 Hello guys, good evening. Can you hear me? OK, so where were we in the last class? Just one minute. Yeah, so I think we in the last class we were talking about optical isomerism, right? Cardinal center, yeah. OK, so we'll continue. Like I said, we discussed about carnal center. We have discussed the phenomenon of optical isomerism. What is the phenomenon of optical isomerism? That when a plane polarized light passes through an organic compound, if it rotates in any direction, clockwise or anticlockwise, we said that the molecule is compound is optically active, correct? And all these things is done in our device called polarimeter. OK, right? If it is clockwise, then we call it as dextrorotatory, OK? That is called as plus. If it is anticlockwise, it is levo-rotatory, OK? That is L minus. This thing we have discussed. Now, these are the practical thing. If I ask you whether the given compound dextrorolivo, you need to perform that experiment and check whether the molecule is rotating in clockwise or anticlockwise or not, OK? Plane polarized light, OK? But theoretically, how do we find out a given compound is optically active? So for that, we have some, you know, a way through which we can find out. So to understand that, we need to understand certain terms, like I said in the last class. And first term, we discussed last class, that is chiral center. Chiral center is what? It is an SP3 hybridized atom with four different groups attached to it, OK? SP3 hybridized atom, four different atoms or groups. If carbon is the atom, is the chiral center, then the carbon is called chiral carbon, OK? The next is one node you write down here. One node you write down. If the molecule has only one chiral center, it is optically active, optically active. Remember, I'm not saying it is a condition we have for a compound to be optically active. Condition is not this. It is observed that if the molecule has only one chiral center, it is optically active, not the condition, OK? So next point you write down, like if it is one, then definitely it is optically active. If it is more than one, like two or three if it is present, then we cannot say with respect to the chiral center, OK? Keep that in mind. I'm repeating this thing because with this particular information, you can solve most of your questions in the test or in the exam, OK? Only one chiral center, optically active. No doubt about it, OK? If it is more than one, then we cannot say. We have some different way for that, OK? Again, one more thing. Presence or absence of chiral center is not any criteria for a molecule to be optically active. Condition or criteria is not chiral center. But we observe this, right? So next point also you write down. Presence, absence of chiral center is not any criteria, criteria for a molecule to be optically active or molecule to be optically active, OK? One exception we have in this. If you talk about tertiary amines, for example, if you have nitrogen atom with three different alkyl group attached to it, like methyl, ethyl, or propyl, R1, R2, R3, are three different alkyl group, OK? Alkyl groups are what? Like methyl, ethyl, propyl, like that, OK? So if you write R in organic chemistry, R always represents alkyl group. How do we get alkyl? You remove one hydrogen from alkane, you will get alkyl, OK? If you write down R dash, R dash or R1, R2, R3, R3, nothing but the alkyl group. But it just means that these two are not the same alkyl group. I mean, suppose if it is methyl, then it is definitely not methyl. It can be anything else, but not methyl, OK? So R basically represents alkyl group in organic chemistry. How do you get alkyl? You have alkyl, remove one hydrogen from that, it becomes alkyl, OK? So exception is what? If nitrogen has three different alkyl group attached and the fourth one is the lone pair, like this. Fourth one is the lone pair. Then obviously this compound, if you look at this, nitrogen has four different groups attached to it. And this nitrogen is also sp3 hybridized. So if you look at the definition of Karel center, it is a Karel center. sp3 hybridized, four different group attach, R1, R2, R3, and this lone pair also. It's all four are different. So logically it should be, it is a Karel center and we have only one Karel center in this. If the molecule has only one Karel center, it must be optically active. I've given you this molecule has only one Karel center. It is optically active, always, right? But this is an exception, OK? This molecule, it should be optically active, but it is not. It is optically inactive, optically inactive. Keep that in mind. Tertiary amines are optically inactive, OK? The reason for this is amine inversion that we'll discuss later, but this term you keep in mind, amine inversion. We'll discuss this later in the organic chemistry, but you keep that in mind if the molecule has only one Karel center, it is always optically active, except this tertiary amines. Is it clear? Yes? OK. So it's very important, like whatever I'm saying, you listen to me very carefully, OK? It's very important. Even a single word I'm using here, it's very important, OK? Like I said, Karel center, only one Karel center, optically active, but condition is not Karel center, OK? Condition is not the Karel center for optical activity, OK? You see this example, do we have a Karel center in this molecule? Do we have a Karel center in this? What is Karel center? Apply the condition for that and check. We do not have. How many of you are saying yes? Take your time, yes, take your time. OK, all of you are saying no. See, this carbon has how many hydrogen? This carbon has how many hydrogen in the terminal one? This carbon has three hydrogen, right? One, two, three. So any two atoms if it is same, then it cannot be Karel carbons. This is not a Karel carbon, definitely, OK? However, it is sp3 hybridized atom, right? Similarly, this carbon is also sp3 hybridized atom, but two atoms that is hydrogen are same here, cannot be Karel. And similarly, this is also not Karel because all three hydrogen atoms are present and it is optically active. But if you look at this carbon here, this has one hydrogen, one BR, one ethyl group and one methyl group. So sp3 hybridized atom with all different, all four atoms of groups are different. So this molecule has one Karel center, which is nothing but this particular carbon atom. Karel center, we represent by star like this. Is this clear? Right? So one Karel center we have into this. And if it has only one Karel center, it is optically active. OK, molecule is optically active. Condition is not that, but yes, we can say. If you have this molecule, do we have Karel center in this? We do not have? Yes, guys, respond first. Do we have Karel center in this? What about it now? Do we have now? We do not have? Do we have now? No. If you place a double bond here, then now we have. Which one? This one, the center one. One is this group, C double bond C. Other one is C C single bond, ethyl group and methyl group. And fourth one is hydrogen. All four groups are different, you check now. Yes or no? I'm talking about this carbon. This is a Karel carbon, this one. Right? This is a Karel carbon. For this one also, Karel carbon is not possible. If I place a double bond here, one more methyl group, do we have a Karel carbon now? Yes or no? Yes, we have. Okay, yes, we have a Karel center. What about it? Do we have a Karel center now? Do we have a Karel center now? Yes, we do not have because this carbon now, it becomes SP, too hybridized, okay? If you add a chlorine here, then, do we have a Karel center? Yes, we have a Karel center and that Karel center is this carbon. Could you find out Karel center if it is given in the book, like in the question? Find out easily. SP3 hybridized, four different atoms are both present. Okay, one example we'll see. If you have a ring structure like this, you see. Okay, here we have CS3 present. Then how do you find out, one chlorine we have here, for example, how do you find out that this molecule has Karel center or not? Okay, so what we do here, obviously you see this carbon has two hydrogen, we cannot think about these two carbon, okay? For this carbon, if you see, this carbon, it has one hydrogen here, one methyl group, but other two groups, we are not sure what is the other two groups we have. So if you have a ring structure, then what do you have to do? You just see this path, like clockwise. We have CS2, CS2, CCL, anti-clockwise. So anti-clockwise and clockwise, if you go, you see at the first, if you go like this, then the third carbon has one chlorine. If you go by this, we have first carbon atom which contains chlorine. So we say this path and this path is different. Hence we have this side different group present, this side different group present. Hence this molecule, this carbon is a Karel carbon. Like this we do. Similarly, if you think of this carbon, this is also a Karel carbon. If you talk about this molecule, here we have CS3. Yes, fine, we have one hydrogen present here. These two group are different, but this path, CCC and this path, CCC are same. So this has no Karel carbon, clear? Means for a ring structure, you just need to follow, go clockwise and anti-clockwise and check whether the path is same or not. Write down, in a ring structure, in a ring structure, in a ring structure, comma, in order to determine, in order to, in order to determine whether the atom is Karel or not, whether the atom is Karel or not, will follow the ring in clockwise and anti-clockwise manner. Will follow the ring in clockwise and anti-clockwise manner. An anti-clockwise manner. If the two paths are different, if the two paths are different, then we say, then we say the carbon atom has different group attached. Hence the carbon atom has different group attached and we can say the carbon is the Karel carbon. And we can say it is a Karel carbon. And we say it is a Karel carbon. Okay, the next term we have, Karelity or Karel molecule. Now, first of all, you must understand this. Karelity has nothing to do with Karel center. First thing is this one, right? Karelity has nothing to do with Karel center. Karelity is different. Karel center is different. Okay, Karelity is what? Karelity is the property of a molecule. It is the property of a molecule. Karelity is the property of the molecule. Which molecule are said to be Karel molecule? The one which shows this property, Karelity property, okay? So when the molecule, molecule which can't be divided into two equal halves is called Karel molecule. Karel molecule. This means what? There is no symmetry in the molecule. Karel molecule, there should not be any symmetry. Okay? Karel molecules, next point, write down. Karel molecules, non superimposable, non superimposable, non superimposable, mirror image of each other. Non superimposable, mirror image of each other. What is superimposable? Superimposability is a 3D phenomenon. If two molecules are superimposable, superimposable, they looks like, they looks like single molecule. Same molecule, single molecule, anything you can say. Karel molecules are non superimposable. Like for example, I'll tell you one example here. Did you copy this, all of you? Yeah, once again. Finished copying this. Now you see superimposability, you can understand like this. Suppose we have a carbon atom and here we have chlorine. This is hydrogen, CH3. And here we have, we can say bromine. Three dimensional figure it is, okay. This means what? This two bond is in the plane of the paper, like this two bond in the plane, right? Or we can say this BRN-CS3 with this carbon, it is present in the plane of the screen, you can say. Okay, in the plane of the screen. CL is coming out of the screen towards me and H is going away from the observer, away from the screen, behind the screen it is going, okay? Understood this? What is this representation means? Tell me, clear? Guys, H is going behind the plane, like behind the screen and CL is coming towards the observer flying away the representation, yeah, the same thing. Okay, so if you take the mirrored image of this one, then what happens you see? If you take the mirrored image, then this carbon will be here and this edge is going into the plane. We have chlorine here, CS3 this side and BR on the top, sideways we are taking. So if you try to place this molecule on this, it is said to be superimposable, then identical atoms or groups overlaps, like suppose what we'll do, we'll just pick this up and we place this over this molecule. This carbon will place over this carbon, right? This carbon will place over this carbon, right? A cube per a correct angle, right? So if you place this carbon, obviously this BR-BR overlaps, CL-CL overlap, right? But this edge will come this side and this CS3 will come this side. So point I'm trying to make that all these identical atoms or groups are not overlapping. And we say these two molecules are non-superimposable. Is it clear? Non-superimposable, tell me. The superimposability is this, it's a 3D phenomenon, okay? If one molecule you place over the other and the identical groups or atoms are overlaps, then the molecule is said to be superimposable. It looks like single molecule, one single molecule is present. You cannot differentiate between the two. Hands are superimposable. Do you have NCRT with you now? Give me one minute. Just a second, guys. Just a second. Just give me one minute, guys. I want you to think this way. Guys, just give me one minute. Don't leave the meeting. I'm just coming back, okay? One link I have to share. Just one second. Hello, can you hear me? Yes. Oh yeah, I'm done. Yeah, actually one urgency has come. Yeah. So I was, yes. So I was talking about superimposability, right? This molecule is non-superimposable because the atoms, the identical atoms or groups are not overlapping actually. Similarly, if you take the example of hand also, it is also non-superimposable because if you place this over this, obviously this thumb is this side and this thumb is this side, okay? So we can place like this only. We cannot flip it. Keep that in mind, okay? Flipping is not allowed here, okay? Flipping is not allowed here. So hands are non-superimposable. Is it clear to all of you? So what we had discussed just now that carol molecules are non-superimposable to their mirror image, okay? So if you have a carol molecule, consider the mirror image of that and check whether it is superimposable or not. For superimposability, we can rotate the molecule this way. Like we can rotate it like this. We cannot flip it. Keep that in mind. Out of the plane rotation is not allowed. So we can rotate the molecule this way. This way we can rotate and we try to place it over this so that identical atoms overlaps each other. But like I said, we cannot flip our hand like this. We cannot flip the molecule this way, okay? Like the car is staring, right? Like that we can rotate the molecule in the same plane we can rotate to check the superimposability. Is it clear? Any doubt in this? Yeah, keep that in mind. Yeah. So next point you write down. Carol molecule, carol molecule is always optically active. Is always optically active, correct? Means carolity is the necessary condition for a molecule to be optically active. Carolity is the necessary condition for a molecule to be optically active. If somebody asks you what is the condition for a molecule to be optically active, right? What is the condition for a molecule to be optically active? Your answer should be the molecule should be carol in nature. What is carol? Carol molecule is a molecule which cannot be divided into two equal halves, okay? Correct? Means there is not any plane of symmetry or center of symmetry present in the molecule, okay? Again, I'm repeating. Carol center and carolity, okay? Are different things. Carol center, carol molecule is completely different. Carolity is the property of carol molecule. Means of the molecule is carol. It means it shows carolity. Carolity is what? We cannot divide the molecule into two equal halves. Clear, no? Fine. One last point in this, you write down. If a molecule is carol, it doesn't mean it has a carol center. If a molecule is carol, it doesn't mean it has a carol center. Any doubt, Illya? Last class, we'll talk about stereo center also. Remember that? Yes. Last class, we'll talk about stereo center. Stereo center and carol center are two different things, again. Carol center is always SP3 hybridized, but stereo center can be SP2 or SP3. Okay? Keep that in mind. Carol center is always SP2, carol center can be SP2 or SP3. Stereo center, sorry, other way, my bad. Stereo center is always SP2 or always SP3. It can be anything, SP2 or SP3, but carol center is always SP3. It cannot be SP2. Stereo center is different. Carol center is different. Right? Stereo center is the center across which if you interchange the atoms or molecules, you will get stereo isomers, cisterns, right? So these are different, different terms we have. You must keep in mind. Now, the representation of molecule, you see two, three types of representation we have. Okay. Two important one we'll discuss over here. Write down the heading, representation of molecule. Representation of molecule. The first one we have, the first one we have a flying wedge representation that we already know. I'll just draw one, flying wedge representation, wedge representation. And this is like the one we have taken the example. Suppose we have carbon, carbon, and then we have one atom attached over here, T. This is Q. This is R coming out of the plane away from towards the observer and going into the plane away from the observer. S. This is flying wedge representation. As you know, this dark wedge, dark wedge is coming out of the plane, off the plane. Like here you can say out of the screen, it is coming towards me, towards you. This one is into the plane, means away from the observer, behind the screen it is going. And this too in the plane of this, in the plane, means in the plane of this screen. Bond angle is 119. This is flying wedge representation. Similarly, the second representation we have, probably this you haven't done before, that we call it as fissure projection. Fissure projection, more important one is this, fissure projection. Fissure projection is, suppose we have a molecule for example, butane 2,3 diol, butane, butane 2,3 diol. Okay, so its structure would be what? Like this, OH, isn't it? This is the structure. How many carol center we have in this? Could you tell me the number of carol center in this molecule? Yes, right? And that is, and that is this carbon. Yeah, that's right. Two carol center we have this carbon and this carbon. Now, since it has two carol center, so we cannot say this molecule is optically active or not. If it is only one, then yes, it is optically active, okay? So if you look at this carbon, what all groups are present over here? One is this, other one is OH, other one is hydrogen here, and other one this bigger group, right? Similarly for this one also you can think. So if you want to represent the, you know, the fissure projection of this molecule, fissure projection will go like this. We'll draw fissure projection this way, okay? In which this is a carol center here, like this line I'll represent it vertically, let's say. This is a carol center, this is another carol center, right? So we can place OH here, other one is H, OH here, other one is H, and these two are CH3. So this is the fissure projection of butane 2,3 diol, which contains two carol carbon here. Now it is not that simple, okay? This molecule is non-planar, you keep that in mind. It is a non-planar molecule. Looks like it is in the plane of the screen, but it is not, okay? Let me explain you this molecule here. This molecule, if you look at the 3D representation of it, this molecule is, can be represented this way. The vertical lines, this one, and this one is going into the plane actually. This carbon, the carol carbon, I'll just write down here, this is the carol carbon, and the vertical bond is going into the plane, so it is a dash here. And this, all the horizontal lines are coming out of the plane towards the observer like this. All the horizontal lines, okay? So it looks like simple and planar, but it is not actually. Tell me, any doubt you have? So this is OH, this is CH3, this is H, this is OH, this is H, and this is CH3. So what do you have to keep in mind? In Fisher projection, the horizontal bond is coming towards you, okay? It's like a V over here, it's like a V over here. Like if you can understand this one, both hydrogen and OH, it is coming towards you, right? Towards you. Same thing we have here, V, and then this is going into the plane, this is going into the plane, and this is along the plane of this, okay? Is this clear, understood this? Yes, tell me guys, clear? So whenever you look at this molecule, always keep that in mind, it is non-planar molecule, looks like planar, but it is not. Yes, all of you, you can type in Y simply if you understood it. Quickly guys, right? So this we call it as Fisher projection, okay? We can have one, this thing also, one carol center also possible. Like for example, you have a compound, say this, this kind of compounds also possible, right? This is suppose OH randomly, I am just placing one atom here, this is CS3, and this is CL suppose we have. This means what? That this carbon atom is a carol center here, carol center. So if you represent the criteria representation of this would be what? Carbon has this OH and H, it is going this way. OH here, H is this, and this two are into the plane. This is CL, this is CS3, this is what it means, okay? We have one more representation formula, and that we call it as saw horse, okay? Saw horse, new man, two more we have. Okay, we'll discuss that later. Which orientation, Anj? Saw horse, new man we'll discuss later, okay? Let it be now, yeah. Which orientation you're talking about? This one, in this molecule. See, it's simple, the horizontal lines are coming towards you, it's not planar. This OH is above the plane of the screen, towards you. This OH is towards you, this H is towards you, this OH is towards you, and this H is towards you. And this vertical line, this one and this one is going into the plane away from you. Means this CS3 is behind the screen, this CS3 is behind the screen. This four atoms are towards you, out of the plane. Towards, out of the plane, towards the observer. That's why you see the horizontal lines are this thick wedge and vertical ones are dash wedge we have. Correct? This two types of representation you must understand, okay? Now, like I said, that the molecule has only one Karel center, it is optically active, okay? DNL configuration we have seen, correct? That is a practical method, okay? You need to perform that experiment to understand the DNL configuration. But theoretically, we can also assign configuration to the optically active compound. And this configuration, we call it as R and S configuration. This is a theoretical method we can discuss. This R can be D also, R can be L also. We have no relation between D, L and R, S configuration. R, S configuration is completely theoretical, okay? And D, L configuration is practical, okay? Practically we need to do that. What is this R, S stands for, okay? R stands for rectus, Latin word rectus we have, which means right. S stands for sinister. Again, a Latin word sinister, which means left, okay? Right and left. So if you want to define or decide here that what is the configuration of the given molecule? It is R or S. Then how do we do it? So to do this R and S configuration, you can have two different representation of molecule, right? First one, if you have flying wedge representation, we have a condition here you need to understand. If you have a flying wedge representation, so in flying wedge what we'll do, the least priority group, just you listen to me for like 10, 15 minutes, you'll understand what I'm talking about. Least priority group must be, must be placed away from the observer. Means it should be into the plane, away from the observer into the plane. Like for example, you see, we have a compound say carbon, okay? So we have here OH, we have CH3, we have CL and we have H. So what we'll do to assign RS configuration if flying wedge is given, okay? Then we'll assign the priority to this atoms or groups attached to it. We know how to assign priority CIP group, okay? So chlorine has the maximum priority one, then we have oxygen two, then we have carbon three, and then we have hydrogen four. So the hydrogen has the least priority, it must be present into the plane, means with the dashed line, it must be there. If it is not, then also we can do, but that we'll discuss later that what to do into that one, okay? This is the first condition we have when we have flying wedge representation, is it clear? Understood what I said? Yeah, if you have Fisher projection, right? In Fisher projection, what happens? For example, you see the least priority group, yes or no, one second, one second. Least priority group must be present the bottom of the vertical line, for example, if you have this one call center, right? So we have H here, chlorine, OH, and CS3. So if you assign the priority again, it is one, it is two, it is three, it is four. So fourth is the least priority, must be present on the bottom of the list, it must be present on the bottom of the list, and the majority must be present on the bottom of the vertical line. This is the condition we have to assign R and S, okay? Yes, did you understand this condition? Just try to understand what I'm telling you. I know you have a lot of doubt and question in your head. Okay, I can tell you once we finish this, you'll understand, okay? Obviously some practice will be required, but all your doubts will be taken care of, just give me some time. Condition, all of you understood? Like in fish operation, what is the condition? And flying fish, what is the condition we have? Yes, could you respond? Are you there? You had your lunch today? All of you had your lunch? Yes, feeling sleepy? Althe is feeling sleepy. Is it? Okay, Oshik is hungry. One pizza, Oshik. Okay, so when we have physical loss, I'll do the, we'll have the party for that. Pizza party we'll have in physical loss. But right now you need to focus on this. Or our next session most probably. So, okay. No, not next session. Next session means your 12th standard. That is the next session. Okay, fine. So all of you understood the condition, right? Now to assign R and S configuration, first of all, you have to assign priority like this. One, two, three, four, like I did. Okay, now after this, what do you need to do? You just need to go from first priority group. You need to go to third priority group. First to third, you need to go. Abduho, there are two ways to go from first to third. Okay? One is what? You can directly go from here to here. Yes, one to three, I can go directly like this. But this is not right. Okay, we won't go like this. We will go from first to third via two. Okay, so first we'll go from first to second, second highest priority. And then second highest priority, third highest priority. Like this will go. So what is the correct path we have here? The correct path that rule that we have here to understand the configuration of the molecule is we'll go from one to two, right? And then two to three, this is the correct path we have. So we have some orientation, you see. If this is clockwise orientation you are getting, one to two, two to three, if it is clockwise, then it's said to be R configuration, okay? If it is anti-clockwise, then it is S configuration. Tell me, did you understand this? Clear? R S configuration. So first of all, you have to assign priority on the group, one, two, three, four. And then first to third you have to go via second one. Clockwise if it is getting, then R, anti-clockwise, then S. You must be thinking of, so what happens if the least priority group is not here? Suppose hydrogen is placed over here, then what happens, right? Correct? The same thing we'll do in this case also, like you see. Condition is the least priority must be in the bottom, which is this, like for this case we have. So one to third we need to go via two. So one to two, two to third. It is what? It is clockwise, so configuration for this is R. Is it clear? Tell me, it's very important. Yes, no doubt. Now let's try this question. You need to assign priority, now you need to tell me the configuration of this molecule. Okay, so you see this. Here we have H, CS3, CL, BR. H, CS3, CL, BR. H, BR, CS3, CL. Okay, CL, H, CS3, BR. Okay, H, CS3, CL, BR. Tell me the configuration, one, two, three, four, five. This is first, second, third, four. RS configuration you need to find out. Yes, done guys. See, I'm doing this. First of all, tell me the priority here. What is the priority of bromine in this? Priority of bromine, you can write down the number. Priority of bromine. We'll compare the priority with atomic number, right? So bromine has the highest priority, one. Then we have chlorine, second priority, and then we have carbon, third priority, and hydrogen, fourth priority. Any doubt in this? Yes, any doubt? Okay, and then what we need to do? We need to go from first to third via second, so this is the path. Clockwise or anticlockwise? Clockwise or anticlockwise? Obviously it is anticlockwise. Anticlockwise is S configuration. Now, we'll do this one in the last. If you assign the priority here, or I'll take one more example, just a second. Before doing this, I'll take one more example. Suppose we have this one. Okay, we have BR here. We have CS3 here. We have CL here and H here. Could you tell me the configuration of this one? R or S? This one first you do. R or S configuration? You're getting R, okay? So the priority would be same. We know the priority. This is the fourth one. This is first priority. This is second priority. This is the third priority. So one to third via two. The path is this. One to third via two. It is clockwise, so R configuration, okay? R configuration. Now, actually, these two are very important, okay? Now you just listen to me what I'm telling you. If you compare this molecule from this, how do you get this molecule from this one? Can we say if we exchange the position of CS3 and BR, exchange CS3 and BR, we'll get this, isn't it? Yes or no? If you can respond, that would be better for me. I can understand. I will understand that you are getting it. Exchange CS3 and BR, you'll get this, fine? Yes. This means if we exchange one pair, one pair if you exchange, you will get configuration, means S converts into R. One pair if you exchange, right? From here, CS3, BR, you'll get this. Now, could you find out the configuration of this one, this molecule? From here to here, we are going. Now we are exchanging the position of what? CL and CS3, correct? Tell me the configuration of this molecule. This is obviously one. This is two, this is three, and this is four. So one, two, three via two. So it is anticlockwise S, okay? Now you see this three molecule here. If you have only one pair exchange from this one, references this molecule you consider, only one pair exchange if you have, then the configuration changes from S to R. If you have only one pair exchange here, configuration changes from R to S, isn't it? Yes, guys, one pair exchange here, only one exchange S converts into R. From here also, CS3, CL, you get exchange, so R becomes S. So what we can conclude here is, the conclusion from this is, like if you have odd number of exchange, odd number of exchange, like one pair is getting exchanged, three pair is getting exchanged, or five, seven like that, odd number of exchange, configuration changes from R to S, or S to R, anything is possible here. Okay, this is the first conclusion we have. What is the second conclusion here? Now, here you see, here you see, from this to this one pair exchange, this to this one pair exchange. So if you compare this to this from first to third, like this to this, then we can have one pair here, one pair total two pair of exchange we have here. If there is exchange in two pair, then the configuration does not change, S is S only. So the second conclusion we have here, odd number of exchange gives you change in configuration, but other hand, if you have even number of exchange, even number of exchange, then there is no change in configuration. Can we say that? Did you understand this? Yes, could you respond? Odd number of exchange, there will be a change in configuration if it is R, R becomes S, S becomes R. But if you have even number of exchange, then there is no change in configuration. So if you get the molecule in which the least priority group is not into the plane of the paper, right? Or not at this position in Fisher, because that is a condition. This position is not present. Then we can use the help, you can take the help of this two conclusion in order to find out the configuration of the molecule. Like for example, you see this. In this molecule, what happens, the second one we have here, if you assign the priority one, two, three and four. So ideally the condition is that this hydrogen should be here and then we can say it is R or S, but that is not the molecule here. Now, to get the configuration of this molecule, what we do, this position, listen to me very carefully here, okay? This position, whatever group is present over here, if it is not the least priority group, then we exchange the position of this group with the group or atom which has the least priority. Which has the least priority. So what we'll do here, since we have CS3 in the bottom and H we have here, so what I'll do, I'll just do the changes over here only, because I don't have that much space, so I'm just doing some change over here. Just you, we'll come back to this question, okay? You can let me know, like the position of these atoms later on, okay? But what I'm doing, in order to get the configuration of this molecule, what I'll do, I'll just change the position of CS3 and H, exchange the position of CS3 and H in fact, okay? So H we'll have here and CS3 we'll have here. And then this CL and BR, you let it be as it is. You know the priority here? Priority is one, two and three. So its priority is one, two and three, so it is what? It is anti-clockwise, it means this is S. If this is S, then what should be this? Yes, what should be this molecule, the priority? If it is S, and if you exchange, we are going ULTA now. Now if you exchange this CS3 and H, we get this. It means only one exchange we have. So if it is S, this must be R. Like this, we'll find out the configuration of this molecule, okay? Or what you can do, you see, suppose you get this two molecule in the exam, right? This two molecule in the exam. If you get only this, then you have to do this one. But suppose all these molecules are given, one and two are given. Then we know the priority of, we know the configuration of this one because hydrogen is in the bottom, it is S. Now how do we get this one from this? By exchanging the position of H and CS3? So one pair we are exchanging, only one exchange is there. So if it is S, this must be R. Any way you can do this. Could you tell me what is the configuration of this molecule, this one? Configuration of this molecule, it is R. I am comparing with this one, you see. With this one, I'm comparing, suppose. We have CS3 in the bottom, fine. H on this side we have. So first of all, what we do, we'll exchange H and BR. So we'll have BR here and H here. And then we'll exchange BR and CL. So from this to this, if you want to go, you have to do two exchange, even number of exchange. So if it is R, this one is also R. Now you can ask your doubt. Now you can ask your doubt. Tell me. Any doubt you have, you can ask. Any doubt, guys? Tell me. No doubt, then I'll move on. Either makes the bottom one least priority or do exchange on existing molecule. Yes, we can do that. We must fulfill the condition that we have. Okay, the group which is present on the bottom of the vertical line, it must have the least priority. If it is not, then we can exchange with the least priority group and then we can find out the configuration. And accordingly, whatever number of exchange we have ordered even, we can say the change in configuration here. Yes, Anusha, go ahead. Okay, fine then, okay. So it's not that tough, okay. Only thing that you can make mistake is in assigning priority. So keep that in mind how to assign priority and then R and S configuration. Again, I'm repeating myself. This R and S has nothing to do with D and L configuration or plus minus configuration. R can be D, R can be L, S can be D, S can be L. Anything is possible. D L plus minus are practical thing which we used to do in polarimeter. R, S is the theoretical one, okay, fine. Now in this, we have two, three terms that we use in the exam also to ask all these questions, okay. So we have two, three terms here, okay. The first term you write down enantiomers, heading right down, enantiomers, enantiomers. These are the compounds, right on. These are the stereo isomers. These are the stereo isomers, stereo isomers which are non-superimposable mirror image of each other, isomers which are non-superimposable, superimposability you know, non-superimposable, mirror image of each other, mirror image of each other are called enantiomers. And this phenomenon is enantiomerism, okay? Like you see the example of cis and trans. We have carbon double bond carbon HCl HCl. If you take mirror image of this carbon double bond carbon, we have H this side, Cl this side, Cl and H this side. So obviously if you place this molecule over this, this is not coming out to be superimposable. Like it is not getting superimposable here. And hence these compounds are enantiomers of each other, okay? Enantiomers. If you take an example of glyceraldehyde, glyceraldehyde the molecule, the formula is this. We have C on the top, we have CH3, CH2 OH and then we have OH and H. This molecule is glyceraldehyde. If you take the mirror image of this, then this carbon will have CH3 OH, CH2 OH, okay? The name of the compound is glyceraldehyde. Remember that. And these two are non superimposable. So enantiomers of each other, right? Now a few properties you must remember here. If you talk about this one, this is optically active. This is optically active, right? So we say enantiomers are optically active, optically active, okay? If this molecule has R configuration, then this one is S or we can say if it is D, then this one will be L. Which one is D? Which one is L? I do not know. We do not know, okay? To understand this D and L configuration, we have to do that experiment, that polarimeter and all. But if it is D, this one is L or if it is L, then this one is D. Both will not have the same configuration. That is important, right? That means what? If it rotates clockwise, PPL, it rotates PPL clockwise, then this will rotate anti-clockwise with the same angle, right? So all enantiomers are optically active. Next point to write down. If we take equal mixture of, if we take equimolar mixture of mixture of both compounds, then the mixture is called, then the mixture is called resmic mixture. Very important term, resmic mixture. Resmic mixture is always optically inactive, optically inactive. What does it mean? Why it is optically inactive? You see, suppose this is A we have and this is B. What we have done, I have taken five moles of A equimolar mixture, right? So five moles A and I have taken five moles B, five moles B. So if you take equimolar mixture of A and B, then A will rotate clockwise, for example, with some angle. B rotates with the same angle, anti-clockwise direction. So we do not observe any deflection in PPL eventually, okay? Suppose the PPL comes like this, this will rotate clockwise like this and when it comes in the contact of B, this will rotate anti-clockwise at the same angle and finally it comes out undeflected. Hence the mixture is optically inactive. However, individually if you see the molecules are optically active in nature. Is it clear? Sometimes in the question, they'll ask you to find, I'll identify the given compounds are enantiomers or not. The best way is what? That you find out the configuration here, right? Find out the configuration. If this configuration is coming out to be R, then this must be S. If it is not S, then these two are not enantiomers of each other. Means they must have opposite configuration. Then only one is clockwise, otherwise anti-clockwise. Okay? Configuration must be opposite and then then obviously the structural formula or the molecular formula must be same. If you have more than one carol center, for example, this compound is given and this compound is given, you need to find out whether it is enantiomers or not. Enantiomers or not. Then we'll find out the configuration here. Suppose it is R, it is S. Then if it is R, this must be S. Both are mirror image. If it is S, this must be R. If this condition is true, right? Then the compounds, the pairs are enantiomers. If not, then it is not enantiomers of each other. So they must have same molecular formula with opposite configuration. In this case, when the opposite configuration we have, these two are called enantiomers. Clear? Now the next one we have, di-stereomers, right? These are the stereoisomers. These are the stereoisomers which are not mirror image of each other. These are the stereoisomers which are not mirror image of each other. Stereoisomers which are not mirror image of each other are called di-stereomers. All geometrical isomers are di-stereomers. Keep that in mind. All GI are di-stereomers because cis and trans cannot be the mirror image of each other. Cis and trans cannot be the mirror image of each other. Right down, di-stereomers in a pair, in a pair, obviously in a pair we have two molecules. So what are the possibility here? Here both molecules can be optically active. One is optically active and other one is optically inactive. This is also possible. Both molecules can be, which I will write down the first sentence only, both molecules can be optically active or optically inactive. That is also possible. Means both can be optically active. Both can be optically inactive. One can be active. One can be inactive. All three possibilities we have. Other important term here is di-stereomers has different physical properties. Means they will have different balling point, different melting point, different refractive index, refractive index. Can be separated by, can be separated by physical methods, physical methods like fractional distillation, fractional distillation, chromatographic crystallization is also possible to separate. Okay, done. O A is optically active, O I is optically inactive. Okay, now you look at this question. We have four different molecules here. Four different molecules we have. H, H, OH, CH3 and then we have CH3 OH. Then we have H, H, the first one is this, second one is this, CH3 OH. Then we have OH CH3, H, H, CH3 CH3 OH and OH. Another one, H, H, OH, OH, CH3 and CH3. These are the four different structures we have. You need to find out the enantiomeric pair and di-stereomeric pair. Which pairs are enantiomers? Which pairs are di-stereomers? Okay, so how do we do this kind of questions? Okay, so enantiomers we know these are non-super impossible mirror image. Okay, non-superimposable mirror image, correct? So if you look at these two molecules, these two molecules are mirror image of each other. They are superimposable or not? If you try to find out, no, they are not superimposable. If you try to place this carbon over it, it is non-superimposable. This OH is coming over. The CH3 is coming over this OH. If you try to rotate it 180 and place, then also it is not superimposable, right? So that way we can say 1 and 2 are enantiomers. 2 and 3 are not mirror image. 1 and 4 are not mirror image. 2 and 4 are not mirror image. So only 1 and 2 are enantiomeric pair we have in this one. Another way is what? You find out the configuration of this particular carbon. I am coming to that. 3 and 4, I am coming to that. Wait, wait. 3 and 4 looks like same molecule, right? 3 and 4 we will discuss once again. One way is mirror image and superimposability you check. Easier way is what? You find out the configuration of this carbon, this carbon, if it is exactly opposite, then also enantiomers. If you find out the configuration here, the priority would be 1 and then this entire group is 2. This entire group is 2. This is 3 and this is 4. This is the priority order. 4 should be down. 4 should be down. I will just exchange this 2 and 4. 1, 2 and 3 it is anti-clockwise S. This one is R configuration. Did you understand this? How this is R the top one? Yes. Tell me guys, configuration of the top one is R. Is it clear? What is the configuration of the bottom one? This we had discussed, it is R now. For the bottom one you see, here the priority is 1. This is the entire group. This one is 2. This is 3 and this is 4. So it is going clockwise. This one is also R. So both configuration is R. Any doubt in this? Tell me. End out. Now you can find out the configuration of this 2 by assigning the priority here. But that is not required because you can compare the 2. How do you get this molecule from this? By exchanging CS3 and OH. So one single exchange we have odd number of exchange. If it is R, this one is S. Same thing we have here. This one is also S. Exactly opposite configuration. Hence 1 and 2 are enantiomers. That also you can do. Could you tell me the configuration of this carbon? Configuration of this carbon? S. Configuration of this carbon? This one is R. It is simple, no? So compare it. This position is all the same here. So this is S and this is S. If you compare this and this carbon, then we have one exchange, OH and CS3. If it is S, this must be R. Could you tell me the configuration of this carbon? If you compare this to CS3 and OH exchange, we have one exchange. So it is R. This one is S. This one is R. Here also, if you see, this is also RS. I'm sorry. This is S, right? Yeah, it is S. Okay. So first of all, which pair are not mirrored image of each other? Die-stereomers, if you try to find out. The pair which is not mirrored image, one and third are not mirrored image of each other. Die-stereomers. Two and third are not mirrored image of each other. Die-stereomers. One and fourth are not mirrored image of each other. Die-stereomers. Two and fourth are not mirrored image of each other. Die-stereomers. One third, one fourth, two third, two fourth. Any doubt? Yes. Which two compound can form resmic mixture? Could you tell me? Which two compound can form resmic mixture? One and two. Yes. If you take equimolar mixture of one and two, it forms resmic mixture. Okay. Only one and two can do that. Okay. You had some confusion. You were some talking about three and four. Are three and four are res, are enantiomers? Three and four? They are mirrored image. They looks like mirrored image. Are they enantiomers? It is superimposable. Very good. You look at this molecule this way. I'm drawing the same molecule. You don't have to draw it again. Okay. The same molecule we have this. Both OH same side and both CS3 same side. You look at this. This was the molecule. Here we have a mirror image. Mirror. This is the mirror. In one of the molecule we have CS3, CS3, OH, OH, H and H. This H will be as it is. This is CS3, CS3, OH, OH. So the configuration of the top one is S and R. So this one is S. This one is R and this one is R. This one is S. You see, this two looks like mirrored image, but they are superimposable. Then this R will be here on the top and this OH will be here. This OH will be here. This CS3 will be here. Rotate this 180 degree clockwise and then you try to place this molecule on this molecule. You'll get it superimposable. Yes. Yes or no? Just imagine this. 180 rotate here and then you place it over the second molecule. It is superimposable. Yes or no? Yes. I'm waiting for your response. Understood all of you? So these two molecules are superimposable to their mirror image. Hence, these are not enhance humans. In fact, they are identical compound actually. These two are identical. See, I'm looking at this molecule from this side. If you look at it from this side, it will be R and then S. The same molecule, if you look at it from this side, then it will be S and then it will be R. Which is nothing but this molecule. These two are identical molecule. See me. Now one more thing. This molecule is optically active or inactive. If you look at this molecule, there is a plane of symmetry present here. Look at this plane. Can we divide the molecule along this plane? Two molecules. So this is actually what? This is the POS, plane of symmetry. And hence we can divide the molecule into two equal halves. Both halves are symmetrical to each other. Hence both molecules are optically inactive. This kind of compounds in which the plane of symmetry present within the molecule. See, the plane of symmetry is passing through the molecule. Within the molecule it is present. This kind of compound, we call it as mesocompound. Mesocompound. We have a plane of symmetry like this. And hence mesocompounds are always optically inactive. Mesocompounds are always optically inactive. Optically inactive because of plane of symmetry present within it. So optically inactive because of the term that we use here because of internal compensation. Internal compensation. Keep this term in mind. Internal compensation. Mesocompounds are optically inactive because of internal compensation. If this rotates clockwise 10 degree, this will rotate anti-clockwise 10 degree. The one that I talked about. Okay, in resmic mixture. See mesocompounds are those compounds in which plane of symmetry present within the molecule. Like you see this molecule you can divide like this. OH is the mirror image of this. Symmetrical. This kind of compounds are mesocompound. The name is mesocompound given. Since we have symmetry present within the molecule, it is optically inactive and it is present within the molecule. So optically inactive because of internal compensation. Internal compensation is the plane of symmetry will pass through the molecule. See the molecule is passing through the plane is passing through the molecule. Half part this side, half part this side. Hence internal compensation. There's one more term like you must be thinking of internal. Right? Isn't it? If there is something called internal, then we must have external thing also. Yes, so external is what I'll tell you. External be here. Don't worry. So external. I have explained that resmic mixture. No way is that. Huh? This one you see. Is there any difference? Just you look at it. What I said, I said we'll take the equimolar mixture of A and B. So this two of your mixture with a two different molecule. We have A and we have B and both are compensating each other. So in this also, we do not have plane of symmetry, but since two different molecules are compensating their optical activity, right? Compensating their optical behavior. A clockwise is the anticlockwise. So this kind of compensation, we call it as external compensation. External compensation. Okay. And both by example, I am here. Resmic mixtures are optically inactive because of external compensation. Mesocompounds are optically active, optically inactive, I'm sorry, because of internal compensation, plane of symmetry present within the molecule. I hope this is clear now. So these terms, internal compensation, external compensation, mesocompounds, resmic mixture, enantiomers, diastereomers, right? Identical compounds are very important. You will get questions on this in the exam, definitely. Tell me any doubt till here. Okay. One last thing we'll see and that is the calculation of stereoisomers. Write down the next point, calculation of stereoisomers. Just we'll use formula like we did in geometrical isomerism. First case, like in case of unsymmetrical molecule, just you need to memorize this formula and use it, that is it. Unsymmetrical molecules, number of optically active form, number of optically active form. Optically active means D and L we are talking about. Optically active forms. I'm assuming it as A, okay? So value of A would be 2 to the power N, okay? A, N is the number of asymmetric carbon atom, means across which the OI is possible. Number of optically inactive, number of optically inactive form, that is M and M will be 0 over here, because unsymmetrical, optically inactive is not possible. Unsymmetrical means optically active. Total number of stereoisomers or optical isomers are A plus M, which is 2 to the power N, and this is the formula we have in the first case. N is the number of asymmetric carbon atom. We'll see one example you'll understand, right? Down N is the number of asymmetric carbon atom, just a second. Just one thing I wanted to ask, have you done a group 13 and 14 P block? Group 13 and 14 P block, have you done? Okay, you start doing it. I'll take one class for that, okay? It's the simple thing you need to memorize things over there. So I'll take only one class, we don't have that much time, okay? Because the conceptual thing I wanted to discuss, because those things that you cannot do on your own, those I want to take first, right? So we'll probably have only one class left for group 13 and 14. Just go through once NCRT, we'll discuss important things in the class only. Okay, clear? Now you see in this one question, if suppose a molecule we have this, like CH3, CHOH and COOH, the name of this compound is lactic acid, lactic acid. You see the number of asymmetrical carbon atom is this carbon atom, SP3 hybridized, all four atoms different. So N value is 1, it is not symmetrical, M is 0. So the total number of isomers is A, optically active form is 2 to the power N, that is 2. So total isomers would be, total isomers would be A plus M, that is 2. Second one, in case of symmetrical molecule, see the molecule is this, no, we do not have any symmetry in this molecule, carbon COOH, CH3, OH and H, we do not have any symmetry in this. So this carbon is the asymmetry carbon we have, carol carbon, N value is 1, since there is no symmetry, so number of mesocompound is 0, optically inactive form is 0, right, mesocompound is 0, optically active form is 2 to the power N, N value is 1, so it is 2. So total isomers would be A plus M. No, there is no optically activity across CH3, Anusha. Asymmetry carbon means carbon atom has all four atoms different. Symmetrical molecule, we have two cases in this, when N is even, so if N is even, number of optically active forms equals to A equals to 2 to the power N minus 1, number of mesoform M equals to 2 to the power N by 2 minus 1, okay. Hence, the total number of isomers, total isomers equals to A plus M, that is 2 to the power N minus 1 plus 2 to the power N by 2 minus 1, okay. Look at this molecule, we have COOH, CH, OH, CH, OH, and COOH, CH, OH, COOH. This molecule we call it as tartaric acid, this is a 2-carol carbon we have, so N value for this molecule is 2, A value would be 2 to the power N minus 1, that is 2, M value would be 2 to the power N by 2 minus 1, that is 1, total optical isomers, okay. And the last one, when N is even, if done, in this only we have the second case, when N is odd, number of optically active form A is equals to 2 to the power N minus 1 minus 2 to the power N minus 1 divided by 2, M value is 2 to the power N minus 1 divided by 2, total isomers A plus M, which is 2 to the power N minus 1. Okay, so for this example you see, we have CH2, OH, CH, OH, CH, OH, CH, OH, CH2, so N value is 3, asymmetric carbon atom, 1, 2, 3, one change I will do, CH2 H5 I will write down here, CH2 H5 OH, N value is 3, so A value number of optically active form 2 to the power 3 minus 1 minus 2 to the power 3 minus 1 divided by 2, so 2 square 4 minus 2, we have 2 here, M value 2 to the power 3 minus 1 divided by 2, which is 2 only, total isomers, okay guys, this is when N is odd, it is we are doing under unsymmetrical one, you are doing in a symmetrical one, yeah, yeah, yeah, that's right, we should write down here CH2 OH only, it is CH2 OH only, okay, fine, so we'll, so we are done with this optical isomerism is done, one thing is left in this chapter that is conformational isomerism, okay, conformational isomers, conformational isomerism, that we'll do in the last, okay, we won't do it now, we'll take a break now, after the break we'll start with hydrocarbon, we'll see preparation of alkane, okay, so after finishing that, then we'll take up conformational isomers, it is not that important for your school exams also, it's not like we won't do, we'll do that later, okay, after finishing hydrocarbon, okay, yeah, so we can take a break now, we'll resume at 6.30, take a break guys, okay,