High school Chemistry Lessons: hydrocarbons

Showing posts with label hydrocarbons. Show all posts

Sunday, March 18, 2018

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

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: 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

■ 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＝CH₂or  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

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

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Monday, December 25, 2017

Chapter 14.15 - Isomerism in Alicyclic compounds

In the previous section, we saw the details about Isomerism. In this section, we will see Isomerism in Alicyclic compounds.

• We have seen three types of structures for hydrocarbons:
(i) Straight chain (ii) Branched chain (iii) Cyclic
• We have seen some basic details about cyclic compounds in our earlier classes. (Details here)
• Now we will analyse them in greater detail. We will do the analysis by taking Cyclopentane as an example:
1. Consider the chain structure in fig.14.94(a) below:

Fig.14.94

• We know that it is the structure of pentane
♦ All the carbon-carbon bonds are single bonds
♦ 3 hydrogen atoms are attached to the carbon atom at the left end
♦ 3 hydrogen atoms are attached to the carbon atom at the right end also
♦ All interior carbon atoms have 2 hydrogen atoms each
2. Let us bend this 'straight chain' into a 'ring' structure. This is shown in fig.14.94(b)
• In fig(b), we have only bend it. The closed ring is not formed yet.
♦ To get a closed ring, the end carbons should meet (ie., bond together)
• Note that in this 'bent state', the 'number of hydrogen atoms carried by each of the carbon atoms' has not changed.
3. Now we will bond the end carbons together. For easy comparison, the 'bent state' and the 'ring'state' are shown side by side in fig.14.95 below

Fig.14.95

• Fig.14.95(a) is the same fig.14.94(b) that we saw above.
4. See the completed ring in fig.14.95(b)
• The end carbons have finally met. That is., a bond is formed between them
• What else do we see?
Ans: Each of the end carbon atoms have lost one Hydrogen atom. So, the molecule as a whole, has lost two hydrogen atoms
• Why did the hydrogen atoms leave?
Ans: Because they are no longer required for octet of the end carbon atoms. A pair of electrons is now shared in the newly formed bond between the end carbon atoms
• When a hydrogen atom leave, it takes it's electron (shown as red dot in the fig.) with it.
♦ The green dot which was in bond with that red dot is now free.
• The same happens in the other end carbon
• The newly freed green dots together form a bond.
• This bond results in the bonding between the end carbon atoms and the ring is completed
• The completed ring in fig.14.95(b) is Cyclopentane
5. Let us write the molecular formulae:
Pentane: C₅H₁₂
Cyclopentane: C₅H₁₀
• So the cyclopentane has two hydrogen atoms less than the pentane
■ This is true for all alkanes:
The number of hydrogen atoms in any cycloalkane will be 2 less than the corresponding alkane
6. At this time we should recall another result that we saw in previous classes:
■ The number of hydrogen atoms in any alkene will be 2 less than the corresponding alkane
7. Combining (5) and (6) we can write:
• The number of hydrogen atoms in a cycloalkane is same as that in the corresponding alkene
This gives an interesting result:
■ An alkene and it's corresponding cycloalkane are isomers
8. By the word 'corresponding', we mean 'same number of carbon atoms'
Example:
• If the cyclic compound is: Cyclopentane
♦ Corresponding alkane is: Pentane
♦ Corresponding alkene is: Pentene
♦ Cyclopentane and Pentene will be isomers
Another example:
• If the cyclic compound is: Cyclopropane
♦ Corresponding alkane is: Propane
♦ Corresponding alkene is: Propene
♦ Cyclopropane and Propene will be isomers

• Thus we saw the relation between a cycloalkane and it's corresponding alkene. For this result, we started out with a cycloalkane.
• Now we will start out another discussion with a cycloalkene. Let us see where we will reach:

1. Consider the chain structure in fig.14.96(a) below:

Fig.14.96

• We know that it is the structure of pentene
♦ One of the carbon-carbon bonds is a double bond
♦ 3 hydrogen atoms are attached to the carbon atom at the left end
♦ 2 hydrogen atoms are attached to the carbon atom at the right end
♦ 1 hydrogen atom is attached to the second carbon atom from the right end
♦ All other interior carbon atoms have 2 hydrogen atoms each
2. Let us bend this 'straight chain' into a 'ring' structure. This is shown in fig.14.96(b)
• In fig(b), we have only bend it. The closed ring is not formed yet.
♦ To get a closed ring, the end carbons should meet (ie., bond together)
• Note that in this 'bent state', the 'number of hydrogen atoms carried by each of the carbon atoms' has not changed.
3. Now we will bond the end carbons together. For easy comparison, the 'bent state' and the 'ring'state' are shown side by side in fig.14.97 below:

Fig.14.97

• Fig.14.97(a) is the same fig.14.96(b) that we saw above.
4. See the completed ring in fig.14.97(b)
• The end carbons have finally met. That is., a bond is formed between them
• What else do we see?
Ans: Each of the end carbon atoms have lost one Hydrogen atom. So, the molecule as a whole, has lost two hydrogen atoms
• Why did the hydrogen atoms leave?
Ans: Because they are no longer required for octet of the end carbon atoms. A pair of electrons is now shared in the newly formed bond between the end carbon atoms
• When a hydrogen atom leave, it takes it's electron (shown as red dot in the fig.) with it.
♦ The green dot which was in bond with that red dot is now free.
• The same happens in the other end carbon
• The newly freed green dots together form a bond.
• This bond results in the bonding between the end carbon atoms and the ring is completed
• The completed ring in fig.14.97(b) is Cyclopentene
5. Let us write the molecular formulae:
• Pentane: C₅H₁₂
♦ Cyclopentane: C₅H₁₀
_{• Pentene: C₅H₁₀}
_{♦ Cyclopentene: C₅H₈}
• So the cyclopentene has two hydrogen atoms less than the pentene
■ This is true for all alkenes:
The number of hydrogen atoms in any cycloalkene will be 2 less than the corresponding alkene
6. At this time we should recall another result that we saw in previous classes:
■ The number of hydrogen atoms in any alkyne will be 2 less than the corresponding alkene
7. Combining (5) and (6) we can write:
• The number of hydrogen atoms in a cycloalkene is same as that in the corresponding alkyne
This gives an interesting result:
■ An alkyne and it's corresponding cycloalkene are isomers
8. By the word 'corresponding', we mean 'same number of carbon atoms'
Example:
• If the cyclic compound is: Cyclopentene
♦ Corresponding alkane is: Pentane
♦ Corresponding alkene is: Pentene
♦ Corresponding alkyne is: Pentyne
♦ Cyclopentene and Pentyne will be isomers
Another example:
• If the cyclic compound is: Cyclopropene
♦ Corresponding alkane is: Propane
♦ Corresponding alkene is: Propene
♦ Corresponding alkyne is: Propyne
♦ Cyclopropene and Propyne will be isomers
• Thus we saw the relation between a cycloalkene and it's corresponding alkyne.
■ We can show the above findings in a simple fig:

Fig.14.98

• The green arrow indicates that Alkenes and Cycloalkanes can be isomers
• The yellow arrow indicates that Alkynes and Cycloalkenes can be isomers
■ In this section, we will be seeing problems related to the green arrow only

But before we see those problems, a more interesting result related to the green arrow is coming up:
1. We know that the pentene itself has isomers. This is based on the position of it's double bond. We saw details here.
Thus we have:
Pent-1-ene: CH₃ㅡCH₂ㅡCH₂ㅡCH＝CH₂
Pent-2-ene: CH₃ㅡCH₂ㅡCH＝CHㅡCH₃
2. The three compounds: cyclopentane, pent-1-ene and pent-2-ene have the same molecular formula C₅H₁₀.
• We can write:
The three compounds: cyclopentane, pent-1-ene and pent-2-ene are isomers
• pent-1-ene and pent-1-ene are chain isomers
3. For higher alkenes like octene, decene etc., more chain isomers are possible.
• All those isomers will be isomers of the corresponding cycloalkanes (cyclooctane, cyclodecane etc.,) also.

Solved example 14.21
Write the IUPAC names and structures of all possible isomers of the compound given below:
CH₃ㅡCH₂ㅡCH＝CH₂
Solution:
1. The given compound is But-1-ene
It has two possible isomers
(i) The chain isomer: But-2-ene
• It is produced by the change in the position of the double bond
• It's structure is: CH₃ㅡCH＝CHㅡCH₃.
(ii) The corresponding cycloalkane: Cyclobutane
• It's structure is shown in the fig.14.99(a) below:

Fig.14.99

2. So we have three compounds:
But-1-ene, But-2-ene and Cyclobutane
• All three of them have the same molecular formula: C₄H₈.
• All three are isomers.

Solved example 14.22
Write the IUPAC names and structures of all possible isomers of the compound shown in fig.14.99(b) above.
Solution:
1. The given compound is Cyclohexane
It has three possible isomers
(i) Hex-1-ene
• It's structure is: CH₃ㅡCH₂ㅡCH₂ㅡCH₂ㅡCH＝CH₂
(ii) Hex-2-ene
• It's structure is: CH₃ㅡCH₂ㅡCH₂ㅡCH＝CHㅡCH₃

(iii) Hex-3-ene
• It's structure is: CH₃ㅡCH₂ㅡCH＝CHㅡCH₂ㅡCH₃
2. So we have four compounds:
Hex-1-ene, Hex-2-ene, Hex-3-ene and Cyclohexane
• All four of them have the same molecular formula: C₆H₁₂
• All four are isomers

In the next section, we will see Reactions of Organic compounds

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Tuesday, December 19, 2017

Chapter 14.11 - Halo functional groups

In the previous section, we saw the nomenclature of organic compounds carrying the Carboxyl group. In this section, we will see the Halo group.

• We know that halogens are one of the several 'families' in the periodic table.
• We have also other families like oxygen family, carbon family, noble gases etc., Details here.
• Some members of the halogen family forms compounds with hydrocarbons. Such compounds are called halo compounds
• We will now see them in detail:
■ The members of the halogen family which form compounds with hydrocarbons are:
Fluorine (F), Chlorine (Cl), Bromine (Br) and Iodine (I)
■ When a halogen become part of the organic compound, it is called a halo group
So we have: Fluoro group (ㅡF), Chloro group (ㅡCl), Bromo group (ㅡBr) and Iodo group (ㅡI)

Now we will see the IUPAC names of halo compounds.
• The rules are very similar to the ones that we saw for branched chain hydrocarbons. Details here.
So we can write:
• The position of carbon atom carrying the halo group+hyphen+name of halo group+word root+suffix
♦ The halo group should be given the lowest possible number
• Let us see a solved example:
Solved example 14.11
Write the IUPAC name of the compounds given in fig.14.76 below:

Fig.14.76

Now we will see the IUPAC names of compounds with more than one halo group.
• The rules are similar to those when more than one branch is present. We saw the details here.
So we can write:
[Position number of carbon atom carrying the group+hyphen+name of halo group]
+hyphen
+[Position number of carbon atom carrying the group+hyphen+name of halo group]
+word root+suffix
• The items inside square brackets are related to halo groups. Those groups should be written in alphabetical order.
♦ But in our present discussion, we will encounter those compounds only in which there is only one type of halo group. So we do not need to worry about alphabetical order.

Solved example 14.12
Write the IUPAC name of the halo compounds given in fig.14.77 below:

Fig.14.77

Solution:
We will write the steps:
Case (i):
1. In the given compound, there are 4 carbon atoms. So the word root is But
There are only single bonds. So the suffix is ane
2. The numbering can be done either from left or right. Both will give the position as 2 for the first halo group
• Thus the first halo group gets number 2 and the second halo group gets number 3
3. Name of the halo group is Chloro
4. Now we can assemble the name. The rule is:
[Position number of carbon atom carrying the group+hyphen+name of halo group]
+hyphen
+[Position number of carbon atom carrying the group+hyphen+name of halo group]
+word root+suffix
• Thus we get:
[2-chloro]
+hyphen
[3-chloro]
+butane
So the name would be: 2-Chloro-3-chlorobutane
But . . .
As we have seen in the case of repeating branches, we can write:
• If a group comes two times, it's name should not be written two times
♦ Instead, the prefix 'di' should be used
• If a group comes three times, it's name should not be written three times
♦ Instead, the prefix 'tri' should be used
• If a group comes four times, it's name should not be written four times
♦ Instead, the prefix 'tetra' should be used
so on . . .
• The positions should be separated by commas
So the name of our compound is:
2,3-Dichlorobutane.

Case (ii):
1. In the given compound, there are 5 carbon atoms. So the word root is pent
There are only single bonds. So the suffix is ane
2. The numbering can be done either from left or right. Both will give the position as 2 for the first halo group
• Thus the first halo group gets number 2 and the second halo group gets number 4
3. Name of the halo group is Chloro
4. Now we can assemble the name. The rule is:
[Position number of carbon atom carrying the group+hyphen+name of halo group]
+hyphen
+[Position number of carbon atom carrying the group+hyphen+name of halo group]
+word root+suffix
• Thus we get:
[2-chloro]
+hyphen
[4-chloro]
+pentane
So the name would be: 2-Chloro-4-chloropentane
But . . .
As we have seen in the previous case, we can write:
• If a group comes two times, it's name should not be written two times
♦ Instead, the prefix 'di' should be used
• If a group comes three times, it's name should not be written three times
♦ Instead, the prefix 'tri' should be used
• If a group comes four times, it's name should not be written four times
♦ Instead, the prefix 'tetra' should be used
so on . . .
• The positions should be separated by commas
So the name of our compound is:
2,4-Dichloropentane.

Solved example 14.13
Write the IUPAC name of the compounds given in fig.14.78 below:

Fig.14.78

Solution:
We will write the steps:
Case (i):
1. In the given compound, there are 3 carbon atoms. So the word root is prop
There are only single bonds. So the suffix is ane
2. The numbering should be done towards the left from the right most carbon atom. This is to ensure that, the carbon atom carrying the groups gets the lowest number.
• Thus the carbon atom carrying the halo groups gets number 2
3. Name of the halo group is Chloro
4. Now we can assemble the name. The rule is:
[Position number of branch+hyphen+name of halo group]
+hyphen
+[Position number of branch+hyphen+name of halo group]
+word root+suffix
• Thus we get:
[2-chloro]
+hyphen
[2-chloro]
+butane
So the name would be: 2-Chloro-2-chlorobutane
But . . .
As we have seen in the previous case, we can write:
• If a group comes two times, it's name should not be written two times
♦ Instead, the prefix 'di' should be used
• If a group comes three times, it's name should not be written three times
♦ Instead, the prefix 'tri' should be used
• If a group comes four times, it's name should not be written four times
♦ Instead, the prefix 'tetra' should be used
so on . . .
• The positions should be separated by commas.
• If the groups are at the same position, the position number should be repeated.
So the name of our compound is:
2,2-Dichlorobutane.

Case (ii):
1. In the given compound, there are 4 carbon atoms. So the word root is but
There are only single bonds. So the suffix is ane
2. The numbering can be done either from left or right. Both will give the position as 2 for the carbon carrying the halo groups
• Thus the carbon atom carrying the halo groups gets number 2
3. Name of the halo group is Chloro
4. Now we can assemble the name. The rule is:
[Position number of branch+hyphen+name of halo group]
+hyphen
+[Position number of branch+hyphen+name of halo group]
+word root+suffix
• Thus we get:
[2-chloro]
+hyphen
[2-chloro]
+propane
So the name would be: 2-Chloro-2-chloropropane
But . . .
As we have seen in the previous case, we can write:
• If a group comes two times, it's name should not be written two times
♦ Instead, the prefix 'di' should be used
• If a group comes three times, it's name should not be written three times
♦ Instead, the prefix 'tri' should be used
• If a group comes four times, it's name should not be written four times
♦ Instead, the prefix 'tetra' should be used
so on . . .
• The positions should be separated by commas.
• If the groups are at the same position, the position number should be repeated.
So the name of our compound is:
2,2-Dichloropropane.

Case (iii):
1. In the given compound, there are 4 carbon atoms. So the word root is but
There are only single bonds. So the suffix is ane
2. The numbering can be done in two ways:
• When numbered from left to right, the position numbers are 2,3,3
♦ This gives a sum of: 2+3+3 = 8
• When numbered from right to left, the position numbers are 2,2,3
♦ This gives a sum of: 2+2+3 = 7
• The numbering from right to left gives a lower sum. So it should be adopted.
• The above rule is known as the sum rule. We will see more details about it in higher classes.
3. Name of the halo group is Bromo
4. Now we can assemble the name. The rule is:
[Position number of branch+hyphen+name of halo group]
+hyphen
+[Position number of branch+hyphen+name of halo group]
+word root+suffix
• Thus we get:
[2-bromo]
+hyphen
[2-bromo]
+hyphen
[3-bromo]
+butane
So the name would be: 2-bromo-2-bromo-3-bromobutane
But . . .
As we have seen in the previous cases, we can write:
• If a group comes two times, it's name should not be written two times
♦ Instead, the prefix 'di' should be used
• If a group comes three times, it's name should not be written three times
♦ Instead, the prefix 'tri' should be used
• If a group comes four times, it's name should not be written four times
♦ Instead, the prefix 'tetra' should be used
so on . . .
• The positions should be separated by commas.
• If the groups are at the same position, the position number should be repeated.
So the name of our compound is:
2,2,3-Tribromobutane.

Now we will see the reverse process. That is., we are given the IUPAC name of a halo compound. We must write the structural formula. We will learn the method by analysing an example:
1. Given IUPAC name is: 2-Chloropropane
2. Consider the word root. It is prop. So there are 3 carbon atoms
3. Consider the suffix. It is ane. So the compound is derived from an alkane
• So all the carbon-carbon bonds are single bonds
4. So we have 3 carbon atoms with single bonds between them. The numbering (from left to right) can also be done at this stage. This is shown in fig.14.79(a) below:

Fig.14.79

5. The halo group is at position 2.
So we attach the ㅡCl to the second carbon atom. This is shown in fig(b)
6. Fill all the valencies of carbon atoms.
• We use hydrogen to fill up the valencies.
• The result is shown in fig.14.79(c) above. So this fig.14.79(c) shows the required structure

Another example:
1. Given IUPAC name is: 2,3-Dibromopentane
2. Consider the word root. It is pent. So there are 5 carbon atoms
3. Consider the suffix. It is ane. So the compound is derived from an alkane
• So all the carbon-carbon bonds are single bonds
4. So we have 5 carbon atoms with single bonds between them. The numbering (from left to right) can also be done at this stage. This is shown in fig.14.80(a) below:

Fig.14.80

5. The halo groups are at positions 2 and 3.
So we attach the ㅡBr to the second and third carbon atoms. This is shown in fig(b)
6. Fill all the valencies of carbon atoms.
• We use hydrogen to fill up the valencies.
• The result is shown in fig.14.80(c) above. So this fig.14.80(c) shows the required structure

■ In the first section in this chapter, we saw how the 'alkyl radical' gets itself attached to an open chain. See details here.
■ In a previous section, we saw how a 'hydroxyl group' that is., 'ㅡOH', gets itself attached to an open chain. See details here.
■ In another previous section, we saw how an 'aldehyde group' that is., 'ㅡCHO', gets itself attached to an open chain. See details here.
■ In yet another previous section, we saw how a 'keto group' that is., 'ㅡCO', gets itself attached to an open chain. See details here.
■ In the just previous section we saw how a 'carboxyl group' that is., 'ㅡCOOH', gets itself attached to an open chain. See details here.
■ Now we will see the same for the halo group:
• Though there are different members (fluoro, chloro etc.,) in the halo group, the bonding details are same for all. So we will take chlorine as an example
• We have seen the bonding between sodium and chlorine to form sodium chloride. it is an ionic bond. Details here.
• We have also seen the covalent bonding between two chlorine atoms. Two chlorine atoms share a pair of electrons between them to achieve octet and thus stability.
• It can be represented as: ClㅡCl. Details here. The two chlorine atoms bonded together by a single covalent bond becomes a chlorine molecule (Cl₂)
• So when a chlorine atom is removed from the chlorine molecule, the remaining portion becomes a chloro group (ㅡCl). It will be unstable and will be looking for an electron.
• If a hydrogen atom is removed from a hydrocarbon, the ㅡCl can take it's place. This is shown in fig.14.81 below:

Fig.14.81

• Thus we get a stable molecule

In the next section, we will see the Alkoxy group.

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