 can you start now yeah so we were discussing about optical isomerism right did you understand this in ansiomerism what is in ansiomerism what is what kind of rotation is allowed and all because you must keep this in mind this flipping of molecule is not allowed right we can rotate the molecule within the same plane but we cannot flip the molecule we cannot rotate it out of the plane okay that's the condition we have always keep this in mind this is very important right so this is one example I have given you so the definition of an ansiomers are those compounds a pair of compounds which are non superimposable mirror image okay hands are not superimposable to you to each other left and right hand you must keep this in mind nothing if you rotate this compound in this direction right after 180 degree of rotation this chlorine will come here and this hydrogen will come here so you'll get the same compound see suppose if you have rotated this molecule in anti-clockwise direction in this direction 180 degree then what happens this chlorine will come here this hydrogen will come here this chlorine will come here and this hydrogen will come here so you'll get the same compound understood so that's the thing okay one more example I'll give you here and that is this is important okay so you must remember the formula also OH H here we have CH2 OH and here we have CHO when you take the mirror image of this right then what happens you just keep one thing in mind the mirror image of this will be this OH H this side we have CHO and here we have CH2 OH this is the mirror image any one of you have doubt in this let me know if you have doubt did you understand this see this this this representation that I have given you now it is a three-dimensional representation correct so the OH and H are respectively coming out of the plane and into the plane correct so when you take the mirror image of this compound in the mirror image also the OH should be coming out of the plate and this H should be going into the plate right that's why I've taken like this you must take care of you must understand the molecule like this OH is coming towards you when you look at the carbon this carbon OH is coming towards you H is going away from you right so when you take the mirror image from this side so in this also the OH will reflect coming towards you and H also reflects going away from you right so that's why here we don't have any change into this but here we have CHO this side and CH2 OH this side okay now the name of this molecule we call it as glyceraldehyde okay the name of this molecule is glyceraldehyde is this name is important formula you must remember glyceraldehyde okay now can you tell me the configuration of this and one more thing you see here we'll talk about the configuration later on these two are enantiomers or not are they super impossible these two molecules are super impossible these two molecules are not super impossible right because when you put this carbon over here OH and H will be fine but this CHO is on the right hand side but here the CHO on the left hand side okay so these two molecules are not at all super impossible so these are not super impossible but mirror image of each other so what we can say they are a pair of enantiomers right these are enantiomers of each other right is it clear these two are enantiomers of each other correct so now you tell me what is the configuration of this molecule and this molecule or you tell me the configuration of this first molecule what is the configuration yesterday only we have discussed oh most of you have did not understand this okay they are not enantiomers they are not super impossible correct correct correct okay so tell me what is the configuration of this molecule and this molecule tell me the configuration R and which one is SL this one is suppose first compound and this one is second first one is what Nikhil Sriramya what is the answer Ishika tell me the answer so we have the first priority you see here this is the highest priority we have here and then second we have this third and fourth correct the fourth is already into the plane of the paper correct so we have to go one to third via two so it is clockwise right right so this one is R and this one since we have exchange or we have swag only one pair odd number of pair the second one is obviously as you can do this in the same way correct so R and S so what we are going to conclude from this example is this molecule and this molecule right so the enantiomers that you have all enantiomers are optically active is this optically active first one is this optically active first one is optically active why there is no plane of symmetry that also you can see or second thing is what there is only one carol carbon this carbon is the carol carbon we have is it this is carol carbon sp3 hybridized for different group carol carbon sp3 hybridized for different group so molecule contains only one carol carbon hence it is optically active right right so this one is optically active this one is also optically active both are optically active in nature but the point here is what if one of the molecule rotates the plane polarized right in clockwise direction then the other one will rotate in anticlockwise direction with same angle right so if one of this molecule has configuration D then other one must be L one is D other one is L Dextro and Levo okay see this D we don't know whether it is D or this is D okay but this is such a relative thing if you write down this molecule as Dextro this must be Levo or if you write down this as Dextro then this must be Levo is it clear understood this tell me first see I have written here D and L but this is just a relative things I have written over here correct it is not necessary that this compound is Dextro only this may be Levo also that we cannot say that we can only say once we see the behavior of this molecule towards PPL okay according to clockwise or anticlockwise rotation and the second shouldn't OHB on the left and HP on the right I'll discuss on this one just give me a second let me finish this okay but if you can explain this you can write down your comment here okay I will explain this one later I'll again explain this over an edge part okay see the point here is what I have written here D and L but we cannot say that this one is will be D only D and L we can only say by looking at the behavior of the molecule towards PPL plane polarized light okay but one thing we can on we can definitely say that if this is D this must be L or if this is L this must be D is it clear right what did you understand Shravan or do I need to explain this see the OH is coming out of the plane correct it is coming towards you and edge is going away from you right so when you when you learn when you keep this mirror this side where OH is coming towards you and edge is going into the plane so in the image also if you imagine OH will be towards you only and edge will be going away from you correct so the point is D and L so here what we can say that another point that we are going to another term one is an insurance we are talking about now this is the two molecule we have correct so what happens if you take equimolar mixture of these two molecules that mixture will be optically active or not if you take equimolar mixture of this two compound the mixture as a whole will be optically active or not tell me optically inactive mega is saying optically inactive Shravan it won't ashrut how it is optically active ashrut is it see what we are taking suppose we have equimolar mixture of this so number of moles of compound one here and equal number of moles of compound two I have taken and when you allow this is the mixture we have right and when we allow PPL to pass through this mixture suppose this molecule will rotate PPL in anti-clockwise direction and this one will rotate PPL in clockwise direction right with the same angle so eventually what happens eventually what happens this PPL or plain polarized light will come out without any deviation right hence it will be optically inactive right so what we can say when you mix the equimolar mixture of this the mixture will be optically inactive and though that kind of mixture we call it as rhythmic mixture right so one note you write down here one note you write down if you take equimolar mixture of if we take equimolar mixture of if you take equimolar mixture of both of these optically active compounds both of these optically active compound then the mixture is known as rhythmic mixture then the mixture is known as rhythmic mixture which is optically inactive in nature which is optically inactive in nature all of you understood this is it clear correct so what is the definition of enantiomers non superimposable mirror image okay now you see one more example i'll tell you i'll give you let me know whether these are enantiomers or not okay and i'm giving you the fissure projection so suppose the molecule is this these two molecules what is the relation between these two are the enantiomers of each other are they mirror image of each other are they mirror image of each other if they are not mirror image then how they are enantiomers if they are not mirror image then how they are enantiomers okay so all of you are saying they are not mirror image they are not enantiomers okay so all of you are saying they are not mirror image of each other So they are not enantiomers. What is your answer Preeti tell me within the plane yes correct. Now you see this is the first molecule if I write down this molecule then what would you say OH, COOH, CH3 and H. These two what is the relation between these two? These two looks like mirror image of each other. Yes or no? Tell me these two look like mirror image of each other? Fine. Correct. These two does not look like mirror image of each other. Right. But and these two are superimposable or not? These two are not superimposable? Are they superimposable? Tell me. See all of you you listen to me very carefully over here. This is very important. Okay that's why I'm taking a lot of time over here. Once you will understand this you will never forget this. Okay. So just concentrate here now. Are they superimposable? Not superimposable. Why they are not superimposable? We can rotate this molecule. You see what I said I am asking you right? What I said within the plane rotation is allowed. Right. So if we rotate this like this here and here 180 degree then these two will be superimposable. Then also it won't be superimposable. There are two things here. You see you might see. Okay. You tell me one thing. When we rotate this molecule by 180 degree in this direction then do we have this horizontal bond that we have here? Are they superimposable to each other? This OH will come over this OH and H will come over this OH. Yes or no? Correct. This COOH will come at the bottom and CS3 will go up and hence this line will not be superimposable to each other. Correct. But what about this OH and H? Are they superimposable to this? So the horizontal one are superimposable to each other. Right. Okay. Now you listen to me. First of all the answer here it will be what? First, second and third. Now you listen to me. Use your brain now. Just you listen to me now. The answer of this question is first and second are enantiomers. First and third are also enantiomers. How I'll tell you. Okay. Here you see about this no superimposable thing here. Like I said this bond is this horizontal bond feature projection is coming towards you. The actual molecule if you draw that will be like this. Here we have OH and here we have H and this COOH is going into the plane with a dotted line. So COOH is here and CS3 is here. Correct. Now when you rotate this molecule, correct, when you rotate this molecule, so whatever the bond is coming towards you, now it is going away from you into the plane. This OH and H is going into the plane now. Right. But here it is what? This OH is coming out of the plane. This hydrogen is also coming out of the plane. But when you rotate this, when you rotate this, then this OH is going into the plane and this H is coming out of the, sorry, into the plane. Correct. Both of this bond is going into the plane now. Okay. So here this OH will be towards you and when you try to superimpose this molecule here after what I did in your rotation, then this OH will be towards you and this OH will be away from you into the plane and this OH is towards you and this H will be away from you. So obviously the vertical line, since we have different molecule here, this line is obviously not superimposable to each other. But horizontal line is also not superimposable to each other. Is it right? Did you understand this? Did you understand this? Clear. So don't have this confusion that this molecule is looking like a planner, but this is actually a 3D molecule, 3D representation. Okay. So what I said, first of all, you have to keep this in mind that this molecule is looking like planner. It is into the plane. Correct? Okay. I'm repeating it again just to listen to me carefully. First of all, the confusion here it is what that this molecule is looking like. I'm coming back to this. I will come back to this and ensure one part again. But first I'm trying to make you understand that how this line are not superimposable to each other. Right. So first of all, you see the confusion that you have is that is because of that this molecule is look like it is into the in the plane of this, you know, the board that you have now. It's it look like planner, but this molecule is not plan. The actual thing is what this horizontal line is coming towards you. Right. And this vertical line, this is going into the plane of this board that we have. First thing is that. Okay. So this molecule is actually like this. It should be V. Right. This corner you understood here. Right. And one line is coming towards you in above this plane. Another line is coming again towards you above this plane. That is H and OH and these two is going into the plane here from into the place you and CS3. Right. So what happens here if you rotate this molecule, this OH is coming towards you when you rotate this to 180 degree, this OH will be now it is going away from you into the plane. Okay. Or you what you what you do, you just look at the molecule from this direction. Suppose this you see this is suppose OH and H we have this bond. I have taken like this is the carbon we have. Now you are looking at the molecule from this direction. Okay. You are looking this molecule from this direction. Suppose. Okay. So this OH and H is coming towards you. The another two molecule is what which is COH and CS3. It is going away from you. It is going away from you. Right. Now when you rotate this molecule like this, then what happens this OH will come here and H will go that side and it will be then away from you. Correct. But here this OH and H is coming towards you and when you rotate this OH and H is going away from you. So these two can't be superimposable to each other. Is it clear now? Is it clear? Tell me. This concept is this is actually very easy to explain in physical class. This is online so that I cannot show you with the help of pen. Okay. But I am trying my best to make you understand. Okay. Try to visualize this. One thing only you have to keep in mind that this molecule, this molecule is not planar. However it looks like. Okay. Don't get confused with it. Whenever you say Fischer projection, always keep this in mind that what is the actual position of this atom or molecule we have. Right. So when you rotate this vertical since we have different molecule not superimposable, horizontal also not superimposable. Okay. We are rotating like this you see. Like this 180 degree. Like this I am rotating you see clockwise. We are rotating this. Okay. I will write down the angle also. 180 degree rotation we have here. Like this we are rotating. Understood. We are rotating like this. This H see this H will come here and this OH will come this side. You have to visualize this actually. Let me know. Okay. If you still have doubt you can ask me in the class in the physical class. I'll explain this. Okay. So the point is here you see now we are coming back to this enantiomer thing here. These two are looking like mirror image of each other but non superimposable. Right. Non superimposable mirror image. That's why that's why what happens. These two are first and third are enantiomers of each other. Okay. Now what happens here you see one thing that you probably didn't take care of in the last slide. The enantiomeric pair must have opposite configuration. Okay. Two molecules if they have same connectivity right or they have similar carbon. Two molecules if they have similar carbon and with opposite configuration then they are enantiomers of each other. This is the another definition we have. Okay. Now when I say that similar carbon what does it mean? Similar carbon means what the Karel carbon must have same atoms or group attached to it. Correct. Like this carbon you see if two molecules you are considering H CS2 OH OH CHO H CS2 OH CHO OH. So these carbon has similar connectivity. These two carbons are similar carbon. Right. Similar carbon. Since the two molecules has similar carbon with opposite configuration then they are said to be enantiomers of each other. Is it clear? Let me know quickly. Guys are you there? I'm not getting any messages. So there are two different way we have to check what to check enantiomeric pair. Okay. What we can do? First thing is what you just check whether they are mirror image or not. Right. And they are superimposable or not. Correct. One thing is this. Another or the best way to check the molecule whether they are superimposable or not is to check the configuration of the molecule and connectivity as well. Okay. Connectivity means what you see again. This carbon has COOH OH CS3 H. This carbon has COOH OH CS3 H. So both the carbon has same connectivity. If these molecules are of opposite configuration then they are enantiomers of each other. Like you try to find out the configuration of this molecule. What is the configuration here we'll get? First is this, second, third and fourth. How do we find out the configuration of this? Can you tell me the configuration of this one? Correct Bharat. Right. If the molecular formula are same we will just find out the configuration R and S. If they have opposite configuration they are enantiomers of each other. So this one what is the configuration we have here? Tell me the configuration of this one. Yeah. This one will be R. What is the rule for that? I can see some of you are making mistake over here. See the condition here is what least priority group must be on the bottom of the vertical line. Correct. So what we have to do we have to swap these two molecules. CS3 would be here and H will be here. So 1, 2 and 3 if CS3 is here then it will be anti-clockwise. So anti-clockwise is S. So this must be R because we have done one swap over here. Odd number of swap, yes odd number of swap gives you opposite configuration. Even number of swap gives you same configuration. So we have swap H and CS3 only one pair of groups and atoms we have swapped here. So whatever the configuration we'll get here 1, 2 and 3 it is anti-clockwise should be S. So this must be R. Right. So this is R. What about this? What about this? In this molecule you see if you compare these two COH and COH position are same. CS3, CH position are same. We have only swapped OH and H. One pair of exchange we have. So if this is R this must be S because we have one pair of exchange here. Right. Now can you find out the configuration of this one? COH will be at this position. How many pair of exchange we have here? We have one pair of exchange. OH and CS3 you see here. OH and CS3 has been exchanged here. Right. So if this is R odd number of exchange this must be S. Correct. So when you see these two molecules you see these two molecules have same molecular formula and their configuration is also same. It means 2 and 3 are identical molecules. 2 and 3 are identical molecules. So if I can say 1 and 3 are enantiomers and 2 and 3 are identical hence 1 and 2 are also enantiomers of each other. One swap. Yes mega correct. Even number of swap if you have configuration will be same odd number of swap will change the configuration from R to S or S to R. Like that. All of you understood this? Yes correct correct. They are identical in Shikha because they have same molecular formula and same configuration that's why they are identical. Same configuration means what? Their behavior will be same towards PPL and molecular formula is also same. So they are exactly identical compound. Only the representation is different. So here we can conclude again here we can conclude again one more thing that to find out enantiomeric pair we can find out the mirror image and then whether they are superimposable or not. That is one way but the best way is what you just check the molecular formula of the two molecule. If the molecular formula are same and their configuration is opposite one is R other one should be S or one is S other one should be R. Same molecular formula with opposite configuration are also a pair of enantiomers correct. So however you see the second molecule does not look like the mirror image of the first then also they are a pair of enantiomers right. So right on this point one and two are enantiomers of each other. One and two are enantiomers of each other. However however they does not look like however they does not look like they does not look like the mirror image of each other. So don't try to check the mirror image just you check the connectivity of the carbon atom means you first check the molecular formula. If we have same molecular formula with opposite configuration they are mirror image of each other. They are sorry they are enantiomers of each other right. How do we identify identical molecule because all these things they are they'll give you two molecules right they'll give you three four molecules and then and the option will be like this one two are enantiomers one two are identical two three are again identical like this they have the they'll give you the options right. So just you check the configuration that would be the best option we have okay. So this is it and this is it for enantiomer thing okay we'll come again back to this. Now the next term that we use here so we have discussed restring mixture and enantiomers okay. So all enantiomers we'll discuss this one now. Next write down di-stereomers di-stereomerism di-stereomerism write down the definition here. Stereo isomers Stereo isomers tautomerism we will not discuss now in general we'll discuss this after GOC okay tautomerism you let it be and tautomerism is there in structural isomerism we're talking about stereo now. So write down di-stereomers whose molecule di-stereomers whose molecules are not mirrored image of each other whose molecules are not mirrored image of each other called di-stereomers called di-stereomers and this phenomenon is di-stereomerism and this phenomenon is di-stereomerism okay. So any stereo isomers if you take okay if they are not mirrored image of each other then they are di-stereomers right they are di-stereomers for example you see are they mirrored image of each other these two molecules these are mirrored image of each other yes or no oh this is not required actually this is not required these two are mirrored image of each other no they're obviously not mirrored image of each other right so these two molecules are not mirrored image of each other but their molecular formula is same that you see see molecular formula is same this one is trans and this one is cis cis and trans isomer so they are what they are GI geometrical isomers right they are geometrical isomers cis and trans isomers they are so those these two molecules with same molecular formula but they are not mirror image of each other hence they are di-stereomers correct so they are di-stereomers of each other so here you see geometrical isomers which can be cis trans is at sin and t right they cannot be means all these and and this is also is this is also defined in a pair okay the pair of molecule which are not mirror image of each other are di-stereomers right like in ensumers also this di-stereomers is also defined for a pair of molecule first of all this thing right so all geometrical isomers all geometrical isomers whether it is cis trans sin and t in z right whatever the geometrical isomers we have these pair of molecule can never be the mirror image of each other mirror image not possible in all these cases however their molecular formula is same so all geometrical isomers are di-stereomers so this is very important okay all geometrical isomers are di-stereomers they have asked this question many times again the second fact to write down here which is again important they have asked this question all these data that I'm giving you they have asked this question in NEGE all these exams okay theoretical question sin and nt what happens in sin and nt sin and nt we define for these molecules c double bond n n double bond n if higher priority group are on the same side sin opposite side nt for example you see n double bond n one side suppose we have a lone pair this side we have OH this side suppose we have CH3 nitrogen will always have a lone pair correct here also it is one lone pair here suppose here we have CH3 CH3 and edge okay you see the higher priority will have this carbon and oxygen first priority first priority same side so this is an example of sin isomer here higher priority on the opposite side so this one is nt sin and nt term we use for this kind of molecule only where we have C double bond n or n double bond n for this kind of molecule we use sin and nt term correct is it clear next point you write down and these points you must remember okay first of all this one geometrical isomer second point you write down dystereomers and when I say dystereomers it means we are talking about a pair dystereomers can be of can be any combination this means what any combination means one of these two pair of molecule can be optically active and other one can be optically inactive another possibility is both are optically active and another possibility is both are optically inactive any combination possible we have here for dystereomers the pair of molecule can be optically active can be optically inactive one can be optically active other one can be optically inactive okay third property you write down here dystereomers have different physical properties have different physical properties physical property means boring point melting point refractive index refractive index density all these dystereomers will have different physical properties and hence they can be separated they can be separated by physical methods physical method for example we have fractional distillation t i o n a l fractional distillation these three four points of dystereomers you must memorize you must remember is it clear can we move on tell me now you see one example we'll see here in the molecule which has carol center okay so heading right down compounds containing compounds containing two carol carbon two carol carbon I'll give you one example here after that again I'll give you one question but first you try to tell me or try to solve this you have to tell me the configuration of each carol carbon I'll give you some random molecule here now in this molecule you see first of all you listen to me here okay I'll just explain you this representation first and then you can do the configuration of this one see like we have we have discussed this one already right at this one correct so this is the same molecule we have here only thing here it is what we have only one carol if you add one molecule like this also you'll get one more carol carbon here so in this also like these two bonds is coming towards you and these two bonds are going away from you the same thing we have there also I'll explain it here you see see suppose one molecule I'll give you like this here we have carbon and here we have carbon this is also fissure projection we have but with two carol carbon so this bond is coming out of the plane towards the observer horizontal bond this bond is also the same horizontal bond coming out of the plane here also we have the same thing horizontal bond coming out of the plane same thing we have here horizontal bond coming out of the plane and these two bonds are into the plane going away from you right this will be in the plane of this this carbon carbon bond will be in the plane of this so same thing we have here this bond is coming towards you this bond is going coming again towards you coming towards you and this is also coming towards you here this whole molecule here so when you have two carol two center here so in this the since this is the carol carbon we have this is also a carol carbon we have so both are carol carbon so four different group attached with this carbon is one two three and this whole group is four this is not the priority I'm just giving you how the groups are attached so one two three and this whole group similarly for this carbon if you select so one two three and this whole group is the fourth understood this tell me tell me guys correct now you try to find out the configuration of this carbon and this carbon find out the configuration of these two carbon atom here this is suppose the first carbon we have and this is the second carbon tell me the configuration now assign the priority according to cip rule I don't know what I have given some random question okay so let them solve also virus is getting r and s the top one is our right the top one is r and the bottom one is s is it worth I don't know I don't know you are getting top one is like all right okay so what it is getting r for the top carbon the bottom carbon is s configuration tell me others need and then is getting s and s yes the least not h the least priority group should be on the bottom of the vertical line all of you are getting different answer most of you are getting s and s okay now you see the priority would be what obviously this one will be the least priority and the maximum will be this one and then two and then three since least priority should be on the bottom of this vertical line so h should be here and this group should be here right so when you exchange these two okay I'm not talking about this molecule now we have exchange here and we'll get three here and four here right then the configuration of this will be what it is anticlockwise s so this molecule when you exchange one pair you are getting s so the original molecule is what original molecule will be r this molecule is r this carbon is r is it clear is it clear the this this the deep yellow color that you have it is the configuration of the actual molecule the purple one is the configuration when you exchange this hydrogen in this whole group what is wow engine correct but the lowest prior that is the constraint we have that is the condition whatever the lowest priority group you have that must be on the bottom of the vertical line if it is not there you have to make some changes by exchanging the atoms or molecules you did not understand the priority order the first atom is carbon obviously this one is fourth one carbon carbon and carbon second atom is oxygen here also oxygen and here it is carbon so obviously this one is third in these two if you compare C double bond O C double bond O also here after that we have H but here we have O so this is the first priority second priority right similarly if you do for this one the priority order will be this that I am right known with the red one so the higher priority will be what is the higher priority this is the fourth one this is the third one in these two we have carbon and then oxygen here we have carbon and carbon so this is first and this is second right so when you exchange these two here so again I'll use the purple one this should be one and this should be four so one two three will be like this it is clockwise so clockwise so it is R right but the actual molecule I'll write down with this deep yellow color that would be S is it clear now yes correct Priti because the actual molecule is this H is here and this whole molecule is here but for our convenience we have swapped these two molecule so when you swap these two atom or molecule like this group like this then the actual molecule is changed right so by swapping these two we got the configuration of this molecule right not this one the new one that you get okay but we have to find out for this molecule right so that one if it is R so this one would be S because we have done one swap is it clear now so we can do if you have more than one carol center like two three four five whatever the number of carol center you will get according to this only you have to find out the configuration okay but in this chapter isomer is a maximum you will get two carol center not more than that okay when you do biomolecules there we have a chain of compound like this there we have more than no two carol center four five carol center will be there okay but there also you don't have to find out the configuration that is not required over fine so you see I'll give you one question onto this the same thing I'll give you one question so let me draw some structure first this is the first one second how does this this you have to identify the pair of enantiomers dyestereomers pair of enantiomers and dyestereomers here what do I do to find out enantiomers and dyestereomers what will do to find out the enantiomeric pair what is the configuration for the first one let's discuss the first one this configuration will be r is it correct this configuration is also r all of you are getting r and r what is the configuration of this what is the configuration here this one you will get s and this also you'll get s okay here the configuration will be this one is s and this one is r this one is r and this one is s how did you find out the configuration of two three four one you have just assigned the priority and then you'll check clockwise anti-clockwise correct okay answer to me first just let me understand one thing first what did you do did you assign priority here on all these molecule and then you find out s and r did you do this tell me what i'm asking did you assign priority and find out for two three and fourth one correct actually you see you don't have to assign priority here in all these molecule just you do that for the first one like in this one you assign the priority you see here what you'll do if you assign the priority here this one will be first right this one will be second third and fourth and then you sap you find out the configuration of these two it is r and r okay once you have done this all the molecule you see the molecular formula is same correct so you don't have to assign the priority again and again in second third and fourth one what you have to see you have to just check the number of swap we have here like from this one first you concentrate on this carbon right how do we get this molecule this part see this is a group right this is the group we have the bottom one this is the group we have right and this group whether you put ch3 on the right hand side or left hand side that won't make any difference on the connectivity of this carbon am i right right whether you put this ch3 here or here this won't make any difference with the correct activity of this carbon atom okay so when you're trying to find out the configuration of this carbon atom just you check how many pair of groups has been exchanged or has been swapped right so from this to this you will get once you exchange this oogen cs3 so odd number of exchange we have if this is r so this would be as correct similarly you compare this one cs3 oh h and this group again this group is same over here you get this molecule or this atom here by the exchange of cs3 and oh h again so again if this is r this one will be s okay so two different molecules have same molecular formula but but exactly opposite configuration r s r s so these two are the pair of enantiomers is it clear first and second are the enantiomeric pair is it clear tell me first point i'm trying to make is what if the if the molecular formula is same you don't have to assign the priority again and again just you do it for one and then you take the number of swapped we have one number of swap you have done accordingly you can find out the configuration one more thing you can do in the another way which is not you know which you should not prefer okay better ways to use the configuration r and s these two molecules you can easily see these two molecules are mirror image of each other right and they are non-superimposable mirror image okay that's why they are enantiomers now similarly if you see the connectivity of this carbon and this carbon is same exactly same oh h left side left side h on the top top cs3 on the right hand side and this group on the bottom so that's why the configuration of this is s so this should be s only right this one if you compare we have swapped what cs3 and oh h then you will get this okay if this is r this one would be s sorry if this is s this one would be r correct similarly you can do uh the configuration of this also you can take right so these two are what three and four are of opposite configuration three and four are of opposite configuration right are they enantiomers third and fourth one that's what I'm asking Bharat opposite configuration we have but they are not they are they are actually you know mirror image of each other they are actually mirror image of each other but they are superimposable these two are non-superimposable non-superimposable so these two since they are superimposable hence they are not enantiomers of each other how they are superimposable that you see well what I said if you remember that we can rotate the molecule in the plane right in the plane rotation is allowed so we'll check the molecule by all these rotation so what happens if you rotate the molecule in clockwise manner 180 degree rotation you'll get the same molecule yes so you rotate this molecule 180 degree and then you can see both molecules are what both molecules are superimposable that's why these are superimposable to each other so only one and two molecules one and two pair are enantiomers right but one two are mirror image of each other three four are mirror image of each other so if I ask you what are di-stereomeric pair we have here di-stereomers one and two since they are mirror image of each other so they are not di-stereomers correct so apart from one two and three four all of the pair if you do if you make that would be what that would be di-stereomeric pair so first and third are di-stereomer first and fourth are di-stereomer second and third second and third are di-stereomer and second and fourth are also di-stereomers to be sure right for di-stereomers you just have to check the mirror image right if they are mirror image of each other they are not di-stereomers if they are not mirror image they are di-stereomers correct for enantiomeric pair you have to find out the configuration if they are exactly opposite configuration with j molecular formula they could be enantiomers of each other but then again provided provided the fact that they must be non superimposable right these two are these two molecules third and fourth you see they have exactly opposite configuration s and r r and s they're exactly opposite configuration but since they are superimposable to each other hence hence they are not di-stereomers right or they are not they are not enantiomers of each other correct what is the relation between three and four what is the relation between three and four three and four are identical compounds right whatever if you see if you rotate this by 180 degree we'll have c s 3 c s 3 this side o h o h this side right so that is nothing but the same molecule that's why they are identical to each other are they three and four i'm talking about three and four are they optically active with one molecule you can actually you know conclude all these concepts that we have studied okay if you can do all these with one two three molecules you will understand the concept of clearly right so what is the optical nature of third and fourth compound are they optically active or not yes right shravan there is a plane of symmetry that's why they are not optically active it is not because of two carol carbon need engine okay i have told you these things many times that presence or absence of carol carbon is no any criteria for a molecule to be optically active correct we have to check plane of symmetry if the molecule has two carol carbon they may or may not be optically active that is not the condition you see here we have clearly a plane of symmetry present here here also we have plane of symmetry understood again one more thing did you understand till here tell me is it clear all of you i am asking you all okay i'm not teaching only three four five persons so all of you please reply these count this kind of compound right in which we have a plane of symmetry we call it as meso compounds right the compound in which we have a plane of symmetry those compounds we call it as a meso compound right generally we define this as two or more carol carbon when the molecule contains two or more carol carbon so since meso compound has plane of symmetry so they are what optically inactive always keep this in mind that meso compounds are optically inactive this is also an example of meso compound and this is also optically inactive can we get a resonant mixture in this is it possible to have a resonant mixture if possible then how how can we get a resonant mixture over here tell me right truthy it is the equimolar mixture of one and two see one and two are enantiomers seven right one and two are enantiomers of each other so if you take equimolar mixture of these two you will get a resonant mixture clear okay now the last part so these are the terms that we use generally in optical isomers that is enantiomers resonant mixture meso compounds di-stereomers okay these are the four things that you must keep in mind okay now we'll see now we'll see the last part of this optical isomerism and that is the calculation of stereo isomers how many stereo isomers are possible calculation of so in this we have a formula and that formula you have to apply to get the answer okay because you cannot you know imagine that structure every time in the exam also because it is a bit tough and there are high probability that you'll make some mistake okay so it is the best way to memorize the formula and get the answer okay so first one first case you write down in case of unsymmetrical molecule unsymmetrical molecule number of optically active means d or l configuration number of optically active form is equals to that will take by a a is the number of optically active form small a and that is equals to 2 to the power n where n is the number of asymmetric carbon atom asymmetric carbon atom right number of mesocompounds number of mesocompound in short we write m is equals to see if the molecule is unsymmetrical then the number of mesocompound will be what zero there is no mesocompound possible because meso is only possible when we have symmetry correct and we are taking the condition of the unsymmetrical molecules so there's no mesocompound possible totally studio isomers isomer will be number of optically active plus the mesocompound which is optically inactive which is equals to 2 to the power n in this case 2 to the power n based on this solve this question example c h 3 c h o h c o o h this is lactic acid formula is lactic acid can you tell me the total number of stereo isomers in this how many stereo center we have here asymmetric carbon atom number of asymmetric carbon atom is only one that is this carbon right carol carbon n value is one since the molecule is unsymmetrical hence the value of mesocompound is zero and a is equals to 2 to the power n that is equals to 2 total stereo isomer is equals to a plus m and that will be 2 only formula you have to keep in mind in case of in case of symmetrical molecules in this we have two possibilities the first one is when n is even even number then the number of optically active form that is d and l is equals to 2 to the power n minus 1 I think this color is not good use this one number of mesoform which is m this is a is equals to 2 to the power n by 2 minus 1 sorry 2 to the power n divided by 2 minus 1 this is the number of mesoform total stereo isomers will be a plus m which is 2 to the power n minus 1 plus 2 to the power n by 2 minus 1 example you see ooc ch oh ch oh oh fine the number of mesocompound the number of optically active compound see they can ask you anything into this they can ask you total number of stereo isomers also they can ask you number of optically active form also they can ask you number of mesocompounds tell me what is the answer yes how many studio center we have this carbon and this carbon and obviously we have a plane of symmetry here is very clear right p os symmetrical molecule it is the molecule is symmetrical so we use this formula right and the number of n is what n values 2 that is even right so a value will be what 2 to the power 2 by 2 2 to the power 2 minus 2 to the power 2 minus 1 is equals to 2 m value is 2 to the power 1 is equals to 1 so total stereo isomers will be 2 plus 1 that is equals to 3 only you have to apply the formula here nothing much okay next one you see case 2 only condition b first was when n is odd this is when n is first was when n is even this is when n is odd here the number of optically active form that will be equals to a which is nothing but 2 to the power n minus 1 minus 2 to the power n minus 1 divided by 2 this whole divided by 2 number of mesoform m that will be equals to 2 to the power n minus 1 divided by total isomer here it will be a plus m so this part will get cancelled total number of isomer really 2 to the power n minus 1 okay the other three formula we have here question you see CH2 OH CH OH CH OH CH OH and CH2 OH tell me number of optically active form number of meso compound total number of isomer of HNAV where are you from of HNAV how there are two studio centers Bharat do we have two studio center in this in this we have two studio center because you see this molecule and this molecule are same okay so i'll just do some change into this question okay this i'll write this one i think this is my mistake COH now it's fine odd one so number of n value is 3 3 now it's 3 1 2 and 3 is 2 to the power 2 minus 2 to the power 1 that will be equals to m is 2 to the power 2 by 2 which is equals to 2 total is your isomer will be 4 number of optically active form is 2 number of meso compound is 2 yeah this is not symmetrical this this question is wrong i guess then or we should have symmetry also correct if it is not symmetrical then we use the formula of unsymmetrical one simply okay we'll use the formula of unsymmetrical one this formula we cannot use okay because this is not symmetry we have to check symmetry also no so you use simply the formula of unsymmetrical one that is 2 to the power 3 you'll get okay i'll give you one i'll just do some change into this for this question the answer will be 2 to the power 3 that will be 8 for this question number of unsymmetrical one okay so if i have to make that question i'll do the change in this one this you let it be as c h 2 oh or here what we do c o h c h 3 and c h 2 h just a second then it will be same no not possible again because we have n should have odd and symmetry also so we'll write c h o h c h o h 1 2 3 this is not there for 3 4 i don't want to use this it's 1 2 3 4 we're getting even it is right it's 1 2 3 and should be odd and symmetry also right we have symmetry in this one and here i put h no it's fine i think is it fine now this one is a carol carbon this one is also a carol carbon again it's not symmetric right for symmetry if i put c s 3 here then we have 4 carol carbon 1 2 3 4 we can draw a molecule like this so where n is odd and the molecule is symmetrical n is odd and molecule is symmetrical i draw that so it is odd this one you check is this a carol carbon yes is this a carol carbon this is not a carol carbon one more change how about this one this is a carol carbon yeah is this fine is this fine you see but we don't have plane of symmetry into this right not possible this one okay let it be i'll just give you one question onto this i'll cross check in the book somewhere if i get i'll give you with even number will have plane of symmetry and is odd plus plane of symmetry let it be i'll just give you some question onto this i have to see then okay yeah yeah yeah we can move forward we are wasting our time i'll give you some example onto this on the group right we'll give you so these are the you know a few uh thing that we have discussed in optical isomerism we have discussed probably everything in optical isomerism our last part of this our last part of this isomerism is we have that is conformational isomerism conformational you know what is conformational isomerism what is conformational isomerism do you know are you there it is not restricted it is unrestricted rotation across sigma bond there is no restriction okay one thing okay just you see this i think i got what where we have we were doing mistake we have ch2 oh edge i got this now this was the first example i took if you remember just give me a minute okay this was the first example we took right here you see the molecule we don't have plane of symmetry here why because first of all you see we have odd number of carbon atoms correct we have odd number of carbon atoms second thing is what you may think where you go get confused here that you may think we have a plane of symmetry like this correct yes this was your doubt right i got this now you you see this you were thinking that we have a plane of symmetry here equal molecule same molecule this side same molecule this side is it tell me fast we'll finish this now but now you see this carbon atom this carbon atom is sp3 hybridized right plane of symmetry means what which divides the molecule into two equal half correct so you see these if you suppose if you divide this molecule also it is not possible that see what happens it is possible that one one possibility is what this group will take this OH and go out and this group will take this over and come this side if you divide the molecule into half right so if you try to get a plane of symmetry here this part is different from this part yes or no because one will have H another will have OH and it is not a straight out also it is tetrahedral so bond angle is 109 degree 28 minute am I clear now it's not Shashant you see it properly like you can say since since you can see that this molecule is a planar molecule that's what the doubt you have you are thinking that this molecule is sp3 but that is not one edge is this side and OH move maybe this side the bond angle between these four bonds are 109 degree 28 minute so if you try to divide the molecule into two equal half however this group and this group are same but one group will have this edge other one will have this OH correct so the two equal two parts are not same not identical so we don't have any plane of symmetry here am I clear now did you understand this right and after all this you can see we have actually five carbon atom into this so odd number of carbon atom how can we have plane of symmetry when we have odd number of atoms present it is only possible in case of even number I hope it is clear now correct now we'll start this conformational iso one more thing you will just write it down here if somebody ask you of the question is number of number of resnick mixture what will be your answer just last thing here number of resnick mixture sushant understood number of resnick mixture will be what that number of optically active form that you get okay first and second so number of optically active form if you mix two optically active form in equimolar form that gives you what one resnick mixture right the two optically active forms gives you one resnick mixture so a optically active form you have that will give you a by two resnick mixture all these molecules where wherever you are getting optically active form whether it is symmetrical or non-unsymmetrical that's not the case wherever you have optically active form the number of resnick mixture will be number of optically active form divided by two because two optically active molecule gives you one resnick mixture so n gives you n by two a gives you a by two is it clear now we are coming back to conformational isomerism yes yes yes obviously if one molecule has if you take the mirror image of that okay one will be d other one will be l obviously when you are talking about resnick mixture overall it is optically inactive so one obviously it will be d other one it will be l okay conformational isomercy write down the different arrangement of the different arrangement atoms in space atoms in space that results from that results from the free rotation free rotation of groups about carbon carbon single bond single bond axis are called conformations conformations conformational isomers conformations or conformational isomers and this phenomenon and this phenomenon is conformational isomerizm confirmational isomerizm okay did you write this now what is conformational isomer you see here suppose we have ethane right so if I write down the formula of ethane that is C2S6 this is the carbon-carbon single bond okay carbon-carbon single bond now in this this is the hydrogen so these are hydrogens actually H H H H we will discuss this in detail also okay because I'm trying to make you understand what is you know what is the conformational isomers okay so this is carbon this is also carbon correct and we have single bond here this bond will be exactly opposite to this this will be exactly opposite to this this will be exactly opposite to this like this it will be okay now the thing is since this carbon carbon has only single bond so you can what you can do you can fix one of the carbon atom and you can rotate the carbon atom in clockwise or anti-clockwise direction whatever you want to do here the way you can rotate this since there is no any pi bond present here correct so there is no restricted rotation so one this you can fix and this you can rotate right around its own axis you can rotate this easily correct so what happens in this there are two three terms we have here right there are two three terms and that terms will discuss first and first we'll see here the first of all you see we have an electron pair here right we have an electron pair here which is at this position we have certain amount of certain strain also in strain in the sense electron electron repulsion whether it is less or more that is a different debate right but since we have bond pair bond pair of electron here so these two electrons will have tendency to repel each other correct and when you rotate the molecule when you rotate one of the carbon atom giving this carbon atom fix anyone you can rotate but other one should be fixed right so if you rotate this carbon atom so the relative position of this bond pair bond pair electron will change yes or no the position of this bond pair of electron and this bond pair of electron will change correct did you understand this tell me fast some what means like the difference whatever it is but yes that will change right so this kind of strain is strain between bond pair of electron this kind of strain we call it as torsional strain torsional strain if this strain is high the stability will be less any kind of repulsive force will reduce the stability of the compound correct so this is one of one kind of repulsive force only right so the repulsion between the bond pair of electron at the adjacent carbon atom is torsional strain okay at the same time what we have we can have repulsion between this hydrogen and this hydrogen atom also what I said that you see if you have these two carbon atom simply I will draw here we have carbon carbon bond hydrogen hydrogen and we have hydrogen hydrogen like this this is nothing but ethane I have written here the simpler not a representation so when these two carbon atom we have so we have certain distance between the bond pair of electron here all these bond pair like these bond pair this bond pair this bond pair this bond pair a certain distance so when you rotate this molecule so obviously the distance between the these two bond pair of electron will change correct and that will affect the repulsive force between these two bond pair of electron also so in the distance increases the torsional is strain which is the strain between the bond pair of electron at the adjacent carbon that will also change correct and when the repulsive force is less stability will be more so torsional strain is what it is the strain or the repulsion between the electron pairs of at adjacent carbon carbon bond right now the second type of strain is what we call it as vendor wall strain vendor wall strain this is between the atoms okay atoms or molecules present at the adjacent carbon atom like we have repulsion between bond pair of electron similarly we have slight repulsion between these hydrogen atom also right so again when you rotate this molecule in clockwise or anticlockwise manner the distance between the hydrogen atom changes and hence the vendor wall strain will decrease again the stability will increase wait wait i'll dictate you the first to try to understand i'll dictate ask you those just let me and let me make you understand this first torsional strain is the strain between the bond pair of electron at adjacent carbon atom vendor wall strain is the strain or the repulsion between the two atoms present at the adjacent carbon atom right so vendor wall strain we are talking about this atoms torsional strain we are talking about the load i'm sorry bond pair of electrons correct so what i am trying to make you understand is once since the rotation is impossible here there's no pi bond so once you rotate one of the carbon atom so the relative distance between the two hydrogen atom and that bond pair of electron will change and that will also affect the stability of the molecule am i right is it clear all of you again you see here this bond and this bond bond and this bond that you have we have a bond angle also here if you extend this bond like this you'll get a bond angle right this is the bond angle which can be anything right so this bond angle we call it as dihedral angle dihedral angle theta okay so once the rotation takes place this dihedral angle will also change and that will change the stability of the compound because the torsional and wander wall strain will be affected in that case is it fine because of the change in this dihedral angle the wander wall and torsional strain will change and hence the stability will get affected at some point of time at some angle the stability will be maximum at some angle the strain will be maximum so stability will be minimum is it clear so now you see what are the conformational isomers like as you keep on rotating one of the carbon atom the dihedral angle will change continuously it can be 1 degree then 2 degree then 4.5 degree 10 degree 20 degree 30 60 90 180 360 like that it may go right so we keep on changing this dihedral angle by rotating the molecule the stability of the molecule changes right so all these conformations all these conformations that is that we get according to this dihedral angle that we call it as conformations or conformational isomers okay so there are actually infinite number of conformational isomers possible what I said that when you change this angle the dihedral angle when you change that will affect the stability of the molecule and at every angle you'll get different different molecules because their stability is different right so like this we can have infinite number of dihedral angle so we can have infinite conformational isomers possible right so if somebody asks you to draw the conformational isomers of ethane right you are not going to draw the infinite number of conformational isomers okay so what we'll draw we'll draw the most stable isomers here right so out of infinite conformational isomers the most stable conformational isomers are called conformers right so right on first of all this point out of infinite conformational isomers infinite conformational isomers the isomers which are more stable which are most stable the isomers which are most stable are called conformers so generally we try to write two most stable conformers or two most stable conformational isomers okay conformers how do we assign stability that we'll see okay but the point is this only conformational isomers is because of the unrestricted rotation of carbon carbon single bond you keep on rotating one of the carbon atom like this around this carbon carbon bond the bond length the length between this hydrogen atom will change the length between this distance between this bond pair of electron will change that will affect the torsional and vendor wall strain and eventually the stability of molecule will get affected right these angle we call it as dihedral angle which at every angle will get one conformational isomers like this at there are infinite angle possible right from 0 to 360 more than that also there are infinite dihedral angle possible and according to each dihedral line corresponding to each dihedral angle will have one conformational isomers so there are infinite conformational isomers possible but out of infinite conformational isomers the isomers which are most stable are called conformers and whenever they ask us to draw the conformers of one particular molecule we'll try to draw the most stable conformers two most stable conformers okay how do we assign or you know generalize the stability of this molecule that we'll see in the next class okay but these two definition you write down torsional strain and vendor wall strain you write down torsional strain write down it is the repulsion between the electron pairs the repulsion between the electron pair at adjacent carbon atom adjacent carbon atom vendor wall strain vendor wall strain it is the repulsion between the atoms present at repulsion between the atoms present at adjacent carbon this vendor wall strain is almost negligible between hydrogen and hydrogen atom both this torsional and vendor wall strain is minimum in is minimum in staggered form is minimum in staggered form what is the staggered form when the dihedral angle when dihedral angle theta is 60 degree these two strain is maximum in eclipsed form these term you have to memorize eclipsed form what is this eclipsed form when dihedral angle is zero right so there are two different forms we have for ethane i'm talking about right two different forms we have for ethane staggered and eclips correct in between these because dihedral angle can be anything right it can be any value from zero to 60 degree from 60 onwards if you go you are again going towards the eclipsed form if you are moving from zero to 60 degree right you are moving towards the staggered form most stable form again to 60 to 120 if you go again you are going back to the original form okay we'll discuss this when you discuss when we'll discuss the new man projection in the next class okay but just i'm giving you some term for today okay so that we can start it from next class staggered and eclipsed theta value if any value we have theta between zero to 60 degree this form we call it as skew or gauche form skew or gauche form okay one more thing the last thing here we have minimum strain in staggered form we have so stability will be maximum here minimum strain maximum stability and here the stability will be minimum is it clear did you understand this so we wind up that last year only okay we have discussed the most important part so that that is useful in hydrocarbon okay little bit more we'll discuss here there are a few things we have into this but that is not required now since this is the these are the portion of this is the angle see dihedral angle just a second dihedral angle is the angle between the two bond present at the adjacent carbon see this carbon hydrogen bond and carbon hydrogen bond that is dihedral angle right dihedral angle is the angle between the two bonds present at the adjacent carbon correct so these are the few you know terms i have given you right these portions are there in 12th class again i'm telling you i have discussed in this like confirmation isomers we have discussed almost 30 percent 30 40 percent we have discussed right but we will not complete it now because we have to finish hydrocarbon first correct so since we didn't have much time today so i have discussed this thing but little bit more i will discuss over here the rest part that you have in this we'll discuss in probably once you finish once we finish hydrocarbon goc right after hydrocarbon we'll finish goc that is more important correct so we'll wind up the class here only okay we'll have another class this week apart from Saturday that will be probably online and we'll let you know on this Wednesday what class you have on Wednesday do you have any class on Wednesday or Wednesday i'm not free Wednesday Thursday maybe you'll have on Friday i'll let you know okay correct any doubt we have can we wind up here yeah hello thank you all and i've given you the you know homework last class to read out hydrogen from the ncrt okay so all of you must finish that yeah i'll let you know regarding Friday Wednesday i have another class so i cannot take that day okay thank you all take care bye