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.
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.
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 SE
♦ S stands for substitution
♦ E stands for electrophilic.
•
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.
Let us see the mechanism of an SE reaction. An SE 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 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.
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'.
Copyright©2023 Higher secondary chemistry.blogspot.com
No comments:
Post a Comment