 see in the molecule if there is ring or double part is present so we will apply the condition of both things double part condition will apply we will also apply the condition of ring if both are satisfied then we can have g i because of double part as well as ring possible okay like this one how many g i possible if it is possible there how many g i possible how many g i possible how many g i possible if a ring contains double bond and the condition of that is satisfied then it is possible like a double bond inside the ring inside the ring both then break the ring if it is able to rotate you know for that the condition of the ring must be 8 or more member of the ring larger ring it is possible it won't break now if you able to rotate about one it is not like that this is not possible for double bond inside the ring the ring must be at least 8 or more larger ring it is possible not that important but one of the examples we see okay so this is geometrical isomerism because of ring 4 g i possible there this will tell me how many g i center how many g i center there g i possible yes or no one this one is or this one is also not question is also not maximum 1 you said all answers what What? What? What? What? 1, 2, 3, 4, 5. It's actually the same. See across it's not possible. That's not possible. Across this? It's not possible. One thing, huh? There's a difference. What? My, my. This is Sv2 hybridized. Oh, yeah. It's not Sv3. If we have one substituent here, there is possible Because this carbon is sp2, not sp3. The condition is at least sp2, at least 2 sp3 hybridizer. This carbon is not sp3, it is sp2. Sir, but that's why this idea is sp3. Which one? This case is possible. I have given you now. So now we are not to sp3. What's sp3? So the carbon is sp2. But it is not dye substituted. Both are dye substituted. This has two hydros, not possible. This two are sp2 hybridized, not possible. Sp3 have a fine, but there is only one sp3. In this one there is no exterior center. I will find out. Dye substituted? Yes. How many sp3 are there? There are all the sp3. Oh wait, I am sorry. Wait, it's a strong sp3, right? Okay. The sp3 is possible. Yeah. Yeah. How? They are all just hybridized everywhere. So they are all just hybridized everywhere. Oh, it has to be dye substituted. Yes. It is dye substituted. So which one is dye substituted? It's those two are substituted. You have to have a hydrogen. Hydrogen and then the other one. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Hydrogen. Sir, can you tell me this? How many stress-enters are there? Yes, I'll check one. How many stress-enters are there? No idea. Across this ring it is possible? Across this ring? No sir. The ring has a condition of 2SP3, have it as a muscle type. But if I place one chlorine here, then it's across DI possible. Because one substance is here. In this question, in this question where there is no chlorine present, only one geographical isomer is there. Only one is serious-enters. How many stress-enters are there? Two. Same thing. Which one is there? I asked how many stress-enters are there? It says there is a chlorine. Then it will be more than 2SP3. If it is chlorine, then two stress-enters, then we can have two across this, two across this, four across this. So now it will be different where this is excited and this is outside. They say it will no longer be two for the ring, two for more than two for the ring. Sir, because even the CL bond can come out of the plane. Because this is possible, across this. So it is more than four, I guess. We have formula for that. In this case, it's not possible you have to draw this structure and find out. Like if it is cis and two structures of this. Sir, DI substitutes mean there have to be two substitutes No, no, no. Two substitutes. Different substitutes. Hydrogen CSP possible. Hydrogen chlorine possible. Chlorine CSP possible. Any two different. Do we have GI possible in this? No. Not possible. Not possible. Along with the ring itself. Across the ring it is possible. But across the double bond. Right, across the ring possible. Across the double bond. Okay, third case you write down. The last one, write down. GI because of double bond inside the ring. There's two points you have to memorize. It is actually experimental, this condition we have. Write down. GI because of double bond inside the ring. Condition is, the ring must be, the ring must be at least eight members. At least eight member ring must be there. More than eight, it's fine. Okay. Next line. For smaller ring, write down next line. For smaller ring. Less than eight. Yes, less than eight. The trans isomers are highly unstable. The trans isomers are highly unstable. And does not exist. For smaller ring, the trans isomers are highly unstable. And does not exist. Due to angle strain. Next point. There's two, three points you have to memorize. Next point write down. For eight, nine and ten member ring. For eight, nine and ten member cycloalkanes. You write down. For eight, nine and ten member cycloalkanes. Cycloalkanes. Alkenes. Alkenes. Can't be alkenes. Yeah, alkenes. For eight, nine and ten member cycloalkanes. Cis form is more stable. Cis form is more stable than the trans form. But for eleven or larger member ring. Trans is more stable. This example you see. How many stereo centers are there? CS3, CH. CS3. That's it. CS3. So why is Cis more stable than eight, nine and ten? Actually there are two three things. One is the diagonal repulsion. Which is more in the case of trans form. Let it be. We'll discuss that. No. Confirmation is more stable. We'll discuss that. Second one is one. Tell me the number of stereo centers. Mayut. How many stereo centers we have here? How many? One. Why one? You like one. That's why it's one. Simply put, we won. Next you. What's your name? Rajdeep, right? Yes. I'll tell you. Will. No idea. What are you doing? What are you doing? I don't get it. You could not even say one single line. How do you say that? Then if I use this concept next class, you'll ask me how it is happening. Action. Here. First one. First one. Two, right? First one, two, fine? Yes sir. Double bond, this one? Ring? No. Not possible. What about this one? Yeah. Okay. Why the ring is not possible? Six things in a double bond. Because this carbon is hot. Yes. We do hybridize. If you have one more substitution here, then because of ring also it is possible. Which is not the case. Right? Across the ring, G.I. is not possible. We have double bond inside the ring, but the ring is not eight member. That is the condition. So because of double bond inside the ring, also G.I. is not possible. So we have G.I. possible because of this double bond and this double bond. Two stereo centers. Yeah. Okay? Now across this possible? Yeah. Stereo center? Yeah. Across this ring? Yeah. Yes. Yes. Because it has two SP2s. SP3? Yes sir. Across this double bond inside the ring? No. It's seven. It's seven? It's not eight. What? Oh, sir. It is eight. I meant to be eight. Okay. So then that also. Double bond inside the ring also. So because of this ring and because of double bond inside the ring also G.I. is not possible. So one, two, three. Across this it is not possible. Three stereo centers. Okay. So you see whenever you get this question you have to apply all the condition. Yeah. Ring, double bond inside the ring and double bond. All those condition we have to. What about this one? Yes. Do we have two separate? Yes. Do we have one substitute? Yes sir. I have one substitute. I have one. I have two. I have two. I have two. I have two. I have two. I have two. I have two. I have two. I have two. I have two. I have two. So again position you and ring. So because of ring G.I. is not possible. Sir, can you show us the stereo structures going through the double bond? Like with the… This one. Yeah. Just put the надо somewhere here. In this one… Yes sir, take this shorter things. See well then suppose your double bond coming towards you, okay. This park can go down, this park can go up and this one starts at bottom. It is difficult to draw this structure. This one is going down, this part is going down, and across this CS3 is going up, this whole part is going down, this part is going down. So this one is also GI possible. One last thing you write down, cumulines, C-U-L-L-E-N-E-S, cumulines are these compounds when we have consecutive double bond, C-double bond, C-double bond, C-double bond, C, right? Suppose we have PQ, so in this case when you have two double bond present, we call it as even cumulines. Number of double bond is 2, 4, 6, then it is even cumulines, right? And even cumulines does not show geometrical isomerism. Whatever the position of this two group we have, suppose P here, Q is here, it does not show geometrical isomerism because this distance, this distance here, P and Q, if this distance is L, right? This structure, if this distance is L, this distance is also L. But this distance is same here because the molecule is non-polar. This is an xy plane, this is yz plane, so it is like this. So I can show you with the hybridization of this also. Orbital diagram we can discuss, but that we will discuss later, okay? So point is when you have two points you have to remember, very important, okay? Even cumulines if you have, even cumulines are non-planar and they does not show geometrical isomerism, right on this point. Even cumulines are non-planar, does not show geometrical isomerism, but they can show optical isomerism, optical isomerism is possible, geometrical isomerism is not possible because this distance is same since the molecule is non-planar, so this part is in different plane, this part is in different plane, so this distance is coming out to be same, okay? One last thing we have left in this, that is number of GI, for that we have two, three formulas, okay? Even cumulines can show GI. No, even cumulines won't show, okay? Sir, how many GI will it show? For that we have formula, we will see the difference of one. Even odd cumulines will be non-planar, right? No, non-planar. Sir, but they are all nice, we do hybridize them. See, this part is in this xy plane, this part is in yz plane, the next one again is in xy plane. That's how it is. We will discuss that later. Anyway, so we have formula of number of GI optical planes, we will discuss that next class, we will finish this and then we will start opticalizing. Okay, thank you.