In the previous section, we saw the details about stability of resonance structures. In this section, we will see resonance effect.
• We know that inorganic molecules like HCl are polar.
♦ The Cl atom being electronegative, pulls the electron density in the covalent bond.
♦ This gives Cl a partial -ve charge. Also H gets a partial +ve charge.
♦ See section 4.10.
• Some organic molecules are also polar. This can be demonstrated using two examples.
Example 1:
In this example, we analyze the resonance structures of aniline. It can be written in 6 steps:
1. The structure I of fig.12.101 below shows aniline. In this structure, all the atoms have octet.
 |
| Fig.12.101 |
• The structure II is obtained by rearranging the electrons present in I
• In I, NㅡC1 is a single bond. But this same NㅡC1 becomes a double bond in II.
• This is because of the movement of electrons indicated by the green curved-arrow.
♦ Two electrons from the lone pair of N moves to NㅡC1
♦ Those two electrons were singly-owned by N.
♦ After the formation of N=C1, the N co-owns those electrons.
♦ So N does not lose any electrons.
• So due to the green curved-arrow,
♦ N does not lose any electrons.
♦ C1 gains two electrons.
• Note the + formal charge of N in II. This can be explained in 3 steps:
(i) An independent N atom will have five electrons around it.
(ii) But in II, there are only four electrons around N.
♦ Two from the bonds with the H atoms.
♦ Two from the double bond.
(iii) That means, N has lost an electron. Thus it gains one +ve formal charge.
• Due to the green curved-arrow in I, C1 now possess two new electrons.
• But C1 already has octet. It does not need any electrons.
• So the magenta curved-arrow comes into play.
♦ Two electrons in C1=C2 moves to C2
♦ They become a lone pair of C2
• Thus the two unwanted electrons gained by C1 through the green curved-arrow, are lost by the action of magenta curved arrow.
• Note the - formal charge of C2 in II. This can be explained in 3 steps:
(i) An independent C atom will have four electrons around it.
(ii) But in II, there are five electrons around C2
♦ Two from the lone pair.
♦ One from C2ㅡC1
♦ One from C2ㅡC3
♦ One from C2ㅡH
(iii) That means, C2 has gained an electron. Thus it gains one -ve formal charge.
2. From the above step (1), we know how structure II is formed. So let us analyze the curved-arrows in structure II
• The C2 in II has two unwanted electrons. So the yellow curved-arrow comes into play.
♦ The lone pair of C2 moves to C2ㅡC3.
♦ Thus C2ㅡC3 becomes C2=C3.
♦ This double bond is shown in III
• When such a double bond is formed, C3 will have two unwanted electrons.
♦ So the cyan curved-arrow comes into play.
♦ Two electrons in C3=C4 moves to C4
♦ They become a lone pair of C4
• Thus the two unwanted electrons gained by C3 through the yellow curved-arrow, are lost by the action of cyan curved arrow.
• Note the - formal charge of C4 in III. This can be explained in 3 steps:
(i) An independent C atom will have four electrons around it.
(ii) But in II, there are five electrons around C4
♦ Two from the lone pair.
♦ One from C4ㅡC3
♦ One from C4ㅡC5
♦ One from C4ㅡH
(iii) That means, C4 has gained an electron. Thus it gains one -ve formal charge.
3. From the above step (2), we know how structure III is formed. We can write similar steps to analyze the curved-arrows in structure III.
• Based on those steps, we would see how structure IV is formed.
• The reader is advised to write those steps in his/her own notebook.
4. In effect, we see that:
• Structure I is neutral.
• Structures II, III and IV are not neutral. They are charged.
♦ This is due to the movement of electrons as indicated by the curved arrows.
♦ Positions with -ve formal charge have high electron densities.
♦ Positions with +ve formal charge have low electron densities.
5. So one contributing structure is neutral while the remaining three contributing structures are charged.
• So the hybrid structure will be charged. That means, in a sample of aniline, the molecules will exist in a charged state. This is just like charged molecules in a sample of HCl.
6. In this example, the electrons are transferred away from the substituent group (ㅡNH2 group).
• If the transfer of electrons is away from an atom or substituent group attached to the conjugated system, it is known as positive resonance effect (+R effect)
• A conjugated system is that system in which single bonds and double bonds occur in an alternating arrangement.
Example 2:
In this example, we analyze the resonance structures of nitrobenzene. It can be written in 6 steps:
1. The structure I of fig.12.102 below shows nitrobenzene. In this structure, all the atoms have octet.
 |
| Fig.12.102 |
• The arrow between N and O indicates that, the bond between N and that O is a coordinate bond. Both electrons in that bond originally belonged to N. However, that coordinate bond has no role in our present discussion.
• The structure II is obtained by rearranging the electrons present in I
• Consider the N=O double bond in I. This same N=O becomes a single bond in II.
• This is because of the movement of electrons indicated by the green curved-arrow.
♦ Two electrons in the N=O moves to O.
♦ They become a lone pair of O
• Due to this movement of electrons, N has lost two electrons. So the magenta curved-arrow comes into play.
♦ Two electrons in C1=C2 moves to NㅡC1
♦ Thus C1=C2 becomes C1ㅡC2 and NㅡC1 becomes N=C1
• Note the - formal charge of O in II. This can be explained in 3 steps:
(i) An independent O atom will have six electrons around it.
(ii) But in II, there are seven electrons around that O
♦ Six from three lone pairs.
♦ One from NㅡO
(iii) That means, O has gained an electron. Thus it gains one -ve formal charge.
• Note the + formal charge of C2 in II. This can be explained in 3 steps:
(i) An independent C atom will have four electrons around it.
(ii) But in II, there are only three electrons around C2.
♦ One from bond with the H atom.
♦ One from C2ㅡC1
♦ One from C2ㅡC3
(iii) That means, C2 has lost an electron. Thus it gains one +ve formal charge.
• Thus the two electrons lost by N through the green curved-arrow, are gained by the action of magenta curved arrow.
2. From the above step (1), we know how structure II is formed. So let us analyze the curved-arrow in structure II
• The C2 in II needs two electrons. So the yellow curved-arrow comes into play.
♦ Two electrons in C3=C4 moves to C2ㅡC3
♦ Thus C3=C4 becomes C3ㅡC4 and C2ㅡC3 becomes C2=C3.
♦ This is shown in III.
• Note the + formal charge of C4 in III. This can be explained in 3 steps:
(i) An independent C atom will have four electrons around it.
(ii) But in III, there are only three electrons around C4.
♦ One from bond with the H atom.
♦ One from C4ㅡC3
♦ One from C4ㅡC5
(iii) That means, C4 has lost an electron. Thus it gains one +ve formal charge.
3.
From the above step (2), we know how structure III is formed. We can
write similar steps to analyze the curved-arrow in structure III.
• Based on those steps, we would see how structure IV is formed.
• The reader is advised to write those steps in his/her own notebook.
4. In effect, we see that:
• Structure I is neutral.
• Structures II, III and IV are not neutral. They are charged.
♦ This is due to the movement of electrons as indicated by the curved arrows.
♦ Positions with -ve formal charge have high electron densities.
♦ Positions with +ve formal charge have low electron densities.
5. So one contributing structure is neutral while the remaining three contributing structures are charged.
•
So the hybrid structure will be charged. That means, in a sample of
nitrobenzene, the molecules will exist in a charged state. This is just like
charged molecules in a sample of HCl.
6. In this example, the electrons are transferred towards the substituent group (ㅡNO2 group).
• If the transfer of electrons is towards an atom or substituent group attached to the conjugated system, it is known as negative resonance effect (-R effect).
◼ Note that:
• In example 1 (+R effect), the C atoms in the chain acquired greater electron densities. This is because of the movement of electrons away from the substituent group.
• In example 2 (-R effect), the C atoms in the chain lost electron
densities. This is because of the movement of electrons towards the
substituent group.
• The atoms or substituent groups which cause +R effect are:
ㅡ halogen, ㅡOH, ㅡOR, ㅡOCOR, ㅡNH2, ㅡNHR, ㅡNR2, ㅡNHCOR
• The atoms or substituent groups which cause -R effect are:
ㅡCOOH, ㅡCHO, >C=O, ㅡCN, ㅡNO2
• As mentioned before, the resonance effect (+R effect and -R effect) occur in conjugated systems. In conjugated systems, single bonds and double bonds in the chain are present in an alternating arrangement.
• Benzene and substituted benzene compounds are examples of a closed chain conjugated system.
• 1,3-butadiene is an example of an open chain conjugated system. It’s condensed formula is shown below:
CH3=CH2ㅡCH2=CH3
◼ Why is it that, the conjugated systems are subjected to +R effect and -R effects ?
• The answer can be written in 3 steps:
1. We know that, in a double bond, there is a 𝜎 bond and a π bond.
• In a 𝜎 bond, there is linear overlap of orbitals. The electrons are locked in position.
• But in a π bond, there is lateral overlap of orbitals. So the electrons are free to move.
2. So the electrons of the π bond move towards an adjacent single bond. That single bond will then become a double bond. We indicated such movements using curved-arrows in the above examples.
• The electrons of the π bond can also move towards an adjacent atom. Those atoms will then become regions of high electron densities. We indicated such movements also using curved-arrows in the above examples.
3. Regions of ‘electron richness’ and ‘electron deficiency’, will make the molecule polar.
In the next section we will see electromeric effect.
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