Class 11 · Chemistry · Chapter 09

Hydrocarbons

Petrol in a tank, wax in a candle, gas in a stove — all of it is just carbon and hydrogen, linked into chains and rings. This chapter teaches you to build, break and burn those chains on paper.

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Chapter Snapshot

The chapter, at a glance

3
Families — alkanes, alkenes & alkynes, plus aromatics
4
Key preparation methods, from Wurtz to Kolbe's electrolysis
7–9
Marks typically weighted in board examinations
1
Single most application-heavy organic chapter of Class 11
Why This Chapter Matters

The fuel behind your organic chemistry score

  • Longest organic chapter in the syllabus — it alone can decide 8–9 marks across board and competitive exams.
  • Mechanism-based questions on electrophilic addition and substitution appear almost every year.
  • Direct foundation for Alcohols, Aldehydes, Ketones and Haloalkanes in Class 12 — the same reagents and mechanisms return.
  • Real-world relevance — petroleum refining and combustion questions are popular in applied/case-based formats.
Where marks concentrate
Reaction mechanismsHigh
Preparation methodsHigh
Aromaticity & benzeneModerate
Key Concept Highlights

Six links in the hydrocarbon chain

Alkanes

Saturated Hydrocarbons

Single-bonded chains; prepared by Wurtz reaction and undergo free-radical substitution.

Alkenes

Unsaturated — Double Bond

Undergo electrophilic addition; Markovnikov's rule decides which carbon gets what.

Alkynes

Unsaturated — Triple Bond

More reactive still; acidic terminal hydrogen enables unique identification tests.

Aromatic

Benzene & Aromaticity

Delocalised π-electrons make benzene resistant to addition but prone to substitution.

Mechanism

Markovnikov & Peroxide Effect

Rules that predict the major product in addition reactions to unsymmetrical alkenes.

Applied

Combustion & Petroleum

How hydrocarbons burn, and how crude oil is fractionated into usable fuels.

Important Formulae & Reactions

The reaction manifold — keep these fired up

CnH2n+2

General formula for alkanes (saturated hydrocarbons)

CnH2n

General formula for alkenes (one double bond)

CnH2n−2

General formula for alkynes (one triple bond)

2R–X + 2Na → R–R + 2NaX

Wurtz reaction for preparing higher alkanes

CH2=CH2 + HBr → CH3–CH2Br

Markovnikov addition of HBr to an unsymmetrical alkene

CnH2n+2 + O2 → CO2 + H2O

Complete combustion of an alkane (balance per given n)

2CH3COONa → C2H6 + 2CO2 + H2 (electrolysis)

Kolbe's electrolytic method for symmetrical alkane synthesis

What You Will Learn

Your chain, link by link

01

Classification & nomenclature of hydrocarbons

Sorting alkanes, alkenes, alkynes and aromatics, and naming each correctly.

02

Preparation methods

Wurtz reaction, Kolbe's electrolysis, and reduction of alkyl halides.

03

Physical & chemical properties of alkanes

Free-radical halogenation and combustion behaviour of saturated chains.

04

Alkenes & electrophilic addition

Markovnikov's rule, the peroxide effect, and key addition reactions.

05

Alkynes & their distinctive reactions

Acidic character, addition reactions, and tests to distinguish terminal alkynes.

06

Aromatic hydrocarbons & benzene

Structure, aromaticity criteria, and electrophilic substitution reactions.

Chapter Resources

Jump straight to what you need

📖Full NotesComplete theory 🎯MCQs PracticeTopic-wise quiz 🗂️PYQ BankPast-year questions 🔥True / FalseQuick-fire check ✍️ExercisesNCERT back-questions
Exam Strategy & Preparation Tips

How to actually score well here

01

Drill Markovnikov and anti-Markovnikov (peroxide effect) side by side — the most commonly confused pair in this chapter.

02

Memorise general formulae (CnH2n+2, CnH2n, CnH2n−2) cold — they underpin nearly every numerical question.

03

Practise combustion equation balancing for at least 3 different hydrocarbons until it's automatic.

04

Draw the mechanism arrows for electrophilic addition/substitution every time — partial credit is real.

05

Keep a name-reaction table (Wurtz, Kolbe, Sabatier-Senderens) with reagents and conditions for quick recall.

06

Revise benzene's structure and resonance carefully — aromaticity-based assertion-reason questions recur often.

Chapter 9 · CBSE · Class XI
⛽

Introduction to Hydrocarbons

Hydrocarbons NCERT Class 11 Chemistry Carbon and Hydrogen Compounds Classification of Hydrocarbons Saturated Hydrocarbons Unsaturated Hydrocarbons Aromatic Hydrocarbons Alkanes Paraffins Methane Ethane Propane Butane Alkyl Groups Chain Isomerism Structural Isomerism Conformations Newman Projection Sawhorse Projection Staggered Conformation Eclipsed Conformation Torsional Strain Hydrogenation Wurtz Reaction Decarboxylation Kolbe Electrolysis Pyrolysis Cracking Aromatization Reforming Isomerisation Combustion of Alkanes Halogenation of Alkanes Free Radical Substitution Free Radical Mechanism Initiation Propagation Termination Alkenes Olefins Ethene Propene Double Bond sp2 Hybridisation Geometrical Isomerism Cis Trans Isomerism Markovnikov Rule Anti Markovnikov Rule Peroxide Effect Kharash Effect Electrophilic Addition Baeyer Test Baeyers Reagent Bromine Water Test Ozonolysis Polymerisation Polyethylene Polypropylene Alkynes Acetylene Ethyne Triple Bond sp Hybridisation Acidic Character of Alkynes Sodium Acetylide Lindlar Catalyst Vicinal Dihalides Dehydrohalogenation Dehalogenation Hydration of Alkynes Aromatic Compounds Arenes Benzene Structure of Benzene Kekule Structure Resonance Resonance Energy Delocalisation Aromaticity Huckel Rule Benzenoid Compounds Non Benzenoid Compounds Nitration Halogenation Sulphonation Friedel Crafts Alkylation Friedel Crafts Acylation Electrophilic Substitution SE Mechanism Sigma Complex Arenium Ion Nitronium Ion Directive Influence Ortho Para Directing Groups Meta Directing Groups Activating Groups Deactivating Groups Combustion of Benzene Benzene Hexachloride Gammaxane Lindane Cyclohexane Petroleum Natural Gas Coal Gas LPG LNG Fractional Distillation of Petroleum Petrochemicals
📋
Table of Contents
13 sections
📖 Introduction
🌟 Importance of Hydrocarbons in Daily Life
🗒️ Natural Sources Of Hydrocarbons
Most naturally occurring hydrocarbons originate from the decomposition of ancient plants and animals buried under the Earth's crust over millions of years under high pressure and temperature.
1. Petroleum (Crude Oil)
fractional distillation.

  • Petroleum Gas (LPG)
  • Petrol (Gasoline)
  • Naphtha
  • Kerosene
  • Diesel
  • Lubricating Oil
  • Paraffin Wax
  • Bitumen
2. Natural Gas
Natural gas is found in the upper layers of petroleum reservoirs and is obtained during drilling of oil wells. It mainly consists of methane with small amounts of ethane, propane and butane.

Natural gas is one of the cleanest fossil fuels because it produces less smoke and fewer pollutants during combustion.
3. Coal
Coal is not a hydrocarbon but a carbon-rich fossil fuel. When coal is heated strongly in the absence of air (destructive distillation), valuable products such as coal gas, coal tar and coke are obtained.
  • Coal Gas: Used as industrial fuel.
  • Coal Tar: Source of aromatic hydrocarbons.
  • Coke: Used in metallurgy.
📌 Fractional Distillation of Petroleum
🗒️ 
Fractional Distillation of Crude Oil Separation of hydrocarbons based on boiling points Cooler (~20°C) Small molecules Low viscosity / High flammability Hotter (~350°C) Large molecules High viscosity / Low flammability Crude Oil Furnace 370°C Refinery Gas (LPG) < 20°C • C₁ – C₄ • Bottled gas, heating Petrol (Gasoline) 20°C – 70°C • C₅ – C₁₀ • Car fuel Naphtha 70°C – 160°C • C₈ – C₁₂ • Chemical production Kerosene (Paraffin) 160°C – 250°C • C₁₀ – C₁₆ • Jet fuel Diesel Oil 250°C – 300°C • C₁₄ – C₂₀ • Diesel engines, trains Fuel Oil 300°C – 370°C • C₂₀ – C₅₀ • Ships, power stations Lubricating Oil & Waxes 370°C – 400°C • C₂₀ – C₅₀ • Engine oils, candles Bitumen (Residue) > 400°C • C₇₀+ • Road surfacing, roofing
🛠️ Application
Applications of Hydrocarbons
  • Manufacture of plastics such as polyethylene, polypropylene and polystyrene.
  • Production of synthetic rubber.
  • Preparation of detergents and surfactants.
  • Manufacture of fertilizers.
  • Preparation of pharmaceuticals.
  • Production of paints and varnishes.
  • Manufacture of dyes and explosives.
  • Preparation of synthetic fibres like nylon and polyester.
  • Used as industrial solvents.
  • Raw material for petrochemical industries.
🗒️ Hydrocarbons As Petrochemical Feedstock
Hydrocarbons are converted into thousands of useful chemicals in petrochemical industries.
Hydrocarbon Product Manufactured
Ethene Polyethylene, Ethanol, Ethylene glycol
Propene Polypropylene, Isopropyl alcohol
Benzene Dyes, Medicines, Detergents
Toluene TNT, Solvents, Paints
Xylene Polyesters, Plastics
🤔 Did You Know?
Why are Hydrocarbons Important in Organic Chemistry?
Hydrocarbons are considered the parent compounds of organic chemistry because almost every organic compound can be obtained by replacing one or more hydrogen atoms with another atom or functional group.

Example:
  • \(\mathrm{CH_4}\)
    → Methane (Hydrocarbon)
  • \(\mathrm{CH_3Cl}\)
    → Chloromethane
  • \(\mathrm{CH_3OH}\)
    → Methanol
  • \(\mathrm{CH_3NH_2}\)
    → Methylamine
  • \(\mathrm{CH_3COOH}\)
    → Acetic acid
Hence, understanding hydrocarbons makes learning all later organic chemistry chapters much easier.
💡 Concept Summary
✏️ Solved Concept Example
Why is ethanol not classified as a hydrocarbon?
  1. 1
    Recall the definition of hydrocarbon.
  2. 2
    Identify the atoms present in ethanol.
  3. 3
    Compare with the definition.
Hydrocarbons contain only carbon and hydrogen atoms. Ethanol has molecular formula
\[ \mathrm{C_2H_5OH} \]
Since oxygen is also present, ethanol is an oxygen-containing organic compound and not a hydrocarbon.
Ethanol is not a hydrocarbon because it contains oxygen in addition to carbon and hydrogen.
⚡ Exam Tip
❌ Common Mistakes
  • Writing ethanol or methanol as hydrocarbons.
  • Considering coal itself as a hydrocarbon.
  • Confusing LPG with natural gas.
  • Writing petrol as a single compound instead of a mixture.
  • Forgetting that petroleum is a mixture of many hydrocarbons.
📋 CBSE Case Study (HOTS)

A petroleum refinery separates crude oil into different fractions. One fraction is used as LPG for cooking, another as petrol for automobiles, while a heavier fraction is used for road construction.

Questions

  1. Which process is used to separate petroleum fractions?
  2. Why do different fractions separate at different heights in the fractionating column?
  3. Name the fraction used in road construction.
  4. Why is methane considered a cleaner fuel than coal?

Answers

  1. Fractional distillation.
  2. Different boiling points of hydrocarbons.
  3. Bitumen.
  4. Methane burns more completely and produces fewer pollutants and less soot.
🗒️ Importnce
  • Definition of hydrocarbon is frequently asked in one-mark questions.
  • Natural sources and applications are common CBSE theory questions.
  • Fractional distillation of petroleum is important for competency-based questions.
  • Hydrocarbons act as the foundation for the complete Organic Chemistry syllabus of Class XI.
  • Memorize examples of fuels and polymers derived from hydrocarbons.
⛽

Classification of Hydrocarbons

📋
Table of Contents
9 sections
🗺️ Overview
Hydrocarbons are classified on the basis of the nature of the carbon-carbon (C–C) bond present in their molecules. The type of bonding determines their physical properties, chemical reactivity, methods of preparation, and industrial applications. Broadly, hydrocarbons are classified into three major categories:
  1. Saturated Hydrocarbons
  2. Unsaturated Hydrocarbons
  3. Aromatic Hydrocarbons

Quick Definition: The greater the number of multiple bonds in a hydrocarbon, the higher is its chemical reactivity. Saturated hydrocarbons are comparatively less reactive, whereas unsaturated hydrocarbons readily undergo addition reactions.
🎨 SVG Diagram
Classification Flow Chart
Hydrocarbons Saturated Only C–C single bonds Max H-atoms bound Unsaturated Double or triple bonds Reactivity: High Aromatic Special cyclic systems Delocalized π electrons Alkanes Open-chain or Aliphatic CₙH₂ₙ₊₂ Single Bonds (C–C) e.g., Methane, Propane Alkenes CₙH₂ₙ (C=C) Alkynes CₙH₂ₙ₋₂ (C≡C) Open-chain or Aliphatic e.g., Ethene, Ethyne Benzene Series Closed Ring (Cyclic) C₆H₆ Ring Base Resonance Stabilized e.g., Benzene, Toluene
📌 Saturated Hydrocarbons (Alkanes)
📘 Definition
Hydrocarbons containing only carbon-carbon single covalent bonds (C–C) are known as saturated hydrocarbons. Since every carbon atom forms the maximum possible number of single bonds, these compounds are said to be saturated with hydrogen atoms.
For open-chain saturated hydrocarbons (alkanes),
CnH2n+2
1
Example
Name Molecular Formula Structure
Methane
\(\mathrm{CH_4}\)
\(\mathrm{CH_4}\)
Ethane
\(\mathrm{C_2H_6}\)
\(\mathrm{CH_3-CH_3}\)
Propane
\(\mathrm{C_3H_8}\)
\(\mathrm{CH_3-CH_2-CH_3}\)
Butane
\(\mathrm{C_4H_{10}}\)
\(\mathrm{CH_3-CH_2-CH_2-CH_3}\)
🔷 Important Characteristics
  • Contain only single covalent bonds.
  • Carbon atoms are sp3 hybridized.
  • All carbon atoms exhibit tetrahedral geometry.
  • Undergo substitution reactions.
  • Comparatively less reactive.
📌 Unsaturated Hydrocarbons
📘 Definition
Hydrocarbons containing one or more carbon-carbon double bonds (C=C) or carbon-carbon triple bonds (C≡C) are called unsaturated hydrocarbons.

These compounds possess fewer hydrogen atoms than the corresponding alkanes because multiple bonds reduce the number of hydrogen atoms attached to carbon.
Alkenes
Hydrocarbons containing at least one carbon-carbon double bond.
General Formula:
\[\boxed{\mathrm{C_n H_{2n}}}\]
Examples:
  • \(\mathrm{CH_{2}=CH_{2}}\)
    (Ethene)
  • \(\mathrm{CH_{3}-CH=CH_{2}}\)
    (Propene)
Alkynes
Hydrocarbons containing at least one carbon-carbon triple bond.
General Formula:
\[\boxed{\mathrm{C_n H_{2n-2}}}\]
Examples:
  • \(\mathrm{HC \equiv CH}\)
    (Ethyne)
  • \(\mathrm{CH_3-C \equiv CH}\)
    (Propyne)
🔷 Characteristics
  • Contain multiple bonds.
  • More reactive than alkanes.
  • Readily undergo addition reactions.
  • Alkenes contain sp2 hybridized carbon.
  • Alkynes contain sp hybridized carbon.
📌 Aromatic Hydrocarbons (Arenes)
📘 Definition
Aromatic hydrocarbons are a special class of cyclic hydrocarbons containing one or more benzene rings with a delocalized cloud of
\(\pi\)
-electrons. Their unusual stability is known as aromaticity.

The word "aromatic" originally referred to pleasant-smelling compounds. However, in modern chemistry, aromatic compounds are identified by their electronic structure rather than their smell.
🔷 Characteristics
  • Contain one or more benzene rings.
  • Possess delocalized
    \(\pi\)
    -electrons.
  • Exceptionally stable.
  • Prefer substitution reactions over addition reactions.
  • Follow Hückel's Rule (
    \(4n+2\)
    \(\pi\)
    -electrons).
2
Example
Compound Molecular Formula
Benzene
\(\mathrm{C_6H_6}\)
Toluene
\(\mathrm{C_7H_8}\)
Naphthalene
\(\mathrm{C_{10}H_8}\)
⚖️ Comparison of Different Classes of Hydrocarbons
Property Alkanes Alkenes Alkynes Aromatic
Main Bond C–C C=C C≡C Delocalized Ring
General Formula
\(\mathrm{C_n H_{2n+2}}\)
\(\mathrm{C_n H_{2n}}\)
\(\mathrm{C_n H_{2n-2}}\)
 No common formula
Hybridisation sp3 sp2 sp sp2
Main Reaction Substitution Addition Addition Electrophilic Substitution
Relative Reactivity Low High Very High Moderate
✏️ Solved Concept Example
Classify the following compounds:
  1. \(\mathrm{CH_3CH_3}\)
  2. \(\mathrm{CH_2=CH_2}\)
  3. \(\mathrm{HC\equiv CH}\)
  4. \(\mathrm{C_6H_6}\)
  1. 1
    Identify the type of carbon-carbon bond.
  2. 2
    Assign the correct hydrocarbon class.
Compound Bond Present Classification
\(\mathrm{CH_3CH_3}\)
 Single Bond Saturated (Alkane)
\(\mathrm{CH_2=CH_2}\)
 Double Bond Alkene
\(\mathrm{HC\equiv CH}\)
 Triple Bond Alkyne
\(\mathrm{C_6H_6}\)
 Aromatic Ring Aromatic Hydrocarbon
⚡ Exam Tip
❌ Common Mistakes
  • Considering every cyclic compound as aromatic.
  • Writing the general formula of alkynes as
    \(\mathrm{C_{n}H_{2n}}\)
    .
  • Confusing alkenes with aromatic compounds because both contain
    \(\pi\)
    -bonds.
  • Assuming aromatic compounds undergo addition reactions like alkenes.
📋 CBSE Competency-Based Case Study (HOTS)

A refinery receives four hydrocarbon samples. Laboratory analysis reveals that Sample A contains only C–C single bonds, Sample B contains one C=C bond, Sample C contains one C≡C bond, and Sample D contains a benzene ring.

Questions

  1. Classify each sample.
  2. Which sample is expected to undergo bromine addition most readily?
  3. Which sample is expected to exhibit aromatic stability?
  4. Which sample generally undergoes substitution reactions instead of addition?

Answers

  1. A: Alkane, B: Alkene, C: Alkyne, D: Aromatic hydrocarbon.
  2. Sample C (Alkyne).
  3. Sample D.
  4. Sample D (Aromatic hydrocarbon).
⛽

Alkanes

📋
Table of Contents
17 sections
🗺️ Overview
Alkanes are saturated, open-chain (acyclic) hydrocarbons containing only carbon-carbon single covalent bonds (C–C) and carbon-hydrogen single covalent bonds (C–H). Since all carbon atoms are connected through single bonds, alkanes contain the maximum possible number of hydrogen atoms attached to carbon. Hence, they are called saturated hydrocarbons.

Alkanes are the simplest members of the hydrocarbon family and form the basis for studying Organic Chemistry. They are also known as the parent hydrocarbons because many other organic compounds are derived from them by replacing one or more hydrogen atoms with other atoms or functional groups.
📘 Definition
🤔 Did You Know?
Why are Alkanes Called Paraffins?
Alkanes are chemically much less reactive than most other classes of organic compounds because they contain only strong sigma (
\(\sigma\)
) bonds and do not possess reactive double or triple bonds.

For this reason, alkanes were earlier called Paraffins, which is derived from the Latin words:
  • Parum = Little
  • Affinis = Affinity
Thus, Paraffin literally means "little affinity", indicating that alkanes have very little tendency to react with common chemical reagents under ordinary conditions.

Alkanes generally do not react with dilute acids, dilute bases, oxidizing agents or reducing agents under normal conditions. However, under suitable conditions (heat, light or catalyst), they undergo reactions such as combustion, halogenation and cracking.
📌 Methane — The First Member of the Alkane Series
🗒️ Natural Occurrence Of Methane
  • Natural gas (major constituent)
  • Coal mines (called Firedamp)
  • Marshy areas (called Marsh Gas)
  • Biogas plants
  • Oil wells
  • Decomposition of organic matter by anaerobic bacteria
🔎 Board Fact
📌 Homologous Series of Alkanes
🔢 General Formula of Alkanes
📐 Derivation of the General Formula
Suppose an alkane contains
\(n\)
 carbon atoms.
  • The first and last carbon atoms each possess three hydrogen atoms.
  • Each intermediate carbon atom possesses two hydrogen atoms.
Therefore,
\[\begin{aligned}\text{Number of Hydrogen atoms}&=3+3+2(n-2)\\&=6+2n-4\\&=2n+2\end{aligned}\]
Hence,
\[\boxed{\mathrm{C_{n}H_{2n+2}}}\]
Verification
\(n\)
 Formula Obtained Compound
1
\(\mathrm{CH_4}\)
 Methane
2
\(\mathrm{C_2H_6}\)
 Ethane
3
\(\mathrm{C_3H_8}\)
 Propane
4
\(\mathrm{C_4H_{10}}\)
 Butane
🗒️ Structure Of Methane
The carbon atom in methane undergoes sp3 hybridisation. The four equivalent hybrid orbitals arrange themselves in space to minimize electron pair repulsion, giving methane a perfectly symmetrical tetrahedral geometry.
  • Hybridisation = sp3
  • Shape = Tetrahedral
  • Bond Angle =
    \(109.5^\circ\)
  • All four C–H bonds are identical.
🎨 SVG Diagram
Structure of Methane - Illustration
Methane (CH₄) — Structure & Geometry The Simplest Saturated Hydrocarbon (Alkane) 2D Structural (Lewis) Formula 4 single covalent C–H bonds in a flat representation C H H H H Valency fulfilled: C (4) + 4×H (1) Shares 4 electron pairs (8 octet electrons for C) 3D Tetrahedral Geometry Real spatial orientation (sp³ hybridization) 109.5° C H H H H Angle: 109.5° Shape: Tetrahedral Hybridization: sp³ Symmetry: Non-polar
🌟 Important Bond Lengths
🏷️ Properties
General Physical Properties of Alkanes
Properties
Color
Colourless and nearly odourless compounds.
Solubility
Insoluble in water but soluble in organic solvents.
Physical State
Lower alkanes are gases, middle members are liquids and higher alkanes are solids.
Boiling Point
Boiling point increases with increase in molecular mass.
Density
Density is lower than water.
Carbon Atoms Physical State
\(\mathrm{C_1-C_4}\)
 Gases
\(\mathrm{C_5-C_{17}}\)
 Liquids
\(\mathrm{C_{18}}\)
and above		 Solids
🔷 Chemical Characteristics of Alkanes
🔷 Characteristics
  • Contain only strong sigma bonds.
  • Least reactive among hydrocarbons.
  • Do not decolourise bromine water or alkaline KMnO4.
  • Undergo combustion to produce carbon dioxide and water.
  • Undergo free-radical substitution with halogens in the presence of sunlight or UV light.
  • Undergo cracking at high temperatures to produce smaller hydrocarbons.
✏️ Example
Solved Example
Determine the molecular formula of an alkane containing eight carbon atoms.
  1. 1
    Recall the general formula.
  2. 2
    Substitute
    \(n=8\)
    .
  1. General formula:
    \[\mathrm{C_{n}H_{2n+2}}\]
  2. Putting \(n=8\),
    \[\mathrm{H=2(8)+2=18}\]
  3. Therefore,
    \[\boxed{\mathrm{C_8H_{18}}}\]
Hence, the required alkane is Octane.
⚡ Exam Tip
❌ Common Mistakes
  • Writing the general formula as
    \(\mathrm{C_{n}H_{2n}}\)
    .
  • Confusing methane with methanol.
  • Assuming alkanes undergo addition reactions.
  • Incorrectly writing the bond angle as
    \(120^\circ\)
    .
  • Forgetting that all bonds in alkanes are sigma (
    \(\sigma\)
    ) bonds.
🗒️ CBSE Competency-Based Case Study (HOTS)

A fuel company is comparing methane, propane and octane for different applications. Scientists observe that methane is obtained from natural gas, propane is the major component of LPG and octane is an important constituent of petrol.

Questions

  1. Which of these compounds belongs to the alkane family?
  2. Why are all of them called saturated hydrocarbons?
  3. Write the molecular formula of propane using the general formula of alkanes.
  4. Why are alkanes called paraffins?

Answers

  1. All three are alkanes.
  2. They contain only carbon-carbon single bonds.
  3. \(\mathrm{C_3H_8}\)
    .
  4. Because they exhibit very little chemical reactivity under ordinary conditions ("little affinity").
⛽

Nomenclature and Isomerism of Alkanes

📋
Table of Contents
16 sections
🗺️ Overview
As the number of carbon atoms increases, the number of possible arrangements of carbon atoms also increases. Consequently, a single molecular formula may represent two or more different compounds having different structures. This phenomenon gives rise to isomerism, one of the most important concepts in Organic Chemistry.

The first three alkanes—methane (
\(\mathrm{CH_4}\)
), ethane (
\(\mathrm{C_2H_6}\)
), and propane (
\(\mathrm{C_3H_8}\)
)—have only one possible structure. However, beginning with butane (
\(\mathrm{C_4H_{10}}\)
), more than one structural arrangement becomes possible.
🗒️ Important Board Point
Isomerism starts from butane (
\(\mathrm{C_4H_{10}}\)
)
📌 Nomenclature of Alkanes
📘 Structural Isomerism
🗒️ Chain Isomerism
Structural isomers that differ in the arrangement or branching of the carbon skeleton are known as chain isomers. This type of isomerism is called chain isomerismbr>
The simplest example is butane (
\(\mathrm{C_4H_{10}}\)
).
Compound Condensed Structural Formula IUPAC Name
Straight Chain
\(\mathrm{CH_3-CH_2-CH_2-CH_3}\)
 Butane (n-Butane)
Branched Chain
\(\mathrm{CH_3-CH(CH_3)-CH_3}\)
 2-Methylpropane (Isobutane)
🎨 SVG Diagram
Representation of Chain Isomerism
n-Butane IUPAC: Butane | Formula: C4H10 Linear Alkane C C C C H H H H H H H H H H 2-Methylpropane Common: Isobutane | Formula: (CH3)3CH Branched Alkane C C C C H H H H H H H H H H
⚖️ Comparison of the Two Isomers
Property n-Butane 2-Methylpropane
Molecular Formula
\(\mathrm{C_4H_{10}}\)
\(\mathrm{C_4H_{10}}\)
Carbon Skeleton Straight Chain Branched Chain
Boiling Point Higher Lower
Reason Larger surface area Compact structure
📌 Classification of Carbon Atoms
📘 Primary (1°) Carbon Atom
📘 Secondary (2°) Carbon Atom
📘 Tertiary (3°) Carbon Atom
📘 Quaternary (4°) Carbon Atom
🎨 SVG Diagram
Classification of Carbon Atoms (Diagram)
Primary (1°) 1 Carbon / R-group Attached C R H H H e.g., Ethanol (CH₃CH₂OH) Secondary (2°) 2 Carbons / R-groups Attached C R¹ H H R² e.g., Isopropanol Tertiary (3°) 3 Carbons / R-groups Attached C R¹ R³ H R² e.g., Isobutane Quaternary (4°) 4 Carbons / R-groups Attached C R¹ R³ R⁴ R² e.g., Neopentane
📘 Alkyl Groups
✏️ Example
Solved Example
How many chain isomers are possible for
\(\mathrm{C_4H_{10}}\)
? Name them.
  1. 1
    Draw all possible carbon skeletons.
  2. 2
    Check whether the molecular formula remains unchanged.
There are two possible arrangements of four carbon atoms.
  1. Straight chain → Butane (n-Butane)
  2. Branched chain → 2-Methylpropane (Isobutane)

Therefore,
\(\mathrm{C_4H_{10}}\)
 has two chain isomers.
⚡ Exam Tip
❌ Common Mistakes
  • Writing structural isomers with different molecular formulae.
  • Confusing branched-chain compounds with cyclic compounds.
  • Assuming every compound with four carbons has only one structure.
  • Writing the alkyl group formula as
    \(\mathrm{C_{n}H_{2n+2}}\)
    instead of
    \(\mathrm{C_{n}H_{2n+1}}\)
    .
  • Identifying the central carbon in neopentane as tertiary instead of quaternary.
📋 CBSE Competency-Based Case Study (HOTS)

A chemist synthesizes two compounds, A and B, each having the molecular formula

\(\mathrm{C_4H_{10}}\)
. Compound A has a higher boiling point than compound B. Both compounds contain only carbon and hydrogen.

Questions

  1. What phenomenon is illustrated here?
  2. Name compounds A and B.
  3. Which compound is expected to have the straight carbon chain?
  4. Write the general formula of the alkyl group obtained from butane after removing one hydrogen atom.

Answers

  1. Chain isomerism (Structural isomerism).
  2. n-Butane and 2-Methylpropane.
  3. n-Butane.
  4. \(\mathrm{C_4H_9}\)
    .
⛽

Preparation of Alkanes

📋
Table of Contents
11 sections
🗺️ Overview
Although petroleum and natural gas are the major commercial sources of alkanes, laboratory and industrial methods are frequently employed to prepare pure alkanes of desired molecular masses. These methods are important not only from the examination point of view but also for understanding organic reaction mechanisms.

CBSE Important: Preparation of alkanes is one of the most frequently asked topics in Class XI examinations. Students should remember the reaction conditions, catalysts, products formed and limitations of each method.
📌 Preparation from Unsaturated Hydrocarbons (Catalytic Hydrogenation)
📘 Definition
Alkenes and alkynes are converted into alkanes by the addition of dihydrogen (
\(\mathrm{H_2}\)
) in the presence of finely divided metal catalysts such as Nickel (Ni), Palladium (Pd) or Platinum (Pt). This process is called catalytic hydrogenation.

Principle

The catalyst adsorbs hydrogen molecules on its surface and weakens the H–H bond. The activated hydrogen atoms then add across the carbon-carbon double or triple bond, converting it into a single bond.

Catalysts Used

Catalyst Reaction Conditions Remarks
Platinum (Pt) Room Temperature Highly efficient but expensive
Palladium (Pd) Room Temperature Fast catalyst
Nickel (Ni) Higher Temperature and Pressure Industrial catalyst
General Reaction
\[ \begin{aligned}\ce{\mathrm{C=C + H_2 \xrightarrow{Ni/Pd/Pt} C-C}\\\\ \mathrm{C\equiv C +2H_2 \xrightarrow{Ni/Pd/Pt} C-C}}\end{aligned} \]
1
Example
(a) Hydrogenation of Ethene
\[\mathrm{CH_2=CH_2 + H_2\xrightarrow{Ni/Pd/Pt}CH_3-CH_3}\]
2
Example
(b) Hydrogenation of Propene
\[\mathrm{CH_3-CH=CH_2+H_2\xrightarrow{Ni}CH_3-CH_2-CH_3}\]
3
Example
(c) Hydrogenation of Propyne
\[\mathrm{CH_3-C\equiv CH+2H_2\xrightarrow{Ni}CH_3-CH_2-CH_3}\]

Reaction Mechanism (Simplified)

  • Hydrogen molecules are adsorbed on the catalyst surface.
  • H–H bond dissociates into hydrogen atoms.
  • Unsaturated hydrocarbon is adsorbed.
  • Hydrogen atoms add across the multiple bond.
  • Alkane is released from the catalyst surface.
🛠️ Applications
  • Manufacture of saturated hydrocarbons.
  • Hardening of vegetable oils into vanaspati ghee.
  • Petroleum refining.
  • Petrochemical industry.
Unsaturated Reactant Alkene (C=C) or Alkyne (C≡C) C C R¹ H R² H sp² / sp Hybridized Carbon Center CATALYTIC HYDROGENATION H₂ (g) Metal Catalyst: Ni, Pd, or Pt Heterogeneous Surface Catalysis Stereochemistry: Syn Addition Saturated Product Alkane (C—C Single Bond) C C R¹ H H R² H H sp³ Hybridized Carbon Center
📌  Preparation from Alkyl Halides

Reduction of Alkyl Halides

Alkyl chlorides, bromides and iodides are reduced by nascent hydrogen (generated from zinc and dilute hydrochloric acid or other reducing agents) to form the corresponding alkanes.

Alkyl fluorides are generally not reduced easily because the C–F bond is extremely strong.
General Reaction
\[ \mathrm{\begin{aligned}\ce{R-X + H_2 &\longrightarrow R-H + HX\\ \small\text{where } X &=\mathrm{Cl,\;Br,\;I}\end{aligned}}} \]
1
Example
\[\mathrm{CH_3Cl+H_2\longrightarrow CH_4+HCl}\]
2
Example
\[\mathrm{C_2H_5Cl+H_2\longrightarrow C_2H_6+HCl}\]
3
Example
\[\mathrm{CH_3CH_2CH_2Cl+H_2\longrightarrow CH_3CH_2CH_3+HCl}\]
👁️
Observation Only the halogen atom is replaced by hydrogen; therefore, the carbon chain length remains unchanged.
📌 Wurtz Reaction
📘 Definition
When an alkyl halide is treated with sodium metal in dry ether, two alkyl groups combine to form a higher alkane. This reaction is called the Wurtz Reaction.

Dry ether is essential because sodium reacts violently with water. Moisture destroys the reaction.
General Reaction
\[ \boxed{\ce{\mathrm{2R-X+2Na\xrightarrow{Dry\ Ether}R-R+2NaX}}} \]
1
Example
Methane Series
\[\mathrm{2CH_3Br+2Na\xrightarrow{Dry\ Ether}CH_3-CH_3+2NaBr}\]
2
Example
Ethane Series
\[\mathrm{2C_2H_5Br+2Na\xrightarrow{Dry\ Ether}C_2H_5-C_2H_5+2NaBr}\]

Reaction Mechanism (Simplified)

  • Sodium removes halogen atoms.
  • Alkyl radicals are produced.
  • Two radicals combine to produce a higher alkane.
🛠️ Applications
  • Preparation of symmetrical alkanes.
  • Preparation of even-carbon alkanes.
  • Laboratory synthesis of higher hydrocarbons.

Limitations of Wurtz Reaction

  • Best suited for symmetrical alkanes.
  • Different alkyl halides produce mixtures.
  • Not suitable for preparing odd-carbon straight-chain alkanes.
  • Tertiary alkyl halides usually undergo elimination instead of coupling.
Alkyl Halide 2 Equivalents (Haloalkanes) R X + X R X = Cl, Br, or I (Halogen) WURTZ REACTION 2 Na in Dry Ether Solvent (Et₂O) Free Radical / Organometallic Mechanism C—C Bond Coupling (Dimerization) Symmetrical Alkane + Sodium Halide Salt R — R + 2 NaX e.g., 2 CH₃Cl → CH₃—CH₃ + 2 NaCl
📌 Preparation from Carboxylic Acids (Decarboxylation)
📘 Definition
The sodium salt of a carboxylic acid, when heated with soda lime (a mixture of sodium hydroxide and calcium oxide), produces an alkane having one carbon atom less than the original acid.

Removal of carbon dioxide from a carboxylic acid or its salt is called decarboxylation.

Composition of Soda Lime

  • Sodium hydroxide (
    \(\mathrm{NaOH}\)
    )
  • Calcium oxide (
    \(\mathrm{CaO}\)
    )
Calcium oxide keeps sodium hydroxide dry and porous, making the reaction more efficient.
General Reaction
\[ \boxed{\ce{\mathrm{RCOONa+NaOH\xrightarrow{CaO,\ \Delta}RH+Na_2CO_3}}} \]
9
Example
\[\mathrm{CH_3COONa+NaOH\xrightarrow{CaO,\ \Delta}CH_4+Na_2CO_3}\]
👁️
Important Observation One carbon atom is removed in the form of carbon dioxide.
Carboxylate Salt Product Carbon Atoms
\(\mathrm{CH_3COONa}\)
\(\mathrm{CH_4}\)
 2 → 1
\(\mathrm{C_2H_5COONa}\)
\(\mathrm{C_2H_6}\)
 3 → 2
\(\mathrm{C_3H_7COONa}\)
\(\mathrm{C_3H_8}\)
 4 → 3
📌 Kolbe's Electrolytic Method
📘 Definition
Electrolysis of an aqueous solution of the sodium or potassium salt of a carboxylic acid produces an alkane containing an even number of carbon atoms. This method is known as Kolbe's Electrolytic Method.
Anode Reaction
\[ \begin{aligned}&\ce{\mathrm{2RCOO^-\rightarrow 2R^\bullet+2CO_2+2e^-}}\\ &\scriptsize\text{The alkyl radicals combine:}\\ &\ce{\mathrm{R^\bullet+R^\bullet\rightarrow R-R}}\end{aligned} \]
Cathode Reaction
\[ \ce{\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}} \]
Overall Reaction
\[ \ce{\mathrm{2CH_3COONa+2H_2O\rightarrow CH_3-CH_3+2CO_2+H_2+2NaOH}} \]
Important Features
  • Produces symmetrical alkanes.
  • Even number of carbon atoms are obtained.
  • Carbon dioxide is evolved at the anode.
  • Hydrogen gas is liberated at the cathode.
⚖️ Comparison of Different Methods of Preparation of Alkanes
Method Starting Compound Main Reagent Carbon Chain
Hydrogenation Alkene / Alkyne
\(\mathrm{H_2}\)
, Ni/Pd/Pt		 Unchanged
Reduction Alkyl Halide Zn/HCl or Hydrogen Unchanged
Wurtz Reaction Alkyl Halide Na / Dry Ether Doubles
Decarboxylation Sodium Carboxylate Soda Lime One Carbon Less
Kolbe Electrolysis Sodium Carboxylate Electrolysis Doubles (Even Number)
✏️ Example
solved Example
Which method would you choose to prepare propane from propene?
  1. 1
    Identify the functional group.
  2. 2
    Select the suitable preparation method.
Propene contains one carbon-carbon double bond. Therefore, catalytic hydrogenation converts it into propane.
\[\mathrm{CH_3CH=CH_2+H_2\xrightarrow{Ni}CH_3CH_2CH_3}\]
Hence, catalytic hydrogenation is the appropriate method.
⚡ Exam Tip
❌ Common Mistakes
  • Writing only one molecule of alkyl halide in Wurtz reaction instead of two.
  • Forgetting that dry ether is essential in Wurtz reaction.
  • Assuming decarboxylation retains the same number of carbon atoms.
  • Confusing Kolbe electrolysis with decarboxylation.
  • Using alkyl fluorides in reduction reactions.
📋 CBSE Competency-Based Case Study (HOTS)

A petrochemical laboratory wants to prepare different alkanes. Sample A is prepared by adding hydrogen to an alkene using nickel catalyst. Sample B is prepared by heating sodium acetate with soda lime. Sample C is obtained by treating ethyl bromide with sodium metal in dry ether.

Questions

  1. Name the three preparation methods used.
  2. Which method decreases the carbon chain by one carbon atom?
  3. Which method is mainly used to prepare symmetrical higher alkanes?
  4. Why is dry ether used in the Wurtz reaction?

Answers

  1. Hydrogenation, Decarboxylation and Wurtz Reaction.
  2. Decarboxylation.
  3. Wurtz Reaction.
  4. Because sodium reacts violently with moisture; dry ether provides an anhydrous medium for the reaction.
⛽

Physical Properties of Alkanes

📋
Table of Contents
16 sections
🗺️ Overview
The physical properties of alkanes are mainly governed by their molecular size, shape, and the nature of the intermolecular forces acting between their molecules. Since alkanes contain only carbon-carbon (C–C) and carbon-hydrogen (C–H) covalent bonds, they are almost non-polar molecules.

Unlike ionic compounds or polar molecules, alkanes do not possess permanent positive and negative poles. Therefore, the only attractive forces acting between alkane molecules are the weak London dispersion forces (also called van der Waals forces).


Important Concept: The physical properties of alkanes depend mainly on the strength of intermolecular van der Waals (dispersion) forces rather than the strength of the C–C or C–H covalent bonds.
🤔 Why are Alkanes Non-Polar?
The electronegativity values of carbon and hydrogen are very close.
Element Electronegativity (Pauling Scale)
Carbon 2.55
Hydrogen 2.20

The electronegativity difference is only
\[2.55-2.20=0.35\]
Because of this very small difference, the C–H bond is almost non-polar. Moreover, the symmetrical arrangement of bonds in alkanes causes the small bond dipoles to cancel each other. Hence, alkanes behave as non-polar molecules.
🎨 SVG Diagram
MOLECULAR POLARITY - Illustration
NCERT CLASS 11 CHEMISTRY • MOLECULAR POLARITY IN HYDROCARBONS C — H Bond Dipole C (2.55) — H (2.20) δ⁻ δ⁺ Electronegativity Diff (Δχ = 0.35) → Very Weak SYMMETRY CANCELS DIPOLE Overall Hydrocarbon Net Dipole Moment (μnet) ≈ 0 D NON-POLAR Intermolecular Forces: Weak Van der Waals
📌 Intermolecular Forces in Alkanes
Since alkanes are non-polar molecules, they do not exhibit hydrogen bonding or permanent dipole-dipole attraction. The only attractive force present between alkane molecules is the London dispersion force (van der Waals force).
🔷 Characteristics of Van der Waals Forces
  • Weakest type of intermolecular force.
  • Present between all molecules.
  • Increase with increase in molecular size.
  • Increase with increase in molecular mass.
  • Increase with increase in surface area of the molecule.
The covalent bonds inside an alkane molecule are very strong, whereas the attractive forces between alkane molecules are weak.
📌 Physical State of Alkanes
🗒️ Consequences For Alkanes
Solvent Solubility of Alkanes Reason
Water Insoluble Water is polar.
Benzene Soluble Both are non-polar.
Carbon Tetrachloride (
\(\mathrm{CCl_4}\)
)		 Soluble Non-polar solvent.
Ether Soluble Organic solvent.
🤔 Why are Alkanes Insoluble in Water?
Water molecules are strongly held together by hydrogen bonding. Alkane molecules cannot form hydrogen bonds with water. Therefore, water molecules prefer to remain associated with one another instead of mixing with alkane molecules.
📌 Boiling Point of Alkanes
🗒️ Svg Image
NCERT CLASS 11 CHEMISTRY • PHYSICAL PROPERTIES OF ALKANES STEP 01 Carbon Chain Length (n) ↑ Increase in Molecular Mass (M) & Size CH₄ → C₂H₆ → C₃H₈ STEP 02 Molecular Contact Surface Area ↑ Stronger Intermolecular Van der Waals Forces London Dispersion Attraction ↑ RESULT Boiling Point (Tb) ↑ MORE ENERGY REQD. Higher Thermal Energy Needed for Vaporization
🔎 Effect of Branching on Boiling Point
🎨 SVG Diagram
Boiling Point - Illustration
NCERT CLASS 11 CHEMISTRY • ISOMERISM & BOILING POINT TRENDS UNBRANCHED ISOMER (e.g., n-Pentane) Straight Chain Extended Molecular Surface Large Surface Area of Contact Stronger Van der Waals Forces Higher Boiling Point (309.1 K) BRANCHING INCREASES Spherical Shape BRANCHED ISOMER (e.g., Neopentane) Branched Chain Compact Reduced Surface Area (Spherical Shape) Weaker Van der Waals Forces Lower Boiling Point (282.5 K)
🗒️ Melting Point Of Alkanes
The melting point generally increases with increase in molecular mass because larger molecules possess stronger intermolecular attractions.

However, unlike boiling point, the melting point does not increase uniformly because crystal packing differs among different alkane molecules.
🗒️ Density Of Alkanes
  • Density increases slightly with molecular mass.
  • All liquid alkanes are lighter than water.
  • Typical density ranges from 0.6–0.8 g mL-1.

Since alkanes are less dense than water, they float on the surface of water.
📝 Summary of Physical Properties
✏️ Example
Solved Example
Arrange the following compounds in increasing order of boiling point:
n-Pentane, 2-Methylbutane and 2,2-Dimethylpropane.
  1. 1
    Identify the degree of branching.
  2. 2
    Greater branching means lower boiling point.
Among the three compounds, 2,2-dimethylpropane is the most branched and n-pentane is the least branched.
\[\boxed{\mathrm{2,2\text{-}Dimethylpropane < 2\text{-}Methylbutane < n\text{-}Pentane }}\]
Thus, the boiling point increases as branching decreases.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming alkanes form hydrogen bonds with water.
  • Believing branched alkanes have higher boiling points than straight-chain alkanes.
  • Confusing intermolecular van der Waals forces with covalent C–C bonds.
  • Assuming all hydrocarbons dissolve in water.
  • Writing alkanes as polar molecules.
📋 CBSE Competency-Based Case Study (HOTS)

Questions

  1. Why does compound A have the highest boiling point?
  2. Which intermolecular force operates between alkane molecules?
  3. Which compound has the smallest surface area?
  4. Why are alkanes insoluble in water?

Answers

  1. Because it has the largest surface area and strongest van der Waals forces.
  2. London dispersion (van der Waals) forces.
  3. Compound C (highly branched).
  4. Because alkanes are non-polar whereas water is polar and hydrogen-bonded.
⛽

Chemical Properties of Alkanes

📋
Table of Contents
9 sections
🗺️ Overview
Alkanes are the least reactive members of the hydrocarbon family because they contain only strong carbon-carbon (C–C) and carbon-hydrogen (C–H) sigma (
\(\sigma\)
) bonds. These bonds possess high bond dissociation energies and are difficult to break under ordinary conditions.

Consequently, alkanes are generally unaffected by dilute acids, alkalis, oxidizing agents and reducing agents at room temperature. However, under suitable conditions such as heat, ultraviolet (UV) light, or in the presence of suitable catalysts, alkanes undergo several important reactions.
Major Chemical Reactions of Alkanes
  • Substitution reactions (Halogenation)
  • Combustion
  • Controlled oxidation
  • Pyrolysis (Cracking)
  • Isomerisation
  • Aromatisation (for higher alkanes)
📌 Substitution Reactions
📘 Definition
A reaction in which one or more hydrogen atoms of an alkane are replaced by another atom or group of atoms without disturbing the carbon skeleton is called a substitution reaction.

The most important substitution reaction of alkanes is halogenation.

Remember: Lower alkanes generally undergo halogenation but do not readily undergo nitration or sulphonation under ordinary laboratory conditions.

📘 Halogenation of Alkanes
Halogenation is the substitution reaction in which one or more hydrogen atoms of an alkane are replaced by halogen atoms such as chlorine or bromine.
Reaction Conditions
  • Ultraviolet light (
    \(h\nu\)
    )
  • Diffused sunlight
  • Heating at about 573–773 K
General Reaction
\[ \ce{\boxed{\mathrm{R-H+X_2\xrightarrow{h\nu\ /\ Heat}R-X+HX}}} \]
where
\(\mathrm{X=Cl,\ Br,\ I,\ F}\)


Successive Chlorination of Methane

During chlorination, hydrogen atoms of methane are replaced one after another by chlorine atoms.
  • 1
    Step 1
    \[\mathrm{CH_4+Cl_2\xrightarrow{h\nu}CH_3Cl+HCl}\]
  • 2
    Step 2
    \[\mathrm{CH_3Cl+Cl_2\xrightarrow{h\nu}CH_2Cl_2+HCl}\]
  • 3
    Step 3
    \[\mathrm{CH_2Cl_2+Cl_2\xrightarrow{h\nu}CHCl_3+HCl}\]
  • 4
    Step 4
    \[\mathrm{CHCl_3+Cl_2\xrightarrow{h\nu}CCl_4+HCl}\]

Products Formed

Compound IUPAC Name Common Name
\(\mathrm{CH_3Cl}\)
 Chloromethane Methyl chloride
\(\mathrm{CH_2Cl_2}\)
 Dichloromethane Methylene chloride
\(\mathrm{CHCl_3}\)
 Trichloromethane Chloroform
\(\mathrm{CCl_4}\)
 Tetrachloromethane Carbon tetrachloride

Relative Reactivity of Halogens

Different halogens react with alkanes at different rates.
\[\boxed{\mathrm{F_2>Cl_2>Br_2>I_2}}\]
Halogen Nature of Reaction
\(\mathrm{F_2}\)
 Very fast and explosive
\(\mathrm{Cl_2}\)
 Fast and useful
\(\mathrm{Br_2}\)
 Slow but selective
\(\mathrm{I_2}\)
 Very slow and reversible

Exam Point: Fluorination is extremely violent and difficult to control, whereas iodination is reversible and therefore does not proceed satisfactorily unless an oxidising agent is present.

Relative Reactivity of Hydrogen Atoms

Hydrogen atoms attached to different types of carbon atoms are replaced at different rates.
\[\boxed{3^\circ>2^\circ>1^\circ}\]
This order is due to the stability of the intermediate free radicals formed during the reaction.
Hydrogen Type Relative Reactivity
Tertiary Hydrogen Highest
Secondary Hydrogen Moderate
Primary Hydrogen Lowest

Iodination of Methane

Iodination is reversible because hydrogen iodide formed during the reaction drives the equilibrium backwards.
\[\mathrm{CH_4+I_2\rightleftharpoons CH_3I+HI}\]
To make the reaction proceed forward, hydrogen iodide is continuously oxidised by an oxidising agent such as iodic acid or nitric acid.
\[\mathrm{HIO_3+5HI\rightarrow 3I_2+3H_2O}\]

Oxidising Agents Used

  • \(\mathrm{HIO_3}\)
  • \(\mathrm{HNO_3}\)

Free Radical Chain Mechanism of Halogenation

Halogenation of alkanes proceeds through a free radical chain mechanism. It consists of three important stages:
  1. Initiation
  2. Propagation
  3. Termination
🔄 Free Radical Chain Mechanism of Halogenation
  • 1
    Initiation
    Definition
    The reaction is initiated by homolysis of chlorine molecule in the presence of light or heat.

    Since the Cl–Cl bond is weaker than C–C and C–H bonds, it breaks most easily.
    \[\boxed{\mathrm{Cl-Cl \xrightarrow[\text{homolysis}]{h\nu} \overset{\bullet}{\mathrm{C}}\mathrm{H_3} +\underset{\text{Chlorine free radicals}}{\mathrm{Cl}}}}\]
    Homolytic Fission
    Each atom takes away one electron from the shared pair, producing free radicals.
  • 2
    Propagation
    This stage is responsible for the formation of the main products. The free radicals produced in the initiation step continuously regenerate new free radicals, thereby sustaining the chain reaction.
    Propagation Step 1
    \[\mathrm{CH_4+\overset{+}Cl\rightarrow \overset{+}{C}H_3+H-Cl}\]
    A chlorine radical abstracts one hydrogen atom from methane to form a methyl radical.
    Propagation Step 2
    \[\mathrm{\overset{\bullet}{C}H_3+Cl_2\rightarrow CH_3Cl+\overset{\bullet}Cl}\]
    The newly formed chlorine radical again attacks another methane molecule, thereby continuing the chain reaction.
    Formation of Higher Chlorinated Products
    \[\mathrm{CH_3Cl+\overset{\bullet}Cl\rightarrow \overset{\bullet}CH_2Cl+HCl }\]
    \[\mathrm{\overset{\bullet}{C}H_2Cl+Cl_2\rightarrow CH_2Cl_2+\overset{\bullet}Cl}\]
    Similarly, trichloromethane and tetrachloromethane are formed by repeated substitution.
  • 3
    Termination
    Termination occurs when two free radicals combine together. Since free radicals are consumed, the chain reaction stops.
    Possible Termination Steps

    (a)

    \[\mathrm{\overset{\bullet}Cl+\overset{\bullet}Cl\rightarrow Cl_2}\]

    (b)

    \[\mathrm{\overset{\bullet}{C}H_3+\overset{\bullet}{C}H_3\rightarrow C_2H_6}\]

    (c)

    \[\mathrm{\overset{\bullet}{C}H_3+\overset{\bullet}Cl\rightarrow CH_3Cl}\]
    Important Observation:
    Formation of ethane (
    \(\mathrm{C_2H_6}\)
    )     as a by-product during chlorination of methane is explained by the combination of two methyl free radicals in the termination step.
🌟 Important Features of Halogenation
⚡ Quick Revision Table
Property Observation
Reaction Type Substitution
Mechanism Free Radical Chain
Initiation Homolysis of Cl₂
Propagation Formation of CH₃Cl and new radicals
Termination Combination of free radicals
Halogen Reactivity
\(\mathrm{F_2>Cl_2>Br_2>I_2}\)
Hydrogen Reactivity
\(3^\circ>2^\circ>1^\circ\)
✏️ Example
Solved Example
Why is ethane formed as a by-product during the chlorination of methane?
  1. 1
    Recall the free radical mechanism.
  2. 2
    Identify the termination step.
During propagation, methyl free radicals (
\(\mathrm{\overset{\bullet}{C}{C}H_3}\)
) are produced. Occasionally, two methyl radicals combine together instead of reacting with chlorine.
\[\mathrm{\overset{\bullet}{C}H_3+\overset{\bullet}{C}CH_3\rightarrow C_2H_6}\]
Therefore, ethane is formed as a by-product during the termination step.
⚡ Exam Tip
❌ Common Mistakes
  • Writing heterolytic cleavage instead of homolytic cleavage during initiation.
  • Using sunlight in the termination step instead of the initiation step.
  • Confusing substitution reactions with addition reactions.
  • Writing iodination as an irreversible reaction.
  • Ignoring the formation of ethane as a by-product.
📋 CBSE Competency-Based Case Study (HOTS)

A laboratory exposes methane and chlorine gas to ultraviolet light. Initially, chloromethane is formed. As the reaction continues, dichloromethane, chloroform and carbon tetrachloride are also produced. A small amount of ethane is detected among the products.

Questions

  1. What type of reaction is taking place?
  2. Which bond undergoes homolytic cleavage during initiation?
  3. Why is ethane formed?
  4. Arrange the halogens in decreasing order of reactivity toward alkanes.

Answers

  1. Free radical substitution (halogenation).
  2. The Cl–Cl bond.
  3. Because two methyl free radicals combine during the termination step.
  4. \(\mathrm{F_2>Cl_2>Br_2>I_2}\)
    .
⛽

Combustion of Alkanes

📋
Table of Contents
16 sections
📘 Definition
📘 Complete Combustion
📘 Definition
When an alkane burns in the presence of sufficient oxygen, complete oxidation takes place.
General Equation
\[ \ce{\boxed{\mathrm{C_{n}H_{2n+2}+\left(\frac{3n+1}{2}\right)O_2\rightarrow nCO_2+(n+1)H_2O+Heat}}} \]
📐 Derivation of the General Combustion Equation
Consider an alkane having molecular formula
\[\mathrm{C_{n}H_{2n+2}}\]
  • There are
    \(n\)
     carbon atoms, therefore
    \(nCO_2\)
     molecules are formed.
  • There are
    \(2n+2\)
     hydrogen atoms, therefore
    \((n+1)H_2O\)
     molecules are formed.
  • Total oxygen atoms required are
    \[2n+(n+1)=3n+1\]
Since each oxygen molecule contains two oxygen atoms,
\[\boxed{\frac{3n+1}{2}}\]
molecules of oxygen are required.
📌 Combustion of Methane
✏️ Examples of Complete Combustion
Ethane
\[\mathrm{2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O}\]
Propane
\[\mathrm{C_3H_8+5O_2\rightarrow 3CO_2+4H_2O}\]
Butane
\[\mathrm{2C_4H_{10}+13O_2 \rightarrow 8CO_2+10H_2O}\]
🤔 Did You Know?
Why are Alkanes Used as Fuels?
Alkanes are excellent fuels because they possess high calorific values and undergo highly exothermic combustion reactions.
Reason Explanation
High Heat Output Large amount of energy is released.
Easy Availability Obtained from petroleum and natural gas.
Clean Burning Complete combustion produces only carbon dioxide and water.
High Energy Density Suitable for domestic and industrial use.
🛠️ Common Fuel Applications
  • Natural gas (Methane)
  • LPG (Propane and Butane)
  • Petrol
  • Diesel
  • Aviation fuel
  • Kerosene
📌 Incomplete Combustion of Alkanes
📎 Side Note
Carbon Black (Soot)
📘 Definition
The finely divided black carbon produced during incomplete combustion is called carbon black.

🛠️ Industrial Applications
  • Manufacture of printer ink.
  • Printing ink.
  • Black pigments.
  • Tyre industry.
  • Rubber reinforcement.
  • Filters.
  • Carbon electrodes.
  • Paints and coatings.
🗒️ Remember
Remember: Carbon black is produced deliberately in industries by controlled incomplete combustion of hydrocarbons.
⚖️ Complete vs Incomplete Combustion
Property Complete Combustion Incomplete Combustion
Oxygen Supply Sufficient Insufficient
Main Products
\(\mathrm{CO_2}\)
and
\(\mathrm{H_2O}\)
\(\mathrm{CO}\)
, Carbon (Soot),
\(\mathrm{H_2O}\)
Flame Blue, non-luminous Yellow, smoky
Heat Produced Maximum Less
Pollution Low High
🚨 Environmental Aspects of Combustion
🚧 Caution
  • Complete combustion produces carbon dioxide, a greenhouse gas.
  • Incomplete combustion produces poisonous carbon monoxide.
  • Soot particles contribute to air pollution and respiratory diseases.
  • Efficient combustion reduces fuel wastage and environmental pollution.
⚡ Quick Revision
Feature Observation
Reaction Type Oxidation
Nature Exothermic
Main Products
\(\mathrm{CO_2}\)
and
\(\mathrm{H_2O}\)
Heat Released Large amount
Fuel Value Very High
Incomplete Combustion Produces CO and Carbon (Soot)
Flame Blue (Complete), Yellow (Incomplete)
✏️ Example
Solved Concept Example
Write the balanced equation for the complete combustion of propane.
  1. 1
    Write molecular formula of propane.
  2. 2
    Apply the general combustion equation.
  3. 3
    Balance oxygen atoms.
Propane has the formula
\[\mathrm{C_3H_8}\]
Applying the combustion reaction,
\[\boxed{\mathrm{C_3H_8+5O_2\rightarrow 3CO_2+4H_2O}}\]
⚡ Exam Tip
❌ Common Mistakes
  • Writing carbon monoxide as the product of complete combustion.
  • Confusing carbon black with carbon dioxide.
  • Forgetting that combustion is highly exothermic.
  • Balancing oxygen atoms incorrectly in combustion equations.
  • Assuming yellow flame indicates complete combustion.
📋 CBSE Competency-Based Case Study (HOTS)

A laboratory compares two burners. Burner A receives sufficient oxygen and produces a blue flame, whereas Burner B receives limited oxygen and burns with a yellow smoky flame. Scientists detect carbon black around Burner B.

Questions

  1. Which burner undergoes complete combustion?
  2. Why is carbon black formed in Burner B?
  3. Name the poisonous gas that may also be produced during incomplete combustion.
  4. Why are alkanes widely used as fuels?

Answers

  1. Burner A.
  2. Because oxygen is insufficient, leading to incomplete combustion.
  3. Carbon monoxide (
    \(\mathrm{CO}\)
    ).
  4. Because they have high calorific values and release a large amount of heat on combustion.
⛽

Controlled Oxidation of Alkanes

📋
Table of Contents
4 sections
📘 Definition
⚖️ Difference from Combustion:
  • Complete combustion converts alkanes completely into carbon dioxide and water.
  • Controlled oxidation stops the oxidation at an intermediate stage, producing useful organic compounds.
✏️ Example
1
Example
(a) Formation of Methanol
\[\boxed{\mathrm{2CH_4+O_2\xrightarrow[\text{High Pressure}]{Catalyst}2CH_3OH}}\]
2
Example
(b) Formation of Formaldehyde (Methanal)
\[\boxed{\mathrm{CH_4+O_2\xrightarrow{Catalyst}HCHO+H_2O}}\]
3
Example
(c) Formation of Acetic Acid
\[\boxed{\mathrm{2CH_3CH_3+3O_2\xrightarrow{Catalyst}2CH_3COOH+2H_2O}}\]
🌟 Industrial Importance
⛽

Isomerisation of Alkanes

📋
Table of Contents
5 sections
📘 Definition
🗒️ Reaction Conditions
  • Anhydrous aluminium chloride (
    \(\mathrm{AlCl_3}\)
    )
  • Hydrogen chloride gas (
    \(\mathrm{HCl}\)
    )
  • Heating
⚗️ General Reaction
\[ \ce{\boxed{\mathrm{n\text{-}Alkane\xrightarrow[\mathrm{HCl}]{Anhydrous\ AlCl_3}Branched\ Alkane}}} \]
✏️ Example
n-Butane
\[ \ce{\mathrm{CH_3CH_2CH_2CH_3\xrightarrow{AlCl_3/HCl}(CH_3)_3CH}} \]
n-Pentane
\[ \ce{\mathrm{CH_3(CH_2)_3CH_3\xrightarrow{AlCl_3/HCl}2\text{-}Methylbutane}} \]
🌟 Importance
🎨 SVG Diagram
Isomerisation of Alkanes
NCERT CLASS 11 CHEMISTRY • ISOMERISATION OF ALKANES REACTANT Straight Chain n-Alkane NCERT Example: n-Hexane (C₆H₁₄) CH₃-(CH₂)₄-CH₃ Anhyd. AlCl₃ / HCl 573 K, 35 atm Isomerisation PRODUCTS Branched Alkanes Major + Minor Isomers 2-Methylpentane CH₃-CH(CH₃)-(CH₂)₂-CH₃ 3-Methylpentane CH₃-CH₂-CH(CH₃)-CH₂-CH₃
⛽

Aromatisation (Reforming)

📋
Table of Contents
4 sections
📘 Definition
📌 Reaction Conditions
✏️ Example
\[ \ce{\boxed{\mathrm{C_6H_{14}\xrightarrow[\mathrm{773\,K}]{Cr_2O_3/Al_2O_3}C_6H_6+4H_2}}} \]
🌟 Industrial Importance
⛽

Reaction of Methane with Steam (Steam Reforming)

📋
Table of Contents
3 sections
📘 Definition
⚗️ Reaction
\[ \ce{\boxed{\mathrm{CH_4+H_2O\xrightarrow[\mathrm{1273\,K}]{Ni}CO+3H_2}}} \]
🌟 Importance
⛽

Pyrolysis (Cracking)

📋
Table of Contents
10 sections
📘 Definition
📌 General Features
✏️ Example

Dehydrogenation

\[ \ce{\mathrm{C_6H_{14}\rightarrow C_6H_{12}+H_2}} \]

Cracking into Alkane and Alkene

\[ \ce{\mathrm{C_6H_{14}\rightarrow C_4H_8+C_2H_6}} \]

Formation of Multiple Smaller Hydrocarbons

\[ \ce{\mathrm{C_6H_{14}\rightarrow C_3H_6+C_2H_4+CH_4}} \]
🗒️ Industrial Cracking Of Petroleum
📘 Definition
Pyrolysis is extensively used in petroleum refineries to convert heavy petroleum fractions into lighter and more valuable fuels such as petrol (gasoline), LPG and petrochemical feedstocks.
1
Example
Dodecane, an important constituent of kerosene, undergoes catalytic cracking.
\[ \ce{\boxed{\mathrm{\underset{Dodecane}{C_{12}H_{26}}\xrightarrow[\mathrm{973\,K}]{Pt/Pd/Ni}\underset{Heptane}{C_7H_{16}}+\underset{Pentene}{C_5H_{10}}}}} \]
Thus,
  • Heptane is produced.
  • Pentene is produced.
🌟 Importance
⚖️ Comparison of Important Reactions of Alkanes
Reaction Main Product Importance
Controlled Oxidation Alcohols, Aldehydes, Acids Industrial chemicals
Isomerisation Branched Alkanes Improves petrol quality
Aromatisation Benzene and Homologues Petroleum refining
Steam Reforming CO + H₂ Hydrogen manufacture
Pyrolysis Lower Alkanes + Alkenes Cracking of petroleum
✏️ Example
Solved Example
Why is pyrolysis important in petroleum refining?
  1. 1
    Recall what pyrolysis does.
  2. 2
    Identify its industrial advantage.
Pyrolysis converts heavy petroleum fractions into lighter hydrocarbons such as petrol, LPG and alkenes. Since lighter fuels are more valuable and widely used, cracking increases the economic value of crude petroleum.
⚡ Exam Tip
❌ Common Mistakes
  • Writing carbon dioxide as the product of controlled oxidation.
  • Confusing aromatisation with isomerisation.
  • Assuming pyrolysis occurs in the presence of air.
  • Forgetting that steam reforming requires a nickel catalyst and high temperature.
  • Believing branched alkanes have lower octane numbers than straight-chain alkanes.
📋 CBSE Competency-Based Case Study (HOTS)

A petroleum refinery processes normal hexane, methane and dodecane in three separate reactors. Reactor A converts hexane into benzene, Reactor B converts methane into carbon monoxide and hydrogen, while Reactor C converts dodecane into heptane and pentene.

Questions

  1. Name the reactions occurring in Reactors A, B and C.
  2. Which reaction is used for industrial manufacture of hydrogen gas?
  3. Which reaction improves petrol quality by producing aromatic hydrocarbons?
  4. Why is cracking economically important?

Answers

  1. Aromatisation, Steam Reforming and Pyrolysis (Cracking).
  2. Steam reforming of methane.
  3. Aromatisation (Reforming).
  4. Because it converts heavy petroleum fractions into lighter, more valuable fuels and petrochemical feedstocks.
⛽

Conformations of Alkanes

📋
Table of Contents
22 sections
🗺️ Overview
One of the unique characteristics of alkanes is the presence of carbon-carbon single (σ) bonds. Since a sigma bond is formed by the head-on overlap of atomic orbitals, its electron cloud is cylindrically symmetrical about the internuclear axis. This symmetry allows one carbon atom to rotate with respect to the other without breaking the C–C bond.

As a result of this rotation, a large number of different three-dimensional arrangements of atoms are possible. These arrangements differ only in the orientation of atoms in space and not in their connectivity.
📘 Definition
🗒️ Origin Of Conformations
The carbon-carbon sigma bond permits rotation because the electron density is uniformly distributed around the bond axis.

Therefore,
  • No bond is broken during rotation.
  • No atoms are added or removed.
  • Only the orientation of attached atoms changes.
Consequently, every alkane containing one or more C–C single bonds can exist in an infinite number of conformations.
📌 Is Rotation Completely Free?
📎 Torsional Strain
The weak repulsive interaction between electron clouds of adjacent bonds during rotation about a sigma bond is called torsional strain.
🗒️ Conformations Of Ethane
Ethane (
\(\mathrm{C_2H_6}\)
) is the simplest alkane that exhibits conformational isomerism.

Each carbon atom is sp3 hybridised and bonded to three hydrogen atoms.

If one carbon atom is kept fixed while the other rotates about the C–C bond, an infinite number of conformations are obtained.

Among these infinite conformations, two are particularly important:
  • Staggered Conformation
  • Eclipsed Conformation
All other conformations lying between these two extremes are called Skew Conformations.
📘 Staggered Conformation
🔷 Characteristics
🔷 Characteristics
  • Maximum separation between hydrogen atoms.
  • Minimum electron cloud repulsion.
  • Minimum torsional strain.
  • Lowest potential energy.
  • Most stable conformation.
📘 Eclipsed Conformation
🔷 Characteristics
🔷 Characteristics
  • Maximum overlap of electron clouds.
  • Maximum torsional strain.
  • Highest potential energy.
  • Least stable conformation.
📘 Skew Conformation
📘 Newman Projection
🎨 SVG Diagram
Newman Projection
NCERT CLASS 11 CHEMISTRY • NEWMAN PROJECTIONS OF ETHANE (C₂H₆) MOST STABLE CONFORMER Staggered Conformation H H H H H H Dihedral Angle (θ = 60°) • Min. Torsional Strain MAXIMUM STABILITY (12.5 kJ/mol Lower Energy) 60° ROTATION C—C Single Bond LEAST STABLE CONFORMER Eclipsed Conformation H H H H H H Dihedral Angle (θ = 0°) • Max. Torsional Strain MINIMUM STABILITY (Maximum Electron Repulsion)
📘 Sawhorse Projection
🎨 SVG Diagram
Sawhorse Projection
NCERT CLASS 11 CHEMISTRY • SAWHORSE PROJECTIONS OF ETHANE (C₂H₆) MOST STABLE CONFORMER Staggered Conformation H H H H H H Dihedral Angle (θ = 60°) • Min. Torsional Strain MAXIMUM STABILITY (Anti-Parallel Bonds) 60° ROTATION C—C Single Bond LEAST STABLE CONFORMER Eclipsed Conformation H H H H H H Dihedral Angle (θ = 0°) • Max. Torsional Strain MINIMUM STABILITY (Parallel Eclipsed Bonds)
⚖️ Relative Stability of Conformations
The stability of different conformations depends upon the magnitude of torsional strain.
Why is the Staggered Form More Stable?
  • Electron clouds are farthest apart.
  • Repulsive interaction is minimum.
  • Potential energy is minimum.
  • Torsional strain is least.
Why is the Eclipsed Form Less Stable?
  • Electron clouds overlap.
  • Repulsive interaction increases.
  • Torsional strain becomes maximum.
  • Potential energy increases.
Important Result
\[\boxed{\text{Staggered} > \text{Skew} > \text{Eclipsed}}\]
(in order of stability)
📌 Dihedral (Torsional) Angle
📌 Potential Energy Diagram of Ethane Rotation
⚖️ Comparison of Newman and Sawhorse Projections
Feature Newman Projection Sawhorse Projection
View Direction Along C–C Bond Oblique View
Front Carbon Point Lower Carbon
Rear Carbon Circle Upper Carbon
Most Useful For Comparing Conformations Visualising Molecular Geometry
👁️ Important Observations
✏️ Example
Solved Example
Why is the staggered conformation of ethane more stable than the eclipsed conformation?
  1. 1
    Compare electron cloud repulsions.
  2. 2
    Compare torsional strain.
In the staggered conformation, adjacent C–H bonds are separated by the maximum possible distance, so electron cloud repulsions are minimum. Consequently, torsional strain and potential energy are minimum, making it the most stable conformation. In the eclipsed conformation, the bonds overlap, increasing repulsion and potential energy.
⚡ Exam Tip
❌ Common Mistakes
  • Confusing conformations with structural isomers.
  • Assuming C–C bonds break during rotation.
  • Writing the eclipsed form as the most stable conformation.
  • Confusing torsional strain with angle strain.
  • Drawing Newman projections with incorrect bond angles.
📋 CBSE Competency-Based Case Study (HOTS)

A student studies the rotation of ethane around its carbon-carbon bond using molecular models. He observes that at one orientation, hydrogen atoms are farthest apart, while at another orientation they are directly aligned behind one another.

Questions

  1. Name the two conformations observed.
  2. Which conformation has minimum torsional strain?
  3. What is the approximate energy difference between these two conformations?
  4. Why are conformers of ethane not isolated under ordinary conditions?

Answers

  1. Staggered and Eclipsed conformations.
  2. Staggered conformation.
  3. Approximately
    \(12.5\ \mathrm{kJ\ mol^{-1}}\)
    .
  4. Because the rotational energy barrier is very small and thermal energy at room temperature allows rapid interconversion between conformers.
⛽

Alkenes

📋
Table of Contents
18 sections
🗺️ Overview
Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond (C=C). The presence of the double bond makes alkenes much more reactive than alkanes because one of the two bonds is a relatively weak pi (
\(\pi\)
) bond, which can be broken easily during chemical reactions.

Alkenes are among the most important organic compounds because they serve as starting materials for the manufacture of plastics, alcohols, synthetic rubber, detergents, pharmaceuticals and numerous industrial chemicals.
📘 Definition
🔢 Formula
General Formula of Alkenes
For open-chain alkenes containing a single double bond, the general molecular formula is
\[ \ce{\boxed{\mathrm{C_{n}H_{2n}}}} \]
📐 Derivation of the General Formula
An alkane containing
\(n\)
carbon atoms has the molecular formula
\[\mathrm{C_{n}H_{2n+2}}\]
Formation of one carbon-carbon double bond removes two hydrogen atoms.
\[\mathrm{C_{n}H_{2n+2}-H_2=C_{n}H_{2n}}\]
Therefore,
\[\boxed{\mathrm{C_{n}H_{2n}}}\]
🤔 Did You Know?
Why are Alkenes Called Olefins?
Alkenes are also known as olefins. The word olefin is derived from the Latin words:
  • Oleum = Oil
  • Facere = To make
The name originated because the first member of the alkene family, ethene (
\(\mathrm{C_2H_4}\)
), reacts with chlorine to form an oily liquid called 1,2-dichloroethane (ethylene dichloride).
\[\boxed{\mathrm{CH_2=CH_2+Cl_2\rightarrow CH_2Cl-CH_2Cl}}\]
Board Fact: "Olefin" literally means "oil-forming compound."
📌 Structure of the Carbon-Carbon Double Bond
🗒️ Formation Of The Pi Bond
Each carbon atom possesses one unhybridised 2p orbital perpendicular to the plane of the molecule.
These two parallel p-orbitals overlap sideways (laterally) to form the pi bond.
Bond enthalpy:
\[\boxed{284\ \mathrm{kJ\ mol^{-1}}}\]
🎨 SVG Diagram
Structure of Ethene
NCERT CLASS 11 CHEMISTRY • STRUCTURE OF ALKENES (ETHENE, C₂H₄) ORBITAL OVERLAP MODEL C=C Double Bond Formation π-cloud (Sideways Overlap) π-cloud (Sideways Overlap) σ bond C C H H H H 1 σ Bond (sp²-sp² axial) | 1 π Bond (unhybridized 2p-2p lateral) GEOMETRY & BOND METRICS Trigonal Planar Hybridization of Carbon: sp² (3 σ bonds + 1 π bond) H—C—H & H—C=C Bond Angles: 116.6° & 121.7° (≈ 120°) C=C Bond Length & Enthalpy: 134 pm | 681 kJ mol⁻¹ Spatial Orientation: Planar (All 6 atoms in 1 plane) RESTRICTED ROTATION C=C Double Bond Prevents Free Rotation
🔎 Hybridisation and Geometry
⚖️ Comparison
Bond Lengths and Bond Enthalpies
Bond Bond Length Bond Enthalpy
C–C Single Bond 154 pm 348 kJ mol-1
C=C Double Bond 134 pm 681 kJ mol-1
σ Bond in Double Bond Part of C=C 397 kJ mol-1
π Bond Part of C=C 284 kJ mol-1
👁️ Important Observation
🤔 Why are Alkenes More Reactive than Alkanes?
The presence of the weak pi bond makes alkenes chemically much more reactive than alkanes.
Reason
  • The π bond is weaker than the σ bond.
  • Its electrons lie above and below the molecular plane.
  • These electrons are more exposed and loosely held.
  • Electrophiles can easily attack the π electron cloud.

Key Concept: The sigma bond usually remains intact during reactions, whereas the pi bond breaks easily and is replaced by two new sigma bonds.
📌 Electrophilic Nature of Reactions
🗺️ Overview
Because the pi bond is rich in electron density, alkenes behave as electron-rich molecules.

Hence, they are readily attacked by electron-deficient species called electrophiles.
📘 Definition
An electrophile is an electron-deficient atom, ion or molecule that seeks electrons and accepts an electron pair during a chemical reaction.
1
Example
  • \(\mathrm{H^+}\)
  • \(\mathrm{Br^+}\)
  • \(\mathrm{Cl^+}\)
  • \(\mathrm{NO_2^+}\)
Therefore, the characteristic reaction of alkenes is electrophilic addition.
🎨 SVG Diagram
Electrophile
NCERT CLASS 11 CHEMISTRY • ELECTROPHILES (ELECTRON-LOVING SPECIES) REACTION MECHANISM CONCEPT Electron Pair Acceptance (Lewis Acids) Nu⁻ Nucleophile (Electron-Rich) e⁻ Pair Transfer E+ Electrophile (Vacant Orbital) Attacks Electron-Dense Sites (Negative Ions or π-Bonds) NCERT CLASSIFICATION Types of Electrophiles 1. Positively Charged Species Possess positive charge and vacant orbitals: H⁺, Cl⁺, Br⁺, NO₂⁺, R₃C⁺ (Carbocation) 2. Neutral Molecules (Incomplete Octet) Neutral species with incomplete central octet: BF₃, AlCl₃, FeCl₃, SO₃, :CH₂ (Carbene) KEY DEFINITION Reagents that ACCEPT an Electron Pair
⚖️ Comparison Between Alkanes and Alkenes
Property Alkanes Alkenes
Nature Saturated Unsaturated
Main Bond C–C C=C
General Formula
\(\mathrm{C_{n}H_{2n+2}}\)
\(\mathrm{C_{n}H_{2n}}\)
Hybridisation sp3 sp2
Geometry Tetrahedral Trigonal Planar
Main Reaction Substitution Addition
Relative Reactivity Low High
🌟 Importance of the Pi Bond
✏️ Example
Solved Example
Why are alkenes more reactive than alkanes although the C=C bond is stronger than the C–C bond?
  1. 1
    Compare σ and π bonds.
  2. 2
    Identify which bond breaks during reactions.
Although the overall bond enthalpy of the carbon-carbon double bond is greater than that of a single bond, the double bond consists of one strong σ bond and one comparatively weak π bond. During chemical reactions, only the weak π bond breaks, while the σ bond remains intact. Therefore, alkenes react more readily than alkanes.
⚡ Exam Tip
❌ Common Mistakes
  • Writing the general formula of alkenes as
    \(\mathrm{C_{n}H_{2n+2}}\)
    .
  • Assuming both σ and π bonds break during reactions.
  • Confusing electrophiles with nucleophiles.
  • Drawing tetrahedral geometry around double-bonded carbon atoms.
  • Assuming free rotation is possible around a carbon-carbon double bond.
📋 CBSE Competency-Based Case Study (HOTS)

A petrochemical plant produces ethene from petroleum. Scientists observe that ethene readily reacts with bromine, hydrogen chloride and chlorine, whereas ethane does not react under similar conditions.

Questions

  1. Why is ethene more reactive than ethane?
  2. Which bond is broken during most reactions of ethene?
  3. Why are alkenes called olefins?
  4. State the hybridisation of each carbon atom in ethene.

Answers

  1. Because ethene contains a weak π bond with exposed electron density.
  2. The π bond.
  3. Because ethene reacts with chlorine to form an oily liquid (1,2-dichloroethane).
  4. Each carbon atom is sp2 hybridised.
⛽

Nomenclature of Alkenes (IUPAC System)

📋
Table of Contents
6 sections
🗺️ Overview
The IUPAC nomenclature of alkenes follows rules similar to those for alkanes, except that the presence of a carbon-carbon double bond (C=C) is indicated by replacing the suffix "-ane" with "-ene".
📘 Definition
🔄 Stepwise Rules for Naming Alkenes
  • 1
    Step 1 : Select the Parent Chain
    Choose the longest continuous carbon chain containing the carbon-carbon double bond.
    The parent chain must include the double bond, even if another longer chain exists without it.
  • 2
    Step 2 : Number the Carbon Chain
    Number the carbon atoms from the end nearest to the double bond.
    The carbon atom where the double bond begins receives the lowest possible number.
    Example
    \[\mathrm{CH_3-CH=CH-CH_2-CH_3}\]
    Numbering starts from the left because the double bond begins at carbon-2.
    Name:
    \[\boxed{\mathrm{Pent\text{-}2\text{-}ene}}\]
  • 3
    Step 3 : Identify the Position of the Double Bond
    The position of the double bond is indicated by the smaller numbered carbon atom.
    Example
    Structure Name
    \(\mathrm{CH_2=CH_2}\)
    		 Ethene
    \(\mathrm{CH_2=CHCH_3}\)
    		 Propene
    \(\mathrm{CH_2=CHCH_2CH_3}\)
    		 But-1-ene
    \(\mathrm{CH_3CH=CHCH_3}\)
    		 But-2-ene
🌟 Importance
🔎 Board Fact
⚖️ Common and IUPAC Names
Formula Common Name IUPAC Name
\(\mathrm{C_2H_4}\)
 Ethylene Ethene
\(\mathrm{C_3H_6}\)
 Propylene Propene
\(\mathrm{C_4H_8}\)
 Butylene Butene
⛽

Isomerism in Alkenes

📋
Table of Contents
14 sections
📘 Definition
📌 Structural Isomerism
🎨 SVG Diagram
Isomers of \(\mathrm{C_4H_8}\)
NCERT CLASS 11 CHEMISTRY • STRUCTURAL & GEOMETRICAL ISOMERS OF BUTENE (C₄H₈) STRUCTURAL ISOMERS (CONSTITUTIONAL) 1-Butene CH₂ CH CH₂ CH₃ Position Isomer (Double Bond at C₁) 2-Methylpropene (Isobutene) CH₂ C CH₃ CH₃ Chain Isomer (Branched Backbone) GEOMETRICAL ISOMERS (STEREOISOMERISM OF 2-BUTENE) C C H₃C CH₃ H H cis-2-Butene (Same Side) C C H₃C H H CH₃ trans-2-Butene (Opposite)
📘 Chain Isomerism
📘 Position Isomerism
📘 Geometrical (Cis-Trans) Isomerism
🤔 Did You Know?
Why is Rotation Restricted Around a Double Bond?
A carbon-carbon double bond consists of one sigma bond and one pi bond.
The pi bond is formed by sideways overlap of p-orbitals.
Rotation about the double bond would destroy this sideways overlap and break the pi bond.
Therefore, free rotation around the double bond is not possible.
🎨 SVG Diagram
Rotation Restricted - Illustration
NCERT CLASS 11 CHEMISTRY • WHY IS ROTATION RESTRICTED AROUND A C=C DOUBLE BOND? 0° ROTATION (STABLE GROUND STATE) Parallel p-Orbitals C C Maximum Sideways Overlap • Intact π-Bond STABLE CONFORMATION (Low Energy) 90° ROTATION Breaks π-Overlap 90° ROTATED STATE (HIGH ENERGY) Perpendicular p-Orbitals NO OVERLAP C C Requires ~264 kJ/mol Energy to Break π-Bond ROTATION PREVENTED AT ROOM TEMP
📘 Definition
⚖️ Comparison
Property Cis Isomer Trans Isomer
Arrangement of Similar Groups Same Side Opposite Side
Dipole Moment Higher Very Low or Zero
Polarity Polar Almost Non-polar
Boiling Point Generally Higher Generally Lower
Melting Point Generally Lower Generally Higher
Stability Slightly Less Stable Generally More Stable
📌 Dipole Moment of But-2-ene
🗒️ Effect Of Geometrical Isomerism On Physical Properties
Property Reason
Boiling Point Cis isomers possess higher dipole-dipole attraction.
Melting Point Trans isomers pack more efficiently in crystals.
Solubility Cis isomers are generally more soluble in polar solvents.
Stability Trans isomers experience less steric repulsion.
✏️ Example
Solved Example
Why does but-2-ene exhibit geometrical isomerism whereas but-1-ene does not?
  1. 1
    Check the groups attached to each doubly bonded carbon.
  2. 2
    Apply the condition for geometrical isomerism.
In but-2-ene, each carbon atom of the double bond is attached to two different groups. Therefore, both cis and trans arrangements are possible.
In but-1-ene, one carbon atom of the double bond is attached to two identical hydrogen atoms. Since one carbon does not have two different substituents, geometrical isomerism is not possible.
⚡ Exam Tip
❌ Common Mistakes
  • Numbering the carbon chain from the wrong end.
  • Ignoring the double bond while selecting the parent chain.
  • Assuming all alkenes exhibit cis-trans isomerism.
  • Confusing chain isomerism with position isomerism.
  • Believing cis isomers are always more stable than trans isomers.
📋 CBSE Competency-Based Case Study (HOTS)

A chemist synthesizes two samples of but-2-ene. Both have the same molecular formula and connectivity but exhibit different melting points, boiling points and dipole moments. One sample has a dipole moment of approximately 0.33 D, whereas the other is almost non-polar.

Questions

  1. What type of isomerism is shown by these compounds?
  2. Which sample has the higher dipole moment?
  3. Why is rotation around the carbon-carbon double bond restricted?
  4. Which isomer generally possesses the higher melting point?

Answers

  1. Geometrical (cis-trans) isomerism.
  2. The cis isomer.
  3. Because rotation would break the π bond formed by sideways overlap of p-orbitals.
  4. The trans isomer.
⛽

Preparation of Alkenes

📋
Table of Contents
10 sections
🗺️ Preparation of Alkenes
Major Methods of Preparation of Alkenes
  • Partial reduction of alkynes
  • Dehydrohalogenation of alkyl halides
  • Dehalogenation of vicinal dihalides
  • Acidic dehydration of alcohols
🗂️ Preparation from Alkynes (Partial Reduction)
(a) Reduction Using Lindlar's Catalyst (Cis-Alkene)
Definition
When an alkyne is treated with one mole of hydrogen in the presence of Lindlar's catalyst, it undergoes partial hydrogenation to form a cis-alkene.
What is Lindlar's Catalyst?
Lindlar's catalyst consists of:
  • Palladium deposited on calcium carbonate or charcoal.
  • The catalyst is partially deactivated (poisoned) using compounds such as quinoline or sulphur compounds.
Purpose of Poisoning
The catalyst activity is reduced so that hydrogenation stops at the alkene stage and does not continue to form an alkane.
General Reaction
\[\boxed{\mathrm{RC\equiv CR'+H_2\xrightarrow{\text{Lindlar Catalyst}}cis\text{-}RCH=CHR'}}\]
Example
\[\boxed{\mathrm{CH_3C\equiv CCH_3+H_2\xrightarrow{\text{Lindlar}}cis\text{-}CH_3CH=CHCH_3}}\]
(b) Reduction with Sodium in Liquid Ammonia (Trans-Alkene)
Definition
Alkynes on reduction with sodium metal in liquid ammonia undergo dissolving metal reduction to produce trans-alkenes.
General Reaction
\[\boxed{\mathrm{RC\equiv CR'\xrightarrow[\mathrm{NH_3(l)}]{Na}trans\text{-}RCH=CHR'}}\]
Example
\[\boxed{\mathrm{CH_3C\equiv CCH_3\xrightarrow[\mathrm{NH_3(l)}]{Na}trans\text{-}CH_3CH=CHCH_3}}\]
📎 Preparation from Alkyl Halides (Dehydrohalogenation)
When an alkyl halide is heated with alcoholic potassium hydroxide, one molecule of hydrogen halide (HX) is eliminated to produce an alkene.
This reaction is called dehydrohalogenation.
Meaning of Dehydrohalogenation
  • Dehydro = Removal of hydrogen
  • Halogenation = Removal of halogen
Therefore, removal of HX produces a double bond.
General Reaction
\[\boxed{\mathrm{R\overset{{\beta}}CH_2\overset{{\alpha}}CH_2X\xrightarrow[\Delta]{Alcoholic\ KOH}RCH=CH_2+HX}}\]
Example
\[\boxed{\mathrm{\overset{\beta}CH_3\overset{\alpha}CH_2Br\xrightarrow[\Delta]{Alcoholic\ KOH}CH_2=CH_2+HBr}}\]
Why is it Called a β-Elimination Reaction?
The carbon atom directly attached to the halogen is called the α-carbon.
The adjacent carbon atom is called the β-carbon.
During the reaction:
  • Halogen leaves from the α-carbon.
  • Hydrogen leaves from the β-carbon.
  • A double bond is formed between α and β carbon atoms.
Factors Affecting Dehydrohalogenation
  • Effect of Halogen:The ease of elimination depends upon the leaving ability of the halogen.
    \[\boxed{\mathrm{I>Br>Cl}}\]
  • Effect of Alkyl GroupThe order of reactivity is:
    \[\boxed{3^\circ>2^\circ>1^\circ}\]
    This is because tertiary carbocations (or transition states) are more stable than secondary and primary ones.
📎 Side Note
Preparation from Vicinal Dihalides (Dehalogenation)
Vicinal dihalides are compounds in which two halogen atoms are attached to adjacent carbon atoms.
On treatment with zinc metal, both halogen atoms are removed simultaneously to produce an alkene.
Definition of Vicinal
The term vicinal means "on neighbouring carbon atoms."
General Reaction
\[\boxed{\mathrm{RCHX-CHXR'+Zn\rightarrow RCH=CHR'+ZnX_2}}\]
Example 1
\[\boxed{\mathrm{BrCH_2CH_2Br+Zn\rightarrow CH_2=CH_2+ZnBr_2}}\]
Example 2
\boxed{\mathrm{CH_3CHBrCH_2Br+Zn\rightarrow CH_3CH=CH_2+ZnBr_2}}
📎 Preparation from Alcohols (Acidic Dehydration)
Alcohols undergo elimination of one molecule of water on heating with concentrated sulphuric acid to form alkenes.
The reaction is called acidic dehydration of alcohols.
Meaning of Dehydration
Removal of one molecule of water (
\(\mathrm{H_2O}\)
).
Reaction Conditions
  • Concentrated sulphuric acid (
    \(\mathrm{H_2SO_4}\)
    )
  • Approximately 443 K (170°C) for ethanol
General Reaction
\[\boxed{\mathrm{RCH_2CH_2OH\xrightarrow[\Delta]{Conc.\ H_2SO_4}RCH=CH_2+H_2O}}\]
Example
\[\boxed{\mathrm{CH_3CH_2OH\xrightarrow[\Delta]{Conc.\ H_2SO_4}CH_2=CH_2+H_2O}}\]
Why is Acidic Dehydration a β-Elimination Reaction?
In alcohols:
  • The carbon attached to the hydroxyl group is the α-carbon.
  • The adjacent carbon is the β-carbon.
During dehydration:
  • The hydroxyl group leaves from the α-carbon.
  • Hydrogen leaves from the β-carbon.
  • A carbon-carbon double bond is formed.
⚖️ Comparison of Different Methods of Preparation of Alkenes
Method Reagent / Catalyst Main Product
Partial Reduction Lindlar Catalyst Cis Alkene
Dissolving Metal Reduction Na / Liquid NH₃ Trans Alkene
Dehydrohalogenation Alcoholic KOH Alkene
Dehalogenation Zinc Dust Alkene
Acidic Dehydration Conc. H₂SO₄ Alkene
✏️ Example
Solved Example
Which reagent would you choose to prepare cis-but-2-ene from but-2-yne? How would you prepare the trans isomer?
  1. 1
    Identify the required stereochemistry.
  2. 2
    Select the appropriate reducing agent.
  • For cis-but-2-ene, use hydrogen in the presence of Lindlar's catalyst.
  • For trans-but-2-ene, reduce the alkyne with sodium in liquid ammonia.
⚡ Exam Tip
❌ Common Mistakes
  • Confusing Lindlar's catalyst with ordinary Pd catalyst (ordinary Pd gives complete reduction to alkane).
  • Using aqueous KOH instead of alcoholic KOH for elimination reactions.
  • Writing dehydration products without removing water.
  • Confusing dehalogenation with dehydrohalogenation.
  • Forgetting that Na/liquid NH₃ produces trans alkenes.
📋 CBSE Competency-Based Case Study (HOTS)

A chemist wants to synthesize different alkenes from various starting materials. One reaction involves but-2-yne and hydrogen in the presence of Lindlar's catalyst. Another uses bromoethane with alcoholic potassium hydroxide, while a third involves ethanol heated with concentrated sulphuric acid.

Questions

  1. Which reaction produces a cis-alkene?
  2. Which reaction is called dehydrohalogenation?
  3. Which reaction is an example of acidic dehydration?
  4. Why is Lindlar's catalyst poisoned?

Answers

  1. Partial reduction using Lindlar's catalyst.
  2. Reaction of alkyl halide with alcoholic KOH.
  3. Heating alcohol with concentrated sulphuric acid.
  4. To stop hydrogenation at the alkene stage and prevent complete reduction to an alkane.
⛽

Physical Properties of Alkenes

📋
Table of Contents
14 sections
🗺️ Overview
The physical properties of alkenes are largely similar to those of alkanes because both are hydrocarbons composed mainly of carbon and hydrogen atoms. However, the presence of a carbon-carbon double bond makes alkenes slightly more polar and chemically more reactive than alkanes.

Most physical properties of alkenes depend upon:
  • Molecular mass
  • Molecular shape
  • Surface area
  • Strength of intermolecular (van der Waals) forces
Key Concept: The carbon-carbon double bond has very little effect on the physical properties of alkenes but greatly influences their chemical properties.
📌 Physical State
📎 Polarity of Alkenes
Alkenes are generally considered non-polar or only slightly polar molecules.
The carbon-carbon double bond produces a small region of higher electron density, making alkenes slightly more polar than alkanes. However, this polarity is not sufficient to make them soluble in water.
Hydrocarbon Relative Polarity
Alkanes Almost Non-polar
Alkenes Slightly More Polar
🗒️ Solubility Of Alkenes
Alkenes follow the general rule: "Like dissolves like."

Since alkenes are essentially non-polar molecules, they dissolve readily in non-polar organic solvents but are almost insoluble in water.
Solvent Solubility Reason
Water Insoluble Water is highly polar.
Benzene Soluble Both are non-polar.
Petroleum Ether Soluble Non-polar solvent.
Carbon Tetrachloride (
\(\mathrm{CCl_4}\)
)		 Soluble Non-polar solvent.
Diethyl Ether Soluble Organic solvent.
Alkenes cannot form hydrogen bonds with water molecules. Consequently, water molecules prefer to remain associated with one another rather than mixing with alkene molecules.
Boiling Point of Alkenes
The boiling point of alkenes increases steadily with increase in molecular size and molecular mass.
Each additional methylene group (
\(\mathrm{-CH_2-}\)
) generally increases the boiling point by approximately
\[\boxed{20-30\ \mathrm{K}}\]
Reason
  • Increase in molecular mass.
  • Increase in surface area.
  • Stronger London dispersion forces.
  • Greater energy required to separate molecules.
🔎 Effect of Branching on Boiling Point
🗒️ Melting Point Of Alkenes
The melting point generally increases with increasing molecular mass.
However, the increase is not perfectly regular because crystal packing differs from one alkene to another.
More symmetrical molecules generally possess higher melting points because they pack more efficiently in the crystal lattice.
📌 Note
🗒️ Intermolecular Forces
The principal intermolecular force present between alkene molecules is the London dispersion (van der Waals) force.
Since alkenes cannot form hydrogen bonds with one another, these weak intermolecular forces determine their boiling points and melting points.
Force Present?
London Dispersion Forces Yes
Hydrogen Bonding No
Strong Dipole-Dipole Attraction Generally No
⚖️ Comparison of Physical Properties of Alkanes and Alkenes
Property Alkanes Alkenes
Nature Saturated Unsaturated
Polarity Almost Non-polar Slightly More Polar
Water Solubility Insoluble Insoluble
Organic Solvent Solubility High High
Main Intermolecular Force London Dispersion London Dispersion
Boiling Point Trend Increases with Molecular Mass Increases with Molecular Mass
Branching Effect Decreases Boiling Point Decreases Boiling Point
⚡ Quick Revision
Property Observation
Colour Colourless
Odour Generally Odourless (Ethene has a faint sweet smell)
Physical State
\(\mathrm{C_2-C_4}\)
: Gases;
\(\mathrm{C_5-C_{18}}\)
: Liquids; Higher members: Solids
Water Solubility Insoluble
Organic Solubility Soluble
Main Force London Dispersion Forces
Boiling Point Increases with Molecular Mass
Density Less than Water
✏️ Example
Solved Example
Why does but-1-ene have a higher boiling point than 2-methylpropene even though both have the molecular formula
\(\mathrm{C_4H_8}\)
?
  1. 1
    Compare molecular shapes.
  2. 2
    Relate surface area to intermolecular forces.
But-1-ene is a straight-chain molecule with a larger surface area than the branched-chain isomer 2-methylpropene. Consequently, stronger London dispersion forces operate between but-1-ene molecules, requiring more energy to separate them. Therefore, but-1-ene has the higher boiling point.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming alkenes are soluble in water because they contain a double bond.
  • Confusing physical properties with chemical reactivity.
  • Writing hydrogen bonding as the principal intermolecular force.
  • Assuming branched alkenes have higher boiling points than straight-chain alkenes.
  • Believing all alkenes possess a characteristic smell.
📋 CBSE Competency-Based Case Study (HOTS)

A laboratory compares ethene, pent-1-ene and 2-methylbut-1-ene. Ethene is found to be a gas at room temperature, while the other two compounds are liquids. Among the liquids, pent-1-ene has a higher boiling point than 2-methylbut-1-ene.

Questions

  1. Why is ethene gaseous at room temperature?
  2. Why are alkenes insoluble in water?
  3. Which intermolecular force is mainly responsible for their boiling points?
  4. Why does pent-1-ene boil at a higher temperature than its branched isomer?

Answers

  1. Because its small molecular size results in very weak London dispersion forces.
  2. Because alkenes are essentially non-polar and cannot form hydrogen bonds with water.
  3. London dispersion (van der Waals) forces.
  4. Because its straight-chain structure provides a larger surface area and stronger intermolecular attractions.
⛽

Chemical Properties of Alkenes

📋
Table of Contents
14 sections
🗺️ Overview
The characteristic chemical behaviour of alkenes arises from the presence of the carbon-carbon double bond (C=C). The double bond consists of one strong sigma (
\(\sigma\)
) bond and one comparatively weak pi (
\(\pi\)
) bond.

Since the electrons of the π bond are loosely held and exposed above and below the plane of the molecule, they are readily attacked by electron-deficient species (electrophiles). Consequently, the most characteristic reaction of alkenes is electrophilic addition.
Important Note
  • Alkanes mainly undergo substitution reactions.
  • Alkenes mainly undergo addition reactions.
  • The π bond breaks during the reaction while the σ bond remains intact.
🗒️ Addition Of Dihydrogen (Catalytic Hydrogenation)
📘 Definition
Alkenes react with one molecule of hydrogen gas in the presence of finely divided nickel, palladium or platinum catalyst to form the corresponding alkanes.
General Reaction
\[ \ce{\boxed{\mathrm{RCH=CHR'+H_2\xrightarrow{Ni/Pd/Pt}RCH_2CH_2R'}}} \]
Example
\[ \ce{\boxed{\mathrm{CH_2=CH_2+H_2\xrightarrow{Ni}CH_3CH_3}}} \]
Industrial Importance
  • Hydrogenation of vegetable oils to manufacture vanaspati ghee.
  • Preparation of saturated hydrocarbons.
  • Petroleum refining.
🗒️ Board Tip
Hydrogenation converts an unsaturated compound into a saturated compound.
🗒️ Addition Of Halogens
Chlorine and bromine add across the carbon-carbon double bond to produce vicinal dihalides.
Iodine normally does not undergo this reaction because the addition is thermodynamically unfavourable.
General Reaction
\[ \ce{\boxed{\mathrm{RCH=CHR+X_2\xrightarrow{}RCHX-CHX R'}}} \]
where
\(X=\mathrm{Cl}\)
or
\(\mathrm{Br}\)
.
Examples
\[ \begin{aligned}\ce{\mathrm{CH_2=CH_2+Br_2}&\mathrm{\rightarrow BrCH_2CH_2Br}}\\\\ \ce{\mathrm{CH_3CH=CH_2+Cl_2}&\mathrm{\rightarrow CH_3CHClCH_2Cl}}\end{aligned} \]

Test for Unsaturation

Bromine Test
  • Bromine solution in carbon tetrachloride is reddish-brown.
  • Unsaturated compounds decolourise bromine solution.
  • Saturated hydrocarbons do not decolourise bromine under ordinary conditions.

Mechanism

The reaction proceeds by an electrophilic addition mechanism through the formation of a cyclic halonium ion (studied in higher classes).
🗒️ Addition Of Hydrogen Halides
📘 Definition
Hydrogen halides such as HCl, HBr and HI add across the carbon-carbon double bond to form alkyl halides.
Reactivity Order
\[ \boxed{HI>HBr>HCl} \]
General Reaction
\[ \ce{\boxed{\mathrm{RCH=CH_2+HX\rightarrow RCHXCH_3}}} \]

Addition of HBr to Symmetrical Alkenes

When both carbon atoms of the double bond are attached to identical groups, only one product is formed.
Example
\[ \begin{aligned}\ce{\mathrm{CH_2=CH_2+HBr}&\mathrm{\rightarrow CH_3CH_2Br}}\\\\ \ce{\mathrm{CH_3CH=CHCH_3+HBr}&\mathrm{\rightarrow CH_3CHBrCH_2CH_3}}\end{aligned} \]

Addition to Unsymmetrical Alkenes (Markovnikov Rule)

Statement of Markovnikov's Rule
During the addition of an unsymmetrical reagent (
\(HX\)
,
\(H_2O\)
,
\(H_2SO_4\)
) to an unsymmetrical alkene, the hydrogen atom attaches to the carbon atom already carrying more hydrogen atoms, while the negative part attaches to the carbon atom carrying fewer hydrogen atoms.
Memory Trick
Rich get Richer
The carbon already rich in hydrogen receives another hydrogen atom.
Example
\[ \ce{\boxed{\mathrm{CH_3CH=CH_2+HBr\rightarrow CH_3CHBrCH_3}}} \]
Major Product: 2-Bromopropane

Mechanism of Markovnikov Addition

Step 1: Formation of Carbocation
The π electrons attack the proton (
\(\mathrm{H^+}\)
) released by HBr.
Two carbocations are theoretically possible.
\[\mathrm{CH_3CH=CH_2+H^+}\]
\[\mathrm{CH_3CH_2CH_2^+}\]
or,
\[\mathrm{CH_3CH^+CH_3}\]
The secondary carbocation is more stable than the primary carbocation.
Step 2 : Attack of Bromide Ion
\[\boxed{\mathrm{CH_3CH^+CH_3+Br^-\rightarrow CH_3CHBrCH_3}}\]
The stability of carbocations follows:
\[\boxed{3^\circ>2^\circ>1^\circ>CH_3^+}\]

Anti-Markovnikov Addition (Peroxide Effect or Kharasch Effect)

📘 Definition
In the presence of organic peroxides, the addition of HBr to an unsymmetrical alkene occurs opposite to Markovnikov's rule

This phenomenon is called:
  • Anti-Markovnikov Addition
  • Peroxide Effect
  • Kharasch Effect
.
Example
\[ \ce{\boxed{\mathrm{CH_3CH=CH_2+HBr\xrightarrow{Peroxide}CH_3CH_2CH_2Br}}} \]
Major Product: 1-Bromopropane
Very Important Board Point

The peroxide effect is observed only with HBr.
It is not observed with HCl or HI.
Reason
The reaction proceeds through a free radical chain mechanism.

Comparison of Markovnikov and Anti-Markovnikov Addition

Feature Markovnikov Addition Anti-Markovnikov Addition
Condition Normal Presence of Peroxide
Mechanism Electrophilic Addition Free Radical Addition
Major Product 2-Bromopropane 1-Bromopropane
Applicable To HX Only HBr
🗒️ Addition Of Sulphuric Acid
Cold concentrated sulphuric acid adds across the double bond according to Markovnikov's rule to produce alkyl hydrogen sulphates.
Reaction
\[ \ce{\boxed{\mathrm{CH_2=CH_2+H_2SO_4\rightarrow C_2H_5HSO_4}}} \]
🌟 Importance
Alkyl hydrogen sulphates can be hydrolysed to obtain alcohols.
🗒️  Addition Of Water (Hydration)
Alkenes react with water in the presence of a few drops of concentrated sulphuric acid to form alcohols.
The reaction follows Markovnikov's rule.
General Reaction
\[ \ce{\boxed{\mathrm{CH_2=CH_2+H_2O\xrightarrow{H^+}CH_3CH_2OH}}} \]
Industrial Importance
  • Manufacture of ethanol.
  • Preparation of industrial alcohols.
🗒️ Oxidation Of Alkenes

Baeyer's Reagent

Cold dilute alkaline potassium permanganate solution oxidises alkenes into vicinal glycols (diols).
The purple colour of
\(\ce{KMnO_4}\)
fades to brown due to the formation of
\(\ce{MnO_2}\)
.
Genearal Reaction
\[ \ce{\boxed{\mathrm{RCH=CHR'+[O]+H_2O\rightarrow RCHOHCHOHR'}}} \]
Examples
\[ \begin{aligned}\ce{\mathrm{CH_2=CH_2+[O]+H_2O}&\mathrm{\rightarrow HOCH_2CH_2OH}}\\\\ \ce{\mathrm{CH_3CH=CH_2+[O]+H_2O}&\mathrm{\rightarrow CH_3CHOHCH_2OH}}\end{aligned} \]

Baeyer's Test

  • Purple KMnO₄ solution becomes colourless.
  • Used to detect unsaturation.
🗒️ Ozonolysis
📘 Definition
Ozone first adds across the carbon-carbon double bond to form an ozonide. The ozonide is then decomposed using zinc and water to produce smaller carbonyl compounds.
General Reaction
\[ \ce{\boxed{\mathrm{Alkene\xrightarrow{O_3}Ozonide\xrightarrow{Zn/H_2O}Aldehydes\ or\ Ketones}}} \]
🌟 Importance
  • Determination of the position of the double bond.
  • Identification of unknown alkenes.
  • Preparation of aldehydes and ketones.
🗒️ Polymerisation
📘 Definition
Polymerisation is the process in which a large number of small alkene molecules (monomers) combine to form a giant molecule called a polymer.
🗒️ Sapcer
🗒️ Ethene Polymerisation
\ce{}\boxed{\mathrm{n(CH_2=CH_2)\xrightarrow{Pressure,\ Catalyst}(-CH_2-CH_2-)_n}}
The polymer formed is called polyethene (polyethylene).

Polypropene Formation

🗒️ Chemical Reaction
\ce{\boxed{\mathrm{n(CH_2=CHCH_3)\rightarrow(-CH_2-CH(CH_3)-)_n}}}
🛠️ Applications
  • Plastic bags
  • Water pipes
  • Milk crates
  • Toys
  • Buckets
  • Electrical insulation
  • Containers

Environmental Concern

Excessive use of polyethylene and polypropylene contributes significantly to plastic pollution because these polymers are non-biodegradable.
📝 Summary of Chemical Reactions of Alkenes
✏️ Example
Solved Example
Propene reacts separately with HBr under ordinary conditions and in the presence of peroxide. Explain the difference in products formed.
  1. 1
    Identify the reaction conditions.
  2. 2
    Apply Markovnikov's rule or peroxide effect.
Under ordinary conditions, HBr adds according to Markovnikov's rule to give 2-bromopropane. In the presence of peroxide, the reaction follows a free radical mechanism (Kharasch effect), giving 1-bromopropane as the major product.
⚡ Exam Tip
❌ Common Mistakes
  • Applying the peroxide effect to HCl or HI.
  • Confusing substitution reactions with addition reactions.
  • Writing iodine as readily adding across the double bond.
  • Ignoring carbocation stability while applying Markovnikov's rule.
  • Confusing Baeyer's test with bromine water test.
  • Assuming ozonolysis always produces only aldehydes.
📋 CBSE Competency-Based Case Study (HOTS)
A chemist investigates an unknown hydrocarbon. The compound rapidly decolourises bromine solution in carbon tetrachloride and also decolourises cold dilute potassium permanganate solution. On ozonolysis followed by treatment with zinc and water, it produces ethanal and methanal.

Questions

  1. What type of hydrocarbon is the compound likely to be?
  2. Which two laboratory tests indicate the presence of unsaturation?
  3. Why is ozonolysis useful in organic chemistry?
  4. What major product is formed when propene reacts with HBr in the presence of peroxide?

Answers

  1. An alkene.
  2. Bromine test and Baeyer's (KMnO₄) test.
  3. It helps determine the position of the carbon-carbon double bond by cleaving it into identifiable carbonyl compounds.
  4. 1-Bromopropane (anti-Markovnikov addition).
⛽

Alkynes

📋
Table of Contents
20 sections
🗺️ Overview
Alkynes are unsaturated hydrocarbons containing at least one carbon-carbon triple bond (
\(\mathrm{C \equiv C}\)
). The triple bond consists of one sigma (
\(\sigma\)
) bond and two pi (
\(\pi\)
) bonds, making alkynes highly reactive towards addition reactions.

Compared with alkanes and alkenes, alkynes contain fewer hydrogen atoms because each triple bond represents two additional degrees of unsaturation. Due to the presence of two weak π bonds, alkynes are important starting materials in organic synthesis and numerous industrial processes.
📘 Definition
🔢 General Formula of Alkynes
📐 Derivation of the Formula
The molecular formula of an alkane is
\[\mathrm{C_{n}H_{2n+2}}\]
Formation of one carbon-carbon triple bond removes four hydrogen atoms.
\[\mathrm{C_{n}H_{2n+2}-2H_2=C_{n}H_{2n-2}}\]
Hence,
\[\boxed{\mathrm{C_{n}H_{2n-2}}}\]
📌 First Stable Member of the Alkyne Series
🔎 Board Fact
⚛️ Structure of Ethyne
Ethyne is the simplest alkyne having the molecular formula
\[\mathrm{HC \equiv CH}\]
Each carbon atom is sp hybridised.
The carbon-carbon triple bond consists of:
  • One strong sigma (
    \(\sigma\)
    ) bond.
  • Two mutually perpendicular pi (
    \(\pi\)
    ) bonds.
Geometry
Property Value
Hybridisation sp
Geometry Linear
Bond Angle
\(180^\circ\)
Carbon Atoms Collinear
🎨 SVG Diagram
Structure of Ethyne (Acetylene)
MOLECULAR STRUCTURE OF ETHYNE (ACETYLENE) NCERT CLASS 11 CHEMISTRY • C₂H₂ (ALKYNES) 180° (Linear) H C C H sp sp C-H σ-bond (sp-1s) 2 × π-bonds (2p-2p unhybridized overlap) 1 × C-C σ-bond (sp-sp overlap) C≡C Bond Length: 120 pm NCERT BONDING SUMMARY: • Total Bonds: 3 σ-bonds + 2 π-bonds • Geometry: Linear (180°) • C-H Bond Length: 106 pm • % s-character: 50%
⚖️ Comparison of Alkanes, Alkenes and Alkynes
Property Alkanes Alkenes Alkynes
Nature Saturated Unsaturated Unsaturated
Main Bond C–C C=C C≡C
General Formula
\(\mathrm{C_{n}H_{2n+2}}\)
\(\mathrm{C_{n}H_{2n}}\)
\(\mathrm{C_{n}H_{2n-2}}\)
Hybridisation sp3 sp2 sp
Geometry Tetrahedral Trigonal Planar Linear
Main Reaction Substitution Addition Addition
🌟 Importance of Alkynes
🛠️ Major Applications of Alkynes
  • Manufacture of plastics.
  • Preparation of synthetic rubber.
  • Production of pharmaceuticals.
  • Preparation of aldehydes and ketones.
  • Manufacture of acetic acid and vinyl compounds.
  • Organic synthesis in laboratories.
🗒️ Acetylene And The Oxyacetylene Flame
The most important industrial alkyne is acetylene (ethyne).
When acetylene burns in pure oxygen, it produces an extremely hot flame known as the oxyacetylene flame.
🔁 Combustion Reaction
\[ \ce{\boxed{\mathrm{2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O+Heat}}} \]
📌 Temperature
🛠️ Applications of Oxyacetylene Flame
  • Arc welding
  • Metal cutting
  • Repair of heavy machinery
  • Steel fabrication
🤔 Did You Know?
Why are Alkynes Reactive?
A carbon-carbon triple bond contains:
  • One strong sigma bond.
  • Two comparatively weak pi bonds.
The two π bonds are rich in electron density and are easily attacked by electrophiles.
Consequently, alkynes readily undergo addition reactions similar to alkenes.
💡 Concept
✏️ Example
Solved Example
Why does ethyne undergo addition reactions more readily than ethane?
  1. 1
    Compare the bonding in ethane and ethyne.
  2. 2
    Identify the role of π bonds.
Ethyne contains a carbon-carbon triple bond consisting of one σ bond and two weak π bonds. The π electrons are loosely held and readily attacked by electrophiles, making ethyne highly reactive towards addition reactions. Ethane contains only strong σ bonds and therefore mainly undergoes substitution reactions instead of addition reactions.
⚡ Exam Tip
❌ Common Mistakes
  • Writing the general formula of alkynes as
    \(\mathrm{C_{n}H_{2n}}\)
    .
  • Confusing ethyne with ethene.
  • Writing sp2 instead of sp hybridisation.
  • Drawing a bent or trigonal planar geometry for alkynes.
  • Assuming acetylene is used directly for welding without oxygen.
📋 CBSE Competency-Based Case Study (HOTS)

A metal fabrication workshop uses cylinders containing acetylene and oxygen for welding steel structures. A chemistry student observes that acetylene belongs to the alkyne family and readily undergoes addition reactions in the laboratory.

Questions

  1. State the IUPAC and common names of the first stable alkyne.
  2. Write the general formula of alkynes containing one triple bond.
  3. What is the hybridisation and geometry of the carbon atoms in ethyne?
  4. Why is the oxyacetylene flame preferred for welding?

Answers

  1. IUPAC name: Ethyne; Common name: Acetylene.
  2. \(\mathrm{C_{n}H_{2n-2}}\)
    .
  3. sp hybridisation with linear geometry (
    \(180^\circ\)
    ).
  4. Because it produces a very high-temperature flame (about
    \(3300^\circ\mathrm{C}\)
    ) suitable for welding and cutting metals.
⛽

Nomenclature and Isomerism of Alkynes

📋
Table of Contents
4 sections
🗺️ Overview
The IUPAC nomenclature of alkynes follows rules similar to those used for alkanes and alkenes. The only difference is that the suffix "-ane" is replaced by "-yne" to indicate the presence of a carbon-carbon triple bond (
\(\mathrm{C \equiv C}\)
).

The carbon chain is always numbered from the end nearest to the triple bond so that the triple bond receives the lowest possible locant (position number).
📘 Definition
🔄 Stepwise Rules for IUPAC Nomenclature of Alkynes
  • 1
    Step 1 : Select the Parent Chain
    Choose the longest continuous carbon chain containing the carbon-carbon triple bond.
    The parent chain must contain the triple bond, even if another longer chain without the triple bond exists.
  • 2
    Step 2 : Number the Parent Chain
    Number the carbon atoms in the parent chain from the end nearest to the triple bond, ensuring the triple bond receives the lowest possible locant.
  • 3
    Step 3 : Write the Name
    Replace the suffix -ane with -yne and indicate the position of the triple bond.
✏️ Example
Structure IUPAC Name
\(\mathrm{HC\equiv CH}\)
 Ethyne
\(\mathrm{CH_3C\equiv CH}\)
 Prop-1-yne (Propyne)
\(\mathrm{HC\equiv CCH_2CH_3}\)
 But-1-yne
\(\mathrm{CH_3C\equiv CCH_3}\)
 But-2-yne
Common and IUPAC Names of Important Alkynes
Molecular Formula Common Name IUPAC Name
\(\mathrm{C_2H_2}\)
 Acetylene Ethyne
\(\mathrm{C_3H_4}\)
 Methyl Acetylene Propyne
\(\mathrm{C_4H_6}\)
 Ethyl Acetylene But-1-yne
🎨 SVG Diagram
Illustrative Examples of IUPAC Nomenclature of Alkynes
IUPAC NOMENCLATURE OF ALKYNES: POSITION ISOMERISM NCERT CLASS 11 CHEMISTRY • NOMENCLATURE COMPARISON Example 1: Terminal Alkyne HC C CH2 CH3 1 2 3 4 Lowest locant rule: Numbering starts from left (1 ≡ 2 - 3 - 4) But-1-yne Example 2: Internal Alkyne CH3 C C CH3 1 2 3 4 Symmetrical triple bond: Either end gives locant 2 (1 - 2 ≡ 3 - 4) But-2-yne NCERT RULE RECAP: • Root Word: "But-" (4-carbon chain) | • Primary Suffix: "-yne" (Triple bond present) | • Locant Rule: Triple bond receives the lowest possible carbon number.
⛽

Isomerism in Alkynes

📋
Table of Contents
11 sections
📘 Definition
📌 Position Isomerism
📎 Chain Isomerism
Chain isomerism arises due to different arrangements of the carbon skeleton while the molecular formula remains the same.
Example: Pentyne (
\(\mathrm{C_5H_8}\)
)
Three structural isomers are possible.
Structure IUPAC Name Type of Isomer
\[\]
 Pent-1-yne Position Isomer
\[ \mathrm{CH_3C\equiv CCH_2CH_3}\]
 Pent-2-yne Position Isomer
\[ \mathrm{HC\equiv CC(CH_3)CH_3}\]
 3-Methylbut-1-yne Chain Isomer
🔗 Relationship Among the Isomers of Pentyne
Compared Compounds Relationship
Pent-1-yne and Pent-2-yne Position Isomers
Pent-1-yne and 3-Methylbut-1-yne Chain Isomers
Pent-2-yne and 3-Methylbut-1-yne Chain Isomers
🤔 Did You Know?
Why Do Alkynes Not Show Geometrical Isomerism?
Geometrical (cis-trans) isomerism requires restricted rotation around a double bond and two different substituents attached to each doubly bonded carbon atom.

In alkynes:
  • The carbon atoms are sp hybridised.
  • The molecule possesses linear geometry (
    \(180^\circ\)
    )    .
  • Each carbon atom of the triple bond is attached to only one substituent.
Therefore, cis-trans arrangements are impossible.
🌟 Important Board Point
⚖️ Comparison of Hydrocarbon Nomenclature
Hydrocarbon Characteristic Bond Suffix Example
Alkane C–C -ane Butane
Alkene C=C -ene But-1-ene
Alkyne C≡C -yne But-1-yne
✏️ Example
Solved Example
1
Question
Write all the structural isomers of molecular formula
\(\mathrm{C_5H_8}\)
belonging to the alkyne family and classify them.
  1. 1
    Construct all possible carbon skeletons.
  2. 2
    Place the triple bond at different positions.
  3. 3
    Identify chain and position isomers.
  • \(\mathrm{HC\equiv CCH_2CH_2CH_3}\)
    — Pent-1-yne.
  • \(\mathrm{CH_3C\equiv CCH_2CH_3}\)
    — Pent-2-yne.
  • \(\mathrm{HC\equiv CC(CH_3)CH_3}\)
    — 3-Methylbut-1-yne.
Pent-1-yne and Pent-2-yne are position isomers, whereas the branched isomer (3-methylbut-1-yne) is a chain isomer of both straight-chain alkynes.
⚡ Exam Tip
❌ Common Mistakes
  • Numbering the carbon chain from the wrong end.
  • Selecting a parent chain that does not include the triple bond.
  • Confusing chain isomerism with position isomerism.
  • Writing the suffix -ene instead of -yne.
  • Assuming alkynes show geometrical (cis-trans) isomerism.
📋 CBSE Competency-Based Case Study (HOTS)

A chemistry student prepares three compounds having the molecular formula

\(\mathrm{C_5H_8}\)
. Two compounds differ only in the position of the carbon-carbon triple bond, whereas the third possesses a branched carbon skeleton.

Questions

  1. Name the three alkynes.
  2. Which two are position isomers?
  3. Which compound is a chain isomer?
  4. Why do these compounds not exhibit geometrical isomerism?

Answers

  1. Pent-1-yne, Pent-2-yne and 3-Methylbut-1-yne.
  2. Pent-1-yne and Pent-2-yne.
  3. 3-Methylbut-1-yne.
  4. Because the carbon atoms of the triple bond are sp hybridised, linear, and each is bonded to only one substituent, making cis-trans arrangements impossible.
⛽

Structure of the Carbon-Carbon Triple Bond

📋
Table of Contents
12 sections
🗺️ Overview
The simplest member of the alkyne family is ethyne (
\(\mathrm{C_2H_2}\)
). The unique properties of alkynes arise from the presence of a carbon-carbon triple bond (
\(\mathrm{C \equiv C}\)
). Understanding the structure of this bond is essential because it explains the geometry, bond strength, bond length and chemical reactivity of alkynes.
📘 Definition
⚛️ Hybridisation in Ethyne
Each carbon atom in ethyne undergoes sp hybridisation.
One 2s orbital mixes with one 2p orbital to produce two equivalent sp hybrid orbitals.
The remaining two p orbitals remain unhybridised.
Formation of sp Hybrid Orbitals
\[\boxed{2s+2p\longrightarrow2(sp)}\]
Each carbon atom therefore possesses:
  • Two sp hybrid orbitals
  • Two unhybridised p orbitals
Key Point
The two sp hybrid orbitals are arranged in opposite directions, producing a linear geometry with a bond angle of
\[\boxed{180^\circ}\]
Formation of Sigma (
\(\sigma\)
) Bonds
The sigma bonds in ethyne are formed by head-on (axial) overlap of atomic orbitals.
Carbon-Carbon Sigma Bond
One sp hybrid orbital of each carbon atom overlaps head-on with an sp hybrid orbital of the other carbon atom to form the carbon-carbon sigma bond.
\[\boxed{sp-sp\longrightarrow C-C\ \sigma\ bond}\]
Carbon-Hydrogen Sigma Bonds
The remaining sp hybrid orbital on each carbon overlaps with the 1s orbital of hydrogen to produce two carbon-hydrogen sigma bonds.
\[\boxed{sp-1s\longrightarrow C-H\ \sigma\ bond}\]
Therefore, ethyne contains:
  • One C–C sigma bond
  • Two C–H sigma bonds
Formation of Pi (
\(\pi\)
) Bonds
Each carbon atom still possesses two unhybridised p orbitals.
These p orbitals are:
  • Mutually perpendicular to each other.
  • Perpendicular to the internuclear axis.
  • Parallel to the corresponding p orbitals of the neighbouring carbon atom.
These parallel p orbitals undergo sidewise (lateral) overlap to produce two independent pi bonds.
\[\boxed{p-p\longrightarrow\pi\ bond}\]
Thus, the carbon-carbon triple bond consists of:
\[\boxed{C\equiv C=1\sigma+2\pi}\]
Bond Type of Overlap Number Present
C–C sp-sp Head-on 1 σ Bond
C–H sp-1s Head-on 2 σ Bonds
C–C p-p Sidewise 2 π Bonds
Total Bonds in Ethyne
  • Three sigma (
    \(\sigma\)
    ) bonds
  • Two pi (
    \(\pi\)
    ) bonds
Geometry of Ethyne
Since the two sp hybrid orbitals are oriented exactly opposite to each other, the molecule possesses a linear geometry.

Therefore,
\[\boxed{\angle H-C\equiv C=180^\circ}\]
All four atoms lie on the same straight line.
🎨 SVG Diagram
sp-hybridization
sp HYBRIDIZATION & ORBITAL OVERLAP IN ETHYNE (C₂H₂) NCERT CLASS 11 CHEMISTRY • CHEMICAL BONDING & HYDROCARBONS H 1s H 1s 2py 2pz C 2py 2pz C π-bond (2py-2py) π-bond (2pz-2pz) σ (sp - 1s) bond σ (sp - 1s) bond σ (sp - sp) bond 180° NCERT HYBRIDIZATION SUMMARY: • 1 s + 1 p orbital mix to form two linearly oriented sp hybrid orbitals (180° apart). • C≡C Triple Bond consists of 1 σ-bond + 2 perpendicular π-bonds.
📌 Electron Cloud Around the Triple Bond
💡 NCERT Concept
⚖️ Bond Length and Bond Strength
The presence of two additional π bonds makes the carbon-carbon triple bond much stronger and shorter than double and single bonds.
Bond Bond Length (pm) Bond Enthalpy (kJ mol-1)
C–C 154 348
C=C 133 681
C≡C 120 823
⚖️ Comparison of Single, Double and Triple Bonds
Property Single Bond Double Bond Triple Bond
Hybridisation sp3 sp2 sp
Geometry Tetrahedral Trigonal Planar Linear
Bond Composition 1σ 1σ + 1π 1σ + 2π
Bond Length (pm) 154 133–134 120
Bond Enthalpy (kJ mol-1) 348 681 823
Relative Strength Lowest Intermediate Highest
🌟 Significance of the Triple Bond
  • Provides high bond strength.
  • Makes the molecule linear.
  • Produces two electron-rich π bonds.
  • Causes alkynes to undergo electrophilic addition reactions.
  • Responsible for the characteristic chemistry of alkynes.
✏️ Example
Solved Example
1
Question
Why is the carbon-carbon triple bond shorter and stronger than the carbon-carbon double bond?
  1. 1
    Compare bond composition.
  2. 2
    Relate bond order to bond strength and bond length.
The carbon-carbon triple bond contains one sigma bond and two pi bonds, whereas the double bond contains only one sigma and one pi bond. The higher bond order (3) increases the attractive force between the two carbon nuclei, making the triple bond stronger (823 kJ mol-1) and shorter (120 pm) than the double bond (681 kJ mol-1, 133 pm).
⚡ Exam Tip
❌ Common Mistakes
  • Writing sp2 instead of sp hybridisation for alkynes.
  • Assuming the triple bond contains three sigma bonds.
  • Drawing a bent geometry instead of a linear geometry.
  • Confusing bond length and bond strength trends.
  • Forgetting that two π bonds are mutually perpendicular.
📋 CBSE Competency-Based Case Study (HOTS)

A student compares ethane, ethene and ethyne using molecular models. The models show that ethane has tetrahedral geometry, ethene has trigonal planar geometry, while ethyne is linear. The bond lengths and bond strengths are also measured experimentally.

Questions

  1. What is the hybridisation of carbon atoms in ethyne?
  2. How many sigma and pi bonds are present in the carbon-carbon triple bond?
  3. Why is the C≡C bond shorter than the C=C bond?
  4. State the bond angle in ethyne.

Answers

  1. sp hybridisation.
  2. One sigma bond and two pi bonds.
  3. Because the bond order is higher (3), resulting in stronger attraction between the nuclei and a shorter bond length.
  4. \(180^\circ\)
    .
⛽

Preparation of Alkynes

📋
Table of Contents
2 sections
🗺️ Overview
Alkynes can be prepared by both industrial methods and laboratory methods. The industrial method mainly produces ethyne (acetylene), whereas laboratory methods generally involve the elimination of hydrogen halides from suitable dihalides.

Important NCERT Methods of Preparation
  • From Calcium Carbide (Industrial Method)
  • From Vicinal Dihalides (Double Dehydrohalogenation)
📌 Preparation from Calcium Carbide (Industrial Method)
📖 Introduction
The most important industrial method for manufacturing ethyne (acetylene) is the reaction of calcium carbide (
\(\mathrm{CaC_2}\)
) with water.

Calcium carbide itself is manufactured from quick lime (calcium oxide) and coke (carbon) in an electric furnace at about 2200–2500 K.
Importance of the Industrial Method
This method is economical and is widely used for the large-scale production of acetylene required in welding, metal cutting and the manufacture of many organic chemicals.
  • 1
    Step 1 : Manufacture of Calcium Carbide
    Preparation of Quick Lime
    Limestone (calcium carbonate) is strongly heated in a lime kiln to produce quick lime.
    Chemical Equation
    \[\boxed{\mathrm{CaCO_3\xrightarrow{\Delta}CaO+CO_2}}\]
    This reaction is called the calcination of limestone.
    Preparation of Calcium Carbide
    Quick lime is heated with coke in an electric furnace.
    \[\boxed{\mathrm{CaO+3C\xrightarrow{2200-2500\,K}CaC_2+CO}}\]
    Why is High Temperature Required?
    The reaction is highly endothermic. Therefore, an electric furnace is used to provide the extremely high temperature needed for the reaction.
  • 2
    Step 2 : Preparation of Ethyne
    Calcium carbide reacts vigorously with water to produce acetylene (ethyne) and calcium hydroxide.
    Chemical Equation
    \[\boxed{\mathrm{CaC_2+2H_2O\xrightarrow{}C_2H_2+Ca(OH)_2}}\]
    The reaction is highly exothermic and produces a large amount of heat.
    Observation
    • Rapid evolution of colourless acetylene gas.
    • Formation of white calcium hydroxide.
    • Heat is liberated.
Flowchart of Industrial Preparation
INDUSTRIAL PREPARATION OF ETHYNE (ACETYLENE, C₂H₂) NCERT CLASS 11 CHEMISTRY • UNIT 13: HYDROCARBONS METHOD 1: CALCIUM CARBIDE ROUTE Step 1: Calcination of Limestone (Kiln, ~1000°C) CaCO₃ (s) ━Δ━→ CaO (s) + CO₂ (g)↑ Step 2: Electric Arc Furnace (~2000°C with Coke) CaO (s) + 3C (Coke) ━2000°C━→ CaC₂ (s) + CO (g)↑ Yields molten Calcium Carbide (CaC₂) Step 3: Hydrolysis Generator (Water Addition) CaC₂ (s) + 2H₂O (l) ━━→ H━C≡C━H (g) + Ca(OH)₂ (aq) [ crude C₂H₂ gas + Slaked Lime byproduct ] METHOD 2: PARTIAL OXIDATION OF METHANE Step 1: Feedstock Preheating & Mixing Natural Gas (CH₄) + Oxygen (O₂) (Ratio ~ 2:1) Step 2: Partial Combustion Reactor (~1500°C) 6CH₄ + 3O₂ ━1500°C━→ 2C₂H₂ + 2CO + 10H₂ High temperature maintained for milliseconds to maximize yield Step 3: Water Quench Chamber Rapid cooling to < 200°C with water sprays Prevents thermal decomposition of C₂H₂ into Carbon & H₂ STAGE A: SCRUBBING & PURIFICATION • Gas washed with acidified CuSO₄ / FeCl₃ solution. • Removes toxic impurities: Phosphine (PH₃) & Hydrogen Sulfide (H₂S). • Passed through dry Lime (CaO) to absorb residual moisture. STAGE B: SAFE STORAGE IN CYLINDERS • Safety Warning: Pure C₂H₂ explodes under pressure (> 2 bar). • Dissolved under pressure (~12 atm) in Acetone absorbed on a porous mass (e.g., asbestos/pumice) inside steel cylinders.
⛽

Preparation from Vicinal Dihalides (Double Dehydrohalogenation)

📋
Table of Contents
9 sections
📘 Definition
🔄 Process
Stepwise Double Dehydrohalogenation
  • 1
    Step 1 : Formation of Vinyl Halide
    Vicinal dihalides react with alcoholic potassium hydroxide to eliminate one molecule of hydrogen halide.
    \[\boxed{\mathrm{BrCH_2CH_2Br\xrightarrow{Alcoholic\ KOH}CH_2=CHBr+HBr}}\]
    The product formed is called a vinyl halide (alkenyl halide).
  • 2
    Step 2 : Formation of Alkyne
    Vinyl halides are comparatively less reactive than ordinary alkyl halides. Therefore, a much stronger base such as sodamide (
    \(\mathrm{NaNH_2}\)
    )     is required for the second elimination.
    \[\boxed{\mathrm{CH_2=CHBr\xrightarrow{NaNH_2}HC\equiv CH+NaBr+NH_3}}\]
  • 3
    Overall Conversion
    \[\boxed{\mathrm{BrCH_2CH_2Br\xrightarrow[\mathrm{NaNH_2}]{Alcoholic\ KOH}HC\equiv CH}}\]
    Thus, a vicinal dihalide is converted into an alkyne through two successive elimination reactions.

Why is Sodamide Used?

Vinyl halides contain the halogen atom directly attached to an sp2-hybridised carbon atom.
The carbon-halogen bond is stronger than that in ordinary alkyl halides.
Hence, ordinary bases such as alcoholic KOH cannot remove the second molecule of hydrogen halide efficiently.
Important NCERT Point
The second dehydrohalogenation requires a much stronger base such as sodamide (
\(\mathrm{NaNH_2}\)
).
💡 Concept
⚖️ Comparison of Preparation Methods
Method Starting Material Reagent Main Product
Industrial Method Calcium Carbide Water Ethyne
Laboratory Method Vicinal Dihalide Alcoholic KOH, NaNH₂ Alkyne
🌟 Industrial Importance of Ethyne
✏️ Example
Solved Example
1
Question
Why is sodamide used instead of alcoholic potassium hydroxide in the second step of converting vicinal dihalides into alkynes?
  1. 1
    Identify the intermediate formed.
  2. 2
    Compare the strength of bases required.
The first elimination with alcoholic potassium hydroxide produces a vinyl halide. Because the carbon-halogen bond in a vinyl halide is stronger (attached to an sp2-hybridised carbon), a much stronger base is required for the second elimination. Therefore, sodamide (
\(\mathrm{NaNH_2}\)
) is used to remove the second molecule of hydrogen halide and produce the alkyne.
⚡ Exam Tip
❌ Common Mistakes
  • Writing oxygen instead of carbon dioxide during calcination of limestone.
  • Using aqueous KOH instead of alcoholic KOH for dehydrohalogenation.
  • Attempting both eliminations using alcoholic KOH alone.
  • Confusing vicinal dihalides with geminal dihalides.
  • Writing calcium hydroxide incorrectly as calcium oxide in the final industrial reaction.
📋 CBSE Competency-Based Case Study (HOTS)

A chemical industry manufactures acetylene using limestone, coke and water. In a laboratory, a student converts 1,2-dibromoethane into ethyne using alcoholic potassium hydroxide followed by sodamide.

Questions

  1. Name the industrial raw material that reacts with water to produce ethyne.
  2. Write the intermediate formed during laboratory preparation.
  3. Why is sodamide required in the second step?
  4. State one industrial use of ethyne.

Answers

  1. Calcium carbide (
    \(\mathrm{CaC_2}\)
    ).
  2. Vinyl bromide (bromoethene).
  3. Because the second elimination from a vinyl halide requires a much stronger base than alcoholic KOH.
  4. Oxyacetylene welding, manufacture of vinyl chloride, or preparation of organic chemicals.
⛽

Physical Properties of Alkynes

📋
Table of Contents
14 sections
🗺️ Overview
The physical properties of alkynes mainly depend upon their molecular mass, length of the carbon chain, and the presence of the carbon-carbon triple bond (
\(\mathrm{C \equiv C}\)
). Since alkynes are only weakly polar, they exhibit physical properties very similar to those of alkanes and alkenes.
📘 Definition
🗒️ Physical State
As the molecular mass increases, intermolecular van der Waals forces become stronger. Consequently, the physical state changes from gases to liquids and finally to solids.
Members Physical State at 298 K Examples
First 3 members Gases Ethyne, Propyne, Butyne
Next 8 members Liquids Pentyne to Dodecyne
Higher members Solids Higher alkynes
NCERT Fact
Increase in molecular size increases the strength of intermolecular forces, leading to a gradual change from gaseous to liquid and finally solid state.
🗒️ Colour And Odour
  • Pure alkynes are colourless.
  • Ethyne (acetylene) possesses a characteristic faint odour.
  • Higher alkynes are generally odourless.
Important Note
Commercial acetylene often has an unpleasant smell because of impurities such as phosphine (
\(\mathrm{PH_3}\)
) and hydrogen sulphide (
\(\mathrm{H_2S}\)
), not because of ethyne itself.
🗒️ Polarity
Alkynes are weakly polar molecules.
Although the carbon atoms of the triple bond are sp hybridised (having 50% s-character), the electronegativity difference between carbon and hydrogen remains very small.
Therefore, alkynes behave almost like non-polar hydrocarbons.
Reason
  • Very small electronegativity difference between carbon and hydrogen.
  • Predominantly covalent carbon-carbon and carbon-hydrogen bonds.
  • Weak intermolecular van der Waals forces.
🗒️ Solubility
Alkynes follow the principle:
"Like dissolves like."
Since alkynes are weakly polar, they dissolve readily in non-polar organic solvents but are almost insoluble in water.
Solvent Solubility Reason
Water Insoluble Water is highly polar.
Diethyl Ether Soluble Weak intermolecular interactions.
Benzene Soluble Both are non-polar.
Carbon Tetrachloride (
\(\mathrm{CCl_4}\)
)		 Highly Soluble Non-polar solvent.
Exam Rule
Non-polar hydrocarbons dissolve in non-polar solvents but not in polar solvents such as water.
🗒️ Density
The density of alkynes gradually increases with increase in molecular mass.
However, all common alkynes remain lighter than water.
Property Observation
Density Trend Increases with molecular mass.
Relative Density Less than water.
🗒️ Melting Point And Boiling Point
Melting point and boiling point increase regularly with increase in the number of carbon atoms.
As the molecular size increases, the surface area available for intermolecular attraction also increases, thereby strengthening van der Waals forces.
Trend
\[ \boxed{ \text{Molecular Mass} \uparrow \Longrightarrow \text{Van der Waals Forces} \uparrow \Longrightarrow \text{Boiling Point} \uparrow } \]
Reason
  • Increase in molecular surface area.
  • Increase in London dispersion forces.
  • Greater energy required to separate molecules.
📝 Summary of Physical Properties
⚖️ Comparison of Alkanes, Alkenes and Alkynes
Property Alkanes Alkenes Alkynes
Nature Non-polar Weakly polar Weakly polar
Water Solubility Insoluble Insoluble Insoluble
Organic Solubility High High High
Density Less than water Less than water Less than water
Trend in Boiling Point Increases with molar mass Increases with molar mass Increases with molar mass
✏️ Example
Solved Example
1
Question
Why do alkynes have higher boiling points as the number of carbon atoms increases?
  1. 1
    Relate molecular mass to intermolecular forces.
  2. 2
    Explain the role of van der Waals forces.
As the number of carbon atoms increases, molecular size and surface area increase. Consequently, London dispersion (van der Waals) forces become stronger. More heat energy is therefore required to separate the molecules, resulting in higher boiling and melting points.
⚡ Exam Tip
❌ Common Mistakes
  • Writing alkynes as completely polar compounds.
  • Assuming alkynes dissolve readily in water.
  • Confusing ethyne with ethene while discussing odour.
  • Stating that boiling point decreases with chain length.
  • Assuming alkynes are denser than water.
📋 CBSE Competency-Based Case Study (HOTS)

A laboratory technician stores ethyne, hexyne and tetradecyne under identical conditions. He observes that ethyne is a gas, hexyne is a liquid, whereas tetradecyne exists as a solid. He also notices that none of these compounds mixes with water but all dissolve readily in benzene.

Questions

  1. Why do higher alkynes exist as solids?
  2. Why are alkynes insoluble in water?
  3. State the intermolecular force responsible for the increase in boiling point.
  4. Why do alkynes dissolve in benzene?

Answers

  1. Because increasing molecular mass strengthens van der Waals forces, leading to solidification.
  2. Because alkynes are weakly polar whereas water is highly polar.
  3. London dispersion (van der Waals) forces.
  4. Because both alkynes and benzene are predominantly non-polar, following the principle "like dissolves like".
⛽

Chemical Properties of Alkynes

📋
Table of Contents
12 sections
🗺️ Overview
The chemical properties of alkynes are mainly governed by the presence of the carbon-carbon triple bond (
\(\mathrm{C \equiv C}\)
). A triple bond contains one sigma (
\(\sigma\)
) bond and two electron-rich pi (
\(\pi\)
) bonds. The weak π bonds are easily attacked by electrophiles, making alkynes highly reactive towards addition reactions.

Unlike alkenes, terminal alkynes (alkynes containing a hydrogen atom directly attached to the triply bonded carbon) also exhibit weak acidic character.
Major Chemical Properties
  • Acidic character
  • Addition reactions
  • Polymerisation
🗒️ Acidic Character Of Alkynes
📖 Introduction
Among hydrocarbons, only terminal alkynes possess appreciable acidic character.
1
Example
\[\mathrm{HC\equiv CH}\]
contains acidic hydrogen atoms attached directly to the sp-hybridised carbon atom.
Terminal alkynes react with sodium metal to form sodium acetylide with evolution of hydrogen gas.
Reaction with Sodium Metal
\[ \ce{\boxed{ 2HC\equiv CH + 2Na \rightarrow 2HC\equiv CNa + H_2 \uparrow }} \]
Evolution of hydrogen gas confirms the acidic nature of terminal alkynes.

Reaction with Sodamide (\(\mathrm{NaNH_2}\))

Sodamide is a very strong base that removes the acidic hydrogen atom of terminal alkynes.
\[ \ce{\boxed{HC\equiv CH+NaNH_2\rightarrow HC\equiv CNa+NH_3}} \]
The product formed is known as sodium acetylide.
🤔 Why are Terminal Alkynes Acidic?
The acidic nature of alkynes is explained on the basis of hybridisation.
Hydrocarbon Hybridisation % s Character
Ethane sp3 25%
Ethene sp2 33%
Ethyne sp 50%
An increase in the percentage of s-character increases the electronegativity of carbon.
Therefore, the sp-hybridised carbon atom attracts the shared electron pair of the C-H bond more strongly, making hydrogen easier to lose as a proton.
Trend of Acidity
\[ \ce{\boxed{\mathrm{HC\equiv CH>CH_2=CH_2>CH_3CH_3}}} \]
Board Point
Only terminal alkynes show acidic character. Internal alkynes do not possess acidic hydrogen atoms.
Comparison of Acidity
Compound Hybridisation Acidic Nature
Ethane sp3 Very Weak
Ethene sp2 Weak
Ethyne sp Highest
🗺️ Addition Reactions
The carbon-carbon triple bond contains two weak π bonds. Therefore, alkynes readily undergo addition reactions.
Since two π bonds are present, alkynes can add two molecules of many reagents.

Important NCERT Point
Most addition reactions of alkynes proceed through the electrophilic addition mechanism.
🔁 1. Addition Of Dihydrogen (Hydrogenation)
Alkynes react with hydrogen in the presence of nickel, platinum or palladium catalyst.
Initially an alkene is produced, which on further hydrogenation gives an alkane.
\[ \begin{aligned}\ce{HC\equiv CH+H_2&\xrightarrow{Ni/Pt/Pd}CH_2=CH_2\\\\ CH_2=CH_2+H_2&\xrightarrow{Ni/Pt/Pd}CH_3CH_3 }\end{aligned} \]
Overall Reaction
\[ \ce{\boxed{HC\equiv CH+2H_2\rightarrow CH_3CH_3}} \]
🔁 2. Addition Of Halogens
Bromine and chlorine add across the triple bond.
First Molecule of Bromine
\[ \ce{\boxed{CH_3C\equiv CH+Br_2\rightarrow CH_3CBr=CHBr}} \]
Second Molecule of Bromine
\[ \ce{\boxed{CH_3CBr=CHBr+Br_2\rightarrow CH_3CBr_2CHBr_2}} \]

Test for Unsaturation

The reddish-brown colour of bromine dissolved in carbon tetrachloride disappears because bromine adds across the multiple bond.
🔁 3. Addition Of Hydrogen Halides
Hydrogen halides (HCl, HBr and HI) add to alkynes according to Markovnikov's rule.
Addition of two molecules finally produces a geminal dihalide, in which both halogen atoms are attached to the same carbon atom.
  • 1
    Step 1
    \[\boxed{HC\equiv CH+HBr\rightarrow CH_2=CHBr}\]
  • 2
    Step 2
    \[\boxed{CH_2=CHBr+HBr\rightarrow CH_3CHBr_2}\]
📘 Definition
A geminal dihalide is a compound in which both halogen atoms are attached to the same carbon atom.
🔁 4. Addition Of Water (Hydration)
Alkynes do not react with water under ordinary conditions.
However, in the presence of dilute sulphuric acid and mercuric sulphate (
\(\mathrm{HgSO_4}\)
), water adds across the triple bond.
Initially an unstable enol is produced, which immediately rearranges into a more stable carbonyl compound by keto-enol tautomerism.
  • 1
    Step 1 : Formation of Enol
    \[\boxed{HC\equiv CH+H_2O\xrightarrow[\mathrm{HgSO_4}]{\mathrm{H_2SO_4}}CH_2=CHOH}\]
  • 2
    Step 2 : Keto-Enol Tautomerism
    \[\boxed{CH_2=CHOH\longrightarrow CH_3CHO}\]
🗒️ NCERT Fact
Hydration of ethyne ultimately produces ethanal (acetaldehyde).
🔁 Polymerisation
Alkynes undergo both linear polymerisation and cyclic polymerisation.
  • 1
    (a) Linear Polymerisation
    Under suitable conditions, numerous ethyne molecules combine to form polyacetylene (polyethyne).
    \[\boxed{n(HC\equiv CH)\longrightarrow (-CH=CH-)_n}\]
    The polymer contains alternating double bonds and behaves as a conducting polymer after suitable doping.
    Applications
    • Conducting polymers.
    • Rechargeable batteries.
    • Electronic devices.
    • Sensors.
  • 2
    (b) Cyclic Polymerisation
    When ethyne is passed through a red-hot iron tube at about 873 K, three molecules combine to form benzene.
    \[\boxed{3HC\equiv CH\xrightarrow{873\,K}C_6H_6}\]
    This reaction is called cyclotrimerisation.
Industrial Importance
The benzene produced serves as an important starting material for dyes, plastics, pharmaceuticals, detergents, explosives and synthetic fibres.
🗒️ Summary of Important Addition Reactions
Reaction Reagent Main Product
Hydrogenation
\(\mathrm{H_2/Ni}\)
 Alkane
Halogenation
\(\mathrm{Br_2}\)
,
\(\mathrm{Cl_2}\)
 Tetrahalide
Hydrogen Halide Addition HCl, HBr, HI Geminal Dihalide
Hydration
\(\mathrm{HgSO_4/H_2SO_4}\)
 Aldehyde or Ketone
Linear Polymerisation Suitable Catalyst Polyacetylene
Cyclic Polymerisation 873 K, Iron Tube Benzene
✏️ Example
Solved Concept Example
2
Question
Why does ethyne react with sodium metal whereas ethene and ethane do not?
  1. 1
    Compare hybridisation.
  2. 2
    Relate s-character to acidity.
The carbon atoms of ethyne are sp hybridised and possess 50% s-character. This makes carbon more electronegative and increases the polarity of the C-H bond. Consequently, the hydrogen atom can be removed as a proton by strong bases such as sodium metal or sodamide. Ethene (sp2) and ethane (sp3) possess lower s-character and therefore do not exhibit appreciable acidic behaviour.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming all alkynes are acidic instead of only terminal alkynes.
  • Confusing geminal dihalides with vicinal dihalides.
  • Writing ethanol instead of ethanal after hydration of ethyne.
  • Ignoring keto-enol tautomerism.
  • Forgetting that alkynes can add two molecules of halogens or hydrogen halides.
  • Writing benzene formation as linear polymerisation instead of cyclic polymerisation.
📋 CBSE Competency-Based Case Study (HOTS)

A research laboratory compares ethane, ethene and ethyne by treating each compound with sodium metal and bromine dissolved in carbon tetrachloride. Only one compound evolves hydrogen gas with sodium, and both unsaturated compounds decolourise bromine solution. The same compound also produces benzene when passed through a red-hot iron tube at 873 K.

Questions

  1. Which hydrocarbon evolves hydrogen gas with sodium? Explain.
  2. Why does bromine solution become colourless in the presence of alkynes?
  3. What is the product obtained after hydration of ethyne using
    \(\mathrm{HgSO_4/H_2SO_4}\)
    ?
  4. Name the reaction in which three molecules of ethyne combine to form benzene.

Answers

  1. Terminal alkyne (ethyne), because its acidic hydrogen attached to an sp-hybridised carbon is replaced by sodium.
  2. Bromine adds across the carbon-carbon triple bond, consuming bromine and removing its reddish-brown colour.
  3. Ethanal (
    \(\mathrm{CH_3CHO}\)
    ), formed through keto-enol tautomerism.
  4. Cyclic polymerisation (cyclotrimerisation).
⛽

Aromatic Hydrocarbons (Arenes)

📋
Table of Contents
10 sections
🗺️ Overview
Aromatic hydrocarbons are a special class of cyclic hydrocarbons that exhibit exceptional stability due to the presence of a delocalised π-electron system. Unlike alkenes and alkynes, aromatic compounds do not readily undergo addition reactions because doing so would destroy their aromatic stability.

These compounds are commonly known as arenes. The simplest and the most important aromatic hydrocarbon is benzene (
\(\mathrm{C_6H_6}\)
), which serves as the parent compound of thousands of aromatic compounds.
📘 Definition
🤔 Did You Know?
Why are They Called Aromatic?
The term "aromatic" originated from the Greek word aroma, meaning pleasant smell.

Many of the first compounds discovered in this family possessed pleasant fragrances. Consequently, chemists grouped them together as aromatic compounds.

Modern chemistry, however, defines aromaticity on the basis of electronic structure rather than smell. Many aromatic compounds are odourless, while some possess unpleasant odours.
🌟 Important Board Point
📘 Benzene – The Parent Aromatic Hydrocarbon
🎨 SVG Diagram
Structure of Benzene
BENZENE (C₆H₆) — STRUCTURE & RESONANCE HYBRID NCERT CLASS 11 CHEMISTRY • UNIT 13: HYDROCARBONS (AROMATICITY) H H H H H H C C C C C C 120° C–C = 139 pm KEKULÉ RESONANCE STRUCTURES (I) (II) NCERT KEY CONCEPTS: • Planar Hexagonal Ring: All 6 Carbon atoms are sp² hybridized with bond angles of 120°. • Intermediate C–C Bond Length: Equal bond length of 139 pm lies between C–C single (154 pm) and C=C double (134 pm) bonds. • Aromaticity (Hückel's Rule): Contains 6 π-electrons satisfying (4n + 2) π rule where n = 1.
💡 Key Concept
🔷 Characteristics of Aromatic Hydrocarbons
🔷 Characteristics
  • Cyclic compounds.
  • Planar molecular structure.
  • Completely conjugated π-electron system.
  • Highly stable because of resonance.
  • Prefer substitution reactions rather than addition reactions.
  • Most contain one or more benzene rings.
🤔 Why Does Benzene Not Readily Undergo Addition Reactions?
Although benzene contains three carbon-carbon double bonds, it behaves differently from alkenes.
Addition reactions would destroy the continuous delocalised π-electron cloud responsible for aromatic stability.
Therefore, benzene generally undergoes electrophilic substitution reactions instead of addition reactions.
⚖️ Comparison
Hydrocarbon Main Reaction
Alkane Substitution
Alkene Addition
Alkyne Addition
Benzene Electrophilic Substitution
🗂️ Classification of Aromatic Hydrocarbons
Aromatic compounds containing one or more benzene rings are known as benzenoid compounds.
Examples
  • Benzene
  • Toluene
  • Xylene
  • Naphthalene
  • Anthracene
Some aromatic compounds do not contain a benzene ring but still possess aromatic character because they satisfy the conditions of aromaticity.
Examples
  • Azulene
  • Tropylium ion
  • Cyclopropenyl cation
⛽

Nomenclature of Aromatic Hydrocarbons

📋
Table of Contents
9 sections
📌 Note
🤔 Did You Know?
Why Does Benzene Give Only One Monosubstituted Product?
All six hydrogen atoms in benzene are chemically equivalent because of resonance and molecular symmetry. Therefore, replacing any one hydrogen atom produces the same compound.
Example
Replacing any hydrogen atom by chlorine always produces only chlorobenzene.
📌 Isomerism in Disubstituted Benzene
🎨 SVG Diagram
Ortho, Meta and Para Positions
G ISOMERISM IN SUBSTITUTED BENZENES (POSITION) NCERT CLASS 11 CHEMISTRY • HYDROCARBONS • ARENES Ortho-Isomer (1,2)-disubstituted Y 1 2 ortho Meta-Isomer (1,3)-disubstituted Y 1 2 3 meta Para-Isomer (1,4)-disubstituted Y 1 2 3 4 para Positions relative to G: C2 & C6 = Ortho | C3 & C5 = Meta | C4 = Para For disubstituted arenes C₆H₄GY, three position isomers exist. The priority group (G) dictates numbering starting from C1.
✏️ Examples of Positional Isomers
Compound IUPAC Name Common Name
\(\mathrm{Cl}\)
at 1,2		 1,2-Dichlorobenzene o-Dichlorobenzene
\(\mathrm{Cl}\)
at 1,3		 1,3-Dichlorobenzene m-Dichlorobenzene
\(\mathrm{Cl}\)
at 1,4		 1,4-Dichlorobenzene p-Dichlorobenzene
⚡ Quick Revision
Feature Aromatic Hydrocarbons
Common Name Arenes
Parent Compound Benzene
Main Reaction Electrophilic Substitution
Ring Type Planar Hexagonal Ring
π Electrons Delocalised
Monosubstituted Products Only One
Disubstituted Isomers Ortho, Meta, Para
✏️ Example
Solved Example
1
Question
Why does benzene produce only one monochloro derivative but three dichloro derivatives?
  1. 1
    Consider the symmetry of benzene.
  2. 2
    Compare mono- and disubstitution.
All six hydrogen atoms in benzene are chemically equivalent because of resonance and symmetry. Therefore, substitution of any one hydrogen atom produces the same monochloro derivative (chlorobenzene). However, when two hydrogen atoms are replaced, the substituents may occupy the 1,2-, 1,3- or 1,4-positions, producing three positional isomers known as ortho, meta and para.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming aromatic compounds always have a pleasant smell.
  • Treating benzene like an alkene and predicting addition reactions under ordinary conditions.
  • Confusing benzenoid and non-benzenoid compounds.
  • Interchanging ortho, meta and para positions.
  • Assuming benzene forms multiple monosubstituted products.
📋 CBSE Competency-Based Case Study (HOTS)

A student chlorinates benzene under suitable laboratory conditions and obtains only one monochloro product. On further chlorination of another benzene derivative containing one chlorine atom, three different dichloro compounds are isolated.

Questions

  1. Why does benzene produce only one monochloro derivative?
  2. Name the three positional isomers obtained after disubstitution.
  3. Why does benzene generally undergo substitution rather than addition reactions?
  4. Differentiate between benzenoid and non-benzenoid aromatic compounds.

Answers

  1. Because all six hydrogen atoms in benzene are chemically equivalent due to resonance and molecular symmetry.
  2. Ortho (1,2), Meta (1,3) and Para (1,4).
  3. Because addition would destroy the aromatic π-electron delocalisation and reduce aromatic stability.
  4. Benzenoid compounds contain one or more benzene rings, whereas non-benzenoid aromatic compounds exhibit aromaticity without containing a benzene ring.
⛽

Structure of Benzene

📋
Table of Contents
17 sections
🗺️ Overview
Benzene is the most important aromatic hydrocarbon and serves as the parent compound of the aromatic family. It was first isolated by Michael Faraday in 1825 from the oily residue obtained during the production of illuminating gas.

The molecular formula of benzene is
\[ \boxed{\mathrm{C_6H_6}} \]
This formula indicates a very high degree of unsaturation because six carbon atoms are associated with only six hydrogen atoms. On the basis of the molecular formula alone, benzene should behave like an alkene or an alkyne. Surprisingly, benzene is extraordinarily stable and undergoes electrophilic substitution reactions instead of addition reactions.
🤔 Did You Know?
Why was the structure of benzene a puzzle?
  • Its molecular formula suggested three carbon-carbon double bonds.
  • It readily formed a triozonide, indicating the presence of three double bonds.
  • It produced only one monosubstituted derivative, proving that all six carbon atoms and all six hydrogen atoms are chemically equivalent.
  • Its chemical behaviour was much more stable than expected for an unsaturated compound.
🏛️ Historical Development of the Structure of Benzene
Year Scientist Contribution
1825 Michael Faraday Isolated benzene.
1865 August Kekulé Proposed the cyclic hexagonal structure with alternating single and double bonds.
Later Linus Pauling and Others Explained the structure using resonance and delocalisation of π electrons.
⚛️ Kekulé Structure of Benzene
In 1865, August Kekulé proposed that benzene consists of a six-membered carbon ring with alternating single and double bonds.

According to his model:
  • Six carbon atoms form a regular hexagonal ring.
  • Adjacent carbon atoms are connected alternately by single and double bonds.
  • Each carbon atom is bonded to one hydrogen atom.
  • Each carbon atom satisfies its valency of four.
🎨 SVG Diagram
Kekulé's Structure
KEKULÉ STRUCTURES OF BENZENE (C₆H₆) Two resonance contributors · August Kekulé, 1865 Kekulé I C C C C C C H H H H H H ═ resonance Kekulé II C C C C C C H H H H H H LEGEND C–C single bond (1.54 Å reference) C=C double bond (1.34 Å reference) Carbon (C) Hydrogen (H) Actual C–C bond: 1.40 Å Bond angles: 120° 6 π electrons delocalized Molecular symmetry: D₆ₕ The true structure is a resonance hybrid — all six C–C bonds are equivalent, intermediate between single and double bond character. Double bonds in Structure I and II are rotated by 60° relative to each other.
✅ Achievement of Kekulé's Structure
The proposed structure successfully explained:
  • Molecular formula
    \(\mathrm{C_6H_6}\)
    .
  • Tetravalency of carbon.
  • Presence of three carbon-carbon double bonds.
⚠️ Limitations of Kekulé's Structure
Although Kekulé's structure explained many properties of benzene, several experimental observations could not be explained.
  • 1. Only One Monosubstituted Product Since all six carbon atoms are equivalent, benzene gives only one monosubstituted derivative such as chlorobenzene.
    Kekulé's structure successfully explains this observation.
  • 2. Two Possible Ortho Dibromo Compounds According to Kekulé's structure, two different ortho-dibromobenzenes should exist.
    • Both bromine atoms attached to doubly bonded carbons.
    • Both bromine atoms attached to singly bonded carbons.
    However, experimentally only one ortho-dibromobenzene is obtained.
  • 3. High Stability of Benzene Ordinary alkenes readily undergo addition reactions.
    Benzene, despite containing three double bonds, prefers substitution reactions and resists addition.
  • 4. Equal Carbon-Carbon Bond Lengths Modern X-ray diffraction studies show that all six carbon-carbon bonds in benzene have exactly the same length:
    \[ \boxed{139\ \text{pm}} \]
    This value lies between:
    • Single bond = 154 pm
    • Double bond = 134 pm
    Kekulé's model predicts alternating long and short bonds, which is incorrect.
📌 Kekulé's Oscillation Hypothesis
🗒️ Resonance And Stability Of Benzene
Modern Valence Bond Theory explains the structure of benzene using the concept of resonance.
The actual benzene molecule is not represented by either Kekulé structure alone. Instead, it is a resonance hybrid of several contributing structures.
Definition of Resonance
When a molecule cannot be represented satisfactorily by a single Lewis structure, it is described by two or more contributing structures. The actual molecule is called the resonance hybrid.
Main Resonating Structures
The two Kekulé structures contribute most significantly to the resonance hybrid.
Representation of Resonance Hybrid
The resonance hybrid is represented by a hexagon containing a circle.
🎨 SVG Diagram
Representation of Resonance Hybrid
RESONANCE STRUCTURE OF BENZENE (C₆H₆) The true structure is a hybrid of two Kekulé contributors RESONANCE CONTRIBUTORS Kekulé I C C C C C C H H H H H H Kekulé II C C C C C C H H H H H H resonance Resonance Hybrid ↓ RESONANCE HYBRID C C C C C C H H H H H H 6 π electrons delocalized all C–C = 1.40 Å BOND KEY C=C double C–C single C–C hybrid π electrons ATOMS Carbon (C) sp² hybrid Hydrogen (H) 1s Bond angles 120° · D₆ₕ symmetry · Bond order 1.5 · Resonance energy ≈ 36 kcal/mol
📌 Orbital Structure of Benzene
👁️ Important Observation
🔍 Bond Length and Bond Strength in Benzene
Bond Bond Length (pm)
C-C Single Bond 154
C=C Double Bond 134
Benzene C-C Bond 139
The equal bond length of 139 pm proves that all carbon-carbon bonds possess identical bond order, intermediate between a single and a double bond.
🗒️ Resonance Energy And Stability
The delocalisation of π electrons lowers the total energy of the molecule.
This extra stability is called resonance energy.
Definition
Resonance energy is the difference between the energy of the actual benzene molecule and that of the most stable contributing Kekulé structure.

Because of resonance energy:
  • Benzene is unusually stable.
  • Addition reactions are difficult.
  • Substitution reactions are favoured.
  • All carbon-carbon bonds become identical.
Kekulé Structure vs Resonance Structure
Property Kekulé Structure Actual Benzene
Double Bonds Alternating Delocalised
C-C Bond Length Two Different Lengths Equal (139 pm)
Stability Ordinary Unsaturated Compound Highly Stable
Main Reaction Addition Expected Substitution Preferred
Electron Distribution Localised Delocalised
⚡ Quick Revision
Feature Benzene
Molecular Formula
\(\mathrm{C_6H_6}\)
Hybridisation sp2
Shape Planar Hexagonal
Bond Angle
\(120^\circ\)
C-C Bond Length 139 pm
Nature of π Electrons Delocalised
Main Reaction Electrophilic Substitution
Reason for Stability Resonance
✏️ Example
Solved Example
1
Question
Why does benzene undergo substitution reactions instead of addition reactions although it contains three double bonds?
  1. 1
    Consider the role of resonance.
  2. 2
    Compare the stability before and after addition.
The six π electrons in benzene are delocalised over the entire ring, giving exceptional resonance stability. Addition reactions would destroy this delocalisation and reduce the stability of the molecule. Therefore, benzene prefers electrophilic substitution reactions, in which the aromatic π-electron system remains intact.
⚡ Exam Tip
❌ Common Mistakes
  • Drawing benzene with fixed alternating double bonds as the actual structure.
  • Assuming the circle inside benzene represents a physical bond.
  • Writing alternating bond lengths instead of equal bond lengths.
  • Confusing oscillation with resonance.
  • Predicting addition reactions as the major reactions of benzene.
📋 CBSE Competency-Based Case Study (HOTS)

A student compares cyclohexene and benzene. Although both compounds contain unsaturation, cyclohexene readily decolourises bromine solution, whereas benzene does not under ordinary conditions. X-ray diffraction further reveals that every carbon-carbon bond in benzene has the same length of 139 pm.

Questions

  1. Why are all carbon-carbon bonds in benzene equal in length?
  2. Why does benzene prefer substitution reactions instead of addition reactions?
  3. What does the circle inside the benzene ring represent?
  4. Name the scientist who proposed the cyclic structure of benzene.

Answers

  1. Because of resonance and complete delocalisation of six π electrons over the ring.
  2. Because addition would destroy aromatic resonance and decrease the stability of the molecule.
  3. It represents six delocalised π electrons spread uniformly over the benzene ring.
  4. August Kekulé (1865).
⛽

Aromaticity

📋
Table of Contents
12 sections
🗺️ Overview
The extraordinary stability of benzene led chemists to recognize a special property known as aromaticity. Initially, only benzene and its derivatives were considered aromatic compounds. Modern chemistry has shown that many cyclic compounds without a benzene ring also exhibit aromatic behaviour, provided they satisfy certain structural and electronic requirements.

Thus, aromaticity is not determined by the presence of a benzene ring alone, but by the arrangement and delocalisation of π electrons within a cyclic molecule.
📘 Definition
🧭 Conditions for Aromaticity
A compound must satisfy all of the following conditions to be aromatic.
  • 1
    1. Planarity
    The molecule must be planar, i.e., all atoms involved in the conjugated ring should lie in the same plane.
    Planarity allows effective sideways overlap of adjacent p orbitals, which is essential for electron delocalisation.
    Why is Planarity Important?
    If the molecule becomes non-planar, the p orbitals cannot overlap continuously and aromaticity is lost.
  • 2
    2. Complete Delocalisation of π Electrons
    Every atom in the ring should possess an unhybridised p orbital so that a continuous cyclic overlap of p orbitals is possible.
    The π electrons should not remain localised between two atoms. Instead, they must be distributed over the entire ring.
    Key Concept
    Continuous overlap of p orbitals produces a delocalised π-electron cloud, which is mainly responsible for the stability of aromatic compounds.
  • 3
    3. Hückel's Rule (
    \(4n+2\)
    Rule)
    The total number of delocalised π electrons present in the cyclic conjugated system must satisfy the expression
    \[\boxed{\pi\ \text{electrons}=4n+2}\]
    where
    \[n=0,1,2,3,\ldots\]
    This relationship is known as Hückel's Rule.
    Allowed Number of π Electrons
    \(n\)
    \(4n+2\)
    		 Aromatic? Example
    0 2 Yes Cyclopropenyl cation
    1 6 Yes Benzene
    2 10 Yes Naphthalene
    3 14 Yes Anthracene (overall aromatic system)
🤔 Did You Know?
Why Does Hückel's Rule Work?
According to Molecular Orbital Theory, cyclic conjugated molecules having
\(4n+2\)
 π electrons possess completely filled bonding molecular orbitals.
This arrangement lowers the total energy of the molecule and provides remarkable stability.
Compounds containing
\(4n\)
 π electrons generally possess partially filled or degenerate orbitals, making them comparatively unstable.
Board Level Understanding
For Class XI, remember that aromatic compounds satisfy
\(4n+2\)
 π electrons, whereas compounds with
\(4n\)
 π electrons are generally not aromatic.
✏️ Examples of Aromatic Compounds
Compound π Electrons Planar Aromatic
Benzene 6 Yes Yes
Naphthalene 10 Yes Yes
Anthracene 14 Yes Yes
Phenanthrene 14 Yes Yes
✏️ Examples of Non-Benzenoid Aromatic Compounds
A benzene ring is not essential for aromaticity. Several compounds without a benzene ring are aromatic because they satisfy all the conditions of aromaticity.
Compound Reason for Aromaticity
Cyclopropenyl Cation 2 π electrons (
\(4n+2\)
,
\(n=0\)
)
Tropylium Ion 6 π electrons
Azulene 10 π electrons
NCERT Note
These compounds are mentioned only to illustrate that aromaticity depends on electron delocalisation rather than the presence of a benzene ring.
⚖️ Aromatic vs Non-Aromatic Compounds
Property Aromatic Compound Non-Aromatic Compound
Structure Cyclic May be cyclic or open chain
Planarity Planar Not necessary
Conjugation Complete May be absent
π Electrons
\(4n+2\)
 Does not satisfy Hückel's rule
Stability Very High Ordinary
Main Reaction Electrophilic Substitution Addition or other reactions
🤔 Did You Know?
How to Check Aromaticity?
Stepwise Method
  1. Check whether the compound is cyclic.
  2. Confirm that the ring is planar.
  3. Verify continuous conjugation (each atom should possess a p orbital).
  4. Count the total number of delocalised π electrons.
  5. Apply Hückel's Rule:
    \[4n+2\]
    If satisfied, the compound is aromatic.
✏️ Example
Solved Example
1
Question
Why is benzene aromatic although it contains three double bonds?
  1. 1
    Verify the three conditions of aromaticity.
  2. 2
    Apply Hückel's Rule.
Benzene is a cyclic and planar molecule in which all six carbon atoms are sp2-hybridised, allowing complete delocalisation of π electrons. It contains six π electrons, which satisfy Hückel's Rule:
\[4n+2=6\]
where
\(n=1\)
. Therefore, benzene is aromatic and exhibits exceptional stability.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming every cyclic compound is aromatic.
  • Ignoring the requirement of planarity.
  • Counting σ electrons instead of π electrons.
  • Applying Hückel's Rule without checking conjugation.
  • Believing that only benzene derivatives can be aromatic.
📋 CBSE Competency-Based Case Study (HOTS)

A chemistry student examines four cyclic compounds. Compound A is cyclic and planar but contains only four π electrons. Compound B is cyclic, planar and contains six delocalised π electrons. Compound C is cyclic but non-planar, whereas Compound D is an open-chain conjugated hydrocarbon.

Questions

  1. Which compound is expected to be aromatic?
  2. State Hückel's Rule.
  3. Why is planarity essential for aromaticity?
  4. Can an aromatic compound exist without a benzene ring? Give one example.

Answers

  1. Compound B, because it is cyclic, planar, fully conjugated and contains six π electrons satisfying Hückel's Rule.
  2. Aromatic compounds must contain
    \((4n+2)\)
    π electrons, where
    \(n=0,1,2,\ldots\)
    .
  3. Planarity allows continuous sideways overlap of p orbitals, resulting in complete delocalisation of π electrons.
  4. Yes. Examples include the tropylium ion, azulene and the cyclopropenyl cation.
⛽

Preparation of Benzene

📋
Table of Contents
8 sections
🗺️ Overview
Benzene is one of the most important industrial organic chemicals. It serves as the starting material for the manufacture of dyes, plastics, detergents, pharmaceuticals, explosives, synthetic fibres and numerous aromatic compounds.
Industrially, benzene is obtained mainly from coal tar and petroleum refining (catalytic reforming). In the laboratory, benzene can be prepared by several convenient methods based on decarboxylation, reduction and polymerisation reactions.
Important Methods of Preparation
  • Commercial isolation from coal tar and petroleum.
  • Cyclic polymerisation (cyclotrimerisation) of ethyne.
  • Decarboxylation of sodium benzoate.
  • Reduction of phenol using zinc dust.
Commercial Sources of Benzene
  • 1. Coal Tar: Coal tar is obtained during the destructive distillation of coal. It contains a large number of aromatic hydrocarbons, from which benzene is separated by fractional distillation.
  • 2. Petroleum: Modern industries obtain most benzene from petroleum by catalytic reforming and steam cracking of petroleum fractions.
Board Note
NCERT mentions coal tar as the traditional commercial source of benzene, whereas present-day industries largely depend on petroleum refining.
🧰 Important Methods of Preparation
  • Cyclic Polymerisation (Cyclotrimerisation) of Ethyne
    Principle
    Three molecules of ethyne combine together at high temperature to form one molecule of benzene.
    The reaction is carried out by passing acetylene through a red-hot iron tube at about 873 K.
    Chemical Equation
    \[\boxed{3HC\equiv CH\xrightarrow[\mathrm{Red\ Hot\ Iron}]{873\,K}C_6H_6}\]
    Reaction Type
    • Cyclic polymerisation
    • Cyclotrimerisation reaction
    Why is it Called Cyclotrimerisation?
    Three (tri) molecules of ethyne combine to form a cyclic six-membered benzene ring.
  • Decarboxylation of Aromatic Acids
    Principle
    The sodium salt of benzoic acid undergoes decarboxylation on heating with soda lime, producing benzene.
    Soda lime is a mixture of:
    • Sodium hydroxide (
      \(\mathrm{NaOH}\)
      )
    • Calcium oxide (
      \(\mathrm{CaO}\)
      )
    Calcium oxide keeps sodium hydroxide dry and porous, improving the efficiency of the reaction.
    Chemical Equation
    \[\boxed{\mathrm{C_6H_5COONa+NaOH\xrightarrow[\mathrm{CaO}]{\Delta}C_6H_6+Na_2CO_3}}\]
    Concept
    The carboxyl group (
    \(\mathrm{-COOH}\)
    ) is removed as carbon dioxide during the reaction. Consequently, the product contains one carbon atom fewer than the original acid.
    Definition
    Decarboxylation is the removal of carbon dioxide (or the carboxyl group) from a carboxylic acid or its salt during heating.
    Concept Behind Decarboxylation
    The reaction proceeds through the removal of the carboxylate group as carbonate, while the aromatic ring gains a hydrogen atom to form benzene.
    General representation:
    \[\boxed{ArCOONa\longrightarrow ArH}\]
    where Ar represents an aromatic ring.
  • Reduction of Phenol
    Principle
    Phenol is converted into benzene by passing its vapours over strongly heated zinc dust.
    Zinc removes the oxygen atom present in the hydroxyl group as zinc oxide.
    Chemical Equation
    \boxed{\mathrm{C_6H_5OH+Zn\xrightarrow{\Delta}C_6H_6+ZnO}}
    Reaction Type
    • Reduction reaction
    • Deoxygenation reaction
    Important Observation
    The hydroxyl group of phenol is removed without disturbing the aromatic benzene ring.
⚖️ Comparison of Laboratory Methods
Method Starting Material Reagent / Condition Main Product
Cyclic Polymerisation Ethyne 873 K, Red-hot Iron Tube Benzene
Decarboxylation Sodium Benzoate Soda Lime Benzene
Reduction Phenol Heated Zinc Dust Benzene
🌟 Industrial Importance of Benzene
✏️ Example
Solved Example
1
Question
Why is soda lime used during the preparation of benzene from sodium benzoate?
  1. 1
    Identify the reaction.
  2. 2
    State the role of soda lime.
Soda lime, a mixture of sodium hydroxide and calcium oxide, promotes the decarboxylation of sodium benzoate. During heating, the carboxyl group is removed as sodium carbonate, producing benzene. Calcium oxide keeps sodium hydroxide dry and porous, making the reaction more effective.
⚡ Exam Tip
❌ Common Mistakes
  • Writing calcium hydroxide instead of calcium oxide in soda lime.
  • Confusing cyclic polymerisation with ordinary polymerisation.
  • Writing carbon dioxide instead of sodium carbonate as the product of decarboxylation.
  • Using zinc metal instead of heated zinc dust in the reduction of phenol.
  • Forgetting that three molecules of ethyne produce one molecule of benzene.
📋 CBSE Competency-Based Case Study (HOTS)

A laboratory prepares benzene by three different methods. In the first experiment, acetylene is passed through a red-hot iron tube. In the second, sodium benzoate is heated with soda lime. In the third, phenol vapours are passed over heated zinc dust.

Questions

  1. Name the reaction involved in converting ethyne into benzene.
  2. What is the role of soda lime in the preparation of benzene?
  3. Which reagent removes oxygen from phenol?
  4. Why is benzene obtained from sodium benzoate said to contain one carbon atom less than the original acid?

Answers

  1. Cyclic polymerisation (cyclotrimerisation).
  2. It brings about decarboxylation of sodium benzoate to form benzene.
  3. Heated zinc dust.
  4. Because the carboxyl group is removed during decarboxylation as sodium carbonate, reducing the carbon count by one.
⛽

Physical Properties of Aromatic Hydrocarbons

📋
Table of Contents
16 sections
🗺️ Overview
The physical properties of aromatic hydrocarbons arise from their planar cyclic structure, delocalised π-electron cloud and predominantly non-polar nature. Although aromatic compounds possess unsaturated carbon atoms, they resemble alkanes in many physical properties because the electron cloud is uniformly distributed over the aromatic ring.
The most common aromatic hydrocarbons include benzene, toluene, xylene, naphthalene and anthracene.
📘 Definition
🗒️ 1. Physical State
The physical state of aromatic hydrocarbons depends primarily upon their molecular mass.
Compound Physical State at 298 K
Benzene Colourless Liquid
Toluene Colourless Liquid
Xylene Colourless Liquid
Naphthalene White Crystalline Solid
Anthracene Crystalline Solid

Lower aromatic hydrocarbons are generally liquids, whereas higher aromatic hydrocarbons containing two or more fused rings are usually solids.
🗒️ 2. Colour And Odour
  • Pure aromatic hydrocarbons are generally colourless.
  • Most possess a pleasant characteristic odour, which led to the historical name "aromatic".
  • The presence of smell is not the criterion for aromaticity.
Important Example
Naphthalene is a white crystalline aromatic hydrocarbon with a characteristic smell. It is commonly used in mothballs for protecting clothes from insects and in toilets as a deodorising agent.
🗒️ 3. Polarity
Aromatic hydrocarbons are non-polar or very weakly polar molecules. Consequently, aromatic hydrocarbons exhibit only weak intermolecular London dispersion (van der Waals) forces.
Concept
Although benzene contains delocalised π electrons, the molecule as a whole remains non-polar because of its highly symmetrical structure.
🗒️ 4. Solubility
Aromatic hydrocarbons are practically insoluble in water because water is highly polar whereas aromatic hydrocarbons are non-polar.
However, they dissolve readily in non-polar organic solvents.
Solvent Solubility Reason
Water Insoluble Polar solvent
Benzene Completely Miscible Non-polar solvent
Diethyl Ether Soluble Weak intermolecular forces
Carbon Tetrachloride (
\(\mathrm{CCl_4}\)
)		 Highly Soluble Non-polar solvent
Chloroform Soluble Organic solvent
Golden Rule
Like dissolves like.

Non-polar aromatic hydrocarbons dissolve in non-polar organic solvents but not in polar solvents like water.
🗒️ 5. Density
Most aromatic hydrocarbons have densities lower than that of water.
Therefore, when mixed with water they float on its surface and form a separate layer.
Property Observation
Density Generally less than water
Layer Formation Upper layer over water
🗒️ 6. Melting Point And Boiling Point
The melting point and boiling point increase gradually with increase in molecular mass.
Larger aromatic molecules possess stronger London dispersion forces because of their greater surface area.
Trend
\[\boxed{ \text{Increase in Molecular Mass} \Longrightarrow \text{Increase in Intermolecular Forces} \Longrightarrow \text{Increase in Melting and Boiling Points} }\]
Example
Naphthalene has a much higher melting point than benzene because it contains two fused benzene rings and therefore stronger intermolecular attractions.
🗒️ 7. Combustion Behaviour
Aromatic hydrocarbons burn with a yellow, luminous and sooty flame.
The high carbon-to-hydrogen ratio causes incomplete combustion, producing tiny particles of carbon (soot).
Reason
  • High percentage of carbon.
  • Incomplete combustion.
  • Formation of unburnt carbon particles.
NCERT Point
The sooty flame is an important experimental characteristic used to distinguish aromatic hydrocarbons from saturated hydrocarbons.
🗒️ 8. Naphthalene And Mothballs
Naphthalene is a fused aromatic hydrocarbon consisting of two benzene rings.
It slowly changes directly from solid to vapour without passing through the liquid state.
This phenomenon is known as sublimation.
Applications of Naphthalene
  • Moth repellent in wardrobes.
  • Protection of woollen clothes.
  • Toilet deodoriser.
  • Intermediate in dye and pharmaceutical industries.
📝 Summary of Physical Properties
⚖️ Comparison of Hydrocarbon Families
Property Alkanes Alkenes Alkynes Aromatic Hydrocarbons
Nature Non-polar Weakly Polar Weakly Polar Non-polar
Water Solubility Insoluble Insoluble Insoluble Insoluble
Organic Solubility High High High High
Combustion Blue Flame Luminous Flame Luminous Flame Sooty Flame
Characteristic Feature Saturated Double Bond Triple Bond Resonance Stability
✏️ Example
Solved Example
1
Question
Why do aromatic hydrocarbons burn with a sooty flame?
  1. 1
    Compare carbon-to-hydrogen ratio.
  2. 2
    Relate it to combustion.
Aromatic hydrocarbons contain a relatively high proportion of carbon. During combustion, complete oxidation does not occur easily, resulting in the formation of fine carbon particles (soot). These glowing carbon particles produce the characteristic yellow, luminous and sooty flame.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming all aromatic compounds are liquids.
  • Confusing aromaticity with pleasant smell.
  • Writing aromatic hydrocarbons as soluble in water.
  • Stating that they burn with a clean blue flame.
  • Ignoring the sublimation property of naphthalene.
📋 CBSE Competency-Based Case Study (HOTS)

A student places benzene, naphthalene and water in separate containers. Benzene forms a separate layer when mixed with water, while naphthalene slowly disappears from an open dish over several days. During combustion, both benzene and naphthalene produce a yellow, smoky flame.

Questions

  1. Why is benzene insoluble in water?
  2. Name the process by which naphthalene disappears on standing.
  3. Why do aromatic hydrocarbons burn with a sooty flame?
  4. State one practical use of naphthalene.

Answers

  1. Because benzene is non-polar whereas water is a polar solvent.
  2. Sublimation.
  3. They possess a high carbon-to-hydrogen ratio, leading to incomplete combustion and formation of carbon (soot).
  4. It is used as a moth repellent for protecting clothes and as a toilet deodoriser.
⛽

Chemical Properties of Aromatic Hydrocarbons (Arenes)

📋
Table of Contents
16 sections
🗺️ Overview
The chemical behaviour of aromatic hydrocarbons is governed by the delocalised π-electron cloud present over the benzene ring. Although benzene contains three π bonds, it is much more stable than ordinary unsaturated hydrocarbons because of aromatic resonance.
Consequently, aromatic hydrocarbons generally undergo electrophilic substitution reactions (ESR) rather than addition reactions. In substitution reactions, one hydrogen atom of the benzene ring is replaced by another atom or group without disturbing the aromatic π-electron system.
📘 Definition
🤔 Did You Know?
Why Does Benzene Prefer Substitution Rather than Addition?
Addition reactions would destroy the delocalised π-electron cloud responsible for aromatic stability.
Substitution reactions replace only one hydrogen atom while preserving the aromatic ring. Therefore, substitution is energetically much more favourable.
💡 Key Concept
🔄 General Mechanism of Electrophilic Substitution
Almost all electrophilic substitution reactions occur through three common steps.
  • 1
    Step 1 : Generation of Electrophile
    A strong electrophile (
    \(E^+\)
    ) is generated using suitable catalysts or reagents.
  • 2
    Step 2 : Attack on Benzene Ring
    The π electrons of benzene attack the electrophile to form a positively charged intermediate called the σ-complex or arenium ion.
  • 3
    Step 3 : Restoration of Aromaticity
    Loss of a proton (
    \(H^+\)
    ) restores the aromatic ring, producing the substituted benzene derivative.
🎨 SVG Diagram
General Mechanism of Electrophilic Substitution - Illustration
Electrophilic Aromatic Substitution (SEAr) General Reaction Mechanism & Intermediate Dynamics STEP 1: Generation of the Strong Electrophile (E+) A Lewis acid catalyst typically activates a precursor to generate a powerful electron-deficient electrophile. E—Y + LA E+ + [Y—LA]– STEP 2: Electrophilic Attack & Formation of the σ-Complex (Wheland Intermediate) The nucleophilic π-system attacks the electrophile. Aromaticity is temporarily lost (Rate-Determining Step). E+ Slow (RDS) Loss of aromaticity H E + H E + H E + Resonance Hybrid + H E STEP 3: Deprotonation & Restoration of Aromaticity A weak base removes the proton from the sp3 carbon, restoring the stable aromatic π-system. + H E :B– Fast Aromaticity regained E + H—B
🔁 1. Nitration Of Benzene
📘 Definition
Nitration is the introduction of a nitro group (
\(\mathrm{-NO_2}\)
) into the benzene ring.
Benzene is heated with a nitrating mixture consisting of concentrated nitric acid and concentrated sulphuric acid.
Chemical Equation
\[ \ce{\boxed{\mathrm{C_6H_6+HNO_3\xrightarrow[\,Conc.\ H_2SO_4\,]{323-333\,K}C_6H_5NO_2+H_2O}}} \]
Role of Concentrated Sulphuric Acid
Concentrated sulphuric acid acts as a stronger acid and generates the active electrophile.
\[ \ce{HNO_3+H_2SO_4\rightarrow NO_2^++HSO_4^-+H_2O} \]
The electrophile responsible for nitration is:
\[\boxed{NO_2^+}\]
Product
Benzene → Nitrobenzene
🔁 2. Halogenation Of Benzene
In halogenation, a hydrogen atom of benzene is replaced by chlorine or bromine in the presence of a Lewis acid catalyst.
Catalysts Used
  • Anhydrous
    \(\mathrm{FeCl_3}\)
  • Anhydrous
    \(\mathrm{FeBr_3}\)
  • Anhydrous
    \(\mathrm{AlCl_3}\)
\[ \ce{\boxed{\mathrm{C_6H_6+Cl_2\xrightarrow[Anhydrous]{FeCl_3}C_6H_5Cl+HCl}}} \]
similarly
\[ \ce{\boxed{\mathrm{C_6H_6+Br_2\xrightarrow[Anhydrous]{FeBr_3}C_6H_5Br+HBr}}} \]

Electrophile

\[ \ce{\boxed{Cl^+ \quad or \quad Br^+}} \]

Product

Chlorobenzene or Bromobenzene
🔁 3. Sulphonation Of Benzene
Sulphonation is the replacement of a hydrogen atom of benzene by a sulphonic acid group.
Benzene is heated with fuming sulphuric acid (oleum).
\[ \ce{\boxed{\mathrm{C_6H_6+SO_3\xrightarrow[Oleum]{}C_6H_5SO_3H}}} \]
or
\[ \ce{\boxed{\mathrm{C_6H_6+H_2SO_4(SO_3)\rightarrow C_6H_5SO_3H+H_2O}}} \]

Electrophile

\[ \ce{\boxed{SO_3 \quad \text{or} \quad SO_3H^+}} \]
🌟 Importance

Sulphonic acids are important intermediates in the manufacture of detergents, dyes and pharmaceuticals.
🔁 4. Friedel Crafts Alkylation
📘 Definition
The reaction is carried out using an alkyl halide in the presence of anhydrous aluminium chloride.
\[ \ce{\boxed{\mathrm{C_6H_6+RCl\xrightarrow[Anhydrous]{AlCl_3}C_6H_5R+HCl}}} \]
Example
\[ \ce{\mathrm{C_6H_6+CH_3Cl\xrightarrow{AlCl_3}C_6H_5CH_3+HCl}} \]

Product

Toluene
\[ \ce{\mathrm{C_6H_6+C_2H_5Cl\xrightarrow{AlCl_3}C_6H_5C_2H_5+HCl}} \]

Product

Ethylbenzene

Electrophile

\[ \ce{\boxed{R^+}} \]
🗒️ Limitations
Polyalkylation may occur because alkyl groups activate the benzene ring.
🔁 5. Friedel Crafts Acylation
📘 Definition
In Friedel-Crafts acylation, an acyl group replaces one hydrogen atom of benzene.
The reaction uses an acyl halide or an acid anhydride in the presence of anhydrous aluminium chloride.
\[ \ce{\boxed{\mathrm{C_6H_6+RCOCl\xrightarrow[Anhydrous]{AlCl_3}C_6H_5COR+HCl}}} \]
1
Example
\begin{aligned}\ce{\mathrm{C_6H_6+CH_3COCl}&\mathrm{\xrightarrow{AlCl_3}C_6H_5COCH_3+HCl}}\\ \ce{\mathrm{C_6H_6+(CH_3CO)_2O}&\mathrm{\xrightarrow{AlCl_3}C_6H_5COCH_3+CH_3COOH}}\end{aligned}

Electrophile

\[ \ce{\boxed{RCO^+}} \]
🗒️ Advantage
Unlike alkylation, acylation generally produces only one substituted product because the acyl group deactivates the aromatic ring.
🔁 Further Electrophilic Substitution
If excess electrophilic reagent is used, additional hydrogen atoms of benzene may also be replaced.
For example, excess chlorine in the presence of anhydrous aluminium chloride converts benzene into hexachlorobenzene.
\[ \ce{\boxed{\mathrm{C_6H_6+6Cl_2\xrightarrow{AlCl_3}C_6Cl_6+6HCl}}} \]
⚖️ Comparison of Electrophilic Substitution Reactions
Reaction Reagent Catalyst Electrophile Main Product
Nitration
\(\mathrm{HNO_3}\)
 Conc.
\(\mathrm{H_2SO_4}\)
\(\mathrm{NO_2^+}\)
 Nitrobenzene
Halogenation
\(\mathrm{Cl_2/Br_2}\)
\(\mathrm{FeCl_3,\ FeBr_3,\ AlCl_3}\)
\(\mathrm{Cl^+,Br^+}\)
 Haloarene
Sulphonation Oleum Heat
\(\mathrm{SO_3}\)
 Benzenesulphonic Acid
Friedel-Crafts Alkylation Alkyl Halide
\(\mathrm{AlCl_3}\)
\(\mathrm{R^+}\)
 Alkylbenzene
Friedel-Crafts Acylation Acyl Halide
\(\mathrm{AlCl_3}\)
\(\mathrm{RCO^+}\)
 Acylbenzene
✏️ Example
Solved Example
2
Question
Why does benzene undergo electrophilic substitution rather than electrophilic addition?
  1. 1
    Consider resonance stability.
  2. 2
    Compare substitution and addition.
The six π electrons in benzene are delocalised over the entire ring, making the molecule exceptionally stable. Electrophilic substitution replaces only one hydrogen atom while preserving aromaticity. In contrast, electrophilic addition would destroy the delocalised π-electron system and significantly reduce the stability of the molecule. Therefore, benzene predominantly undergoes electrophilic substitution reactions.
⚡ Exam Tip
❌ Common Mistakes
  • Writing addition instead of substitution for benzene.
  • Using aqueous instead of anhydrous
    \(\mathrm{AlCl_3}\)
    .
  • Forgetting that concentrated
    \(\mathrm{H_2SO_4}\)
    generates the nitronium ion.
  • Confusing alkylation with acylation.
  • Ignoring the role of Lewis acids in halogenation.
📋 CBSE Competency-Based Case Study (HOTS)

A student performs five reactions on benzene. In one experiment, benzene is heated with concentrated nitric acid and concentrated sulphuric acid. In another, chlorine is passed through benzene in the presence of anhydrous ferric chloride. Later, benzene is treated with methyl chloride and acetyl chloride separately in the presence of anhydrous aluminium chloride.

Questions

  1. Name the electrophile generated during nitration.
  2. Why is anhydrous
    \(\mathrm{FeCl_3}\)
    used in chlorination?
  3. Differentiate between Friedel-Crafts alkylation and acylation.
  4. Why does benzene undergo substitution instead of addition?

Answers

  1. The electrophile is the nitronium ion,
    \(\mathrm{NO_2^+}\)
    .
  2. It acts as a Lewis acid catalyst and generates the electrophile
    \(\mathrm{Cl^+}\)
    .
  3. Alkylation introduces an alkyl group (
    \(\mathrm{R}\)
    ), whereas acylation introduces an acyl group (
    \(\mathrm{RCO}\)
    ).
  4. Substitution preserves the aromatic resonance of benzene, whereas addition destroys the delocalised π-electron system.
⛽

Mechanism of Electrophilic Substitution Reactions (SE Reaction)

📋
Table of Contents
14 sections
🗺️ Overview
The most characteristic reactions of benzene and other aromatic hydrocarbons are electrophilic substitution reactions (SE reactions). In these reactions, a hydrogen atom attached to the aromatic ring is replaced by an electrophile without destroying the aromaticity of the benzene ring.

Unlike alkenes, which undergo electrophilic addition, benzene undergoes substitution because the aromatic ring possesses exceptional resonance stability. Any reaction that destroys this aromaticity is energetically unfavorable.
📘 Definition
🗒️ General Mechanism Of Electrophilic Substitution
Almost all electrophilic substitution reactions of benzene follow the same three fundamental steps.
  1. Generation of the electrophile (
    \(E^+\)
    )
  2. Formation of σ-complex (Arenium ion)
  3. Removal of proton and restoration of aromaticity
🗒️ Step 1 : Generation Of Electrophile (\(E^+\))
An electrophile is an electron-deficient species that seeks electrons. Since benzene possesses a π-electron cloud, it attracts electrophiles.
Most electrophiles are generated in situ using strong acids or Lewis acids.

Role of Lewis Acid (\(\mathrm{AlCl_3}\))

Anhydrous aluminium chloride is a Lewis acid because it can accept an electron pair. It reacts with halogens, alkyl halides and acyl halides to produce strong electrophiles.
  • 1
    (a) During Chlorination
    \[\boxed{\mathrm{Cl_2+AlCl_3\rightarrow Cl^++[AlCl_4]^-}}\]
    Electrophile
    \[\boxed{Cl^+}\]
  • 2
    (b) During Friedel-Crafts Alkylation
    \[\boxed{\mathrm{CH_3Cl+AlCl_3\rightarrow CH_3^++[AlCl_4]^-}}\]
    Electrophile
    \[\boxed{R^+}\]
  • 3
    (c) During Friedel-Crafts Acylation
    \[\boxed{\mathrm{CH_3COCl+AlCl_3\rightarrow CH_3CO^++[AlCl_4]^-}}\]


    The electrophile produced is called the acylium ion.
    \[\boxed{RCO^+}\]
🗒️ Generation Of Nitronium Ion During Nitration
In nitration, the active electrophile is the nitronium ion (
\(\mathrm{NO_2^+}\)
).
It is produced by the reaction between concentrated nitric acid and concentrated sulphuric acid.

Step 1 : Proton Transfer

\[ \ce{\mathrm{HNO_3+H_2SO_4\rightleftharpoons H_2NO_3^++HSO_4^-}} \]

Step 2 : Formation of Nitronium Ion

\[ \ce{\boxed{\mathrm{H_2NO_3^+\rightarrow NO_2^++H_2O}}} \]
Important Concept
Sulphuric acid behaves as the acid by donating a proton, whereas nitric acid behaves as a base by accepting the proton.
Thus, the first step is simply an acid-base reaction.
🗒️ Step 2 : Formation Of σ Complex (Arenium Ion)
The electrophile attacks the π-electron cloud of benzene to form a positively charged intermediate known as the σ-complex, arenium ion or Wheland intermediate.

During this step:
  • One carbon atom changes from sp2 to sp3 hybridisation.
  • The aromatic ring temporarily loses its aromatic character.
  • The positive charge is delocalised by resonance over the ring.
🗒️ Why is the σ-Complex Unstable?
One carbon atom becomes sp3 hybridised, interrupting the continuous overlap of p orbitals. Consequently, aromaticity is temporarily lost, making this intermediate highly unstable.

Resonance Stabilisation of the Arenium Ion

Although the σ-complex is unstable, the positive charge is delocalised over several carbon atoms through resonance.
This resonance stabilisation lowers the energy of the intermediate and allows the reaction to proceed.
👁️
Key Observation The arenium ion is resonance stabilised but is still much less stable than benzene because complete aromaticity has been lost.
🗒️ Step 3 : Removal Of Proton And Restoration Of Aromaticity
Finally, the σ-complex loses a proton (
\(H^+\)
).
The proton is removed by:
  • \([AlCl_4]^-\)
    during halogenation, alkylation and acylation.
  • \(HSO_4^-\)
    during nitration.
Loss of the proton restores the delocalised π-electron cloud and aromaticity.
\[ \ce{\boxed{\sigma\text{-Complex}\rightarrow Substituted\ Benzene+H^+}} \]
Most Important Point
The restoration of aromaticity is the major driving force for electrophilic substitution reactions.

Energy Profile of Electrophilic Substitution<

Formation of the σ-complex requires considerable energy because aromaticity is temporarily destroyed.
The final step releases energy because aromaticity is restored.
🗒️ Addition Reactions Of Benzene
Although benzene generally resists addition reactions, under vigorous conditions it undergoes addition because aromaticity can be overcome by supplying sufficient energy.
  • 1
    1. Catalytic Hydrogenation
    Hydrogen gas adds across the benzene ring in the presence of nickel catalyst under high temperature and pressure to produce cyclohexane.
    \[\boxed{\mathrm{C_6H_6+3H_2\xrightarrow[\text{High Pressure}]{Ni,\ Heat}C_6H_{12}}}\]
    Reaction Type
    • Addition reaction
    • Catalytic hydrogenation
  • 2
    2. Addition of Chlorine (Formation of Benzene Hexachloride)
    In the presence of ultraviolet light, benzene undergoes photochemical addition of chlorine.
    \[\boxed{\mathrm{C_6H_6+3Cl_2\xrightarrow{h\nu}C_6H_6Cl_6}}\]
    The product is called:
    • Benzene Hexachloride (BHC)
    • Hexachlorocyclohexane (HCH)
    • \(\gamma\)
      -isomer is called Gammaxane (Lindane).
    Applications
    Lindane was formerly used as an agricultural insecticide. Its use is now restricted in many countries because of environmental persistence and toxicity.
🗒️ Combustion Of Benzene
Like other hydrocarbons, benzene undergoes complete combustion in excess oxygen to produce carbon dioxide and water.
\[\boxed{\mathrm{2C_6H_6+15O_2\rightarrow 12CO_2+6H_2O}}\]
or equivalently,
\[\boxed{\mathrm{C_6H_6+\dfrac{15}{2}O_2\rightarrow 6CO_2+3H_2O}}\]
Because of its relatively high carbon content, benzene burns with a yellow, luminous and sooty flame.
General Combustion Equation for Hydrocarbons
For any hydrocarbon having the formula
\(\mathrm{C_{x}H_y}\)
:
\[\boxed{\mathrm{C_{x}H_y+\left(x+\frac{y}{4}\right)O_2\rightarrow xCO_2+\frac{y}{2}H_2O}}\]
Shortcut
Remember the coefficient of oxygen:
\[\boxed{x+\frac{y}{4}}\]
📝 Summary of Important Reactions of Benzene
✏️ Example
Solved Example
1
Question
Why is the σ-complex considered the rate-determining intermediate during electrophilic substitution?
  1. 1
    Consider aromaticity.
  2. 2
    Examine the stability of the intermediate.
Formation of the σ-complex requires temporary destruction of the aromatic π-electron system, making this step energetically difficult. Consequently, it has the highest activation energy and therefore becomes the slowest (rate-determining) step. Subsequent loss of a proton rapidly restores aromaticity.
⚡ Exam Tip
❌ Common Mistakes
  • Writing the hydrogenation product as
    \(\mathrm{C_6H_{14}}\)
    instead of cyclohexane
    \(\mathrm{C_6H_{12}}\)
    .
  • Confusing the σ-complex with the final product.
  • Assuming aromaticity is retained throughout the reaction.
  • Forgetting that sulphuric acid generates the nitronium ion.
  • Writing aqueous
    \(\mathrm{AlCl_3}\)
    instead of anhydrous
    \(\mathrm{AlCl_3}\)
    .
📋 CBSE Competency-Based Case Study (HOTS)

During a laboratory experiment, a student nitrates benzene using a mixture of concentrated nitric acid and concentrated sulphuric acid. In another experiment, benzene is hydrogenated using nickel catalyst under high pressure. Finally, benzene is exposed to chlorine under ultraviolet light.

Questions

  1. Name the electrophile generated during nitration.
  2. Why is the σ-complex unstable?
  3. Which compound is formed on catalytic hydrogenation of benzene?
  4. Name the product obtained when benzene reacts with chlorine in the presence of ultraviolet light.

Answers

  1. Nitronium ion,
    \(\mathrm{NO_2^+}\)
    .
  2. Because one carbon atom becomes sp3-hybridised, interrupting π-electron delocalisation and temporarily destroying aromaticity.
  3. Cyclohexane,
    \(\mathrm{C_6H_{12}}\)
    .
  4. Benzene hexachloride (BHC), also called hexachlorocyclohexane; its
    \(\gamma\)
    -isomer is known as Lindane (Gammaxane).
⛽

Directive Influence of Functional Groups in Monosubstituted Benzene

📋
Table of Contents
10 sections
🗺️ Overview
When a monosubstituted benzene undergoes another electrophilic substitution reaction, the incoming electrophile does not attack all positions of the benzene ring equally. Instead, the substituent already present on the ring influences both:
  • The position where the new group enters.
  • The rate of the substitution reaction.
This phenomenon is known as the directive influence of substituents.
📘 Definition
🗒️ Possible Products Of Further Substitution
Consider chlorination of toluene.
The incoming chlorine atom may theoretically enter at three different positions.
  • Ortho (1,2)
  • Meta (1,3)
  • Para (1,4)
Experimentally, these products are not formed in equal amounts. The major product depends upon the substituent already attached to the benzene ring.
🗂️ Types / Category
Types of Directing Groups
Substituents are broadly classified into two categories.
  1. Ortho- and Para-directing groups
  2. Meta-directing groups
  • 1. Ortho- and Para-Directing Groups Groups that direct an incoming electrophile mainly to the ortho and para positions are called ortho- and para-directing groups.
    Most of these groups donate electron density to the benzene ring through resonance (+R or +M effect), thereby increasing the electron density at the ortho and para positions.
    Definition
    An ortho-para directing group increases the electron density mainly at the ortho and para positions, making these positions more reactive towards electrophilic attack.
    Directive Influence of the Hydroxyl Group (-OH)
    Phenol is an excellent example of an ortho-para directing compound.
    The oxygen atom possesses lone pairs of electrons, which participate in resonance with the benzene ring.
    As a result:
    • Electron density increases at the ortho positions.
    • Electron density also increases at the para position.
    • These positions become highly attractive to electrophiles.
    Important Concept
    The hydroxyl group exhibits:
    • −I effect (electron withdrawing through sigma bonds)
    • +R (+M) effect (electron donation through resonance)
    The resonance effect is much stronger than the inductive effect. Therefore, the overall electron density increases at the ortho and para positions.
    Activating Groups
    Groups that donate electron density into the benzene ring increase the rate of electrophilic substitution. Such groups are called activating groups.
    Group Reason Directive Nature
    \(\mathrm{-OH}\)
    		 Strong +R effect Ortho/Para
    \(\mathrm{-NH_2}\)
    		 Strong +R effect Ortho/Para
    \(\mathrm{-NHR}\)
    		 +R effect Ortho/Para
    \(\mathrm{-NHCOCH_3}\)
    		 +R effect Ortho/Para
    \(\mathrm{-OCH_3}\)
    		 +R effect Ortho/Para
    \(\mathrm{-CH_3}\)
    		 Hyperconjugation + +I effect Ortho/Para
    \(\mathrm{-C_2H_5}\)
    		 +I effect Ortho/Para
    Why are Halogens an Exception?
    Halogens exhibit a unique behaviour.
    • They are deactivating because of their strong −I effect.
    • However, they are still ortho-para directing because they possess lone pairs that participate in resonance.
    Important Exception
    Halogens = Deactivating but Ortho-Para Directing
    Halogen Inductive Effect Resonance Effect Overall Nature
    F −I +R o/p directing, deactivating
    Cl −I +R o/p directing, deactivating
    Br −I +R o/p directing, deactivating
    I −I +R o/p directing, deactivating
  • 2. Meta-Directing Groups Groups that direct the incoming electrophile mainly towards the meta position are called meta-directing groups.
    These substituents withdraw electron density from the benzene ring by strong −I and/or −R effects.

    As a consequence:
    • Electron density at ortho and para positions decreases significantly.
    • The meta position becomes relatively richer in electron density.
    • The electrophile attacks predominantly at the meta position.
    Directive Influence of the Nitro Group (-NO₂)
    The nitro group is one of the strongest electron-withdrawing substituents.
    It withdraws electrons by:
    • Strong −I effect
    • Strong −R effect
    Consequently:
    • Ortho and para positions become electron deficient.
    • The meta position remains comparatively electron rich.
    • Electrophilic substitution occurs predominantly at the meta position.
    Important Observation
    Nitrobenzene reacts much more slowly than benzene because the nitro group strongly deactivates the aromatic ring.
    Deactivating Groups
    Groups that decrease the electron density of the benzene ring and reduce the rate of electrophilic substitution are called deactivating groups.
    Group Reason Directive Nature
    \(\mathrm{-NO_2}\)
    		 −I, −R Meta
    \(\mathrm{-CN}\)
    		 −I Meta
    \(\mathrm{-CHO}\)
    		 −I, −R Meta
    \(\mathrm{-COR}\)
    		 −I, −R Meta
    \(\mathrm{-COOH}\)
    		 −I, −R Meta
    \(\mathrm{-COOR}\)
    		 −I, −R Meta
    \(\mathrm{-SO_3H}\)
    		 −I Meta
⚖️ Comparison of Directing Groups
Property Ortho/Para Directors Meta Directors
Electron Density Increase Decrease
Main Effect +R or +I −I and/or −R
Reaction Rate Usually Faster Usually Slower
Nature Activating (except halogens) Deactivating
Major Product Ortho & Para Meta
🗒️ Golden Exception
Halogens = Ortho-Para Directing but Deactivating
✏️ Example
Solved Example
1
Question
Predict the major product obtained when phenol undergoes nitration.
  1. 1
    Identify the substituent already present.
  2. 2
    Determine its directing influence.
  3. 3
    Predict the major products.
The hydroxyl group exhibits a strong +R effect and increases the electron density at the ortho and para positions. Therefore, nitration of phenol predominantly produces ortho-nitrophenol and para-nitrophenol, with the para product usually predominating because of lower steric hindrance.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming all electron-withdrawing groups are ortho-para directing.
  • Forgetting the exception shown by halogens.
  • Confusing activating groups with ortho directors only (they usually direct to both ortho and para positions).
  • Ignoring resonance while predicting the major product.
  • Thinking the incoming electrophile decides the orientation instead of the existing substituent.
📋 CBSE Competency-Based Case Study (HOTS)

A student performs nitration on three aromatic compounds: phenol, chlorobenzene and nitrobenzene. The products obtained are analysed to determine the orientation of substitution.

Questions

  1. Which compound mainly forms ortho and para products because of resonance donation?
  2. Which compound mainly gives the meta product?
  3. Which compound is deactivating but still directs substitution to the ortho and para positions?
  4. Why does the nitro group direct substitution to the meta position?

Answers

  1. Phenol.
  2. Nitrobenzene.
  3. Chlorobenzene (and all haloarenes).
  4. Because the nitro group strongly withdraws electrons by both the −I and −R effects, reducing electron density at the ortho and para positions and making the meta position relatively more favourable for electrophilic attack.
· Updated
NCERT · Class XI Chemistry · Chapter 9

Hydrocarbons — The Chemistry of Carbon & Hydrogen

A complete self-contained learning engine covering alkanes, alkenes, alkynes and aromatic hydrocarbons — with a rule-based AI solver, formulas, exam-smart tips, common pitfalls, concept-wise practice questions with full solutions, and interactive modules.

CONCEPT 1 Alkanes — The Saturated Hydrocarbons

Alkanes (CnH2n+2) contain only carbon–carbon and carbon–hydrogen single (sigma) bonds. Every carbon is sp³ hybridised, giving free rotation about C–C bonds and a family of conformations.

Preparation Routes
1
Wurtz Reaction: 2R–X + 2Na dry ether → R–R + 2NaX. Works best for symmetrical alkanes with an even number of carbons; unsuitable for making alkanes with odd carbon count from a single halide.
2
Kolbe's Electrolysis: Electrolysis of an aqueous solution of a sodium/potassium salt of a carboxylic acid. At the anode, RCOO⁻ loses an electron, decarboxylates to R•, and two radicals combine: 2RCOO⁻ → R–R + 2CO₂ + 2e⁻.
3
Decarboxylation: Sodium salt of a carboxylic acid heated with soda lime (NaOH + CaO) gives an alkane with one carbon less: R–COONa + NaOH CaO, Δ → R–H + Na₂CO₃.
4
From Grignard Reagent: Grignard reagents react with any compound containing an active/acidic hydrogen (water, alcohols, amines) to give an alkane: R–MgX + H–OH → R–H + Mg(OH)X.
5
Catalytic Hydrogenation: Unsaturated hydrocarbons add H₂ over finely divided Ni, Pd or Pt (Sabatier–Senderens reaction) to give the corresponding alkane.
Key Reactions of Alkanes
  • Free-radical halogenation: proceeds via initiation (homolysis of X₂ by light/heat), propagation (H-abstraction, then X-abstraction), and termination (radical–radical combination). Reactivity order of C–H bonds: 3° > 2° > 1°, but this is offset by the greater number of 1° hydrogens available, so product ratios must be calculated statistically.
  • Combustion: complete combustion gives CO₂ + H₂O and releases heat; incomplete combustion (limited O₂) gives carbon black/CO — the basis of the general formula CnH2n+2 + (3n+1)/2 O₂ → nCO₂ + (n+1)H₂O.
  • Isomerism: alkanes from butane onward show chain isomerism (n-butane vs isobutane); the number of possible isomers rises sharply with carbon count.
  • Conformations of ethane: rotation about the C–C bond gives a continuum of conformations; the two extremes are staggered (most stable, minimum torsional strain) and eclipsed (least stable, maximum torsional strain), related by a Newman projection dihedral angle of 60°.

CONCEPT 2 Alkenes & Alkynes — The Unsaturated Hydrocarbons

Alkenes (CnH2n) contain one C=C double bond (one σ + one π, sp² carbons); alkynes (CnH2n−2) contain one C≡C triple bond (one σ + two π, sp carbons). The π electron cloud makes both families electron-rich and reactive toward electrophiles.

Alkene Preparation
  • Dehydration of alcohols: conc. H₂SO₄/H₃PO₄, Δ; follows Zaitsev's rule (more substituted alkene is the major product).
  • Dehydrohalogenation of alkyl halides: alcoholic KOH, Δ; also follows Zaitsev's rule.
  • Partial hydrogenation of alkynes: H₂/Pd-BaSO₄ (Lindlar's catalyst) gives the cis-alkene; Na/liq. NH₃ gives the trans-alkene.
Alkyne Preparation
  • From calcium carbide: CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂ (industrial route to ethyne).
  • Double dehydrohalogenation: vicinal or geminal dihalides treated with excess alcoholic/molten KOH or NaNH₂.
Markovnikov & Anti-Markovnikov Addition (HX to Alkenes)
1
Markovnikov's rule (normal, no peroxide): the electrophile H⁺ adds to the carbon that already carries more hydrogens, generating the more stable carbocation (3° > 2° > 1°); the halide ion then attacks this carbocation.
2
Peroxide (Kharasch) effect: with HBr only, in the presence of peroxides, the mechanism switches to free-radical addition — Br• adds first to the terminal carbon (less hindered), giving the anti-Markovnikov product. This effect is not observed with HCl or HI because of bond-energy/reactivity mismatches in the propagation step.
Other Signature Reactions
  • Ozonolysis: alkene + O₃ then Zn/H₂O cleaves the C=C bond to give two carbonyl compounds — a powerful tool to deduce the position of a double bond in an unknown alkene.
  • Oxidation: cold dilute KMnO₄ (Baeyer's reagent) gives a vicinal diol and decolourises purple colour — a classic unsaturation test.
  • Acidic hydrogen of terminal alkynes: the ≡C–H hydrogen in terminal alkynes is weakly acidic (sp carbon holds electrons closer to nucleus); reacts with Na, NaNH₂, or ammoniacal AgNO₃/Cu₂Cl₂ to give substituted metal acetylides — this test distinguishes terminal from internal alkynes.
  • Polymerisation: alkenes undergo addition polymerisation (e.g. ethene → polythene) under high pressure with suitable catalysts.

CONCEPT 3 Aromatic Hydrocarbons — Benzene & Electrophilic Substitution

Benzene (C₆H₆) is a planar, cyclic, fully conjugated ring with delocalised π electrons over all six sp² carbons, represented as a resonance hybrid of two Kekulé structures. Its unusual stability (resonance energy ≈ 152 kJ/mol) makes it undergo substitution rather than addition under normal conditions.

Criteria for Aromaticity (Hückel's Rule)
  • The ring system must be cyclic and planar, allowing continuous overlap of p-orbitals.
  • Every ring atom must have a p-orbital available for delocalisation (complete conjugation).
  • The total number of delocalised π electrons must equal 4n + 2 (n = 0, 1, 2…) — benzene has 6 π electrons (n = 1).
General Mechanism of Electrophilic Aromatic Substitution (EAS)
1
Generation of the electrophile (E⁺) using a catalyst — e.g. Br₂/FeBr₃ for bromination, conc. HNO₃/H₂SO₄ for nitration, conc. H₂SO₄ (fuming, SO₃) for sulfonation, R–Cl/anhyd. AlCl₃ for Friedel–Crafts alkylation, RCOCl/anhyd. AlCl₃ for Friedel–Crafts acylation.
2
The electrophile attacks the π cloud, forming a resonance-stabilised, non-aromatic arenium ion (σ-complex) — the rate-determining, high-energy intermediate.
3
A base (often the counter-anion) removes the H⁺ from the sp³ carbon, restoring full aromaticity and giving the substituted product.
Directive Influence of Substituents
  • Activating, ortho/para-directing: –OH, –NH₂, –OR, –NHCOCH₃, alkyl groups (–CH₃ etc.) — these donate electron density into the ring by resonance or hyperconjugation, stabilising the arenium ion most at the ortho/para positions.
  • Deactivating, meta-directing: –NO₂, –CN, –SO₃H, –CHO, –COOH, –COR — these withdraw electron density by resonance, and would place a positive charge on the substituent-bearing carbon if attack occurred at ortho/para, so meta attack is favoured.
  • Deactivating but ortho/para-directing (the odd one out): halogens (–F, –Cl, –Br, –I) — inductively electron-withdrawing (net deactivating, slower than benzene) yet still ortho/para-directing because their lone pairs can be donated by resonance into the ring at those positions.
Carcinogenicity & Toxicity

Polynuclear (fused-ring) aromatic hydrocarbons such as benzo[a]pyrene, produced during incomplete combustion of organic matter (tobacco smoke, coal, diesel exhaust), are known carcinogens — a key reason NCERT flags this chapter's environmental-health relevance.

🧮 Tool 1 — Index of Hydrogen Deficiency (Degree of Unsaturation) Calculator

Enter atom counts for a hydrocarbon (optionally with halogens/N/O). IHD = (2C + 2 + N − H − X) / 2. Oxygen does not affect the formula.

Result will appear here.

🧮 Tool 2 — Markovnikov / Anti-Markovnikov Product Predictor

Choose an unsymmetrical alkene, a hydrogen halide, and whether peroxide is present, and the solver walks through the correct mechanism.

Result will appear here.

🧮 Tool 3 — Alkane Synthesis Route Advisor

Tell the solver what starting material you have on hand; it recommends the matching named reaction and shows the steps.

Result will appear here.

General Molecular Formulas

Alkanes: CnH2n+2
Alkenes: CnH2n
Alkynes: CnH2n−2
Cycloalkanes: CnH2n
Arenes (benzene series): CnH2n−6

Index of Hydrogen Deficiency

IHD = (2C + 2 + N − H − X) / 2

Each ring or π bond (double bond) contributes 1 to IHD; a triple bond contributes 2. Oxygen is ignored in the formula.

Combustion of Alkanes

CnH2n+2 + (3n+1)/2 O₂ → n CO₂ + (n+1) H₂O

Wurtz Reaction

2R–X + 2Na → R–R + 2NaX

Kolbe's Electrolysis

2RCOO⁻ → R–R + 2CO₂ + 2e⁻ (at anode)

Decarboxylation (Soda Lime)

RCOONa + NaOH → R–H + Na₂CO₃

Hückel's Rule (Aromaticity)

π-electron count = 4n + 2 (n = 0, 1, 2, …)

Must also be cyclic, planar, and fully conjugated.

Number of Chain (Structural) Isomers

C4H10 → 2   |   C5H12 → 3   |   C6H14 → 5   |   C7H16 → 9

No simple closed-form formula exists — these must be memorised or systematically enumerated.

✅ Remembering Zaitsev's Rule

"The rich get richer" — in an elimination reaction, the double bond forms toward the carbon that already has more alkyl groups attached, giving the more substituted, more stable alkene as the major product.

✅ Spotting the Peroxide Effect Instantly

Peroxide effect (anti-Markovnikov) applies only to HBr. Remember it as "Br only, because of Bond energy": H–Br's bond is weak enough for Br• to form easily, but the propagation step is thermodynamically unfavourable for HCl and kinetically unfavourable for HI.

✅ Ortho/Para vs Meta Directors — One-Line Test

Draw the resonance structures of the arenium ion for attack at ortho, meta, and para. If the substituent's atom directly attached to the ring has a lone pair or a δ⁻, it stabilises ortho/para attack (o,p-director). If that atom carries a δ⁺ or multiple bond to a more electronegative atom (C=O, N=O, S=O), it destabilises ortho/para attack, so meta wins.

✅ Halogens Are the Exception, Not the Rule

Only halogens are simultaneously deactivating (inductive effect dominates reactivity) and ortho/para-directing (resonance donation dominates orientation). This split behaviour is a favourite exam trap — memorise it as a standalone case.

✅ Distinguishing Terminal from Internal Alkynes

Use ammoniacal AgNO₃ or Cu₂Cl₂: terminal alkynes (with ≡C–H) give a precipitate (silver/copper acetylide); internal alkynes give no reaction, since they have no acidic hydrogen on the sp carbon.

✅ Ozonolysis as a Detective Tool

To find where a double bond sits in an unknown alkene, mentally "cut" the C=C bond and cap each end with an oxygen (=O). Matching the resulting two carbonyl fragments against the ozonolysis products tells you exactly where the double bond was.

✅ Free Radical Halogenation — Statistics Beat Reactivity

3° C–H bonds are most reactive per hydrogen, but a molecule usually has far more 1° hydrogens. Always multiply (relative reactivity) × (number of that type of H) for each type before deciding the major monohalogenated product.

❌ Confusing IHD contribution of rings vs π bonds

Students often forget that a ring counts exactly like a π bond in the IHD formula — both add 1. A cyclohexene (one ring + one double bond) has IHD = 2, not 1. Always count rings and π bonds separately, then add.

❌ Applying the peroxide effect to HCl or HI

A very common slip is writing an anti-Markovnikov product for HCl/peroxide or HI/peroxide. The Kharasch effect is exclusive to HBr — with HCl the radical chain is endothermic in one step, and with HI it is endothermic in the other, so free-radical addition does not proceed to completion for either.

❌ Treating halogens as meta directors because they deactivate

Deactivating and meta-directing are not synonyms. Halogens deactivate the ring (slower reaction than benzene) yet still direct to ortho/para (resonance donation of lone pairs). Don't assume "deactivating ⇒ meta" — check the mechanism, not just the label.

❌ Writing Wurtz reaction for odd-carbon or unsymmetrical targets

Wurtz coupling of two different alkyl halides gives a statistical mixture of three products (symmetrical + symmetrical + cross product), which is inefficient and rarely the intended synthetic route. Wurtz is reliable only for making a symmetrical alkane with an even number of carbons from a single alkyl halide.

❌ Forgetting Lindlar's catalyst gives the cis-alkene (not trans)

Students frequently reverse these: Lindlar's catalyst (Pd/BaSO₄, poisoned) → cis-alkene via syn addition of H₂ from the catalyst surface. Na/liquid NH₃ (dissolving metal reduction) → trans-alkene via a radical-anion mechanism. Mixing these up is one of the most tested errors in this chapter.

❌ Assuming benzene undergoes addition reactions like alkenes

Because benzene contains "double bonds," students sometimes predict addition reactions (like with Br₂ water) as the default. In reality, aromatic stabilisation (resonance energy) makes substitution the preferred pathway under normal conditions — addition happens only under forcing conditions (e.g. Cl₂/UV light, H₂/Ni at high pressure).

Q1. A student attempts Wurtz coupling using a 1:1 mixture of CH₃CH₂Br and CH₃CH₂CH₂Br with sodium metal. List every possible product and explain why this method is a poor choice for preparing pure 2-methylbutane... wait, predict all hydrocarbon products formed.

Concept building+
1
Two different alkyl radicals are generated on the sodium surface: C₂H₅• (from ethyl bromide) and C₃H₇• (n-propyl, from 1-bromopropane).
2
Radicals combine randomly in three possible pairings: C₂H₅• + C₂H₅• → butane (C₄H₁₀); C₃H₇• + C₃H₇• → hexane (C₆H₁₄); C₂H₅• + C₃H₇• → pentane (C₅H₁₂), the cross product.
3
Conclusion: a statistical mixture of butane, hexane, and pentane is obtained (roughly 1:1:2 by simple statistics), which is why Wurtz coupling of two different halides is synthetically inefficient — separating three volatile alkanes is impractical, so this route is reserved for single, symmetrical alkyl halides.

Q2. Monochlorination of 2-methylpropane (isobutane) under UV light gives two possible products. Using bond-dissociation-based reactivity (3° : 1° ≈ 5 : 1 per hydrogen) and the number of each type of hydrogen, calculate the expected percentage of each monochlorinated product.

Numerical reasoning+
1
2-Methylpropane, (CH₃)₃CH, has 9 primary (1°) hydrogens and 1 tertiary (3°) hydrogen.
2
Weighted contribution: 1° → 9 × 1 = 9; 3° → 1 × 5 = 5. Total weighted units = 14.
3
Percentage of 1-chloro-2-methylpropane (from 1° attack) = 9/14 × 100 ≈ 64.3%. Percentage of 2-chloro-2-methylpropane, i.e. tert-butyl chloride (from 3° attack) = 5/14 × 100 ≈ 35.7%.
4
Conclusion: even though each individual 3°-H is five times more reactive, the large statistical excess of 1°-H means the primary chloride is still the major product by mass.

Q3. Sodium propanoate is electrolysed (Kolbe's method) alongside sodium butanoate in the same cell. Predict all alkane products and identify which one is the "cross" product of the mixed Kolbe electrolysis.

Application+
1
At the anode, both carboxylates lose an electron and decarboxylate: CH₃CH₂COO⁻ → CH₃CH₂• + CO₂, and CH₃CH₂CH₂COO⁻ → CH₃CH₂CH₂• + CO₂.
2
Ethyl radicals combine to give butane (C₄H₁₀); propyl radicals combine to give hexane (C₆H₁₄); and the cross-combination of an ethyl radical with a propyl radical gives pentane (C₅H₁₂) — this pentane is the cross product.
3
Conclusion: as with mixed Wurtz reactions, mixed Kolbe electrolysis gives a mixture of three alkanes, so it is only a clean method when a single carboxylate salt is electrolysed.

Q4. An unknown alkene X, on ozonolysis followed by Zn/H₂O workup, gives only one product: propanone (acetone), (CH₃)₂C=O, with no other carbonyl fragment detected. Deduce the structure of X and explain the reasoning.

Deductive reasoning+
1
Since ozonolysis of an unsymmetrical alkene normally cleaves it into two different carbonyl fragments, obtaining only one type of product means the alkene must have been symmetrical about the double bond — both carbons of C=C bear identical substituent sets.
2
Each C=O in the product traces back to one carbon of the original double bond. Since both fragments are (CH₃)₂C=O, both alkene carbons must each bear two methyl groups.
3
Structure of X: (CH₃)₂C=C(CH₃)₂, i.e. 2,3-dimethylbut-2-ene — a tetrasubstituted, symmetrical alkene, consistent with getting a single identical carbonyl product on cleavage.

Q5. But-1-yne and but-2-yne are both passed separately through ammoniacal silver nitrate solution. Predict and explain the observation in each case, and describe how this test would be used to distinguish an equimolar mixture of the two.

Application+
1
But-1-yne, CH≡C–CH₂–CH₃, has a terminal ≡C–H bond. This hydrogen is acidic (the sp carbon holds bonding electrons closer to the nucleus), so it is displaced by Ag⁺ to form a white precipitate of silver salt, AgC≡C–CH₂CH₃.
2
But-2-yne, CH₃–C≡C–CH₃, has no hydrogen directly on either sp carbon (both are substituted by methyl groups), so no acidic proton is available — no precipitate forms; the solution remains unreacted.
3
Conclusion: passing the mixture through ammoniacal AgNO₃ selectively precipitates the but-1-yne as its silver acetylide, leaving but-2-yne in solution — filtration separates the two, confirming this reagent's diagnostic value for terminal alkynes.

Q6. 3-methylbut-1-ene is treated with HBr, once with no peroxide and once with peroxide present. Give the major product in each case with full mechanistic reasoning, including any possible rearrangement.

High-order / rearrangement+
1
Without peroxide (ionic, Markovnikov): H⁺ adds to the terminal =CH₂ carbon (C1), generating a secondary carbocation at C2: (CH₃)₂CH–CH⁺–CH₃.
2
This secondary cation is adjacent to a carbon bearing a hydrogen that can undergo a 1,2-hydride shift to form the more stable tertiary cation: (CH₃)₂C⁺–CH₂–CH₃.
3
Br⁻ then attacks this rearranged tertiary cation, giving 2-bromo-2-methylbutane as the major (rearranged) product — a classic carbocation-rearrangement trap.
4
With peroxide (radical, anti-Markovnikov): Br• adds first to the less-hindered terminal carbon (C1), giving the more stable secondary radical at C2: (CH₃)₂CH–CH•–CH₃. Radical intermediates do not undergo hydride shifts the way carbocations do, so no rearrangement occurs; H is abstracted from HBr to complete the chain, giving 1-bromo-3-methylbutane as the major product.

Q7. Nitrobenzene is treated with a nitrating mixture (conc. HNO₃/H₂SO₄) to introduce a second nitro group. Predict the position of the new substituent and justify it by comparing the arenium-ion resonance structures for ortho, meta, and para attack.

Mechanistic reasoning+
1
–NO₂ is strongly electron-withdrawing by resonance: the N atom bears a formal positive charge and is doubly bonded to one oxygen, so it cannot donate electron density into the ring.
2
For ortho or para attack, one resonance structure of the arenium ion places the positive charge on the ring carbon directly bonded to –NO₂ — this is severely destabilised, since it forces two positive charges (ring cation + NO₂'s own δ⁺ nitrogen) close together.
3
For meta attack, none of the resonance structures place positive charge on the carbon bearing –NO₂, so this pathway avoids the adjacent-positive-charge penalty and is comparatively more stable.
4
Conclusion: the second nitro group enters predominantly at the meta position, giving 1,3-dinitrobenzene as the major product.

Q8. Explain, using Hückel's rule, why cyclooctatetraene (an 8-membered ring with alternating double bonds) is not aromatic even though it is fully conjugated, while the cyclopentadienyl anion (a 5-membered ring bearing a negative charge) is aromatic.

Conceptual extension+
1
Cyclooctatetraene has 8 π electrons (4 double bonds). Testing Hückel's rule 4n+2: solving 4n+2 = 8 gives n = 1.5, which is not a whole number, so the electron count fails the rule.
2
In practice, cyclooctatetraene also adopts a non-planar, "tub-shaped" geometry, breaking continuous p-orbital overlap — it fails both the electron-count and the planarity criteria, so it behaves as a simple polyene, not an aromatic system.
3
The cyclopentadienyl anion has 5 ring carbons, each sp² and each contributing one p-orbital; the ring carries 6 π electrons in total (4 from two double bonds + 2 from the lone pair of the carbanion). Testing 4n+2 = 6 gives n = 1, a whole number.
4
Conclusion: the cyclopentadienyl anion is planar, fully conjugated, and satisfies 4n+2 with n = 1, so it is aromatic — despite being an ion rather than a neutral hydrocarbon, demonstrating that Hückel's rule applies to any cyclic conjugated π system, not only neutral arenes.

Q9. Toluene (methylbenzene) is subjected to Friedel–Crafts acylation with CH₃COCl/anhyd. AlCl₃. Predict the major product and explain why acylation, unlike alkylation, does not suffer from polysubstitution or rearrangement problems.

Synthesis reasoning+
1
–CH₃ is an activating, ortho/para-directing group (electron donation by hyperconjugation), so the incoming acylium ion, CH₃CO⁺, attacks predominantly at the para position (less steric hindrance than ortho, next to the existing methyl group).
2
Major product: 1-methyl-4-acetylbenzene (para-methylacetophenone / 4′-methylacetophenone).
3
Why no polysubstitution: the newly installed acetyl group, –COCH₃, is strongly electron-withdrawing (carbonyl carbon is δ⁺), which deactivates the ring toward further electrophilic attack — the product is less reactive than the starting toluene, so the reaction self-limits to mono-substitution.
4
Why no rearrangement: the acylium ion, CH₃C≡O⁺ ↔ CH₃–C⁺=O, is resonance-stabilised and does not rearrange the way carbocations formed in Friedel–Crafts alkylation often do — this is precisely why acylation is the preferred route when a straight-chain substituent is required.

Classify Each Hydrocarbon

Click a compound, then click the family you think it belongs to. Cards turn green for correct, red for incorrect.

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Electrophilic Substitution — Directing Effect Simulator

Pick a substituent already on the benzene ring, and the simulator will show whether it is activating/deactivating and where the next electrophile will attack.

Select a substituent above to see the prediction.

Reagent & Reaction Flashcards

Click any card to flip it and reveal the reaction / rule.

Rapid-Fire Concept Quiz

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    hydrocarbons — Learning Resources

    🧠 Practice MCQs
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    Hydrocarbons-Exercises

    Frequently Asked Questions

    Hydrocarbons are organic compounds made up only of carbon and hydrogen atoms. They are classified into alkanes, alkenes, alkynes and aromatic hydrocarbons.

    Hydrocarbons are classified into saturated hydrocarbons (alkanes), unsaturated hydrocarbons (alkenes and alkynes), and aromatic hydrocarbons based on the type of carbon-carbon bonds present.

    Alkanes contain only single C-C bonds, alkenes contain at least one C=C double bond, and alkynes contain at least one C=C triple bond.

    Benzene is highly stable because of resonance and the delocalisation of six p-electrons over the ring, making it an aromatic compound.

    Benzene prefers electrophilic substitution because addition reactions destroy its aromatic stability, whereas substitution preserves the delocalised p-electron system.

    Electrophilic substitution reactions involve the replacement of a hydrogen atom on the benzene ring by an electrophile. Common examples include nitration, halogenation, sulphonation and Friedel-Crafts reactions.

    The substituent already attached to the benzene ring determines whether the incoming electrophile enters the ortho, meta or para position during electrophilic substitution.

    Activating groups such as –OH, –NH2 and alkyl groups are generally ortho-para directing, while deactivating groups like –NO2, –COOH and –SO3H are meta directing. Halogens are an important exception because they are deactivating but ortho-para directing.

    Hydrocarbons are one of the most important chapters in Organic Chemistry and form the basis for reaction mechanisms, named reactions and advanced organic chemistry questions in CBSE Board, JEE Main and NEET examinations.

    These notes cover classification, nomenclature, preparation, physical and chemical properties, reaction mechanisms, electrophilic substitution, directive effects, solved examples, diagrams, NCERT solutions, exam tips and competency-based questions.

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