In the previous section, we saw the explanation for the unusual stability of benzene. In this section, we will see aromaticity.
Suppose that, we are given a compound. We must be able to check whether it is an aromatic compound or not. For that, first we must learn about conjugated systems.
In the previous chapter, we saw a definition for conjugated systems [see step (6) just above fig.12.102 in section 12.16].
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We wrote that:
A conjugated system is that system in which single bonds and double bonds occur in an alternating arrangement
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Now we will see a more general definition. A general definition will be more convenient because, it will be applicable in all cases. It can be written in 5 steps:
1. Consider the benzene molecule. It is a ring structure. All six molecules which make up the ring, are C atoms.
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Each of those six C atoms has a p orbital.
2. Consider the naphthalene molecule. It is a ring structure. All ten molecules which make up the ring, are C atoms.
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Each of those ten C atoms has a p orbital.
3. The definition for conjugated system is:
If every atom which make up the ring has a p orbital, then that system is called a conjugated system.
4. We have seen heterocyclic compounds [see fig.12.21(ii) in section 12.3]. In such compounds, there will be other atoms like O or N in the ring. Such systems can also be considered as conjugated systems if the hetero atom (hetero atom is an atom other than C) also have a p orbital. The definition written in (3) is valid for heterocyclic compounds also.
5. What is the significance of every atom which make up the ring, “having a p orbital”?
The answer is simple:
If p orbitals are present in every atom in the ring, then lateral bonding is possible for the entire area of the ring. The whole ring will become very stable.
◼ So to determine whether a system is conjugated, we need to acquire a special skill:
The skill to check each atom in the ring and tell whether that atom has a p orbital.
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The following four steps will help us to acquire this skill.
1. An atom which is sp3 hybridized will not have a p orbital.
♦ One s orbital and three p orbitals mix together to give four sp3 hybrid orbitals.
♦ No p orbitals will be available after the hybridization.
2. An atom which is sp hybridized will have two p orbitals.
♦ One s orbital and one p orbital mix together to give two sp hybrid orbitals.
♦ Two p orbitals will be available after the hybridization.
♦ Two p orbitals are not acceptable. There must be only one p orbital.
3. An atom which is sp2 hybridized will have a p orbital.
♦ One s orbital and two p orbitals mix together to give three sp2 hybrid orbitals.
♦ One p orbital will be available after the hybridization.
4. So if the atom that we check is sp2 hybridized, then it will have a single p orbital.
◼ So the next skill that we need to acquire is this:
The skill to check each atom in the ring and tell the hybridization of that atom.
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The following two steps will help us to acquire this skill:
1.Find the steric number (SN) of the atom.
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SN of an atom is the sum of two numbers:
(a) The number of atoms attached to the atom under consideration.
(b) The number of lone pairs of electrons possessed by the atom under consideration.
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Based on the SN, we can determine the hybridization using the following chart:
♦ If SN is 4, then the atom is sp3 hybridized.
♦ If SN is 3, then the atom is sp2 hybridized.
♦ If SN is 2, then the atom is sp hybridized.
2. The above step gives us an easy method to determine hybridization of any atom.
However, there is an exception. This can be written in four steps:
(i) Consider the atom whose hybridization is to be determined.
(ii) Check whether that atom satisfies two conditions:
(a) The atom has one or more lone pairs of electrons.
(b) The atom is attached to another atom which is sp2 hybridized.
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If both (a) and (b) are satisfied, then the hybridization of the atom under consideration will be sp2
(iii) While using the SN method, the hybridization may work out to be sp3. But if both (a) and (b) are satisfied, then we must use sp2 only.
(iv) This exception is due to resonance effect. We will see more details about it in higher classes.
◼ Let us see an example where we apply the above two steps:
The example can be written in 8 steps:
(i) Consider the C atom marked as ‘2’ in fig.13.105 below:
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| Fig.13.105 Cyclopentadienyl anion |
(ii) This C atom is attached to two other C atoms. Also it is attached to one H atom. So in total, this C atom is attached to three atoms.
(iii) There are no lone pairs for this C atom.
(iv) So SN = (3+0) = 3
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Since SN is 3, this C atom is sp2 hybridized.
(v) Consider the C atoms 3, 4 and 5. Each of them are attached to three atoms. Also, lone pairs are absent. So these C atoms have SN value 3 and thus they are sp2 hybridized.
(vi) We can write: The C atoms 2, 3, 4 and 5 in fig.13.105 are sp2 hybridized.
(vii) Consider the C atom marked as ‘1’ in the fig.105. This C atom has a lone pair of electrons. Also, this C atom is attached to another C atom which is sp2 hybridized.
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So this C atom is sp2 hybridized.
(viii) We can write: All C atoms in fig.13.105 are sp2 hybridized.
◼ Let us see another example:
The example can be written in 12 steps:
(i) Consider the C atom marked as ‘2’ in fig.13.106(a) below:
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| Fig.13.106 Cycloheptatrienyl cation |
(ii) This C atom is attached to two other C atoms. Also it is attached
to one H atom. So in total, this C atom is attached to three atoms.
(iii) There are no lone pairs for this C atom.
(iv) So SN = (3+0) = 3
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Since SN is 3, this C atom is sp2 hybridized.
(v) Consider the C
atoms 3, 4, 5, 6 and 7. Each of them are attached to three atoms. Also, lone
pairs are absent. So these C atoms have SN value 3 and thus they are
sp2 hybridized.
(vi) We can write: The C atoms 2, 3, 4, 5 and 6 in fig.13.106(a) are sp2 hybridized.
(vii)
Consider the C atom marked as ‘1’ in the fig.106(a). This C atom has a
+ve charge.
(viii) This C atom has no lone pair of electrons. So we cannot use the exception rule.
(ix) In this situation, we draw a Lewis structure related to this C atom.
♦ C1 is shown to have a +ve charge.
♦ An independent C atom has four valence electrons.
♦ Since now there is a +ve charge, there must be three electrons around C1
♦ Let us distribute those three electrons:
✰ One among the three, is for C2
✰ One among the three, is for C7
✰ The remaining one is for an H atom.
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Thus we get the Lewis structure in fig.13.106(b)
(x) Based on the Lewis structure, we can write:
C1 is attached to three atoms.
(xi) Since C1 is attached to three atoms, the hybridization of C1 is sp2
(xii) We can write: All C atoms in fig.13.106(a) are sp2 hybridized.
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For a molecule to be aromatic, it must be a conjugated system. After acquiring the above skills, we will be able to tell whether a given system is a conjugated system or not.
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It can be summarized in 2 steps:
1. First we check each atom in the system and note down it’s hybridization.
2. If all atoms in the system are sp2 hybridized, then that system is a conjugated system.
"Being a conjugated system" is only one of the conditions for aromaticity that a molecule must satisfy. Now let us see another condition. It can be written in 4 steps:
1. π electrons are those electrons which reside in the p-orbitals of the atom. They are called π electrons because, they take part in the π bonds.
2. We must count the number of all the π electrons in a conjugated system. Let this number be N
3. The number N must satisfy the equation: N = 4n+2
♦ Where n is an integer (n = 0, 1, 2, 3, . . .)
◼ This rule is known as Hückel rule.
4. To apply this rule, we use the following 5 steps:
(i) Count the number N
(ii) Equate it to 4n+2
(iii) solve the equation for n
(iv) If n is an integer, then the given compound satisfies the rule.
(v) If n is a fraction, then the given compound does not satisfy the rule.
Let us see some examples:
Example I:
Fig.13.107(a) below shows cyclohexane. All atoms in the ring are sp3 hybridized. So there are no p orbitals. Consequently, there are no π electrons. So cyclohexane does not satisfy Hückel rule.
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| Fig.107 |
Example II:
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This can be written in two steps:
(i) Fig.13.107(b) above shows benzene. All six C atoms in the ring are sp2 hybridized. So all atoms will have one p orbital each. So there are six p orbitals. In each of those six p orbitals, there is one electron. So in total, there are six π electrons. That means, N = 6
(ii) Now we equate this value of N to (4n+2). We get: 6 = 4n + 2
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Solving for n, we get: n = 1
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‘1’ is an integer. So benzene satisfies Hückel rule.
Example III:
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This can be written in two steps:
(i) Fig.13.108(a) below shows
naphthalene.
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| Fig.13.108 |
• Using the skills that we acquired in this section, we will be able to write:
All ten C atoms in the rings are sp2 hybridized.
• So all atoms will have one p orbital each. So there are ten p orbitals. In each of those ten p orbitals, there is one electron. So in total, there are ten π electrons.
• That means, N = 10
(ii) Now we equate this value of N to (4n+2). We get: 10 = 4n + 2
• Solving for n, we get: n = 2
• ‘2’ is an integer. So naphthalene satisfies Hückel rule.
Example IV:
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This can be written in two steps:
(i) Fig.13.108(b) above shows
anthracene.
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Using the skills that we acquired in this section, we will be able to write:
All fourteen C atoms in the rings are sp2 hybridized.
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So all atoms
will have one p orbital each. So there are fourteen p orbitals. In each of
those fourteen p orbitals, there is one electron. So in total, there are fourteen
π electrons.
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That means, N = 14
(ii) Now we equate this value of N to (4n+2). We get: 14 = 4n + 2
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Solving for n, we get: n = 3
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‘3’ is an integer. So anthracene satisfies Hückel rule.
Example V:
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This can be written in two steps:
(i) Fig.13.108(c) above shows
phenenthrene.
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Using the skills that we acquired in this section, we will be able to write:
All fourteen C atoms in the rings are sp2 hybridized.
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So all atoms
will have one p orbital each. So there are fourteen p orbitals. In each of
those fourteen p orbitals, there is one electron. So in total, there are fourteen
π electrons.
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That means, N = 14
(ii) Now we equate this value of N to (4n+2). We get: 14 = 4n + 2
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Solving for n, we get: n = 3
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‘3’ is an integer. So phenanthrene satisfies Hückel rule.
Example VI:
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This can be written in two steps:
(i) An earlier fig.13.105 above shows
cyclopentadienyl anion.
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We saw that:
All five C atoms in the ring are sp2 hybridized.
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So all atoms
will have one p orbital each. So there are five p orbitals.
♦ The p orbitals of C2, C3, C4 and C5, will contain one electron each
♦ The p orbital of C1 will contain the lone pair.
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So in total, there are (4+2) = 6
π electrons.
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That means, N = 6
(ii) Now we equate this value of N to (4n+2). We get: 6 = 4n + 2
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Solving for n, we get: n = 1
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‘1’ is an integer. So cyclopentadienyl anion satisfies Hückel rule.
Example VII:
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This can be written in two steps:
(i) An earlier fig.13.106 above shows
cycloheptatrienyl cation.
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We saw that:
All seven C atoms in the ring are sp2 hybridized.
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So all atoms
will have one p orbital each. So there are seven p orbitals.
♦ The p orbitals of C2, C3, C4, C5, C6 and C7 will contain one electron each
♦ The p orbital of C1 will contain zero electrons.
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So in total, there are (6+0) = 6
π electrons.
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That means, N = 6
(ii) Now we equate this value of N to (4n+2). We get: 6 = 4n + 2
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Solving for n, we get: n = 1
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‘1’ is an integer. So cycloheptatrienyl cation satisfies Hückel rule.
Now we can write the conditions for aromaticity.
For a molecule to be considered as aromatic, the following four conditions must be satisfied:
1. The molecule must be planar.
2. The molecule must have a ring structure.
3. The molecule must be a conjugated system (complete delocalization of electrons).
4. The molecule must satisfy Hückel rule.
• The discussion in this section helps us to acquire the required skills to check whether a given molecule is aromatic or not.
In the next section, we will see preparation of benzene.
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