 So, in the bromination of propane, a whopping 97% of my products is in the form of 2-brome propane and only 3% of it is in the form of 1-brome propane. Now this is actually really interesting because if we compare it with the corresponding chlorination reaction, then in chlorination, both of these products are more 50-50. The 2-chloro product is only slightly higher at 55%. Now do note that in both these cases, the 2-degree product is the major product and this is because it's much easier to break these 2-degree carbon hydrogen bonds compared to breaking these 1-degree hydrogen bonds as this will lead to the formation of a more stable 2-degree radical compared to a 1-degree radical. In fact, it's been experimentally found out that while it requires 98 kcal to break each of these bonds, it only requires 95 kcal per mole to break these bonds. So therefore in chlorination, even though it's more likely for the chlorine radicals that are formed to collide against these 1-degree hydrogen atoms as they are more in number, but because it's much easier to break these bonds, it's much easier to form a 2-degree radical compared to forming a 1-degree radical. So therefore ultimately, a greater number of 2-degree radicals are formed in the course of the reaction, which ultimately makes the 2-chloro product the major product, right? Now even in bromination, we have the same number of 1-degree and 2-degree hydrogen atoms compared to chlorination and even out here it requires the exact same 95 kcal to break these bonds and 98 kcal per mole to break these ones. So even in bromination, we should expect the 1-brom product and the 2-brom product to be in the same 45-to-55 ratio, right? So what's really going on out here? Let's find out. Now before we move forward, I'll give you a little hint. The only thing that's different between these two reactions is actually the nature of the bonds that are being formed rather than the nature of the bonds that are being broken, right? So if we dig in deeper, that in chlorination and even in bromination, even though the energy that is required to break these carbon hydrogen bonds is the same in both the cases, but the energy that is released is different in both these cases, right? HCl is a much stronger bond compared to HBr. 103 kcal per mole of energy is released during the formation of HCl, but only 87 kcal is released by the formation of HBr. So therefore, in the formation of 1-degree radicals via chlorination, while 98 kcal is required to break these bonds, 103 kcal is released by the formation of these bonds. So ultimately in this process, 5 kcal of energy will be released, right? Similarly, in the formation of 2-degree radicals, 8 kcal of energy will be released. It's slightly higher as breaking these 2-degree carbon hydrogen bonds requires lesser energy, but ultimately formation of these carbon radicals via chlorination is exothermic, right? It's exothermic. On the other hand, if you look at bromination, while 98 kcal is required to break this bond, only 87 kcal is released by the formation of an HBr bond. So this reaction as well as this one, both of these reactions are endothermic, right? So while formation of these carbon radicals via chlorination is exothermic, in case of bromination it's endothermic, and this and this as we'll see in this video, for the high selectivity of bromination compared to chlorination. Now to understand the relationship between endothermicity or exothermicity with selectivity, we need to look at the energy profile of the reactions. Now I'm sure you must have come across these kind of diagrams, these energy diagrams earlier, but let's do a quick recap of the most essential points. So let's say we have some reactants A2 and B2 and these reactants react to form some products let's say two times of AB. Now chemical reactions most of the time don't happen automatically, but instead the first need to cross an energy barrier which we call the activation energy. So let's say if we have some A2 molecules and some B2 molecules in a container, then these molecules will not spontaneously convert into my products. But only if they collide with the proper orientation and proper energy, only then some new bonds can get broken, some new bonds can get formed and all bonds broken and only then we'll get our products, right? Now this state in which new bonds are getting formed and all bonds are broken, this is the most energetically unfavorable state these reactants can find themselves in and this is what we call the transition state. So once the energy of the reactants, in other words we can say that when the energy of the reactants reaches the energy of the transition state, the reaction trips over to form the products. Now it turns out that in exothermic reactions like this one in which we have a chlorine radical abstracting a hydrogen atom from propene to form a first degree radical, in this kind of reactions in which there's a decrease in energy, in such reactions the transition state of the reaction is achieved early into the reaction. On the other hand in case of endothermic reactions in which there's an increase in energy going from the reactant to the products, in endothermic reactions the transition state of the reaction is achieved late into the reaction. So what does being early and being late mean and what are the consequences? To understand let's think about a chlorine radical trying to abstract a hydrogen atom from a carbon hydrogen bond. In fact let me first talk about a bromine radical, let me take a bromine radical it will be much easier to explain and then we will come to the chlorine radical. Now if you recollect chemical bonds are not static, right? They keep vibrating so this carbon hydrogen bond is actually vibrating. So this bonds like a vibrating and as it vibrates as these atoms move further away. See every single time I try like forcing myself to do this, okay, okay so this is like this is a very bad representation of a bond vibrating. But let's imagine this bond vibrates and if it vibrates it's going to move further away, right? So if it moves further away then this will actually start developing this radical character, right? It won't be like fully formed radicals but there will be some radical character that's getting developed. Now if there weren't any bromine radicals or anything else then this bond will vibrate it will develop some radical character it can never like totally break because if it breaks then we have two reactive species and both of them will combine with each other. So basically the bond will keep vibrating. However, in the presence of some external reactive agent like say a bromine radical there can be a new bond formation between these two electrons, right? So this is how the bond is getting abstracted. I get you get the picture of how a new bond formation happens. Now what happens is that if we do promenation, let's come back to promenation. So if we do promenation these bromine hydrogen bonds that will get formed, these bromine hydrogen bonds are not as strong as these carbon hydrogen bonds, right? That's why this reaction was endothermic. The amount of energy that was released by this bromine hydrogen bond was lower than the energy that was required to break this bond. So these are like weaker bonds. So therefore a bromine radical will only abstract a hydrogen atom from a carbon hydrogen bond. It will only abstract that when this bond has like vibrated far enough, when this bonds like really, really weak, only then the bromine will even start to interacting with this hydrogen atom, okay? So this means that in this reaction, this transition state which we have talked about earlier, this transition state is reached right before the radical formation, right? So we can in fact go ahead and draw it out here. This transition state is actually formed right before the radical formation. The radical, right before the radicals almost form, that's where you get your transition state. So that's why this transition state is late. And one important thing that you should also note out here is that the energy of this transition state is pretty similar to the energy of the product that we get because the radical is almost formed, right? When the bromine radical starts interacting with it, okay? Now on the other hand, if you come to chlorination, as you have seen, chlorine hydrogen bonds, if you talk about chlorination, then this chlorine hydrogen bonds are far stronger than carbon hydrogen bonds. The reaction is in fact exothermic. So therefore, because chlorine forms really strong bonds with hydrogen, so it doesn't even wait for this bond to vibrate and reach its breaking point, it pretty much starts reacting with this bond right away. So therefore, the transition state in chlorination is early and because the chlorine radical pretty much starts reacting right away, so the energy of this transition state is pretty close to the energy of the reactants, okay? So therefore, generally, generally, let me get rid of all this stuff if you have understood all this. So generally, in exothermic reactions, the transition state is early and more importantly, the energy of the transition state, the energy of the transition state is closer to that of the reactants, to that of reactants rather than the products. On the other hand, in endothermic reactions, because the transition state is late, it's formed just before the products get formed, so the energy of this transition state, energy of transition state is closer to products, okay? So now if we look at chlorination, so in chlorination, we had chlorine radicals abstracting hydrogen atoms. We get either a first degree hydrogen atom or a second degree hydrogen atom. And chlorination we have seen was exothermic. So it's exothermic. This is our propane and because it's exothermic, so the energy of the products will be lower. So this is the most stable product. This is energy. So this is out here while the first degree product will be less stable. So it's going to be out here, right? Now if we draw the energy diagram, the transition state of both of these reactions are going to be early because both of them are exothermic. So the energy of both of these transition states, the energy is going to resemble that of the reactants. So the energy of the transition state of both these parts are going to be pretty similar. Right. Pretty similar energy of TS. On the other hand, if we talk about, if we talk about bromination, bromination was endothermic. It was endothermic. So this time, if you look, if we draw our energy diagram again, so we have energy versus time. So we have our propane out here and it's an endothermic reaction, right? So these things are going to be higher up out here. So and the first degree radical is like less stable. So it's going to be higher in energy compared to the second degree radical. So let's say that their energies are out here. Now because this is endothermic, so the transition state is going to resemble the products, right? So in this case, the transition state, the energy of the transition state will resemble the energy of this one. While in this case, the energy of the transition state will resemble this. So the energy profile is going to look like this. Right. Now if you like, let's, let's look at our chlorination and bromination once more. So in chlorination, because it was exothermic, so the chlorine radicals pretty much started reacting with the, started like reacting with the carbon-hydrogen bond right away. So the transition state resembled that of the reactants. So in both these pathways, the both these pathways have pretty similar transition state. But in bromination, the transition state is late. The bromine radical starts react, starts interacting with the carbon-hydrogen bond right before the radical formation. So the transition state was late. So as you can see, the activation energies in both these reactions are pretty close to each other. Right. While the activation energies out here are quite different. So therefore in chlorination, both of these reactions, because they have pretty much similar activation energies, so both of these reaction formation of the first degree as well as the second degree radical will have the same, not the same, but similar rate. While in bromination, there is going to be a sharp difference in rates between these two reactions. Because the second degree radical formation is energetically more favorable, so there will be a much lower activation energy, which will make the formation of second degree radical much faster compared to the first degree radical. So therefore in chlorination, this was supposed to be the conclusion. So therefore in bromination, 97% of our products is in the form of two-brom-propane. Bromination is much more selective towards the second-degree product. While in case of chlorination, it's more 50-50, it's 45% and 55%.