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NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons

NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons

Edited By Shivani Poonia | Updated on Jul 17, 2025 09:23 AM IST

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 WallahAakash Unacademy

NEET Scholarship Test Kit (Class 11): Narayana | Physics Wallah Aakash | ALLEN

This Story also Contains
  1. Download PDF of NCERT Solutions for Class 11 Chemistry Chapter 9
  2. NCERT Solutions for Class 11 Chemistry Hydrocarbons- Exercise Questions
  3. Class 11 Chemistry NCERT Chapter 9: Higher Order Thinking Skills (HOTS) Questions
  4. Approach to Solve Questions of Chapter 9
  5. Topics and Subtopics of NCERT Class 11 Chemistry Chapter 9
  6. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  7. NCERT Solutions for Class 11 Chemistry
  8. NCERT Class 11 Solutions
  9. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons
NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons

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.

Also Read

Download PDF of NCERT Solutions for Class 11 Chemistry Chapter 9

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.

Download PDF

NCERT Solutions for Class 11 Chemistry Hydrocarbons- Exercise Questions

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 ClCl bond as
Homolytic clevage

(ii) Propagation-
Chlorine-free radicals attack methane molecules and generate methyl-free radicals as
Free radicals
This methyl radical reacts with other molecules of chlorine ( ClCl ) to form methyl chloride and liberate chlorine-free radicals.
Methyl chloride
Then the chlorine free radical reacts with methyl chloride and this way propagation occurs.
Chlorine free radical

(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.

Etane

Question 9.2(a) Write IUPAC names of the following compounds

CH3CH=C(CH3)2

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.

CH3CH=C(CH3)2
The IUPAC name of the given compound is 2-methylbut-2-ene

Question 9.2(b) Write IUPAC names of the following compounds

CH2=CHCCCH3

Answer:

Both alkene and alkyne are present but the priority will go to the alkene group.

CH2=CHCCCH3
The IUPAC name of the compound is Pent-1-ene-3-yne

Question 9.2(c) Write IUPAC names of the following compounds

1, 3-Butadiene

Answer:

1, 3-Butadiene

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 :

 4-phenylbut-1-ene

Answer:

 4-phenylbut-1-ene

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 :

 2-methylphenol

Answer:

Here there are two functional groups hydroxy and methyl. The priority will go to the hydroxy group.

 2-methylphenol

The IUPAC name of the compound is 2-methylphenol

Question 9.2 (f) Write IUPAC names of the following compounds :

5-(2-methyl propyl)-decane

Answer:

5-(2-methyl propyl)-decane

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 :

4-ethylDeca-1, 4, 8-triene

Answer:

4-ethylDeca-1, 4, 8-triene
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.

Question 9.3 For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of the double or triple bond as indicated :

C4H8 ( one double bond)

Answer:

C4H8 the following isomers are possible with one double bond ;

But-1-eneBut-1-ene
But-2-eneBut-2-ene
2-Methylprop-1-ene2-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.
ozonolysis
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.
Ozonolysis
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.
ozonolysis
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

ozonolysis
Thus, by removing the ozone from ozonide, we can get the parent alkene structure.

Alkene

Question 9.6 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’.

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 = YC=CY
There are eight CH sigma bonds, which means eight hydrogen atoms in the structure A. Also, three CC bonds that indicates the presence of four carbon atoms in A.
Now, by combining all the observations, the structure of the A would be

Alkene
The ozonolysis reaction is shown below-
ozonolysis

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
ozonolysis
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
(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( O2)
CH3(CH2)2CH3(g)+13O2(g)8CO2(g)+10H2O(g)+Heat

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( O2 ). Combustion reaction of the Pentene-
CH3(CH2)3CH3(g)+15O2(g)10CO2(g)+10H2O(g)+Heat

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(O2) that gives carbon dioxide, water molecules and produces some amount of heat.

Structure of hexyne

The reaction

2C6H10+17O212CO2+10H2O+heat

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( O2 ) to give carbon dioxide, water molecules and produces some amount of heat.

Structure of toluene

C6H5(CH3)+9O27CO2+4H2O+heat

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-
 hex-2-ene
Now, the Geometrical, cis and trans isomers -
is 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.
Benzene
Each carbon of benzene is in sp2 hybridization. Two sp2 hybrid orbitals of each carbon overlap with the adjacent carbon atoms' orbital, resulting in six C-C sigma bonds that all are in a hexagonal plane and the remaining sp2 orbital overlaps with the s-orbitals of hydrogen and forms C-H sigma bonds.

In benzene, six C-H sigma bonds present. And the remaining unhybridised p-orbital (which are perpendicular to the plane) formed π-bond by lateral overlapping. The possibility of forming π -bond is six (C1C2,C3C4,C6C1,/C2C3,C4C6,C6C1).
There are 6 π electrons, whichare delocalised and move freely about the six carbon nuclei, and the presence of these delocalised π electrons in benzene makes it more stable.

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

  1. The cyclic compound should be a planner
  2. The complete(continuous) delocalisation of π -electrons in the ring
  3. Follow the Huckel rule- it states that the ring has (4n+2) π -electrons, where n =integer(n = 0, 1, 2, 3....)

Question 9.12 (i) Explain why the following systems are not aromatic?

Ethylpiperidine

Answer:

Ethylpiperidine

Not an aromatic compound because the  pi-electrons in the ring are not in complete conjugation. And it is a non-planner structure.

Question 9.12(ii) Explain why the following systems are not aromatic?

Cyclopentene

Answer:

cyclopentene

There is no complete conjugation of a π electron in the ring as there is sp3 hybridized carbon between them. And also it does not obey huckle rule [(4n+2) π ]electron.

Question 9.12(iii) Explain why the following systems are not aromatic?

cyclooctane

Answer:

cyclooctane

It disobeys the Hückel rule of (4n+2) π electrons. According to this rule, it should have 2, 6, 10 .... number of π electrons but it has 8 π electrons. Also, this structure becomes non-planar.

Question 9.13(i) How will you convert benzene into

p-nitrobromobenzene

Answer:

Bromination of a benzene ring in the presence of anhydrous FeCl3 and Br2 gives bromobenzene. Then, treating bromobenzene with conc. nitric acid in the presence of sulphuric acid gives p-nitrobenzene. Halogens are ortho and para-directing groups.
Bromination

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 (AlCl3) gives m-nitrochlorobenzene. Here, Cl will attach at meta position as NO2 is a meta directing group.
nitrochlorobenzene.

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 HCl. When methylbenzene reacts with conc. nitric acid and sulphuric acid it gives a mixture of para and ortho products of nitrotoluene, which on distillation gives p-nitrotoluene.
Alkylation of benzene

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.
Benzene

Question 9.14 In the alkane H3C- CH2 -C(CH3 )2-CH2 -CH(CH3)2 , identify 1°,2°,3° carbon atoms and give the number of H atoms bonded to each one of these.

Answer:

Pentane
10 carbon are those that are directly bonded with only one carbon atom; the given structure has five 10 carbon atoms and 15 hydrogens attached to it.
20 carbon are those that are connected with only two carbon atoms. In the above structure, there are two 20 carbons present, and four hydrogens attached to them.
30 are those that connect with three carbon atoms. In the above structure, 1 carbon atom is 30 carbon atom and only one hydrogen is attached to it.

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.

Question 9.16 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.

Answer:

Addition of HBr to propene-
In this addition, an electrophile H+ attacks the double bond of the alkene to form 10 and 20 carbocations-
 propene
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)
Propane

Question 9.17 Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?

Answer:

Ortho-xylene has two resonant structures, so,
xylene
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, n-hexane and ethyne in decreasing order of acidic behaviour. Also, give a reason for this behaviour.

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.

Benzene

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  pi-electrons are delocalised above and below the ring. Thus, it is an electron-rich species. In nucleophilic substitution, the attacking species is electron-rich in nature, so it becomes challenging to attack benzene because there will be repulsion. On the other hand, when the attacking agent is an electrophile, which is electron-deficient, they are easily attracted by the benzene.

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.
ethyne

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 NaNH2/liq. NH3. Cyclic polymerisation of ethyne gives us benzene.


ethene

Question 9.20(iii) How would you convert the following compounds into benzene?

Hexane

Answer:

The cyclisation of hexane in the presence of Cr2O3 produces cyclohexane, which on aromatisation gives benzene.
hexane

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;
2-methylbutane
As per the above structure, the following alkene compounds produce 2-methylbutane by hydrogenation are
2-methylbutane 2-methylbutane 2-methylbutane

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.
NO2 is an electron-withdrawing group; it deactivates the benzene ring towards electrophiles by decreasing the electron density of the ring.
Decreasing order of their reactivity with an electrophile(E+)
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, pH3CC6H4NO2,pO2 NC6H4NO2.

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 >pCH3C6H4NHO2>pO2NC6H4NO2

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, NO2+ is used as an electrophile. Here methyl group is electron electron-donating group, and the nitro group is electron electron-withdrawing group. So, the benzene ring attached with the CH3 group has high electron density, and the ring that is attached with the nitro group has the least electron density. Hence, toluene undergoes nitration most easily.

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 (FeCl3), stannic chloride (SnCl4) , BF3 etc. can be used instead of aluminium chloride.

Question 9.25 Why is Wurtz reaction not preferred for the preparation of alkanes containing an odd number of carbon atoms? Illustrate your answer by taking one example.

Answer:

Wurtz reaction-
Bromo methane
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.

Butane
All the products in the mixture have nearly the same boiling point. So, the separation will be difficult.

Class 11 Chemistry NCERT Chapter 9: Higher Order Thinking Skills (HOTS) Questions

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 m-xylene with HNO3,H2SO4 followed by oxidation, 4-nitrobenzene-1, 3-dicarboxylic acid is obtained as the major product.
Statement (II) : CH3 group is o/p-directing whileNO2 group is m-directing group.

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 CH3 group is o/p directing while NO2 group is meta directing.

Hence, the correct answer is option (3).

Question: Given below are two statements :
Statement I : Ozonolysis followed by treatment with Zn,H2O of cis-2-butene gives ethanal.
Statement II : The production obtained by ozonolysis followed by treatment with Zn,H2O of 3, 6-dimethyloct-4-ene has no chiral carbon atom.
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) Br2/Fe;Cl2,Δ; alc. KOH

2) Cl2/Fe; Br2/ anhy. AlCl3; aq. KOH

3) Br2/ anhy. AlCl3;Cl2,Δ; aq. KOH

4) Cl2/ anhy. AlCl3;Br2/Fe; alc. KOH

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).

Approach to Solve Questions of Chapter 9

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.

Topics and Subtopics of NCERT Class 11 Chemistry Chapter 9

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

What Extra Should Students Study Beyond NCERT for JEE/NEET?

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.

NCERT Solutions for Class 11 Chemistry

Below are the chapter-wise solutions to help you excel in exams.

NCERT Class 11 Solutions

The hyperlinks of the NCERT solution of class 11 are given below.

NCERT Books and NCERT Syllabus

Students can refer to the links given below for the NCERT books and Syllabus.

Frequently Asked Questions (FAQs)

1. What is hydration of alkynes

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.

2. What is Markovnikov’s rule in hydrocarbons Class 11?

Markovnikov’s Rule states that in the addition of HX to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogen atoms. 

3. Explain Aromaticity and Hückel's Rule.

 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:

  1. It must be cyclic.
  2. It must be planar.
  3. It must have a continuous ring of p-orbitals (i.e., be fully conjugated).
  4. It must have (4n+2) pi (π) electrons, where 'n' is a non-negative integer (0, 1, 2, 3...). For n=1, it's 6 pi electrons (as in benzene).
4. What is the significance of isomerism in hydrocarbons?

 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.

5. What types of reactions do hydrocarbons typically undergo?

 Hydrocarbons can undergo a variety of reactions, including combustion, substitution, and addition reactions. 

Articles

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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