Chapter 15: Haloalkanes and Haloarenes

Introduction

Overview

• Haloalkanes: Compounds formed by replacing one or more hydrogen atoms in alkanes with halogen atoms (-F, -Cl, -Br, -I).
• Haloarenes: Compounds where a hydrogen atom of the benzene ring is replaced by a halogen atom, general formula C₆H₅X.
• Haloalkanes have halogen attached to sp³ hybridized carbon atoms (alkyl group).
• Haloarenes have halogen attached to sp² hybridized carbon atoms (aryl group).
• Many halogenated organic compounds occur naturally and have clinical uses.

Solved Examples

• Ex 1: Identify the type of compound: CH₃Cl.
  • CH₃Cl is a haloalkane with halogen on sp³ carbon.
• Ex 2: Identify the compound type: C₆H₅Br.
  • C₆H₅Br is a haloarene with halogen on sp² carbon of benzene ring.

Practice Set

• Q1: What is the hybridization of carbon bonded to halogen in haloalkanes?
  • A. sp
  • B. sp²
  • C. sp³
  • D. None
• Q2: Which of the following is a haloarene?
  • A. CH₃Cl
  • B. C₆H₅Cl
  • C. C₂H₅Br
  • D. CH₂Cl₂
• Q3: Haloalkanes differ from haloarenes in that halogen is attached to:
  • A. sp carbon in haloalkanes
  • B. sp³ carbon in haloarenes
  • C. sp³ carbon in haloalkanes
  • D. sp² carbon in haloalkanes

Answer Key

• Q1: C - Haloalkanes have halogen attached to sp³ carbon.
• Q2: B - C₆H₅Cl is a haloarene (halogen on benzene ring).
• Q3: C - Haloalkanes have halogen on sp³ carbon; haloarenes on sp².

Memory Aids

• Haloalkanes = Halogen on sp³ carbon (alkyl).
• Haloarenes = Halogen on sp² carbon (aryl ring).
• Remember: "Alkane = sp³, Arene = sp²" for halogen attachment.

Classification

Types of Halo Compounds

• Classified by number of halogen atoms: Mono-, Di-, Tri-halo compounds.
• Classified by carbon-halogen bond hybridization: sp³ (alkyl) or sp² (aryl/vinyl).
• Monohalo compounds:
  • sp³ C–X: Alkyl halides (e.g., CH₃Cl)
  • sp² C–X: Vinylic halides (e.g., CH₂=CHCl)
• Dihalo compounds:
  • sp³ C–X: Allylic halides (halogen on carbon adjacent to double bond)
  • sp² C–X: Aryl halides (halogen on aromatic ring)
• Trihalo compounds:
  • sp³ C–X: Benzylic halides (halogen on benzylic carbon)
• Examples and structures illustrate each category.

Solved Examples

• Ex 1: Classify CH₂=CHCl.
  • Monohalo compound with sp² C–X bond → Vinylic halide.
• Ex 2: Identify type of C₆H₅CH₂Cl.
  • Trihalo compound with sp³ C–X bond → Benzylic halide.

Practice Set

• Q1: What type of halide is CH₃CH₂Cl?
  • A. Vinylic halide
  • B. Alkyl halide
  • C. Aryl halide
  • D. Allylic halide
• Q2: Which halide has halogen attached to sp² carbon?
  • A. Alkyl halide
  • B. Vinylic halide
  • C. Benzylic halide
  • D. Allylic halide
• Q3: C₆H₅CH₂Cl is an example of:
  • A. Allylic halide
  • B. Benzylic halide
  • C. Aryl halide
  • D. Vinylic halide

Answer Key

• Q1: B - CH₃CH₂Cl is an alkyl halide (sp³ C–X).
• Q2: B - Vinylic halides have halogen on sp² carbon.
• Q3: B - C₆H₅CH₂Cl is a benzylic halide (sp³ C–X).

Memory Aids

• Mono = single halogen; Di = two; Tri = three halogens.
• sp³ C–X: Alkyl, allylic, benzylic halides.
• sp² C–X: Vinylic, aryl halides.
• Benzylic halides have halogen on carbon next to aromatic ring.

Nomenclature

Common and IUPAC Naming of Alkyl Halides

• Common names: Alkyl group name + halide (e.g., ethyl bromide).
• IUPAC names: Halosubstituted hydrocarbons (e.g., bromoethane).
• Monohalogen benzene derivatives: Common and IUPAC names are identical.
• Dihalogen benzene derivatives: Common prefixes o-, m-, p- correspond to IUPAC locants 1,2-; 1,3-; 1,4- respectively.

Solved Examples

• Ex 1: Name CH₃CH₂Br
  • Common: Ethyl bromide
  • IUPAC: Bromoethane
• Ex 2: Name C₆H₄Br₂ with bromines at 1,3 positions
  • Common: m-Dibromobenzene
  • IUPAC: 1,3-Dibromobenzene

Practice Set

• Q1: What is the IUPAC name of CH₃CH₂Cl?
  • A. Ethyl chloride
  • B. Chloroethane
  • C. Ethane chloride
  • D. Chloromethane
• Q2: The common name for 1,4-dibromobenzene is:
  • A. o-Dibromobenzene
  • B. m-Dibromobenzene
  • C. p-Dibromobenzene
  • D. 1,4-Dibromobenzene
• Q3: Which prefix corresponds to 1,2-dibromobenzene in common naming?
  • A. o-
  • B. m-
  • C. p-
  • D. None

Answer Key

• Q1: B - Chloroethane is the correct IUPAC name for CH₃CH₂Cl.
• Q2: C - p-Dibromobenzene is the common name for 1,4-dibromobenzene.
• Q3: A - o- prefix corresponds to 1,2-dibromobenzene.

Memory Aids

• o- = 1,2- (ortho)
• m- = 1,3- (meta)
• p- = 1,4- (para)
• Alkyl halides: Alkyl name + halide (common), Halosubstituted hydrocarbon (IUPAC)

Dihaloalkanes

Classification and Naming

• Dihaloalkanes with same halogen atoms are named as alkylidene or alkylene dihalides.
• Geminal halides: Both halogen atoms on the same carbon (gem-dihalides).
• Vicinal halides: Halogen atoms on adjacent carbons (vic-dihalides).
• Common names: Gem-dihalides = alkylidene halides; Vic-dihalides = alkylene dihalides.
• IUPAC names: Both are named as dihaloalkanes with locants.

Solved Examples

• Ex 1: Name CH₃CHCl₂
  • Common: Ethylidene chloride
  • IUPAC: 1,1-Dichloroethane
• Ex 2: Name CH₂ClCH₂Cl
  • Common: Ethylene dichloride
  • IUPAC: 1,2-Dichloroethane

Practice Set

• Q1: What is the IUPAC name of CH₃CHCl₂?
  • A. 1,2-Dichloroethane
  • B. 1,1-Dichloroethane
  • C. Ethylene dichloride
  • D. Ethylidene chloride
• Q2: Which is a geminal dihalide?
  • A. CH₂ClCH₂Cl
  • B. CH₃CHCl₂
  • C. CH₃CH₂Cl
  • D. CH₃CH₂Br
• Q3: The common name for 1,2-dichloroethane is:
  • A. Ethylidene chloride
  • B. Ethylene dichloride
  • C. 1,1-Dichloroethane
  • D. 1,2-Dichloroethane

Answer Key

• Q1: B - 1,1-Dichloroethane is the IUPAC name for CH₃CHCl₂.
• Q2: B - CH₃CHCl₂ is a geminal dihalide (both Cl on same carbon).
• Q3: B - Ethylene dichloride is the common name for 1,2-dichloroethane.

Memory Aids

• Geminal = same carbon (gem-dihalide = alkylidene halide)
• Vicinal = adjacent carbons (vic-dihalide = alkylene dihalide)
• IUPAC uses locants for exact positions (e.g., 1,1- or 1,2-)

Nature of C — X Bond

Bond Polarity and Electronegativity

• Halogen atoms (X) are more electronegative than carbon (C).
• Carbon-halogen bond in alkyl halides is polar covalent.
• Carbon atom carries partial positive charge (δ⁺), halogen carries partial negative charge (δ^-).
• Electron density shifts from carbon to halogen, making the bond polar.

Solved Examples

• Ex 1: Identify the polarity in CH₃Cl.
  • Cl is more electronegative than C.
  • Bond is polar with C as δ⁺ and Cl as δ^-.

Practice Set

• Q1: Which atom in alkyl halides carries a partial positive charge?
  • A. Halogen
  • B. Carbon
  • C. Hydrogen
  • D. Oxygen
• Q2: What type of bond is formed between carbon and halogen in alkyl halides?
  • A. Ionic
  • B. Nonpolar covalent
  • C. Polar covalent
  • D. Metallic
• Q3: In the C—X bond, the arrow indicating electron density points from:
  • A. Halogen to Carbon
  • B. Carbon to Halogen
  • C. Hydrogen to Carbon
  • D. Carbon to Hydrogen

Answer Key

• Q1: B - Carbon carries partial positive charge due to electronegativity difference.
• Q2: C - The bond is polar covalent because electrons are shared unevenly.
• Q3: B - Electron density shifts from carbon to more electronegative halogen.

Memory Aids

• Remember: "C is δ⁺, X is δ^-" in alkyl halides.
• Polar bond due to electronegativity difference between C and X.

Bond Length and Bond Enthalpy

Trend with Halogen Size

• As halogen size increases (F to I), bond length increases.
• Bond enthalpy decreases with increasing bond length.
• Fluorine forms shortest and strongest C—F bond.
• Iodine forms longest and weakest C—I bond.

Solved Examples

• Ex 1: Compare bond enthalpy of C—F and C—I bonds.
  • C—F bond enthalpy = 452 kJ/mol (strongest)
  • C—I bond enthalpy = 234 kJ/mol (weakest)

Practice Set

• Q1: Which C—X bond has the shortest bond length?
  • A. C—Cl
  • B. C—I
  • C. C—F
  • D. C—Br
• Q2: What happens to bond enthalpy as bond length increases?
  • A. Increases
  • B. Decreases
  • C. Remains same
  • D. First increases then decreases
• Q3: Arrange the following bonds in order of increasing bond length: C—Br, C—F, C—I.
  • A. C—F < C—Br < C—I
  • B. C—I < C—Br < C—F
  • C. C—Br < C—I < C—F
  • D. C—F < C—I < C—Br

Answer Key

• Q1: C - Fluorine forms shortest bond due to smallest atomic size.
• Q2: B - Bond enthalpy decreases as bond length increases.
• Q3: A - Bond length increases from F to I.

Memory Aids

• Bond length: F < Cl < Br < I
• Bond enthalpy: F > Cl > Br > I

Dipole Moment Variation

Effect of Halogen Size and Electronegativity

• Dipole moment depends on bond polarity and bond length.
• From F to I, halogen size increases, electronegativity decreases.
• Dipole moment varies irregularly due to these opposing effects.
• CH₃Cl has highest dipole moment among CH₃F, CH₃Cl, CH₃Br, CH₃I.

Solved Examples

• Ex 1: Identify molecule with highest dipole moment among CH₃F, CH₃Cl, CH₃Br, CH₃I.
  • CH₃Cl has dipole moment 1.860 D (highest).

Practice Set

• Q1: Which molecule has the highest dipole moment?
  • A. CH₃F
  • B. CH₃Cl
  • C. CH₃Br
  • D. CH₃I
• Q2: Dipole moment generally depends on:
  • A. Only electronegativity
  • B. Only bond length
  • C. Both electronegativity and bond length
  • D. Neither electronegativity nor bond length
• Q3: Why does dipole moment vary irregularly from F to I?
  • A. Because electronegativity and size change oppositely
  • B. Because all halogens have same electronegativity
  • C. Because bond length remains constant
  • D. Because carbon electronegativity changes

Answer Key

• Q1: B - CH₃Cl has highest dipole moment due to optimal electronegativity and size.
• Q2: C - Dipole moment depends on both electronegativity difference and bond length.
• Q3: A - Opposite trends in electronegativity and size cause irregular variation.

Memory Aids

• Dipole moment peaks at intermediate halogen size (Cl).
• Remember: Dipole moment ≠ directly proportional to electronegativity alone.

Methods of Preparation of Haloalkanes

From Alcohols

• Alcohols (R 6OH) react with halogenating agents to form haloalkanes (R 6X).
• Alkyl chlorides (R 6Cl) formed using HCl with anhydrous ZnCl₂ catalyst for 1 6 and 2 6 alcohols; tertiary alcohols react directly with concentrated HCl.
• Alkyl bromides (R 6Br) formed by reacting alcohols with NaBr and H₂SO₄.
• Phosphorus halides (PX₃, X = Cl, Br) convert alcohols to alkyl halides.
• Red phosphorus with halogens (Br₂, I₂) also converts alcohols to alkyl halides.
• Other reagents: PCl₅, SOCl₂ with pyridine convert alcohols to alkyl chlorides.
• Good yields of alkyl iodides (R 6I) obtained by heating alcohols with sodium or potassium iodide in 95% ortho phosphoric acid.
• Reactivity order with haloacid: tertiary > secondary > primary alcohols.

Solved Examples

• Ex 1: Convert 2 6propanol to 2 6bromopropane using NaBr and H₂SO₄.
• Step 1: Mix 2 6propanol with NaBr and H₂SO₄.
• Step 2: Reaction proceeds to form 2 6bromopropane and water.

Practice Set

• Q1: Which reagent converts tertiary alcohols to alkyl chlorides at room temperature?
  • A. HCl with ZnCl₂
  • B. Concentrated HCl
  • C. NaBr and H₂SO₄
  • D. PCl₅
• Q2: What is the order of reactivity of alcohols with haloacids?
  • A. 1 6 > 2 6 > 3 6
  • B. 3 6 > 2 6 > 1 6
  • C. 2 6 > 3 6 > 1 6
  • D. 1 6 > 3 6 > 2 6
• Q3: Which reagent is used to obtain good yields of alkyl iodides from alcohols?
  • A. NaI in 95% ortho phosphoric acid
  • B. NaBr and H₂SO₄
  • C. PCl₅
  • D. SOCl₂ with pyridine

Answer Key

• Q1: B - Tertiary alcohols react directly with concentrated HCl at room temperature.
• Q2: B - Reactivity order is tertiary > secondary > primary alcohols.
• Q3: A - Heating alcohols with NaI in 95% ortho phosphoric acid gives good yields of alkyl iodides.

Memory Aids

• Remember: 3 6 alcohols react fastest with haloacids.
• Use ZnCl₂ catalyst for 1 6 and 2 6 alcohols with HCl.
• NaI + 95% H₃PO₄ for alkyl iodides.

From Hydrocarbons

Free Radical Halogenation

• Methane (CH₄) reacts with Cl₂ under UV light or heat to form chloromethanes.
• Chlorine splits into radicals that replace hydrogen atoms stepwise.
• Products: CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄ plus HCl byproduct.
• Ease of hydrogen replacement: tertiary > secondary > primary.

Addition to Alkenes

• Alkenes react with hydrogen halides (HX) and halogens (X₂).
• HX addition follows Markovnikov's rule: halogen attaches to carbon with fewer hydrogens.
• In presence of peroxide, HBr adds via Anti-Markovnikov's rule.
• Halogens add across double bonds to form vic-dihalides.

Solved Examples

• Ex 1: Chlorination of methane under UV light forms CH₃Cl.
• Step 1: UV light splits Cl₂ into radicals.
• Step 2: Cl radical replaces H in CH₄ forming CH₃Cl.

Practice Set

• Q1: What initiates free radical halogenation of methane?
  • A. Heat or UV light
  • B. Catalyst ZnCl₂
  • C. Concentrated HCl
  • D. Peroxide
• Q2: Which rule does HX addition to alkenes usually follow?
  • A. Anti-Markovnikov
  • B. Markovnikov
  • C. Zaitsev
  • D. Saytzeff
• Q3: In presence of peroxide, HBr adds to alkenes following?
  • A. Markovnikov's rule
  • B. Anti-Markovnikov's rule
  • C. Free radical substitution
  • D. Electrophilic addition

Answer Key

• Q1: A - Heat or UV light initiates radical formation.
• Q2: B - HX addition follows Markovnikov's rule normally.
• Q3: B - HBr adds via Anti-Markovnikov's rule in presence of peroxide.

Memory Aids

• Free radical halogenation needs UV or heat.
• Markovnikov: halogen to carbon with fewer H.
• Anti-Markovnikov: halogen to carbon with more H (only HBr + peroxide).

Halogen Exchange

Finkelstein Reaction

• Alkyl halide (R 6X, X = Cl or Br) reacts with NaI in dry acetone.
• Halogen X is replaced by iodine forming alkyl iodide (R 6I).
• NaCl or NaBr precipitates out, driving reaction forward (Le Chatelier's principle).

Swarts Reaction

• Alkyl halide (RX, X = Cl or Br) reacts with AgF.
• Halogen replaced by fluorine forming alkyl fluoride (RF).
• Byproduct: silver halide (AgX).
• Other reagents: HgF₂, CoF₃, SbF₅ can also be used.

Additional Notes

• Relative rates of hydrogen abstraction by Cl₂: 3 6:2 6:1 6 = 5:3.8:1.
• By Br₂ at 400 K: 1600:82:1.
• Direct iodination is reversible; requires oxidising agents (HNO₃, HIO₃).

Solved Examples

• Ex 1: Convert R 6Cl to R 6I using NaI in dry acetone (Finkelstein reaction).
• Step 1: Mix R 6Cl with NaI in dry acetone.
• Step 2: R 6I forms, NaCl precipitates.

Practice Set

• Q1: In Finkelstein reaction, which solvent is used?
  • A. Water
  • B. Dry acetone
  • C. Ethanol
  • D. Benzene
• Q2: Swarts reaction replaces halogen with?
  • A. Chlorine
  • B. Bromine
  • C. Fluorine
  • D. Iodine
• Q3: Which reagent is NOT used in Swarts reaction?
  • A. AgF
  • B. HgF₂
  • C. NaI
  • D. CoF₃

Answer Key

• Q1: B - Dry acetone is solvent in Finkelstein reaction.
• Q2: C - Swarts reaction replaces halogen with fluorine.
• Q3: C - NaI is not used in Swarts reaction; it is used in Finkelstein reaction.

Memory Aids

• Finkelstein: halogen exchange with NaI in dry acetone.
• Swarts: halogen replaced by fluorine using AgF or similar.
• Direct iodination needs oxidising agents.

Preparation of Haloarenes

From Hydrocarbons by Electrophilic Substitution

• Aryl chlorides and bromides are prepared by electrophilic substitution of arenes with Cl₂ or Br₂.
• Lewis acid catalysts like Fe or FeCl₃ are used to activate halogens.
• Reaction occurs in the dark to prevent side reactions.
• Example: Toluene reacts with X₂ (Cl₂/Br₂) in presence of Fe catalyst to give ortho- and para-halotoluene.
• Ortho- and para-isomers differ in melting points and can be separated easily.
• Iodination requires oxidizing agents (HNO₃, HIO₄) to oxidize HI formed; reaction is reversible.
• Fluoro compounds are not prepared by this method due to fluorine's high reactivity.

Solved Examples

• Ex 1: Toluene + Br₂ / Fe → o-bromotoluene + p-bromotoluene
• Step 1: Add Br₂ and Fe catalyst to toluene in dark.
• Step 2: Electrophilic substitution occurs at ortho and para positions.
• Step 3: Separate isomers by melting point difference.

Practice Set

• Q1: Which catalyst is used for electrophilic halogenation of benzene?
  A. FeCl₃
  B. NaCl
  C. HCl
  D. CuCl
• Q2: What is the major product when toluene reacts with Cl₂ in presence of Fe?
  A. Meta-chlorotoluene
  B. Ortho- and para-chlorotoluene
  C. Benzyl chloride
  D. Chlorobenzene
• Q3: Why are fluoroarenes not prepared by electrophilic substitution?
  A. Fluorine is too reactive
  B. Fluorine is unreactive
  C. Fluorine forms unstable intermediates
  D. Fluorine reacts only with alkanes

Answer Key

• Q1: A. FeCl₃ - It acts as a Lewis acid catalyst activating halogens.
• Q2: B. Ortho- and para-chlorotoluene - Electrophilic substitution occurs at ortho and para positions due to methyl group activation.
• Q3: A. Fluorine is too reactive - High reactivity prevents controlled substitution.

Memory Aids

• Fe/FeCl₃ catalyzes Cl₂/Br₂ substitution on benzene rings.
• Methyl group directs substitution to ortho and para positions.
• Fluorine too reactive for electrophilic substitution on arenes.

From Benzene Diazonium Chloride

Key Reactions and Mechanism

• Start with aniline (C₆H₅–NH₂).
• React with NaNO₂ and HCl at 273–278 K to form benzene diazonium chloride (C₆H₅–N₂⁺Cl⁻).
• Benzene diazonium chloride is a versatile intermediate for halogenation.
• Balz-Schiemann reaction: React with NaNO₂ and HBF₄ to produce fluorobenzene.
• Sandmeyer's reaction: React with Cu₂Cl₂/HCl to give chlorobenzene + N₂ gas.
• Similarly, react with Cu₂Br₂/HBr to give bromobenzene + N₂ gas.

Solved Examples

• Ex 1: Preparation of fluorobenzene
  Step 1: Aniline + NaNO₂/HCl → Benzene diazonium chloride (273–278 K)
  Step 2: Benzene diazonium chloride + NaNO₂/HBF₄ → Fluorobenzene + N₂
• Ex 2: Preparation of chlorobenzene
  Step 1: Aniline + NaNO₂/HCl → Benzene diazonium chloride
  Step 2: Benzene diazonium chloride + Cu₂Cl₂/HCl → Chlorobenzene + N₂

Practice Set

• Q1: What is the temperature range for diazotization of aniline?
  A. 273–278 K
  B. 300–310 K
  C. 250–260 K
  D. 280–290 K
• Q2: Which reagent is used in Balz-Schiemann reaction to produce fluorobenzene?
  A. Cu₂Cl₂
  B. NaNO₂ and HBF₄
  C. HCl
  D. Cu₂Br₂
• Q3: What gas is released during Sandmeyer's reaction?
  A. CO₂
  B. N₂
  C. H₂
  D. O₂

Answer Key

• Q1: A. 273–278 K - Low temperature is essential for stable diazonium salt formation.
• Q2: B. NaNO₂ and HBF₄ - Used in Balz-Schiemann reaction to replace diazonium group with fluorine.
• Q3: B. N₂ - Nitrogen gas is released as a byproduct in Sandmeyer's reaction.

Memory Aids

• Benzene diazonium chloride is key intermediate for halogenation of benzene.
• Balz-Schiemann reaction produces fluorobenzene using HBF₄.
• Sandmeyer's reaction replaces diazonium group with Cl or Br releasing N₂.

Physical Properties

General Characteristics

• Alkyl halides are colourless when pure.
• Bromides and iodides develop colour on exposure to light.
• Many volatile halogen compounds have a sweet smell.
• Methyl chloride, methyl bromide, ethyl chloride, and some chlorofluoromethanes are gases at room temperature; higher members are liquids or solids.

Boiling Points and Intermolecular Forces

• Halogen derivatives have stronger intermolecular forces (dipole-dipole and van der Waals) than parent hydrocarbons due to greater polarity and molecular mass.
• Boiling points of chlorides, bromides, and iodides are higher than hydrocarbons of similar molecular mass.
• For the same alkyl group, boiling points decrease in order: RI > RBr > RCl > RF due to increasing van der Waals forces with halogen size.
• Boiling points of isomeric haloalkanes decrease with increased branching.
• Boiling points of isomeric dihalobenzenes are nearly the same; para-isomers have higher melting points due to better crystal lattice symmetry.

Density and Solubility

• Bromo, iodo, and polychloro derivatives are heavier than water.
• Density increases with the number of carbon atoms, halogen atoms, and atomic mass of halogen atoms.
• Example densities (g mL^-1): n-C₃H₇Cl = 0.89, n-C₃H₇Br = 1.335, n-C₃H₇I = 1.747.
• Haloalkanes are very slightly soluble in water but more soluble in organic solvents due to similar intermolecular attractions.

Photochemical Decomposition of Alkyl Iodides

• Alkyl iodides become violet or brown on standing due to weak C–I bond.
• They decompose in sunlight, releasing iodine which dissolves in alkyl iodide, imparting a dark colour.

The reaction: 2R–I 1hv1 R1R + I₂

Solved Examples

• Ex 1: Explain why alkyl iodides develop colour on standing.
  - C–I bond is weak.
  - Sunlight breaks 2R–I into R–R and I₂.
  - Iodine dissolves in alkyl iodide, causing violet/brown colour.
• Ex 2: Arrange the boiling points of RI, RBr, RCl, and RF.
  - Order: RI > RBr > RCl > RF due to increasing van der Waals forces with halogen size.

Practice Set

• Q1: Which alkyl halide is a gas at room temperature?
  A. n-C₃H₇Cl
  B. Methyl bromide
  C. n-C₃H₇I
  D. Ethyl iodide
• Q2: What causes the higher boiling points of alkyl halides compared to hydrocarbons?
  A. Lower molecular mass
  B. Stronger intermolecular forces
  C. Less polarity
  D. Smaller size
• Q3: Why do para-isomers of dihalobenzenes have higher melting points?
  A. Less symmetry
  B. Better crystal lattice fitting
  C. Lower molecular mass
  D. Higher branching

Answer Key

• Q1: B - Methyl bromide is a gas at room temperature due to low molecular weight.
• Q2: B - Alkyl halides have stronger dipole-dipole and van der Waals forces, raising boiling points.
• Q3: B - Para-isomers fit better in crystal lattice due to symmetry, increasing melting points.

Memory Aids

• Boiling point order: RI > RBr > RCl > RF (increasing halogen size increases van der Waals forces).
• Alkyl iodides decompose under light: 2R–I 1hv1 R1R + I₂.
• Para-isomers have higher melting points due to symmetry.

Chemical Reactions

Reactions of Haloalkanes

Nucleophilic Substitution Reactions

• C—X bond is highly polarised due to electronegativity difference (X > C).
• Polarity causes nucleophilic substitution reactions.
• General mechanism: Nucleophile attacks δ+ carbon, displaces X⁻.
• Common nucleophilic substitutions:
  • RX + KOH (aq) → R–OH (alcohol)
  • RX + NaOR' → R–OR' (ether, Williamson synthesis)
  • RX + NaSH → R–SH (thioalcohol)
  • RX + KCN → R–CN (alkyl cyanide)
• Ambident nucleophiles (e.g., CN⁻, NO₂⁻) can attack through two sites, giving different products.
• Mechanisms: SN1 (two-step, carbocation intermediate) and SN2 (one-step, backside attack).
• Reactivity order: R–I > R–Br > R–Cl > R–F.
• Primary allylic and benzylic halides react faster in SN1 due to resonance-stabilized carbocations.
• Optical isomerism arises from chiral centers (carbon with four different groups).
  • Dextrorotatory (+) rotates plane-polarized light right.
  • Laevorotatory (−) rotates left.
  • Racemisation: conversion of enantiomer to racemic mixture (optically inactive).
  • Meso compounds have internal symmetry and are optically inactive.

Solved Examples

• Ex 1: SN2 reaction of CH₃Br with OH⁻ gives CH₃OH with inversion of configuration.
• Ex 2: 2-Bromobutane + alc. KOH → 2-butene (major) + 1-butene (minor) via elimination.

Practice Set

• Q1: Which halide is most reactive in SN2?
  • A. R–F
  • B. R–Cl
  • C. R–Br
  • D. R–I
• Q2: What product forms when RX reacts with KCN?
  • A. Alcohol
  • B. Alkyl cyanide
  • C. Ether
  • D. Thioalcohol
• Q3: Which mechanism involves carbocation intermediate?
  • A. SN1
  • B. SN2
  • C. E2
  • D. E1

Answer Key

• Q1: D. R–I; Iodides have weakest C–X bond, highest reactivity.
• Q2: B. Alkyl cyanide; KCN substitutes CN group forming R–CN.
• Q3: A. SN1; involves carbocation intermediate in two-step mechanism.

Memory Aids

• SN2: One step, backside attack, inversion of configuration.
• SN1: Two steps, carbocation intermediate, racemisation possible.
• Saytzeff’s rule: Major alkene has more alkyl substituents.
• Ambident nucleophiles: CN⁻ (C or N), NO₂⁻ (O or N).

Elimination Reactions

Elimination of HX from Haloalkanes

• Haloalkanes with β-hydrogen undergo dehydrohalogenation with alc. KOH on heating.
• Reaction forms alkenes, KX, and H₂O.
• Major alkene follows Saytzeff’s rule: more substituted alkene is favored.

Solved Examples

• Ex 1: 2-Bromobutane + alc. KOH → 2-butene (major) + 1-butene (minor)

Practice Set

• Q1: What is the major product of elimination from 2-bromopentane?
  • A. 1-pentene
  • B. 2-pentene
  • C. Pentane
  • D. 3-pentene
• Q2: Which reagent is used for dehydrohalogenation?
  • A. Aqueous KOH
  • B. Alcoholic KOH
  • C. NaOH (solid)
  • D. HCl
• Q3: Saytzeff’s rule states that the major alkene formed has?
  • A. Least alkyl groups
  • B. Most alkyl groups
  • C. Equal alkyl groups
  • D. No alkyl groups

Answer Key

• Q1: B. 2-pentene; more substituted alkene is major product.
• Q2: B. Alcoholic KOH; used for elimination reactions.
• Q3: B. Most alkyl groups; Saytzeff’s rule favors more substituted alkene.

Memory Aids

• Elimination requires β-hydrogen and base (alc. KOH).
• Saytzeff’s rule: More substituted alkene is major product.

Reactions with Metals

Reactions of Haloalkanes with Metals

• RX + Mg (dry ether) → R–MgX (Grignard reagent)
• Grignard reagents react with proton donors to form RH.
• RX + Li (dry ether) → RLi (organolithium compound)
• 2 RX + 2 Na (dry ether) → R–R + 2 NaX (Wurtz reaction)

Solved Examples

• Ex 1: Preparation of Grignard reagent from bromobenzene and Mg.
• Ex 2: Wurtz reaction: 2 CH₃Cl + 2 Na → C₂H₆ + 2 NaCl.

Practice Set

• Q1: Grignard reagent is formed by reaction of RX with?
  • A. Na
  • B. Mg
  • C. Li
  • D. K
• Q2: Wurtz reaction joins two alkyl halides using?
  • A. Mg
  • B. Li
  • C. Na
  • D. K
• Q3: Organolithium compounds are formed by reaction of RX with?
  • A. Na
  • B. Mg
  • C. Li
  • D. K

Answer Key

• Q1: B. Mg; Grignard reagents form with magnesium in dry ether.
• Q2: C. Na; Wurtz reaction uses sodium metal.
• Q3: C. Li; Organolithium compounds form with lithium.

Memory Aids

• Grignard reagent: RX + Mg → R–MgX.
• Wurtz reaction: 2 RX + 2 Na → R–R + 2 NaX.
• Organolithium: RX + Li → RLi.

Reactions of Haloarenes

Nucleophilic Substitution Reactions

• Haloarenes less reactive than alkyl halides due to:
  • Partial double bond character of C—X (resonance with benzene).
  • Less polar C—X bond (sp² C more electronegative than sp³ C).
  • Phenyl cation unstable; SN1 mechanism not favored.
  • Electron-rich nucleophiles repel electron-rich aromatic ring.
• Replacement of Cl by OH requires harsh conditions (NaOH, 623 K, 300 atm, acid work-up).
• Electron withdrawing groups (-NO₂) at ortho/para increase reactivity.

Solved Examples

• Ex 1: Chlorobenzene + NaOH (623 K, 300 atm) → Phenol + NaCl
• Ex 2: 2-Chloronitrobenzene + NaOH → 2-Nitrophenol

Practice Set

• Q1: Why are haloarenes less reactive in nucleophilic substitution?
  • A. Strong C—X bond
  • B. Partial double bond character and resonance
  • C. High polarity of C—X bond
  • D. Stable phenyl cation
• Q2: Which group increases reactivity of haloarenes?
  • A. Electron donating
  • B. Electron withdrawing (-NO₂)
  • C. Alkyl
  • D. Halogen
• Q3: Conditions to replace Cl by OH in chlorobenzene?
  • A. NaOH, room temp
  • B. NaOH, 623 K, 300 atm, acid work-up
  • C. NaOH, 100 K
  • D. NaOH, 500 atm

Answer Key

• Q1: B; Resonance and partial double bond reduce reactivity.
• Q2: B; Electron withdrawing groups activate haloarenes.
• Q3: B; High temp, pressure, and acid needed for substitution.

Memory Aids

• Haloarenes: C—X partial double bond, less reactive.
• Electron withdrawing groups increase nucleophilic substitution.
• Harsh conditions needed for OH substitution.

Electrophilic Substitution Reactions

• Halogen I-effect withdraws electrons, deactivating ring.
• Halogen +M effect increases electron density at ortho/para.
• Haloarenes are ortho/para directing.
• Halogenation: Haloarene + Cl₂ + FeCl₃ → dihalobenzene.
• Nitration: Haloarene + HNO₃/H₂SO₄ → o- and p-halonitrobenzene.
• Sulphonation: Haloarene + H₂SO₄ → o- and p-halo benzene sulfonic acid.
• Friedel-Crafts alkylation: Haloarene + alkyl halide + AlCl₃ → o- and p-haloalkylbenzene.
• Friedel-Crafts acylation: Haloarene + acid chloride + AlCl₃ → o- and p-acetophenone.

Solved Examples

• Ex 1: Chlorobenzene + Cl₂/FeCl₃ → 1,4-dichlorobenzene (major) + 1,2-dichlorobenzene (minor)
• Ex 2: Chlorobenzene + HNO₃/H₂SO₄ → 1-chloro-4-nitrobenzene (major) + 1-chloro-2-nitrobenzene (minor)
• Ex 3: Chlorobenzene + CH₃Cl/AlCl₃ → 1-chloro-4-methylbenzene (major) + 1-chloro-2-methylbenzene (minor)

Practice Set

• Q1: Haloarenes are less reactive due to?
  • A. Strong C—X bond
  • B. Partial double bond character and resonance
  • C. High polarity
  • D. Stable carbocation
• Q2: Halogen directs substitution to?
  • A. Meta
  • B. Ortho and para
  • C. Only para
  • D. Only ortho
• Q3: Major product of chlorobenzene + Cl₂/FeCl₃ is?
  • A. 1,2-dichlorobenzene
  • B. 1,3-dichlorobenzene
  • C. 1,4-dichlorobenzene
  • D. 1,5-dichlorobenzene

Answer Key

• Q1: B; Resonance and partial double bond reduce reactivity.
• Q2: B; Halogens are ortho/para directing.
• Q3: C; Para is major due to less steric hindrance.

Memory Aids

• Halogen I-effect deactivates, +M effect directs ortho/para.
• Halogenation needs FeCl₃ catalyst.
• Friedel-Crafts reactions give ortho/para products.

Polyhalogen Compounds

Overview

• Carbon compounds containing more than one halogen atom.
• Examples: CHCl₃ (chloroform), CH₂Cl₂ (dichloromethane).

Solved Examples

• Ex 1: Identify a polyhalogen compound with three chlorine atoms.
  • Answer: CHCl₃ (chloroform).

Practice Set

• Q1: Which of the following is a polyhalogen compound?
  • A. CH₄
  • B. CHCl₃
  • C. C₂H₆
  • D. C₂H₄
• Q2: Polyhalogen compounds contain how many halogen atoms?
  • A. One
  • B. Two or more
  • C. None
  • D. Exactly one
• Q3: Which is an example of a polyhalogen compound?
  • A. CH₃Cl
  • B. CH₂Cl₂
  • C. CH₄
  • D. CCl₄

Answer Key

• Q1: B - CHCl₃ contains multiple halogen atoms.
• Q2: B - Polyhalogen compounds have two or more halogen atoms.
• Q3: B - CH₂Cl₂ is a polyhalogen compound.

Memory Aids

• Polyhalogen = multiple halogens attached to carbon.
• Common examples: chloroform (CHCl₃), dichloromethane (CH₂Cl₂).

Dichloromethane (Methylene Chloride), CH₂Cl₂

Uses

• Widely used as a solvent.
• Used as paint remover.
• Used as propellant in aerosol sprays.

Harmful Effects

• Harms the human central nervous system.
• Low exposure: impaired hearing and vision.
• High exposure: dizziness, nausea, tingling, numbness in fingers and toes.
• Direct skin contact: intense burning and mild redness.
• Eye contact: can burn the cornea.

Solved Examples

• Ex 1: Name a use of dichloromethane.
  • Answer: Solvent or paint remover.

Practice Set

• Q1: What is a common use of CH₂Cl₂?
  • A. Fuel
  • B. Solvent
  • C. Food additive
  • D. Fertilizer
• Q2: Which harmful effect is caused by high exposure to CH₂Cl₂?
  • A. Improved vision
  • B. Dizziness and nausea
  • C. Increased energy
  • D. Skin whitening
• Q3: Direct contact of CH₂Cl₂ with eyes causes?
  • A. Cornea burn
  • B. Eye whitening
  • C. No effect
  • D. Eye color change

Answer Key

• Q1: B - CH₂Cl₂ is widely used as a solvent.
• Q2: B - High levels cause dizziness and nausea.
• Q3: A - Direct eye contact can burn the cornea.

Memory Aids

• CH₂Cl₂ = paint remover and solvent.
• Harmful to CNS; avoid skin and eye contact.

Trichloromethane (Chloroform), CHCl₃

Uses

• Major use: production of freon refrigerant R-22.
• Solvent for resins, rubbers, oils, fats, alkaloids, iodine.
• Preparation of chloretone (drug) and chloropicrin (insecticide).
• Used to preserve anatomical specimens.

Harmful Effects

• Oxidised to phosgene (COCl₂), a poisonous gas causing liver and kidney damage.
• Inhalation depresses central nervous system causing dizziness, fatigue, headache.

Solved Examples

• Ex 1: Write the reaction of chloroform with oxygen under light.
  • 2 CHCl₃ + O₂ 1hv1 COCl₂ + 2 HCl

Practice Set

• Q1: Chloroform is mainly used to produce?
  • A. Freon refrigerant R-22
  • B. Paint remover
  • C. Fertilizer
  • D. Fuel
• Q2: Phosgene gas formed from chloroform causes damage to?
  • A. Heart
  • B. Liver and kidney
  • C. Lungs only
  • D. Brain
• Q3: Inhalation of chloroform vapors causes?
  • A. Increased energy
  • B. CNS depression, dizziness, fatigue
  • C. Skin rash
  • D. No effect

Answer Key

• Q1: A - Major use is freon refrigerant production.
• Q2: B - Phosgene damages liver and kidney.
• Q3: B - Vapors depress CNS causing dizziness and fatigue.

Memory Aids

• Chloroform 1hv1 Phosgene + HCl (photochemical reaction).
• Phosgene is highly toxic; avoid inhalation.

Triiodomethane (Iodoform), CHI₃

Uses

• Used as antiseptic in wound dressing (liberates iodine).
• Used as methylating agent in organic synthesis.
• Replaced due to objectionable smell by other iodine formulations.

Solved Examples

• Ex 1: Why is iodoform replaced in antiseptics?
  • Due to its strong, unpleasant smell.

Practice Set

• Q1: Iodoform is used as?
  • A. Solvent
  • B. Antiseptic
  • C. Fuel
  • D. Fertilizer
• Q2: Iodoform liberates which element?
  • A. Chlorine
  • B. Iodine
  • C. Fluorine
  • D. Bromine
• Q3: Why is iodoform replaced in medical use?
  • A. Toxicity
  • B. Objectionable smell
  • C. Instability
  • D. High cost

Answer Key

• Q1: B - Used as antiseptic.
• Q2: B - Liberates iodine.
• Q3: B - Replaced due to bad smell.

Memory Aids

• Iodoform = antiseptic with iodine release.
• Smell is main drawback.

Tetrachloromethane (Carbon Tetrachloride), CCl₄

Uses

• Manufacture of refrigerants and aerosol propellants.
• Synthesis of chlorofluorocarbons and pharmaceuticals.
• Cleaning agent and spot remover.
• Fire extinguisher (pyrene).

Harmful Effects

• Causes liver cancer.
• Common effects: dizziness, nausea, vomiting.
• Can cause permanent nerve damage leading to coma or death.
• May irritate eyes.
• Depletes ozone layer.

Solved Examples

• Ex 1: Name a harmful effect of CCl₄.
  • Liver cancer in humans.

Practice Set

• Q1: Carbon tetrachloride is used as?
  • A. Fertilizer
  • B. Fire extinguisher
  • C. Food additive
  • D. Fuel
• Q2: Which is a harmful effect of CCl₄?
  • A. Eye whitening
  • B. Liver cancer
  • C. Increased energy
  • D. Skin tanning
• Q3: CCl₄ causes ozone layer depletion because?
  • A. It is stable in atmosphere
  • B. It reacts with ozone directly
  • C. It releases chlorine radicals in stratosphere
  • D. It absorbs UV rays

Answer Key

• Q1: B - Used as fire extinguisher (pyrene).
• Q2: B - Causes liver cancer.
• Q3: C - Releases chlorine radicals causing ozone depletion.

Memory Aids

• CCl₄ = toxic, causes liver cancer and ozone depletion.
• Used in fire extinguishers and solvents.

Freons

Preparation

• Chlorofluorocarbon compounds of methane and ethane with chlorine and fluorine.
• Example: CF₂Cl₂ (Freon-12).
• Manufactured by Swarts reaction: CCl₄ + 2AgF₂ 1 CCl₂F₂ + 2AgCl.

Properties

• Extremely stable, unreactive, non-toxic, non-corrosive gases.
• Easily liquefiable.

Uses

• Aerosol propellants.
• Refrigeration and air conditioning.

Harmful Effects

• Stable freons diffuse unchanged into stratosphere.
• Initiate free radical chain reactions with ozone.
• Cause depletion of ozone layer.

Solved Examples

• Ex 1: Write the Swarts reaction for Freon-12 preparation.
  • CCl₄ + 2AgF₂ 1 CCl₂F₂ + 2AgCl

Practice Set

• Q1: Freons are used as?
  • A. Fertilizers
  • B. Aerosol propellants
  • C. Food preservatives
  • D. Fuels
• Q2: Freons cause ozone depletion because?
  • A. They react with oxygen
  • B. They release chlorine radicals in stratosphere
  • C. They absorb UV light
  • D. They are unstable
• Q3: Freon-12 is prepared by?
  • A. Reaction of methane with chlorine
  • B. Swarts reaction of CCl₄ with AgF₂
  • C. Direct fluorination of ethane
  • D. Electrolysis of chlorofluorocarbons

Answer Key

• Q1: B - Used as aerosol propellants.
• Q2: B - Release chlorine radicals causing ozone depletion.
• Q3: B - Prepared by Swarts reaction.

Memory Aids

• Freons = stable chlorofluorocarbons used in refrigeration.
• Cause ozone depletion via chlorine radicals.

p, p’-Dichlorodiphenyl Trichloroethane, DDT

Preparation

• Condensation of chloral with chlorobenzene in presence of concentrated H₂SO₄.

Properties

• Powerful insecticide.
• Effective against mosquitoes (malaria) and lice (typhus).

Harmful Effects

• Insects developed resistance.
• Highly toxic to fish.
• Chemically stable and fat soluble.
• Not rapidly metabolised by animals.
• Accumulates in fatty tissues, builds up over time.

Solved Examples

• Ex 1: How is DDT prepared?
  • By condensation of chloral with chlorobenzene using conc. H₂SO₄.

Practice Set

• Q1: DDT is used as?
  • A. Fertilizer
  • B. Insecticide
  • C. Solvent
  • D. Fuel
• Q2: Which is a harmful effect of DDT?
  • A. Rapid metabolism in animals
  • B. High toxicity to fish
  • C. Water soluble
  • D. No resistance in insects
• Q3: DDT accumulates in?
  • A. Blood
  • B. Fatty tissues
  • C. Bones
  • D. Muscles

Answer Key

• Q1: B - DDT is a powerful insecticide.
• Q2: B - Highly toxic to fish.
• Q3: B - Deposited in fatty tissues.

Memory Aids

• DDT = stable, fat soluble insecticide accumulating in animals.
• Resistance developed by insects reduces effectiveness.

Polyhalogen Compounds Table

Summary Table

• Lists important polyhalogen compounds with formula and uses.
• Examples: Chloretone (drug), Phosgene (poison gas), Freon (refrigerant), DDT (insecticide).
• Shows diverse applications in medicine, industry, and agriculture.

Solved Examples

• Ex 1: Name a polyhalogen compound used as insecticide.
  • Answer: DDT.

Practice Set

• Q1: Which compound is used as a refrigerant?
  • A. DDT
  • B. Phosgene
  • C. Freon
  • D. Chloretone
• Q2: Phosgene is used as?
  • A. Poisonous gas
  • B. Fertilizer
  • C. Solvent
  • D. Insecticide
• Q3: Chloretone is a?
  • A. Hypnotic drug
  • B. Insecticide
  • C. Refrigerant
  • D. Solvent

Answer Key

• Q1: C - Freon is used as refrigerant.
• Q2: A - Phosgene is a poisonous gas.
• Q3: A - Chloretone is a hypnotic drug.

Memory Aids

• Polyhalogen compounds have diverse uses: drugs, insecticides, refrigerants, gases.

Distinction between Chloroform and Iodoform

Key Difference

• Iodoform on heating with AgNO₃ gives yellow precipitate of AgI.
• Chloroform does not give this test due to its stable nature.

Solved Examples

• Ex 1: Which compound gives yellow precipitate with AgNO₃?
  • Iodoform (CHI₃).

Practice Set

• Q1: Which compound gives yellow precipitate with AgNO₃?
  • A. Chloroform
  • B. Iodoform
  • C. Dichloromethane
  • D. Carbon tetrachloride
• Q2: Why does chloroform not give yellow precipitate with AgNO₃?
  • A. It is unstable
  • B. It is stable
  • C. It reacts slowly
  • D. It forms colorless precipitate
• Q3: Yellow precipitate formed is?
  • A. AgCl
  • B. AgI
  • C. AgBr
  • D. AgF

Answer Key

• Q1: B - Iodoform gives yellow AgI precipitate.
• Q2: B - Chloroform is stable, no precipitate formed.
• Q3: B - Yellow precipitate is AgI.

Memory Aids

• Iodoform test: yellow AgI precipitate with AgNO₃.
• Chloroform is stable, no precipitate.