In the previous section, we saw the details about inductive effect. In this section, we will see resonance.
We have seen some basics about resonance structures in a previous chapter (see section 4.9). Now we will see resonance in organic compounds. Some basics can be written based on some examples.
Example 1:
This can be written in 6 steps:
1. Consider the benzene molecule in fig.12.89(I) below.
• It has single and double bonds in an alternating arrangement.
♦ C2ㅡC3, C4ㅡC5, C6ㅡC1, are single bonds.
♦ C1ㅡC2, C3ㅡC4, C5ㅡC6, are double bonds.
 |
| Fig.12.89 |
2. Consider bond lengths in general:
♦ The length of a CㅡC single bond is 154 pm
♦ The length of a C=C double bond is 134 pm
• So we would expect both these lengths in a benzene molecule.
• But in reality, all CㅡC bonds in benzene molecule are of the same length, which is: 139 pm
• This 139 pm is an intermediate value of 154 pm and 134 pm.
• So the structure (I) is not acceptable.
3. Consider the molecule in fig.12.89(II). It is the mirror image of the molecule in (I).
♦ The single bonds in (I) are double bonds in (II)
♦ The double bonds in (I) are single bonds in (II)
• We can draw the Lewis structure of benzene in this way also. All the atoms have octet.
• But this structure is also not acceptable because, in reality, all bond lengths are the same.
4. Next we consider the energy of molecule (energy stored in the bonds of the molecule)
• Both I and II are energetically same.
♦ We cannot say that I is more stable than II.
♦ We cannot say that II is more stable than I.
5. So it is clear that, benzene cannot be represented by any one Lewis structure.
• We will have to draw both I and II.
◼ We can write:
♦ I and II are resonance structures of benzene.
♦ The actual benzene molecule is a hybrid of I and II.
6. Let us see the transfer of electrons taking place during resonance. It can be written in 6 steps:
(i) In structure I of fig.12.89(a) below, all the atoms have octet.
• The structure II is obtained by rearranging the electrons present in I
 |
| Fig.12.89(a) |
(ii) In I, C1=C2 is a double bond. But this same C1=C2 becomes a single bond in II.
• This is because of the movement of electrons indicated by the green curved-arrow.
♦ Two electrons from the C1=C2 moves to C2ㅡC3
♦ Those two electrons were co-owned by C1 and C2
• So due to the green curved-arrow,
♦ C1 loses both those electrons.
♦ C2 is not affected.
• We will soon see how C1 makes up for this loss.
(iii) Due to the green curved-arrow, C3 now possess two new electrons.
• But C3 already has octet. It does not need any electrons.
• So the magenta curved-arrow comes into play.
♦ Two electrons in C3=C4 moves to C4ㅡC5
• Thus the two unwanted electrons gained by C3 through the green curved-arrow, are now lost.
• Recall why the green curved-arrow did not affect C2. In the same way, the magenta curved-arrow will not affect C4
(iv) But the magenta curved-arrow will give two unwanted electrons to C5
• So the cyan curved-arrow comes into play.
♦ Two electrons in C5=C6 moves to C6ㅡC1
• Thus the two unwanted electrons gained by C5 through the magenta curved-arrow is now lost.
• Recall why the green curved-arrow did not affect C2.
• Also recall why the magenta curved-arrow did not affect C4.
• In the same way, the cyan curved-arrow will not affect C6
(v) The cyan curved-arrow will give two electrons to C1.
• Thus the electrons initially lost by C1 due to the green arrow, are regained.
(vi) Thus the three curved-arrows show us how the structure II is obtained.
• Even after structure II is obtained, the movement of electrons will continue. This is indicated by the three curved-arrows in II.
• Thus the benzene molecule resonates between I and II. The bonds are neither single nor double.
Example 2:
This can be written in 7 steps:
1. Consider the nitromethane (CH3NO2) molecule in fig.12.90(I) below.
• It has two nitrogen-oxygen bonds.
♦ One nitrogen-oxygen bond is a single bond.
♦ The other nitrogen-oxygen bond is a double bond.
 |
| Fig.12.90 |
2. Consider bond lengths in general:
♦ The length of a NㅡO single bond is 145 pm.
♦ The length of a N=O double bond is 115 pm.
• So we would expect both these lengths in a nitromethane molecule.
• But in reality, both the nitrogen-oxygen bonds in nitromethane are of the same length, which is:125 pm.
• This 125 pm is an intermediate value of 145 pm and 115 pm.
• So the structure (I) is not acceptable.
3. Consider the molecule in fig.12.90(II). It is the mirror image of the molecule in (I).
♦ The NㅡO single bond in (I) is a N=O double bond in (II)
♦ The N=O double bond in (I) is a NㅡO single bond in (II)
• We can draw the Lewis structure of nitromethane in this way also. All the atoms have octet.
• But this structure is also not acceptable because, in reality, both the nitrogen-oxygen bond lengths are the same.
4. Next we consider the energy of molecule (energy stored in the bonds of the molecule)
• Both I and II are energetically same.
♦ We cannot say that I is more stable than II.
♦ We cannot say that II is more stable than I.
5. So it is clear that, nitromethane cannot be represented by any one Lewis structure.
• We will have to draw both I and II.
◼ We can write:
♦ I and II are resonance structures of nitromethane.
♦ The actual nitromethane molecule is a hybrid of I and II.
6. The +ve sign given to N and -ve sign given to O are formal charges.
• We have seen how to calculate formal charges in section 4.4.
7. Let us see the significance of the curved arrows in fig.12.90 above. It can be written in 5 steps:
(i) The green and magenta curved-arrows show how resonance occurs.
(ii) The green curved-arrow in I shows that, the two electrons in the bond are shifted to the O atom.
• When this shift occurs, the double bond becomes a single bond.
• The two electrons become a lone pair of O. They are no longer available to N. So in effect, the N has now sextet.
(iii) Now the magenta curved-arrow comes into play. Two lone pair electrons of bottom O are shifted to become a double bond. Thus the N atom regains it's octet.
(iv) Thus the two curved-arrows show us how the structure II is obtained.
• Even after structure II is obtained, the movement of electrons will continue. This is indicated by the two curved-arrows in II.
• Thus the molecule resonates between I and II. The bonds are neither single nor double.
Example 3:
This can be written in 7 steps:
1. Consider the acetate ion (CH3CO2-) ion in fig.12.91(I) below.
• It has two carbon-oxygen bonds.
♦ One carbon-oxygen bond is a single bond.
♦ The other carbon-oxygen bond is a double bond.
 |
| Fig.12.91 |
2. Consider bond lengths in general:
♦ The length of a CㅡO single bond is different from the length of a C=O double bond.
• So we would expect two different carbon-oxygen bond lengths in an acetate ion.
• But in reality, both the carbon-oxygen bonds in acetate ion are of the same length.
• This 'same length' is an intermediate value of the two bond lengths.
• So the structure (I) is not acceptable.
3. Consider the molecule in fig.12.91(II). It is the mirror image of the molecule in (I).
♦ The CㅡO single bond in (I) is a C=O double bond in (II)
♦ The C=O double bond in (I) is a CㅡO single bond in (II)
• We can draw the Lewis structure of acetate ion in this way also. All the atoms have octet.
• But this structure is also not acceptable because, in reality, both the carbon-oxygen bond lengths are the same.
4. Next we consider the energy of molecule (energy stored in the bonds of the molecule)
• Both I and II are energetically same.
♦ We cannot say that I is more stable than II.
♦ We cannot say that II is more stable than I.
5. So it is clear that, acetate ion cannot be represented by any one Lewis structure.
• We will have to draw both I and II.
◼ We can write:
♦ I and II are resonance structures of acetate.
♦ The actual acetate molecule is a hybrid of I and II.
6. The -ve sign given to O is the formal charge.
• We have seen how to calculate formal charges in section 4.4.
7. Let us see the significance of the curved arrows in fig.12.91 above. It can be written in 5 steps:
(i) The green and magenta curved-arrows show how resonance occurs.
(ii) The green curved-arrow in I shows that, the two electrons in the bond are shifted to the O atom.
• When this shift occurs, the double bond becomes a single bond.
• The two electrons become a lone pair of O. They are no longer available to C. So in effect, the C has now sextet.
(iii) Now the magenta curved-arrow comes into play. Two lone pair electrons of bottom O are shifted to become a double bond. Thus the C atom regains it's octet.
(iv) Thus the two curved-arrows show us how the structure II is obtained.
• Even after structure II is obtained, the movement of electrons will continue. This is indicated by the two curved-arrows in II.
• Thus the molecule resonates between I and II. The bonds are neither single nor double.
Based on the above examples, we can write 5 important points related to resonance.
1. We see that, resonance is due to the movement of electrons.
• Always, those electrons in double bonds are being moved. The reason can be written in steps:
(i) The two electrons in a single bond will not be able to move because, single bond is a sigma bond.
♦ The sigma bond is a linear overlap of orbitals.
♦ Thus the electrons are locked in position.
♦ So we do not see any movement of electrons of single bonds.
(ii) In a double bond, there will be a sigma bond and a pi bond.
♦ The pi bond occurs due to the lateral overlap of orbitals.
♦ The electrons in a pi bond can move.
2. When the electrons are locked in a bond, we say that, those electrons are localized.
• When electrons move away from a bond, they become unpaired electrons of an atom. In such a situation, the electrons appear as clouds. We then say that, those electrons are delocalized.
• When delocalization of electrons take place, the negative charge is spread out on a larger space in the molecule. This gives greater stability to the molecule.
3. All resonance structures must be identical as far as the positions of the atoms are concerned.
• Changes in the positions of atoms are not allowed.
• Only electrons are allowed to move.
4. All resonance structures must be Lewis structures. Each atom in the structure must have octet.
• The only exception allowed is for the carbocation. (carbon with 6 electrons)
5. When there is resonance, a single Lewis structure will not be sufficient to describe a molecule.
• We will need to show all the possible Lewis structures.
• The actual molecule will be a hybrid of the various resonance structures.
• The hybrid will have an average characteristics of the individual resonance structures.
We saw that a molecule can have two or more resonance structures. Our next aim is to compare the stability of those resonance structures. We will see it in the next section.
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