 In this video, we are going to look at how hyperconjugation affects the stability of a benzyl cation. So, we have these methyl groups that are attached at various positions of a benzyl cation and we are going to see how it affects the stability of the cation. So, if you look at this particular methyl group, it's attached to the para position of a benzyl cation and as you can see we have this double bond that's connected to an sp3 carbon atom, right? Now, this sp3 carbon also has this hydrogen that's attached to it and as you've seen in such a scenario, the electrons in this sigma bond, in this carbon hydrogen sigma bond can actually get delocalized, right? So, the electrons in this sigma bond can be pushed into the benzene ring and this is what we call sigma pi resonance or more commonly we call it the hyperconjugating effect, right? So, this will lead to the formation of a negative charge on the carbon atom. This alkyl group is going to act as an electron donor. It's going to increase the electron density of the benzene ring. So, this alkyl group shows what we call the plasage effect. Now, the lone pair that's formed out here is again connected to a double bond. So, we can have resonance out here too and these pi electrons. These pi electrons will now land over this carbon atom, right? Now, we can keep delocalizing these electrons further and ultimately this negative charge will land over here. So, as you can see, just like in case of resonance, this methyl group, this methyl group will bring about a lone pair of electrons, will bring about negative charge at only these very specific positions, right? Now, in one of the resonating structures, when the lone pair lands over here, so when the lone pair is out here, you can see that it's right next to the empty orbital of this carbon atom, right? A positively charged carbon always signifies an empty orbital and whenever we have a lone pair right next to an empty orbital, both of this can overlap and this is going to create a pi bond right over here. It's going to remove the positive charge from carbon and create a double bond right over here, right? So, what's essentially happening is that this methyl group, by a hyper conjugation, is going to help in removing the positive charge from the carbon atom. It's going to help in completing the octet of this carbon and in this way it's going to help in stabilizing the cation, right? Now, if there was no methyl group out here, this positive charge would be stabilized only by resonance from the benzene ring. So, if this methyl group was not there, this positive charge would have delocalized only over the benzene ring. But because of this methyl group, this positive charge can spread out even more. It can spread out even to this methyl group, right? So therefore, presence of a methyl group at the para position is going to add to the stability of the benzyl cation, right? Let us now look at what happens if we have a CH2 group at the meta position. Even out here, we can have sigma pi resonance. So the electrons on this sigma bond can shift over here while these pi electrons can move over to this carbon atom and this will bring about a negative charge on this carbon, right? So CH3 even out here can show a plus H effect. Let us now try to delocalize this lone pair that's created out here. Now, this lone pair is connected to this double bond. So this lone pair can go over here while this pi electrons can move over here, right? So this will bring about a negative charge on this carbon atom and if we keep delocalizing the negative charge will ultimately land up over here, right? So as you can see just like in resonance, if we put an electron donating group at the meta position, the negative charges will only land up at these ortho and para positions at this ortho and para of the benzene ring, right? So the negative charges or we should say the lone pair of electrons will only land at these positions. Now, as you can see, if we put a methyl group at meta, we won't get a negative charge. We won't get a lone pair right under the empty orbital of this carbon atom, right? So there can't be any pi bond formation out here. So the presence of a methyl group at meta will not be able to stabilize this cation. It won't be able to delocalize this positive charge over this methyl group via hyper conjugation, right? This cation can only get stabilized via resonance from the benzene ring. This positive charge can only get delocalized over the benzene ring, but it can never get delocalized over this CH3 group, right? So CH3 even though it's a plus H group, it's an electron donor, but because it's not going to add to the stability of this cation, so hyper conjugating effects of an alkyl group at meta are actually not that important. So you should not give much importance to the hyper conjugating effects if we have such groups attached at the meta position. Let us now look at what happens if we have a CH3 group at the ortho position. In fact, why don't you give this a thought? Why don't you pause the video and try and see if this alkyl group at the meta position can stabilize the cation. Now, even out here, we can have sigma pi resonance and if you keep moving these electrons, if you keep growing your resonating structures, you'll see that these lone pair of electrons, these negative charges will get created at this ortho and para position of the alkyl group, right? Now, even out here, because we have an empty orbital right next to a lone pair, so even out here, we can have a pi bond formation. So just like in case of para, this methyl group will actively help in stabilizing the cation, right? So therefore, just like in case of para, if this methyl group was not present out here, this cation would only have been stabilized by resonance from the benzene ring. This positive charge would only have gotten delocalized over the benzene ring, but presence of a methyl group at ortho is going to help in delocalizing the positive charge even more, is going to bring it even on this methyl group, right? So to summarize, presence of a CH3 group at this para position is going to bring about negative charges right under the cation, is going to bring about lone pair of electrons right under the cation, in both para and ortho, which is going to help in delocalizing this positive charge even more, it's going to bring about the positive charge even on these methyl groups, but if we put a CH3 group at meta, because there won't be a lone pair right under this carbon atom, so this methyl group will not be able to delocalize this cation, so it won't be able to add to the stability of the benzyl cation, right? So therefore, between a, b and c, a and c is going to be more stable compared to b, right? Now, how do we compare the stability between a and c? Well, purely based on hyper conjugation, both a and c are going to have exactly the same stability, putting a methyl group at the para position or the ortho position brings about negative charges exactly at identical positions, both of these bring lone pair of electrons right under the cation and both of these can stabilize the cation via hyper conjugation, so considering only hyper conjugating effects, a and c will have exactly the same stability. However, do note that besides hyper conjugation, there are also inductive effects that are at play out here, right? This alkyl group can also donate electron cloud to the benzene ring via induction, this carbon has these hydrogen atoms that are attached to it and as you must know, carbon is more electronegative than hydrogen, so it can pull these electrons in these sigma bonds towards itself, right? So, this will bring about a negative charge on this carbon atom as some slight negative charge on this carbon. It's going to increase the electron density of this carbon which can then push the electrons in the surrounding sigma bonds and in this way, this methyl group can also push electrons to the benzene cation via induction, right? It's going to push electrons via induction, so it can also show a plus i effect. Similarly, even out here, this methyl group can show a plus i. It's also showing plus h, let me write that down too. It's showing both plus h and plus i. The plus h effects of both these groups will be exactly the same, but their ability to stabilize this cation via induction will be very different, right? Induct effects if you remember are distance dependent and because I have a methyl group, I have an inductive group placed closer out here, so its ability to stabilize the cation via induction will be much higher compared to putting the methyl group far away at the para position, right? So therefore, between A and C, C is going to be more stable compared to A followed by B, right? Now, a quick note out here before I end the video. Even out here, when I put a methyl group at the meta position, even out here the inductive effects are at play and in fact, whenever we have such electron donating groups at meta, just like in case of resonance, these inductive effects turn out to be more important compared to hyper conjugation. Keep this in mind and see you in the next video.