Chapter 13.20 - Electrophilic Substitution Reactions of Arenes

In the previous section, we saw aromaticity. In this section, we will see preparation of benzene. Later in this section, we will see physical and chemical properties of benzene.

Preparation of Benzene

For the industrial preparation of Benzene, we use coal tar. Coal tar is a thick black liquid, which is obtained as a by product while processing tar. The coal tar thus obtained is subjected to fractional distillation. Benzene is collected from the upper part of the fractionating column.

For the preparation of benzene in the lab, we can use any of the three methods explained below:
I. Cyclic polymerization of ethyne.
We have already seen this process in a previous section [see fig.13.94 in section 13.16]

II. Decarboxylation of aromatic acids.
• This can be explained in 3 steps:
1. In a decarboxylation reaction, a -COOH group or a -COONa group is removed from the molecule. One H atom replaces the removed group.
2. Consider the reaction shown in fig.13.109 below.

Fig.13.109

• It is clear that, on the reactant side, a -COONa group is occupying the position of an H atom.
• This reactant is the sodium salt of benzoic acid.
3. When this salt is heated with soda lime (soda lime is a mixture of NaOH and CaO), the -COONa group will be removed. An H atom will take it’s place. Thus we get benzene.

III. Reduction of phenol.
• This can be explained in 2 steps:
1. Consider the reaction shown in fig.13.110 below.

Fig.13.110

• It is clear that, on the reactant side, a -OH group is occupying the position of an H atom.
• This reactant is phenol.
2. When phenol vapours is passed over heated zinc dust, it get reduced to benzene.
• We can see that, the -OH group is removed. One H atom will take it’s place. Thus we get benzene.

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Physical properties of aromatic compounds

This can be written in 5 steps:
1. Aromatic compounds are non-polar molecules.
2. They are colorless liquids or colorless solids.
3. They have as a characteristic aroma.
• We are familiar with naphthalene balls.
   ♦ Naphthalene has a unique smell. It also has moth repellent property.
   ♦ So naphthalene balls can be used in toilets.
   ♦ They can also be used for preserving clothes.
4. Aromatic compounds are immiscible with water. But they are readily miscible with organic solvents.
5. Aromatic compounds burn with sooty flame. A flame with black carbon particles is called sooty flame. Some images can be seen here.

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Chemical properties of aromatic compounds

• We have to learn about two chemical properties of aromatic compounds:
(i) Electrophilic substitution reaction
(ii) Addition reaction

• First we will see electrophilic substitution reaction.
• An electrophile is a species which tries to gain an electron pair. It will enter into reaction with other species which can donate an electron pair.
• A chlorine atom which has lost an electron is an example for an electrophile. We know that, an ordinary chlorine atom will be trying to gain one electron to attain octet. If a chlorine atom has already lost an electron, it will be trying to gain an electron pair to attain octet.
• An electrophile is represented by the symbol $E^{\oplus}$.
   ♦ The '+' sign indicates that:
         ✰ The species is +ve charged.
         ✰ And the species is trying to gain electrons.
• In an electrophilic substitution reaction,
   ♦ An atom or group is removed from a molecule.
   ♦ An electrophile takes the place of that atom or group.
• An electrophilic substitution reaction is represented by the symbol S_E
   ♦ S stands for substitution
   ♦ E stands for electrophilic.

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• Recall that in section 13.17, we wrote this:
Benzene prefer substitution reaction rather than addition reaction. This is to preserve the ring structure. See fig.13.98 in section 13.17. 

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Let us see the mechanism of an S_E reaction. An S_E reaction proceeds in three steps:
(a) Generation of the electrophile ($E^{\oplus}$)
(b) Formation of carbocation intermediate
(c) Removal of proton from carbocation intermediate

• First we will see the generation of $E^{\oplus}$. This can be written in 4 steps:
1. In the fig.13.111 below, a Cl₂ molecule reacts with anhydrous AlCl₃ molecule.
(Anhydrous means, it contains no water. In many reactions, it is important to use anhydrous substances because water molecules if present, will enter into the reaction, giving unwanted results)

Fig.13.111

 
• The Cl₂ molecule undergoes heterolytic cleavage.
   ♦ The left Cl atom has lost two electrons.
         ✰ It is an electrophile.
         ✰ It is represented as $Cl^{\oplus}$.
   ♦ The right Cl atom attaches to the AlCl₃ to give tetrachloroaluminum ion.
• Note the Lewis structure of tetrachloroaluminum ion. The Al atom has four electrons around it. We see an extra yellow dot for Al. This electron was brought by the fourth Cl atom. Thus the molecule attains a -ve charge. We write it as [AlCl₄]^-.
2. We have seen how the electrophile $Cl^{\oplus}$ is formed. Now we will see how another electrophile $R^{\oplus}$ is formed.
• In the fig.13.112 below, a chloromethane molecule reacts with anhydrous AlCl₃ molecule.

Fig.13.112

• The chloromethane molecule undergoes heterolytic cleavage.
   ♦ The CH₃ group has lost two electrons.
         ✰ It is an electrophile.
         ✰ Since it is an alkyl group, it is represented as $R^{\oplus}$.
   ♦ The Cl atom attaches to the AlCl₃ to give tetrachloroaluminum ion.
3. Next we will see how another electrophile $RC^{\oplus} O$ is formed.
• In the fig.13.113 below, an acetyl chloride (CH₃COCl) molecule reacts with anhydrous AlCl₃ molecule.

Fig.13.113

• The acetyl chloride molecule undergoes heterolytic cleavage.
   ♦ The CH₃CO group has lost two electrons.
         ✰ A group with a 'double bonded O atom' and and an alkyl group is called an acyl group.
         ✰ It is represented as $RCO$
         ✰ When electrons are lost from this group, we represent it as $RC^{\oplus} O$
         ✰ It is an electrophile.
         ✰ It is called acylium ion.
   ♦ The Cl atom attaches to the AlCl₃ to give tetrachloroaluminum ion.
4. The last electrophile that we have to learn about, is the nitronium ion ($N^{\oplus} O_2$). It’s formation can be written in 5 steps:
(i) Fig.13.114 below shows the reaction between nitric acid (HNO₃) and sulphuric acid (H₂SO₄)

Fig.13.114

◼ Let us become familiarized with the charges in each of the reactants and products.
• Consider HNO₃ on the reactant side.
     An independent N atom has five electrons around it. But in HNO₃, it has only four electrons. So it has a +ve formal charge.  
     An independent O atom has six electrons around it. But in HNO₃, one of the O atoms has seven electrons. So it has a -ve formal charge.  
• On the product side, consider the protonated nitric acid.
     An independent O atom has six electrons around it. But in the protonated nitric acid, the left O atom has only five electrons. So it has a +ve formal charge.  
     The charges of the other O atom and the N atom are already explained when we considered HNO₃ on the reactant side.
• On the product side, consider the HSO₄^-.
     An independent O atom has six electrons around it. But in HSO₄^-, the left side O atom has seven electrons. So it has a -ve formal charge.  
(ii) Now we will see how the products are formed.
• The H₂SO₄ undergoes heterolytic cleavage. Both electrons in the O-H bond is retained the O atom. That is why we see a green dot for the O atom in HSO₄^-.
• The detached H atom now has zero electrons. The left side O atom of the nitric acid, donates both the electrons required for bond formation. The H atom thus gets attached to the nitric acid molecule. It is called protonated nitric acid because, the nitric acid has gained a proton. Recall that, a H atom with zero electrons is a proton.
• Thus stage I is complete.
(iii) The protonated nitric acid formed in stage I is very unstable. It decomposes into nitronium ion. This happens in stage II. Fig.13.115 below shows the process.

Fig.13.115

◼ Let us become familiarized with the charge in the nitronium ion.
     An independent N atom has five electrons around it. But in HNO₃, it has only three electrons. So it has a +ve formal charge.
(iv) Now we will see how the products are formed.
• The protonated nitric acid undergoes heterolytic cleavage. Both electrons in the N-O bond are retained by the O atom. Those two electrons help to form an independent water molecule.
• The right side of the heterolytic cleavage becomes the nitronium ion.
(v) We are familiar with acid-base reactions. But here, we see a reaction between two acids. The sulphuric acid is more powerful than nitric acid. So the nitric acid serves as the base. We know that acids are proton donors. Indeed we see in stage I that, the proton is transferred from the sulphuric acid to nitric acid.

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We have completed a discussion on 'generation of the electrophile. It is the first step in the reaction mechanism of electrophilic substitution reaction. In the next section, we will see the second step, which is 'formation of carbocation intermediate'.

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Chapter 13.3 - Preparation of Alkanes

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

We have to learn three methods used for preparing alkanes.
I. Preparation of alkanes from unsaturated hydrocarbons.
This can be written in 6 steps:
1. We know that unsaturated hydrocarbons contain double or triple bonds.
• If enough dihydrogen is supplied, those double or triple bonds can be converted into single bonds. Thus we can prepare alkanes. This process is known as hydrogenation.
2. This process requires the presence of finely divided catalysts like platinum, palladium or nickel.
• If platinum or palladium is used, room temperature will be sufficient to carry out the reaction.
• If nickel is used, high temperature and pressure will be required.
3. When catalysts are present, the H-H bond in dihydrogen breaks.
• The H atoms get absorbed on the surface of the catalysts.
• These H atoms are then transferred to the unsaturated hydrocarbons.
4. The chemical equation for the hydrogenation of ethene is:
$\rm{CH_2 = CH_2~+~H_2~ \color {green}{\xrightarrow[{}]{Pt/Pd/Ni}} ~ CH_3 - CH_3}$
5. The chemical equation for the hydrogenation of propene is:
$\rm{CH_3 - CH = CH_2~+~H_2~ \color {green}{\xrightarrow[{}]{Pt/Pd/Ni}} ~ CH_3 - CH_2 - CH_3}$
6. The chemical equation for the hydrogenation of propyne is:
$\rm{CH_3 - C ≡ CH~+~2H_2~ \color {green}{\xrightarrow[{}]{Pt/Pd/Ni}} ~ CH_3 - CH_2 - CH_3}$

II. Preparation of alkanes from alkyl halides
Method A:
This can be written in 4 steps:
1. Alkyl halides can be reduced using zinc and dilute hydrochloric acid to give alkanes.
• As an example, we can write the chemical equation for the reduction of chloromethane:
$\rm{CH_3 - Cl~+~H_2~ \color {green}{\xrightarrow[{}]{Zn,\;H^+}} ~ CH_4~+~HCl}$
• It is clear that the C-Cl bond in chloromethane undergoes fission. We have seen the mechanism of this fission in the previous chapter. [see fig.12.68 of section 12.10]
• We will see more details about this type of reactions in later sections.
2. Note that, in this method, we are dealing with alkyl halides.
Halogens are: F, Cl, Br and I
• But for the preparation of alkanes by this method, we cannot use alkyl fluorides. Only chlorides, bromides and iodides can be used. This is because, F being highly electronegative, the C-F bond will be very strong and hence difficult to break.
3. The chemical equation for the reduction of chloroethane is:
$\rm{CH_3 - CH_2 - Cl~+~H_2~ \color {green}{\xrightarrow[{}]{Zn,\;H^+}} ~ CH_3 - CH_3~+~HCl}$
4. The chemical equation for the reduction of 1-chloropropane is:
$\rm{CH_3 - CH_2 - CH_2 - Cl~+~H_2~ \color {green}{\xrightarrow[{}]{Zn,\;H^+}} ~ CH_3 - CH_2 - CH_3~+~HCl}$

Method B:
This can be written in 4 steps:
1. Alkyl halides can be treated with sodium metal in dry ether solution to give higher alkanes. (The ether solution should be dry because, if moisture is present, the sodium metal will react with that moisture)
• This reaction is known as Wurtz reaction. 
• As an example, we can write the chemical equation for the reaction between bromomethane and sodium metal:
$\rm{CH_3 - Br~+~2Na~+~Br - CH_3~ \color {green}{\xrightarrow[{}]{dry~ether}} ~ CH_3 - CH_3~+~2NaBr}$
2. Let us write the chemical equation when bromoethane is used:
$\rm{CH_3 - CH_2 - Br~+~2Na~+~Br - CH_2 - CH_3~ \color {green}{\xrightarrow[{}]{dry~ether}} ~ CH_3 - CH_2 - CH_2 - CH_3~+~2NaBr}$
3. In both the examples that we have seen, we have used the same type of alkyl halide.
    ♦ In the first example, we used bromomethane only.
    ♦ In the second example, we used bromoethane only.
◼ Now the question arises:
What will happen if we take different alkyl halides?
• The answer can be written in 5 steps:
(i) Suppose that, we take two different alkyl halides, say CH₃Br and CH₃CH₂Br
Then we will get three different higher alkanes as explained below:
(ii) Two molecules of CH₃Br will react with two sodium atoms to give one molecule of CH₃-CH₃.
• This is the same reaction that we wrote in step (1)
(iii) Two molecules of CH₃-CH₂-Br will react with two sodium atoms to give one molecule of CH₃-CH₂-CH₂-CH₃.
• This is the same reaction that we wrote in step (2)
(iv) One molecule of CH₃Br and one molecule of CH₃-CH₂-Br will react with two sodium atoms to give one molecule of CH₃-CH₂-CH₃.
• The equation is:
$\rm{CH_3 - Br~+~2Na~+~Br - CH_2 - CH_3~ \color {green}{\xrightarrow[{}]{dry~ether}} ~ CH_3 - CH_2 - CH_3~+~2NaBr}$
(v) So from (ii), (iii) and (iv), it is clear that, the product mixture will contain three different alkanes.
• The boiling points of these alkanes are close to each other. So it will be difficult to separate them.
• So we never take different alkyl halides as reactants in Wurtz reaction.
4. We just saw that, the same type of alkyl halide is taken in Wurtz reaction. Based on this information, we can obtain an interesting result. It can be written in steps:
(i) If the alkyl halide taken has an even number of C atoms, then the product alkane will also have an even number of C atoms.
• This is because, an even number added to another even number will always give an even number.
(ii) If the alkyl halide taken has an odd number of C atoms, then also, the product alkane will have an even number of C atoms.
• This is because, an odd number added to another odd number will always give an even number.
(iii) So we can write:
Wurtz reaction can be used to produce higher alkanes having even number of C atoms.

III. Preparation of alkanes from carboxylic acids
Method A:
This can be written in 2 steps:
1. Sodium salt of a suitable carboxylic acid is heated with soda lime (soda lime is a mixture of NaOH and CaO).
• During this reaction, one C atom, two O atoms and the Na atom is removed from the salt molecule.
• Thus the salt molecule gets converted to an alkane molecule.
2. Since one C atom and two O atoms are removed, we can say that, one carbon dioxide molecule is removed.
• This process of elimination of carbon dioxide from carboxylic acid is known as decarboxylation.
• As an example, we can write the chemical equation when sodium ethanoate (the sodium salt of ethanoic acid) is heated:
$\rm{CH_3 COO^- Na^+~+~NaOH~ \color {green}{\xrightarrow[{△}]{CaO}} ~ CH_4~+~Na_2 CO_3}$

Solved example 13.6
Sodium salt of which acid will be needed for the preparation of propane? Write the chemical equation of the reaction.
Solution:
1. During decarboxylation, one C atom is removed from the carboxylic acid.
• So we must consider that acid which has one C atom more than the number of C atoms in propane.
• That means, we must consider butanoic acid.
2. We can write:
Sodium salt of butanoic acid is needed for the preparation of propane.
3. The chemical equation is:
$\rm{CH_3 - CH_2 - CH_2 COO^- Na^+~+~NaOH~ \color {green}{\xrightarrow[{△}]{CaO}} ~ CH_3 - CH_2 - CH_3~+~Na_2 CO_3}$

Method B:
This can be written in 2 steps:
1. In an aqueous solution, the sodium salt of the carboxylic acid will dissociate as shown below:
$\rm{CH_3 COO^- Na^+~ \rightleftarrows  ~ CH_3 COO^- ~+~Na^+}$
• $\rm{CH_3 COO^- }$ is acetate ion.
• This dissociation process can be represented using Lewis structures as shown in fig.13.26 below:

Fig.13.26

• If this aqueous solution is subjected to electrolysis, the acetate ion will move towards the anode.
2. Consider the Lewis structure of acetate ion.
• We see that, all atoms have octet.
• But for oxygen, only the six brown electrons are original. The blue electron is acquired from the Na atom.
• So the species as a whole acquire a -ve charge. It is called the acetate ion
3. This acetate ion moves towards the anode.
• At the anode, it donates one of the brown electrons possessed by the right side O atom. The result is shown in fig.13.27(a) below:

Fig.13.27

4. So the acetate ion has lost the -ve charge. It is no longer an ion.
• But an unpaired electron is created. Thus the acetate ion has become acetate free radical.
• We have seen homolytic cleavage and formation of free radicals in the previous section [see fig.12.74 of section 12.11]
5. This free radical undergoes homolytic cleavage as shown in fig.13.27(b) above.
• As a result of this cleavage, two species are formed.
    ♦ The species on the left side is methyl free radical. It is shown in fig.c
    ♦ The species on the right side is shown in fig.d
6. So starting with an acetate ion, we obtained a methyl radical.
• A large number of acetate ions will be reaching the anode. Each of them will give a methyl free radical.
• We can group them into pairs. So each group will contain two methyl free radicals.
• Those two radicals can form a bond using the lone electrons. This will result in an ethane molecule. It is shown in fig.13.28(a) below:

Fig.13.28

7. We saw that, after the homolytic cleavage there will be two species.
• We saw that, the left side species (fig.13.27.a) will help to obtain ethane. Now we will see the right side species (fig.13.27.b).
• If we examine this species carefully, we can see that, the lone electrons can rearrange to form a double bond. This will result in a CO₂ molecule.
• This is shown in fig.13.28(b) above.
8. The above steps help us to understand the processes taking place at the anode. Those processes can be written in a condensed form as shown in fig.13.29 below:

Fig.13.29

9. Next we will see the reactions taking place at the cathode. It can be written in steps:
(i) Consider the Lewis structure of water molecule. It is shown in fig.13.30(a) below:

Fig.13.30

• If this water molecule is near the cathode, it can receive an electron from the cathode.
• Then one of the O-H bonds will undergo homolytic cleavage. This is shown in fig.b
(ii) The left portion is a hydrogen free radical. It is very unstable. It is shown in fig.c
(iii) The right portion is also unstable. It is shown in fig.d
(iv) But the right portion obtains one electron from the cathode. Thus it becomes the hydroxyl ion. It is shown in fig.e
• Note that in fig.e, the O atom has only the six brown electrons originally. The red electron is received from some other source. In our present case, it is received from the cathode. So the structure as a whole has a negative charge.
(v) So starting with a water molecule, we obtained a hydrogen free radical.
• A large number of water molecules will be present in the solution. Each of them will give a hydrogen radical.
• We can group them into pairs. So each group will contain two hydrogen free radicals.
• Those two radicals can form a bond using the lone electrons. This will result in a dihydrogen molecule.
(vi) The above 5 steps help us to understand the processes taking place at the cathode. Those processes can be written in a condensed form as shown below:
At cathode:
2H₂O + 2e^- ⟶ 2OH^- + 2 H•
H• + H• ⟶ H₂ ↑
10. This method is known as Kolbe's electrolytic method.
11. This method cannot be used to prepare methane. The reason an be written in 3 steps:
(i) In fig.13.28(a), we see that, two identical alkyl free radicals combine together to form an alkane molecule.
(ii) Each of the alkyl free radicals will contain at least one C atom. So the resulting alkane will obviously contain more than one C atom.
(iii) Methane has only one C atom. So methane cannot be prepared using this method.
12. This method for preparation will always give alkanes having even number of C atoms. The reason can be written in 4 steps:
(i) In fig.13.28(a), we see that, two identical alkyl free radicals combine together to form an alkane molecule.
(ii) If the free radical has an even number of C atoms, then the product alkane will also have an even number of C atoms.
• This is because, an even number added to another even number will always give an even number.
(iii) If the free radical taken has an odd number of C atoms, then also, the product alkane will have an even number of C atoms.
• This is because, an odd number added to another odd number will always give an even number.
(iv) So we can write:
Kolbe's electrolytic method will always give alkanes having even number of C atoms.

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• We have completed a discussion on the various methods for preparation of Alkanes.
• Those methods can be written in the form of a flow chart as shown in fig.13.31 below:

Fig.13.31

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In the next section we will see physical properties of alkanes.

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