 In the previous videos, we discussed various nucleophilic substitution reactions and we talked about how the rate of an SN1 reaction depends on the stability of the carbocation intermediate. During the substitution, there is an intermediate formed and the more stable it is, the faster the reaction is. So, before going further, I think it's important that we recapitulate or we revise or we recall how exactly a carbocation is stabilized. It was way back that we studied the various electronic effects that stabilize a carbocation. A carbocation needs electron density. It's electron deficient. So, the groups attached to it must donate electron density to it to make it stable. Let's take an example to quickly recall what we studied. So, if I'm given the following substrates and someone comes and asks me which of the following would actually react the fastest via an SN1 mechanism, I would wonder that if I want to make them react via SN1 mechanism, I'd have to make a carbocation and the carbocation that's most stable will be the fastest to form and the one that's the fastest to form will give me the product the most quickly. So, for that I'll have to break the CCl bond and see which carbocation would be the most stable. So, let's do that. These are the carbocations so formed. Hey, why don't you write these down and try comparing the stability yourself before we do it together. Talking about the first one, I can see something here. I can see how the positive charge is alternate to a pi bond. Is there a possibility of resonance? Yes, there is. This pi bond would try and stabilize it. Once it does that, we get the next canonical structure. Can there be more resonance? Yes. How do we do that? We'll have to remember that we have to move in the same direction. I'm going anti-clockwise. Let's just keep doing that. And once we keep doing that, we get these structures. So, the hybrid would look something like this. Now, what are the things that I can see here? I know there's resonance but is there any aromatic character to it? The ring is cyclic. All the carbon atoms are sp2 hybridized. So, it's planar. The total number of pi electrons would be 2 which satisfies the Huckel rule. Taking n equal to 0, we get the total pi electron value to be 2. So, this ring is in fact aromatic. Was it aromatic previously? No, it wasn't. As soon as it got the positive charge, it became aromatic. Let's take up the next one. Can I see a possibility of resonance? The positive charge is alternate to a pi bond. So, yes, there is a possibility of resonance. Try drawing the resonating structures yourself and then we'll do it together. These are the resonating structures for the given carbocation. So, there's resonance that's stabilizing this carbocation. Cool? But is there aromaticity? Had the bond not broken, it was still aromatic. Benzene ring was still there. So, it's not the formation of carbocation that has made it aromatic. It was already aromatic. Carbocation is actually stabilized by a resonance. So, there's a difference. In the previous case, it became aromatic due to the formation of carbocation. Here, it already was and we can say that the previous carbocation was more stable than this one. Let's go to the next one. If we look at this carbocation, there's no pi bond, lone pair of electrons or negative charge. Alternate to this positive to help it get resonance stabilized. So, there's no resonance possibility. Can I see something else? Alpha hydrogen? Yeah, there is one alpha hydrogen. So, there's a possibility of hyper conjugation. So, it is stabilized via hyper conjugation. Cool? What about the next one? Does this carbocation have alpha hydrogens? If yes, how many? If you're thinking these hydrogens are alpha hydrogens, they are not. The carbon next to the positively charged carbon is the alpha carbon and it has already formed four bonds and none of them is with the hydrogen atom. So, there's no alpha hydrogen here and since there's no alpha hydrogen, there's no possibility of hyper conjugation. I cannot see any pi bond also alternate to this. So, there's no resonance also. So, what am I left with? Inductive effect. There's a plus i effect. These methyl groups are good donors. They donate electron density via sigma bonds. Okay. So, what have I seen until now? That the first one is stabilized by aromaticity, the second one by resonance and the third one has hyper conjugation due to one alpha hydrogen while the fourth one is stabilized via inductive effect provided by the methyl groups that are present close to it and I already know how aromaticity plays the strongest role. So, the first one is the most stable carbocation while the last one is the least stable carbocation. Coming back to the first question, which of the following substrates would react the fastest via SN1 mechanism? Well, the one that's forming the most stable carbocation. This will be fastest. So, the substrate that actually reacts the fastest via SN1 mechanism would be this one. Now that we have recalled how different electronic factors help us stabilize the carbocation and how it affects the rate of an SN1 mechanism, in the next video we'll be talking about why and how do we rearrange a carbocation.