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 8.5 - Details and Nomenclature of Cyclic hydrocarbons

In the previous section, we saw the classification and nomenclature of hydrocarbons. In this section we will learn about Cyclic hydrocarbons.

The hydrocarbons that we saw so far were 'chains'. For example: 
• In propane, the three carbon atoms bond together to form a chain. 
• In pentene, the five carbon atoms bond together to form a chain.

Now we consider a different case. 
• If the last carbon atom in the chain is connected back to the first carbon atom, it becomes a 'closed chain'. Then it will resemble a ring. So they are called cyclic or ring hydrocarbons. 
• According to the IUPAC rules, the prefix 'cyclo'  is used in naming them. 
The fig.18.16 below shows some cyclic hydrocarbons.

Fig.8.16

Let us see each of them in detail:
1. In fig.18.16(a) we have a ring with 3 carbon atoms. So the root name is 'prop'
• All the carbon-carbon bonds are single bonds. So the suffix is 'ane'
• Thus we get: prop + ane → propane
• But since it is a cyclic hydrocarbon, the suffix 'cyclo' must be given. 
• So the final name is cyclo + propane → cyclopropane
■ Let us consider the number of hydrogen atoms:
• In cyclopropane, each carbon atom is bonded to 2 hydrogen atoms. The molecular formula is C₃H₆
• We can see that all hydrogen and carbon atoms have attained octet
• Compare cyclopropane with open chain propane:
    ♦ Open chain propane has a molecular formula C₃H₈. 
    ♦ So when the 'open chain' become 'cyclic', two hydrogen atoms are not required. Why is this so?
Ans: We have seen the structural formula of open chain propane. (Details here). 
• There is a hydrogen on the extreme left, and another hydrogen on the extreme right. 
• These two are not required when the 'open chain' becomes 'cyclic'. 
• This is because, the last carbon atom cycles back to the first carbon atom. The two of them share a pair to form a single bond. Thus two hydrogen atoms are no longer required

2. In fig.b we have a ring with 4 carbon atoms. So the root name is 'but'
• All the carbon-carbon bonds are single bonds. So the suffix is 'ane'
• Thus we get: but + ane → butane
• But since it is a cyclic hydrocarbon, the suffix 'cyclo' must be given. 
• So the final name is cyclo + butane → cyclobutane
■ Let us consider the number of hydrogen atoms.
• In cyclobutane, each carbon atom is bonded to 2 hydrogen atoms. The molecular formula is C₄H₈
• We can see that all hydrogen and carbon atoms have attained octet
• Compare cyclobutane with open chain butane:
    ♦ Open chain butane has a molecular formula C4H10. 
    ♦ So when the 'open chain' become 'cyclic', two hydrogen atoms are not required. Why is this so?
• The answer is same as the one we saw above in the case of cyclopropane. The last carbon atom cycles back to the first carbon atom. The two of them share a pair to form a single bond. Thus two hydrogen atoms are no longer required.

■ So we have seen the details about cyclopropane (fig.a) and cyclobutane (fig.b). Based on them the details about cyclopentane (fig.c) and cyclohexane (fig.d) can be written. The reader may write them in his/her own notebooks.
3. Next we will see fig.e. We have a ring with 4 carbon atoms. So the root name is 'but'
• There is one carbon-carbon double bond. So the suffix is 'ene'
• Thus we get: but + ene → butene
• But since it is a cyclic hydrocarbon, the suffix 'cyclo' must be given. 
• So the final name is cyclo + butene → cyclobutene
■ Let us consider the number of hydrogen atoms:
• In cyclobutene, 2 carbon atoms are bonded to 2 hydrogen atoms each. The other 2 carbon atoms are bonded to one hydrogen atom each. The molecular formula is C₄H₆
• We can see that all hydrogen and carbon atoms have attained octet
• Compare cyclobutene with open chain butane:
    ♦ Open chain butane has a molecular formula C₄H₁₀. 
    ♦ So when the 'open chain' become 'cyclic', and also with atleast one double bond, four hydrogen atoms are not required. Why is this so?
Ans: We have seen the structural formula of butane (Details here)
• There are 3 hydrogen atoms on the extreme left, and another 3 hydrogen atoms on the extreme right.
• Out of them two are not required from either ends, when the 'open chain' becomes 'cyclic'. 
• This is because, the last carbon atom cycles back to the first carbon atom. The two of them form a double bond, sharing two pairs of electrons. 
• Thus, a total of four hydrogen atoms are no longer required

4. Next we will see fig.f. We have a ring with 5 carbon atoms. So the root name is 'pent'
• There is one carbon-carbon double bond. So the suffix is 'ene'
• Thus we get: pent + ene → pentene
• But since it is a cyclic hydrocarbon, the suffix 'cyclo' must be given. 
• So the final name is cyclo + pentene → cyclopentene
■ Let us consider the number of hydrogen atoms.
• In cyclopentene, 2 carbon atoms are bonded to 2 hydrogen atoms each. The other 2 carbon atoms are bonded to one hydrogen atom each. The molecular formula is C₅H₈
• We can see that all hydrogen and carbon atoms have attained octet
• Compare cyclopentene with open chain pentane:
    ♦ Open chain pentane has a molecular formula C₅H₁₂
    ♦ So when the 'open chain' become 'cyclic', and also with atleast one double bond, four hydrogen atoms are not required. Why is this so?
• The answer is same as the one we saw above in the case of cyclobutene. The last carbon atom cycles back to the first carbon atom. The two of them share two pairs to form a double bond. Thus four hydrogen atoms are no longer required.

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■ From the above discussion, it is clear that, a minimum of two carbon atoms is required to form a cyclic hydrocarbon.

■ In general we can say this about the number of hydrogen atoms in a cyclic hydrocarbon:

• If the cyclic hydrocarbon has all the carbon-carbon bonds as single bonds:
    ♦ 2 less than the corresponding alkane 
• If the cyclic hydrocarbon has one carbon-carbon bonds as double bond:
    ♦ 4 less than the corresponding alkane
    ♦ 2 less than the corresponding alkene

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So we have completed the discussion on cyclic hydrocarbons. We will now see some solved examples based on the discussions that we had so far in this chapter.

Solved example 8.3

Write the chemical formula of the compounds that are missing in the homologous series given below:

Solution:
(i). A series:
1. When the the number of carbon atoms is 3, number of hydrogen atoms is 8 
2. When the the number of carbon atoms is 4, number of hydrogen atoms is 10
3. So it is clear that, the general formula for this series is C_nH_2n+2.
4. The missing compounds are those with number of carbon atoms (n) equal to 2 and 5.
5. The number of hydrogen atoms when n = 2 is ( 2 × 2 + 2) = 6. So the compound is C₂H₆.
6. The number of hydrogen atoms when n = 5 is ( 2 × 5 + 2) = 12. So the compound is C₅H₁₂.
(ii). B series:
1. When the the number of carbon atoms is 3, number of hydrogen atoms is 6 
2. When the the number of carbon atoms is 6, number of hydrogen atoms is 12
3. So it is clear that, the general formula for this series is C_nH_2n.
4. The missing compounds are those with number of carbon atoms (n) equal to 4, 5 and 7.
5. The number of hydrogen atoms when n = 4 is ( 2 × 4) = 8. So the compound is C₄H₈.
6. The number of hydrogen atoms when n = 5 is ( 2 × 5) = 10. So the compound is C₅H₁₀.
7. The number of hydrogen atoms when n = 7 is ( 2 × 7) = 14. So the compound is C₇H₁₄.
(iii). C series:
1. When the the number of carbon atoms is 3, number of hydrogen atoms is 4 
2. When the the number of carbon atoms is 4, number of hydrogen atoms is 6
3. So it is clear that, the general formula for this series is C_nH_2n-2.
4. The missing compounds are those with number of carbon atoms (n) equal to 2, 5 and 6.
5. The number of hydrogen atoms when n = 2 is ( 2 × 2 - 2) = 2. So the compound is C₂H₂.
6. The number of hydrogen atoms when n = 5 is ( 2 × 5 - 2) = 8. So the compound is C₅H₈.
7. The number of hydrogen atoms when n = 6 is ( 2 × 6 - 2) = 10. So the compound is C₆H₁₀.
The completed table is given below:

Note:
• For the A series we found that general formula is C_nH_2n+2. Also successive compounds differ by CH₂. So it is the Alkane series
• For the B series we found that general formula is C_nH_2n. Also successive compounds differ by CH₂. So it is the Alkene series
• For the C series we found that general formula is C_nH_2n-2. Also successive compounds differ by CH₂. So it is the Alkyne series

Solved example 8.4
Given below are the chemical formulae of some hydrocarbons:
C₃H₆, C₂H₆, C₃H₄
(a) Represent their structure
(b) Based on the structure, classify them into alkane, alkene and alkyne.
Solution:
Part (a):
1. C₃H₆ : We have discussed about the structure of this hydrocarbon in detail here. 
2. C₂H₆ : We have discussed about the structure of this hydrocarbon in detail here.
3. C₃H₄ : We have discussed about the structure of this hydrocarbon in detail here.
Part (b):
1. There is one double bond in C₃H₆. So it is an alkene
2. All the carbon-carbon bonds in C₂H₆ are single bonds. So it is an alkane
3. There is one triple bond in C₃H₄. So it is an alkyne

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In the next section, we will learn about position of atoms in a three dimensional space. 

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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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Chapter 8.3 - Alkynes Homologous series

In the previous section, we saw the properties of the alkene homologous series. In this section we will see one more homologous series.

In the previous section we started with C₂H₄, and saw that it was the first member of alkene series. In this section we will start with another compound C₂H₂. It is a molecule of 'ethyne'. We want to know how the atoms are arranged in the molecule. Let us find out:
Step 1: In C₂H₂, there are two carbon atoms. The first step is to connect those two, using a single bond. It is shown in fig.8.13(a) below:

Fig.8.13

Step 2: • Consider the carbon atom on the left in fig.(a). It is in single bond with the right side carbon atom. So it has acquired one electron through this single bond. Three more electrons are required. 
• There are not enough hydrogen atoms to supply these 3 electrons. There are only a total of two hydrogen atoms. 
• So we will equally share the 2 hydrogen atoms and see what happens. This step is shown in fig.8.13(b)
Step 3: In the fig.(b), the two hydrogen atoms have attained octet. But both the carbon atoms still need 3 electrons each. This can be supplied by changing the single bond between the carbon atoms to a triple bond. This is shown in fig.(c). Now all the atoms have attained octet. 
■ With the above three steps, we get the final arrangement of atoms in a molecule of C₂H₂. The electron dot diagram is shown in the fig.8.14 below:

Fig.8.14

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Now we will consider another example: C₃H₄. We will write the required steps as in the previous example.

Step 1: In C₃H₄, there are 3 carbon atoms. The first step is to connect the three using a single bond. It is shown in fig.8.15(a) below:

Fig.8.15

Step 2: • Consider the left most carbon atom. It is in single bond with the middle carbon atom. So it has acquired one electron through this single bond. Three more electrons are required.
• If we bond it to three hydrogen atoms, the middle carbon atom and the right side carbon atom will be left with only one hydrogen. Because there are only 4 hydrogen atoms in total. We cannot share 1 hydrogen atoms among 2 carbon atoms. So we can not give 3 hydrogen atoms to the left side carbon.
• Also, we cannot share the total 4 hydrogen atoms among the total 3 carbon atoms equally. 
• So we will give importance to symmetry and see what happens. That is., we will give 2 electrons to the middle carbon atom. And one hydrogen atom each to the carbon atoms on left and right. This step is shown in fig.8.15(b)
Step 3: In fig.(b), the requirements of all the hydrogen atoms are satisfied. There is also symmetry. But consider the left most carbon atom. It needs two more electrons. So change the single bond between it and the middle carbon atom to a triple bond. This step is shown in fig.c.
Step 4: • In fig.c, all the hydrogen atoms has octet
• The left most carbon atom has octet
• But the middle carbon atom has two excess electrons. [3 single bonds and 1 triple bond]
• Also the right most carbon atom is in shortage of two electrons
• So remove two hydrogen from the middle carbon
• Bond this removed hydrogen atoms to the right most carbon atom.
• This step is shown in fig.d. Now all atoms have octet

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So we determined the arrangement of atoms in the following two molecules:

(i) Molecule with 2 carbon atoms and 2 hydrogen atoms
(ii) Molecule with 3 carbon atoms and 4 hydrogen atoms
• A pattern is beginning to emerge. The next two cases will be:
(iii) Molecule with 4 carbon atoms and 6 hydrogen atoms
(iv) Molecule with 5 carbon atoms and 8 hydrogen atoms
• Note that number of carbon atoms is increasing by 1. Also, number of hydrogen atoms is increasing by 2
• It is like a series. Let us draw the arrangement for the first six members of the series:

With the above table, the analysis about the series becomes easy. Let us learn the properties of the series:
Property 1:
• The 1^st member of the series has 2 carbon atoms
And 2 hydrogen atoms
• The 2^nd member of the series has 3 carbon atoms
And 4 hydrogen atoms
• The 3^rd member of the series has 4 carbon atoms
And 6 hydrogen atoms
So on...
■ So there exists a definite relation between the two quantities below:
• No. of carbon atoms
• No. of hydrogen atoms
■ If the no. of carbon atoms is 'n', then the no. of hydrogen atoms will be (2n-2)
So we can represent the members of the series by a general formula: C_nH_(2n-2)
Property 2:
• Consider the second row. Consider the item in the 'Condensed formula' column in this row. We have: CH = C – CH₃
• Now consider the third row, same column. We have: CH = C – CH₂ – CH₃
The difference is a CH₂ group
• Now consider the fourth row, same column. We have:   CH = C – CH₂ – CH₂ – CH₃
Here also, the difference between the previous third row is a CH₂ group
• We will find the same difference through out the series. That is:
■ Successive members differ by a CH₂ group

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In the previous sections, we saw that a series of compounds having the above two properties is called a homologous series. So this series is also a homologous series. The name given to this series is: Alkynes.

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In the next section, we will learn about hydrocarbons. 

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