In the previous section, we saw the nomenclature of unsaturated hydrocarbons. In this section, we will see functional groups.
• We know that hydrocarbons consist of carbon and hydrogen only. But in some hydrocarbons, one or more hydrogen atoms will be removed and another atom will take it's place.
• Let us see some examples:
■ Consider Propane: CH3 ㅡCH2 ㅡCH3
• One hydrogen atom in the end carbon can be replaced by a chlorine atom. Then we get:
CH3 ㅡCH2 ㅡCH3 ㅡ Cl.
♦ This new compound is called Chloropropane
• One hydrogen atom in the end carbon can be replaced by a fluorine atom. Then we get:
CH3 ㅡCH2 ㅡCH3 ㅡ F.
♦ This new compound is called Fluoropropane
• In the above examples, one hydrogen atom was replaced by another 'single atom'
We will see a case when one hydrogen atom is replaced by a 'group of atoms':
■ Consider Methane: CH4
• One hydrogen atom can be replaced by a (OH) group. Then we get:
CH3 ㅡOH.
♦ This new compound is called Methanol
Similarly, one hydrogen in ethane (CH3 ㅡCH3) can be replaced by a (OH) group. Then we get:
CH3 ㅡCH2 ㅡOH.
♦ This new compound is called Ethanol
Another example:
■ Consider Methane: CH4
• One hydrogen atom can be replaced by a (NH3) group. Then we get:
CH3 ㅡNH3.
♦ This new compound is called Methanamine
• Similarly, one hydrogen in ethane (CH3 ㅡCH3) can be replaced by a (NH3) group. Then we get:
CH3 ㅡCH2 ㅡNH3.
♦ This new compound is called Ethanamine
• The chemical and physical properties of propane are quite different from those of chloropropane or fluoropropane
• The chemical and physical properties of methane are quite different from those of methanol or methanamine
• The chemical and physical properties of ethane are quite different from those of ethanol or ethanamine
■The presence of certain atoms or groups imparts certain characteristic properties to compounds. They are called functional groups.
We will now discuss about the various functional groups and their nomenclature
Hydroxyl Group (ㅡOH)
• The IUPAC names of compounds with hydroxyl function group end in 'ol'. Such compounds are called alcohols.
♦ In any alcohol, there will be one or more ㅡOH group
■ The naming of alcohols is done by the following procedure:
• Remove the 'e' from the name of the corresponding alkane
♦ Put 'ol' in it's place
• So we can write:
Alkane - e + ol → Alkanol
• Thus we get:
♦ Methane - e + ol → Methanol
♦ Ethane - e + ol → Ethanol
♦ Propane - e + ol → Propanol
so on . . .
But this is not the only IUPAC rule related to the nomenclature of alcohols. Let us analyse more:
Consider the two compounds given in fig.14.54 below:
• In the first compound, there are 3 carbon atoms, 8 hydrogen atoms and 1 oxygen atom
♦ So it's molecular formula is: C3H8O
• In the second compound, there are 3 carbon atoms, 8 hydrogen atoms and 1 oxygen atom
♦ So it's molecular formula is also: C3H8O
■ Though the two compounds have the same molecular formula, they have different physical and chemical properties. This is due to the difference in the position of the functional group. So we need a method to give differnt names for such compounds.
■ The method specified by the IUPAC is very similar to the one we used to name branched alkanes. We saw it in an earlier section of this chapter. Details here.
Let us learn the naming procedure by taking the first compound in fig.14.54 as an example:
Step 1:
• Identify the 'longest chain'. It should be considered as the main chain.
♦ The 'longest chain' is the chain with the maximum number of carbon atoms
• In our present case, the longest chain is highlighted in red color in the fig.14.55(a) below:
• Note that, in our present case, there is not much difficulty in finding the 'longest chain containing carbon atoms'.
♦ This is because, there are no 'branches containing carbon atoms'.
• We will see such cases in higher classes. There we will see hydrocarbons in which there are both of the following two items attached to the 'main chain':
(i) Branches with carbon atoms
(ii) Functional groups
• In fact for our present discussion, we will be dealing with those cases in which only functional groups are attached to the main chain
♦ So there will not be any difficulty in identifying the 'main chain'
Step 2:
• Identify the position at which the hydroxyl group is attached.
• For that, we have to 'give numbers' to the carbon atoms in the main chain.
• Two possible ways of numbering are shown in fig 14.55(b) and (c) above. Let us analyse them:
(i) In fig(b), the numbering is done from left to right
(ii) In fig(c), the numbering is done from right to left.
(iii) So there are two possible ways to do the numbering.
♦ But only one is acceptable. Because, the numbering should satisfy a condition:
■ The 'carbon atom carrying the hydroxyl group' should get the lowest possible number
• Let us see how this condition applies to our case:
♦ In fig(b) the 'carbon atom carrying the group gets the number 3
♦ In fig(c) the 'carbon atom carrying the group gets the number 1
• So the numbering done in fig(c) is correct.
♦ Position of the group is 1
• Thus in this step 2, we accomplish two things:
(a) We finalize the number of carbon atoms in the 'main chain'
(b) We identify the position of the group
Step 3:
• From the 'finalized number of carbon atoms in the main chain', write down the word root.
♦ If the finalized number is 1, then word root is meth
♦ If the finalized number is 2, then word root is eth
♦ If the finalized number is 3, then word root is prop
♦ If the finalized number is 4, then word root is but
So on . . .
• The suffix is also written down in this step:
♦ The suffix is ane if all the carbon-carbon bonds in the molecule are single bonds
♦ The suffix is ene if any carbon-carbon double bonds are present in the molecule
♦ The suffix is yne if any carbon-carbon tripple bonds are present in the molecule
• In our present case, 'finalized number of carbon atoms in the main chain' is 3
♦ So the word root is prop
• All the carbon-carbon bonds in the molecule are single bonds
♦ So the suffix is ane
Step 4:
• In this step, we assemble the final name of the alcohol using the information obtained in the above three steps
■ The assembling is done as follows:
alkane minus 'e' +hyphen+position number of group+hyphen+ 'ol'
• In our present case, the IUPAC name is:
propane minus 'e' +hyphen+1+hyphen+ 'ol'
This gives: propan-1-ol.
Now we will name the second compound in fig.14.54 above
Step 1:
• Identify the 'longest chain'. It should be considered as the main chain.
♦ The 'longest chain' is the chain with the maximum number of carbon atoms
• In our present case, the longest chain is highlighted in red color in the fig.14.56(a) below:
Step 2:
• Identify the position at which the hydroxyl group is attached.
• For that, we have to 'give numbers' to the carbon atoms in the main chain.
• Two possible ways of numbering are shown in fig 14.56(b) and (c) above. Let us analyse them:
(i) In fig(b), the numbering is done from left to right
(ii) In fig(c), the numbering is done from right to left.
(iii) So there are two possible ways to do the numbering.
♦ But only one is acceptable. Because, the numbering should satisfy a condition:
■ The 'carbon atom carrying the hydroxyl group' should get the lowest possible number
• Let us see how this condition applies to our case:
♦ In fig(b) the 'carbon atom carrying the group gets the number 2
♦ In fig(c) the 'carbon atom carrying the group gets the number 2
• So the numbering done in both figs,(b)&(c) is correct.
♦ Position of the group is 2
• Thus in this step 2, we accomplish two things:
(a) We finalize the number of carbon atoms in the 'main chain'
(b) We identify the position of the group
Step 3:
• From the 'finalized number of carbon atoms in the main chain', write down the word root.
♦ If the finalized number is 1, then word root is meth
♦ If the finalized number is 2, then word root is eth
♦ If the finalized number is 3, then word root is prop
♦ If the finalized number is 4, then word root is but
So on . . .
• The suffix is also written down in this step:
♦ The suffix is ane if all the carbon-carbon bonds in the molecule are single bonds
♦ The suffix is ene if any carbon-carbon double bonds are present in the molecule
♦ The suffix is yne if any carbon-carbon tripple bonds are present in the molecule
• In our present case, 'finalized number of carbon atoms in the main chain' is 3
♦ So the word root is prop
• All the carbon-carbon bonds in the molecule are single bonds
♦ So the suffix is ane
Step 4:
• In this step, we assemble the final name of the alcohol using the information obtained in the above three steps
■ The assembling is done as follows:
alkane minus 'e' +hyphen+position number of group+hyphen+ 'ol'
• In our present case, the IUPAC name is:
propane minus 'e' +hyphen+2+hyphen+ 'ol'
This gives: propan-2-ol.
Solved example 14.6
Write the IUPAC name of the compound shown in fig.14.57(a) below:
Solution:
The correct method of numbering is shown in fig.14.57(c)
We get the IUPAC name as: pentan-2-ol
The reader may write the steps in his/her own notebooks
Now we will see the reverse process. That is., we are given the IUPAC name of an alcohol. We must write the structural formula. We will learn the method by analysing an example:
1. Given IUPAC name is: Hexan-2-ol
2. Consider the word root. It is hex. So there are 6 carbon atoms
3. Consider the suffix. It is an. So the alcohol is derived from an alkane
• So all the carbon-carbon bonds are single bonds
♦ alkane: ane minus e gives an
♦ alkene: ene minus e gives en
♦ alkyne: yne minus e gives yn
4. So we have 6 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.14.58(a) below:
5. The hydroxyl group is at position 2.
So we attach the ㅡOH 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.14.58(c) above. So this fig.14.58(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. Now we will see how a 'hydroxyl group' that is., 'ㅡOH', gets itself attached to an open chain:
1. We have seen the covalent bonds in water molecule. The bonds are between oxygen atom and two hydrogen atoms. See fig.3.19 for details.
• So any one water molecule will be as shown in fig.14.59 below:
• Note that oxygen has 6 electrons (the 6 cyan dots around the 'O' on the right side in the above fig.) in it's outermost shell. So it needs two more electrons.
• This is achieved by entering into covalent bonds with two hydrogen atoms.
2. So the oxygen atom has attained octet configuration.
♦ In other words, valency of oxygen atom is satisfied
(ii) Any hydrogen atom that we take, has two electrons around it.
• So every hydrogen atom has attained octet configuration.
♦ In other words, valency of hydrogen atoms are satisfied
3. But if we remove one hydrogen atom, the situation will change. The 'new form' of water will be as shown in fig.14.60 below:
• One hydrogen atom has left the water molecule.
• While leaving, it took it's electron also with it.
4. Now the oxygen atom is in need of one electron to fill it's valency.
• So the remaining portion, which is OH as a whole, will change to a 'reactive group of atoms'.
• They are said to be a 'reactive group' because, they tend to enter into reactions and obtain one electron from other sources.
5. Such reactive groups are called radicals.
• The radical formed by the removal of one hydrogen atom from a water molecule is called hydroxyl radical or hydroxyl group. It is represented as: ㅡOH
• Note that, here we saw 'how ㅡOH is formed from a water molecule'. There are other methods also by which ㅡOH is formed. We will see them when we discuss about reactions of organic compounds in the next chapter. But whatever be the method of formation, the structure of ㅡOH will be as shown in fig.14.60 above.
6. Now we will see how this group gets itself attached to the 'straight chain' hydrocarbons.
Consider fig.14.61 below:
• On the left most end, we have a stable 'straight chain' molecule.
8. But a hydroxyl group enters into reaction with this stable molecule.
• As a result, a hydrogen atom is removed from the chain and the group takes it's place.
9. The newly formed molecule is stable because, all the valencies are satisfied as shown in fig.14.62 below:
In fig.a, note the two points:
(i) Any carbon atom that we take has the required 8 electrons around it.
• So every carbon has attained octet configuration.
♦ In other words, valency of carbon atoms are satisfied
(ii) Any hydrogen atom that we take has two electrons around it.
• So every hydrogen atom has attained octet configuration.
♦ In other words, valency of hydrogen atoms are satisfied
(ii) The single oxygen atom has eight electrons around it.
• So it has also attained octet configuration.
♦ In other words, valency of oxygen atom is satisfied
10. The same structure in fig.14.62(a) is drawn again in fig.14.62(b).
• But a portion is specially marked within dotted lines. This is our 'portion of interest'.
• This is where the hydroxyl group got itself attached to the 'straight chain'.
• Thus an 'alcohol' is formed.
In the next section, we will see Aldehyde group.
• We know that hydrocarbons consist of carbon and hydrogen only. But in some hydrocarbons, one or more hydrogen atoms will be removed and another atom will take it's place.
• Let us see some examples:
■ Consider Propane: CH3 ㅡCH2 ㅡCH3
• One hydrogen atom in the end carbon can be replaced by a chlorine atom. Then we get:
CH3 ㅡCH2 ㅡCH3 ㅡ Cl.
♦ This new compound is called Chloropropane
• One hydrogen atom in the end carbon can be replaced by a fluorine atom. Then we get:
CH3 ㅡCH2 ㅡCH3 ㅡ F.
♦ This new compound is called Fluoropropane
• In the above examples, one hydrogen atom was replaced by another 'single atom'
We will see a case when one hydrogen atom is replaced by a 'group of atoms':
■ Consider Methane: CH4
• One hydrogen atom can be replaced by a (OH) group. Then we get:
CH3 ㅡOH.
♦ This new compound is called Methanol
Similarly, one hydrogen in ethane (CH3 ㅡCH3) can be replaced by a (OH) group. Then we get:
CH3 ㅡCH2 ㅡOH.
♦ This new compound is called Ethanol
Another example:
■ Consider Methane: CH4
• One hydrogen atom can be replaced by a (NH3) group. Then we get:
CH3 ㅡNH3.
♦ This new compound is called Methanamine
• Similarly, one hydrogen in ethane (CH3 ㅡCH3) can be replaced by a (NH3) group. Then we get:
CH3 ㅡCH2 ㅡNH3.
♦ This new compound is called Ethanamine
• The chemical and physical properties of propane are quite different from those of chloropropane or fluoropropane
• The chemical and physical properties of methane are quite different from those of methanol or methanamine
• The chemical and physical properties of ethane are quite different from those of ethanol or ethanamine
■The presence of certain atoms or groups imparts certain characteristic properties to compounds. They are called functional groups.
We will now discuss about the various functional groups and their nomenclature
Hydroxyl Group (ㅡOH)
• The IUPAC names of compounds with hydroxyl function group end in 'ol'. Such compounds are called alcohols.
♦ In any alcohol, there will be one or more ㅡOH group
■ The naming of alcohols is done by the following procedure:
• Remove the 'e' from the name of the corresponding alkane
♦ Put 'ol' in it's place
• So we can write:
Alkane - e + ol → Alkanol
• Thus we get:
♦ Methane - e + ol → Methanol
♦ Ethane - e + ol → Ethanol
♦ Propane - e + ol → Propanol
so on . . .
But this is not the only IUPAC rule related to the nomenclature of alcohols. Let us analyse more:
Consider the two compounds given in fig.14.54 below:
 |
| Fig.14.54 |
♦ So it's molecular formula is: C3H8O
• In the second compound, there are 3 carbon atoms, 8 hydrogen atoms and 1 oxygen atom
♦ So it's molecular formula is also: C3H8O
■ Though the two compounds have the same molecular formula, they have different physical and chemical properties. This is due to the difference in the position of the functional group. So we need a method to give differnt names for such compounds.
■ The method specified by the IUPAC is very similar to the one we used to name branched alkanes. We saw it in an earlier section of this chapter. Details here.
Let us learn the naming procedure by taking the first compound in fig.14.54 as an example:
Step 1:
• Identify the 'longest chain'. It should be considered as the main chain.
♦ The 'longest chain' is the chain with the maximum number of carbon atoms
• In our present case, the longest chain is highlighted in red color in the fig.14.55(a) below:
 |
| Fig.14.55 |
♦ This is because, there are no 'branches containing carbon atoms'.
• We will see such cases in higher classes. There we will see hydrocarbons in which there are both of the following two items attached to the 'main chain':
(i) Branches with carbon atoms
(ii) Functional groups
• In fact for our present discussion, we will be dealing with those cases in which only functional groups are attached to the main chain
♦ So there will not be any difficulty in identifying the 'main chain'
Step 2:
• Identify the position at which the hydroxyl group is attached.
• For that, we have to 'give numbers' to the carbon atoms in the main chain.
• Two possible ways of numbering are shown in fig 14.55(b) and (c) above. Let us analyse them:
(i) In fig(b), the numbering is done from left to right
(ii) In fig(c), the numbering is done from right to left.
(iii) So there are two possible ways to do the numbering.
♦ But only one is acceptable. Because, the numbering should satisfy a condition:
■ The 'carbon atom carrying the hydroxyl group' should get the lowest possible number
• Let us see how this condition applies to our case:
♦ In fig(b) the 'carbon atom carrying the group gets the number 3
♦ In fig(c) the 'carbon atom carrying the group gets the number 1
• So the numbering done in fig(c) is correct.
♦ Position of the group is 1
• Thus in this step 2, we accomplish two things:
(a) We finalize the number of carbon atoms in the 'main chain'
(b) We identify the position of the group
Step 3:
• From the 'finalized number of carbon atoms in the main chain', write down the word root.
♦ If the finalized number is 1, then word root is meth
♦ If the finalized number is 2, then word root is eth
♦ If the finalized number is 3, then word root is prop
♦ If the finalized number is 4, then word root is but
So on . . .
• The suffix is also written down in this step:
♦ The suffix is ane if all the carbon-carbon bonds in the molecule are single bonds
♦ The suffix is ene if any carbon-carbon double bonds are present in the molecule
♦ The suffix is yne if any carbon-carbon tripple bonds are present in the molecule
• In our present case, 'finalized number of carbon atoms in the main chain' is 3
♦ So the word root is prop
• All the carbon-carbon bonds in the molecule are single bonds
♦ So the suffix is ane
Step 4:
• In this step, we assemble the final name of the alcohol using the information obtained in the above three steps
■ The assembling is done as follows:
alkane minus 'e' +hyphen+position number of group+hyphen+ 'ol'
• In our present case, the IUPAC name is:
propane minus 'e' +hyphen+1+hyphen+ 'ol'
This gives: propan-1-ol.
Now we will name the second compound in fig.14.54 above
Step 1:
• Identify the 'longest chain'. It should be considered as the main chain.
♦ The 'longest chain' is the chain with the maximum number of carbon atoms
• In our present case, the longest chain is highlighted in red color in the fig.14.56(a) below:
 |
| Fig.14.56 |
• Identify the position at which the hydroxyl group is attached.
• For that, we have to 'give numbers' to the carbon atoms in the main chain.
• Two possible ways of numbering are shown in fig 14.56(b) and (c) above. Let us analyse them:
(i) In fig(b), the numbering is done from left to right
(ii) In fig(c), the numbering is done from right to left.
(iii) So there are two possible ways to do the numbering.
♦ But only one is acceptable. Because, the numbering should satisfy a condition:
■ The 'carbon atom carrying the hydroxyl group' should get the lowest possible number
• Let us see how this condition applies to our case:
♦ In fig(b) the 'carbon atom carrying the group gets the number 2
♦ In fig(c) the 'carbon atom carrying the group gets the number 2
• So the numbering done in both figs,(b)&(c) is correct.
♦ Position of the group is 2
• Thus in this step 2, we accomplish two things:
(a) We finalize the number of carbon atoms in the 'main chain'
(b) We identify the position of the group
Step 3:
• From the 'finalized number of carbon atoms in the main chain', write down the word root.
♦ If the finalized number is 1, then word root is meth
♦ If the finalized number is 2, then word root is eth
♦ If the finalized number is 3, then word root is prop
♦ If the finalized number is 4, then word root is but
So on . . .
• The suffix is also written down in this step:
♦ The suffix is ane if all the carbon-carbon bonds in the molecule are single bonds
♦ The suffix is ene if any carbon-carbon double bonds are present in the molecule
♦ The suffix is yne if any carbon-carbon tripple bonds are present in the molecule
• In our present case, 'finalized number of carbon atoms in the main chain' is 3
♦ So the word root is prop
• All the carbon-carbon bonds in the molecule are single bonds
♦ So the suffix is ane
Step 4:
• In this step, we assemble the final name of the alcohol using the information obtained in the above three steps
■ The assembling is done as follows:
alkane minus 'e' +hyphen+position number of group+hyphen+ 'ol'
• In our present case, the IUPAC name is:
propane minus 'e' +hyphen+2+hyphen+ 'ol'
This gives: propan-2-ol.
Solved example 14.6
Write the IUPAC name of the compound shown in fig.14.57(a) below:
 |
| Fig.14.57 |
The correct method of numbering is shown in fig.14.57(c)
We get the IUPAC name as: pentan-2-ol
The reader may write the steps in his/her own notebooks
Now we will see the reverse process. That is., we are given the IUPAC name of an alcohol. We must write the structural formula. We will learn the method by analysing an example:
1. Given IUPAC name is: Hexan-2-ol
2. Consider the word root. It is hex. So there are 6 carbon atoms
3. Consider the suffix. It is an. So the alcohol is derived from an alkane
• So all the carbon-carbon bonds are single bonds
♦ alkane: ane minus e gives an
♦ alkene: ene minus e gives en
♦ alkyne: yne minus e gives yn
4. So we have 6 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.14.58(a) below:
 |
| Fig.14.58 |
So we attach the ㅡOH 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.14.58(c) above. So this fig.14.58(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. Now we will see how a 'hydroxyl group' that is., 'ㅡOH', gets itself attached to an open chain:
1. We have seen the covalent bonds in water molecule. The bonds are between oxygen atom and two hydrogen atoms. See fig.3.19 for details.
• So any one water molecule will be as shown in fig.14.59 below:
 |
| Fig.14.59 |
• This is achieved by entering into covalent bonds with two hydrogen atoms.
2. So the oxygen atom has attained octet configuration.
♦ In other words, valency of oxygen atom is satisfied
(ii) Any hydrogen atom that we take, has two electrons around it.
• So every hydrogen atom has attained octet configuration.
♦ In other words, valency of hydrogen atoms are satisfied
3. But if we remove one hydrogen atom, the situation will change. The 'new form' of water will be as shown in fig.14.60 below:
 |
| Fig.14.60 |
• While leaving, it took it's electron also with it.
4. Now the oxygen atom is in need of one electron to fill it's valency.
• So the remaining portion, which is OH as a whole, will change to a 'reactive group of atoms'.
• They are said to be a 'reactive group' because, they tend to enter into reactions and obtain one electron from other sources.
5. Such reactive groups are called radicals.
• The radical formed by the removal of one hydrogen atom from a water molecule is called hydroxyl radical or hydroxyl group. It is represented as: ㅡOH
• Note that, here we saw 'how ㅡOH is formed from a water molecule'. There are other methods also by which ㅡOH is formed. We will see them when we discuss about reactions of organic compounds in the next chapter. But whatever be the method of formation, the structure of ㅡOH will be as shown in fig.14.60 above.
6. Now we will see how this group gets itself attached to the 'straight chain' hydrocarbons.
Consider fig.14.61 below:
 |
| Fig.14.61 |
8. But a hydroxyl group enters into reaction with this stable molecule.
• As a result, a hydrogen atom is removed from the chain and the group takes it's place.
9. The newly formed molecule is stable because, all the valencies are satisfied as shown in fig.14.62 below:
 |
| Fig.14.62 |
(i) Any carbon atom that we take has the required 8 electrons around it.
• So every carbon has attained octet configuration.
♦ In other words, valency of carbon atoms are satisfied
(ii) Any hydrogen atom that we take has two electrons around it.
• So every hydrogen atom has attained octet configuration.
♦ In other words, valency of hydrogen atoms are satisfied
(ii) The single oxygen atom has eight electrons around it.
• So it has also attained octet configuration.
♦ In other words, valency of oxygen atom is satisfied
10. The same structure in fig.14.62(a) is drawn again in fig.14.62(b).
• But a portion is specially marked within dotted lines. This is our 'portion of interest'.
• This is where the hydroxyl group got itself attached to the 'straight chain'.
• Thus an 'alcohol' is formed.
In the next section, we will see Aldehyde group.
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