In the previous section, we saw the naming procedure for unsaturated hydrocarbons. In this section, we will see a reverse process. That is., we will be given the IUPAC name of an alkene or an alkyne. We must draw it's structure.
We will learn the process with the help of an example:
Given: IUPAC name is hex-3-ene
We will write the procedure in steps:
1. The word root is hex. So there will be 6 carbon atoms.
• Draw these 6 carbon atoms with out any bonds between them.
• Number the carbon atoms from left to right
• This is shown in fig.14.50(a) below:
2. The suffix is ene. So there will be a double bond.
• The position of the bond is given as 3
• This indicates that, the bond is in between carbon atoms 3 and 4
♦ The bond cannot be in between carbon atoms 2 and 3
♦ This is because, the IUPAC name would then be hex-2-ene
• So draw a double bond between carbon atoms 3 and 4
♦ All other carbon-carbon bonds can be shown as single bonds
• This is shown in fig.14.50(b)
3. Fill all the valencies of carbon atoms.
• Since it is a hydrocarbon, only carbon and hydrogen will be present.
♦ So we use hydrogen to fill up the valencies.
• The result is shown in fig.14.50(c) above. So this fig.14.50(c) shows the required structure
Another example:
Given: IUPAC name is but-1-yne
We will write the procedure in steps:
1. The word root is but. So there will be 4 carbon atoms.
• Draw these 4 carbon atoms with out any bonds between them.
• Number the carbon atoms from left to right
• This is shown in fig.14.51(a) below:
2. The suffix is yne. So there will be a triple bond.
• The position of the bond is given as 1
• This indicates that, the bond is in between carbon atoms 1 and 2
♦ There is no other possibility because, 1 is the lowest possible number
• So draw a triple bond between carbon atoms 1 and 2
♦ All other carbon-carbon bonds can be shown as single bonds
• This is shown in fig.14.51(b)
3. Fill all the valencies of carbon atoms.
• Since it is a hydrocarbon, only carbon and hydrogen will be present.
♦ So we use hydrogen to fill up the valencies.
• The result is shown in fig.14.51(c) above. So this fig.14.51(c) shows the required structure
• Note that, there is only one hydrogen atom for the first carbon and no hydrogen for the second carbon. Reader may write the reason for this in his/her own notebooks.
Now we will see some solved examples based on what we have discussed in this section and the previous section
Solved example 14.4:
Write the IUPAC name of the compound shown in fig.14.52(a) below:
Solution:
1. Two possible methods of 'numbering of carbon atoms' are shown in figs.(b) and (c)
• In fig.(b), the numbering is done from left to right
♦ In this, the 'carbon atoms linked to the double bond' gets 4 and 5
• In fig.(c), the numbering is done from right to left
♦ In this, the 'carbon atoms linked to the double bond' gets 2 and 3
2. According to the IUPAC rules, 'carbon atoms linked to the double bond' should get the lowest number. In our present problem,
• The numbering in fig.(b) gives 4 and 5
• The numbering in fig.(c) gives 2 and 3
3. The numbering in fig.(c) gives the lowest number 2. So it is the correct method of numbering.
4. Now we can assemble the name. The rule for assembling is:
Word root+hyphen+position of double bond+hyphen+suffix
• As there are 6 carbon atoms, the word root is hex
• As one carbon-carbon bond is a double bond, the suffix is ene
So we get: hex-2-ene.
Solved example 14.5
Given: IUPAC name is hex-3-yne. Draw the structure
Solution:
We will write the procedure in steps:
1. The word root is hex. So there will be 6 carbon atoms.
• Draw these 6 carbon atoms with out any bonds between them.
• Number the carbon atoms from left to right
• This is shown in fig.14.53(a) below:
2. The suffix is yne. So there will be a triple bond.
• The position of the bond is given as 3
• This indicates that, the bond is in between carbon atoms 3 and 4
♦ The bond cannot be in between carbon atoms 2 and 3
♦ This is because, the IUPAC name would then be hex-2-yne
• So draw a triple bond between carbon atoms 3 and 4
♦ All other carbon-carbon bonds can be shown as single bonds
• This is shown in fig.14.53(b)
3. Fill all the valencies of carbon atoms.
• Since it is a hydrocarbon, only carbon and hydrogen will be present.
♦ So we use hydrogen to fill up the valencies.
• The result is shown in fig.14.53(c) above. So this fig.14.53(c) shows the required structure
• Note that, there is no hydrogen for the third and fourth carbons. Reader may write the reason for this in his/her own notebooks.
In the next section, we will see functional groups.
We will learn the process with the help of an example:
Given: IUPAC name is hex-3-ene
We will write the procedure in steps:
1. The word root is hex. So there will be 6 carbon atoms.
• Draw these 6 carbon atoms with out any bonds between them.
• Number the carbon atoms from left to right
• This is shown in fig.14.50(a) below:
 |
| Fig.14.50 |
• The position of the bond is given as 3
• This indicates that, the bond is in between carbon atoms 3 and 4
♦ The bond cannot be in between carbon atoms 2 and 3
♦ This is because, the IUPAC name would then be hex-2-ene
• So draw a double bond between carbon atoms 3 and 4
♦ All other carbon-carbon bonds can be shown as single bonds
• This is shown in fig.14.50(b)
3. Fill all the valencies of carbon atoms.
• Since it is a hydrocarbon, only carbon and hydrogen will be present.
♦ So we use hydrogen to fill up the valencies.
• The result is shown in fig.14.50(c) above. So this fig.14.50(c) shows the required structure
Another example:
Given: IUPAC name is but-1-yne
We will write the procedure in steps:
1. The word root is but. So there will be 4 carbon atoms.
• Draw these 4 carbon atoms with out any bonds between them.
• Number the carbon atoms from left to right
• This is shown in fig.14.51(a) below:
 |
| Fig.14.51 |
• The position of the bond is given as 1
• This indicates that, the bond is in between carbon atoms 1 and 2
♦ There is no other possibility because, 1 is the lowest possible number
• So draw a triple bond between carbon atoms 1 and 2
♦ All other carbon-carbon bonds can be shown as single bonds
• This is shown in fig.14.51(b)
3. Fill all the valencies of carbon atoms.
• Since it is a hydrocarbon, only carbon and hydrogen will be present.
♦ So we use hydrogen to fill up the valencies.
• The result is shown in fig.14.51(c) above. So this fig.14.51(c) shows the required structure
• Note that, there is only one hydrogen atom for the first carbon and no hydrogen for the second carbon. Reader may write the reason for this in his/her own notebooks.
Now we will see some solved examples based on what we have discussed in this section and the previous section
Solved example 14.4:
Write the IUPAC name of the compound shown in fig.14.52(a) below:
 |
| Fig.14.52 |
1. Two possible methods of 'numbering of carbon atoms' are shown in figs.(b) and (c)
• In fig.(b), the numbering is done from left to right
♦ In this, the 'carbon atoms linked to the double bond' gets 4 and 5
• In fig.(c), the numbering is done from right to left
♦ In this, the 'carbon atoms linked to the double bond' gets 2 and 3
2. According to the IUPAC rules, 'carbon atoms linked to the double bond' should get the lowest number. In our present problem,
• The numbering in fig.(b) gives 4 and 5
• The numbering in fig.(c) gives 2 and 3
3. The numbering in fig.(c) gives the lowest number 2. So it is the correct method of numbering.
4. Now we can assemble the name. The rule for assembling is:
Word root+hyphen+position of double bond+hyphen+suffix
• As there are 6 carbon atoms, the word root is hex
• As one carbon-carbon bond is a double bond, the suffix is ene
So we get: hex-2-ene.
Solved example 14.5
Given: IUPAC name is hex-3-yne. Draw the structure
Solution:
We will write the procedure in steps:
1. The word root is hex. So there will be 6 carbon atoms.
• Draw these 6 carbon atoms with out any bonds between them.
• Number the carbon atoms from left to right
• This is shown in fig.14.53(a) below:
 |
| Fig.14.53 |
• The position of the bond is given as 3
• This indicates that, the bond is in between carbon atoms 3 and 4
♦ The bond cannot be in between carbon atoms 2 and 3
♦ This is because, the IUPAC name would then be hex-2-yne
• So draw a triple bond between carbon atoms 3 and 4
♦ All other carbon-carbon bonds can be shown as single bonds
• This is shown in fig.14.53(b)
3. Fill all the valencies of carbon atoms.
• Since it is a hydrocarbon, only carbon and hydrogen will be present.
♦ So we use hydrogen to fill up the valencies.
• The result is shown in fig.14.53(c) above. So this fig.14.53(c) shows the required structure
• Note that, there is no hydrogen for the third and fourth carbons. Reader may write the reason for this in his/her own notebooks.
In the next section, we will see functional groups.
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