Chapter 13.16 - Polymerisation of Alkynes

In the previous section, we saw the preparation of alkynes. We also saw some chemical properties of alkynes. In this section, we will see two more chemical properties.

Addition of water

Let us see the reaction between an alkyne and water. It can be written in 5 steps:
1. Alkynes react with water. Mercuric sulphate and concentrated sulphuric acid is also required for the reaction. The reaction mixture should be warmed to a temperature of 333 K. Carbonyl compounds will be obtained as products. An example is shown in fig.13.92 below:

 

Fig.13.92

 

• 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 one of the two C atoms of the ethyne molecule. This is shown in fig.a.
• The H⁺ ion does not have any electrons. Both electrons required for the bond, are obtained by breaking the triple bond in the alkyne. 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 other C atom.
• Both electrons required for the bond, are supplied by this -ve part. They are the green and yellow electrons.
4. The alcohol thus formed will undergo isomerisation. As a result we get ethanal. This is shown in fig.b.
• Isomerisation is a reaction in which a molecule gets transformed into it's isomer.
• We saw a type of isomerisation in an earlier section [see fig.13.38 in section 13.6].
5. In this example, we see that, the two species H⁺ and OH^- get added to the original alkyne molecule.
• So we can write:
The reaction between alkynes and water is an addition reaction.

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Let us see another example. It is shown in fig.13.93 below. It can be written in 6 steps.

 

Fig.13.93

1. 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 propyne molecule. This is shown in fig.a.
• The H⁺ ion does not have any electrons. Both electrons required for the bond, are obtained by breaking the triple bond in the alkyne. 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. The alcohol thus formed will undergo isomerisation. As a result we get propanone. This is shown in fig.b.
6. In this example, we see that, the two species H⁺ and OH^- get added to the original alkyne molecule.
• So we can write:
The reaction between alkynes and water is an addition reaction.

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Polymerization of alkynes

• First we will see linear polymerization. It can be written in 6 steps:
1. Consider the molecule of ethyne ($\rm{CH ☰ CH}$).
• We know that in ethyne, the triple bond is necessary to satisfy the valencies of C and H atoms.
2. If we break the triple bond, the structure would look like this: $\rm{-CH = CH -}$
• 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:
$\rm{-CH = CH - CH = CH - CH = CH - CH = CH -}$
• The process of making the joined structure is known as polymerization. We saw this in the case of alkenes also.
4. In our present polymerization, we subject ethyne molecules to high pressure and high temperature. Presence of a suitable catalyst is also necessary.
• Due to the high pressure and temperature, ethyne molecules will change into $\rm{-CH = CH -}$
• 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,
   ♦ Ethyne is the monomer.
   ♦ The polymer obtained has a common name: polyacetylene.
        ✰ It's IUPAC name is polyethyne.
5. Under special conditions, polyethyne conducts electricity. Thin films of this polymer can be used as electrodes in batteries. These films are good conductors. They are lighter and cheaper than metal conductors. Some images can be seen here.
6. The joined structure in step (3) can be written in short form as: $\rm{-(CH = CH - CH = CH )_n -}$
• So the process of this polymerization to make polyethyne can be written as:
$\rm{n(CH ☰ CH)~ \color {green}{\xrightarrow[{\text{Catalyst}}]{{\text{High temp./pressure}}}} ~ -(CH = CH - CH = CH )_n -}$

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Now we will see cyclic polymerization. It can be written in 4 steps:

1. We have already seen the details about the structure of the benzene ring [see fig.12.53 of section12.8].
• We see that, there are six CH groups in benzene. Can three ethyne molecules join together to form a benzene ring?
2. Fig.13.94(a) below shows three ethyne molecules aligned together in favorable positions.

Fig.13.94

• If two electrons in the triple bonds can shift, new single bonds will be formed between the three ethyne molecules. The shifting of electrons are indicated by the three curved red arrows in fig.b.
3. This is a cyclic polymerization. For this polymerization to take place, we allow the ethyne gas to pass through red hot iron tube kept at 873 K.
4. So we have an easy method to prepare benzene. Once we obtain benzene, we can make a variety of useful products like benzene derivatives, dyes, drugs etc.,
• This method helps us to enter the world of aromatic compounds from the world of aliphatic compounds.
• We know that:
    ♦ Aliphatic compounds have an open chain structure.
    ♦ Aromatic compounds have a closed chain structure.
• So we enter the world of closed chain structures from the world of open chain structures.

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Let us see a solved example.

Solved example 13.14
How will you convert ethanoic acid into benzene ?
Solution:
1. We can start from benzene and think in a reverse direction.
• To obtain benzene, we must have ethyne (CH☰CH). Because, ethyne when passed through a red hot iron tube will give benzene. See fig.13.94 above.
• This process is marked as the last step (X) in fig.13.95 below.
• So our aim must be to convert the given ethanoic acid to ethyne.

Fig.13.95

2. Ethyne can be obtained from alkenyl halides. We saw this in the preparation of alkynes. See the topic at the beginning of the previous section.
• In our present case, the alkenyl halide CH₂=CHBr can be used to obtain ethyne.
• This process is marked as (IX) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the alkenyl halide CH₂=CHBr.
3. Alkenyl halides can be obtained from vicinal dihalides. We saw this in the preparation of alkynes. See the topic at the beginning of the previous section.
• In our present case, CH₂Br-CH₂Br can be used to obtain the alkenyl halide CH₂=CHBr.
• This process is marked as (VIII) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the vicinal dihalide CH₂Br-CH₂Br.
4. The above vicinal dihalide can be prepared by the addition reaction between ethene and Br₂. We saw this in the chemical properties of alkenes. See fig.13.68 of section 13.11.
• This process is marked as (VII) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to ethene.
5. Ethene can be obtained from the alkyl halide CH₃-CH₂Cl. We saw this in the preparation of alkenes from alkyl halides. See fig.13.65 of section 13.10.
• This process is marked as (VI) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the alkyl halide CH₃-CH₂Cl.
6. The above alkyl halide can be prepared from the alkane CH₃-CH₃. We saw this in the addition reaction of alkanes. See substitution reactions in section 13.4.
• This process is marked as (V) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the alkane CH₃-CH₃.
7. The above alkane can be obtained using CH₃Cl by Wurtz reaction, We saw this in the preparation of alkanes from alkyl halides. See section 13.3.
• This process is marked as (IV) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the alkyl halide CH₃Cl.
8. The above alkyl halide can be obtained from CH₄. We saw this in the chemical properties of alkanes. See substitution reactions in section 13.4.
• This process is marked as (III) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the alkane CH₄.
9. CH₄ can be obtained from the sodium salt of ethanoic acid. We saw this in the preparation of alkanes. See preparation of alkanes from carboxylic acids in section 13.3.
• This process is marked as (II) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the sodium salt of the ethanoic acid.
10. The preparation of the above sodium salt is a simple process.
• Ethanoic acid is an acid. It will react with the base NaOH to give the sodium salt.
• This process is marked as (I) in fig.13.95 above.
• So we have worked in the reverse order from benzene to ethanoic acid.

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In the next section we will see aromatic hydrocarbons.

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