 addition of in cyclic alkene, addition of Br2 in cyclic alkene forms, forms enantiomeric pair, forms enantiomeric pair. So, since alkene you always get enantiomers, Br2 with CCl4 this reaction takes place in the inert solvent carbon tetrachloride. So, the addition of Br was exciting. So, why is it? Why is it synubitrile and anti over there, what is the? Because it is see it is trans. Sir, but like what is the mechanism? You will get the same thing here, see I guess we are Cs3 here, Cs3 here, Cs3 here. Now, this Br minus will attack from this side and this goes up, because this methyl group no, this side attack is acyclic leader. So, back side side is attack, it goes up. So, both side we have to attach or if it attacks from this side, the attack will be preferred over here, because you see here methyl group here this side and this ring opens up like this. So, this attached from this side, you see Br group is attacking from this side and this bond comes over here. So, we have two Br molecules from this side. So, but then if it attacks from that side, you can still open up that bond with Br. This one, this one. So, if you attack from this side, this side is not possible because we are behind here. So, this bond gets open and when you convert, this is Sahar's position and when you convert this into fission, you will get both Br group on the same side, because the carbon atom and this carbon atom the configuration must be same. So, you must have done the. So, why does it have to be this side? No, you see this is Sahar's, this is fission. So, we are converting this into this. So, when you convert this, you get this only to have the same configuration in this carbon atom. Now, the point is you should know the conversion of Sahar's into this Sahar's into fission. You must have done this. And this you have to do, then only you will get two Br on the same side. So, it is not like I am attaching Br from the same side in this carbon atom. These two carbon Br minus will attack, you will get two products, Sahar's protection and you convert that into this fission, you will get this configuration. This conversion, you know, right? That is what you have to do. Here, if you do, you will get both Br on the opposite side. You must have the same configuration of this carbon atom and this carbon atom. You must have discussed this into Newman also, Newman into fission, right? All these three conversion. So, all these things when you draw the structure and then you convert this into fission, you will get the same molecule like that. But to memorize this, like I said, cat and tail, you have to remember, right? This is addition to your transformation, it is true. Cyclic alkene always gives you enantiomeric sphere. Okay, you must do this first. Next write down addition of hydrogen halide. Now, look at enantiomeric sphere. One we are, here we write down into the plane, out. Hydrogen halide, write down hydrogen halide reacts with alkene. So, this is a whole new reaction. Next, next reaction. Hydrogen halide, HX. Hydrogen halide reacts with alkene and forms alkide halide. And forms alkide halide. This reaction, this reaction follows Marconi-Coff addition. What is Marconi-Coff rule? It forms, it follows Marconi-Coff rule. Write down next line. In this reaction, we do not use aqueous solution of HX, hydrogen halide. We do not use aqueous solution of HX because it leads to the formation of alcohol. See, actually what happens, suppose you have an alkene, CH2, double bond CH2. When it reacts with HI, the product, the first one here, you get CH2 and CH3. And then in the next step, I- that acts as a nucleophile, attach with these two carbon atom. And the product is what? CH3, CH2. This is the product. Now, we do not use H2O here. If you use H2O, then what happens in this step? H2O takes us as a nucleophile because H2O is a solvent, its concentration is more and oxygen has don't bear on it. So, instead of I- this move will attack onto this carbon. And finally, H plus comes out and it leads to the formation of what? CS3, CS2, OH alcohol. That is why we do not use water for this reaction. We only use HI for this. What is the product in this reaction? CS3, CH, double bond CH2 with HI. Tell me the product. Now in this one, the major product is CS3, CS2, OH alcohol. Okay, write down next one. The addition of HBr hydrogen bromide, addition of HBr in presence of peroxide. Only HBr. Only HBr, yes. This one? No, it's not. It is in a vertical reaction. So, which one is major? This one is major. So, in presence of? Addition of HBr in presence of peroxide follows Malkonikov rule, anti-Malkonikov, sorry, follows anti-Malkonikov rule and this is known as peroxide effect. Peroxide effect or we also call it as peroxide effect, k-h-a-r-a-s-c-h. Peroxide effect and garage effect, it is applicable only for HBr. Okay, only for HBr. So, in this molecule, if you add HBr with H2O2, what is the product we get? So, C-H3, C-H2, C-H2Br, plus C-H3, C-H3Br, the major one is, oh, we are only sorry. And this is neither. So, why is it only with Br? Why does it only work with Br? Because in this presence of this, this HBr has some bond with and bond strength, right? This reaction, why is this one is major? Because it follows free radical mechanism, right on this also. Why it forms this? Because it follows free radical mechanism. C-H2O2, then also the major product is this one. It won't change the product. Okay, because anti-Malkonikov rule is very relevant, HBr. Okay. Next, we'll write down this reaction follows the formation of carbocation, the same one. This reaction follows. Here you see carbocation is forming, right? So, it follows the formation of carbocation. Hence, wherever possible, we'll do the rearrangement. Okay, tell me what is the product in this reaction? C-H3, C-H3 is the product in this reaction. So, here also it's anti-Malkonikov. So, when you shift... So also, what if I have a RBr? Instead of H, I have RBr. And I still follow on anti-Malkonikov. And it's not a RBr. RBr is not full boiling. For RBr, we have its LL, we have different reactions. RBr we cannot add. Cargo-roving bond won't break it. Here, FBr... See, what happens here? Hydrogen, it attacks onto this HBr. This pi-electron cannot take RBr from LK. Get on it and go to this big one there. And actually over here, we'll get a cargo breakdown. That is not possible here for Rx. So, what is the number of breaks which you also see in the FBr? Cargo breakdown we get. So, we always have it in order to get more stable time. So, just cargo breakdown is what degree? Just cargo breakdown is 2 degree. 2 degree for more stable time. Sir, you've been a donor shift. Yes. Right on the problem. Which one? Just cargo breakdown. Right. So... So... Cargo breakdown is wrong. Yes. Okay. First question, this electron pair comes over here and this takes H plus from this, right? Yes. So, the cargo breakdown we get is CS3C... CS3... CH positive and CS3. Now, this is alpha-hydrogen is what? 3. Now, to stabilize this, we have rearrangement of cargo breakdown. So, one of the methyl group from the resin carbon and this bond pair of electron and rearrange itself onto this carbon atom. So, 1, 2, methyl shift. Let's give CS3... C CS3... CH CS3... CS3 positive charge. This is the most stable cargo breakdown we have. If you compare the alpha-hydrogen here, 3 plus 3 plus 1, that is 7. Here it was C. It's more stable. Now, in this, C L minus will attach over here. So, the tertiary alkynide we'll get as a final product here. C CS3... CH CS3... CS3... This is the product. Okay. Now, in this one, what happens? Cargo breakdown is this? Yes. Next, what happens? So, you can apply the dancingness contract. Okay. Now, after this, we have ring contraction. Since the positive charge present on the ring itself, so, we have ring contraction here. Ring contraction, how do we do? Suppose it is first, second, third. Right. And this one is fourth. It is not required here. So, we can take this bond pair. This bond pair of electron, we can say it comes over here. And then, this third carbon attached with the second and first carbon. Right. So, you see here, we have 1, 2, 3, 3 membered ring. First one is this. Second, third one is this. And third, first one has one methyl group. So, this is methyl. And first are attached. First has fourth. Right. So, this we have two methyl group present here, with one positive charge on it. This is the electron. Here, we have positive charge. And this is the fourth carbon. Now, you see. You understand this. Fine. Understood this. This pair of electron comes out of third carbon. And third and first attached. So, 1, 2, 3. 1, 2, 3. 1 attached with fourth. So, why did we number that? So, how can we do that? How can we do that? You can start from the first carbon atom from here. How can we do that? How can we do that? You always get this only. Numb ring, but it changes. Probably, it will be this one. You can number like the way you want. Okay. There is no condition for that. No, you know, constant for that. 1, 2, 3, 4. We are just numbing it so that we can write down all these, you know, molecules or groups which is attached to the respective carbon atom. Now, this is, you see, 3-member ring with this carbon as positive charge. So, this is stabilized through. That same reason is the most stable carbocation we have. Nucleophile is what? Cl minus. So, this Cl minus will attack on to this positive charge carbon atom. The product of this reaction is we have CS3. Here, we have 2 methyl group. This one included. See, all these things, they won't ask you. You have to identify it. Since the point is the reaction involves formation of carbocation. So, wherever the carbocation forms, you have to form the most stable, possible carbocation. Okay. So, what all factors we have, everything will apply. Hyperconjugation, resonance, aromaticity, anything will apply. Right. Now, in this one, what is the product we get? So, whether we see it as a this electron there, I'll just try it or then we'll just first try it. The objective is to get most stable carbocation. You missed one thing. Hinging expansion. Yes. Hinging expansion. This is a positive over here. This is after shifting also. Wherever it is, it will be wrong. Every step, you should get more stable carbocation. It's so good. I thought you can only shift one. How can you shift again? If you shift, you get more stable here. And this thing will get expanded. No, but it's not shifting. It's expansion. Hinging expansion after this. We have right right. It's the number of atoms. The degree is right. You should find it. But the angle is different. Angle is different. Angle is different. Ring is expansion. But you said every step, you have to get the most stable one. You are taking only hyper one degree. I'm telling you what. This is four-member ring. We're converting it to five-member. Sir, but then even if that happens next step, you said every step, you should get something more stable. No, we'll see how. But we can't take it to them. We'll see what happens after that. Sir, but that's what. Before Hinging expansion takes place, this step, before that, we have to get more stable carbocation. Yes, we should do that. We'll see that. We'll see that. So probably we'll get this. Angle strain is down. Angle strain is down. This is positive charge. And this is... We'll get this right. How many alpha iron you have here? Three. Three alpha iron. Now in one, two hydrides, yes. There's hydrogen. This comes over here. So we'll get... Sir, the carbocation of the other carb is more stable. Sir, the carbocation on the second carb. More stable. Five. No, five. No, no, no. You see, one possibility is this. Another possibility is water. We'll get positive charge here, right? Five. Okay, how many alpha iron you have? Five. This is more stable. Sir, you'll just say it or just match that? Yes. Sir, there's no expansion. No, because see, first of all, the reaction is, what's this? This carbocation is more stable than this one. This will form. Now, when shifting is there, and it's going towards the less stable state, that's why it won't work. So this is a more stable carbocation that forms in the first step itself, right? This gives you minor product. This gives you major product. CL minus expansion. Sir, but if it went there, can you draw the expansion? Expansion, I think. Yeah. I'll draw that. So for this, suppose, this carbocation... Wait, so it won't go? No, it won't go. No, it won't go because this one is more stable. Wait, so why, why is it more stable? That's why I have a hyzer joint. That one is 3. Oh, man. If it forms here, right? So in the next step, this will shift over here, and then we'll have expansion, right? Yes. So how do we write? First here, second, third, fourth and fifth. Yes, sir. Right? So we'll have this thing a little bit. Just you draw a five-member ring. Five-member ring. And first carbon, suppose it is this. So first carbon has one minute height and one minute room, right? Yes. Okay. First, attached with five, suppose this one talks over here. So this is five. This is two, three and four. Correct? Where we'll have the positive charge? Two. On a second carbon. Below positive charge here. So again, we can shift. Again, we can shift this here. We'll get positive charge and two in a time. But this carbon cation won't work because this one is more stable. Okay, so we'll have some more reactions and we'll face this next class. So most of the time, also, we'll have some more reactions. Okay, so anyway. Sir, sir, sir. In the, the, this compound. I don't know why. I don't know why I have six of them. No, no, it's three only. We have positive charge here. Yeah, I get it. One here and two here. I think it's something else.