Azo Coupling Reactions and Dye Chemistry

Complete guide to coupling reactions, azo dye formation, mechanism, synthetic applications, and textile chemistry for JEE

The Hook: The Colorful World of Azo Dyes

Connect: Real Life → Chemistry

Look around you right now - your clothes, food packaging, cosmetics, even your pen ink likely contain azo dyes!

Mind-blowing statistics:

  • 60-70% of all synthetic dyes are azo compounds
  • Methyl Orange (pH indicator): Azo dye you use in chemistry lab
  • Sunset Yellow (E110): Food coloring in soft drinks, candies
  • Tartrazine (E102): Yellow food dye in chips, beverages
  • Congo Red: Biological stain for detecting amyloid proteins

Historical importance:

  • 1856: First synthetic dye (Mauveine) discovered by William Perkin
  • 1884: Paul Griess discovered azo coupling
  • Revolutionary: Freed textile industry from natural dyes (expensive, rare)

The chemistry: All azo dyes contain the -N=N- (azo) linkage connecting aromatic rings, formed through coupling reactions of diazonium salts!

JEE Question: Why do azo compounds show intense colors? How does pH affect their color?


The Core Concept

What is Azo Coupling?

Azo coupling is an electrophilic aromatic substitution reaction where a diazonium salt (electrophile) attacks an activated aromatic compound (nucleophile).

General Reaction:

$$\boxed{[\text{Ar-N}_2^+]\text{X}^- + \text{Ar'-H} \rightarrow \text{Ar-N=N-Ar'} + \text{HX}}$$

Product: Azo compound (contains -N=N- group)

Key requirement: Ar’-H must be highly activated (electron-rich)

JEE Weightage
Coupling Reactions: 2-3 questions in JEE Main, 1-2 in JEE Advanced Focus areas: Mechanism, conditions, dye structures, pH dependence, color chemistry

Coupling Partners

What Couples with Diazonium Salts?

Requirement: Aromatic ring must be highly activated (electron-rich)

Suitable coupling partners:

1. Phenols and Naphthols

Why reactive?

  • -OH group is strongly activating (+R effect)
  • Increases electron density on ring (especially ortho/para)
  • Ring is electron-rich enough for electrophilic attack

Common partners:

  • Phenol (C₆H₅OH)
  • β-Naphthol (2-naphthol)
  • α-Naphthol (1-naphthol)
  • Substituted phenols

Conditions: Alkaline medium (pH 8-10)

Why alkaline?

  • Phenol exists as phenoxide ion (C₆H₅O⁻)
  • Phenoxide is MORE activated than phenol
  • Better nucleophile
$$\text{C}_6\text{H}_5\text{OH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{O}^- + \text{H}_2\text{O}$$

2. Aromatic Amines

Why reactive?

  • -NH₂ or -NR₂ groups are strongly activating (+R effect)
  • Lone pair delocalizes into ring
  • Increases electron density

Common partners:

  • Aniline (C₆H₅NH₂)
  • N,N-Dimethylaniline (C₆H₅N(CH₃)₂)
  • N-Methylaniline
  • Naphthylamines

Conditions: Weakly acidic medium (pH 4-5)

Why acidic?

  • In strong acid: -NH₂ becomes -NH₃⁺ (deactivating)
  • In strong base: Diazonium salt decomposes
  • Weak acid: Balance between activation and stability
pH Control is Critical!

For phenols: pH 8-10 (alkaline)

  • Forms phenoxide ion (C₆H₅O⁻)
  • More activated, better coupling

For amines: pH 4-5 (weakly acidic)

  • Keeps -NH₂ free (not protonated)
  • Maintains diazonium salt stability

Wrong pH consequences:

Phenol in acidic medium:

  • No phenoxide ion
  • Less activated
  • Poor/no coupling

Amine in alkaline medium:

  • Diazonium salt decomposes to phenol
  • No coupling

JEE Memory: “Phenols are Proud (alkaline pH)” “Amines are Acidic (acidic pH)”

Interactive Demo: Visualize Coupling Reactions

See how diazonium salts couple with phenols and amines to form azo dyes.


Mechanism of Azo Coupling

Step-by-Step Mechanism

Example: Benzenediazonium chloride + Phenol (in NaOH)

Step 1: Formation of phenoxide ion

$$\text{C}_6\text{H}_5\text{OH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{O}^-\text{Na}^+ + \text{H}_2\text{O}$$

Step 2: Electrophilic attack by diazonium ion

         N≡N⁺
          |
          ⊕  ←  attacks para position
         / \
  HO-⟨⟩-O⁻

Diazonium cation (N₂⁺) acts as electrophile, attacks electron-rich ortho/para position

Step 3: Formation of σ-complex (arenium ion)

     O⁻           N≡N⁺
      |            |
      ⟨⟩⁺  ←  sigma complex
      |
      H

Step 4: Deprotonation (rearomatization)

     O⁻               O⁻
      |                |
      ⟨⟩-N=N-⟨⟩  +  H⁺

Final product: p-Hydroxyazobenzene (yellow-orange dye)

$$\text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_4\text{-OH}$$
Regioselectivity: Ortho vs Para

Q: Why does coupling occur predominantly at para position?

Answer:

Para position is favored:

  1. Steric factors: Less crowded than ortho
  2. Electronic factors: Both ortho and para are activated, but para is more accessible

For phenoxide ion:

  • Major product: p-Hydroxyazobenzene (para coupling)
  • Minor product: o-Hydroxyazobenzene (ortho coupling, some steric hindrance)

For naphthols:

  • β-Naphthol: Couples at α-position (position 1, adjacent to OH)
  • More complex regioselectivity

JEE Note: Unless specified, assume para coupling for phenol derivatives


Important Azo Dyes and Their Preparation

1. Methyl Orange (Acid-Base Indicator)

Structure:

(CH₃)₂N-⟨⟩-N=N-⟨⟩-SO₃H

Full name: 4-Dimethylaminoazobenzene-4’-sulfonic acid

Preparation:

Step 1: Diazotization of sulfanilic acid

$$\text{H}_2\text{N-C}_6\text{H}_4\text{-SO}_3\text{H} \xrightarrow{\text{NaNO}_2/\text{HCl}, 0-5°\text{C}} [\text{⁺N}_2\text{-C}_6\text{H}_4\text{-SO}_3^-]$$

(Sulfanilic acid exists as zwitterion)

Step 2: Coupling with N,N-dimethylaniline

$$[\text{⁺N}_2\text{-C}_6\text{H}_4\text{-SO}_3^-] + (\text{CH}_3)_2\text{N-C}_6\text{H}_5 \xrightarrow{\text{pH 4-5}}$$ $$(\text{CH}_3)_2\text{N-C}_6\text{H}_4\text{-N=N-C}_6\text{H}_4\text{-SO}_3\text{H}$$

Properties:

  • In acid (pH < 3.1): Red (protonated form)
  • In base (pH > 4.4): Yellow (deprotonated form)
  • Transition: pH 3.1-4.4 (orange)

Use: Acid-base indicator for titrations

Color change:

$$\text{Red (acidic)} \rightleftarrows \text{Orange} \rightleftarrows \text{Yellow (basic)}$$
Why Does Methyl Orange Change Color?

In acidic medium (pH < 3.1):

The -N=N- group gets protonated (or -N(CH₃)₂ gets protonated):

$$(\text{CH}_3)_2\text{N-C}_6\text{H}_4\text{-N=N-C}_6\text{H}_4\text{-SO}_3\text{H} \xrightarrow{+\text{H}^+}$$ $$(\text{CH}_3)_2\text{NH}^+-\text{C}_6\text{H}_4\text{-N=N-C}_6\text{H}_4\text{-SO}_3\text{H}$$

Protonated form: RED (extended conjugation, lower energy gap)

In basic medium (pH > 4.4):

Unprotonated form:

$$(\text{CH}_3)_2\text{N-C}_6\text{H}_4\text{-N=N-C}_6\text{H}_4\text{-SO}_3^-$$

Deprotonated form: YELLOW (different conjugation, higher energy gap)

JEE Key: Color depends on extent of conjugation and electron distribution!

Related: Acid-base indicators

2. Orange II (Food Dye, E111)

Structure:

HO-⟨⟩-N=N-⟨⟩-SO₃Na

Preparation:

Step 1: Diazotization

$$\text{H}_2\text{N-C}_6\text{H}_4\text{-SO}_3\text{Na} \xrightarrow{\text{NaNO}_2/\text{HCl}} [\text{⁺N}_2\text{-C}_6\text{H}_4\text{-SO}_3\text{Na}]$$

Step 2: Coupling with β-naphthol

$$[\text{⁺N}_2\text{-C}_6\text{H}_4\text{-SO}_3\text{Na}] + \text{β-naphthol} \xrightarrow{\text{NaOH}}$$ $$\text{HO-naphthyl-N=N-C}_6\text{H}_4\text{-SO}_3\text{Na}$$

Color: Orange-red

Use: Textile dye, food coloring (now banned in many countries)

3. Congo Red (Biological Stain)

Structure: Contains two azo groups (-N=N-)

NH₂-⟨⟩-N=N-⟨naphthyl⟩-N=N-⟨⟩-NH₂

(with sulfonic acid groups)

Preparation:

  • Requires benzidine (diamine)
  • Two-step diazotization and coupling
  • Complex synthesis

Properties:

  • In acid: Blue-violet
  • In base: Red
  • Transition pH: 3.0-5.0

Uses:

  • pH indicator
  • Biological stain for amyloid proteins
  • Detects cellulose fibers

4. Aniline Yellow (Para-aminoazobenzene)

Structure:

H₂N-⟨⟩-N=N-⟨⟩

Preparation:

Step 1: Diazotization of aniline

$$\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}} [\text{C}_6\text{H}_5\text{N}_2^+]\text{Cl}^-$$

Step 2: Coupling with aniline (in weakly acidic medium)

$$[\text{C}_6\text{H}_5\text{N}_2^+]\text{Cl}^- + \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{pH 4-5}} \text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_4\text{-NH}_2$$

Color: Bright yellow

Note: This is an amine-amine coupling (both rings have NH₂)


Why Are Azo Compounds Colored?

The Science of Color

Color in azo compounds comes from:

1. Extended Conjugation

Azo group (-N=N-) extends conjugation:

⟨⟩-N=N-⟨⟩

Electrons delocalized over:

  • Benzene ring 1
  • Azo group (N=N)
  • Benzene ring 2

Extended π-system allows lower energy transitions

2. Chromophore and Auxochrome

Chromophore: The azo group (-N=N-)

  • Responsible for light absorption
  • Creates basic color

Auxochrome: Substituents like -OH, -NH₂, -SO₃H

  • Modify/intensify color
  • Shift absorption wavelength

Example:

  • Azobenzene (C₆H₅-N=N-C₆H₅): Yellow (λmax ≈ 440 nm)
  • p-Hydroxyazobenzene: Orange-red (λmax shifted, deeper color)
  • p-Aminoazobenzene: Yellow-orange (λmax modified)

3. Light Absorption

Visible light range: 400-700 nm

Azo compounds absorb:

  • π → π* transitions
  • n → π* transitions (from N lone pairs)

Wavelength absorbed determines color seen:

Color AbsorbedColor Seen (Complementary)
Violet (400nm)Yellow-green
Blue (480nm)Yellow
Green (530nm)Red
Yellow (580nm)Blue
Orange (610nm)Blue-green
Red (680nm)Green

Most azo dyes absorb in blue-green region → appear red/orange/yellow

JEE Concept: Structure-Color Relationship

Q: How does adding -OH group to azobenzene change its color?

Answer:

Azobenzene: C₆H₅-N=N-C₆H₅ (Yellow, λmax ≈ 440 nm)

p-Hydroxyazobenzene: HO-C₆H₄-N=N-C₆H₅ (Orange-red, λmax ≈ 480 nm)

Reason:

1. Extended conjugation:

  • -OH group has +R effect
  • Lone pair on oxygen delocalizes into ring
  • Further extends π-conjugation
  • Lower energy gap (HOMO-LUMO)

2. Bathochromic shift (red shift):

  • Absorption shifts to longer wavelength
  • Blue light absorbed instead of violet
  • Complementary color changes: yellow → orange/red

3. Intensified color:

  • -OH is auxochrome
  • Increases molar absorptivity (ε)
  • Deeper, more intense color

JEE Key: More conjugation = Longer wavelength absorption = Deeper color

Related: Conjugation and aromaticity


Conditions for Successful Coupling

Essential Requirements

1. Activated aromatic ring (coupling partner)

  • Must have strong EDG (-OH, -NH₂, -NR₂)
  • Electron density high enough for electrophilic attack

2. Correct pH

  • Phenols: pH 8-10 (alkaline)
  • Amines: pH 4-5 (weakly acidic)

3. Temperature

  • Usually 0-5°C (like diazonium salt formation)
  • Low temperature maintains diazonium salt stability

4. Freshly prepared diazonium salt

  • Never store
  • Use immediately after preparation
  • Decomposes on warming

What Does NOT Couple?

❌ Benzene: Not activated enough

  • No EDG
  • Too low electron density
  • No coupling occurs

❌ Nitrobenzene: Deactivated ring

  • -NO₂ is EWG
  • Decreases electron density
  • No coupling

❌ Chlorobenzene: Weakly deactivated

  • -Cl has weak -I effect
  • Not activated enough

❌ Toluene: Only weakly activated

  • -CH₃ has weak +I effect
  • Usually insufficient for coupling

✓ Must have strong EDG: -OH, -OR, -NH₂, -NHR, -NR₂


Test for Azo Dye Formation

Laboratory Test

Procedure:

Step 1: Prepare diazonium salt

$$\text{Aniline} + \text{NaNO}_2 + \text{HCl} \xrightarrow{0-5°\text{C}} \text{Benzenediazonium chloride}$$

Step 2: Add coupling partner

  • For phenol: Add NaOH, then add phenol
  • For amine: Adjust pH to 4-5, add amine

Step 3: Observe

  • Immediate appearance of bright color (orange, red, yellow)
  • Indicates azo dye formation

Example:

Benzenediazonium chloride + β-naphthol (in NaOH):

Observation: Brilliant orange-red color appears immediately

Product: 1-(Phenylazo)-2-naphthol (Orange II precursor)

JEE Lab Question Pattern

Typical question: “A colorless solution A is treated with NaNO₂ and HCl at 0-5°C. The resulting solution is added to an alkaline solution of β-naphthol, and a brilliant orange color appears. Identify A.”

Solution:

Analysis:

  1. Forms diazonium salt (NaNO₂/HCl, 0-5°C) → Must be aromatic primary amine
  2. Couples with β-naphthol in alkaline medium → Confirms diazonium salt
  3. Orange color appears → Azo dye formed

Compound A: Aniline (C₆H₅NH₂) or substituted aniline

Reactions:

$$\text{A} \xrightarrow{\text{NaNO}_2/\text{HCl}} [\text{Ar-N}_2^+]\text{Cl}^-$$ $$[\text{Ar-N}_2^+]\text{Cl}^- + \text{β-naphthol} \xrightarrow{\text{NaOH}} \text{Orange azo dye}$$

This is a classic identification test for aromatic primary amines!

Related: Properties of amines


Applications of Azo Compounds

1. Textile Dyes

Direct dyes:

  • Azo compounds with -SO₃H groups
  • Water-soluble
  • Directly dye cotton, wool

Examples:

  • Congo Red
  • Direct Blue
  • Chrysophenine

Process:

  1. Dissolve dye in water
  2. Immerse fabric
  3. Dye molecules adsorb onto fibers
  4. Fix with salt or heat

2. Food Colorings

Common food azo dyes:

DyeE NumberColorUse
TartrazineE102YellowSoft drinks, chips
Sunset YellowE110OrangeDesserts, candies
Allura RedE129RedBeverages, sweets
AmaranthE123Red(Banned in USA)

Health concerns:

  • Some linked to hyperactivity in children
  • Regulations vary by country
  • Many being replaced with natural dyes

3. pH Indicators

Azo dyes that change color with pH:

Methyl Orange:

  • Red (pH < 3.1) ↔ Yellow (pH > 4.4)
  • Used in acid-base titrations

Methyl Red:

  • Red (pH < 4.4) ↔ Yellow (pH > 6.2)
  • More sensitive range than methyl orange

Congo Red:

  • Blue (pH < 3.0) ↔ Red (pH > 5.0)
  • Used in biological staining

4. Biological Stains

Congo Red:

  • Stains amyloid proteins (disease marker)
  • Used in Alzheimer’s research

Sudan dyes:

  • Stain lipids (fats)
  • Used in histology

5. Industrial Applications

Paper industry:

  • Colored papers, cardboard

Leather industry:

  • Dye leather goods

Cosmetics:

  • Hair dyes
  • Makeup products

Printing inks:

  • Newspaper, magazines

Common Mistakes to Avoid

Mistake #1: Wrong pH for Coupling

Wrong: “Couple phenol with diazonium salt in acidic medium”

Correct: “Phenol requires ALKALINE medium (pH 8-10)”

Reason:

  • Phenol (C₆H₅OH) is weak acid (pKa ≈ 10)
  • In base: Forms phenoxide (C₆H₅O⁻)
  • Phenoxide is much more activated (better coupling)

Wrong: “Couple aniline with diazonium salt in alkaline medium”

Correct: “Aniline requires WEAKLY ACIDIC medium (pH 4-5)”

Reason:

  • Strong base: Diazonium salt decomposes
  • Strong acid: Aniline protonated to -NH₃⁺ (deactivated)
  • Weak acid: Balance between both

JEE Memory: Phenols = alkaline, Amines = acidic

Mistake #2: Expecting Benzene to Couple

Wrong: “Benzene will couple with benzenediazonium chloride”

Correct: “Benzene is NOT activated enough to undergo coupling”

Requirements for coupling:

  • Must have strong EDG (-OH, -NH₂, -NR₂)
  • Benzene has no activating groups
  • Electrophilic substitution requires very strong electrophile (diazonium ion is moderate)

Only highly activated rings couple: ✓ Phenol, naphthols ✓ Aniline, N,N-dimethylaniline ✓ Substituted phenols/amines with EDG

✗ Benzene ✗ Toluene (too weakly activated) ✗ Chlorobenzene (deactivated) ✗ Nitrobenzene (strongly deactivated)

Mistake #3: Temperature Control

Wrong: “Warming speeds up coupling reaction”

Correct: “Keep at 0-5°C to prevent diazonium salt decomposition”

If temperature rises:

$$[\text{Ar-N}_2^+]\text{Cl}^- \xrightarrow{>10°\text{C}} \text{Ar-OH} + \text{N}_2↑$$

Diazonium salt decomposes to phenol!

JEE Note: Coupling must be done immediately after diazotization, keeping cold throughout


Practice Problems

Level 1: Foundation (NCERT)

Problem 1: Basic Coupling

Q: What happens when benzenediazonium chloride is treated with phenol in NaOH solution?

Solution:

Reaction:

Step 1: Formation of phenoxide

$$\text{C}_6\text{H}_5\text{OH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{O}^-\text{Na}^+ + \text{H}_2\text{O}$$

Step 2: Coupling

$$[\text{C}_6\text{H}_5\text{N}_2^+]\text{Cl}^- + \text{C}_6\text{H}_5\text{O}^-\text{Na}^+ \rightarrow$$ $$\text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_4\text{-OH} + \text{NaCl}$$

Product: p-Hydroxyazobenzene (orange-yellow azo dye)

Structure:

⟨⟩-N=N-⟨⟩-OH

Observation: Orange-yellow color appears

Conditions:

  • Alkaline medium (NaOH)
  • Temperature: 0-5°C
  • Coupling at para position (major product)
Problem 2: Methyl Orange Preparation

Q: Write the preparation of methyl orange from sulfanilic acid.

Solution:

Step 1: Diazotization of sulfanilic acid

Sulfanilic acid exists as zwitterion: ⁺H₃N-C₆H₄-SO₃⁻

$$\text{H}_2\text{N-C}_6\text{H}_4\text{-SO}_3\text{H} + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0-5°\text{C}}$$ $$[\text{Cl-}^+\text{N}_2\text{-C}_6\text{H}_4\text{-SO}_3\text{H}]$$

(Sulfanilic acid diazonium chloride)

Step 2: Coupling with N,N-dimethylaniline

$$[\text{⁺N}_2\text{-C}_6\text{H}_4\text{-SO}_3\text{H}] + (\text{CH}_3)_2\text{N-C}_6\text{H}_5 \xrightarrow{\text{pH 4-5}}$$ $$(\text{CH}_3)_2\text{N-C}_6\text{H}_4\text{-N=N-C}_6\text{H}_4\text{-SO}_3\text{H}$$

Product: Methyl orange (orange dye, used as indicator)

Conditions:

  • Step 1: 0-5°C, acidic
  • Step 2: pH 4-5 (weakly acidic for amine coupling)

Properties:

  • Red in acid (pH < 3.1)
  • Yellow in base (pH > 4.4)

Level 2: JEE Main

Problem 3: Identifying Coupling Partner

Q: Which of the following will couple with benzenediazonium chloride?

(A) Benzene (B) Nitrobenzene (C) Phenol (in NaOH) (D) Toluene

Solution: (C) Phenol (in NaOH)

Analysis:

(A) Benzene:

  • No activating group
  • Not electron-rich enough
  • No coupling

(B) Nitrobenzene:

  • -NO₂ is strong EWG (deactivating)
  • Ring is electron-deficient
  • No coupling

(C) Phenol (in NaOH):

  • Forms phenoxide ion (C₆H₅O⁻)
  • -O⁻ is very strong activating group
  • Highly electron-rich ring
  • Coupling occurs

(D) Toluene:

  • -CH₃ is weak activating group
  • Not activated enough for coupling with diazonium salt
  • No coupling

Correct answer: (C)

Requirement: Must have strong EDG (-OH, -NH₂, -NR₂) for coupling

Problem 4: pH Dependence

Q: Explain why coupling of benzenediazonium chloride with: (a) Phenol is done in alkaline medium (b) Aniline is done in weakly acidic medium

Solution:

(a) Phenol in alkaline medium (pH 8-10):

Reason:

In alkaline medium:

$$\text{C}_6\text{H}_5\text{OH} + \text{OH}^- \rightarrow \text{C}_6\text{H}_5\text{O}^- + \text{H}_2\text{O}$$

Phenoxide ion (C₆H₅O⁻) formed:

  • -O⁻ is much stronger activating group than -OH
  • Increases electron density on ring significantly
  • Better nucleophile for electrophilic substitution
  • Faster, more efficient coupling

In acidic medium:

  • Phenol remains as -OH (weaker activator)
  • Coupling is slow or doesn’t occur

(b) Aniline in weakly acidic medium (pH 4-5):

Reason:

Balance required:

If too acidic (pH < 3):

$$\text{C}_6\text{H}_5\text{NH}_2 + \text{H}^+ \rightarrow \text{C}_6\text{H}_5\text{NH}_3^+$$

Anilinium ion (-NH₃⁺) is deactivating → No coupling

If alkaline (pH > 7):

$$[\text{C}_6\text{H}_5\text{N}_2^+]\text{Cl}^- + \text{OH}^- \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{N}_2↑$$

Diazonium salt decomposes → No coupling

At pH 4-5 (weakly acidic):

  • Aniline remains as -NH₂ (activated)
  • Diazonium salt remains stable
  • Both coupling partners available
  • Successful coupling occurs

JEE Key: pH must be optimized for both partners!

Level 3: JEE Advanced

Problem 5: Multi-step Dye Synthesis

Q: Starting from benzene, outline the preparation of Orange II dye (structure: HO-naphthyl-N=N-C₆H₄-SO₃Na).

Solution:

Retrosynthetic analysis: Orange II ← β-Naphthol + Sulfanilic acid diazonium salt

Route 1: Prepare sulfanilic acid diazonium salt

Step 1: Benzene → Nitrobenzene

$$\text{C}_6\text{H}_6 \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{C}_6\text{H}_5\text{NO}_2$$

Step 2: Nitrobenzene → Aniline

$$\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Sn/HCl}} \text{C}_6\text{H}_5\text{NH}_2$$

Step 3: Aniline → Sulfanilic acid

$$\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{conc. H}_2\text{SO}_4, 180°\text{C}} \text{H}_2\text{N-C}_6\text{H}_4\text{-SO}_3\text{H}$$

(Sulfonation)

Step 4: Diazotization

$$\text{H}_2\text{N-C}_6\text{H}_4\text{-SO}_3\text{H} \xrightarrow{\text{NaNO}_2/\text{HCl}, 0-5°\text{C}} [\text{⁺N}_2\text{-C}_6\text{H}_4\text{-SO}_3^-]$$

Route 2: β-Naphthol (given or prepared from naphthalene)

Step 5: Coupling

$$[\text{⁺N}_2\text{-C}_6\text{H}_4\text{-SO}_3\text{Na}] + \text{β-naphthol} \xrightarrow{\text{NaOH, 0-5°\text{C}}}$$ $$\text{HO-naphthyl-N=N-C}_6\text{H}_4\text{-SO}_3\text{Na}$$

Product: Orange II (orange-red azo dye)

Conditions summary:

  • All steps at controlled temperatures
  • Coupling in alkaline medium (phenol derivative)
  • 0-5°C for diazonium salt stability

JEE Insight: Multi-step synthesis combining several concepts!

Problem 6: Structure-Color Relationship

Q: Compound A has molecular formula C₁₂H₁₁N₃ and shows the following properties:

  • Gives orange color with HCl
  • Reacts with acetyl chloride to give a product with molecular formula C₁₄H₁₃N₃O
  • Contains azo linkage

Identify compound A and explain the color change.

Solution:

Analysis:

Molecular formula: C₁₂H₁₁N₃

  • Contains 3 nitrogen atoms
  • One is likely azo group (-N=N-)
  • Other is likely -NH₂ group

Clue 1: Azo linkage present

  • Must have -N=N-

Clue 2: Reacts with acetyl chloride

  • Has free -NH₂ group (forms amide)
  • C₁₂H₁₁N₃ + CH₃COCl → C₁₄H₁₃N₃O
  • Gain: C₂H₂O (acetyl group, -COCH₃)

Clue 3: Orange color with HCl

  • Protonation changes color (typical for azo dyes)

Structure deduction:

Must be: H₂N-C₆H₄-N=N-C₆H₅

Compound A: p-Aminoazobenzene (para-phenylazoaniline)

Verification:

  • Formula: C₁₂H₁₁N₃ ✓
  • Has -NH₂ (reacts with acetyl chloride) ✓
  • Has azo group ✓

Acetylation:

$$\text{H}_2\text{N-C}_6\text{H}_4\text{-N=N-C}_6\text{H}_5 + \text{CH}_3\text{COCl} \rightarrow$$ $$\text{CH}_3\text{CO-NH-C}_6\text{H}_4\text{-N=N-C}_6\text{H}_5$$

(C₁₄H₁₃N₃O) ✓

Color change explanation:

In neutral form: Yellow-orange

  • Normal conjugation

With HCl (protonated): Orange-red

$$\text{H}_2\text{N-C}_6\text{H}_4\text{-N=N-C}_6\text{H}_5 + \text{HCl} \rightarrow$$ $$[\text{H}_3\text{N}^+-\text{C}_6\text{H}_4\text{-N=N-C}_6\text{H}_5]\text{Cl}^-$$
  • Protonated -NH₃⁺ is electron-withdrawing
  • Changes conjugation pattern
  • Shifts absorption wavelength
  • Deeper color (orange → red)

JEE Key: Protonation affects conjugation and color in azo dyes!


Quick Revision Box

TopicKey PointsConditions
Coupling partnersPhenols, naphthols, aromatic aminesMust have strong EDG
pH for phenolspH 8-10 (alkaline)Forms phenoxide (more activated)
pH for aminespH 4-5 (weakly acidic)Keeps NH₂ free, N₂⁺ stable
Temperature0-5°CPrevents decomposition
RegioselectivityPara position (major)Less steric hindrance
Color originExtended conjugationπ → π* transitions
Chromophore-N=N- (azo group)Absorbs visible light
Auxochrome-OH, -NH₂, -SO₃HModifies/intensifies color
Methyl OrangeRed (acid) ↔ Yellow (base)pH 3.1-4.4

Connection to Other Topics

Prerequisites:

Related Topics:

Applications:

  • Textile chemistry
  • Analytical chemistry (indicators)
  • Organic synthesis
  • Pharmaceutical chemistry

Summary

Key Takeaways

1. Coupling Reaction:

General reaction:

$$[\text{Ar-N}_2^+]\text{Cl}^- + \text{Ar'-H} \rightarrow \text{Ar-N=N-Ar'} + \text{HCl}$$

Product: Azo compound (colored dyes)

2. Coupling Partners:

Must be highly activated: ✓ Phenols, naphthols → pH 8-10 (alkaline) ✓ Aromatic amines → pH 4-5 (weakly acidic)

Won’t couple: ✗ Benzene (not activated) ✗ Nitrobenzene (deactivated) ✗ Toluene (too weakly activated)

3. pH Control (Critical!):

Phenols:

  • Alkaline (pH 8-10)
  • Forms phenoxide (C₆H₅O⁻)
  • More activated, better coupling

Amines:

  • Weakly acidic (pH 4-5)
  • Keeps -NH₂ free (not -NH₃⁺)
  • Maintains diazonium salt stability

4. Important Azo Dyes:

Methyl Orange:

  • From sulfanilic acid + N,N-dimethylaniline
  • pH indicator (red ↔ yellow)
  • pH range: 3.1-4.4

Orange II:

  • From sulfanilic acid + β-naphthol
  • Orange-red textile dye

5. Color Chemistry:

Why colored?

  • Extended conjugation (π-system)
  • Chromophore: -N=N- group
  • Auxochromes: -OH, -NH₂, -SO₃H

Structure affects color:

  • More conjugation → Deeper color
  • EDG → Bathochromic shift (red shift)
  • EWG → Hypsochromic shift (blue shift)

6. JEE Strategy:

✓ Remember pH requirements (phenols = alkaline, amines = acidic) ✓ Only strongly activated rings couple ✓ Temperature: Always 0-5°C ✓ Know methyl orange preparation ✓ Understand color-structure relationship

7. Applications:

  • Textile dyes (60-70% of all dyes)
  • Food colorings (E numbers)
  • pH indicators (methyl orange, congo red)
  • Biological stains
  • Industrial applications

“Azo coupling transforms diazonium salts into the vibrant world of colored compounds - from textiles to pH indicators, it’s chemistry that colors our world!”

Congratulations! You’ve completed the nitrogen compounds chapter. Master these concepts to excel in JEE organic chemistry!