In the previous section, we saw the nomenclature of organic compounds carrying the alkoxy group. In this section, we will see the 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 sections, we saw that the halo group can be represented as 'ㅡF, ㅡCl, ㅡBr or ㅡI'.
And the alkoxy group can be represented as ㅡOㅡR
■ Now, our present amino group is represented as: 'ㅡNH2'
■ 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 the alkoxy group (ㅡOㅡR), the bond which connects the group to a main chain starts from the oxygen atom
• In our present amino group (ㅡNH2), the bond which connects the group to a main chain starts from the nitrogen 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 keto group end in amine.
• Such compounds are called amines.
• In any amine, there will be a ㅡNH2 group
■ The naming of amines is done by the following 2 steps:
1. Remove the 'e' from the name of the corresponding alkane
2. Put 'amine' in it's place
• So we can write:
Alkane - e + amine → Alkanamine
• Thus we get:
♦ Methane - e + amine → Methanamine
♦ ethane - e + amine → ethanamine
♦ Propane - e + amine → propanamine
so on . . .
Consider the fig.14.54 that we saw in a previous section (alcohols). 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 an interior position.
• Accordingly we gave two distinct names: Propan-1-ol and propan-2-ol
■ In our present discussion on amines also, we encounter a similar situation.
• This is because, the ㅡNH2 group can be anywhere in the interior of the structure
♦ Obviously, the C that carries the group should get the lowest possible number
• Let us see a solved example:
Solved example 14.14
Write the IUPAC name of the compounds shown in fig.14.84 below:
Solution:
We will write the steps:
Case (i):
1. In the given compound, there are 2 carbon atoms. So the corresponding alkane is ethane
• Remove the 'e' at the end.
2. Put 'amine' in it's place. We get:
ethane - e + amine= ethanamine
3. The lowest position number is 1 from right. So we get: Ethanan-1-amine
Case (ii):
1. In the given compound, there are 3 carbon atoms. So the corresponding alkane is propane
• Remove the 'e' at the end.
2. Put 'amine' in it's place. We get:
propane - e + amine = propanamine
3. The lowest position number is 1 from right. So we get: Propan-1-amine
Case (iii):
1. In the given compound, there are 3 carbon atoms. So the corresponding alkane is propane
• Remove the 'e' at the end.
2. Put 'amine' in it's place. We get:
propane - e + amine = propanamine
3. The lowest position number is 2 from left as well as right. So we get: propan-2-amine
Case (iv):
1. In the given compound, there are 6 carbon atoms. So the corresponding alkane is hexane
• Remove the 'e' at the end.
2. Put 'amine' in it's place. We get:
hexane - e + amine = hexanamine
3. The lowest position number is 3 from right. So we get: hexan-3-amine
Solved example 14.15
We cannot write 'Propan-3-amine'. Give reason.
Solution:
1. The structure of propan-1-amine is: CH3ㅡCH2 ㅡCH2ㅡNH2.
• In this case we begin the numbering from the right most carbon atom
2. Then we would be inclined to think about propan-3-amine as: NH2 ㅡCH2 ㅡCH2ㅡCH3.
• But this is wrong. In this case the numbering should begin from the left most carbon atom.
• So we cannot write 'Propan-3-amine'
Now we will see the reverse process. That is., we are given the IUPAC name of an amine. We must write the structural formula. We will learn the method by analysing an example:
1. Given IUPAC name is: Hexan-2-amine
2. Consider the word root. It is hex. So there are 6 carbon atoms
3. Consider the suffix. It is an. So the ketone 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. This is shown in fig.14.85(a) below:
5. The functional group is at the position 2. So we can attach the amino group to the carbon atom at position 2. This is shown in fig.14.85(b)
6. Now fill all the valencies of carbon atoms.
• We use hydrogen to fill up the valencies.
• The result is shown in fig.14.85(c)
■ In the first section in this chapter, we saw how the 'alkyl radical' gets itself attached to an open chain. See details here.
■ In another previous section we saw how a 'hydroxyl group' that is., 'ㅡOH', gets itself attached to an open chain. See details here.
■ In yet 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 amino group:
• We have seen the covalent bonding between two nitrogen atoms. Two nitrogen atoms share three pairs of electrons between them to achieve octet and thus stability.
• It can be represented as: N☰N. Details here. The two nitrogen atoms bonded together by a triple bond becomes as Nitrogen molecule (N2)
• So a nitrogen atom require three electrons to satisfy the valency. In ammonia molecule (NH3) three hydrogen atoms form bonds with nitrogen.
• So when a hydrogen atom is removed from the ammonia molecule, the remaining portion becomes an amino group (ㅡNH2). It will be unstable and will be looking for an electron.
• If a hydrogen atom is removed from a hydrocarbon, the ㅡNH2 can take it's place. This is shown in fig.14.86 below:
• Thus we get a stable molecule
In the next section, we will see Isomers.
• 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 sections, we saw that the halo group can be represented as 'ㅡF, ㅡCl, ㅡBr or ㅡI'.
And the alkoxy group can be represented as ㅡOㅡR
■ Now, our present amino group is represented as: 'ㅡNH2'
■ 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 the alkoxy group (ㅡOㅡR), the bond which connects the group to a main chain starts from the oxygen atom
• In our present amino group (ㅡNH2), the bond which connects the group to a main chain starts from the nitrogen 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 keto group end in amine.
• Such compounds are called amines.
• In any amine, there will be a ㅡNH2 group
■ The naming of amines is done by the following 2 steps:
1. Remove the 'e' from the name of the corresponding alkane
2. Put 'amine' in it's place
• So we can write:
Alkane - e + amine → Alkanamine
• Thus we get:
♦ Methane - e + amine → Methanamine
♦ ethane - e + amine → ethanamine
♦ Propane - e + amine → propanamine
so on . . .
Consider the fig.14.54 that we saw in a previous section (alcohols). 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 an interior position.
• Accordingly we gave two distinct names: Propan-1-ol and propan-2-ol
■ In our present discussion on amines also, we encounter a similar situation.
• This is because, the ㅡNH2 group can be anywhere in the interior of the structure
♦ Obviously, the C that carries the group should get the lowest possible number
• Let us see a solved example:
Solved example 14.14
Write the IUPAC name of the compounds shown in fig.14.84 below:
 |
| Fig.14.84 |
We will write the steps:
Case (i):
1. In the given compound, there are 2 carbon atoms. So the corresponding alkane is ethane
• Remove the 'e' at the end.
2. Put 'amine' in it's place. We get:
ethane - e + amine= ethanamine
3. The lowest position number is 1 from right. So we get: Ethanan-1-amine
Case (ii):
1. In the given compound, there are 3 carbon atoms. So the corresponding alkane is propane
• Remove the 'e' at the end.
2. Put 'amine' in it's place. We get:
propane - e + amine = propanamine
3. The lowest position number is 1 from right. So we get: Propan-1-amine
Case (iii):
1. In the given compound, there are 3 carbon atoms. So the corresponding alkane is propane
• Remove the 'e' at the end.
2. Put 'amine' in it's place. We get:
propane - e + amine = propanamine
3. The lowest position number is 2 from left as well as right. So we get: propan-2-amine
Case (iv):
1. In the given compound, there are 6 carbon atoms. So the corresponding alkane is hexane
• Remove the 'e' at the end.
2. Put 'amine' in it's place. We get:
hexane - e + amine = hexanamine
3. The lowest position number is 3 from right. So we get: hexan-3-amine
Solved example 14.15
We cannot write 'Propan-3-amine'. Give reason.
Solution:
1. The structure of propan-1-amine is: CH3ㅡCH2 ㅡCH2ㅡNH2.
• In this case we begin the numbering from the right most carbon atom
2. Then we would be inclined to think about propan-3-amine as: NH2 ㅡCH2 ㅡCH2ㅡCH3.
• But this is wrong. In this case the numbering should begin from the left most carbon atom.
• So we cannot write 'Propan-3-amine'
Now we will see the reverse process. That is., we are given the IUPAC name of an amine. We must write the structural formula. We will learn the method by analysing an example:
1. Given IUPAC name is: Hexan-2-amine
2. Consider the word root. It is hex. So there are 6 carbon atoms
3. Consider the suffix. It is an. So the ketone 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. This is shown in fig.14.85(a) below:
5. The functional group is at the position 2. So we can attach the amino group to the carbon atom at position 2. This is shown in fig.14.85(b)
6. Now fill all the valencies of carbon atoms.
• We use hydrogen to fill up the valencies.
• The result is shown in fig.14.85(c)
■ In the first section in this chapter, we saw how the 'alkyl radical' gets itself attached to an open chain. See details here.
■ In another previous section we saw how a 'hydroxyl group' that is., 'ㅡOH', gets itself attached to an open chain. See details here.
■ In yet 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 amino group:
• We have seen the covalent bonding between two nitrogen atoms. Two nitrogen atoms share three pairs of electrons between them to achieve octet and thus stability.
• It can be represented as: N☰N. Details here. The two nitrogen atoms bonded together by a triple bond becomes as Nitrogen molecule (N2)
• So a nitrogen atom require three electrons to satisfy the valency. In ammonia molecule (NH3) three hydrogen atoms form bonds with nitrogen.
• So when a hydrogen atom is removed from the ammonia molecule, the remaining portion becomes an amino group (ㅡNH2). It will be unstable and will be looking for an electron.
• If a hydrogen atom is removed from a hydrocarbon, the ㅡNH2 can take it's place. This is shown in fig.14.86 below:
 |
| Fig.14.86 |
In the next section, we will see Isomers.
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