Chapter 15: Haloalkanes and Haloarenes

Introduction

Haloalkanes and Haloarenes

• Haloalkanes: Alkanes with one or more hydrogens replaced by halogen atoms (-F, -Cl, -Br, -I).
• Haloarenes: Benzene derivatives where a hydrogen atom is replaced by a halogen; general formula C₆H₅X.
• Haloalkanes have halogen attached to sp³ hybridized carbon atoms.
• Haloarenes have halogen attached to sp² hybridized carbon atoms of the aromatic ring.
• Many halogenated organic compounds occur naturally and have clinical uses.

Solved Examples

• Ex 1: Identify the hybridization of carbon in chloromethane (CH₃Cl).
  Step 1: Carbon in methane is sp³ hybridized.
  Step 2: Replacement of H by Cl does not change hybridization.
  Answer: Carbon is sp³ hybridized.
• Ex 2: Determine if chlorobenzene is a haloalkane or haloarene.
  Step 1: Cl attached to benzene ring carbon (sp² hybridized).
  Answer: Chlorobenzene is a haloarene.

Practice Set

• Q1: Which hybridization is present in haloalkanes?
  A. sp
  B. sp²
  C. sp³
  D. sp³d
• Q2: The general formula of haloarenes is:
  A. C_nH_2n+1X
  B. C₆H₅X
  C. C_nH_2nX
  D. C_nH_2n-2X
• Q3: Which of the following is a haloalkane?
  A. Chlorobenzene
  B. Bromoethane
  C. Iodobenzene
  D. Fluorobenzene

Answer Key

• Q1: C. sp³ - Haloalkanes have halogen attached to sp³ carbon atoms.
• Q2: B. C₆H₅X - Haloarenes have halogen substituted on benzene ring.
• Q3: B. Bromoethane - An alkane derivative with halogen attached to sp³ carbon.

Memory Aids

• Haloalkanes = Halogen on sp³ carbon (alkane).
• Haloarenes = Halogen on sp² carbon (aromatic ring).
• General formula: Haloalkanes vary; Haloarenes = C₆H₅X.

Classification

Types of Haloalkanes and Haloarenes

• Classified by number of halogen atoms: Mono-, Di-, Tri-halo compounds.
• Classified by carbon-halogen bond hybridization: sp³ or sp².
• Monohalo compounds:
  • sp³ C–X: Alkyl halides (e.g., chloromethane)
  • sp² C–X: Vinylic halides (e.g., vinyl chloride)
• Dihalo compounds:
  • sp³ C–X: Allylic halides
  • sp² C–X: Aryl halides (e.g., dichlorobenzene)
• Trihalo compounds:
  • sp³ C–X: Benzylic halides (e.g., trichloromethane)

Solved Examples

• Ex 1: Classify CH₂=CHCl.
  Step 1: Halogen attached to sp² carbon of alkene.
  Answer: Vinylic halide (monohalo, sp² C–X).
• Ex 2: Classify CCl₃H.
  Step 1: Three halogens attached to sp³ carbon.
  Answer: Trihaloalkane (benzylic halide).

Practice Set

• Q1: Which is a vinylic halide?
  A. CH₃Cl
  B. CH₂=CHCl
  C. C₆H₅Cl
  D. CH₂Cl–CH₃
• Q2: Dichlorobenzene is classified as:
  A. Alkyl halide
  B. Allylic halide
  C. Aryl halide
  D. Benzylic halide
• Q3: Trichloromethane (CHCl₃) is an example of:
  A. Monohaloalkane
  B. Dihaloalkane
  C. Trihaloalkane
  D. Haloarene

Answer Key

• Q1: B. CH₂=CHCl - Halogen on sp² carbon of alkene (vinylic halide).
• Q2: C. Aryl halide - Halogen attached to aromatic ring carbon.
• Q3: C. Trihaloalkane - Three halogens on sp³ carbon.

Memory Aids

• Mono = one halogen; Di = two; Tri = three halogens.
• sp³ C–X: Alkyl, allylic, benzylic halides.
• sp² C–X: Vinylic and aryl halides.
• Benzylic halides have halogen on carbon adjacent 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: Use prefixes o- (ortho), m- (meta), p- (para).
  • IUPAC: Use numbers 1,2-; 1,3-; 1,4- to indicate positions.

Solved Examples

• Ex 1: Name CH₃CH₂Br
  • Common: Ethyl bromide
  • IUPAC: Bromoethane
• Ex 2: Name C₆H₄Br₂ with bromines at 1,4 positions
  • Common: p-Dibromobenzene
  • IUPAC: 1,4-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 1,3-dibromobenzene is:
  • A. o-Dibromobenzene
  • B. m-Dibromobenzene
  • C. p-Dibromobenzene
  • D. 3,5-Dibromobenzene
• Q3: Which is the correct IUPAC name for CH₃C₆H₄Br with Br at position 2?
  • A. 2-Bromotoluene
  • B. o-Bromotoluene
  • C. Bromobenzene
  • D. 1-Bromotoluene

Answer Key

• Q1: B - Chloroethane is the correct IUPAC name; common name is ethyl chloride.
• Q2: B - m-Dibromobenzene corresponds to 1,3-dibromobenzene.
• Q3: A - 2-Bromotoluene is the correct IUPAC name indicating bromine at position 2.

Memory Aids

• o-, m-, p- prefixes correspond to 1,2-; 1,3-; 1,4- positions on benzene ring.
• Common name = Alkyl group + halide; IUPAC = Halosubstituted hydrocarbon.

Dihaloalkanes

Classification and Naming

• Dihaloalkanes contain two halogen atoms of the same type.
• Common names:
  • Geminal (gem-) dihalides: Both halogens on the same carbon; called alkylidene halides.
  • Vicinal (vic-) dihalides: Halogens on adjacent carbons; called alkylene dihalides.
• IUPAC names: Both types named as dihaloalkanes with position numbers (e.g., 1,1-dichloroethane, 1,2-dichloroethane).

Solved Examples

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

Practice Set

• Q1: What is the common name for 1,1-dibromoethane?
  • A. Ethylene dibromide
  • B. Ethylidene dibromide
  • C. Ethane dibromide
  • D. Ethyl dibromide
• Q2: Which compound is a vicinal dihalide?
  • A. 1,1-Dichloropropane
  • B. 1,2-Dichloropropane
  • C. 2,2-Dichloropropane
  • D. 3,3-Dichloropropane
• Q3: Identify the IUPAC name of ethylene dichloride.
  • A. 1,1-Dichloroethane
  • B. 1,2-Dichloroethane
  • C. 2,2-Dichloroethane
  • D. 1,3-Dichloroethane

Answer Key

• Q1: B - Ethylidene dibromide is the common name for 1,1-dibromoethane (geminal).
• Q2: B - 1,2-Dichloropropane is vicinal (halogens on adjacent carbons).
• Q3: B - 1,2-Dichloroethane is the IUPAC name for ethylene dichloride (vicinal).

Memory Aids

• Geminal = same carbon (alkylidene halide); Vicinal = adjacent carbons (alkylene dihalide).
• IUPAC uses numbering to specify halogen positions precisely.

Nature of C — X Bond

Bond Polarity and Electronegativity

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

Solved Examples

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

Practice Set

• Q1: Which atom in alkyl halide carries partial positive charge?
  • A) Halogen
  • B) Carbon
  • C) Hydrogen
  • D) Oxygen
• Q2: The polarity of C—X bond arises due to:
  • A) Equal electronegativity
  • B) Halogen more electronegative than carbon
  • C) Carbon more electronegative than halogen
  • D) Ionic bond formation
• Q3: In CH₃Br, the partial charges are:
  • A) C δ⁻, Br δ⁺
  • B) C δ⁺, Br δ⁻
  • C) Both neutral
  • D) Both positive

Answer Key

• Q1: B — Carbon carries partial positive charge due to lower electronegativity.
• Q2: B — Halogen is more electronegative, causing bond polarity.
• Q3: B — Carbon is δ⁺ and bromine is δ⁻ in polar bond.

Memory Aids

• Remember: "Electronegativity difference causes polarity".
• Halogen atoms pull electron density towards themselves.
• Carbon always δ⁺ in C—X bond.

Bond Length and Enthalpy

Trend with Halogen Size

• Bond length increases with halogen size: F < Cl < Br < I.
• Bond enthalpy decreases as bond length increases.
• Shorter bonds are stronger; longer bonds are weaker.
• Example: C—F bond length = 139 pm, enthalpy = 452 kJ/mol.
• C—I bond length = 214 pm, enthalpy = 234 kJ/mol.

Solved Examples

• Ex 1: Compare bond strength of C—Cl and C—I.
  • C—Cl bond is shorter and stronger than C—I bond.

Practice Set

• Q1: Which bond has the highest bond enthalpy?
  • A) C—F
  • B) C—Cl
  • C) C—Br
  • D) C—I
• Q2: As halogen size increases, bond length:
  • A) Decreases
  • B) Increases
  • C) Remains same
  • D) Varies irregularly
• Q3: Which bond is weakest?
  • A) C—F
  • B) C—Cl
  • C) C—Br
  • D) C—I

Answer Key

• Q1: A — C—F bond has highest bond enthalpy (452 kJ/mol).
• Q2: B — Bond length increases with halogen size.
• Q3: D — C—I bond is weakest due to longest bond length.

Memory Aids

• Bond length ↑ → Bond enthalpy ↓
• Fluorine forms shortest, strongest bonds.
• Iodine forms longest, weakest bonds.

Dipole Moment Trends

Variation from F to I

• Dipole moment depends on bond polarity and bond length.
• Electronegativity decreases from F to I.
• Bond length increases from F to I.
• Dipole moments vary irregularly due to opposing effects.
• Example values (Debye): CH₃F (1.85), CH₃Cl (1.86), CH₃Br (1.75), CH₃I (1.55).

Solved Examples

• Ex 1: Explain why CH₃Cl has higher dipole moment than CH₃F.
  • CH₃Cl has longer bond length compensating lower electronegativity difference.

Practice Set

• Q1: Dipole moment is a measure of:
  • A) Bond length only
  • B) Charge separation in molecule
  • C) Molecular weight
  • D) Number of bonds
• Q2: Which molecule has highest dipole moment?
  • A) CH₃F
  • B) CH₃Cl
  • C) CH₃Br
  • D) CH₃I
• Q3: Dipole moment variation from F to I is:
  • A) Regular increase
  • B) Regular decrease
  • C) Irregular due to size and electronegativity
  • D) Constant

Answer Key

• Q1: B — Dipole moment measures charge separation.
• Q2: B — CH₃Cl has highest dipole moment (1.860 D).
• Q3: C — Variation is irregular due to opposing factors.

Memory Aids

• Dipole moment = Electronegativity difference × Bond length.
• Opposite trends of size and electronegativity cause irregular dipole moments.
• CH₃Cl shows maximum dipole moment among alkyl halides.

Methods of Preparation of Haloalkanes

From Alcohols

• Primary and secondary alcohols react with HCl in presence of ZnCl₂ catalyst to form alkyl chlorides.
• Tertiary alcohols react with concentrated HCl by simple shaking at room temperature.
• Alkyl bromides are prepared by reacting alcohols with NaBr and H₂SO₄.
• Phosphorus halides (PX₃, X = Cl, Br) convert alcohols to alkyl halides (R–X).
• 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–I) obtained by heating alcohols with NaI or KI in 95% ortho phosphoric acid.
• Reactivity order with haloacid: tertiary > secondary > primary alcohols.

Solved Examples

• Ex 1: Convert 2° alcohol to alkyl chloride using HCl and ZnCl₂.
• Step 1: Mix 2° alcohol with HCl and anhydrous ZnCl₂.
• Step 2: Reaction proceeds via SN1/SN2 mechanism forming alkyl chloride.

Practice Set

• Q1: Which reagent converts tertiary alcohol to alkyl chloride at room temperature?
  A. HCl + ZnCl₂
  B. Concentrated HCl
  C. NaBr + H₂SO₄
  D. PCl₅
• Q2: Which reagent is used to prepare alkyl bromide from alcohol?
  A. NaI + H₃PO₄
  B. NaBr + H₂SO₄
  C. SOCl₂ + pyridine
  D. HCl + ZnCl₂
• Q3: Order of reactivity of alcohols with haloacid is:
  A. 1° > 2° > 3°
  B. 3° > 2° > 1°
  C. 2° > 3° > 1°
  D. 1° > 3° > 2°

Answer Key

• Q1: B - Tertiary alcohols react with concentrated HCl directly at room temperature.
• Q2: B - NaBr with H₂SO₄ converts alcohols to alkyl bromides.
• Q3: B - Reactivity order is tertiary > secondary > primary alcohols.

Memory Aids

• Use ZnCl₂ catalyst for 1° and 2° alcohols with HCl.
• Tertiary alcohols react directly with concentrated HCl.
• Phosphorus halides (PX₃) are versatile for halide formation.

From Hydrocarbons

Free Radical Halogenation

• Methane reacts with Cl₂ under UV light or heat to form chloromethanes.
• Chlorine splits into radicals initiating substitution of H by Cl.
• Products: CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄ plus HCl byproduct.
• Ease of H replacement: tertiary > secondary > primary.

Addition to Alkenes

• Alkenes react with HX (X = Cl, Br, I) adding across double bond.
• Markovnikov's rule: H adds to carbon with more H, halogen to carbon with fewer H.
• In presence of peroxide, HBr adds via Anti-Markovnikov rule.
• Halogens (Br₂) add to alkenes forming vicinal dihalides.

Solved Examples

• Ex 1: Chlorination of methane under UV light.
• Step 1: UV light splits Cl₂ into radicals.
• Step 2: Cl radicals replace H atoms in methane stepwise.
• Ex 2: Addition of HBr to propene in presence of peroxide.
• Step 1: Peroxide initiates radical addition.
• Step 2: Br adds to less substituted carbon (Anti-Markovnikov).

Practice Set

• Q1: Which condition initiates free radical halogenation?
  A. Heat or UV light
  B. Acid catalyst
  C. Base catalyst
  D. High pressure
• Q2: In addition of HBr to alkene with peroxide, halogen adds to:
  A. More substituted carbon
  B. Less substituted carbon
  C. Both carbons equally
  D. No addition occurs
• Q3: Order of ease of H abstraction by Br₂ at 400 K is:
  A. 1° > 2° > 3°
  B. 3° > 2° > 1°
  C. 2° > 3° > 1°
  D. Equal for all

Answer Key

• Q1: A - Heat or UV light initiates radical formation.
• Q2: B - Peroxide causes Anti-Markovnikov addition; halogen adds to less substituted carbon.
• Q3: B - Br₂ abstracts 3° H fastest, then 2°, then 1°.

Memory Aids

• Free radical halogenation needs UV or heat.
• Markovnikov addition: halogen to more substituted carbon.
• Peroxide effect reverses Markovnikov rule for HBr.

Halogen Exchange

Finkelstein Reaction

• Alkyl halide (R–X, X = Cl or Br) reacts with NaI in dry acetone.
• NaI replaces Cl or Br with I forming alkyl iodide (R–I).
• NaCl or NaBr precipitates out, driving reaction forward (Le Chatelier’s principle).

Swarts Reaction

• Alkyl halide (R–X, X = Cl or Br) reacts with AgF to form alkyl fluoride (R–F).
• Silver halide (AgX) precipitates as byproduct.
• Other reagents: HgF₂, CoF₃, SbF₅ can also be used.

Solved Examples

• Ex 1: Convert R–Cl to R–I using NaI in dry acetone.
• Step 1: Mix R–Cl with NaI in dry acetone.
• Step 2: R–I forms; NaCl precipitates out.
• Ex 2: Prepare alkyl fluoride from R–Br using AgF.
• Step 1: React R–Br with AgF.
• Step 2: R–F forms; AgBr precipitates.

Practice Set

• Q1: In Finkelstein reaction, solvent used is:
  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 salt precipitates in Finkelstein reaction?
  A. NaI
  B. NaCl or NaBr
  C. AgX
  D. KI

Answer Key

• Q1: B - Dry acetone is used to precipitate NaCl/NaBr.
• Q2: C - Swarts reaction replaces Cl or Br with fluorine.
• Q3: B - NaCl or NaBr precipitates, driving reaction forward.

Memory Aids

• Finkelstein: Halogen exchange using NaI in dry acetone.
• Swarts: Halogen replaced by fluorine using AgF.
• Precipitation of salt drives equilibrium forward.

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 dark to prevent side reactions.
• Example: Toluene reacts with X₂ (Cl₂/Br₂) + Fe catalyst to give ortho- and para-halotoluene.
• Ortho- and para-isomers differ in melting points, allowing easy separation.
• Iodination is reversible; requires oxidising agents (HNO₃, HIO₄) to oxidise HI formed.
• 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: Fe activates Br₂ forming Br⁺.
• Step 2: Electrophilic substitution at ortho and para positions relative to CH₃.
• Step 3: Products separated by melting point difference.

Practice Set

• Q1: Which catalyst is used for electrophilic halogenation of benzene?
  A. FeCl₃ B. AlCl₃ C. Cu D. Zn
• Q2: Why are fluoroarenes not prepared by direct halogenation?
  A. Low reactivity B. High reactivity of F₂ C. No catalyst D. Fluorine is inert
• Q3: Which oxidising agent is used to oxidise HI during iodination?
  A. HNO₃ B. H₂SO₄ C. HCl D. NaOH

Answer Key

• Q1: A. FeCl₃ - Lewis acid catalyst activates halogen for electrophilic substitution.
• Q2: B. High reactivity of F₂ - Fluorine reacts violently, unsuitable for direct halogenation.
• Q3: A. HNO₃ - Strong oxidising agent to convert HI back to I₂.

Memory Aids

• FeCl₃ activates Cl₂/Br₂ for aromatic substitution.
• Toluene directs halogenation to ortho and para positions.
• Iodination needs oxidising agent to prevent reversal.
• Fluorine too reactive for direct aromatic substitution.

From Benzene Diazonium Chloride

Key Reactions Using Benzene Diazonium Chloride

• Start with aniline (C₆H₅–NH₂).
• React with NaNO₂ and HCl at 273–278 K to form benzene diazonium chloride (C₆H₅–N₂⁺Cl⁻).
• Balz-Schiemann reaction: Benzene diazonium chloride + NaNO₂ + HBF₄ → Fluorobenzene + N₂ + BF₃.
• Sandmeyer's reaction: Benzene diazonium chloride + CuCl + HCl → Chlorobenzene + N₂ + CuCl₂.
• Similarly, with CuBr + HBr → Bromobenzene + N₂.
• These reactions replace –NH₂ with halogens or fluorine on benzene.

Solved Examples

• Ex 1: Preparation of chlorobenzene from aniline
• Step 1: Aniline + NaNO₂ + HCl (273 K) → Benzene diazonium chloride
• Step 2: Benzene diazonium chloride + CuCl + HCl → Chlorobenzene + N₂

Practice Set

• Q1: What is the temperature range for diazotization?
  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?
  A. HBF₄ B. HCl C. HBr D. HNO₃
• Q3: What is the gaseous product released in Sandmeyer's reaction?
  A. CO₂ B. N₂ C. H₂ D. O₂

Answer Key

• Q1: A. 273–278 K - Low temperature needed to stabilize diazonium salt.
• Q2: A. HBF₄ - Used to form fluorobenzene in Balz-Schiemann reaction.
• Q3: B. N₂ - Nitrogen gas is released during Sandmeyer's reaction.

Memory Aids

• Diazotization at low temperature (273–278 K) forms diazonium salt.
• Balz-Schiemann reaction replaces –N₂⁺ with F.
• Sandmeyer's reaction replaces –N₂⁺ with Cl or Br using Cu salts.
• Nitrogen gas (N₂) is always released in these reactions.

Physical Properties

General Characteristics

• Pure alkyl halides are colourless; bromides and iodides develop colour on light exposure.
• 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 dipole-dipole and van der Waals forces than parent hydrocarbons due to polarity and higher molecular mass.
• Boiling points of alkyl halides 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.
• Isomeric dihalobenzenes have nearly same boiling points; 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 more carbon atoms, halogen atoms, and heavier halogens.
• 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 breaking under sunlight.
• Photochemical reaction: 2R–I 1hv1 R–R + I₂
• Liberated iodine dissolves in alkyl iodide, imparting a dark colour.

Solved Examples

• Ex 1: Arrange the boiling points of CH₃Cl, CH₃Br, and CH₃I.
• Step 1: Identify halogen size and mass: I > Br > Cl.
• Step 2: Boiling points increase with halogen size due to van der Waals forces.
• Answer: CH₃I > CH₃Br > CH₃Cl.
• Ex 2: Explain why para-dihalobenzenes have higher melting points than ortho- and meta-isomers.
• Step 1: Para-isomers are more symmetrical.
• Step 2: Symmetry allows better packing in crystal lattice.
• Answer: Higher melting points due to better lattice fitting.

Practice Set

• Q1: Which alkyl halide is a gas at room temperature?
  • A. n-Butyl chloride
  • B. Methyl bromide
  • C. Ethyl iodide
  • D. Propyl chloride
• Q2: Order the boiling points of the following: CH₃Cl, CH₃Br, CH₃I, CH₃F.
  • A. CH₃F < CH₃Cl < CH₃Br < CH₃I
  • B. CH₃I < CH₃Br < CH₃Cl < CH₃F
  • C. CH₃Cl < CH₃F < CH₃Br < CH₃I
  • D. CH₃Br < CH₃Cl < CH₃F < CH₃I
• Q3: What causes the violet/brown colour in aged alkyl iodides?
  • A. Formation of alkyl radicals
  • B. Dissolved iodine from photochemical decomposition
  • C. Reaction with water
  • D. Oxidation by air

Answer Key

• Q1: B. Methyl bromide is a gas at room temperature due to low molecular weight.
• Q2: A. Boiling points increase with halogen size: F < Cl < Br < I.
• Q3: B. Violet/brown colour is due to iodine formed by photochemical decomposition dissolving in alkyl iodide.

Memory Aids

• Boiling point order: RI > RBr > RCl > RF (increasing van der Waals forces).
• Para-isomers have higher melting points due to symmetry and better crystal packing.
• Alkyl iodides decompose under light: 2R–I 1hv1 R–R + I₂, causing colour change.

Chemical Reactions

Reactions of Haloalkanes

• Haloalkane reactions: (i) Nucleophilic substitution (ii) Elimination (iii) Reaction with metals

Nucleophilic Substitution Reactions

Key Points

• C–X bond is highly polar due to electronegativity difference; polarity decreases F > Cl > Br > I.
• Polarity causes nucleophilic substitution reactions.
• Ambident nucleophiles (e.g., CN^–, NO₂^–) can attack through different atoms, forming different products.
• Two mechanisms: SN1 (unimolecular) and SN2 (bimolecular).
• SN1: two-step, carbocation intermediate, rate depends on carbocation stability, weak nucleophiles preferred, racemization occurs.
• SN2: one-step, nucleophile attacks opposite leaving group, rate depends on steric hindrance, strong nucleophiles preferred, inversion of configuration.
• Reactivity order: R–I > R–Br > R–Cl > R–F.
• Primary allylic and benzylic halides show higher SN1 reactivity due to resonance stabilization.

Solved Examples

• Ex 1: SN2 reaction of CH₃Br with OH^– gives CH₃OH with inversion of configuration.
• Ex 2: SN1 reaction of (CH₃)₃CBr forms carbocation intermediate, then reacts with H₂O to give (CH₃)₃COH racemic mixture.

Practice Set

• Q1: Which halide undergoes SN1 fastest? A) CH₃Cl B) (CH₃)₃CBr C) CH₃Br D) C₆H₅Cl
• Q2: In SN2 reaction, nucleophile attacks from: A) Same side as leaving group B) Opposite side C) Any side D) No attack
• Q3: Ambident nucleophile CN^– can form: A) Alkyl cyanide B) Alkyl isocyanide C) Both A and B D) None

Answer Key

• Q1: B; tertiary halides form stable carbocations, favoring SN1.
• Q2: B; SN2 involves backside attack causing inversion.
• Q3: C; CN^– attacks via C or N forming cyanides or isocyanides.

Memory Aids

• SN1: 1 step rate, carbocation intermediate, racemization.
• SN2: 2 molecules rate, backside attack, inversion.
• Reactivity order: I > Br > Cl > F.
• Ambident nucleophiles attack via different atoms.

Elimination Reactions

Key Points

• Haloalkanes with β-hydrogen undergo dehydrohalogenation with alcoholic KOH on heating.
• Elimination removes H from β-carbon and halogen from α-carbon forming alkene.
• Saytzeff’s rule: major alkene has maximum alkyl substitution.

Solved Examples

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

Practice Set

• Q1: Product of elimination from 2-chloropropane with alc. KOH? A) Propene B) Propane C) Propan-2-ol D) No reaction
• Q2: According to Saytzeff’s rule, major product is the alkene with: A) Least alkyl groups B) Most alkyl groups C) No double bond D) Triple bond
• Q3: Which reagent favors elimination over substitution? A) Dilute KOH B) Alc. KOH C) H₂SO₄ D) NaCl

Answer Key

• Q1: A; elimination forms propene.
• Q2: B; more substituted alkene is major.
• Q3: B; alcoholic KOH favors elimination.

Memory Aids

• Elimination: β-H and halogen removed, forms alkene.
• Saytzeff’s rule: major alkene = most substituted.
• Alc. KOH → elimination; aq. KOH → substitution.

Reactions with Metals

Key Points

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

Solved Examples

• Ex 1: C₂H₅Br + Mg → C₂H₅MgBr (Grignard reagent).
• Ex 2: 2 C₂H₅Br + 2 Na → C₄H₁₀ + 2 NaBr (Wurtz reaction).

Practice Set

• Q1: Grignard reagent is formed by reaction of alkyl halide with: A) Na B) Mg C) Li D) K
• Q2: Wurtz reaction joins: A) Two alkyl halides B) Alkyl halide and metal C) Two metals D) Alkyl halide and water
• Q3: Organolithium compounds are formed by reaction of alkyl halide with: A) Li B) Na C) Mg D) K

Answer Key

• Q1: B; Grignard reagents form with Mg.
• Q2: A; Wurtz reaction couples two alkyl halides.
• Q3: A; organolithium formed with Li.

Memory Aids

• Grignard: R–X + Mg → R–MgX.
• Wurtz: 2 R–X + 2 Na → R–R + 2 NaX.
• Organolithium: R–X + Li → R–Li.

Nucleophilic Substitution in Haloarenes

Key Points

• Haloarenes less reactive than haloalkanes due to partial double bond character of C–X bond (resonance).
• C–X bond less polar; sp² carbon more electronegative than sp³.
• 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).
• Electron withdrawing groups (-NO₂) at ortho/para increase reactivity.

Solved Examples

• Ex 1: Chlorobenzene + NaOH (623 K, 300 atm) → Phenol.
• Ex 2: 2,4-Dinitrochlorobenzene + NaOH → 2,4-Dinitrophenol.

Practice Set

• Q1: Why are haloarenes less reactive in nucleophilic substitution? A) Strong C–X bond B) Resonance stabilization C) Less polar bond D) All
• Q2: Replacement of Cl by OH in chlorobenzene requires: A) Room temp B) High temp and pressure C) Acidic medium D) No reaction
• Q3: Electron withdrawing groups on haloarenes: A) Decrease reactivity B) Increase reactivity C) No effect D) Make it explosive

Answer Key

• Q1: D; all factors reduce reactivity.
• Q2: B; harsh conditions needed.
• Q3: B; increase reactivity by stabilizing intermediate.

Memory Aids

• Haloarenes: C–X bond partial double bond, less reactive.
• OH substitution needs harsh conditions.
• -NO₂ groups increase reactivity.

Electrophilic Substitution in Haloarenes

Key Points

• Halogen's inductive effect (–I) decreases electron density; haloarenes less reactive than benzene.
• Halogen's resonance effect (+M) increases electron density at ortho and para positions.
• Haloarenes are deactivating but ortho/para directing.
• Halogenation with Cl₂/FeCl₃ gives dihalobenzenes (major para, minor ortho).
• Nitration with conc. HNO₃/H₂SO₄ gives o- and p-halonitrobenzenes.
• Sulphonation with conc. H₂SO₄ gives o- and p-halo benzene sulfonic acids.
• Friedel-Crafts alkylation/acylation gives o- and p-haloalkyl/acetophenones.

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: Halogens in haloarenes are: A) Activating B) Deactivating C) Neutral D) Explosive
• Q2: Major product in chlorobenzene nitration is: A) Ortho B) Meta C) Para D) None
• Q3: Friedel-Crafts alkylation of chlorobenzene gives: A) Meta product B) Ortho and para products C) Only para D) No reaction

Answer Key

• Q1: B; halogens deactivate benzene ring.
• Q2: C; para product is major due to sterics.
• Q3: B; both ortho and para products formed.

Memory Aids

• Halogens: –I effect deactivates, +M effect ortho/para directs.
• Electrophilic substitution: major para product.
• Friedel-Crafts: ortho/para substitution.

Polyhalogen Compounds

Definition and Examples

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

Solved Examples

• Ex 1: Identify the halogen atoms in CHI₃.
  Answer: Iodine atoms (3 atoms).

Practice Set

• Q1: Which of the following is a polyhalogen compound?
  A. CH₄
  B. CHCl₃
  C. C₂H₆
  D. C₂H₅OH
• Q2: Polyhalogen compounds contain how many halogen atoms?
  A. One
  B. Two or more
  C. None
  D. Only fluorine
• Q3: Which of the following is NOT a polyhalogen compound?
  A. CCl₄
  B. CHBr₃
  C. CH₃Cl
  D. CF₂Cl₂

Answer Key

• Q1: B - CHCl₃ contains multiple halogen atoms.
• Q2: B - Polyhalogen compounds have two or more halogen atoms.
• Q3: C - CH₃Cl has only one halogen atom, not polyhalogen.

Memory Aids

• Polyhalogen = multiple halogen atoms attached to carbon.
• Common halogens: Cl, Br, I, F.
• Look for compounds with 2 or more halogen atoms for polyhalogen.

Dichloromethane (Methylene Chloride), CH₂Cl₂

Uses

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

Harmful Effects

• Harms the human central nervous system.
• Low exposure causes slight hearing and vision impairment.
• High exposure causes dizziness, nausea, tingling, numbness in fingers and toes.
• Direct skin contact causes burning and redness.
• Contact with eyes can burn the cornea.

Solved Examples

• Ex 1: Name one use of CH₂Cl₂.
  Answer: Solvent or paint remover.

Practice Set

• Q1: What is a common use of dichloromethane?
  A. Fuel
  B. Solvent
  C. Food additive
  D. Fertilizer
• Q2: Which system does CH₂Cl₂ harm?
  A. Digestive
  B. Central nervous
  C. Respiratory
  D. Circulatory
• Q3: What effect does high exposure to CH₂Cl₂ cause?
  A. Improved vision
  B. Dizziness and numbness
  C. Increased appetite
  D. Skin whitening

Answer Key

• Q1: B - Dichloromethane is widely used as a solvent.
• Q2: B - It harms the central nervous system.
• Q3: B - High exposure causes dizziness and numbness.

Memory Aids

• CH₂Cl₂ = solvent + paint remover + aerosol propellant.
• Harmful to CNS; watch for dizziness and skin burns.

Trichloromethane (Chloroform), CHCl₃

Uses

• Major use: production of freon refrigerant R-22.
• Solvent for resins, rubbers, oils, fats, alkaloids, iodine.
• Used in 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.
• Inhaling vapours depresses central nervous system causing dizziness, fatigue, headache.

Solved Examples

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

Practice Set

• Q1: What is the major use of chloroform?
  A. Fuel
  B. Refrigerant production
  C. Food preservative
  D. Fertilizer
• Q2: Phosgene gas formed from chloroform causes damage to?
  A. Heart
  B. Liver and kidney
  C. Lungs
  D. Brain
• Q3: Which of the following is a photochemical reaction?
  A. CHCl₃ + O₂ 1hv1 COCl₂ + HCl
  B. CH₄ combustion
  C. NaCl dissolution
  D. H₂O electrolysis

Answer Key

• Q1: B - Chloroform is used to produce freon refrigerant R-22.
• Q2: B - Phosgene causes liver and kidney damage.
• Q3: A - Chloroform oxidation under light is a photochemical reaction.

Memory Aids

• Chloroform 1hv1 phosgene + HCl (toxic).
• Used in refrigerant production and solvents.

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?
  Answer: Due to its bad smell.

Practice Set

• Q1: Iodoform is mainly used as?
  A. Solvent
  B. Antiseptic
  C. Fuel
  D. Fertilizer
• Q2: Iodoform liberates which element for antiseptic action?
  A. Chlorine
  B. Iodine
  C. Fluorine
  D. Bromine
• Q3: Why is iodoform replaced in medical use?
  A. Toxicity
  B. Objectionable smell
  C. Expensive
  D. Unstable

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.
• Used as cleaning agent and spot remover.
• Used as fire extinguisher (pyrene).

Harmful Effects

• Causes liver cancer in humans.
• 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 use of CCl₄ in fire safety.
  Answer: Fire extinguisher (pyrene).

Practice Set

• Q1: Carbon tetrachloride is used as?
  A. Food additive
  B. Fire extinguisher
  C. Fertilizer
  D. Fuel
• Q2: Which organ is mainly affected by CCl₄ toxicity?
  A. Heart
  B. Liver
  C. Brain
  D. Lungs
• Q3: What environmental harm does CCl₄ cause?
  A. Acid rain
  B. Ozone depletion
  C. Global warming
  D. Water pollution

Answer Key

• Q1: B - Used as fire extinguisher.
• Q2: B - Causes liver cancer.
• Q3: B - Depletes ozone layer.

Memory Aids

• CCl₄ = fire extinguisher + liver toxin + ozone depleter.
• Watch for dizziness and nerve damage symptoms.

Freons

Definition and Preparation

• Chlorofluorocarbon compounds of methane and ethane with chlorine and fluorine.
• Example: CF₂Cl₂ (Freon-12).
• Prepared by Swarts reaction: CCl₄ + 2 AgF₂ 11 CCl₂F₂ + 2 AgCl.

Properties and Uses

• Extremely stable, unreactive, non-toxic, non-corrosive gases.
• Easily liquefiable.
• Used as aerosol propellants, refrigerants, and in air conditioning.

Harmful Effects

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

Solved Examples

• Ex 1: Write the Swarts reaction for Freon-12.
  Answer: CCl₄ + 2 AgF₂ 11 CCl₂F₂ + 2 AgCl

Practice Set

• Q1: Freons are mainly used as?
  A. Fertilizers
  B. Refrigerants
  C. Fuels
  D. Solvents
• Q2: Freons cause environmental harm by?
  A. Acid rain
  B. Ozone depletion
  C. Water pollution
  D. Soil erosion
• Q3: Which reaction is used to prepare Freon-12?
  A. Swarts reaction
  B. Wurtz reaction
  C. Friedel-Crafts
  D. Cannizzaro

Answer Key

• Q1: B - Used as refrigerants.
• Q2: B - Cause ozone layer depletion.
• Q3: A - Prepared by Swarts reaction.

Memory Aids

• Freons = stable CFCs used in cooling.
• Swarts reaction replaces Cl with F.
• Environmental hazard: ozone depletion.

p, p’-Dichlorodiphenyl Trichloroethane (DDT)

Preparation and Properties

• Prepared by condensation of chloral with chlorobenzene in presence of conc. H₂SO₄.
• Powerful insecticide effective against mosquitoes and lice.

Harmful Effects

• Insects have developed resistance to DDT.
• Highly toxic to fish.
• Chemically stable and fat soluble.
• Not rapidly metabolised; accumulates in fatty tissues.
• Bioaccumulation leads to increased concentration over time.

Solved Examples

• Ex 1: Name the reactants for DDT preparation.
  Answer: Chloral and chlorobenzene.

Practice Set

• Q1: DDT is mainly used as?
  A. Fertilizer
  B. Insecticide
  C. Solvent
  D. Fuel
• Q2: Which is a harmful effect of DDT?
  A. Rapid metabolism
  B. Resistance in insects
  C. Water purification
  D. Increased plant growth
• Q3: DDT accumulates in?
  A. Muscle tissue
  B. Fatty tissue
  C. Bone
  D. Blood plasma

Answer Key

• Q1: B - Used as insecticide.
• Q2: B - Insects developed resistance.
• Q3: B - Deposited in fatty tissues.

Memory Aids

• DDT = insecticide, fat soluble, bioaccumulates.
• Resistance and toxicity to aquatic life are key concerns.

Polyhalogen Compounds Table

Summary

• Table lists important polyhalogen compounds with formula and uses.
• Examples include chloretone (drug), phosgene (poison gas), freon (refrigerant), tear gas, insecticides, antiseptics, anesthetics, fire extinguishers, solvents.
• Shows diverse applications in medicine, industry, and agriculture.

Solved Examples

• Ex 1: Identify the compound used as a fire extinguisher.
  Answer: Carbon tetrachloride (pyrene).

Practice Set

• Q1: Which compound is used as an insecticide?
  A. Phosgene
  B. DDT
  C. Freon
  D. Chloretone
• Q2: Phosgene is used as?
  A. Refrigerant
  B. Poisonous gas
  C. Solvent
  D. Antiseptic
• Q3: Which compound is a hypnotic drug?
  A. Chloretone
  B. DDT
  C. Freon
  D. Phosgene

Answer Key

• Q1: B - DDT is an insecticide.
• Q2: B - Phosgene is a poisonous gas.
• Q3: A - Chloretone is a hypnotic drug.

Memory Aids

• Match compound to use: DDT-insecticide, Phosgene-poison, Freon-refrigerant.
• Remember chloretone as drug, pyrene as fire extinguisher.

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: What precipitate forms when iodoform reacts with AgNO₃?
  Answer: Yellow AgI precipitate.

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 AgI precipitate?
  A. It is unstable
  B. It is stable
  C. It contains iodine
  D. It reacts with water
• Q3: The yellow precipitate formed is?
  A. AgCl
  B. AgBr
  C. AgI
  D. AgF

Answer Key

• Q1: B - Iodoform gives yellow AgI precipitate.
• Q2: B - Chloroform is stable and does not react.
• Q3: C - Yellow precipitate is silver iodide (AgI).

Memory Aids

• Iodoform test: Yellow AgI precipitate with AgNO₃.
• Chloroform is stable; no precipitate formed.