Course Content
CHAPTER 3: CLASSIFICATION OF ELEMENTS
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
0/6
CHAPTER 7: EQUILIBRIUM
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
0/7
CHAPTER 10: S-BLOCK ELEMENTS
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
0/7
CHAPTER 12: CHARACTERIZATION OF ORGANIC COMPOUND
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
0/13
Class 11th Chemistry Online Class For 100% Result
About Lesson

Alkanes can be classified as:

  • Primary alkanes(10)
  • Secondary alkanes (20)
  • Tertiary alkanes (30)
  1. Primary alkane: Alkane in which a Carbon atom attached to one or none alkyl group  is   =  10

       CH3-CH3  (ethane)

              10

 2. Secondary Alkane: In which a carbon is linked to two alkyl groups is =20

     Example in propane:- CH3-CH2-CH3 

                                        20

       3.Tertiary Alkane: Alkane in which a carbon atom is linked to three alkyl groups is =30

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

Preparation of alkanes

  1. Naturally, they are synthesised by decomposition of plants and waste matter.
  2. In laboratory: They are prepared from :
  • From unsaturated hydrocarbons
  • From alkyl halides
  • By reduction of alkyl halides
  • By use of Grignard reagent
  • From carboxylic acids

 (a) From unsaturated Hydrocarbons:-The method involved is by hydrogenation that is addition of H2. It is also called as reduction reaction.

The general reaction involved:

 A Special reaction occurs where we use Raney Ni which is actually activated nickel   (alloy of 50 % Ni, 50 % Al). But the reaction is assigned a special name that is Sabatier Sanderson Reduction. The reaction is given below:

 Similarly this reaction can be carried out for alkynes and the same name is assigned:  

Hydrocarbons Notes and Solution Class 11 Chemistry

 

(b)From Alkyl halides (RX)

These alkyl halides are formed by replacing Hydrogen by halogen (X).

Example: RCl or CH3Cl etc

       The methods that can be used for preparing alkanes by using alkyl halides are:

  • Wurtz reaction
  • Reduction of alkyl halide
  • By use of Grignard reagent

Wurtz reaction: in this two molecules of alkyl halide react with sodium in presence of dry ether as shown below:

Hydrocarbons Notes and Solution Class 11 Chemistry

Please note the dry ether is used so as to keep sodium dry so that it does not catch fire in    moist   conditions.

             Another example:

Limitation of this reaction: is that we need to take same alkyl halides otherwise we get mixtures of alkanes and due to almost same boiling points etc, they are difficult to separate. Another limitation of this is that by this method we can prepare alkanes only with even number of Carbon atoms.

(c)Reduction of alkyl halides: – It can be done by using various Reducing agents like:

  • With Zn and HCl

  CH3CH2Cl + H2  à CH3CH3    +        HCl

(Ethyl Chloride)      (Ethane)        Hydrogen chloride

  In this the Zn and HCl both react and produce nascent Hydrogen to carry out reduction.                     

                                (Zn/HCl)

CH3CH2X           + 2H2   –>     CH3CH3  + HX          

(Ethyl halide)                        (Ethane)       (Hydrogen Halide)             

  • Pt or  Pd can be used as reducing agent

                                        (Pt/Pd)

CH3CH2Br          +       H2    –>  CH3CH3    + HBr

(Ethyl Bromide)                        (Ethane)    (Hydrogen Bromide)

  • With use of reducing agent HI in presence of Red Phosphorous.

                        (Red P)

CH3CH2X + HI    –>       CH3CH3  +    I2 +            H2

(Ethyl Halide)                (Ethane)  (Iodine)      Hydrogen Gas

The Role of Red P is to remove I2 from reaction so as to get maximum yield of alkane because it is a reversible reaction.

 Red P + I2 –>      PI3

                          Phosphorous Iodide

  • By use of Reducing agent Zn and Cu  couple
Hydrocarbons Notes and Solution Class 11 Chemistry

 With use of very strong Reducing agent like LiAlH4 ,NaBH4

(d) By use of Grignard reagent (RMgX)

Hydrocarbons Notes and Solution Class 11 Chemistry

(e) From Carboxylic acids (RCOOH)

  • By Decarboxylation: It is removal of of CO2 In this first carboxylic acid, is made to react with sodium metal then followed by its reaction with sodium hydroxide to get the desired alkane.
Hydrocarbons Notes and Solution Class 11 Chemistry

Kolbe’s electrolysis Reaction: In this electric current is passed through metal carboxylic salt

Reduction of carboxylic acid: In this carboxylic acid is reduced in presence of hydrogen iodide as shown :

Exercise Files
No Attachment Found
No Attachment Found
Wisdom Academy