 So, what do you think will happen to the stability of a benzyl cation if I add an electron with drawing group to it? So, if I add something like an aldehyde group which is an electron with drawing group and we simply call them EWGs, what do you think will happen to the stability of this cation if I have such groups attached to it? Is it going to make it more stable or is it going to make it less stable? Well, an electron with drawing group is going to withdraw electrons from my system, right? So, what I have out here is a positively charged system. This has a positive charge, which is already electron deficient. It has a fewer number of electrons compared to protons. From this positively charged system, we are withdrawing more electrons, right? So this is going to make this system even more positive. It's going to make it even more positive, right? Now, anything that has a charge, either a positive charge or a negative charge, is already unstable and making it even more positive will make it even more unstable, right? So, this is a no-brainer. Adding an electron with drawing group to a cation, to a benzyl cation will definitely make it more unstable. However, it turns out that the amount of instability that is added actually depends upon the position on which we are adding this electron with drawing groups. It turns out that adding this electron with drawing groups at the ortho position, the ortho position and the para position, it turns out that adding electron with drawing groups out here in these positions make it much more unstable compared to adding it at this meta position. Adding an electron with drawing group at meta makes a benzyl cation more unstable but not as unstable as putting it in either say ortho or para. So what's really going on out here? Let's find out. Let us start by analyzing what happens to this benzyl cation if I add an electron with drawing group at the para position. So this is an electron with drawing group and as you can see there is this pi bond out here that is in conjugation with this pi bond, right? So we can have resonance out here and this electron with drawing group is going to withdraw electrons. These pi electrons are going to move over here while these pi electrons will move over to the oxidatum, right? So this will lead to the formation of a new resonating structure that's going to look like this. So the pi bond moves over here leaving a plus one formal charge on this carbonatum while these pi electrons move over to oxygen and this oxygen also had some lone pair of electrons which I did not draw earlier. So this will now have an extra set of electrons giving it a formal charge of minus one. Now this positive charge is then again directly connected to a pi bond. So therefore these pi electrons can now shift over here leading to the formation of a new resonating structure and if we keep continuing drawing our resonating structures, if we keep doing that, if you draw all the resonating structures that will be added to the benzyl cation due to the presence of this electron withdrawing group. If you look at all these resonating structures then you realize that the electron withdrawal by this aldehyde group, by this electron withdrawing group creates a positive charge on the benzene ring but this positive charge only gets formed over these very specific positions, right? So we can say that due to this electron withdrawing group, there's a positive charge that gets developed on the benzene ring but this positive charge gets developed only at these positions, right? Okay, let us now see what happens if instead of adding this electron withdrawing group at the para position, let's see what happens if I add it to the meta position. Where do we think the positive charges will get formed in this scenario? You can pause the video for a moment and try to come up with your own answer. Well out here, we have these pi electrons that are in conjugation with this pi bond. So this can move over here. So this will lead to the formation of a plus charge out here and a double bond out here and a negative charge over here and then this pi electrons can move over here and this will lead to the formation of another plus charge over here. Do remember these positions and then this can move over here and this will lead to the formation of another plus charge out here. Then this can move over here and this electrons can move over here and this will lead to the formation of this other resonating structure, right? This will lead to the formation of this. So as you notice, the positive charge got developed here, and here, and here, right? In fact, if you look closely, you'll realize that with respect to this electron-withdrawing group, these positive charges get developed at the ortho and para position, right? So with respect to this electron-withdrawing group, these positive charges are getting developed at the ortho and the para of this particular group, right? Similarly, if you look at where the positive charges get developed, if I add my electron-withdrawing group at the ortho position of this benzyl cation, then if you draw your resonating structures, you'll see that these positive charges, they get developed at the ortho and para of this electron-withdrawing group. Now if you compare these three structures, you'll see that in both para as well as in ortho, we have these positive charges that get formed closest to each other, right? We have this positive charge out here and another positive charge on this carbon atom. So we have two positive charges that are placed closer in ortho and para. In fact, the relative placement of this positive with all the other positive is exactly the same in ortho and para. But in meta, these positive charges are relatively placed further apart, right? We don't have a positive charge that's coming directly below this positive on this carbon atom, like in the case of para and ortho. So these positive charges are relatively spread out further apart, right? Now we all know that if we take two positive charges, two like charges together and if we just leave it, they are just going to move away, right? So their natural position or their more stable position is being far away from each other and if we bring them close, we actually need to spend some energy in doing that. So therefore, if we have like charges that are placed close to each other, they are going to have much higher energy, much higher potential energy and therefore they are going to be much more unstable compared to like charges being far away from each other, right? So therefore, because in ortho and para, because placing electron withdrawing groups at the para and the ortho position creates a positive charge directly under this cation, it brings like charges close to each other. So therefore, the system becomes much more unstable compared to putting an electron withdrawing group at meta. So to summarize, adding electron withdrawing groups to an benzyl cation will remove electrons from it, thereby making it even more positive and therefore even more unstable. So all of this in which we have an electron withdrawing group attached, all of these are going to be less stable compared to the benzyl cation itself, right? So one is going to be the most stable amongst all of this. Now putting an electron withdrawing group at the ortho and para positions creates positive charges and one of these positive charges comes directly under this cation that we had out here in both ortho and para, but in meta, in meta this does not happen. So meta becomes relatively more stable compared to ortho and para. From a purely resonance point of view, both ortho and para will have exactly the same energy, but do remember that there are also other electronic forces at play like the inductive effect, which actually causes a difference in stability of the ortho and para form. But that's a topic for a different day. Purely from a resonance point of view, ortho and para will have the same energy and the same stability as the relative placement or the relative positions of these positive charges in both ortho and para is exactly the same.