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Ever wondered what powers your car, cooks your food, and heats your home? The answer lies in hydrocarbons! The hydrocarbons are the compounds that consist of carbon and hydrogen only. They play a vital role in our day-to-day lives as they are the primary source of energy. Petrol and diesel are hydrocarbons that are used as fuel for vehicles, LPG is also a hydrocarbon that is used for cooking. This is one of the important chapters in organic chemistry.
JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy
NEET Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | ALLEN
The important topics like the preparation and properties of these compounds, are well discussed in this chapter. The NCERT solutions for class 11 chemistry chapter 9 are prepared by our subject experts to help you master the topics. These NCERT solutions offer detailed solutions with simple explanations that will help you develop a clear understanding of the topics. The higher-order thinking skills (HOTS) questions are included to improve your critical thinking. We have also highlighted some points that will help you build a good approach to solve the questions. Students can also go through the hydrocarbons class 11 notes available on our website.
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You can download the NCERT solutions for class 11 chemistry chapter 9 PDF here. Click on the icon below to get detailed solutions to all the questions that are given in the textbook. The hydrocarbons class 11 questions and answers are too important for the exams.
The questions are solved with simple explanations. You can also go through the hydrocarbons class 11 notes for learning the concepts.
Question 9.1 How do you account for the formation of ethane during the chlorination of methane?
Answer:
Chlorination of methane occurs by a free radical mechanism and it takes place in three steps-
(i) Initiation-
First, homolytic cleavage of the
(ii) Propagation-
Chlorine-free radicals attack methane molecules and generate methyl-free radicals as
This methyl radical reacts with other molecules of chlorine (
Then the chlorine free radical reacts with methyl chloride and this way propagation occurs.
(iii) Termination-
Ethane will be produced as a final product in this step. When two methyl free radicals react with each other, ethane will be formed.
Question 9.2(a) Write IUPAC names of the following compounds
Answer:
The parent chain has 4 carbons and the double bond is at C-2, so it but-2-ene and also methyl group is also present C-2.
The IUPAC name of the given compound is 2-methylbut-2-ene
Question 9.2(b) Write IUPAC names of the following compounds
Answer:
Both alkene and alkyne are present but the priority will go to the alkene group.
The IUPAC name of the compound is Pent-1-ene-3-yne
Question 9.2(c) Write IUPAC names of the following compounds
Answer:
The IUPAC name of the compound is 1, 3-Butadiene or Buta-1, 3-diene. It is a symmetrical compound so the numbering cam be done from either end.
Question 9.2(d) Write IUPAC names of the following compounds :
Answer:
The parent chain will be the aliphatic carbon and benzene will be the side group. The IUPAC name of the compound is 4-phenylbut-1-ene.
Question 9.2(e) Write IUPAC names of the following compounds :
Answer:
Here there are two functional groups hydroxy and methyl. The priority will go to the hydroxy group.
The IUPAC name of the compound is 2-methylphenol
Question 9.2 (f) Write IUPAC names of the following compounds :
Answer:
The parent chain has ten carbons so decane (longest chain). The IUPAC name of the compound is 5-(2-methylpropyl)-decane.
Question 9.2(g) Write IUPAC names of the following compounds :
Answer:
The parent chain has ten carbons and the numbering starts where the alkene give the least number. The IUPAC name of the compound is 4-ethyl-1,4,8-deca triene.
Answer:
![]() | But-1-ene |
![]() | But-2-ene |
![]() | 2-Methylprop-1-ene |
Question 9.4(i) Write IUPAC names of the products obtained by the ozonolysis of the following compounds :
Pent-2-ene
Answer:
Ozonolysis of Pent-2-ene gives two product,s both are aldehyde compounds. In this process, the ozone molecules attach to the double bond of the molecule and break it into two products.
The IUPAC name of the compounds-(i) ethanal (ii) propanal
Question 9.4 (ii) Write IUPAC names of the products obtained by the ozonolysis of the following compounds :
3,4-Dimethylhept-3-ene
Answer:
Ozonolysis of 3,4-dimethylhept-3-ene gives two products of keto-compounds. The ozone molecules attach to the double bond.
The IUPAC name of the compound is -(i) Butan-2-one (ii) Pentan-2-one
Question 9.4(iii) Write IUPAC names of the products obtained by the ozonolysis of the following compounds:
2-Ethylbut-1-ene
Answer:
Ozonolysis of 2-ethylbut-1-ene gives two products, one is a keto compound and another is an aldehyde.
.
The IUPAC name of one compound is Pentan-3-one and the other compound is methanal.
Question 9.4 (iv) Write IUPAC names of the products obtained by the ozonolysis of the following compounds :
1-Phenylbut-1-ene
Answer:
Ozonolysis of 1-Phenylbut-1-ene gives two aldehydes. One is aromatic in nature and the other is an aliphatic aldehyde.
The IUPAC name of the compound is-(i) Benzaldehyde (ii) propanal
Question 9.5 An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3- one. Write structure and IUPAC name of ‘A’.
Answer:
In the process of ozonolysis, an ozonide, a cyclic ring structure intermediate, is formed, which undergoes cleavage to give the product. The compound A produces pentan-3-one and ethanal. So, the possible structure of A should be
Thus, by removing the ozone from ozonide, we can get the parent alkene structure.
Answer:
As per the given data, compound A on ozonolysis gives two moles of aldehyde, having a molar mass of 44u. It indicates that the compound is symmetrical. So, the possible general structure of
A =
There are eight
Now, by combining all the observations, the structure of the A would be
The ozonolysis reaction is shown below-
The atomic mass of ethanal is 44 u.
Question 9.7 Propanal and pentane-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?
Answer:
In the process of ozonolysis, an ozonide (a cyclic ring structure intermediate) is formed, which undergoes cleavage to give the product. The parent compound produces pentan-3-one and propanal. So, the possible structure should be
Here in the above structure, the right side will give Pent-3-one and the left-hand side will give propanal.
Thus, by removing the ozone from ozonide, we can get the parent alkene structure.
Therefore, the structure of the parent alkene is
(3-ethyl-3-hexene)
Question 9.8(i) Write chemical equations for the combustion reaction of the following hydrocarbons:
Butane
Answer:
Combustion is a chemical reaction in which a substance reacts with oxygen releasing energy in the form of heat and light. It is commonly seen in burning fuels like wood, petrol or gas.
Combustion means the reaction of a compound with dioxygen(
Question 9.8(ii) Write chemical equations for combustion reaction of the following hydrocarbons:
Pentene
Answer:
Combustion means the reaction of a compound with dioxygen(
Question 9.8 (iii) Write chemical equations for combustion reaction of the following hydrocarbons:
Hexyne
Answer:
Combustion means the reaction of the given compound with dioxygen(
Structure of hexyne
The reaction
Question 9.8(iv) Write chemical equations for combustion reaction of the following hydrocarbons:
Toluene
Answer:
Combustion is the reaction of the given compound with dioxygen(
Structure of toluene
Question 9.9 Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?
Answer:
The structure of the hex-2-ene is shown here-
Now, the Geometrical, cis and trans isomers -
Cis-isomer has a higher boiling point than trans-isomers due to more dipole-dipole interactions between the molecules. The cis-form is more polar than the trans-form because it has a higher dipole moment. The trans-form is almost non-polar.
Question 9.10 Why is benzene extraordinarily stable though it contains three double bonds?
Answer:
Benzene is a hybrid of various resonating structures.
Each carbon of benzene is in
In benzene, six C-H sigma bonds present. And the remaining unhybridised
There are 6
Question 9.11 What are the necessary conditions for any system to be aromatic?
Answer:
The necessary conditions for any system to be aromatic are
Question 9.12 (i) Explain why the following systems are not aromatic?
Answer:
Not an aromatic compound because the
Question 9.12(ii) Explain why the following systems are not aromatic?
Answer:
There is no complete conjugation of a
Question 9.12(iii) Explain why the following systems are not aromatic?
Answer:
It disobeys the Hückel rule of (4n+2)
Question 9.13(i) How will you convert benzene into
p-nitrobromobenzene
Answer:
Bromination of a benzene ring in the presence of anhydrous
Question 9.13 (ii) How will you convert benzene into
m- nitrochlorobenzene
Answer:
Benzene on treatment with conc. nitric acid and sulphuric acid give nitrobenzene, which on further treatment with chlorine in the presence of anhydrous aluminium chloride (
Question 9.13(iii) How will you convert benzene into
p - nitrotoluene
Answer:
Alkylation of benzene in the presence of anhydrous aluminium chloride gives methylbenzene and
Question 9.13 (iv) How will you convert benzene into
acetophenone
Answer:
Benzene on reacting with an acyl chloride in the presence of anhydrous aluminium chloride, gives acetophenone with hydrochloric acid as a by-product.
Answer:
Question 9.15 What effect does branching of an alkane chain has on its boiling point?
Answer:
With an increase in the branching of the alkane, the boiling point of the alkane decreases. Alkanes experience intermolecular van der Waals forces. The strong is the force, strong will be the boiling point. When we increase the branching, the surface area of the molecule decreases; as a result, the van der Waals force also decreases.
Answer:
Addition of
In this addition, an electrophile
A secondary carbocation is more stable than a primary carbocation. Thus bromide ion attacks the carbocation to form 2-bromopropane as a major product. (This mechanism is followed by Markovnikov's rule)
Answer:
Ortho-xylene has two resonant structures, so,
Since all three products, methylglyoxal, 1,2-methylglyoxal and glyoxal cannot be obtained from either of the two structures (i and ii). Hence, we can say that o-xylene is a resonant hybrid of two Kekule structures (I and II)
Question 9.18 Arrange benzene,
Answer:
The acidic character of a species is defined on the basis of the ease with which it can lose its H+ ions.
The acidic character decreases in the order: Ethyne > Benzene > Hexane.
As the s-character decreases, carbon electronegativity decreases. Higher the s-character, more will be the negativity so the negative charge developing on carbon will be more stabilized. So the H that is attached to the alkyne will leave more readily.
Question 9.19 Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
Answer:
In benzene, the
Question 9.20 (i) How would you convert the following compounds into benzene?
Ethyne
Answer:
Benzene can be formed by the cyclic polymerisation of ethyne. Ethyne on passing through a red-hot tube(made of iron) at 873K forms benzene. Three molecules of ethyne polymerise to form benzene.
Question 9.20(ii) How would you convert the following compounds into benzene?
Ethene
Answer:
By converting ethene to ethyne by reacting with bromine in the presence of carbon tetrachloride. And then heating in the presence of alc. KOH followed by
Question 9.20(iii) How would you convert the following compounds into benzene?
Hexane
Answer:
The cyclisation of hexane in the presence of
Question 9.21 Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
Answer:
Structures of all the alkenes that on hydrogenation, give 2-methylbutane.
The general structure of the 2-methylbutane is shown here;
As per the above structure, the following alkene compounds produce 2-methylbutane by hydrogenation are
Question 9.22 (a) Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E +
Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene
Answer:
Electrophiles are electron-deficient species, so they want a nucleophile that donates electrons to them. The higher the electron density on a benzene ring, the higher the reactivity towards electrophiles.
Decreasing order of their reactivity with an electrophile(
Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene
Question 9.22 (b) Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E +
Toluene,
Answer:
Electrophiles are electron-deficient species, so they want a nucleophile that donates electrons to them. The higher the electron density on a benzene ring, the higher the reactivity towards electrophiles.
Since the methyl group is an electron-donating group, it increases the electron density on the benzene ring. The more the number of electron withdrawing groups, the less reactive towards electrophiles.
Therefore, the decreasing order of reactivity towards electrophile -
toluene
Question 9.23 Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most easily and why?
Answer:
Nitration occurs by an electrophilic substitution reaction, in which an electron-rich species is attacked by an electron-deficient molecule known as an electrophile. In nitration,
Question 9.24 Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
Answer:
Lewis acid like anhydrous ferric chloride
Answer:
Wurtz reaction-
Wurtz reaction is not preferred for the preparation of alkanes containing an odd number of carbon atoms because if we take two dissimilar alkyl halides as reactants, the product will be a mixture of alkanes and alkene may also form due to free radical mechanism. example- bromomethane and iodoethane.
All the products in the mixture have nearly the same boiling point. So, the separation will be difficult.
The hydrocarbons class 11 Important questions with answers are provided here. These questions will help you tackle complex problems.
Question: Given below are two statements :
Statement (I) : On nitration of
Statement (II) :
In light of the above statements, choose the correct answer from the options given below:
1) Both Statement I and Statement II are false
2) Statement I is false but Statement II is true
3) Both Statement I and Statement II are true
4) Statement I is true but Statement II is false
Answer:
Statement-I
Statement-II
Hence, the correct answer is option (3).
Question: Given below are two statements :
Statement I : Ozonolysis followed by treatment with
Statement II : The production obtained by ozonolysis followed by treatment with
In the light of the above statements, choose the correct answer from the options given below :
(1) Both Statement I and Statement II are true
(2) Statement I is false but Statement II are true
(3) Statement I is true but Statement II is false
(4) Both Statement I and Statement II are false
Answer:
Statement I: Correct statement
Statement II: Incorrect statement because the product has a chiral center.
Hence, the correct answer is option (3).
Question: Choose the correct set of reagents for the following conversion.
1)
2)
3)
4)
Answer:
The first step is electrophilic aromatic substitution, in which bromination of ethylbenzene is done in the presence of iron. The second step is the free radical halogenation of the above obtained product, followed by an elimination reaction in the presence of alcoholic KOH, and the obtained product is 4-bromostyrene.
Hence, the correct answer is option (1).
A structured approach that works well for both theory-based and numerical problems is given below. It is advisable to follow the NCERT solutions for class 11 chemistry chapter 9 hydrocarbons for deep learning.
1. Understand the basic concepts
Students can start by identifying the key topics covered: Types of hydrocarbons- alkanes, alkenes, alkynes, aromatic hydrocarbons, nomenclature rules, Isomerism, preparation and reactions of hydrocarbons, and their physical and chemical properties.
You can make the use of class 11 hydrocarbons notes available on our website.
2. Categorize the question
Learn to categorise questions into conceptual/theoretical questions, structural drawing, reaction-based questions, IUPAC nomenclature or numerical questions. This will help you prepare for them accordingly.
3. Attempt questions wisely.
While solving questions, it is very important to read the question carefully and identify exactly what is being asked. Relate it to the concept learned.
The hydrocarbons class 11 questions and answers are well explained in our exemplar solutions.
4. Try to solve the question in a stepwise manner
While solving questions of organic chemistry, it is very important to note down the information given and you can also use bullet points for clarity when writing answers.
5) Practice more
Students can refer to the solved examples in the textbook and then solve the in-text questions. Students can also access the NCERT solutions for class 11 chemistry PDF for quick revision of the concepts.
Given below are topics that are covered in the NCERT textbook. Learn these concepts through class 11 hydrocarbons notes available on our website.
9.1 Classification
9.2 Alkanes
9.2.1 Nomenclature and Isomerism
9.2.2 Preparation
9.2.3 Properties
9.2.4 Conformations
9.3 Alkenes
9.3.1 Structure of Double Bond
9.3.2 Nomenclature
9.3.3 Isomerism
9.3.4 Preparation
9.3.5 Properties
9.4 Alkynes
9.4.1 Nomenclature and Isomerism
9.4.2 Structure of Triple Bond
9.4.3 Preparation
9.4.4 Properties
9.5 Aromatic Hydrocarbon
9.5.1 Nomenclature and Isomerism
9.5.2 Structure of Benzene
9.5.3 Aromaticity
9.5.4 Preparation of Benzene
9.5.5 Properties
9.5.6 Directive influence of a functional group in monosubstituted benzene
9.6 Carcinogenicity and Toxicity
Given below is a comparison table highlighting what to study beyond the NCERT for JEE. For detailed learning refer the NCERT solutions for class 11 chemistry chapter 9 hydrocarbons.
Below are the chapter-wise solutions to help you excel in exams.
The hyperlinks of the NCERT solution of class 11 are given below.
Students can refer to the links given below for the NCERT books and Syllabus.
The addition of H₂O in the presence of HgSO₄ and H₂SO₄ to form enols, which quickly tautomerize to aldehydes or ketones. Ethyne gives acetaldehyde while other alkynes give ketones.
Markovnikov’s Rule states that in the addition of HX to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogen atoms.
Aromaticity is a property of cyclic, planar molecules with a conjugated pi electron system that gives them enhanced stability compared to their acyclic counterparts. This special stability is due to the delocalization of pi electrons in a closed ring.
Hückel's Rule (4n+2 Rule) is used to predict if a monocyclic compound is aromatic. A compound is aromatic if it meets the following criteria:
Isomerism in hydrocarbons refers to compounds that have the same molecular formula but different structural arrangements. It is significant because it leads to variations in physical and chemical properties, influencing reactivity and usability in different applications. You can learn more topic like this from our NCERT solutions for class 11 chemistry chapter 9 PDF.
Hydrocarbons can undergo a variety of reactions, including combustion, substitution, and addition reactions.
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Exam Date:22 July,2025 - 28 July,2025
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