 Now, what are the conditions we have for a molecule to show geometrical isomerism, okay? Write on condition for a molecule to show geometrical isomerism. Condition for a molecule to show geometrical isomerism. See, geometrical isomerism is possible in three different cases. One is because of double bond, another one is because of ring and the third one is double bond inside the ring, got it? Okay? In different cases, we have GI, geometrical isomerism because of double bond, because of ring, because of double bond inside the ring. For all these possibilities, they are different, different cases, conditions we have to satisfy. Like for the first one, the condition for a double bond to show geometrical isomerism is write down, the first one GI because of double bond. The condition is the double bonded carbon atom, write down, the double bonded carbon atom must not have, must not have identical atoms or groups attached to it, identical atoms or groups attached to it, okay? So if you have any carbon-carbon double bond and if this carbon atom, the double bonded carbon atom contains identical group, where on this carbon or this carbon? This is the identical group, right? Same group present here. Then this double bond won't show geometrical isomerism, okay? It must be different. So in this case, if you have C double bond, C, PQ, PQ, here we have GI possible. One is cis, other one is trans, two GI possible. Why in this case it is not possible? Because even if it rotates, then also the relative position is same. There is no difference in the, you know, distance between the two atoms. AQ if you place here, P here, then also the molecule is same, right? That's why it is not so geometrical isomerism. So condition is the double bonded carbon atom must have two different atoms or groups attached to it. Which, there can be more than one geometrical isomer. More than one means? More than two. More than two geometrical atoms. How? In the molecule, suppose you have three double bonds. Oh, okay. I'll show you the example, okay? That's possible. Example you see? And one more thing, number of double bonded carbon atom, right, number of sp2 hybridized carbon atom, double bonded carbon atom, across which the GI is possible, that we call it as stereocentrum, okay? Right on the stereocentrum, it is the sp2 or sp3 hybridized carbon atom across which GI is possible. In case of ring, sp3 possible. Okay. Tell me this example. Do we have GI possible in this one? No. Right? No. Sir, so what do you mean by a cross which isomerism? This is the star bond, this is the star bond across this. So there will be two stereocentrum. The stereocentrum can be, you can also think about this double bond is the stereocentrum. Because across this GI. What about this one? CS3, CH, double bond CH2. Which GI is possible? No. Why? The second carbon has identical atoms. What about this one? CS3CH, double bond CH2, CS3. Yeah. Possible? Yeah. Number of stereocentrum is what? Number of stereocentrum. Two. This is the 11 bond group. Versus two carbon groups. Only one stereocentrum. No, double bond? What about the triangle? These, across these two carbon atoms we have GI possible. No, SP2, this is the SP200 carbon atom, so across this double bond GI is possible. GI is not possible across carbon atom because of double bond. So these two, GI is possible. So GI is possible in this one. This is the ampoule you see. GI is possible in this. Do we have GI possible in this one? Possible or no? Yes or no? Yes. Possible? Fourth one? No, this one, this one. Fourth one. This one, fourth one. Across this GI is possible. Because this carbon in this carbon has one hydrogen in CS35, one hydrogen in this entire group, right? Two different groups. This is one of the stereo center we have. And anyone in the center is present. GI is possible. So here one of the carbons that's part of a stereo center is the thing attached to another stereo center. Across this, it's trans-possible. Across this, it's trans-possible. Across this, it's trans-possible. So this is not the stereo center, but these two are the stereo center. But in general, it's yes. Yes. If one of the stereo center is there, GI is possible. You can draw this structure like this also. Suppose this C double bond C is this one here. CS3 is this side. H here. And suppose one hydrogen is this side. And this is CH2. CH double bond CH. CH double bond CH. This is the same problem. Anyway, it's possible. Across this possible. This is what you can also draw across this one. But here it is not possible. Okay. Fifth one. Fifth one. What here? Possible. When ring is there, then what we'll do? Can't give up. See, even if you add it the other way, nothing will happen. Sir, can I get to that carbon atom here for the same things? See, actually, when you have ring like this. See, here the condition is satisfying. Yeah. To find out this, what we'll do? Suppose you have to find out this side, what the group is attached. So you have to start from here and go this way. This side. So it is CH2, CH2, C. Right? Yes. No matter how. Clockwise it's not possible. Where will you go? We have to find out. Sir, it's not like that. Yes, so that too will go. Let me go from this side first. The other side too will go. One side will be complete. We'll go both way. Will it go like this or like this? Because we need to find out what group is attached to this side and what group is attached to this side. So, what we need is the oscillation. What we need is oscillation. Okay? Now, if these two paths are different, suppose, then it means two different groups are attached on this carbon atom. Method is the best. No logic here. Because that's the method. Other way, if there is a ring attached like this, it will go this way. Right? What you are getting? CH2 here. CH2 here and then carbon. So it means this side we have CH2, CH2, C attached to this. Or this means CH2, CH2, C. Okay? Now, this side if it was CH2, CH2, C again. So what it attached to? CH2, CH2, C. Ironical group attached, it means across this GI is not possible. Basically, if carbon is same here, it means the ironical group is attached. GI is not possible. So in this case, no GI knows to the center. If there was suppose... No reason. Yeah. So then how do you draw the two different structures? You don't have to draw anything. Suppose, my example here. Thank you very much. Suppose we have chlorine here. Okay? When you see, when you go this way, at the first carbon, we have one chlorine attached. And when you go this way, at the first carbon, we do not have chlorine attached. This only means that this path and this path is different. It means two different groups are attached. So GI is possible. You don't have to find out what group is attached to. Priority. This side will be more... This will be more... But you do not have to find out cis and trans isomer. They can ask you whether GI is possible or not. Like suppose, if they give you this compound in mid-width somewhere, and they ask you how many stereocenters are there. For example, this example is... CS3, C2, H, double bond, C, H, single bond, C, H. And there is a ring. How many stereocenters are there? Across this, GI is possible. And across this also GI is possible. Because this path and this path is different. They add two stereocenters. And first carbon, we have this group. Here we have two hydrogens. Understood this? So because of double bond, the condition is the carbon atom which are bonded with the double bond must have different group attached with it. Second one that I don't... GI because of ring. We already did it. No, ring. It's not ring. Right on GI because of ring. Condition is right on... At least two sp3 hybridized carbon atom at least two sp3 hybridized carbon atom in the ring, at least two sp3 hybridized carbon atom in the ring must be dye substituted with two different atoms on group. With two different atoms on group. Dye substituted with two different atoms on group attached. Now suppose if you have a compound like this. Okay. And here we have CS3. Here we have CS3. We do not have double bond in this compound. Only ring is there. Okay. This molecule will show geometrical isomerism and there is only one steve center which is ring in this case. Across ring GI is possible. Why because the two sp3 hybridized carbon atom has two different group attached with it. That's why it is GI possible because of ring. Correct? Up sp3. Okay. In this, do we have GI possible? If yes, then how many geometrical isomers are there? M e is a methyl. But it's not dye substituted. No, but this is symmetric. However, it's oxygen. Yeah, this is symmetric. No, it's an incident also because it's not dye substituted. It's only like three to six. What is the hydrogen? There is a hydrogen besides the methyl. Sir, it's not. Why is it not? How do we rotate it? How do we rotate it? Why do we rotate it? Oh, yeah. Here it is, I think. Possible or not? Yes, possible. Possible. How many GI possible? Six. Six. Six. Six. Six. Six. Six. Six. Six. Six. Six. The condition is, the ring must be dye substituted, at least two SG-3 hybridized carbon. Yeah, one sector. One sector? Okay. Yeah. So it is at least two. So more than two also, it's five. Okay? This carbon is SP-3? SP-3, SP-3. One is methyl, other group is what? high-tides. So two different groups are there? Yes. GI possible number of this entry is one. That is ring. Correct? So now, when you draw the structure, can draw various structure of this. The first one is when all the methyl group is coming out of the plane towards the observer. This is one possibility. What about this one? Sir, here there is only one secondary. Yes. Sir, if we have a double bond coming out of the ring, then we can have a center. Double bond in the ring. Coming out of the ring. Then we have to see across the double bond also. Yes. Double bond condition is satisfied. If it is there, then possible. Sir, here we have one more extra. This question, do you think the ring has a center? No, that carbon is not SQ3 otherwise. Ring was a carbon. Sir, if I look at that, like when two things coming out of the plane and one going in and two going in and one coming out, I can... Same. Yeah, same. So, just do that. Yeah. Okay. There are two structures possible here. See, this one is the same thing. Yeah, but you just rotate the planes. When you look at the structure this way, the two methyl groups coming towards you, right? If you look at from this side, this is the structure. If you look at from the opposite side, this is the structure. Okay. These two are same structure. Even these two are the same structure. So, only two GI possible. Either this two or this two. Correct? Understood.