The Hook: The Gateway to Azo Dyes and Medicines
Azo dyes - the most widely used synthetic dyes in textiles, food, and cosmetics - are all prepared through diazonium salts!
Real-world applications:
- Sunset Yellow (food coloring): Made via diazonium chemistry
- Methyl Orange (pH indicator): Diazonium coupling product
- Prontosil (first antibiotic drug, 1932): Azo compound from diazonium salt
- Congo Red (biological stain): Used in medical diagnostics
The synthetic power: Diazonium salts are like molecular transformers - they allow us to replace the -NH₂ group in aniline with almost any functional group:
$$\text{C}_6\text{H}_5\text{-NH}_2 \xrightarrow{\text{via diazonium}} \text{C}_6\text{H}_5\text{-X}$$Where X = -F, -Cl, -Br, -I, -CN, -OH, -H, etc.
JEE Question: Why are aliphatic diazonium salts unstable but aromatic ones stable?
The Core Concept
What are Diazonium Salts?
Diazonium salts contain the -N₂⁺ (diazonium) group attached to an aryl or alkyl group.
General formula: R-N₂⁺X⁻ or Ar-N₂⁺X⁻
Structure:
$$\text{R-N≡N}^+ \text{X}^-$$Key features:
- Nitrogen-nitrogen triple bond
- Positive charge on terminal nitrogen
- Counter ion (usually Cl⁻, HSO₄⁻, BF₄⁻)
Preparation: Diazotization
The Diazotization Reaction
Best for: Aromatic primary amines ONLY
General Reaction:
$$\boxed{\text{Ar-NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0-5°\text{C}} [\text{Ar-N}_2^+]\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}}$$Conditions:
- Temperature: 0-5°C (ice-cold)
- Acid: HCl or H₂SO₄
- Reagent: Sodium nitrite (NaNO₂)
Mechanism of Diazotization
Step 1: Generation of nitrous acid
$$\text{NaNO}_2 + \text{HCl} \rightarrow \text{HNO}_2 + \text{NaCl}$$Step 2: Formation of nitrosonium ion (electrophile)
$$\text{HNO}_2 + \text{H}^+ \rightarrow \text{H}_2\text{O-N=O}^+ \rightarrow \text{H}_2\text{O} + \text{N≡O}^+$$(Nitrosonium ion: NO⁺)
Step 3: Electrophilic attack on amine
$$\text{C}_6\text{H}_5\text{-NH}_2 + \text{NO}^+ \rightarrow \text{C}_6\text{H}_5\text{-NH-N=O}^+ \text{(N-nitrosoanilinium ion)}$$Step 4: Deprotonation
$$\text{C}_6\text{H}_5\text{-NH-N=O}^+ \xrightarrow{-\text{H}^+} \text{C}_6\text{H}_5\text{-N=N-OH}$$Step 5: Tautomerization
$$\text{C}_6\text{H}_5\text{-N=N-OH} \rightarrow \text{C}_6\text{H}_5\text{-NH-N=O}$$Step 6: Protonation and dehydration
$$\text{C}_6\text{H}_5\text{-N=N-OH} \xrightarrow{+\text{H}^+, -\text{H}_2\text{O}} [\text{C}_6\text{H}_5\text{-N≡N}^+]\text{Cl}^-$$Final product: Benzenediazonium chloride
Temperature must be 0-5°C
If temperature > 10°C:
- Diazonium salt decomposes
- Forms phenol + nitrogen gas $$[\text{C}_6\text{H}_5\text{-N}_2^+]\text{Cl}^- \rightarrow \text{C}_6\text{H}_5\text{-OH} + \text{N}_2↑ + \text{HCl}$$
Why stable at low temperature?
- At 0-5°C, decomposition is slow (kinetic stability)
- Can be used immediately for further reactions
- Storage: Use immediately or keep cold
JEE Note: Diazonium salts are never isolated - used in situ (in the same reaction mixture)
Examples of Diazotization
Example 1: Benzenediazonium chloride
$$\text{C}_6\text{H}_5\text{-NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0-5°\text{C}} [\text{C}_6\text{H}_5\text{-N}_2^+]\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}$$Example 2: p-Toluenediazonium chloride
$$\text{p-CH}_3\text{-C}_6\text{H}_4\text{-NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}, 0-5°\text{C}} [\text{p-CH}_3\text{-C}_6\text{H}_4\text{-N}_2^+]\text{Cl}^-$$Example 3: With H₂SO₄
$$\text{C}_6\text{H}_5\text{-NH}_2 + \text{NaNO}_2 + \text{H}_2\text{SO}_4 \xrightarrow{0-5°\text{C}} [\text{C}_6\text{H}_5\text{-N}_2^+]\text{HSO}_4^-$$Stability of Diazonium Salts
Aromatic Diazonium Salts
Stable at 0-5°C (can be used for further reactions)
Why stable?
- Resonance stabilization with benzene ring
- Electron-withdrawing effect of N₂⁺ delocalized
Resonance:
N≡N⁺ N=N⁺ N=N⁺
| || ||
⟨⟩ ←→ ⟨⟩⁺ ←→ ⟨⟩⁺
Positive charge partially delocalized → more stable
Interactive Demo: Visualize Diazonium Chemistry
See the formation and reactions of diazonium salts step-by-step.
Aliphatic Diazonium Salts
Unstable even at 0°C (decompose immediately)
Why unstable?
- No resonance stabilization (no aromatic ring)
- Positive charge localized on nitrogen
- Extremely unstable
Decomposition:
$$\text{R-N}_2^+\text{Cl}^- \rightarrow \text{R}^+ + \text{N}_2 + \text{Cl}^-$$(Carbocation, very reactive)
Further reaction:
$$\text{R}^+ + \text{H}_2\text{O} \rightarrow \text{R-OH} + \text{H}^+$$Question type: “What happens when ethylamine reacts with NaNO₂/HCl?”
Wrong answer: “Forms ethyldiazonium chloride (stable)”
Correct answer:
- Forms ethyldiazonium chloride (momentarily)
- Immediately decomposes to ethanol + N₂
Key: Only aromatic diazonium salts are stable enough for synthetic use!
Related: Properties of amines
Reactions of Diazonium Salts
graph TD
A[Diazonium Salt] --> B[Replacement by Halogen]
A --> C[Replacement by CN]
A --> D[Replacement by OH]
A --> E[Replacement by H]
A --> F[Coupling Reactions]
B --> B1[Sandmeyer Cl, Br]
B --> B2[Gattermann Cl, Br]
B --> B3[Balz-Schiemann F]
C --> C1[Sandmeyer CN]
D --> D1[Hydrolysis]
E --> E1[Reduction]
F --> F1[Azo Dyes]Type 1: Sandmeyer Reactions
Named after: Swiss chemist Traugott Sandmeyer (1884)
Catalyst: Cu₂X₂ (cuprous salts) or CuX
General Reaction:
$$\boxed{[\text{Ar-N}_2^+]\text{X}^- \xrightarrow{\text{Cu}_2\text{X}_2 \text{ or CuX}} \text{Ar-X} + \text{N}_2↑}$$Sandmeyer Reaction: Chlorination
Reagent: CuCl or Cu₂Cl₂
$$[\text{C}_6\text{H}_5\text{-N}_2^+]\text{Cl}^- \xrightarrow{\text{CuCl}} \text{C}_6\text{H}_5\text{-Cl} + \text{N}_2↑$$Product: Chlorobenzene
Example:
$$[\text{p-CH}_3\text{-C}_6\text{H}_4\text{-N}_2^+]\text{Cl}^- \xrightarrow{\text{CuCl}} \text{p-CH}_3\text{-C}_6\text{H}_4\text{-Cl}$$(p-Chlorotoluene)
Sandmeyer Reaction: Bromination
Reagent: CuBr or Cu₂Br₂
$$[\text{C}_6\text{H}_5\text{-N}_2^+]\text{Cl}^- \xrightarrow{\text{CuBr}} \text{C}_6\text{H}_5\text{-Br} + \text{N}_2↑$$Product: Bromobenzene
Sandmeyer Reaction: Cyanation
Reagent: CuCN (cuprous cyanide)
$$[\text{C}_6\text{H}_5\text{-N}_2^+]\text{Cl}^- \xrightarrow{\text{CuCN}} \text{C}_6\text{H}_5\text{-CN} + \text{N}_2↑$$Product: Benzonitrile
Importance: -CN can be converted to -COOH or -CH₂NH₂
Further reactions:
$$\text{C}_6\text{H}_5\text{-CN} \xrightarrow{\text{H}_3\text{O}^+} \text{C}_6\text{H}_5\text{-COOH}$$(Benzoic acid)
$$\text{C}_6\text{H}_5\text{-CN} \xrightarrow{\text{LiAlH}_4} \text{C}_6\text{H}_5\text{-CH}_2\text{-NH}_2$$(Benzylamine)
“Sandmeyer Substitutes with Copper Catalysts”
Sandmeyer reactions:
- Ar-N₂⁺ → Ar-Cl (CuCl)
- Ar-N₂⁺ → Ar-Br (CuBr)
- Ar-N₂⁺ → Ar-CN (CuCN)
All require cuprous (Cu⁺) salts!
JEE Pattern: Questions often ask “How to convert aniline to chlorobenzene?” Answer: Aniline → Diazonium salt → Sandmeyer (CuCl) → Chlorobenzene
Type 2: Gattermann Reactions
Named after: German chemist Ludwig Gattermann (1890)
Difference from Sandmeyer: Uses Cu powder + HX instead of CuX
General Reaction:
$$\boxed{[\text{Ar-N}_2^+]\text{X}^- \xrightarrow{\text{Cu + HX}} \text{Ar-X} + \text{N}_2↑}$$Gattermann Reaction: Chlorination
Reagent: Cu + HCl
$$[\text{C}_6\text{H}_5\text{-N}_2^+]\text{Cl}^- \xrightarrow{\text{Cu + HCl}} \text{C}_6\text{H}_5\text{-Cl} + \text{N}_2↑$$Product: Chlorobenzene
Gattermann Reaction: Bromination
Reagent: Cu + HBr
$$[\text{C}_6\text{H}_5\text{-N}_2^+]\text{Cl}^- \xrightarrow{\text{Cu + HBr}} \text{C}_6\text{H}_5\text{-Br} + \text{N}_2↑$$Product: Bromobenzene
Comparison:
| Feature | Sandmeyer | Gattermann |
|---|---|---|
| Reagent | CuCl, CuBr, CuCN | Cu + HCl, Cu + HBr |
| Products | Ar-Cl, Ar-Br, Ar-CN | Ar-Cl, Ar-Br (no CN) |
| Condition | Cuprous salt | Copper powder + acid |
| Scope | Wider (includes CN) | Limited (Cl, Br only) |
JEE Note: For -CN, only Sandmeyer works (CuCN)!
Both give same products for Cl/Br, just different catalysts
Type 3: Balz-Schiemann Reaction (Fluorination)
For: Introducing fluorine (F) into aromatic ring
Reagent: Fluoroboric acid (HBF₄)
Steps:
Step 1: Form diazonium fluoroborate
$$[\text{C}_6\text{H}_5\text{-N}_2^+]\text{Cl}^- + \text{HBF}_4 \rightarrow [\text{C}_6\text{H}_5\text{-N}_2^+]\text{BF}_4^- + \text{HCl}$$Step 2: Dry and heat
$$[\text{C}_6\text{H}_5\text{-N}_2^+]\text{BF}_4^- \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{-F} + \text{N}_2↑ + \text{BF}_3$$Product: Fluorobenzene
Special features:
- Diazonium fluoroborate is solid (can be isolated)
- More stable than diazonium chloride
- Heating causes decomposition to fluorobenzene
Q: Why can’t we use CuF for fluorination (like Sandmeyer)?
Answer:
- CuF is unstable and difficult to prepare
- Sandmeyer/Gattermann don’t work for F
- Balz-Schiemann is only practical method for Ar-F
JEE Memory:
- For Cl, Br: Sandmeyer or Gattermann
- For CN: Only Sandmeyer (CuCN)
- For F: Only Balz-Schiemann (HBF₄)
“Fluorine is Fancy - needs Balz-Schiemann!”
Type 4: Replacement by Iodine
Reagent: Potassium iodide (KI)
No catalyst needed!
$$\boxed{[\text{C}_6\text{H}_5\text{-N}_2^+]\text{Cl}^- + \text{KI} \rightarrow \text{C}_6\text{H}_5\text{-I} + \text{N}_2↑ + \text{KCl}}$$Product: Iodobenzene
Why no catalyst?
- I⁻ is a good nucleophile
- Direct substitution occurs
- No need for copper catalysts
Comparison:
| Halogen | Reagent | Method |
|---|---|---|
| F | HBF₄, heat | Balz-Schiemann |
| Cl | CuCl or Cu/HCl | Sandmeyer/Gattermann |
| Br | CuBr or Cu/HBr | Sandmeyer/Gattermann |
| I | KI (no catalyst) | Direct substitution |
Type 5: Replacement by Hydroxyl Group (Hydrolysis)
Reagent: Water (warm) or dilute acid
$$\boxed{[\text{C}_6\text{H}_5\text{-N}_2^+]\text{Cl}^- \xrightarrow{\text{H}_2\text{O, warm}} \text{C}_6\text{H}_5\text{-OH} + \text{N}_2↑ + \text{HCl}}$$Product: Phenol
Conditions:
- Warm the diazonium salt solution
- Or add dilute H₂SO₄ and heat
Example:
$$[\text{p-CH}_3\text{-C}_6\text{H}_4\text{-N}_2^+]\text{Cl}^- \xrightarrow{\text{H}_2\text{O, warm}} \text{p-CH}_3\text{-C}_6\text{H}_4\text{-OH}$$(p-Cresol)
Importance: One of the best methods to prepare phenols!
Alternate pathway:
$$\text{Aniline} \rightarrow \text{Diazonium salt} \rightarrow \text{Phenol}$$Q: How to convert benzene to phenol?
Method 1 (via diazonium):
- Benzene → Nitrobenzene (HNO₃/H₂SO₄)
- Nitrobenzene → Aniline (Sn/HCl)
- Aniline → Diazonium salt (NaNO₂/HCl, 0-5°C)
- Diazonium salt → Phenol (H₂O, warm)
Method 2 (industrial):
- Benzene → Cumene → Phenol + Acetone (Cumene process)
JEE prefers: Diazonium route (shows reaction sequence)
Related: Oxygen compounds
Type 6: Replacement by Hydrogen (Reduction)
Reagent: Hypophosphorous acid (H₃PO₂) or C₂H₅OH/Zn dust
$$\boxed{[\text{C}_6\text{H}_5\text{-N}_2^+]\text{Cl}^- \xrightarrow{\text{H}_3\text{PO}_2} \text{C}_6\text{H}_6 + \text{N}_2↑}$$Product: Benzene
Also works:
$$[\text{C}_6\text{H}_5\text{-N}_2^+]\text{Cl}^- \xrightarrow{\text{C}_2\text{H}_5\text{OH/Zn}} \text{C}_6\text{H}_6 + \text{N}_2↑$$Application: Deamination
This reaction removes the amino group!
Example:
$$\text{Aniline} \rightarrow \text{Diazonium salt} \rightarrow \text{Benzene}$$Uses:
- Remove unwanted -NH₂ after directing a substitution
- Prepare hydrocarbons from amines
Problem: Prepare 1,3-dibromobenzene from benzene
Cannot do directly: Br is ortho/para directing, gives 1,2- and 1,4-dibromobenzene
Strategy: Use -NH₂ as meta-directing group (as -NH₃⁺)
Step 1: Benzene → Nitrobenzene → Aniline
Step 2: Aniline + Br₂ → 2,4,6-Tribromoaniline (ortho/para directing)
Wait, that’s not what we want!
Better strategy:
Use -NH₂ after protonation as -NH₃⁺ (meta-directing):
Actually, this is complex. Better approach:
Step 1: Introduce -NO₂ (meta-directing) Step 2: Brominate → m-bromonitrobenzene Step 3: Reduce NO₂ → m-bromoaniline Step 4: Diazotize and brominate → 3,5-dibromoaniline Step 5: Diazotize and reduce → 1,3-dibromobenzene
Key: Diazonium salt helps remove -NH₂ after using it for directing!
Related: Electrophilic aromatic substitution
Type 7: Coupling Reactions (Azo Compounds)
Most important for dyes!
General Reaction:
$$\boxed{[\text{Ar-N}_2^+]\text{Cl}^- + \text{Ar'-OH or Ar'-NR}_2 \xrightarrow{\text{pH 8-10}} \text{Ar-N=N-Ar'}}$$Product: Azo compound (bright colored dyes)
Reaction partners:
- Phenols (in alkaline medium)
- Aromatic amines (in weakly acidic medium)
Details in: Coupling Reactions
Summary of Replacements
| Group | Reagent | Product | Method |
|---|---|---|---|
| -F | HBF₄, heat | Ar-F | Balz-Schiemann |
| -Cl | CuCl or Cu/HCl | Ar-Cl | Sandmeyer/Gattermann |
| -Br | CuBr or Cu/HBr | Ar-Br | Sandmeyer/Gattermann |
| -I | KI | Ar-I | Direct substitution |
| -CN | CuCN | Ar-CN | Sandmeyer only |
| -OH | H₂O, warm | Ar-OH | Hydrolysis |
| -H | H₃PO₂ or C₂H₅OH/Zn | Ar-H | Reduction |
| -N=N-Ar’ | Ar’-OH or Ar’-NR₂ | Azo dye | Coupling |
“Fancy Colored Brainy Iodine Chemistry Helps Create Beautiful Dyes”
- Fancy → F (Balz-Schiemann)
- Colored → Cl (Sandmeyer/Gattermann)
- Brainy → Br (Sandmeyer/Gattermann)
- Iodine → I (KI)
- Chemistry → CN (Sandmeyer)
- Helps → H (H₃PO₂ reduction)
- Create → Coupling
- Beautiful → (also for Br)
- Dyes → -OH (hydrolysis to phenol)
JEE Strategy: Know all 8 transformations by heart!
Synthetic Importance of Diazonium Salts
Why Diazonium Salts are Important
1. Convert -NH₂ to various groups:
- Cannot directly substitute -NH₂ in aniline
- Must go via diazonium salt
- Allows access to many derivatives
2. Introduce groups that can’t be added directly:
Example: Preparing fluorobenzene
- Cannot do: C₆H₆ + F₂ (too reactive, violent)
- Must use: C₆H₆ → C₆H₅NH₂ → C₆H₅N₂⁺ → C₆H₅F
3. Prepare phenols easily:
- Direct hydroxylation of benzene is difficult
- Via diazonium: C₆H₆ → C₆H₅NH₂ → C₆H₅N₂⁺ → C₆H₅OH
4. Form azo dyes:
- Coupling reactions give colored compounds
- Textile dyes, food colors, indicators
Common Synthetic Transformations
Benzene → Chlorobenzene:
$$\text{C}_6\text{H}_6 \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Sn/HCl}} \text{C}_6\text{H}_5\text{NH}_2$$ $$\xrightarrow{\text{NaNO}_2/\text{HCl}, 0-5°\text{C}} [\text{C}_6\text{H}_5\text{N}_2^+]\text{Cl}^- \xrightarrow{\text{CuCl}} \text{C}_6\text{H}_5\text{Cl}$$Benzene → Phenol:
$$\text{C}_6\text{H}_6 \rightarrow \text{C}_6\text{H}_5\text{NO}_2 \rightarrow \text{C}_6\text{H}_5\text{NH}_2 \rightarrow [\text{C}_6\text{H}_5\text{N}_2^+]\text{Cl}^- \xrightarrow{\text{H}_2\text{O, warm}} \text{C}_6\text{H}_5\text{OH}$$Aniline → Benzonitrile → Benzoic acid:
$$\text{C}_6\text{H}_5\text{NH}_2 \rightarrow [\text{C}_6\text{H}_5\text{N}_2^+]\text{Cl}^- \xrightarrow{\text{CuCN}} \text{C}_6\text{H}_5\text{CN} \xrightarrow{\text{H}_3\text{O}^+} \text{C}_6\text{H}_5\text{COOH}$$Common Mistakes to Avoid
Wrong: “Diazonium salts are stable at room temperature”
Correct: “Aromatic diazonium salts are stable only at 0-5°C”
If temp > 10°C:
$$[\text{Ar-N}_2^+]\text{Cl}^- \rightarrow \text{Ar-OH} + \text{N}_2↑$$JEE Trap: Questions may not mention temperature - always assume 0-5°C for diazotization!
Wrong: “All diazonium salts are stable at 0-5°C”
Correct: “Only AROMATIC diazonium salts are stable”
Aliphatic diazonium salts:
- Decompose immediately even at 0°C
- Cannot be used for further reactions
- Give alcohol + N₂
JEE Note: Diazotization is synthetically useful only for aromatic amines!
Wrong: “Use CuCN for chlorination”
Correct: “Use CuCl for Cl, CuBr for Br, CuCN for CN”
Also wrong: “Use Sandmeyer for fluorination”
Correct: “Fluorination requires Balz-Schiemann (HBF₄)”
Summary:
- Cl, Br, CN → Sandmeyer (CuX)
- F → Balz-Schiemann (HBF₄)
- I → KI (no catalyst)
Practice Problems
Level 1: Foundation (NCERT)
Q: How will you convert aniline to chlorobenzene?
Solution:
Step 1: Diazotization
$$\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0-5°\text{C}} [\text{C}_6\text{H}_5\text{N}_2^+]\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}$$Step 2: Sandmeyer reaction
$$[\text{C}_6\text{H}_5\text{N}_2^+]\text{Cl}^- \xrightarrow{\text{CuCl}} \text{C}_6\text{H}_5\text{Cl} + \text{N}_2↑$$Final product: Chlorobenzene
Alternative for step 2: Gattermann reaction (Cu + HCl)
Q: Write the reaction for conversion of benzenediazonium chloride to phenol.
Solution:
$$[\text{C}_6\text{H}_5\text{N}_2^+]\text{Cl}^- \xrightarrow{\text{H}_2\text{O, warm}} \text{C}_6\text{H}_5\text{OH} + \text{N}_2↑ + \text{HCl}$$Conditions:
- Warm water (or dilute H₂SO₄)
- Temperature > 10°C
Product: Phenol
Observation: Nitrogen gas evolution (effervescence)
JEE Note: This is hydrolysis of diazonium salt
Level 2: JEE Main
Q: What reagent is used to convert benzenediazonium chloride to: (a) Fluorobenzene (b) Benzonitrile (c) Iodobenzene
Solution:
(a) Fluorobenzene:
- Reagent: HBF₄, then heat
- Method: Balz-Schiemann reaction $$[\text{C}_6\text{H}_5\text{N}_2^+]\text{Cl}^- \xrightarrow{\text{HBF}_4} [\text{C}_6\text{H}_5\text{N}_2^+]\text{BF}_4^- \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{F}$$
(b) Benzonitrile:
- Reagent: CuCN
- Method: Sandmeyer reaction $$[\text{C}_6\text{H}_5\text{N}_2^+]\text{Cl}^- \xrightarrow{\text{CuCN}} \text{C}_6\text{H}_5\text{CN} + \text{N}_2↑$$
(c) Iodobenzene:
- Reagent: KI (potassium iodide)
- Method: Direct substitution (no catalyst) $$[\text{C}_6\text{H}_5\text{N}_2^+]\text{Cl}^- \xrightarrow{\text{KI}} \text{C}_6\text{H}_5\text{I} + \text{N}_2↑$$
JEE Memory: F needs fancy method (Balz-Schiemann), I is independent (no catalyst)
Q: How will you prepare benzoic acid from aniline?
Solution:
Retrosynthetic analysis: C₆H₅COOH ← C₆H₅CN ← C₆H₅N₂⁺Cl⁻ ← C₆H₅NH₂
Forward synthesis:
Step 1: Diazotization
$$\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}, 0-5°\text{C}} [\text{C}_6\text{H}_5\text{N}_2^+]\text{Cl}^-$$Step 2: Sandmeyer cyanation
$$[\text{C}_6\text{H}_5\text{N}_2^+]\text{Cl}^- \xrightarrow{\text{CuCN}} \text{C}_6\text{H}_5\text{CN} + \text{N}_2↑$$Step 3: Hydrolysis of nitrile
$$\text{C}_6\text{H}_5\text{CN} \xrightarrow{\text{H}_3\text{O}^+, \Delta} \text{C}_6\text{H}_5\text{COOH}$$Final product: Benzoic acid
Key: Diazonium → CN → COOH (useful chain extension)
Level 3: JEE Advanced
Q: Starting from benzene, how will you prepare: (a) m-Dibromobenzene (b) p-Bromoaniline
Solution:
(a) m-Dibromobenzene:
Strategy: Use -NO₂ as meta-directing group
Step 1: Nitration
$$\text{C}_6\text{H}_6 \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{C}_6\text{H}_5\text{NO}_2$$Step 2: Bromination (meta-directed)
$$\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Br}_2/\text{Fe}} \text{m-BrC}_6\text{H}_4\text{NO}_2$$Step 3: Reduce to amine
$$\text{m-BrC}_6\text{H}_4\text{NO}_2 \xrightarrow{\text{Sn/HCl}} \text{m-BrC}_6\text{H}_4\text{NH}_2$$Step 4: Diazotize and brominate
$$\text{m-BrC}_6\text{H}_4\text{NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}} [\text{m-BrC}_6\text{H}_4\text{N}_2^+]\text{Cl}^-$$ $$\xrightarrow{\text{CuBr}} \text{1,3-dibromobenzene}$$(b) p-Bromoaniline:
Strategy: Protect NH₂, brominate, deprotect
Step 1: Benzene → Aniline (via nitrobenzene)
Step 2: Acetylation (protect)
$$\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{(\text{CH}_3\text{CO})_2\text{O}} \text{C}_6\text{H}_5\text{NHCOCH}_3$$Step 3: Bromination (para-directed)
$$\text{C}_6\text{H}_5\text{NHCOCH}_3 \xrightarrow{\text{Br}_2} \text{p-BrC}_6\text{H}_4\text{NHCOCH}_3$$Step 4: Deprotection (hydrolysis)
$$\text{p-BrC}_6\text{H}_4\text{NHCOCH}_3 \xrightarrow{\text{H}_3\text{O}^+} \text{p-BrC}_6\text{H}_4\text{NH}_2$$Product: p-Bromoaniline
JEE Insight: Different strategies for meta vs para substitution!
Related: Aromatic substitution
Q: Explain why aromatic diazonium salts are more stable than aliphatic diazonium salts.
Solution:
Aromatic diazonium salts (e.g., C₆H₅N₂⁺Cl⁻):
Stability factors:
1. Resonance stabilization:
N≡N⁺ N=N⁺ N=N⁺
| || ||
⟨⟩ ←→ ⟨⟩⁺ ←→ ⟨⟩⁺
Positive charge delocalized into benzene ring → stabilized
2. Electron-withdrawing effect:
- N₂⁺ group withdraws electrons from ring
- Ring stabilizes positive charge through resonance
Aliphatic diazonium salts (e.g., CH₃N₂⁺Cl⁻):
Instability factors:
1. No resonance:
- No aromatic ring to delocalize charge
- Positive charge localized on terminal N
2. Rapid decomposition:
$$\text{R-N}_2^+ \rightarrow \text{R}^+ + \text{N}_2$$Carbocation forms → very unstable
3. Strong driving force:
- Formation of N₂ gas (very stable, triple bond)
- Large entropy increase (gas evolution)
- Thermodynamically favorable
Conclusion:
| Feature | Aromatic | Aliphatic |
|---|---|---|
| Resonance | Yes (stabilizing) | No |
| Stability at 0-5°C | Stable | Unstable |
| Synthetic use | Yes | No |
| Decomposition | Slow (at low temp) | Instantaneous |
JEE Key: Only aromatic diazonium salts are synthetically useful!
Quick Revision Box
| Transformation | Reagent | Method | Product |
|---|---|---|---|
| Ar-NH₂ → Ar-N₂⁺ | NaNO₂/HCl, 0-5°C | Diazotization | Diazonium salt |
| Ar-N₂⁺ → Ar-F | HBF₄, heat | Balz-Schiemann | Aryl fluoride |
| Ar-N₂⁺ → Ar-Cl | CuCl or Cu/HCl | Sandmeyer/Gattermann | Aryl chloride |
| Ar-N₂⁺ → Ar-Br | CuBr or Cu/HBr | Sandmeyer/Gattermann | Aryl bromide |
| Ar-N₂⁺ → Ar-I | KI | Direct substitution | Aryl iodide |
| Ar-N₂⁺ → Ar-CN | CuCN | Sandmeyer | Aryl cyanide |
| Ar-N₂⁺ → Ar-OH | H₂O, warm | Hydrolysis | Phenol |
| Ar-N₂⁺ → Ar-H | H₃PO₂ | Reduction | Arene |
Connection to Other Topics
Prerequisites:
- Preparation of Amines - Getting aniline
- Properties of Amines - Reaction with HNO₂
- Haloalkanes - Substitution reactions
Related Topics:
- Coupling Reactions - Azo dye formation
- Phenols - Preparation via diazonium
- Aromatic Chemistry - Directing effects
Applications:
- Dye synthesis
- Pharmaceutical intermediates
- Aromatic substitution control
Summary
1. Preparation (Diazotization):
Reaction:
$$\text{Ar-NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0-5°\text{C}} [\text{Ar-N}_2^+]\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}$$Critical conditions:
- Temperature: 0-5°C (ice-cold)
- Only for aromatic primary amines
- Never isolate - use immediately
2. Stability:
Aromatic: Stable at 0-5°C (resonance stabilization) Aliphatic: Unstable even at 0°C (no resonance)
3. Replacement Reactions:
| Product | Reagent | Method |
|---|---|---|
| Ar-F | HBF₄, Δ | Balz-Schiemann |
| Ar-Cl | CuCl | Sandmeyer |
| Ar-Br | CuBr | Sandmeyer |
| Ar-I | KI | Direct |
| Ar-CN | CuCN | Sandmeyer |
| Ar-OH | H₂O, warm | Hydrolysis |
| Ar-H | H₃PO₂ | Reduction |
4. Sandmeyer vs Gattermann:
Sandmeyer: CuCl, CuBr, CuCN (wider scope) Gattermann: Cu/HCl, Cu/HBr (Cl, Br only)
5. Synthetic Importance:
- Convert -NH₂ to any group
- Prepare phenols easily
- Introduce halogens (especially F)
- Form azo dyes (coupling)
6. JEE Strategy:
✓ Always mention 0-5°C for diazotization ✓ For F: Use Balz-Schiemann (only method) ✓ For CN: Use Sandmeyer with CuCN (not Gattermann) ✓ For I: KI works directly (no catalyst) ✓ Remember: Aromatic only for stable salts
“Diazonium salts are the Swiss Army knife of organic synthesis - one intermediate, countless products!”
Next, explore coupling reactions to understand how diazonium salts create the vibrant world of azo dyes!