Chapter 13 - Production of Metals

In the previous section, we completed a discussion on Electrochemistry. In this chapter, we will see production of metals.
■ If we examine the history of human civilisation, we can see that:
• Gold and copper were the early metals to be discovered by man. They were discovered in the stone age itself. 
• Isn't it amazing that those expensive metals were discovered in very early stages of human civilisation?
• In fact it is not so amazing because, gold, though scarce, is found in pure state in nature. 
• They are found in pure state because, the reactivity of gold is very low. 
    ♦ It does not react with other elements. So it is often found as gold nuggets.
■ Similar is the case of copper. 
• Though reactivity of copper is greater than gold, it has low reactivity when compared to other metals. 
• We have seen the reactivity series in the previous chapter.
■As human civilisation progressed, methods were discovered to combine copper with other metals like tin and zinc. 
• The alloy of copper and zinc is called brass. 
• The alloy of copper and tin is called bronze.
• The use of 'tools made of bronze' became widespread. Thus began the 'bronze age'
■ As the civilisation progressed further, methods were discovered to isolate a much more reactive metal – Iron. 
• Soon the 'bronze age' came to an end and 'iron age' began.
■ The metals which are more reactive than iron (aluminium, potassium, sodium etc.,) could be discovered only much later, with the application of electricity.
• So we see that, the metals were discovered in the order of their reactivity.

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• Today various metals are widely used in our day to day life
• The unique properties pocessed by each metal makes it useful for particular applications. 
For example:
• Copper and aluminium have high electrical conductivity.
    ♦ High electrical conductivity means, less resistance to the flow of elecricity.
    ♦ So the wastage of electric energy in the form of heat energy will be very low
    ♦ So copper and aluminium are used for making electric wires.
• Aluminium has high thermal conductivity
    ♦ High thermal conductivity means, less resistance to the flow of heat.
    ♦ So aluminium is used for making cooking utensils. 
    ♦ Because, heat will be readily transferred to the food to be cooked.
• Some metals are malleable. That is., they can be pressed or hammered into different shapes with out breaking.  For example, a piece of hot iron can be hammered into a thin sheet.
• Some metals are ductile. That is., they can be lengthened into thin wires by the application of a tensile force. Copper, aluminium and steel are examples of metals pocessing ductility. 

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So we find that metals are a very important part of our day to day life. We will now try to find out how they are produced.
We have seen that:
• 'Less reactive metals' are found in pure form. Examples are paltinum and gold 
• 'More reactive metals' are found along with impurities. Examples are copper, aluminium, iron etc.,
■ So we have to learn about the 'more reactive metals' in some detail. We will write it in steps:
1. We know that, in certain localities, there will be an abundance of certain metals.
• Those metals and the impurities will be in a 'combined state'.
• The combination of 'pure metal' and 'impurities' is called mineral. 
2. We have to remove the impurities from the mineral to obtain the pure metal. 
• It may not be easy to remove those impurities. 
• We will have to take the mineral to a lab or an industrial factory.
3. But that is not all. Some times, it will involve very expensive processes to remove those impurities.
Consider an example: 
(i) A certain 'quantity of pure metal' is obtained after removing all impurities from a mineral sample. 
(ii) The total money spent for 'obtaining the mineral sample' and the 'purification process' is Rs. 'x'
(iii) The value of the 'quantity of pure metal' obtained is Rs.'y'
(iv) If y is greater than x, there is profit
(v) If y is less than x, then there is loss
4. We would not want to lose money. 
• So it is important to make the 'selection of the mineral' carefully.
Let us see an example:
    ♦ Bauxite (Al₂O₃.2H₂O) is a mineral containing aluminium 
    ♦ Cryolite (Na₃AlF₆) is a mineral containing aluminium 
    ♦ Clay (Al₂O₃.2SiO₂.2H₂O) is a mineral containing aluminium 
5. Which one of them would we use to obtain aluminium?
• Based on theoretical and experimental results, scientists have given us clear instructions about which one to use.
• The mineral that we would use is bauxite. Because it is easier to obtain aluminium from it.
■ A mineral from which a metal is economically, easily and quickly extracted is called the ore of the metal
• Thus we see that all minerals of a metal cannot be it's ores
6. So, for a mineral should possess the following properties to acquire the status of an ore:
• The mineral should be available in abundance in any particular locality
    ♦ That locality should be easily accessible
• It must be easy to separate the metal from the mineral
• The mineral should have a high metal content. 
    ♦ That is., if we take a sample of the mineral, a greater portion of that sample must be the metal.
■ If the mineral possess the above properties, it will be called an ore

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Following are some of the metals and their ores:
• Metal: Aluminium
    ♦ Ore: Bauxite
    ♦ Chemical formula: Al₂O₃.2H₂O 
• Metal: Iron
    ♦ Ores: Haematite, Magnetite
    ♦ Chemical formula: Fe₂O₃, Fe₃O₄ 
• Metal: Copper
    ♦ Ores: Copper pyrites, Cuprite
    ♦ Chemical formula: CuFeS₂, Cu₂O 
• Metal: Zinc
    ♦ Ores: Zinc blende, Calamine
    ♦ Chemical formula: ZnS, ZnCO₃ 

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Metallurgy is a branch of science and technology which deals with the properties of metals, their production and purification.
There are three important stages in the production of metals. They are:
1. Concentration of ores
2. Extraction of the metal from concentrated ore
3. Refining of the metal  
We will now see each stage in detail:

1. Concentration of ores

• The process of removing the impurities from the ore obtained from the earth's crust is termed concentration of ore. 
• The impurities are called gangue.
The method used for concentration will depend on:
(a)  Nature of the ore
(b) Nature of the impurities
• What ever be the method used for concentration, the first step is to powder the ore. 
    ♦ This is known as pulverisation. 
• The ore obtained from the mine will be in the form of lumps of various sizes. 
• But when pulverisation is done, the ore will be turned into smaller particles of uniform size. 
• This makes it easier to remove the impurities.

Levigation or Hydraulic washing

• When the impurities are lighter and the metal particles are heavier, we use Levigation method.
• Levigation is also known as Hydraulic washing. 
• Consider the fig.13.1 below:

Fig.13.1

We can write the steps:
1. The powdered ore is supplied from the top into a conical tank. 
2. The conical tank is full of water which is supplied from the bottom. 
3. The lighter impurities will float at the top portion while the heavier metal particles will settle down. 
4. The impurities are collected from the top and the metal particles are collected from the bottom. 
• Large quantities of impurities can be removed by this method. We can write a summary:
When the impurities are lighter and the ore particles are heavier, the lighter impurities are removed by washing in a current of water. 
• Examples:
    ♦ If the impurities are oxides, this method can be used
    ♦ Some ores of gold can be concentrated by this method

Froth floatation

• This method is used when impurities are heavier and ore particles are lighter. 
• Usually, sulphide ores are concentrated by this method. 
• Consider fig.13.2 below. We can write the steps:

Fig.13.2

1. A mixture of the powdered ore, water and pine oil is supplied to the bottom portion of a tank
• Pine oil is an oil obtained from pine trees
2. This mixture is agitated by rotating a paddle. Compressed air is also supplied to the bottom of the tank at the same time
• As a result, froth is formed. 
3. The sulphide impurities are wetted by water. They are heavier and so sinks to the bottom.
• The ore particles are wetted by pine oil. They stick to the froth and floats on the top. 
4. The froth can be skimmed off from the top of the tank. 
• From this froth, the ore particles can be easily obtained. 
• Copper pyrites is concentrated by this method.

Magnetic separation

• This method is used when any one (and only one) of the following two conditions are satisfied:
    ♦ The ore is magnetic
    ♦ The impurity is magnetic
• Note that, for this method, only one should be magnetic. If both the ore and impurities are magnetic, the method will not work.
We will now write the steps:
1. In fig.13.3 below, the powdered ore is fed onto a conveyor belt. 
• The roller on the left side is magnetic.

Fig.13.3

2. When the powdered ore falls over, the magnetic component (which may be either the ore or the impurity), will be attracted towards the roller. 
• So it get separated from the other component. 
3. This method is used in the following cases:
(i) Powdered magnetite. It is an iron ore
The iron, which is the magnetic component can be easily separated
(ii) Powdered tin stone. It is an ore of tin
• Tin is non magnetic. 
• But the impurity is tin tungstate. It is magnetic.

Leaching

• In this method, the powdered ore is added to a solvent. The solvent should satisfy both the conditions given below:
    ♦ The ore must dissolve in it
    ♦ The impurity must not dissolve in it
Let us write the steps:
1. The ore dissolves and forms a solution. 
2. The insoluble impurities are filtered off.
3. Now we have the ore in dissolved form. 
• So next step is to obtain the ore from the solution. 
• For this, a suitable chemical reaction is used
4. Bauxite, the ore of aluminium is concentrated by this method

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So we have seen the basic details about 'concentration of ore'. We will now see some solved examples:
Solved example 13.1
(a) In a powdered ore, the ore is of high density and the impurities present is of low density. Which method would you use for concentration of the powdered ore?
(b) In a powdered ore, the ore is magnetic in nature and the impurities present is non magnetic in nature. Which method would you use for concentration of the powdered ore? 
(c) In a powdered ore, the ore is of low density and the impurities present is of high density. Which method would you use for concentration of the powdered ore?
(d) In a powdered ore of aluminium, the aluminium gets dissolved in a solution and the impurities is insoluble in the same solution. Which method would you use for concentration of the powdered ore?
Solution:
(a) Levigation (hydraulic washing)
• Because, the denser ore will settle down in water and the lighter impurities will float above the water
(b) Magnetic separation
• Because, the ore, which is the magnetic component, will get separated from the non magnetic impurities
(c) Froth flotation
• Because, the ore which is lighter, will rise to the top along with the froth. The heavier impurities will sink to the bottom. Thus they will get separated.
(d) Leaching
• Because aluminium will dissolve while the impurities will not. So the impurities can be filtered off. The useful aluminium can be recovered from the solution using a suitable chemical reaction.

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In the next section, we will see the second stage after 'concentration of ore'. 

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Chapter 15.1 - Combustion and Thermal cracking

In the previous section, we completed a discussion on substitution, addition and polymerisation reactions. In this section we will see more such reactions.

Combustion of Hydrocarbons

• Kerosene, Petrol, LPG etc., are hydrocarbons. 
• They are used as fuels because, they produce heat when they burn in the presence of oxygen. 
• Let us see an example:
When methane burns, it combines with oxygen in the air to form CO₂ and H₂O along with heat and light. The equation is:
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g) + Heat

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When hydrocarbons burn, they combine with oxygen in the air to form CO₂ and H₂O along with heat and light. This process is called combustion.

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Two more examples are given below:
Combustion of ethane:
2C₂H₆ (g) + 7O₂ (g) → 4CO₂ (g) + 6H₂O (g) + Heat
Combustion of butane:
2C₄H₁₀ (g) + 13O₂ (g) → 8CO₂ (g) + 10H₂O (g) + Heat

Thermal cracking

• Consider a hydrocarbon molecule having a large number of carbon molecules. For example hexane.
• It contains 6 carbon atoms. It can be considered a heavy molecule when compared to smaller molecules like ethane or propane
■ Can we break down a hexane molecule to make two new compounds?
• For example, if we break the hexane just after the second carbon atom, we get two sets:
CH₃ㅡCH₂│ㅡCH₂ㅡCH₂ㅡCH₂ㅡCH₃
• The first set having two carbon atoms and the second set having 4 carbon atoms
    ♦ So the first set having two carbon atoms can become ethane
    ♦ and the second set having 4 carbon atoms can become butane
• But there may not be enough hydrogen atoms to give two new alkanes.
• We know the general formulae:
Alkanes: C_nH_2n+1
Alkenes: C_nH_2n
Alkynes: C_nH_2n-1.
• So the alkenes and alkynes need lesser number of hydrogen atoms to form molecules.
■ Thus, when we break down the heavy hydrocarbons, we get a mixture of alkanes and alkenes

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■ But why would we want to break down the heavy ones and obtain lighter ones?
• The answer is that, many valuable hydrocarbons can be obtained in this way.
• For example, if we can obtain large quantities of butane, we will be able to produce LPG on a large scale because, butane is a main constituent of LPG
• Another application is in 'pollution control'
• We have seen that plastics are polymers of hydrocarbons. If those polymer chains in plastic wastes can be broken down to harmless hydrocarbons, pollution can be controlled to some extent
■ Thus we see that breaking down heavier hydrocarbons is an important requirement.  So how do we achieve it?
• Answer is that, we need to apply large quantities of heat. Application of pressure may also be required in some cases. 
• In any case, the heating should be done in the absence of air. Other wise oxygen will enter into reaction with the hydrocarbon.

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The process of heating some hydrocarbons with high molecular masses in the absence of air to form  hydrocarbons with lower molecular masses is called thermal cracking.

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Let us see some examples:
• Propane is one of the simplest hydrocarbons which has the capacity to undergo thermal cracking. The equation is:
CH₃ㅡCH₂│ㅡCH₃ (Propane) + Heat → CH₂＝CH₂ (Ethene) + CH₄ (Methane).
■ But when hydrocarbons with large number of carbon atoms undergo thermal cracking, there are different possibilities for the 'cleavage of the carbon chain' to occur.
• Possibilities for butane:
(i) CH₃│ㅡCH₂ㅡCH₂ㅡCH₃ (Butane) + Heat 
→ CH₄ (Methane) + CH₂＝CHㅡCH₃ (Propene) 
(ii) CH₃ㅡCH₂│ㅡCH₂ㅡCH₃ (Butane) + Heat 
→  CH₃ㅡCH₃ (Ethane) + CH₂＝CH₂ (Ethene)
• Possibilities for hexane:
(i) CH₃│ㅡCH₂ㅡCH₂ㅡCH₂ㅡCH₂ㅡCH₃ (Hexane) + Heat 
→ CH₄ (Methane) + CH₂＝CHㅡCH₂ㅡCH₂ㅡCH₃ (Pentene)  
(ii) CH₃ㅡCH₂│ㅡCH₂ㅡCH₂ㅡCH₂ㅡCH₃ (Hexane) + Heat 
→ CH₃ㅡCH₃ (Ethane) + CH₂＝CHㅡCH₂ㅡCH₃ (Butene) 
(iii) CH₃ㅡCH₂ㅡCH₂│ㅡCH₂ㅡCH₂ㅡCH₃ (Hexane) + Heat 
→ _{CH3ㅡCH2ㅡCH3} (Propane) + CH₂＝CHㅡCH₃ (Propene)
■ We can see a pattern in the cleavage of hexane:
• The numbers are:
1+5. This gives meth + pent
2+4. This gives eth + but
3+3. This gives prop+ prop
• In the product side, the first one is always an alkane
• The second one is always an alkene
• The hydrogen atoms are shared suitably between the product alkane and product alkene   
■ So we see that there are different possibilities when a heavy hydrocarbon is broken down. Then how can we predetermine what products to get?
• The answer is that, the products depend on the temperature and pressure applied during the process.
• Scientists and engineers have ready catalogues which shows 'how much pressure and temperature' should be applied in order to get any particular products

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Now we will see some solved examples based on what we have discussed so far in this chapter
Solved example 15.1
Fill in the blanks:
(i) CH≡CH + H₂ →  _____
Solution:
This is an addition reaction. The triple bond will become a double bond. The new hydrogen atoms will supply the required electrons. So the product is C₂H₄. Thus in the blank space, we write: C₂H₄
(ii)  CH₃Cl + Cl₂ → _____ + HCl
Solution:
On the product side we see an HCl molecule. That means, One H is displaced from CH₃Cl. So it is a substitution reaction. One H will be replaced by Cl. So in the blank space, we write: CH₂Cl₂.
(iii) ____  →  ㅡ[CH₂ㅡCH₂]_nㅡ
Solution:
• On the product side we see square brackets with subscript 'n'. So it is a polymerisation reaction
• Within the square brackets we have 'CH₂ㅡCH₂'. So the polymer is: polythene 
• This is ethene, which has changed the double bond to single bond
• That means, ethene is the monomer. Thus in the blank space, we write: 
CH₂＝CH₂ or  C₂H₄.
(iv) ____ + H₂ →  CH₃ㅡCH₃
Solution:
• One of the reactants is H₂. And the only product is CH₃ㅡCH₃.
• So the other reactant should be having two H atoms less than CH₃ㅡCH₃.
• When we remove two H atoms from CH₃ㅡCH₃, we get ethene.
• Thus in the blank space, we write: CH₂＝CH₂

Solved example 15.2
Match columns A, B and C suitably

	Reactants (A)	Products (B)	Name of the reaction (C)
1	CH3ㅡCH3 + Cl2	CO2 + H2O	Addition reaction
2	C2H6 + O2	CH2＝CH2	Thermal cracking
3	CH2＝CH2	CH2＝CH2 + CH4	Substitution reaction
4	CH3ㅡCH2ㅡCH3	CH3ㅡCH2Cl + HCl	Polymerisation
5	CH≡CH + H2	ㅡ[CH2ㅡCH2]nㅡ	Combustion

Solution:
(i) Row 1, column 1:
• In this reaction we get chloroethane and HCl as products. It is a substitution reaction
• So the matching will be: Row 1 → Row 4 → Row 3
(ii) Row 2, column 1:
• Reaction with oxygen is combustion. Two of the products are carbondioxide and water
• So the matching will be: Row 2 → Row 1 → Row 5
(iii) Ethene will undergo polymerisation to become polythene
• So the matching will be: Row 3 → Row 5 → Row 4
(iv) Propane will undergo thermal cracking to give methane and ethene
• So the matching will be: Row 4 → Row 3 → Row 2
(v) Ethyne will undergo addition reaction to give ethene.
• So the matching will be: Row 5 → Row 2 → Row 1

Solved example 15.3
Given below are two chemical equations:
(a) CH₂＝CH₂ + H₂ → A
(b) A + Cl₂ → B + HCl 
Identify the compounds 'A' and 'B'. Name these reactions
Solution:
1. Reaction in (a) is an addition reaction. The unsaturated ethene will become saturated ethane. 
So A is CH₃ㅡCH₃ (ethane) 
2. In reaction (b), ethane and chlorine undergo substitution reaction. 
So B is CH₃ㅡCH₂Cl (chloroethane)

Solved example 15.4
Write the chemical formula of propane. Write the names and structural formulae of two compounds that may be formed during it's substitution reaction with chlorine
Solution:
1. The chemical formula of propane is: CH₃ㅡCH₂ㅡCH₃
2. When propane reacts with chlorine, substitution reaction takes place in stages:
Stage 1: CH₃ㅡCH₂ㅡCH₃ + Cl₂ →CH₃ㅡCH₂ㅡCH₂Cl (chloropropane) + HCl
Stage 2: _{CH3ㅡCH2ㅡCH2Cl}  + Cl₂ →CH₃ㅡCHClㅡCH₂Cl (1,2-dichloropropane) + HCl.
• The chemical formula of chloropropane is: C₃H₇Cl
It's structural formula is shown in fig.15.19(a) below.
• The chemical formula of 1,2-dichloropropane is: C₃H₆Cl₂
It's structural formula is shown in fig.15.19(a) below:

Fig.15.19

Solved example 15.5
Complete the equation of the following chemical reaction:
CH₃ㅡCH₂ㅡCH₂ㅡCH₃ + __O₂ →_____ + _____
Name the reaction
Solution:
1. This is the reaction between a hydrocarbon and oxygen. So it is the combustion reaction of a hydrocarbon
2. The hydrocarbon is butane. We are asked to write:
• The number of oxygen molecules required and
• The products with number of molecules of each
3. The products of the combustion of a hydrocarbon are CO₂ and H₂O
• When we write the balanced equation of the reaction, we will get the number of molecules of each:
2C₄H₁₀ + 13O₂ →8CO₂ + 10H₂O

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In the next section, we will see some important Organic compounds

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Chapter 15 - Chemical Reactions of Organic compounds

In the previous section, we completed a discussion on nomenclature of hydrocarbons. In this section we will discuss some chemical reactions involving hydrocarbons.
• In organic chemistry, we deal with the following three topics:
    ♦ Study of different hydrocarbons
    ♦ Study of different compounds obtained from hydrocarbons
    ♦ Study of the different chemical reactions in which hydrocarbons are reactants or products
• Many substances that we use in our day to day life, and also many substances which are essential in scientific and engineering fields, are contributions of organic chemistry.
• Medicines, polymers, fuels etc., are examples  
• Such substances are obtained through different chemical reactions. 
• In this chapter we will discuss some of those basic chemical reactions.

Substitution reaction

First let us see the reaction between methane (CH₄) and chlorine in the presence of sunlight. This reaction can be analyzed in stages:
• Note that, the reaction will take place in the presence of sunlight because, sunlight is required to break the bonds between atoms
Stage 1:
• We know that, in methane, there are 4 hydrogen atoms around a carbon atom. 
• In stage 1, one of those hydrogen atoms will be removed and a chlorine atom will take it's place. 
• So the original methane will become chloromethane. 
• The removed hydrogen atom will combine with a chlorine atom to become hydrogen chloride (HCl).
• This is shown in fig.15.1 below:

Fig.15.1

Stage 2:
• In stage 2, one hydrogen atom from the newly formed chloromethane will be removed and a chlorine atom will take it's place. 
• So the original chloromethane will become dichloromethane. The prefix 'di' is used to indicate two chlorine atoms in the molecule. 
• So in dichloromethane, there are 2 hydrogen atoms and 2 chlorine atoms around a carbon atom. 
• The removed hydrogen atom will combine with a chlorine atom to become hydrogen chloride (HCl).
• This is shown in fig.15.2 below:

Fig.15.2

 Stage 3:
• In stage 3, one hydrogen atom from the newly formed dichloromethane will be removed and a chlorine atom will take it's place. 
• So the original dichloromethane will become trichloromethane. The prefix 'tri' is used to indicate three chlorine atoms in the molecule. Trichloromethane is also called chloroform 
• So in trichloromethane, there are 1 hydrogen atom and 3 chlorine atoms around a carbon atom. 
• The removed hydrogen atom will combine with a chlorine atom to become hydrogen chloride (HCl).
• This is shown in fig.15.3 below:

Fig.15.3

Stage 4:
• In stage 4, one remaining hydrogen atom from the newly formed trichloromethane will be removed and a chlorine atom will take it's place. 
• So the original trichloromethane will become tetrachloromethane. The prefix 'tetra' is used to indicate four chlorine atoms in the molecule. Tetrachloromethane is also called carbon tetrachloride 
• So in tetrachloromethane, there 4 chlorine atoms around a carbon atom. There are no hydrogen atoms.
• The removed hydrogen atom will combine with a chlorine atom to become hydrogen chloride (HCl). 
• This is shown in fig.15.4 below:

Fig.15.4

• The above figs., shows the structural formulae. The same reactions can be written using molecular formulae as follows:
Stage 1: CH₄ + Cl₂ → CH₃Cl + HCl
Stage 2: CH₃Cl + Cl₂ → CH₂Cl₂ + HCl
Stage 3: CH₂Cl₂ + Cl₂ → CHCl₃ + HCl
Stage 4: CHCl₃ + Cl₂ → CCl₄ + HCl
• In the above reaction, in each stage,  one hydrogen atom is replaced by a chlorine atom.

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Reactions in which an atom or a group in a compound is replaced by another atom or a group are called substitution reactions.

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• Substitution reactions between ethane and chlorine can be written as follows:
Stage 1: C₂H₆ + Cl₂ → C₂H₅Cl + HCl
Stage 2: C₂H₅Cl + Cl₂ → C₂H₄Cl₂ + HCl
Stage 3: C₂H₄Cl₂ + Cl₂ → C₂H₃Cl₃ + HCl
Stage 4: C₂H₃Cl₃ + Cl₂ → C₂H₂Cl₄ + HCl
Stage 5: C₂H₂Cl₄ + Cl₂ → C₂HCl₅ + HCl
Stage 6: C₂HCl₅ + Cl₂ → C₂Cl₆ + HCl

Addition reaction

• We know that ethane has all single bonds. It is a saturated compound
• We also know that in ethene, one double bond is present. It is an unsaturated compound.
• They are shown in fig.15.5 below:

Fig.15.5

• Fig.15.5(a) is an ethane molecule. Fig.(b) is an ethene molecule
• When unsaturated compounds take part in chemical reactions, they tend to form saturated compounds. Let us check if this is true for ethene:
• Ethene reacts with hydrogen at high temperature in the presence of nickel catalyst. 
• The product formed is ethane, which is a saturated compound. The reaction is shown below:

Fig.15.6

• Note that, the two carbon atoms no longer need to maintain a double bond between them, as the two new hydrogen atoms will supply the required electrons
• Using molecular formulae, it can be written as:
C₂H₄ + H₂ → C₂H₆

Another example:

Fig.15.7

• In the above reaction, we have propene and chlorine on the left side. 
• Propene is an unsaturated compound. It reacts with chlorine to form 1,2 - Dichloropropane, which is a saturated compound
• This can be written in another form also:
CH₃ㅡCH＝CH₂ + Cl₂ → C₃H₆Cl₂
• Note that, the two carbon atoms no longer need to maintain a double bond between them, as the two new chlorine atoms will supply the required electrons
• We have already seen in the previous chapter, how a chloro group gets itself attached to a carbon chain (Details here)

Some more examples:
(i) CH₂＝CH₂ + Cl₂ → C₂H₄Cl₂
The structural formulae are shown below:

Fig.15.8

The product is Dichloroethane. Note that, it has all single bonds

(ii) CH₂＝CH₂ + HCl → C₂H₅Cl
The structural formulae are shown below:

Fig.15.9

The product is Chloroethane. Note that, it has all single bonds

(iii) CH₃ㅡCH＝CH₂ + H₂ → C₃H₈
The structural formulae are shown below:

Fig.15.10

The product is propane. Note that, it has all single bonds

(iv) CH₃ㅡCH＝CHㅡCH₃ + HBr → C₄H₉Br
The structural formulae are shown below:

Fig.15.11

The product is 2-bromobutane. 
• Note that, the new hydrogen and bromine atoms cannot take up 'any position they like' in the But-2-ene. 
    ♦ The new hydrogen atom has to be attached to one of the two carbon atoms in the double bond in But-2-ene
    ♦ The new bromine atom has to be attached to the other carbon atom in the double bond in But-2-ene

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■ So we find that, compounds with double bond changes to compounds with all single bonds. 
• In a similar way, compounds with triple bonds can also change to compounds with all single bonds.
• This will take place in stages. 
    ♦ In the first stage, the triple bond in an alkyne changes to a double bond. 
    ♦ In the second stage, the double bond changes to a single bond. 
• The example of ethyne is given below:
Stage 1: Ethyne changes to ehene:
CH ☰ CH + H₂ → C₂H₂ 
Stage 2: Ethene changes to ethane:
CH₂＝CH₂ + H₂ → C₂H₆

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Reactions in which unsaturated organic compounds with double bond or triple bond react with other molecules to form saturated compounds are called addition reactions.

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Polymerisation

• From the discussion on addition reaction above, we find that the unsaturated compounds have a tendency to become saturated compounds.
• In addition reactions, the original unsaturated molecule combine with a 'molecule of another compound' to become saturated. 
    ♦ This 'another compound' can be hydrogen, chlorine, hydrogen chloride, hydrogen bromide etc.,
• Now we will see the change from unsaturated to saturated, with out any help from 'another compound'. 
    ♦ That is., the molecules of some unsaturated compound will combine among themselves to form saturated compounds. 
■ Let us see an example. 
• Consider an ethene molecule. It wants to become a saturated molecule by combining with another ethene molecule. Is it possible? Let us see:
We will write the steps:
1. In fig.15.12(a) below, two individual ethene molecules are shown. 
• All the valency requirements are satisfied and each of them are stable individually.

Fig.15.12

2. Now, each of them individually decides to discard the double bond and 'maintain a single bond only' between their two carbon atoms. The result is shown in fig.15.12(b) 
3. But as we can see, there are unsatisfied bonds. They are shown inside small red circles. 
• There are a total of four unsatisfied bonds. They are called 'unsatisfied' because, such bonds have only one electron. 
• This is shown more clearly in the electron dot diagram in fig.15.13(a) below:

Fig.15.13

• Consider any one carbon atom in fig.15.7(a). There are only 7 electrons around it: 
    ♦ 5 green and two red.
• So one more electron is required. 
4. Consider fig.15.13(b)
• The two molecules have decided to combine together. 
• So the two carbon atoms in the middle will bond together. Thus a larger molecule is formed. 
• The result is shown in fig.15.14(a) below:

Fig.15.14

5. In fig.14.14(a), the larger molecule consists of two simpler molecules of ethene. 
• Even when this larger molecule is formed from two smaller molecules, the outer two carbon atoms still need one electron each. We can represent it as: ㅡ[CH₂ㅡCH₂]₂ㅡ 
• 'CH₂ㅡCH₂' is one ethene molecule. Putting it inside square brackets and giving a subscript '2' indicates that two ethene molecules are combined together inside the square brackets.
• The 'ㅡ' symbol at both ends indicate that, even after the combination, unsatisfied bond exists at the ends
6. Obviously, another two such simple molecules of ethene can be attached at the ends. This is shown by the ellipses in fig.15.8(b). 
• The resulting larger molecule can be represented as: ㅡ[CH₂ㅡCH₂]₄ㅡ .
7. In this way a long chain can be formed with a large number of ethene molecules.
• The structure of the chain is shown in fig.15.15 below:

Fig.15.15

• This structure can be represented as: ㅡ[CH₂ㅡCH₂]_nㅡ 
    ♦ Where 'n' is a very large number

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Polymerisation is the process in which a large number of simple molecules combine under suitable conditions to form complex molecules. 
• The complex molecules thus formed are called polymers.
• The simple molecules which combine in this manner are called monomers.

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• We use a number of polymers in our daily life. 
    ♦ Some of them are natural polymers
    ♦ Others are man-made polymers
■ In the example that we saw above, the monomer is ethene
■ The IUPAC name of the polymer is polyethene
• It is commonly known as polythene. 
• It is also known as polyethylene. Since ethylene was the common name for ethene before IUPAC names were implemented.
• It is the most common type of plastic, and is used for making plastic bags, bottles etc.,

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■ The IUPAC name of another important polymer is: Poly(1-chloroethene)
• From this IUPAC name, we get the following information:
(i) The monomer is written inside the brackets. It is: 1-chloroethene
(ii) '1-chloroethene' indicates that it is an alkene. In this case, the alkene is obviously ethene.
(iii) One hydrogen atom in the ethene is replaced by a chlorine atom. Hence the name: 1-chloroethene
(iv) So the monomer is: CH₂＝CHCl. It's structure is shown in fig.15.16(a) below.
• But the double bond between the two carbon atoms breaks and become a single bond. 
    ♦ So it can be represented as: ㅡ[CH₂ㅡCHCl]ㅡ
    ♦ It's structure is shown in fig.15.16(b) below.
• Now it is ready to form the chain. The chain can be represented as: ㅡ[CH₂ㅡCHCl]_nㅡ
    ♦ It's structure can be represented as shown in fig.15.16(c) below:

Fig.15.16

[We have already seen in the previous chapter, how a chloro group gets itself attached to a carbon chain (Details here). We saw it again above in this section in addition reaction. So the electron dot diagram of the chain is not drawn here. However, the reader may draw it in his/her notebooks] 
• The monomer '1-chloroethene', or simply 'chloroethene', is commonly known as vinylchloride
• So when the monomer vinylchloride becomes a polymer, it's common name will be polyvinylchloride.
• It is popularly known as PVC, and is used for making pipes, electric cable insulation etc.,

■ The IUPAC name of another important polymer is: Poly(1,1,2,2-tetrafluoroethene)
• From this IUPAC name, we get the following information:
(i) The monomer is written inside the brackets. It is: 1,1,2,2-tetrafluoroethene
(ii) '1,1,2,2-tetrafluoroethene' indicates that it is an alkene. In this case, the alkene is obviously ethene.
(iii) One ethene molecule has four hydrogen atoms. All those hydrogen atoms are replaced by fluorine atoms. Hence the name: 1,1,2,2-tetrafluoroethene
(iv) So the monomer is: CF₂＝CF₂. It's structure is shown in fig.15.17(a) below.
• But the double bond between the two carbon atoms breaks and become a single bond. 
    ♦ So it can be represented as: ㅡ[CF₂ㅡCF₂]ㅡ
    ♦ It's structure is shown in fig.15.17(b) below.
• Now it is ready to form the chain. The chain can be represented as: ㅡ[CF₂ㅡCF₂]_nㅡ
    ♦ It's structure can be represented as shown in fig.15.17(c) below:

Fig.15.17

[We have already seen in the previous chapter, how a chloro group gets itself attached to a carbon chain (Details here). We saw it again above in this section in addition reaction. Fluorine and chlorine belongs to the same halogen family. Their valencies are the same. So the electron dot diagram of the chain is not drawn here. However, the reader may draw it in his/her notebooks] 

• The monomer 1,1,2,2-tetrafluoroethene, is commonly known simply as tetrafluoroethene

• When the monomer tetrafluoroethene becomes a polymer, that polymer is commonly know as teflon.
• It is used for coating the inner surface of non-stick cookware.

■ The IUPAC name of yet another important polymer is: Poly(propene)
• From this IUPAC name, we get the following information:
(i) The monomer is written inside the brackets. It is: propene
(ii) We know that 'propene' is an alkene. It's formula is: CH₃ㅡCH＝CH₂
(iii) It's structure is shown in fig.15.18(a) below.
• But the double bond between the two carbon atoms breaks and become a single bond. 
But it cannot be represented as: ㅡ[CH₃ㅡCHㅡCH₂]ㅡ
Let us analyse the reason:
(i) Till now we have seen monomers with upto two carbon atoms only. 
(ii) But our present monomer propene has 3 carbon atoms. 
iii) So while writing the structure, the carbon atoms on either sides of the double bond should come in a straight line. 
(iv) This is because, the 'unsatisfied bonds' will be starting from the carbon atoms on either sides of the double bond
(v) The third carbon atom should be written above or below the line. This is shown in fig.15.18(b) below:

Fig.15.18

[As in the previous two cases, the reader may draw the electron dot diagram in his/her own notebooks]
• The propene is commonly known as propylene
• So when the monomer propylene becomes a polymer, it's common name will be polypropylene
• It is used for making table tops and other utensils which require thermal resistance.
■ It is interesting to note that rubber (natural and man-made) is a polymer. The monomer in this case is isoprene
• Isoprene is an alkene having two double bonds. We will see it's polymer structure in higher classes

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In the next section, we will see some more reactions.

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