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📋 Table of Contents

9.1 Classification of Hydrocarbons

Definition

Hydrocarbons are compounds containing only carbon and hydrogen. They are the most important sources of energy (LPG, CNG, petrol, diesel, kerosene) and are used in manufacture of polymers, dyes, drugs and solvents.

LPG = Liquefied Petroleum Gas  ·  CNG = Compressed Natural Gas  ·  LNG = Liquefied Natural Gas

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AI Image: Classification of Hydrocarbons – Mind Map
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A complete hierarchical tree / mind map showing the classification of hydrocarbons into saturated, unsaturated, and aromatic, with sub-types and examples.
🤖 AI Image Prompt
A clean professional chemistry mind map on a pure white background showing the classification of hydrocarbons. Root node: "Hydrocarbons" in a dark navy rectangle. Three main branches: (1) "Saturated — Alkanes (CnH2n+2)" in teal, sub-branches: "Open chain (alkanes)" e.g. CH4, C2H6 and "Cyclic (cycloalkanes)" e.g. Cyclopropane. (2) "Unsaturated" in purple, sub-branches: "Alkenes (CnH2n)" e.g. C2H4, "Alkynes (CnH2n-2)" e.g. C2H2, "Dienes" e.g. Buta-1,3-diene. (3) "Aromatic" in dark green, sub-branches: "Benzenoid" e.g. Benzene, Toluene, Naphthalene and "Non-benzenoid" e.g. Tropone. All connecting arrows in amber/gold. Clean sans-serif fonts. White background only. No decorative patterns.

Fig. 9.0 — Classification tree of hydrocarbons

Type General Formula Bonds Present Examples
Alkanes (Saturated) CnH2n+2 C–C and C–H single bonds only CH4, C2H6, C3H8
Cycloalkanes CnH2n C–C single bonds in a ring Cyclopropane, Cyclohexane
Alkenes (Unsaturated) CnH2n At least one C=C double bond C2H4, C3H6
Alkynes (Unsaturated) CnH2n–2 At least one C≡C triple bond C2H2, C3H4
Aromatic Variable Delocalised π electrons; (4n+2)π rule Benzene C6H6, Toluene, Naphthalene

9.2 Alkanes

Definition

Alkanes are saturated open-chain hydrocarbons with only C–C and C–H single bonds. General formula: CnH2n+2. Earlier called paraffins (Latin: parum = little, affinis = affinity) because they are largely inert.

First member: Methane (CH4) — found in coal mines and marshy places.

9.2.1 Nomenclature and Isomerism

Methane, ethane, and propane each have only one possible structure. From butane onwards, multiple structural isomers exist.

IUPAC Nomenclature Rules
Step Rule Example
1 Find the longest continuous carbon chain (parent chain) For CH₃–CH₂–CH(CH₃)–CH₃
Longest chain = 4 carbons → butane
2 Number the chain to give lowest numbers to substituents Number from either end, choose direction giving smaller locant
3 Name and number the substituents (alkyl groups) –CH₃ at position 2 → 2-methyl
4 Use prefixes for multiple identical substituents di-, tri-, tetra-, penta-, hexa-
5 Arrange substituents in alphabetical order ethyl comes before methyl (ignore prefixes di, tri, etc.)
6 Write complete name: substituents + parent alkane 2-methylbutane
Important Alkyl Groups (–R)
Formula Name Structure (SVG) Alternative Name
–CH₃ Methyl CH₃–
–CH₂CH₃ Ethyl CH₃CH₂–
–CH₂CH₂CH₃ Propyl CH₃CH₂CH₂– n-propyl
–CH(CH₃)₂ Isopropyl CH₃ CH CH₃ sec-propyl
–CH₂CH₂CH₂CH₃ Butyl CH₃CH₂CH₂CH₂– n-butyl
–CH(CH₃)CH₂CH₃ sec-Butyl CH₃ CH CH₂ CH₃ secondary butyl
–C(CH₃)₃ tert-Butyl CH₃ C CH₃ CH₃ tertiary butyl
Problem 9.1

Question: Write structures of different chain isomers of alkanes corresponding to the molecular formula C₆H₁₄. Also write their IUPAC names.

Solution (5 Isomers)
  1. CH₃–CH₂–CH₂–CH₂–CH₂–CH₃
    n-Hexane (Hexane)
  2. CH₃–CH(CH₃)–CH₂–CH₂–CH₃
    2-Methylpentane
  3. CH₃–CH₂–CH(CH₃)–CH₂–CH₃
    3-Methylpentane
  4. CH₃–CH(CH₃)–CH(CH₃)–CH₃
    2,3-Dimethylbutane
  5. CH₃–C(CH₃)₂–CH₂–CH₃
    2,2-Dimethylbutane
Problem 9.2

Question: Write structures of different isomeric alkyl groups corresponding to the molecular formula C₅H₁₁. Write IUPAC names of alcohols obtained by attachment of –OH groups at different carbons of the chain.

Solution
Alkyl Group Structure (C₅H₁₁) Corresponding Alcohol IUPAC Name
CH₃–CH₂–CH₂–CH₂–CH₂– CH₃–CH₂–CH₂–CH₂–CH₂–OH Pentan-1-ol
CH₃–CH(CH₂–CH₂–CH₃)– CH₃–CH(OH)–CH₂–CH₂–CH₃ Pentan-2-ol
CH₃–CH₂–CH(CH₂–CH₃)– CH₃–CH₂–CH(OH)–CH₂–CH₃ Pentan-3-ol
CH₃–CH(CH₃)–CH₂–CH₂– CH₃–CH(CH₃)–CH₂–CH₂–OH 3-Methylbutan-1-ol
CH₃–CH₂–CH(CH₃)–CH₂– CH₃–CH₂–CH(CH₃)–CH₂–OH 2-Methylbutan-1-ol
CH₃–C(CH₃)(CH₂–CH₃)– CH₃–C(CH₃)(OH)–CH₂–CH₃ 2-Methylbutan-2-ol
CH₃–C(CH₃)₂–CH₂– CH₃–C(CH₃)₂–CH₂–OH 2,2-Dimethylpropan-1-ol
CH₃–CH(CH₃)–CH(CH₃)– CH₃–CH(CH₃)–CH(OH)–CH₃ 3-Methylbutan-2-ol
Important IUPAC Nomenclature Rules

1. Lowest Sum Rule

Number the parent carbon chain from the end that gives the lowest possible locants (numbers) to the substituents.

Example:
CH₃–CH(CH₃)–CH₂–CH(C₂H₅)–CH₂–CH₃
✓ Correct: 4-Ethyl-2-methylhexane (positions 2,4)
✗ Wrong: 3-Ethyl-5-methylhexane (positions 3,5)

2. Alphabetical Arrangement

When multiple substituents are present, list them alphabetically. Prefixes like sec- and tert- are ignored, but iso- and neo- are considered.

Key Points:
sec- and tert-ignored for alphabetical order
iso- and neo-considered for alphabetical order
di-, tri-, tetra-ignored for alphabetical order

3. Multiple Same Substituents

Use di-, tri-, tetra- etc. for multiple identical substituents. List all position numbers.

Example:
CH₃–C(CH₃)₂–CH₂–C(CH₃)₂–CH₃
Name: 2,2,4,4-Tetramethylpentane

4. Longest Chain Priority

Always select the longest continuous carbon chain as the parent chain, even if it's not straight.

Steps:
1. Find longest chain → parent alkane
2. Number from end giving lowest locants
3. Name substituents alphabetically
4. Use prefixes for multiple same groups
IUPAC Naming Examples

Example 1 3-Methylpentane

3-Methylpentane molecular structure showing skeletal formula with teal-highlighted carbon 3 and CH₃ branch
Formula: C₆H₁₄
Chain: 5C (pentane)
Substituent: CH₃ at position 3 (highlighted in teal)
IUPAC Name: 3-methylpentane

Example 2 2,4-Dimethylpentane

2,4-Dimethylpentane molecular structure showing skeletal formula with violet-highlighted carbons 2 and 4 with CH₃ branches
Formula: C₇H₁₆
Chain: 5C (pentane)
Substituents: CH₃ at positions 2 and 4 (highlighted in violet)
IUPAC Name: 2,4-dimethylpentane

🔑 IUPAC Naming Key Points:

  • Longest Chain: Always identify the longest continuous carbon chain as the parent name
  • Numbering: Number from the end that gives substituents the lowest possible numbers
  • Substituent Position: Use numbers to indicate where branches are attached
  • Multiple Substituents: Use prefixes (di-, tri-, tetra-) for multiple identical groups
  • Alphabetical Order: List different substituents alphabetically (ignore prefixes)
Molecular Formula No. of Isomers IUPAC Names
C4H10 2 Butane (n-butane); 2-Methylpropane (isobutane)
C5H12 3 Pentane; 2-Methylbutane (isopentane); 2,2-Dimethylpropane (neopentane)
C6H14 5 Hexane; 2-Methylpentane; 3-Methylpentane; 2,3-Dimethylbutane; 2,2-Dimethylbutane
C7H16 9
C10H22 75
Structural Isomers - AI Generated Images

C₄H₁₀ Butane Isomers — 2 isomers

1. n-Butane
Linear
n-Butane skeletal line formula diagram
IUPAC: Butane • Common: n-butane
b.p.: 272 K (−1°C) • Density: 0.58 g/mL
2. 2-Methylpropane
Branched
2-Methylpropane T-shaped skeletal structure
IUPAC: 2-methylpropane • Common: isobutane
b.p.: 261 K (−12°C) • Density: 0.55 g/mL

C₅H₁₂ Pentane Isomers — 3 isomers

1. n-Pentane
Linear
n-Pentane linear chain
IUPAC: Pentane
b.p.: 309 K (36°C)
2. 2-Methylbutane
1° Branch
2-Methylbutane with single branch
IUPAC: 2-methylbutane
Common: isopentane
3. 2,2-Dimethylpropane
Quaternary
2,2-Dimethylpropane cross-shape
IUPAC: 2,2-dimethylpropane
Common: neopentane

📋 Complete AI Generation Specifications:

🎨 Visual Requirements:
  • Style: Clean skeletal line formulas
  • Lines: Bold black, 2.5-3px thickness
  • Dots: Small circles marking carbon positions
  • Colors: Teal (#14B8A6), Violet (#7C3AED), Rust (#B91C1C)
  • Background: Pure white (#FFFFFF)
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🔍 Key Observations

  • Boiling Point Trend: As branching increases, boiling point decreases due to reduced surface area and weaker van der Waals forces
  • Stability: More branched alkanes are generally more stable due to hyperconjugation
  • Isomer Count: Number of isomers increases rapidly with molecular size (C₁₀H₂₂ has 75 isomers!)
  • Naming Priority: Find longest chain first, then number to give substituents lowest locants

Types of Carbon Atoms

Type Attached to how many C atoms Symbol Example in 2-Methylpropane
Primary (1°) 1 (or 0 as in methane) Terminal –CH3 groups
Secondary (2°) 2 –CH2– groups in chain
Tertiary (3°) 3 –CH– at branch point
Quaternary (4°) 4 4° (neo) –C– with no H; e.g. in neopentane

Molecular Structure Comparison

Three-dimensional representation showing tetrahedral geometry and linear chain structure

3D Molecular Structure of Methane and Butane

💡 Methane Key Points

  • Simplest alkane with sp³ hybridization
  • Perfect tetrahedral symmetry
  • All C-H bonds are equivalent
  • Natural gas main component

🔗 Butane Key Points

  • First alkane showing structural isomerism
  • Has two forms: n-butane and isobutane
  • Used in LPG (liquefied petroleum gas)
  • Shows conformational isomerism

9.2.2 Preparation of Alkanes

Petroleum and natural gas are the main sources of alkanes. However, they can be prepared in the laboratory by the following methods:

1 From Unsaturated Hydrocarbons (Hydrogenation)

Dihydrogen gas adds to alkenes and alkynes in the presence of finely divided catalysts like platinum (Pt), palladium (Pd), or nickel (Ni). This process is called catalytic hydrogenation.

Reaction Conditions
  • Pt / Pd: Catalyze at room temperature.
  • Raney Ni: Requires relatively higher temperature (523–573 K) and pressure.
CH₂=CH₂ + H₂
Pt/Pd/Ni
CH₃–CH₃ Ethene to Ethane
CH₃–C≡CH + 2H₂
Pt/Pd/Ni
CH₃–CH₂–CH₃ Propyne to Propane

2 From Alkyl Halides

(i) Reduction with Zinc

Alkyl halides (except fluorides) on reduction with zinc and dilute hydrochloric acid give alkanes.

CH₃–Cl + H₂
Zn, H⁺
CH₄ + HCl Chloromethane to Methane
C₂H₅–Cl + H₂
Zn, H⁺
C₂H₆ + HCl Chloroethane to Ethane
(ii) Wurtz Reaction

Alkyl halides on treatment with sodium metal in dry ether solution give higher alkanes (usually with even number of carbon atoms).

2CH₃–Br + 2Na
dry ether
CH₃–CH₃ + 2NaBr Bromomethane to Ethane
2C₂H₅–Br + 2Na
dry ether
C₄H₁₀ + 2NaBr Bromoethane to n-Butane
Note on Wurtz Reaction

If a mixture of two different alkyl halides is used, a mixture of three different alkanes is formed, which is difficult to separate. Hence, it is not preferred for preparing alkanes with odd number of carbons.

3 From Carboxylic Acids

(i) Decarboxylation

Sodium salts of carboxylic acids on heating with soda lime (mixture of NaOH and CaO in 3:1 ratio) give alkanes containing one carbon less than the parent acid.

CH₃COONa + NaOH
CaO, Δ
CH₄ + Na₂CO₃ Sodium Acetate to Methane
(ii) Kolbe's Electrolytic Method

An aqueous solution of sodium or potassium salt of a carboxylic acid on electrolysis gives an alkane containing even number of carbon atoms at the anode.

2CH₃COONa + 2H₂O
Electrolysis
C₂H₆ + 2CO₂ + H₂ + 2NaOH
Reaction Mechanism (At Electrodes)
At Anode (+): 2CH₃COO⁻ → 2CH₃COO• + 2e⁻
2CH₃COO• → 2•CH₃ + 2CO₂
2•CH₃ → C₂H₆ (Ethane)
At Cathode (-): 2H₂O + 2e⁻ → 2OH⁻ + H₂•
2H• → H₂ (Gas)
Preparation Summary at a Glance
Starting Material Method / Reagents Key Product Note
Alkenes / Alkynes H₂ / Pt, Pd or Ni (Hydrogenation) Simple addition of hydrogen
Alkyl Halides Zn / dil. HCl (Reduction) Replacement of Halogen by H
Alkyl Halides Na / dry ether (Wurtz) Chain doubling (Even carbons)
Carboxylic acid salt NaOH + CaO, Heat (Decarboxylation) One carbon less than parent
Carboxylic acid salt Electrolysis (Kolbe's) Even number of carbons

9.2.3 Properties of Alkanes

Physical Properties

Physical Properties Data for Alkanes
Name Formula Molar Mass (g/mol) Melting Point (K) Boiling Point (K) Density (g/cm³) at 293K State at 298K
Methane CH₄ 16.04 90.7 111.7 0.000 67* Gas
Ethane C₂H₆ 30.07 90.4 184.6 0.001 26* Gas
Propane C₃H₈ 44.10 85.5 231.1 0.001 83* Gas
Butane C₄H₁₀ 58.12 134.6 272.4 0.579 Gas
2-Methylpropane C₄H₁₀ 58.12 114.7 261.0 0.549 Gas
Pentane C₅H₁₂ 72.15 143.3 309.1 0.626 Liquid
2-Methylbutane C₅H₁₂ 72.15 113.1 300.9 0.620 Liquid
2,2-Dimethylpropane C₅H₁₂ 72.15 256.4 282.5 0.614 Liquid
Hexane C₆H₁₄ 86.18 178.5 341.9 0.659 Liquid
Decane C₁₀H₂₂ 142.28 243.3 447.1 0.730 Liquid
Eicosane C₂₀H₄₂ 282.55 309.6 617 0.789 Solid

* Gas density at STP (273K, 1 atm)

📈 Boiling Point Trends

Effect of Chain Length:
  • Increases with molecular size due to stronger van der Waals forces
  • More surface contact in longer chains
  • Linear relationship with number of carbons
Effect of Branching:
  • Decreases with branching
  • Spherical shape reduces surface contact
  • Example: n-pentane (309K) > 2-methylbutane (301K) > neopentane (283K)

❄️ Melting Point Trends

Odd-Even Effect:
  • Even-numbered alkanes have higher m.p. than adjacent odd-numbered ones
  • Better packing in crystal lattice
  • More symmetric molecular arrangement
Branching Effect:
  • Highly branched alkanes (like neopentane) have higher m.p.
  • Spherical molecules pack more efficiently
  • Symmetry enhances crystal stability

🔬 General Physical Characteristics

Polarity & Bonding:
  • Almost non-polar molecules
  • C–H electronegativity difference = 0.4
  • Weak van der Waals forces only
  • No hydrogen bonding
Solubility & Miscibility:
  • Insoluble in water (polar)
  • Soluble in non-polar solvents (benzene, CCl₄)
  • "Like dissolves like" principle
  • Lower density than water (float on water)
Physical State Classification:
C₁–C₄: Gases at 298K
C₅–C₁₇: Liquids at 298K
C₁₈+: Solids at 298K

Chemical Properties

General Reactivity of Alkanes

Alkanes are generally unreactive towards ionic reagents due to:

  • Strong C–C and C–H σ bonds
  • Low polarity of bonds
  • Absence of functional groups

However, they undergo free radical reactions at high temperature or in presence of UV light.

Reaction Type Conditions Chemical Equations (Balanced) Key Observations
1. Substitution (UV) or
573–773 K
CH₄ + Cl₂ CH₃Cl + HCl
Sequence:
CH₄ → CH₃Cl → CH₂Cl₂ → CHCl₃ → CCl₄
Halogenation: F₂ > Cl₂ > Br₂ > I₂
Iodination: Reversible. Requires oxidizing agent (HIO₃ or HNO₃).
Others: Alkanes also undergo Nitration & Sulphonation at high temp.
Oxidation & Combustion Processes
2. Combustion Air / O₂, Heat
CH₄ + 2O₂ CO₂ + 2H₂O
General:
CₙH₂ₙ₊₂ + (3n+1)/2 O₂ → nCO₂ + (n+1)H₂O
ΔH: Highly exothermic
• Incomplete combustion → Carbon Black (C)
3. Controlled Oxidation Limit O₂, Catalyst
2CH₄ + O₂ [Cu/523K/100atm] → 2CH₃OH (Methanol)
CH₄ + O₂ [Mo₂O₃/Δ] → HCHO + H₂O (Methanal)
2C₂H₆ + 3O₂ [(CH₃COO)₂Mn/Δ] → 2CH₃COOH + 2H₂O
3° H Atom: Oxidized by KMnO₄ to alcohols.
• Highly specific to catalyst.
Mn-acetate gives Ethanoic acid.
KMnO₄ oxidizes 2-methylpropane to 2-methylpropan-2-ol.
Structural & Thermal Transformations
4. Isomerisation AlCl₃ / HCl n-alkanes branched alkanes (Isomers)
n-Hexane → 2-methylpentane + 3-methylpentane
• Major products are reported.
• Increases performance of fuels.
Mechanism of Isomerisation
5. Aromatisation Cr₂O₃/V₂O₅/Mo₂O₃, 773K, 10-20 atm n-C₆H₁₄ Benzene (C₆H₆) + 4H₂
n-C₇H₁₆ Toluene (C₆H₅CH₃)
• Also called Reforming.
• Alkane must have ≥ 6 carbons.
• Involves Dehydrogenation & cyclization.
Aromatization Cyclization Diagram
Industrial Processes & Cracking
6. Reaction with Steam Ni, 1273 K CH₄ + H₂O CO + 3H₂ • Industrial method to produce H₂ gas (Syngas).
7. Pyrolysis Δ (973 K for Alkanes)
C₆H₁₄ C₆H₁₂ + C₄H₈ + C₂H₆ + C₂H₄ + CH₄
Higher Alkanes:
C₁₂H₂₆ [Pt/Pd/Ni] → C₇H₁₆ + C₅H₁₀ + other products
Cracking of large molecules.
• Dodecane example (Kerosene oil) shows its utility.
Pyrolysis Branching Products
Free Radical Mechanism: Halogenation of Methane

Step 1: Initiation

Cl₂ → 2Cl• (chlorine radicals)

UV light breaks Cl–Cl bond homolytically to produce highly reactive chlorine free radicals.

Bond enthalpy: Cl–Cl = 242 kJ/mol

Step 2: Propagation

(a) CH₄ + Cl• → •CH₃ + HCl
(b) •CH₃ + Cl₂ → CH₃Cl + Cl•

Chain reaction: Each step produces a new radical to continue the cycle.

Rate-determining step: (a)
C–H bond enthalpy: 413 kJ/mol

Step 3: Termination

(a) Cl• + Cl• → Cl₂
(b) •CH₃ + •CH₃ → C₂H₆
(c) •CH₃ + Cl• → CH₃Cl

Chain termination: Free radicals combine to form stable molecules.

Formation of ethane confirms
free radical mechanism

📊 Factors Affecting Halogenation Rate

Reactivity of Halogens:
F₂ > Cl₂ > Br₂ > I₂
  • F₂: Explosive reaction
  • Cl₂: Controlled reaction in sunlight
  • Br₂: Requires higher temperature
  • I₂: No reaction under normal conditions
H-atom Replacement Order:
Tertiary (3°) > Secondary (2°) > Primary (1°)
  • 3° C–H: More reactive (lower bond strength)
  • 2° C–H: Moderate reactivity
  • 1° C–H: Less reactive (stronger bond)

9.2.4 Conformations of Ethane

Key Definitions

Conformations (Conformers / Rotamers): Different spatial arrangements of atoms in a molecule that can be interconverted by rotation around a C–C single bond. Alkanes can have infinite conformations, but rotation is NOT completely free — hindered by torsional strain (1–20 kJ/mol).

Torsional strain: Weak repulsive interaction between electron clouds of adjacent C–H bonds that resist rotation.

Dihedral angle (torsional angle): The angle of rotation about the C–C bond.

Eclipsed Conformation

H atoms on adjacent carbons are as close together as possible (0° dihedral angle). Maximum electron cloud repulsion → maximum torsional strainleast stable.

Staggered Conformation

H atoms are as far apart as possible (60° dihedral angle). Minimum repulsion → minimum torsional strainmost stable. Preferred conformation.

Any intermediate conformation = Skew conformation. Energy difference between eclipsed and staggered = 12.5 kJ/mol (small enough that free rotation occurs at room temperature).

Representation Methods

Sawhorse Projection

View along C–C axis. C–C drawn as a long diagonal line. Front carbon at lower end, rear carbon at upper end. Each C has 3 lines for H at 120°.

Newman Projection

View at C–C bond head-on. Front carbon = central point; rear carbon = circle. Each has 3 H-bonds at 120°. Eclipsed: H overlaps; Staggered: H at 60° to each other.

Sawhorse projections of ethane showing eclipsed and staggered conformations

Fig. 9.2: Sawhorse projections of ethane (eclipsed and staggered)

Newman projections of ethane showing eclipsed and staggered conformations

Fig. 9.3: Newman projections of ethane (eclipsed and staggered)

💡
Stability Rule Staggered > Skew > Eclipsed (in order of decreasing stability). Staggered = minimum torsional strain = preferred. Ethane conformers CANNOT be isolated (energy barrier too small at room temp).

9.3 Alkenes

Definition

Alkenes are unsaturated hydrocarbons containing at least one C=C double bond. General formula: CnH2n. Also called olefins (oil-forming) because ethylene forms an oily liquid with Cl2. First stable member: Ethene (C2H4).

9.3.1 Structure of the Double Bond

Orbital picture of ethene depicting sigma bonds only

Fig. 9.4: Orbital picture of ethene depicting sigma bonds only

Orbital picture of ethene showing pi bond formation, pi cloud, bond angles and bond length

Fig. 9.5: Orbital picture of ethene showing pi bond, pi cloud, bond angles and bond length

9.3.2 Nomenclature

Structure IUPAC Name
CH3–CH=CH2 Propene
CH3–CH2–CH=CH2 But-1-ene
CH3–CH=CH–CH3 But-2-ene
CH2=CH–CH=CH2 Buta-1,3-diene
CH2=C(CH3)–CH3 2-Methylprop-1-ene
CH2=CH–CH(CH3)–CH3 3-Methylbut-1-ene

9.3.3 Isomerism in Alkenes

Alkenes show both structural isomerism and geometrical isomerism.

Structural Isomerism (C₄H₈ — 3 isomers)

As in alkanes, ethene (C2H4) and propene (C3H6) have only one structure, but alkenes higher than propene show multiple structures.

Structure Name Type of Isomerism
CH2=CH–CH2–CH3 But-1-ene Position isomers (I vs II)
CH3–CH=CH–CH3 But-2-ene Position isomers (I vs II)
CH2=C(CH3)2 2-Methylprop-1-ene Chain isomers (I or II vs III)
Problem 9.9 (NCERT Style)

Write structures and IUPAC names of structural isomers corresponding to C5H10.

  1. CH2=CH–CH2–CH2–CH3   →   Pent-1-ene
  2. CH3–CH=CH–CH2–CH3   →   Pent-2-ene
  3. CH3–C(CH3)=CH–CH3   →   2-Methylbut-2-ene
  4. CH3–CH(CH3)–CH=CH2   →   3-Methylbut-1-ene
  5. CH2=C(CH3)–CH2–CH3   →   2-Methylbut-1-ene

Geometrical (cis-trans) Isomerism

Definition

Occurs when rotation about C=C is restricted. If each doubly bonded carbon has two different groups, two spatial arrangements are possible:

  • cis: identical groups on the SAME side of the double bond
  • trans: identical groups on OPPOSITE sides

Condition for cis-trans isomerism: Both doubly bonded carbons must each carry two DIFFERENT substituents. If either carbon carries two identical groups → no cis-trans isomerism.

cis-trans isomers are stereoisomers with the same structural formula but different spatial arrangement. Rotation about C=C is restricted, so interconversion does not happen by simple bond rotation.

General Representation

Doubly bonded carbon atoms satisfy the remaining valencies by joining with two atoms/groups. If the two groups attached to each carbon are different, they can be represented by YX C = C XY type structure.

This YX C = C XY arrangement can be represented in space in the following two ways:

C C X Y X Y (a) cis-config
C C X Y Y X (b) trans-config

(a) and (b) represent the two different spatial arrangements that give geometrical isomers.

Geometrical isomerism is commonly represented by these patterns:

  • XYC=CXY: shows cis and trans forms (example: but-2-ene)
  • XYC=CXZ and XYC=CZW: can also show geometrical isomerism when each C has two different groups.
cis-But-2-ene C C CH₃ H CH₃ H
μ = 0.33 D · bp 277 K
Polar · Higher mp
trans-But-2-ene C C CH₃ H H CH₃
μ ≈ 0 D · bp 274 K
Non-polar · Higher mp
cis vs trans — Key Differences
  • Polarity: cis is more polar (bond dipoles don't cancel); trans ≈ non-polar (dipoles cancel)
  • Boiling point: cis > trans (higher polarity → stronger intermolecular forces)
  • Melting point of solids: trans > cis (more symmetric → better packing in crystal)
  • Dipole moment: cis-but-2-ene = 0.33 D; trans-but-2-ene ≈ 0 D
Problem 9.10 (NCERT Style)

Draw cis and trans isomers and write IUPAC names:

  1. CHCl=CHCl   →   cis-1,2-Dichloroethene and trans-1,2-Dichloroethene
  2. C2H5C(CH3)=C(CH3)C2H5   →   cis-3,4-Dimethylhex-3-ene and trans-3,4-Dimethylhex-3-ene
Problem 9.11 (NCERT Style)

Which compounds show cis-trans isomerism?

  1. (CH3)2C=CH–C2H5   Does not show
  2. CH2=CBr2   Does not show
  3. C6H5CH=CH–CH3   Shows
  4. CH3CH=CClCH3   Shows

Rule check: cis-trans isomerism appears only if each doubly bonded carbon has two different substituents.

9.3.4 Preparation of Alkenes

Alkenes can be prepared by controlled elimination or partial reduction methods. Each process is shown below with reaction conditions, mechanism clue, and reaction-specific SVG representation.

1. From Alkynes (Partial Reduction)

Alkynes on reduction with one mole of H2 give alkenes. Below, expanded structural formulas are used (NCERT style) for clear product configuration.

  • Pd/C (partially deactivated, Lindlar type) gives cis-alkene.
  • Na/liquid NH3 gives trans-alkene.
Alkene Preparation from Alkynes - Reactions 9.30-9.31

Fig. 9.30, 9.31: Partial reduction of alkynes to cis/trans alkenes

Concept check: Propene formed here does not show geometrical isomerism because one double-bond carbon has two identical H atoms.

2. From Alkyl Halides (Dehydrohalogenation)

Alkyl halides on heating with alcoholic KOH eliminate HX and form alkene. This is a β-elimination reaction.

Alkene Preparation from Alkyl Halides - Reaction 9.34

Fig. 9.34: Dehydrohalogenation of alkyl halides

  • Base removes H from β-carbon; halide leaves from adjacent α-carbon.
  • Because H and X are eliminated from adjacent carbons, it is called β-elimination.
3. From Vicinal Dihalides (Dehalogenation)

Vicinal dihalides (halogens on adjacent carbons) react with Zn metal to eliminate ZnX2 and form alkenes. This is called dehalogenation.

Alkene Preparation from Vicinal Dihalides - Reactions 9.35-9.36

Fig. 9.35, 9.36: Dehalogenation of vicinal dihalides

4. From Alcohols by Acidic Dehydration

On heating alcohols with concentrated H2SO4, one molecule of water is removed and alkene is formed. This is called acidic dehydration of alcohols (also a β-elimination process).

Alkene Preparation from Alcohols - Reaction 9.37

Fig. 9.37: Acidic dehydration of alcohols

  • Common laboratory condition: conc. H2SO4 and heat.
  • Water elimination converts alcohol to corresponding alkene.
Quick Comparison
  • Partial reduction (alkynes): addition pathway, controlled to stop at C=C.
  • Dehydrohalogenation / Dehalogenation / Dehydration: elimination pathways to create C=C.
  • All methods are frequently tested through reagent-condition matching and product prediction.

9.3.5 Properties of Alkenes

A. Physical Properties

Alkenes as a class resemble alkanes in physical properties, with the following important exceptions based on types of isomerism and differences in polar nature:

Physical State and Boiling Points
  • First three members (ethene, propene, but-1-ene): gases at room temperature
  • Next fourteen members (approximately C₅ to C₁₈): liquids
  • Higher members (C₁₉ and above): solids
Odour and Colour
  • Ethene (C₂H₄): Odourless in pure form, but has characteristic sweet smell in commercial preparations
  • All other alkenes: Colourless liquids or solids with no characteristic odour
Solubility and Boiling Point Trends
  • Solubility in water: Insoluble in water (due to non-polar C–C double bond)
  • Solubility in non-polar solvents: Soluble in non-polar organic solvents such as benzene, petroleum ether, carbon tetrachloride, etc.
  • Boiling point increase with chain length: For every additional –CH₂– unit in a straight chain alkene, the boiling point increases by approximately 20–30 K
  • Effect of branching on boiling point: Like alkanes, straight-chain alkenes have higher boiling points than their isomeric branched-chain counterparts (due to greater surface area and stronger intermolecular van der Waals forces)

B. Chemical Properties

Alkenes are the rich source of loosely held π electrons due to the electron-rich C=C double bond. This makes them highly reactive. The π electrons are:

The reactions of alkenes are predominantly addition reactions in which electrophiles add on to the carbon–carbon double bond to form addition products. Some reagents also add by free radical mechanism. Additionally, oxidation and ozonolysis reactions are quite prominent in alkene chemistry.

1. ADDITION OF DIHYDROGEN (HYDROGENATION)
Hydrogenation of Alkenes

General Reaction:

R₂C=CR₂ + H₂ —(Ni, Pd, or Pt catalyst)→ R₂CH–CHR₂

Example:

CH₂=CH₂ + H₂ —(Ni/Pt/Pd)→ CH₃–CH₃

Mechanism: This is a catalytic hydrogenation process involving:

  • Formation of π-complex between alkene and catalyst surface
  • Addition of H atoms from the catalyst surface to the C=C double bond
  • Release of saturated alkane product

Key Points:

  • Reaction requires finely divided metal catalysts: Ni (nickel powder), Pt (platinum black), or Pd (palladium)
  • The catalyst acts as a surface on which both H and C=C can attach, facilitating the addition
  • Used primarily to prepare alkanes (saturated hydrocarbons) from alkenes
2. ADDITION OF HALOGENS (Br₂ AND Cl₂)
Addition of Halogens to Alkenes

General Reaction:

R₂C=CR₂ + X₂ R₂CX–CXR₂ (vicinal dihalide)

Example:

CH₂=CH₂ + Br₂ —(CCl₄)→ CHBr–CHBr (1,2-dibromoethane)

Specific Halogen Behaviour:

  • Bromine (Br₂): Readily adds to alkenes at room temperature in CCl₄ solution. The characteristic reddish-orange colour of bromine water (or CCl₄ solution) is rapidly discharged when added to an alkene, providing the brown/orange decolourisation test for unsaturation
  • Chlorine (Cl₂): Also adds readily, but reaction is more vigorous
  • Iodine (I₂): Does NOT add to alkenes under normal conditions due to thermodynamic considerations (the reaction is reversal)

Mechanism (Electrophilic Addition):

  • Step 1: Br₂ (or Cl₂) approaches the π-electron cloud of the C=C bond
  • Step 2: Formation of a bromonium (or chloronium) ion intermediate as one Br atom forms a bond with one carbon, displacing the other Br as Br⁻ ion
  • Step 3: The bromide ion attacks the bromonium ion from the opposite side (backside attack)
  • Result: Anti-stereochemistry (if applicable); vicinal dihalide is formed

Reactivity Order and Applications:

  • This decolourisation of Br₂/CCl₄ solution is a qualitative test for C=C unsaturation
  • The test distinguishes between alkenes (which decolourise Br₂) and alkanes (which do not)
3. ADDITION OF HYDROGEN HALIDES (HCl, HBr, HI)
Addition of HX to Alkenes

General Reaction:

R₂C=CR₂ + HX R₂CH–CXR₂ (alkyl halide)

Example:

CH₂=CH₂ + HBr CH₃–CH₂Br (bromoethane)

Reactivity of Hydrogen Halides:

HI > HBr > HCl

(Order reflects H–X bond strength and ease of protonation: HI is easiest to dissociate, HCl is hardest)

⚠️ MARKOVNIKOV'S RULE (1869)

Statement: In the addition of an unsymmetrical reagent (H–X, where H ≠ X) to an unsymmetrical alkene, the negative/halide part (X⁻) attaches to the carbon atom which possesses the lesser number of hydrogen atoms.

Alternatively: The H attaches to the C with more H, and X attaches to the C with fewer H.

Mechanistic Explanation of Markovnikov Rule:

ELECTROPHILIC ADDITION MECHANISM

  • Step 1 (Protonation): H⁺ from HX attacks the π-electron cloud of the C=C double bond. The proton can add to either carbon, forming TWO possible carbocation intermediates
  • Step 2 (Carbocation Stability): The carbocation that forms depends on H⁺ attack position:
    • If H⁺ adds to the carbon with MORE H: a less substituted (primary or secondary) carbocation forms
    • If H⁺ adds to the carbon with FEWER H: a more substituted (secondary or tertiary) carbocation forms
  • Step 3 (Major Product Formation): The more stable carbocation (more substituted) forms preferentially because it is formed faster. Stability order: 3° > 2° > 1°
  • Step 4 (Nucleophilic Attack): The halide ion (X⁻) attacks the carbocation from behind, neutralizing the positive charge and forming the C–X bond
  • Result: The major product is the one from the more stable carbocation intermediate

Detailed Example: Addition of HBr to Propene:

  • Markovnikov pathway (MAJOR):
    • H⁺ adds to terminal carbon (C-3, which has 2 H atoms)
    • Forms secondary carbocation: CH₃–CH⁺–CH₃ (2°)
    • Br⁻ attacks → product: 2-Bromopropane (isopropyl bromide)
  • Anti-Markovnikov pathway (MINOR):
    • H⁺ adds to internal carbon (C-2, which has 1 H atom)
    • Forms primary carbocation: CH₃–CH₂–CH₂⁺ (1°) — LESS STABLE
    • This pathway is much slower
    • Product: 1-Bromopropane (trace/negligible amount)
📄 NCERT Mechanism Diagram Placeholder (9.34)
Image Placeholder: Shows step-by-step mechanism with:
  • Formation of secondary carbocation (CH₃–CH⁺–CH₃)
  • Nucleophilic attack by Br⁻ to give major product
  • Formation of primary carbocation as less stable pathway
3a. ADDITION OF HBr — ANTI-MARKOVNIKOV (KHARASH EFFECT / PEROXIDE EFFECT)
Kharash Effect / Peroxide Effect

Discovery: Kharash and F.R. Mayo (1933) at the University of Chicago observed that HBr addition to unsymmetrical alkenes in the presence of peroxide takes place contrary to Markovnikov's rule.

General Reaction:

R₂C=CR₂ + HBr —(peroxide)→ R₂C(Br)–CR₂H (ANTI-Markovnikov product)

Example:

CH₃–CH=CH₂ + HBr —(peroxide)→ CH₃–CH₂–CH₂Br (1-Bromopropane — major)

Compare with Markovnikov (without peroxide): CH₃–CHBr–CH₃ (2-Bromopropane)

Key Observation: This effect is observed ONLY with HBr, not with HCl or HI!

Why Only HBr?

  • HBr (363.7 kJ/mol): H–Br bond strength is moderate — can be easily cleaved by free radicals to produce Br• radicals, which then add to the C=C bond
  • HCl (430.5 kJ/mol): H–Cl bond is TOO STRONG — cannot be cleaved easily by free radicals, so anti-Markovnikov addition does not occur
  • HI (296.8 kJ/mol): H–I bond is too WEAK — I• free radicals combine to form I₂ instead of adding to the C=C double bond

Mechanism: FREE RADICAL CHAIN MECHANISM

Initiation Step

Peroxide (R-O-O-R) —(heat)→ 2 R-O• (free radicals)

Propagation Steps

(i) R-O• + H-Br → R-O-H + Br•
(ii) Br• + C=C → (C-C)• (secondary free radical, more stable)
(iii) (C-C)• + H-Br → C-C-Br + Br•

Step (iii) regenerates Br•, continuing the chain reaction

Why Anti-Markovnikov?

  • The Br• radical adds to the alkene carbon with MORE hydrogens
  • This forms a secondary free radical (more substituted, more stable) intermediate
  • The H-Br then attacks this intermediate to give the anti-Markovnikov product
📄 NCERT Free Radical Mechanism Placeholder
Image Placeholder: Shows:
  • Peroxide decomposition to form R-O• free radicals
  • Three propagation steps with Br• and C=C interactions
  • Secondary free radical formation as key intermediate (more stable)
  • Final attack of H-Br on secondary radical giving anti-Markovnikov product
4. ADDITION OF COLD CONCENTRATED SULPHURIC ACID
Addition with H₂SO₄

Reaction:

R₂C=CR₂ + H₂SO₄ (conc., cold) → R₂CH–OSOH + Additional H₂SO₄ reaction

Example:

CH₃–CH=CH–CH₂ + H₂SO₄ → Alkyl hydrogen sulphate

Key Features:

  • Reaction follows Markovnikov's rule
  • Acidic H₂SO₄ acts as electrophile source (H⁺)
  • Forms intermediate alkyl hydrogen sulphate (ROSO₃H)
  • Further reaction possible with water to form alcohol
5. ADDITION OF WATER (HYDRATION / HYDROLYSIS)
Hydration of Alkenes

General Reaction:

R₂C=CR₂ + H₂O —(dilute H₂SO₄ or H₃PO₄ catalyst)→ R₂CH–OHCR₂ (alcohol)

Example:

CH₂=CH₂ + H₂O —(H₂SO₄ catalyst)→ CH₃–CH₂OH (ethanol)

Reaction Conditions:

  • Catalyst: Dilute H₂SO₄ or H₃PO₄ (acidic catalyst required)
  • Water requirement: Few drops of concentrated H₂SO₄ mixed with water and alkene
  • Temperature: Usually at moderate temperature

Markovnikov's Hydration:

In the hydration of unsymmetrical alkenes, the OH group attaches to the carbon with fewer H atoms (Markovnikov rule applies). For example:

CH₃–CH=CH₂ + H₂O → CH₃–CHOH–CH₃ (secondary alcohol — major) + CH₃–CH₂–CH₂OH (primary — minor)

Mechanism (Electrophilic Addition):

Hydration Mechanism Steps

  • Step 1: H⁺ (from H₂SO₄) protonates the more substituted carbon of the C=C bond, forming a carbocation
  • Step 2: Water molecule attacks the carbocation (nucleophilic attack)
  • Step 3: Deprotonation of the resulting oxonium ion yields the alcohol product
📄 NCERT Hydration Mechanism Placeholder (9.46)
Image Placeholder: Shows:
  • H⁺ protonation of C=C to form carbocation
  • Water molecule attacking carbocation
  • Oxonium ion intermediate formation
  • Deprotonation to form alcohol product
  • Example: CH₃–CH=CH₂ conversion to CH₃–CHOH–CH₃

Industrial Importance:

  • Major method for large-scale production of ethanol from ethene
  • Produces secondary and tertiary alcohols from alkenes
6. OXIDATION OF ALKENES
6a. Oxidation with Baeyer's Reagent (Cold Dilute KMnO₄)

Reagent: Cold, dilute, aqueous KMnO₄ solution (Baeyer's reagent), at ~273 K (0°C)

Reaction:

R₂C=CR₂ + [O] (from dilute KMnO₄) → R₂C(OH)–C(OH)R₂ (vicinal diol / glycol)

Example:

CH₂=CH–CH=CH₂ + KMnO₄/(cold, dil.) → CH(OH)–CH(OH)–CH(OH)–CH(OH) (But-1,2,3,4-tetraol)

Products:

  • The C=C bond is hydroxylated (OH groups add to the double bond carbons)
  • Two OH groups attach to adjacent carbons (vicinal position) → vicinal diol or glycol
  • The purple colour of KMnO₄ solution is discharged/decolourised

Test for Unsaturation:

The rapid decolourisation of purple/pink KMnO₄ solution (colour change to colourless or brown) is used as a qualitative test for C=C unsaturation in organic compounds.

Stereochemistry:

  • Addition of OH groups occurs with syn-stereochemistry (both OH groups add from the same side of the C=C bond)
6b. Oxidation under Acidic Conditions (KMnO₄/H⁺ or K₂Cr₂O₇/H⁺)

Reagent: Acidified KMnO₄ or K₂Cr₂O₇ solutions under heating

Reaction:

• Oxidizing C=C bond cleavage occurs
• Products depend on structure of alkene

Product Formation:

  • From terminal alkenes (R–CH=CH₂): Carboxylic acid (RCOOH) + CO₂
  • From internal alkenes (R–CH=CH–R'): Ketones (RCOR') and/or carboxylic acids

Examples:

CH₃–CH=CH–CH₃ + KMnO₄/H⁺ → CH₃–CO–CH₃ (acetone/propanone) + no other products
 
CH₃–CH=CH₂ + KMnO₄/H⁺ → CH₃–CHO (acetaldehyde) + HCOOH (formic acid)
or further oxidation: CH₃–COOH + CO₂

7. OZONOLYSIS OF ALKENES
Detection of Double Bond Position by Ozonolysis

Definition: Ozonolysis is the reaction of alkenes with ozone (O₃) followed by decomposition of the ozonide with reducing agents (e.g., Zn/H₂O) to isolate aldehydes and/or ketones.

General Reaction:

R₂C=CR₂ + O₃ → [R₂C–O–O–CR₂] (ozonide, primary adduct)
              ↓ (Zn/H₂O reduction)
         R₂C=O + R₂C=O (carbonyl compounds)

Example 1: Propene

CH₃–CH=CH₂ + O₃ → [Ozonide]
[Ozonide] + Zn/H₂O → CH₃–CHO (Ethanal/Acetaldehyde) + HCHO (Formaldehyde/Methanal)

Example 2: 2-Methylpropene

(CH₃)₂C=CH₂ + O₃ → [Ozonide]
[Ozonide] + Zn/H₂O → (CH₃)₂C=O (Propanone/Acetone) + HCHO (Formaldehyde)

Key Points:

  • The C=C double bond is cleaved and each carbon forms a C=O (carbonyl) bond
  • Primary carboxylic acid → aldehyde (under Zn/H₂O reduction)
  • Secondary doubly bonded carbon → ketone

Mechanism (Overview):

  • Step 1: Cycloaddition — Ozone (1,3-dipole) undergoes 1,3-cycloaddition to the C=C double bond, forming a primary ozonide (1,2,4-trioxolane)
  • Step 2: Rearrangement — The primary ozonide rearranges through a Criegee mechanism to form a more stable secondary ozonide
  • Step 3: Decomposition — Zn (reducing agent) and water cleave the O–O bonds of the ozonide, releasing carbonyl compounds (aldehydes/ketones)
📄 NCERT Ozonolysis Mechanism Placeholder (9.51-9.52)
Image Placeholder: Shows:
  • 1,3-cycloaddition of ozone to C=C double bond
  • Primary ozonide (1,2,4-trioxolane) formation
  • Criegee rearrangement to secondary ozonide
  • Zn/H₂O decomposition to form aldehyde/ketone
  • Example structures showing ozonide intermediates

Practical Application: Structure Determination

Ozonolysis is highly useful for determining the position of double bonds in unknown alkenes.

  • By identifying the carbonyl products (aldehydes/ketones), one can work backwards to determine the structure of the original alkene
  • Each carbonyl compound tells us the number of carbons and functional groups on each side of the double bond

Note on Alternative Workup:

  • If H₂O₂ is used instead of Zn/H₂O, aldehydes are oxidized to carboxylic acids
  • If no workup agent is used and only aqueous workup: aldehydes + hydration may occur
8. POLYMERISATION OF ALKENES
Addition Polymerisation of Alkenes

Definition: Polymerisation is the process in which a large number of simple unsaturated molecules (monomers) combine together through repeated addition reactions to form a single giant molecule (polymer).

General Reaction:

n(CH₂=CHR) —(High T, High P, catalyst)→ –(CH₂–CHR)–n (Polymer)

Examples of Addition Polymers:

1. Polythene (Polyethylene)

n(CH₂=CH₂) –(CH₂–CH₂)–n

  • Most common polymer; used for plastic bags, squeeze bottles, etc.
  • Produced by polymerisation under high temperature (100–300°C) and high pressure (1000–2000 atm)

2. Polypropylene (Polypropene)

n(CH₃–CH=CH₂) (Propene/Propylene) –(CH₃–CH–CH₂)–n

  • Strong polymer with diverse applications

3. Polyvinyl Chloride (PVC)

n(CH₂=CHCl) –(CH₂–CHCl)–n

Reaction Conditions:

  • Temperature: High temperature (100–300°C)
  • Pressure: High pressure (1000–2000 atmospheres)
  • Catalyst: Peroxides (for initiating free radical polymerisation) or Ziegler-Natta catalysts (for stereospecific polymerisation)

Mechanism: Free Radical Addition Polymerisation

Polymerisation Steps

  • Initiation: Peroxide decomposes under heat to form free radicals (R•), which attack the first monomer molecule to form a growing radical chain
  • Propagation: The growing radical chain successively adds monomer units through repeated addition to the C=C double bond
  • Termination: Two radical chains combine or a radical loses an electron via hydrogen abstraction, ending chain growth

Industrial und Commercial Importance:

  • Polythene bags and squeeze bottles — extremely widespread household items
  • Refrigerator bottles and containers — due to chemical resistance
  • Radio and TV cabinets — due to insulating properties
  • Plastic pipes — used in construction and plumbing
  • Textile fibres — synthetic fabrics
  • Protective coatings — due to durability

Environmental Note:

While addition polymers are extremely useful, their excessive use of fossil fuel-derived monomers and their persistence in the environment have raised concerns about waste management and sustainability. However, these polymers remain indispensable materials for modern society.


9.4 Alkynes

Definition

Alkynes are unsaturated hydrocarbons with at least one C≡C triple bond. General formula: CnH2n–2. First stable member: Ethyne (C2H2) — commonly called acetylene. Used for oxyacetylene welding.

9.4.2 Structure of the Triple Bond (Ethyne)

🖼️
Fig. 9.6 – Orbital Picture of Ethyne (σ and π overlaps)
📄 NCERT Page 315
(a) σ-bond picture: H–C–C–H linear arrangement, showing sp hybrid orbitals (1s on H, sp on each C) forming two C–H and one C–C σ bonds. All atoms in a straight line.
(b) π-bond picture: Two sets of p orbitals (2px and 2py) on each C overlap laterally to form two π bonds. The 2px orbitals overlap in the plane of the paper (π bond), and 2py orbitals overlap perpendicular to the paper (second π bond).

Fig. 9.6 — Orbital picture of ethyne: (a) sigma overlaps forming the C–C and C–H bonds; (b) two pi bonds formed by lateral overlap of p orbitals

9.4.1 Nomenclature and Isomerism

Suffix: –yne replaces –ane. In common system: derivatives of acetylene (e.g., methylacetylene).

n Formula Common Name IUPAC Name
2 C2H2 Acetylene Ethyne
3 C3H4 Methylacetylene Propyne
4 C4H6 Ethylacetylene But-1-yne
4 C4H6 Dimethylacetylene But-2-yne

9.4.3 Preparation of Alkynes

# Method Reaction
1 From Calcium Carbide CaCO3 →(Δ) CaO + CO2;  CaO + 3C → CaC2 + CO;  CaC2 + 2H2O → Ca(OH)2 + C2H2↑ (Industrial method)
2 From Vicinal Dihalides R–CHX–CHX–R' + alc. KOH → alkenyl halide → + NaNH2 → alkyne (two steps of dehydrohalogenation)

9.4.4 Properties of Alkynes

First 3 = gases; next 8 = liquids; higher = solids. All colourless; ethyne has characteristic odour. Weakly polar. Lighter than water; immiscible with water; soluble in organic solvents.

A. Acidic Character

KEY CONCEPT

Terminal alkynes (H attached to triply bonded C) are acidic — they react with strong bases like Na metal and NaNH2 to liberate H2.

Reason: sp hybridised C (50% s character) is most electronegative → attracts C–H shared electrons more strongly → H is released as H⁺ more easily than in alkenes (sp²) or alkanes (sp³).

HC≡CH + Na → HC≡C⁻Na⁺ + ½H2    (Monosodium ethynide)
HC≡C⁻Na⁺ + Na → Na⁺C≡C⁻Na⁺ + ½H2    (Disodium ethynide)
CH3–C≡C–H + NaNH2 → CH3–C≡C⁻Na⁺ + NH3    (Sodium propynide)

Note: But-2-yne (internal alkyne, CH3–C≡C–CH3) is NOT acidic — no H on triply bonded C!

Acidity order:
HC≡CH
>
H₂C=CH₂
>
CH₃–CH₃
 (sp > sp² > sp³ carbon)

B. Addition Reactions of Alkynes

Alkynes add up TWO molecules of the reagent (compared to one for alkenes). Addition in unsymmetrical alkynes follows Markovnikov rule.

Reaction Reagent Product (Step 1 → Step 2)
(i) + H2 Pt/Pd/Ni HC≡CH + H2 → H2C=CH2 (Ethene) → + H2 → CH3–CH3 (Ethane)
(ii) + X2 Br2/CCl4 HC≡CH + Br2 → CHBr=CHBr (1,2-dibromo) → + Br2 → CHBr2–CHBr2 (1,1,2,2-tetrabromo)
(iii) + HX HCl, HBr, HI HC≡CH + HBr → CH2=CHBr (Bromoethene) → + HBr → CH3–CHBr2 (gem-dihalide: 1,1-dibromoethane)
(iv) + H2O HgSO4/dil. H2SO4, 333 K HC≡CH + H2O → CH2=CHOH (vinyl alcohol) → isomerisation → CH3CHO (Ethanal/Acetaldehyde)
CH3–C≡CH + H2O → CH3–CO–CH3 (Propanone/Acetone)
(v) Polymerisation Suitable catalyst Linear: → Polyacetylene (conducting polymer used in batteries)
Cyclic: 3 C2H2 → Benzene (red hot iron tube, 873 K)
Gem vs Vicinal Dihalide
  • Gem dihalide: two halogens on the same carbon (from HX + alkyne)
  • Vicinal dihalide: two halogens on adjacent carbons (from X2 + alkene)
Cyclic Polymerisation — Benzene from Ethyne
3 HC≡CH —(red hot Fe tube, 873 K)→ C6H6 (Benzene)

This is the best route from aliphatic to aromatic compounds.

🎨
AI Image: Cyclic Polymerisation of Ethyne to Benzene
📄 AI Generate (NCERT p.318)
Reaction diagram showing 3 molecules of ethyne (HC≡CH) combining in a red hot iron tube at 873 K to form benzene ring.
🤖 AI Image Prompt
A clean chemistry reaction diagram on pure white background. Shows 3 structural formulas of ethyne (H–C≡C–H) on the left side with a large arrow pointing right, labelled "Red hot iron tube, 873 K" above the arrow. On the right side, show the benzene ring structure (hexagon with a circle inside, in amber/gold colour). Below the benzene, write "Benzene (C₆H₆)". The three ethyne molecules should be arranged diagonally or in a triangular arrangement to suggest they come together. All molecular structures drawn clearly with proper bond lines. Professional chemistry diagram style. White background only. No text other than the labels mentioned.

Cyclic polymerisation of ethyne → benzene (aromatisation route from aliphatic to aromatic)


9.5 Aromatic Hydrocarbons (Arenes)

Aromatic hydrocarbons are also called arenes (from Greek aroma = pleasant smell). Most contain benzene ring. Benzenoids contain benzene ring; non-benzenoids don't (e.g. Tropone).

Benzene
C₆H₆
Toluene
CH₃
Methylbenzene
Naphthalene
C₁₀H₈ · Two fused rings

9.5.2 Structure of Benzene

Historical Facts
  • Benzene isolated by Michael Faraday (1825)
  • Molecular formula C6H6 — indicates high degree of unsaturation
  • Benzene forms a triozonide → 3 double bonds present
  • Benzene gives only one monosubstituted product → all 6 C and H atoms are identical
  • Kekulé (1865): proposed cyclic structure with alternating single and double bonds
  • Problem with Kekulé: predicts 2 ortho-disubstituted isomers, but only ONE is observed → solved by "oscillating double bonds"
  • Modern explanation: resonance (delocalisation)

Electronic Structure of Benzene

Kekulé A
Structure A
Kekulé B
Structure B
Resonance Hybrid (C)
Actual structure
C–C = 139 pm (all equal)
🖼️
Fig. 9.7 (a), (b), (c), (d) – Orbital Overlap in Benzene & Delocalised π Cloud
📄 NCERT Pages 320–321
Fig 9.7(a): p orbitals overlapping as in Kekulé structure A (C1–C2, C3–C4, C5–C6 π bonds).
Fig 9.7(b): p orbitals overlapping as in Kekulé structure B (C2–C3, C4–C5, C6–C1 π bonds).
Fig 9.7(c): Actual equal overlap — each C's p orbital overlaps equally with both neighbours; dashed lines showing equal probability.
Fig 9.7(d): The resulting delocalised π electron cloud — two doughnut-shaped lobes, one above and one below the plane of the hexagonal ring.

Fig. 9.7 — Orbital overlap and delocalised π electron cloud in benzene: (a) Kekulé A, (b) Kekulé B, (c) equal overlap (actual), (d) doughnut-shaped π clouds above and below the ring

KEY CONCEPT

Why benzene prefers substitution over addition:

Benzene's delocalised π electrons are attracted more strongly by the 6 carbon nuclei than localised electrons between 2 carbons. This extra stability (resonance energy) means benzene resists addition reactions (which would destroy delocalisation) and instead undergoes electrophilic substitution (which retains the aromatic ring).

X-Ray data: all C–C = 139 pm (intermediate between single 154 and double 134). No pure double bonds → explains resistance to addition.

9.5.3 Aromaticity

KEY CONCEPT

Hückel's Rule of Aromaticity

A compound is aromatic if it satisfies ALL THREE conditions:

  1. Planarity: the molecule is flat (planar)
  2. Complete delocalisation of π electrons in the ring
  3. (4n + 2) π electrons in the ring where n = 0, 1, 2, 3, … (Hückel Rule)
Compound n value π electrons Aromatic?
Benzene n=1 ✅ Yes
Cyclopentadienyl anion (C5H5⁻) n=1 ✅ Yes
Cycloheptatrienyl cation (C7H7⁺) n=1 ✅ Yes
Naphthalene n=2 10π ✅ Yes
Anthracene / Phenanthrene n=3 14π ✅ Yes
Cyclobutadiene 4π (4n, n=1) ❌ Anti-aromatic
Cyclohexadiene (non-planar ring) Not fully delocalised ❌ Not aromatic

9.5.4 Preparation of Benzene

Method Reaction
Commercial Isolated from coal tar (fractional distillation)
Cyclic polymerisation of ethyne 3 HC≡CH →(Fe, 873 K)→ C6H6
Decarboxylation of benzoic acid C6H5COONa + NaOH →(CaO/Δ)→ C6H6 + Na2CO3
Reduction of phenol C6H5OH + Zn →(Δ)→ C6H6 + ZnO

9.5.5 Electrophilic Substitution Reactions of Benzene

Definition

Arenes undergo electrophilic substitution (SE) reactions in which an electrophile (E⁺) replaces a hydrogen atom on the ring, retaining aromaticity. Three steps: (a) Generation of E⁺, (b) Formation of σ-complex (arenium ion), (c) Removal of H⁺ to restore aromaticity.

Reaction Reagents & Conditions Electrophile Generated Product Eq. No.
1. Nitration Conc. HNO3 + Conc. H2SO4 (nitrating mixture), 323–333 K NO2⁺ (Nitronium ion) Nitrobenzene (C6H5NO2) 9.72
2. Halogenation Cl2 or Br2, anhydrous FeCl3/FeBr3/AlCl3 (Lewis acid) Cl⁺ or Br⁺ Chlorobenzene / Bromobenzene 9.73
3. Sulphonation Fuming H2SO4 (oleum), heat SO3 (electrophilic) Benzene sulphonic acid (C6H5SO3H) 9.74
4. Friedel-Crafts Alkylation Alkyl halide (R–X) + anhydrous AlCl3 R⁺ (carbocation) Alkylbenzene (e.g. Toluene with CH3Cl) 9.75–9.76
5. Friedel-Crafts Acylation Acyl halide (R–COX) or acid anhydride + AlCl3 RCO⁺ (Acylium ion) Acylbenzene (e.g. Acetophenone with CH3COCl) 9.77–9.78

Mechanism of Electrophilic Substitution (SE) — Detailed

(a) Generation of Electrophile E⁺:
For nitration: HNO3 + H2SO4 → H2NO3⁺ + HSO4⁻ → H2O + ⁺NO2 (Nitronium ion)
For halogenation: Cl–Cl + AlCl3 → Cl⁺ + [AlCl4]⁻
For F-C alkylation: CH3Cl + AlCl3 → CH3⁺ + [AlCl4]⁻

(b) Formation of σ-complex (Arenium ion):
Benzene + E⁺ → Arenium ion (σ-complex) — one C becomes sp³, aromaticity LOST; ion stabilised by resonance over remaining π system.

(c) Removal of H⁺ (Restoration of Aromaticity):
[AlCl4]⁻ or [HSO4]⁻ removes H⁺ from sp³ carbon → aromaticity restored → substituted benzene + HCl/H2SO4 + AlCl3 (Lewis acid regenerated)

Addition Reactions (Under Special Conditions)

Hydrogenation: Benzene + 3H2 —(Ni, Δ, high P)→ Cyclohexane 9.80
Chlorination: Benzene + 3Cl2 —(UV, 500 K)→ BHC (Benzene Hexachloride, C6H6Cl6, Gammaxane) 9.81
🎨
AI Image: Mechanism of Electrophilic Substitution (Nitration of Benzene)
📄 AI Generate (NCERT p.323)
Step-by-step mechanism showing: (1) generation of NO₂⁺, (2) attack on benzene ring to form arenium ion (σ-complex) with sp³ carbon shown, (3) loss of H⁺ to restore benzene ring with NO₂ group.
🤖 AI Image Prompt
A clean three-step chemistry mechanism diagram on a pure white background showing electrophilic nitration of benzene. Step 1 (top): Shows HNO3 + H2SO4 producing nitronium ion (NO2+) + HSO4-. Step 2 (middle): Shows benzene ring (hexagon with circle) being attacked by NO2+ to form the sigma complex/arenium ion — show the benzene ring with one sp3 carbon (indicated by a small H on top of that carbon), NO2 attached, and a positive charge distributed over the ring shown by resonance brackets. Step 3 (bottom): Shows the arenium ion losing H+ to HSO4- to form nitrobenzene (benzene ring with NO2 group) + H2SO4 + restored aromaticity. Each step has a curved arrow showing electron movement. All drawn in black ink on white background. Professional organic chemistry mechanism style. No extra text or decorative elements.

Mechanism of electrophilic nitration of benzene: generation of NO₂⁺ → σ-complex (arenium ion) → loss of H⁺ → nitrobenzene

9.5.6 Directive Influence of Substituents in Monosubstituted Benzene

Definition

When monosubstituted benzene undergoes further substitution, the incoming group is directed to specific positions (ortho, meta or para) based on the nature of the substituent already present — not on the nature of the incoming group.

Ortho & Para Directing Groups (Activating)

Mechanism: These groups donate electrons into the ring via +R or +I effect → electron density increases at o- and p-positions → electrophile attacks o- and p-positions.

Examples: –OH, –NH2, –NHR, –NR2, –NHCOCH3, –OCH3, –CH3, –C2H5

Halogens are o- and p-directing but moderately deactivating (strong –I but +R effect increases density at o/p)

These groups make benzene ring MORE reactive toward electrophilic substitution.

Meta Directing Groups (Deactivating)

Mechanism: These groups withdraw electrons from ring via –R or –I effect → electron density decreases more at o- and p-positions than at m-position → electrophile attacks the relatively electron-rich meta position.

Examples: –NO2, –CN, –CHO, –COR, –COOH, –COOR, –SO3H

These groups make benzene ring LESS reactive toward electrophilic substitution.

📝 Examples of Directive Effects

Phenol (–OH group): Resonance structures show high electron density at C-2, C-4, C-6 (ortho and para positions) → attack at o- and p-positions → o-nitrophenol + p-nitrophenol (major products)

Nitrobenzene (–NO2 group): –NO2 withdraws electrons strongly (–I and –R) → electron density decreases at o- and p-positions more than m → electrophile attacks m-position → m-dinitrobenzene (major)

Toluene (–CH3 group): +I effect of alkyl group + hyperconjugation → increased electron density at o- and p-positions → o-nitrotoluene + p-nitrotoluene (major)

🎨
AI Image: Resonance Structures of Phenol – Showing o- and p-Directing Nature
📄 AI Generate (NCERT p.324)
Five resonance structures of phenol connected by double-headed arrows, showing how electron density from the lone pair on oxygen delocalises into the ortho and para positions of the benzene ring.
🤖 AI Image Prompt
A clean chemistry resonance diagram on a pure white background showing five resonance structures of phenol (C6H5OH). Structure I: normal phenol with lone pairs on oxygen and the benzene ring represented as a hexagon with alternating bonds. Structure II: curved arrow from O lone pair into ring, positive charge on O, negative charge at ortho carbon (C2). Structure III: negative charge moves to para carbon (C4), positive on O. Structure IV: different ortho position (C6). Structure V: another resonance form. All five structures connected by double-headed arrows. Below each structure, label the position of the partial negative charge (ortho-2, para-4, ortho-6). At the bottom, add the summary: "High electron density at ortho (2,6) and para (4) positions → o- and p-substitution". Black structural drawing on white background. Professional organic chemistry style.

Resonance structures of phenol: electron density builds up at o- and p-positions → ortho and para directing

Substituent already on ring Directing Effect Activating/Deactivating? Effect on ring reactivity
–OH, –NH2, –OCH3, –CH3 o/p directing Activating Ring MORE reactive
–F, –Cl, –Br, –I (halogens) o/p directing Moderately deactivating Ring less reactive
–NO2, –CN, –COOH, –CHO, –SO3H m directing Strongly deactivating Ring much less reactive

9.6 Carcinogenicity and Toxicity

Important

Benzene and polynuclear hydrocarbons (containing more than two fused benzene rings) are toxic and possess carcinogenic (cancer-causing) properties.

They are formed during incomplete combustion of organic materials like tobacco, coal and petroleum. They enter the human body, undergo biochemical reactions, damage DNA and cause cancer.

Carcinogenic Hydrocarbon Notes
1,2-Benzanthracene Polynuclear aromatic hydrocarbon; fused ring system
3-Methylcholanthrene Very potent carcinogen
1,2-Benzpyrene Found in cigarette smoke
1,2,5,6-Dibenzanthracene Strong carcinogen
9,10-Dimethyl-1,2-benzanthracene Potent carcinogen

Use of polythene bags and polypropylene is also a concern — excessive use raises environmental issues.


Comprehensive Comparison: Alkanes vs Alkenes vs Alkynes

Property Alkanes Alkenes Alkynes
General Formula CnH2n+2 CnH2n CnH2n–2
Hybridisation sp³ sp² (at C=C) sp (at C≡C)
Bond angle 109.5° 120° 180°
C–C bond length 154 pm (single) 134 pm (double) 120 pm (triple)
Molecular shape Tetrahedral at each C Planar around C=C Linear around C≡C
Characteristic reactions Substitution (free radical) Addition (electrophilic) Addition (electrophilic); Acidic character
Test for unsaturation No reaction with Br2/KMnO4 Decolourises Br2/KMnO4 Decolourises Br2/KMnO4
Acidic character None None Terminal H is acidic (sp C)
Isomerism Chain, position Chain, position, geometrical (cis-trans) Chain, position
First stable member Methane (CH4) Ethene (C2H4) Ethyne (C2H2)

✏️ Practice Questions

Q1
How do you account for the formation of ethane as a by-product during chlorination of methane?
Q2
An alkene A on ozonolysis gives a mixture of ethanal and pentan-3-one. Write the structure and IUPAC name of A.
Q3
An alkene A contains three C–C bonds, eight C–H σ bonds and one C–C π bond. On ozonolysis it gives two moles of an aldehyde of molar mass 44 u. Write the IUPAC name of A.
Q4
Addition of HBr to propene gives 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction gives 1-bromopropane. Explain.
Q5
Why is benzene extraordinarily stable though it contains three double bonds?
Q6
What are the necessary conditions for any system to be aromatic?
Q7
Draw the cis and trans structures of hex-2-ene. Which isomer will have higher boiling point and why?
Q8
Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Give reason.
Q9
Why is the Wurtz reaction not preferred for preparation of alkanes with an odd number of carbon atoms?
Q10
Out of benzene, m-dinitrobenzene and toluene, which will undergo nitration most easily and why?
Q11
Write all the alkenes which on hydrogenation give 2-methylbutane.
Q12
Methane cannot be prepared by Kolbe's electrolytic method. Why?

🎯 Important Exam Points – Quick Reference

CONCEPTGeneral formulas: Alkanes CnH2n+2, Alkenes CnH2n, Alkynes CnH2n–2.
REACTIONWurtz reaction: always gives even-number carbon alkane from same alkyl halide. Mixed halides → 3 products (not useful).
CONCEPTKolbe's method: methane CANNOT be prepared. Decarboxylation: alkane has 1 C less than acid.
CONCEPTStaggered conformation of ethane is MORE stable than eclipsed. Energy difference = 12.5 kJ/mol. Conformers CANNOT be isolated.
CONCEPTMarkovnikov's rule: H adds to C with more H atoms (negative part to C with fewer H). Mechanistically: more stable carbocation forms faster.
CONCEPTAnti-Markovnikov (Kharash effect): only with HBr, only in presence of peroxide. Free radical mechanism. HCl/HI do NOT show peroxide effect.
CONCEPTAcidity of terminal alkynes: sp C (50% s) > sp² C (33% s) > sp³ C (25% s). HC≡CH > H₂C=CH₂ > CH₃–CH₃.
CONCEPTHückel Rule: aromatic if (4n+2) π electrons + planar + complete delocalisation. Benzene = 6π (n=1), Naphthalene = 10π (n=2).
CONCEPTo/p directors (activating): –OH, –NH₂, –CH₃, –OCH₃. o/p directors (deactivating): halogens. m directors (deactivating): –NO₂, –CN, –COOH, –CHO, –SO₃H.
CONCEPTLindlar's catalyst → cis-alkene from alkyne. Na in liquid NH₃ → trans-alkene from alkyne.
REACTIONBaeyer's test: cold dil. KMnO₄ decolourised by alkenes/alkynes (not alkanes). Also: reddish-orange Br₂/CCl₄ decolourised.
REACTIONOzonolysis: identifies position of double bond. Work backwards from carbonyl fragments to get original alkene.
REACTIONAromatisation (Reforming): n-alkane (≥C6) + V/Mo/Cr oxide, 773K → benzene/homologues.
MCQEthane by-product in chlorination of methane → formed in chain termination step: •CH₃ + •CH₃ → C₂H₆.
MCQWhy benzene prefers substitution over addition → resonance stabilisation/aromaticity would be destroyed by addition; retained by substitution.