 Okay. Hello, everyone. Please type in your name so that I can understand that we're out there. Okay, just know Shreya, Shreyas, R.M.A.N. Wait, God, I'll tell you. Shree Ramya, Shree S.Ramesh. Which class you are in, Shree S.Ramesh? Okay. Anyways, see, today we are going to discuss some basic concepts of general organic chemistry. Okay, this is only for a school exam. Basic level will do. Okay. I've already started this in 11th class, HSR and PSHSR, but at Ajay Nagar, I will start soon. Okay. So again, all these things we'll discuss in the class and we'll go into so detail up to Ajay's advanced level. Okay. So today's class, we'll just go through the basic concepts of this so that when we discuss this again to the class and we'll go to the advanced level concept, you can understand it properly. Okay. That is the purpose of this class we have. Okay. So actually, there are few electronics effects. If I talk about a chapter we are going to study today, that is general organic chemistry. Okay. Now, in this, there are a few things that I will not discuss today, which usually I start from. Right. But I will not discuss it today. We'll discuss this in the class when we take the physical class or offline class. Okay. So those four, like, especially for Ajay Nagar, guys, if you are, you just try to understand the concept we'll study. Okay. It's not like complete thing we are doing today. Okay. Since many of you are not here today in the session. Okay. So for Ajay Nagar, I will discuss this thing again into the physical class also. You don't worry with that. Okay. So a few things I am not discussing it here. Right. That I will discuss in the class. But again, the basic concepts that we have already done in chemical bonding and that is resonance. So we'll start today's class. We'll start from the concept of resonance. Okay. Once we have already done this particular concept in chemical bonding, what resonance is and how do we draw the various resonating structures? Okay. Okay. So the point is, this actually the part of GOC, if you remember, I told you this thing in the class when we were doing chemical bonding that we will discuss this again into GOC. Okay. So here we are going to discuss resonance again. So first of all, you see what is resonance. So the definition of this is what it is the phenomenon. It is the phenomenon in which two or more molecules in which two or more structures. Two or more structures can be drawn can be drawn for a given compound for a given compound. But none of them, whatever structures we draw, but none of them represents the actual molecule represents the actual molecule. Okay. The structure we draw here, all these structures, like I said, two or more structures, right? So all these structures are called resonating structures. Okay. So next time you write down all these structures are known as resonating structures. Resonating structure or canonical structure, we also call it as canonical structures. So from now onwards, these resonating structure will use RS to write down the resonating because every time we have to write down this particular thing resonating structures here and there. So we'll use RS as resonating structure. Okay. Like you see examples, if I write down the structure of benzene, benzene, the structure is this. Okay. And we can also draw another structure as this. These two are called resonating structure of benzene. If you look at this molecule, this molecule looks like similar molecule, but they are not similar. Okay. They are different molecules because if you do the numbering over here, one, two, three, four, five and six. Here if you do in a similar way, one, two, three, four, five and six. Okay. So both these is structure you see here the double bond we have between one and six carbon atom. Here we have one and two carbon atom. Here you see one or two carbon atom we have single bond. Right. So both these structures are different structures. Right. And both these structures, we call it as resonating structure or canonical structure. The actual structure of benzene is not this or this. These two are not, these two does not represent the actual structure of benzene. Then what is the actual structure of benzene we have? You must have seen the structure of benzene in the book it is written like this, a circle in between. Okay. So this is what this is the actual structure of benzene or true structure of benzene. And this is structure we call it as hybrid structure, hybrid structure. Why hybrid structure? Because in any molecules where resonance is possible, this is nothing but you see one single molecule benzene we can represent in two different form. Right. So this phenomenon is resonance. Right. But none of these molecule represents the actual structure of benzene. So actual structure of benzene is this. Now what is this actual structure or we also call it as hybrid structure. Hybrid structure means what these two see this structure has certain properties. This is structure also has certain properties. Okay. So the actual benzene if you take that will contain the properties of this structure also and this structure also. So when you combine these two molecule, you'll get a hybrid molecule. That hybrid molecule is a hybrid structure. Okay. So actually when you are able to draw a resonating structure of any compound, so all those resonating structures collectively or combine the hybrid structure. The hybrid structure of that molecule. Right. And none of these resonating structure represents the actual molecule. Actual molecule is the hybrid of all these resonating structure. Right. Now, if you talk about this benzene molecule here, you see what is this circle that I put here in between like on this molecule. And what this circle means actually. You see here, in these two molecule, we have double bond between first and second, third and fourth, fifth and sixth carbon atom. Here we have second and third, fourth and fifth, first and sixth carbon atom. So actually what we say that in the actual benzene molecule, this double bond is this pi electron, which has this double bond, this pi electron delocalized over the entire molecule through resonance. Okay. Now the point is here, if these two structure you compare first and second carbon has double bond, first and second carbon has single bond. So if I ask you what is the bond length or bond order of C1 and C2 carbon in benzene molecule. Right. Point here is what, if I ask you this question, what is the bond order of C1 C2 carbon in benzene molecule. Right. C1 and C2 carbon in benzene molecule. Right. So from this structure, it will be two. From this structure, it should be one. Right. But since the actual benzene molecule is this, with these two, we cannot give this answer because these are not the real structure. The real structure is this. Now this actual molecule you have to consider for this answer. Now in this, what happens, the bond order or number of bond between first and second carbon will be somewhere in between one and two. It won't be two exactly. It won't be one exactly. It will be somewhere in between one and two. Right. So that's what the resonating resonance we have. Okay. The meaning of here is what, here is what, here we have three pi bond with six pi electron we have through resonance. Resonance what happens, this six pi electron delocalized over the entire molecule like this and it continuously moves around this molecule like this. We have electron cloud which continuously moves over this cyclic structure. And this movement of the electrons is nothing but resonance we have. Okay. So resonance is what? Resonance is nothing but is a phenomenon by which two or more structure can be drawn for a given compound. Now how this structure draws by the movement of electrons, pi electrons. Okay. So basically this kind of resonating structure gives a kind of stability also to the compound. We'll talk on this stability thing later on. Okay. But resonance is this only. Now when you draw this molecule, right, actual molecule is this, this pi electron is continuously moving over the entire molecule like this. Now the point is what, if the actual structure is this, then why we draw these two structures? Whenever you see any reactions, they must have drawn this structure or this structure not this one. They never draw the actual structure of benzene while they represent the reaction of benzene. The reason behind this is what? In this structure, which is the actual structure of benzene, the electrons keeps on moving over the entire molecule. These are not static electrons, right. And to represent any reaction, we must require static electrons like this or this, right. So we use any of these molecule to represent the reaction of benzene. But actual molecule of benzene is this, which is due to resonance. Understood this? So these two are resonating structure or canonical structure. This is hybrid structure or we also call it as resonance hybrid. Okay. The resonating structure also call it as resonance hybrid. Now, there are some condition of resonance. Okay. That condition you have to keep in mind. And if on that condition only, the resonance is possible. Otherwise, it won't be possible. Okay. So next thing we'll see condition of resonance. There are some more examples in this we'll see. But after this condition, we'll see some more examples under this. Okay. So first of all, you write down condition of resonance. The first condition is it involves all of you write down these conditions. It involves delocalization of pi electrons of pi electrons. All resonating structures, all resonating structures must follow octet rule. Second one, resonating structure, resonating structure, which has more covalent bond, more covalent bond are more stable. Third one, third one you write down. All resonating structure, resonating structure must be planar. Planarity is the necessary condition for any molecule to show resonance. Okay. If the molecules are not planar resonating structure, we cannot draw. Next one, delocalization of electrons takes place the orbital and which is only possible when the molecule is planar. That's what it is. Fifth one, molecules must have must have conjugated system. What is conjugated system? We'll discuss now. Molecules must have conjugated system. What all systems are said to be conjugated? First one, all these are examples of conjugated system. If you have carbon, C double bond, C single bond, carbon with positive charge. This is a conjugated system. The meaning of this is what? We should have a double bond, then a single bond and then a positive charge or vacant P orbital. This is the meaning of this particular thing we have. We'll see the example of this also. Then you will understand that what is the meaning of this. First of all, you write down this. This is one type of conjugated system. Another one, if you have C double bond, C single bond, C and negative charge on it conjugated system. If you have C double bond, C single bond, C and lone pair on it conjugated system. If you have double bond, single bond and double bond conjugated system. We have positive charge, single bond and lone pair if you have then conjugated system. Negative charge, single bond and vacant D orbital conjugated system. If you have double bond, single bond and free radical conjugated system. All these are examples of conjugated system. On the basis of this, if I ask you some question here, which of these are conjugated system? First example you see. If I write down this one, C double bond C double bond C and double bond C. Is this a conjugated system? Double bond O with lone pair on it. Is this a conjugated system? C double bond C, single bond C and triple bond C. Is this a conjugated system or not? See, the first one we have here. If you compare this, it has a pi bond, right? Then we have sigma and then we have pi. We have one more pi here, but that is not our concern. Our concern is what? If you have pi sigma pi, so we have here pi sigma pi, so this is a conjugated system. Similarly, this one you see, we have a pi bond, then we have a sigma bond here and then we have a lone pair, right? So pi sigma lone pair. Where is this? See here, pi sigma and lone pair. This is also a conjugated system. When you consider this, pi sigma and again we have pi here. So this is also a conjugated system. One note you write down here, sp3 hybridized carbon, hybridized carbon never takes part in resonance. Okay, sp3 hybridized carbon will never take part in resonance. Next one. Next you write down characteristics, resonance or resonating structure. Anything you can write. Characteristics of resonance or resonating structure. First of all, all resonating structures are imaginary. They are not real structures. All resonating structures are imaginary. Second one, all Rs contributes to real structure or resonance hybrid. Next one, write down. Among all structures, real hybrid or resonance hybrid, both are same thing, real hybrid will have least energy. It is most stable. Most stable. That is why the actual molecule exists in real hybrid form. Okay? Most stable. What is this Rs contributors? I will discuss this. Okay? First you write down all these properties. Real hybrid, real hybrid will use RH for this term, real hybrid. Okay? Real hybrid Rs is more stable. Means for one compound, if you can draw three resonating structure and for other compound, if you can draw five resonating structure, then the compound with five resonating structure will be more stable. That is the meaning of this. Last point you write down into this. In some cases, some cases equal contributors are present, equal contributors are present. What is this contributor? I will discuss just now. Just wait for some time. Contributors are present, equal contributors are present. In this case, in this case, the stability is especially very high, very high if the total number of Rs is more, is less. Even if the total number of Rs is less. I will discuss this contributing part, contributors part now. I will see if I draw one molecule. Now we will take the example here. Suppose the molecule we have now and this one I have already discussed in chemical bonding. I will discuss now also. CO3 2 minus, carbonate ion. Okay? So carbonate ion, if you draw the structure of this, the structure will be like this. You see C double bond O, single bond O minus and single bond O minus. This is the structure. Now, this structure, how we get this structure that you can understand by Lewis acid, sorry, Lewis structure of this. Okay? Lewis dot structure you can draw, you will get this. Now you see here, is this a conjugated system? You see? Pi, sigma, negative charge. It is a conjugated system. Pi, sigma, negative charge. Means if it is a conjugated system, then we can draw the resonating structure of this. And how do we draw the resonating structure? For that, what happens? This negative charge comes over here onto this bond and this pi bond goes onto this carbon atom. Okay? Now if you numbering this oxygen as 1, it is 2, it is 3 and 4. Right? So when this electron pair comes over here, we will get a double bond here and this pi electron goes here, we will get negative charge here. So the structure will be, now it is C carbon O negative double bond O, single bond O minus. Another structure also we can draw and that structure is this. This negative charge comes over here because here also you see pi, sigma, negative charge. Negative charge electron comes over here and this pi electron goes onto this oxygen and the structure will be this C single bond O minus double bond O, single bond O minus. Now if you see the position of double bond in all these molecules, this is first oxygen, second, third and fourth, first, second, third and fourth. Right? First of all, all these structures are, all these structures are resonating structure. Okay? All these are resonating structure and since these are resonating structure, so these are not real structure. One more thing I will tell you here, all these resonating structures are separated by this double headed arrow. Right? Resonating structure of any molecule if you draw, those resonating structure will be separated by this double headed arrow. Why this double headed arrow we are using? Because the meaning of this is what? This structure can be converted into this and this also can be converted into this, both way it can go. Right? That's why we are using double headed arrow. Anyway, so this is what the resonating structures we have in just a second. Okay? So these are resonating structures and these are not real structures. Okay? So if I draw the real structure of this molecule carbonate and that is resonating resonance hybrid or real hybrid. Okay? So that will be like this. You see carbon oxygen and oxygen. So we'll draw like this carbon oxygen here, oxygen here and here we have again oxygen. Right? Now you see here one thing that in the first compound, this one, we have double bond between this and this carbon. If I numbering this first, let me write down the number here. One, two, three and four. Right? So here we have double bond and all these molecules we have single bond. Right? Between C2 and oxygen O3, we have single bond in the first, double bond in the second and single bond in the last. Similarly, in this also two and four also. So all these molecule what we say, these won't have any double bond or single bond, but they have a bond which is in between double and single bond. And we draw this delocalization of electron like this with this dotted line. And then here also we have the dotted line. Okay? Means what this pi bond is distributed uniformly over this oxygen atom. One more thing. If I ask you about the charge on oxygen atom, you see on the first oxygen, there is no charge in the first structure. But in the last two structure, we have negative, negative charge. Right? So this two negative, which is here, this two negative charge is equally distributed over this oxygen atom. So how many oxygen atom we have? We have three oxygen atom, two charge distributed among three oxygen atom. So here in this resonance hybrid, the charge on each oxygen atom will be minus two by three, minus two by three and minus two by three. So this structure is the actual structure of carbonate ion. This is resonance hybrid or real structure. Now this resonance hybrid is the, will have the property which has, which is given by all these resonating structure. Means few property will be contributing by this first structure, few property will be contributing by this second structure and few will be contributing by this third structure. So all these resonating structures will contribute into the properties of resonance hybrid. Okay? See, the reason of the charge minus two by three is what? We have, we have minus two total charge. Right? But in all these structure, what happens? This pi electron is delocalizing over the entire molecule. You see here, first of all, in the first structure, we have double bond between first and second. Second structure, we have double bond between second and third. And third one, second and fourth. So this means what? All these bond will have double bond in each of these structure. Right? It means what? That this pi bond is distributing over the entire molecule. It is, it is between one and two also, two and three also and three and four and two and four also. Right? So the point is this charge, which is minus two charge is distributed equally over the three oxygen atom. Right? The total charge is minus two distributed among three oxygen atom. So each oxygen atom will have charge minus two by three. Okay? Each oxygen atom will have charge minus two by three. Now the point is this structure is the real structure. And this structure will have property given by all these resonating structures. Right? Means suppose this has supposed 10 properties I can write for this one. So it is possible that few properties will be given by this molecule. Few will be given by this molecule. Few will be given by this molecule. So actually this structure is the, is the combined structure of all these resonating structures. Right? That's why the properties of resonance hybrid is given by resonance is given by its resonating structure only. Right? So since these resonating structure contributes into the property of resonance hybrid. So we also call it as contributors. All these resonating structures also call it as contributors. Now in this it is possible that one resonating structure will contribute maximum other contributes minimum. Right? Or another possibility is what all resonating structures contribute equally. Right? So when all the resonating structures right on this point when all resonating structures resonating structures contributes equally resonance hybrid resonance hybrid then they called as as equal contributors equal contributors. Like the example I have taken just now this is the case of equal contributors. All these resonating structures contributes equally. Whenever you have uniform distribution of pi electrons they will have equal contributors. Okay? If one resonating structures contributes more than the other then it is known as major contributors. Right? So if one of the resonating structures contributes equally then equal contributors if one of the resonating structure contributes more than the other structures then that particular structure we call it as major contributors and all other one are the minor contributors. Okay? One RS contributes more than the other RS then it is major contributors and others are minor contributors. Right on this and others are minor contributors. Okay? Now we will see again some example here. Next example you see we have to draw the resonating structure of it. CH2 double bond CH single bond CH2 plus. Is this a conjugated system? Tell me. Yes or no? Is this a conjugated system? Is this a conjugated system? Yes. This is a conjugated system because we have pi sigma and positive charge. Okay? Pi sigma and positive charge. Now when you look at this molecule I will just draw the exact structure of this first because it is the first example here we are discussing. So here you see this three carbon are bonded like this carbon, carbon and carbon. And this carbon has two hydrogen. This carbon has two hydrogen and there is a pi bond which forms by the lateral overlap of this p orbital of these two carbon. Right? This is the lateral overlap we have and pi bond forms and here we have one p orbital which is vacant. That's why this carbon has positive charge on it. Positive charge means what? It has lose its electron. Right? So that's why this orbital is vacant now. There is no electron present into this orbital. And that's why when this electron pair jumps over here it can accept this electron and will have a pi bond over here. So here you see when you draw the resonating structure of this that will be this bond pair of electrons jumps over here. Right? And we get the resonating structure and since these, I'm going to draw the resonating structure here. So I have given this arrow, double headed arrow to separate these two resonating structure. Now the resonating structure of this will be, this carbon, single bond carbon, single bond carbon. Right? This has two hydrogen like this. This carbon has two hydrogen. Everything will be as it is. Only change will be what? This bond pair will shift here between these two carbon atoms. That is the only change. Every other thing will be same. Now you see here since this carbon which has one electron here and one electron here and both this electron, this one electron jump over here and now this bond pair of electron is shared between these two carbon atoms. Right? This one and these two carbon atoms. So obviously one electron, this carbon has lost. Right? This carbon loses its electron. So that's why here we have the positive charge on this carbon atom. Right? And the pi bond shifted over here. Right? So these two are the resonating structure of this molecule. Two RS we have. Two resonating structure possible. But again in the exam you don't have to draw this orbital thing. Right? This orbital you don't have to draw. Then what do we write here? Suppose the compound is this which is given in this. So we have a double headed arrow, this pi electron shifted over here and then we'll write CH2 single bond CH double bond CH2. And since this electron pair, one of this carbon, one of this carbon shifted over here. Right? So this carbon loses its electron. So here we have the positive charge. And this carbon already has one positive charge and one extra electron. So this electron will neutralize this positive charge. We don't have any charge on this carbon atom. Right? One more thing you must keep in mind. Number of charge present in one resonating structure will always be same on all other resonating structures. Like you see these two are RS of each other. This contains one positive charge. So another RS must contains one positive charge. Right? That is the thing we have. Understood? So this is the orbital diagram I have drawn here to make you understand. But this you don't have to draw in the exam. Just you do like this, shifting and you'll get that. So if I ask you how many RS possible for this? Total RS resonating structure is two. Right? This also they ask sometimes in the exam. For this molecule how many RS resonating structure we can draw? Right? Another one you see. If one more example we'll see here. Oh, before that we'll see one more thing here. Now, since I said that these two are resonating structure. These two are not the actual molecule. So if I ask you to draw the actual molecule, how do you draw? So first of all, you just write down all those sigma bonds that will be CH2, single bond CH, single bond CH2. Now this you see this pi electron is shifting is distributed among these high carbon, right? Three carbon. If it is one, two and three, this is also one, two and three. So pi electron is between one and two and here between two and three. So here we have the partial double bond characteristics. It is not the actual double bond we have here in the actual structure. These two carbon and these two carbon will have partial double bond characteristics and that will denote by this dotted line. This dotted line means that this has partial double bond character. This bond is not exactly equals to carbon-carbon single bond. Not exactly equals to carbon-carbon double bond. It is somewhere in between single and double bond. Now here you see the positive charge in the first structure has at third carbon and here it is at first carbon. So these two carbon will have slight positive charge. So del plus, del plus. This is the actual structure or resonance hybrid. Resonance hybrid. So when you take this molecule, actually the structure of the molecule will be this. This we write to for our own convention, for our own ease, like to represent the reaction and all. Okay. Now, if I give you one more molecule here, CH2 double bond CH, single bond CH, double bond CH, single bond CH2 positive charge. Is this a conjugated system? Is this a conjugated system? This molecule is a conjugated system or not? You see the rules actually. What is the condition we have for resonating structure? It must have conjugated system, right? So you see here, it is pi, sigma and positive charge. So this is a conjugated system. So you can draw the resonating structure of this. Since the resonance is possible into this. Now, how do we draw the resonating structure? This pi electron jumps over here, right? And we get what every other thing will be same. CH2 double bond CH, single bond CH, single bond CH, single bond CH. Single bond CH, single bond CH, double bond CH2. Now you see here, this CH2 takes this electron pair, neutral, and this carbon loses its electron, right? So this carbon, the third carbon will have the positive charge. So here we'll have the positive charge. This is the one resonating structure we have. Now in this, another resonating structure is also possible. Because you see again this pi, sigma and pi. If this bond, here also pi, sigma and positive charge we have. If this bond comes over here, we'll get this structure only, first structure only, right? So this is first structure, this is second structure. Now when this pi bond comes over here, we can draw the third structure here, which is nothing but CH2, CH, double bond CH, single bond CH, and double bond CH2, and here we'll have the positive charge. So this is the third structure we have. So these three structures are known as resonating structure. So for this molecule, we have total three resonating structure possible, right? We have total three resonating structure possible. So when you compare the stability of this compound, the first one and the second one, right? So since the second one contains more number of resonating structure, we have three resonating structure here, and here we have only two. So we can say the second one is more stable than the first one. Is it clear? This is a pi, sigma, pi conjugated system. In this, yeah, in this, yeah, you can see, you see here, you'll get the same thing actually. Suppose pi, sigma, pi conjugation, right? So if this pi bond you shift here, right? So obviously this pi bond has to shift onto this carbon atom. That is obviously true because otherwise this carbon will have five bonds, right? So this pi bond goes here, this pi bond goes here. Here we'll have double bond, here we'll have double bond, and here we'll have the positive charge, which is nothing but this structure. So with this also, you'll get this only. Anyone you can take. Understood? Now you'll see the another one. This one you tell me how many resonating structure possible? Possible or not? Yes or no? Tell me. Possible, right? Because again, we have pi, sigma, positive charge or pi, sigma, pi, right? So when you draw the resonating structure here, so one step to draw the first resonating structure or one resonating structure, we just shift one pi bond here, right? So here we'll have negative charge, this negative positive forms a bond, and here we'll have the positive charge. So we'll get this double bond here, positive charge here, and this double bond will be as it is. Now in this again, two ways possible. Either this pi bond comes here, then we'll get the same structure this, and the second one is what? This pi bond goes here. So we'll get the structure then, which is this. Positive charge here, double bond, double bond, and positive charge here. Sorry. This is a positive charge here we have. So in this you see again, there are three resonating structure possible. If we draw the real hybrid of this molecule, the actual structure, actual structure will be like this. You see, this is first, second, third, fourth, and fifth carbon. So from fifth to first, the pi electron is delocalizing, right? From here to here it is moving, pi electron, right? So from here, this is one, two, three, four, and five. Five to one pi electron is delocalized. So just write down the partial double bond like this. Means pi electron is delocalized from this first carbon to fifth carbon. Now we have positive charge where we are getting positive charge. At first carbon, then third carbon and fifth carbon. So we'll have del plus positive charge in the first, del plus on third, and del plus on fifth. This is the actual structure of this molecule. Is it clear? Can you draw the resonating structure of this molecule? First you check whether resonance is possible or not. If it is possible, then draw the resonating structure and tell me how many resonating structure possible. Tell me three. Only three are there. All of you are getting four. Is it four? Let's see. The first structure we'll draw here. So for that this pi electron goes here. Here we have the positive charge. So the first structure will be this double bond CH2, positive charge on this carbon atom, and this double bond will be as it is. Okay, Neha is getting four. Okay, we'll see. This pi electron again comes here. Double bond we have here, positive charge, double bond here, and double bond CH2. Again, this pi electron comes over here, double bond, double bond, positive charge, and double bond CH2. And the next structure will be what? This pi electron comes over here, double bond, double bond, and then CH2 plus. So five resonating structure possible, five RS total, including this. The given molecule is also a resonating structure. You see in all these we are considering the given molecule also. Understood? You're talking about carbonate, Margava. So here you didn't understand this. Y minus 2 by 3. Okay, see again I will explain this quickly. Or I'll go back to the last slide. This one, right? See, you tell me one thing. This molecule has total two negative charge. Right? Okay, just one second. Let's do it again. This molecule has total two negative charge. And this negative charge is delocalizing over oxygen atom. You see here, when this negative charge comes over here, this oxygen will have the negative charge. And when this comes over here, this oxygen will have the negative charge. So in all these three structure, you see this minus two charge is moving on to these oxygen atom continuously one by one. Right? So what we can say that total charge is minus two and this is distributed among three oxygen atom. Right? So each oxygen atom will have charge of what? Minus two by three. It is something like this. If I have 15 rupees and if I distribute this 15 rupees into three people equally, then each one of these will have what? Five and five. Right? Which is nothing but 15 by three. So what we are doing here, we have total minus two charge and we have distributed this minus two charge onto three oxygen atom equally. So each oxygen atom will have then minus two by three charge. That is what we are doing. Okay? Now coming back to this, if you have to draw the resonance hybrid of this molecule, resonance hybrid will be what? You see, we have a benzene ring draw all the sigma bonds and CH2. First of all, you see in this one, two, three, four, five, six, say this numbering has nothing to do with the nomenclature thing. Okay? I am just randomly giving these carbon atom some number. Okay? So don't get confused with the nomenclature part. Okay? So here you have seen all these resonating structure, the positive charge present at second carbon, then fourth carbon and then sixth carbon. Second carbon, fourth carbon and sixth carbon and then on the carbon which is outside the ring. Okay? So and this pi electron is delocalizing over the entire molecule. So here you see we have the double bond, partial double bond characteristics among this molecule, which we can show by this circle, partial double bond characteristics. And this carbon also joined by partial double bond characteristics. So this is the dotted line. This has partial positive charge here. And then at the second carbon, partial positive charge, then at the fourth carbon, partial positive charge, sixth carbon, partial positive charge. So this structure is the real structure we have or we also call it as resonance hybrid. Yeah, no problem, Sridham. Understood. So this is how we draw the resonating structure or various resonating structure possible. Okay? One last example we'll see, and then we'll see the r effect and plus r effect and minus r effect. Okay? One last example we'll see. Two question I'll give you. Tell me how many rs possible for this? We have negative charge here. And the another one is this negative charge, double bond and double bond. Tell me how many structure possible resonating structures? Okay. See, here this is negative charge, right? So this negative charge comes over here and this pi electron will go onto this carbon atom. So the resonating structure here will be this. We have double bond here and negative charge here. So only two rs possible into this because all other carbon you see, these are sp3, sp3, sp3 hybridized and we know sp3 hybridized carbon does not take part in resonance. Okay? So only two rs possible. In this one, this negative charge comes over here and this pi electron goes onto this carbon atom. So the resonating structure will be like this. Here we have double bond. Here we have negative charge and this double bond will be as it is. The other one you see, this negative charge comes over here. This pi electron goes here and we have this. Here we have double bond. This double bond will be as it is and again negative charge will be here. Okay? Another resonating structure possible into this. When this negative charge comes over here, this pi electron goes here. So we have double bond here. This double bond will be as it is and here we have the negative charge resonating structure possible into this. This comes over here and this pi electron goes here. Negative charge, double bond and double bond. Just a second. Here we have the double bond. Okay? So now you see if it is first, second, third, fourth and fifth carbon. So here the negative charge at first carbon. Here we have third carbon. Here we have fifth carbon. Here we have second carbon and here we have fourth carbon. So basically in this molecule, the negative charge is present in one by one. It is present at all the carbon atoms, right? So when the charge present at all the atoms in the ring, one by one in the resonating structure, then it means the charge is equal distributed over the entire molecule. Okay? It is equally distributed over the entire molecules or the carbon atom. So if you draw the resonating structure or resonance hybrid of this molecule, the resonance hybrid will be like this. This is the ring, right? And this ring has partial double bond characteristics because it is equally distributed double bond characteristics. And each of these carbon atom has one, one negative charge all this resonating structure, which means the charge is equally distributed over the five carbon atoms. And here it will be on each carbon atom, it will be minus one by five. Here it will be minus one by five, minus one by five, minus one by five, minus one by five, and minus one by five. This is what the ideal structure we have. Understood this? Why minus one by five? Because we have five carbon atom, right? And minus one charge is distributed among the five carbon atoms equally in all these resonating structures. Is it clear? RS for the previous one, you can draw it easily. Six member ring is this. Six member ring is this. And this minus one charge we put number over here, one, two, three and four, right? So minus one, this minus charge present on first and fourth carbon, right? So you write down here, del negative first and fourth, del negative. This is just a mistake. It is not double bond will not appear. Love here. I'll just make it correct. It's not a mistake. I have drawn double bond over there. Let me correct this first. You see this one. What happens in this? This pi electron comes over here on the third carbon and this negative charge comes over here. So here we'll have the double bond, right? This is a single bond we have. Here we have double bond. Pi electron comes over here on third carbon. So here we have the negative charge. This is the another resonating structure we have. So the point here is that if you have to draw the resonance hybrid of this one the negative charge present at first and third carbon. So del negative one first, del negative one second and pi electron delocalize from first to third carbon, right? From here to here. So we have a partial double bond characteristics like this from first to third. That is it. First to third, they're partial double bond characteristics and then negative charge, partial negative on this, del negative one first, del negative one third. That is it. Now you understood this first one? Shifting off negative charge connection in the middle. Nothing. You just keep on shifting this negative charge because you see with practice you will understand that how it is going on. Numbering you do on the carbon atom you will understand this. You see if you put the number here it is one, two, three, four and five, right? So third carbon. Here we have first carbon, third carbon. Now again pi sigma negative charge conjugation resonance structure possible. So it's here and you'll have then one, two, three, four, five. Just cross checking whether conjugation is there or not. When you get this structure again you see conjugation is possible. Negative charge sigma and pi. So this will come here. This will go here. You'll get this. Again one, two, three, four and five. Again you see negative charge, sigma and pi conjugation we have. This will come over here. This will go here. You'll get this. Again you see one, two, three, four and five. In this also you see, in this also there is conjugation. Pi sigma negative charge. But when this comes here and this goes here it is nothing but the first structure we have if you see. That's why further resonating structure is not possible. Did you get this? Yes or no? Okay. Now next one you see in this resonance we have two types of effect that is plus R effect plus R or minus R effect. Like yesterday we have done in the class plus i and minus i. Okay. So plus R also known as plus M. Minus R can also be written as minus M where this M means mesomeric effect. Mesomeric effect. So this resonance effect and mesomeric effect are the same thing. Plus M and minus M. Right? Plus R means what? See first of all whatever examples we have done like yesterday class we have discussed i effect okay not for Rajinagar I will discuss this i effect in Rajinagar also later on but i effect I have discussed yesterday and there if you remember I told you that there's only sigma electrons involved. Right? But the difference in R effect is what or mesomeric effect is what here we have pi electrons involved that's why you see always pi electrons shift from one point to another point or one carbon to another carbon. So here we have pi bond, pi electrons evolve, involve and there we have sigma electrons involved. That's the difference we have in these two. Right? That one is distance dependent this one is not distance dependent. Okay? So here you see if I write what is plus R effect plus R group plus R or plus M are the same thing plus R group are electrons releasing group like we have plus i and minus R or minus M group are electrons withdrawing group withdrawing group. Okay? That withdraws electron from the ring. Now you see here if I write down two examples how do you know that the group attached to the ring is electron withdrawing or electron releasing. That's the thing. Suppose we have this OH and the second one if I write down C double bond O and OH. Right? So how do you understand that the group attached to the ring it is electron releasing electron withdrawing. Okay? So only two things you have to keep in mind. Okay? If the group attached to the ring contains any negative charge or lone pair then it is electron releasing. Okay? If the group attached to the ring contains any negative charge or lone pair of electron then it is electron releasing. Now here you see when OH is attached with the ring so it has two lone pair so it can donate this lone pair to this carbon atom and this pi electron comes over here. This is nothing but resonance now. Okay? So the resonating structure we draw here will be like this. Oxygen attached with double bond now because this bond, comes into bond pair OH. Now when this oxygen loses its electron into the bond pair so it will have partial it will have positive charge onto it. And now this pi electron comes over here so on this carbon will have negative charge and then all other bonds will be as it is. This is one resonating structure. Another resonating structure we can draw when this negative charge comes over here and this pi electron goes here because this is a conjugated system. So we will get this double bond OH positive charge on oxygen double bond we have here double bond we will have here and here we have the negative charge. In this again this negative charge comes over here double bond goes here and this will be double bond double bond negative charge double bond OH plus and the last structure will be when this negative charge comes over here this pi electron goes here and we get this double bond, double bond, double bond and OH. Is it clear? So this group you see OH is electron releasing. That's why it is plus R group OH. Which has lone pair or negative charge present on the atom with which it is attached to the ring is known as electron releasing group or plus R group. Understood this? Write down this point somewhere if the group attached to the ring contains contains lone pair or negative charge on it then it behaves as electron releasing group electron releasing group. So this is how you can understand okay if lone pair or negative charge present. Now coming back to this example so this is the example of plus R effect. Now in this example you see this carbon does not have any lone pair on it. So when group attached to the ring or the atom with which the group attached to the ring that atom and any other atom if it contains any unsaturation or double or triple bond. Then the group behaves as electron with drawing group and this is minus R effect. Now how it is electron with drawing you see this has a double bond between first and second atom has double bond. So this electron goes here because it is pi sigma pi conjugation again this pi electron goes here when this pi electron jumps onto this carbon atom. So the another resonating structure you will draw that will be double bond C O minus and OH here we have the positive charge because the pi electron goes onto this carbon atom. Then double bond and double bond. Another structure is what when this pi electron comes over here so we will get double bond here all other thing will be as it is O minus OH double bond here here we will have the positive charge. Another one you see when this pi electron comes over here double bond, double bond positive charge here double bond CO minus OH now the last thing is what when this negative charge comes over here this pi electron goes here and we get double bond double bond double bond C double bond O OH So this is the resonating structure again we have due to minus R effect electron with drawing nature is it clear? Yes can we move ahead tell me okay so this is like for you to understand we will discuss some more example of this into the class right we will move again we move ahead in hyper conjugation and all so this is actually the plus Rn minus R effect okay we will do some more example some more example good examples on this into the in the class right now we are moving ahead into hyper conjugation next you write down the another electronic effects we have that is hyper conjugation write down into this takes place hyper conjugation takes place what is hyper conjugation I will tell you it takes place only in the case of Keen means double bond if it is there or free radical for a Rajinagar student you must have you must get confused with this what is free radical and all so don't get confused with all this just you write it down when I will discuss this in the class you will understand this okay this is not important now I am just giving you the information here okay so carbocation alkene and free radical next slide write down carbon ion does not take part does not take part carbon ion does not take part in hyper conjugation what is hyper conjugation I will tell you again first you write down these information what is the condition for hyper conjugation we have that you write down condition you write down condition for hyper conjugation write down sp3 hybridized carbon sp3 hybridized carbon with at least one hydrogen with at least one hydrogen must be connected with must be connected with a positive charge carbon or double bonded carbon or free radical or free radical now you see what hyper conjugation is suppose if I write down this example carbon has three hydrogen here and this this is connected with carbon cf2 positive charge condition you must read it is given that sp3 hybridized carbon with at least one hydrogen must be connected with positive charge carbon double bonded or free radical right so you see this one this is positive charge carbon this carbon is sp3 hybridized right which has three hydrogen the condition is at least one hydrogen it should have right this is alpha carbon also so in the condition you can also write it down sp3 hybridized alpha carbon that would be better hybridized alpha carbon so this is carbocation and this position is the alpha position this is alpha carbocation so when sp3 hybridized alpha carbon contain at least one hydrogen atom then hyper conjugation is possible first thing is this when sp3 hybridized alpha carbon contains at least one hydrogen then hyper conjugation is possible now what is hyper conjugation you see this bond pair here because of this bond we have a bond pair here because this positive charge present here this electron pair is dragged by this carbon atom towards its side and the structure that you get here we call it as hyper conjugative structure and that structure will be hc here we have h plus because the bond pair of electron which has one hydrogen also comes over here double bond ch2 and this hydrogen will be as it is right now this in this it is also possible that either this bond pair comes over here or this one comes over here or this one comes over here any one of these bond pair can come so in all these possibilities the possible hyper conjugative structure will be this and the last one will be ch2 this structure will get when this bond pair comes over here this structure will get when this bond pair comes over here so these are the hyper conjugative structure now the point here you have to note is what write down this h plus iron h plus iron without any bond without any bond over there present over there hence we also called hyper conjugation called hyper conjugation as no bond resonance as no bond resonance see this bond pair comes over here but h plus is not going anywhere it is there only within the molecule actually in this what happens this bond breaking and bond making takes place simultaneously sometime this comes over here this bond break and then this comes over here this bond will be as it is this comes over here this bond will be as it is so any of these structures are possible or these structures are inter convertible into each other and this we call it as hyper conjugative structure ok now if you have to find out for a given molecule number of hyper conjugative structure number of hyper conjugating structure for this what we write for this one you see for this one you see the total hyper conjugating structure will be one three and four these three structures these three structures you see it is because of these three hydrogen atom number of alpha hydrogen atom so number of hyper conjugating structure formula you can write number of alpha hydrogen plus one ok number of alpha hydrogen plus one is it clear number of hyper conjugating structure is equals to number of alpha hydrogen plus one the formula we have one more point you write down here more hyper conjugating structure more hyper conjugating structure more will be the will be the stability and more hyper conjugating structure means what more alpha hydrogen so to sum up all this if number of alpha hydrogen is more then more will be the stability of the compound ok like you see on the basis of this if I give you one question you have to compare the stability of these molecules stability order C carbon positive CH3 CH3 and CH3 this is three degree carbocation then we have CH positive CH3 CH positive CH3 CH3 CH2 positive and then we have CH3 positive okay in these four structure if you compare the stability how many alpha hydrogen are present here alpha hydrogen are those hydrogen which is attached with the alpha carbon right alpha carbon is the carbon which is adjacent to the functional group or any carbocation, carbonyl whatever it is you see these three are alpha carbon we have so total number of alpha hydrogen is 3 plus 3 plus 3 that is 9 similarly here the alpha hydrogen is 6 alpha hydrogen is 3 and there is no alpha hydrogen right so more number of alpha hydrogen more will be the hyper conjugating structure will be the stability so stability order is maximum for this then we have this is it clear okay so this is what hyper conjugation is the last part we will discuss now and that is electromeric effect electromeric effect is the only temporary effect now in this one you see this effect involves this effect involves complete transfer of transfer of shared pair of electron to one of the atoms joined by a multiple bond okay the effect involves the complete transfer of shared pair of electrons to one of the atoms joined by multiple bond under the influence of under the influence of any attacking reagent so this kind of shifting of or transfer of electron takes place under the influence of attacking reagent okay next slide and write down as soon as attacking reagent removed the original condition is restored means whatever the molecule was initially will get the same molecule is restored that's why it is temporary effect okay that's why it is temporary effect this kind of effect only observes in the molecule contains multiple bond in the molecule contains multiple bond either double or triple multiple bond there now in this one if I take the example here suppose the compound is this C double bond O right so in this case what happens is oxygen is more electronegative so obviously any nucleophile if it attacks this pile electron will shift towards the more electronegative element which is nothing but oxygen so we get here C positive and O negative charge on it right this is how shifting takes place now this electron I will give you some example here but before that we will see the types of electromagnetic effect this electromagnetic effect also have or also two types of electromagnetic effects we have the first one will write plus E effect plus E effect you write down when transfer of electron takes place towards the attacking reagent when the transfer of electron takes place towards the attacking reagent meaning of this is what suppose we have carbon-carbon double bond right and suppose H plus is attacking onto this carbon atom so suppose if this bond shift over here right and this H plus attached onto this carbon atom only means the attachment or these H plus will attack onto the carbon atom where the bond pair of electron is shifting so this kind of attack or reaction it comes under electro plus E effect or positive electromagnetic effect the product that you get here this carbon will have positive charge because the bond pair shifted towards this and this carbon will have one hydrogen with two bond as it is right so this kind of effect is electromagnetic effect when the bond pair of electron shifted towards the carbon atom where the attacking reagent where the attack is being is taking place right so this is plus E effect now the next one is negative electromagnetic effect which we also call it as minus E effect negative electromagnetic effect when this is possible when the transfer of electron of electron takes place away from the attacking reagent for example you see if I take this C double bond O carbonyl group and the nucleophile is Cn minus this Cn minus is attacking onto this group right so obviously since oxygen is more electronegative this bond pair will shift towards this and since this is negatively charged so this will attack onto the positive side oxygen will take the bond pair of electron so this will never attack onto the oxygen but this will attack onto the positive side so you see this attack takes place away from the shifting of electrons the product here would be C with this cyanide group and O minus simply one thing you can keep in mind here that attacking reagent will not attack will not attach where on the atom on which the electron is transferred that's the meaning simply we have in minus E effect now when I say this temporary effect why it is temporary now you remove this cyanide group from here again you will get C double bond O the same thing that's why it is temporary effect in previous example also you remove H plus H that is what we have the temporary nature of this effect you remove attacking reagent you will get the original compound as it is is it clear so the last thing we will discuss today is carbines see I am just discussing the basics of all these concepts application part again we will see we are not completing the portion here and for Rajasri Nagar students if you are getting 50% of this also this helps you a lot when I discuss these things into the class so next to write down the one intermediate we have and that we call it as carbine the last thing for today I will take 5-10 minutes more carbines see like carbocation carbion free radical carbines are also the intermediates one of the intermediates we have and similarly we have nitrenes also so that also we will discuss nitrenes carbine is nothing but methylene group carbine is methylene only which is nothing but CH2 with one lone pair methylene is CH2 one lone pair ok write down into this these are the these are the neutral covalent compounds neutral covalent compound of carbon having two bond pair and one lone pair you see one hydrogen to one pair and one lone pair ok one lone pair of electrons so if I write down this one C6H5 CH one lone pair on it again you see this is what one hydrogen from here has been replaced by this phenyl group so we call it as phenyl carbine the name of this is phenyl carbine if I write down any alkyl group here RCR one lone pair on it this is dialkyl carbine because the hydrogen is replaced by two alkyl group dialkyl carbine now this carbine exist in two carbine exist in two types two types of carbines are there so if I write down singlet carbine or only singlet singlet in this type write down in this type in this type both unpaired electrons both unpaired electrons go into the both unpaired electrons goes into the one orbital one orbital we have opposite spin first you write down this I will explain what this means opposite spin and it forms it forms bent structure bent structure you see this one singlet carbine is this C with two hydrogen and the two electron that you have unpaired electron is present in one orbital one p orbital one is opposite spin and non spin clockwise one electron is clockwise other one is anti clockwise now you see here there are few properties of this it is less stable less stable means in comparison to other carbine that we will see in some time it is less stable since both electrons are unpaired now so it is diamagnetic and we also call it as hot methylene just the name methylene now if you calculate the spin multiplicity of it spin multiplicity the formula is two summation s plus one now how do we calculate this two open bracket one electron is clockwise other one is anti clockwise for clockwise plus half plus one electron anti clockwise and for that is minus half bracket close plus one this is nothing but the s summation of s so this will get cancelled and will get one now you see since this spin multiplicity is one that's why we are calling it as singlet carbine because of spin multiplicity right how do we calculate this formula of spin multiplicity is two summation of s plus one s how do we calculate one electron clockwise so one into plus half for that plus one electron anti clockwise one into minus half for that plus half minus half gets cancelled we are left with one so spin multiplicity is one that's why it is singlet carbine now we will do the last thing in this which is triplet when we calculate the spin multiplicity of this we will get three we will do that okay first we write down in this type the electron electron goes into different orbital orbital same spin it has linear structure linear structure sp hybridized sp hybridized this one is sp2 hybridized it is sp2 hybridized okay so the last thing you see if I draw the structure of this we have hc h and the two electron will be in two different orbital with one one one electron each of same spin okay now the two things here since the electrons are unpaired now so it is what it is paramagnetic in nature the previous one was diamagnetic it is paramagnetic right the name of this is cold methylene just opposite of that and it is more stable than the singlet one since the two electron is in different orbital less repulsion will be here okay when you calculate the spin multiplicity of this multiplicity of this the formula is again same two summation of s plus one but here what happens both electron has clockwise right so two into plus half I have written it together only or we can also write one into plus half then again one into plus half plus one so when you solve this the spin multiplicity you will get as okay and since this is three that's why we are calling it as triplet the name of this is triplet so this is it for this carbene we have two types of carbene singlet and triplet okay so today I have given you the basic information of basic concepts of GOC okay application part will discuss this in class again few things are left application part but the theory the basic thing definition we have covered with few examples okay so I hope you have understood this okay next class whenever we have will continue some application part of it and then we will see some more examples okay thank you any doubt you have you can ask me