Chapter 14.6 - Structure from names of Alkenes and Alkynes

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:

Fig.14.50

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:

Fig.14.51

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.

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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

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:

Fig.14.53

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.

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In the next section, we will see functional groups.

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Chapter 14.5 - IUPAC Nomenclature of Unsaturated hydrocarbons

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:

Fig.14.43

• 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

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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:

Fig.14.44

• 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:

Fig.14.45

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

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1. The second compound is shown in fig.14.46(a) below:

Fig.14.46

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

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The same steps can be adopted for alkynes also. The fig.14.47 below shows two alkynes. 

Fig.14.47

• 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

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Another example:
1. An alkene is shown in fig.14.48(a) below:

Fig.14.48

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:

Fig.14.49

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

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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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Chapter 8.4 - Classification and Nomenclature of Hydrocarbons

In the previous section, we completed the discussion on alkynes. So far we have seen three homologous series. Alkanes, alkenes and alkynes. A very large number of compounds are present in each of those series. We can note one peculiarity in all those compounds. That is., they contain only two elements: Carbon and Hydrogen.
• The compounds which contain only carbon and hydrogen are called Hydrocarbons.
• So any compound that we pick from the alkanes, alkenes or alkynes will be a hydrocarbon.

Classification of hydrocarbons

If in a hydrocarbon, all the ‘carbon-carbon bonds’ are single bonds, that hydrocarbon will come under the category of Saturated hydrocarbons.
Let us analyse the reason for such a name:
• Take a hydrocarbon in which all the carbon-carbon bonds are single bonds
• In that hydrocarbon, pick any carbon atom
• Each of it’s 4 valence electrons will be locked in a single bond. 
[This is because, a hydrocarbon will contain only carbon and hydrogen, and if the carbon-carbon bond is a single bond, the other bonds with hydrogen atoms will also be single bonds, as hydrogen can enter into single bond only]
• Any carbon atom that we pick, all around it, there will be single bonds only. This is a kind of 'saturation.'.
■ So we can write the definition of a saturated hydrocarbon:
Any hydrocarbon is a saturated hydrocarbon if it satisfies the following condition:
• All carbon-carbon bonds should be single bonds

The opposite of saturation is ‘unsaturation’. So The other category is: Unsaturated hydrocarbons. We can write it’s definition easily:
■ Any hydrocarbon is an unsaturated hydrocarbon if it satisfies any one of the following conditions:
• Any one carbon-carbon bond is a double bond
• Any one carbon-carbon bond is a triple bond

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Based on the above definitions, we can write this:
■ Alkanes come under the category of Saturated hydrocarbons
■ Alkenes and Alkynes come under the category of unsaturated hydrocarbons
We can draw a flow chart like diagram:

Nomenclature of hydrocarbons

We have seen that a large number of hydrocarbons are present in the three homologous series. It is difficult to find appropriate names for each of them. IUPAC has put forward some rules for the naming of hydrocarbons.

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IUPAC (International Union of Pure and Applied Chemistry) is an international organisation that strives to carry forward the new trends in the field of chemical sciences. Thus the developments in chemistry can be effectively utilised for the service of mankind. This organisation was founded in 1919. It’s headquarters is in Zurich, Switzerland. IUPAC takes the lead role in naming of elements and compounds. It standardises the atomic weights and physical constants. It also recognises new terms in chemistry.

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Let us see the rules:
• Consider a group of hydrocarbons. We want to name each one of them.
• What is the first difference that we notice among them?
Ans: It is of course, the number of carbon atoms. Different hydrocarbons will be having different number of carbon atoms. So we give a root name based on this number. The table below shows the root name:
C₁ - Meth
C₂ - Eth
C₃ - Prop
C₄ - But
C₅ - Pent
C₆ - Hex
C₇ - Hept
C₈ - Oct
C₉ - Non
C₁₀ - Dec 
• Let us see a few examples which show the application of the above root names:
    ♦ If a hydrocarbon has 3 carbon atoms, it’s root name will be ‘Prop’
    ♦ If a hydrocarbon has 8 carbon atoms, it’s root name will be ‘Oct’
• It may be noted that, from 'C₅' onwards, the root names can be connected to the names of ‘polygons’ that we see in maths classes. 
    ♦ A pentagon has 5 sides
    ♦ A hexagon has 6 sides
    ♦ A heptagon has 7 sides
So on...
• Now we know how to fix root names. But another problem arises:
Compounds with same number of carbon atoms are present in the three homologous series. 
• For example, C₃H₈ belongs to the alkane series. C₃H₆ belongs to the alkene series. C₃H₄ belongs to the alkyne series.
    ♦ They all have the same number of carbon atoms which is 3
    ♦ So they all will have the same root name ‘prop’
    ♦ In fact, any number from C₂ in the alkane series, will have a ‘cousin’ in alkene and alkyne series
Thus it is clear that, a root name is not enough.
• To solve this problem, we use a suffix after the root name. This suffix is derived from the name of the series.
    ♦ If the hydrocarbon is from the alkane series, the suffix will be ‘ane’  
    ♦ If the hydrocarbon is from the alkene series, the suffix will be ‘ene’
    ♦ If the hydrocarbon is from the alkyne series, the suffix will be ‘yne’
• So how do we apply the suffix? We will see with the help of an example:
(i) Pick any one hydrocarbon from the alkane series. Say C₄H₁₀
(ii) It has 4 carbon atoms. So the root name is ‘But’
(iii) It is from the alkane series. So the suffix will be ‘ane’
(iv) Now we can assemble the name:
Root name + suffix ⇒ But + ane → Butane
(v) Butane is the name of C₄H₁₀. It’s cousins in the alkene and alkyne series are C₄H₈ and C₄H₆ respectively.
Their names will be:
But + ene → Butene
But + yne → Butyne

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So we now know how to name the hydrocarbons. The names of hydrocarbons upto C₁₀, in the three homologous series are given here.

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We may get a different type of problem. We will see it as solved examples:
Solved example 8.1
Write the IUPAC name of C₅H₁₀
Solution:
1. In this problem, only the molecular formula is given. Structural formula or condensed formula are not given. 
2. So we do not know whether it is an alkane, alkene or alkyne. But we can find out.
3. Number of carbon atoms ‘n’ = 5
4. Number of hydrogen atoms = 10. This is ‘2n’
5. So the general formula is C_nH_2n. It belongs to the alkene series. Now we can name it. 
6. The root name is ‘pent’  (since number of carbon atoms = 5)
7. The suffix is ‘ene’. (since it belongs to the alkene series)
8. So the name is pent + ene → pentene

Solved example 8.2
Write the IUPAC name of C₃H₄
Solution:
1. In this problem, only the molecular formula is given. Structural formula or condensed formula are not given. 
2. So we do not know whether it is an alkane, alkene or alkyne. But we can find out.
3. Number of carbon atoms ‘n’ = 3
4. Number of hydrogen atoms = 4. This is ‘2n-2’
5. So the general formula is C_nH_2n-2. It belongs to the alkyne series. Now we can name it. 
6. The root name is ‘prop’  (since number of carbon atoms = 3)
7. The suffix is ‘yne’. (since it belongs to the alkyne series)
8. So the name is prop + yne → propyne

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In the next section, we will learn about Cyclic hydrocarbons. 

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