NIOS Lesson 24 - HYDROCARBONS

 You have studied in the previous lesson that hydrocarbons are the compounds containing carbon and hydrogen. You also know that they are classified as aliphatic, alicyclic and aromatic hydrocarbons. 

They constitute a very important class of organic compounds and are widely used as fuels, lubricants and dry-cleaning agents. They are also used as important ingredients in medicines and in dyes.

Petroleum and coal are the major sources of various types of hydrocarbons. The products obtained from fractional distillation of petroleum and destructive distillation of coal are used almost in every sphere of life. 

Hydrocarbons are considered to be the parent organic compounds, from which other organic compounds can be derived by replacing one or more hydrogen atoms with different functional groups.

In this lesson, you will study about the preparation, important physical and chemical properties of hydrocarbons.

Hydrocarbons are organic compounds made up of only carbon (C) and hydrogen (H) atoms. They are broadly classified based on the type of bonds and structure:

🔹 Major Classification of Hydrocarbons

• Alkanes

  • Saturated hydrocarbons (only single C–C bonds).
  • General formula: ( C_nH_{2n+2} ).
  • Example: Methane (CH₄), Ethane (C₂H₆).

• Alkenes

  • Unsaturated hydrocarbons with at least one double bond (C=C).
  • General formula: ( C_nH_{2n} ).
  • Example: Ethene (C₂H₄), Propene (C₃H₆).

• Alkynes

  • Unsaturated hydrocarbons with at least one triple bond (C≡C).
  • General formula: ( C_nH_{2n-2} ).
  • Example: Ethyne (C₂H₂), Propyne (C₃H₄).

• Aromatic hydrocarbons

  • Contain one or more benzene rings (cyclic, conjugated π-electrons).
  • Example: Benzene (C₆H₆), Toluene (C₆H₅CH₃).

🔹 Subdivision Based on Structure

• Aliphatic hydrocarbons: Straight-chain, branched, or non-aromatic cyclic compounds (alkanes, alkenes, alkynes).
• Aromatic hydrocarbons: Compounds with benzene-like rings.

Hydrocarbons can be classified in different ways depending on their structure and bonding:

🔹 1. Based on Presence of Aromatic Ring

• Aliphatic hydrocarbons: Do not contain aromatic rings.
  • Subdivided into alkanes, alkenes, and alkynes.
• Aromatic hydrocarbons: Contain benzene-like rings with delocalized π-electrons.

🔹 2. Based on Structure

• Acyclic hydrocarbons: Open-chain hydrocarbons (straight or branched).
• Cyclic hydrocarbons: Carbon atoms form closed rings.
  • Alicyclic hydrocarbons: Non-aromatic rings (e.g., cyclohexane).
  • Aromatic hydrocarbons: Benzene and related compounds.

🔹 3. Based on Saturation

• Saturated hydrocarbons: Only single bonds (alkanes, cycloalkanes).
• Unsaturated hydrocarbons: One or more double/triple bonds (alkenes, alkynes).

🔹 4. Special Classes

• Cycloalkanes: Saturated hydrocarbons with ring structures.
• Polycyclic hydrocarbons: Multiple fused rings (e.g., naphthalene).
• Heteroaromatic hydrocarbons: Aromatic rings containing atoms other than carbon (like nitrogen in pyridine).

It shows hydrocarbons branching into Aliphatic (alkanes, alkenes, alkynes) and Aromatic (benzene rings), further divided into Acyclic (open-chain), Cyclic (alicyclic, aromatic), and then into Saturated, Unsaturated, Polycyclic, and Heteroaromatic categories.

OBJECTIVES

􀁺 list different methods of preparation of alkanes;

􀁺 explain the reasons for variation in physical properties of alkanes;

􀁺 draw the conformations of ethane and compare their relative stability;

􀁺 describe different chemical properties of alkanes;

􀁺 list different methods of preparation of alkenes;

􀁺 explain the physical properties of alkenes;

􀁺 describe the chemical properties of alkenes;

􀁺 list different methods of preparation of alkynes;

􀁺 explain physical and chemical properties of alkynes;

􀁺 discuss the cause of greater reactivity of alkenes and alkynes over alkanes;

􀁺 distinguish alkanes, alkenes and alkynes;

􀁺 list various fractions obtained by destructive distillation of coal;

􀁺 explain the stability of various aromatic compounds using resonance;

􀁺 state Huckel rule and its use;

􀁺 describe methods of preparation, physical properties and chemical properties of benzene;

􀁺 list various uses of hydrocarbons; and

􀁺 explain the term carcinogenicity and Toxicity.

24.1 ALKANES (PARAFFINS)

Alkanes are saturated hydrocarbons. They are very less reactive towards various reagents; hence, they are also referred to as paraffins (parum means little, affins means affinity).

24.1.1 Methods of Preparation

Some important methods of preparation of alkanes are as follows:

1. From Haloalkanes (Alkyl Halides): Monohaloalkanes can be converted to alkanes by following three methods:

a) By reduction of haloalkanes: The replacement of halogen atom of haloalkanes with hydrogen is called the reduction and can be carried out by the following reagents:

The Grignard’s reagents are used to prepare various compounds like Compounds hydrocarbons, ethers, alcohols and carboxylic acids. It reacts with the compounds containing active hydrogen and forms alkanes. An easily replaceable hydrogen atom present in the compounds is called active hydrogen. An active hydrogen is present in (i) alcohols and (ii) water and (iii) acids.

c) By Wurtz Reaction : In this reaction, an alkyl halide reacts with sodium metal in the presence of dry ether and forms the higher alkanes.

2. From Unsaturated Hydrocarbons: The unsaturated hydrocarbons (i.e. alkenes and alkynes) can be converted to alkanes by the addition of hydrogen in the presence of a catalyst like nickel, platinum or palladium.

This reaction is also called hydrogenation and is used to prepare vegetable ghee from edible oils (by converting unsaturated fats to saturated ones.)

3. From Alcohols, Aldehydes and Ketones: Alcohols, aldehydes and ketones on reduction with HI, in presence of red phosphorus, give alkanes. The general reactions are as shown below.

4. From Carboxylic Acids: Carboxylic acids can produce alkanes in a number

of ways as shown below:

i) Heating with soda lime:

In this reaction, an alkane with one carbon less than those present in the

parent carboxylic acid is obtained.

ii) By Reduction of carboxylic acid:

24.1.2 Physical Properties of Alkanes

Physical State: The physical state of alkanes depends upon the intermolecular

forces of attraction present between molecules which in turn, depend upon the

surface area of the molecules. As the molecular mass of the alkanes increases,

their surface area also increases, which in turn, increases the intermolecular

forces of attraction, and accordingly, the physical state of alkanes changes from

gaseous to liquid, and then to solid. The alkanes containing 1 to 4 carbon atoms

are gases, whereas those containing 5 to 17 carbon atoms are liquids, and the still

higher ones are solids. In the case of isomeric alkanes, the straight chain alkanes

will have maximum surface area, and hence, stronger intermolecular force of

attraction. As the branching increases, surface area decreases. Hence the

intermolecular forces of attraction decrease. Let us consider the isomers of pentane

Amongst these three isomeric compounds, neopentane will have the weakest intermolecular forces of attraction due to the smallest surface area of its molecules.

Density: The density of alkanes increases with the increase in molecular mass which increases with the increase in the number of carbon atoms. All alkanes are lighter than water i.e. their density is less than 1.0 g/cm3. The maximum density in the case of alkanes is 0.89 g cm3. The lower density of alkanes than water is due to the absence of strong intermolecular attractions in alkanes.

Boiling Point: The boiling points of alkanes also increase with the increase in the molecular mass. In straight chain alkanes, the increase in boiling points due to the increase in surface area of the molecules. Branching in a chain reduces the surface area and therefore, decreases the boiling point of alkanes. Thus, in the above example, isopentane and neopantane have a lower boiling point than pentane.

Melting Point: Similar to the boiling points, the melting points of alkanes also increase with the increase in their molecular mass, but there is no regular variation in melting point. The melting points of alkanes depend not only upon the size and shape of the molecules, but also on the arrangement (i.e. the packing) of the molecules in the crystal lattice.

In alkanes, each carbon atom is sp3 hybridized which results in a bond angle of 109°28′. In straight chain hydrocarbons the carbon atoms are arranged in a zigzag way in the chain. If the molecule contains an odd number of carbon atoms, then the two terminal methyl groups lie on the same side. So, the interaction between the alkane molecules, with odd number of carbon atoms, is less than the molecule with even number of carbon atoms, in which terminal methyl groups lie on the opposite sides.

In the above structures, we find that alkanes containing even number of carbon atoms are more symmertical and can be more closely packed as compared with alkanes containing odd number of carbon atoms and can be more closely packed.

Van der Waal’s force of attraction is stronger, due to which they have higher melting points. Therefore, the alkanes with odd number of carbon atoms have lower melting point than those having even number of carbon atoms.

24.1.3 Conformations of Ethane

You have studied section 25.3.2 that electronic displacements affect the physical and chemical properties of organic compounds. You will now study how the forces present within the molecules affect their structures and stabilities. In fact, these interactions make some geometric arrangements of atoms energetically more favorable than others.

alkanes containing odd number of carbon atoms and can be more closely packed.

Van der Waal’s force of attraction is stronger, due to which they have higher melting points. Therefore, the alkanes with odd number of carbon atoms have lower melting point than those having even number of carbon atoms.

24.1.3 Conformations of Ethane

You have studied section 25.3.2 that electronic displacements affect the physical and chemical properties of organic compounds. You will now study how the forces present within the molecules affect their structures and stabilities. In fact, these interactions make some geometric arrangements of atoms energetically more favorable than others.

The groups bonded through a sigma bond can easily rotate with respect to each other. i.e. the two — CH3 groups in ethane can rotate with respect to each other.

The different arrangement of atoms resulting from such a rotation are called conformations and each such specific conformation is called a conformer (from conformational isomer).

The conformational isomers can be represented in the following two ways:

(i) Sawhorse representations

(ii) Newman projections

The Sawhorse representations show the carbon–carbon bond from an oblique angle and indicate the spatial arrangement of all C — H bonds.

The Newman projections are easier to draw and, in such drawings, the relative

positions of atoms are easily visualized. Therefore, we will use Newman projections

to study the conformations of ethane.

Several conformations of ethane are possible. But, there are two extreme possibilities.

These are discussed below:

(i) In this conformation all the six C — H bonds are as far away as possible.

This conformation is called staggered conformation and is shown below:

(ii) Another conformation in which all the six C — H bonds are as close as

possible is shown below:

Eclipsed conformation of ethane

This is called eclipsed conformation. The three rear hydrogens are drawn little more rotated than the perfectly eclipsed positions to make them visible in the structure.

Remember that there are infinite number of possible conformations in between the staggered and the eclipsed conformations. All these conformations originate by the rotation of the C — C bond.

The staggered conformation is the most slable conformation whereas the eclipsed conformation is the least stable conformation of ethane. The eclipsed conformation

has about 12 kJ mol–1 higher energy than the staggered conformation.

Conformation means the shape, structure, or spatial arrangement of something — often used in biology to describe the body structure of animals, and in chemistry to describe the different 3D arrangements of atoms in a molecule due to rotation around single bonds.

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📘 General Definitions

• Adaptation: The act of conforming or producing conformity.
• Structure: The arrangement or formation of parts into a whole.
• Animal shape: The physical build or proportions of an animal, often judged in dog or livestock shows.
• Molecular arrangement: Any spatial arrangement of atoms in a molecule that can be obtained by rotation around a single bond. Merriam Webster Cambridge Dictionary

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🧪 Chemistry Context

• Conformations in alkanes: Molecules like ethane can rotate around their C–C single bonds, producing different spatial arrangements.
• Conformers: These are specific conformations that can interconvert by rotation.
• Eclipsed conformation: Hydrogen atoms on adjacent carbons are as close as possible, leading to higher energy due to repulsion.
• Staggered conformation: Hydrogen atoms are as far apart as possible, making this arrangement more stable. 

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🐕 Biology/Animal Science Context

• Dog shows: Animals are judged on their conformation, meaning their physical structure and how well it matches breed standards.
• Livestock: Used to evaluate body proportions, muscle distribution, and overall health. 

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📊 Comparison Table

Context	Meaning of Conformation	Example
General	Adaptation or arrangement of parts	Embryo forming into a whole
Chemistry	Spatial arrangement of atoms due to bond rotation	Staggered vs. eclipsed ethane
Biology	Shape/structure of animals	Dog judged in conformation show

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⚠️ Key Notes

• In chemistry, conformations are dynamic and can change with rotation around single bonds.
• In biology, conformation is static and refers to physical structure.
• The term is specialized, so its meaning depends heavily on the field of use

In chemistry, conformations are the different 3D shapes a molecule can adopt due to rotation around single (σ) bonds, without breaking any bonds. These conformations—like staggered and eclipsed—differ in stability and energy, and their analysis is key to understanding molecular behavior.

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🔬 What Are Conformations?

• Definition: Spatial arrangements of atoms in a molecule caused by rotation around single bonds.
• Conformational isomers (or conformers): Different shapes of the same molecule that interconvert freely by bond rotation.
• Key difference: Unlike configurations (which require breaking bonds to change), conformations change only by rotation. 

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⚡ Types of Conformations

• Staggered conformation

  • Atoms/groups are as far apart as possible.
  • Most stable due to minimized electron repulsion.
  • Example: Ethane has hydrogens at 60° dihedral angles. 

• Eclipsed conformation

  • Atoms/groups line up directly behind each other.
  • Least stable due to torsional strain (repulsion between electron clouds).
  • Example: Ethane hydrogens overlap at 0° dihedral angle. 

• Anti conformation

  • In larger molecules like butane, two bulky groups are opposite (180° apart).
  • Lowest energy arrangement. 

• Gauche conformation

  • Bulky groups are 60° apart.
  • Less stable than anti due to steric hindrance. 

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📊 Energy & Stability

Conformation	Dihedral Angle	Relative Stability	Reason
Staggered	60°	Most stable	Minimal torsional strain
Eclipsed	0°	Least stable	Maximum torsional strain
Anti	180°	Very stable	Bulky groups far apart
Gauche	60°	Moderately stable	Steric hindrance between bulky groups

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🧩 Visualization Methods

• Newman projection: View straight down a bond axis to see relative positions of substituents.
• Sawhorse projection: Angled view showing both carbons and attached groups. 

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⚠️ Key Notes

• Barrier to rotation: For ethane, ~12 kJ/mol separates staggered and eclipsed forms, meaning molecules rotate freely at room temperature. 
• Conformational analysis helps predict reactivity, stability, and interactions in organic molecules.
• Larger molecules (like butane, cyclohexane) show more complex conformations due to steric effects.

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Would you like me to expand into cyclohexane conformations (chair, boat, twist-boat) which are classic examples in organic chemistry, or keep the focus on simple alkanes like ethane and butane?

24.1.4 Chemical Properties of Alkanes

1. Halogenation reactions: The chemical reactions in which a hydrogen atom of an alkane is replaced by a halogen atom are known as halogenation. Alkanes react with chlorine in the following way.

Chlorination of methane takes place via the free radical mechanism. When the

reaction mixture is exposed to sunlight, chlorine molecules absorb energy from

sunlight and get converted to free radicals i.e. chlorine atoms with an unpaired

electron (Cl)

•

. The chlorine radicals then combine with methane and form methyl radical [CH3]

•

. The methyl radical further reacts with chlorine molecule and

produces chloromethane. This reaction continuously takes place till it is stopped Compounds

or the reactants completely react to form the products. The free radical mechanism

involves the following three steps.

(i) Chain Initiation Step: It involves the formation of free radicals.

24.1.4 Uses of Alkanes

Alkanes are used as fuel gases, solvents, drycleaning agents, lubricants and in ointments (paraffin wax). Methane is used for illuminating and domestic fuel and also for the production of other organic compounds such as haloalkanes, methanol, formaldehyde and acetylene. Propane is used as a fuel, refrigerant and as raw material in the petrochemical industry. Butane and its isomer–isobutane, are the major constituents of LPG.

INTEXT QUESTIONS 24.1

1. List four important uses of hydrocarbons.

They are used as fuels and to prepare detergents, dyes, drugs, explosives etc.

Hydrocarbons are used to prepare some important organic compounds like

alcohols, aldehydes, carboxylic acids etc.

2. What is Grignard’s reagent in a molecule?

The alkyl magnesium halides (R-MgX) are called Grignard’s reagent.

3. What is an active hydrogen in a molecule?

Easily replaceable hydrogen present in a molecule is called active hydrogen.

4. What makes the physical properties of various hydrocarbons different?

The physical properties of hydrocarbons differ from one another due to difference in molecular mass, surface area, intermolecular force of attraction.

5. Name two alkanes which are gases and two alkanes which are liquids at room temperature.

Methane and ethane are gases, pentane and hexane are liquids.

6. Name three isomers of pentane.

Three isomers of pentane are: n-pentane, isopentane and neopentane.

7. Which one has higher b.p. n-butane or n-pentane? Explain.

n-pentane has higher boiling point than n-butane.

8. Write the balanced chemical equation for the complete combustion of propane.

n-pentane has higher boiling point than n-butane.

24.2 Alkenes

These are unsaturated hydrocarbons containing at least one double bond between two carbon atoms. The hydrocarbons of this class are also called olefines (olefiant = oil forming).

24.2.1 Methods of Preparation

In the laboratory, alkenes are generally prepared either from haloalkanes (alkyl

halides) or alcohols.

1. From Haloalkanes: Halaoalkanes are converted to alkenes by dehydrohalogenation. The process of removal of halogen acid like HCl, HBr or HI from the adjacent carbon atoms of alkyl halides, when reacted with alcoholic solution of potassium hydroxide, is called dehydrohalogenation.

The major product is formed according to the Saytzeff’s Rule.

Saytzeff’s Rule: It states that when an alkyl halide reacts with alcoholic solution of potassium hydroxide and if two alkenes are possible, then the one which is more substituted, will be the major product. In the above example,

but-2-ene is the major product because it contains two alkyl groups attached to

the –C=C– group.

24.2.2 Physical Properties of Alkenes

Some important physical properties of alkanes are as follows:

Physical State : Unbranched alkenes containing upto four carbon atoms are gases and containing five to sixteen carbon atoms are liquids while those with more than 16 carbon atoms are solids.

Boiling Points : The boiling points of alkenes increase with molecular mass as is

shown in Table 24.1.

Table 24.1 : Boiling points of Alkenes

The increase in boiling point can be attributed to the van der Waals forces which

increases with number of carbon atoms of the alkene. The branched chain alkenes

have lower boiling points than those of straight chain isomers.

Melting Point : In alkenes, there is increase in the melting point with the increase

in molecular mass. In the case of isomeric alkenes, the cis and trans isomers have

different melting points.

24.2.3 Chemical Properties of Alkenes

1. Addition Reactions : The chemical reactions in which a molecule adds to

another molecule are called an addition reaction. These reactions are

characteristic of unsaturated compounds like alkenes and alkynes. The

following reactions illustrate the addition reactions of alkenes.

(i) Addition of Hydrogen : Addition of hydrogen to unsaturated

hydrocarbons takes place in the presence of a catalyst like Ni, Pt or Pd.

In case of unsymmetrical alkenes (which contain unequal number of Hatoms attached to the carbon atoms of the double bonds), the addition of HX takes place according to the Markownikoff’s rule. This rule states that in the addition of halogen acids to unsymmetrical alkenes, the halogen of HX goes to that carbon atom of C = C bond which

already has less H-atoms attached to it. In other words, hydrogen atom of HX goes to the carbon atom with more number of H-atoms Compounds

attached to it.

Mechanism of Electrophilic Addition: You have studied earlier that the electron cloud of the pi bond is present above and below the plane of the molecule in alkenes. Various electron seeking species and reagents thus react with the alkenes. For example, H+ of HX(HBr) can

add to the double bond to yield a carbocation.

The carbocation being highly reactive reacts with the halide ion in the second step to yield an alkyl halide (alkyl bromide).

In case, the starting alkene is unsymmetrical e.g. propene, then in the first step of formation of a carbocation, there are two possibilities of attachment of H+ of HX which are shown below:

The carbocation being highly reactive reacts with the halide ion in the second step to yield an alkyl halide (alkyl bromide).

In case, the starting alkene is unsymmetrical e.g. propene, then in the first step of formation of a carbocation, there are two possibilities of attachment of H+ of HX which are shown below:

This would lead to the formation of two carbocations as shown above.

The two possible carbocations have different stabilities i.e. the secondary carbocation (II) is more stable than the primary carbocation (I). 

Therefore, the secondary carbocation (II) is formed preferentially in the first step. Further reaction, i.e. attack of Br– on the carbocation, thus yields 2-bromopropane as the major product.

Thus, the above explanation describes for the formation of 2- bromopropane as the major product as per the Markownikoff’s rule.

If the addition of HBr is carried out in the presence of peroxides such as benzoyl peroxide, then the reaction takes place contrary to Markownikoff’s rule. This is also known as Anti-Markownikoff’s addition or peroxide effect.

2. Oxidation: The oxidation of alkenes can be done by using different oxidizing agents like KMnO4, oxygen and ozone.

(i) Oxidation with KMnO4

Alkenes are unsaturated hydrocarbons having Pi (π)-bond(s) between the carbon atoms, so they are easily oxidized by cold dilute alkaline solution of KMnO4.

(ii) Oxidation with Oxygen : Ethene on oxidation with oxygen in the

presence of silver (Ag) gives epoxyethane. The reaction is shown below:

(iii) Combustion: The oxidation reaction, in which carbon dioxide and water are formed along with the liberation of heat and light, is called combustion.

(iv) Oxidation with Ozone : Ozone adds to the alkene forming ozonide.

The ozonide when further reacted with water in the presence of zinc dust, forms aldehydes or ketones, or both.

This process of addition of ozone to an unsaturated hydrocarbon followed by hydrolysis is called ozonolysis.

Ozonolysis can be used for the determination of the position of double bonds in alkenes by analyzing the products formed i.e. aldehydes and ketones. This is explained below.

When but-1-ene is oxidized with ozone and the ozonide formed is hydrolysed, we get one mole of propanal and one mole of methanal, showing that the double bond is between carbon atom 1 and 2. Whereas but-2-ene on oxidation with ozone, followed by hydrolysis, gives two

moles of ethanal, showing that the double bond is present between carbon atoms 2 and 3 as shown below.

When but-1-ene is oxidized with ozone and the ozonide formed is hydrolysed, we get one mole of propanal and one mole of methanal, showing that the double bond is between carbon atom 1 and 2. Whereas but-2-ene on oxidation with ozone, followed by hydrolysis, gives two

moles of ethanal, showing that the double bond is present between carbon atoms 2 and 3 as shown below.

24.2.4 Uses of Alkenes

Ethene is used for making mustard gas, which is a poisonous gas used in warfare.

It is also used for artificial ripening of fruits, as a general anaesthetic and for producing other useful materials such as polythene, ethanal, ethylene glycol (antifreeze), ethylene oxide (fumigant) etc.

INTEXT QUESTIONS 24.2

1. Which one has higher boiling point: cis but-2-ene or trans but-2-ene?

Trans-2-butene has higher boiling point than cis-isomer.

2. Name the products formed when ethene is oxidized with cold alkaline solution of KMnO4. 

Ethane-1, 2-diol

3. Write the conditions for hydrogenation of alkenes.

Hydrogen in presence of catalist Ni, Pt or Pd

4. What happens when ethene reacts with oxygen at 575 K in presence of Ag?

Epoxyethane is produced.

24.3 ALKYNES

These are also unsaturated hydrocarbons which contain atleast one triple bond

between two carbon atoms. Some examples are as follows:

24.3.1 Preparation of Ethyne (Acetylene) Compounds

Some important methods for preparation of ethyne are explained below.

1. From Calcium Carbide: Ethyne can be prepared in the laboratory, as well as

on a large scale, by the action of water on calcium carbide.

Ethyne prepared by this method generally contains the impurities of hydrogen sulphide and phosphine due to the impurities of calcium sulphide and calcium phosphide in calcium carbide.

2. Preparation of Ethyne from Dihaloalkanes

Ethyne can be prepared by refluxing geminal dihaloalkanes (having both halogens

attached to the same carbon atom) or vicinal dihaloalkanes (having halogen atoms

attached to the adjacent carbon atoms) with alcoholic solution of KOH.

3. Preparation of higher alkynes : Higher alkynes can be prepared by the reaction

of alkynides of lower alkynes with primary alkyl halides.

24.3.2 Physical Properties of Alkynes

1. First three members of alkynes are gases, the next eight members are liquids

and members having more than twelve carbon atoms are solids.

2. They are colourless and odourless, except ethyne which has a garlic odour.

3. The melting points, boiling points and densities of alkynes increase with the

increasing molar mass. In alkynes, there are π(pi)-electrons due to which

these molecules are slightly polar. So charge separation takes place in alkynes,

and hence dipoles are formed. The presence of dipoles increases the inter

molecular force of attraction, and hence the boiling points of alkynes are

higher than those of the corresponding alkanes.

4. Alkynes are very slightly soluble in water and soluble in acetone.

As alkynes have 50% s- character, they are the most acidic in nature. An sphybridized

carbon atom is more electronegative than sp2 or sp3 carbon atoms.

Due to greater electronegativity of sp hybridized carbon atom in ethyne, hydrogen

atom is less tightly held by the carbon and hence, it can be removed as a proton

(H+) by a strong base like sodium metal and sodamide. The following reactions

24.3.5 Uses of Alkynes

Ethyne (acetylene) is used for producing oxyacetylene flame (2800ºC) which is used for for welding and cutting of iron and steel. It is also used for artificial ripening of fruits and vegetables. It also finds use in the production of a number of other organic compounds such as ethanal, ethanoic acid, ethanol, synthetic rubbers and synthetic fibre orlon.

24.3.6 Distinction Between Alkanes, Alkenes and Alkynes

The folloiwng table shows different tests for distinction between alkanes, alkenes and alkynes :

Table 24.3 : Tests for identification of alkanes, alkenes and alkynes

INTEXT QUESTIONS 24.3

1. How is ethyne prepared from calcium carbide?

Calcium carbide is reacted with water to prepare ethyne.

2. Give one reaction to confirm the acidic nature of ethyne.

Reaction with sodium metal confirms the acidic nature of ethyne.

3. What is the percentage of s-character in ethane, ethene and ethyne?

The s-character in : Ethane = 25%,

Ethene = 33%,

Ethyne = 50%

24.4 AROMATIC HYDROCARBONS Compounds

Till now, we have explained various methods of preparation of aliphatic hydrocarbons. Now, we shall deal with an aromatic hydrocarbon (benzene) in detail. It is one of the major components obtained by the destructive distillation of coal as shown in Fig. 24.1

24.4.1 Structure of Benzene

The molecular formula of benzene is C6H6 which indicates that benzene is an unsaturated hydrocarbon. The unsaturation in benzene can be verified by the following reactions.

(i) Benzene undergoes the addition of H2 in the presence of Ni or Pt as catalyst.

Benzene does not respond to the tests of unsaturation which are shown by alkenes and alkynes i.e., both the alkenes and the alkynes decolourize bromine water and alkaline solution of potassium permanganate (Bayer’s Reagent). However, benzene undergoes substitution reactions.

Kekule Structure : A ring structure for benzene was proposed by Kekule in 1865. According to him, six carbon atoms are joined to each other by alternate single and double bonds to form a hexagon ring. As the proposed structure of benzene has three double bonds, so its properties should resemble with the properties of alkenes. But the chemical properties of benzene are different from alkenes.

As Kekule’s structure contains three single bonds and three double bonds, one may expect that in benzene there should be two different bond lengths i.e. 154 pm for C-C single bond and 134 pm for C=C double bond. But the experimental studies show that benzene is regular hexagon with an angle of 1200 and all the carbon-carbon bond lengths are equal i.e. 139 pm.

If Kekule’s structure is to be taken as a true structure, then benzene should form only one monosubstitution product and two ortho distubstitution products, shown below as (a) and (b).

In structure (a), the two halogen atoms are on the doubly bonded carbon atoms, whereas in structure (b), the two halogen atoms are on singly bonded carbon atoms. As per the Kekule’s structure these two isomers (a and b) should exist and show different properties. But, in reality, only one ortho disubstituted product exists. In order to explain this, Kekule proposed a dynamic equilibrium between the two structures.

Kekule’s structure does not explain the stability of benzene and its some unusual reactions. Resonance can explain the unusual behaviour of benzene. Let us now study about resonance.

Resonance: The phenomenon by virtue of which a single molecule can be Compounds

represented in two or more structures is called resonance. The actual structure

is the resonance hybrid of all the canonical or resonating structure. (see lesson 25)

Heat of hydrogenation data provides proof for resonance stabalization in benzene.

The heat of hydrogenation is the amount of heat liberated when hydrogen

is added to one mole of an unsaturated compound in the presence of a

catalyst.

and the actual structure is intermediate of these two forms. This can be represented as III where the circle inside the ring indicates the equivalence of the carboncarbon bonds. The carbon-carbon bond length has been found to be 139 pm.

which is same for all the carbon-carbon bonds in benzene. Also, this value of bond length is intermediate between the typical C—C single (154 pm) and C=C double bond (134 pm) lengths.

Molecular Orbital Picture of Benzene

All the six carbon atoms of benzene are sp2 hybridised. All C—C—C bond angles are 120° and there is a p orbital on each carbon atom. All the six p-orbitals are perpendicular to the plane of the six-membered carbon ring. The overlap of these p orbitals leads to a delocalised electron cloud above and below the place of the carbon ring. This is shown below in IV and V.

24.4.2 Aromaticity

So far you have studied that benzene

􀁺 is a cyclic conjugated molecule.

􀁺 is unusually stable.

􀁺 is planar in nature and its all C — C bond lengths are equal.

􀁺 can be represented as a resonance hybrid of two structures.

􀁺 undergoes substitution reactions rather than addition reactions.

Though the above properties indicate that benzene is aromatic in nature. But to

complete the argument for its aromatic nature, we have to check whether it follows

Huckel’s rule or not. According to Huckel rule – a molecule is aromatic only if

it has a planar, monocyclic system of conjugated 4n + 2 π electrons where n = 0,

1, 2, 3, …. Thus, molecules with 2, 6, 10, 14 π … electrons can be aromatic.

24.4.3 Physical Properties of Aromatic Hydrocarbons

1. Benzene and its homologues are colourless liquids having a characteristic odour.

2. They are immiscible in water but are miscible in all proportions with organic solvents such as alcohol, ether, petrol, etc. They dissolve fats and many other organic substances.

3. Most of the aromatic hydrocarbons are lighter than water.

4. Their boiling points show a gradual increase with increasing molecular mass e.g. benzene (b.p. 353 K), toluene (b.p. 383 K) and ethylbenzene (b.p. 409 K) and so on.

24.4.4 Chemical Properties of Aromatic Hydrocarbons

Aromatic hydrocarbons generally undergo electrophilic substitution reactions in which hydrogen atom of the aromatic ring is replaced by an electrophile. Such reactions are discussed below in detail taking benzene as an example.

(i) Halogenation : The reaction in which a hydrogen atom of benzene is replaced

by a halogen atom is called halogenation of benzene. Halogenation takes place in

the presence of iron, or ferric halides (FeX3, where X = Cl or Br).

24.4.5 Directive Influence of Functional Groups

In case of substituted aromatic compounds, the functional group(s) already present directs the next incoming group to a particular position in the aromatic ring. It is called directive influence of the group already attached to the benzene ring. For example, phenol on chlorination gives a mixture of ortho – chlorophenol and para– chlorophenol as – OH groups is an ortho and para directing group.

24.4.6 Carcinogenicity & Toxicity

There exist several aromatic compounds-many of them being very important for the life while there are some others which are harmful. A large number of them are toxic in nature. For example, benzene is carcinogenic in nature.

Another such hydrocarbon is benzo[α] pyrene which has been found in cigarette smoke and in the exhaust from automobiles. This compound is also carcinogenic and can cause skin cancer in mice.

24.4.7 Uses of Aromatic Hydrocarbons

Benzene is used as a solvent for several organic compounds and thus, acts as a medium for carrying out synthetic reactions. It is the basic aromatic hydrocarbon and can be converted to other organic compounds by carrying out substitution in the benzene ring. Toluene, a higher homologue of the benzene, finds its uses for dry-cleaning, as a solvent, and as a starting material for the manufacture of dyes, drugs, explosive (trinitrotoluene, T.N.T.), benzaldehyde, benzoic acid etc.

INTEXT QUESTIONS 24.4

1. What is the value of resonance energy of benzene?

The resonance energy of benzene is 150.3 KJ mol–1.

2. Name the product formed when :

(i) benzene reacts with chlorine in the presence of light.

(i) Benzene hexachloride (BHC).

(ii) phenol reacts with chlorine in the presence of FeCl3.

(ii) o-Chlorophenol and p-chlorophenol.

(iii) nitrobenzene reacts with chlorine in the presence of FeCl3.

(iii) m-Chloronitrobenzene.

3. Classify the following into o-and p- or m-directing groups:

WHAT YOU HAVE LEARNT

􀁺 Alkanes can be prepared by (i) the reduction of haloalkanes, (ii) action of water or alcohol on Grignard’s reagent, (iii) Wurtz reaction and (iv) hydrogenation of unsaturated hydrocarbons.

􀁺 Physical properties of hydrocarbons depend on the intermolecular forces of attraction. which in turn depend upon the shapes of molecules and their surface area.

􀁺 The melting points of hydrocarbons depends upon the symmetry of the molecules i.e. hydrocarbons with even number of carbon atoms are more symmetrical and have higher melting points.

􀁺 Rotation about carbon-carbon single bond leads to various conformations of

a molecule. Ethane has many conformations out of which the staggered conformation is the most stable one and the eclipsed conformation is the least stable one.

􀁺 Alkenes can be prepared by dehydrohalogenation of alkyl halides and by dehydration of alcohols.

􀁺 Alkenes and alkynes undergo addition reactions e.g. addition of hydrogen, halogens, halogen acids, water, sulphuric acid etc. due to the presence of carbon-carbon double or triple bonds.

􀁺 Addition of halogen acids and other unsymmetrical reagents to unsymmetrical

alkenes and alkynes takes place according to the Markownikoff’s rule.

􀁺 Alkenes undergo polymerization on heating under pressure.

􀁺 All hydrocarbons (saturated as well as unsaturated) form CO2 and H2O on Compounds

combustion and liberate energy.

􀁺 An alkaline solution of KMnO4 can oxidize alkenes and alkynes forming different products such as carboxylic acids, aldehydes and/or ketones and carbon dioxide.

􀁺 Ozone can oxidize unsaturated hydrocarbons (alkenes and alkynes) forming ozonides which when further reacted with water in the presence of zinc dust either form aldehydes or ketones or both.

􀁺 Reaction of ozone with alkenes can be used to determine the position of double bond.

􀁺 Ethyne can be prepared by the action of water on calcium carbide and by dehydrogenation of dihaloalkanes.

􀁺 Alkynes are acidic in nature due to sp-hybridization of carbon atoms. Because sp hybridized carbon atoms are more electronegative than sp2 and sp3, and the bond between C—H is weakened. Hence, hydrogen atoms in alkynes can be replaced by certain metal atoms.

􀁺 Alkanes, alkenes and alkynes can be distinguished by using:

a) Br2 dissolved in carbon tetrachloride.

b) Ammoniacal solution of AgNO3

c) Ammoniacal solution of Cu2Cl2

d) Alkaline solution of KMnO4

􀁺 Benzene is obtained by destructive distillation of coal.

􀁺 A ring structure of benzene was suggested by Kekule. Actual structure of benzene is the resonance hybrid of the canonical structures.

􀁺 Aromatic hydrocarbons undergo substitution reactions i.e. the reactions in which hydrogen atom of hydrocarbons is replaced by another atom or group of atoms. Halogenation, sulphonation, nitration and Friedel Craft’s reaction are substitution reactions of benzene.

􀁺 The position of second substituent on a benzene ring depends upon the nature

of the group already present.

TERMINAL EXERCISE

1. What happens when : (Write chemical equations)

(i) Iodoethane is heated with HI in the presence of red phosphorus.

(ii) 2-Chlorobutane reacts with sodium metal.

(iii) Ethyl magnesium bromide is reacted with methyl alcohol (methanol).

(iv) 2-Chloropropane reacts with alcoholic solution of KOH.

(v) 1,1-Dichloroethane reacts with alcoholic solution of KOH.

Here’s a clear breakdown of the chemical reactions you asked about, expressed as equations and explained step by step.

1

Iodoethane with HI and Red Phosphorus

Heating iodoethane with HI in presence of red phosphorus produces ethane.

C2H5I + 2HI → C2H6 + I2

• Red phosphorus regenerates HI during the reaction

• Ethane is formed as the main product

2

2-Chlorobutane with Sodium Metal

This is a Wurtz reaction, forming a higher alkane.

2 C4H9Cl + 2 Na → C8H18 + 2 NaCl

• Sodium removes chlorine atoms

• Octane (C8H18) is formed

3

Ethyl Magnesium Bromide with Methanol

Grignard reagent reacts with alcohol to form alkane.

C2H5MgBr + CH3OH → C2H6 + MgBrOCH3

• Ethyl group abstracts hydrogen from methanol

• Ethane is formed

4

2-Chloropropane with Alcoholic KOH

Alcoholic KOH induces elimination (dehydrohalogenation).

CH3-CHCl-CH3 + KOH(alc) → CH2=CH-CH3 + HCl

• Hydrogen and chlorine are eliminated

• Propene is formed

5

1,1-Dichloroethane with Alcoholic KOH

Double elimination produces ethyne (acetylene).

CH3-CHCl2 + 2KOH(alc) → HC≡CH + 2KCl + 2H2O

• Two molecules of KOH remove two HCl units

• Ethyne is formed

These reactions illustrate substitution, elimination, and organometallic chemistry in action. Would you like me to also explain the mechanisms behind each step — for example, how the Grignard reagent abstracts a proton or how alcoholic KOH favors elimination over substitution? That would give you a deeper understanding of why these products form.

2. Give reasons for the following:

(i) The boiling point of neopentane is less than that of n-pentane.

(ii) Stability of benzene ring

(iii) Boiling points of hydrocarbons decrease with the increase in branching.

Here are the reasons explained clearly:

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(i) Boiling point of neopentane vs n-pentane

• Neopentane is highly branched and has a nearly spherical shape.
• This reduces the surface area available for intermolecular van der Waals forces.
• n-Pentane, being linear, has a larger surface area and stronger intermolecular attractions.
• Hence, neopentane boils at a lower temperature than n-pentane.

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(ii) Stability of benzene ring

• The benzene ring is stabilized by resonance: six π-electrons are delocalized over six carbon atoms.
• This delocalization creates a uniform electron cloud above and below the ring, lowering its energy.
• The ring follows Hückel’s rule (4n+2 π electrons, with n=1), making it aromatic and exceptionally stable compared to hypothetical localized structures.

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(iii) Effect of branching on boiling points of hydrocarbons

• In branched hydrocarbons, molecules are more compact and spherical.
• This reduces the contact area between molecules, weakening van der Waals forces.
• As a result, branched isomers have lower boiling points than their straight-chain counterparts.
• The more the branching, the lower the boiling point.

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So, the key theme across these points is intermolecular forces and resonance stabilization. Would you like me to also illustrate this with molecular structure diagrams (like neopentane vs n-pentane, benzene resonance forms) so you can visualize why these differences occur?

3. How will you prepare the following?

(i) Ethane from ethene

(ii) Ethene from ethanol

(iii) Cyclohexane from benzene

(iv) Methane from sodium acetate

(v) Butane from bromoethane

Here’s a clear breakdown of how each compound can be prepared using standard organic chemistry reactions.

1

Ethane from Ethene

Convert ethene to ethane by catalytic hydrogenation.

• Pass ethene gas over finely divided nickel catalyst

• Maintain temperature around 300°C

• Add H₂ gas to saturate the double bond

• Product: ethane

2

Ethene from Ethanol

Dehydrate ethanol to form ethene.

• Heat ethanol with concentrated H₂SO₄ at ~170°C

• Acid acts as a dehydrating agent

• Elimination of water gives ethene

3

Cyclohexane from Benzene

Hydrogenate benzene to form cyclohexane.

• Pass benzene vapor with excess H₂ gas

• Use nickel/platinum catalyst

• Maintain high temperature and pressure

• Product: cyclohexane

4

Methane from Sodium Acetate

Decarboxylate sodium acetate with soda lime.

• Mix sodium acetate with soda lime (NaOH + CaO)

• Heat strongly in a test tube

• Decarboxylation releases methane gas

5

Butane from Bromoethane

Use Wurtz reaction to couple bromoethane molecules.

• React bromoethane with sodium metal

• Use dry ether as solvent

• Two ethyl radicals couple to form butane

These are classic lab-scale methods for preparing hydrocarbons. Each reaction highlights a fundamental transformation: hydrogenation, dehydration, decarboxylation, and coupling.

4. What happens when (write balanced chemical equations):

(i) Hydrochloric acid is added to ethene.

(ii) Hydrobromic acid (HBr) is added to propene in the presence of benzoyl peroxide.

(iii) Benzene reacts with chloromethane in the presence of anhydrousAlCl3.

(iv) Br2 is added to ethyne.

(v) Methane is oxidized with oxygen in the presence of copper at 475K and a high pressure of 120 atm.

Here’s a breakdown of each reaction with the balanced chemical equations:

(i) Hydrochloric acid + ethene

Ethene undergoes electrophilic addition with HCl to form chloroethane.

$$C{H}_2=C{H}_2+HCl\to C{H}_3-C{H}_{2}Cl$$

(ii) Hydrobromic acid + propene with benzoyl peroxide

This is the Kharasch effect (anti-Markovnikov addition) due to free radical mechanism initiated by benzoyl peroxide. The product is 1-bromopropane.

$$C{H}_3-CH=C{H}_2+HBr\underset{\text{benzoyl peroxide}}{\overset{}{\to }}C{H}_3-C{H}_2-C{H}_{2}Br$$

(iii) Benzene + chloromethane with AlCl3

This is a Friedel–Crafts alkylation reaction. Benzene reacts with chloromethane in the presence of anhydrous AlCl₃ to form toluene.

$$C_6{H}_6+C{H}_{3}Cl\underset{{\text{AlCl}}_3}{\overset{}{\to }}{C}_6{H}_{5}C{H}_3+HCl$$

(iv) Bromine + ethyne

Ethyne undergoes halogen addition. First, dibromoethene forms; with excess Br₂, tetrabromoethane is produced.

$$CH\equiv CH+B{r}_2\to CHBr=CHBr$$

$$CHBr=CHBr+B{r}_2\to CHB{r}_2-CHB{r}_2$$

(v) Methane oxidation with oxygen at 475K, 120 atm, copper catalyst

This is catalytic oxidation of methane to methanol.

$$C{H}_4+\frac{1}{2}{O}_2\underset{\text{Cu, 475K, 120 atm}}{\overset{}{\to }}C{H}_{3}OH$$

✅ Each reaction highlights a different type of organic transformation: electrophilic addition, free radical addition, Friedel–Crafts alkylation, halogen addition, and controlled oxidation.

Would you like me to also sketch the reaction mechanisms (step-by-step electron flow) for these, or keep it at the balanced equations level?

https://copilot.microsoft.com/shares/AU3VrZd2FEXtjHiGgzPGm

5. How are the following conversions carried out?

(i) Ethyne to ethane

(ii) Benzene to nitrobenzene

(iii) Ethyl alcohol (ethanol) to ethene

(iv) Ethyne to ethanedioic acid

(v) Benzene to o-nitrochlorobenzene.

These conversions are classic organic transformations, each requiring specific reagents and conditions. Here’s a clear step-by-step guide for carrying them out:

These conversions are classic organic transformations, each requiring specific reagents and conditions. Here’s a clear step-by-step guide for carrying them out:

1

Hydrogenate Ethyne

Recommended

Convert ethyne to ethane by catalytic hydrogenation.

CH≡CH + 2H₂ → CH₃-CH₃

• Pass ethyne (C₂H₂) with hydrogen gas

• Use Ni/Pd catalyst at room temperature

• Ethene forms first, then further hydrogenation gives ethane

2

Nitrate Benzene

Safety

Introduce a nitro group into benzene.

C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O

• Mix benzene with conc. HNO₃ and conc. H₂SO₄

• Maintain temperature around 330K

• Produces nitrobenzene

3

Dehydrate Ethanol

Remove water from ethanol to form ethene.

CH₃CH₂OH → CH₂=CH₂ + H₂O

• Heat ethanol with conc. H₂SO₄ at 443K

• Alternatively pass vapors over heated Al₂O₃

• Collect ethene gas

4

Oxidize Ethyne

Strong oxidation of ethyne yields ethanedioic acid.

CH≡CH → HOOC-COOH

• Treat ethyne with alkaline KMnO₄ solution

• Reaction proceeds to form oxalic acid

5

Prepare o-Nitrochlorobenzene

Sequential substitution reactions on benzene.

C₆H₆ → C₆H₅Cl → o-ClC₆H₄NO₂ + p-ClC₆H₄NO₂

• Step 1: Chlorinate benzene with Cl₂/FeCl₃ → chlorobenzene

• Step 2: Nitrate chlorobenzene with conc. HNO₃/H₂SO₄

• Chlorine directs substitution to ortho and para positions

• Mixture contains o- and p-nitrochlorobenzene

https://copilot.microsoft.com/shares/4xfSQZkaUg1tFTdUnxFDd

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Each conversion highlights a different reaction type: hydrogenation, electrophilic substitution, dehydration, oxidation, and directed substitution. These are foundational transformations in organic chemistry, often used to build more complex molecules.

Would you like me to also explain the mechanistic pathways (electron flow and intermediates) for these reactions, so you can see why each reagent works the way it does?

Here’s how each of the requested conversions is carried out, with the reagents and conditions clearly laid out:

(i) Ethyne → Ethane

• Ethyne is hydrogenated in the presence of a finely divided nickel or palladium catalyst.

• First, ethene forms; further hydrogenation gives ethane.

$$CH\equiv CH+2H_2\underset{\text{Ni/Pd}}{\overset{}{\to }}C{H}_3-C{H}_3$$

(ii) Benzene → Nitrobenzene

• Benzene undergoes nitration with a nitrating mixture (conc. HNO₃ + conc. H₂SO₄).

• The nitronium ion ($N{O}_2^{+}$) acts as the electrophile.

C_6H_6 + HNO_3 \;\;\xrightarrow[\text{conc. H_2SO_4}]{}\;\; C_6H_5NO_2 + H_2O

(iii) Ethanol → Ethene

• Ethanol undergoes acid-catalyzed dehydration.

• Conc. H₂SO₄ or heated Al₂O₃ is used.

CH_3CH_2OH \;\;\xrightarrow[\text{conc. H_2SO_4, 443K}]{}\;\; CH_2=CH_2 + H_2O

(iv) Ethyne → Ethanedioic acid (Oxalic acid)

• Ethyne reacts with alkaline potassium permanganate (KMnO₄).

• Oxidation yields ethanedioic acid.

$$CH\equiv CH\underset{{\text{KMnO}}_4\mathrm{/}\text{alkaline}}{\overset{}{\to }}HOOC-COOH$$

(v) Benzene → o-Nitrochlorobenzene

• Step 1: Chlorination of benzene with Cl₂ in the presence of FeCl₃ gives chlorobenzene.

$$C_6{H}_6+C{l}_2\underset{{\text{FeCl}}_3}{\overset{}{\to }}{C}_6{H}_{5}Cl+HCl$$

• Step 2: Nitration of chlorobenzene with conc. HNO₃ + conc. H₂SO₄ gives o- and p-nitrochlorobenzene (due to chlorine being an ortho/para directing group).

C_6H_5Cl + HNO_3 \;\;\xrightarrow[\text{conc. H_2SO_4}]{}\;\; o\text{-}ClC_6H_4NO_2 + p\text{-}ClC_6H_4NO_2

1

Hydrogenate Ethyne

Convert ethyne to ethane via catalytic hydrogenation.

CH≡CH + 2H₂ → CH₃-CH₃

• Use Ni or Pd catalyst

• Add 2 moles of H₂

• Product: ethane

2

Nitrate Benzene

Introduce nitro group to benzene.

C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O

• Mix conc. HNO₃ + conc. H₂SO₄

• Generate NO₂⁺ electrophile

• Product: nitrobenzene

3

Dehydrate Ethanol

Remove water from ethanol to form ethene.

CH₃CH₂OH → CH₂=CH₂ + H₂O

• Heat with conc. H₂SO₄ at 443K

• Alternative: Al₂O₃ catalyst

• Product: ethene

4

Oxidize Ethyne

Strong oxidation yields oxalic acid.

CH≡CH → HOOC-COOH

• Use alkaline KMnO₄

• Break triple bond completely

• Product: ethanedioic acid

5

Form o-Nitrochlorobenzene

Sequential chlorination and nitration of benzene.

C₆H₆ → C₆H₅Cl → o-ClC₆H₄NO₂ + p-ClC₆H₄NO₂

• Step 1: Chlorinate with Cl₂/FeCl₃ → chlorobenzene

• Step 2: Nitrate with conc. HNO₃/H₂SO₄

• Product: o- and p-nitrochlorobenzene

Each conversion demonstrates a classic organic transformation: hydrogenation, electrophilic substitution, dehydration, oxidation, and directed substitution. Would you like me to also explain the mechanistic pathways (electron flow, intermediates) for these reactions?

6. You are provided with three gas jars containing ethane, ethene and ethyne.

Give the suitable chemical tests to identify the three hydrocarbons.

To distinguish between ethane (C₂H₆), ethene (C₂H₄), and ethyne (C₂H₂), you can use simple chemical tests that exploit their different bonding structures:

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🔬 Chemical Tests

• Ethane (saturated hydrocarbon, alkane)

  • Test: Pass the gas through bromine water or acidified potassium permanganate solution.
  • Observation: No decolorization occurs because ethane has only single bonds and does not react under normal conditions.

• Ethene (unsaturated hydrocarbon, alkene)

  • Test: Bubble the gas through bromine water.
  • Observation: The orange color of bromine water disappears rapidly due to addition across the double bond.
  • Similarly, with acidified KMnO₄, the purple color is discharged (Baeyer's test).

• Ethyne (unsaturated hydrocarbon, alkyne)

  • Test 1: Like ethene, it also decolorizes bromine water and acidified KMnO₄ because of the triple bond.
  • Test 2 (to distinguish from ethene): Pass ethyne into ammoniacal silver nitrate (AgNO₃) or ammoniacal cuprous chloride (Cu₂Cl₂) solution.
    • Observation: Formation of a white precipitate of silver acetylide or red precipitate of copper acetylide confirms ethyne.

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🧾 Summary Table

Hydrocarbon	Key Bond Type	Test with Bromine Water	Test with KMnO₄	Special Test
Ethane	Single bond (alkane)	No reaction	No reaction	None
Ethene	Double bond (alkene)	Decolorizes	Decolorizes	None
Ethyne	Triple bond (alkyne)	Decolorizes	Decolorizes	Forms precipitate with ammoniacal AgNO₃ or Cu₂Cl₂

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Would you like me to also explain the mechanism of these reactions (addition vs substitution) so you can see why the tests work, or keep it focused on just the practical identification?

7. What is ozonolysis? How is it used to determine the position of a double bond?

🌐 What is Ozonolysis?

Ozonolysis is a reaction in which ozone (O₃) cleaves the double bond (C=C) of an alkene. The process breaks the carbon–carbon double bond and forms carbonyl compounds (aldehydes, ketones, or carboxylic acids depending on conditions).

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⚗️ Reaction Steps

1. Addition of ozone: Ozone adds across the double bond to form an unstable intermediate called an ozonide.
2. Cleavage: The ozonide is then decomposed (usually by reduction with zinc + water or dimethyl sulfide).
3. Products: The double bond is split, and each carbon of the double bond becomes part of a carbonyl group.

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🔍 How it Determines the Position of a Double Bond

• When an unknown alkene undergoes ozonolysis, the products reveal where the double bond was located.
• Example:
  • Ethene (CH₂=CH₂) → Ozonolysis gives two molecules of formaldehyde (HCHO).
  • Propene (CH₃–CH=CH₂) → Ozonolysis gives acetaldehyde (CH₃CHO) and formaldehyde (HCHO).
  • But-2-ene (CH₃–CH=CH–CH₃) → Ozonolysis gives two molecules of acetaldehyde (CH₃CHO).

By analyzing the carbonyl compounds formed, chemists can deduce the exact position of the double bond in the original molecule.

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🧾 Summary Table

Alkene	Ozonolysis Products	Double Bond Position
Ethene	2 × Formaldehyde	Between two CH₂ groups
Propene	Acetaldehyde + Formaldehyde	Between CH₃–CH and CH₂
But-2-ene	2 × Acetaldehyde	Between two CH groups

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Would you like me to also sketch out a reaction mechanism diagram showing how ozone attacks the double bond and forms the ozonide before cleavage? That visual often makes the process much clearer.

8. Give reasons for the following:

(i) Alkanes do not undergo addition reactions like alkenes and alkynes.

(ii) Ethyne is more acidic than ethane.

(iii) Ethene undergoes polymerization but not ethane. 

(iv) Benzene undergoes electrophilic substitution reactions.

Here’s a clear explanation of each point:

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(i) Alkanes do not undergo addition reactions like alkenes and alkynes

• Alkanes contain only C–C single bonds (sigma bonds), which are very stable and have no regions of high electron density.
• Alkenes and alkynes have double or triple bonds with π-electrons that are more reactive and easily attacked by electrophiles.
• Since alkanes lack π-bonds, they generally undergo substitution reactions rather than addition.

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(ii) Ethyne is more acidic than ethane

• In ethyne, the carbon is sp-hybridized, meaning it has 50% s-character. Electrons are held closer to the nucleus, making the C–H bond more polar and easier to lose a proton (H⁺).
• In ethane, the carbon is sp³-hybridized with only 25% s-character, so the C–H bond is less polar and less acidic.
• Thus, ethyne has a lower pKa and is more acidic.

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(iii) Ethene undergoes polymerization but not ethane

• Ethene has a double bond with reactive π-electrons that can open up and link with other ethene molecules to form long chains (polymers like polyethylene).
• Ethane has only single bonds, which are stable and cannot easily break to form new bonds for polymerization.
• Therefore, ethene polymerizes, while ethane does not.

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(iv) Benzene undergoes electrophilic substitution reactions

• Benzene is stabilized by aromatic resonance, where π-electrons are delocalized over the ring.
• Addition reactions would disrupt this aromatic stability, which is energetically unfavorable.
• Instead, benzene reacts via electrophilic substitution, where the aromatic ring remains intact while a hydrogen atom is replaced by an electrophile.

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Would you like me to also illustrate these differences with reaction mechanism diagrams (like showing ethene polymerization vs. ethane stability, or benzene substitution vs. addition)? That could make the reasoning even clearer.

Here’s the visual set of reaction mechanism diagrams you asked for — they’re ready now. They neatly illustrate why alkanes resist addition, why ethyne is more acidic, how ethene polymerizes, and why benzene prefers substitution over addition.

Looking at them side by side makes the contrasts much clearer:

• The alkane vs. alkene panel shows how π-bonds enable addition.
• The ethyne vs. ethane panel highlights proton loss due to higher s-character.
• The ethene vs. ethane panel captures polymerization versus stability.
• The benzene panel emphasizes preservation of aromaticity through substitution.

Would you like me to also walk you through step-by-step mechanisms (with curved arrow notation) for one of these reactions, say benzene electrophilic substitution or ethene polymerization? That way you can see exactly how electrons move during the process.