 So, today let's discuss if the leaving group type would affect the rate of an SN2 reaction. But before we begin, why don't we just recall the mechanism of an SN2 reaction. In this reaction, the attacking nucleophile Y- attacks the carbon attached to the leaving group and the leaving group X- leaves. All of this happens in a single step. So, Y- attacking and X- leaving happens simultaneously and therefore the more stable the leaving group, the faster it would leave and the more would be the rate of the reaction. So yes, the type of the leaving group is going to affect the rate of an SN2 reaction. Let's take up a few problems to understand this better. In this question, we are given three reactions and we have to compare the rates of these three reactions towards an SN2 mechanism. Okay, just look at these reactions carefully and think what are the common things here. Here the substrate apart from the leaving group in each case is the same. The attacking nucleophile is also the same in each case. But in the first case, this nucleophile replaces Br- while in the second case it replaces F- and in the third case it has to replace Cl- right. So what do we compare here? We talked about how the more stable the leaving group, the faster it would leave and the more would be the rate of that reaction, correct? How do I compare the stability of these anions? Well, we use the electronic effects. Remember the electronic effects that we studied about in the previous units? We use them. First, we look at the atom that has the charge. If the atoms are different, we compare them based on the sizes or the electronegativities. If the atom containing the charge in each case is the same, then we go further thinking about resonance, inductive effect, etc., etc. So here, in this case Br- fluoride ion and Cl- need to be compared. What do we know about them? Well, they belong to the same group. Good. What else? If we look at the size, Br- is the largest of them all, followed by the Cl- followed by the fluoride ion. So if the sizes are in this order, let's think a little more deeply. The larger the size of the ion, the more spread out is the charge. And the more spread out the charge, the lesser is the charge per unit area, the lesser the potential energy and the more stable the ion is, correct? So the stability order would be Br- is more stable than the Cl- followed by the fluoride ion. So who leaves the most quickly? It's the Br- ion out of all three and therefore the first reaction would happen the most quickly out of all three and it has the fastest rate. And if we are asked to compare the order of the rates, we say that the rate of the first reaction would be greater than the third one followed by the second one. Let's take up another problem, shall we? And this question, one equivalent of cyanide ion has to attack and replace the leaving group via SN2 mechanism. What do the options tell me? Well, either the bromide ion leaves as in the first case or the iodide ion leaves as in the second case or the both may leave. What could happen? What would be the role of one equivalent? One equivalent cyanide or CN- means it can replace either of the two potential leaving groups here. It would replace either the iodine as iodide ion or the bromine as bromide ion. And how do we know which one is going to go away? Why don't you try it yourself and then we'll do it together. If we think about the size, the iodide ion is much larger than the bromide ion. The larger the size of the ion, the more spread out the charges, the lesser is the charge per unit area and the more stable the ion is. So the iodide ion is more stable than the bromide ion. So who would want to go? The iodide ion. So the answer to this question would be option B. Let's take up another question. Okay, this question asks us to arrange the following in the increasing order of reactivity towards the same nucleophile during an SN2 reaction. Ooh, some things to focus on. We see that the attacking nucleophile is the same and it's happening via SN2. What else is similar in each case? It's the substrate apart from the leaving group, right? Only the leaving group in each case would be different. Think about which of the three leaving groups would be the most stable, okay? In each case, the negative charge resides on the oxygen atom, correct? The negative charge is on the same atom. Ooh, we use the electronic effects. In the first case, the ethyl group attached to the oxygen atom don't need electron density on the oxygen atom via plus-i effect. The oxygen atom is already electron dense. It already has a negative charge plus somebody is pushing more electron density on it. That would make it unstable, right? In order to help it, somebody has to come and take that electron density away rather than pushing more and that is what is happening in the third case. If you look carefully, the negative charge is alternate to the pi bond and there is a possibility of resonance. And due to this resonance, this negative charge spreads out more, right? It is stabilized more, so which of the three leaving groups would be the most stable? The third one would be the most stable one while the first one would be the least stable one. And therefore, the reactivity towards the nucleophile during an SN2 reaction would be CH3CH2OCH2CH3 would react the slowest while CH3CH2OCOCH3 would react the fastest.