 So, we have been treating alkenes with hydrogen halides and we have already figured out what happens when an alkene is treated with HBr. Romine attaches to the more substituted carbon, it is called a Marconikov product. Today we will be talking about what happens if we add peroxide to this mixture and heat it. Would the product still be the same? If not, why and what could be the possible mechanism? So, let us begin. So what reagents are we adding? We are adding HBr, any alkyl or aryl peroxide would work. Here I am adding benzoil peroxide and we heat this mixture, right? What happens when this peroxide is heated? Well, heat or strong heat can lead to bond cleavage and what bond would break this one? How? Well, the bond has an oxygen atom on either side. So there is no electronegativity difference. So this bond cleaves homolithically and both the oxygen atoms get one electron each. This results in the formation of free radicals. Free radicals are really, really reactive intermediates. So once this free radical forms, it goes out looking for stability and it sees HBr. If it gets this edge, it would form a stable molecule that is benzoic acid. So that is what it does. It helps break the HBr bond, takes away edge and we are left with bromine atom. Now this bromine would go to the alkene and try and react with it. Let's see how. This bromine atom would help break the Cc pi bond. Since pi bonds are weaker than the sigma bonds, this is the bond that breaks. When it breaks, each carbon would get one electron each, right and then there are two possibilities. Either the bromine attaches to the terminal carbon and we get this radical or it attaches to the central carbon and we get this radical. So which of the two radicals is more likely to be formed? Well, the left one is a two degree radical, while the right one is a one degree radical. The more stable the intermediate, the more are the chances for the reaction to be driven in its formation. So I'll have to figure out which of the two radicals is more stable. It's the left one. It's the two degree radical since it has more electron donating groups attached. That is, it has electron donating groups present on either side. But what about the one degree one? Will it not form? Well, it would but in lesser amount. So the two degree radical would give the major product while the one degree one would give the minor product. Let's proceed. So now these three radicals intend to form stable neutral species. For that, they would want this HBr bond to break. They are looking for this HBr atom would be asked to go away. What happens when the left radical does that? Well, it forms CH3, CH makes a bond with this H and we get CH2Br here plus bromine atom. And what happens with the right free radical? CH3, CHBr, CH2 and it forms a bond with this H plus Br atom. What are these Br atoms going to do? Well, they are going to react with two other molecules of alkenes and this reaction continues in the same manner. So what is the major product? The major product would be the one that comes from the most stable free radical and the major product in this case would be CH3, CH2, CH2Br while CH3, CHBr, CH3 would be the minor product. So this major product is not the one that we got when we reacted alkene with HBr. This product is called antimalconic of product, the one where bromine attaches itself to the less substituted carbon and how does this reaction actually stop? We don't really pay heed to this termination step as it leads to the formation of various minus products because all these free radicals would be constantly colliding with other molecules or other radicals and when do they stop doing that? When they collide with other radicals and form neutral molecules. So what could be the other side minor products? They could be two Br atoms could just collide and form bromine. This Br could also react with this free radical that we got there, the two degree free radical and from this as one of the products but this all would be minor. So we don't pay much attention to the termination process. What we really need to focus on here is how the major product changes if we add peroxide to the same reaction mixture that we were dealing with initially and we heat this mixture. Why did it happen? Well the entire mechanism was different when we just treated it with HBr, we got the marconic of product as the major one because this was an electrophilic addition reaction wherein this HBr broke down as H plus and Br minus ions, the electron dense pi bond goes for this proton, takes up this proton, forms the most stable carbocation as the major product and this most stable carbocation is later attacked by Br minus and we get the major product, right? While if we treat the same alkene with HBr in the presence of peroxide and heat this mixture we get the anti-marconic of product as the major product because the mechanism is entirely different. It is a free radical mechanism where bromine attaches itself first and we get a more stable radical that drives the reaction into the formation of this product as the major product. In the first case we have carbocation as the intermediate while in the second case we have a free radical as the intermediate.