In the previous section, we saw the nomenclature of organic compounds carrying the halo group. In this section, we will see the Alkoxy 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))
• We have seen that the carboxyl group can be represented in two ways (see fig.14.72 (a) & (b))
• In the previous section, we saw that the halo group can be represented as 'ㅡF, ㅡCl, ㅡBr or ㅡI'.
■ Now, our present alkoxy group is represented as: 'ㅡOㅡR'
• We know what 'R' is. It stands for alkyl radicals like methyl radical, ethyl radical etc., We saw it in a previous section of this chapter. Details here.
• So 'ㅡOㅡR' is a combination of oxygen and alkyl radicals. Hence the name alkoxy.
■ 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 the carboxyl group (ㅡCOOH), the bond which connects the group to a main chain starts from the carbon atom
• In the halo group, the bond which connects the group to a main chain starts from halogen atom
• In our present alkoxy group (ㅡOㅡR), the bond which connects the group to a main chain starts from the oxygen atom
• 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 ㅡOㅡR group end in an ethane.
♦ This end part indicates 'the main alkane to which the group is attached'.
• The beginning part indicates the name of the group
Let us analyse this with an example:
■ Consider the compound: CH3ㅡOㅡCH2ㅡCH3.
• In this, which is the ㅡOㅡR?
♦ It seems that 'ㅡOㅡCH2ㅡCH3' at the right side is the ㅡOㅡR
♦ Where R is the ethyl radical 'ㅡCH2ㅡCH3'
• But 'CH3ㅡOㅡ' on the left side is also an ㅡOㅡR.
♦ Because the methyl radical 'ㅡCH3' is attached to O
• So we have a situation:
♦ There are alkyl groups on either side of O.
♦ which one belongs to the ㅡOㅡR group? and
♦ which one is the main alkane?
■ According to the IUPAC rule:
• The longest alkyl group should be taken as the main alkane
• The other alkyl group should be the 'R' of the ㅡOㅡR group
■ Thus in our present case,
• The end part of the name = main alkane = longest alkyl group = ethane
• The beginning part = the other alkyl group = methoxy
• So the IUPAC name is: Methoxyethane
Another example:
Consider the compound: CH3ㅡCH2ㅡCH2ㅡCH2ㅡOㅡCH2ㅡCH3.
1. The end part of the name = main alkane
= longest alkyl group = butane (4 carbon atoms on the left side of O)
2. The beginning part = the other alkyl group = ethoxy
3. So the IUPAC name is: Ethoxybutane.
■ Compounds with the alkoxy group (ㅡOㅡR) are called ethers.
Now we will see the reverse process. That is., we are given the IUPAC name of an ether. We must write the structural formula. We will learn the method by analysing an example:
1. Given IUPAC name is: Methoxyethane
2. The end part of the name = main alkane = ethane
• So the 'ethyl radical' is the main part. It should be written on one side (either left or right) of 'O'
♦ We will write it on the right side of O
3. The beginning part = the other alkyl group = methoxy
• So methyl radical is the R of the ㅡOㅡR.
♦ We will write it on the left side of O
4. Thus we get: CH3ㅡOㅡCH2ㅡCH3.
Another example:
1. Given IUPAC name is: Ethoxyethane
2. The end part of the name = main alkane = ethane
• So the 'ethyl radical' is the main part. It should be written on one side (either left or right) of 'O'
♦ We will write it on the right side of O
3. The beginning part = the other alkyl group = ethoxy
• So ethyl radical is the R of the ㅡOㅡR.
♦ We will write it on the left side of O
4. Thus we get: CH3ㅡCH2ㅡOㅡCH2ㅡCH3.
• Note that on both sides, we have the ethyl radicals
One more example:
1. Given IUPAC name is: Methoxypropane
2. The end part of the name = main alkane = propane
• So the 'propyl radical' is the main part. It should be written on one side (either left or right) of 'O'
♦ We will write it on the right side of O
3. The beginning part = the other alkyl group = methoxy
• So methyl radical is the R of the ㅡOㅡR.
♦ We will write it on the left side of O
4. Thus we get: CH3ㅡOㅡCH2ㅡCH2ㅡCH3.
■ 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.
so on . . .
■ Now we will see the same for the alkoxy group:
In the ㅡOㅡR group, a bond is starting out from the oxygen atom. This indicates that the O is in deed of an electron. This will be clear when we draw the electron dot diagram of the group. It is shown in the fig.14.82 below:
• In the above fig.14.82, a methyl radical (CH3) is attached to an oxygen atom.
• The valency needs of all H atoms are satisfied. Since they have one red dot and one green dot
• The valency needs of C atom is satisfied. Since it has 3 red plus 4 green plus 1 cyan, giving a total of 8 dots
• Oxygen has it's own 6 cyan dots plus the green dot of C. This gives a total of only 7. So one more electron is required. Thus the group as a whole requires one more electron
• Note: In the fig.14.82, The oxygen atom is attached to a methyl radical. It is the 'R' in 'ㅡOㅡR.' Instead of methyl, we can show ethyl, propyl etc., also
• This need will be satisfied when the group gets itself attached to another alkyl radical as shown in fig.14.83 below:
• Thus we get a stable molecule
In the next section, we will see Amino 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))
• We have seen that the carboxyl group can be represented in two ways (see fig.14.72 (a) & (b))
• In the previous section, we saw that the halo group can be represented as 'ㅡF, ㅡCl, ㅡBr or ㅡI'.
■ Now, our present alkoxy group is represented as: 'ㅡOㅡR'
• We know what 'R' is. It stands for alkyl radicals like methyl radical, ethyl radical etc., We saw it in a previous section of this chapter. Details here.
• So 'ㅡOㅡR' is a combination of oxygen and alkyl radicals. Hence the name alkoxy.
■ 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 the carboxyl group (ㅡCOOH), the bond which connects the group to a main chain starts from the carbon atom
• In the halo group, the bond which connects the group to a main chain starts from halogen atom
• In our present alkoxy group (ㅡOㅡR), the bond which connects the group to a main chain starts from the oxygen atom
• 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 ㅡOㅡR group end in an ethane.
♦ This end part indicates 'the main alkane to which the group is attached'.
• The beginning part indicates the name of the group
Let us analyse this with an example:
■ Consider the compound: CH3ㅡOㅡCH2ㅡCH3.
• In this, which is the ㅡOㅡR?
♦ It seems that 'ㅡOㅡCH2ㅡCH3' at the right side is the ㅡOㅡR
♦ Where R is the ethyl radical 'ㅡCH2ㅡCH3'
• But 'CH3ㅡOㅡ' on the left side is also an ㅡOㅡR.
♦ Because the methyl radical 'ㅡCH3' is attached to O
• So we have a situation:
♦ There are alkyl groups on either side of O.
♦ which one belongs to the ㅡOㅡR group? and
♦ which one is the main alkane?
■ According to the IUPAC rule:
• The longest alkyl group should be taken as the main alkane
• The other alkyl group should be the 'R' of the ㅡOㅡR group
■ Thus in our present case,
• The end part of the name = main alkane = longest alkyl group = ethane
• The beginning part = the other alkyl group = methoxy
• So the IUPAC name is: Methoxyethane
Another example:
Consider the compound: CH3ㅡCH2ㅡCH2ㅡCH2ㅡOㅡCH2ㅡCH3.
1. The end part of the name = main alkane
= longest alkyl group = butane (4 carbon atoms on the left side of O)
2. The beginning part = the other alkyl group = ethoxy
3. So the IUPAC name is: Ethoxybutane.
■ Compounds with the alkoxy group (ㅡOㅡR) are called ethers.
Now we will see the reverse process. That is., we are given the IUPAC name of an ether. We must write the structural formula. We will learn the method by analysing an example:
1. Given IUPAC name is: Methoxyethane
2. The end part of the name = main alkane = ethane
• So the 'ethyl radical' is the main part. It should be written on one side (either left or right) of 'O'
♦ We will write it on the right side of O
3. The beginning part = the other alkyl group = methoxy
• So methyl radical is the R of the ㅡOㅡR.
♦ We will write it on the left side of O
4. Thus we get: CH3ㅡOㅡCH2ㅡCH3.
Another example:
1. Given IUPAC name is: Ethoxyethane
2. The end part of the name = main alkane = ethane
• So the 'ethyl radical' is the main part. It should be written on one side (either left or right) of 'O'
♦ We will write it on the right side of O
3. The beginning part = the other alkyl group = ethoxy
• So ethyl radical is the R of the ㅡOㅡR.
♦ We will write it on the left side of O
4. Thus we get: CH3ㅡCH2ㅡOㅡCH2ㅡCH3.
• Note that on both sides, we have the ethyl radicals
One more example:
1. Given IUPAC name is: Methoxypropane
2. The end part of the name = main alkane = propane
• So the 'propyl radical' is the main part. It should be written on one side (either left or right) of 'O'
♦ We will write it on the right side of O
3. The beginning part = the other alkyl group = methoxy
• So methyl radical is the R of the ㅡOㅡR.
♦ We will write it on the left side of O
4. Thus we get: CH3ㅡOㅡCH2ㅡCH2ㅡCH3.
■ 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.
so on . . .
■ Now we will see the same for the alkoxy group:
In the ㅡOㅡR group, a bond is starting out from the oxygen atom. This indicates that the O is in deed of an electron. This will be clear when we draw the electron dot diagram of the group. It is shown in the fig.14.82 below:
 |
| Fig.14.82 |
• The valency needs of all H atoms are satisfied. Since they have one red dot and one green dot
• The valency needs of C atom is satisfied. Since it has 3 red plus 4 green plus 1 cyan, giving a total of 8 dots
• Oxygen has it's own 6 cyan dots plus the green dot of C. This gives a total of only 7. So one more electron is required. Thus the group as a whole requires one more electron
• Note: In the fig.14.82, The oxygen atom is attached to a methyl radical. It is the 'R' in 'ㅡOㅡR.' Instead of methyl, we can show ethyl, propyl etc., also
• This need will be satisfied when the group gets itself attached to another alkyl radical as shown in fig.14.83 below:
 |
| Fig.14.83 |
In the next section, we will see Amino group.
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