🧪
Exercise 9.1Classification, Preparation and Reactions of Amines
Q1
Classify the following amines as primary, secondary or tertiary: (i) (C₂H₅)₂CHNH₂ (ii) (C₂H₅)₂NH (iii) Ph₃N (iv) C₆H₅NHCH₃ (v) C₂H₅NHC₂H₅
Solution
Primary: –NH₂ (N attached to one C)
Secondary: –NHR (N attached to two C)
Tertiary: –NR₂ (N attached to three C)
(i) (C₂H₅)₂CHNH₂:
Nitrogen is bonded to only one carbon (the CH) via NH₂ → Primary amine
(ii) (C₂H₅)₂NH:
Nitrogen is bonded to two C₂H₅ groups → Secondary amine
(iii) Ph₃N (triphenylamine):
Nitrogen is bonded to three phenyl groups → Tertiary amine
(iv) C₆H₅NHCH₃ (N-methylaniline):
N is bonded to phenyl (C₆H₅) and CH₃ → Secondary amine
(v) C₂H₅NHC₂H₅ (diethylamine):
N is bonded to two ethyl groups → Secondary amine
Q2
Write structural formulae and IUPAC names of all the isomeric amines corresponding to the molecular formula C₃H₉N.
Solution
Molecular formula C₃H₉N — degree of unsaturation = 0 → saturated amines
Primary amines (–NH₂):
1. CH₃CH₂CH₂NH₂ — propan-1-amine (n-propylamine)
2. (CH₃)₂CHNH₂ — propan-2-amine (isopropylamine)
Secondary amines (–NH–):
3. CH₃NHCH₂CH₃ — N-methylethanamine (methylethylamine)
Tertiary amines (–N<):
4. (CH₃)₃N — N,N-dimethylmethanamine (trimethylamine)
Total: 4 isomers
Q3
How will you convert benzene to aniline?
Solution
Benzene → Aniline (C₆H₅NH₂):
Step 1 — Nitration:
C₆H₆ + HNO₃ → (conc. H₂SO₄, 55°C) → C₆H₅NO₂ + H₂O
(Benzene → Nitrobenzene)
Step 2 — Reduction:
C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O
Reducing agents used:
• Fe/HCl (Bechamp reduction) — most common
• Sn/conc. HCl
• H₂/Pd or Ni (catalytic hydrogenation)
Or in acidic medium:
C₆H₅NO₂ + 3Fe + 6HCl → C₆H₅NH₂ + 3FeCl₂ + 2H₂O
Final product: Aniline (phenylamine)
Q4
How will you convert aniline to phenol?
Solution
Aniline → Phenol via diazonium salt:
Step 1 — Diazotisation:
C₆H₅NH₂ + NaNO₂ + HCl → (0–5°C) → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
(Aniline → Benzene diazonium chloride)
Must be done at 0–5°C to prevent decomposition of diazonium salt.
Step 2 — Hydrolysis:
C₆H₅N₂⁺Cl⁻ + H₂O → (heat) → C₆H₅OH + N₂↑ + HCl
(Benzene diazonium chloride → Phenol)
Or using dilute H₂SO₄ and warming:
C₆H₅N₂⁺HSO₄⁻ + H₂O → C₆H₅OH + N₂↑ + H₂SO₄
Q5
Give the IUPAC name of the product formed when aniline reacts with: (i) Acetic anhydride (ii) Bromine water
Solution
(i) Aniline + Acetic anhydride [(CH₃CO)₂O]:
C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH
Product: N-phenylacetamide (acetanilide)
IUPAC name: N-phenylacetamide
(Acylation/acetylation reaction)
(ii) Aniline + Bromine water (Br₂/H₂O):
The amino group activates the ring strongly toward electrophilic substitution.
All three positions (2,4,6) are substituted:
C₆H₅NH₂ + 3Br₂ → 2,4,6-tribromoaniline + 3HBr
Product: 2,4,6-tribromoaniline
IUPAC name: 2,4,6-tribromoaniline
Characteristic white precipitate forms immediately.
Q6
Arrange the following in order of decreasing pKb values (increasing base strength): aniline, methylamine, dimethylamine.
Solution
Base strength (stronger base → lower pKb):
Aliphatic amines are more basic than aniline because:
• In aniline, the lone pair on N is delocalised into the benzene ring (resonance)
• This reduces availability of lone pair for accepting H⁺
• In alkylamines, the lone pair is freely available
Among dimethylamine vs methylamine:
• In aqueous solution: dimethylamine is more basic than methylamine
(inductive effect of two CH₃ > one CH₃, and solvation)
Actual pKb values:
Dimethylamine: pKb ≈ 3.27 (strongest base, lowest pKb)
Methylamine: pKb ≈ 3.36
Aniline: pKb ≈ 9.40 (weakest base, highest pKb)
Decreasing pKb order (i.e., increasing base strength):
Aniline > Methylamine > Dimethylamine
So base strength: Dimethylamine > Methylamine > Aniline
Q7
How are diazonium salts prepared from aromatic primary amines? Write the equation for the reaction of benzene diazonium chloride with (i) HBF₄ (ii) CuBr (iii) CuCN (iv) H₃PO₂
Solution
Preparation of diazonium salts (diazotisation):
ArAr-NH₂ + NaNO₂ + 2HCl → (0–5°C) → ArN₂⁺Cl⁻ + NaCl + 2H₂O
For aniline:
C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
(0–5°C; above 5°C the diazonium salt decomposes)
Reactions of benzene diazonium chloride:
(i) With HBF₄ (Balz-Schiemann reaction):
C₆H₅N₂⁺Cl⁻ + HBF₄ → C₆H₅N₂⁺BF₄⁻ → (heat) → C₆H₅F + N₂ + BF₃
Product: Fluorobenzene
(ii) With CuBr (Sandmeyer reaction):
C₆H₅N₂⁺Cl⁻ + CuBr → C₆H₅Br + N₂ + CuCl
Product: Bromobenzene
(iii) With CuCN (Sandmeyer reaction):
C₆H₅N₂⁺Cl⁻ + CuCN → C₆H₅CN + N₂ + CuCl
Product: Benzonitrile (phenyl cyanide)
(iv) With H₃PO₂ (hypophosphorous acid):
C₆H₅N₂⁺Cl⁻ + H₃PO₂ + H₂O → C₆H₆ + N₂ + HCl + H₃PO₃
Product: Benzene (replacement of –N₂⁺ by –H)
Q8
Write a short note on Gabriel synthesis.
Solution
Gabriel synthesis is a method for preparing pure primary aliphatic amines.
Starting material: Phthalimide (cyclic imide from phthalic anhydride + NH₃)
Steps:
Step 1 — Deprotonation of phthalimide:
Phthalimide + KOH → Potassium phthalimide (K⁺[phthalimide]⁻)
The N–H is acidic due to electron-withdrawing carbonyl groups.
Step 2 — N-alkylation:
K-phthalimide + R–X → N-alkylphthalimide + KX
(The nitrogen is alkylated by alkyl halide via SN2)
Step 3 — Hydrolysis:
N-alkylphthalimide + H₂O/NaOH (or hydrazine) → R–NH₂ + phthalic acid (or phthalhydrazide)
Advantages:
• Gives only primary amines (no secondary or tertiary)
• Clean, predictable yield
Limitation:
• Only works for primary amines
• Cannot use aryl or vinyl halides (poor SN2 reactivity)
The textbook is solved. The exam won't ask the textbook.
Board papers twist Amines into exam-pattern questions the exercises never show. Practice unlimited chapter-wise tests with step-wise solutions — verify your WhatsApp number and your free trial starts right here.
CBSE · CLASS 12
🎟️14-day free trial · code PRAC auto-applied
Mobile Number
+91
OTP will be sent to your WhatsApp
⭐ Trusted by 12,000+ students across India