 So, what we have out here are different cations, different benzyle cations to be precise, to which we have these functional groups that are attached, right? Now, we need to arrange these cations in either increasing or decreasing order of stability, and this might look really daunting at first, we have so many groups that are attached, but trust me, it's actually not that hard, we just need to analyze individually what each of these groups does to the benzyle cation. So, let's start by looking at this particular group. So, it's an OH group which has these lone pair of electrons that are attached to it. Now, these lone pair of electrons are connected to a double bond, so there can be resonance out here, these electrons can get pushed into the benzene ring, and this is going to increase the electron density of my system, right? It's going to bring about a negative charge into my benzene ring. So, this OH is what we call a placer group. It increases the electron density of the system via resonance, so it's placer. Similarly, if you look at this particular group, even this has these lone pair of electrons connected to a double bond, so we can have resonance even out here, so even this is going to increase the electron density of the system, so even this is going to be a placer group. Now, if you come to this particular group, this carbon atom doesn't have any lone pair of electrons, right? So, there can't be any placer out here. However, there is these pi electrons, this pi bond between carbon and oxygen. Now, we have learned that if we have a pi bond, then we can actually move this pi bond to the more electronegative element, right? So, if we do that, there will be an empty orbital over this carbon atom, and this can then pull electrons from my system. So, this particular group is an electron withdrawing group. It is going to decrease the electron density of the system. It is going to bring about a positive charge into my benzene ring. So, this is what we call a minus R group, right? It decreases electron density via resonance. Similarly, if you look at this particular group, even this has a C double bond O, so even this can withdraw electrons from my benzene ring. So, even this is a minus R group. Let's now look at this particular CHC group that's attached out here. Now, carbon doesn't have any lone pair out here, but we have these hydrogen connected to the carbon atom, right? Now, whenever we have an SP3 carbon connected to a double bond, and if this SP3 carbon has a hydrogen attached to it, then this sigma bond, the electrons in this carbon hydrogen sigma bond can actually be pushed into the system via an effect which we know as hyperconjugation. So, hyperconjugation is also going to push electrons into the benzene ring and it's going to bring about these negative charges into my system, right? So, therefore, this CHC group is also an electron donor. It donates electrons via hyperconjugation, so we call it a plus H group. Similarly, in CD3, D is nothing but deuterium, which is an isotope of hydrogen. So, even in CD3, we are going to have a similar effect. The electrons in this carbon deuterium sigma bond can be pushed into the system. So, even this is going to act as a plus H group, right? Now, that we have established the characteristics of each of these functional groups, let us see how they affect the stability of this cation. Now, as we must know, whenever we have this electron donating or electron withdrawing group attached to a benzene ring, they only affect the electron density at these very specific positions, right? If you draw the resonating structures arising out of this electron donation via this OH group, then this will lead to the formation of a double bond out here and these pi electrons will shift over here creating a negative charge out here, right? Now, these pi electrons can resonate further and this will bring about a negative charge at this position and if we keep delocalizing this electron, then the negative charge will move further over here, right? So, therefore, this electron donation by the OH group brings about a negative charge only at these very specific positions. It brings about a negative charge only at the ortho and para of the OH group, right? Now, as you can see, there is a negative charge right under this carbocation. In fact, this negative charge arises due to the presence of a lone pair of electron out here and this lone pair can overlap with the m2 orbital of this cation thereby stabilizing the cation, right? So, therefore, these plaza groups, they bring about a negative charge directly under this cation and the same goes even for these plaza groups. Even out here, these sigma pi resonance will ultimately result in negative charges over the ortho and para of this CHC group. So, even this, even these are going to bring about a lone pair of electrons right here which can overlap with the m2 orbital of this carbocation thereby stabilizing it, right? So, these plaza, these plaza and plaza groups are going to stabilize mycation. However, if you look at these minus r groups, minus r withdraws electrons from the ring, right? So, this is in fact going to bring about a positive charge on my benzene ring and if you keep growing your resonating structures, you will see that even out here, this positive charge gets developed very specifically only at this ortho and para of the electron withdrawing group, right? So, they bring about these positive charges. Let me denote the positive charges in red. So, they bring about these positive charges at these ortho and para positions of the electron withdrawing group. Now, as you can see in both these structures, we are getting a positive charge right under another positive charge. So, this is not good. It's going to create a repulsion out here. It's going to make the system more unstable. So, therefore, the electron withdrawing groups in 3 and 4 are going to destabilize the benzylcatine, right? So, stability wise, 3 and 4 are going to be the least stable amongst all of this. Now, as you know, resonance effects are stronger compared to hyperconjugating effects because we actually need to break sigma bonds in hyperconjugation, which is difficult to break compared to the pi bonds in case of resonance. So, therefore, electron push via resonance is much easier compared to hyperconjugation. So, therefore, stability imparted in 1 and 2 is going to be greater than 5 and 6, right? And of course, these are going to be more stable compared to 3 and 4. Now, between 1 and 2, you can see that the lone pair on this oxygen atom can also be involved in resonance out here with this C double bond O, right? So, this lone pair is going to be less available for donation to the benzene ring. So, this particular group is going to be a weaker placer compared to OH, right? So, therefore, this electron donation from OH is going to be better compared to this. So, this OH is going to stabilize the cation more compared to this. So, therefore, between 1 and 2, 1 is going to be more stable compared to 2, right? Now, between 5 and 6, a carbon hydrogen bond is actually much easier to break compared to a carbon deuterium bond because a deuterium is heavier compared to hydrogen and chemical bonds actually are not static but they keep vibrating. So, you can think of these bonds vibrating like a spring. So, therefore, because hydrogen is lighter, it can vibrate more. This makes a carbon hydrogen bond longer and weaker compared to a carbon deuterium bond. So, it's easier to break. So, therefore, the plus H effect of a CH3 group is in fact higher than that of CD3 as CH3 bonds are easier to break. So, therefore, the electron push by CH3 into the system is going to be more compared to CD3 which will make it much better at stabilizing this cation compared to CD3, right? So, therefore, between 5 and 6, 5 is going to be definitely more stable compared to 6, right? Now, if you compare between 3 and 4, both of these are electron withdrawing groups and if you think about how this electron withdrawal happens, if you think about the mechanism then this happens when we shift these pi electrons to the more electronegative element, right? This can create an empty orbital over this carbon atom which can then pull electron from the benzene ring thereby decreasing the electron density of the benzene ring. Now, if you analyze these two carbon atoms, if you analyze the electron deficiency of these two carbon atoms you'll realize that because there's a CH3 group attached out here and CH3 groups as you know are electron donating groups they can donate electrons either via induction or via hyper conjugation so this CH3 can donate some of its electrons to this electron deficient carbon atom, right? So, this is actually going to lower the deficiency, the electron deficiency of this carbon atom compared to the carbon out here as we only have a hydrogen out here which doesn't show any kind of electron donation So, therefore, if you think about these two groups as a whole you'll realize that because there's a CH3 group attached out here so this carbon is actually not as deficient as this carbon atom, right? So, therefore, the ability to withdraw electrons from the benzene ring will be lower for this particular group when we have a CH3 attached compared to this group, right? So, an LDID group is a stronger minus R group compared to a C double bond O with a CH3 attached, right? Now, because minus R out here destabilizes the system so this because it's a stronger minus R group it's going to destabilize my ket ion more compared to this one, right? So, between 3 and 4, 4 is going to be more stable compared to 3, right?