 Hello, guys. Good evening. Hello, guys. Are you there? Can you see their screen now? Good evening, guys. Who are not there? Please type in your name. Please type in your name in the chat box. Can you see their screen now? Hello, Aretu. Hello, Aretya. All of you can see their screen. Can we start now? So, guys, we are going to start today optical isomerism. And probably in this two-hour session, we'll finish optical isomerism. So, first of all, you see, optical isomerism is what? How do we define the phenomenon called optical isomerism? So, the optical isomerism is actually the behavior of a molecule towards plain polarized light. How a molecule can behave towards plain polarized light? What is plain polarized light that also we'll see? If the behavior of the molecule towards plain polarized light, we observe. If the molecule can rotate the plain polarized light in any direction, whether it is clockwise or anti-clockwise, if the rotation is possible, it means that the molecule is optically active. What is optically active molecule? All those molecules which can rotate the plain polarized light in any direction. That we call it as optically active compound. When there is no deflection in plain polarized light, when there is no deflection, when there is no rotation, then the molecule is said to be optically inactive. So, this is what? When I say the behavior of a molecule towards plain polarized light, obviously from this you won't understand what is PPL, what is behavior, what we are talking about actually. But the basic definition of optically active compound is the compound which can rotate plain polarized light in any direction called optically active. So, now we'll start with the definition of optical isomerism. So, optical isomerism is defined as the compounds having, all of you write down, the compounds having similar physical and chemical properties, similar physical and chemical properties, compounds having similar physical and chemical properties. Their behavior towards plain polarized light, but their behavior towards plain polarized light is different, are called optical isomers. And this phenomenon is optical isomerism. Hi everyone Siddharth, Nikita, Shaughnak, Siddhi, Shreya. So, the behavior or the definition of optical isomerism is what? Compounds having same physical and chemical properties, but behavior towards plain polarized light is different, are called optical isomers and this phenomenon is optical isomerism. Now, what is plain polarized light? So, first of all we'll see what is plain polarized light that we are discussing. What is plain polarized light? Light you see when you have a light source, suppose you'll have a light source that light source will emit light in all direction actually. In this direction, this direction, every possible direction, the light source emits light. So, what we do? This is a white light, light source which emits white light basically. Now, this light we allow to pass through a prism that we call it as Nicole prism. This is Nicole prism. Nicole prism in tablet is not working. I don't know. I have to see the settings any less. Nicole prism, right? Now, when this light passes through this plain Nicole prism, all the lights, we can say polarized in one direction. Which is not exactly true. Why I'll tell you, but when the light passes through the plain polarized light, we get the light which is parallel to each other or all the lights are there in one plane. You see, here the lights are in all the direction, but here the lights are in one direction along a same plane, all the lights along a same plane. So, what you can understand? Hello, Chirag. Hello, Arya. So, what is the meaning of this? You can understand actually Nicole prism. There is, you know, in Nicole prism, there is a slit. Very ultra fine slits present in the Nicole prism, right? Ultra fine slits present in the Nicole prism like this. Very fine slits present in the Nicole prism. All those lights which are parallel to these slits, basically slits means we have a gap in between, right? So, suppose the lights are going in this direction, we have horizontal light, suppose we have that horizontal light will, you know, collide with this bar that you have over here and reflects back or will reflects in the other direction. Only those lights which are parallel to these slits will pass through the slit and hence like on the other side of the prism will get the light which is parallel to each other, okay? And this light we call it as plain polarized light because all the lights are in one plane. So, we call it as plain polarized light. In short, we call it as PPL, okay? Now, what we have to do here? The organic compound which whose behavior you have to, you know, understand whether it is optically active or not. Now, this plain polarized light, we allow this light to pass through the organic compound, whatever organic compound we have, we allow this light to pass through the organic compound. Suppose this is the organic compound we have. When this light passes through the organic compound, there are three possible observations we may have, right? One observation is what? The light will deflect and the deflection of light may be in clockwise direction like this or it may be in anti-clockwise direction. This is one type of behavior and other behavior is what? There is no deflection in the light at all. The light comes out from the organic compound as it went into the organic compound, right? There is no deflection in the organic compound. So, when there is no deflection like in this case, the given organic compound is said to be optically inactive, optically inactive. When there is no deflection, it is optically inactive. When the light deflects in any direction, it is optically active OA in short, I am writing it down. Got it? Any doubt, let me know. If you have any doubt, let me know. Till here, any doubt. Any doubt, Shreya, Tridiv, Aditya, Aniruddh, Siddharth, Sharnak, Chirag, Arya, Nikita. Any doubt, let me know till here. Behavior of light towards plain polarized light we are observing, right? Now, the assumption is what? If the PPL that is plain polarized light, if the PPL rotates clockwise or towards light, the PPL rotates clockwise. The compound is said to be dextrorotatory. The compound is said to be dextrorotatory or in short we call it as d, only d, small d. Or we also write it as plus. Plus means dextrorotatory, d, clockwise rotation, okay? When PPL rotates clockwise, it is dextrorotatory, small d or plus. If the PPL rotates anti-clockwise, anti-clockwise, it is called it as levorotatory, A-E-V-O, levorotatory, that is small l or we call it as minus. All these are experimental things, plus and minus. If I ask you, if I give you a compound and ask you whether it is dextrorivo, you have to perform this experiment. You have to perform this experiment and then you have to check whether the light is rotating clockwise or anti-clockwise. Then only you can say it is dextroro or levo. Listen to me very carefully here that dextroro and levorotatory compounds is experimental. There is no theoretical way so that we can say that the given compound is dextroro or levo or plus or minus. Now suppose if I give you one compound or if you get a question in the exam and the question is the given compound is optically active, yes or no. Then what do you have to do? Till now, we haven't seen any theoretical method to understand whether a given compound is optically active or not. You have to perform this experiment and you have to see or observe the deflection in PTA. If yes deflection is there, then it is optically active, otherwise inactive. Obviously this method you cannot use in the exam. The examiner will not give you that compound. It is in the question paper. You don't have all those facilities so that you can perform this experiment. So obviously there must be a theoretical way to find out the compound whether it is optically active or not. All these are experimental. Again I am repeating for a given compound DNL there is no other way to find out but there is no other theoretical way to find out DNL nature of any compound. In the book for some compounds it is written, it is Dextro, it is Levo. That is based on this experiment. We are not going to perform this experiment in the exam so we must have some other way or theoretical way to find out optically active nature of the compound. We are moving towards that theoretical method. So before moving, before doing all those methods, before understanding all those methods, there are few terms we have to understand. And all these terms are very important. All of you understand it carefully. The first term we are going to understand here is chiral centre. Write down chiral centre. I will dictate guys, all of you write down definitions in everything in your notebook. See this pen tablet is not working smoothly so I am just struggling a bit. That is why I am trying to dictate you otherwise I would have written things on the screen. Whatever is important urgent that I will definitely write but when I am dictating you must write it down. Now the definition of chiral centre is what? Write down all of you. An SP3 hybridised atom, an SP3 hybridised atom having four different atoms or groups attached to it. An SP3 hybridised atom having four different atoms or groups attached to it is called chiral centre. So chiral centre, the condition is what? It must be SP3 hybridised. And when SP3 hybridised, the geometry is tetrahedral. For suppose if I write down this molecule, carbon has OH. Then we have suppose CH3 and one more atom is suppose deuterium is attached to it. Now when you talk about this carbon, first of all the condition is what? The carbon must be or the atom must be SP3 hybridised. So this carbon atom is yeah, it is SP3 hybridised. And there are four different atoms or groups attached with it, OH, H, CH3 and D. Since there are four different atoms or groups attached to it, this carbon is chiral carbon. And chiral carbon we denote by star. Star means this carbon is the chiral carbon. So any centre or any atom which is SP3 hybridised having four different atoms or groups attached to it is called chiral centre or chiral atom. If carbon is that atom, then we call it as chiral carbon. Write down this point. If carbon is the atom, then we call it as chiral carbon. Then we call it as chiral carbon. Okay, one very important note we have here. All of you write down this note. All of you write down this note. Write down presence or absence of chiral centre. Is not any criteria, absence of chiral centre is not any criteria for a molecule to be optically active. Presence or absence of chiral centre is not any criteria for a molecule to be optically active. Next line. But if the molecule has only one chiral centre, then it is said to be optically active. When the molecule has only one chiral centre, then it is said to be optically active. Did you understand this point? For a given molecule, if you have to find out whether it is optically active or not, then presence or absence of chiral centre is not the criteria. But if you find out that or if you see that there is only one chiral centre in the molecule, then you can say that the molecule is optically active. Okay, molecule is optically active. But still, we have an example where there is no chiral carbon or chiral atom present, then also the molecule is optically active. Okay, we have that example also. That is why I am saying that presence or absence of chiral centre is not any criteria for a molecule to be optically active. What observation we have that if the molecule has only one chiral centre, then it is optically active. This is our observation. It is not the condition. It is not the criteria. Chiral centre is not the criteria for a molecule to be optically active. Very important point this one. Okay, now in this, we have one exception also. Write down the exception and what is that exception you see? Tertiary amines. Tertiary amines are what? When the nitrogen has three alkyl group attached and I am taking three different alkyl group R1, R2, R3. And apart from these three alkyl group, we will have one lone pair also present onto the nitrogen atom. Lone pair on nitrogen atom. So you see this nitrogen atom is also sp3 hybridized. Four different groups are there, R1, R2, R3 and a lone pair. According to the definition of optical activity, if the molecule has only one chiral centre, it is optically active. So according to that logic you see, this molecule should also be optically active because it has only one chiral centre which is the nitrogen here and four different groups attached. But this molecule or tertiary amine is optically inactive. It is an exception. Tertiary amine is optically inactive. This is an exception we have. Got it? Is it clear guys? Is it clear guys? Let me know. Any doubt, any confusion? You can write your doubt over here. Can you tell me how many chiral centre are present in this molecule? If present then. Bromy, question number one. Do we have any chiral centre present here? How many chiral centre we have in these two molecules? Tell me this is first and this is second. Do we have chiral centre in first compound? First compound we do not have any chiral centre. Now you see this. This carbon is sp3, fine. But we have three hydrogen present. It cannot be chiral carbon. Condition is what? sp3 hybridised and four different atoms are grown. So this is sp3, sp3 but we have two hydrogen here, three hydrogen here, not possible. This carbon is also sp3 but three hydrogen here, not possible. But when you see this carbon atom, it has one hydrogen present here. It has one hydrogen present here. And one group is this, ethyl group. This whole group is ethyl group. Another group is methyl group. Other one is hydrogen and other one is bromine. So yes, carbon is sp3 hybridised and we have four different groups attached. Hence, this carbon is the chiral, right? Only one chiral carbon we have. This carbon you see in the ring, we cannot have chiral carbon because all the carbon atoms are sp2 hybridised, okay? Sp2 hybridised carbon or any atom can never be chiral centre or chiral atom because the condition is the atom must be sp3 hybridised, okay? So this carbon you see, yes, this carbon is sp3 hybridised but since it contains two hydrogen atoms, so it is also not a chiral carbon. This carbon is again an sp3 carbon and it has four different groups attached to it. One is CS3, other one is etch, then cl and other one is the whole group. So yeah, this carbon is a chiral carbon and we have one chiral carbon also present in this compound. Understood? Now see, like I discussed and said that presence or absence of chiral centre is not any criteria. It is not any criteria for a molecule to be optically active. Then what is the criteria and condition for a molecule to show optical activity, right? So that condition or that criteria is chirality. Next heading you write down, chiral molecule. No aditya, second carbon from the left, you are talking about this carbon, okay? You are talking about this carbon, this carbon is sp3 hybridised but it contains two hydrogen atoms. Since two hydrogen atoms we have, that's why it is not a chiral carbon because we must have different atoms or groups attached to it. That is the condition. So this carbon is not chiral carbon, got it? Got it guys? Yeah. Another condition for a molecule to show optical activity is chirality or chiral molecule. One thing you always keep in mind that this chirality has nothing to do with chiral centre, okay? Chiral centre is different, chirality is different, okay? Both are not at all, you know, same or they have nothing to do with each other. Both term has nothing to do with each other, right? The first point in this you write down, any molecule which cannot be divided into two equal half, any molecule which cannot be divided into two equal half, two equal half is called chiral molecule, is called chiral molecule. Chiral, chirality is the property of a molecule, chirality is the property of a molecule by which, by which it cannot be divided into two equal half. Chirality is the property of a molecule by which it cannot be divided into two equal half. Next line, chirality and chiral centre, chirality and chiral centre are two different terms, are two different terms and they have nothing to do with each other and they have nothing to do with each other. Always keep this in mind that chiral centre and chirality are not at all same, they are completely different from each other, right? Next line, chirality if the molecule is not symmetric, yeah, yeah, right, right. I am coming to that Aniruddha, I am coming to that. Just give me a few more minutes, I am coming to that point. Next point you write down, chirality is the main criteria, chirality is the main criteria, comma, necessary and sufficient condition. Chirality is the main criteria, comma, necessary and sufficient condition for a molecule to be, a molecule to be optically active. Next line, if the molecule is chiral, it does not mean that it has a chiral centre. Not necessary. If the molecule is chiral, it does not mean that it has a chiral centre. Next slide. Parallel molecules are non-superimposable to their mirror image. Parallel molecules are non-superimposable to their mirror image. Superimposability, write down the definition next slide. Superimposability is a 3D phenomenon. If two molecules are superimposable, if we place one molecule onto another, two molecules are superimposable and if we place one molecule onto another, it looks like a single molecule. All these 4-5 points that I have given you, all these points are very important. Now, again I am telling you the necessary and sufficient condition for a molecule to be optically active that it should be chiral in nature. Now, what is chiral? Chiral molecule are those molecules in which there is no kind of symmetry. If there is any kind of symmetry present, then the molecule cannot be chiral and it won't be optically active. How do we decide that the molecule is chiral in nature? We have to check the plane of symmetry or center of symmetry. How to check that? That also we will see with some example but to decide for a given molecule is chiral or not, we will check what kind of symmetry we have if it is there. If the symmetry is not there any kind of mainly plane of symmetry and center of symmetry, if the symmetry is not there then the molecule is chiral and it is optically active. If any kind of symmetry is present, then the molecule is said to be chiral. If there is any kind of symmetry present in the molecule, then it is said to be chiral and it is optically inactive. This is said to be a chiral and it is optically inactive. One last time in this, if the molecule has only one chiral center, if the molecule has only one chiral center, optically active. Only one chiral center, optically active. Remember, if you have two, three chiral center present then we cannot say whether it is optically active or not. If only one chiral center, it is optically active. Having said that presence or absence of chiral center is not any criteria, must keep in mind. The only criteria for a molecule to be optically active is chirality, the molecule must be chiral molecule. How do we decide chirality? There should not be any kind of symmetry present in the molecule. When I say symmetry means there should not be any plane of symmetry or center of symmetry. How to do that? We will discuss. Now there is one more term and then we will move on to discuss or to understand if the given molecule is optically active. There is one more term you write down and that is stereocenter. Last class also I have given you the definition of it. I am discussing this here also because we should know the difference between stereocenter and chiral center. Any doubt till here? Aniruddha, Shreya, Aditya, Arya, Sridheep, Nikita, Siddharth. Any doubt tell me till here? Is it clear? Now next I write down stereocenter like I discussed in the last class. It is a double bond or ring or chiral center. It is a double bond or ring or chiral center due to which it is a double bond or ring or chiral center. Due to which stereoisomerism is possible. Stereo center, stereogenic center, both are same thing. It is a double bond or ring or chiral center due to which stereoisomerism is possible. Similarly important point next one you write down. Stereocenter may be sp2 or sp3 hybridized. Stereocenter may be sp2 or sp3 hybridized. You see chiral center is what? The hybridization of chiral center is always sp3. But stereocenter can be sp2 or sp3. We can conclude from this that all chiral centers are stereocenter. Write down all chiral centers are stereocenter. But the converse of this is not true. But the converse of this is not true. This is the difference between stereocenter and chiral center. Stereocenter can be sp2 also but chiral center cannot be sp2. Now you see this molecule. Do we have any chiral center in this CH3? Tell me in the first one, do we have chiral center present in the first one? In the first one, yes. We have chiral center in the first one because you see first of all we have to check or we have to find sp3 hybridized carbon. This carbon is sp3 hybridized with chlorine and hydrogen. Two different atoms we have here. Now we have to decide what are the other two groups attached with this carbon. For that what we have to do? We have to go clockwise and anti-clockwise. If the path is same in both directions it means the same atoms or groups attached with that carbon. If the path is different it means we have different groups attached. So when you go in this direction you see we have carbon-carbon double bond but such carbon-carbon double bond is not present here. It means this way and this way the path is different and hence two different groups attached with this carbon atom and hence this carbon atom is the chiral carbon. Similarly here both paths are the same. There is no chiral carbon in this compound. Only one chiral carbon present here and since we have only one chiral carbon present optically active this compound will be. Since only one chiral carbon is present. For any molecule if you have to decide whether it is optically active or not you can go through with presence or absence of chiral centre. If there is only one chiral centre yes, optically active. If there is no casual centre we cannot say whether it is optically active. If more than one chiral centre then also we cannot say whether it is optically active or not. In that case when we have more than one chiral centre or there is no chiral centre in these two cases we have to do, we have to check chirality, whether the molecule is chiral or not, okay. And how do we check chirality? We will check plane of symmetry or center of symmetry, any kind of symmetry we will check. Got it? The stereo center is this. Now we have representation of molecule, okay. Write down the heading representation of molecule, representation of molecule. Now you see D and L configuration are experimental, correct? I told this in the beginning only. D and L are experimental, right? Then what is the theoretical way to find out the behavior of a molecule towards PPL? That is chirality we have discussed. Now one compound can rotate the plane if it is optically active, can rotate the plane polarized light in clockwise direction and one can rotate in anticlockwise direction, right? So how do we know that the given compound rotates clockwise or anticlockwise, that is what we are going to understand now, okay? And what do we do for that? Means the behavior is to be a behavior towards PPL, how the molecule behaves towards PPL, whether it rotates clockwise or anticlockwise, how do we understand that? That we are going to understand now. And for that, you should know one or two representation of molecule, how the molecule represents, how do we write down the structure of a molecule, okay? So there are two representation we have that we are going to discuss here. The one is flying wedge representation. The first one you write down, flying wedge. You must know this, flying wedge. We also call it as wedge dash representation, flying wedge representation. And the other one is fissure, fissure projection, fissure projection. These two we have to discuss. Now you see flying wedge representation is what? Suppose we have a molecule, right, carbon. And it is attached with four different atoms of which, suppose O H we have. And it is also attached with, this is dash actually, hydrogen. Then we have chlorine. And then we have uterium, right? So this dark wedge, we call it as dark wedge is this. And this is what? This is the dash, which dash representation. You must have seen this kind of bond, like, which is connected with the, you know, the dot line. And one kind of bond which is connected with this thick line, which is this, this kind of bond you must have seen, okay? So what is the meaning of these two bonds actually? This bond which is represented by the dark wedge, it is coming towards the observer. It means what you see, I am looking at this molecule, I am looking towards my screen, okay? You are also looking towards your screen. So this carbon O H bond is coming towards me, or it's coming towards you. It is coming out of the, no, the screen towards you, right? So this dark wedge means the bond is coming towards the observer, right? So write down this dark wedge means this kind of wedge means the bond is coming towards the observer. Dark wedge means write down the bond is towards the observer and the dash representation that we have, it means the bond is going into the plate, means this hydrogen, which is connected with a dash bond here, right? It is going away from me, it is going away from you, right? Into the plane of the paper or into the plane of this, right? So this is the meaning of flying wedge representation, okay? Flying wedge representation. Got it? Flying wedge, any doubt in this? Flying wedge, any doubt? Guys, tell me, any doubt? Flying wedge dash bond means away from the observer, dark wedge means towards the observer, that is it, okay? Now fissure projection is what? You must have seen this kind of structure like this, like the structure of glucose I will draw here, you must have seen this kind of bond, no, horizontal bonds like this, okay? Where we have your CH co-present, co-edge, then edge, co-edge, like this structure. The point is, you must have seen this kind of structure, whether we have two Karel center or one this kind of structure you must have seen, right? So this representation is fissure projection, first of all. The cross representation we have is fissure projection. Now what is the meaning of fissure projection, okay? So first of all you see, if I take one example, this one suppose, here we have CH2OH, this is hydrogen, this is chlorine and this is methyl CH3. Do we have any Karel center in this molecule? Do we have any Karel center in this molecule? Yes, the center one, right? This is the Karel center, right? This is the Karel center and we have only one Karel center, right? That's what the thing here. Now what is the meaning of this fissure projection? In fissure projection what happens? The horizontal bond, this bond is coming towards the observer, horizontal bond right down the two points here in fissure projection, horizontal bond is coming towards the observer and vertical bond is going away from the observer into the plane and vertical bond is going away from the observer into the plane. The horizontal bond is what? It is coming towards the observer, horizontal bond is coming towards the observer out of the plane and vertical bond is going into the plane of the into the plane away from the observer. Okay. So now if I represent a three dimensional view of this molecule for a better you know picture what we'll write since the horizontal bond is coming towards the observer means this bond we must write with a dark wedge like this. This is CH2OH, this is CH3, CH3, this is chlorine and this is hydrogen. Okay. This is second, did you understand this, did you understand guys that's what we are representing Shreya. Line is into the plane that's why the dashed line and horizontal line coming out of the plane is that's what you know dark wedge that we have. Okay. So you can imagine a V shape okay that you know which is coming towards you and another V that is going away from you like that the open end. Okay. So that's the thing I can explain in the class I think you can understand from this any one of you have doubt here picture and all why in the plane because carbon is sp3 you know which one is co-planar here which one is co- there's no planar the carbon atom is sp3 hybridized the molecule is not at all planar the mutual bond angle is 109 degree got it the mutual bond angle is 109 degree atoms or groups in fissure projection when they we have only one sp3 hybridized carbon it's not at all planar they are in different different plane what Nikita I didn't get what about those not going away or towards see see first of all this is this molecule this molecule you see it looks like planar this molecule when you look at this culture like this it looks like planar in a plane fissure but actual thing or actual representation is this it is not at all planar these two atoms are gross which is on the horizontal line is actually coming towards you right and these two atoms are gross which is on the vertical line it is going away from you into the screen we have only four groups here Nikita two is coming towards us and two is going away from us only four we have this you let it be flying wedge you let it be you don't compare this and this now first you try to understand this fissure flying wedge is this got it these two points you must keep in mind fissure projection looks like planar because we draw like this we draw across like this okay but it is not at all planar understood understood guys tell me any doubt and we move on my you see like I said that the molecule the behavior of a molecule can either be D or L dextrored levo but dextrored levo which that's representation yes we show that to bond but it is not planar it is actually you know see there are only two things we can say one is going towards away from us and other one is coming towards us right this is only two possibilities with flying wedge we only represent these two and other two we say that it is in us different plane not in the plane of these two that's the meaning okay these two are in different plane these two will be in different plane but since we have only two possibilities we can show one is coming away from us and one is going away from and one is going away from us and one is coming towards us there's no other third there's no other you know possibility away and towards us that's what we can say so in flying wedge other two bonds are not in the same plane but when I say it is not it is planar it does not mean that these two are in the same plane the point is the plane of this and this and these two will be different and these two will also not be in the same plane it's a very simple logic the bond angle between these two is 109 degree right so it is tetrahedral it cannot be in the same plane but the molecule to be in the same plane they must have linear they must be linear 180 degree now you see like I said dnl representation or configuration of a molecule is experimental right to find dnl you have to perform that experiment the first one that I discussed correct but that we cannot do in the exam so we have different terms which we use to you know define the behavior of a molecule towards speed towards plane polarized light and those term we call it as r and s configuration rs configuration we can find out theoretically dnl is experimentally we cannot find it out we have to perform that experiment so whenever you get the question in the exam you have to find out r or s configuration correct so how to find out r and s configuration that you write down write down the heading rs configuration you write down rs configuration so in this you see we have we have one rule actually one condition we have to apply and that is it this is again rs configuration we have to understand okay so there are actually two things okay so in Fisher prediction what we do and in West dash representation what we have to write the first of all we see we can also convert Fisher into which dash and vice versa also this conversion is possible we can convert Fisher prediction into which dash or which dash into Fisher okay so that conversion we won't understand now that will do in the class okay that will be you know physical class will be better for you to understand so that conversion we are skipping now okay what constraint we have and what we have to how we have to you know assign rs configuration in both type of projection that will understand today okay now suppose the molecule is this the molecule I am taking is this carbon with this carbon we have chlorine this is chlorine and this is hydrogen and here we have CH3 and here we have this is the molecule obviously the molecule is optically active because it has only one kind of center right now to assign rs configuration in this which dash representation the condition is what first of all the first step is step one write down is to assign priority assign priority assign priority to the atoms or groups attached to atoms or groups attached to the carol carbon the carol carbon or carol atom now the condition in Fisher prediction is what write down the condition condition you write down in in sorry in which dash representation in which dash representation the least priority group must be into the plane must be into the plane the least priority group must be into the plane that's the condition right the group or atom which has the least priority that must be with this dash bond that's the condition we have right now the third point is what we have to go or we have to trace this line trace the circle you write down trace the circle and how do we trace the circle we have to go from first priority group to the third priority group first to third via to write first to third via to we have to complete the circle like this condition is what the least priority group must be on the bottom of the yes yes yes correct and it will priority will assign according to CIP rule okay will assign the priority the least priority must must be going into the plane I'm talking about which dash now Fisher prediction will discuss okay in which dash the least priority group must be into the plane then we have to trace the circle we have to go from first priority to third priority via second what are CIP rules Shreya we are not going to discuss here last class I have discussed you go through the notes if not then you can ask me personally I'll tell you right once the IP rule again there are five six rules so I cannot dictate all those only thing is what here actually here I'll tell you how to assign priority atomic number you have to compare but I'll suggest you to write down the rules from your friends notes right so condition is I have given you first to third via to now you see to assign RS configuration the fourth thing is what if this circle is clockwise right if this circle is clockwise clockwise then the configuration is our if it is anti-clockwise anti-clockwise then the configuration is s if it is clockwise if it is clockwise once a configuration is R anti-clockwise configuration is s now you see we are we are going to assign the configuration of this molecule correct first of all we have to assign priority Shreya priority will assign by the comparison of atomic number okay so we'll compare the atomic number of all these elements if the groups are attached then we compare the atomic number of first atom here it is carbon right so which one has the highest atomic number obviously bromine so bromine has the highest priority then we have chlorine second then we have carbon third and in the last we have hydrogen fourth that is what the condition we have if the like the least priority group must be into the plane of the paper right so it is already there if it is not there then what we have to do that also we'll see don't think about that now okay what we have to do we have to go one to third first to third priority via second right like this we have to go so when you see this circle is what it is clockwise connect it is clockwise clockwise configuration is R so this compound has R configuration any doubt dies let me know guys if you have any doubt fast is it clear all of you understood where is the deep what I did first of all I assigned priority one two three four then what we have to do we have to go from one to two to three right this is the condition we have from first priority to third priority group I have to go via two right since via two condition we have so we cannot go like this right so first then second and then third this rotation you see this rotation is what it is clockwise so when it is clockwise it is R okay s and plus minus again we cannot say a little plus minus like I said it is experimental D L what I said it is experimental the one which is D that must be plus the one which is L that must be minus okay yeah that's what so plus and minus we cannot say by looking at the molecule there is no theoretical method to find out plus or minus configuration don't worry with that plus and minus they are not going to ask they are they will ask you in the exam the configuration whether it is R or S no it does matter that that's the condition share the fourth priority group must be in the into the plane of the paper must be into the that's the condition I have given you for our for flying which representation or which dash representation got it now I assume that all of you have written this part this part right so I will just erase this first now you see very important you know point we are going to conclude here I have seen this molecule where configuration we have already known which is our we have just now we have concluded that now what we are doing I am just swapping chlorine and CH3 right swap CL and CH3 what we get you see we have carbon chlorine is coming towards the observer which is swapped by CS3 so now this CH3 is coming towards the observer towards us chlorine is here bromine is here bromine is here and we have and we have hydrogen present here right no that's what that's the only representation it should be in the into the plane of the paper right whether you write between bromine and CS3 or bromine or chlorine or CS3 or chlorine that does not make any difference it should be into the plane that is it right so now when you exchange one pair or when you swap CS3 and CL then what happens again what we have to do again we have to assign the priority so we'll assign the priority first bromine is 1 chlorine is 2 methyl is 3 and hydrogen is 4 since the least priority is into the plane so fine that's what the condition we have to follow that's what the condition we have to follow right so we have to go from 1 to 3rd via 2 this is the rotation we have and this rotation is what this rotation is anti clockwise and anti clockwise is S configuration got it got it now next is what again we are swapping chlorine and bromine suppose swap chlorine and bromine when we swap chlorine and bromine what we get you see carbon as CS3 as it is CS3 as it is here we have bromine here we have chlorine and hydrogen we have here again what we do we'll assign the priority so priority is 1 here 2 3 4 1 2 3 via 2 like this we have to go now you see it is clockwise so clockwise is what R R configuration how many of you understood this how many of you quickly you tell me it's already 8 we have half an hour more Fisher also I have to discuss Nikita understood okay wait wait on the road so what what you can observe from these three example you see initially the configuration is R correct when we swap one pair when we swap one pair the configuration becomes S again if you swap one pair from here the configuration becomes R right but when you compare this first and third first and third molecule you compare they have same configuration right and to get third from first what you have to do you have to exchange first of all you have to exchange CS3 and CL and then in CL and VR means when you have to form third from first you have to swap two pair right one is CH3 CH3 and CL you have to swap first and then you have to swap what CL and VR correct now next what we can say if you have to a pair if you swap to pair then the configuration remains same if you swap one pair then the configuration will change from R to S and S to R right so how we can generalize this we can say when we have odd number of exchange means the exchange number is odd then the configuration change from R changes from R to S and S to R but if you have even number of exchange even number of swap then the configuration won't change did you get this right write down this point if you have odd number of exchange if you have odd number of exchange the configuration changes from R to S or S to R this is very important point when the configuration changes from R to S or S to R did you write this point when there is odd number of exchange the configuration changes from R to S and S to R next point if we have even number of exchange like two pair four pair six pair if you are getting exchanged if we have even number of exchange then the configuration remains same it won't change is it clear did you understand this even number of exchange no change in configuration it remains same odd number of exchange the configuration changes from R to S or S to R is it clear all of you tell me fast now you see in all these molecule first second and third the least priority is into the plane of the paper right so what we do what we will do when the least priority is not into the plane of the paper like suppose if I write this example carbon is attached with chlorine here we have hydrogen here we have CH3 here we have bromine can you tell me the configuration of this what is the configuration of this molecule only condition for least priority group is it must be into the plane of the paper that is it that's the only thing you have to keep in mind odd number of configuration changes even number configuration won't change tell me the answer see here what we do when we assign the priority okay when we assign the priority this will be 1 this will be 2 this is 3 and this is 4 now the condition for flying which representation is what the least priority must be into the plane of the paper which is not here right which is not here so what we have to do here this edge or the least priority will swap with the group or atom which is present into the plane of the paper that's the condition we have to satisfy the least priority must be into the plane right so we have to satisfy that condition so what we do we'll swap this edge and and the group which is present into the plane of the paper that is CH3 in this case when H and CS3 we swap so we'll get what we'll get this molecule yes or no when we swap H and CS3 we'll get this molecule right which is our very which whose configuration is our first we swap this find out the configuration if it is our and we know this molecule we get by odd number of exchange from this molecule so if it is our odd number of exchange we have this must be as got it is it clear now we'll see Fisher projection how to assign RS in Fisher projection we have the same thing all conditions are same or Fisher projection only condition is what like we have in in which dash we have the condition that the least priority must be in the into the plane of the paper in Fisher projection the condition is what yes area when the least priority group is into the plane of the paper then only you will get the right configuration you can determine that's not an issue but that will be wrong okay condition we have to satisfy then we'll determine the RS configuration okay so that's the condition in which dash representation similarly in Fisher projection all conditions are same we have to assign priority we have to go 1 to 3rd via to everything clockwise is our anti-clockwise is as everything is same but there we have condition similarly here also we have one condition and the condition you write down condition is what the least priority group write down the condition in Fisher projection the least priority group must be must be on the bottom of the vertical line the least priority group must be on the bottom of the vertical line got it that's the condition now you see in this one you have to assign RS configuration this molecule suppose we have CH2 OH this is CL and this is CH3 tell me the configuration RS right it is up on like priorities what loading is first CH2 OH is 2 this is 3 and this is 4 the least priority always on the bottom of the vertical line so here it is present we have to go 1 to 2 1 to 3rd via 2 like this it is clockwise configuration is R here also the same thing we have even number of swap configuration remains same odd number of swap configuration will change right can you tell me what is the configuration of this molecule configuration of this molecule you tell me see triple bound CH then we have CH3 and we have CH double bound CH2 here we have CH2 CH3 the second one is what the second one is S yeah right whenever you have this kind of ring smaller ring will have the higher priority we have discussed this how to assign priority in this kind of molecule if you see that smaller ring will have the higher priority this is 1 this is 2 this is 3 and this is 4 right 4th one is here and second one is here so we'll exchange these two when you exchange these two will have 1 2 and 3 1 2 and 3 is clockwise clockwise is R so this one will be S here you see again triple bond will have the highest priority then we have double bond and then we have single bond and then we have only CS3 here again we'll exchange these two this one will also be S is it S no it is not S it will be R I think right so two here three here so one two and three it is anti clockwise S so this one will be R tell me is it clear understood all of you understood any doubt on the another term write down enantiomers enantiomers definition of enantiomers write down these are the compounds which are non superimposable which are non super impossible mirror image of each other these are the compounds which are non superimposable mirror image of each other are called enantiomers non superimposable mirror image of each other okay enantiomers are what non superimposable mirror image of each other non superimposable means what the molecule next you write down the next term or one example we'll see and then we'll see the next up like you see the example of glyceroldehyde we have C OH CHO and CH2 OH now the name of this compound is glyceroldehyde write down the name glyceroldehyde this is glyceroldehyde right now the mirror image of this if you see mirror image you are taking the mirror image that would be this see first of all the bond which is going into the plane and out of the plane that won't change because we have mirrored this side right we have mirrored placed adjacent to this molecule right so this two bond will not change OH will be coming out of the plane H will be going into the plane this side we have CHO and this side we have CH2 OH now you see this molecule these are non superimposable mirror image non super impossible how you take this molecule and place onto this molecule the OH OH will coincide H H will coincide but this CH2 OH will be placed on CHO and this CHO will be placed on CH2 OH just keep this is put take this molecule put it over here then find OH and H will coincide but CH2 OH and CHO will not coincide here that's why this molecule are non superimposable right and they are non superimposable mirror image hence they are enantiomers of each other enantiomers of each other got it now can you tell me the configuration of these two molecule what is the configuration of the both molecule what is the configuration of first one tell me what is the configuration of the second one tell me two different answer most of you are getting R anyone is getting S anyone is getting S most of you are getting R R is the right answer R is the right answer you see here again the first priority is this OH second will be what second will be CHO third will be this and fourth will be this again it is which that's representation so we have to go 1 2 third via 2 and this orientation or this rotation is clockwise this configuration is R when this is R when you compare this will have odd number of swap here this must be S right so what we can conclude here the molecule which are enantiomers of each other they must have same molecular formula but different configuration right right on this to the molecules which are enantiomers of each other the molecules which are enantiomers of each other they must have same molecular formula but different configuration but it understood guys same molecular formula but different configuration any doubt guys any doubt did you understand this wait are they there wait wait wait so enantiomers if you have to find out right what we'll see the molecule must have same molecular formula but they must have opposite configuration right this is very important now this molecule since both are optically active one so one will be D one will be dextro but other one will be levo right we do not know whether this one is dextro or this one is dextro but if it is D this one is L if it is L or if it is D this one is L understood that we can only decide by the experiment or the rotation of plane polarized like right so we cannot say that the which one is D which one is L for that we have to perform the experiment but one thing is very sure that if this one is D this will be L or if this one is D this is this will be L right so they won't ask you the question of dextro and levo right but they'll ask you R and S they can ask you enantiomers right now these two are optically active compound if this rotates the plane polarized like clockwise by some angle theta this will also rotate the plane polarized light by the same angle theta but the direction will be anti-clockwise because one is R other one is S one is D so other one is L so rotation will be opposite rotation of plane polarized light for this will be clockwise suppose for this will be anti-clockwise angle will be same right so when we take the equimolar mixture of these two right then that compound is said to be the rest mixture right and suppose we have ten moles of this and ten moles of this then whatever rotation this molecule will bring into the plane polarized light the same rotation this molecule will bring but in opposite direction so eventually the ray will come out without any deflection and hence the mixture will be what hence the mixture will be optically inactive so this kind of mixture or equimolar mixture we call it as Resnick mixture right down right down this particular definition of Resnick mixture is the equimolar mixture Resnick mixture definition you write down it is the equimolar mixture of the two compounds which are enantiomers of each other did you write this did you write this Resnick mixture it is the equimolar mixture of the two compounds which are enantiomers of each other equimolar mixture of the two compounds which are enantiomers of each other but it any doubt guys till here any doubt all of you answer fast and you know that is experimentally have to perform the experiment okay so that you don't have to bother with because you are not they are not going to ask you that question you have to perform that experiment in the lab okay you cannot do that at home right so don't worry with that relative things we can say we can say if it this is D this one is L if this is L this one is D but to know exactly which one is D or L either you should refer some book whether the experimental thing is written or you have to perform the experiment right one last time you write down that is die stereomers die stereomers are the stereo isomers definition you write down die stereomers are the stereo isomers die stereomers are the stereo isomers whose molecules are not mirror image of each other stereo isomers or simply you can write down di-stereo-mers or di-stereo-mers which are not mirror image of each other, simply like this is better, di-stereo-mers or di-stereo-mers which are not mirror image of each other, okay. The two molecules which have same molecular formula but they are not mirror image of each other are called di-stereo-mers, okay. All the geometrical isomers, cis-trans, syn-NT, easy are not easily cannot say, easily cannot say, write down all geometrical isomers in bracket you write down cis-trans, all geometrical isomers in bracket you write down cis-trans are geometrical isomers of each other. You can draw one cis and trans molecule which has the same molecular formula, you must see they cannot be the mirror image of each other under any circumstances, okay. So all the molecules, all the geometrical isomers in bracket you write down cis-trans are di-stereo-mers of each other. One important point, the last point for today you write down, di-stereo-mers can be of any combination, the last point for today, di-stereo-mers can be of any combination, can be of any combination. Any combination means what, when we explain di-stereo-mers we explain in a pair, right, for a pair of molecule, we say these two pairs are di-stereo-mers of each other, these two pairs are enantiomers of each other, right. So any combination means what, both pair or both molecule in a pair can be optically active, right, both can be, both can be optically inactive or one can be, one can be optically active and other optically inactive. Any of these three combinations possible for di-stereo-mers, this on the basis of this also they have asked question and the first point that I have given you geometrical isomers, cis-trans are di-stereo-mers of each other on the basis of that also they have asked question, okay. Any doubt till here, let me know, let me know if you have any doubt. So we'll, we'll, we'll wind up the class here, okay. You can solve the questions of studio isomerism, it's given in any of the books, I'll share the assignment with you tomorrow not today, okay. But you can revise all the things that we have discussed, every each and every line is very important, I have been very specific, right, towards, you know, the each and every concept, okay. I hope all of you have written all the, you know, the small, small concept that I have talked today, okay. You must revise that, solve any book if you have, solve questions of studio isomerism, you will, you can, you can solve those questions, I, I assume that, right. So I will share the assignment on this, okay, tomorrow then you can solve that also. No doubt, so can you wind up the class here, okay guys, so, okay guys, let's see you in the next class, take care, bye-bye, thank you for joining.