Course Content
Section Name Topic Name 3 Classification of Elements and Periodicity in Properties 3.1 Why do we Need to Classify Elements ? 3.2 Genesis of Periodic Classification 3.3 Modern Periodic Law and the present form of the Periodic Table 3.4 Nomenclature of Elements with Atomic Numbers > 100 3.5 Electronic Configurations of Elements and the Periodic Table 3.6 Electronic Configurations and Types of Elements: s-, p-, d-, f – Blocks 3.7 Periodic Trends in Properties of Elements
Section Name Topic Name 7 Equilibrium 7.1 Equilibrium in Physical Processes 7.2 Equilibrium in Chemical Processes – Dynamic Equilibrium 7.3 Law of Chemical Equilibrium and Equilibrium Constant 7.4 Homogeneous Equilibria 7.5 Heterogeneous Equilibria 7.6 Applications of Equilibrium Constants 7.7 Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G 7.8 Factors Affecting Equilibria 7.9 Ionic Equilibrium in Solution 7.10 Acids, Bases and Salts 7.11 Ionization of Acids and Bases 7.12 Buffer Solutions 7.13 Solubility Equilibria of Sparingly Soluble Salts
Section Name Topic Name 10 The s-Block Elements 10.1 Group 1 Elements: Alkali Metals 10.2 General Characteristics of the Compounds of the Alkali Metals 10.3 Anomalous Properties of Lithium 10.4 Some Important Compounds of Sodium 10.5 Biological Importance of Sodium and Potassium 10.6 Group 2 Elements : Alkaline Earth Metals 10.7 General Characteristics of Compounds of the Alkaline Earth Metals 10.8 Anomalous Behaviour of Beryllium 10.9 Some Important Compounds of Calcium 10.10 Biological Importance of Magnesium and Calcium
Section Name Topic Name 12 Organic Chemistry – Some Basic Principles and Techniques 12.1 General Introduction 12.2 Tetravalence of Carbon: Shapes of Organic Compounds 12.3 Structural Representations of Organic Compounds 12.4 Classification of Organic Compounds 12.5 Nomenclature of Organic Compounds 12.6 Isomerism 12.7 Fundamental Concepts in Organic Reaction Mechanism 12.8 Methods of Purification of Organic Compounds 12.9 Qualitative Analysis of Organic Compounds 12.10 Quantitative Analysis
Class 11th Chemistry Online Class: Elevate Your CBSE Board Success
About Lesson


  They are the hydrocarbons  with C = C (double bond) with one   bond and other  π bond.

  • Their General formula  Cn H2n
  • They are oil like – called olefins.
  • They are reactive due to π bond and common reaction they undergo in addition Reaction.
  • The lowest member of alkenes is ethene (C2H4).
  • The structure of alkene is:
Hydrocarbons Notes and Solution Class 11 Chemistry
Hydrocarbons Notes and Solution Class 11 Chemistry

Nomenclature:  Primary suffix + ene

            CH2  =  CH2          CH3 – CH =  CH2     CH3  CH = CH – CH3

            Ethene                      Propene                    But-2-ene

             CH2  =  CH  –  CH  =  CH2

            But –  1 , 3 – diene

ISOMERISM: There are two types of isomerism.

  • Structural isomerism
  • Stereoisomerism

The structural isomerism shown by them are :

  • Chain
  • Position

           Chain:       CH3  CH2  CH2  CH = CH2 (pentene)

 CH3 –  C  –  CH  =  CH2 (3-methyl butene)


  • Position: CH3  CH2  –  CH2  –  CH  =  CH2


                      CH3 –  CH2  –  CH  –  CH  –  CH3


Geometrical Isomerism –  The spatial arrangement of atoms is going to be different in it.

            It is caused due to restricted rotation about C = C bond .

             Condition Required to show this geometrical isomerism :

  1. Presence of double
  2. Both atom attached to C should be different.
Hydrocarbons Notes and Solution Class 11 Chemistry
Hydrocarbons Notes and Solution Class 11 Chemistry

      Example: but-2 ene

CH3  CH  =  CH  –  CH3

The cis and trans have some chemical  properties but different physical properties as arrangement of atoms is different. Like cis is more polar than trans.

                                           (Cis) but 2 ene                            Trans but-2 ene

                                           Dipole moment   =  0.35D         Dipole moment is zero

Please note

  • Because of different polarites the Cis have higher values of boiling point than Trans.
  • Trans will have higher melting point whereas Cis will have lower melting point.
  • Trans has symmetrical structure and will fit better in crystal lattice.
  • Stability of trans isomer is more than Cis as Cis itself suffer steric hindrence .

If all 4 groups attached  are different then we have E and Z system of configuration. In this numbering is done according to masses  instead of cis and trans ,we have Z= zussmen form  E= entgagen form.

Hydrocarbons Notes and Solution Class 11 Chemistry
Hydrocarbons Notes and Solution Class 11 Chemistry

Stability of Alkenes

 It is explained on the basis of amount of energy evolved when 1 mole of alkene is hydrogenated

  CH2  =  CH2  + H -> CH3 – CH3                   change in energy(H) =  -ve

 Ethene    hydrogen       ethane

Example: In cis butene and trans butene, the energy difference is 5kJ/mol, as cis has energy 20kJ/mol and trans has energy 115kJ/mol. Therefore, both on hydrogenation forms butane CHCH2CH2CH3.

Trans isomer is more stable as  initially it is at lower energy.So, We can say that trans  is 5KJ more stable than the Cis isomer.

Please keep in mind :“ More alkylated alkenes are more stable”

 Due to hyperconjugation, max conjugated  structures are formed and more  it is stable as shown below :

Preparation  of Alkenes

  1. From alkyl halides: (RX) by the process of  dehydrohalogenation, it is also called  Beta elimination .

In this case it has two beta carbons so, we make use of special rule that is Saytzeff’s rule.

 Saytzeff’s rule: In case of reacting alcoholic  KOH with unsymmtrical halide , that

  alkene is preffered which is maximum alkylated. So , in above  example But – 2 – ene is preferred .

                    If we see ease of  elemination the order is :

                                           I  >  Br  >  Cl  >  F

The better the leaving group more easily the reaction occurs .The reactivity of alkanes towards this reaction is : tertiary > secondary > primary

2.From dihalogen derivatives

Let us take an example of  Vicinal dihalide that is :

 In this renewal of halogen is taking place therefore the process is called dehalogenation .

 3. From alkynes : by the process hydrogenation.The reaction involved is given below: 

  1. From Alcohols :

   It is prepared by Dehyration Reaction of alcohols.

The ease of Reaction is: 30>   20>    10

                    Another example is:

  1. Kolbe’s electrolytic reduction

In this electric current is passed through potassium salt of carboxylic acid and we get alkene as shown below.

 Physical Properties Of alkenes

  1. Existence: Lower members are colourless (Ethene is with sweet smell and other alkenes are odourless).
  • C5 to C18  are   
  • > C18 are     
  1. Melting Point: – Alkenes have higher melting point than alkanes because π e cloud will be more polarized. If we compare melting point of Cis and Trans then, Trans possess high melting point then Cis because Trans have symmetrical structure so, they fit better in lattice.
  2. Boiling Point: They show regular graduation.
  • With increase in Carbon atoms, Vander wall force increases so, boiling point also increases.
  • Boiling point decreases with increase in branching. As surface area reduces vander wall force decreases, therefore boiling point also decrease.
  • Out of cis and trans, trans has low boiling point and cis possess high boiling point due to polarity.
  1. Dipole moment: Cis have certain dipole moment, but Trans have zero dipole moment as both the dipoles are in opposite direction therefore, they cancel each other effect.

In unsymmetrical Trans µ ≠ O it will have certain value due to difference in the type of groups attached.

  1. Solubility: – Alkenes have greater solubility then alkanes.
  • Solubility increase with increase in C atoms.
  • Lower members are sparingly soluble but soluble in organic solvents like Benzene, ether, Carbon tetra chloride etc.
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