Chapter 13 — Organic Nitrogen Compounds Exercise Solutions
This chapter requires you to know basicity order, tests, and diazonium salt reactions thoroughly.
Key Concepts Quick Recap
| Term | Definition |
|---|
| Primary amine (1°) | One alkyl/aryl group on N: RNH₂ |
| Secondary amine (2°) | Two groups on N: R₂NH |
| Tertiary amine (3°) | Three groups on N: R₃N |
| Basicity | Due to lone pair on N; ability to accept a proton |
| Carbylamine reaction | Test for 1° amines: RNH₂ + CHCl₃ + 3KOH → RNC (isocyanide, foul smell) |
| Diazotisation | ArNH₂ + NaNO₂ + HCl (0–5 °C) → ArN₂⁺Cl⁻ |
| Coupling reaction | Diazonium salt + phenol/amine → azo dye (orange/red) |
| Sandmeyer's reaction | Replace N₂⁺ with Cl, Br, CN using CuCl, CuBr, CuCN |
| Balz-Schiemann reaction | ArN₂⁺BF₄⁻ →(heat) ArF + N₂ + BF₃ (introduces F) |
Basicity Comparison
| Type | Order in Water |
|---|
| Aliphatic amines | 2° > 1° > 3° > NH₃ |
| Aromatic amines | Aniline << all aliphatic amines |
| Reason for 2° > 1° | More alkyl groups donate electrons; more +I effect |
| Reason for 3° < 2° | Steric effect and solvation: 3° amine cation poorly solvated |
| Reason for aniline << aliphatic | Lone pair on N delocalised into ring → less available for protonation |
I. Choose the Best Answer (MCQs)
1. The carbylamine reaction is used to identify: (a) Secondary amines (b) Tertiary amines (c) Primary amines (d) All amines
Answer(c) Primary amines Primary amines react with CHCl₃/KOH to form isocyanides (carbylamines) with a very foul smell. Secondary and tertiary amines do NOT give this test.
2. Diazotisation must be carried out at: (a) 100 °C (b) 50 °C (c) Room temperature (d) 0–5 °C
Answer(d) 0–5 °C Diazonium salts are unstable and decompose rapidly above 5 °C. Ice-cold conditions are necessary to maintain the salt long enough for use.
3. The basicity of amines in water follows the order (aliphatic): (a) 1° > 2° > 3° (b) 3° > 2° > 1° (c) 2° > 1° > 3° (d) All equal
Answer(c) 2° > 1° > 3° In water, solvation of the ammonium ion is important. 3° amines form poorly solvated cations (less H for H-bonding) → lower basicity than expected.
4. The coupling reaction of diazonium salt with β-naphthol produces: (a) A yellow azo dye (b) An orange-red azo dye (c) A green azo dye (d) No colour
Answer(b) An orange-red azo dye ArN₂⁺ + β-naphthol (in alkaline solution) → orange-red azo compound with the −N=N− (azo) linkage.
5. Which of the following introduces a −CN group into the benzene ring via diazonium salt? (a) Gattermann's reaction (b) Balz-Schiemann reaction (c) Sandmeyer's reaction with CuCN (d) Wurtz reaction
Answer(c) Sandmeyer's reaction with CuCN ArN₂⁺ + CuCN → ArCN + N₂. Similarly: CuCl gives ArCl, CuBr gives ArBr.
6. Why is aniline less basic than methylamine? (a) N−H bond in aniline is stronger (b) Lone pair on N in aniline is delocalised into benzene ring (c) Aniline has more electrons (d) Benzene ring donates electrons to N
Answer(b) Lone pair delocalised into benzene ring In aniline, the lone pair of N is in conjugation with the π electrons of the benzene ring (resonance). This delocalisation makes the lone pair less available for protonation → weaker base.
7. Hinsberg's test distinguishes primary, secondary, and tertiary amines using: (a) CHCl₃/KOH (b) Benzenesulphonyl chloride (c) HNO₂ (d) CuCN
Answer(b) Benzenesulphonyl chloride (C₆H₅SO₂Cl)
- 1° amines react to give sulphonamide soluble in NaOH.
- 2° amines react to give sulphonamide insoluble in NaOH.
- 3° amines do not react.
II. Short Answer Questions
Q1. Explain the carbylamine reaction with an equation. What class of amines give this test?
Answer
The carbylamine (isocyanide) test is used to identify primary amines (1°).
Reaction: Primary amine + chloroform + alcoholic KOH → isocyanide (foul smell)
C₂H₅NH₂ + CHCl₃ + 3KOH →(heat) C₂H₅NC + 3KCl + 3H₂O
(ethylamine) (ethyl isocyanide — very foul smell)
Also: C₆H₅NH₂ + CHCl₃ + 3KOH → C₆H₅NC + 3KCl + 3H₂O
Secondary and tertiary amines do NOT give this reaction.
Q2. Why does aniline not undergo Friedel-Crafts reactions?
Answer
Aniline (C₆H₅NH₂) reacts with the Lewis acid catalyst (AlCl₃) used in Friedel-Crafts reactions:
C₆H₅NH₂ + AlCl₃ → C₆H₅NH₂·AlCl₃ (salt complex)
The lone pair on N coordinates to AlCl₃, forming a complex that deactivates the amine's ring-activating effect. The nitrogen's lone pair is no longer available to donate to the ring → electrophilic aromatic substitution becomes very difficult.
Q3. Write the reactions of diazonium salts to replace N₂⁺ with various groups (Sandmeyer's and related reactions).
Starting material: C₆H₅N₂⁺Cl⁻ (benzenediazonium chloride)
Replace N₂⁺ with Cl: + CuCl → C₆H₅Cl + N₂ (Sandmeyer's)
Replace N₂⁺ with Br: + CuBr → C₆H₅Br + N₂ (Sandmeyer's)
Replace N₂⁺ with CN: + CuCN → C₆H₅CN + N₂ (Sandmeyer's)
Replace N₂⁺ with F: + HBF₄ → ArN₂⁺BF₄⁻ →(heat) ArF + N₂ + BF₃ (Balz-Schiemann)
Replace N₂⁺ with OH: + H₂O(heat) → C₆H₅OH + N₂ + HCl
Replace N₂⁺ with NO₂: + CuNO₂ → C₆H₅NO₂ + N₂
Replace N₂⁺ with I: + KI → C₆H₅I + N₂ (no Cu needed for I)
Replace N₂⁺ with H: + H₃PO₂(hypophosphorous acid) → C₆H₆ + N₂ (deamination)
Q4. Explain the synthesis of an azo dye by the coupling reaction.
Answer
Coupling reaction = diazonium salt + an activated aromatic compound (phenol or amine) in weakly alkaline/acidic medium.
Step 1 — Diazotisation:
C₆H₅NH₂ + NaNO₂ + HCl →(0–5 °C) C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
Step 2 — Coupling:
C₆H₅N₂⁺Cl⁻ + C₆H₅OH(alkaline) → C₆H₅−N=N−C₆H₄−OH + HCl
(para-hydroxyazobenzene — orange dye)
Azo dyes have the −N=N− chromophore. They are used as fabric dyes, food colorants, and indicators.
Worked Examples
Example 1 — IUPAC names of amines
CH₃NH₂ → methanamine (methylamine) — 1°
(CH₃)₂NH → N-methylmethanamine (dimethylamine) — 2°
(CH₃)₃N → N,N-dimethylmethanamine (trimethylamine) — 3°
C₆H₅NH₂ → aniline (benzenamine) — aromatic 1°
C₆H₅NHCH₃ → N-methylaniline — aromatic 2°
Example 2 — Basicity comparison with reasoning
Compare: (CH₃)₃N vs (CH₃)₂NH vs CH₃NH₂ vs NH₃ vs C₆H₅NH₂
In gas phase (no solvation): 3° > 2° > 1° > NH₃ (purely based on +I induction)
In water (solvation matters):
2° > 1° > 3° > NH₃ > C₆H₅NH₂
pKb values (approx.):
CH₃NH₂ 3.38
(CH₃)₂NH 3.27 (most basic)
(CH₃)₃N 4.19 (less basic due to poor solvation)
C₆H₅NH₂ 9.37 (very weak — N lone pair delocalised into ring)
Example 3 — Identifying amine class using Hinsberg's test
Hinsberg's reagent: C₆H₅SO₂Cl (benzenesulphonyl chloride)
With primary amine:
RNH₂ + C₆H₅SO₂Cl → RNH−SO₂C₆H₅ (sulphonamide soluble in NaOH) ✓
With secondary amine:
R₂NH + C₆H₅SO₂Cl → R₂N−SO₂C₆H₅ (insoluble in NaOH) ✓
With tertiary amine:
R₃N + C₆H₅SO₂Cl → No reaction (no N−H to react) ✓
Result: If soluble in NaOH after reaction → 1°
If insoluble → 2°; If no reaction → 3°
Example 4 — Diazotisation step-by-step
C₆H₅NH₂ + NaNO₂ + 2HCl →(0–5 °C) C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
Steps:
1. Dissolve aniline in HCl.
2. Cool to 0–5 °C using ice bath.
3. Add NaNO₂ solution dropwise.
4. Test with starch-iodide paper (turns blue → confirms diazonium salt formed).
5. Use diazonium solution immediately or store at low temperature.
Conversion Pattern
For diazonium conversions: 1. Diazotise the primary aromatic amine (NaNO₂ + HCl, 0–5 °C). 2. Choose the replacement reagent for the desired group. 3. Write the product and N₂ as a by-product.
Example: Aniline → iodobenzene
C₆H₅NH₂ →(NaNO₂/HCl, 0°C) C₆H₅N₂⁺Cl⁻ →(KI) C₆H₅I + N₂ + KCl
Common Mistakes
- Forgetting that diazotisation requires 0–5 °C — the salt decomposes above this.
- Wrong basicity order: in water, 2° > 1° > 3° (not 3° > 2° > 1°).
- Applying carbylamine test to secondary amines — only 1° amines give it.
- Missing N₂ gas as a by-product in diazonium replacement reactions.
- Not explaining why aniline is less basic (must mention resonance/delocalisation).
Exam Checklist
- Amine class (1°, 2°, 3°) is correctly identified.
- Basicity order is correct for the medium (water vs gas phase).
- Diazotisation conditions (0–5 °C, NaNO₂, HCl) are clearly stated.
- Replacement reagent and product are correct.
- Azo dye structure shows the −N=N− linkage.
