 So, today we will be discussing the effect of leaving group and the solvent on the rate of an SN1 reaction but before we proceed let us just recall the mechanism of an SN1 reaction. During this reaction the bond between the carbon and the leaving group breaks, a carbocation forms and the leaving group leaves. In the next step the attacking nucleophile attacks and we get the desired product. Step one is the slowest step. It is the rate determining step. So, we need to focus on this bond breaking. If the carbocation and the leaving group so formed are stable this bond breaks easily and this step is fast. So, to increase the rate of the reaction we have to focus on the stability of the carbocation and the leaving group. We have already discussed the stability of carbocation in the previous videos. So, today let us bring our focus on the leaving group. More stable the leaving group faster is the rate. Let us take an example to dig a little deeper. So, let us compare the reactivity of these two substrates undergoing nucleophilic substitution via SN1 mechanism. In both the cases this part is the same right. When the bond breaks we get the same carbocation in either case and the leaving group is Cl- in the first case while it is Br- in the second case. Since the carbocation formed is the same in each case we need to focus on what is different. It is the leaving groups. So, let us bring all our focus on comparing the stability of Cl- and Br-. In the periodic table both chlorine and bromine belong to the same group while the chlorine atom resides in the third period while the bromine atom resides in the fourth period. Bromine is larger in size than the chlorine atom right. Therefore, the bromide ion would also be larger in size as compared to the chloride ion. The negative charge will spread out more in case of the bromide ion and the charge per unit area would be less. More spread out the charge lesser is the potential energy more is the stability. Therefore, bromide ion would be more stable than the chloride ion. Coming back to our question we see that the second substrate reacts faster than the first one for an Sn1 reaction. Let us take another example shall we. We need to compare the reactivity of these three substrates for the nucleophilic substitution reaction unimolecular. Why don't you try this problem yourself and then we will do it together. If we look carefully in each case the substrate has only one thing different which is the leaving group. The rest of the structure is same in each case right. So, if I try and break these bonds I will get OH- F- and I- respectively as the leaving groups while the carbocation formed is the same in each case. Here I- has the largest size the charge spreads out well in this case the charge per unit area is less and it is the most stable out of all. Let's now focus on the other two OH- and F-. In the first case the negative charge is present on the oxygen atom while in the second case it's on the fluorine atom. Both oxygen and fluorine belong to the same period. There is not much difference in their size so we will compare their electronegativity here. Fluorine is more electronegative than oxygen. It has a higher tendency to keep electron density to itself right. So fluoride ion would be much more stable as compared to the hydroxide or OH- ion. So the leaving group stability is I- is more stable than F- which is more stable than OH-. So the CI bond breaks the most easily so which of the following would react the fastest via SN1. It would be the third one and the one that reacts the slowest would be the first one. Let's take up another problem. Here we have two substrates both undergo SN1 reaction and we have to compare the rates of the two reactions in this case. If we look carefully the substrate is the same in each case. The leaving group is also the same in each case. Interesting isn't it? What do I compare then? We already know that the strength of nucleophile doesn't really matter in the SN1 reaction because it's not a part of the redetermining step. What is different here? The mixture looks a little different. In the first case there's 80% ethanol and 20% water mixture. In the second case I have 20% ethanol and 80% water in a mixture. It has something to do with the polarity. More water would mean more polarity but how is it helping my reaction? Let's find out. Since the substrate is same in each case the carbocation form would also be the same. The leaving group is the same as well I-. I know water molecule is a polar molecule. Oxygen has a partial negative charge while hydrogen has a partial positive charge. The partial negative charge would be attracted towards the carbocation and this would lead to the stability of the carbocation. The more the amount of this electrostatic interaction or attraction the more stabilized the carbocation is. So if there are more number of water molecules there would be more attraction the carbocation will be stabilized further. So if I increase the polarity of the solvent the stability of the carbocation intermediate increases. Hey the rate of the reaction is directly proportional to the stability of carbocation. Oh so more the polarity of the solvent more is the stability of the carbocation and faster would be the rate in that case. This is why the solvent is the hidden factor. It does not only participate in the reaction but it helps stabilize the carbocation and more stable the carbocation more easily the bond between the carbon and the leaving group breaks and faster is the reaction. So cool. So in the question given to us the reaction mixture having a more polar solvent will be more reactive towards an SM1 mechanism.