Chapter 9: Hydrocarbons

Introduction

Key Points

• Hydrocarbons are organic compounds made up only of carbon (C) and hydrogen (H) atoms.
• Main natural sources: natural gas, petroleum, and coal.

Solved Examples

• Ex 1: Identify the elements in hydrocarbons.
  Answer: Carbon and Hydrogen only.

Practice Set

• Q1: What elements do hydrocarbons contain?
  A. Carbon and Oxygen
  B. Carbon and Hydrogen
  C. Hydrogen and Oxygen
  D. Carbon, Hydrogen, and Oxygen
• Q2: Which of the following is a major source of hydrocarbons?
  A. Water
  B. Natural gas
  C. Air
  D. Salt
• Q3: Hydrocarbons are primarily composed of which two elements?
  A. Carbon and Nitrogen
  B. Carbon and Hydrogen
  C. Hydrogen and Oxygen
  D. Nitrogen and Oxygen

Answer Key

• Q1: B - Hydrocarbons contain only carbon and hydrogen atoms.
• Q2: B - Natural gas is a main source of hydrocarbons.
• Q3: B - Hydrocarbons are composed of carbon and hydrogen only.

Memory Aids

• Hydrocarbons = Hydro (H) + Carbon (C) only.
• Sources: Natural gas, Petroleum, Coal (NPC).

Classification

Types of Hydrocarbons

• Based on carbon-carbon bonds, hydrocarbons are classified into:
• Saturated: Only single bonds (C–C, C–H).
• Unsaturated: Contain double or triple bonds (C=C, C≡C).
• Aromatic: Special cyclic compounds with resonance (e.g., benzene).

• Saturated Hydrocarbons:
• Alkanes: Open chain with single bonds.
• Cycloalkanes: Closed ring with single bonds.
• Unsaturated Hydrocarbons: Contain multiple bonds (double/triple).
• Aromatic Hydrocarbons: Cyclic compounds with special stability due to resonance.

Solved Examples

• Ex 1: Classify C₂H₆.
  Answer: Saturated alkane (ethane, single bonds only).
• Ex 2: Classify C₂H₄.
  Answer: Unsaturated alkene (contains a double bond).

Practice Set

• Q1: Which hydrocarbon is saturated?
  A. Ethene
  B. Ethyne
  C. Ethane
  D. Benzene
• Q2: Which hydrocarbon contains a triple bond?
  A. Ethane
  B. Ethene
  C. Ethyne
  D. Cyclohexane
• Q3: Aromatic hydrocarbons are characterized by:
  A. Only single bonds
  B. Multiple bonds without resonance
  C. Cyclic structure with resonance
  D. Open chain with double bonds

Answer Key

• Q1: C - Ethane is saturated with single bonds only.
• Q2: C - Ethyne contains a triple bond.
• Q3: C - Aromatic hydrocarbons have cyclic structure with resonance.

Memory Aids

• Saturated = Single bonds only (alkanes, cycloalkanes).
• Unsaturated = Double or triple bonds (alkenes, alkynes).
• Aromatic = Cyclic with resonance (benzene ring).

Alkanes

General Properties

• Alkanes are saturated hydrocarbons with formula C_nH_2n+2.
• Examples: Methane (CH₄), Ethane (C₂H₆), Propane (C₃H₈).

Structure

• Carbon atoms are sp³ hybridized, forming tetrahedral geometry.
• All C–C and C–H bonds are sigma (σ) bonds.
• Bond lengths: C–C = 1.54 Å, C–H = 1.09 Å; bond angle = 109° 28'.

Isomerism

• Exhibit chain isomerism.
• First three members (methane, ethane, propane) have no isomers.
• Number of isomers increases with carbon atoms: Butane (2), Pentane (3), Hexane (5), Heptane (9), Octane (18), Decane (75).

Solved Examples

• Ex 1: Identify isomers of pentane.
  Answer: n-pentane (straight chain), iso-pentane (one branch), neo-pentane (more branched).

Practice Set

• Q1: What is the general formula of alkanes?
  A. C_nH_2n B. C_nH_2n+2 C. C_nH_2n-2 D. C_nH_n
• Q2: Which alkane has 5 isomers?
  A. Butane B. Pentane C. Hexane D. Heptane
• Q3: What type of isomerism do alkanes exhibit?
  A. Position B. Chain C. Functional D. Geometrical

Answer Key

• Q1: B - Alkanes have the formula C_nH_2n+2.
• Q2: C - Hexane has 5 isomers.
• Q3: B - Alkanes show chain isomerism.

Memory Aids

• Alkanes = Saturated hydrocarbons, formula C_nH_2n+2.
• Isomer count increases with carbon number; no isomers for C1-C3.

Preparation

Hydrogenation of Unsaturated Hydrocarbons

• Alkenes/alkynes + H₂ (Pt, Pd, Ni catalyst) → Alkanes.

Reduction of Alkyl Halides

• Alkyl halides (except fluorides) + Zn + dilute HCl → Alkanes.

Wurtz Reaction

• Alkyl halides + 2Na (dry ether) → Higher alkanes (even carbon number) + 2NaX.

From Carboxylic Acids

• Decarboxylation: Sodium salts of carboxylic acids + soda lime (NaOH + CaO) → Alkane (one C less) + CO₂.

• Kolbe’s Electrolysis: Electrolysis of sodium/potassium salts of carboxylic acids → Alkane with even number of carbons.

Solved Examples

• Ex 1: Write reaction for hydrogenation of ethene.
  CH₂=CH₂ + H₂ (Pt) → CH₃–CH₃
• Ex 2: Write Wurtz reaction for chloromethane.
  2CH₃Cl + 2Na → C₂H₆ + 2NaCl

Practice Set

• Q1: Which catalyst is used in hydrogenation?
  A. Fe B. Ni C. Cu D. Zn
• Q2: Wurtz reaction is used to prepare alkanes with:
  A. Odd number of carbons B. Even number of carbons C. Any number of carbons D. Only methane
• Q3: Decarboxylation removes which group?
  A. Hydroxyl B. Carboxyl C. Carbonyl D. Amino

Answer Key

• Q1: B - Nickel is a common hydrogenation catalyst.
• Q2: B - Wurtz reaction forms alkanes with even carbon numbers.
• Q3: B - Decarboxylation removes the carboxyl group.

Memory Aids

• Hydrogenation uses Pt, Pd, Ni catalysts.
• Wurtz reaction joins alkyl halides using sodium in dry ether.
• Decarboxylation removes CO₂ from carboxylic acids.

Physical Properties

States and Solubility

• C₁ to C₄: gases; C₅ to C₁₇: liquids; C₁₈ and above: solids at 298 K.
• Colourless, odourless, insoluble in water, soluble in non-polar solvents.

Boiling and Melting Points

• Boiling points increase with molecular mass.
• Branching decreases boiling points.
• Alkanes with even carbon numbers have higher melting points than adjacent odd-numbered alkanes due to symmetry.

Solved Examples

• Ex 1: State physical state of hexane at room temperature.
  Answer: Liquid (C₅ to C₁₇ are liquids).

Practice Set

• Q1: Which alkane is a gas at room temperature?
  A. Hexane B. Methane C. Octadecane D. Nonane
• Q2: Alkanes are soluble in:
  A. Water B. Polar solvents C. Non-polar solvents D. Acids
• Q3: Branching in alkanes causes:
  A. Increase in boiling point B. Decrease in boiling point C. No change D. Increase in melting point

Answer Key

• Q1: B - Methane is a gas at room temperature.
• Q2: C - Alkanes dissolve in non-polar solvents.
• Q3: B - Branching lowers boiling points.

Memory Aids

• First 4 alkanes are gases; next are liquids; long chains are solids.
• Branching lowers boiling points due to less surface contact.

Chemical Properties

Reactivity

• Generally inert to acids, bases, oxidizing and reducing agents.
• Undergo halogenation under UV light and heat.
• Halogenation rate: F₂ > Cl₂ > Br₂ > I₂.
• Hydrogen replacement rate: 3° > 2° > 1° hydrogens.
• Fluorination is violent; iodination is slow and reversible with oxidizing agents.

Halogenation Mechanism

• Chain initiation: UV light splits Cl₂ into 2 Cl radicals.
• Chain propagation: Cl radical abstracts H from alkane forming alkyl radical + HCl; alkyl radical reacts with Cl₂ forming alkyl halide + Cl radical.
• Chain termination: Radicals combine to form stable molecules.

Other Reactions

• Combustion: Complete → CO₂ + H₂O; Incomplete → C (soot) + H₂O.
• Controlled oxidation forms methanol, formaldehyde, ethanoic acid.
• Isomerisation converts n-butane to iso-butane.
• Aromatisation converts hexane to benzene.
• Reaction with steam produces CO and H₂.
• Pyrolysis/cracking breaks large hydrocarbons into smaller ones.

Solved Examples

• Ex 1: Write the product of methane chlorination.
  CH₄ + Cl₂ → CH₃Cl + HCl (under UV light)

Practice Set

• Q1: Which halogen reacts fastest with alkanes?
  A. I₂ B. Br₂ C. Cl₂ D. F₂
• Q2: Rate of hydrogen replacement in alkanes is:
  A. 1° > 2° > 3° B. 3° > 2° > 1° C. 2° > 3° > 1° D. Equal for all
• Q3: What is the main step in chain propagation during halogenation?
  A. Radical formation B. Radical combination C. Hydrogen abstraction D. Radical termination

Answer Key

• Q1: D - Fluorine reacts fastest but is violent.
• Q2: B - Tertiary hydrogens are replaced fastest.
• Q3: C - Hydrogen abstraction by halogen radical is key propagation step.

Memory Aids

• Halogenation proceeds via free radical mechanism: initiation, propagation, termination.
• Fluorination is uncontrollable; iodination is slow and reversible.
• Substitution rate: 3° > 2° > 1° hydrogens.

Conformations

Definition

• Conformational isomers arise from rotation about C–C single bonds.
• Also called rotational isomers.

Conformations of Ethane

• Infinite conformations possible by rotation of one methyl group around C–C bond.
• Eclipsed: Hydrogens align, causing maximum repulsion; least stable.
• Staggered: Hydrogens are maximally apart; most stable, 12.5 kJ/mol more stable than eclipsed.
• Skew (Gauche): Intermediate conformations between eclipsed and staggered.

Solved Examples

• Ex 1: Identify the most stable conformation of ethane.
  Answer: Staggered conformation.

Practice Set

• Q1: Which conformation has highest energy?
  A. Staggered B. Eclipsed C. Skew D. Chair
• Q2: Rotation about which bond causes conformational isomerism?
  A. C=O B. C–C single bond C. C≡C D. C–H
• Q3: Energy difference between eclipsed and staggered ethane is approximately:
  A. 5 kJ/mol B. 12.5 kJ/mol C. 20 kJ/mol D. 50 kJ/mol

Answer Key

• Q1: B - Eclipsed conformation has highest energy due to repulsion.
• Q2: B - Rotation about C–C single bond causes conformational isomerism.
• Q3: B - Energy difference is about 12.5 kJ/mol.

Memory Aids

• Staggered = stable; eclipsed = unstable; skew = intermediate.
• Rotation about C–C single bond changes conformation.

Alkenes Structure

Explanation

• Alkenes have molecular formula C_nH_2n with a carbon-carbon double bond (C=C).
• In ethylene, sp² orbitals of carbons overlap axially forming a sigma (σ) bond.
• Remaining sp² orbitals overlap with hydrogen 1s orbitals.
• Pure p orbitals on carbons overlap sideways forming a pi (π) bond.
• Double bond consists of one σ and one π bond.

Solved Examples

• Ex 1: Describe bonding in ethene.
  • Each carbon is sp² hybridized.
  • σ bond formed by axial overlap of sp² orbitals.
  • π bond formed by lateral overlap of p orbitals.

Practice Set

• Q1: What type of hybridization is present in alkene carbons?
  • A. sp
  • B. sp²
  • C. sp³
  • D. dsp²
• Q2: The pi bond in alkenes is formed by overlap of?
  • A. sp² orbitals
  • B. p orbitals
  • C. s orbitals
  • D. sp orbitals
• Q3: Which bond in alkene allows free rotation?
  • A. Sigma bond
  • B. Pi bond
  • C. Both sigma and pi bonds
  • D. None

Answer Key

• Q1: B - Alkene carbons are sp² hybridized.
• Q2: B - Pi bond is formed by lateral overlap of p orbitals.
• Q3: A - Sigma bond allows free rotation; pi bond restricts rotation.

Memory Aids

• Double bond = 1 σ + 1 π bond.
• sp² hybridization in alkene carbons.
• Pi bond formed by sideways p orbital overlap.

Alkenes Isomerism

Explanation

• Chain isomerism: Different carbon chain arrangements (e.g., 1-pentene and 2-methyl-1-butene).
• Position isomerism: Double bond position differs (e.g., 1-butene vs 2-butene).
• Ring chain isomerism: One is open chain, other is cyclic (e.g., propene and cyclopropane).
• Geometrical isomerism: Due to restricted rotation around C=C, cis (same side) and trans (opposite side) forms exist.
• Trans isomers are generally more stable than cis due to less steric crowding.

Solved Examples

• Ex 1: Identify isomers of butene.
  • 1-butene: double bond at carbon 1.
  • 2-butene: double bond at carbon 2 (cis and trans forms).

Practice Set

• Q1: Which is an example of chain isomerism?
  • A. 1-butene and 2-butene
  • B. 1-pentene and 2-methyl-1-butene
  • C. Cis-butene and trans-butene
  • D. Propene and cyclopropane
• Q2: Geometrical isomerism arises due to?
  • A. Free rotation around C=C
  • B. Restricted rotation around C=C
  • C. Different molecular formula
  • D. Different carbon chain
• Q3: Which is more stable?
  • A. Cis-isomer
  • B. Trans-isomer
  • C. Both equally stable
  • D. None

Answer Key

• Q1: B - Chain isomerism involves different carbon chains.
• Q2: B - Geometrical isomerism due to restricted rotation around double bond.
• Q3: B - Trans isomers are more stable due to less steric hindrance.

Memory Aids

• Cis = same side; Trans = opposite side.
• Chain isomers differ in carbon skeleton.
• Position isomers differ in double bond location.

Alkenes Preparation

Explanation

• Partial hydrogenation of alkynes: Using Lindlar’s catalyst (Pd on CaCO₃ poisoned with Pb), alkynes convert to cis-alkenes.
• Birch reduction: Sodium in liquid ammonia converts alkynes to trans-alkenes.
• Dehydrohalogenation of alkyl halides: β-elimination using alcoholic KOH removes HX forming alkenes.
• From vicinal dihalides: Zn reduces vicinal dihalides to alkenes.
• Dehydration of alcohols: Heating alcohols with conc. H₂SO₄ removes water forming alkenes.
• Saytzeff rule: More substituted alkene is major product.

Solved Examples

• Ex 1: Partial hydrogenation of ethyne.
  • C₂H₂ + H₂ (Lindlar’s catalyst) → C₂H₄ (cis-ethene)
• Ex 2: Dehydrohalogenation of 1-bromoethane.
  • CH₃CH₂Br + KOH (alc) → CH₂=CH₂ + KBr + H₂O

Practice Set

• Q1: Lindlar’s catalyst produces which alkene?
  • A. Trans-alkene
  • B. Cis-alkene
  • C. Alkane
  • D. Alkyne
• Q2: Dehydration of alcohols requires?
  • A. Dilute H₂SO₄
  • B. Conc. H₂SO₄ and heat
  • C. KOH and heat
  • D. Na and NH₃
• Q3: In dehydrohalogenation, which carbon loses hydrogen?
  • A. α-carbon
  • B. β-carbon
  • C. γ-carbon
  • D. None

Answer Key

• Q1: B - Lindlar’s catalyst gives cis-alkenes.
• Q2: B - Dehydration requires conc. H₂SO₄ and heat.
• Q3: B - Hydrogen is removed from β-carbon in dehydrohalogenation.

Memory Aids

• Lindlar’s catalyst → cis-alkene.
• Birch reduction → trans-alkene.
• Saytzeff rule → more substituted alkene major product.

Alkenes Physical Properties

Explanation

• First 3 alkenes are gases; next 14 are liquids; higher are solids.
• All alkenes are colourless and odourless except ethene (faint sweet smell).
• Insoluble in water; soluble in non-polar solvents.
• Boiling points increase with molecular size; straight chains have higher boiling points than branched isomers.

Solved Examples

• Ex 1: Boiling point trend.
  • Butene (C₄H₈) boiling point higher than 2-methylpropene due to straight chain.

Practice Set

• Q1: Which alkene is a gas at room temperature?
  • A. Ethene
  • B. Hexene
  • C. Decene
  • D. Dodecene
• Q2: Alkenes are generally soluble in?
  • A. Water
  • B. Polar solvents
  • C. Non-polar solvents
  • D. None
• Q3: Boiling point increases with?
  • A. Branching
  • B. Molecular size
  • C. Decrease in chain length
  • D. None

Answer Key

• Q1: A - First three alkenes including ethene are gases.
• Q2: C - Alkenes dissolve in non-polar solvents.
• Q3: B - Boiling point increases with molecular size.

Memory Aids

• Alkenes: gases (C1-C3), liquids (C4-C17), solids (higher).
• Insoluble in water, soluble in non-polar solvents.
• Straight chain > branched in boiling point.

Markovnikov’s and Anti-Markovnikov’s Rule

Explanation

• Markovnikov’s rule: In addition to unsymmetrical alkenes, the negative part of the reagent attaches to carbon with fewer hydrogens.
• Anti-Markovnikov’s rule (Peroxide effect): In presence of peroxide, HBr adds opposite to Markovnikov’s rule.
• Peroxide effect observed only with HBr, not HCl or HI.

Solved Examples

• Ex 1: Addition of HBr to propene.
  • Without peroxide: Br attaches to carbon with fewer H (Markovnikov product).
  • With peroxide: Br attaches to carbon with more H (Anti-Markovnikov product).

Practice Set

• Q1: Markovnikov’s rule applies to addition of?
  • A. HCl
  • B. HBr without peroxide
  • C. HBr with peroxide
  • D. HI
• Q2: Anti-Markovnikov addition occurs in presence of?
  • A. Heat
  • B. Peroxide
  • C. Acid
  • D. Base
• Q3: Peroxide effect is observed with?
  • A. HCl
  • B. HBr
  • C. HI
  • D. All HX

Answer Key

• Q1: B - Markovnikov’s rule applies to HBr without peroxide.
• Q2: B - Peroxide causes Anti-Markovnikov addition.
• Q3: B - Peroxide effect only with HBr.

Memory Aids

• Markovnikov: Negative part attaches to carbon with fewer H.
• Anti-Markovnikov: Peroxide reverses addition in HBr.
• Peroxide effect only with HBr.

Dienes and Polymerization

Explanation

• Dienes: Alkenes with two double bonds.
• Types of dienes:
  • Isolated dienes: double bonds separated by ≥1 single bond.
  • Conjugated dienes: double bonds separated by exactly one single bond.
  • Cumulated dienes: double bonds adjacent on same carbon.
• Polymerization: Alkenes like propene polymerize under heat, pressure, and catalyst to form polymers (e.g., polypropylene).
• Polymer structure: Repeating units with opened double bonds linked in chains.

Solved Examples

• Ex 1: Polymerization of propene.
  • Many propene molecules (CH₃-CH=CH₂) link to form polypropylene.
  • Requires heat, pressure, and catalyst.

Practice Set

• Q1: Dienes have how many double bonds?
  • A. One
  • B. Two
  • C. Three
  • D. None
• Q2: Conjugated dienes have double bonds separated by?
  • A. No bond
  • B. One single bond
  • C. Two single bonds
  • D. Adjacent carbons
• Q3: Polymerization of propene produces?
  • A. Polyethylene
  • B. Polypropylene
  • C. Polystyrene
  • D. Polyvinyl chloride

Answer Key

• Q1: B - Dienes have two double bonds.
• Q2: B - Conjugated dienes have double bonds separated by one single bond.
• Q3: B - Propene polymerizes to polypropylene.

Memory Aids

• Dienes = 2 double bonds.
• Conjugated = alternate double bonds.
• Polymerization opens double bonds to form chains.

Alkynes

Definition and Characteristics

• Alkynes are unsaturated hydrocarbons with the general formula C_nH_2n-2.
• Contain a carbon-carbon triple bond (≡), e.g., acetylene (C₂H₂, HC≡CH) and propyne (HC≡CCH₃).
• More acidic than alkenes and alkanes due to sp-hybridised carbon atoms which are more electronegative.

Isomerism

• Ethyne does not show isomerism.
• Other alkynes exhibit chain, position, and functional isomerism.
• Examples include:
  • Position isomers: 1-Pentyne and 2-Pentyne (same chain, different triple bond position).
  • Chain isomers: 1-Pentyne and 3-Methyl-1-butyne (different carbon chain arrangement).
  • Functional isomers: 1-Pentyne and 1,3-Pentadiene (alkyne vs diene).
  • Ring-chain isomers: Propyne (linear) and cyclopropene (ring) with same formula but different structures.

Structure

• In acetylene, sp-hybridised orbitals of carbons overlap axially to form a C–C σ-bond.
• Each carbon's other sp orbital overlaps with hydrogen 1s orbitals forming two C–H σ-bonds.
• Remaining p-orbitals overlap sideways forming two C–C π-bonds.

Preparation of Acetylene

• From calcium carbide:
  • Heat CaCO₃ → CaO + CO₂
  • CaO + C → CaC₂
  • CaC₂ + H₂O → Ca(OH)₂ + C₂H₂
• From vicinal dihalides:
  • Vicinal dihalide + alcoholic KOH → alkene
  • Alkene + NaNH₂ → acetylene (C₂H₂)

Solved Examples

• Ex 1: Write the reaction for acetylene preparation from calcium carbide.
  • CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂

Practice Set

• Q1: What is the general formula of alkynes?
  • A. C_nH_2n+2
  • B. C_nH_2n
  • C. C_nH_2n-2
  • D. C_nH_n
• Q2: Which type of isomerism is not shown by ethyne?
  • A. Chain isomerism
  • B. Position isomerism
  • C. Functional isomerism
  • D. All of the above
• Q3: Which compound is a ring-chain isomer of propyne?
  • A. 1-Pentyne
  • B. Cyclopropene
  • C. 1,3-Pentadiene
  • D. 3-Methyl-1-butyne

Answer Key

• Q1: C - Alkynes have the formula C_nH_2n-2 due to one triple bond.
• Q2: D - Ethyne does not show chain, position, or functional isomerism.
• Q3: B - Cyclopropene is a ring-chain isomer of propyne (same formula, different structure).

Memory Aids

• Alkynes = C_nH_2n-2, triple bond present.
• Ethyne = no isomers; others show chain, position, functional isomers.
• Ring-chain isomers: linear alkynes vs cyclic alkenes.

Physical Properties

General Properties

• First three alkynes are gases; next eight are liquids; higher members are solids.
• All alkynes are colourless and odourless except ethyne which has a characteristic odour.
• Insoluble in water; soluble in non-polar solvents.
• Melting point, boiling point, and density increase with molar mass.

Acidic Nature

• Acetylene reacts with sodium in liquid ammonia to form sodium acetylide and hydrogen gas:
  • CH≡CH + 2Na → 2NaC≡CH + H₂
• Acidic character order: CH≡CH > CH₂=CH₂ > CH₃–CH₃.
• —C≡CH group is acidic due to 50% s-character of sp-hybridised carbon, facilitating proton release.

Solved Examples

• Ex 1: Write the reaction of acetylene with sodium in liquid ammonia.
  • CH≡CH + 2Na → 2NaC≡CH + H₂

Practice Set

• Q1: Which alkyne member is a gas at room temperature?
  • A. C₅H₈
  • B. C₂H₂
  • C. C₁₀H₁₈
  • D. C₁₂H₂₂
• Q2: Why are alkynes more acidic than alkenes?
  • A. Due to sp³ hybridisation
  • B. Due to sp² hybridisation
  • C. Due to sp hybridisation with higher s-character
  • D. Due to presence of double bond
• Q3: What is the product when acetylene reacts with sodium in liquid ammonia?
  • A. Sodium ethoxide
  • B. Sodium acetylide
  • C. Sodium hydroxide
  • D. Sodium chloride

Answer Key

• Q1: B - First three alkynes including acetylene (C₂H₂) are gases.
• Q2: C - sp hybridisation has 50% s-character, increasing acidity.
• Q3: B - Acetylene reacts with sodium to form sodium acetylide.

Memory Aids

• Acidity order: alkyne > alkene > alkane.
• sp hybridised carbons increase acidity due to higher s-character.
• Acetylene + Na → sodium acetylide + H₂.

Chemical Reactions

Addition Reactions

• Hydrogenation:
  • C₂H₂ + H₂ (Ni, heat) → C₂H₄ → C₂H₆
• Halogenation:
  • C₂H₂ + Cl₂ (CCl₄) → addition of Cl atoms
  • C₂H₂ + HBr → stepwise addition of Br atoms
• Hydration:
  • C₂H₂ + H₂O + H₂SO₄ (HgSO₄) → acetaldehyde (via tautomerism)

Polymerisation

• Linear polymerisation: multiple acetylene molecules form polyacetylene (polyethyne).
• Cyclic polymerisation: 3 acetylene molecules react in hot iron tube to form benzene.

Solved Examples

• Ex 1: Write the product of acetylene hydrogenation.
  • C₂H₂ + 2H₂ (Ni) → C₂H₆

Practice Set

• Q1: What is the product of acetylene hydrogenation?
  • A. Ethylene
  • B. Ethane
  • C. Ethanol
  • D. Acetaldehyde
• Q2: Which catalyst is used for acetylene hydration?
  • A. Nickel
  • B. Mercury sulfate
  • C. Platinum
  • D. Iron
• Q3: What is formed by cyclic polymerisation of acetylene?
  • A. Polyacetylene
  • B. Benzene
  • C. Ethane
  • D. Cyclohexane

Answer Key

• Q1: B - Acetylene hydrogenates to ethane with Ni catalyst.
• Q2: B - HgSO₄ catalyses hydration to acetaldehyde.
• Q3: B - Cyclic polymerisation forms benzene.

Memory Aids

• Hydrogenation: acetylene → ethylene → ethane (Ni catalyst).
• Hydration uses HgSO₄ catalyst to form acetaldehyde.
• 3 acetylene molecules → benzene (cyclic polymerisation).

Aromatic Hydrocarbons

Overview

• Aromatic hydrocarbons are called arenes; benzene is the simplest arene.
• General formula: C_nH_2n-6m, where n ≥ 6, m = number of benzene rings.

Nomenclature

• Aromatic compounds consist of nucleus (benzene ring) and side chain (alkyl or aliphatic groups attached).
• Benzene ring carbons are sp² hybridised with alternating double bonds or delocalised π-electrons.
• Monosubstituted derivatives: one hydrogen replaced by alkyl/group.
• Disubstituted derivatives: substituents at ortho (1,2), meta (1,3), para (1,4) positions.
• Phenyl derivatives: side chains with functional groups attached to benzene.

Isomerism

• All six hydrogens in benzene are equivalent; only one monosubstituted derivative possible.
• Position isomerism: two substituents form ortho, meta, para isomers.
• Three identical substituents yield three position isomers.
• Two same and one different substituent: six isomers; all different: ten isomers.
• Functional isomerism: e.g., benzyl alcohol and o-cresol (same formula, different structure).
• Tautomerism: keto-enol isomerism in acetophenone due to α-hydrogen mobility.

Solved Examples

• Ex 1: Identify the type of isomerism in benzyl alcohol and o-cresol.
  Solution: Functional isomerism due to different functional groups (-CH₂OH vs -OH).
• Ex 2: How many position isomers are possible for disubstituted benzene?
  Solution: Three (ortho, meta, para).

Practice Set

• Q1: What is the general formula of arenes?
  A. C_nH_2n B. C_nH_2n-6m C. C_nH_2n+6 D. C_nH_n
• Q2: Which positions are called meta in disubstituted benzene?
  A. 1,2 B. 1,3 C. 1,4 D. 2,4
• Q3: How many isomers are possible when two substituents are same and one is different?
  A. 3 B. 6 C. 9 D. 10

Answer Key

• Q1: B - Arenes follow C_nH_2n-6m formula.
• Q2: B - Meta positions are 1,3.
• Q3: B - Six isomers possible with two same and one different substituent.

Memory Aids

• Ortho = adjacent carbons (1,2), Meta = separated by one carbon (1,3), Para = opposite carbons (1,4).
• Functional isomers have same formula but different functional groups.
• Tautomerism involves keto-enol interconversion.

Structure of Benzene

Bonding and Hybridisation

• Each carbon is sp² hybridised.
• Two sp² orbitals form σ bonds with adjacent carbons; third forms σ bond with hydrogen.
• Unhybridised p-orbitals overlap to form delocalised π-electrons above and below the ring plane.
• All bonds lie in the same plane.
• High unsaturation and unusual stability due to resonance.

Solved Examples

• Ex 1: Explain why benzene is more stable than cyclohexatriene.
  Solution: Resonance delocalisation of π-electrons in benzene stabilises it.

Practice Set

• Q1: What is the hybridisation of carbon atoms in benzene?
  A. sp B. sp² C. sp³ D. sp²d
• Q2: What type of bond forms between carbon and hydrogen in benzene?
  A. π bond B. σ bond C. Ionic bond D. Hydrogen bond
• Q3: Why is benzene unusually stable?
  A. Due to single bonds B. Due to resonance C. Due to ionic bonds D. Due to hydrogen bonding

Answer Key

• Q1: B - Carbon atoms in benzene are sp² hybridised.
• Q2: B - σ bonds form between carbon and hydrogen.
• Q3: B - Resonance delocalisation provides stability.

Memory Aids

• Benzene ring is planar with delocalised π-electrons.
• Resonance structures show bond equalisation.

Huckel Rule of Aromaticity

Criteria for Aromaticity

• Ring must be planar.
• Complete delocalisation of π-electrons in the ring.
• Number of π-electrons must be (4n + 2), where n = 0,1,2,...

Examples

• Benzene (6 π-electrons, n=1) is aromatic.
• Cyclopentadienyl anion and cycloheptatrienyl cation (6 π-electrons) are aromatic.
• Naphthalene (10 π-electrons, n=2) and anthracene/phenanthrene (14 π-electrons, n=3) are aromatic.

Comparison of Aromatic, Anti-aromatic and Non-aromatic

• Aromatic: Cyclic, planar, delocalised π-electrons, follow (4n+2) rule, e.g., benzene.
• Anti-aromatic: Cyclic, planar, delocalised π-electrons, follow 4n rule, e.g., cyclopentadienyl cation.
• Non-aromatic: Non-planar or non-cyclic, no delocalisation, e.g., cyclooctatetraene.

Solved Examples

• Ex 1: Determine if cyclopentadienyl anion is aromatic.
  Solution: It has 6 π-electrons (4n+2, n=1), planar and cyclic, so aromatic.

Practice Set

• Q1: What is the number of π-electrons in benzene?
  A. 4 B. 6 C. 8 D. 10
• Q2: Which rule defines aromaticity?
  A. (4n) B. (4n+2) C. (2n+1) D. (3n+2)
• Q3: Which molecule is anti-aromatic?
  A. Benzene B. Cyclopentadienyl cation C. Naphthalene D. Anthracene

Answer Key

• Q1: B - Benzene has 6 π-electrons.
• Q2: B - Huckel's rule is (4n+2) π-electrons.
• Q3: B - Cyclopentadienyl cation follows 4n rule, anti-aromatic.

Memory Aids

• Aromatic: planar, cyclic, (4n+2) π-electrons.
• Anti-aromatic: planar, cyclic, 4n π-electrons.
• Non-aromatic: non-planar or non-cyclic.

Preparation of Benzene

Methods

• Cyclic polymerisation of ethyne: 3 CH≡CH molecules heated at 873 K in red-hot iron tube form benzene.
• Decarboxylation of aromatic acids: Benzene carboxylate salt + NaOH + CaO heat → benzene + CO₂ + Na₂CO₃.
• Reduction of phenol: Phenol + Zn heat → benzene + ZnO.

Solved Examples

• Ex 1: Write the reaction for benzene preparation by decarboxylation.
  Solution: C₆H₅COONa + NaOH + CaO → C₆H₆ + Na₂CO₃ + CO₂.

Practice Set

• Q1: Which reagent is used in cyclic polymerisation of ethyne?
  A. Zn B. CaO C. Red-hot iron tube D. NaOH
• Q2: What is the product of phenol reduction with zinc?
  A. Cyclohexane B. Benzene C. Phenol D. Benzene oxide
• Q3: Decarboxylation of benzoic acid salt produces?
  A. Toluene B. Benzene C. Phenol D. Benzaldehyde

Answer Key

• Q1: C - Red-hot iron tube is used for ethyne polymerisation.
• Q2: B - Phenol reduces to benzene with zinc.
• Q3: B - Decarboxylation yields benzene.

Memory Aids

• Ethynes polymerise at high temperature to benzene.
• Decarboxylation removes CO₂ from aromatic acid salts.
• Phenol reduction with Zn yields benzene.

Physical Properties

Key Points

• Aromatic hydrocarbons are non-polar.
• Colourless liquids or solids with characteristic aroma.
• Immiscible with water, miscible with organic solvents.
• Burn with sooty flame.
• Boiling point increases with molecular size due to van der Waals forces.
• Melting point depends on size and symmetry; p-isomer of xylenes has highest melting point.

Solved Examples

• Ex 1: Why does boiling point increase with molecular size?
  Solution: Increased van der Waals forces with larger molecules.

Practice Set

• Q1: Aromatic hydrocarbons are generally:
  A. Polar B. Non-polar C. Ionic D. Metallic
• Q2: Which is miscible with water?
  A. Benzene B. Ethanol C. Toluene D. Xylene
• Q3: Which isomer of xylene has highest melting point?
  A. Ortho B. Meta C. Para D. None

Answer Key

• Q1: B - Aromatic hydrocarbons are non-polar.
• Q2: B - Ethanol is miscible with water, arenes are not.
• Q3: C - Para-xylene has highest melting point due to symmetry.

Memory Aids

• Non-polar, aromatic smell, sooty flame.
• Boiling point ↑ with size; melting point depends on symmetry.

Chemical Properties

Reactivity

• Aromatic hydrocarbons are less reactive than alkenes/alkynes due to π-electron delocalisation.
• Undergo electrophilic substitution reactions (EAS) replacing H on benzene ring.
• Can undergo addition and oxidation under special conditions.

Solved Examples

• Ex 1: Name the product of nitration of benzene.
  Solution: Nitrobenzene (benzene with -NO₂ group).
• Ex 2: What catalyst is used in Friedel-Crafts alkylation?
  Solution: AlCl₃.

Practice Set

• Q1: Which reaction is typical for benzene?
  A. Addition B. Electrophilic substitution C. Free radical substitution D. Polymerisation
• Q2: Which reagent is used for sulphonation of benzene?
  A. HNO₃ B. H₂SO₄ C. SO₃ in H₂SO₄ D. AlCl₃
• Q3: Friedel-Crafts alkylation requires?
  A. RCl + AlCl₃ B. RCOCl + AlCl₃ C. HNO₃ + H₂SO₄ D. Cl₂ + FeCl₃

Answer Key

• Q1: B - Benzene undergoes electrophilic substitution.
• Q2: C - Sulphonation uses SO₃ in H₂SO₄.
• Q3: A - Friedel-Crafts alkylation uses alkyl halide and AlCl₃.

Memory Aids

• Benzene prefers substitution over addition due to π-electron stability.
• Common EAS: nitration, sulphonation, Friedel-Crafts alkylation/acylation.

Directive Influence of a Functional Group in Mono-Substituted Benzene

Ortho and Para Directing Groups

• Groups with +M effect increase electron density at ortho and para positions.
• Electrophiles attack electron-rich ortho and para positions.
• Examples: —CH₃, —C₂H₅, —OCH₃, —NH₂, —NHR, —NHCOCH₃, —OH, —F, —Cl, —Br, —I.

Meta Directing Groups

• Groups with -M effect decrease electron density at ortho and para positions.
• Electrophiles attack meta position.
• Examples: —NO₂, —CHO, —COR, —COOH, —COOR, —SO₃H, —CN.
• Halogens are ortho/para directing but deactivating.

Solved Examples

• Ex 1: Predict the major product when chlorobenzene undergoes nitration.
  Solution: Nitro group attaches at meta position due to -M effect of Cl.

Practice Set

• Q1: Which group directs electrophiles to ortho and para positions?
  A. —NO₂ B. —OH C. —COOH D. —CN
• Q2: Halogens are:
  A. Ortho/para directing and activating B. Meta directing and deactivating C. Ortho/para directing and deactivating D. Meta directing and activating
• Q3: Which group is meta directing?
  A. —NH₂ B. —CH₃ C. —NO₂ D. —OH

Answer Key

• Q1: B - —OH is ortho/para directing (+M effect).
• Q2: C - Halogens are ortho/para directing but deactivating.
• Q3: C - —NO₂ is meta directing (-M effect).

Memory Aids

• +M groups: activate ring, direct ortho/para.
• -M groups: deactivate ring, direct meta.
• Halogens: special case, ortho/para directing but deactivating.

Carcinogenicity and Toxicity

Overview

• Benzene and polynuclear hydrocarbons with more than two fused benzene rings are toxic and carcinogenic.
• These compounds form during incomplete combustion of organic materials like tobacco, coal, and petroleum.
• They enter the human body, undergo biochemical reactions, damage DNA, and cause cancer.

Solved Examples

• Ex 1: Identify a carcinogenic compound formed from incomplete combustion of coal.
  Answer: Benzo[a]pyrene (a polynuclear hydrocarbon with five fused benzene rings).

Practice Set

• Q1: Which of the following is a carcinogenic polynuclear hydrocarbon?
  A. Benzene
  B. Naphthalene
  C. Benzo[a]pyrene
  D. Toluene
• Q2: Polynuclear hydrocarbons are primarily formed by:
  A. Complete combustion of organic materials
  B. Incomplete combustion of organic materials
  C. Photosynthesis
  D. Electrolysis of water
• Q3: What is the primary biological effect of carcinogenic hydrocarbons?
  A. DNA repair
  B. DNA damage leading to cancer
  C. Protein synthesis
  D. Cell respiration

Answer Key

• Q1: C. Benzo[a]pyrene - It is a known carcinogenic polynuclear hydrocarbon formed during incomplete combustion.
• Q2: B. Incomplete combustion of organic materials - This process forms toxic polynuclear hydrocarbons.
• Q3: B. DNA damage leading to cancer - Carcinogens damage DNA causing mutations and cancer.

Memory Aids

• Polynuclear hydrocarbons = multiple fused benzene rings > 2 rings.
• Incomplete combustion = source of carcinogens like benzo[a]pyrene.
• Carcinogens damage DNA → cancer.