In the previous section, we saw the nomenclature of organic compounds carrying the hydroxyl group. In this section, we will see the aldehyde group.
• We have seen that the hydroxyl group is represented as 'ㅡOH'. But the aldehyde group can be represented in two ways:
• We can use either (a) or (b) in the above fig.14.63 to represent the aldehyde 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
• 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 aldehyde group end in al.
• Such compounds are called aldehydes.
• In any aldehyde, there will be a ㅡCHO group
■ The naming of aldehydes is done by the following 2 steps:
1. Remove the 'e' from the name of the corresponding alkane
2. Put 'al' in it's place
• So we can write:
Alkane - e + al → Alkanal
• Thus we get:
♦ Methane - e + al → Methanal
♦ ethane - e + al → ethanal
♦ Propane - e + al → propanal
so on . . .
■ But there are some important points to note:
• The 'newly attaching aldehyde 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 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 methanal.
• Because, One hydrogen atom is removed from Methane and the ㅡCHO group takes it's place to give the aldehyde.
• But when the group takes it's place, the total number of C atoms will become 2.
♦ So it should be named as Ethanal.
• That means, the compound methanal cannot exist.
• However, methanal does exist. It's structure is shown in fig.14.64(a) below.
■ So how is methanal formed?
• We know that, the ㅡCHO group will be looking for an electron.
♦ This electron is supplied by a hydrogen atom.
• Thus we get the methanal shown in fig.14.64(a) above.
♦ The portion shown within the magenta lines is the aldehyde group.
■ How is the compound in fig(b) formed?
• It was originally CH4. One of it's H atoms was removed and the ㅡCHO group took it's place.
♦ Now there are a total of two C atoms, and so it is ethanal
■ How is the compound in fig(c) formed?
• It was originally CH3ㅡCH3. One of it's H atoms was removed and the ㅡCHO group took it's place.
♦ Now there are a total of three C atoms, and so it is propanal
■ The compounds in fig.64 above can be represented in another way also:
• Methanal: HㅡCHO
• Ethanal: CH3ㅡCHO
• Propanal: CH3ㅡCH2ㅡCHO
• Butanal: CH3ㅡCH2ㅡCH2ㅡCHO
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 aldehyde group, we do not have to learn about such change in positions.
• That is., In all the aldehydes that we see in the present discussion, the ㅡCHO group will be at the end (or at the beginning).
♦ So we will not see names like Propan-1-al, propan-2-al etc., in this discussion.
Solved example 14.7
Write the IUPAC name of the compound given below:
CH3ㅡCH2ㅡCH2ㅡCH2ㅡCHO
Solution:
• As mentioned above, in the present discussion on aldehydes, we will not encounter those aldehydes where the ㅡCHO group takes different positions.
• We will see those aldehydes 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 'al' in it's place. We get:
Pentane - e + al = Pentanal
Now we will see the reverse process. That is., we are given the IUPAC name of an aldehyde. We must write the structural formula. We will learn the method by analysing an example:
1. Given IUPAC name is: Propanal
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 ㅡCHO 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 ㅡCHO
5. Fill all the valencies of carbon atoms.
• We use hydrogen to fill up the valencies.
• The result is: CH3ㅡCH2ㅡCHO
■ In the first section in this chapter, we saw how the 'alkyl radical' gets itself attached to an open chain. See details here.
■ In the previous section we saw how a 'hydroxyl group' that is., 'ㅡOH', gets itself attached to an open chain. See details here.
■ Now we will see the same for the aldehyde group:
• We have seen the structure of ㅡOH group in the previous section. It is shown again in fig.14.65 (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 O atom.
• Now consider fig.14.65(b). It shows the bonding details in the ㅡCHO group
♦ All the valency needs of oxygen atoms is satisfied (6 cyan dots plus two green dots, giving a total of 8 dots.
♦ Those of hydrogen is also satisfied (1 red dot plus 1 green dot, giving a total of 2 dots
♦ Those of carbon is not fully satisfied. (4 green plus 2 cyan plus 1 red dots 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.14.66 below:
• Thus we get a stable molecule
In the next section, we will see the Ketone group.
• We have seen that the hydroxyl group is represented as 'ㅡOH'. But the aldehyde group can be represented in two ways:
 |
| Fig.14.63 |
■ 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
• 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 aldehyde group end in al.
• Such compounds are called aldehydes.
• In any aldehyde, there will be a ㅡCHO group
■ The naming of aldehydes is done by the following 2 steps:
1. Remove the 'e' from the name of the corresponding alkane
2. Put 'al' in it's place
• So we can write:
Alkane - e + al → Alkanal
• Thus we get:
♦ Methane - e + al → Methanal
♦ ethane - e + al → ethanal
♦ Propane - e + al → propanal
so on . . .
■ But there are some important points to note:
• The 'newly attaching aldehyde 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 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 methanal.
• Because, One hydrogen atom is removed from Methane and the ㅡCHO group takes it's place to give the aldehyde.
• But when the group takes it's place, the total number of C atoms will become 2.
♦ So it should be named as Ethanal.
• That means, the compound methanal cannot exist.
• However, methanal does exist. It's structure is shown in fig.14.64(a) below.
 |
| Fig.14.64 |
• We know that, the ㅡCHO group will be looking for an electron.
♦ This electron is supplied by a hydrogen atom.
• Thus we get the methanal shown in fig.14.64(a) above.
♦ The portion shown within the magenta lines is the aldehyde group.
■ How is the compound in fig(b) formed?
• It was originally CH4. One of it's H atoms was removed and the ㅡCHO group took it's place.
♦ Now there are a total of two C atoms, and so it is ethanal
■ How is the compound in fig(c) formed?
• It was originally CH3ㅡCH3. One of it's H atoms was removed and the ㅡCHO group took it's place.
♦ Now there are a total of three C atoms, and so it is propanal
■ The compounds in fig.64 above can be represented in another way also:
• Methanal: HㅡCHO
• Ethanal: CH3ㅡCHO
• Propanal: CH3ㅡCH2ㅡCHO
• Butanal: CH3ㅡCH2ㅡCH2ㅡCHO
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 aldehyde group, we do not have to learn about such change in positions.
• That is., In all the aldehydes that we see in the present discussion, the ㅡCHO group will be at the end (or at the beginning).
♦ So we will not see names like Propan-1-al, propan-2-al etc., in this discussion.
Solved example 14.7
Write the IUPAC name of the compound given below:
CH3ㅡCH2ㅡCH2ㅡCH2ㅡCHO
Solution:
• As mentioned above, in the present discussion on aldehydes, we will not encounter those aldehydes where the ㅡCHO group takes different positions.
• We will see those aldehydes 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 'al' in it's place. We get:
Pentane - e + al = Pentanal
Now we will see the reverse process. That is., we are given the IUPAC name of an aldehyde. We must write the structural formula. We will learn the method by analysing an example:
1. Given IUPAC name is: Propanal
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 ㅡCHO 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 ㅡCHO
5. Fill all the valencies of carbon atoms.
• We use hydrogen to fill up the valencies.
• The result is: CH3ㅡCH2ㅡCHO
■ In the first section in this chapter, we saw how the 'alkyl radical' gets itself attached to an open chain. See details here.
■ In the previous section we saw how a 'hydroxyl group' that is., 'ㅡOH', gets itself attached to an open chain. See details here.
■ Now we will see the same for the aldehyde group:
• We have seen the structure of ㅡOH group in the previous section. It is shown again in fig.14.65 (a) below:
 |
| Fig.14.65 |
♦ This need is indicated by the absence of a dot at the end of the bond line attached to the O atom.
• Now consider fig.14.65(b). It shows the bonding details in the ㅡCHO group
♦ All the valency needs of oxygen atoms is satisfied (6 cyan dots plus two green dots, giving a total of 8 dots.
♦ Those of hydrogen is also satisfied (1 red dot plus 1 green dot, giving a total of 2 dots
♦ Those of carbon is not fully satisfied. (4 green plus 2 cyan plus 1 red dots 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.14.66 below:
 |
| Fig.14.66 |
In the next section, we will see the Ketone group.
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