In the previous section, we saw the nomenclature of organic compounds carrying the keto group. In this section, we will see the carboxylic acid group.
• We have seen that the hydroxyl group is represented as 'ㅡOH'. (see details)
• We have seen that the aldehyde group can be represented in two ways (see fig.14.63)
• We have seen that the keto group can be represented in two ways (see fig.14.67 (a) & (b))
• Now we will see that the carboxylic acid group can also be represented in two ways:
• We can use either (a) or (b) in the above fig.14.72 to represent the carboxylic acid group
■ Note that:
• In the hydroxyl group (ㅡOH), the bond which connects the group to a main chain starts from the oxygen atom
• In the aldehyde group (ㅡCHO), the bond which connects the group to a main chain starts from the carbon atom
• In our present carboxylic acid group (ㅡCOOH), the bond which connects the group to a main chain starts from the carbon atom
• Also there is a double bond within the group. We will see all the bonding details at the end of this section. At present, we will see the nomenclature.
• The IUPAC names of compounds with ㅡCOOH group end in oic acid.
• Such compounds are called carboxylic acids.
• In any carboxylic acid, there will be a ㅡCOOH group
■ The naming of carboxylic acids is done by the following 2 steps:
1. Remove the 'e' from the name of the corresponding alkane
2. Put 'oic acid' in it's place
• So we can write:
Alkane - e + oic acid → Alkanoic acid
• Thus we get:
♦ Methane - e + oic acid → Methanoic acid
♦ ethane - e + oic acid → ethanoic acid
♦ Propane - e + oic acid → propanoic acid
so on . . .
■ But there are some important points to note:
• The 'newly attaching carboxylic acid group' has one C atom of it's own.
• So when it is attached, the total number of carbon atoms will increase by one.
• This 'total number' should be considered while naming the compound.
• This is a general rule for any functional group which carries a C atom of it's own. We applied it earlier for aldehyde group and keto group
We can write:
■ If the functional group contains a carbon atom that carbon atom should be treated as part of the main chain.
■ Then we will be wondering about methanoic acid.
• Because, One hydrogen atom is removed from Methane and the ㅡCOOH group takes it's place to give the carboxylic acid.
• But when the group takes it's place, the total number of C atoms will become 2.
♦ So it should be named as Ethanoic acid.
• That means, the compound methanoic acid cannot exist.
• However, methanoic acid does exist. It's structure is shown in fig.14.73(a) below:
■ So how is methanoic acid formed?
• We know that, the ㅡCOOH group will be looking for an electron.
♦ This electron is supplied by a hydrogen atom.
• Thus we get the methanoic acid shown in fig.14.73(a) above.
♦ The portion shown within the magenta polygon is the carboxylic acid group.
■ How is the compound in fig(b) formed?
• It was originally CH4. One of it's H atoms was removed and the ㅡCOOH group took it's place.
♦ Now there are a total of two C atoms, and so it is ethanoic acid
■ How is the compound in fig(c) formed?
• It was originally CH3ㅡCH3. One of it's H atoms was removed and the ㅡCOOH group took it's place.
♦ Now there are a total of three C atoms, and so it is propanoic acid
■ The compounds in fig.73 above can be represented in another way also:
• Methanoic acid: HㅡCOOH
• Ethanoic acid: CH3ㅡCOOH
• Propanaoic acid: CH3ㅡCH2ㅡCOOH
Consider the fig.14.54 that we saw in the previous section. It is shown again below:
• In both compounds, there are 3 carbon atoms, 8 hydrogen atoms and 1 oxygen atom.
♦In fig(a), the hydroxyl group is at the end position.
♦ But in fig(b), it is at another position.
• Accordingly we gave two distinct names: Propan-1-ol and propan-2-ol
■ But in our present discussion on carboxylic acid group, we do not have to learn about such change in positions.
• That is., In all the carboxylic acids that we see in the present discussion, the ㅡCOOH group will be at the end (or at the beginning).
♦ So we will not see names like Propan-1-oic acid, propan-2-oic acid etc., in this discussion.
Solved example 14.10
Write the IUPAC name of the compound given below:
CH3ㅡCH2ㅡCH2ㅡCH2ㅡCOOH
Solution:
• As mentioned above, in the present discussion on aldehydes, we will not encounter those carboxylic acids where the ㅡCOOH group takes different positions.
• We will see those carboxylic acids only where the group is at the end (or at the beginning).
♦ So we do not need to give position numbers.
We will write the steps:
1. In the given compound, there are 5 carbon atoms. So the corresponding alkane is Pentane
• Remove the 'e' at the end.
2. Put 'oic acid' in it's place. We get:
Pentane - e + oic acid = Pentanoic acid
Now we will see the reverse process. That is., we are given the IUPAC name of a carboxylic acid. We must write the structural formula. We will learn the method by analysing an example:
1. Given IUPAC name is: Propanoic acid
2. Consider the word root. It is prop. So there are 3 carbon atoms
3. Consider the suffix. It is an. So the aldehyde 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 3 carbon atoms with single bonds between them.
• Note that this 'total 3 numbers' is including the C atom in the ㅡCOOH group.
• Also, we do not need to write the numbering. Because the group will be at the end (or the beginning).
• So we can write:
C ㅡC ㅡCOOH
5. Fill all the valencies of carbon atoms.
• We use hydrogen to fill up the valencies.
• The result is: CH3ㅡCH2ㅡCOOH
■ 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 the just previous section we saw how a 'keto group' that is., 'ㅡCO', gets itself attached to an open chain. See details here.
■ Now we will see the same for the carboxylic acid group:
• We have seen the structure of ㅡCHO group (keto group) in a previous section. It is shown again in fig.14.74 (a) below:
• From fig(a), it is clear that the group is in need for an electron.
♦ This need is indicated by the absence of a dot at the end of the bond line attached to the C atom.
• Now consider fig.14.74(b). It shows the bonding details in the ㅡCOOH group
♦ All the valency needs of top oxygen atom is satisfied (6 cyan dots plus two green dots, giving a total of 8 dots.
♦ Those of bottom oxygen atom is satisfied (6 cyan dots plus one green plus one red dot, giving a total of 8 dots.
♦ Those of hydrogen is also satisfied (1 red dot plus 1 cyan dot, giving a total of 2 dots
♦ Those of carbon is not fully satisfied. (4 green plus 3 cyan give a total of only 7)
• The need for one electron is indicated by the absence of a dot at the end of the bond line attached to the C atom.
♦ Thus the group as a whole, is in need for one electron.
• This need will be satisfied when the group gets itself attached to a hydrocarbon chain as shown in fig.14.75 below:
• Thus we get a stable molecule
In the next section, we will see the Halo group.
• We have seen that the hydroxyl group is represented as 'ㅡOH'. (see details)
• We have seen that the aldehyde group can be represented in two ways (see fig.14.63)
• We have seen that the keto group can be represented in two ways (see fig.14.67 (a) & (b))
• Now we will see that the carboxylic acid group can also be represented in two ways:
 |
| Fig.14.72 |
■ Note that:
• In the hydroxyl group (ㅡOH), the bond which connects the group to a main chain starts from the oxygen atom
• In the aldehyde group (ㅡCHO), the bond which connects the group to a main chain starts from the carbon atom
• In our present carboxylic acid group (ㅡCOOH), the bond which connects the group to a main chain starts from the carbon atom
• Also there is a double bond within the group. We will see all the bonding details at the end of this section. At present, we will see the nomenclature.
• The IUPAC names of compounds with ㅡCOOH group end in oic acid.
• Such compounds are called carboxylic acids.
• In any carboxylic acid, there will be a ㅡCOOH group
■ The naming of carboxylic acids is done by the following 2 steps:
1. Remove the 'e' from the name of the corresponding alkane
2. Put 'oic acid' in it's place
• So we can write:
Alkane - e + oic acid → Alkanoic acid
• Thus we get:
♦ Methane - e + oic acid → Methanoic acid
♦ ethane - e + oic acid → ethanoic acid
♦ Propane - e + oic acid → propanoic acid
so on . . .
■ But there are some important points to note:
• The 'newly attaching carboxylic acid group' has one C atom of it's own.
• So when it is attached, the total number of carbon atoms will increase by one.
• This 'total number' should be considered while naming the compound.
• This is a general rule for any functional group which carries a C atom of it's own. We applied it earlier for aldehyde group and keto group
We can write:
■ If the functional group contains a carbon atom that carbon atom should be treated as part of the main chain.
■ Then we will be wondering about methanoic acid.
• Because, One hydrogen atom is removed from Methane and the ㅡCOOH group takes it's place to give the carboxylic acid.
• But when the group takes it's place, the total number of C atoms will become 2.
♦ So it should be named as Ethanoic acid.
• That means, the compound methanoic acid cannot exist.
• However, methanoic acid does exist. It's structure is shown in fig.14.73(a) below:
 |
| Fig.14.73 |
• We know that, the ㅡCOOH group will be looking for an electron.
♦ This electron is supplied by a hydrogen atom.
• Thus we get the methanoic acid shown in fig.14.73(a) above.
♦ The portion shown within the magenta polygon is the carboxylic acid group.
■ How is the compound in fig(b) formed?
• It was originally CH4. One of it's H atoms was removed and the ㅡCOOH group took it's place.
♦ Now there are a total of two C atoms, and so it is ethanoic acid
■ How is the compound in fig(c) formed?
• It was originally CH3ㅡCH3. One of it's H atoms was removed and the ㅡCOOH group took it's place.
♦ Now there are a total of three C atoms, and so it is propanoic acid
■ The compounds in fig.73 above can be represented in another way also:
• Methanoic acid: HㅡCOOH
• Ethanoic acid: CH3ㅡCOOH
• Propanaoic acid: CH3ㅡCH2ㅡCOOH
Consider the fig.14.54 that we saw in the previous section. It is shown again below:
 |
| Fig.14.54 |
♦In fig(a), the hydroxyl group is at the end position.
♦ But in fig(b), it is at another position.
• Accordingly we gave two distinct names: Propan-1-ol and propan-2-ol
■ But in our present discussion on carboxylic acid group, we do not have to learn about such change in positions.
• That is., In all the carboxylic acids that we see in the present discussion, the ㅡCOOH group will be at the end (or at the beginning).
♦ So we will not see names like Propan-1-oic acid, propan-2-oic acid etc., in this discussion.
Solved example 14.10
Write the IUPAC name of the compound given below:
CH3ㅡCH2ㅡCH2ㅡCH2ㅡCOOH
Solution:
• As mentioned above, in the present discussion on aldehydes, we will not encounter those carboxylic acids where the ㅡCOOH group takes different positions.
• We will see those carboxylic acids only where the group is at the end (or at the beginning).
♦ So we do not need to give position numbers.
We will write the steps:
1. In the given compound, there are 5 carbon atoms. So the corresponding alkane is Pentane
• Remove the 'e' at the end.
2. Put 'oic acid' in it's place. We get:
Pentane - e + oic acid = Pentanoic acid
Now we will see the reverse process. That is., we are given the IUPAC name of a carboxylic acid. We must write the structural formula. We will learn the method by analysing an example:
1. Given IUPAC name is: Propanoic acid
2. Consider the word root. It is prop. So there are 3 carbon atoms
3. Consider the suffix. It is an. So the aldehyde 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 3 carbon atoms with single bonds between them.
• Note that this 'total 3 numbers' is including the C atom in the ㅡCOOH group.
• Also, we do not need to write the numbering. Because the group will be at the end (or the beginning).
• So we can write:
C ㅡC ㅡCOOH
5. Fill all the valencies of carbon atoms.
• We use hydrogen to fill up the valencies.
• The result is: CH3ㅡCH2ㅡCOOH
■ 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 the just previous section we saw how a 'keto group' that is., 'ㅡCO', gets itself attached to an open chain. See details here.
■ Now we will see the same for the carboxylic acid group:
• We have seen the structure of ㅡCHO group (keto group) in a previous section. It is shown again in fig.14.74 (a) below:
 |
| Fig.14.74 |
♦ This need is indicated by the absence of a dot at the end of the bond line attached to the C atom.
• Now consider fig.14.74(b). It shows the bonding details in the ㅡCOOH group
♦ All the valency needs of top oxygen atom is satisfied (6 cyan dots plus two green dots, giving a total of 8 dots.
♦ Those of bottom oxygen atom is satisfied (6 cyan dots plus one green plus one red dot, giving a total of 8 dots.
♦ Those of hydrogen is also satisfied (1 red dot plus 1 cyan dot, giving a total of 2 dots
♦ Those of carbon is not fully satisfied. (4 green plus 3 cyan give a total of only 7)
• The need for one electron is indicated by the absence of a dot at the end of the bond line attached to the C atom.
♦ Thus the group as a whole, is in need for one electron.
• This need will be satisfied when the group gets itself attached to a hydrocarbon chain as shown in fig.14.75 below:
 |
| Fig.14.75 |
In the next section, we will see the Halo group.
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