 heading right down all of you, stereo isomerism. We have two types, configurational and conformational. We are talking about configurational first and in that also we are discussing geometrical isomerism. Okay, so stereo isomerism is what? Stereo isomerism is, you know, it is due to the different position of atoms or groups in the space. Okay, so stereo isomerism is defined as, or right on this way, isomers, which has the same structural formula, which has the same structural formula, different relative arrangements, different relative arrangements of atoms or group, atoms or group in space. Or we can also say the 3D arrangement is different. Okay, called, called stereo isomers, isomers and this phenomenon is stereo isomerism. One second to copy this and just connect the charger. Give me just one minute. Automers will discuss Pradyum, not now, but we'll discuss later, let it be. Okay, for now you let it be. Okay, so stereo isomerism is because of different arrangement in space of atoms or groups, right? It is basically because of different relative arrangements, the repulsion between the atoms or groups, you know, it either decreases or increases, which affects the overall stability and property of the molecule and hence they are isomers, correct? So in this, if you see, if you talk about geometrical isomerism, okay, so geometrical isomerism is because of restricted rotation, right down, geometrical isomerism, isomerism GI is because of, because of restricted rotation across the double bond, the double bond, double bond, ring, or double bond inside the ring, double bond, ring, double bond inside the ring. So what do you mean by this? Because it means that we can have three different conditions for geometrical isomerism to possible. Geometrical isomerism is possible because of double bond, because of a ring and because of double bond inside the ring, these three different conditions we have, okay? So, look at this example. Suppose if I write down this molecule, CHCH, CH3H, CH3H, and CHCH, CH3H, another one, C double bond C, CH3H, CH3H, and C double bond C, at CH3, at CH3. Do we have any difference in this two compound and in this two compound? Do we have any difference in the first one? The first one you see, the molecule is identical. It's the same molecule, we don't have any difference because we can always rotate this sigma bond and we'll get this molecule, okay? But here the rotation is hindered because if you want to, if you fix this carbon and want to rotate this carbon, so you have to break this bond. And for isomers, we cannot break any bond. We can change the position, right? But the number of bond must be same. So if you want to rotate the second carbon, you have to break the pi bond which forms over there, okay? And hence we say we have restricted rotation across a double bond which results into the different properties of this two molecule. So these two are isomers actually. What kind of isomers? Geometrical isomers because the relative distance here and here is different. Correct? Understood. What is the name of this compound? IUPAC name? The second one, second one. Second pair that we have. Tell me, is it but one in, how? What is the position of double bond there? Right, it is but two in, correct? Both molecule is but two in, yes? So now here what happens when the isomers or the concept of isomerism came into the picture, then the whole objective of IUPAC nomenclature has got defeated. Why? Because the purpose of IUPAC nomenclature is that each and every molecule should have a unique name. Correct? Should have unique name. But these two compounds are different compounds different in their properties, melting point, et cetera, but they have the same name that is but two in, okay? Now to overcome this drawback, we have given different terms to this kind of compound that we use only in case of isomers, right? Those terms are what? Those terms are, if you talk about geometrical isomers, can be defined as cisterns. I'll write it on the next page, wait. Write down the name of this type of compound. We have cis and trans nomenclature system. We have E and Z nomenclature system. We have syn and NT nomenclature system. Cisterns, easy, syn and T. Okay? Cisterns we use, you see here, we have carbon-carbon double bond. One attached here, second group here, three, third group here and fourth group here. So out of this four group, right? If only two are different, right? We must have two different group attached to this position, right? One, two, three, four. Only two should be different over here, not all four. So in that case, we use cis and trans isomers. Cis and trans we use when two atoms or groups are different out of four. With example, you'll understand. E and Z, write down, when more than two atoms or groups are different, syn and NT we use when we have carbon-nitrogen double bond or nitrogen-nitrogen double bond. In these two cases, we use this syn and NT, basically for nitrogen-containing compound. Okay? So for cis and trans, what we can write, if identical group are on the same side, so cis, if identical group on the opposite side, then trans, copy this. We'll do some example here. Like the first one, if you take CH3, C double bond, C, CH3, H, H. So since anyone, if you see identical group, same side, so it is cis. This one's the same side, cis only. If you talk about this one, C double bond, C, H, CL, CL, H, identical group, you can take other chlorine or hydrogen on the opposite side, it is trans. Okay? If you have this, C double bond, C, H, CL, BR, and H. Now, this one is, again, it is highly substituted only, but since we have three different groups attached over here, so we can preferably write down E or Z isomers for this. Right? Preferably we'll write ENZ isomers. How do we write ENZ? We'll see that later. But if you want to write down cis and trans, then we can write down cis with respect to the identical group, which is hydrogen here. This cis. Got it? Write down ENG system of nomenclature. Write down ENZ system of nomenclature. Nomenclature we use when the alkene is highly substituted. Okay? When the alkene is highly substituted. Highly substituted means when the double bonded carbon atom has different group attached. If three groups are different, then also we can consider this as substituted alkene, highly substituted alkene. But if all the four groups are different, it is obviously highly substituted alkene. And in this case, cis and trans, obviously we cannot define because all the four groups are different. There's no identical group present. So in this case, what we do, we can write down the name of the compound using ENZ system, right? So to define ENZ, what we need to do, we need to assign priority on this group, which is attached to the double bonded carbon atom. So A and B will compare its priority, suppose it is one and this is two. Its priority is one and this is two. Suppose I'm getting this priority. Then what we'll write? If higher priority group on the same side, same side, then it is Z isomer. If higher priority are on the opposite side, it is E isomer, okay? And write down, this has nothing to do with the cis and trans system of nomenclature. Done? This has nothing to do with the cis and trans system of nomenclature. This means what? E can be cis, E can be trans also. Similarly, Z, it can be cis and trans both. So it's not like Z is cis and E is trans. Don't confuse with that. Okay, Z can be cis or trans, E can be cis or trans, anything is possible. Now the whole question is, how do we assign the priority? Because if you are not able to assign the priority, you won't be able to find out the E and Z system of nomenclature. So first thing is to assign priority here. Okay, any doubt here? Tell me, to assign priority, what we'll do, we'll have a rule given. We have CIP rule. The rule is CIP ruled. So it is called, it is based on a scientist name, three scientist name, Khan. We have ingold and prelog, rule we have, CIP. Khan ingold, prelog. Four or five rules we have that you have to keep in mind. First one is higher atomic number has higher priority. Higher priority. The second one is in case of isotopes, isotopes we compare mass number. So based on these two rules you see, suppose you need to assign priority on chlorine, chlorine, bromine and iodine. Iodine has highest atomic number, the priority order is this. Isotope if you take, tritium, deuterium and proteum. Tritium has highest mass number, the order of priority is this. If you talk about O18 and O16, O18 has more priority. CL37 and CL35, 37 has more priority. Done, copy guys. Okay, next one is if a group is attached, all right, I don't like this. In case of a group, we compare the atomic number of first atom, first atom. For example, chlorine, SO3H, OOH, NHCH2, NHCH3 and COH. So if you want to compare all these atoms or groups, we'll compare the atomic number of the first atom. Like for example, chlorine is atom only, atomic number is 17, it is sulfur, 16, it is oxygen here, eight, nitrogen, seven, carbon, six. So order of priority is this, done. Sometimes what happens, the first atom is same, like see this one. We have CH2, CH3, other one is suppose we have, wait a second, we have suppose CH3, other one is CH2, CH3 and other one is COOH. If these three groups you need to compare, first atom is carbon only. So obviously with first atom, we cannot compare that which one is higher priority, all are carbon. If we cannot decide on the basis of first atom, then we'll go for the second atom, which is hydrogen here, we don't have any choice. Here the second atom is carbon. Chain will follow, here it is oxygen. So obviously the priority of oxygen is maximum, then carbon and then hydrogen. Right down, next rule, the lone pair is considered as the least priority, lone pair is considered as the least priority. Now you see this example and how do we name this compound, ENC. We have carbon-carbon double bond OH, just tell me ENZ nomenclature. Try this for question, done, okay. So first one will assign priority O and C, O is higher priority, one, two. CL has one, this is two. Higher priority same side, it is Z isomer. Carbon, carbon, carbon, chlorine. Chlorine is the more, this is two. This carbon, carbon we cannot define. So hydrogen and carbon is one, it is two, again Z isomer. It is deuterium one, hydrogen two, chlorine one. So it is also Z. This two if you want to compare, smaller ring will have the higher priority. This one and then this. Check your answer. Yes, the smaller one has more priority. I'll show you how, but you can also keep that in mind. See, how do we do this? The ring, the last example that we have. You can go either clockwise or anti-clockwise here. Suppose I'm going clockwise. First, second, third, fourth, fifth, six atom we have like this. Here also we'll go clockwise. One, two, three, four, five. So first atom if you compare, it is only carbon and carbon, we cannot do. Second atom, oxygen, oxygen, cannot do. Third atom, carbon, carbon, cannot do. Fourth atom, carbon, carbon, cannot do. What is the fifth atom in both group? What is the fifth atom in both group? Ring one. The first one on the top we have carbon, but the bottom we have oxygen, right? Oxygen has more priority. Hence the below one, smaller group has higher priority than the upper one. Clear? Understood? Few more examples you see. Tell me this too. Z and E. Okay. See nitrogen here, the fourth group we have, which is this lone pair. Here also we have one lone pair this side, and one is this side. So obviously, lone pair has the least priority. Higher priority opposite side, E. One, one, it is Z. Okay? We can write E and Z, but since nitrogen is present here, so preferably we can write this as anti and sin. This is also right. Correct? Understood? So now, this is the nomenclature system we have discussed for geometrical isomerism. What nomenclature? How do we do the naming of the compounds? Now, what is the condition and what condition the compound can show geometrical isomerism? So write down the heading condition for GI, geometrical isomerism. Geometrical isomerism is possible under three different conditions. Okay? First one, geometrical isomerism, because of double bond, because of ring, and the last one we have, because of double bond inside the ring. In this, the first one you write down, GI because of double bond, due to double bond. Condition for a molecule having double bond to show geometrical isomerism is, like suppose we have this molecule C double bond C, we have A, B and PQ. So atoms are grouped, which is attached to the double bonded carbon atom must not be same. That is the condition for a double bond to show geometrical isomerism. Write down, for a double bond to show geometrical isomerism, for a double bond to show geometrical isomerism, the atoms or groups, the atoms or groups attached to the double bonded carbon atom attached to the double bonded carbon atom must be different, right? So this means what? For this double bond to show GI, A must not be equal to B or identical to B and P must not be identical to Q. This is the condition we have for GI due to this double bond. A can be equal to P or Q. B can also be equal to P or Q. So you must take care of this thing that here, this two group must be different. A, B will compare and PQ will compare. Any one of these two are same, P equals to Q or A equals to B, the double bond won't show GI. I'm giving you some example. Just tell me that which compound can show GI? First one, second one, third one, this one. How many compounds can show GI here? Only two. Others can also answer. First one, GI possible? Because we have both hydrogen atom here, no GI possible for the first one. Second one, here it is fine, but here we have two hydrogen. So no GI possible for the two hydrogen. Across this GI possible, yes. Across this also GI possible. So two GI possible here in the third one. Okay, the double bond across which geometrical isomerism possible, we call it as stereo center. So we have two stereo center in this SC in this molecule. This double bond here, it is fine, but if you compare this side, okay, if you have a ring structure given like this, then what you do, you just go like this, anticlockwise and clockwise you move. If you move anticlockwise, you are getting CH2, CH2 and C. This is, we have CH2, CH2 and C. We can assume this side, we have CH2, CH2, C present. These are the method to find out GI possible or not. If you go clockwise, then again we are getting the same thing, CH2, CH2, C. And hence we say both path are same here, identical. Both path are identical, same, and hence we can conclude that the identical group is attached both side on the bottom and on the top. Hence GI is not possible here, no GI. If you compare this one, then this path is different because we have chlorine at first carbon. This side, we do not have chlorine at first carbon. So we have two GI possible here. Number of GI, how to find out, we'll see that. Concern here is the molecule will show GI. This GI is because of double bond. So number of stereocenter here, if I write down, SC equals to one. Tell me, how many of you understood this? Is it clear, guys? No, no, no, not four. Only two, for one molecule. Number of GI, just try to be able to discuss that later. No, no, no, you see, if the ring is attached to a double bond, no. You don't have to do anything. You try to move anti-clockwise, see what all groups are present, and clockwise, see what all groups are present. If both groups are same like we have here, then it means both sides we have same group attached, and hence GI is not possible. So basically you need to just check the path is different or not. Clockwise or anti-clockwise, that is it. Okay, so condition for double bond is this. Similarly, we have condition for a ring to show geometrical isomerism. Second write down, second point. Geometrical isomerism due to ring. Nation is at least two sp3 hybridized atom, at least two we have. Two sp3 hybridized atom must be dye substituted with two different group. For example, you see, we have this molecule. CS3 attach here, H we have here, CS3 we have here, and H we have here. Okay, so this carbon you see, this carbon is sp3 hybridized, this carbon is sp3 hybridized, and two different groups attached to this carbon atom, hence this will show GI. Yes, GI possible. Number of studio center is one, and we say this geometrical isomerism is possible because of ring, ring is the studio center we have here. And it out, SC is the studio center, okay? SC is the studio center, yeah. Second one you see, tell me in this one, GI possible or not? Tell me, is it possible? Yes, obviously hydrogen atom is present over there. We do not represent that. Yes guys, answer. Okay, you see here, this molecule, along with this carbon atom, we have, along with Smithile, we have hydrogen present here, that is understood, and one hydrogen we have here. Similarly, here also we have hydrogen present, and here also we have hydrogen present. All these carbon atom are sp3 hybridized, because sp3, two sp3 hybridized carbon atom is disubstituted. So in this case, yes, GI possible. Number of studio center is one. In this case also, condition is at least two. So more than two also finds. Yes, GI possible. The number of studio center is again, the ring that is one. In this also, yes, GI possible. The number of studio center is still the ring that is one. Few more example you see. GI possible or not? And also find out the number of studio center. Yes, tell me. In the first one it is possible, geometrical isomerism? No. If I write down this one then, yes or no? Then you double bond, right? Okay. So in that case, we have this double bond. Is it this double bond? Is it possible with this double bond? No. Why? Because this is equal. So double bond won't show. This molecule will not show GI, right? Further. If you add here CS3 then, then it shows. Number of studio center? Number of studio center is one. That is this ring, right? What if I place a double bond here? GI possible? How? Yes. Yes. In this case also, GI is not possible because the carbon becomes sp2 now. We need sp3 hybridized carbon atom. So this will just remove. Answer for this molecule is yes, GI possible. Obviously across this GI not possible because we have hydrogen here. Here we have one hydrogen present. So because of ring GI possible, because of double bond GI possible, overall yes, geometrical isomerism possible. Number of studio center is two over here. Number of studio center is two. Here also because of this larger ring, GI possible, right? So yes, GI possible. Number of studio center is one here. That is a hexagon. Number of studio center is one and that is the ring we have. Understood all of you? Now the third condition we have that is geometrical isomerism due to double bond inside the ring. It's a factual, basically you have to keep this in mind. GI due to double bond inside the ring. So for this one, the condition for this type of geometrical isomerism is that the ring must contains at least at least eight or more, means the ring size must be at least eight membered. Okay, that is the condition we have. Like you see this example, this molecule is cis-cyclo-octene because the hydrogen atom present like this way. Here one hydrogen atom is this side, another one is this side. So this is trans-cyclo-octene. Trans-cyclo-octene. What happens if the ring size is less than eight member ring? So in that case, what happens? The trans isomers are highly unstable. Two points to write down in notes. First of all, you copy down this and then I dictate. Write down, for a smaller ring, a smaller ring means smaller than eight member. For a smaller ring, the trans isomers, the trans isomers are highly unstable. Trans isomers are highly unstable due to strain. The trans isomers are highly unstable due to strain and hence does not exist, right? That's why seven and lesser carbon atom present in the ring, then double bond inside the ring does not show any kind of geometrical isomerism. You simply ignore that. Okay. Second point to write down. For eight, nine and 10 and 10 member ring, for eight, nine and 10 membered, 10 membered cycloalkene you write down. For eight, nine and 10 membered cycloalkene, cis isomer is more stable than trans. Cis isomer is more stable than trans. Cis isomer is more stable than trans. Next slide. For 11 and larger member ring, for 11 and larger member ring, trans is more stable. The last two points you must keep in mind, it's a factual, okay, experimental based fact we have. So you have to memorize it. Should I repeat again? Cis is more stable because in trans what happens, you see? This hydrogen, this hydrogen, we have a repulsion here. Because of this dihedral repulsion, trans becomes less stable than cis. Cis we do not have any repulsion, you see here. Tell me GI possible or not, and number of stereo center. Larger member ring, 11 or more. See, trans is always more stable because we have the two groups present at more distance, the distance is more between the two identical group. The trans is generally more stable than cis, okay, because of less repulsion over there, because the identical groups are far away from each other relatively. Actually in 11 member ring, we don't have that repulsion that I have shown you just now. Larger ring, more space, it can accommodate the hydrogen atom over there, and hence the repulsion is less. Which one shows GI? If GI possible, number of stereo center, then tell me the answer. First one, GI possible, you can type Y or N. First one, any one condition is there, any one, we consider this yes, right? So across this double bond, yes, GI possible, because of ring, because of ring, Y or N, because of double bond inside the ring, N, so yes, GI possible, but the number of stereo center is one here. Because of this double bond, this double bond, yes. Because of ring, because of ring, yes. Because of double bond inside the ring, yes. Double bond inside the ring, yes, right. Yes, GI possible, and SC equals to three. This one, because of ring, it is possible because this is also sp3 hybridized, nitrogen is also sp3 hybridized with a lone pair on it. So yes, GI possible, that is because of ring, SC is one. One very important example we have, in J, they have asked this question. See this kind of compound, when we have continuous double bond like this, you see. C double bond, C, double bond, C, like this, and it goes on. Right, this kind of compound, the common name is cumulines, or allines also we call it as, C-U-M-U-L-L-E-N-E-S, cumulines, common name, right. So when we have even number of double bond, like double C double bond, C double bond, C double bond, C like this, it goes, right. So if you have even number of double bond, even number of double bond, then the molecule is non-planar. Even number of double bond, we call it as even cumulines, right. And this kind of compound does not show geometrical isomerism, okay. I'll write down here CH3, CH3, CH, CH, and in this one, it looks like trans, but GI is not possible in this one because it has been observed that in even cumulines, since this is in different plane, this is in different plane, non-planar molecule, then the distance between, distance between this methyl group and this methyl group, it's same only. The distance difference is not there. And hence the relative repulsion is also same. If it is L, this is also L. Hence the relative repulsion is same. Relative repulsion is same, correct. So it does not show GI. Property will be same in that case, okay. Keep that in mind. Even cumulines is non-planar and it does not show geometrical isomerism. If this groups are different, it can show optical isomerism, right. Non-planar does not show GI can show optical isomerism, why? So in optical isomerism, we'll discuss this also. Must remember this, keep this in mind, it's very important. Yeah, once again, Auro, correct. So here the condition of GI is done over. We are now left with properties. Anusha will discuss this. I'll tell you this after the break, you can copy this down, okay. We'll continue this after the break. I just said one thing. This, I'll do this one minute, I'll take. When you have continuous double bond like this, this kind of compound, we call it as cumulines, general name, okay. When you have even number of double bond like we have in this case, then this cumuline is non-planar. How and why? We'll discuss in optical isomerism, okay. But don't get confused that this CSC, CSC same side, so it is cis and this one is trans, no. Since it is non-planar, so distance between the methyl group here and here is same. So relative repulsion is also same. There's no difference in this two molecules as far as the repulsion is concerned, correct. And hence, there is no GI possible for this two molecules, correct, that is what it means. Keep this in mind in optical isomerism also, we'll see this. So this even number of cumulines does not show GI, but it can show optical isomerism. Take a break, we'll resume at six, 38, 38. Okay, take a break, heading right down, properties. See, when you talk about boiling point or suppose dipole moment, so dipole moment is directly proportional to the boiling point, okay. It is also directly proportional to the solubility, heat of combustion, heat of combustion, heat of hydrogenation, density, refractive index, okay. So dipole moment is associated with all these properties of the molecule, okay. Second one, if you talk about thermal stability, thermal stability and melting point, melting point depends upon packing, okay. And melting point of trans isomer is more. Because in trans we have better packing, hence we have more melting point. Now only these two you need to memorize. Now, how do we compare the dipole moment, correct? Copy down this first. So we know dipole moment is a vector quantity. No, by mistake I wrote 638, it was 633 actually. Anyway, just now we started, did you copy? Okay, I'll go back. So dipole moment, how do we compare? Dipole moment is represented by mu and it is a vector quantity directed from positive to negative, dipole moment positive to negative. Look at this example. We have C double bond C, CH3, H, CH3, H, okay. So actually what happens when a bond is polar, like suppose hydrogen-chlorine bond, it is more electronegative, so delta negative and delta positive. So because of this electronegativity difference, we'll have some charge developed over here and because of this charge, we'll have a dipole moment from positive to negative direction. That is what the meaning we have. So whenever we have electronegative, so whenever we have electronegativity difference, some charge develops, that leads to polarity in the bond and then in the molecule, okay. If you look at this carbon atom, which is attached here, this one is sp2, this one is also sp2. And this carbon here and here, these two are sp3 hybridized carbon. We know that electronegativity of sp2 is more than sp3. So what happens after some time because of electronegativity difference, this would be delta negative, this would be delta positive, delta negative and delta positive, right. And hence we have a net dipole moment in this direction because of this bond positive to negative, positive to negative, right. It's a vector quantity, so it is there in this direction, it is there in this direction. And hence we have a net dipole moment in this direction, new net, which is obviously not equal to zero and you see the molecule is polar. We don't have to count or calculate the dipole moment, exact value, we just need to check whether it is zero or not. This is cis, correct? This one was cis. If you talk about trans, then what happens you see? CH3, C double bond C, CH3, H, H. So here we have dipole moment positive to negative this direction, right. And then we have positive negative this direction, sorry. Equal dipole moment in opposite direction, this two will cancel out, okay. And the dipole moment here mu two, here suppose it is mu one, right. Obviously it is getting canceled over here. Here it is not getting canceled. So we can say mu one is greater than mu two. Cis has more dipole moment here than trans. When cis has more dipole moment, means the boiling point of cis is more, the solubility of cis is more, heat of hydrogenation more, heat of combustion more, okay, density more, activity index more, anything. Did you understand? So here the cis has more dipole moment than trans, but we cannot generalize this. We cannot say a cis will have, will always have more dipole moment, okay. Look at this example. Suppose we have CH3, C double bond C, CL, H, and H. C double bond C, CL, H, CH3, and H. Right? Positive negative, this is the direction of dipole moment. Positive negative because chlorine is more electronegative, this is the direction of dipole moment. This one is obviously cis. Why cis? Because both hydrogen on the same side, okay. This one is trans because hydrogen on the opposite side. Positive negative and positive negative. This two will add up. We'll get mu two here, we'll get mu one. Here the direction of dipole moment, net direction will be this, this direction, but obviously this two is getting add up. This one is obviously more than this. So in this case, dipole moment of cis is less than the dipole moment of trans. So we cannot generalize this that the dipole moment of cis is always more than trans. It depends upon what is the bonding, what is the structure, what is the molecule we have. Tell me which one is E and which one is Z, left and right. E or Z, which one is Z? First one is Z and second one is E, right? Okay, so we can also say this is Z and this is E. Okay, what if I place here instead of H, if I place here C, H3, then what happens? E or Z, the second one, it's E. Okay, this one is E. If I ask you cis and trans, it is cis, okay, cis. So always keep this in mind, cis can be anything, you see. Here cis is Z, here cis is E. So don't get confused that cis is always Z and trans is always E. This is not true, okay? This can be cis can be Z, cis can be E and trans can be Z also, depending upon what is the molecule we have, correct? Just I'll remove this and this also. We had H over here, correct? Okay, now the last one in geometrical isomerism, that how to calculate the number of GIs? For this, we have formula that we need to follow, okay? Logically, also you can draw the structure and find out, but that logic won't work. It won't, I should not say it won't work, but when you get some bigger molecule, then with logic, you'll get confused. It is difficult to get answer from the logic, okay? Because you need to draw the structure and really very complex if you are going to draw all the structure to count the number of GIs, okay? The best is to memorize the formula. What formula we have? So we'll count the number of stereocentre basically, first of all, we'll count the number of stereocentre, that is N. If N is equals to one, then the number of GI, geometrical isomerism is two to the power one, that is two. Cis-trans-sin-n-t-easy, we can say, okay? If N is greater than one, greater than one, then the number of GI, then the number of GI would be two to the power N, if the ends of the Pauline are different. Earlier, what does it mean? First of all, you copied on the formula. It is equals to two to the power N minus one plus two to the power P minus one, if ends are same, once again. This is the formula we have, copied on this. Okay, done, copied. Now, how do we use this formula you see? Suppose we have a molecule, say CH3CH, double bond CHCH3. GI possible, obviously the number of stereocentre is one, so number of GI is equals to the formula is two to the power one, that is two. Which means we get across this double bond, we can draw cis, we can draw trans. Hence two isomers possibly, okay? Look at this question. We have CH3, CH double bond, CH single bond, CH double bond, CH single bond, CL. Ends of this molecule is different. You see here CL and CL and CS3 we have, means ends are different over here. If it is CS3, then ends are same. Number of stereocentre is one and two. N value is two. If ends are different, the number of GI would be, the formula we have, two to the power N, where N is greater than one, that is two to the power two, four. Means the possibility is what? Across this double bond, cis, trans, cis, cis, cis, cis, trans, trans, cis, trans, trans. Hence we are getting four. So four possible combination we have. One is cis, other one is trans, one is cis, other one is cis, one is cis, one is weight. This one is trans, this one is cis, this one is trans, this one is trans. These are the four possible combinations we have. Hence answer is four. So like this also you can analyze that for a given molecule, how many cis, trans structure you can draw. What happens if this is equal, CS3, CS3? Here it is CS3, here it is CS3. Then one of this four possibilities that we have that you need to eliminate because both will give the same thing because ends are similar. Once you draw the structure you will get another example 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 here. CH3CH, double bond CH, single bond CH, double bond, CH, single bond CH, double bond CH2, number of GI here. CS3C, double bond CH, double bond CH, single bond CH, double bond CH, CH3. This one we have done, no, should I try? Try this three, 16, 8 and 3, okay. So number of stereo center here, one, two, three and four. Ends are different, number of GI equals to two to the power four that is 16. Number of stereo center, one, two, this is not the stereo center because we have CH2 here. So number of GI would be two, ends are different. So two to the power two that is four. One, two, ends are same. N value is two, P value is N by two, that is one. So number of GI would be two to the power N minus one plus two to the power P minus one, that is three. Formula you check or oh, N is events of P is equals to N by two, check the formula. No, previous one that ends are not same. You see it ends is this, this one you check. Done, what is the answer? No, methyl ethyl is different, no, so they are different. It is the group attached to the double bonded carbon atom. This carbon you need to check which group is attached. Correct, yeah. So how many stereo center here? One, two and three. Ends are same, N value is three. So P value is what? N plus one by two, that is two. Number of GI would be two to the power N minus one plus two to the power P minus one. Two to the power two plus two to the power two minus one right here. Four plus two, six we are getting here. One, two, three and four. Did you write the formula or check that once? N value is four, no, two to the power N minus one plus two to the power P minus one it is. Yes, see this, check the last one. I'll come back to the previous one. Ends are same, N value is four. What is the number of GI when the ends are same or it'll be? Number of GI is the same formula that you have written just now. Two to the power N minus one plus two to the power P minus one, correct? Yeah, what is N value? Four, if N is four, could you tell me what is the value of P? See the, check the formula or first and tell me. If N is four, what is the value of P? I don't think you have written this, wait. Did you copy this? That's what I said, copy down first. So P value is four by two, we'll substitute it here. Two to the power four minus one. Can you read it out? Okay, so this is, here it is geometrical isomerism is done over here.