In the previous section, we saw the properties of a homologous series. We also saw the Alkane series. In this section we will see another homologous series.
In the previous section we started with CH4, and saw that it was the first member of a series. In this section we will start with another compound C2H4. It is a molecule of 'ethene'. We want to know how the atoms are arranged in the molecule. Let us find out:
Step 1: In C2H4, there are two carbon atoms. The first step is to connect those two, using a single bond. It is shown in fig.8.10(a) below:
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.
• If we bond it to three hydrogen atoms, the right side carbon atom will be left with only one hydrogen. Because there are only 4 hydrogen atoms in total. So we can not give 3 hydrogen atoms to the left side carbon.
• So we will equally share the 4 hydrogen atoms and see what happens. This step is shown in fig.8.10(b)
Step 3: In the fig.(b), all the hydrogen atoms have attained octet. But both the carbon atoms still need one electron each. This can be supplied by changing the single bond between the carbon atoms to a double 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 C2H4. The electron dot diagram is shown in the fig.8.11 below:
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 three hydrogen. Because there are only 6 hydrogen atoms in total. We cannot share 3 hydrogen atoms among 2 carbon atoms. So we can not give 3 hydrogen atoms to the left side carbon.
• So we will equally share the 6 hydrogen atoms and see what happens. This step is shown in fig.8.12(b)
Step 3: In fig.(b), the requirements of all the hydrogen atoms are satisfied. But consider the left most carbon atom. It needs one more electron. So change the single bond between it and the middle carbon atom to a double 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 an excess electron. [3 single bonds and 1 double bond]
• Also the right most carbon atom is in shortage of one electron
• So remove one hydrogen from the middle carbon
• Bond this removed hydrogen atom to the right most carbon atom.
• This step is shown in fig.d. Now all atoms have octet
(ii) Molecule with 3 carbon atoms and 6 hydrogen atoms
• A pattern is beginning to emerge. The next two cases will be:
(iii) Molecule with 4 carbon atoms and 8 hydrogen atoms
(iv) Molecule with 5 carbon atoms and 10 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 tabulate 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 1st member of the series has 2 carbon atoms
And 4 hydrogen atoms
• The 2nd member of the series has 3 carbon atoms
And 6 hydrogen atoms
• The 3rd member of the series has 4 carbon atoms
And 8 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
So we can represent the members of the series by a general formula: CnH2n
Property 2:
• Consider the second row. Consider the item in the 'Condensed formula' column in this row. We have: CH2 = CH – CH3
• Now consider the third row, same column. We have: CH2 = CH – CH2 – CH3
The difference is a CH2 group
• Now consider the fourth row, same column. We have: CH2 = CH – 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
In the next section, we will see one more homologous series.
In the previous section we started with CH4, and saw that it was the first member of a series. In this section we will start with another compound C2H4. It is a molecule of 'ethene'. We want to know how the atoms are arranged in the molecule. Let us find out:
Step 1: In C2H4, there are two carbon atoms. The first step is to connect those two, using a single bond. It is shown in fig.8.10(a) below:
 |
| Fig.8.10 |
• If we bond it to three hydrogen atoms, the right side carbon atom will be left with only one hydrogen. Because there are only 4 hydrogen atoms in total. So we can not give 3 hydrogen atoms to the left side carbon.
• So we will equally share the 4 hydrogen atoms and see what happens. This step is shown in fig.8.10(b)
Step 3: In the fig.(b), all the hydrogen atoms have attained octet. But both the carbon atoms still need one electron each. This can be supplied by changing the single bond between the carbon atoms to a double 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 C2H4. The electron dot diagram is shown in the fig.8.11 below:
 |
| Fig.8.11 |
Now we will consider another example: C3H6. We will write the required steps as in the previous example.
Step 1: In C3H6, there are 3 carbon atoms. The first step is to connect the three using a single bond. It is shown in fig.8.12(a) below: |
| Fig.8.12 |
• If we bond it to three hydrogen atoms, the middle carbon atom and the right side carbon atom will be left with only three hydrogen. Because there are only 6 hydrogen atoms in total. We cannot share 3 hydrogen atoms among 2 carbon atoms. So we can not give 3 hydrogen atoms to the left side carbon.
• So we will equally share the 6 hydrogen atoms and see what happens. This step is shown in fig.8.12(b)
Step 3: In fig.(b), the requirements of all the hydrogen atoms are satisfied. But consider the left most carbon atom. It needs one more electron. So change the single bond between it and the middle carbon atom to a double 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 an excess electron. [3 single bonds and 1 double bond]
• Also the right most carbon atom is in shortage of one electron
• So remove one hydrogen from the middle carbon
• Bond this removed hydrogen atom to the right most carbon atom.
• This step is shown in fig.d. Now all atoms have octet
So we determined the arrangement of atoms in the following two molecules:
(i) Molecule with 2 carbon atoms and 4 hydrogen atoms(ii) Molecule with 3 carbon atoms and 6 hydrogen atoms
• A pattern is beginning to emerge. The next two cases will be:
(iii) Molecule with 4 carbon atoms and 8 hydrogen atoms
(iv) Molecule with 5 carbon atoms and 10 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 tabulate the arrangement for the first six members of the series:
Property 1:
• The 1st member of the series has 2 carbon atoms
And 4 hydrogen atoms
• The 2nd member of the series has 3 carbon atoms
And 6 hydrogen atoms
• The 3rd member of the series has 4 carbon atoms
And 8 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
So we can represent the members of the series by a general formula: CnH2n
Property 2:
• Consider the second row. Consider the item in the 'Condensed formula' column in this row. We have: CH2 = CH – CH3
• Now consider the third row, same column. We have: CH2 = CH – CH2 – CH3
The difference is a CH2 group
• Now consider the fourth row, same column. We have: CH2 = CH – 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
In the previous section, 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: Alkenes.
In the next section, we will see one more homologous series.
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