In the previous section, we completed a discussion on substitution, addition and polymerisation reactions. In this section we will see more such reactions.
• 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
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.
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
• 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: CnH2n+1
Alkenes: CnH2n
Alkynes: CnH2n-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
■ 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.
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.
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
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
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:
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
In the next section, we will see some important Organic compounds
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
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.
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: CnH2n+1
Alkenes: CnH2n
Alkynes: CnH2n-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
■ 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.
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.
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
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 |
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
In the next section, we will see some important Organic compounds
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