 In this video we are going to see how the stability of a benzyl cation if I have a benzene ring along with a CH2 attached then this is what we call a benzyl group. So I want to see how the stability of a benzyl cation changes if I add different functional groups like an alcoholic group or maybe an aldehyde group to different positions of this benzyl cation. So let's say I take an alcoholic group and attach it to different positions of this benzyl cation. So I want to see how the stability of these cations differ from a simple benzyl cation. Now we even have special names for each of these positions. This is what we call an ortho position ortho while this is meta and this is the para position. Now a quick note out here it doesn't really matter in which direction I put the wedge group towards my right or towards my left both of these are going to be ortho because remember that in a benzene ring these pi electrons are in resonance with each other right. So therefore a more realistic scenario of this benzyl cation is actually going to look something like this in which I have a partial double bond distributed throughout the benzene ring right. So therefore it doesn't really matter if I put my OH group to this side or to the other side I can simply rotate this to get back my other orientation right. So it doesn't actually matter in which direction you put these OH groups. Anyways let's start with our analysis. Let's start by first analyzing the cation in which we had an OH group attached at the para position. So out here we have an oxygen with this lone pair of electrons and this is connected to a pi bond right. So we can have resonance out here this lone pair can go over here while these pi electrons can move over here and this will lead to the formation of a new resonating structure that's going to look like this right. Now this lone pair of electrons that are now on carbon can again resonate with this pi bond so it can move out here while these pi electrons can move over to this carbon atom right. So this will lead to our next resonating structure that's going to look like this and the moment you have this the moment you have a lone pair that's directly attached to an empty orbital the moment you have such a scenario both of these orbitals can overlap and this will lead to the formation of a pi bond right. Now as you can see this will remove the formal charge from the carbon atom but more importantly what's happening if you look at this molecule closely more importantly what's happened is that this carbon atom now has a complete octet right. It has a complete octet in fact if you count the total number of shared electrons out here so we'll have one two three four five six seven eight so it has eight electrons in fact if you look at all these atoms all these carbon atoms as well as this oxygen atom you'll see that all of these atoms actually have a complete octet out here even this oxygen atom if you count the total number of shared electrons so it's going to be one two three four five six seven eight so even this oxygen atom has a complete octet so therefore if we put some electron donating groups like OH electron donors these have electrons that they can give they can share and if we put such electron donating groups at the para position then this lone pair of electron can also get involved in resonance and ultimately stabilize the cation right. Now one important thing that I'd like to add out here and this might also be a source of some confusion is that you might argue that hey if this oxygen donates electrons then this oxygen is going to become positively charged right. If you look at this canonical structure at this resonating form in which all the atoms have a complete octet the oxygen atom is has a positive charge right. So you might argue that hey this is in fact not a stable structure because we all know that having a positive charge on a highly electronegative element like oxygen is not energetically favorable so you might argue that hey this lone pair should not even delocalize much right. Now that's a very valid confusion but what I'd like to add is that do remember that these formal charges that we assign to our atoms may not always represent the actual electron density that is present within the molecule. What do I mean by this? Let me explain. Let's say I take a hydronium ion H3O+. Now I'm sure you must have come across this in your textbooks H plus in water actually exist in the form of H3O+. So it's a really strong acid. Now if we draw the Cauchy-Louis structure of H3O+, it's going to look something like this. Now you can see that this oxygen atom even though it has a complete octet it only has one two three four five electrons five valence electrons and a neutral oxygen has six valence electrons. So therefore this oxygen should be positively charged right. However if you think about it even more deeply then you realize that this oxygen is a highly electronegative element right. On top of that we are putting a positive charge on oxygen. So this is going to make it even more electronegative. So it's going to actually pull all these electrons all these surrounding electrons towards itself and ultimately if we look at the electron density if you look at the actual electron density of hydronium ion then you realize that this positive charge actually gets spread out to these hydrogen atoms while oxygen in fact is slightly negative. So therefore formal charges and the actual electron density may not always mean the same thing and even out here even though it looks like that this oxygen has a plus one positive charge but in reality this oxygen is not going to be that positive because it's going to pull all the electron cloud that's around it towards itself right. So this positive charge is in fact going to spread out to other surrounding atoms. So the bottom line is we can always have an oxygen atom a positively charged oxygen atom provided that its octet is complete. Anyways moving ahead let us see what happens if I put an OH group at the meta position instead of para. So again we can have a resonance this lone pair can shift over here this pi electrons can move over here and this will lead to the formation of a new resonating structure that's going to look like this but what's going to happen next. I have this lone pair on the carbon atom and I have a pi bond out here. So what's going to happen is that this lone pair is going to move out here while these pi electrons are going to move here right. So this will lead to the formation of a new resonating structure and this can resonate further and ultimately let's let me move this. I'll get back to OH once again right. Now as you can see if I put this OH at the meta position the lone pair of the OH does get involved in resonance but in none of these resonating structures can it form a pi bond with this empty orbital of the carbon atom. This is because in none of the resonating structures these pi electrons will come directly below this CH2 plus group so it can't form a pi bond like in this one. In other words if we draw all the resonating structures of this particular molecule then we'll get a resonating structure in which all the atoms have a complete octet and this will greatly stabilize my molecule my cation but if I put an OH at the meta position then I won't be able to find a resonating structure in which all the atoms have a complete octet. So therefore putting an OH at the para position will stabilize a bensile cation while putting it at meta won't make much of a difference. So finally what about the ortho position? Do you think putting an OH at ortho will stabilize my bensile cation or do you think it won't make much of a difference? You can pause the video and try to come up with your own answer. Well let's draw the resonating structures right. So this lone pair can resonate it can go out here while this pi electrons can move over here and this will lead to the formation of a new resonating structure that's going to look like this. This can then further resonate and this can go on and finally you can see I can get a lone pair directly under the empty orbital of this carbon atom right. So the moment this happens both of this can overlap both of this can form a bond out here so just like putting an OH at para even out here I'll have a resonating structure in which all the atoms have a complete octet right. So therefore stability wise putting an OH group at ortho and para makes it more stable compared to putting it at the meta position right. But how do we determine the stability between ortho and para which is going to be more stable. Now first and foremost you should remember that resonance is distance independent it doesn't matter if I put an OH group out here or out here in both these cases the lone pair of electron will delocalize and stabilize the cation. So resonance is distance independent it doesn't matter if the OH group is closer or further. However what does depend on distance is the inductive effect. Now as you might recollect inductive effect arises due to difference in electronegativity across these sigma bonds and because oxygen is a highly electronegative element the OH group as a whole is a good minus i group it pulls electrons in in sigma bonds across sigma bonds towards itself. So therefore having a minus i group at the ortho position which is closer to this positively charged cation will relatively destabilize the system compared to putting the OH group far away at the para position. Inductive effect decreases exponentially with distance so the effect of this OH group at para on this carbon atom will be very minimal right. So to summarize putting an OH group at the ortho position as well as the para position stabilizes a positively charged benzyl cation via resonance while putting an OH at meta does not or we should say cannot. So therefore ortho and para will be more stable than the meta out here. Moreover between ortho and para because resonance is distance independent it doesn't matter how close or far you are from this cation. So resonance isn't a factor when it comes to stability between ortho and para but what does play a role is the inductive effect and because we have a good minus i group in OH placed closer at the ortho position compared to para. So therefore this OH group will destabilize the ortho more compared to para via induction right. So therefore para is going to be more stable compared to ortho and between the meta and the benzyl cation addition of an OH group doesn't impart any extra stability via resonance but do remember that OH is a minus i group so it can pull electrons via induction and this will destabilize the cation right. So therefore putting an OH at the para will make it even less stable than the benzyl cation itself.