Chapter 9 - Hydrocarbons

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NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

NCERT Solutions for Class 11 Chemistry Chapter 9 (Hydrocarbons) help students understand the main types of hydrocarbons, including alkanes, alkenes, alkynes, and aromatic compounds. This chapter covers their structure, properties, preparation, and chemical reactions. Hydrocarbons are found in fuels such as natural gas and petrol and are used to make many everyday products.

NCERT Solutions for Class 11 Chemistry Chapter 9 by ALLEN explain important reactions, acidity trends, organic conversions, and the structure of benzene. The solutions include answers to NCERT textbook questions to help students understand the chapter, practise questions, and prepare for CBSE exams, JEE, and NEET.

1.0Class 11 Chemistry Chapter 9 Hydrocarbons: Key Concepts

This chapter explains the types of hydrocarbons, their preparation, properties, and chemical reactions. The key concepts include:

Alkanes: Learning about ethane conformations using Sawhorse and Newman projections, the Wurtz reaction, and free radical halogenation.
Alkenes: Understanding cis-trans isomerism, electrophilic addition reactions, and Markovnikov’s and anti-Markovnikov’s rules.
Ozonolysis: Learning how ozonolysis helps identify the position of double and triple bonds in hydrocarbons.
Alkynes: Understanding the acidic nature of terminal alkynes and how alkynes can form benzene through cyclic polymerisation.
Aromatic Hydrocarbons: Learning about the structure of benzene and Hückel’s rule for identifying aromatic compounds.
Electrophilic Substitution Reactions: Studying nitration, sulphonation, halogenation, and Friedel-Crafts alkylation and acylation of benzene.
Directive Influence: Understanding how groups already attached to a benzene ring affect where a new group enters, such as the ortho, meta, or para position.
Carcinogenicity and Toxicity: Learning about harmful polynuclear hydrocarbons and their effects on human health and the environment.

2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 9

NCERT EXERCISE

How do you account for the formation of ethane during chlorination of methane? Ans. Chlorination of methane proceeds via a free radical chain mechanism. The whole reaction takes place in the given three steps. Step 1: Initiation: The reaction begins with the homolytic cleavage of $Cl - Cl$ bond as:

$Cl - Cl h ν \dot{C} l + \dot{C} l$

Chlorine free radicals Step 2: Propagation: In the second step, chlorine free radicals attack methane molecules and break down the C-H bond to generate methyl radicals as

$CH_{4} + \dot{C} l h ν \dot{C} H_{3} + H - \dot{C} l$ Methane These methyl radicals react with other chlorine free radicals to form methyl chloride along with the liberation of a chlorine free radical.

$\dot{C} H_{3} + Cl - Cl \to CH_{3} - Cl + Cl$ Methyl chloride Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While HCl and $CH_{3} Cl$ are the major products formed, other higher halogenated compounds are also formed as:

$CH_{3} + Cl - \dot{C} l \to \dot{C} H_{2} Cl + HCl$

$\dot{C} H_{2} Cl + Cl - Cl \to CH_{2} Cl_{2} + \dot{C} l$

Step 3: Termination: Formation of ethane is a result of the termination of chain reactions taking place as a result of the consumption of reactants as:

$\dot{C} l + \dot{C} l \to Cl + Cl$

$H_{3} \dot{C} + \dot{C} H_{3} \to H_{3} C - CH_{3}$

Hence, by this process, ethane is obtained as a by-product of chlorination of methane.

Write IUPAC names of the following compounds:

Ans.

For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated: (a) $C_{4} H_{8}$ (one double bond) (b) $C_{5} H_{8}$ (one triple bond)

Ans.

(a) The following structural isomers are possible for $C_{4} H_{8}$ with one double bond: (I) $H_{2} C 1 = C 2 H - C 3 H_{2} - C 4 H_{3}$ (II) $C 1 H_{3} - C 2 H = C 3 H - C 4 H_{3}$ (III) $C 1 H_{2} = CH_{3} ∣ C 2 - C 3 H_{3}$

The IUPAC name of Compound (I) is But-1-ene, Compound (II) is But-2-ene, and Compound (III) is 2-Methylprop-1-ene.

(b) The following structural isomers are possible for $C_{5} C_{8}$ with one triple bond:

(i) $H C 1 \equiv C 2 - C 3 H_{2} - C 5 H_{3}$

(ii) $H_{3} C 1 - C 2 \equiv C 3 - C 4 H_{2} - C 5 H_{3}$

(iii) $H_{3} C 4 - CH_{3} ∣ C 3 H - C 2 \equiv C 1 H$

The IUPAC name of Compound (I) is Pent-1-yne, Compound (II) is Pent-2-yne, and Compound (III) is 3-Methylbut-1-yne.

Write IUPAC names of the products obtained by the ozonolysis of the following compounds: (i) Pent-2-ene (ii) 3,4-Dimethyl-hept-3-ene (iii) 2-Ethylbut-1-ene (iv) 1-Phenylbut-1-ene

Ans.

(i) Pent-2-ene undergoes ozonolysis as:

The IUPAC name of Product (I) is ethanal and Product (II)is propanal. (ii) 3, 4-Dimethylhept-3-ene undergoes ozonolysis as:

The IUPAC name of Product (I)is butan-2-one and Product (II)is Pentan-2-one.

(iii)2-Ethylbut-1-ene undergoes ozonolysis as:

The IUPAC name of Product (I)is pentan-3-one and Product (II)is methanal.

(iv) 1-Phenylbut-1-ene undergoes ozonolysis as:

The IUPAC name of Product (I)is benzaldehyde and Product (II)is propanal.

5. An alkene 'A' on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of 'A'. Ans.

During ozonolysis, an ozonide having a cyclic structure is formed as an intermediate which undergoes cleavage to give the final products. Ethanal and pentan-3-one are obtained from the intermediate ozonide. Hence, the expected structure of the ozonide is:

This ozonide is formed as an addition of ozone to 'A'. The desired structure of 'A' can be obtained by the removal of ozone from the ozonide. Hence, the structural formula of 'A' is:

$H_{3} C 1 - CH_{2} - CH_{3} ∣ C 2 H = C 3 - C 4 H_{2} - C 5 H_{3}$

The IUPAC name of 'A' is 3-Ethylpent-2-ene.

An alkene 'A' contains three C - C, eight $C - H σ$ bonds and one $C - C π$ bond. 'A' on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of 'A'. Ans. As per the given information, 'A' on ozonolysis gives two moles of an aldehyde of molar mass 44 u. The formation of two moles of an aldehyde indicates the presence of identical structural units on both sides of the double bond containing carbon atoms. Hence, the structure of 'A' can be represented as:

XC = CX

There are eight $C - H σ$ bonds. Hence, there are 8 hydrogen atoms in 'A'. Also, there are three C-C bonds. Hence, there are four carbon atoms present in the structure of 'A'. Combining the inferences, the structure of 'A' can be represented as:

'A' has 3 C-C bonds, 8 C-H $σ$ bonds, and one $C - C π$ bond. Hence, the IUPAC name of 'A' is But-2-ene. Ozonolysis of 'A' takes place as:

The final product is ethanal with molecular mass $= [(2 \times 12) + (4 \times 1) + (1 \times 16)] = 44 u$

Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene? Ans. As per the given information, propanal and pentan-3-one are the ozonolysis products of an alkene. Let the given alkene be 'A'. Writing the reverse of the ozonolysis reaction, we get:

The products are obtained on the cleavage of ozonide 'X'. Hence, 'X' contains both products in the cyclic form. The possible structure of ozonide can be represented as:

Now, 'X' is an addition product of alkene 'A' with ozone. Therefore, the possible structure of alkene 'A' is: 8. Write chemical equations for combustion reaction of the following hydrocarbons: (i) Butane (ii) Pentene (iii) Hexyne (iv) Toluene Ans. Combustion can be defined as a reaction of a compound with oxygen.

Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why? Ans. Hex-2-ene is represented as:

CH_{3} - CH = CH - CH_{2} - CH_{2} - CH_{3}

Geometrical isomers of hex-2-ene are:

The dipole moment of cis-compound is a sum of the dipole moments of $C - CH_{3}$ and $CCH_{2} CH_{2} CH_{3}$ bonds acting in the same direction.

The dipole moment of trans-compound is the resultant of the dipole moments of $CCH_{3}$ and $C - CH_{2} CH_{2} CH_{3}$ bonds acting in opposite directions. Hence, cis-isomer is more polar than trans-isomer. The higher the polarity, the greater is the intermolecular dipole-dipole interaction and the higher will be the boiling point. Hence, cis-isomer will have a higher boiling point than trans-isomer.

Why is benzene extra ordinarily stable though it contains three double bonds? Ans. Benzene is a hybrid of resonating structures given as:

All six carbon atoms in benzene are $sp^{2}$ hybridized. The two $sp^{2}$ hybrid orbitals of each carbon atom overlap with the $sp^{2}$ hybrid orbitals of adjacent carbon atoms to form six sigma bonds in the hexagonal plane. The remaining $sp^{2}$ hybrid orbital on each carbon atom overlaps with the s-orbital of hydrogen to form six sigma C-H bonds. The remaining unhybridized p-orbital of carbon atoms has the possibility of forming three $π$ bonds by the lateral overlap of $C_{1} - C_{2}, C_{3} - C_{4}, C_{5} - C_{6}$ , or $C_{2} - C_{3}, C_{4} - C_{5}, C_{6} - C_{1}$ ,.

The six $π$ 's are delocalized and can move freely about the six carbon nuclei. Even after the presence of three double bonds, these delocalized $π$ -electrons stabilize benzene.

11.What are the necessary conditions for any system to be aromatic? Ans. A compound is said to be aromatic if it satisfies the following three conditions: (i) It should have a planar structure. (ii) The $π$ -electrons of the compound are completely delocalized in the ring. (iii) The total number of $π$ -electrons present in the ring should be equal to $(4 n + 2)$ , where $n = 0, 1, 2 \dots$ etc. This is known as Huckel's rule.

Explain why the following systems are not aromatic?

Ans.

For the given compound, the number of $π$ -electrons is six. But only four $π$ -electrons are present within the ring. Also there is no conjugation of $π$ -electrons within the ring and the compound is not planar in shape. Hence, the given compound is not aromatic in nature. (ii)

For the given compound, the number of $π$ -electrons is four. By Huckel's rule,

4 n + 2 = 4 4 n = 2 n = \frac{1}{2}

For a compound to be aromatic, the value of n must be an integer $(n = 0, 1, 2 \dots)$ , which is not true for the given compound. Hence, it is not aromatic in nature.

(iii) For the given compound, the number of $π$ -electrons is eight. By Huckel's rule,

4 n + 2 = 8 4 n = 6 n = \frac{3}{2}

For a compound to be aromatic, the value of n must be an integer $(n = 0, 1, 2 \dots)$ . Since the value of n is not an integer, the given compound is not aromatic in nature.

How will you convert benzene into (i) p-nitrobromobenzene (ii) m-nitrochlorobenzene (iii) p -nitrotoluene (iv) acetophenone

Ans.

(i) Benzene can be converted into p-nitrobromobenzene as:

(ii) Benzene can be converted into m-nitrochlorobenzene as:

(iii) Benzene can be converted into p-nitrotoulene as:

(iv) Benzene can be converted into acetophenone as:

14. In the alkane $H_{3} C - CH_{2} - C (CH_{3})_{2} - CH_{2}$ $- CH (CH_{3})_{2}$ , identify $1^{\circ}, 2^{\circ}, 3^{\circ}$ carbon atoms and give the number of H atoms bonded to each one of these. Ans.

1° carbon atoms are those which are bonded to only one carbon atom, i.e., they have only one carbon atom as their neighbour. The given structure has five 1° carbon atoms and fifteen hydrogen atoms are attached to it. 2° carbon atoms are those which are bonded to two carbon atoms, i.e., they have two carbon atoms as their neighbours. The given structure has two 2° carbon atoms and four hydrogen atoms are attached to it. 3° carbon atoms are those which are bonded to three carbon atoms, i.e., they have three carbon atoms as their neighbours. The given structure has one 3° carbon atom and only one hydrogen atom is attached to it.

What effect does branching of an alkane chain has on its boiling point? Ans. Alkanes experience inter-molecular Van der Waals forces. The stronger the force, the greater will be the boiling point of the alkane. As branching increases, the surface area of the molecule decreases which results in a small area of contact. As a result, the Van der Waals force also decreases which can be overcome at a relatively lower temperature. Hence, the boiling point of an alkane chain decreases with an increase in branching.
Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism. Ans. Addition of HBr to propene is an example of an electrophilic substitution reaction. Hydrogen bromide provides an electrophile, $H^{+}$ . This electrophile attacks the double bond to form 1° and 2° carbocations as shown:

Secondary carbocations are more stable than primary carbocations. Hence, the former predominates since it will form at a faster rate. Thus, in the next step, Br- attacks the carbocation to form 2 - bromopropane as the major product.

This reaction follows Markovnikov's rule where the negative part of the addendum is attached to the carbon atom having a lesser number of hydrogen atoms. In the presence of benzoyl peroxide, an addition reaction takes place anti to Markovnikov's rule. The reaction follows a free radical chain mechanism as:

Secondary free radicals are more stable than primary radicals. Hence, the former predominates since it forms at a faster rate. Thus, 1-bromopropane is obtained as the major product.

In the presence of peroxide, Br free radical acts as an electrophile. Hence, two different products are obtained on addition of HBr to propene in the absence and presence of peroxide.

Write down the products of ozonolysis of 1, 2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene? Ans. o-xylene has two resonance structures:

All three products, i.e., methyl glyoxal, 1, 2-dimethylglyoxal, and glyoxal are obtained from two Kekule structures. Since all three products cannot be obtained from any one of the two structures, this proves that o-xylene is a resonance hybrid of two Kekule structures (I and II).

Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour. Ans. Acidic character of a species is defined on the basis of ease with which it can lose its H-atoms. The hybridization state of carbon in the given compound is:<img

As the s-character increases, the electronegativity of carbon increases and the electrons of C-H bond pair lie closer to the carbon atom. As a result, partial positive charge of Hatom increases and $H^{+}$ ions are set free. The s -character increases in the order:

sp^{3} < sp^{2} < sp

Hence, the decreasing order of acidic behaviour is Ethyne > Benzene > Hexane.

Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty? Ans. Benzene is a planar molecule having delocalized electrons above and below the plane of ring. Hence, it is electron-rich. As a result, it is highly attractive to electron deficient species i.e., electrophiles. Therefore, it undergoes electrophilic substitution reactions very easily. Nucleophiles are electron-rich. Hence, they are repelled by benzene. Hence, benzene undergoes nucleophilic substitutions with difficulty.
How would you convert the following compounds into benzene? (i) Ethyne (ii) Ethene (iii) Hexane

Ans.

#Please refer attached PDF for the next set of questions and solutions.

3.0NCERT Class 11 Biology – Chapter-Wise Solutions & Explanations

Study Class 11 Chemistry with NCERT Solutions containing chapter-wise answers to textbook questions. Revise important topics, solve practice questions, and prepare for school exams.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Some Basic Concepts of Chemistry
Chapter 2	Structure of Atom
Chapter 3	Classification of Elements and Periodicity in Properties
Chapter 4	Chemical Bonding and Molecular Structure
Chapter 5	Chemical Thermodynamics
Chapter 6	Equilibrium
Chapter 7	Redox Reactions
Chapter 8	Organic Chemistry – Some Basic Principles and Techniques

4.0Key Features of NCERT Solutions for Class 11 Chemistry Chapter 9

Reaction Mechanisms: Step-by-step explanations of electrophilic addition and electrophilic substitution reactions to help students understand how these reactions occur.
Conversion Problems: Clear steps for solving organic conversion questions, such as converting calcium carbide into nitrobenzene.
Stability of Carbocations: Understanding how carbocation stability helps predict the major product in addition reactions.
Practical Applications: Learning about cracking and reforming of alkanes and their use in the petroleum industry.
Prepared by ALLEN Faculty: Solutions prepared by ALLEN subject experts to help students understand important concepts and prepare for CBSE exams, JEE, and NEET.

Chapter-wise Class 11 Chemistry NCERT Solutions

Chapter 1 Some Basic Concepts of Chemistry

Chapter 2 Structure of Atom

Chapter 3 Classification of Elements and Periodicity in Properties

Chapter 4 Chemical Bonding and Molecular Structure

Chapter 5 Chemical Thermodynamics

Chapter 6 Equilibrium

Chapter 7 Redox Reactions

Chapter 8 Organic Chemistry – Some Basic Principles and Techniques

Chapter 9 Hydrocarbons

Subject-wise Class 11 NCERT Solutions

Class-wise NCERT Solutions

Study Material by ALLEN: JEE/NEET Topper's First Choice

JEE Main Previous Year Papers

JEE Main Sample Papers

JEE Main Mock Test

JEE Adv Sample Papers

JEE Adv Previous Year Papers

JEE Advanced Mock Test

NEET Previous Year Papers

NEET Sample Papers

NEET Mock Test

NEET Revision Notes

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NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

NCERT Solutions for Class 11 Chemistry Chapter 9 (Hydrocarbons) help students understand the main types of hydrocarbons, including alkanes, alkenes, alkynes, and aromatic compounds. This chapter covers their structure, properties, preparation, and chemical reactions. Hydrocarbons are found in fuels such as natural gas and petrol and are used to make many everyday products.

NCERT Solutions for Class 11 Chemistry Chapter 9 by ALLEN explain important reactions, acidity trends, organic conversions, and the structure of benzene. The solutions include answers to NCERT textbook questions to help students understand the chapter, practise questions, and prepare for CBSE exams, JEE, and NEET.

1.0Class 11 Chemistry Chapter 9 Hydrocarbons: Key Concepts

This chapter explains the types of hydrocarbons, their preparation, properties, and chemical reactions. The key concepts include:

Alkanes: Learning about ethane conformations using Sawhorse and Newman projections, the Wurtz reaction, and free radical halogenation.
Alkenes: Understanding cis-trans isomerism, electrophilic addition reactions, and Markovnikov’s and anti-Markovnikov’s rules.
Ozonolysis: Learning how ozonolysis helps identify the position of double and triple bonds in hydrocarbons.
Alkynes: Understanding the acidic nature of terminal alkynes and how alkynes can form benzene through cyclic polymerisation.
Aromatic Hydrocarbons: Learning about the structure of benzene and Hückel’s rule for identifying aromatic compounds.
Electrophilic Substitution Reactions: Studying nitration, sulphonation, halogenation, and Friedel-Crafts alkylation and acylation of benzene.
Directive Influence: Understanding how groups already attached to a benzene ring affect where a new group enters, such as the ortho, meta, or para position.
Carcinogenicity and Toxicity: Learning about harmful polynuclear hydrocarbons and their effects on human health and the environment.

2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 9

NCERT EXERCISE

How do you account for the formation of ethane during chlorination of methane? Ans. Chlorination of methane proceeds via a free radical chain mechanism. The whole reaction takes place in the given three steps. Step 1: Initiation: The reaction begins with the homolytic cleavage of $Cl - Cl$ bond as:

$Cl - Cl h ν \dot{C} l + \dot{C} l$

Chlorine free radicals Step 2: Propagation: In the second step, chlorine free radicals attack methane molecules and break down the C-H bond to generate methyl radicals as

$CH_{4} + \dot{C} l h ν \dot{C} H_{3} + H - \dot{C} l$ Methane These methyl radicals react with other chlorine free radicals to form methyl chloride along with the liberation of a chlorine free radical.

$\dot{C} H_{3} + Cl - Cl \to CH_{3} - Cl + Cl$ Methyl chloride Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While HCl and $CH_{3} Cl$ are the major products formed, other higher halogenated compounds are also formed as:

$CH_{3} + Cl - \dot{C} l \to \dot{C} H_{2} Cl + HCl$

$\dot{C} H_{2} Cl + Cl - Cl \to CH_{2} Cl_{2} + \dot{C} l$

Step 3: Termination: Formation of ethane is a result of the termination of chain reactions taking place as a result of the consumption of reactants as:

$\dot{C} l + \dot{C} l \to Cl + Cl$

$H_{3} \dot{C} + \dot{C} H_{3} \to H_{3} C - CH_{3}$

Hence, by this process, ethane is obtained as a by-product of chlorination of methane.

Write IUPAC names of the following compounds:

Ans.

For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated: (a) $C_{4} H_{8}$ (one double bond) (b) $C_{5} H_{8}$ (one triple bond)

Ans.

(a) The following structural isomers are possible for $C_{4} H_{8}$ with one double bond: (I) $H_{2} C 1 = C 2 H - C 3 H_{2} - C 4 H_{3}$ (II) $C 1 H_{3} - C 2 H = C 3 H - C 4 H_{3}$ (III) $C 1 H_{2} = CH_{3} ∣ C 2 - C 3 H_{3}$

The IUPAC name of Compound (I) is But-1-ene, Compound (II) is But-2-ene, and Compound (III) is 2-Methylprop-1-ene.

(b) The following structural isomers are possible for $C_{5} C_{8}$ with one triple bond:

(i) $H C 1 \equiv C 2 - C 3 H_{2} - C 5 H_{3}$

(ii) $H_{3} C 1 - C 2 \equiv C 3 - C 4 H_{2} - C 5 H_{3}$

(iii) $H_{3} C 4 - CH_{3} ∣ C 3 H - C 2 \equiv C 1 H$

The IUPAC name of Compound (I) is Pent-1-yne, Compound (II) is Pent-2-yne, and Compound (III) is 3-Methylbut-1-yne.

Write IUPAC names of the products obtained by the ozonolysis of the following compounds: (i) Pent-2-ene (ii) 3,4-Dimethyl-hept-3-ene (iii) 2-Ethylbut-1-ene (iv) 1-Phenylbut-1-ene

Ans.

(i) Pent-2-ene undergoes ozonolysis as:

The IUPAC name of Product (I) is ethanal and Product (II)is propanal. (ii) 3, 4-Dimethylhept-3-ene undergoes ozonolysis as:

The IUPAC name of Product (I)is butan-2-one and Product (II)is Pentan-2-one.

(iii)2-Ethylbut-1-ene undergoes ozonolysis as:

The IUPAC name of Product (I)is pentan-3-one and Product (II)is methanal.

(iv) 1-Phenylbut-1-ene undergoes ozonolysis as:

The IUPAC name of Product (I)is benzaldehyde and Product (II)is propanal.

5. An alkene 'A' on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of 'A'. Ans.

During ozonolysis, an ozonide having a cyclic structure is formed as an intermediate which undergoes cleavage to give the final products. Ethanal and pentan-3-one are obtained from the intermediate ozonide. Hence, the expected structure of the ozonide is:

This ozonide is formed as an addition of ozone to 'A'. The desired structure of 'A' can be obtained by the removal of ozone from the ozonide. Hence, the structural formula of 'A' is:

$H_{3} C 1 - CH_{2} - CH_{3} ∣ C 2 H = C 3 - C 4 H_{2} - C 5 H_{3}$

The IUPAC name of 'A' is 3-Ethylpent-2-ene.

An alkene 'A' contains three C - C, eight $C - H σ$ bonds and one $C - C π$ bond. 'A' on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of 'A'. Ans. As per the given information, 'A' on ozonolysis gives two moles of an aldehyde of molar mass 44 u. The formation of two moles of an aldehyde indicates the presence of identical structural units on both sides of the double bond containing carbon atoms. Hence, the structure of 'A' can be represented as:

XC = CX

There are eight $C - H σ$ bonds. Hence, there are 8 hydrogen atoms in 'A'. Also, there are three C-C bonds. Hence, there are four carbon atoms present in the structure of 'A'. Combining the inferences, the structure of 'A' can be represented as:

'A' has 3 C-C bonds, 8 C-H $σ$ bonds, and one $C - C π$ bond. Hence, the IUPAC name of 'A' is But-2-ene. Ozonolysis of 'A' takes place as:

The final product is ethanal with molecular mass $= [(2 \times 12) + (4 \times 1) + (1 \times 16)] = 44 u$

Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene? Ans. As per the given information, propanal and pentan-3-one are the ozonolysis products of an alkene. Let the given alkene be 'A'. Writing the reverse of the ozonolysis reaction, we get:

The products are obtained on the cleavage of ozonide 'X'. Hence, 'X' contains both products in the cyclic form. The possible structure of ozonide can be represented as:

Now, 'X' is an addition product of alkene 'A' with ozone. Therefore, the possible structure of alkene 'A' is: 8. Write chemical equations for combustion reaction of the following hydrocarbons: (i) Butane (ii) Pentene (iii) Hexyne (iv) Toluene Ans. Combustion can be defined as a reaction of a compound with oxygen.

Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why? Ans. Hex-2-ene is represented as:

CH_{3} - CH = CH - CH_{2} - CH_{2} - CH_{3}

Geometrical isomers of hex-2-ene are:

The dipole moment of cis-compound is a sum of the dipole moments of $C - CH_{3}$ and $CCH_{2} CH_{2} CH_{3}$ bonds acting in the same direction.

The dipole moment of trans-compound is the resultant of the dipole moments of $CCH_{3}$ and $C - CH_{2} CH_{2} CH_{3}$ bonds acting in opposite directions. Hence, cis-isomer is more polar than trans-isomer. The higher the polarity, the greater is the intermolecular dipole-dipole interaction and the higher will be the boiling point. Hence, cis-isomer will have a higher boiling point than trans-isomer.

Why is benzene extra ordinarily stable though it contains three double bonds? Ans. Benzene is a hybrid of resonating structures given as:

All six carbon atoms in benzene are $sp^{2}$ hybridized. The two $sp^{2}$ hybrid orbitals of each carbon atom overlap with the $sp^{2}$ hybrid orbitals of adjacent carbon atoms to form six sigma bonds in the hexagonal plane. The remaining $sp^{2}$ hybrid orbital on each carbon atom overlaps with the s-orbital of hydrogen to form six sigma C-H bonds. The remaining unhybridized p-orbital of carbon atoms has the possibility of forming three $π$ bonds by the lateral overlap of $C_{1} - C_{2}, C_{3} - C_{4}, C_{5} - C_{6}$ , or $C_{2} - C_{3}, C_{4} - C_{5}, C_{6} - C_{1}$ ,.

The six $π$ 's are delocalized and can move freely about the six carbon nuclei. Even after the presence of three double bonds, these delocalized $π$ -electrons stabilize benzene.

11.What are the necessary conditions for any system to be aromatic? Ans. A compound is said to be aromatic if it satisfies the following three conditions: (i) It should have a planar structure. (ii) The $π$ -electrons of the compound are completely delocalized in the ring. (iii) The total number of $π$ -electrons present in the ring should be equal to $(4 n + 2)$ , where $n = 0, 1, 2 \dots$ etc. This is known as Huckel's rule.

Explain why the following systems are not aromatic?

Ans.

For the given compound, the number of $π$ -electrons is six. But only four $π$ -electrons are present within the ring. Also there is no conjugation of $π$ -electrons within the ring and the compound is not planar in shape. Hence, the given compound is not aromatic in nature. (ii)

For the given compound, the number of $π$ -electrons is four. By Huckel's rule,

4 n + 2 = 4 4 n = 2 n = \frac{1}{2}

For a compound to be aromatic, the value of n must be an integer $(n = 0, 1, 2 \dots)$ , which is not true for the given compound. Hence, it is not aromatic in nature.

(iii) For the given compound, the number of $π$ -electrons is eight. By Huckel's rule,

4 n + 2 = 8 4 n = 6 n = \frac{3}{2}

For a compound to be aromatic, the value of n must be an integer $(n = 0, 1, 2 \dots)$ . Since the value of n is not an integer, the given compound is not aromatic in nature.

How will you convert benzene into (i) p-nitrobromobenzene (ii) m-nitrochlorobenzene (iii) p -nitrotoluene (iv) acetophenone

Ans.

(i) Benzene can be converted into p-nitrobromobenzene as:

(ii) Benzene can be converted into m-nitrochlorobenzene as:

(iii) Benzene can be converted into p-nitrotoulene as:

(iv) Benzene can be converted into acetophenone as:

14. In the alkane $H_{3} C - CH_{2} - C (CH_{3})_{2} - CH_{2}$ $- CH (CH_{3})_{2}$ , identify $1^{\circ}, 2^{\circ}, 3^{\circ}$ carbon atoms and give the number of H atoms bonded to each one of these. Ans.

1° carbon atoms are those which are bonded to only one carbon atom, i.e., they have only one carbon atom as their neighbour. The given structure has five 1° carbon atoms and fifteen hydrogen atoms are attached to it. 2° carbon atoms are those which are bonded to two carbon atoms, i.e., they have two carbon atoms as their neighbours. The given structure has two 2° carbon atoms and four hydrogen atoms are attached to it. 3° carbon atoms are those which are bonded to three carbon atoms, i.e., they have three carbon atoms as their neighbours. The given structure has one 3° carbon atom and only one hydrogen atom is attached to it.

What effect does branching of an alkane chain has on its boiling point? Ans. Alkanes experience inter-molecular Van der Waals forces. The stronger the force, the greater will be the boiling point of the alkane. As branching increases, the surface area of the molecule decreases which results in a small area of contact. As a result, the Van der Waals force also decreases which can be overcome at a relatively lower temperature. Hence, the boiling point of an alkane chain decreases with an increase in branching.
Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism. Ans. Addition of HBr to propene is an example of an electrophilic substitution reaction. Hydrogen bromide provides an electrophile, $H^{+}$ . This electrophile attacks the double bond to form 1° and 2° carbocations as shown:

Secondary carbocations are more stable than primary carbocations. Hence, the former predominates since it will form at a faster rate. Thus, in the next step, Br- attacks the carbocation to form 2 - bromopropane as the major product.

This reaction follows Markovnikov's rule where the negative part of the addendum is attached to the carbon atom having a lesser number of hydrogen atoms. In the presence of benzoyl peroxide, an addition reaction takes place anti to Markovnikov's rule. The reaction follows a free radical chain mechanism as:

Secondary free radicals are more stable than primary radicals. Hence, the former predominates since it forms at a faster rate. Thus, 1-bromopropane is obtained as the major product.

In the presence of peroxide, Br free radical acts as an electrophile. Hence, two different products are obtained on addition of HBr to propene in the absence and presence of peroxide.

Write down the products of ozonolysis of 1, 2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene? Ans. o-xylene has two resonance structures:

All three products, i.e., methyl glyoxal, 1, 2-dimethylglyoxal, and glyoxal are obtained from two Kekule structures. Since all three products cannot be obtained from any one of the two structures, this proves that o-xylene is a resonance hybrid of two Kekule structures (I and II).

Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour. Ans. Acidic character of a species is defined on the basis of ease with which it can lose its H-atoms. The hybridization state of carbon in the given compound is:<img

As the s-character increases, the electronegativity of carbon increases and the electrons of C-H bond pair lie closer to the carbon atom. As a result, partial positive charge of Hatom increases and $H^{+}$ ions are set free. The s -character increases in the order:

sp^{3} < sp^{2} < sp

Hence, the decreasing order of acidic behaviour is Ethyne > Benzene > Hexane.

Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty? Ans. Benzene is a planar molecule having delocalized electrons above and below the plane of ring. Hence, it is electron-rich. As a result, it is highly attractive to electron deficient species i.e., electrophiles. Therefore, it undergoes electrophilic substitution reactions very easily. Nucleophiles are electron-rich. Hence, they are repelled by benzene. Hence, benzene undergoes nucleophilic substitutions with difficulty.
How would you convert the following compounds into benzene? (i) Ethyne (ii) Ethene (iii) Hexane

Ans.

#Please refer attached PDF for the next set of questions and solutions.

3.0NCERT Class 11 Biology – Chapter-Wise Solutions & Explanations

Study Class 11 Chemistry with NCERT Solutions containing chapter-wise answers to textbook questions. Revise important topics, solve practice questions, and prepare for school exams.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Some Basic Concepts of Chemistry
Chapter 2	Structure of Atom
Chapter 3	Classification of Elements and Periodicity in Properties
Chapter 4	Chemical Bonding and Molecular Structure
Chapter 5	Chemical Thermodynamics
Chapter 6	Equilibrium
Chapter 7	Redox Reactions
Chapter 8	Organic Chemistry – Some Basic Principles and Techniques

4.0Key Features of NCERT Solutions for Class 11 Chemistry Chapter 9

Reaction Mechanisms: Step-by-step explanations of electrophilic addition and electrophilic substitution reactions to help students understand how these reactions occur.
Conversion Problems: Clear steps for solving organic conversion questions, such as converting calcium carbide into nitrobenzene.
Stability of Carbocations: Understanding how carbocation stability helps predict the major product in addition reactions.
Practical Applications: Learning about cracking and reforming of alkanes and their use in the petroleum industry.
Prepared by ALLEN Faculty: Solutions prepared by ALLEN subject experts to help students understand important concepts and prepare for CBSE exams, JEE, and NEET.

Chapter-wise Class 11 Chemistry NCERT Solutions

Chapter 1 Some Basic Concepts of Chemistry

Chapter 2 Structure of Atom

Chapter 3 Classification of Elements and Periodicity in Properties

Chapter 4 Chemical Bonding and Molecular Structure

Chapter 5 Chemical Thermodynamics

Chapter 6 Equilibrium

Chapter 7 Redox Reactions

Chapter 8 Organic Chemistry – Some Basic Principles and Techniques

Chapter 9 Hydrocarbons

Subject-wise Class 11 NCERT Solutions

Class-wise NCERT Solutions

Study Material by ALLEN: JEE/NEET Topper's First Choice

JEE Main Previous Year Papers

JEE Main Sample Papers

JEE Main Mock Test

JEE Adv Sample Papers

JEE Adv Previous Year Papers

JEE Advanced Mock Test

NEET Previous Year Papers

NEET Sample Papers

NEET Mock Test

NEET Revision Notes

Frequently Asked Questions

According to NCERT Solutions Class 11 Chemistry Chapter 9, what is the difference between alkanes, alkenes, and alkynes?

What is the difference between Markovnikov’s rule and the anti-Markovnikov effect, as explained in NCERT Solutions Class 11 Chemistry Chapter 9?

What is the difference between cis and trans isomers in alkenes according to NCERT Solutions Class 11 Chemistry Chapter 9?

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NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

1.0Class 11 Chemistry Chapter 9 Hydrocarbons: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 9

NCERT EXERCISE

3.0NCERT Class 11 Biology – Chapter-Wise Solutions & Explanations

4.0Key Features of NCERT Solutions for Class 11 Chemistry Chapter 9

Frequently Asked Questions

According to NCERT Solutions Class 11 Chemistry Chapter 9, what is the difference between alkanes, alkenes, and alkynes?

What is the difference between Markovnikov’s rule and the anti-Markovnikov effect, as explained in NCERT Solutions Class 11 Chemistry Chapter 9?

What is the difference between cis and trans isomers in alkenes according to NCERT Solutions Class 11 Chemistry Chapter 9?

Where can students find NCERT Solutions Class 11 Chemistry Chapter 9 for exam preparation?

Is Class 11 Chemistry Chapter 9 important for board and competitive exams?

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NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

1.0Class 11 Chemistry Chapter 9 Hydrocarbons: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 9

NCERT EXERCISE

3.0NCERT Class 11 Biology – Chapter-Wise Solutions & Explanations

4.0Key Features of NCERT Solutions for Class 11 Chemistry Chapter 9

Frequently Asked Questions

According to NCERT Solutions Class 11 Chemistry Chapter 9, what is the difference between alkanes, alkenes, and alkynes?

What is the difference between Markovnikov’s rule and the anti-Markovnikov effect, as explained in NCERT Solutions Class 11 Chemistry Chapter 9?

What is the difference between cis and trans isomers in alkenes according to NCERT Solutions Class 11 Chemistry Chapter 9?

Where can students find NCERT Solutions Class 11 Chemistry Chapter 9 for exam preparation?

Is Class 11 Chemistry Chapter 9 important for board and competitive exams?

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NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

1.0Class 11 Chemistry Chapter 9 Hydrocarbons: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 9NCERT EXERCISE

3.0NCERT Class 11 Biology – Chapter-Wise Solutions & Explanations

4.0Key Features of NCERT Solutions for Class 11 Chemistry Chapter 9

Frequently Asked Questions

According to NCERT Solutions Class 11 Chemistry Chapter 9, what is the difference between alkanes, alkenes, and alkynes?

What is the difference between Markovnikov’s rule and the anti-Markovnikov effect, as explained in NCERT Solutions Class 11 Chemistry Chapter 9?

What is the difference between cis and trans isomers in alkenes according to NCERT Solutions Class 11 Chemistry Chapter 9?

Where can students find NCERT Solutions Class 11 Chemistry Chapter 9 for exam preparation?

Is Class 11 Chemistry Chapter 9 important for board and competitive exams?

Join ALLEN!

NCERT Solutions Class 11 Chemistry Chapter 9 - Hydrocarbons

1.0Class 11 Chemistry Chapter 9 Hydrocarbons: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 9NCERT EXERCISE

3.0NCERT Class 11 Biology – Chapter-Wise Solutions & Explanations

4.0Key Features of NCERT Solutions for Class 11 Chemistry Chapter 9

2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 9

NCERT EXERCISE

2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 9

NCERT EXERCISE