 In the previous video we started talking about how the leaving group type affects the rate of an SN2 reaction. Let's take up a few more examples to understand the concept better. In this question we have to compare the rates of the following substrate towards an SN2 reaction, assuming they are attacked by the same nucleophile. What do we see in this question? We see how the leaving group is different in each case but the carbon that is being attacked is the same in each case. Right, let's take a step further and start comparing which of the following leaving groups would be the most stable. In each case the negative charge resides on the oxygen atom. What else is similar? Well, I see O minus attached to S double bond O double bond O in either case. Right, and there is a possibility of resonance since the negative charge is alternate to the pi bond. So there is the same resonance that's happening in each case. Wherein the negative is able to undergo conjugation with this oxygen and it can also undergo conjugation with this oxygen. And we get these three resonating structures. So this negative gets stabilized via resonance in each case and in a similar manner. How do I compare the stability of these leaving groups then? We think about these attached things in each case. In the first case the CH3 group attached has a donating effect, the plus I effect. It pushes electron density towards this group via the sigma bonds. However, in the second case the CF3 group has three fluorine atoms attached to the carbon atom which are electronegative. So overall this CF3 group has a withdrawing effect. It withdraws electron density via sigma bond and it shows a minus I effect. Well, when we are talking about stabilizing an anion, we see that the anion stabilizes when it is attached to an electron withdrawing group that takes away electron density from it. Right, what about the third one? A benzene ring is attached. Benzene, the carbon has sp2 hybridization. The more the percentage is character, the more the electronegativity and the more is the ability to pull electron density via the sigma bond. So the benzene ring also depicts a minus I effect. What now? Well, the first one would be the least stable. Why? Because in the other two cases the electron density is being pulled while in the first case it is being pushed on an anion. How do I compare the other two? Sp2 hybridized carbon is electronegative. Fluorine is also electronegative. But even if an sp2 hybridized carbon is electronegative it has a tendency to withdraw electron density. Would it be anything when compared to the most electronegative atom of the periodic table? I don't think so. So CF3 where the carbon is attached to three strongly withdrawing fluorine atoms would be more electron withdrawing as compared to the benzene ring. And therefore the first one would be the least stable leaving group while the second one would be the most stable. Comparing the reactivity of these three substrates towards an SN2 reaction I see that the second one would react the fastest followed by the third and the first one would be the slowest to react. Let's take up something different now. Here we need to compare the reactivity of these two substrates towards an SN2 reaction. Let's do it. The attacked carbon would be the same here as well. So why don't you try comparing the stability of the leaving group? In each case and then we'll do it together. In both the cases the negative charge resides on oxygen atom. Okay, so we need to take help of the electronic effects. We can do that. Hey, in both these cases the negative charge is alternate to a pi bond. There is a possibility of resonance in each case. Let's try drawing the resonating structures for the first anion. When we draw these resonating structures we see how the negative charge travels through the entire benzene ring. So it is stabilized. Now let's try drawing the resonating structures for the second one. The negative charge is being shared between these two oxygens. Okay, there is resonance in both the cases. We may feel that since the first anion has more resonating structures it would be a little more stable than the second one. But if we look carefully we see how during this resonance the negative charge moves from the oxygen atom to the carbon atom. While in the second case the negative charge moves from the oxygen to another oxygen atom. Well, oxygen is more electronegative than carbon. Even though the first case has more resonating structure the second case has a symmetrical resonance. There the negative charge moves from oxygen to another oxygen atom. And oxygen being more electronegative than carbon can hold this negative charge in a much more stable way. Correct? And therefore the second leaving group is more stable than the first one. And the rates would be the second substrate would react faster than the first one.