 Yeah. Hello guys, can you hear me? Okay. So today we are going to start a new chapter in organic chemistry that is isomerism. Okay. So isomerism the first part that is structural isomerism we have already done. Okay. So if you look at the classification of isomerism, there's a quick introduction of this and then we'll move on directly into optical isomerism. Okay. So isomerism, if you see, it is the property of a molecule. Okay. Property of a molecule. For example, suppose if I take one example of C2 at 6 O. If I ask you to draw the structure of C2 at 6 O, we can draw two possible structure for this. One is you can draw CH3 OCH3 and other one is CH3 CH2 OH only single bond is possible in this. Why because the degree of unsaturation if you count for this one. DOU, degree of unsaturation if you remember it is C plus one minus H plus X minus N divided by two. You've done this in first chapter if you remember. Okay. So C plus one is carbon atom we have two, two plus one minus H is six. There's no halogen, there's no nitrogen divided by two. So when you solve this you're getting zero. So degree of unsaturation is zero. That means for this molecule we do not have any single bond double bond ring structure possible and hence we can draw these two possible structure. Isn't it right? Right. One structure is this other one is this. Now what is this? What is the name of this compound? Could you tell me? IUPEC name? What is the IUPEC name of this compound? It is ethanol. Right. It is an alcohol. Right. It is an alcohol because the functional group OH is present. This we call it as ethanol. This one is ether and we call it as dimethyl ether. The name of this compound is dimethyl ether. Okay. So whether you write this alcohol that is ethanol or dimethyl ether, yeah that also you can write. That's fine. Methoxymethane also you can say. So the point is this molecule and this molecule has same molecular formula and that molecular formula is C2H6O. Right. So with the same molecular formula, two different compounds we can draw. One is alcohol, other one is ether. Obviously their chemical and physical properties are different. Okay. Because they have different functional group. One is alcohol, second one is ether. So this kind of phenomenon we call it as isomerism. What phenomenon? A given molecule with a given molecular formula can exist in two different form. This phenomenon is isomerism. Okay. And these two molecules are called isomers of each other. What kind of isomers? Since they have different functional group, so they are functional isomers. Functional isomers. Isomers we always define in a pair. Right. We can say these two are isomers. We cannot say this molecule is isomer. That is not possible. Okay. Functional isomers why? Because the functional group are different. OH and O. Okay. So isomerism is what the compound having same molecular formula but different structural formula and different physical and chemical properties are called isomers. Okay. So isomers are compounds with same molecular formula, different structural formula and different physical and chemical properties are called isomers. And this phenomenon is isomerism. And this phenomenon is isomerism. So this is what we had discussed already in the previous class of isomerism. Okay. Now when you see the classification of isomerism, first of all it is classified into two categories. One is structural. Structural we also call it as constitutional. Structural or constitutional. And the second one is stereo. Structural and stereo isomerism. Further, this is structural isomerism is classified into different different categories. The first one under structural is chain isomerism, then positional, and then we have functional. We have done this I'm just giving a quick recap, and then after this we'll move on to optical. Additional, functional, metamerism, metamerism, and then we have tautomerism, tautomerism, and then we have ring chain. So if you remember in this we have discussed all type, but only this tautomerism is left isn't it. So we'll discuss tautomerism also in the last of this chapter. We have discussed this part already so we are not going to discuss this again you will have to revise this. Okay, now stereo further classified into two categories. Further classified into two categories. That is a configurational one second configurational right down this side. And the second part is confirmation so conformational and configurational. Okay. So conformational we have different different form configurational also classified into two categories further. So also has two part. One is geometrical, and other one is optical isomanism. Okay, so we had discussed geometrical as well isn't it. So once easy sin anti we had discussed yes or no. Yes. So, yeah, do you see practical we have almost done few things are still left, but it is done for grade 11. Once we start a reaction mechanism, there will take up all those things which is left. Okay, have we done the first part geometrical in normal classes, could you tell me. Okay. Anyway, so we'll start from the beginning only then. Okay, we'll do structure also don't worry. But we'll start today with this part. Okay, geometrical and optical will start. No problem. We'll do the entire chapter again. Okay, those who has already done in KVP classes. Okay, that's fine they can revise this, but today we'll start with stereo isomerism here. Okay, confirmation and configuration will do. And then in the last will go towards the structural isomerism. Correct. So first we are going to discuss is geometrical isomerism. Okay, so write down the heading. One more thing. These two are different from each other. It's not like if you do not understand this you won't understand this one. Okay, so we can start with this and then the last will come to this point here we have different definitions. Okay, that we can do is we'll do that don't worry. Okay, so we'll start with stereo isomerism heading all of you write down that is geometrical isomerism. See for isomerism, the concept is that only same molecular formula and different properties. Whether it is geometrical, whether it is optical or any other kind of isomerism molecular formula must be same. Okay, molecular formula must be same. Now you see this example will take one example and then we'll discuss other things. Yes, we'll do everything in the chapter will do optical will do geometrical will do confirmation will do everything. Okay, we are starting today with geometrical is it fine with all of you. Once we finish all this will do a structural isomerism as well. Correct. Okay, so suppose I'm writing down to example here. One is CH single bond CH. Here we have CS3H here we have CH3H. And another one is CHCH CH CH CH CH CH CH CH CH CH CH CHCH this is one pair, another one see double bond C, CH3 H3 H, CH3 H, CH double bond C, HCH3 and CH3 H. What is the difference in this two molecule? This two means I am talking about, this two molecule do you have any difference and what is the difference or similarity in this two molecule? These two molecules are same or different. These two molecules are same or different. These two molecules are different molecules. This is different molecule but this one is identical molecule. There is no difference in this two. Why? Because we do not have double bond present here. If double bond is not present, then you can easily rotate this carbon across this single bond. Rotation is there so you can rotate this and you will get this molecule. But here the rotation is not possible because you see the double bond forms this way. We have a sigma bond like this. The orbital overlaps will form a sigma bond and double bond forms by lateral overlapping like this one. This forms the double bond. Now once you try to rotate this, obviously this pi bond you need to break and that we cannot do in isomerism. Hence what we say that across a double bond, the rotation is hindered across the double bond. We can also say triple bond. Same logic. We can also say ring structure. The rotation is hindered and hence these two molecules are different. But this here, the rotation is easily possible across a single bond. Hence these two molecules are identical. Now this kind of compound where the properties differs because of double bond or triple bond, not triple bond, double bond or ring. This kind of molecules we call it as geometrical isomers. For example, you see these two molecules, both CS3 are on the same side. Both CS3 are on the opposite side. So relative stability, if you see, this one is more stable than this because we have less repulsion over here. This kind of phenomenon, we call it as geometrical isomerism. Write down the definition, all of you. Write down the isomers, geometrical isomers. Write down isomers which has the same structural formula, same structural formula, but differ in the arrangement of atoms or groups. Atoms or groups in space. Yes, I am writing. Can't you see the screen? Guys, the screen is visible. Arujita, you can raise wine. It's fine for all of us. Atoms or groups in space due to restricted rotation, due to restricted rotation around the double bond are called geometrical isomers. And this phenomenon is geometrical isomerism. We can have here a double bond or ring also we can write because across ring also GI possibly are called geometrical isomers. According to this, you see this two compound has different properties. They have the same molecular formula. You see here these two have the same molecular formula, but different properties. So these two are different molecules like I have written over here. Could you tell me the IUPEC name of this compound? IUPEC name of this compound? It is but 2 in. Yes, this molecule is but 2 in. What is the name of this compound? But 2 in. Now the whole purpose of IUPEC non nomenclature got defeated over here because the purpose of the naming here is what? That each and every molecule should have its own name. A specific name must be given to each molecule. That is the whole purpose of IUPEC nomenclature. So when it comes to isomers like this, the purpose of the whole objective of IUPEC nomenclature got defeated here. So we have to find out some way by which we can differentiate. Suppose if I ask you to draw the structure of but 2 in some of you will draw this, some of you will draw this and both are different molecules. So we have to specify that which structure we need to draw or we are talking about. So to overcome this difficulty here, we have given a term here and that term is cis trans, easy and syn anti nomenclature. So to define or to write down the name of geometrical isomers, we have different naming system, naming systems of geometrical isomers, GI. GI means geometrical isomers. Three types we have mainly. One is cis and trans. Second one is ENZ. Third one is syn and anti. Syn and anti is used when we have nitrogen present in the molecule. It is this nomenclature, this naming system we use when CN double bond is there, in case of this we use or NN double bond is there. In case of this also, we use syn and anti system of nomenclature. Cis trans and easy and syn anti. These three nomenclature we have. When we use cis and when we use trans and when we use ENZ. ENZ is generally used in more substituted alkene. More substituted alkene. More substituted means what? The double bonded carbon at least have three different groups. This means write down the double bonded carbon at least have three different groups. Cis and trans we use when the alkene is not highly substituted means the double bonded carbon has only two groups different on both double bonded carbon at it. Two different groups here, totally if you see. Copy this down. Yeah. So one by one we will discuss. First of all we will discuss cis and trans. So pretty simple. Suppose we have a molecule like this C double bond C. Now here we can have any group attached on this carbon atom. We can have any group attached to it. So total four we have here. Four group possible. Atoms across one, two, three, four position we have. So out of this four, if three groups are different, three atoms are different, then we use preferably we use ENZ system. If out of the four, if three groups are different, if only two groups are different we have, then we use cis and trans system. So here you see suppose we have a group attached to it. Same example. I'll take CH3, H and H CH3 like this. Another one if I write down CH3, both CH3 on the same side and both H on the same side. So you see here cis trans isomers or cis isomers we define when identical group group are present on on the same side. That is cis. Trans means when identical groups groups are present on the opposite side. Opposite side of the double bond actually. Okay. So you see in this one, CS3 and CS3 are identical group. It is opposite side. So it is trans isomers. H and H also on the opposite side. Trans CS3 CS3 on the same side. So it is cis isomers. So if you write the name of this compound, it is obviously but 2 in, but this name would be since it is cis. So it is cis but 2 in. This one is trans but 2 in. So if I ask you to draw the structure of cis but 2 in, cis but 2 but 2 in and trans but 2 in, it is clearly an understanding and you can understand this at which structure you need to draw. Prokola is talking about this four group. One, two, three, four positions we have no. Out of four if three are different. One, two, three, four. Out of four if three are different. We'll discuss this. Easy and Z and Z will discuss. Correct. So it is cis and this is trans. So cis and trans we can define only when identical group are present on the both double-bonded carbon atom. Like another example you see, we have a carbon-carbon double bond with C, O, O, H and H. This is also C, O, O, H and H. Suppose we have this and another one in which both C, O, O, H is present on the opposite side. H, C, O, O, H it is C, O, O, H and H. So this one is obviously, it is cis the first one and this is trans. Okay. Cis molecule, this we call it as malic acid. You see the relative positions changes, the compound itself changes. It is malic acid. It is fumaric acid. So relation between malic and fumaric is they are geometrical isomers of each other. If you have this compound, C double bond C, C, O, O, H, H instead of C, O, H we have C6, H5 here and this is H, C6, H5 like this. Okay. Name of this compound, if this C, O, H you replace by phenyl group, name of this compound is synnamic acid. Right. This one is also transform of synnamic acid. So this is the, you know, point I was talking about when at least three groups are different. You see one, two, three different groups we have. Then preferably we can say, we can define this as ENZ isomers also, but ENZ we'll discuss a bit later. Suppose if you have to assign cis and trans over here, right, cis and trans over here, then we can define cis and trans with respect to always identical group that is hydrogen here. Hydrogen present on the same side. So we can say it is cis if we have to define cis and trans. Hydrogen on the opposite side, it is trans. But preferably for this molecule, we define ENZ isomers. How do we define ENZ? We'll discuss that a bit later. Okay. Now you see what is the condition for a double bond to show geometrical isomers. Okay. So condition for, if you talk about GI, GI is possible under three conditions because of double bond, we can say GI possible, right? Because of ring GI possible. And the third condition is because of double bond inside the ring, double bond inside the ring. This three condition you have to take care of. When you talk about a double bond, we're talking about this only now, ring and this will see a bit later. Double bond you see the condition is the two atoms, atoms or groups attached to the attached to the double bonded carbon atom must be different. This means what? Suppose we have a molecule C double bond C. And here I'm writing down a term A and B here. And here suppose P and Q are attached like this. So condition for double bond is what? This A and B must be different. It cannot be the same atoms or molecules or groups. A and B must be different. P and Q also different here. So condition for a double bond to show GI is A must not equal to B and P must not equal to Q. This is the condition for a double bond to show GI. A can be equal to P. B can be equal to Q. But this two are not equal, must not equal. Clear? Now the second type of naming is E and Z isomers. E and Z isomers if you see, we have a molecule say carbon-carbon double bond A, B, P and Q. So first of all E and Z isomers we define when out of the four atoms or groups at least three are different or four are different, then we can define E and Z. Because if any two of these groups or atoms are equal or identical, then with respect to that atoms or groups, we can always define cis and trans. But when we have highly substituted alkene, highly substituted alkene means suppose we have ethene. Ethene is this H, H, H and H. Now if you remove one of the hydrogen atom and place CS3 here, this we call it as substituted alkene. Another one if you substitute here, it is more substituted than the previous one. This one is even more substituted and this one is highly substituted alkene. So when we have highly substituted alkene, then we use E and Z, a technique for the nomenclature of the geometrical isomers. So when we use E and when we use Z. So to define this E and Z, first what we do, first we assign priority on these atoms or groups. Suppose this has higher priority than the other one. We assign priority. It depends on the molecule. If all the four groups are different, you cannot even define cis and trans. But if any two are same, then you can also write down cis and trans. That won't be wrong, but preferably we'll write E and Z for that. Okay, that's always better to write E and Z. Cis and trans with respect to the atom which is identical on both carbon atoms. Okay, so to define E and Z isomers, what we'll do, we'll assign priority on the atoms or groups attached to the double bonded carbon atom, these two carbon atoms. This carbon atom has these two groups attached. So we'll assign priority on these two. Suppose it is first priority. This is a second priority. Means this one is higher priority. This one is lower priority. Similarly, this one is higher priority. This one is lower priority. Like this, we assign priority. Now Z isomers is the one in which the higher priority groups or atoms are present on the same side of double bond. Same side of double bond. Okay, if you talk about E isomer, higher priority group or atom are present on the opposite side of double bond, opposite side of double bond. So if you talk about this one, same priority on the same side, so this is Z isomer we can say. We'll do some examples first to copy this down. So basically if you know how to assign priority, we can say Z and E isomers easily we can define. But the question is how to assign priority here, correct? So to assign priority, we have a rule that we call it as CIP rule. The name CIP is based on the name of a scientist, of three scientists actually, that is Khan Ingold and the third one is Prelog. So Khan Ingold Prelog are the name of the scientist. These three scientists has given this rule. We just need to follow the rules to assign priority on the atoms or groups attached to the double bonded carbon atom. First rule is what? Higher atomic number, see atomic number, not the atomic mass means higher priority. So suppose iodine, bromine, chlorine, fluorine attached, so priority order would be this higher atomic number. In case of isotopes, in case of isotope, we'll see mass number, we'll see mass number. So hydrogen isotope you see, T, D, H, the priority order. Oxygen if you see, O18 is higher priority than O16. If you talk about chlorine, chlorine 37, higher priority than chlorine 35. Done. Do you have bio exam tomorrow? Okay, so who all are writing? Okay, so some of you have requested to finish the class a bit early. Okay, is it fine with you if I finish it by seven? Yes, so we'll finish it by seven. We'll take just five minutes of break in between. Like break, you won't take much because we need to finish many things. So five minutes break, we'll finish it by seven. Okay, so this is the thing we have. Now the third thing is third priority, third rule you write down. If groups are attached, are attached, then we'll compare the priority, then we'll assign priority, priority on the basis of, on the basis of atomic number first atom, of first atom. See this example. Suppose we have chlorine, we have SO3H, we have OH, we have NHCH3, we have COH. These groups are there. Then how do you assign priority to this? We'll compare the, no, we won't do like that. We'll compare the priority by comparing the atomic number of the first atom, like we have chlorine, first atom is sulfur, first atom is oxygen. One second Mr. I'll go back. First atom is nitrogen, first atom is carbon. Chlorine, we know atomic number is 17, sulfur it is 16, oxygen it is 8, nitrogen it is 7 and carbon it is 6, atomic number. More atomic number, more will be the priority, priority order is this. So if groups are attached, we'll compare the atomic number of the first atom. I'll go back one second. I'm so in the third point, so what is the fourth word? In, if groups are attached, this one. Yes, thank you sir. If groups are attached, done Frida. Now, obviously if you have different groups attached with different atom here, like chlorine, sulfur, oxygen, if different atoms easily we can compare. Sometimes what happens? Sometimes we have the same atom attached, like for example, suppose one group is CS3 attached to the double bonded carbon atom, other one is CH2, CS3 attached to the double bonded carbon atom, other one is COOH attached to the double bonded carbon atom. So in this case, obviously we compare first atom. So first atom is same only here, that is carbon, we cannot compare with respect to the first atom, then we'll compare the second atom in the chain. So carbon, we don't have the chain, we have hydrogen only, so we'll compare hydrogen with carbon. In the chain we'll compare. So carbon, carbon chain we have, carbon, oxygen we have here. So carbon, here we have carbon, here we have oxygen. So priority order would be like this. Oxygen, carbon and hydrogen, this is the priority order. That's what I said before. If the first atom is same, then we cannot decide priority on the basis of first atom, then we compare the second atom. Second atom in this carbon, we have hydrogen and we go along the chain, the carbon chain that we have. Here we don't have chain, we don't have any choice. So we'll compare hydrogen here. Here we have carbon first atom, second atom is again carbon. So carbon and hydrogen we compare then. Obviously carbon is more, so this one is higher priority than this. If we compare the second atom here, that is oxygen, oxygen and carbon, obviously oxygen has more atomic number. Hence this is the priority order. Yes, so again I'll write down here, how do we do it? We have carbon with hydrogen attached, and here we have carbon with hydrogen, then again carbon with hydrogen like this. Here we have carbon and then oxygen, oxygen and etch. So first atom, we cannot decide with respect to the first atom. That is carbon, carbon and carbon only. So leave the first atom. Now you compare the second atom. Second atom we compare only in the chain, carbon-carbon chain. So second atom here is carbon. So we'll see this carbon atom. But the second atom here is hydrogen because we do not have any choice. We'll take this hydrogen. So carbon-hydrogen, carbon is higher priority, this second atom here is oxygen. Oxygen is higher priority than this, order is this. Is it clear? So these are the trick that we use. It is just the method we have. We do not have any logic here. Just the method we use in order to assign priority. You have to do it this way only. Only one complication you may have when we have double or triple bond present. In that case also, we have a hypothetical method that we use. I'll give you one example here. Not very common, but you should know, sometimes they ask this. Okay, next point you write down. A double triple bonded atom bonded atom is considered equivalent to two or three such atoms. Two or three such atoms. With example, you'll understand what is the meaning here. Suppose you have to compare C double bond O and Cn, C triple bond N. These two we need to compare. Since we have a double bond here, so it is a hypothetical method. It's not logically correct. We have a double bond here. So what we assume in this, so in this we assume that the carbon is attached with two oxygen and each oxygen also attached with two carbon, because we have a double bond here. So carbon is attached with two oxygen and each oxygen also attached with two carbon. So you see one oxygen with one carbon this side, one more carbon we have here, one more carbon we have here. This is the meaning here. Hypothetical method. It doesn't happen practically. Similarly here, we have a triple bond. It means carbon is attached with three nitrogen and each nitrogen attached with three carbon. So one carbon we have here, another two is this, another two is this and another two is this. This is what the meaning we have. So if you want to compare these two, we'll compare the first atom carbon and carbon. We cannot decide this. When we compare the second atom, we have oxygen and we have nitrogen. So obviously oxygen is more, so priority order is this. So first, second and then third, like this we go. Is it always two C attached? Means no. When we have a double bond, then we say CO double bond we have. It means each carbon has two oxygen, each oxygen has two carbon. That is what it means. Yes, so number of oxygen and nitrogen we are not considering. We are considering the atom we have here and their atomic number, that is it. Each O will not have two C. Like I said, yeah, we can assume that in order to compare the priority. Each oxygen has two carbon, each carbon has two oxygens. So first atom is carbon, second atom is oxygen, third atom is carbon. Carbon, second atom is nitrogen, third atom is carbon, like that. Once again, just so these are not very common. It's not like you will get this always, like very often in the exam, but yes, rarely you get, but you should know this. Right. Now, one more example we have here. Like sometimes what happens, the carbon atom contains a ring attached to it. I'll explain this with an example. Suppose we have carbon, carbon double bond, sorry, carbon, carbon double bond, and this carbon atom has a ring attached to it. This is the ring, for example, oxygen atom. And here also we have a ring, five member ring like this. So obviously this site, it's pretty clear. Gromine has more atomic number. So it is higher priority one and this is two. The problem is how to assign priority in this ring. Okay. How to assign priority in this ring. So consider this ring. This ring is nothing but this. And this ring you have to open. Like suppose this carbon I'll write down here, attached with two oxygen. So we'll write down O this side. We'll write down O this side. And then O with carbon, carbon, O with carbon, carbon. How many carbon atoms do we have? We have three carbon atoms, right? Like you start from this, oxygen, carbon, carbon, carbon, oxygen, carbon. Like this, like this is the way we have. It's not like we have some logic. This is the method we use in order to assign priority. How did we write? We start from this carbon, this carbon, carbon, oxygen, carbon, carbon, carbon, oxygen, carbon. We'll write this. Again, other way. Like you need to write down clockwise, anti-globally. Both way you need to travel. When you start from this side, carbon, oxygen. Then again, we have carbon, carbon, carbon, oxygen, carbon. This is the first ring. Now similarly, we'll write down the smaller one also, this ring. This is the ring we have. We'll start from this one, carbon. I'm going clockwise. So we have oxygen, carbon, carbon, oxygen, and carbon. Anti-glockwise, oxygen, carbon, carbon, oxygen, carbon. Like this, you open it and then you compare the two atoms like this. I'll tell you how. First atom is carbon. We cannot decide the signal. Second atom you see we have only oxygen. Again, oxygen, oxygen, we cannot decide ignored. Third atom is again carbon. We cannot decide ignored. Fourth atom is again carbon. We cannot decide ignored. Fifth atom if you see, here we have carbon, but here we have oxygen. Here we can easily compare. Correct. So this is oxygen here and this is carbon here. So we can compare oxygen, carbon. Oxygen is more, you know, has more atomic weight and this has higher priority than the other one. This is one. This is two. Higher priority group on the same side, it is Z. Tell me. So if you have ring, you have to do like this. Both way you need to write clockwise, anti-glockwise and compare. Higher priority, same side, we have Z. Higher priority, opposite side, we have E. I'll write down. I'll write down. This Z you just let it be. I'll write down how it is Z. Here I have written, I guess you see. See, I've written this. Z isomer, when higher priority are present on the same side or double bond. Z, opposite side and E. Okay. So you see, both higher priority, same side, hence it is Z. Yeah. So when you have a compound like this, a smaller ring will have the higher priority in this kind of pattern. Okay. A smaller ring will have higher priority. Yeah. I'll explain. So what you need to do, you write down the two ring which is attached like this and open up the ring, like this carbon right down here and then you start going clockwise. So oxygen, carbon, carbon, oxygen, carbon, right down here. Oxygen, carbon, carbon, carbon, oxygen, carbon, done. Then anti-glockwise, oxygen, carbon, carbon, carbon, oxygen, carbon. This is what we have. Same thing you need to do for this group. Carbon, oxygen, carbon, carbon, carbon, oxygen, and we have this. Then you start comparing one by one. First atom, same, we cannot do. Second atom, oxygen, second atom, oxygen, we cannot do. Third atom, carbon, carbon, cannot do. Carbon, carbon, cannot do. Carbon, oxygen, hence here we can compare. Oxygen has more atomic number. This has higher priority than this, hence one and two. Tell me, clear? 12 of you? It's pretty simple. Logic we do not have. Just a method you need to open the ring and then compare one by one. This is also very rare. You'll get questions like this because yes, in JEE sometimes they have asked these questions. So must understand this. Now, one last thing in this you write down. Number five and number six, you write down the number here. If lone pair is present, it has the least priority. It has the least priority. Like for example, we have carbon, nitrogen, double bond, OH, CH3, H. This nitrogen you see, it has one lone pair present on it. So in this case also, geometrical isomerism possible. Priority, if you check carbon, hydrogen, carbon is higher priority. Oxygen is always have higher priority because lone pair only. Higher priority opposite side, this is E. And we also call it as Nt because we have carbon-nitrogen bond. So when nitrogen is present, sin and Nt we can define. So higher priority opposite side, Nt isomer. Clear? Understood? Okay. Now we'll see some examples. The logic and the theory we have. You need to see in these molecules, geometrical isomerism is possible or not. H2C, double bond, H2. Copy down all of you first. Geometrical isomerism is possible or not. CH, double bond, CH2, CH3, CH3, CH double bond, CH, CH3. Try this. Just need to find out, GI is possible or not. Condition, I have already told you. What is the condition for a double bond to show GI? Okay. So now you see the condition for a double bond to show GI is what? Again, I'll repeat. If A and B is there, P and Q is there. What is the condition? A does not equals to B. P does not equals to Q. Isn't it? Identical group or atom should not present on the double bonded carbon atom. Isn't it? Yeah. So do we have GI possible in the first one? Yes or no? Why or N you can type? No. No GI. Okay. In this one, do we have GI possible? No. Because left side to it's fine, but here we have both atoms hydrogen only, so not possible. In this one, yes, we have possible, right? And we have two isomers possible, cis and trans for this. Remember you let it be. In this one, yes, possible. And in the last one, is it possible? Last one, yes, possible. Last one possible, not across this, across this, it is not possible. No GI, but across this double bond, it is possible. Any one bond is satisfying, so across this double bond is possible. So overall the molecule will show. Okay. One more thing. The number of double bond across which the GI is possible is called stereo center. Okay. Stereo center is the number of bond across which the GI is possible. We can have double bond. We can have ring also. We can count in GI. Okay. I'll write down here. So in this one, if I ask you, number of GI, the first one, obviously there's no GI present, there's no stereo center present. In this, we have one stereo center. This is star mark. We represent the stereo center. In this, how many stereo center we have? How many stereo center we have in this? Two, because across this possible, across this also it is possible. In this one, we have only one stereo center. This is not possible. If any one stereo center is there, molecule will show GI. So these are the some very basic examples we have. Now for J point of view, if you look at this kind of example, see, sometimes what happens, we have a carbon double bond C, CS3H. And with this carbon, we have a ring present here. Like this, we have a ring. Carbon atom present in the ring. In this molecule, what happens? So when the ring is present, then what we'll do in order to find out whether the molecule will show GI or not. Okay. So if you want to understand what group is attached this side and what group is attached this side, obviously, this is fine. Two different we have. Everything is based on this. To understand what group we have this side, you just go like this. Like this, you need to go. The white one you consider. What you get? You get CH2, CH2 and C. You'll get CH2, CH2 and C. This means on this side, CH2, CH2, C group is attached. Okay. Similarly, if you go other way, like suppose you need to understand which group is attached this side. So you need to go like this, the green one. This gives again CH2, CH2 and C. Since we have identical group, it means this side and this side, we have identical group present and hence the molecule won't show GI. I'll give you another example. See this one. We have a ring, double bond, C, CH3H. So this obviously it won't show, but when I place this chlorine here, you don't have to write this actually. You just see this path, both path must be different. If you go anticlockwise or clockwise, the path must be different. This path at the first carbon we have CL and when you go other way, this path, the chlorine is present at the second carbon. It means both clockwise and anticlockwise we have different path. It means a different group attached over there. Did you understand this? Both way you need to go. If both paths are different, it means different group attached both side. This will be a stereo center and yes, GI possible across this. Could you tell me this one? Yes or no? Yes, possible? If I attach, possible. If I attach chlorine here, then yes guys, tell me. All of you must respond. What about this? Not possible? What about this? Possible? What about this? What I will say that? What is the route? Tell me. Previous one. This one. Yeah, same thing. No, you see the same. If you go like this, then at the first carbon, you are having two chlorine. If you go like this, first carbon, we have one chlorine and one bromine. Means the path are different. Hence, yes, possible. Right? See, it's the same logic. Like I'm not giving you any different logic in all these compounds. Let's take another example. Suppose you have a four-member ring. And we have a double bond here. Carbon, we have CH3 and D present here. Now, we need to find out in this one, do we have GI possible or not? Could you tell me? Yes, it is possible or not? Not possible, correct? Because what we'll do? From here, we'll go like this. So, C, C, C, we are getting. If you go like this, again the same atom we are getting. The same path, so it's not possible. What if I attach one chlorine here? Shraddha, tell me. No, very good. How it is possible, Ritu? You see, this path, we have chlorine at the second carbon. This way also chlorine at the second carbon. So, it's not possible. Paths are same. Oh, yeah, fine. What about this one? If I attach one chlorine here, okay? What if I attach one chlorine here? Not possible because this path, again, the path is same. This way, this way, path is same. What if I attach one chlorine here? Again, not possible. What if I attach one CS3 here? Possible, right? So, like this, you will get the question, okay? A question could be anything. It's very dynamic, right? I cannot give you the, you know, like this type or that type. But you should know the logic if any group is attached, a ring is attached, then what you should do to find out whether the given molecule is, you know, does show geometrical isonarism or not. Is it clear with you? Okay? So, now we had discussed that across a double bond, how we can find out GIA is possible or not. But like I said, here if you see, if I go back and show you the slide, that one slide I show you, that I have shown you that there are three possibilities under which the geometrical isonarism is possible. You see this, right? Double bond, ring and double bond inside the ring. So, for double bond, we understood the condition. This was the condition we have. So, what is the condition for a ring to show geometrical isonarism? Okay? So, right? Next heading, write down. Okay, one more thing I forgot to tell you. Stereocenter you must have written. So, write down. Stereocenter can be sp2 or sp3 hybridized, anything possible. Stereocenter could be sp2 or sp3. Sp2 or sp3, anything. No, no, I'm telling you. I'm telling you. Stereocenter sp2 be hosatta and sp3 also it's possible. So, both it is possible. Like obviously, if you have a double bond, then sp2. We'll see the ring valla. In that case, it is sp3. Stereocenter can be sp2 also or can be sp3 also. That you write down. Till now we have discussed sp2. No, in triple bond, it's not possible. See, I'll tell you what is happening. See, triple bond won't show geometrical isonarism. What is happening here? Yes, right? What is happening here? You see we have h and cs3, correct? We have cs3 and h. So, when you change the position of cs3 and h, then the relative distance between the methyl group is decreasing or changing. That will affect the stability of molecule. But when it is a triple bond, then what you will change? Nothing you can change, right? We have only one atom. It's linear. So, you can arrange any, like the way you want, you can arrange this in space. You can also write down this way. Like, you can arrange any which way you want, right? So, like, arrangement is same only in this space. Hence, across triple bond is not possible. That time in the first slide, I have written triple bond because we have hindered rotation across this. When you rotate, when you try to fix this and rotate this, then again, this pi bond, you have to break. So, rotation is hindered, but GI is not possible across this. Okay. Now, you write down geometrical isomerism because of ring. Geometrical isomerism because of ring. So, condition is what? Condition you have to keep in mind. There are two key points in this. Write down. At least two sp3 hybridized, at least two sp3 hybridized atom must be disubstituted, must be disubstituted. Okay. I'll give you one example. Sp3 is very important, right? It's not sp2, it's sp3. That's why I have, you know, given you the note there that stereo center could be sp2 or sp3 hybridized. Okay. So, look at this example. Suppose we have a ring, three-membered ring. If you have CS3 here and H here and CS3 here and H here, right? So, first of all, this is sp3 hybridized. So, two atoms are sp3 hybridized and they are disubstituted also, means disubstitution means this two group must be different atom or group at sp3 hybridized carbon. This must be different. Then only GI possible. So, we can say in this compound because of ring GI possible and the stereo center is nothing but the ring over here, right? And we say geometrical isomerism is possible because of ring here. Just the condition you need to apply. We don't have logic how it happened and all. The way or the rule is given in the book that you have to keep in mind. Okay. Obviously, there are, you know, like, you know, research behind this to give this particular rule. We have a lot of research. So, we don't have to go all the history like how it happened and how do we get to this particular thing that is not at all required. Okay. Now, suppose you have a compound like this, four-membered ring I'm talking about here. One CS3 is attached here and another CS3 is attached here. Yes or no? Okay. Why no? Yes. Why it is yes? Where it is sp2 hybridized? GI3. It is not sp2 and it is disubstituted also. I haven't drawn this but you have to consider this H is there. No. Yes, right. Yes, clear. And both are what? Both are sp3 hybridized. Isn't it? This is sp3 and this is sp3. So, we can say GI is possible because of ring. Number of stereo center is one. Any doubt? This one? Possible? Yes. Now, possible? Yes. Condition is at least two. So, previous one was also fine. GI possible. This one is also fine. GI possible here because the condition is at least two. We have three over here. So, more than two also fine. Serious one. Tell me the number of stereo center in this. How many stereo center we have here? If GI possible, number of stereo center, number of stereo center. Okay. So, in this one, if you see, because we have an identical good present over here, so GI is possible. Yes, it is possible but only because of ring. Number of stereo center is one over here. That is the ring over here. Okay. In this one, this won't show because of ring it is possible. Double bond it is possible. So, yes, GI possible. Number of stereo center is two. This one? This is a studio center. This is a studio center. Like, studio center is not this. It is the ring only the studio center. But these two are sp3 hybridized. Right? So, yes, GI possible. Number of stereo center is one over here. That is this ring. Any doubt? Suppose if you place a double bond here, GI possible? Yes or no? Yes, possible because at least two again we have sp3 hybridized. What about this? GI possible? No, it's not possible because this two become sp2 but the condition is sp3 hybridized. Okay. Any doubt till here? Now, the third condition, write down, geometrical isomerism due to double bond inside the ring. This is the least important but still the condition you should know. Write down the note over here. The condition is the ring must be must be at least at least eight members. Right? At least eight members should be present in the ring. Then only it is possible. Otherwise, double bond inside the ring won't show geometrical isomerism. Okay. Example you see, just one or two examples we have not much. Suppose we have a double bond here and we have a double bond here. Okay. So, in this one, since the number of carbon atoms you see in the ring, one, two, three, four, five, six, seven, eight. Then we can say we have GI possible in this and the stereo center is this double bond. So, double bond inside the ring show geometrical isomerism. Okay. In this, you see both hydrogen is there on this side. So, it is cis form and in this case one hydrogen is this side. Another one is this side. It is a trans form we have. Just one condition you have to keep in mind. Nothing much we have. One note you write down here. For eight, nine and 10 membered ring, cis form is more stable than trans. This is the experimental fact we have. For 11 or larger membered ring, 11 or more membered ring, trans is more stable. That's these two points you have to keep in mind. Nothing else. So, three conditions we have for any compound to show geometrical isomerism. Double bond ring and double bond inside the ring. Now, just two, three more examples. You need to tell me, GI is possible or not? If possible, then how many stereo center? CH3, CH, double bond CH. The second one is CH3, CH, double bond CH, double bond CH, double bond CH2. Tell me. Okay. The first one you see, if any one condition is there, then GI is possible. Like you see, across this double bond, GI is possible. Anything is possible. Like in a given compound, we have to consider all three conditions. So, we have this molecule. Yes, GI possible. Number of stereo center you see, SC would be one over here. That is a double bond. Okay. If I place one chlorine suppose over here, then number of SC will be what? Number of stereo center in that case would be two. One is this double bond and other one is ring. Okay. Double bond inside the ring won't show GI because the ring is not eight member. Okay. Just we'll remove this. Across this, yes possible. Because of ring, yes possible. Because of double bond inside the ring, yes possible. This one, not possible. Yes, GI possible. Number of SC would be, isn't it? Yes. Second one. This one is obviously SP3 hybridized. This one is also SP3 hybridized with one lone pair. So, yes, because of ring, GI possible. And number of SC stereo center is one. Okay. Understood. Just a second guys. Okay. Now, one very important example we have. We call it as ALINs. Okay. I'll show you. It's very important. GI also, they have asked this question many times. If you have a compound like this, C double bond, C double bond, C. I would request all of you now, just for now, you just memorize this. We'll discuss the logic of this later when we do optical isomerism. Just you do not get confused when you look at this structure. So, this type of compound, this type of compound, it won't show geometrical isomerism. No GI possible here. Keep that in mind because it looks like the, you know, the distance is different. But here, what happens this distance between this two CS3 group and this distance is exactly same. It is a non-planar compound. It does not look like, but this distance is exactly same. Hence, it does not show geometrical isomerism. All the repulsion, hysteric hindrance, everything is exactly same. That's why GI is not possible in this, but it shows optical isomerism. How and like, how it's possible. We'll discuss this in optical isomerism, but must keep this in mind. Don't get confused that it is Cicertrans type. So, it's GI possible. Okay. Note it down. Jay may have asked this question many times. In general, this kind of compound where we have continuous double bond, we call it as cumulines. This is cumulines, double bond, double bond, double bond, double bond like this. And when we have even number of double bond in cumulines, even number of double bond, even number of double bond means it is non-planar and optically active. Just write it down and memorize it. Okay. We'll see this later. Fine. So, we'll take a break now, guys. Five minutes break and then we'll start after this. Take a break. Five minutes. Okay. So, we discussed this. So, this one is important for Jay's point of view. Also, Jay, they have asked this question many times in the exam. Neat. They've also asked in Neat as well. Okay. Now, we see the physical properties of geometrical isomerism. Okay. Two things are left property and there is the number of GI. How do we count? Okay. Properties of GI, geometrical isomerism. If they ask you to find out by dipole moment, oh, sorry, boiling point, first of all. The boiling point is directly proportional to the dipole moment. If you talk about dipole moment, you see this example. We have C double bond C, CH3H and here we have CH3H. Cis or trans? This one? Cis. And this one is trans. Correct? So, if you look at the charge separation in this molecule, this carbon atom here, it is sp2 hybridized and this one is sp3. So, we know sp2 hybridized is more electronegative than sp3. So, this carbon will drag the electron pair towards this side. So, we have the charge separation dipole moment like this, from positive to negative. Right? From positive to negative. So, basically, it is in this direction. If you talk about the second part, the right side of it, it is like this. Again, it is sp2, sp3. So, dipole moment is like this in this direction. From positive to negative, always in chemistry. Right? So, here it is positive to negative dipole moment. So, this dipole moment is in this direction, basically. Dipole moment is a vector quantity. So, we have a net dipole moment, which is the resultant of this two in this direction, mu. So, we don't have to find out, calculate the dipole moment. We can simply say, this molecule had some dipole moment. It is not equals to zero. And hence, the molecule is polar also. It is in, it is sp3. So, from positive to negative, the dipole moment in chemistry. Physics, it is ultra. Physics, it is negative to positive. Here, we have positive to negative. Okay? So, this is for cis. We have some net dipole moment. Similarly, here also, if you think, positive to negative like this and positive to negative like this. So, this two will cancel out. And mu is equals to zero for this, for trans. So, we can say, cis will have more dipole moment than trans. Write it down. The dipole moment of cis is more than to that of trans. For this exam, oh, wait, one second. Don't write this particular line. Okay? This is the molecule, which has more dipole. Cis and trans, we cannot say. One example, I'll show you. If you take this example here, did you copy all of you this? Don't write cis and trans exam, just to copy down this. You can write down this one. This is second. And then you can write down dipole moment for the first one is greater than to that of the second one. It's a reference, basically, Kensho. There's no, like, we have pressure, we take internal pressure and fetus will take external pressure like that. So, here also, reference we have. Based on that reference, we got this. It is defined like this only. Here you see this example. Again, one more example I'll show you. CH3C, double bond C, CL, H, and H. GI possible in this? Is it possible? Here also, it is possible. Could you tell me E or G it is? This one, if you think of, this has higher priority. This has higher priority, higher priority, same side. It is Z. If you want to say cis, this one is cis. This one is E and trans. With respect to hydrogen, we can say cis or trans. Overall, it is Z or E. Here, if you think of the dipole moment, right? So, dipole moment in this molecule from positive to negative like this it goes and here from positive to negative like this it goes. Here you see positive to negative like this, positive to negative like this. You see, these two are aligned in the same direction. So, here also we have mu does not equal to 0 for trans, mu does not equal to 0 for cis. Which one has more dipole moment? Could you tell me? The right one, correct? So, this is one, this is second. So, you see in this case the trans one will have more dipole moment, isn't it? So, we cannot say cis always have more dipole moment than trans. It depends upon what molecule we have. Is it clear? So, now with dipole moment, you see what all things we can conclude. Like I already told you, dipole moment boiling point is directly proportional. Okay? Dipole moment is also directly proportional to solubility. And all these questions they have asked. Okay? Solubility. One second Shradha, I will go back. Solubility and dipole moment directly proportional. We can also write down heat of combustion, heat of combustion directly proportional. Hydrogenation directly proportional. Dipole moment is more than density is more. Like dipole moment is more than refractive index. This you need to memorize all this property. Okay? Melting point. Okay, this is one thing. Last two property we have, that is about the thermal stability, stability melting point. It is more for, more for trans isomers. Because in trans, the packing is better than cis and hence the melting point is more. Right? Trans the packing is better than cis and hence the melting point is more. I'll go back once. Okay Shradha, let me know once you're done. Done? Okay. Okay. Done all of you? Okay. Now the last thing here we need to understand is the calculation of geometrical isomers. Okay? You need to memorize the formula based on condition. Geometrical isomers. See, when the number of stereo center is one, when the number of stereo center equals to, suppose n I am assuming in general, expression I'll write down. Suppose the number of stereo center is n. So in this first case, if n is equals to one, then the number of GI would be two to the power n, that is two, cis and trans, only possible. Like you see, if you have a compound like this, trans, cis, two isomers possible. Second case, if n is greater than one, if n is greater than one, then we have two possibility here. Number of GI equals to two to the power n, if when the ends are different. What do you mean by this? I'll explain with example just to copy down the formula first. Different ends two to the power n. Two to the power n minus one plus two to the power p minus one, when the ends are same, where p is equals to n by two, if n is even, and p is equals to n plus one by two, if n is odd. This is what you have to memorize. And then we can solve the questions. Copy down this first. Yeah, done. Okay, what do you mean by this end are same or different? We'll discuss this for example. Suppose the first question we have, we have CH3, CH, double bond CH, single bond CH, double bond CHCl. Tell me how many studio center we have. We have two studio center, clearly. This is the one. And another one is this. Here you see the ends here, we have CH3 and here we have CH. So we can say here the end are different, different. So number of GI, the formula you see, it is two to the power n, n value is two here, two to the power two. Hence four is the answer. Actually what happens across both the double bond, we can have cis trans possible. Here also we have cis trans possible. Hence four possible combination we have totally see cis cis cis trans cis trans trans trans like that. If the ends are same, then one of the cis over here will be same as this. So that you need to ignore in that case. Understood this? So like that we can think about it and then we can write down the answer. But like this, if you do cis cis trans trans cis trans, you need to draw the structure and then compare. Okay, that takes a lot of time. That's why in this particular thing, when you have to find out the number of GIs, okay, geometrical isomers, you just see the molecule, look at the condition and apply formula. Okay, another one you see, we have C6H5, CH double bond, CH single bond, CH double bond, CH single bond, CH double bond, CH single bond, CH double bond, CH CL. Tell me the number of GI. How many GI is possible for this? Four, number of studio center is one, two, three and four. Four studio center ends are different, right? Hence two to the power four, 16 is the answer. If you have this, CH3, CH double bond, CH single bond, CH double bond, CH, CH3. Tell me in this one, number of GI. N value is two. Okay, yes. N value is two. Number of GI would be ends are same. Two to the power N minus one plus two to the power P minus one. Could you tell me the value of P here? P is equals to what? N by two or N plus one by two. It is N by two, right? Because N is even. So it is one. And hence this one would be two to the power one plus two to the power zero. That is what three we got. Clearly you can logically you can understand the sys of this and sys of this will be same, correct? Sys across this double bond and sys across this double bond would be same, right? So you need to count it only once. So sys trans here and trans only here, correct? Three. This one you check. CH3, CH double bond, CH single bond, CH double bond, CH single bond, CH double bond, CH CH3. N value is three. One, two, three. N value is three. Ends are same. So number of GI would be two to the power N minus one plus two to the power P minus one. What is P value? N plus one by two since N is odd. So P value is four, sorry, two. P value is two. When you substitute it here, two to the power three minus one plus two to the power two minus one. So four plus two, that is six. Understood? So like this, we can find out the number of geometrical isomers. So the first part of this studio isomerism is done. That is a geometrical isomerism. Okay. Very important. This one is because in 12th grade, we have a chapter called coordination compound. There also we have the application of geometrical isomerism and optical isomerism. Okay. Now we'll start with optical isomerism.