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 present in hydrocarbons.
  Answer: Carbon and hydrogen only.

Practice Set

• Q1: What elements constitute hydrocarbons?
  A. C and O
  B. C and H
  C. H and O
  D. C, H, and O

• Q2: Which of the following is a major source of hydrocarbons?
  A. Water
  B. Natural gas
  C. Air
  D. Salt

• Q3: Hydrocarbons are primarily found in:
  A. Minerals
  B. Petroleum
  C. Metals
  D. Salts

Answer Key

• Q1: B - Hydrocarbons contain only carbon and hydrogen atoms.

• Q2: B - Natural gas is a main source of hydrocarbons.

• Q3: B - Petroleum is a primary source of hydrocarbons.

Memory Aids

• Hydrocarbons = C + H 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 (C=C) or triple (C≡C) bonds.

• Aromatic: Special cyclic compounds with delocalized π electrons.

Subtypes

• Alkanes: Saturated hydrocarbons with open chains.

• Cycloalkanes: Saturated hydrocarbons with carbon atoms in rings.

Solved Examples

• Ex 1: Classify CH4.
  Answer: Saturated alkane (methane, single bonds, open chain).

• Ex 2: Classify C6H6.
  Answer: Aromatic hydrocarbon (benzene ring with delocalized electrons).

Practice Set

• Q1: Which hydrocarbon contains only single bonds?
  A. Alkene
  B. Alkyne
  C. Alkane
  D. Aromatic

• Q2: Cycloalkanes are:
  A. Unsaturated open chain hydrocarbons
  B. Saturated ring hydrocarbons
  C. Aromatic compounds
  D. Unsaturated ring hydrocarbons

• Q3: Which of the following is an aromatic hydrocarbon?
  A. Ethane
  B. Benzene
  C. Ethene
  D. Cyclohexane

Answer Key

• Q1: C - Alkanes have only single bonds.

• Q2: B - Cycloalkanes are saturated hydrocarbons with ring structures.

• Q3: B - Benzene is a classic aromatic hydrocarbon.

Memory Aids

• Saturated = single bonds only (alkanes, cycloalkanes).

• Unsaturated = double/triple bonds (alkenes, alkynes).

• Aromatic = special cyclic with delocalized π electrons (benzene).

Alkanes

Structure

• General formula: CnH2n+2 (n = 1, 2, 3...)

• Carbon atoms are sp3 hybridized, forming tetrahedral geometry

• All C–C and C–H bonds are sigma (σ) bonds

• C–C bond length: 1.54 Å; C–H bond length: 1.09 Å

• Bond angles approximately 109°28'

Isomerism

• Exhibit chain isomerism

• First three alkanes (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)

Preparation

By hydrogenation of unsaturated hydrocarbons

• Alkenes and alkynes react with H2 in presence of Pt, Pd, or Ni catalysts

• Double/triple bonds convert to single bonds forming alkanes

Reduction of alkyl halides

• Alkyl halides (except fluorides) reduced by Zn and dilute HCl to form alkanes

Wurtz reaction

• Alkyl halides react with sodium metal in dry ether to form higher alkanes

• Used to prepare alkanes with even number of carbons

From carboxylic acids

• Decarboxylation: Sodium salts of carboxylic acids heated with soda lime produce alkanes with one less carbon atom

Kolbe’s electrolysis

• Electrolysis of aqueous sodium or potassium salts of carboxylic acids produces alkanes with even number of carbons

Physical Properties

• 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 points increase with molecular mass; decrease with branching

• Alkanes with even number of carbons have higher melting points due to symmetry

Chemical Properties

• Generally inert to acids, bases, oxidizing and reducing agents

• Undergo halogenation under UV light or heat

• Rate of halogenation: F2 > Cl2 > Br2 > I2

• Rate of hydrogen replacement: 3° > 2° > 1°

• Fluorination is violent; iodination is slow and reversible, requires oxidizing agents

Mechanism of Halogenation

• Chain initiation: UV light splits Cl2 into 2 Cl radicals

• Chain propagation: Cl radical abstracts H from alkane forming alkyl radical and HCl; alkyl radical reacts with Cl2 to form alkyl halide and Cl radical

• Chain termination: radicals combine to form stable molecules

Other Reactions

• Combustion: Complete (CO2 + H2O), incomplete (C + H2O)

• Controlled oxidation: partial oxidation to methanol, formaldehyde, acetic acid

• Isomerisation: n-butane to iso-butane using AlCl3/HCl

• Aromatisation: n-hexane to benzene using catalysts at high T and P

• Reaction with steam: methane + steam → CO + H2 (Ni catalyst)

• Pyrolysis/cracking: breaking large hydrocarbons into smaller ones

Conformations

Conformational Isomerism

• Different spatial arrangements due to rotation about C–C single bond

• Called conformational or rotational isomers

Conformations of Ethane

• Infinite conformations possible by rotation about C–C bond

• Eclipsed: hydrogens aligned, least stable

• Staggered: hydrogens maximally apart, most stable (12.5 kJ/mol more stable than eclipsed)

• Skew (gauche): intermediate conformations between eclipsed and staggered

Alkenes Structure

Orbital Overlap

• Alkene carbon atoms are sp2 hybridised.

• One sp2 orbital from each carbon overlaps axially to form a sigma (σ) bond.

• Remaining sp2 orbitals form sigma bonds with hydrogen atoms.

• Unhybridised p orbitals on each carbon overlap sideways to form a pi (π) bond.

• Double bond = 1 σ + 1 π bond; π bond restricts rotation.

Solved Examples

Ex 1

• Explain bonding in ethene: Two sp2 orbitals overlap to form σ bond; p orbitals form π bond.

Practice Set

• Q1: What type of hybridisation 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 sideways

  • C. s orbitals

  • D. sp orbitals

• Q3: Which bond restricts rotation in alkenes?

  • A. Sigma bond

  • B. Pi bond

  • C. Hydrogen bond

  • D. Ionic bond

Answer Key

• Q1: B - Alkene carbons are sp2 hybridised.

• Q2: B - Pi bond is formed by sideways overlap of p orbitals.

• Q3: B - Pi bond restricts rotation around double bond.

Memory Aids

• Alkene carbons: sp2 hybridised = 3 sigma bonds + 1 pi bond.

• Pi bond = sideways p orbital overlap, no free rotation.

Alkenes Isomerism

Types of Isomerism

• Chain isomerism: Different carbon chain arrangements (e.g., 1-pentene and 2-methyl-1-butene).

• 

• Position isomerism: Double bond at different positions (e.g., 1-butene and 2-butene).

• 

• Ring chain isomerism: Open chain vs cyclic compounds (e.g., propene and cyclopropane).

• 

• Geometrical isomerism: Cis (same side) and trans (opposite side) around double bond.

• Trans isomers are generally more stable due to less steric hindrance.

• 

Solved Examples

Ex 1

• Identify isomers of butene: 1-butene (double bond at C1), 2-butene (double bond at C2).

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 isomers differ in:

  • A. Position of double bond

  • B. Arrangement around double bond

  • C. Carbon chain length

  • D. Functional group

• Q3: Which is more stable?

  • A. Cis-isomer

  • B. Trans-isomer

  • C. Both equal

  • D. Depends on temperature

Answer Key

• Q1: B - Chain isomerism involves different carbon chains.

• Q2: B - Geometrical isomers differ in spatial arrangement around double bond.

• Q3: B - Trans-isomers are more stable due to less steric hindrance.

Memory Aids

• Chain isomers: different carbon skeletons.

• Position isomers: double bond shifts.

• Geometrical isomers: cis (same side), trans (opposite side).

Alkenes Preparation

Methods

• Partial hydrogenation of alkynes: Using Lindlar's catalyst yields cis-alkenes; Birch reduction (Na/NH3) yields trans-alkenes.

• 

• Dehydrohalogenation of alkyl halides: β-elimination with alcoholic KOH and heat removes HX to form alkene.

• 

• From vicinal dihalides: Reduction with Zn removes halogens to form alkene.

• Dehydration of alcohols: Heating with conc. H2SO4 removes water to form alkene.

• 

• Saytzeff rule: More substituted alkene formed preferentially.

Solved Examples

Ex 1

• Partial hydrogenation: HC≡CH + H2 (Lindlar's catalyst) → H2C=CH2 (cis-alkene)

Practice Set

• Q1: Lindlar's catalyst produces:

  • A. Trans-alkene

  • B. Cis-alkene

  • C. Alkane

  • D. Alkyne

• Q2: Dehydrohalogenation involves removal of:

  • A. Hydrogen and halogen from same carbon

  • B. Hydrogen and halogen from adjacent carbons

  • C. Two hydrogens

  • D. Two halogens

• Q3: In dehydration of alcohols, the major product follows:

  • A. Less substituted alkene

  • B. More substituted alkene

  • C. Equal mixture

  • D. No alkene formed

Answer Key

• Q1: B - Lindlar's catalyst yields cis-alkenes.

• Q2: B - β-elimination removes H and halogen from adjacent carbons.

• Q3: B - Saytzeff rule favors more substituted alkene.

Memory Aids

• Lindlar's catalyst = partial hydrogenation → cis-alkene.

• Dehydrohalogenation = β-elimination with KOH.

• Saytzeff rule = more substituted alkene major product.

Alkenes Physical Properties

Key Points

• 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 chain alkenes have higher boiling points than branched isomers.

• 

Solved Examples

Ex 1

• Compare boiling points: n-butene > isobutene due to less branching.

Practice Set

• Q1: Ethene is:

  • A. Colourless and odourless

  • B. Colourless with faint sweet smell

  • C. Coloured gas

  • D. Solid at room temperature

• Q2: Alkenes are generally:

  • A. Soluble in water

  • B. Insoluble in water

  • C. Ionic compounds

  • D. Polar solvents

• Q3: Boiling point trend in alkenes is:

  • A. Branched > straight chain

  • B. Straight chain > branched

  • C. No difference

  • D. Random

Answer Key

• Q1: B - Ethene has faint sweet smell.

• Q2: B - Alkenes are insoluble in water.

• Q3: B - Straight chain alkenes have higher boiling points.

Memory Aids

• Ethene smell: faint sweet.

• Alkenes: non-polar, water insoluble.

• Boiling point: straight > branched.

Markovnikov and Anti-Markovnikov Rule

Rules

• Markovnikov's rule: In addition of HX to unsymmetrical alkene, H attaches to carbon with more H atoms; X attaches to carbon with fewer H atoms.

• 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 secondary carbon (Markovnikov product).

• Addition of HBr to propene with peroxide: Br attaches to primary carbon (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 with:

  • A. HCl

  • B. HBr with peroxide

  • C. HI

  • D. HBr without peroxide

• Q3: In Markovnikov addition, the halogen attaches to carbon with:

  • A. More hydrogen atoms

  • B. Fewer hydrogen atoms

  • C. Equal hydrogen atoms

  • D. No hydrogen atoms

Answer Key

• Q1: B - Markovnikov's rule applies to HBr without peroxide.

• Q2: B - Anti-Markovnikov addition occurs with HBr in presence of peroxide.

• Q3: B - Halogen attaches to carbon with fewer hydrogen atoms.

Memory Aids

• Markovnikov: "Rich get richer" (H to carbon with more H).

• Anti-Markovnikov: peroxide reverses addition in HBr.

• Peroxide effect only with HBr.

Dienes and Polymerization

Dienes Classification

• 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 polymerize by opening double bonds under heat, pressure, and catalyst.

• Example: Propene polymerizes to polypropylene.

• Polymer structure: repeating units with side groups (e.g., CH3 in polypropylene).

  

Solved Examples

Ex 1

• Polymerization of propene: n CH3-CH=CH2 → (−CH−CH2−)n with CH3 side group.

Practice Set

• Q1: Conjugated dienes have double bonds separated by:

  • A. No bond

  • B. One single bond

  • C. Two single bonds

  • D. Adjacent carbons

• Q2: Polymerization of alkenes requires:

  • A. Heat, pressure, catalyst

  • B. Cold temperature

  • C. No catalyst

  • D. Light only

• Q3: Polypropylene is formed from:

  • A. Ethene

  • B. Propene

  • C. Butene

  • D. Benzene

Answer Key

• Q1: B - Conjugated dienes have double bonds separated by one single bond.

• Q2: A - Polymerization requires heat, pressure, and catalyst.

• Q3: B - Polypropylene is polymer of propene.

Memory Aids

• Dienes: isolated (separated), conjugated (one bond), cumulated (adjacent).

• Polymerization: double bonds open and link to form long chains.

Alkynes

Definition and General Properties

• Alkynes are unsaturated hydrocarbons with the general formula CnH2n-2.

• They contain at least one carbon-carbon triple bond (≡).

• Examples: Ethyne (acetylene) C2H2, Propyne (methyl acetylene) C3H4.

• Alkynes have sp-hybridised carbons, making them more acidic than alkenes and alkanes.

Isomerism

• Ethyne shows no isomerism.

• Other alkynes exhibit chain, position, and functional isomerism.

• Ring-chain isomerism example: Propyne (alkyne) and cyclopropene (cycloalkene).

Structure

• Each carbon in acetylene is sp-hybridised.

• One sp orbital from each carbon overlaps axially to form a C–C σ bond.

• Remaining sp orbitals overlap with hydrogen 1s orbitals forming two C–H σ bonds.

• Two unhybridised p orbitals on each carbon overlap sideways forming two C–C π bonds.

Preparation of Acetylene

• From calcium carbide (CaC2): CaC2 + 2H2O → C2H2 + Ca(OH)2

• Calcium carbide is prepared by heating CaO with carbon.

• From vicinal dihalides: Vicinal dihalide + 2KOH (alc) → alkene; then treated with NaNH2 to form acetylene.

Physical Properties

• First three alkynes are gases; next eight are liquids; higher members are solids.

• All 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.

Acidity

• Alkynes are acidic due to sp-hybridised carbon with 50% s-character.

• Acetylene reacts with sodium in liquid ammonia: 2Na + C2H2 → 2NaC≡CH + H2

• Acidic strength order: Alkyne (CH≡CH) > Alkene (CH2=CH2) > Alkane (CH3–CH3).

Chemical Reactions

• Addition Reactions:

  • Hydrogenation: C2H2 + 2H2 (Ni, heat) → C2H6

  • Halogenation: C2H2 + Cl2 → C2H2Cl2

  • Hydrohalogenation: Stepwise addition of HBr to form bromoalkenes and dibromoalkanes.

  • Hydration: C2H2 + H2O (H2SO4, HgSO4) → Acetaldehyde (via tautomerism).

• Polymerisation:

  • Linear polymerisation forms polyacetylene.

  • Cyclic polymerisation of 3 acetylene molecules forms benzene.

Solved Examples

Ex 1

• Write the product when acetylene reacts with excess hydrogen in presence of Ni catalyst.

• Solution: C2H2 + 2H2 → C2H6 (ethane)

Ex 2

• Explain the formation of sodium acetylide from acetylene.

• Solution: C2H2 + 2Na → 2NaC≡CH + H2

Practice Set

Q1

• Which of the following is the molecular formula of alkynes?

• A. CnH2n+2

• B. CnH2n

• C. CnH2n-2

• D. CnHn

Q2

• What is the product when acetylene reacts with water in presence of H2SO4 and HgSO4?

• A. Ethanol

• B. Acetaldehyde

• C. Ethane

• D. Ethene

Q3

• Which of the following shows ring-chain isomerism?

• A. 1-Pentyne and 2-Pentyne

• B. Propyne and Cyclopropene

• C. 1-Pentyne and 3-Methyl-1-butyne

• D. 1,3-Pentadiene and 1-Pentyne

Answer Key

• Q1: C. Alkynes have formula CnH2n-2 due to one triple bond.

• Q2: B. Hydration of acetylene forms acetaldehyde via tautomerism.

• Q3: B. Propyne (alkyne) and cyclopropene (cycloalkene) are ring-chain isomers.

Memory Aids

• Alkyne formula: "2n minus 2" hydrogens.

• Acidity order: Alkyne > Alkene > Alkane due to s-character.

• Ring-chain isomerism: Linear alkynes vs cyclic alkenes with same formula.

• Preparation of acetylene: Calcium carbide + water or vicinal dihalides + strong base.

Aromatic Hydrocarbons

Overview

• Aromatic hydrocarbons, also called arenes, contain benzene rings.

• General formula: CnH2n-6m, where n ≥ 6, m = number of benzene rings.

Nomenclature

• Nucleus: Benzene ring with sp2 hybridised carbons and alternating double bonds.

• 

• Side Chain: Alkyl or aliphatic groups replacing hydrogen on benzene.

• Monosubstituted derivatives: One hydrogen replaced by a substituent.

• Disubstituted derivatives: Positions relative to first substituent: ortho (1,2), meta (1,3), para (1,4).

• 

Isomerism

• All six hydrogens in benzene are equivalent; only one monosubstituted derivative possible.

• Position isomerism: Two substituents give ortho, meta, para isomers.

• 

• Three identical substituents yield three isomers; mixed substituents increase isomer count.

• 

• Functional isomerism: Example: Benzyl alcohol vs o-Cresol (C7H8O).

• 

• Tautomerism: Keto-enol tautomerism in acetophenone.

• 

Structure of Benzene

• Each carbon is sp2 hybridised; forms σ bonds with adjacent carbons and hydrogen.

• Unhybridised p orbitals overlap to form delocalised π-electron cloud above and below the plane.

• High unsaturation and resonance stability.

• 

Huckel Rule of Aromaticity

• Criteria for aromaticity:

  • Planar cyclic structure.

  • Complete delocalisation of π-electrons.

  • Contains (4n + 2) π-electrons, n = 0,1,2...

• 

• Comparison of aromatic, anti-aromatic, and non-aromatic compounds based on structure and electron count.

• 

Preparation of Benzene

• Cyclic polymerization of ethyne at 873 K in red-hot iron tube.

• Decarboxylation of aromatic acids with NaOH and CaO.

• Reduction of phenol with zinc and heat.

• 

Physical Properties

• 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.

Chemical Properties

• Less reactive than alkenes/alkynes due to delocalised π-electrons.

• Undergo electrophilic substitution reactions (EAS).

• Can undergo addition and oxidation under special conditions.

• 

• Common EAS reactions: nitration, sulphonation, Friedel-Crafts acylation and alkylation, halogenation.

• Oxidation breaks side chains to CO2 and H2O.

• Addition under UV light; reduction to cyclohexane with H2/Ni.

• 

Directive Influence of Functional Groups

• Ortho and para directing groups: Groups with +M effect increase electron density at ortho and para positions, favoring electrophilic attack there.

• Examples: —CH3, —C2H5, —OCH3, —NH2, —OH, halogens (F, Cl, Br, I).

• 

• Meta directing groups: Groups with -M effect decrease electron density at ortho and para positions, directing electrophiles to meta position.

• Examples: —NO2, —CHO, —COR, —COOH, —COOR, —SO3H, —CN.

• 

• Halogens are ortho/para directing but deactivating due to electronegativity.

Solved Examples

• Ex 1: Identify the directing effect of —OH group in phenol.
  Answer: —OH is an ortho/para directing group due to +M effect increasing electron density at these positions.

• Ex 2: Predict the major product of nitration of toluene.
  Answer: Nitro group substitutes at ortho and para positions relative to —CH3 group (ortho/para directing).

• Ex 3: How many position isomers are possible for disubstituted benzene with two different substituents?
  Answer: Three isomers (ortho, meta, para).

Practice Set

• Q1: Which of the following is an ortho/para directing group?
  A. —NO2
  B. —CH3
  C. —COOH
  D. —CN

• Q2: How many π-electrons are present in benzene according to Huckel's rule?
  A. 4
  B. 6
  C. 8
  D. 10

• Q3: Which reaction converts phenol to benzene?
  A. Decarboxylation
  B. Reduction with Zn
  C. Friedel-Crafts alkylation
  D. Nitration

Answer Key

• Q1: B —CH3 is ortho/para directing due to +M effect.

• Q2: B Benzene has 6 π-electrons fitting (4n+2) with n=1.

• Q3: B Reduction of phenol with Zn produces benzene.

Memory Aids

• Huckel's rule: Aromatic if π-electrons = 4n + 2.

• Ortho/para directors: Groups with lone pairs or alkyl groups (+M effect).

• Meta directors: Electron withdrawing groups (-M effect).

• Halogens: Ortho/para directing but deactivating.

• Electrophilic substitution preferred over addition in benzene due to resonance stability.

Carcinogenicity and Toxicity

Overview

• Benzene and polynuclear aromatic hydrocarbons (PAHs) with more than two fused benzene rings are toxic and carcinogenic.

• PAHs form during incomplete combustion of organic materials such as tobacco, coal, and petroleum.

• These compounds enter the human body, undergo biochemical transformations, damage DNA, and induce cancer.

Solved Examples

• Ex 1: Identify a source of carcinogenic PAHs.
  Answer: Incomplete combustion of tobacco produces carcinogenic PAHs.

• Ex 2: Explain the effect of PAHs on human DNA.
  Answer: PAHs metabolize into reactive intermediates that bind DNA, causing mutations leading to cancer.

Practice Set

• Q1: Which of the following is a carcinogenic compound?
  A. Benzene
  B. Methane
  C. Ethanol
  D. Water

• Q2: Polynuclear aromatic hydrocarbons are primarily formed by:
  A. Complete combustion of organic matter
  B. Incomplete combustion of organic matter
  C. Photosynthesis
  D. Electrolysis of water

• Q3: The mechanism by which PAHs cause cancer involves:
  A. Direct DNA repair
  B. DNA damage through reactive metabolites
  C. Protein synthesis enhancement
  D. Increased oxygen transport

Answer Key

• Q1: A. Benzene - It is a toxic and carcinogenic aromatic hydrocarbon.

• Q2: B. Incomplete combustion of organic matter - This process forms PAHs.

• Q3: B. DNA damage through reactive metabolites - PAHs metabolize into compounds that bind and damage DNA.

Memory Aids

• PAHs = Multiple fused benzene rings → Toxic + Carcinogenic.

• Incomplete combustion = Source of PAHs (think tobacco smoke, coal burning).

• PAHs → Metabolized → DNA damage → Cancer.