In the previous section, we saw the details about nucleophiles and electrophiles. In this section, we will see electron movement in organic reactions.
• The movement of electrons in organic chemical reactions can be shown by curved-arrow notation. This can be explained using some examples.
Example 1:
This can be written in 3 steps:
1. Fig.12.86 below shows the reaction between OH- ion and CH3Br to produce methanol and Br- ion.
• We see that, all atoms on reactant side and product side have octet.
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| Fig.12.86 |
2. Consider the green curved arrow.
♦ It starts from the two yellow dots on the top of the O atom of the OH- ion.
♦ It ends at the C of CH3Br.
• That means, C gets two electrons of O.
• The C atom uses those two electrons to form the bond with O. This can be seen in the methanol molecule on the product side.
• So the green curved arrow helps us to track the movement of electrons.
3. Consider the magenta curved arrow.
♦ It starts from the bond between C and Br.
♦ It ends at the Br.
• That means, Br gets the two electrons in that bond.
• The Br then leaves with the two newly acquired electrons. It has octet. But the octet is attained using the one extra electron. So it becomes Br-.
• So the magenta curved arrow also helps us to track the movement of electrons.
Example 2:
• Each curved arrow indicates the movement of two electrons.
• If only one electron is moved, we use half headed curved arrows. Such arrow heads have the shape of fish hooks. In this example, we will see such a case. It can be written in steps:
1. Fig.12.87 below shows the dissociation of CH3Cl.
• We see that, all atoms on reactant side have octet.
• But the species on the product side do not have octet. They have unpaired electrons.
 |
| Fig.12.87 |
♦ It starts from the bond between C and Cl.
♦ It ends at the C of CH3Cl.
• That means, C gets one electron from the bond.
♦ The resulting species is: $\mathbf{\rm{H_3\overset{.}{C}}}$
♦ The dot indicates unpaired electron.
• So the green half headed curved arrow helps us to track the movement of one electron.
3. Consider the magenta half headed curved arrow.
♦ It starts from the bond between C and Cl.
♦ It ends at the Cl of CH3Cl.
• That means, Cl gets one electron from the bond.
♦ The resulting species is: $\mathbf{\rm{\overset{.}{Cl}}}$
♦ The dot indicates unpaired electron.
• So the magenta half headed curved arrow also helps us to track the movement of one electron.
• We will see more applications of curved arrows in later sections.
• At present we will see electron displacement effects in covalent bonds. It can be written in 3 steps:
1. Consider a sample of a pure substance like CH3Br. It is written as 'pure sample' because, the sample contains only CH3Br molecules.
• Consider any one molecule in that sample. There will be displacement of electrons in the C-Br bond.
• This is because the Br, being more electronegative, pulls the electrons in the bond.
• Thus we see that, displacement of electrons can occur even in pure samples where there is no external influence.
2. Now consider the reaction that we saw in fig.12.86 above. We see that the displacement of electrons occur as shown by the green and magenta curved arrows.
• Here, the sample is not pure. It contains OH- ions in addition to the CH3Br molecules.
• We can write: The displacement of electrons occurred due to the action of OH- ions, which is the attacking reagent.
3. So we see two cases:
(i) Displacement of electrons can occur in pure samples even when there is no external influence.
(ii) Displacement of electrons can also occur due to the action of an attacking reagent.
• Our next aim is to study the various types of electron displacements in detail.
Inductive effect
This can be explained in 9 steps:
1. Consider a covalent bond between two different atoms.
• If one of those two atoms is more electronegative than the other, there will be a shift of electron density towards the more electronegative atom.
• Such a bond is called a polar covalent bond. We have seen the details in an earlier section 4.10.
2. Fig.12.88 below shows the structure of chloro ethane.
• The C-Cl bond in this molecule is a polar covalent bond.
♦ The Cl atom gains some -ve charge (δ-).
♦ The C1 atom gains some +ve charge (δ+).
 |
| Fig.12.88 |
3. The shift of electron density is indicated by arrows. The arrow points in the direction in which the shift occurs.
♦ This is shown in fig.12.88(b) above.
4. We have seen how C1 acquires the +ve charge.
• Now, due to this +ve charge, C1 is able to pull the electrons in the C1C2 bond.
• So C2 loses some electron density and thus acquires a small +ve charge.
• The +ve charge acquired by C2 is smaller when compared to the δ+ of C1.
♦ So the +ve charge of C2 is denoted as δδ+.
◼ So we can write:
The polar covalent bond C1Cl, induces polarity in the adjacent bonds.
5. We know that all the bonds in CH3CH2Cl are σ bonds. So we can write a definition for inductive effect. It can be written in 3 steps:
(i) Consider a covalent bond (σ bond) which does not have any electronegative atoms on either of it’s ends.
(ii) Even though it does not have any electronegative atom, it get polarized. This is due to the polarization in an adjacent σ bond.
(iii) This effect is known as inductive effect.
6. Inductive effect is passed on to subsequent bonds also.
• But the effect decreases rapidly with the increase in the number of intervening bonds.
• For example, in fig.12.88, suppose that, it is chloro pentane instead of chloro ethane. Then the chain will be:C5ㅡC4ㅡC3ㅡC2ㅡC1ㅡCl
7. The C1ㅡCl bond is the original polar covalent bond.
• The other bonds C1ㅡC2, C2ㅡC3, C3ㅡC4, C4ㅡC5 will also undergo polarization. But that polarization will be due to inductive effect.
♦ C1ㅡC2 will experience the highest inductive effect.
✰ Because, it is nearest to the polar bond.
✰ It has zero number of intervening bonds.
♦ C2ㅡC3 will experience a lesser inductive effect.
✰ It has one intervening bond.
♦ C3ㅡC4 will experience a still lesser inductive effect.
✰ It has two intervening bonds.
♦ C4ㅡC5 will experience a still lesser inductive effect.
✰ It has three intervening bonds.
• If the number of intervening bonds is greater than three, the inductive effect is considered to be negligible.
8. In the above steps, we saw that inductive effect is caused due to the electron pulling by Cl atom. Atoms similar to Cl, which have high electronegativity, will cause the such inductive effect.
• The reverse can also happen. This can be explained in 4 steps:
(i) Suppose that, in fig.12.88 above, instead of CH3CH2Cl, we have CH3CH2Z. Where Z is a highly electropositive atom.
(ii) Then the Z will push the electron density towards C1. The C1 will acquire a small -ve charge (δ-)
(iii) Due to this newly acquired -ve charge, the C1 will push the electrons in the C1ㅡC2 bond. Thus the C2 will also acquire a small -ve charge (δδ-)
This is shown in fig.12.88(c) above.
(iv) The C1ㅡZ is the actual polar covalent bond. The polarization in C1ㅡC2 is due to inductive effect.
9. We see that:
• Electronegative atoms like Cl can cause inductive effect due to their ability to withdraw electrons.
• Electropositive atoms can also cause inductive effect due to their ability to donate electrons.
◼ Based on this, the substituents are classified into two groups:
(i) Electron-withdrawing group
(ii) Electron-donating group
• Halogens, groups like nitro (ㅡNO2), cyano (ㅡCN), carboxy (ㅡCOOH), ester (ㅡCOOR) etc., fall in the electron-withdrawing group.
• Groups like methyl (ㅡCH3), ethyl (ㅡCH2CH3) fall in the electron-donating group.
• Recall what we saw in section 12.11:
The alkyl groups attached to the 'C atom with unpaired electron' helps to stabilize the radical. (step 9, below fig.12.76 in section 12.11)
• Now we are in a position to give an explanation for this 'stabilizing action'. It can be written in 4 steps:
Groups like methyl (ㅡCH3), ethyl (ㅡCH2CH3) fall in the electron-donating group. 2. The ethyl free radical has one methyl group attached to the 'C atom with unpaired electron'.
• This methyl group donates electron density to the 'C atom with unpaired electron'.
• So that C atom will not feel that much necessity to acquire a new electron.
• As a result, the radical as a whole, will get some stability.
3. The isopropyl free radical has two methyl groups attached to the 'C atom with unpaired electron'.
• Those two methyl groups donate electron densities to the 'C atom with unpaired electron'.
• So that C atom will not feel that much necessity to acquire a new electron.
• As a result, the radical as a whole, will get some stability.
• Since two methyl groups donate electron densities, the stability will be greater than that in ethyl free radical.
4. The tertiary butyl free radical has three methyl groups attached to the 'C atom with unpaired electron'.
• Those three methyl groups donate electron densities to the 'C atom with unpaired electron'.
• So that C atom will not feel that much necessity to acquire a new electron.
• As a result, the radical as a whole, will get some stability.
• Since three methyl groups donate electron densities, the stability will be greater than that in isopropyl free radical.
Now we will see some solved examples
Solved example 12.16
Which bond is more polar in the following pairs of molecules:
(a) H3CㅡH, H3CㅡBr
(b) H3CㅡNH2, H3CㅡOH
(b) H3CㅡOH, H3CㅡSH
Solution:
Part (a):
• We have to compare two bonds: CㅡH and CㅡBr
• CㅡBr will be more polar because, Br is more electronegative than H
Part (b):
• We have to compare two bonds: CㅡN and CㅡO
• CㅡO will be more polar because, O is more electronegative than N
Part (c):
• We have to compare two bonds: CㅡO and CㅡS
• CㅡO will be more polar because, O is more electronegative than S
Solved example 12.17
In which CㅡC bond of CH3CH2CH2Br, the inductive effect is expected to be the least?
Solution:
• We can expand the given formula as: CH3ㅡCH2ㅡCH2ㅡBr
• C atoms are to be numbered (starting from the substituent Br) from right to left.
• C1ㅡBr is the original polar covalent bond.
• C1ㅡC2 will experience inductive effect.
♦ This bond has zero number of intervening bonds.
♦ So this bond will experience the maximum inductive effect .
• C2ㅡC3 will experience inductive effect.
♦ This bond has one intervening bond.
♦ So this bond will experience the least inductive effect.
In the next section, we will see resonance in organic molecules.
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