Chapter 13.1 - Extraction and Refining

In the previous section, we completed a discussion on 'concentration of ore'. In this section, we will see the next stage.
The second stage after concentration of ore is:
■ Extraction of metal from concentrated ore
There are two stages in it:
(i) Conversion of the concentrated ore into it's oxide
(ii) Reduction of the oxide
• We will now see them in detail:
For the first stage, any one of the following two methods can be used. 
(a) Calcination
(b) Roasting
The choice depends on the type of the ore
(a) Calcination:
1. The concentrated ore is heated to a high temperature in the absence of air
• But the temperature should not exceed the melting point of the ore
2. So the impurities which have lower melting points will melt and then evaporate away.
• The most common impurities which are removed in this way are water and organic matter
3. When the impurities are removed, the useful ore remains. 
• These are the carbonates and hydroxides of the metal. 
4. Due to the heat, these carbonates and hydroxides decompose to form the oxides of the metal. 
• When oxides of the metal are formed, they are ready to be passed on to the next stage
5. Since there is no air, oxygen will not take part in this decomposition reaction.
Example: Zinc carbonate is converted into Zinc oxide
(b) Roasting:
1. The concentrated ore is heated to a high temperature in a current of air
• But the temperature should not exceed the melting point of the ore
• Because of the current of air, oxygen will also take part in the reaction
2. The impurities which have lower melting points will melt and then evaporate away.
• The most common impurities which are removed in this way are water and organic matter
3. When the impurities are removed, the useful ore remains. 
• These remaining ore is converted into oxides because of the presence of oxygen
• When oxides of the metal are formed, they are ready to be passed on to the next stage

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■ At this stage we will draw a flow chart to show various steps that we have seen so far:

Fig.13.4

• As we learn more details, we will fill up the missing spaces in the above chart.
• Now we move to the next step in 'extraction of metals', which is: 

Reduction of the oxide

• We have seen the details about reducing agents and reduction reactions here. 
• We have seen some redox reactions in the previous chapter on 'Reactivity' also. Details here. 
• When we 'reduce' the oxide of a metal, we will get the original metal. 
• Suitable reducing agents can be used for this purpose. 

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Before we proceed further, let us discuss the reason for saying:
'reduce' the oxide of a metal.
The following steps will help us to understand the reason:
1. Let 'X' be a metal. So we can denote it's oxide as 'XO', where 'O' stands for oxygen.
2. Thus 'XO' is a compound. The individual elements in that compound are:
• The metal 'X'
• The oxygen 'O'
3. Now, X is a metal. Metals are electron donors. 
• So in the compound XO, the metal X will be having a positive oxidation state. Let it be +m
• And O will be having a negative oxidation state. Let it be -n
4. So we can write: X^+m O^-n.
• Mg⁺² O^-2 is an example
5. Now we allow this X^+m O^-n to react with 'another compound'.
This 'another compound' must be able to donate electrons
6. The electrons thus donated will be received by X^+m .
• Then X^+m will become X⁰ or simply X
• The oxidation number is reduced from +m to zero
7. This X will no longer attach itself to O
• It becomes independent and we get the pure metal X
8. So we see that some 'reduction' is indeed involved
• The oxidation number decreases from +m to zero
• That is why we say:
'reduce' the oxide of a metal to get the pure metal  
9. The 'another compound' that we saw in step (5) above is the reducing agent.
10. Electricity is the strongest reducing agent. It has an unlimited supply of electrons. Also, it's energy level can be increased by increasing the voltage.

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Now we continue our original discussion:
Some examples of the application of reducing agents are:
• Carbon monoxide is used as the reducing agent to extract iron from haematite

    ♦ The equation is: Fe₂O₃ (s) + 3CO (s) → 2Fe (s) + 3CO₂ (g)

    ♦ In Fe₂O₃, The Fe exists as positive ion. It receive electrons from the reducing agent CO. 

    ♦ The Fe ion thus becomes Fe atom

• Carbon is used as reducing agent to extract zinc from zinc oxide.

    ♦ The equation is: ZnO (s) + C (s) → Zn (s) + CO (g)

    ♦ In ZnO, The Zn exists as positive ion. It receive electrons from the reducing agent C. 

    ♦ The Zn ion thus becomes Zn atom

• Electricity is the strongest reducing agent. This electricity is required to extract highly reactive  metals like sodium, potassium and calcium from their oxides.
■ So we completed the second stage also. The next stage is:

Refining of metals

• The metal obtained by the reduction process that we discussed above, will still contain some impurities. 
• Such impurities include small quantities of other metals and may be some non metals. 
• We must remove them also to get the pure metal. 
■ Following are some of the methods that can be used. The choice of the method will depend on the nature of the metals and the impurities present in them.
(a) Liquation
(b) Distillation
(c) Electrolytic refining
• Before seeing each of them in detail, we will complete the chart. 
• This will give us a better understanding about our exact position in the whole discussion. 
• The completed chart is shown in fig.13.5 below:

Fig.13.5

• Now we can continue the discussion. We were discussing the last step, which is 'Refining of the metal'.
• So we are in the bottom most green box. We can use any one of the three. The choice depends on the type of metal and the impurities.
(a) Liquation
(b) Distillation
(c) Electrolytic refining

Liquation

• This method is used when both the following condition are satisfied:
    ♦ The pure metal to be extracted, has a low melting point
    ♦ The impurities are metals having high melting points 
• Consider fig.13.6 below. We can write the steps:

Fig.13.6

1. The impure metal is supplied from the top of a specially shaped funnel
• This funnel is kept in an inclined position
2. The funnel is heated from it's bottom side
3. The pure metal which has low melting point will melt and will flow down along the inclined surface.
• It are collected at the bottom
4. The impurities having high melting points will be left behind
• Tin and lead, which have low melting points can be refined using this method
5. Note that, the structure in which heat is produced, is made of bricks
• But they are not ordinary bricks. They are 'refractory bricks'. They can withstand very high temperatures.

Distillation

• This method is used when both the following condition are satified:
    ♦ The pure metal to be extracted, has a low boiling point
    ♦ The impurities are metals having high boiling points 
• We can write the steps:
1. The impure metal is heated in a retort. 
2. The pure metal alone will boil and turn into vapours
3. These vapours are collected and condensed to get the pure metal.
4. The impurities will be left behind
• Zinc, cadmium and mercury have low boiling points. So they can be refined by this method

Electrolytic refining

• In this method, we use electrolysis to obtain the metal. 
• We have seen the details about electrolysis in the previous chapter. (Details here). 
• For electrolysis, we need an electrolyte. For our present case, we use a solution of the 'salt of the metal' as the electrolyte. 
■ A small piece of the pure metal is used as the negative electrode. 
■ A piece of the impure metal is used as the positive electrode.
• We want to know why pure metal is used as negative and the impure metal is used as positive electrodes.
• We will see the reason when we analyse the reaction that takes place when we turn on the switch. 
We will write it in steps:
1. We want to refine the impure piece. 
• For that, the pure metal should separate from the impurities and dissolve in the solution. 
2. How does such a dissolution occur?
Ans: The positive metal ions get separated from the impure piece and goes into the solution.
3. Now we want these positive ions to be converted into pure metal. 
• For that, the positive ions move to the negative electrode. 
• There they receive electrons and will be converted into pure metal atoms. 
4. So the formation of the 'new pure metal' occurs at the negative electrode. 
• This newly formed pure metal should stick to the already available pure metal. Then only we can use it. 
■ So the already available pure metal piece is used as the negative electrode. 
5. The reaction taking place at the negative electrode is reduction. 
• So the negative electrode is the cathode. 
• The impure piece is the positive electrode and it is the anode

The chemistry of metallurgy

Consider the table given below. The metals are arranged in the decreasing order of their reactivity.
Also they are arranged in four groups. 

Let us write the details:
1. The five members of the top most group are very reactive. 
• We will need electricity to extract them from their compounds. These compounds are first heated and melted into the liquid state. 
Then electrolysis is carried out
2. The five members of the second group are extracted by reducing their oxides. 
• Depending on the metal, carbon or carbon monoxide can be used as the reducing agent.
3. There is only one member in the third group. It is copper. 
• It has a lower reactivity than the first and second groups. It can be extracted more easily than the metals above it. 
• First it's sulphide is converted to oxide and then the oxide is reduced.
4. The fourth group has two members. 
• Their reactivity is very low. So they do not form any compounds. 
• Thus they are found in free state in nature.

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We can write a summary:
1. Metals have a tendency to lose electrons.
• So they are seen as positive ions in their compounds
2. The positive ions must be converted to ordinary atoms to obtain the pure metal
• So we must supply electrons to the positive ions
3. The positive ions receive these electrons and get reduced to the original atoms
■ So in short, if we are asked to choose between oxidation and reduction:
We must choose reduction to extract metal from it's compounds.

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So we have seen the basics of metallurgy and the various stages involved in the extraction of metals.  In the next section, we will see how these stages are applied in the case of iron. 

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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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