 So, what is the conjugated system? This is the most important one, if you want to know whether the resonance is possible or not, the system must be conjugated, molecule ions must be conjugated. The first type of conjugated, there are six, five or six different types of conjugated system. Five or six different types of conjugated, so we will see one by one also. The first type is we will have a lone pair present on the first atom, right, we will have a pi bond, then sigma bond and then we will have a vacant orbital, right on this way, pi sigma vacant orbital, this is vacant orbital. The example of this you see, suppose we have C H 2 double bond, C H single bond, C H 2 positive charge, okay. Positive charge means what? It is one orbital is vacant because it has, the electron is not present on the carbon atom, right. So, this positive charge, you can also write this or when you write this sigma pi sigma positive charge, that is also the same type of conjugated system. Both are same, this one or this one. Means when you write positive charge, it means one orbital is what? One orbital is vacant, that is what it means, okay. So, pi sigma positive charge is first type of conjugated system. Now, when you draw the orbital diagram of this which is not required, so I will not draw it always, because the first one I will just make you understand this orbital diagram. This pi orbital is vacant, right. Here we have a p orbital and here we have a p orbital. Can you see all the p orbitals are parallel, right and this forms lateral overlapping, it forms pi bond, right. So, what happens if this is the orbital is vacant, so it can take it to electron into this orbital. This orbital, this orbital can also overlap, right. So, this is the orbital of C H 2. So, this gives you the another resonating structure which is C H 2 C H C H 2 and this orbital will be what? Empty and have one orbital, okay. So, this is the resonating structure. Pi is double bond. So, what will happen double bond? So, that structure is also negated out of type 1, right. No, not type 1. No. You have negative, pi sigma negative. So, pi sigma lone well. So, some of that double bond oxygen has a empty. Well, well, well, yeah, this one, this one. Pi sigma here we do not have positive charge, do we have? Yeah. You do not have positive charge. So, would you say that that O has a like a vacant orbital because it can have a orbital? No, no, no, it is not. It has one lone vector. Exterior vacant orbital means it is electron deficient. Vacant orbital means what? It has lost its electron, right. So, this has negative charge means it has extra electron present. So, is it always free that it has to be positive because what about if it is a vacant orbital? No, it is not always. It is not always because there are some molecules which are electron deficient, neutral but electron deficient. That is an example. That is what I have given this. Right, this one is the better one. This also is the same thing but this representation is better. We will do that example first. Understood this one? So, when you draw the resonating structure of this, simply what you do? This pi electron shift over here and it gives CS2 positive charge C H double bond C H. This is the RSOS. Orbital diagram is not required. Directly they are shift electron and right. Can you draw the resonating structure in this? Check also if whether resonance is possible or not. No, I have given you these ions, molecules. You have to write down the RS of this if it is possible. If it is possible. See, first of all how do we check the system must be conjugated, planar to conjugated. So, at least you see these three carbons, one, two and three, these carbons P2 hybridized. Okay, planar. And it is pi, sigma and vacant orbital. On this positive charge means vacant orbital on the carbons right. Pi, sigma, positive charge, type one conjugation. So, when you draw the resonating structure, this pi electron comes over here and the RS will be. We have a pi bond here in most structure. So, that can shift to all six pieces right. No. See at this side we do not have resonance. Is that we are conjugated? From here to here delocalization is possible. This part is not involved because all are SP3 hybridized carbons. This side we do not have conjugation. Okay. So, if you draw the hybrid of this, the real structure will be like this. From this carbon to this carbon, the pi electron will be. Delta positive on this carbon and delta positive on this one. This is the hybrid of this. Because this part is not involved in resonance. Now, the RS on this is what? Three RS possible. Is it clear? It is hybrid. Pi electron delocalized. See the first we have pi electron here. Second we have pi electron here. We have pi electron here also and here also. Positive charge here, this carbon, this carbon, this carbon. Delta positive. Delta positive. Is this a conjugated system? Last one. See it does happen. Is this a conjugated system? No. Boron has a vacant orbit. Boron has a vacant orbit. When there is three bond, that is what is electron deficient. It can accept clear of electron and behaves as the Lewis acid. Okay. You can draw this. How many electrons Boron has? Five. Right? So, one has two and two has two. Two p one. So, diagram of two s one will be this. One, two and three. One vacant orbit will be this. Three bond, one electron pair makes a bond with this and two hydrogen atom overlap. So, one orbital vacant will be this. That is why when you write pi sigma positive charge is not always true. But when you write pi sigma vacant will be, okay, vacant orbital in this case is possible. Correct? So, what is the resonating structure? This comes over here and R s will be C h two positive charge C h double bond B h. Is it right? Yeah. Is it right? Yeah. It is not because we will have one negative charge. See the R s, all R s will have same charge. Neutral is the molecule will be neutral. Whatever structure you are drawing, resonating structure that must be neutral. This is a neutral molecule, right? So, this cannot be the charged one because Boron takes extra electrons. So, must have a negative charge. It loses electron, positive charge on it. All these ions you see, one positive charge here, she also one positive charge. This is what the same thing. Here we have one positive charge and one positive charge. One positive charge, one positive charge, one positive charge, one positive charge. So, charge must be same in all R s. If you are not getting same charge, it means you must have done something wrong. So, is this a different type? No, no, no. The pi sigma vacant free, same type. But minus also. Minus because of extra electron. That is why we have a negative charge. See the conjugation is this. Pi sigma vacant free, that is what the question given. Pi sigma vacant free is type 1. Is it clear? One more question you see. How many R s possible? What is that now? So, what if I see this little bit? So, can you draw it? For this one. For this one. What is what when this comes for this carbon atom? Further you see this is also in conjugation. This comes over here. So, there is no further resonance possible. So, we call this as 3. The given one also R s. That also will be included. Because the resonance is possible. So, we always give a resonating instance. Real hybrid we do not draw. Sir, there are only 2. No, the given one we also count. Got it. Now, the second type you write down. Second type of conjugated system, type 2. We will have pi sigma and lone pair. Pi sigma and lone pair. Example. First one, C h 2 double bond C h single bond C h 2 negative. This comes over here in this book. R s of this is C h 2 lone pair negative C h double bond C h 2. Okay. Oxygen. It is not. Nothing you will get. After some time you will realize, okay, nothing is coming. If you want to talk, we do not do GOC things. This is oxygen. This is also a wrong value. I said this is also a wrong value. Possible resonance in second one. Because oxygen has 2 lone pairs. 2 lone pairs. So, it satisfies this condition, right. Pi sigma and lone pair. So, 1 lone pair takes part in resonance. Only 1. 2 lone pairs. Because then oxygen becomes highly unstable. Oxygen 1 positive charge. See, when you draw the resonating structure, this comes over here and this pi electron shifts onto this carbon. So, it is R s is what? Double bond, lone pair negative charge and positive charge. Okay. So, this oxygen is highly unstable. We can draw the resonating structure, but oxygen is highly unstable. One more lone pair on this oxygen we have, but this will not take part in resonance further. Okay. Because over positive charge is already unstable. It is electron deficient now. So, it holds that lone pair of electron very, you know, formally. So, that further the tendency to lose electron is very less. Okay. This, any compound or element which is present in the ring, which has more than 1 lone pair. So, only 1 lone pair takes part in resonance. This property we will use later on. Okay. When we do the aromaticity, we will discuss over there also. In this, do we have resonance possible? How many resonating structure you will get? Only one. No, like a whole ring. Draw the resonating structure. Tell me how many R s possible? Only this in one hole there. No, it can be like the lone pair can go around the road. Which one? The second one, even if it is the second one, it can take part in the reaction. Could you get any more resonating structures? No, that won't take part. Even if it goes, could you get more structures? If you want to draw, if conjugation is there, then you can draw. But here there isn't, right? Yes, it's not there. So, what will be the third one of the lone pair? How many R s you are getting? I don't know. Why? First one is this. Is this right? Okay. Further you see, this is also resonance, this part. Conjugation. I will draw this side. Now this is also resonance, sorry, in conjugation. This comes over here. And this is pi electron straight up to this oxygen. 5 or 6, 6 R s possible. 2 by 1 plus 4, 6. Sir, the last one is the same thing. No, it's not the same. Double bond here, sir. Double bond on this side. This comes over here. This comes over here. Where? I mean before. Draw two more then. You get 6. After you shifted it towards that side, you got it in short. Where? Sir, from the second one to the third one. Yes. You shifted that double bond. See, electron all the way. Okay, so what is the double bond? This comes over here. And this pi electron under this carbonate. This carbonate is negatively transferred. Now again it comes over here, this pi electron. Okay, last one, fourth one. How many R s possible? No, it's the same thing. 5. So 5 possible. This one I have drawn here. This one I have drawn here. So this is one only, so 5 R s possible. 5 R s. Okay, see on the ring the negative charge is coming on alternate carbon if you see. What is this position on the ring? If you have a benzene ring like this. What is this position? You know these two positions? This is called ortho position. Yeah, ortho position. This is ortho, this is also ortho. We have two ortho positions. We have one para position. And two meta positions. So it can't come to an end. Yeah, that's all. So one thing is very important. In this position we got ortho, para and meta position in benzene ring. Okay. When you have OH present on the ring, molecule is this. And any electrophile reacts with this ring. The reaction of an electrophile and phenol. OH on the ring is phenol. Okay. That electrophile is what? Electrophile is positively charged electron-deficient species. So chances for this electrophile to attach it will be at ortho or para position. Why? Because we are getting negative charge more electron density at ortho and para position. Oh, because of resonance. Resonance we are getting negative charge on ortho and para position. That's why E plus will attach at ortho and para position. Both fraud you'll get. One will be major. Other one will be minor. Okay. That we'll discuss later. So what do major B also? Depends. Because... It's like two ortho is not para. Correct. But sometimes what happens will have a hydrogen bonding here. So in that case, ortho is more stable. In general, here we have steric hindrance. It makes the ortho position less stable. So para is more stable in that case. So that depends. I'm not talking about which one is major or minor. But you'll get ortho and para substituted product. Not meta substituted product. Got it? What do you want? Ortho is this solution. So what is this? This two positions. Okay, those two. Para... So any group is attached here. So these two positions are ortho position. These two positions are meta position. And this one is para. So we have two ortho, two meta, and one para position in a ring. Correct. One more thing I'll tell you which is not required here. All electron releasing group. See, OH group is attached on the ring. Correct. And OH group, you see it is releasing electron. Correct. It is releasing to the ring. So all electron releasing group are ortho para directly. Because in case of electron releasing nature, we'll have more electron density at ortho and para position. Right. So this group, increases electron density at ortho and para position. Hence we call it as ortho para directing group. Right. Ortho para directing groups are activating group also. Okay. So are there any meta? Yes, no. Meta activating, we have only one that is not meta activating. We have deactivating but ortho para director is halogen. Meta activating, we do not have. Let it be. We won't be confused. No, because it's two positions. What, what, what? If a nucleophile was the nucleophilic ring, then it should attack the ring. All meta? All meta, because nucleophilic is negatively charged. Yes, but then what about the oxygen when the electron goes away? That has a positive charge. In that case, this group should be electron withdrawing. Electron, and talking about electron releasing group. Correct. So, but how is oxygen electron? Because oxygen was blown. But it's highly electronegative. That's what. If you draw this structure, this becomes unstable. But if somebody asks you whether this has electron with drawing nature or releasing nature, it is electron releasing in nature. Because oxygen is more electronegative than this carbon. This sigma bond is shifted towards oxygen atom due to eye effect. But this effect is plus M if it is more dominant. Plus M. Mezomeric effect. So this is plus M. Plus M. What's my name? I'm coming to that. Sir, but in this case, the most stable ones are the ones in which there is no charge in it. So it's just the ring. Most stable one is the one which is neutral. The one where oxygen is not positive. So that have the most character. It'll have the most character in that. Then the other one. Then the other one. This one and this one. Okay. Got it? So only one thing you see, RS, we have seen how to draw the resonating structure. Two ortho, two metal and one para position. When you have an electron releasing group present on the ring, it is ortho para directing group for an electrophile. We'll discuss this later also. I'll give you one. Now about this one, how many RS possible? Five. Five RS possible? Yes, five RS possible. And when you draw the RS for this, resonating structure, you'll get this negative charge on each carbon atom one by one. The first RS is this. First, second, third, fourth, fifth. And all this RS you see, the negative charge is one by one is present on each carbon atom. Yes. Your first, third, fifth. So negative charge is equally distributed. Resonance hybrid for this structure is pi electron distributed equally. And each carbon atom will have minus one by five charge. The actual. So this is the resonance hybrid, radial structure. And this kind of molecule in which the electrons pairs are distributed equally we call it as aromatic compound. Okay, so this is an aromatic compound. Sir, sir, will this compound try and how will it become neutral? It can't bond with one by four by five electron. Why don't you want to try it? It's negative, right? No, no. But negative, but it is stable because it is an aromatic compound. Sir, aromatic compound. I think it doesn't smell. What? Sir, aromatic has nothing to do with smell. It is related only with stability of the iron or molybians. Oh, that's how it smells. And that aroma is a different thing. It is aromatic. It's not that popular. It is a factor by which the molecule iron gets extra stability. Okay? One by five is the charge of each carbon atom. See, this minus one charge is distributed around five carbon atoms. So one carbon atom will have minus one by five electron. So aromatic molecule will be more stable compared to a non-aromantic molecule. Yes, obviously. Always. Always. There are... aromatic compounds are actually more stable. Most stable, in fact. Okay? We have one more factor here which we call as equal dancing resonance. So in case of dancing resonance, we have the stability over part of the stability of aromatic mixture. Only one condition, but that is just an exception. We will see that. So that is more stable than aromatic. Dancing resonance. The definition of aromatic is the most stable. There we go. It is generally... See, the point is like I said, in resonance we have only five electrons and lone pairs are involved. But in dancing resonance sigma electrons are involved. That is one special kind of resonance. Okay? In that case, that stability is more than the aromatic amount. In that also, we have one factor. So I am not going into that. We will discuss that later. But apart from dancing resonance, when the compound is aromatic, it is the most stable compound. Okay? So we will see the factors affecting stability of compounds or ions in which we have aromaticity, one of the factor. We have non-aromatic, one of the factor, anti-aromatic, one of the factor, plus M, minus M, plus H, minus H, plus I, minus I. All those factors we have. So when acids congregate, bases aromatic, they will become more stable. More stable and it is a very good acid. Okay? Two acids you have. When the conjugate base of one is aromatic, other one is not aromatic. So we can say the previous one, the first one is more acidic than the second one. Okay? So this is the second type of conjugated system that is pi, sigma, and lone pair.