Chemistry | Class 12 | NCERT Comprehensive Edition

HALOALKANES & HALOARENES

Every concept. Every mechanism. Every reaction. Decoded.

0. Overview

NCERT Line 1: A haloalkane (alkyl halide) contains a carbon–halogen (C‑X) bond where X = F, Cl, Br or I.
NCERT Line 2: A haloarene (aryl halide) has the halogen attached directly to an aromatic ring.

📐 FORMULA CHEAT SHEET – MUST MEMORIZE

General Formula: CₙH₂ₙ₊₁X (monohaloalkane)
Lucas Test: ROH + HCl/ZnCl₂ → RCl + H₂O
Free Radical: RH + X₂ ──hν──► RX + HX
Markovnikov: CH₃CH=CH₂ + HBr → CH₃CHBrCH₃
Anti-Markovnikov: CH₃CH=CH₂ + HBr + ROOR → CH₃CH₂CH₂Br
Sandmeyer: ArN₂⁺Cl⁻ + CuCl → ArCl + N₂↑
Wurtz: 2R-X + 2Na ──dry ether──► R-R + 2NaX
Grignard: R-X + Mg ──dry ether──► R-MgX

PART 1 – INTRODUCTION & CLASSIFICATION

Board Weightage: ★★★☆☆

What are Haloalkanes?

  • Haloalkanes (also called alkyl halides) are organic compounds containing a carbon-halogen (C-X) bond
  • The halogen atom (X) can be fluorine (F), chlorine (Cl), bromine (Br), or iodine (I)
  • General formula: CₙH₂ₙ₊₁X for monohaloalkanes
  • Example: CH₃Cl (chloromethane), C₂H₅Br (bromoethane)

What are Haloarenes?

  • Haloarenes (also called aryl halides) have the halogen atom directly attached to an aromatic ring
  • The halogen is bonded to an sp² hybridized carbon of the benzene ring
  • Example: C₆H₅Cl (chlorobenzene), C₆H₅Br (bromobenzene)

Functional Group

  • Functional group: -X (halogen atom)
  • The halogen is the characteristic functional group in both haloalkanes and haloarenes

General Formula

  • Monohaloalkanes: CₙH₂ₙ₊₁X
  • Dihaloalkanes: CₙH₂ₙX₂
  • Trihaloalkanes: CₙH₂ₙ₋₁X₃
  • Polyhaloalkanes: CₙH₂ₙ₊₁₋ₓXₓ

Classification of Haloalkanes

Based on Number of Halogen Atoms

Monohaloalkanes

Contain one halogen atom

Example: CH₃CH₂Cl (chloroethane)

Dihaloalkanes

Contain two halogen atoms

Geminal: CH₂Cl₂ (both on same carbon)

Vicinal: CH₂ClCH₂Cl (adjacent carbons)

Trihaloalkanes

Contain three halogen atoms

Example: CHCl₃ (chloroform)

Polyhaloalkanes

Contain four or more halogen atoms

Example: CCl₄ (carbon tetrachloride)

Based on Carbon Atom Bearing Halogen

Primary (1°) Haloalkanes

Halogen attached to a primary carbon

Example: CH₃CH₂Cl (1-chloroethane)

Secondary (2°) Haloalkanes

Halogen attached to a secondary carbon

Example: CH₃CHClCH₃ (2-chloropropane)

Tertiary (3°) Haloalkanes

Halogen attached to a tertiary carbon

Example: (CH₃)₃CCl (2-chloro-2-methylpropane)

Classification of Haloarenes

  • Classified based on the position of halogen on the benzene ring
  • Ortho: Halogens at positions 1 and 2
  • Meta: Halogens at positions 1 and 3
  • Para: Halogens at positions 1 and 4
📝 NCERT Examples:
  • CH₃CH₂Cl – primary chloride
  • (CH₃)₂CHBr – secondary bromide
  • (CH₃)₃CCl – tertiary chloride

NOMENCLATURE (IUPAC)

Board Weightage: ★★★★☆

Rules for Haloalkanes

  1. Select the longest continuous carbon chain containing the halogen
  2. Number the chain to give the halogen the lowest possible number
  3. Name the halogen as a prefix: fluoro, chloro, bromo, iodo
  4. For multiple halogens, use prefixes: di-, tri-, tetra-
  5. List substituents alphabetically (ignoring prefixes like di-, tri-)

Examples

Formula IUPAC Name Common Name
CH₃CH₂Cl Chloroethane Ethyl chloride
CH₃CHClCH₃ 2-Chloropropane Isopropyl chloride
CH₂ClCH₂Cl 1,2-Dichloroethane Ethylene dichloride
CH₃CH(Br)CH₂Cl 2-Bromo-1-chloropropane -
⚠️ Common Mistake: Forgetting to order substituents alphabetically – e.g., writing "bromo-chloro-" instead of "chloro-bromo-".

Board Naming Questions

Q1. Write the IUPAC name of CH₃CH₂CH₂Cl
Answer: 1-Chloropropane

Q2. Write the IUPAC name of CH₃CHClCH₂CH₃
Answer: 2-Chlorobutane

Q3. Write the IUPAC name of CH₃CH(Br)CH₃
Answer: 2-Bromopropane
🔴 REACTION SHEET – PART 1

No reactions to memorize in this section

  • Only structures, classification, and nomenclature
  • Focus on identifying 1°, 2°, 3° haloalkanes
  • Practice IUPAC naming

PART 2 – PREPARATION OF HALOALKANES FROM ALCOHOLS

Board Weightage: ★★★★★

1. Reaction with HX (Hydrogen Halides)

General Reaction: R-OH + HX → R-X + H₂O

Conditions:

  • HCl, HBr, or HI
  • Anhydrous ZnCl₂ (Lucas Reagent) for HCl
  • Conc. H₂SO₄ for HBr and HI

Lucas Test

🧠 Memory Trick:
  • Lucas Reagent: Conc. HCl + ZnCl₂
  • Used to distinguish between 1°, 2°, and 3° alcohols
  • 3° alcohols: Immediate turbidity (forms alkyl halide, immiscible)
  • 2° alcohols: Turbidity within 5 minutes
  • 1° alcohols: No turbidity at room temperature

Mechanism:

  • 3° alcohols: SN1 mechanism (carbocation formation)
  • 1° and 2° alcohols: SN2 mechanism
Examples:
  • CH₃CH₂OH + HCl/ZnCl₂ → CH₃CH₂Cl + H₂O
  • (CH₃)₃COH + HCl/ZnCl₂ → (CH₃)₃CCl + H₂O

2. Reaction with SOCl₂ (Thionyl Chloride)

General Reaction: R-OH + SOCl₂ → R-Cl + SO₂ + HCl

Conditions:

  • Anhydrous conditions
  • Pyridine can be used as base
🧠 Advantages:
  • Gaseous byproducts (SO₂ and HCl) escape
  • Pure product obtained
  • No rearrangement
Example: CH₃CH₂OH + SOCl₂ → CH₃CH₂Cl + SO₂↑ + HCl↑

3. Reaction with PCl₃ (Phosphorus Trichloride)

General Reaction: 3R-OH + PCl₃ → 3R-Cl + H₃PO₃

Conditions:

  • Heat
  • Anhydrous conditions
Example: 3CH₃CH₂OH + PCl₃ → 3CH₃CH₂Cl + H₃PO₃

4. Reaction with PCl₅ (Phosphorus Pentachloride)

General Reaction: R-OH + PCl₅ → R-Cl + POCl₃ + HCl

Conditions:

  • Heat
  • Anhydrous conditions
Example: CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl

5. Reaction with PBr₃ (Phosphorus Tribromide)

General Reaction: 3R-OH + PBr₃ → 3R-Br + H₃PO₃

Conditions:

  • Heat
  • Anhydrous conditions
Example: 3CH₃CH₂OH + PBr₃ → 3CH₃CH₂Br + H₃PO₃

6. Reaction with PI₃ (Phosphorus Triiodide)

General Reaction: 3R-OH + PI₃ → 3R-I + H₃PO₃
🧠 Note: PI₃ is usually generated in situ by adding red phosphorus to iodine
Example: 3CH₃CH₂OH + PI₃ → 3CH₃CH₂I + H₃PO₃
🔴 REACTION SHEET – PREPARATION FROM ALCOHOLS
  1. ROH + HCl/ZnCl₂ → RCl + H₂O (Lucas Reagent)
  2. ROH + SOCl₂ → RCl + SO₂ + HCl (Thionyl chloride)
  3. 3ROH + PCl₃ → 3RCl + H₃PO₃
  4. ROH + PCl₅ → RCl + POCl₃ + HCl
  5. 3ROH + PBr₃ → 3RBr + H₃PO₃
  6. 3ROH + PI₃ → 3RI + H₃PO₃

PART 2 (CONTINUED) – PREPARATION FROM ALKANES (FREE RADICAL HALOGENATION)

Board Weightage: ★★★★★ (Extremely Important)

General Reaction

RH + X₂ ──hν/Δ──► RX + HX

Where: RH = Alkane, X₂ = Cl₂ or Br₂, RX = Haloalkane, HX = HCl or HBr

Why is Sunlight Required?

  • Sunlight (hν) or heat (Δ) provides energy to break the X-X bond
  • This bond breaks homolytically (each atom gets one electron)
  • Free radicals are formed: Cl• or Br•

What is a Free Radical?

  • A neutral species containing an unpaired electron
  • Highly reactive because it tries to pair its unpaired electron
  • Examples: Cl•, CH₃•, Br•

Mechanism of Free Radical Halogenation

🧠 Memory Trick: I-P-T = "I Prefer Tea"
  • I = Initiation
  • P = Propagation
  • T = Termination
STEP 1: INITIATION
↓
Cl₂ + hν → Cl• + Cl•
(Homolytic cleavage - each atom gets 1 electron)
↓
STEP 2: PROPAGATION
↓
Cl• + CH₄ → CH₃• + HCl
(Chlorine radical attacks methane)
↓
CH₃• + Cl₂ → CH₃Cl + Cl•
(New chlorine radical formed - chain continues)
↓
STEP 3: TERMINATION
↓
Cl• + Cl• → Cl₂
CH₃• + Cl• → CH₃Cl
CH₃• + CH₃• → C₂H₆
↓
REACTION STOPS

Stage 1 – Initiation

Definition: Formation of free radicals

Reaction: Cl₂ ──hν──► Cl• + Cl•

This is called homolytic cleavage
Board Definition: Homolytic cleavage is the equal breaking of a covalent bond in which each bonded atom takes one electron, producing free radicals.
💡 Important Note: Sunlight (hν) or heat (Δ) provides the energy needed to break the Cl-Cl bond. Without light/heat, no reaction occurs.

Stage 2 – Propagation

This is the main step. Two reactions occur repeatedly.

Step 1: Cl• + CH₄ ──► CH₃• + HCl
Chlorine radical removes one hydrogen atom. Methane becomes a methyl radical.

Step 2: CH₃• + Cl₂ ──► CH₃Cl + Cl•
Product: Chloromethane. A new chlorine radical is formed.
🧠 Why is it called Chain Reaction? Because one free radical produces another free radical. The reaction continues automatically until radicals are removed.
⚠️ Exam Alert: Propagation step is the most important. Remember that the chlorine radical is regenerated, which allows the chain to continue.

Stage 3 – Termination

Definition: Two free radicals combine together. No new radical is produced. Reaction stops.

Reactions:
  1. Cl• + Cl• ──► Cl₂
  2. CH₃• + Cl• ──► CH₃Cl
  3. CH₃• + CH₃• ──► C₂H₆ (Ethane is formed)

Chlorination of Methane

📝 Complete Series (Must Memorise):
CH₄
 ↓
CH₃Cl (Chloromethane)
 ↓
CH₂Cl₂ (Dichloromethane)
 ↓
CHCl₃ (Trichloromethane/Chloroform)
 ↓
CCl₄ (Tetrachloromethane)

Reactions:

  1. CH₄ + Cl₂ ──hν──► CH₃Cl + HCl
  2. CH₃Cl + Cl₂ ──hν──► CH₂Cl₂ + HCl
  3. CH₂Cl₂ + Cl₂ ──hν──► CHCl₃ + HCl
  4. CHCl₃ + Cl₂ ──hν──► CCl₄ + HCl

Bromination of Alkanes

Reaction: RH + Br₂ ──hν──► RBr + HBr
Example: CH₄ + Br₂ ──hν──► CH₃Br + HBr

Chlorination vs Bromination

Feature Chlorination Bromination
Speed Faster Slower
Selectivity Less selective More selective
Products Mixture of products Major product predominates
Control Less controlled Better controlled

Why is Bromination More Selective?

  • Bromination has a higher activation energy
  • Therefore, it reacts mainly with the most stable free radical
  • Major product predominates
  • Chlorination reacts rapidly and forms many products

Stability of Free Radicals

🧠 Most Important Order:

3° > 2° > 1° > CH₃•

Reason: Alkyl groups donate electrons by the +I (inductive) effect and hyperconjugation, stabilizing the free radical.

Mnemonic: 3 > 2 > 1 > M (Methyl)

Example: Monochlorination of Propane

Reaction: CH₃CH₂CH₃ + Cl₂ ──hν──► ?

Possible Products:

  1. 1-Chloropropane: CH₃CH₂CH₂Cl (Primary)
  2. 2-Chloropropane: CH₃CHClCH₃ (Secondary)

📝 Quick Summary: Radical Stability

Order: 3° > 2° > 1° > CH₃•

Reason: Alkyl groups donate electrons by +I effect and hyperconjugation, stabilizing the free radical.

Application: More stable radical = Major product

Major Product: 2-Chloropropane (because secondary radical is more stable than primary radical)
⚠️ Exam Tip: When asked to predict the major product in free radical halogenation, always consider the stability of the intermediate radical. The more stable the radical, the more likely it will form.

Conditions Required

Reagent Condition
Cl₂ Sunlight (hν) or Heat (Δ)
Br₂ Sunlight (hν) or Heat (Δ)

Without light, no reaction occurs.

NCERT Points to Remember

  • Chlorination and bromination proceed by a free radical chain mechanism
  • UV light or heat is essential to generate free radicals
  • Homolytic bond cleavage occurs in the initiation step
  • Bromination is more selective than chlorination
  • Radical stability order: 3° > 2° > 1° > CH₃•

Board PYQs

Q1. Why is sunlight necessary during chlorination of methane?
Answer: Sunlight provides energy for the homolytic cleavage of the Cl–Cl bond, producing chlorine free radicals that initiate the reaction.

Q2. Name the three steps involved in free radical halogenation.
Answer:
  1. Initiation
  2. Propagation
  3. Termination


Q3. Define homolytic cleavage.
Answer: Homolytic cleavage is the equal breaking of a covalent bond such that each atom takes one electron, resulting in the formation of free radicals.

Q4. Why is bromination more selective than chlorination?
Answer: Bromination has a higher activation energy and therefore occurs through the formation of the more stable free radical, making it more selective than chlorination.
🔴 REACTION SHEET – PREPARATION FROM ALKANES
  1. RH + Cl₂ ──hν/Δ──► RCl + HCl
  2. RH + Br₂ ──hν/Δ──► RBr + HBr

Initiation:

  • Cl₂ ──hν──► Cl• + Cl•

Propagation:

  • Cl• + CH₄ ─► CH₃• + HCl
  • CH₃• + Cl₂ ─► CH₃Cl + Cl•

Termination:

  • Cl• + Cl• ─► Cl₂
  • CH₃• + Cl• ─► CH₃Cl
  • CH₃• + CH₃• ─► C₂H₆

Methane Chlorination Series:

  • CH₄ + Cl₂ ──hν──► CH₃Cl + HCl
  • CH₃Cl + Cl₂ ──hν──► CH₂Cl₂ + HCl
  • CH₂Cl₂ + Cl₂ ──hν──► CHCl₃ + HCl
  • CHCl₃ + Cl₂ ──hν──► CCl₄ + HCl

PART 2 (CONTINUED) – PREPARATION FROM ALKENES

Board Weightage: ★★★★★ (One of the most important topics)

Introduction

  • Alkenes contain a carbon-carbon double bond (C=C)
  • The π-bond is weaker than the σ-bond
  • Therefore, the π-bond breaks easily
  • Halogens or hydrogen halides add across the double bond
  • This is called an Addition Reaction

General Reaction

Alkene + HX ──► Haloalkane

where HX = HCl, HBr, or HI

Why are Alkenes More Reactive?

Double bond
 ↓
Contains π bond
 ↓
π bond is weak
 ↓
Breaks easily
 ↓
Electrophilic addition occurs

Addition of Hydrogen Halides (HX)

General Reaction: RCH=CH₂ + HX ──► Haloalkane

HX may be: HCl, HBr, or HI

Examples:

  1. Ethene + HCl: CH₂=CH₂ + HCl ──► CH₃CH₂Cl (Chloroethane)
  2. Ethene + HBr: CH₂=CH₂ + HBr ──► CH₃CH₂Br (Bromoethane)
  3. Ethene + HI: CH₂=CH₂ + HI ──► CH₃CH₂I (Iodoethane)

Unsymmetrical Alkenes

This is where Markovnikov Rule comes into the picture.

What is an Unsymmetrical Alkene? An alkene in which both carbon atoms have different numbers of hydrogen atoms.
Example: CH₃CH=CH₂ (Propene)
  • One carbon: CH₃ (3 H)
  • Other carbon: CH₂ (2 H)
  • Different → Unsymmetrical

🔴 MARKOVNIKOV RULE ⭐⭐⭐⭐⭐

Most important theory of this chapter.

Definition

When an unsymmetrical reagent (HX) adds to an unsymmetrical alkene:

  • Hydrogen goes to the carbon already containing more hydrogen atoms
  • Halogen goes to the carbon having fewer hydrogen atoms
🧠 Easy Memory Trick:

"Rich Become Richer"

Hydrogen-rich carbon → Gets another hydrogen

Example: Propene

CH₃CH=CH₂

Hydrogen count:

  • CH₃ = 3 H
  • CH = 1 H
  • CH₂ = 2 H

Carbon with more hydrogen: CH₂

Therefore:

  • H goes to CH₂
  • Br goes to middle carbon
Product: CH₃CHBrCH₃ (2-Bromopropane)

Mechanism of Markovnikov Addition

STEP 1: PROTONATION
↓
CH₃CH=CH₂ + H⁺ → ?
(Double bond attacks H⁺)
↓
Two possible carbocations form:
1. Primary carbocation (CH₃CH₂CH₂⁺) - UNSTABLE
2. Secondary carbocation (CH₃CH⁺CH₃) - STABLE ✓
↓
STEP 2: NUCLEOPHILIC ATTACK
↓
Br⁻ attacks the MORE STABLE carbocation
↓
CH₃CH⁺CH₃ + Br⁻ → CH₃CHBrCH₃
↓
PRODUCT: 2-Bromopropane
Step 1: Double bond attacks H⁺. Carbocation forms.

Two possibilities:
  1. Carbocation 1: Primary Carbocation (Unstable)
  2. Carbocation 2: Secondary Carbocation (More stable)

Reaction proceeds through secondary carbocation.

Step 2: Br⁻ attacks.

Final product: 2-Bromopropane

Why is Markovnikov Product Major?

Because reaction proceeds through the most stable carbocation

🧠 Stability:

3° > 2° > 1° > CH₃⁺

Exactly like free radicals.

📝 Quick Summary: Markovnikov Rule

Rule: "Rich get Richer" - Hydrogen goes to carbon with more H atoms

Mechanism: Carbocation formation → Most stable carbocation forms → Nucleophile attacks

Key Point: Stability determines the product

Another Example: But-1-ene

CH₂=CHCH₂CH₃ + HCl
Major product: CH₃CHClCH₂CH₃ (2-Chlorobutane)

Board Question

Q. State Markovnikov Rule.
Answer: When an unsymmetrical reagent adds to an unsymmetrical alkene, the negative part attaches to the carbon atom carrying fewer hydrogen atoms, while hydrogen attaches to the carbon atom carrying more hydrogen atoms.

🔴 ANTI-MARKOVNIKOV RULE (PEROXIDE EFFECT) ⭐⭐⭐⭐⭐

Also called Peroxide Effect

Definition

In the presence of Organic Peroxides (ROOR), the addition of HBr takes place opposite to Markovnikov Rule.

Therefore:

  • Br attaches to carbon having more hydrogen atoms

Important Condition

🧠 Works ONLY with HBr
  • Not with HCl ❌
  • Not with HI ❌

Board Favourite Question

Q. Why does peroxide effect occur only with HBr?
Answer: The free radical mechanism is energetically favourable only for HBr. It does not occur with HCl or HI because the reaction steps become energetically unfavourable.

Example: Propene

CH₃CH=CH₂ + HBr + Peroxide
Product: CH₃CH₂CH₂Br (1-Bromopropane)

Comparison

Condition Reaction Product
Without Peroxide CH₃CH=CH₂ + HBr CH₃CHBrCH₃ (2-Bromopropane)
With Peroxide CH₃CH=CH₂ + HBr + ROOR CH₃CH₂CH₂Br (1-Bromopropane)
🧠 Memory Trick:

Peroxide → Breaks → Free Radicals → Free Radical Mechanism → Anti-Markovnikov

Why only HBr?

🧠 Bond Energies:
  • HCl: Very strong bond → Reaction not favourable
  • HBr: Perfect bond strength → Reaction proceeds
  • HI: Very weak → Chain reaction not maintained

Hence, only HBr shows peroxide effect.

Mechanism of Peroxide Effect

Three steps (exactly like free radical chlorination):

STEP 1: INITIATION
↓
ROOR → 2RO•
(Peroxide breaks into radicals)
↓
RO• + HBr → ROH + Br•
(Bromine radical formed)
↓
STEP 2: PROPAGATION
↓
Br• + CH₃CH=CH₂ → CH₃CH•CH₂Br
(Br adds to carbon with MORE H)
↓
CH₃CH•CH₂Br + HBr → CH₃CH₂CH₂Br + Br•
(Br radical regenerated - chain continues)
↓
STEP 3: TERMINATION
↓
Br• + Br• → Br₂
R• + Br• → RBr
↓
REACTION STOPS

Initiation

ROOR ──► 2RO•
Then: RO• + HBr ──► ROH + Br•

Propagation

Br• + CH₃CH=CH₂ ──► CH₃CH•CH₂Br
Then: CH₃CH•CH₂Br + HBr ──► CH₃CH₂CH₂Br + Br•

Again, Br radical is regenerated.

Termination

Br• + Br• ──► Br₂
or
R• + Br• ──► RBr

Reaction stops.
⚠️ Critical Exam Point: Notice that in the propagation step, Br• adds to the carbon with MORE hydrogen atoms (the CH₂ end), not the carbon with fewer H atoms. This is the OPPOSITE of Markovnikov's rule!

Addition of Halogens

Very Important

Alkenes react with Cl₂ or Br₂ to form Vicinal Dihalides.

Examples:

  1. Ethene + Br₂: CH₂=CH₂ + Br₂ ──► CH₂BrCH₂Br (1,2-Dibromoethane)
  2. Ethene + Cl₂: CH₂=CH₂ + Cl₂ ──► CH₂ClCH₂Cl (1,2-Dichloroethane)

Test for Unsaturation

This reaction is used in laboratory.

🧠 Bromine water test:
  • Brown colour
  • Alkene added
  • Brown colour disappears

Reason: Br₂ is consumed. Hence, presence of double bond.

NCERT Important Notes

  • Addition to alkenes is an electrophilic addition reaction
  • Markovnikov's rule applies to addition of HX to unsymmetrical alkenes
  • Peroxide effect is shown only by HBr
  • Addition of Cl₂ or Br₂ forms vicinal dihalides
  • Bromine water is decolourised by alkenes because bromine adds across the double bond

Comparison Table

Feature Markovnikov Anti-Markovnikov
Reagent HX HBr + ROOR
Mechanism Carbocation Free radical
Br goes to Carbon with fewer H Carbon with more H
Major Product (Propene) 2-Bromopropane 1-Bromopropane
Condition Normal Presence of peroxide

Board PYQs

Q1. State Markovnikov's rule.
Answer: When an unsymmetrical reagent (HX) adds to an unsymmetrical alkene, hydrogen attaches to the carbon atom already containing more hydrogen atoms, while the halide attaches to the carbon atom containing fewer hydrogen atoms.

Q2. Why does peroxide effect occur only with HBr?
Answer: The free radical chain mechanism is energetically favourable only for HBr. It is not favourable for HCl or HI.

Q3. What is the major product formed when propene reacts with HBr in the presence of peroxide?
Answer: 1-Bromopropane

Q4. What is formed when ethene reacts with bromine?
Answer: 1,2-Dibromoethane, a vicinal dihalide.
🧠 MEMORY TRICKS
  1. Mechanism: I – P – T → I Prefer Tea (I = Initiation, P = Propagation, T = Termination)
  2. Radical Stability: 3 > 2 > 1 > M (Tertiary > Secondary > Primary > Methyl)
  3. Methane Chlorination Series: CH₄ → CH₃Cl → CH₂Cl₂ → CHCl₃ → CCl₄
  4. Markovnikov Rule: Rich get Richer (Carbon with more H → Gets another H)
  5. Peroxide Effect: P = Peroxide = Position Reversed (Normal product becomes reversed)
  6. Exception: Only HBr (Mnemonic: "Bromine Believes in Peroxide.")
🔴 REACTION SHEET – PREPARATION FROM ALKENES
  1. CH₂=CH₂ + HCl → CH₃CH₂Cl
  2. CH₂=CH₂ + HBr → CH₃CH₂Br
  3. CH₂=CH₂ + HI → CH₃CH₂I
  4. CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (Markovnikov)
  5. CH₃CH=CH₂ + HBr + ROOR → CH₃CH₂CH₂Br (Anti-Markovnikov)
  6. CH₂=CH₂ + Br₂ → CH₂BrCH₂Br
  7. CH₂=CH₂ + Cl₂ → CH₂ClCH₂Cl
📝 Exam Tip:
  • If you see: HBr + Peroxide (ROOR) → immediately think Anti-Markovnikov
  • HCl or HI, even with peroxide → No peroxide effect; use Markovnikov's rule

PART 3 – PREPARATION OF HALOARENES

Board Weightage: ★★★★★ (Very Important)

Introduction

Haloarenes cannot be prepared by free radical halogenation like alkanes because:

  • The benzene ring is highly stable due to resonance
  • Addition reactions would destroy aromaticity
  • Therefore, benzene undergoes electrophilic substitution instead of addition
🧠 Exam Point: Benzene prefers substitution over addition to preserve aromaticity.

METHOD 1 – DIRECT HALOGENATION OF BENZENE

General Reaction

C₆H₆ + X₂ ──FeX₃/AlX₃──► C₆H₅X + HX

where X = Cl or Br

This is an Electrophilic Aromatic Substitution (EAS) reaction.

A. Chlorination of Benzene ⭐⭐⭐⭐⭐

Reaction: C₆H₆ + Cl₂ ──FeCl₃/AlCl₃──► C₆H₅Cl + HCl

Product: Chlorobenzene

Catalyst Used

  • FeCl₃
  • AlCl₃

Both are Lewis acids.

Role of FeCl₃

FeCl₃ reacts with Cl₂ to generate the electrophile (Cl⁺).

Cl₂ + FeCl₃ → Cl⁺ + FeCl₄⁻

The Cl⁺ ion attacks the benzene ring.

Why is FeCl₃ Needed?

Without FeCl₃

  • Cl₂ is not electrophilic enough
  • Benzene does not react easily

With FeCl₃

  • Cl⁺ is produced
  • Reaction becomes fast

Mechanism of Chlorination

Step 1 – Electrophile Formation:
Cl₂ + FeCl₃ ──► Cl⁺ + FeCl₄⁻

Step 2 – Attack on Benzene:
The π electrons of benzene attack Cl⁺.
A positively charged intermediate (σ-complex or arenium ion) is formed.

Step 3 – Loss of H⁺:
Hydrogen leaves. Aromaticity is restored.

Final product: Chlorobenzene
🧠 Memory Trick:

EAS = Electrophile → Attacks → Substitution → Not Addition

B. Bromination of Benzene ⭐⭐⭐⭐⭐

Reaction: C₆H₆ + Br₂ ──FeBr₃──► C₆H₅Br + HBr

Product: Bromobenzene

Catalyst: FeBr₃

Mechanism is exactly the same as chlorination.

C. Iodination of Benzene

Reaction: C₆H₆ + I₂ ──Oxidising agent──► C₆H₅I

Why oxidising agent?

Because HI formed during the reaction reverses the reaction.

Common oxidising agents:

  • HNO₃
  • HIO₃

They remove HI, shifting the reaction forward.

D. Fluorination

Direct fluorination is not preferred.

Reason:

  • Fluorine is extremely reactive
  • Reaction becomes violent and difficult to control
  • Therefore, fluorobenzene is generally prepared by the Balz–Schiemann reaction

PREPARATION FROM DIAZONIUM SALTS

Very Important for Board Exams

What is a Diazonium Salt?

General Formula: ArN₂⁺X⁻

Example: C₆H₅N₂⁺Cl⁻ (Benzenediazonium chloride)

Preparation of Benzenediazonium Chloride

Aniline
↓
NaNO₂ + HCl
0–5°C
↓
C₆H₅N₂⁺Cl⁻
Reaction: C₆H₅NH₂ + NaNO₂ + 2HCl ──0-5°C──► C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
🧠 Important:
  • Temperature must be maintained at 0-5°C
  • Higher temperatures cause decomposition

🔴 SANDMEYER REACTION ⭐⭐⭐⭐⭐

Definition

The Sandmeyer reaction is used to replace the diazonium group with various halogens using copper salts as catalysts.

General Reaction

ArN₂⁺X⁻ + CuX ──► ArX + N₂↑

Specific Reactions

  1. Sandmeyer Chlorination: C₆H₅N₂⁺Cl⁻ + CuCl ──► C₆H₅Cl + N₂↑
  2. Sandmeyer Bromination: C₆H₅N₂⁺Cl⁻ + CuBr ──► C₆H₅Br + N₂↑
  3. Sandmeyer Cyanation: C₆H₅N₂⁺Cl⁻ + CuCN ──► C₆H₅CN + N₂↑

Mechanism

  1. Diazonium salt reacts with Cu(I) salt
  2. Free radical mechanism
  3. Nitrogen gas is evolved
  4. Aryl halide is formed

Board Question

Q. What is the Sandmeyer reaction?
Answer: The Sandmeyer reaction is a chemical reaction used to synthesize aryl halides from aryl diazonium salts using copper(I) halides as catalysts.

🔴 GATTERMANN REACTION ⭐⭐⭐⭐

Definition

The Gattermann reaction is similar to the Sandmeyer reaction but uses Cu powder and HX instead of CuX.

General Reaction

ArN₂⁺X⁻ + Cu + HX ──► ArX + N₂↑

Specific Reactions

  1. Gattermann Chlorination: C₆H₅N₂⁺Cl⁻ + Cu + HCl ──► C₆H₅Cl + N₂↑
  2. Gattermann Bromination: C₆H₅N₂⁺Cl⁻ + Cu + HBr ──► C₆H₅Br + N₂↑

🔴 BALZ–SCHIEMANN REACTION ⭐⭐⭐⭐⭐

Definition

The Balz–Schiemann reaction is used to prepare aryl fluorides from diazonium salts.

General Reaction

ArN₂⁺BF₄⁻ ──Heat──► ArF + N₂↑ + BF₃

Specific Reaction

C₆H₅N₂⁺BF₄⁻ ──Heat──► C₆H₅F + N₂↑ + BF₃

Why is this important?

  • Direct fluorination of benzene is not possible
  • This is the only practical method to prepare fluorobenzene

Board Question

Q. How is fluorobenzene prepared?
Answer: Fluorobenzene is prepared by the Balz–Schiemann reaction. Benzenediazonium fluoroborate is heated to give fluorobenzene, nitrogen gas, and boron trifluoride.

NCERT Points to Remember

  • Diazonium salts are prepared at 0-5°C
  • Sandmeyer reaction uses CuX (X = Cl, Br, CN)
  • Gattermann reaction uses Cu powder + HX
  • Balz–Schiemann reaction is used for fluorination
  • All these reactions evolve nitrogen gas
🔴 REACTION SHEET – PREPARATION OF HALOARENES

Direct Halogenation:

  1. C₆H₆ + Cl₂ ──FeCl₃──► C₆H₅Cl + HCl
  2. C₆H₆ + Br₂ ──FeBr₃──► C₆H₅Br + HBr
  3. C₆H₆ + I₂ ──HNO₃──► C₆H₅I (oxidising agent required)

Diazonium Salt Preparation:

  1. C₆H₅NH₂ + NaNO₂ + 2HCl ──0-5°C──► C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O

Sandmeyer Reaction:

  1. C₆H₅N₂⁺Cl⁻ + CuCl ──► C₆H₅Cl + N₂↑
  2. C₆H₅N₂⁺Cl⁻ + CuBr ──► C₆H₅Br + N₂↑
  3. C₆H₅N₂⁺Cl⁻ + CuCN ──► C₆H₅CN + N₂↑

Gattermann Reaction:

  1. C₆H₅N₂⁺Cl⁻ + Cu + HCl ──► C₆H₅Cl + N₂↑
  2. C₆H₅N₂⁺Cl⁻ + Cu + HBr ──► C₆H₅Br + N₂↑

Balz–Schiemann Reaction:

  1. C₆H₅N₂⁺BF₄⁻ ──Heat──► C₆H₅F + N₂↑ + BF₃

PART 4 – PHYSICAL PROPERTIES

Board Weightage: ★★★☆☆

State

  • Lower alkanes (C₁-C₄): Gases
  • C₅-C₁₀: Liquids
  • Higher: Solids

Colour

  • Colourless

Odour

  • Sweetish characteristic odour

Solubility

  • Insoluble in water
  • Soluble in organic solvents (ether, alcohol, chloroform)

Boiling Point

  • Increases with molecular weight
  • For isomeric haloalkanes: 1° < 2° < 3° (due to surface area)
  • Order: RI > RBr > RCl > RF (for same alkyl group)

Melting Point

  • Generally increases with molecular weight
  • Symmetrical molecules have higher melting points

Density

  • Most haloalkanes are denser than water
  • Order: RI > RBr > RCl > RF

Dipole Moment

  • CH₃F > CH₃Cl > CH₃Br > CH₃I (due to electronegativity difference)

Comparison Table

Compound State BP (°C) MP (°C) Density (g cm⁻³)
CH₃Cl Gas -24 -97 0.91
C₂H₅Br Liquid 38 -115 1.34
(CH₃)₃CCl Liquid 71 -50 1.19

Physical Properties of Haloarenes

  • Generally liquids (C₆-C₁₀) or solids (higher)
  • Higher boiling points than corresponding alkanes due to polarity
Compound BP (°C) MP (°C) Density (g cm⁻³)
Chlorobenzene 131 -45 1.11
Bromobenzene 156 -73 1.50
Iodobenzene 184 -84 1.93

PART 5 – CHEMICAL REACTIONS OF HALOALKANES

Board Weightage: ★★★★★ (Highest weightage)

Nucleophilic Substitution Reactions

SN1 Mechanism (Substitution Nucleophilic Unimolecular)

STEP 1: IONIZATION
↓
R-X → R⁺ + X⁻
(Carbocation formed - SLOW step)
↓
STEP 2: NUCLEOPHILIC ATTACK
↓
R⁺ + Nu⁻ → R-Nu
(Nucleophile attacks carbocation - FAST step)
  • Occurs with: Tertiary halides (3°)
  • Rate depends on: Only substrate concentration (Rate = k[RX])
  • Mechanism: Two-step process with carbocation intermediate
  • Stereochemistry: Racemisation occurs (50% retention, 50% inversion)
  • Solvent: Polar protic solvents favour SN1
Example: (CH₃)₃C-Br + OH⁻ → (CH₃)₃C-OH + Br⁻

SN2 Mechanism (Substitution Nucleophilic Bimolecular)

ONE STEP: BACKSIDE ATTACK
↓
Nu⁻ + R-X → [Nu---R---X]‡ → Nu-R + X⁻
(Transition state with pentacoordinate carbon)
↓
PRODUCT WITH INVERSION
  • Occurs with: Primary (1°) and secondary (2°) halides
  • Rate depends on: Both substrate and nucleophile (Rate = k[RX][Nu⁻])
  • Mechanism: One-step process with transition state
  • Stereochemistry: Inversion of configuration (Walden inversion)
  • Solvent: Polar aprotic solvents favour SN2
Example: CH₃CH₂-Br + OH⁻ → CH₃CH₂-OH + Br⁻

Comparison: SN1 vs SN2

Feature SN1 SN2
Substrate Tertiary (3°) Primary (1°), Secondary (2°)
Rate Law Rate = k[RX] Rate = k[RX][Nu⁻]
Mechanism Two-step (carbocation) One-step (transition state)
Stereochemistry Racemisation Inversion
Nucleophile strength Not important Important

Stereochemical Aspects of Nucleophilic Substitution

Board Weightage: ★★★★★ (Very Important)

Optical Activity

Compounds that rotate the plane of plane-polarised light are called optically active compounds.

  • Dextrorotatory (+): Rotates plane to the right (clockwise)
  • Laevorotatory (-): Rotates plane to the left (anticlockwise)

Chirality and Enantiomers

Chiral molecules: Molecules that are non-superimposable on their mirror images (like left and right hands).

Achiral molecules: Molecules that are superimposable on their mirror images.

Enantiomers: Stereoisomers that are non-superimposable mirror images of each other.

Racemic mixture: Equal mixture of two enantiomers (optically inactive, denoted as dl or ±).

Asymmetric Carbon (Stereocentre)

A carbon atom attached to four different groups is called asymmetric carbon or stereocentre.

Example: Butan-2-ol is chiral (has asymmetric carbon), while propan-2-ol is achiral (no asymmetric carbon).

Retention, Inversion and Racemisation

When a bond to an asymmetric carbon is broken during a reaction, three outcomes are possible:

  • Retention of configuration: Spatial arrangement remains the same
  • Inversion of configuration: Spatial arrangement is inverted (like umbrella turning inside out)
  • Racemisation: 50:50 mixture of both configurations (optically inactive)

Stereochemistry of SN1 and SN2 Reactions

SN2 Reactions

  • Nucleophile attacks from the side opposite to the leaving group
  • Result: Inversion of configuration (Walden inversion)
  • Example: (–)-2-bromooctane + NaOH → (+)-octan-2-ol

SN1 Reactions

  • Carbocation intermediate is planar (sp² hybridised, achiral)
  • Nucleophile can attack from either side
  • Result: Racemisation (50:50 mixture of retention and inversion)
  • Example: (–)-2-bromobutane + NaOH → (±)-butan-2-ol

📌 Important Board Points:

  • SN2 → Inversion of configuration
  • SN1 → Racemisation
  • Optical activity is due to molecular asymmetry (chirality)
  • Enantiomers have identical physical properties except optical rotation

Elimination Reactions

β-Elimination (Dehydrohalogenation)

  • Reagent: Alcoholic KOH
  • Product: Alkene

Zaitsev Rule

The more substituted alkene is the major product.

Hoffmann Rule

With bulky bases, the less substituted alkene is the major product.

Reactions with Metals

Wurtz Reaction

2R-X + 2Na ──dry ether──► R-R + 2NaX

Fittig Reaction

2Ar-X + 2Na ──dry ether──► Ar-Ar + 2NaX

Wurtz-Fittig Reaction

R-X + Ar-X + 2Na ──dry ether──► R-Ar + 2NaX

Organometallic Compounds

Grignard Reagent

R-X + Mg ──dry ether──► R-MgX
  • Preparation: Alkyl halide + magnesium in dry ether
  • Importance: Used in organic synthesis to form C-C bonds

Reduction

Reducing Agents

  • Zn + HCl
  • LiAlH₄
  • NaBH₄
  • H₂/Pd
R-X + 2[H] ──► R-H + HX
🔴 REACTION SHEET – CHEMICAL REACTIONS OF HALOALKANES

Nucleophilic Substitution:

  1. R-Cl + NaOH ──► R-OH + NaCl (SN1/SN2)
  2. R-Cl + NaCN ──► R-CN + NaCl
  3. R-Cl + NaNH₂ ──► R-NH₂ + NaCl
  4. R-Cl + AgNO₂ ──► R-NO₂ + AgCl

Elimination:

  1. R-CH₂-CH₂-Cl + KOH (alcoholic) ──► R-CH=CH₂ + KCl + H₂O

With Metals:

  1. 2R-Cl + 2Na ──dry ether──► R-R + 2NaCl (Wurtz)
  2. 2Ar-Cl + 2Na ──dry ether──► Ar-Ar + 2NaCl (Fittig)
  3. R-Cl + Ar-Cl + 2Na ──dry ether──► R-Ar + 2NaCl (Wurtz-Fittig)
  4. R-Cl + Mg ──dry ether──► R-MgCl (Grignard)

Reduction:

  1. R-Cl + Zn + HCl ──► R-H + ZnCl₂
  2. R-Cl + LiAlH₄ ──► R-H
  3. R-Cl + H₂ ──Pd──► R-H + HCl

Complete Nucleophilic Substitution Reactions

Important reagents and their substitution products:

1. With NaOH/KOH (Formation of Alcohol)

Reaction: R-X + NaOH ──► R-OH + NaX

Example: CH₃CH₂Cl + NaOH ──► CH₃CH₂OH + NaCl

Product: Alcohol

2. With H₂O (Formation of Alcohol)

Reaction: R-X + H₂O ──► R-OH + HX

Example: CH₃CH₂Cl + H₂O ──► CH₃CH₂OH + HCl

Product: Alcohol

3. With NaOR′ (Formation of Ether)

Reaction: R-X + NaOR′ ──► R-OR′ + NaX

Example: CH₃CH₂Cl + NaOCH₃ ──► CH₃CH₂OCH₃ + NaCl

Product: Ether

4. With NaI (Formation of Alkyl Iodide)

Reaction: R-X + NaI ──► R-I + NaX

Example: CH₃CH₂Cl + NaI ──► CH₃CH₂I + NaCl

Product: Alkyl iodide

5. With NH₃ (Formation of Primary Amine)

Reaction: R-X + NH₃ ──► R-NH₂ + HX

Example: CH₃CH₂Cl + NH₃ ──► CH₃CH₂NH₂ + HCl

Product: Primary amine

6. With R′NH₂ (Formation of Secondary Amine)

Reaction: R-X + R′NH₂ ──► R-NHR′ + HX

Example: CH₃CH₂Cl + CH₃NH₂ ──► CH₃CH₂NHCH₃ + HCl

Product: Secondary amine

7. With R′R′′NH (Formation of Tertiary Amine)

Reaction: R-X + R′R′′NH ──► R-NR′R′′ + HX

Example: CH₃CH₂Cl + (CH₃)₂NH ──► CH₃CH₂N(CH₃)₂ + HCl

Product: Tertiary amine

8. With KCN (Formation of Nitrile)

Reaction: R-X + KCN ──► R-CN + KX

Example: CH₃CH₂Cl + KCN ──► CH₃CH₂CN + KCl

Product: Nitrile (cyanide)

9. With AgCN (Formation of Isonitrile)

Reaction: R-X + AgCN ──► R-NC + AgX

Example: CH₃CH₂Cl + AgCN ──► CH₃CH₂NC + AgCl

Product: Isonitrile (isocyanide)

Note: AgCN gives isonitrile, KCN gives nitrile

10. With KNO₂ (Formation of Alkyl Nitrite)

Reaction: R-X + KNO₂ ──► R-ONO + KX

Example: CH₃CH₂Cl + KNO₂ ──► CH₃CH₂ONO + KCl

Product: Alkyl nitrite

11. With AgNO₂ (Formation of Nitroalkane)

Reaction: R-X + AgNO₂ ──► R-NO₂ + AgX

Example: CH₃CH₂Cl + AgNO₂ ──► CH₃CH₂NO₂ + AgCl

Product: Nitroalkane

Note: AgNO₂ gives nitroalkane, KNO₂ gives alkyl nitrite

12. With R′COOAg (Formation of Ester)

Reaction: R-X + R′COOAg ──► R′COOR + AgX

Example: CH₃CH₂Cl + CH₃COOAg ──► CH₃COOCH₂CH₃ + AgCl

Product: Ester

13. With LiAlH₄ (Formation of Hydrocarbon)

Reaction: R-X + LiAlH₄ ──► R-H + LiX + AlH₃

Example: CH₃CH₂Cl + LiAlH₄ ──► CH₃CH₃ + LiCl + AlH₃

Product: Hydrocarbon (alkane)

14. With R′-M⁺ (Formation of Alkane - Coupling)

Reaction: R-X + R′-M⁺ ──► R-R′ + M⁺X⁻

Example: CH₃CH₂Cl + CH₃MgBr ──► CH₃CH₂CH₃ + MgBrCl

Product: Alkane (coupling product)

📌 Important Notes:

  • AgCN vs KCN: AgCN gives isonitrile (R-NC) while KCN gives nitrile (R-CN)
  • AgNO₂ vs KNO₂: AgNO₂ gives nitroalkane (R-NO₂) while KNO₂ gives alkyl nitrite (R-ONO)
  • LiAlH₄: Strong reducing agent, converts haloalkanes to alkanes
  • Grignard (R-MgX): Acts as a source of R⁻ nucleophile for coupling reactions
  • Ammonolysis: Reaction with NH₃ gives primary amine, with secondary amines gives tertiary amines

PART 6 – CHEMICAL REACTIONS OF HALOARENES

Board Weightage: ★★★★☆

Why Haloarenes are Less Reactive

Reasons:

  1. Resonance: The lone pair on halogen participates in resonance with the benzene ring
  2. Partial double bond character: C-X bond acquires partial double bond character
  3. sp² carbon: The carbon is sp² hybridized (more electronegative than sp³)
  4. Bond length: C-X bond in haloarenes is shorter than in haloalkanes
🧠 Memory Trick:

R-S-B (Resonance, sp², Bond length) = Less reactive

Electrophilic Substitution

Haloarenes undergo electrophilic substitution reactions, but more slowly than benzene because halogens withdraw electron density by the -I effect.

Why halogen is ortho/para directing?

Although halogen is electron-withdrawing by inductive effect, it donates electron density to the ring by resonance through its lone pair. This resonance stabilizes the sigma complex formed during attack at ortho and para positions more than at meta position. Thus, haloarenes are deactivated yet ortho/para directing.

NCERT reactions to remember:

  1. Halogenation: C₆H₅Cl + Cl₂ ──anhyd. FeCl₃──► o-dichlorobenzene + p-dichlorobenzene + HCl
  2. Nitration: C₆H₅Cl + HNO₃ ──conc. H₂SO₄──► o-nitrochlorobenzene + p-nitrochlorobenzene + H₂O
  3. Sulphonation: C₆H₅Cl + H₂SO₄ ──heat──► o-chlorobenzenesulphonic acid + p-chlorobenzenesulphonic acid + H₂O
Exam note: In these reactions, para product is generally formed in greater amount than ortho product because substitution at para position faces less steric hindrance.

📌 Important Note:

Haloarenes generally do not undergo Friedel-Crafts reaction easily. The ring is deactivated by the halogen and the halogen can also form a complex with Lewis acid catalysts such as AlCl₃.

Nucleophilic Substitution

Haloarenes are less reactive towards nucleophilic substitution.

Reasons for low reactivity:

  1. Resonance effect: Lone pair on halogen conjugates with benzene ring, giving C-X bond partial double bond character
  2. Hybridisation difference: Carbon is sp² (more electronegative) in haloarenes vs sp³ in haloalkanes
  3. Bond length: C-X bond is shorter in haloarenes (169 pm vs 177 pm), harder to break
  4. Instability of phenyl cation: Phenyl cation is not stabilized by resonance, SN1 ruled out
  5. Electron repulsion: Nucleophile repelled by electron-rich arene ring

Conditions required:

  • High temperature and pressure
  • Strong electron-withdrawing groups (NO₂) ortho/para to halogen

Dow's Process (Replacement by hydroxyl group)

Chlorobenzene can be converted to phenol by heating with aqueous NaOH at 623K and 300 atm pressure.

C₆H₅Cl + NaOH ──623K, 300atm──► C₆H₅ONa + NaCl C₆H₅ONa + HCl ──► C₆H₅OH + NaCl

Effect of NO₂ group

The presence of electron-withdrawing NO₂ group at ortho and para positions increases reactivity of haloarenes towards nucleophilic substitution. The NO₂ group withdraws electron density from the ring and stabilizes the carbanion intermediate through resonance. No effect is observed at meta position.

Benzyne Mechanism

Under extreme conditions, elimination-addition occurs via benzyne intermediate.

Reaction with Metals

Haloarenes undergo sodium-metal coupling reactions in dry ether to form larger hydrocarbons.

Wurtz-Fittig Reaction

A mixture of an alkyl halide and an aryl halide reacts with sodium in dry ether to give an alkylarene.

R-X + Ar-X + 2Na ──dry ether──► R-Ar + 2NaX

Example: CH₃Cl + C₆H₅Cl + 2Na ──dry ether──► C₆H₅CH₃ + 2NaCl

Fittig Reaction

Two molecules of an aryl halide react with sodium in dry ether to give a biaryl compound.

2Ar-X + 2Na ──dry ether──► Ar-Ar + 2NaX

Example: 2C₆H₅Cl + 2Na ──dry ether──► C₆H₅-C₆H₅ + 2NaCl

📌 Coupling Shortcut:

  • Wurtz: alkyl + alkyl → alkane
  • Fittig: aryl + aryl → biaryl
  • Wurtz-Fittig: alkyl + aryl → alkylarene
🔴 REACTION SHEET – REACTIONS OF HALOARENES

Electrophilic Substitution:

  1. C₆H₅Cl + Cl₂ ──anhyd. FeCl₃──► o-C₆H₄Cl₂ + p-C₆H₄Cl₂ + HCl (Halogenation)
  2. C₆H₅Cl + HNO₃ ──conc. H₂SO₄──► o-NO₂C₆H₄Cl + p-NO₂C₆H₄Cl + H₂O (Nitration)
  3. C₆H₅Cl + H₂SO₄ ──heat──► o-HO₃SC₆H₄Cl + p-HO₃SC₆H₄Cl + H₂O (Sulphonation)

Nucleophilic Substitution:

  1. C₆H₅Cl + NaOH ──623K, 300atm──► C₆H₅OH + NaCl (Dow's process)

With Metals:

  1. R-X + Ar-X + 2Na ──dry ether──► R-Ar + 2NaX (Wurtz-Fittig)
  2. 2Ar-X + 2Na ──dry ether──► Ar-Ar + 2NaX (Fittig)

📌 Important Board Points:

  • Haloarenes are less reactive than haloalkanes due to resonance and sp² hybridisation
  • Halogen is ortho/para directing but deactivating in electrophilic substitution
  • NO₂ at ortho/para positions increases reactivity towards nucleophilic substitution
  • Electrophilic substitution in haloarenes mainly includes halogenation, nitration and sulphonation
  • Dow's process requires extreme conditions (623K, 300 atm)
  • Haloarenes react with sodium in dry ether to give Fittig and Wurtz-Fittig coupling products

PART 7 – POLYHALOGEN COMPOUNDS

Board Weightage: ★★★☆☆

Chloroform (CHCl₃)

  • Preparation: Chlorination of methane or by haloform reaction
  • Properties: Sweet-smelling liquid, heavy
  • Uses: Previously used as anaesthetic, now limited
  • Harmful effects: Carcinogenic, liver damage

Carbon Tetrachloride (CCl₄)

  • Preparation: Exhaustive chlorination of methane
  • Properties: Colourless liquid, non-flammable
  • Uses: Solvent, fire extinguisher (now limited)
  • Harmful effects: Liver and kidney damage, ozone depletion

Freons (CCl₂F₂, CCl₃F)

  • Preparation: Manufactured via halogen exchange
  • Properties: Colourless gases, non-toxic
  • Uses: Refrigerants, propellants
  • Harmful effects: Ozone-depleting (banned by Montreal Protocol)

DDT (C₁₄H₉Cl₅)

  • Preparation: Reaction of chloral with chlorobenzene
  • Properties: White crystalline solid
  • Uses: Potent insecticide (banned)
  • Harmful effects: Bioaccumulation, environmental persistence

Iodoform (CHI₃)

  • Preparation: Haloform reaction of methyl ketones with I₂/NaOH
  • Properties: Yellow crystalline solid, characteristic odour
  • Uses: Antiseptic
🔴 REACTION SHEET – POLYHALOGEN COMPOUNDS
  1. CH₄ + 3Cl₂ ──hν──► CCl₄ + 6HCl
  2. CH₃COCH₃ + 3I₂ + 4NaOH ──► CHI₃ + CH₃COONa + 3NaI + 3H₂O (Iodoform test)

📚 FINAL REVISION SHEET

30-Second Revision – Haloalkanes & Haloarenes:

• Haloalkane = C-X (X = F,Cl,Br,I). Classification: primary/secondary/tertiary; mono-/di-/poly-.
• SN1 (tert), SN2 (prim/sec), E1/E2 for elimination, Wurtz, Grignard, reduction.
• Preparation: from alcohols (HX), alkanes (radical), alkenes (addition).
• Haloarenes less reactive; prepare via direct halogenation, Sandmeyer, Gattermann, Balz-Schiemann.
• Polyhalogen compounds – chloroform, CCl₄, freons, DDT, iodoform – know preparation & hazards.

Key Reactions to Memorise

From Alcohols:

  1. ROH + HCl/ZnCl₂ → RCl + H₂O
  2. ROH + SOCl₂ → RCl + SO₂ + HCl
  3. 3ROH + PCl₃ → 3RCl + H₃PO₃

From Alkanes:

  1. RH + Cl₂ ──hν──► RCl + HCl
  2. Mechanism: Initiation → Propagation → Termination

From Alkenes:

  1. CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (Markovnikov)
  2. CH₃CH=CH₂ + HBr + ROOR → CH₃CH₂CH₂Br (Anti-Markovnikov)

Haloarenes:

  1. C₆H₆ + Cl₂ ──FeCl₃──► C₆H₅Cl + HCl
  2. C₆H₅N₂⁺Cl⁻ + CuCl → C₆H₅Cl + N₂↑ (Sandmeyer)
  3. C₆H₅N₂⁺BF₄⁻ ──Heat──► C₆H₅F + N₂↑ (Balz-Schiemann)

Memory Tricks Summary

  1. I-P-T: Initiation-Propagation-Termination (Free radical mechanism)
  2. 3 > 2 > 1 > M: Radical stability order
  3. Rich get Richer: Markovnikov rule
  4. Peroxide = Position Reversed: Anti-Markovnikov (only HBr)
  5. R-S-B: Resonance, sp², Bond length (Haloarenes less reactive)

Board Exam Tips

  1. Always write balanced chemical equations
  2. Mention mechanisms for 5-mark questions
  3. Draw structures clearly
  4. Remember exceptions (peroxide effect only with HBr)
  5. Practice numerical problems on physical properties

🔄 IMPORTANT CONVERSIONS (Maximum 2-Step Reactions)

Board Weightage: ★★★★★ (Very Important)

Alkane to Alkene

Ethane → Ethene:

  1. CH₃CH₃ + Cl₂ ──hν──► CH₃CH₂Cl + HCl (Chlorination)
  2. CH₃CH₂Cl + KOH (alc) ──Δ──► CH₂=CH₂ + KCl + H₂O (β-Elimination)

Alkene to Alkane

Ethene → Ethane:

  1. CH₂=CH₂ + HBr ──► CH₃CH₂Br
  2. CH₃CH₂Br + Zn + H⁺ ──► CH₃CH₃ + ZnBr₂ (Reduction)

Direct: CH₂=CH₂ + H₂ ──Ni──► CH₃CH₃

Alcohol to Alkene

Ethanol → Ethene:

  1. CH₃CH₂OH + PCl₅ ──► CH₃CH₂Cl + POCl₃ + HCl
  2. CH₃CH₂Cl + KOH (alc) ──Δ──► CH₂=CH₂ + KCl + H₂O

Direct: CH₃CH₂OH ──conc. H₂SO₄, 170°C──► CH₂=CH₂ + H₂O

Alcohol to Alkane

Ethanol → Ethane:

  1. CH₃CH₂OH + SOCl₂ ──► CH₃CH₂Cl + SO₂ + HCl
  2. 2CH₃CH₂Cl + 2Na ──dry ether──► CH₃CH₂CH₂CH₃ + 2NaCl (Wurtz)

Alcohol to Haloalkane

Ethanol → Chloroethane: CH₃CH₂OH + HCl/ZnCl₂ ──► CH₃CH₂Cl + H₂O

Ethanol → Bromoethane: CH₃CH₂OH + HBr ──► CH₃CH₂Br + H₂O

Ethanol → Iodoethane: CH₃CH₂OH + PI₃ ──► CH₃CH₂I + H₃PO₃

Haloalkane to Alcohol

Chloroethane → Ethanol: CH₃CH₂Cl + KOH (aq) ──► CH₃CH₂OH + KCl

Haloalkane to Amine

Chloroethane → Ethylamine: CH₃CH₂Cl + NH₃ (excess) ──► CH₃CH₂NH₂ + HCl

Haloalkane to Nitrile

Chloroethane → Ethanenitrile: CH₃CH₂Cl + KCN ──► CH₃CH₂CN + KCl

Nitrile to Amine

Ethanenitrile → Ethylamine: CH₃CH₂CN + 4[H] ──Ni──► CH₃CH₂CH₂NH₂

Haloalkane to Grignard

Chloroethane → Ethylmagnesium chloride: CH₃CH₂Cl + Mg ──dry ether──► CH₃CH₂MgCl

Grignard to Alcohol

Ethylmagnesium chloride → Ethanol: CH₃CH₂MgCl + H₂O ──► CH₃CH₂OH + Mg(OH)Cl

Benzene to Chlorobenzene

C₆H₆ + Cl₂ ──FeCl₃──► C₆H₅Cl + HCl

Chlorobenzene to Phenol

  1. C₆H₅Cl + NaOH ──623K, 300atm──► C₆H₅ONa + NaCl (Dow's process)
  2. C₆H₅ONa + HCl ──► C₆H₅OH + NaCl

Aniline to Chlorobenzene (Sandmeyer)

  1. C₆H₅NH₂ + NaNO₂ + HCl ──0-5°C──► C₆H₅N₂⁺Cl⁻ + 2H₂O
  2. C₆H₅N₂⁺Cl⁻ + CuCl ──► C₆H₅Cl + N₂↑ + CuCl₂

Aniline to Bromobenzene (Sandmeyer)

  1. C₆H₅NH₂ + NaNO₂ + HCl ──0-5°C──► C₆H₅N₂⁺Cl⁻ + 2H₂O
  2. C₆H₅N₂⁺Cl⁻ + CuBr ──► C₆H₅Br + N₂↑ + CuBr₂

Aniline to Fluorobenzene (Balz-Schiemann)

  1. C₆H₅NH₂ + NaNO₂ + HCl ──0-5°C──► C₆H₅N₂⁺Cl⁻ + 2H₂O
  2. C₆H₅N₂⁺Cl⁻ + HBF₄ ──► C₆H₅N₂⁺BF₄⁻ ──Δ──► C₆H₅F + N₂↑ + BF₃

Aniline to Iodobenzene

  1. C₆H₅NH₂ + NaNO₂ + HCl ──0-5°C──► C₆H₅N₂⁺Cl⁻ + 2H₂O
  2. C₆H₅N₂⁺Cl⁻ + KI ──► C₆H₅I + N₂↑ + KCl

Ethane to Ethanol

  1. CH₃CH₃ + Cl₂ ──hν──► CH₃CH₂Cl + HCl
  2. CH₃CH₂Cl + KOH (aq) ──► CH₃CH₂OH + KCl

Ethanol to Ethanoic Acid

  1. CH₃CH₂OH ──K₂Cr₂O₇/H₂SO₄──► CH₃CHO
  2. CH₃CHO ──K₂Cr₂O₇/H₂SO₄──► CH₃COOH

Ethene to Ethanol

  1. CH₂=CH₂ + HBr ──► CH₃CH₂Br
  2. CH₃CH₂Br + KOH (aq) ──► CH₃CH₂OH + KBr

Direct: CH₂=CH₂ + H₂O ──H⁺──► CH₃CH₂OH

Ethene to 1,2-Dibromoethane

CH₂=CH₂ + Br₂ ──CCl₄──► CH₂BrCH₂Br

1,2-Dibromoethane to Ethyne

CH₂BrCH₂Br + 2KOH (alc) ──Δ──► HC≡CH + 2KBr + 2H₂O

Alkene to Haloalkane

Ethene → Chloroethane: CH₂=CH₂ + HCl ──► CH₃CH₂Cl

Ethene → Bromoethane: CH₂=CH₂ + HBr ──► CH₃CH₂Br

Haloalkane to Alkene

Chloroethane → Ethene: CH₃CH₂Cl + KOH (alc) ──Δ──► CH₂=CH₂ + KCl + H₂O

Haloalkane to Alkane

Chloroethane → Ethane: CH₃CH₂Cl + Zn + H⁺ ──► CH₃CH₃ + ZnCl₂

Benzene to Bromobenzene

C₆H₆ + Br₂ ──FeBr₃──► C₆H₅Br + HBr

Benzene to Iodobenzene

C₆H₆ + I₂ ──HNO₃──► C₆H₅I + HI

Ethanol to Diethyl Ether

2CH₃CH₂OH ──conc. H₂SO₄, 140°C──► CH₃CH₂OCH₂CH₃ + H₂O

Ethanol to Acetaldehyde

CH₃CH₂OH ──K₂Cr₂O₇/H₂SO₄──► CH₃CHO + H₂O

Ethanol to Ethane

  1. CH₃CH₂OH + PCl₅ ──► CH₃CH₂Cl + POCl₃ + HCl
  2. CH₃CH₂Cl + 2[H] ──Zn/HCl──► CH₃CH₃ + HCl

Ethyne to Ethene

HC≡CH + H₂ ──Lindlar catalyst──► CH₂=CH₂

Ethyne to Ethane

HC≡CH + 2H₂ ──Ni──► CH₃CH₃

Ethene to Ethyne

  1. CH₂=CH₂ + Br₂ ──CCl₄──► CH₂BrCH₂Br
  2. CH₂BrCH₂Br + 2KOH (alc) ──Δ──► HC≡CH + 2KBr + 2H₂O

1-Bromopropane to Propene

CH₃CH₂CH₂Br + KOH (alc) ──Δ──► CH₃CH=CH₂ + KBr + H₂O

2-Bromopropane to Propene

CH₃CHBrCH₃ + KOH (alc) ──Δ──► CH₃CH=CH₂ + KBr + H₂O

Chloroform to Dichloromethane

CHCl₃ + Zn + HCl ──► CH₂Cl₂ + ZnCl₂

Carbon Tetrachloride to Chloroform

CCl₄ + Zn + HCl ──► CHCl₃ + ZnCl₂

Phenol to Chlorobenzene

  1. C₆H₅OH + PCl₅ ──► C₆H₅Cl + POCl₃ + HCl

Phenol to Bromobenzene

  1. C₆H₅OH + PBr₃ ──► C₆H₅Br + H₃PO₃

📝 Quick Tips:

  • Alkane → Alkene: Chlorination + β-elimination (alcoholic KOH)
  • Alkene → Alkane: HX addition + Zn/H⁺ reduction or direct H₂/Ni
  • Alcohol → Alkene: Conc. H₂SO₄ at 170°C (direct)
  • Haloalkane → Alcohol: Aqueous KOH (SN1/SN2)
  • Haloalkane → Alkene: Alcoholic KOH (elimination)
  • Sandmeyer: Diazonium + CuX → Aryl halide
  • Wurtz: 2R-X + 2Na → R-R

🎯 Best of Luck for Your Exam!

Study hard, stay focused, and ace your chemistry exam!