Question 13.1: How do you account for the formation of ethane during chlorination of methane?

NCERT Solutions-Hydrocarbons Organic Chemistry for Class 11

by

NCERT Solutions-Hydrocarbons Organic Chemistry for Class 11

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Answer 13.1:

The methane chlorination process works through a free radical chain mechanism.

Step 1: Initiation:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

The result begins with the hemolytic cleavage of Cl – Cl bond as: 

Step 2: Propagation:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

In this propagation step, chlorine-free radicals attack methane molecules, and the C-H bond collapses to form methyl radicals as:

When these methyl radicals react with other chlorine-free radicals, methyl chloride is formed next to the liberation of a chlorine-free radical.

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Step 3: Termination:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Ethane formation is the result of the termination of the chain reactions resulting from the ingestion of reactants as:

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

Question 13.2:

Write the IUPAC names of the following compounds :

(i) CH3 CH = C (CH3)2

(ii) CH2 = CH – C  C – CH3

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(iii)

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(iv)

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(v)

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(vi)

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(vii)

Answer 13.2:

(i)

2-Methylbut-2-ene is the required IUPAC name

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(ii)

Pen-1-ene-3-yne is the required IUPAC name

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(iii) 1, 3-Butadiene or Buta-1,3-diene is the required IUPAC name

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(iv) 4-Phenyl but-1-ene is the required IUPAC name

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(v)

2-Methyl phenol is the required IUPAC name

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(vi)

5-(2-Methylpropyl)-decane is the required IUPAC name

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(vii)

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

4-Ethyldeca-1, 5, 8-triene is the required IUPAC name

Question 13.3:

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 :

(a) C4H8 (one double bond)

(b) C5H8 (one triple bond)

Answer 13.3:

(a) The resulting structural isomers with one double bond are probable for C4H8:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

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 subsequent structural isomers are probable for C5H8 with one triple bond:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

The IUPAC name of

Compound (I) is Pent-1-yne,

Compound (II) is Pent-2-yne, and

Compound (III) is 3-Methylbut-1-ene.

Question 13.4:

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

Answer 13.4:

(i) Ozonolysis of Pent-2-ene is shown as:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

The following Product IUPAC names are:

Product (I) ethanal, and

Product  (II) propanal.

(ii) Ozonolysis of 3, 4-Dimethylhept-3-ene is shown as:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

The product names subsequent to IUPAC are:

Product (I) butan-2-one, and

Product (II) Pentan-2-one

(iii) Ozonolysis of 2-Ethylbut-1-ene is shown as:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

The subsequent IUPAC names of the products are:

Product (I) pentan-3-one, and

Product (II) methanal

(iv) Ozonolysis of 1 – Phenylbut-1-ene is shown as

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

The subsequent IUPAC names of the products are :

Product (I) benzaldehyde, and

Product (II) propanal

Question 13.5:

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

Answer 13.5:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

In the course of ozonolysis, an ozonide is produced as an intermediate that has a cyclic structure; it undergoes cleavage to provide the final products. Ethanal and pentan-3-one are obtained from the ozonide intermediates. So, the possible ozonide structure will be:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

This ozonide is produced by adding ozone to ‘ A ‘ as a transitional product that forms final products undergoing cleavage. The preferred ‘ A ‘ arrangement may be obtained by removing ozone from the ozonide. Therefore ‘ A ”s structural formula is:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

As an outcome, ‘A’ is 3-Ethylpent-2-ene is the IUPAC name of alkene.

Question 13.6:

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 13.6:

From the information given, it gives two moles of an aldehyde of molar mass 44 u when ‘ A ‘ undergoes ozonolysis. The formation of an aldehyde’s two moles suggests that the presence of similar structural units on both sides of the C-C ÿ bond is present. Consequently, the ‘ A ‘ structure can be represented as shown below:

We know XC = CX

Because there are eight C – H σ bonds, 8 hydrogen atoms are present in ‘ A. ‘ There are also three C – C bonds, so the ‘ A ‘ structure contains four carbon atoms.

The combination of all these inferences can represent the structure of ‘ A ‘ as shown below:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

‘A’ has 3 C–C bonds, 8 C–H σ bonds, and one C–C π bond.

Therefore, the IUPAC name of ‘A’ is But-2-ene.

After Ozonolysis of ‘A’ takes place,

The end product is ethanol with molecular weight = 44 u

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Question 13.7:

Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?

Answer 13.7:

From the given information, the two ozonolysis products of an alkene are pentan-3-one and propanal.

Suppose the alkene given is A. The reverse of the ozonolysis reaction is what we get,

Question 13.1: How do you account for the formation of ethane during chlorination of methane?
Question 13.1: How do you account for the formation of ethane during chlorination of methane?

The products obtained by ozonide cleavage, let it be ‘ X. ‘ Since ‘ X ‘ contains both products in cyclic form, the possible ozonide structure may be represented as shown below:

Because ‘ X ‘ is an additional component of the ozone alkene ‘ A. ‘ Therefore, the potential structure of the given ‘ A ‘ alkene will be:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Question 13.8:

Write chemical equations for the combustion reaction of the following hydrocarbons:

(i) Butane, (ii) Pentene, (iii) Hexyne,  (iv) Toluene

Answer 13.8:

Combustion reactions may be defined as an oxygen or oxygen reaction of a compound.

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Question 13.9:

Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?

Answer 13.9:

Hex-2-ene is represented as shown below:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Geometrical isomers of hex-2-ene are as follows:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

A cis compound’s dipole moment is equal to the sum of the C – CH3 bond’s dipole moments, and the C – CH2CH3 bonds that both act in the same direction.

On the other hand, a trans-compound dipole moment is the result of the dipole moments of C – CH3 bonds and C – CH2CH3 bonds both acting in opposite directions

As cis-isomer is more polar than trans-isomer. Thus, the higher the polarity, the greater the intermolecular dipole-dipole interaction and the need for more heat to break the bonds. The boiling will be higher, therefore.

Therefore, cis-isomer of a compound will have a higher boiling point than trans-isomer of that compound.

Question 13.10:

Why is benzene extraordinarily stable though it contains three double bonds?

Answer 13.10:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Benzene is a hybrid of the resonating structures, and it is shown as:

All six carbon atoms in benzene are hybridized to sp2.In benzene, each carbon atom’s two sp2 hybrid orbital overlaps with its adjacent carbon atoms ‘ sp2 hybrid orbital, forming a six-sigma bond in the hexagonal plane. The remaining hybrid orbital sp2 on each carbon atom overlaps with the hydrogen atom’s s-orbital to form six sigma C – H bonds. Now, the rest are unhybridized p-orbital of carbon atoms that will have the possibility of forming three C-C π bonds by the lateral overlap of

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

These six π-bonds are delocalized and can move about the six-carbon nuclei freely. Therefore, due to the delocalization of these π-bonds, benzene is stabilized even after the presence of three C-C π -bonds.

Question 13.11:

What are the necessary conditions for any system to be aromatic?

Answer 13.11:

The essential conditions for any aromatic system are as follows:

(i) Firstly, the compound arrangement or structure should be planar.

(ii) The n-electrons are completely delocalized in the ring of a compound

(iii) The sum of n-electrons in the ring of a compound must be equal to (4n + 2)π,

Where n = 0, 1, 2, ….. and so on. This rule is called Huckel’s rule.

Question 13.12: Explain why the following systems are not aromatic?

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(i)

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(ii)

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(iii)

Answer 13.12:

(i)

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Due to the presence of a sp3-hybridized carbon, the system is not planar. It does contain six n-electrons but the system is not fully conjugated since all the six n-electrons do not form a single cyclic electron cloud that surrounds all the atoms of the ring. Therefore, it is not an aromatic compound.

(ii)

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Due to the presence of sp3– hybridized carbon, the system is not planar. Further, it contains only four n-electrons, therefore, the system is not aromatic because it does not contain planar cyclic cloud having (4n+2) n-electrons.

(iii)

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Cyclo-octatetraene is not planar but is tub-shaped. It is, therefore, a non-planar system having 8 n-electrons. Therefore, the molecule is not aromatic as it does not contain a planar cyclic cloud having (4n + 2) n-electrons.

Question 13.13:

How will you convert benzene into:

(i) p – nitrobromobenzene

(ii) m-nitrochlorobenzene

(iii) p -nitrotoluene

(iv) acetophenone

Answer 13.13:

(i) Benzene converted to p – nitrobromobenzene

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(ii) Benzene converted to m-nitrochlorobenzene

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(iii) Benzene converted to p –nitrotoluene

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(iv) Benzene converted to acetophenone

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Question 13.14:

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 13.14:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Primary carbon atoms (1o): Carbon atoms bonded to a single atom of carbon are called primary atoms of carbon. The arrangement given has corresponding five 1o carbon atoms and fifteen H atoms.

Secondary carbon atoms (2o): The secondary carbon atom is called a carbon atom bonded to a double carbon atom. The arrangement given has two corresponding 2o carbon atoms and four H atoms.

Tertiary carbon atoms (3o): The term tertiary carbon atom is used to refer to carbon atoms bonded to three carbon atoms. The arrangement given has one corresponding 3 degree carbon atom and a single H atom.

Question 13.15:

What effect does branching of an alkane chain has on its boiling point?

Answer 13.15:

Alkanes encounter Van-der Waals forces between molecules. The higher the alkane’s power, the greater is the boiling point.

As the molecule branching increases, the surface area decreases, which leads to a small contact area. As a result, the force of the Van-der Waals (or intermolecular force) decreases too. Those forces can be overcome very easily at a relatively lower temperature. Thus, the boiling point of an alkane chain decreases as branching increases.

Question 13.16:

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 13.16:

The addition of HBr to propene is an example of an electrophilic substitution reaction.

Being an acid, the hydrogen bromide provides an electrophile, H+. This electrophile attacks the propene double bond to form carbocations of 1 ° and 2 °, as shown below:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Secondary carbocations are stable in comparison with primary carbocations. The secondary carbocations therefore predominate, as they form at a faster rate than primary carbocations. Thus, Br– attacks the primary carbocation to form 2-bromopropane as the main product in the next step.

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

This reaction follows the rule of Markovnikov, in which the negative part of the addendum is attached to the carbon atom with fewer hydrogen atoms than other carbon atoms present in the compound.

While, 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 shown below:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Secondary free radicals are stable in comparison with primary radicals. The secondary radical therefore predominates, for it forms at a faster rate than the primary radical. Therefore 1 – bromopropane is obtained as the main product.

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

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

Question 13.17:

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

Answer 13.17:

o-xylene has two resonance structures, which are as follows:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

All three products are obtained from two Kekule structures of o-xylene, i.e., methyl glyoxal, 1, 2-demethylglyoxal, and glyoxal. As all three products can not be obtained from either of the two structures, this proves that o-xylene is a resonance hybrid of two Kekule (I and II) structures.

Question 13.18:

Arrange benzene, n-hexane and ethyne in increasing order of their acidic behavior. Also, give a reason for this behavior.

Answer 13.18:

Acidic character of a species is defined on the basis of the ease with which it can lose its H– atoms.

The hybridization state of carbon in the given compound is:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

As the s – character decreases, carbon electronegativity decreases and C – H bond pair electrons lie away from the carbon atom. As a result, H– atom partially positive charge increases, and H+ ions are set free.

The s–character decreases in the order:

sp > sp2 > sp3

Hence, the increasing order of acidic behavior is Hexane < Benzene < Ethyne.

Question 13.19:

Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?

Answer 13.19:

Benzene is a planar molecule with electrons delocalized under and above the ring plane. Hence, it is a material rich in electrons. As a consequence, electron-deficient species, i.e., electrophiles, are extremely attractive.

Benzene, therefore, very easily undergoes electrophilic substitution reactions. Nucleophiles, on the other hand, are also species that are rich in electrons. Therefore, benzene is repelled as compared to electrophiles. Thus, benzene suffers from difficulty with nucleophilic substitutions.

Question 13.20:

How would you convert the following compounds into benzene?

(i) Ethyne     (ii) Ethene   (iii) Hexane

Answer 13.20:

(i)Benzene from Ethyne:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(ii)Benzene from Ethene:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

(iii) Hexane to Benzene

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Question 13.21:

Write structures of all the alkenes which on hydrogenation give 2-methylbutane.

Answer 13.21:

The basic structure of 2-methylbutane is shown below:

On the basis of the above structure, various alkenes that will give 2-methylbutane on hydrogenation are shown below:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Question 13.22:

Arrange the following set of compounds in order of their increasing relative reactivity with an electrophile, E+

(a) p-nitrochlorobenzene, Chlorobenzene, 2,4-dinitrochlorobenzene,

(b) p – H3C – C6H4 – NO2, Toluene, p-O2N – C6H4 – NO2.

Answer 13.22:

Electrophiles are reagents that participate in a reaction by accepting a pair of electrons to bind to nucleophiles.

The higher the density of electrons on a benzene ring, the more reactive the compound is to an electrophile, E+ (Electrophilic reaction).

(a) The electron density of the aromatic ring decreases due to the presence of an electron-withdrawing group (i.e., NO2 – and Cl –) which deactivates the aromatic ring.

Since Cl – group is less electron-withdrawing (due to the inductive effect) than NO2 – group (due to the resonance effect), the increasing order of reactivity is as follows:

2, 4 – dinitrochlorobenzene < p – nitrochlorobenzene < Chlorobenzene

(b) While NO2– group is electron-withdrawing, CH3– is an electron-donating group.

Toluene, therefore, has the maximum density of electrons and is most easily attacked by E+. Since NO2– is an electron-removing group. Therefore, when the number of NO2 substitutes is higher, the order is the following.:

p-O2N – C6H4 – NO2 < p – H3C – C6H4 – NO2  < Toluene.

Question 13.23:

Out of benzene, m–dinitrobenzene and toluene, state the increasing order of nitration. Justify your answer?

Answer 13.23:

The ease of nitration depends on the presence of electron density on the compound to form nitrates. Nitration reactions are examples of electrophilic substitution reactions where a nitronium ion (NO2–) attacks an electron-rich species.

Now NO2– is electron-withdrawing and CH3– group is electron-donating. Since, m– Dinitrobenzene will have the least electron density. Hence, it will undergo nitration with difficulty. Therefore, toluene will have the maximum electron density among the three compounds, followed by benzene. Hence, the increasing order of nitration is as follows:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Question 13.24:

List the names of some Lewis acid which can be used during ethylation of benzene in a Friedel-Craft alkylation reaction.

Answer 13.24:

The reaction of benzene to the presence of Lewis acids (AlCl3) with an acyl halide or acid anhydride yields acyl benzene (or benzene ring). A Friedel-Craft alkylation reaction is called such a reaction. The reaction occurs in the presence of a Lewis acid.

In the Friedel-Craft alkylation reaction, any Lewis acid, such as anhydrous AlCl3, FeCl3, SnCl4, BF3 etc., may be used during the ethylation of benzene.

Question 13.25:

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

Answer 13.25

For the synthesis of symmetrical alkanes (i.e. alkanes with an even number of carbon atoms), the Wurtz reaction is limited. Two similar alkyl halides are taken as reactants in the reaction, then an alkane is formed, which contains double the number of carbon atoms.

Example:

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

Wurtz reaction can not be used to produce unsymmetric alkanes (i.e. alkanes with an odd number of carbon atoms), because if two dissimilar alkyl halides are used as reactants, a mixture of alkanes is obtained as the items. Since the reaction involves free radical species, there is also a side reaction to creating an alkene.

Question 13.1: How do you account for the formation of ethane during chlorination of methane?

The boiling points of alkanes (obtained in the mixture) are very close. Hence, it becomes difficult to separate them.

The important topics covered in NCERT Solutions for Class 11 Chemistry Chapter 13 are:

  • Classification of hydrocarbons
  • Nomenclature, preparation, properties and isomerism of:
  1. Alkanes
  2. Alkenes
  3. Alkynes
  4. Aromatic hydrocarbons
  • Carcinogenicity and toxicity

The topics given in the Class 11 CBSE Syllabus are the basics of the topics to be taught in Class 12. Also, students are advised to solve and go through this NCERT Solutions for Class 11 Chapter 13 to determine their strengths and weaknesses and plan their studies accordingly. Along with NCERT questions, students should try to solve the previous years’ questions as well as the CBSE Sample Papers to get acquainted with the latest exam pattern and marking scheme.

Frequently Asked Questions on NCERT Solutions for Class 11 Chemistry Chapter 13

Q1

Why should we follow NCERT Solutions for Class 11 Chemistry Chapter 13?

NCERT Solutions for Class 11 Chemistry Chapter 13 is the best reference material that offers complete and quality information about different Math concepts. The questions that have been given in the solutions have been solved in an easy-to-remember format, which further helps students to understand and remember the answers clearly. So, it’s clear that NCERT Textbooks for Class 11 are essential reference books to score high marks in the examinations. To score good marks, practising these solutions for Class 11 Chemistry can help to a great extent.

Q2

How do ANAND CLASSES (A School Of Competitions) NCERT Solutions for Class 11 Chemistry Chapter 13 help the students in preparing for the exams?

Our solution module uses various examples and diagrams to explain the questions wherever necessary. For CBSE board students aiming at securing an excellent score, solving NCERT Solutions for Class 11 is a must. These NCERT Solutions for Class 11 Chemistry Chapter 13 help the students in gaining a better knowledge of the topics covered. Solving the questions from each exercise will ensure that the students score high marks in the exams.

Q3

What are the topics covered under NCERT Solutions for Class 11 Chemistry Chapter 13?

NCERT Solutions for Class 11 Chemistry Chapter 13 includes the following topics:
1. Classification
2. Alkanes
Nomenclature and Isomerism
Preparation
Properties
Conformations
3. Alkenes
Structure of Double Bond
Nomenclature
Isomerism
Preparation
Properties
4. Alkynes
Nomenclature and Isomerism
Structure of Triple Bond
Preparation
Properties
5. Aromatic Hydrocarbon
Nomenclature and Isomerism
Structure of Benzene
Aromaticity
Preparation of Benzene
Properties
Directive Influence of a Functional Group in Monosubstituted Benzene
6. Carcinogenicity and Toxicity

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.

...

CBSE Class 11 Chemistry Syllabus

CBSE Class 11 Chemistry Syllabus is a vast which needs a clear understanding of the concepts and topics. Knowing CBSE Class 11 Chemistry syllabus helps students to understand the course structure of Chemistry.

Unit-wise CBSE Class 11 Syllabus for Chemistry

Below is a list of detailed information on each unit for Class 11 Students.

UNIT I – Some Basic Concepts of Chemistry

General Introduction: Importance and scope of Chemistry.

Nature of matter, laws of chemical combination, Dalton’s atomic theory: concept of elements,
atoms and molecules.

Atomic and molecular masses, mole concept and molar mass, percentage composition, empirical and molecular formula, chemical reactions, stoichiometry and calculations based on stoichiometry.

UNIT II – Structure of Atom

Discovery of Electron, Proton and Neutron, atomic number, isotopes and isobars. Thomson’s model and its limitations. Rutherford’s model and its limitations, Bohr’s model and its limitations, concept of shells and subshells, dual nature of matter and light, de Broglie’s relationship, Heisenberg uncertainty principle, concept of orbitals, quantum numbers, shapes of s, p and d orbitals, rules for filling electrons in orbitals – Aufbau principle, Pauli’s exclusion principle and Hund’s rule, electronic configuration of atoms, stability of half-filled and completely filled orbitals.

UNIT III – Classification of Elements and Periodicity in Properties

Significance of classification, brief history of the development of periodic table, modern periodic law and the present form of periodic table, periodic trends in properties of elements -atomic radii, ionic radii, inert gas radii, Ionization enthalpy, electron gain enthalpy, electronegativity, valency. Nomenclature of elements with atomic number greater than 100.

UNIT IV – Chemical Bonding and Molecular Structure

Valence electrons, ionic bond, covalent bond, bond parameters, Lewis structure, polar character of covalent bond, covalent character of ionic bond, valence bond theory, resonance, geometry of covalent molecules, VSEPR theory, concept of hybridization, involving s, p and d orbitals and shapes of some simple molecules, molecular orbital theory of homonuclear diatomic molecules(qualitative idea only), Hydrogen bond.

UNIT V – Chemical Thermodynamics

Concepts of System and types of systems, surroundings, work, heat, energy, extensive and intensive properties, state functions. First law of thermodynamics – internal energy and enthalpy, measurement of U and H, Hess’s law of constant heat summation, enthalpy of bond dissociation, combustion, formation, atomization, sublimation, phase transition, ionization, solution and dilution. Second law of Thermodynamics (brief introduction)
Introduction of entropy as a state function, Gibb’s energy change for spontaneous and nonspontaneous processes.
Third law of thermodynamics (brief introduction).

UNIT VI – Equilibrium

Equilibrium in physical and chemical processes, dynamic nature of equilibrium, law of mass action, equilibrium constant, factors affecting equilibrium – Le Chatelier’s principle, ionic equilibrium- ionization of acids and bases, strong and weak electrolytes, degree of ionization,
ionization of poly basic acids, acid strength, concept of pH, hydrolysis of salts (elementary idea), buffer solution, Henderson Equation, solubility product, common ion effect (with illustrative examples).

UNIT VII – Redox Reactions

Concept of oxidation and reduction, redox reactions, oxidation number, balancing redox reactions, in terms of loss and gain of electrons and change in oxidation number, applications of redox reactions.

UNIT VIII – Organic Chemistry: Some basic Principles and Techniques

General introduction, classification and IUPAC nomenclature of organic compounds. Electronic displacements in a covalent bond: inductive effect, electromeric effect, resonance and hyper conjugation. Homolytic and heterolytic fission of a covalent bond: free radicals, carbocations, carbanions, electrophiles and nucleophiles, types of organic reactions.

UNIT IX – Hydrocarbons

Classification of Hydrocarbons
Aliphatic Hydrocarbons:
Alkanes – Nomenclature, isomerism, conformation (ethane only), physical properties, chemical reactions.
Alkenes – Nomenclature, structure of double bond (ethene), geometrical isomerism, physical properties, methods of preparation, chemical reactions: addition of hydrogen, halogen, water, hydrogen halides (Markovnikov’s addition and peroxide effect), ozonolysis, oxidation, mechanism of electrophilic addition.
Alkynes – Nomenclature, structure of triple bond (ethyne), physical properties, methods of preparation, chemical reactions: acidic character of alkynes, addition reaction of – hydrogen, halogens, hydrogen halides and water.

Aromatic Hydrocarbons:

Introduction, IUPAC nomenclature, benzene: resonance, aromaticity, chemical properties: mechanism of electrophilic substitution. Nitration, sulphonation, halogenation, Friedel Craft’s alkylation and acylation, directive influence of functional group in monosubstituted benzene. Carcinogenicity and toxicity.

To know the CBSE Syllabus for all the classes from 1 to 12, visit the Syllabus page of CBSE. Meanwhile, to get the Practical Syllabus of Class 11 Chemistry, read on to find out more about the syllabus and related information in this page.

CBSE Class 11 Chemistry Practical Syllabus with Marking Scheme

In Chemistry subject, practical also plays a vital role in improving their academic scores in the subject. The overall weightage of Chemistry practical mentioned in the CBSE Class 11 Chemistry syllabus is 30 marks. So, students must try their best to score well in practicals along with theory. It will help in increasing their overall academic score.

CBSE Class 11 Chemistry Practical Syllabus

The experiments will be conducted under the supervision of subject teacher. CBSE Chemistry Practicals is for 30 marks. This contribute to the overall practical marks for the subject.

The table below consists of evaluation scheme of practical exams.

Evaluation SchemeMarks
Volumetric Analysis08
Salt Analysis08
Content Based Experiment06
Project Work04
Class record and viva04
Total30

CBSE Syllabus for Class 11 Chemistry Practical

Micro-chemical methods are available for several of the practical experiments. Wherever possible such techniques should be used.

A. Basic Laboratory Techniques
1. Cutting glass tube and glass rod
2. Bending a glass tube
3. Drawing out a glass jet
4. Boring a cork

B. Characterization and Purification of Chemical Substances
1. Determination of melting point of an organic compound.
2. Determination of boiling point of an organic compound.
3. Crystallization of impure sample of any one of the following: Alum, Copper Sulphate, Benzoic Acid.

C. Experiments based on pH

1. Any one of the following experiments:

  • Determination of pH of some solutions obtained from fruit juices, solution of known and varied concentrations of acids, bases and salts using pH paper or universal indicator.
  • Comparing the pH of solutions of strong and weak acids of same concentration.
  • Study the pH change in the titration of a strong base using universal indicator.

2. Study the pH change by common-ion in case of weak acids and weak bases.

D. Chemical Equilibrium
One of the following experiments:

1. Study the shift in equilibrium between ferric ions and thiocyanate ions by increasing/decreasing the concentration of either of the ions.
2. Study the shift in equilibrium between [Co(H2O)6] 2+ and chloride ions by changing the concentration of either of the ions.

E. Quantitative Estimation
i. Using a mechanical balance/electronic balance.
ii. Preparation of standard solution of Oxalic acid.
iii. Determination of strength of a given solution of Sodium hydroxide by titrating it against standard solution of Oxalic acid.
iv. Preparation of standard solution of Sodium carbonate.
v. Determination of strength of a given solution of hydrochloric acid by titrating it against standard Sodium Carbonatesolution.

F. Qualitative Analysis
1) Determination of one anion and one cation in a given salt
Cations‐ Pb2+, Cu2+, As3+, Al3+, Fe3+, Mn2+, Ni2+, Zn2+, Co2+, Ca2+, Sr2+, Ba2+, Mg2+, NH4 +
Anions – (CO3)2‐ , S2‐, NO2‐ , SO32‐, SO2‐ , NO ‐ , Cl‐ , Br‐, I‐, PO43‐ , C2O2‐ ,CH3COO‐
(Note: Insoluble salts excluded)

2) Detection of ‐ Nitrogen, Sulphur, Chlorine in organic compounds.

G) PROJECTS
Scientific investigations involving laboratory testing and collecting information from other sources.

A few suggested projects are as follows:

  • Checking the bacterial contamination in drinking water by testing sulphide ion
  • Study of the methods of purification of water.
  • Testing the hardness, presence of Iron, Fluoride, Chloride, etc., depending upon the regional
    variation in drinking water and study of causes of presence of these ions above permissible
    limit (if any).
  • Investigation of the foaming capacity of different washing soaps and the effect of addition of
    Sodium carbonate on it.
  • Study the acidity of different samples of tea leaves.
  • Determination of the rate of evaporation of different liquids Study the effect of acids and
    bases on the tensile strength of fibres.
  • Study of acidity of fruit and vegetable juices.

Note: Any other investigatory project, which involves about 10 periods of work, can be chosen with the approval of the teacher.

Practical Examination for Visually Impaired Students of Class 11

Below is a list of practicals for the visually impaired students.

A. List of apparatus for identification for assessment in practicals (All experiments)
Beaker, tripod stand, wire gauze, glass rod, funnel, filter paper, Bunsen burner, test tube, test tube stand,
dropper, test tube holder, ignition tube, china dish, tongs, standard flask, pipette, burette, conical flask, clamp
stand, dropper, wash bottle
• Odour detection in qualitative analysis
• Procedure/Setup of the apparatus

B. List of Experiments A. Characterization and Purification of Chemical Substances
1. Crystallization of an impure sample of any one of the following: copper sulphate, benzoic acid
B. Experiments based on pH
1. Determination of pH of some solutions obtained from fruit juices, solutions of known and varied
concentrations of acids, bases and salts using pH paper
2. Comparing the pH of solutions of strong and weak acids of same concentration.

C. Chemical Equilibrium
1. Study the shift in equilibrium between ferric ions and thiocyanate ions by increasing/decreasing
the concentration of eitherions.
2. Study the shift in equilibrium between [Co(H2O)6]2+ and chloride ions by changing the
concentration of either of the ions.

D. Quantitative estimation
1. Preparation of standard solution of oxalic acid.
2. Determination of molarity of a given solution of sodium hydroxide by titrating it against standard
solution of oxalic acid.

E. Qualitative Analysis
1. Determination of one anion and one cation in a given salt
2. Cations – NH+4
Anions – (CO3)2-, S2-, (SO3)2-, Cl-, CH3COO-
(Note: insoluble salts excluded)
3. Detection of Nitrogen in the given organic compound.
4. Detection of Halogen in the given organic compound.

Note: The above practicals may be carried out in an experiential manner rather than recording observations.

We hope students must have found this information on CBSE Syllabus useful for their studying Chemistry. Learn Maths & Science in interactive and fun loving ways with ANAND CLASSES (A School Of Competitions) App/Tablet.

Frequently Asked Questions on CBSE Class 11 Chemistry Syllabus

Q1

How many units are in the CBSE Class 11 Chemistry Syllabus?

There are 9 units in the CBSE Class 11 Chemistry Syllabus. Students can access various study materials for the chapters mentioned in this article for free at ANAND CLASSES (A School Of Competitions).

Q2

What is the total marks for practicals examination as per the CBSE Class 11 Chemistry Syllabus?

The total marks for the practicals as per the CBSE Class 11 Chemistry Syllabus is 30. It includes volumetric analysis, content-based experiment, salt analysis, class record, project work and viva.

Q3

Which chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry?

The organic chemistry chapter carries more weightage as per the CBSE Syllabus for Class 11 Chemistry.

© 2025 Anand Classes • Published by Anand Technical Publishers
Privacy Policy Terms Contact

Sitemap