diazonium salt.
⚗ Example: C₆H₅NH₂ → C₆H₅N₂⁺Cl⁻.
🎯 Reaction used for azo dye synthesis and coupling reactions.
✨ Section C (Q22–Q28) – Short Answer (3 marks each)
Q22. Explain Hinsberg’s test.
🟦 Reagent: benzenesulphonyl chloride (C₆H₅SO₂Cl).
➤ Primary amines → soluble sulphonamide (dissolves in alkali).
➤ Secondary amines → insoluble sulphonamide.
➤ Tertiary amines → no reaction.
✅ Thus, it distinguishes 1°, 2°, 3° amines.
Q23. Write the steps of Hofmann bromamide reaction with example.
🧪 Step 1: Amide + Br₂/NaOH → N-bromoamide.
🧪 Step 2: Base-induced rearrangement → isocyanate.
🧪 Step 3: Hydrolysis of isocyanate → primary amine (1C less).
🎯 Example: CH₃CONH₂ → CH₃NH₂.
Q24. Write chemical equation for diazotisation of aniline and its coupling with phenol.
⚗ Aniline + NaNO₂/HCl (0–5 °C) → C₆H₅N₂⁺Cl⁻.
➜ Coupling with phenol in alkaline medium → azo dye (p-hydroxyazobenzene).
✅ Used in dye industry.
Q25. Give reasons:
(i) Aniline is less basic than aliphatic amines.
(ii) Tertiary amines cannot be prepared by Gabriel synthesis.
(iii) Diazonium salts are used in dyeing.
🟦 (i) Resonance delocalises lone pair in aniline.
🟩 (ii) Gabriel method only gives primary amines.
🟨 (iii) They couple with aromatic rings to form coloured azo dyes.
Q26. Write short note on carbylamine test with example.
🧪 Primary amine + CHCl₃ + alcoholic KOH → isocyanide (foul smell).
➜ Example: C₆H₅NH₂ + CHCl₃ + 3KOH → C₆H₅NC + 3KCl + 3H₂O.
✅ Test for primary amines.
Q27. Compare basic strengths of NH₃, CH₃NH₂ and (CH₃)₂NH in aqueous medium.
🔷 NH₃: weak base.
🔶 CH₃NH₂: +I effect of CH₃ increases basicity.
🟩 (CH₃)₂NH: stronger base due to more +I and effective H-bonding.
🎯 Order: (CH₃)₂NH > CH₃NH₂ > NH₃.
Q28. Write the difference between aliphatic and aromatic amines with examples.
🟦 Aliphatic: nitrogen attached to alkyl group (e.g., CH₃NH₂).
🟩 Aromatic: nitrogen attached to aryl group (e.g., C₆H₅NH₂).
📌 Aromatic amines show resonance → less basic.
✨ Section D (Q29–Q30) – Case-Based (4 marks each)
Q29.
Read the passage and answer:
Gabriel phthalimide synthesis is an important method for preparing pure primary amines. However, aromatic amines cannot be prepared by this method.
(i) Write the reaction of Gabriel phthalimide synthesis.
(ii) Why is it preferred over direct ammonolysis of alkyl halides?
(iii) Why are aryl halides not suitable for this method?
Answer 29
🟦 (i) C₆H₄(CO)₂N⁻K⁺ + R–X → C₆H₄(CO)₂N–R → (hydrolysis) → RNH₂ + phthalic acid.
🟩 (ii) Ammonolysis of alkyl halides → mixture of 1°, 2°, 3° amines. Gabriel synthesis → only pure 1° amine.
🟨 (iii) Aryl halides: resonance → C–X bond has partial double bond character → resistant to nucleophilic substitution.
Q30.
Read the passage and answer:
Aniline is an important aromatic amine used in dye industry. It undergoes electrophilic substitution reactions like bromination and nitration.
(i) Write the reaction of bromination of aniline.
(ii) Write the reaction of nitration of aniline in acidic medium.
(iii) Explain why nitration of aniline gives meta-product predominantly.
Answer 30
🟦 (i) Aniline + Br₂ (aq) → 2,4,6-tribromoaniline (white ppt).
🟩 (ii) Aniline + HNO₃/H₂SO₄ → p-nitroaniline + o-nitroaniline (minor) + m-nitroaniline (major).
🟨 (iii) In acidic medium, –NH₂ group is protonated to –NH₃⁺ (deactivating, meta-directing). Hence, m-product predominates.
✨ Section E (Q31–Q33) – Long Answer (5 marks each, with OR)
Q31.
Describe the mechanism of Hofmann bromamide reaction with example.
OR
Explain the steps involved in diazotisation of aniline and its subsequent reactions with different reagents.
Answer 31
🟦 Hofmann Bromamide Reaction
RCONH₂ + Br₂ + NaOH → N-bromoamide.
Rearrangement → isocyanate intermediate.
Hydrolysis → R–NH₂ (1 carbon less).
✅ Example: CH₃CONH₂ → CH₃NH₂.
🟩 OR – Diazotisation
Aniline + NaNO₂ + HCl (0–5 °C) → C₆H₅N₂⁺Cl⁻.
Reactions:
➤ With H₂O → Phenol.
➤ With CuCl → Chlorobenzene.
➤ With HBF₄ → Fluorobenzene.
➤ With H₃PO₂ → Benzene.
Q32.
(i) Why do amines act as bases?
(ii) Compare basic strength of aliphatic and aromatic amines in aqueous solution.
(iii) Explain order of boiling points of primary, secondary and tertiary amines.
OR
Discuss carbylamine reaction, Hinsberg test and acylation of amines with examples.
Answer 32
🟦 (i) Amines: lone pair on N available → accept protons → behave as bases.
🟩 (ii) Aliphatic amines > NH₃ > Aromatic amines (due to +I effect of alkyl groups and resonance in aromatic amines).
🟨 (iii) Boiling points: 1° > 2° > 3° (due to hydrogen bonding capacity).
🟧 OR
Carbylamine: R–NH₂ + CHCl₃ + alc. KOH → R–NC (foul smell, test for 1° amines).
Hinsberg: 1° amines → soluble sulphonamide, 2° → insoluble, 3° → no reaction.
Acylation: R–NH₂ + CH₃COCl → RNHCOCH₃ (amide).
Q33.
(i) Write the structure and uses of diazonium salts.
(ii) Explain coupling reactions with phenol and aniline.
(iii) Why are diazonium salts unstable above 5 °C?
OR
Explain preparation and uses of aniline.
Answer 33
🟦 (i) Structure: Ar–N₂⁺Cl⁻ (benzene diazonium chloride).
Uses: azo dye synthesis, intermediate in halogenation, cyanation, reduction.
🟩 (ii) Coupling reactions:
C₆H₅N₂⁺Cl⁻ + C₆H₅OH → p-hydroxyazobenzene (dye).
C₆H₅N₂⁺Cl⁻ + C₆H₅NH₂ → p-aminoazobenzene.
🟨 (iii) Diazonium salts unstable > 5 °C because N₂ gas escapes, leaving unstable carbocation.
🟧 OR – Aniline
Preparation: Nitrobenzene + Sn/HCl → Aniline.
Uses: manufacture of dyes, rubber chemicals, drugs, agrochemicals.
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