Chapter 12.7 - Nomenclature When More Than One Functional Group Of The Same Type Are Present

In the previous section, we saw how the compounds with functional groups are named. In this section, we will see the nomenclature when more than one functional group of the same type are present.

Nomenclature of organic compounds having more than one functional group of the same type

We have seen that, when more than one branch of the same type is present, we use di, tri, tetra etc., We can apply the same method here also. Let us see some examples. We will learn some new rules also while analyzing those examples.

Example 1
Write the IUPAC name of CH₂(OH)CH₂(OH)
Solution:
1. We are given the condensed formula. Let us draw the complete structural formula. It is shown in fig.12.45(a) below:

Fig.12.45

2. We want to write the IUPAC name of the molecule in fig.12.45(a). It can be done in 4 steps:
(i) First we make a list of the functional groups in the given molecule. We get: ㅡOH and ㅡOH
All functional groups are of the same type. So there is no need to assign any priority
(ii) The IUPAC name of the molecule will be based on the ㅡOH group.
    ♦ When the component obtained from ㅡOH is the suffix, we use ‘-ol’  
    ♦ When the component obtained from ㅡOH is the prefix, we use ‘hydroxy-’
• Here, suffix is to be used because, the name is going to be based on ㅡOH
• So the main component of the final name will be: ethanol
(iii) Now we have to show that there are two ㅡOH groups.
So we write: ethan-diol
(iv) Also, we have to show the positions. So we write: ethan-1,2-diol
This is the IUPAC name.
(Note that in this example, the numbering from either sides will give the same result)

Example 2
Write the IUPAC name of CH₂=CH-CH=CH₂
Solution:
1. We are already given the structural formula. But we have to number the C atoms properly. This is shown in fig.12.45(b) above.
2. We want to write the IUPAC name of the molecule in fig.12.45(b). It can be done in 4 steps:
(i) First we make a list of the functional groups in the given molecule. We get: C=C and C=C
All functional groups are of the same type. So there is no need to assign any priority
(ii) The IUPAC name of the molecule will be based on the C=C group.
• We know that, for this double bond, the 'ane' changes to 'ene'
• So the main component of the final name will be: butene
(iii) Now we have to show that there are two C=C groups.
So we write: but-diene
(iv) Also, we have to show the positions. So we write: buta-1,3-diene
This is the IUPAC name.
(The double bonds should get the lowest number possible. In this example, the numbering from either sides will give the same result)

Example 3
Write the IUPAC name of the compound shown in fig.12.46(a) below:

Fig.12.46

Solution:
• We want to write the IUPAC name of the molecule in fig.12.46(a). It can be done in 4 steps:
(i) First we make a list of the functional groups in the given molecule. We get: C=C There is only one functional group. So there is no need to assign any priority
(ii) The IUPAC name of the molecule will be based on the C=C group.
We know that, for this double bond, the 'ane' changes to 'ene'
• So the main component of the final name will be: hexene
• The numbering is done so as to include the double bond. If we do the number in an exact horizontal manner, we will get a longer chain. But such a numbering is not allowed. It is compulsory to select that chain which contains the double bond.
(iii) Now we have to show the position of the double bond. We get: hex-1-ene.
(iv) Also, we have to include the ethyl group at position 2.
So we write: 2-Ethylhex-1-ene
This is the IUPAC name.

Example 4
Write the IUPAC name of the compound shown in fig.12.46(b) above.
Solution:
• We want to write the IUPAC name of the molecule in fig.12.46(b). It can be done in 5 steps:
(i) First we make a list of the functional groups in the given molecule. We get: C=C and C=C. There is only one type of functional group. So there is no need to assign any priority.
(ii) The IUPAC name of the molecule will be based on the C=C group.
We know that, for this double bond, the 'ane' changes to 'ene'
• So the main component of the final name will be: hexene
• The numbering is done so as to include both the double bonds. If we do the number in an exact horizontal manner, we will get a longer chain. But such a numbering is not allowed. It is compulsory to select that chain which contains all the double bonds.
(iii) Now we have to show that there are two C=C groups.
So we write: hexa-diene
(iv) Also, we have to show the positions. So we write: hexa-1,4-diene
(v) Now we have to include the ethyl group at position 2.
So we write: 2-Ethylhexa-1,4-diene
This is the IUPAC name.

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• Examples 3 and 4 are similar except that, there is an extra double bond in example 4.
• The names are also similar:
   ♦ 2-Ethylhex-1-ene
   ♦ 2-Ethylhexa-1,4-diene
• Note that:
   ♦ When there is only one double bond, we write 'hex'
   ♦ When there is more than one double bond, we write 'hexa'
• Another example is: 'pent' and 'penta'

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Now we will see some solved examples.

Solved example 12.10
Write the IUPAC names of the compounds shown in figs.12.47 (a) and (b) below:

Fig.12.47

Solution:
Part (a):
• We want to write the IUPAC name of the molecule in fig.12.47(a). It can be done in 4 steps:
(i) First we make a list of the functional groups in the given molecule. We get: >C=O and >C=O
All functional groups are of the same type. So there is no need to assign any priority
(ii) The IUPAC name of the molecule will be based on the >C=O group.
    ♦ When the component obtained from >C=O is the suffix, we use ‘-one’  
    ♦ When the component obtained from >C=O is the prefix, we use ‘oxo-’
• Here, suffix is to be used because, the name is going to be based on >C=O
• So the main component of the final name will be: hexanone
(iii) Now we have to show that there are two >C=O groups.
So we write: hexane-dione
(iv) Also, we have to show the positions. So we write: Hexane-2,4-dione
This is the IUPAC name.
(Note that in this example, the numbering should be done from left to right. This is to give the functional groups, the lowest possible numbers) 

Part (b):
• We want to write the IUPAC name of the molecule in fig.12.47(b). It can be done in 7 steps:
(i) First we make a list of the functional groups in the given molecule. We get: C=C and C☰C
(ii) Next we compare the list with table 1. We see that, C=C gets top priority. So it is the principal functional group. The C☰C is the subordinate functional group.
(iii) Imagine that, the C☰C is not present. Then the IUPAC name of the molecule will be based on the C=C group.
• So the main component of the final name will be: hexene
(iii) Now we have to show that there are two C=C groups.
So we write: hexa-diene
(iv) Also, we have to show the positions. So we write: Hexa-1,3-diene
(v) Now we have to account for the C☰C group.
So we include 'yne' in the name.
(vi) The C☰C group is at position 5. So we write: Hexa-1,3-diene-5-yne
This is the IUPAC name.
(vii) Priority between double and triple bonds:
• Alphabetically, 'ene' comes before 'yne'. So 'ene' should be written first.
• While numbering, the lowest possible number should be given to the double bonds.

Solved example 12.11
Derive the structure of (i) 2-Chlorohexane, (ii) Pent-4-en-2-ol, (iii) 3-Nitrocyclohexene, (iv) cyclohex-2-en-1-ol, (v) 6-Hydroxy-heptanal.
Solution:
Part (i): 2-Chlorohexane
1. The suffix 'ane' indicates that the compound contains only single bonds. It is an alkane.
• 'hex' indicates that, there are six C atoms.
• So we can draw the skeletal structure as shown in fig.12.48(a) below:

Fig.12.48

2. 'Chloro' indicates the functional group: ㅡCl
• This is the only one functional group. The numbering must be in such a way that, the ㅡCl gets lowest number.
• So '2-Chloro' indicates that, the ㅡCl is at the second position.
• We can add this information to fig.a. We get the modified fig.b
3. Now we fill up the valencies of all C atoms. We get the final structure in fig.c

Part (ii): Pent-4-en-2-ol
1. The suffix '-ol' indicates that one ㅡOH group is present. It is an alcohol.
• It is the suffix. So it is the principal group.
• The numbering must be in such a way that, the ㅡOH gets lowest number
• So '-2-ol' indicates that, the ㅡOH is at the second position.
2. 'Pent' indicates that, there are five C atoms. So we can draw the skeletal structure as shown in fig.12.49(a) below:

Fig.12.49

3. 'en' indicates that there is one double bond.
'-4-en' indicates that, the double bond is between positions 4 and 5
We can add this information to fig.a. We get the modified fig.b
4. Now we fill up the valencies of all C atoms. We get the final structure in fig.c

Part (iii): 3-Nitrocyclohexene
1. The suffix 'ene' indicates that the compound contains one double bond. It is an alkene.
• 'hex' indicates that, there are six C atoms.
• 'cyclo' indicates that, it is a cyclic alkene.
• So we can draw the skeletal structure as shown in fig.12.50(a) below:

Fig.12.50

• Note that, the numbering is done in such a way that, the double bond gets the lowest numbers 1 and 2
2. 'Nitro' indicates that there is one ㅡNO₂ functional group.
• '3-Nitro' indicates that the ㅡNO₂ group is at position 3
• So we get the final structure shown in fig.b
3. From the tables-1 and 2 of the previous section, we know that, the double bond gets priority over the ㅡNO₂ group. We see that, ㅡNO₂ which is in table-2, will always be a subordinate group. That is why, 'ene' is the suffix.

Part (iv): cyclohex-2-en-1-ol
1. The suffix '-ol' indicates that one ㅡOH group is present. It is an alcohol.
• It is the suffix. So it is the principal group.
• The numbering must be in such a way that, the ㅡOH gets lowest number
• So '-1-ol' indicates that, the ㅡOH is at the first position
2. 'Cyclohex' indicates that, it is a cyclic structure with six C atoms.
3. 'en' indicates that, it contains a double bond.
'2-en' indicates that, the double bond is between 2 and 3.
4. Based on the above information, the structure will be as shown in fig.12.50(c) above.
5. From the table-1 of the previous section, we know that, the ㅡOH group gets priority over the double bond. That is why, it is given the lowest number.

Part (v): 6-Hydroxy-heptanal
1. The suffix 'al' indicates that, it is an aldehyde. It has a ㅡCHO group.
• This group is always at the end of the chain. So it do not require any number. (Details here)
2. 'hept' indicates that there are seven C atoms in the chain.
• The 'an' in 'heptan' indicates that, it is an alkane. All bonds are single bonds.
• So the skeletal structure will be as shown in fig.12.51(a) below:

Fig.12.51

• Note that, the seven C atoms include the C atom coming from the ㅡCHO group.   
3. '6-Hydroxy' indicates that, there is a ㅡOH group at the sixth C atom.
• So the skeletal structure can be modified as shown in fig.12.51(b) above.
4. Now we fill up the valencies of all C atoms. We get the final structure in fig.c

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In the next section, we will see nomenclature of substituted Benzene compounds.

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