Chapter 13.13 - Ozonolysis And Polymerisation of Alkenes

In the previous section, we completed a discussion on anti Markovnikov rule. In this section, we will see a few more chemical properties of alkenes.

Addition of sulphuric acid

Let us see the reaction between an alkene and sulphuric acid. It can be written in 3 steps:
1. Alkenes react with cold concentrated sulphuric acid. An example is shown in fig.13.78 below:

Fig.13.78

• This reaction can be explained in 3 steps:
(i) The sulphuric acid molecule undergoes heterolysis. This is indicated by the blue dashed curve.
• We get two parts:
    ♦ H⁺ ion, which is the +ve part.
    ♦ OSO₂OH^- ion, which is the -ve part.
(ii) The H⁺ ion thus produced, will get attached to the left side C atom of the ethene molecule.
• The H⁺ ion does not have any electrons. Both electrons required for the bond, are obtained by breaking the double bond in ethene. That is why both electrons in the bond are shown in red color.
(iii) When the H⁺ ion leaves the sulphuric acid molecule, the remaining portion will have an extra electron. So it will have a -ve charge.
• This remaining portion is the -ve part of the addendum. This -ve part (OSO₂OH^-) gets attached to the right side C atom.
• Both electrons required for the bond, are supplied by this -ve part. They are the green and yellow electrons.
2. Let us see another example. It is shown in fig.13.79 below. The equation is written in a condensed form:

Fig.13.79

• Here we see that, the -ve part gets attached to the C atom with the least number of H atoms.
• So we can write:
The reaction between alkenes and sulphuric acid obeys Markovnikov’s rule.
3. In the above examples, we see that, the two species H⁺ and OSO₂OH^- get added to the original alkene molecule.
• So we can write:
The reaction between alkenes and sulphuric acid is an addition reaction.

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Addition of water

Let us see the reaction between an alkene and water. It can be written in 3 steps:
1. Alkenes react with water. A few drops of concentrated sulphuric acid is also required for the reaction. An alcohol will be obtained as product. An example is shown in fig.13.80 below:

Fig.13.80

• First the water molecule splits to give two parts:
    ♦ H⁺ ion, which is the +ve part.
    ♦ OH^- ion, which is the -ve part.
2. The H⁺ ion thus produced, will get attached to the first C atom of the 2-Methylpropene.
• The H⁺ ion does not have any electrons. Both electrons required for the bond, are obtained by breaking the double bond in the alkene. That is why both electrons in the bond are shown in red color.
3. When the H⁺ ion leaves the water molecule, the remaining portion will have an extra electron. So it will have a -ve charge.
• This remaining portion is the -ve part of the addendum. This -ve part (OH^-) gets attached to the second C atom.
• Both electrons required for the bond, are supplied by this -ve part. They are the green and yellow electrons.
4. Here we see that, the -ve part gets attached to the C atom with the least number of H atoms.
• So we can write:
The reaction between alkenes and water obeys Markovnikov’s rule.
5. In this example, we see that, the two species H⁺ and OH^- get added to the original alkene molecule.
• So we can write:
The reaction between alkenes and water is an addition reaction.

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Oxidation

Let us see how alkenes undergo oxidation. It can be written in 2 steps:
1. Alkenes are oxidised when they react with cold, dilute, aqueous solution of potassium permanganate (Baeyer’s reagent).
• The product will be a vicinal glycol.
    ♦ Glycol is an alcohol in which two -OH groups are present.
    ♦ The two -OH groups must be attached to two different C atoms.
    ♦ If the two -OH groups are attached to two adjacent C atoms, then it is called a vicinal glycol.
• Two examples are shown in the fig.13.81 below:

Fig.13.81

• We see that, the -OH groups get attached to the C atoms on either sides of the double bond.
2. During the reaction, the KMnO₄ gets reduced. So the purple color gradually fades away.
• Because of this decolourisation, this reaction can be used as a test for unsaturation. This can be explained in three steps:
(i) If double or triple bonds (unsaturation) are present, then those bonds will break to form single bonds. This will lead to the reduction of KMnO₄ and the subsequent decolourisation.
(ii) If there is no unsaturation, no decolourisation will occur.
(iii) So decolourisation is an indication of unsaturation.

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Let us see another type of oxidation reaction. It can be explained in two steps:
1. Alkenes are oxidised when they react with acidic potassium permanganate or acidic potassium dichromate.
2. The product is ketone and/or acid. It depends on the nature of the alkene and the experimental conditions.
• Two examples are shown in fig.13.82 below:

Fig.13.82

• Recall the structure of organic acids containing -COOH group that we saw in previous classes (Fig.14.73 in an earlier chapter)

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Ozonolysis

This can be explained in 7 steps:
1. First, the alkene reacts with ozone (O₃)
• The product formed is called an ozonide.
• Two examples are shown in fig.13.83 below:

Fig.13.83

2. In the second step, the ozonide is cleaved into smaller molecules:
• Cleavage of the ozonides in fig.13.83 are shown in fig.13.84 below.
(the cleavages are indicated by dashed cyan curves)
• The cleavage is accomplished by using Zn + H₂O

Fig.13.84

3. In fig.13.84(a) we see that, cleavage of the ozonide gives ethanal and methanal.
• Both are aldehydes.
• Recall the structure of organic compounds containing -CHO group that we saw in previous classes (Fig.14.64 in an earlier chapter)
4. In fig.13.84(b), we see that, cleavage of the ozonide gives propan-2-one and methanal.
• Propan-2-one is a ketone.
• Recall the structure of organic compounds containing -CO- group that we saw in previous classes (Fig.14.67 in an earlier chapter)
5. If we ignore the intermediate product (ozonide), we can write four points:
(i) The original alkene is cleaved at the double bond.
(ii) The double bond requirement of the C atoms are then satisfied by a aldehyde group or keto group.
(iii) If the C atom contains one or more H atoms, we get an aldehyde (-CHO)
(iv) If the C atom contains no H atoms, we get a ketone (-CO-)
6. So the process of ozonolysis acts as a scissors which cuts the alkene. The cut is made at the double bond.
• By examining the products, we will be able to determine the location of the double bond in the original alkene, alkynes or other unsaturated compounds.
7. In the product side in fig.13.84 above, we see that one O atom is missing in both examples. We will see the role of this O atom in higher classes.

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Polymerization

This can be written in 7 steps:
1. Consider the molecule of ethene ($\rm{CH_2 = CH_2}$).
• We know that in ethene, the double bond is necessary to satisfy the valencies of C and H atoms.
2. If we break the double bond, the structure would look like this: $\rm{-CH_2 -CH_2 -}$
• The ‘${}-{}$’ on the sides indicate that, the structure is looking for electrons.
3. Consider the structure shown in the above step (2).
• If there are a large number of such structures, they can join together to satisfy the valencies. The joined structure would look like this:
$\rm{-CH_2 - CH_2 - CH_2 - CH_2 - CH_2 - CH_2 - CH_2 - CH_2 -}$
• The process of making the joined structure is known as polymerization.
4. In polymerization, we subject ethene molecules to high pressure and high temperature. Presence of a suitable catalyst is also necessary.
• Due to the high pressure and temperature, ethene molecules will change into $\rm{-CH_2 -CH_2 -}$
• These structures will join together to form a large molecule.
   ♦ The large molecule is called a polymer.
   ♦ The simple molecule from which polymer is obtained is called a monomer.
• In our present case,
   ♦ Ethene is the monomer.
   ♦ The polymer obtained has a common name: polythene.
   ♦ Polythene is used to make polythene bags. Some images can be seen here.
5. The joined structure in step (3) can be written in short form as: $\rm{-(CH_2 -CH_2 )_n -}$
• So the process of polymerization to make polythene can be written as:
$\rm{n(CH_2 = CH_2)~ \color {green}{\xrightarrow[{\text{Catalyst}}]{{\text{High temp./pressure}}}} ~ -(CH_2 -CH_2 )_n -}$
6. Another example is shown in fig.13.85 below:

Fig.13.85

• In this case,
   ♦ The monomer is propene.
   ♦ The polymer obtained is polypropene.
   ♦ Another name of polypropene is polypropylene.
• Polypropene is used for making moulded articles. Some images can be seen here.
7. Polymers have become a common commodity in our every day life.
• But they are not biodegradable. So the used polymer materials will remain as such for centuries. Toxic materials gradually leak from those waste materials and cause pollution of the soil and water.

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We have completed a discussion on alkenes. In the next section we will see alkynes.

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