 Let us now see what happens if I add chlorine to propane instead of methane and heat the mixture up. What are the kind of product or maybe products will we get in this scenario? Let's explore. Now as you have seen in the previous video, this kind of reaction actually starts with an initiation step in which a few chlorine molecules in the presence of heat or even UV rays, UV rays, a few chlorine molecules in the presence of heat or UV rays can actually break apart in the form of chlorine radicals, right? Now the electrons in this bond are attracted to the nucleus of two identical chlorine atoms. So when this bond breaks, one of the electron stays with one chlorine atom while the other electron stays with the other chlorine atom. So therefore this bond breaking is represented by this single-headed arrows. Now the chlorine radical that is formed, it only has 1, 2, 3, 4, 5, 6, 7 electrons in its outer most shell, so its octet is not complete, so it's pretty reactive and it can in fact react with the other chlorine radical and both of these can combine to give me back my chlorine molecule. So there's actually a reversible reaction that's going on out here. Now one of the other things that might happen is that this chlorine radical instead of colliding and reacting with another chlorine radical, it can actually go ahead and collide with other propane molecules that are present in the mixture. So this chlorine radical can collide against this hydrogen of a propane molecule and because it's very reactive it can actually react with this hydrogen atom. It can in fact go ahead and abstract this hydrogen from a propane molecule. Now if this happens, then one of the electrons in this bond has to be with hydrogen so that both of these electrons can combine to form a chlorine hydrogen bond, a chlorine hydrogen bond while the other electron will be left over the carbon atom, right? So therefore whenever a chlorine radical abstracts a hydrogen from propane, then this will lead to the formation of a new radical, a new carbon radical, right? Let me in fact go ahead and draw a more condensed form of this radical. So I'll be left with nothing but a CH2CH2CH3 radical, right? A CH2CH2CH3 radical. Now this chlorine radical instead of colliding and abstracting this hydrogen atom, it can go ahead and maybe abstract this hydrogen or even this one. So this chlorine radical can collide against any of these hydrogen atoms and abstract it. So if it goes ahead and say abstracts this hydrogen from a propane molecule, then even in this case I'll be left with a CH2CH2CH3 radical, right? So if this chlorine radical abstracts this hydrogen or this one or even this one from a propane molecule, then in all these cases I'll be left with a CH2CH2CH3 radical, right? However if the chlorine radical goes ahead and collides against this hydrogen atom, if it abstracts this hydrogen from propane, then this time I'll be left with a different radical. I'll be in fact left with a CH3CHCH3 radical, right? Now this is clearly a different radical. This is in fact what we call a second degree carbon radical, second degree because the carbon on which the radical is formed is attached to two other carbon atoms, while the radical out here is called a first degree carbon radical as the carbon on which this radical is formed is first degree as it's attached to only one other carbon atom. So therefore depending on the hydrogen which this chlorine radical abstracts from a propane molecule, we can get the formation of different carbon radicals, right? Now do you think we can get the formation of some other different radical besides this by the abstraction of a hydrogen atom from a propane molecule? What do you say? You can pause the video and think about this for a moment. Now if this chlorine radical goes ahead and say abstracts this hydrogen atom from propane, then even in this case I'll be left with a CH3CHCH3 radical, so I'll be left with a second degree radical. And if it goes ahead and abstracts say any of these hydrogen atoms from this end, then in all these cases I'll be left with a CH3CH2CH2 radical, right? Now this is nothing but equal to the first degree radical that we had out here. So therefore a chlorine radical can abstract any of these hydrogen atoms from a propane molecule but in all these cases I'll be left with either a first degree radical or a second degree carbon radical, right? Now these new carbon radicals that are formed these are also very reactive and these can then go ahead and react with any undissociated chlorine molecules that are left in the mixture. This first degree carbon radical can react with any undissociated chlorine, it can in fact abstract a chlorine atom from unreacted chlorine molecules that are present in the mixture and this will ultimately lead to the formation of one chloropropane along with the liberation of a new chlorine radical, right? Similarly the second degree radical can go ahead and abstract a chlorine atom from any undissociated chlorine that is left in the mixture and this will lead to the formation of two chloropropane along with the formation of a chlorine radical out here also, right? Now these chlorine radicals that are formed can then again go ahead and react with other propane molecules and this will lead to the formation of some more one chloropropane and two chloropropane along with the liberation of some more chlorine radicals and this will then go ahead and react with even more propane molecules and this will set off a chain reaction which will ultimately result in almost all the propane molecules getting chlorinated in the form of one chloropropane and two chloropropane, right? So to summarize whenever we add chlorine to propane and what I have drawn out here is the bond line formula of propane so whenever we add chlorine to propane in the presence of heat or even UV light a few of these chlorine molecules actually break apart in the form of chlorine radicals. Now these chlorine radicals that are formed are very reactive and these can react with the propane molecules that are present in the mixture they can in fact go ahead and abstract any of these hydrogens that are attached to a propane molecule. Now if this chlorine radical abstracts any of these hydrogen atoms that are present at the end so if it abstracts a hydrogen atom from either this end or the other end then in all these cases I will be left with a first degree carbon radical a first degree carbon radical either at this end or at the other end all of these will be first degree but if the chlorine radical abstracts any of these hydrogen atoms so if it abstracts any of these hydrogen from propane then this time will get a different second degree carbon radical. Now these carbon radicals that are formed these are also very reactive and these can then go ahead and react with any undissociated chlorine molecules that are present in the mixture and this ultimately leads to the formation of our products one chloropropane and two chloropropane well now that we know that the products are going to be one chloropropane and two chloropropane then the next question that automatically arises is how much of it should be in the form of one chloropropane and how much of it should be in the form of two chloropropane is it going to be 50 50 or maybe something else let's find out more in the next video