 So, what we have out here this time are different anions, different benzyl anions to which we have these different functional groups that are attached at these various positions, right? Now we need to arrange these anions in increasing order of stability. So we need to arrange them in such a way so that we have the least stable anion first and we move on to the most stable anion, right? So how do we do that? Let's start by analyzing what each of these groups does to the stability of each of these individual anions. So out here I have an OH group that has this lone pair of electrons and whenever we have such a system, whenever we have a lone pair attached to a double bond then we can have resonance out here, right? So this OH group is going to push electrons into the system via resonance. So this is what we call a plus R group, right? Similarly, if you look at this NH2, even this has a lone pair attached to a double bond. So even this can push this lone pair into the benzene ring. So even this is going to act as a plus R group, right? Now if you look at this OH group, this OH also has lone pair of electrons and even this can push this lone pair into the system. So even this can act as a plus R group, right? However, whenever we have an OH attached to a meta, then these kind of resonance effects that's imparted by OH is actually not that important, right? Now we have talked about this a lot in our previous videos, but let me go ahead and show you what I mean by this. So if you draw the resonating structures out here, so if you push this lone pair of electron out here, then these pi electrons are going to move over to this carbon atom, right? So we are going to have a negative charge over this carbon. Now if we keep doing my resonance, if I keep drawing my resonating structures, then this lone pair of electrons will move over here while this pi electrons will move over to this carbon atom, right? So the negative charge is going to move over to this carbon atom. And if I keep pushing this electrons further, then ultimately this lone pair is going to land up over here, right? So as you can see, this OH group brings about this negative charge at this ortho and para positions of this OH group, right? So when we have an OH at this meta position, it brings about negative charge at these very specific positions only. Let us now compare what happens when I have an OH at the para position. So this lone pair of electrons can be pushed into the benzene ring and this will create a negative charge on this carbon atom. And if I keep drawing my resonating structures, you will see that the lone pair ends over here and if I keep delocalizing this pi electrons further, it will ultimately land over here, right? So just like putting an OH out here, even in this case, there's a negative charge that will get developed at this ortho and para position at this ortho and para of my OH group, right? So the presence of an electron donating group will bring about a negative charge only at these very specific positions out here, right? So even out here, there will be negative charges that will get developed over these positions of the benzene ring. Similarly, if you look at this NH2 group, the negative charges will get developed only at these positions, right? Now as you can see when I have this OH group at these para positions, there will be a negative charge that is going to get developed right under this carbon anion, right under this negative charge and this is going to create a repulsion between these two charges, which is going to destabilize the anion further, right? However, when we look at these two cases, because we have an OH at meta out here, so the charges gets developed only at these positions. So we will not have a negative charge right under this carbon anion. So even though this OH is acting as a placer group, but it's not going to destabilize this anion as much as putting an OH out here, right? In fact, whenever we have an OH at the meta, because the resonance is not affecting the stability much, so a more important factor out here will actually be the inductive effect of this OH group. A more important factor will in fact be the minus i effect of this OH group. Because oxygen is a highly electronegative element, so it can also pull electrons via induction and out here because resonance doesn't affect the stability of the anion much, so the more dominating factor out here actually turns out to be the inductive effect. Similarly, even out here, the placer effect of this NH2 group is not going to contribute much to the stability of the anion and the more dominating factor will be the minus i. Nitrogen is also more electronegative than these carbon atoms, so it can pull electrons via induction, so the more dominating factor even out here will be the minus i of the NH2, right? Let us now look at these two NNs. Let's start by looking at this particular group. So out here, I have a cyanide group and if you look at it closely, then it's actually C triple bond N and this carbon atom now does not have any lone pair of electrons that it can donate to the benzene ring, right? So it definitely is not a placer group. However, we have these pi electrons and because nitrogen is more electronegative than carbon, so we can actually shift these pi electrons towards this nitrogen atom, right? So this can create an empty orbital over this carbon which can then pull electrons from the benzene ring towards itself. So a cyanide group is a minus r group. It will actually withdraw electrons from the benzene ring and actually bring about a positive charge on the benzene ring, right? Now just like in case of placer, if you draw the resonating structures out here, you will see that this placer charge will only land at this ortho and para position of this cyanide group, right? So attaching a Cn to a benzene ring is going to bring about positive charges at these positions, right? Now because there will be a positive charge under this carbon anion and a positive charge on a carbon atom actually represents an empty orbital, so therefore in one of the resonating structures when we have a placer charge at the para position because a placer charge on a carbon atom represents an empty orbital and a negative charge on a carbon atom is due to a lone pair of electrons. So whenever we have a lone pair directly attached to an empty orbital, so there can be a pi bond that's formed out here, right? So this is going to remove the negative charge from this carbon atom. It's going to help in delocalizing this negative charge. So therefore the presence of an electron withdrawing group at para is going to stabilize this carbon anion via resonance. The negative charge can get delocalized over the cyanide group. Now if you look at this particular group, this NO2 group, even this is a minus a group. We have a nitrogen double bond oxygen and because oxygen is more electronegative than nitrogen so we can push this pi electrons out here which will create an empty orbital over this nitrogen atom which can then withdraw electrons from the benzene ring, right? So NO2 is also an electron withdrawing group. It's a minus R group and NO2 if you remember is also one of the strongest minus R groups that's out there, right? The minus R of NO2 is stronger, will be stronger than that of the Cm. However because NO2 is attached at meta, so this time these positive touches will only get developed at these positions. So we won't have a positive charge right under this carbon anion and more importantly we won't have an empty orbital right under the lone pair of this carbon atom. So there can't be any pi bond formation out here. So this NO2 group cannot actually help in delocalizing this negative charge from the carbon atom via resonance, right? So the minus R effect of NO2 when you place it at meta isn't of much importance what will be more important will be the inductive effect of this NO2 group, right? Now nitrogen is more electronegative than carbon and this nitrogen also has these oxygen atoms that are attached which can pull electrons from this nitrogen making this nitrogen even more electron deficient. So this NO2 group is also a very strong minus R group, right? So this NO2 can also pull electrons via induction. Now because we have a carbon anion out here, so therefore these electron donating groups they are going to push electrons into an already negatively charged system making it even more unstable. So presence of electron donating groups at 1 and 2 is going to make it really unstable but if you look at all these anions all of them have electron withdrawing groups that are attached to it, right? So these electron withdrawing groups are going to remove some of these negative charge from this carbon anion either via resonance or induction. So therefore these groups 3, 4, 5 and 6 these anions are going to be more stable compared to 1 and 2, right? Now between 1 and 2 because oxygen is more electronegative than nitrogen it loves electrons more. So it's going to pull these electron clouds, these lone pair of electrons more towards itself. It doesn't want to donate these electrons as much as nitrogen which makes OR a relatively weaker placer group compared to NH2, right? Now because NH2 is the stronger placer group out here so it can donate electrons more thereby making this anion more unstable compared to putting an OH out here, right? So therefore between 1 and 2 2 is going to be more unstable compared to 1, right? Now between 3, 4, 5 and 6 you can see that you can see that in 6 there's a minus r effect that's going on. These lone pair on this carbon atom can get delocalized over this cyanogrupe, right? Now because resonance involves the actual delocalization of charge so resonance effects are much stronger compared to these inductive effects. Electron withdrawal via resonance is going to stabilize this anion much more compared to electron withdrawal via induction, right? So therefore 6 is going to be the most stable of all of these, right? Now between 3, 4 and 5 if you compare their minus i strength then the minus i of NO2 will be greater than that of OH which in turn will be greater than that of NH2, right? This is because in NO2 if you look at the structure of NO2 you have a nitrogen atom that has a positive formal charge on top of it and it also has these electronegative oxynatoms that are attached to it. So all of these oxynatoms are going to pull electrons towards itself making this nitrogen even more positive. So this particular nitrogen is going to have a really high electronegativity and it is going to pull electrons much more via induction compared to OH and NH2, right? Now between OH and NH2 because oxygen is more electronegative than nitrogen, so it's going to pull electrons across a sigma bond much more compared to nitrogen so OH will have a stronger minus i compared to NH2, right? Now because electron withdrawal by induction will also remove some of these negative charges from this carbon atom and this electron withdrawal via induction will be highest in case of NO2 followed by OH and NH2. So therefore between 3, 4 and 5, 5 is going to be more stable followed by 3 followed by 4, right?