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:
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:
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
(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 1st member of the series has 2 carbon atoms
And 2 hydrogen atoms
• The 2nd member of the series has 3 carbon atoms
And 4 hydrogen atoms
• The 3rd 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: CnH(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
In the next section, we will learn about hydrocarbons.
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 |
• 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 |
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 |
• 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
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:
Property 1:
• The 1st member of the series has 2 carbon atoms
And 2 hydrogen atoms
• The 2nd member of the series has 3 carbon atoms
And 4 hydrogen atoms
• The 3rd 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: CnH(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
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.
In the next section, we will learn about hydrocarbons.
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