In the previous section, we saw the naming procedure for saturated hydrocarbons. In this section, we will see unsaturated hydrocarbons.
We have seen the basics of nomenclature of unsaturated hydrocarbons in a previous chapter. Let us refresh our memory:
• Alkenes and Alkynes come under the category of unsaturated hydrocarbons.
♦ C4H10 is an alkane.
♦ C4H8 is an alkene
♦ C4H6 is an alkyne
• Note that all of them have 4 carbon atoms. But the 'number of hydrogen atoms' is different.
• We know that the valencies in saturated hydrocarbons (alkanes) are satisfied by single bonds.
♦ We saw an example in fig.14.1 earlier in this chapter.
• We know 'how the valencies are satisfied (with the help of double bonds) in alkenes'. See fig.8.11.
♦ A larger molecule is shown in fig.14.43(a) below:
• We know 'how the valencies are satisfied (with the help of triple bonds) in alkynes'. See fig.8.14.
♦ A larger molecule is shown in fig.14.43(b) above
■ Note that:
• Which ever carbon atom we take from the above fig.14.43(a) or (b), there will be 4 pairs of electrons around it.
♦ So there will be eight electrons available to any carbon atom that we take
• Which ever hydrogen atom we take, there will be 1 pair of electrons around it.
♦ So there will be two electrons available to any hydrogen atom that we take
Now we will see their nomenclature:
■ For our present discussion on 'nomenclature of alkenes and alkynes', we will be considering only 'straight chains'.
• That means, there will not be any 'branches'.
• So for our present discussion on alkenes and alkynes, we do not need to worry about 'highlighting the main chain' as we did in the case of alkanes in the previous sections.
■ Consider the two alkenes in fig.14.44 below:
• The first compound has 4 carbon atoms and 8 hydrogen atoms.
♦ So it's molecular formula is C4H8
• The second compound has 4 carbon atoms and 8 hydrogen atoms.
♦ So it's molecular formula is also C4H8
■ But the structure of each is different from the other. This is due to the difference in the position of the double bond.
• So it is clear that, each of them should be given distinct names.
• Let us see how this is done. We will analyse it in steps:
1. The first compound is shown in fig.14.45(a) below:
2. 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 1 and 2
• In fig.(c), the numbering is done from right to left
♦ In this, the 'carbon atoms linked to the double bond' gets 3 and 4
3. 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 1 and 2
• The numbering in fig.(c) gives 3 and 4
4. The numbering in fig.(b) gives the lowest number 1. So it is the correct method of numbering.
5. Now we can assemble the name. The rule for assembling is:
Word root+hyphen+position of double bond+hyphen+suffix
• As there are 4 carbon atoms, the word root is but
• As one carbon-carbon bond is a double bond, the suffix is ene
So we get: But-1-ene
1. The second compound is shown in fig.14.46(a) below:
2. 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 2 and 3
• 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
3. 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 2 and 3
• The numbering in fig.(c) gives 2 and 3
4. The numbering by both the methods gives the same numbers 2 and 3. So the lowest number is 2.
5. Now we can assemble the name. The rule for assembling is:
Word root+hyphen+position of double bond+hyphen+suffix
• As there are 4 carbon atoms, the word root is but
• As one carbon-carbon bond is a double bond, the suffix is ene
So we get: But-2-ene
The same steps can be adopted for alkynes also. The fig.14.47 below shows two alkynes.
• The first compound has 4 carbon atoms and 6 hydrogen atoms.
♦ So it's molecular formula is C4H6
• The second compound has 4 carbon atoms and 6 hydrogen atoms.
♦ So it's molecular formula is also C4H6
■ But the structure of each is different from the other. This is due to the difference in the position of the triple bond.
• So it is clear that, each of them should be given distinct names.
■ By following the same steps that we saw for alkenes, we will get:
• Name of the first compound: But-1-yne
• Name of the first compound: But-2-yne.
The reader may write the steps in his/her own notebooks
Another example:
1. An alkene is shown in fig.14.48(a) below:
2. 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 3 and 4
• 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
3. 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 3 and 4
• The numbering in fig.(c) gives 2 and 3
4. The numbering in fig.(c) gives the lowest number 2. So it is the correct method of numbering.
5. Now we can assemble the name. The rule for assembling is:
Word root+hyphen+position of double bond+hyphen+suffix
• As there are 5 carbon atoms, the word root is pent
• As one carbon-carbon bond is a double bond, the suffix is ene
So we get: Pent-2-ene.
One more example:
1. An alkyne is shown in fig.14.49(a) below:
2. 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 triple bond' gets 4 and 5
• In fig.(c), the numbering is done from right to left
♦ In this, the 'carbon atoms linked to the triple bond' gets 1 and 2
3. According to the IUPAC rules, 'carbon atoms linked to the triple 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 1 and 2
4. The numbering in fig.(c) gives the lowest number 1. So it is the correct method of numbering.
5. Now we can assemble the name. The rule for assembling is:
Word root+hyphen+position of double bond+hyphen+suffix
• As there are 5 carbon atoms, the word root is pent
• As one carbon-carbon bond is a triple bond, the suffix is yne
So we get: Pent-1-yne
In the next section, we will see the reverse process. That is., we will be given the IUPAC name of an alkene or an alkyne. We must draw it's structure.
We have seen the basics of nomenclature of unsaturated hydrocarbons in a previous chapter. Let us refresh our memory:
• Alkenes and Alkynes come under the category of unsaturated hydrocarbons.
♦ C4H10 is an alkane.
♦ C4H8 is an alkene
♦ C4H6 is an alkyne
• Note that all of them have 4 carbon atoms. But the 'number of hydrogen atoms' is different.
• We know that the valencies in saturated hydrocarbons (alkanes) are satisfied by single bonds.
♦ We saw an example in fig.14.1 earlier in this chapter.
• We know 'how the valencies are satisfied (with the help of double bonds) in alkenes'. See fig.8.11.
♦ A larger molecule is shown in fig.14.43(a) below:
 |
| Fig.14.43 |
♦ A larger molecule is shown in fig.14.43(b) above
■ Note that:
• Which ever carbon atom we take from the above fig.14.43(a) or (b), there will be 4 pairs of electrons around it.
♦ So there will be eight electrons available to any carbon atom that we take
• Which ever hydrogen atom we take, there will be 1 pair of electrons around it.
♦ So there will be two electrons available to any hydrogen atom that we take
Now we will see their nomenclature:
• That means, there will not be any 'branches'.
• So for our present discussion on alkenes and alkynes, we do not need to worry about 'highlighting the main chain' as we did in the case of alkanes in the previous sections.
■ Consider the two alkenes in fig.14.44 below:
 |
| Fig.14.44 |
♦ So it's molecular formula is C4H8
• The second compound has 4 carbon atoms and 8 hydrogen atoms.
♦ So it's molecular formula is also C4H8
■ But the structure of each is different from the other. This is due to the difference in the position of the double bond.
• So it is clear that, each of them should be given distinct names.
• Let us see how this is done. We will analyse it in steps:
1. The first compound is shown in fig.14.45(a) below:
 |
| Fig.14.45 |
• In fig.(b), the numbering is done from left to right
♦ In this, the 'carbon atoms linked to the double bond' gets 1 and 2
• In fig.(c), the numbering is done from right to left
♦ In this, the 'carbon atoms linked to the double bond' gets 3 and 4
3. 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 1 and 2
• The numbering in fig.(c) gives 3 and 4
4. The numbering in fig.(b) gives the lowest number 1. So it is the correct method of numbering.
5. Now we can assemble the name. The rule for assembling is:
Word root+hyphen+position of double bond+hyphen+suffix
• As there are 4 carbon atoms, the word root is but
• As one carbon-carbon bond is a double bond, the suffix is ene
So we get: But-1-ene
1. The second compound is shown in fig.14.46(a) below:
 |
| Fig.14.46 |
• In fig.(b), the numbering is done from left to right
♦ In this, the 'carbon atoms linked to the double bond' gets 2 and 3
• 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
3. 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 2 and 3
• The numbering in fig.(c) gives 2 and 3
4. The numbering by both the methods gives the same numbers 2 and 3. So the lowest number is 2.
5. Now we can assemble the name. The rule for assembling is:
Word root+hyphen+position of double bond+hyphen+suffix
• As there are 4 carbon atoms, the word root is but
• As one carbon-carbon bond is a double bond, the suffix is ene
So we get: But-2-ene
The same steps can be adopted for alkynes also. The fig.14.47 below shows two alkynes.
 |
| Fig.14.47 |
♦ So it's molecular formula is C4H6
• The second compound has 4 carbon atoms and 6 hydrogen atoms.
♦ So it's molecular formula is also C4H6
■ But the structure of each is different from the other. This is due to the difference in the position of the triple bond.
• So it is clear that, each of them should be given distinct names.
■ By following the same steps that we saw for alkenes, we will get:
• Name of the first compound: But-1-yne
• Name of the first compound: But-2-yne.
The reader may write the steps in his/her own notebooks
Another example:
1. An alkene is shown in fig.14.48(a) below:
 |
| Fig.14.48 |
• In fig.(b), the numbering is done from left to right
♦ In this, the 'carbon atoms linked to the double bond' gets 3 and 4
• 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
3. 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 3 and 4
• The numbering in fig.(c) gives 2 and 3
4. The numbering in fig.(c) gives the lowest number 2. So it is the correct method of numbering.
5. Now we can assemble the name. The rule for assembling is:
Word root+hyphen+position of double bond+hyphen+suffix
• As there are 5 carbon atoms, the word root is pent
• As one carbon-carbon bond is a double bond, the suffix is ene
So we get: Pent-2-ene.
One more example:
1. An alkyne is shown in fig.14.49(a) below:
 |
| Fig.14.49 |
• In fig.(b), the numbering is done from left to right
♦ In this, the 'carbon atoms linked to the triple bond' gets 4 and 5
• In fig.(c), the numbering is done from right to left
♦ In this, the 'carbon atoms linked to the triple bond' gets 1 and 2
3. According to the IUPAC rules, 'carbon atoms linked to the triple 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 1 and 2
4. The numbering in fig.(c) gives the lowest number 1. So it is the correct method of numbering.
5. Now we can assemble the name. The rule for assembling is:
Word root+hyphen+position of double bond+hyphen+suffix
• As there are 5 carbon atoms, the word root is pent
• As one carbon-carbon bond is a triple bond, the suffix is yne
So we get: Pent-1-yne
In the next section, we will see the reverse process. That is., we will be given the IUPAC name of an alkene or an alkyne. We must draw it's structure.
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