 So, whenever we add chlorine to propane in the presence of heat or even UV light, we get a mixture of one chloropropane and two chloropropane as our products, right? Now, in this video, we are going to look at how much of this is in the form of one chloropropane and how much of it is in the form of two chloropropane. Is it 50-50 or something else? Let's find out. Now, to understand the ratio in which these products are formed, let's do a very quick recap. Let's look at how this reaction happens in the first place. So, in the presence of heat or even UV light, a few of these chlorine molecules that we take, a few of these chlorine molecules actually break apart in the form of chlorine radicals and these chlorine radicals that are formed are very reactive. They can react with the propane molecules that is present in the mixture. They can in fact go ahead and abstract these hydrogen atoms that are attached to a propane molecule. Now, if this chlorine radical collides against any of these hydrogen atoms that are present at the end, if it abstracts any of these hydrogen atoms from either this end or the other end, then in all these cases, it will lead to the formation of a new first-degree carbon radical. But if the chlorine radical goes ahead and collides against any of these hydrogen atoms, if it abstracts any of these hydrogen from propane, then this will lead to the formation of different second-degree carbon radicals. Right? Now, this chlorine radical can abstract any of these hydrogen atoms from a propane molecule. It can collide against any of these hydrogen atoms. But all of these will ultimately lead to the formation of either a first-degree radical or a second-degree carbon radical. Right? Now, these carbon radicals that are formed, these two are unstable and they are very reactive. And these can then go ahead and react with any undissuscitated chlorine molecules that are present in the mixture. And this ultimately leads to the formation of our products, one chloropropane and two chloropropane. So therefore, these radicals that are formed are the precursors to my products. Right? Now, if you look at it purely statistically, and if we assume that this chlorine radical is equally likely to abstract any of these hydrogen atoms from a propane molecule. So therefore, because there are eight hydrogen atoms in propane, so the probability of a chlorine radical abstracting any one of these hydrogen atoms, the probability of abstracting any one of these hydrogen atoms from propane will be one out of eight. Right? However, because the abstraction of all of these hydrogen atoms, all of these six hydrogen atoms that are attached at the end, because the abstraction of all of these hydrogen atoms will lead to the formation of a first-degree radical, either at this end or at the other end. So therefore, the probability, the total probability of forming a first-degree carbon radical will be six times, six times of one by eight. So it's six by eight, which is nothing but 0.75. Right? Now, on the other hand, because the abstraction of only these two hydrogens, because the abstraction of only these two hydrogens will lead to the formation of a second-degree radical, so therefore, the probability of forming a second-degree carbon radical will be two times of one by eight. So it's two by eight, which is nothing but 0.25. So therefore, we can now go ahead and say that of all the radicals that are formed, 75% of it should be in the form of first-degree radicals, while only 25% should be in the form of second-degree radicals. Right? Now, because these radicals ultimately lead to the formation of my products, so therefore, purely statistically, the amount of one chloropropane that we should get should also be 75%, while that of two chloropropane should be 25%. Right? However, when we actually carry out this reaction, we find that in reality, the amount of one chloropropane that is formed is actually not 75%, but it's in fact closer to 45%, while that of two chloropropane is 55%. So even though purely statistically, one chloropropane should have been the major product, however, it turns out that in reality, chlorination of propane actually yields a two-chloropropane as my major product. So therefore, chlorination of propane seems to somehow favor or we can say select the two-chloroproduct over the one-chloroproduct. Now if ultimately, the products that are formed are 45% and 55%, then this means that the amount of first-degree radical and the second-degree radical that is formed should also be 45% and 55%, right? So therefore, in reality, the probability of forming a first-degree radical is not 75%, it's in fact lower than that, it's equal to 0.45, right? Similarly, the probability of forming a second-degree radical is not 0.25, it's much higher than that, it's going to be equal to 0.55. So what's really going on out here? How do we make sense of these new probabilities? Now if you think about it, abstraction of any of these six hydrogen atoms from a propane molecule will always lead to the formation of a first-degree radical, while the abstraction of these hydrogen atoms will always lead to the formation of a second-degree radical. So if you think about it, the only way the probability of forming a first-degree radical is lower than 0.75, the only way it's possible is if this factor, which is nothing but the probability of abstracting a first-degree hydrogen atom from a propane molecule, it's only possible if this factor, which would assume to be 1 by 8 for all the hydrogen atoms, if this factor is lower than 1 by 8, right? Similarly, the only way the probability of forming a second-degree radical is greater than 0.25, the only way it's possible is if this factor, which is the probability of abstracting a second-degree hydrogen, is greater than 1 by 8, right? So in reality, the probability of abstracting a first-degree hydrogen and a second-degree hydrogen are not the same. We can in fact go ahead and calculate these new probabilities using our new data. So if the probability, if the actual probability of forming a first-degree radical is 0.45, then because this has to be equal to 6 times the probability of abstracting a first-degree hydrogen from a propane molecule, so therefore if you do this calculation, the probability, the actual probability of abstracting a first-degree hydrogen from propane will be 0.45 divided by 6, which will come out to be equal to 0.075. Now if you compare, you will see that this value is less than 1 by 8, 1 by 8 turns out to be 0.125. Similarly, the probability of forming a second-degree radical in reality is 0.55, so 0.55 should be equal to two times of the actual probability of abstracting a second-degree hydrogen atom from propane. So if you do our calculations again, this will be equal to 0.55 divided by 2. So this will come out to be equal to 0.275. So in reality, the probability of a chlorine radical abstracting a second-degree hydrogen from propane is much higher than the probability of a chlorine radical abstracting a first-degree hydrogen from propane, right? In fact, if you divide these probabilities, then we will see that the probability of abstracting a second-degree hydrogen is in fact 3.66 times greater than the first-degree hydrogen atom. So therefore a chlorine radical is much more selective in abstracting a second-degree hydrogen from a propane molecule. It's in fact 3.66 times more selective over a first-degree hydrogen. And this selectivity ultimately results in a greater than expected second-degree radicals which ultimately leads to the formation of two chloropropane as my major product. So the big question that we now have is why does a chlorine radical seem to prefer abstracting a second-degree hydrogen over a first-degree hydrogen? What do you think is the answer? Well, if you look at a second-degree radical, we will realize that it has these six alpha hydrogens, six hydrogens that are directly attached to this radical. And if you recollect from our videos on electronic effects, then you must remember that in such cases when we have these hydrogen atoms directly attached to a radical, in such cases the hydrogen atom can actually stabilize the radical by an effect known as hyperconjugation. So a second-degree radical has six alpha hydrogens that can stabilize it via hyperconjugation. But if you look at a first-degree radical, we will realize that it only has these two hydrogens that can stabilize it via hyperconjugation. So therefore a second-degree radical is more stable than a first-degree radical. So it's actually much easier to break this carbon-hydrogen bond as it will lead to the formation of a relatively more stable second-degree radical compared to breaking these hydrogen bonds as this will lead to the formation of relatively less stable first-degree radicals. In fact, it has been experimentally found out that while it takes 98 kilocalories to break these first-degree carbon-hydrogen bonds, it only takes 95 kilocalories to break these second-degree carbon-hydrogen bonds. So therefore the probability of a chlorine radical abstracting each of these hydrogen atoms is not same. A chlorine radical is in fact 3.66 times more selective in abstracting a second-degree hydrogen atom compared to a first-degree hydrogen. So to summarize in the chlorination of propene, even though we have more number of first-degree hydrogen atoms which this chlorine radical can collide against, but because it's much easier to break these carbon-hydrogen bonds. So therefore in reality the amount of 2-chloropropene that we get is much higher than 25%, it's actually 55% while that of 1-chloropropene is 45%. Now if you have stuck around till the end of this video, let me ask you a question. If instead of chlorination I did bromination of propene, then how much of it do you think should be in the form of 1-bromopropene and how much of it in the form of 2-bromopropene? In fact I want you to actually pause the video and think about this for a moment. Well, one might expect that just like in chlorination, the amount of 1-bromoproduct that is formed should be 45% while that of 2-bromoproduct should be 55%. But in reality it turns out that the 2-bromoproduct that is formed is actually a whopping 97% while that of 1-bromopropene is only 3%. Now that's really interesting, right? Bromination of propene seems to be even more selective towards the second-degree product compared to chlorination. So what's really going on out here? Let's find out in the next video.