Haloarenes: Aromatic Halogen Compounds

Master aryl halides - preparation, properties, nucleophilic substitution, and electrophilic substitution for JEE Chemistry

The Hook: Why Doesn’t Chlorobenzene React Like Chloroethane?

Connect: Real Life → Chemistry

DDT (dichlorodiphenyltrichloroethane), once the world’s most famous insecticide, contains aromatic chlorine atoms. These C-Cl bonds are so stable that DDT persists in the environment for decades! In contrast, alkyl chlorides are reactive and decompose quickly.

Here’s the JEE question: Why is chlorobenzene 1000 times less reactive than chloroethane in nucleophilic substitution? And why does chlorobenzene direct incoming electrophiles to ortho and para positions, even though it’s a deactivating group?


The Core Concept

What are Haloarenes?

Haloarenes (Aryl Halides) are compounds where halogen is directly attached to benzene ring.

General structure:

$$\boxed{\text{Ar-X}}$$

where Ar = aromatic ring, X = F, Cl, Br, I

Examples:

  • Chlorobenzene: C₆H₅Cl
  • Bromobenzene: C₆H₅Br
  • p-Dichlorobenzene: Cl-C₆H₄-Cl
Haloarenes vs Haloalkanes

Key Difference:

Haloalkanes: X attached to sp³ carbon

  • Example: CH₃CH₂Cl
  • Reactive in nucleophilic substitution

Haloarenes: X attached to sp² carbon of benzene

  • Example: C₆H₅Cl
  • Unreactive in nucleophilic substitution (under normal conditions)

Why the difference?

  1. Resonance: C-X bond has partial double bond character
  2. Hybridization: sp² carbon holds electrons more tightly than sp³
  3. Stability: Breaking C-X disrupts aromatic resonance
JEE Weightage
Haloarenes: 2-3 questions in JEE Main, 2-3 in JEE Advanced Focus areas: Difference from alkyl halides, preparation methods, SNAr mechanism, directing effects

Why Haloarenes are Unreactive

Reason 1: Resonance Stabilization

Lone pair on halogen delocalizes into benzene ring.

Resonance structures:

    Cl                 Cl⁺              Cl⁺              Cl⁺
     |                  |                |                |
     ⚬        ↔        ⚬        ↔      ⚬        ↔      ⚬
                      (-)              (-)              (-)

Consequence:

  • C-Cl bond has partial double bond character
  • Bond length: 169 pm (vs 177 pm in alkyl chlorides)
  • Stronger and shorter → harder to break

Reason 2: sp² Hybridization

In haloarenes:

  • Carbon is sp² hybridized
  • Higher s-character (33%) vs sp³ (25%)
  • Electrons held more tightly
  • C-X bond is stronger

Reason 3: Stability of Phenyl Cation

If nucleophilic substitution occurred:

$$\text{C}_6\text{H}_5\text{-Cl} \rightarrow \text{C}_6\text{H}_5^+ + \text{Cl}^-$$

Phenyl cation (C₆H₅⁺):

  • Extremely unstable
  • Cannot be stabilized by resonance
  • Positive charge on sp² carbon (not favorable)
  • Does NOT form under normal conditions
Memory Trick: Why No Substitution

“RIP: Resonance, Inert, Phenyl cation”

Resonance makes C-X bond stronger Inert sp² carbon (high s-character) Phenyl cation too unstable

JEE Tip: This explains why chlorobenzene doesn’t react with aq. NaOH, but alkyl chlorides do!


Preparation of Haloarenes

Method 1: From Benzene (Electrophilic Halogenation)

Direct halogenation:

$$\boxed{\text{C}_6\text{H}_6 + \text{X}_2 \xrightarrow{\text{Lewis acid}} \text{C}_6\text{H}_5\text{X} + \text{HX}}$$

Catalyst required:

  • For Cl₂: FeCl₃ or AlCl₃
  • For Br₂: FeBr₃ or AlBr₃
  • For I₂: HNO₃ (oxidizing agent)

Mechanism: Electrophilic aromatic substitution

Step 1: Generation of electrophile

$$\text{Cl}_2 + \text{FeCl}_3 \rightarrow \text{Cl}^+ + [\text{FeCl}_4]^-$$

Step 2: Electrophilic attack

         Cl⁺
    ⚬  →  ⚬⁺  →  ⚬
               H        -H⁺      Cl
          (σ-complex)

Example:

$$\text{C}_6\text{H}_6 + \text{Cl}_2 \xrightarrow{\text{FeCl}_3} \text{C}_6\text{H}_5\text{Cl} + \text{HCl}$$

Method 2: Sandmeyer Reaction (Most Important for JEE)

From diazonium salts:

$$\boxed{\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{CuX} \rightarrow \text{C}_6\text{H}_5\text{X} + \text{N}_2 + \text{CuCl}}$$

Works for: X = Cl, Br

Example:

$$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{CuCl/HCl} \rightarrow \text{C}_6\text{H}_5\text{Cl} + \text{N}_2 \uparrow$$

Advantages:

  • Excellent yields
  • Specific substitution position
  • Works when direct halogenation would give wrong isomer

Method 3: Gattermann Reaction

Alternative to Sandmeyer:

$$\boxed{\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{HX/Cu} \rightarrow \text{C}_6\text{H}_5\text{X} + \text{N}_2}$$

Difference from Sandmeyer:

  • Uses HX instead of CuX
  • Slightly lower yields
  • Cheaper reagents

Method 4: For Iodobenzene

Iodine doesn’t work with Lewis acids (weak electrophile)

Method A: From diazonium salt + KI

$$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{KI} \rightarrow \text{C}_6\text{H}_5\text{I} + \text{N}_2 + \text{KCl}$$

Method B: Oxidative iodination

$$\text{C}_6\text{H}_6 + \text{I}_2 \xrightarrow{\text{HNO}_3} \text{C}_6\text{H}_5\text{I}$$

HNO₃ oxidizes I₂ to I⁺ (electrophile)

Method 5: For Fluorobenzene

Balz-Schiemann Reaction:

$$\boxed{\text{C}_6\text{H}_5\text{N}_2^+\text{BF}_4^- \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{F} + \text{N}_2 + \text{BF}_3}$$

Unique feature: Diazonium fluoroborate is dry and isolable

JEE Question: Which Method?

Q: How would you prepare m-chloronitrobenzene from benzene?

Analysis:

  • Need meta-substitution
  • Direct chlorination of nitrobenzene gives meta (NO₂ is m-director)

Correct sequence:

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: Chlorination

$$\text{C}_6\text{H}_5\text{NO}_2 + \text{Cl}_2 \xrightarrow{\text{FeCl}_3} m\text{-ClC}_6\text{H}_4\text{NO}_2$$

Why this order?

  • NO₂ is meta-directing
  • Chlorination after nitration gives meta-product

Wrong order: Chlorination first would give o/p mixture, then nitration gives mixture!

JEE Tip: Order of substitution matters for getting correct isomer!


Reactions of Haloarenes

Type 1: Nucleophilic Substitution (Very Difficult)

Under normal conditions: Haloarenes DO NOT undergo nucleophilic substitution.

Harsh conditions required:

  • Very high temperature (300-400°C)
  • Very high pressure
  • Strong base/nucleophile

Example (industrial):

$$\text{C}_6\text{H}_5\text{Cl} + \text{NaOH} \xrightarrow{623\text{ K, 300 atm}} \text{C}_6\text{H}_5\text{OH} + \text{NaCl}$$

Activated Nucleophilic Aromatic Substitution (SNAr)

When haloarene has electron-withdrawing groups at ortho/para positions:

Mechanism: Addition-Elimination (not SN1 or SN2!)

Example: 2,4-dinitrochlorobenzene

    NO₂                    NO₂                    NO₂
     |                      |                      |
Cl—⚬—NO₂  +  OH⁻  →  Cl—⚬—NO₂  →  HO—⚬—NO₂  +  Cl⁻
                         |
                         OH
                    (Meisenheimer
                     complex)

Conditions for SNAr:

  1. Electron-withdrawing groups (NO₂, CN, COR) at o/p positions
  2. Groups stabilize negative charge in intermediate
  3. Room temperature reaction possible!
JEE Concept: Activating Groups for SNAr

Electron-withdrawing groups activate SNAr:

Order of activation:

$$\boxed{\text{-NO}_2 > \text{-CN} > \text{-COCH}_3 > \text{-CHO} > \text{-COOH}}$$

Position matters:

  • o/p to halogen → highly activated
  • meta to halogen → less effective

Most reactive: 2,4,6-trinitrochlorobenzene (picryl chloride)

  • Can react with water at room temperature!

Example:

  • 2,4-dinitrochlorobenzene: reacts easily
  • 3,5-dinitrochlorobenzene: much slower (meta positions)
  • Chlorobenzene: no reaction under normal conditions

JEE Strategy: Count NO₂ groups at o/p positions to predict reactivity!

Type 2: Electrophilic Substitution

Haloarenes undergo typical electrophilic aromatic substitution reactions.

Halogen effect:

  • Deactivating: Withdraws electrons by inductive effect
  • ortho/para directing: Donates electrons by resonance

Nitration

$$\text{C}_6\text{H}_5\text{Cl} + \text{HNO}_3 \xrightarrow{\text{H}_2\text{SO}_4} o\text{-ClC}_6\text{H}_4\text{NO}_2 + p\text{-ClC}_6\text{H}_4\text{NO}_2$$

Products: ortho + para mixture (para usually major)

Sulfonation

$$\text{C}_6\text{H}_5\text{Br} + \text{H}_2\text{SO}_4 \xrightarrow{\text{heat}} o/p\text{-BrC}_6\text{H}_4\text{SO}_3\text{H}$$

Friedel-Crafts Alkylation

$$\text{C}_6\text{H}_5\text{Cl} + \text{CH}_3\text{Cl} \xrightarrow{\text{AlCl}_3} o/p\text{-ClC}_6\text{H}_4\text{CH}_3$$

Note: Slower than benzene (deactivating effect)

Friedel-Crafts Acylation

$$\text{C}_6\text{H}_5\text{Br} + \text{CH}_3\text{COCl} \xrightarrow{\text{AlCl}_3} o/p\text{-BrC}_6\text{H}_4\text{COCH}_3$$
JEE Trap: Directing Effect

Common mistake: “Cl is electron-withdrawing, so it’s meta-directing”

Correct: Halogen is ortho/para directing BUT deactivating

Why ortho/para?

  • Resonance effect (electron donation by lone pair) > Inductive effect
  • Lone pair on X can donate to ortho/para positions

Why deactivating?

  • Inductive withdrawal of electrons from ring
  • Makes ring less nucleophilic overall
  • Reaction is slower than benzene

Memory: “Halogens are Odd: Ortho/para but Deactivating”

Type 3: Reduction

With Ni/H₂:

$$\text{C}_6\text{H}_5\text{Cl} + \text{H}_2 \xrightarrow{\text{Ni, } \Delta} \text{C}_6\text{H}_6 + \text{HCl}$$

Catalytic hydrogenolysis: Removes halogen

Type 4: Reaction with Metals

Wurtz-Fittig Reaction:

$$\text{C}_6\text{H}_5\text{Br} + \text{CH}_3\text{Br} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5\text{-CH}_3 + 2\text{NaBr}$$

Fittig Reaction:

$$2\text{C}_6\text{H}_5\text{Br} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5\text{-C}_6\text{H}_5 + 2\text{NaBr}$$

Product: Biphenyl

Formation of Grignard Reagent:

$$\text{C}_6\text{H}_5\text{Br} + \text{Mg} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5\text{MgBr}$$

Phenyl magnesium bromide - useful in synthesis


Physical Properties

1. Physical State

  • Chlorobenzene, bromobenzene: colorless liquids
  • Iodobenzene: colorless to brown liquid
  • p-Dichlorobenzene: white crystals (mothballs!)

2. Odor

  • Characteristic aromatic odor
  • p-Dichlorobenzene: distinctive mothball smell

3. Solubility

  • Insoluble in water (non-polar)
  • Soluble in organic solvents (benzene, ether, alcohol)

4. Boiling Points

  • Higher than corresponding alkyl halides
  • Due to stronger C-X bond and aromatic π-π interactions
CompoundBP (°C)
Fluorobenzene85
Chlorobenzene132
Bromobenzene156
Iodobenzene188

5. Density

  • Denser than water
  • Chlorobenzene: 1.11 g/mL

Uses and Applications

Chlorobenzene

  • Solvent in organic synthesis
  • Intermediate for phenol, aniline production
  • DDT synthesis (insecticide production)

Bromobenzene

  • Grignard reagent preparation
  • Pharmaceutical intermediate

Iodobenzene

  • Organic synthesis - iodine source
  • Reagent in organometallic chemistry

p-Dichlorobenzene

  • Mothballs and air fresheners
  • Deodorant blocks
  • Insecticide

Common Mistakes to Avoid

Mistake #1: Treating Like Alkyl Halides

Wrong: “Chlorobenzene + NaOH → Phenol (like alkyl halides → alcohols)”

Correct: Haloarenes do NOT undergo simple nucleophilic substitution

Exception: With activating groups (NO₂ at o/p) or very harsh conditions

JEE Tip: Always note if halogen is on aromatic ring or aliphatic chain!

Interactive Demo: Visualize Haloarene Resonance Structures

See how halogen lone pairs delocalize into the aromatic ring.

Mistake #2: Wrong Directing Effect

Wrong: “Cl deactivates, so it’s meta-directing”

Correct: Halogens are ortho/para directing (despite being deactivating)

Reason: Resonance donation > Inductive withdrawal for directing effect

JEE Rule: Only -NO₂, -CN, -SO₃H, -COOH, -CHO, -COR are meta-directing

Mistake #3: Ignoring Substitution Order

Wrong: Starting synthesis without planning substitution sequence

Correct: Use directing effects strategically

Example: To make m-bromoaniline:

  • Wrong: Br first (o/p director) → NH₂ second gives o/p-isomers
  • Correct: NO₂ first (m-director) → Br second → reduce NO₂ to NH₂

JEE Strategy: Plan backwards from target molecule!


Practice Problems

Level 1: Foundation (NCERT)

Problem 1: Reactivity Comparison

Q: Why is chlorobenzene less reactive than chloroethane toward nucleophilic substitution?

Answer:

Three reasons:

  1. Resonance: Lone pair on Cl delocalizes into benzene ring

    • C-Cl bond has partial double bond character
    • Stronger and harder to break
  2. Hybridization: sp² carbon (vs sp³ in chloroethane)

    • Higher s-character
    • Holds electrons more tightly
  3. Phenyl cation instability: C₆H₅⁺ is extremely unstable

    • Cannot form under normal conditions

Conclusion: All three factors make C-Cl bond unreactive in chlorobenzene.

Problem 2: Preparation Method

Q: How will you prepare chlorobenzene from benzene?

Solution:

Method: Direct chlorination

$$\text{C}_6\text{H}_6 + \text{Cl}_2 \xrightarrow{\text{FeCl}_3} \text{C}_6\text{H}_5\text{Cl} + \text{HCl}$$

Mechanism: Electrophilic aromatic substitution

  • FeCl₃ generates Cl⁺ electrophile
  • Cl⁺ attacks benzene ring
  • H⁺ is eliminated

Level 2: JEE Main

Problem 3: SNAr Reactivity

Q: Arrange in order of increasing reactivity toward nucleophilic substitution: (a) Chlorobenzene (b) 2,4-Dinitrochlorobenzene (c) 2,4,6-Trinitrochlorobenzene

Solution:

Order: (a) < (b) < (c)

Reasoning:

(a) Chlorobenzene: No activating groups

  • Extremely unreactive (needs 300°C, high pressure)

(b) 2,4-Dinitrochlorobenzene: Two NO₂ at o and p

  • Moderately reactive (room temperature with strong base)

(c) 2,4,6-Trinitrochlorobenzene: Three NO₂ (all o/p)

  • Very reactive (reacts with water!)

Key: More NO₂ groups at o/p positions → higher reactivity

Problem 4: Directing Effects

Q: What are the major products when chlorobenzene undergoes nitration?

Solution:

$$\text{C}_6\text{H}_5\text{Cl} + \text{HNO}_3 \xrightarrow{\text{H}_2\text{SO}_4} ?$$

Products:

  • o-Chloronitrobenzene (ortho)
  • p-Chloronitrobenzene (para) - major

Ratio: ~30% ortho, 70% para (para favored due to less steric hindrance)

No meta product - Cl is o/p director!

Level 3: JEE Advanced

Problem 5: Multi-step Synthesis

Q: How will you prepare m-chloroaniline from benzene?

Solution:

Strategy: Need meta-substitution, so use meta-director first

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: Chlorination (NO₂ is meta-directing)

$$\text{C}_6\text{H}_5\text{NO}_2 + \text{Cl}_2 \xrightarrow{\text{FeCl}_3} m\text{-ClC}_6\text{H}_4\text{NO}_2$$

Step 3: Reduction of NO₂ to NH₂

$$m\text{-ClC}_6\text{H}_4\text{NO}_2 \xrightarrow{\text{Sn/HCl}} m\text{-ClC}_6\text{H}_4\text{NH}_2$$

Answer: Three-step synthesis

Why this order?

  • NH₂ is o/p director (would give wrong isomer if done first)
  • NO₂ is meta director (gives correct position)
  • Reduce NO₂ to NH₂ at the end

JEE Tip: For meta-substitution, use NO₂, CN, or COOH as temporary director!

Problem 6: Mechanism Understanding

Q: Explain why 2,4-dinitrochlorobenzene reacts with NaOH at room temperature but chlorobenzene doesn’t.

Solution:

2,4-Dinitrochlorobenzene: Undergoes SNAr mechanism

Mechanism:

Step 1: Nucleophilic addition (rate-determining)

    NO₂                     NO₂
     |                       |
Cl—⚬—NO₂  +  OH⁻  →  Cl—⚬—NO₂
                          |
                          OH⁻
                    (Meisenheimer
                     complex)

Step 2: Elimination of Cl⁻

    NO₂                    NO₂
     |                      |
Cl—⚬—NO₂  →  HO—⚬—NO₂  +  Cl⁻
     |
     OH⁻

Why this works:

  • NO₂ groups are electron-withdrawing
  • Stabilize negative charge in intermediate by resonance
  • Lower activation energy

Chlorobenzene:

  • No electron-withdrawing groups
  • Cannot stabilize negative intermediate
  • Phenyl cation too unstable
  • Needs extreme conditions (300°C, 300 atm)

Conclusion: Activating groups make ALL the difference!


Quick Revision Box

TopicKey PointsJEE Rule
StructureX directly on benzenesp² C-X bond
UnreactiveResonance + sp² + unstable C⁺1000× slower than alkyl
PreparationSandmeyer (best), direct halogenationFrom diazonium salt
SNArNeeds NO₂ at o/pAddition-elimination
Directingortho/para but deactivatingResonance > Inductive
ReductionNi/H₂ removes XGives benzene
Wurtz-FittigAr-X + R-X + NaGives Ar-R
UsesSolvents, intermediates, DDTp-DCB = mothballs

Connection to Other Topics

Prerequisites:

Related Topics:

Applications:


Teacher’s Summary

Key Takeaways

1. Haloarenes vs Haloalkanes (MOST IMPORTANT DISTINCTION)

  • Haloarenes: X on aromatic ring, unreactive in nucleophilic substitution
  • Haloalkanes: X on aliphatic chain, reactive

2. Why Unreactive:

  • Resonance: C-X partial double bond
  • sp² carbon: Higher s-character
  • Phenyl cation: Extremely unstable

3. Preparation Methods:

  • Direct halogenation: Benzene + X₂/Lewis acid
  • Sandmeyer: Best method, from diazonium salt
  • For F: Balz-Schiemann (diazonium fluoroborate)

4. Activated Nucleophilic Substitution (SNAr):

  • Requires electron-withdrawing groups at o/p
  • Addition-elimination mechanism
  • Meisenheimer complex intermediate

5. Electrophilic Substitution:

  • ortho/para directing (resonance donation)
  • Deactivating (inductive withdrawal)
  • Slower than benzene

6. Synthesis Strategy:

  • For meta: Use NO₂ or other meta-director first
  • For ortho/para: Use o/p-director first
  • Plan backward from target!

“Haloarenes are unreactive in substitution but follow normal electrophilic substitution - understand WHY for JEE success!”

Next: Study polyhalogen compounds to learn about CHCl₃, CCl₄, DDT, and CFCs!