Chapter 13.10 - Preparation Of Alkenes

In the previous section, we completed a discussion on the isomerism in alkenes. In this section, we will see preparation of alkenes.

We will see four different methods for the preparation of alkenes.
A. From alkynes
B. From alkyl halides
C. From vicinal dihalides
D. From alcohols by acidic dehydration

A. From alkynes
This can be written in 10 steps:
1. In this method, we supply dihydrogen gas, which reacts with the alkyne.
• The dihydrogen gas gives H atoms which can be used to satisfy the valencies of the triple bonded C atoms.
• The triple bond thus becomes a double bond. The equation is shown in fig.13.63 below:

Fig.13.63

• In the above equation,
   ♦ R represents an alkyl group.
   ♦ R¹ represents another alkyl group.
   ♦ $\rm{CH_3 - C ≡ C_2 H_5}$ is an example of such an alkyne.
2. We must supply only calculated amounts of dihydrogen.
• If excess dihydrogen is available, the double bonds will become single bonds. This will result in alkanes instead of alkenes.
3. When H atoms are added to an alkyne, we say that the alkyne is reduced to an alkene. The reason can be written in steps:
(i) The C atom makes a new C-H bond with the incoming H atom.
(ii) The incoming H atom has one electron available. This electron forms a part of the new C-H bond.
(iii) The two electrons in the C-H bond will be attracted towards the C atom. This is because, C is more electronegative than H.
(iv) So the C atom gains some extra negative charge. That is, the C atom is reduced.
(v) Since the two triple bonded C atoms in the alkyne are reduced in this way, we say that, the alkyne is reduced.
4. In our present case, the reduction of the alkyne is partial.
• If complete reduction occur, the alkyne will become an alkane.
• As mentioned in (2), the complete reduction is prevented by supplying only calculated amounts of dihydrogen.
5. For the reaction to take place, we must provide palladised charcoal as a catalyst.
• This palladised charcoal must be partially deactivated with poisons like sulphur compounds or quinoline.
• Partially deactivated palladised charcoal is known as Lindlar's catalyst.
6. As a result of this reaction, we get cis-Alkenes. We saw this in fig.13.63 above.
7. If instead of palladised charcoal, we use sodium in liquid ammonia, we get trans alkene. The equation is shown in fig.13.64 below:

Fig.13.64

8. If the alkyne selected is ethyne, we will get ethene. The equation is:
$\rm{CH ≡ CH~+~H_2~ \color {green}{\xrightarrow[{}]{Pd/C}} ~ CH_2 = CH_2}$
9. If the alkyne selected is propyne, we will get propene. The equation is:
$\rm{CH_3 - C ≡ CH~+~H_2~ \color {green}{\xrightarrow[{}]{Pd/C}} ~ CH_3 - CH = CH_2}$
• The propene thus obtained will not show cis/trans isomerism. The reason can be written in 2 steps:
(i) Consider the right C atom involved in the double bond.
(ii) Two identical atoms (H atoms) are attached to this C atom. So the compound cannot show cis/trans isomerism.
10. Based on the above steps, we will now write a summary about the process:
Alkynes on partial reduction with calculated amount of dihydrogen in the presence of palladised charcoal partially deactivated with poisons like sulphur compounds or quinoline give alkenes.

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B. From alkyl halides
This can be written in 4 steps:
1. In this method, the alkyl halide is heated with alcoholic potash.
• Alcoholic potash is prepared by dissolving potassium hydroxide in an alcohol like ethanol.
2. During the reaction, one molecule of halogen acid will be eliminated from the alkyl halide.
• When such a molecule is eliminated, the remaining atoms in the alkyl halide rearrange to form an alkene. An example is shown in fig.13.65 below:

Fig.13.65

 
• We know that, X represents a halogen atom (Cl, Br, I). In the above fig., we see that, an H atom and the X atom are removed from the original alkyl halide. (HX represents a molecule of halogen acid)
• Since halogen acid is being removed, this reaction is called dehydrohalogenation.
3. In the original alkyl halide, the C atom to which X is attached is the 𝛼 carbon atom.
• The C atom next to the 𝛼 carbon atom is called β carbon atom.
• Since the H atom is removed from the β carbon atom, this reaction is a β-elimination reaction.
4. Now we will see the rate of this reaction. It can be written in steps:
(i) Suppose that the alkyl part ‘R’ is fixed. Then the possible original alkyl halides that we can take are: R-Cl, R-Br and R-I
• It is found that:
   ♦ R-I will give more alkene in unit time than R-Cl and R-Br.
   ♦ R-Br will give more alkene in unit time than R-Cl.
• So we can write:
For halogens, the rate of the reaction decreases in the order: I > Br > Cl
(ii) Suppose that, the halogen X is fixed. Then the possible original alkyl halides that we can take are: Primary alkyl halide, secondary alkyl halide and tertiary alkyl halide.
• It is found that:
   ♦ Tertiary alkyl halide will give more alkene in unit time than primary and secondary.
   ♦ Secondary alkyl halide will give more alkene in unit time than primary.
• So we can write:
For alkyl groups, the rate of the reaction decreases in the order: tertiary > secondary > primary

• Consider the C atom which is holding the X atom. If this C atom is attached to only one other C atom, then that molecule as a whole, is a primary alkyl halide. • Consider the C atom which is holding the X atom. If this C atom is attached to two other C atoms, then that molecule as a whole, is a secondary alkyl halide. • Consider the C atom which is holding the X atom. If this C atom is attached to three other C atoms, then that molecule as a whole, is a tertiary alkyl halide. • The above classification is similar to what we saw in 1o, 2o and 3o carbon atoms [see step (11) below fig.13.8 in section 13.1].

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C. From vicinal dihalides
This can be written in 2 steps:
1. Dihalides are compounds which contain two halogen atoms.
• If in a dihalide, the two halogen atoms are attached to two adjacent C atoms, then that dihalide is called a vicinal dihalide.
2. In this process, a vicinal dihalide is treated with zinc metal.
• When the two X atoms are removed, the remaining atoms in the dihalide rearrange to form an alkene.
• The two X atoms combines with Zn to form ZnX₂.
• Two examples are written below:
CH₂Br-CH₂Br + Zn ⟶ CH₂=CH₂ + ZnBr₂
CH₃CHBr-CH₂Br + Zn ⟶ CH₃CH=CH₂ + ZnBr₂

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D. From alcohols by acidic dehydration
This can be written in 2 steps:
1. We know that, alcohols are the hydroxy derivatives of alkanes [see solved example 13.2 in section 13.1].
• So alcohols can be represented as R-OH
   ♦ R is an alkyl group.
2. Consider the C atom of the R-OH which holds the OH group.
   ♦ One valency of that C will be satisfied by another C atom.
   ♦ Two valencies will be satisfied by two H atoms.
   ♦ The fourth valency will be satisfied by the OH group.
• If we can remove the OH group and an additional H atom, the atoms in the R-OH will have to rearrange and become an alkene.
3. To remove the OH group and the H atom, we heat the alcohol with concentrated sulphuric acid. The equation is shown in fig.13.66 below:

Fig.13.66

• Since H and OH are removed, we say that, a molecule of water is removed.
• Since a molecule of water is removed, it is a dehydration reaction.
• Since the dehydration is achieved in the presence of an acid, this reaction is called acidic dehydration reaction.
4. From fig.13.66, it is clear that, the H atom belonging to the β carbon atom is being removed. So this reaction is an example of β-elimination reaction.

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In the next section we will see properties of alkenes.

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