 write down it is a delocalization of pi electrons or lone pair. It is the delocalization of pi electrons or lone pair. Next line, it is the method of representation of method of representation of one molecule or ion into more than one structural unit into more than one structural unit in which more than one structural unit in which the position of pi electrons or lone pair in which the position of pi electrons or lone pair is different. So, resonance in a single line you can say it is delocalization of pi electrons or lone pair. For example, C H 3 C double bond O O minus negative charge and minus means what we have three lone pairs on this oxygen atom. In this oxygen atom we have two lone pairs present. So, the resonating structure of this ion is separated by this error, double parent error. This error we use to represent the R S resonating structure. Now, when you drop the resonating structure of this, this lone pair forms a bond pair of electron and this pi electron shifts onto this oxygen atom and it gives what? C H 3 C single bond O negative charge and double bond. This is the resonating structure. Another resonating structure we can take the example of benzene. This is also a resonating structure because the two molecules are different. We have single bond between first and second carbon in double bond here. 3 4 5 and 6. This is the resonance. Now, like I said, all resonating structure or the different representation that we have, all these different representations are called as resonating structures or canonical structures or we also call it as contributors. Resonating, canonical or contributor. Resonating structure will take the reference of R S. R S means resonating structure. Contributor means also the same thing because all these R S contributes into the real structure. We know all these resonating structures are what? These are imaginary structures. These are not real structures. The real structure is the combination of all these resonating structures and that we call it as resonating resonance hybrid or resonating hybrid. So, resonating hybrid or the resonance hybrid for this structure is this one. Benzene and this pi electron is delocalized over the entire ring. This is the real structure, hybrid and radial structure, resonance hybrid or radial structure. For this, the resonance hybrid is what? C X 3, C O O and the pi electron delocalized from this oxygen to this oxygen delta negative and this is the hybrid or radial structure. All these things we have discussed already in chemical. Now, the next question is in which of the molecule the resonance is possible? When you draw the resonating structure, first of all you should understand that whether the resonance is possible or not. So, for that we have certain conditions that condition must be satisfied by the molecule or ion then only the resonance is possible. So, what are those conditions that we should understand? Different types of conditions we have different different types of conjugated system. We have in all these conjugated system the resonance is possible. The first condition I will just tell you one by one. Do not write now I will just tell you. The first condition is molecule must be planet. Planet molecule is required because delocalization means what? It is the shifting of electron pair. This electron pair maybe it is a pi electron or lone pair. So, benzene example if you take this benzene has a p orbital here. So, there is a p orbital here also we have a p orbital and all these orbitals are parallel. It is parallel and that is why the shifting of electron pair is possible. If it is not parallel not planar then one p orbital like this another one like this. So, overlap is not possible. When p orbital faces each other then only overlap is possible and pi bond forms. That is why a planar molecule is required. If I draw this structure you see this structure if you rotate this structure this way it means when these two p orbital overlaps it forms a pi bond. We will get a pi bond here. When these two overlaps we will get a pi bond here and when these two overlaps we will get a pi bond. So, if I say these two are overlapping. So, why not these two orbital can overlap? These two may also overlap because both are parallel. So, we can have overlapping of these two this two this two this two all overlap is possible. That is why this pi electron continuously is continuously shift from one point to another and this delocalization of electron is nothing but. The first condition is planar molecule point is required because then only the orbitals will be facing each other and then only shifting of pi electrons. Got it? So, this means that the electron keeps moving in a circle. Yes. But when it is next to one of the same molecules won't the attraction with the nucleus be strong so just stay there. No, not possible. Attraction of nucleus is everywhere wherever the pi electron we have attraction will be same at every point. Because of this delocalization of electron it gives stability. I did the center of two C atoms the force will be weak. That will attract this sigma electrons not pi electrons. Sigma electrons will attract but that is also that much attraction is not possible because both carbon atom has same electron. Sorry same electronegativity. So, that would attract the sigma electrons also. But whatever attraction is there it is there for all the electrons but that won't affect the shifting of electrons into different orbitals. Got it? So, first of all when you solve questions in organic chemistry you should have this idea that in the given molecule whether the resonance is possible. So, how do we know that the resonance is possible? For that we have certain conditions one I have told you that molecule must be planar. Because in planar molecule only the orbital will be parallel to each other and then shifting of electron possible. So, condition and characteristics of resonance and characteristics of resonance and characteristics of resonance and resonating structures. See one more thing I will tell you I said that this resonating structure R s are contributors why these are contributors because this molecule also contributes into the real structure and this also contributes into the real structure. Some property this hybrid structure will have the property from this molecule and this molecule both. When the hybrid molecule takes property or the property of hybrid molecule will be because of its all resonating structure. Now, if this two both R s are equally stable right there are two possibilities one is what if only two R s is possible for example this one right. So, this two can contribute equally into the real structure this is one possibility another possibility is what one contributes more than the other right. So, in these two cases we have two types of contributors also one is equal contributor another one is unequal contributors correct equal contributors are those R s which contributes equally into the real structure right means half property is the half property is equal contributors correct. So, how will they not be equal contributors? I am not saying they are not. No sir, but then we are saying in some cases there is no problem. Now, we will discuss that I mean we cannot finish resonance in five minutes I am just giving you an idea what are possibility we can have. So, it is like if you have equal contributors you will have half property from here and half property from land one what do you mean by half property. Half property means they contribute equally suppose if this structure has ten properties you talk about venting point you talk about boundary you talk about bond is ten bond land ok. So, it has certain number of properties we can define what is the bond length of this C 1 C 2 carbon bond order of C 1 C 2 carbon it is two right what is the bond order of this C 1 C 2 carbon 1. So, this wants that the bond order in real structure should be one C 1 C 2, but this wants the bond structure should be two. So, the actual bond structure you see 1.5 that is what it means it will it will be in between the average of the two right in case of equal contributors, but when the contribution is not equal suppose now I am assuming this is these two are unequal contributors right then which one will contribute more that depends upon the stability of the contributors. If this one is more stable then this structure wants that the bond order of C 1 C 2 carbon should be two right and this wants what but since it is more stable. So, we can say the bond order of C 1 C 2 carbon should be closer to two because it is more stable it contributes more. So, we can say that the bond order of C 1 C 2 carbon will be in the range of 1.5 to 2 exact calculation of bond order is not possible in case of unequal contributors. Equivalently we can take average, but unequal contributors we cannot find out the exact bond order of the bond right any bond, but we can give you a range we can write a range of it that right like in this case if this one is more stable then obviously the bond order should be 1.5 to 2 we can say 1 to 1.5 will be there if this one is more stable understood. So, these are the application of resonance with resonance what we all thinks we can conclude the first application is to find out the bond order of the bond in a molecule and if you have the idea of bond order we can find out what bond strength we can find out bond length because you know the bond order and bond strength are directly proportional bond order and bond length are inversely proportional all these idea we can have. So, one by one we will see that but first of all we write down the condition and characteristics of resonance the first condition is the molecule must be the molecule must be planar the molecule must be planar the position of atoms next one the position of atoms and the position of atoms and sigma electrons are fixed position of atoms and sigma electrons are fixed. Sigma electron does not take part in resonance it is always pi electrons are lone pairs but in IE effect we have sigma electrons right sigma electron does not take part in resonance right next point to write down it is independent of distance independent of distance like IE effect it depends upon distance right it is independent of distance in resonance next one in resonance pi electrons and lone pair pi electrons and lone pair are interconvertible lone pair can form pi bond and pi electrons can form can convert into lone pairs rs all rs are separated by all rs are separated by this double heredero right all rs are separated by double heredero right on resonating structure rs does not explain all the properties of the molecule or ion the resonating structure does not explain all the properties of molecule or ion all resonating structures are imaginary or hypothetical structures these are not real structures all resonating structures are imaginary or hypothetical structures for molecule to show resonance molecule or ion to show resonance so resonance it must be conjugated it must be conjugated the system must be conjugated ok if conjugated system is there a system is molecule or ion it must be conjugated and planar that is the two condition we have two sure resonance now what is a conjugated system what is one minus is it possible to both the o's to have carbon bond in C to have one minus charge no its not possible but c1 won't be stable now its won't be stable because of charge it is unstable carbon cannot have 5 bar This is the most important one, if you want to know whether the resonance is possible or not, the system must be conjugated, molecule or ions must be conjugated. The first type of conjugated, there are 6, 5 or 6 different types of conjugated system we have, 5 or 6 different type of conjugated system, we will see one by one also. The first type is, we will have a lone pair present on the first atom, we will have a pi bond, then sigma bond and then we will have a vacant orbital, write down 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, positive charge means what? It is one orbital is vacant because it has, the electron is not present on the carbon atom, one electron is lost. So, this positive charge you can also write this or when you write this sigma pi sigma positive charge, that is also same type of conjugated system, both are same, this one or this, means when you write positive charge, it means one orbital is what? One orbital is vacant, that is what it means. 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, just the first one I will just make you understand this orbital diagram. This pi orbital is vacant, here we have a p orbital and here we have a p orbital. You see all the p orbital are parallel and this forms lateral overlapping, it forms pi upon. So, what happens if this is the orbital is vacant? So, it can take the two electron into this orbital, this orbital can also overlap, C H 2. So, this gives you another resonating structure, which is C H 2 C H C H 2 and this orbital will be what? Empty and have one orbital. So, this is the resonating structure. So, pi means double model, so overlap is double model. So, that structure is also an example of type 1. No, not type 1, you have negative, pi sigma negative. So, pi sigma lone pair. So, that double model oxygen has an empty structure. Yeah, this one, this one, pi sigma here we do not have positive charge. So, would you say that that O has a vacant orbital, because it can have orbital. No, no, no it is not, it has one lone pair, extra here vacant orbital means it is an electron deficient. Vacant orbital means what? It has lost its electron, right. So, this has negative charge means it has extra electron present. So, does it always mean 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 is shipped over here and it is here, CS2 positive charge, CN double bond, this is the RSOS. Orbital diagram is not required, ok. Directly you can shift electron and right. Can you draw the resonating structure in this? Check also if whether resonance is possible or not. 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, let us just see these three carbon, one, two and three, these carbons are sp2 hybridized, ok planar. And it is pi, sigma and vacant orbital, on this positive charge means vacant orbital on the carbon atom, right. Pi, sigma positive charge, type 1 conjugation. So, when you draw the resonating structure, this pi electron comes over here and the RS will be, we have a pi electron here in most structure, ok. So, that can shift to all 6 phase. No, see at this side, this side we do not have resonance, it is not conjugated. From here to here delocalization is possible, this part is not involved because all are sp3 hybridized. This side we do not have conjugation, ok. 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 like this. 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? 3 RS possible. Is it clear? This one is hybrid. Pi electrons delocalize. 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 we have this carbon, this carbon, this carbon, this carbon. Delta positive. Is this a conjugated system? Last one. So, it does have a vacant orbit. Is this a conjugated system? Boron has one vacant orbit. Boron has a vacant orbit. When there is 3 bonds, that is what is electron deficient, it can accept clear of electron and behaves as the noise has to be. You can draw this. How many electrons Boron has? 5. So, 1 is 2, 2 is 2, 2 is 1, 2 is 2, 2 is 2, 2 is 2. So, diagram of 2 is 1 will be this. 1, 2 and 3, 1 vacant orbital beam. 3 bonds, 1 electron pair makes a bond with this and 2 hydrogen atom overlap. So, 1 orbital vacant beam. So, that is why when you write pi sigma positive charge is not always true, but when you write pi sigma vacant beam, vacant orbital in this case is possible. So, what is the resonating structure? This comes over here and R s will be C h 2 positive charge C h double bond B h. Is it right? Yeah. It is not because we will have one negative charge here. See the R s, all R s will have same charge. When neutral, the molecule will be neutral. Whatever structure you are drawing resonating structure, that must be neutral. This is a neutral molecule. So, this cannot be the charged one because Boron takes extra electrons, so must have one negative charge. It loses electron, positive charge one. All these ions you see, one positive charge here, here also one positive charge. This is what is the same thing. Here we have 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. The pi sigma vacant beam, same type. So, but minus also. Minus because of x y electron, that is why we have a negative charge. See the conjugation is this, pi sigma vacant beam, that is what the question given. Pi sigma vacant beam type one. Is it clear? One more question you see. How many R s possible? What is that? What is the value? It is a little bit. So, can you draw it? For this one. So, what is what when this comes on this terminal? The terminal you see, this is also in conjugation. This comes over here. There is no further resonance possible. So, we call this as 3. 3, the given one also R s, that also we include. Because the resonance is possible, so we always give a resonating distance, real hybrid we do not draw. No, the given one will also come. Got it? Got it. Okay. 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. For 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. There is no content in this. After some time you will realize, okay, nothing is coming. If you want to talk, we do not do GOC then. It is not in the syllabus, it is not in the book. What is that round thing? It is oxygen. This is also round by the way. I said this is also round by the way. Possible resonance in second one? Oxygen has two lone pairs. Two lone pairs. So it satisfies this condition, right? Pi sigma and lone pair. So one lone pair takes part in resonance, only one. Two lone pair involved only one. Because then oxygen becomes highly unstable. Oxygen one positive charge. See when you draw the resonating structure, this comes over here and this pi electron shift onto this carbon. Because R 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 won't take part in resonance further. Okay. Because over positive charge already unstable. It is electron deficient now. So it holds that lone pair of electron very, you know, formally. 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 one lone pair. So only one lone pair takes part in resonance. This property we'll use later on. Okay. When we do the aromaticity, we'll discuss over there also. In this, do we have resonance possible? How many resonating structure you'll get? Only one. No, like a whole ring. Draw the resonating structure. The resonating structure. Tell me how many are as possible? Only this in one hole. No. The lone pair can go around the hole. No. And not come. It's not like... Yeah, yeah. Five. Five. Six. Seven. Okay, sir. In the second one, even if the second one could take part in the reaction, could you get any more resonating structures? No. That won't take part. But if it could, 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 are as you're getting? I'm not getting any. I'm only one. Why? First one is this. The answer is this. Is this right? Yeah. Okay. Further you see, this is also a resonance, this part. Conjugation. I'll draw this side. Now this is also a resonance. Sorry, in conjugation. This comes over here. And this is pi electron strip, not this oxygen. Five or six. Six are as possible. Two and plus four. Six. Sir, the last one is the same thing as the other one. No, it's not the same. No, no, no. Double bond is the same. Double bond will be on this side. It's supposed to come here. Where? Down. Right. What's the number? What's the number? It's supposed to come here. This comes over here. Where? I mean before start. Draw two more then. Six. Six. You know, after you shifted it towards that side, you got it in short. Where? So from the second one to the third one. See. You shifted that double bond. See, electron all the way. This comes over here. Okay. And this pi electron under this carbonate. This carbonate is negatively charged. Now again it comes over here, it's pi electron. Okay. Last one. Fourth one. How many are as possible? What's the difference between the second one of that and the first one? What? No, it's the same thing, no? Yeah. So five possible. Yeah. So this one I have drawn here. This one I have drawn here. So this is one only. So five are as possible. Five are as possible. 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 Beijing ring like this, what is this position? You know these two positions? This is called ortho position. Yeah, ortho. This is ortho. This is also ortho. We have two ortho positions. We have one para position. And two metaposition. So it can't come here? Yeah, that's all. So one thing is very important. In this position we call it ortho, para and metaposition in Beijing ring. Okay. So 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 we do? Be ortho. Depends. Because there are two ortho and not para. Correct. But sometimes what happens will have hydrogen bonding here. So in that case, ortho is more stable. In general, here we have steric hindrance. That 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 it? This two positions. Okay, both two. Para is... 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-para position. Hence we call it as ortho-paradirecting group. Right? Ortho-paradirecting groups are activating group also. Okay? Sir, is there any meta-activating group? Yes, no. Meta-activating, we have only one that is... Not meta-activating, we have deactivating but ortho-paradirecting is halogenist. Meta-activating we do not have. Let it be. Let it be. Oh, because it's too positionless. What? If a nucleophile was... If there's a nucleophile in the area, 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. So it will be more attached to the water. In that case, this group should be electron withdrawing group. And talking about electron-relasing group. So, but how is oxygen electron? Because oxygen has two pairs. But it's highly electronegative. That's what. If you draw this structure, this becomes unstable. But if somebody asks you whether this has electrons with drawing nature or releasing nature, it is electron-relasing in nature. Because oxygen is more electronegative than this car. But this sigma bond is shifted towards oxygen atom. Due to eye effect. But this effect is plus-sum effect which is more dominating. Plus-sum. Mesomeric effect. So this is plus-sum. Plus-sum. I'm coming to that. So in this case, the most stable ones are the ones in which there's no charge in it. So it's just the ring. The most stable one is the one which is neutral. The one where oxygen is not positive. So will that have the most character? It will have the most character. Then the other one. Then the other one. This one. And this one. Okay. Got it? So only one thing you see, RSVF's seen how to draw the resonating structure. Two ortho, two metaheism and para-position. When you have an electron-relasing group present on the ring, it is ortho-paradirecting group for an electrophile. Okay? We'll discuss this later also. I'll give you one here. Now, about this one, how many RS possible? 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. So negative charge is equally distributed. The resonance hybrid for this structure is pi electron distributed equally. And each carbon atom will have minus one by five charge. The actual. Okay? 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, we'll just compound try and how will it become neutral? It can't bond with one by four by five electron. Why do we need to do it alone? It's negative, right? But negative, but it is stable because it's an aromatic compound. Sir, aromatic compound. Sir, I don't think it does smell well. It is related only with the stability of the ion or the molecule. Sir, I thought it's minus one by five. That aroma is a different thing. It is aromatic. It's not that popular. It is a factor by which the molecule ion gets extra stability. Okay? So what's my one by five? One by five is the charge on each carbon atom. See, this minus one charge is distributed around five carbon atoms. So one carbon atom will have minus one by five. An aromatic molecule will be more stable compared to a non-aromatic 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 the stability of aromatic. Only one condition, but that is just an exception. We have, we'll see that. So that is more stable than that aromatic? Dancing resonance. So the definition of aromatic is the most stable? Take that. There we go. It is generally, see, see the point is, like I said, in resonance we have only pi 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, the stability is more than the aromatic compound. In that also, we have one factor. So I'm not going into that. We'll discuss that later. But apart from dancing resonance, when the compound is aromatic, it is the most stable compound. Okay? So we'll 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 conjugate bases aromatic, then it will become more stable. More stable and it is a very good acid. More acidic. Yes, 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. Okay? So this is the second type of conjugated system that is pi, sigma and lone pair. The third type you write down, the third type of conjugated system is pi, sigma and free radical. Pi, sigma and free radical. One electron, right? One dot means one single electron. Example of this is CH2, double bond CH, single bond CH2 and free radical. Okay? So in this we draw the resonating is like this. So this pi electron, one electron comes over here, it is in half arrow, because homolysis it is. One electron is taken up by this carbon, another electron is taken up by this carbon. Okay? And this free radical, free radical combines, forms a double bond and the structure is CH2 radical, CH double bond, CH2. Got it? Any hours possible here? Sir, sir. Right? All is possible. Free radical is any carbon atom, which has three bond pair and one unpaired electron. So its number of electron is what? Seven. Seven electron. But it is not, it is electron deficient, but cannot behave in Lewis acid, because more than eight electron is not possible. If you have, if you remove this electron also, it becomes carbon atom. Right? If you have one electron radical, add one more electron, it becomes a carbon atom. So it is an intermediate, which forms during the process of a reaction. Generally free radical forms in those reactions where we have pure sunlight, hatching heat. Okay? So in carbon-carbon bonds, homolysis takes place. Homolysis means when you have a carbon-carbon bond, it has two electron on it. Homolysis and heterolysis, two types of bond cleavage we have. Right? So one heterolysis is when the two bond pair is taken up by only one carbon atom. Okay? C negative C positive, like this, which is in this case not possible, because one of the atom must be more electronegative. Like suppose we have carbon and chlorine bond, carbon-carbon bond. Chlorine is more electronegative. This bond pair is taken up by this chlorine and it forms C plus the carbocation and Cl minus. Okay? So this is heterolysis. Now when you break this bond, homolysis, one electron is taken up by this carbon, one by this carbon. This forms a radical link, this carbon-carbon radical link. And it is possible when you use sunlight generally. Okay? So free radical is, what free radical is, is an intermediate, which has three bond pair and one unpaired electron, paramagnetic energy. Got it? Type IV, we have pi sigma pi. This is type IV. Okay? Alternate pi bond, CH2, double bond CH, single bond CH, double bond CH2. Its RS is what, this pi electron comes over here and this pi electron on this carbon. So they get CH2, positive charge, CH double bond, CH2, CH2, negative. So this is going to have a lesser stability. In comparison to this, no, if you compare the stability of this two, right, if you compare the stability of this two, this one is more stable. More stable. Two pi bonds. But while reaction, it won't give reactions with this. The addition takes place with this carbon and that carbon. We'll discuss that in alkene chapter. Okay? Reaction of HX on alkene. Okay? It gives conjugated dienes, it is. What is the name of this compound? 13 butadiene. 13 butadiene. Okay? So 13 butadiene gives you one-four addition, not one-two addition. The reason we'll discuss in alkene. Okay? This is the type IV system here. What about this one? In this, do we have resonance possible? Do we have as possible in this? Sir, that one is not possible. It's not possible. Why? It's possible. It's possible. But I see him up high. So how do you know if something is playing? All carbon atom are sp2 hybridized. So in that one? No. So sp2 means? Every time? sp3. Sp3. Sp3. So it's a diagram, plane and the jubilee. No. It will be in the same plane. Carbon, three-part in the same. Sp1. Sp3. Sp3. Sp3. Sp3. Sp3. Sp3. Sp3. Sp1. Sp3. Sp3. Sp3. Sp3. Sp4. Sp4, yes. It's very. Is it that the other structure also should have the other structure? In this it is not possible. But if you have further conjugations in this slide, then also it is possible, of course. Sir, this is also still resonant. It is resonant. See resonating is such that you can draw till conjugation is there. Like I affect me what we say it is distance dependent effect because sigma electrons are involved. Here distance factor is not there. But it requires only conjugation. So in the molecule we can draw resonating structure till conjugation is there. Like here you see this part is not in conjugation, sp3 hybridized. So we can draw resonating structure of this one from this carbon to this carbon. This is not involved in this. So this resonating structure is what when you draw it, we will get a double bond here, negative charge here and positive charge. For this one if you draw this pi electron comes over here and this pi electron jumps onto this oxygen. This is CH2 CH single bond C4 minus H double bond and positive charge. Can we draw this way? This pi electron comes over here and this pi electron comes over here. Reverse. That is what I am asking. Can we draw this way? This comes over here and this comes over here. You can draw. Draw the name. Second name also. Yes that will be a different resonating structure. But that is the least, not stable at all. Because you see oxygen is more electronegative. So it has tendency to attract the electron. That is why the best way to draw the RS is the shifting of the electron towards oxygen. Not away from oxygen. That is what, see, you can draw any structure. Whether it is stable or not, that is a different event. But whether you can draw it or not, you can draw it. But usually what we say, usually we will draw the structure and we shift the electron towards the mode electronegative. Second one sir. You can have more red double bond, single bond, positive. That is the type one that you can draw. Where? No, no. See what you can do. You can do one thing. This comes over here and this comes over here. This gives you the same. Sir, sir, can you? Why? This is this. What do you do? Whatever you want to do, you will get this one. Draw it and see. See, what you can do, you can shift this here. So we will have a double bond here. And this is positive charge. This positive negative makes a bond. Carbon-access carbon with positive negative charge is never stable. Carbon-carbon double bond, if you have, you see this. Carbon-carbon double bond. If you want to change this structure into this, see single bond, see positive, negative like this. For this, you have to break this pi bond. Correct? And to break this pi bond, you have to provide a lot of energy. Right? This is the higher energy molecule right? No. That you can do. It's the same thing. That you can do, it's not different. If you draw, if you, as you go over here, same thing. Understood this? So is that a structure where minus is on, where the plus is on? Generally, we don't consider that. But if they ask you to draw, you have to draw. Shifting of electron, you can do. Molecule won't do. Molecule, though, automatically the shifting of electron takes. Listen, we are not, see, benzene if you are taking. Right? We are trying to study the structure of benzene by shifting the electron. But when benzene exists, the electron is already delocalized in its own way. So when you draw, if you want to draw, you can draw it this way, reverse. But what molecule will do? Oxygen takes electron. So it goes this way. Yes, you can draw, but molecule won't do that. That's why that structure, you won't do that. Okay? Understood this? Now apart from these four, there are two more conjugated systems possible, which is not that important. I'll just write down the type. Type 5 and type 6, you write it down. Type 5, we have a vacant orbital p, sigma bond, and a lone pair. Vacant orbital, sigma and lone pair. Okay? And type 6, we have a lone pair, p orbital lone pair, sigma and vacant p. Vacant p and lone pair. One example we have already done, p of 3. It's lone pair in p orbital. Lone pair in p orbital, sigma bond vacant d. Vacant p, sigma bond lone pair. Example is this. P orbital has vacant p orbital, we know this. Vacant p and fluorine has lone pair on it. So back bonding is possible. This is back bonding. That's also resonance. Back bonding is nothing but resonance. Each one. Each one of us. It's possible. Same thing. Sir, isn't back bonding just a double bond? No, it's not. It has double bond characteristics. It has pi bond characteristics. Which forms when the sigma bond has been formed. After that, the central atom, if it has vacant orbital, and the other atom has the lone pair, and if science is comparable, then this has tendency to donate electrons. That is back bonding. After the sigma bond is formed, the tendency to donate electrons is back bonding. But that BCL3, that bonding is not that strong. Because chlorine may have 3p orbital. Or here we have 2p. 2p 3p overlap is not that great. For bromine also, even lesser is stable. For BF3, the back bonding is strongest. So why is this different from vacant? Where we have vacant orbital? Like this is vacant p. The first one has the lone pair. Got it? Vacant d, the example is this. We have CH2, lone pair on it, single bond Cl. Chlorine is the element of third period, right? So it has vacant orbital. Vacant d orbital. So in this, the resonance is possible. CH2 double bond Cl, negative, here we have positive. Not that important, but in this case, what? All these are the 6 different types of conjugated system, where the resonance is possible. Now like I said, the resonating structures or contributors, we have 2 types of contributors. So write down next, contributors or resonating structures. We have 2 types. One is equal contributor. Other one is unequal contributor. The equal contributor example we have, see both molecules are identical, right? This one, equal contributor. Even if you draw the benzene ring, this is also equal contributor, benzene ring. The stability of RS in case of equal contributor is generally more than the stability of unequal contributor. Stability will see that comparison, but unequal contributor, for example is CH2, negative, C double bond O, H. What is the conjugated system? So in benzene ring, can't we apply pi sigma pi N? Will it end up to be the same thing? Yes, it should. You can draw. Okay? And when you shift this here, this becomes negative, this becomes positive. Double bond. Here we have positive, here you have negative. This negative comes over here, this pi electron goes here. Double bond, positive. What's the one in the middle? We have like positive and negative, won't that be a separate estimate? Yeah. Which one? You're talking about this one. Double bond, positive and negative. Yeah, that one. This one? Yeah. No, it is also not. Why? See, this looks like different structures, right? Yeah. Positive, negative charge here is highly unstable. Okay? And when this comes over here, see, you can take it this way also. We don't draw like this, okay? We just continuously shift this electron there like this. This comes over here, this comes over here, this comes over here. And we'll get this. Yeah, so we'll, we don't write this, actually. Okay? So it converts into this, and these two are the resonating structures. So when you draw the hybrid of it, you'll draw the entire molecule. Sir, so there are only two resonating structures. No. See, if they ask you, you know, if they ask you how many are as possible for benzene ring. So we'll contribute, we'll count all these structures. But this two structure, if you're talking about, I'm talking about this two structure. This to the contribution of this, and this is same in the real structure. This two are equal contributors to these two structures. Okay? This one, the stability is lesser than this one. Even any charge species is lesser stable than these two structures. Right? So that is the stability. But if you talk about the stability of these two, it is same and its contribution is also same. So which type is that one? This one. Yeah. So it will be pi sigma lone pair, or pi sigma, any type you will consider. So we can start at one type of conjugation and then end up with a different conjugation. Yes. In between, you can have any kind of conjugation. If it is conjugation, shifting of it like this. Understood this? Yeah. Now this molecule you see, this comes over here. CH2 double bond CO minus CS3. It is unequal contributed. Okay? The contribution is not same. Even if take this example, double bond O, this part is not involved in resonance. This part is not involved in resonance. Why? Or a conjugation? Why? Why? Why? Why? Why? Why? Why? Why? Why? Why? Why? Why? Why? First one, why is it here? Huh? First one. This part is not involved in resonance. They have conjugation only here, not this part. So it is unequal consequences. When contributing, when the resonance is there in the entire molecule, then H1 wasn't actually associated with it. It's not the parent chain. Parent chain is this only. Parent chain is this only. Now one thing you see, like in equal contributed structures, the stability is same, exactly same. And here the contribution is also same. But here, since the stability is different, so contribution is also different. And that's why it is unequal contributors. So in this case, we can talk about major contributors and major and minor contributors. Major contributors is that structure, which is the most stable structure among the resonating structures. And it contributes maximum into the real structure. Minor contributors, so stability is less. And here major and minor is not possible because all RS has the same stability. All are equal contributors. Sir, but in that one, like in this one, I have to add one. No, no, no. If you draw this, it's fine. But these two are equal contributors. We'll have that. Basically, we'll have six, seven different rules to find out the stability that we'll do now. So next slide down. Like I said, one of the structures is major contributors and other one is minor contributors. So which one is major or minor that depends upon the stability of the RS. To find out the stability of RS, we have six or seven different rules, three different rules we have. So according to that rule, we'll find out the stability of RS. And this is the stability of RS comparison of the same molecules or ions. We cannot compare the RS of this and this. Same molecule to RS we are comparing. So write down the, write down next, factors affecting the stability of RS, resonating structure. Factors affecting the relative stability of RS. So we'll compare the relative stability of RS in the same manner. First we have to apply rule one, then rule two and rule three like this behind. So rule one in this, you write down, the structure which has more number of pi bond is more stable. The structure which has more number of pi bond is more stable. So first example in this, suppose the molecule is this. Can you draw the RS in this? Can you draw RS in this one? Suppose this is A and this is B. Which one is more stable? A or B? A. Because it has more number of pi. So stability of A is more than the amount of B. What is the RS for this one? This comes over here. Second one has more number of pi bond. So B is more stable. Just you have to follow the rules in this. First rule if fails then we use the second rule, third rule and so on. Now the property of this is what? Suppose you have to draw the RS of this two molecule. So RS will have a structure like this, single bond O and this pi bond is delocalized from this oxygen to this carbon. So we will have a pi bond characteristic here. Pi bond characteristic here. Now which one is more stable here? A is more stable. If the question is the bond order of this, C1 C2 carbon if you want to compare it, the bond order of C1 C2 carbon. So we can definitely say that the bond order of C1 C2 carbon is closer to 2 since this is more stable. This contributes more into the real structure. So it is closer to 2 and we can say the bond order of C1 C2 carbon lies in this range. And if they ask the question for equal contributors, remember we cannot find out the exact value of bond order. We can give this range. The option also will be in range only. Whether it is closer to 1 or closer to 2. Got it? First of all, the bond order here is what? 1. Here it is 2. So it lies in this range 1 to 2. And since this one is more stable, which has bond order 2, so we can say that bond order is closer to 2. So it must be more than 1.5 but less than 2. Because if both are equal, then bond order is what? 1.5, average of the two. Since it is not equal, so this is more stable. So closer to 2. Understood this? Rule number 2 you write down. And remember always from these rules, we will compare the relative stability of RS of a given molecule. We cannot compare the RS of the two different molecules. C is this C1, C2 bond order. C1, C2 bond order. C1, C2 bond order. Okay? We will do this bond order part also later on. Okay, I am just giving you an idea. Rule number 2 you write down. When the number of pi points are equal, when the number of pi points are equal, then stability decreases with charge separation. And stability decreases with charge separation. Charge separation means when you have one molecule is charged, another one is neutral. So neutral one is more stable in the other way we can see. So for this the example is, you see this one, benzene pair will have OH like this, again. And its RS is what? We have done this already, this comes over here and this comes over here. So its resonating structure is OH positive. Another example if you see, NH2 nitrogen has one lone pair. It is also electron releasing. Okay? This electron pair comes over here, this pi electron shift on to the star one. Now in this two example if you compare, how many pi bonds we have here? Three. How many pi bonds we have here? Three. So first rule we cannot compare, then we will have a charge separation here in this, plus minus. Here it is, no charge, right? So this one is more stable, A is more stable than B. Got it? Similarly in second one, again A is more stable than B. There is no charge separation. So by charge separation you mean like presence of the charge. Yes, charge present. Because that in the next. Oh, what all it is neutral? Yeah, so it is like the next resonating structure of that the minus moves one more down. Yes, yes, yes. So both of those we cannot compare using the second one. With this rule we cannot compare. For that we have another one. Okay. Next time, rule number three. Which is the actual physical distance between the charges doesn't matter. That is also a factor here. We will see that. Another rules we have. But that's not this rule. No. Like if you want to compare from rule one and rule two, another R is if you draw. These two structure we cannot compare with rule one and two. For that we have another rule. Okay. Next, rule number three. Charge separate is just the presence of charge. Charge present, that's all. Rule number three, write it down. When the molecule has equal number of pi bonds. When the molecule has equal number of pi bonds and charge separation. Equal charge separation and equal number of pi bonds. Then stability is directly proportional to the number of atoms having complete octet. Number of atoms having complete octet. Example you write down BX2. In this resonating, R is possible. Resonance in this is possible. Now it's possible. Okay. So when you draw the R's of this. Pi, sigma, vacant B. Boron and vacant B. This comes over here. BH2 double bond. CH. Single bond of CH. Here we have the negative charge. Now when you look at these two structures. Okay. How many pi bonds in the two structures we have? One. What is the charge separation? One negative charge. So we cannot do this from the first and second rule. Correct? Now this boron has how many electrons? Six electrons? Carbon has eight. Chlorine has more than eight. Right? It has expanded octet. Now this chlorine has eight. Carbon has eight. Even boron has eight electrons. Right? All three atoms has complete octet here. That's why the B is what? B is more stable than that. Okay. So charge separation is same number of pi bonds. It's same. We'll count the atoms as complete octet. That's what we see. So what if in like, we have like a molecule and one resonating structure, oxygen doesn't have a complete octet and the other carbon doesn't. If you have two different atoms with incomplete octets, then how would you choose? You see, we cannot see like this. If you can give me the structure, then I can tell you. Because we'll see the number of pi bonds first, charge separation. Then we'll see this rule. We'll see the other factors also, whether the compound is aromatic or not. There are many things. Like, you cannot say, what will happen if it's like this? We'll have to see how it will be like this. So this is the given list that you gave. That's in the order, decreasing order. It's in the crazy order. Mostly when you do some practice, you will understand what rule you have to apply. So we have rules for that rule. Follow that rule. Rule one, rule two, rule three. So we have to go in order. Yes. Apart from one or two rules, like in case of ring or aromatic compound, we'll see that in the line. But mostly for first four or five rules, we'll follow the same one. Okay? Rule number one, one more example, we'll see. Tell me in this one. This one is more stable. A or B? Yes. B is more stable, right? One by one. One by one. Directly we can say B. Or if you apply rules, this rule also. This carbon does not have offset here. But this carbon has offset. Right? And that is why B is also more stable. But then in the first one, offset has come here. Huh? Here also has come here. Okay. Next one. Rule number. What is rule number four or five? Four. Four. Right down. Right on. If the first three rule fails now, this carbon has six electrons here. Two bond and one hydrogen. Six electrons. This carbon has eight electrons. So complete opposite. And that is true for oxygen also. Okay? That's why. Second one is more stable. Right down. If the first three rules fails, the first three rules fails, then the negative charge, then the negative charge is more stable on, more stable on, more electronegative atom. And the positive charge is more stable on, more stable on, stable on. Have you got it, sir? Yes. Okay. More stable on less electronegative. Yes. Yes. Less electronegative element. Okay? So, for example, sir, all are more electropositive. Negative charge, more electronegative. Yes. More electropositive, you can say. And positive charge, less electronegative or more electropositive. Okay? For example, you see this. Correct? Which one? Superstition. Can you draw eyes into this one? Draw the eyes into this one. This is the four. It is. Yes. This is the same as the last one. Yes. No, it is the same. Yes. It is the same. Yes. It is the same. Yes. So, sir, I have failed. Maybe that the rest of it is the same. Yes. It is the same. Yes. It is the same. Yes. Anything else? I don't know. This is the number of I-1 same. The charge is also same. The octet is also fine here. The octet is also fine here. So, first one is more stable. First one is more stable. Why? Because the charge on more electronegative. It is more stable than that. This one, can you draw the RS for this? Three. Three. One is this? Yes. One you can draw with this. Another one is what? Okay. So, this one is A. Which one is A? This one is C. The second one will be B. Which one is more stable? Yes. Second one. Yes. Second one. Yes. Because octet is more stable. Yes. Three is stable. Sir, but then even... But then when carbon was charged... Yeah, but then what? Deteletivity. Oxygen and nitrogen. Here we have... Oxygen and carbon then? Sorry, this is negative, this is like my usual circle, I don't know, I don't know. It's looling. Not required. Okay. Ah, one more in there. Good night. It's such cold. She's like good to me. Good night. She's like good to me. That's all I wanted to say. This is okay. You said don't. Why do you always want to be a person happy? More energy. Oh, minus sign. You can. I have given you these three circles. Ah, this is given. So, but there will be one more where it's single point O and double point C, double point N. Five. You said AC, man. So is it? A is most stable. Yeah, AC. Then C, SMB. Oxygen, nitrogen and carbon. Electronic activity order. Order of stability is A maximum. So, but if we draw the structure that I also said, how will we compare this idea? What? So, we draw the structure that I also said. What was there? Single point O. And double point C, double point N. Yeah. What's that? Leason. Less number of time. Only two. I do it. Done. Write down problem of five. Eight. Seven. Seven. Like in the STL, five points C. So, it should be more stable than A and B. Which one, which one? C. C. No, here also you have three five points. Two plus one B. Three two plus one. One triple point is two five points, two five points and one there. Three. Okay, next rule number five we write down. Stability will be more. stability will be more when the opposite charges are closer, the light charges are far apart. You can take the example of phenol, you can draw a structures, this one you see, A, B, A, B, C, which one is the most stable, A, B, C, A, A, B, C, A, B, C, A, B, C, A, B, C, A, B, C, A, B, C, A, C, A, B. ABC again. You can close up. ABC is ordering. Same order. Next, we'll write down. We'll do the same. Yes, sir. Thanks. Thanks. Thank you. Sir, isn't there some benzene which is attached to the center also? Some measuring structure where the center is connected and the center point are connected. That is, we have learned those at a price. What is the price? Write down next. Stability of carbocation. That is plus i effective. Carbocation increases when it is surrounded to, stability of carbocation increases when it is surrounded to more number of alkyl group. Sir, is that the primary tertiary? Yes. RS will be what this comes over here. RS, pi, sigma, positive charge, type 1. And it gives you this. Which one is more stable? First one. First one because it is 3 degree? One degree. One degree. First one is more stable. And it is exactly opposite to negative charges there. One more. Which one is more stable? Negative charge. RS is this comes over here. This comes over here. First one because it is 3 degree. And this is? 2 degree. Right? So this is more stable. Carbocation. Next write down. One more. Rule number 7. The structure which has the structure which has more benzenoid form is more stable. The structure which has more benzenoid form is more stable. More benzenoid. Benzenoid form is more stable. This rule we call it as Fry's rule. Fry's rule. Fry's rule. More benzenoid form, more stable. Sir, these are all these rules. No, this is not in the order. It is different. This is for cyclic aromatic component. Till rule number 6. 6 it is in the order. This is for the specific compound. When you have cyclic ring like this. This is one structure. And when you draw the resonating structure of this, you will get. So in this, A is more stable than B. Why? Because A has two benzenoid form. This is the benzen thing. This is also a benzen thing. This is 5. 5. 5. 5. 5. 5. 5. This is also a benzen thing. This 5 bond is shared for the two days. So this two benzen ring you can see. A is more stable than B. More benzenoid form. So it has two benzenoid form here. Second one is not shared. Second one we have only one. Here we have one and two. Two benzenoid forms. This molecule we call it as A3. So they shared. Superb. How did they share it to bond? So why is it the first one? Now it's stable. One. Yeah. And two benzenoid. This 5 bond is shared to the two. Okay, okay. One and two. Here we have only one. This is not a benzenoid. Sorry. First one. This molecule and this A. So you cannot have more than two, right? Two benzenoid. Yes. Other insects are also possible. Okay. And this we cannot have. A is more than B. Last rule right now. Aromaticity. Aromatic compound. It's the most stable. Then we have non-aromatic. Anti-aromatic. Okay. See this aromaticity will discuss separately, okay. I'll just discuss one thing here. You have to compare the stability of RS. Okay. Aromatic compounds were those compounds. 4n plus. Which are conjugated, cyclic, planar, 4n plus 2, pi electrons. Where n equals 2, any number, 0, 1, 2 and so on. Okay. If this condition is satisfied, then the compound is aromatic. Conjugated, cyclic, planar, 4n plus 2, pi electrons. Anti-aromatic. All conditions are true. But it should have 4n pi electrons. And it equals to 0, 1, 2 and 2. So odd pi bonds. So the aromatic. Odd number of pi bonds. Odd number of pi bonds is true. So this is pi electrons. Circular is 2 and plus 1, pi bonds. Oh, by 2 here. Okay. First one is, see here we have, here we have 3 pi bonds. Correct? Here we have 2 and this is 2. But again, this rule 1, we won't apply here. Because if this compound, if you see how many pi electrons it has, 1, 2, 3, 4. And lone pair is also counted pi bond, because it can form a pi bond here. Right? So 1, 2 and 6. So it has 6 pi electrons. For this one it is 1, 2, 3, 4. 4 pi electrons. Okay? So obviously this one is, this one is aromatic compound. Right? For n is equals to 1 it is aromatic. For n is equals to 1 it is anti-aromatic. And this one is non-aromatic compound. No, no, please. It's only for rain. Cyclic. Conjugated, cyclic, plain. It's only for cyclic compounds. Non-aromatic related. We will discuss in aromatic city center. Okay? We will discuss. Okay? Aromatic city we again discuss. Okay, separately we will discuss this. But obviously this one is aromatic, this one is anti-aromatic, and this one is non-aromatic. Okay? Usually the compounds in the league, it's not non-aromatic, but it's stability of non-aromatic compounds and the normal compounds is almost similar. Okay? Aromaticity gives extra stability to the molecule at normal condition. Anti-aromaticity is unstable. Unstable, sorry. It is unstable molecule at normal condition. Right? So obviously this one is most stable and this one is least stable because the order is this aromatic, non-aromatic and anti-aromatic. Sir, when did you know how to apply this rule and then you got to know? When you solve some questions and you look at the molecules, you'll understand this molecule is aromatic. So it's all about practice. Like I said, the first molecule has three. See, when aromaticity is there, it dominates all other factors. So aromatic compounds, always you have to consider the most stable compound except dancing resonance that we'll discuss later, which is just one case. But in general, aromatic compounds are most stable. So whenever the compound is aromatic, you have to take that compound as the most stable compound. Sir, so this is the order. What is the problem? As if you have to start, you have to start. No, no, no. So one, two, three, four, five and six, you can consider in an order. But seventh and eighth one, seventh one is simply a different benzenoid form. You'll understand. When you look at the structure like this, you can understand that this is the benzenoid form rule you have to apply. No. This and this will, no, dominate all of them. No, when you have aromaticity, it's also more stable. Aromaticity overbars everything. It dominates everything. Sir, what about in other factors what are the orders? Rule one, two, six is in the order. Till five, it's in the order. Sixth one is that one, no. Carbocation stability plus i minus i. So that is, again, different. But one to five is in order. When you solve some question, you'll understand what rule you have to apply. There's no doubt in that. So don't get worried about that. So aromaticity, these two rules we apply when we have cyclic compounds like this, where you can have the possibility of aromatic compounds. Aromatic compounds is only possible for cyclic systems. Open chain rules, seven and eight cannot apply. This order they have asked many times in the exam. Needless to say, many times they have asked this order. Aromaticity, like I said, we'll discuss this separately also. What are the conditions? Like this rule we call it as the molecule which has this electron. This rule we call it as Huckel's rule. It is Huckel's rule. 4n plus 2 pi electron, Huckel's rule. There are other molecules which does not follow Huckel's rule, but it's still aromatic. So all those things we'll discuss separately in aromaticity. So this is the rules we have to follow to find out the stability of resonating structure. Now to find out the bond order like I said, bond order depends upon the stability of resonating structure. The RS which is more stable contributes more and according to that we'll have the range of bond order. So bond order, we have already discussed in chemical bonding that it has fractional as well as integral value. Both bond order are possible. MOT may. So if the fractional bond order is possible in the molecule which has resonance, in the molecule which has resonance, how does it work? Sir, you're supposed to ask for a break. He's under under pressure, he can't think anything. Okay, we'll take a break. I have ligation of resonance, let's start. It's not fully, the entire organic mixture based on resonance, so it'll be very important to take the last balance. I'll open up. I'll open up. See? It's going. Look at this board. Then what's the point of the timing? Oh! That's so good. That's beautiful. Then you take one more.