 Sit down, sit down. What's wrong? I have to put on our school uniform. Just a minute. Get him out of here. I said, you're the principal. You're the principal. You're the principal. You're the principal. Why are you doing this? Why are you doing this? Why are you doing this? Why are you doing this? I'm glad. Who's this? You go ahead. You'll get a... prepared right now. You're the principal. You're the principal. You're the principal. You're the principal. You're the principal. ... substance. the application of resonance. Aromaticity, stability of intermediates, stability of intermediates. Okay. The first one right on comparison of bond order. It's fine for you, no? No. Right on the first one, comparison of bond order. Okay. This is not working. No, it's not. This is not the balance. I know it's coming. This is going through. Switch off the fan. Switch off the fan. Switch off the fan. Comparison of bond order is what? Bond order is the number of bond present, is the number of bond present between the two atom in a molecule. Number of bond present between the two atom in a molecule. It can be fractional value of bond order is possible. But in case of resonance, when resonance is possible, then fractional value of bond order is there. Right? Internal value will have when there is no resonance. Okay? CH3, CH2, CH3, CH3, CH2, CH3. All the carbon-carbon bond, if you see, it has bond order 1. Right? Because there is no resonance. Okay? When there is no resonance, integral value we have, otherwise we'll have the fractional value. Now calculation of bond order, we know also one thing. Here, that bond order is inversely proportional to bond length and directly proportional to bond strength. All these comparisons we can do if you know the bond order. Okay? Calculation of bond order in case of equal resonance. In case of equal resonance. Like for example, you see the first example, hc double bond O, O minus, and its resonating structure is hc O minus double bond O. Okay? The bond order in this case will write number of sigma bonds, number of pi bonds divided by number of sigma bonds. This is the formula. Sigma plus pi by sigma. Okay? See, in this molecule, the resonance is possible from this oxygen to this oxygen. Right? Right? Only in this part. So we'll count the sigma bond and pi bond in this part. Where the resonance is possible. How many sigma bonds we have here? Two. Two sigma bonds? Two. Two. How many pi bonds? Divided by? Sir, in the formula there is number of bonds by sigma bonds. Yes. Huh? In the formula there is number of bonds by sigma bonds. No. The back bonding is a weird stuff. No. That's some of the answers. Yes. How many sigma bonds? How many sigma bonds? No. Number of sigma bonds plus number of bonds. How many sigma bonds we have here? One plus two. One plus one. Two sigma bonds? And one five. Two plus two. Yeah. Yeah. Sir. Yeah. Sir, how number of sigma bonds plus number of pi was equal to number of bonding and electrons minus number of antibodies? It is not that. In that case, we'll get integral value. No. Not integral value. No. No. But that is different quality if we have two and two like that. Same atom principle. We're talking about this molecule's shape of mind. So here we'll take, we have resonance values of this formula. We have another formula for this also. We can also write a mission of number of bonds in all rs. But this one is the easier one. The first one. The mission of number of bonds in all rs is divided by the number of rs. There's an ending in scripture. Like in this one you see, suppose you have to count the number, the bond order of this bond. Bond order of this carbon oxygen bond you have to find out. Bond order of this carbon oxygen you have to find out. How many rs we have? One plus one, two? Two. Right? So we have two rs. And how many bonds? Two bonds here and one bond here. Two plus one? Three. Again it is one. Okay? But for this formula you have to draw the resonating structure. Bond resonating structure. But here the advantage is what? With this only you can say one plus two sigma plus one number of sigma plus number of pi by number of s. Second formula is number of bonds in all rs. How many rs we have? Two. And we have to count the bond order of any carbon oxygen bond. Suppose this one we are taking. C1, C2. C1 and C2. We have two bonds here and one bond here. Two plus one by number of rs. Sir, is hydrogen bonding a sigma bond or a pi bond? Hydrogen bonding. It has a sigma bond characteristic. No. Because resonance is possible here. So it is only that part in which the resonance is possible. For the first method, it will be counted as hydrogen bond. You have resonance here. So number of sigma bond plus pi. This will take only that part in which the resonance is possible. Sir, but in the first case we consider the bond and then the second one will be the same. Why did we consider like both? We are considering both oxygen only. How many of you understood this? See, we are taking this portion because resonance is possible here only. This edge we are not considering. So how many sigma bonds we have? One plus one, two sigma bonds. This we let it be. So two sigma bonds, how many pi bonds? One pi bond. So two plus one by two. Sir, I didn't get the second one. I got the first one. The formula is exactly same. Either you can use this one or this one. Look at the same answer. But the disadvantage of this is you have to draw the rs, the entire rs. Then you have to substitute here because the number of rs you know. But if you are using this formula, you can directly say with this one number of sigma and pi by sigma. Now when you draw the rs, suppose one molecule has five rs. Six rs. And you have to draw all those rs and then you see how many rs possible and then you have to write this. So better use this formula, sigma and pi. Got it? Okay. Another example right down. CO32 minus. What is the bond order of carbon-oxygen bond? CO32 minus. Is it... So with this formula, how you are getting ? Okay, how many of you understood the second formula? No, that I didn't understand. Why? Sir, how do you understand one? See, this thing I have explained five times. You don't concentrate, that's the problem is. It's not a rocket science. One simple thing is just not to take one minute to understand you. You don't concentrate, that's the problem. If you want me to do this entire formula, this two formula in the entire class, how would I finish this? Tell me that. If there are something which is there to understand, then I can understand, you can take time, we'll discuss all those. Just one simple formula you are not getting. How would you prepare for J, E and K? Problem is you are not concentrating. When I explain, you'll talk over there. And then later on you realize when I write down the answer, how did you get the answer? When I am discussing something, you should concentrate. I have explained this thing twice. Four or five times I have explained. See, number of bonds in all Rs. First of all, you have to draw the resonating structure. We have two resonating structures. Like I said in the beginning, you have to only concentrate the part of the molecule in which the resonance is possible. Now here also you see, in this part, the resonance is possible only here. This part is not in resonance. So we won't take this into the found. Take it? This is one thing. Now we have two Rs. Formula is this. Number of bonds in all Rs. Now how many bonds you'll have here? Number of bonds will what? Suppose we have to find out the bond order. First of all, you try to understand. Bond order between two atoms will be fine. It's not defined for a molecule. Molecules are bond running order. It's a bond order between the two atoms of a molecule. Like you can find out the bond order of this carbon. This carbon-oxygen bond. You can find out the bond order of carbon-carbon bond. Carbon-oxygen bond like this. So we are taking only one bond here. So I'm taking this C1 in this oxygen bond. This carbon-oxygen bond. This carbon-oxygen bond. Which will be same for this carbon-oxygen bond and this carbon-oxygen bond. So any one of this bond you can take because it is equal resonance it won't affect the answer. So I'm taking this bond. How many bonds we have here? Two. How many bonds we have here? One. So summation of number of bonds in all Rs. We have two Rs. Two bond here. One bond here. Two plus one divided by number of Rs. Two of C1. Sir, but in the first method we counted the bonds of both oxygens. Because we are here the formula is that's what I said the advantage of this formula is what you don't have to draw the entire Rs. of this structure. You have this structure. Let it be. How many since we have resonance possible from here to here? So we will count the sigma bond in this part and pi bond in this part. So we have two sigma bond and one pi bond. Two plus one. So one point five is the bond for both oxygens. Both, you know both carbon-oxygen bond. See, it's the same thing. When you have this molecule since it is the equal resonance so hybrid is what? This molecule wants to give this bond on an S2. This wants to give one. Half will be given by this. Half will be given by this. So average is what? Two plus one by two. Because it is equal resonance this and this contributes equally. AB contributed equally or AB contributed equally? So you will get the average of the two. So the average of two plus one by two that is one point five. So in case of equal resonance you can find out the average. It is nothing but the average. If you look at the formula it is nothing but the average. We have to find out the bond. So this is the average value. This is the average value. That's why we have number of RS. Bond divided by RS. It's average only. Understood this. Correct? Yes. Okay. Now what is the bond order here? O by two. Carbon-oxygen bond. Which formula we should use? This one or this one? First one. The first one may be you will have number of sigma bond. You see first of all here is the resonance possible this part also and this part also. Okay. Only this one. Abhiyam. Resonance possible this part and this part. How many sigma bond we have? One, two and three sigma bond. Three sigma bond plus one pi bond. Four pi z. With this also you can do. You will get three resonating structures. First you will get this. Now it's over here. C O minus double bond O. O minus. And another one from this is what? When this comes here this goes here. C O minus double bond O. O minus. To use the second formula what you have to do? You have to take any one of the carbon-oxygen bond. This one. Okay. So total bond is what? One plus one plus one. Four. One plus one plus two. Four divided by R s is what? Three. Four pi z. So any one formula. I would suggest this one. Okay. Now this one you tell me. For this one, third one. Resonance is possible in this molecule, entire molecule, right? So it is what? We have a sigma bond is three. One pi bond by three. Four pi z. What about this one? This part is not in resonance. Okay. So it is three by two? Clear. Okay. So equal contributing structure you won't get. It's very easy and simple. You have to find out the average. Okay. Mostly you'll get unequal contributing structure. So next I write down unequal contributing, unequal contributing resonating structure. In this case, we cannot find out, we cannot find out the exact value of bond order. The exact value of bond order. But we can find out the range of the bond order. Exact value they won't ask you also in the exam. For unequal contributing. Okay. So in this one example you see. Range of bond order. We can say bond order lies in this range. Okay. So in this example I have already taken this example. CH2 negative C double bond O. Okay. I'll write down H in this form. Okay. Now the resonating structure of this is what? CH2 double bond O minus and H. These are equal contributing? Yes or no? How do we find out? Equal contributing or not? So they should be a part of it. No, no. How do we find out equal contributing structure? If the stability is same. Correct? Are they have same stability? No. No. Because for this we use the rules of relative stability of resonating structure. For this one is what? This one is less stable. Because negative charge is also equal. So now when you have the rules, you will have the idea that what rules you have to apply. It's not so important that rule one, rule two, rule three. Then directly you can apply. Here you see equal pi bond, oxygen on negative charge, carbon negative. This is more stable. Now when this is more stable, so this contributes more into the structure. Correct? Now if you have to find out the bond order, bond order of carbon-carbon bond for this quality. Okay? Exact value we cannot find out. But we can say the bond order of carbon-carbon bond lies in this way. First of all this one to two. Because here we have one, here we have two. Right? So this one is more stable, so this contributes more. Right? So we can further write that the bond order lies in this way. 1.5 and 2. Right? Understood this? So we can write down the range. We cannot write down the exact value in case of unequal contributing structure. Okay? Now what is the bond order of carbon-oxygen bond? What is the range for this? 1 and 1. It is 1 to 1.5. Clear? So for this you should know the rules which RS is more stable. Accordingly you can write down the range of bond order. Got it? I will write down some examples. Okay, one example write down. Sir, if the contribution is like 60 percent, 40 percent, then the bond order we can read. No, that is difficult to judge. Especially 40 or 70, 30. That's why we write down range. We don't write exact value. Okay? Next one you see. Polo benzene. Have you heard about polo benzene? Yes. What is polo benzene? Suppose it is bond A and bond B, carbon-chlorine bond. Right, carbon-chlorine bond. Tell me the bond order of A and B. Bond order comparison of A and B. What is that? Is that hexane? So what is the name of this compound? This one is chloro benzene. That's... This one is what? Cyclohexane. Cyclohexane. Chlorocyclohexane. The first one doesn't have the double bond. Yes, the first one doesn't have the double bond. It is cyclohexane and chlorine. B is more stable. B is more stable. Why? Because it is more stable. Here we have resonance. Do we have resonance here? Do we have resonance here? Do we have resonance here? No. Here we have resonance, you know, 5 sigma lone pair conjugate system. So we can draw the resonating structure, right? So when you draw the resonating structure, we'll have partial double bonds characteristics here. That's why negative double bonds and double bonds cl positive. Right? So partial double bond characteristics we have between carbon and chlorine. That's why the second bond order is mode. Bond order of B is mode. Can you just say how it's aromatic? It's the most stable. No, no, we are talking about this bond. Aromatic is this ring. We are talking about carbon and chlorine bond. Can't we say the whole thing has more ring structures in it? So that's more stable. No, it's stability. It's stability of both. See. Of the whole thing, right? No, no. Both of them aren't measurements of each other. See, the point is this is the different molecule. This is the different molecule. So what is stability if you ask me? This is aromatic compound because ring is more stable. Obviously this one is more stable. But here we are trying to find out the bond order of carbon-chlorine bond. Like I said, bond order between two. Atom will say in a molecule. So we're talking about bond order of this bond and this bond. Since here we do not have resonance, bond order of this is what we can write. Bond order is one, exactly one. But here it will be between one and two. Obviously the bond order of B is more than that. That's why aryl halide, chlorine on benzene, that we call it as aryl halide. When you remove one hydrogen from chlorine, it becomes aryl. Aryl halide is stronger than the normal alkyl halide because of resonance. Because of this resonance, it develops a partial double bond characteristics and it's difficult to break. So in comparison, alkyl halide and aryl halide, aryl halide is more stable. Carbon-herogen bond is more stable than carbon-herogen bond in alkyl halide. It is stable. First one, carbon-chlorine bond is more stronger. If the second one is less stable, so won't it bond to change back because of force structure? That is a different thing. But it has some partial double bond characteristics. Because of resonance. Because this delocalized will be there. Continuously there. If you draw the actual structure, the actual structure is this. Partial double bond characteristics. That's why the second one bonds. And we know the order of bond length and bond strength. The order of bond strength is directly proportional and bond length is inversely proportional. So any one of these questions if they ask in the exam, you can compare this. Okay? This question will tell me. Bond order of alpha-beta bond. This is alpha. This is beta. And this is bond. Alpha means this carbon-carbon bond. Beta is this. Gamma is this. So it's a bond stability. What is the order? Bond order of beta is the highest beta and alpha is gamma. Oh, it's alpha, beta, gamma, not normal. Beta, alpha, gamma. They are not normal clature. No, no, no. It just represents this carbon-carbon bond is alpha. Carbon-carbon bond is beta. Carbon-carbon bond is gamma. Like here we have carbon-carbon bond is A and carbon-carbon bond is B. Which one is the most? Beta. Beta. What is the bond order of this? And what about this alpha? One to two. One to two. One to two. Right? And this one is? One. Right? So beta is max. Then alpha. And then gamma. Okay? What about this one? CS3. Because CS2 double bond, CH. OH. And CS3, CH2, OH. Carbon-oxygen. Alpha, carbon-oxygen. Beta, alpha-beta comparison. Alpha is green. Between what? And you don't even care about him. Fine. He is not paired there for draw in the exam. draw in the exam you have to know this. When oxygen bond is 2 bonds it has 2 lone pairs. It is conjugated, right? This term is conjugated. This term has partial double bond characteristics and that's why the bond order of alpha is more than to that of beta. Alpha and beta. So this has more resonating structures in this. So if we do the dimensions by number of resonating structures this one will have less. Alpha is less. See we have this lone pair and this lone pair and because of this resonance it develops a partial level bond characteristics, right? So here the bond order of beta is what? Less than 2 but greater than 1. Okay? Here alpha corner is what? It is 1 here. In this molecule there is no resonance. No resonance when I say it means this lone pair is not involving resonance. Why? Because there is conjugation but because of this group CS3 there is a steric repulsion here and because of this steric repulsion this group changes its plane so the entire molecule is not plane at all. Okay? So this CS3, one of the CS3 will go into the plane. You know this bond? It is coming out of the plane and this wedge is into the plane. This is CS3. This is CS3. See because of these two group present we will have steric repulsion here, right? Steric repulsion means hindrance. Both this CS3 and CS3 are pushing each other. So because of this hindrance the stability decreases, right? And to minimize this hindrance what happens? The plane of this molecule will change. It rotates to minimize the repulsion and adjust the steric repulsion to the repulsion here. This group and CS3 and CS3 repulsion. See why doesn't that CS3 and that carbon, that CS2 not that much to repulsion? If you have this group present, O C double bonds is bulky group, right? And if any group is present here suppose cyclopropyl like this then since these two groups are big so we will have steric repulsion. Both groups will push each other, right? And because of this steric repulsion the stability decreases usually. So here also we will have the steric repulsion, right? See this question if you get in the exam directly you will do it wrong. If you don't do it then you won't know. That's how it is, okay? Organic chemistry is like that only. You will see many things where the basic logic won't apply. You cannot say if you can ask where we consider this steric repulsion. Right? You can ask this question. So I can say when there is bulky group attached, right? So because of this steric repulsion the stability changes and you should know when you have to apply. Or if any group present and this thing is present this will have the steric repulsion. And to minimize this repulsion it changes its plane and you don't have to do anything into it. It adjusts in such a way so that the repulsion is minimized, right? So what have you conjugated? What will happen in rotation? The plane of this and this one changes. So one is like this and another one is like this. So it's not parallel orbit. No conjugation. Sir is steric repulsion the same as steric hindrance? Yes, steric crowding, steric repulsion, steric hindrance or the same thing. Okay? Static crowding, steric repulsion, steric hindrance or the same thing. So because of this steric hindrance this NH3 hole twice changes its plane and this becomes non-planar now, that Taramali. And hence the resonance is not possible. Write down this example properly. Okay? So alpha is one, beta is this. So the bond order of beta is more than that of... Sir, there was only one bond there and the right one was very stiff as a bunch. Which one? The bond of the bond. This entire thing will be under repulsion. Only this bond was there. That's what the difference between the two bonds. Okay. Thank you. Write down next, second application mesomeric effect. This is the second electronic effect. Resonance is also an electronic effect. Resonance, mesomeric effect, the second electronic effect after conducting effect. Mesomeric effect, we also call it as resonance effect. Both are same thing. Resonance effect. Like inductive effect will have plus i and minus i. What is plus i? What is plus i? Electron releasing. Electron releasing. So here also we have plus m and minus m. Plus means electron releasing. Okay? So you can write this as plus m or minus m effect. Plus m is nothing but plus r and minus m is nothing but minus r. Both are same. Okay? So plus m groups or what? These are the groups which can release electron pair. Electron pair donor. Okay? Plus m groups are electron pair donor. And what is the electron pair involved into this? Either pi bond or pi bond or donor pair. Because we know in mesomeric effect resonance, sigma bond does not involve, right? Pi electrons or donor pair. Okay? So plus m groups are those groups which has at least one lone pair present on the first atom. These are the groups which has at least one lone pair present on the first atom. First atom which, like suppose you have a group. Benzene pair, whatever group is attached. Suppose NH2. Right? NH2 has one lone pair. This is the first atom. Means the group that is attached with the benzene ring. The first atom of that group or the electron which attached with the benzene ring. If that atom contains at least one lone pair, it has plus m effect. Okay? It's very important. Plus m effect. I'll give you the examples. But this is a general thing you should memorize. Okay? The first atom or the atom of the group which is attached with the ring. That atom must have at least one lone pair. Then it behaves as electron pair. So the previous question, wouldn't there be mesomeric effect also? Where? The previous question. This one? Yes sir. So it's the same. Whatever resonating is that we are drawing. We have done the RS for this, no? Yes sir. So this is mesomeric effect because it is releasing electron to the ring. The structure that you draw, that is mesomeric effect. So what about the chlorobenzene? I'll come to that. I'll come to that. Wait. It's a plus m effect means that it has more tendency to resonate. No, no, no. Plus c. First of all, this effect will define for a group which is attached on any conjugated ring. Conjugation should be there because for resonance, conjugation is the basic requirement. Correct? So plus m effect is defined for a group and these are the groups which can release electron to the ring. Correct? And it can release electron only when it has extra electron or available electron. Available electron means what? Lone pair of electron. Bond pair so it cannot release. Right? So this has at least one lone pair so it can release this electron pair to this ring and this pi electron strip over here and then further you can draw the resonating effect. Correct? So plus m group are those groups which can release electron pair to the ring. Right? Release, not withdraw. Okay? Condition is what? The atom must has at least one lone pair. Now what are the examples we have? We can have OH plus m group because oxygen has how many electrons? Two lone pairs on this. Right? It can release electron. O CS3 plus m effect. Okay? Next is NH2 plus m effect and R2. What is R? UL-alkyl group. CS3, methyl, methyl, propyl, whatever. Okay? And R2. We can write O C double bond O plus m. Light oxygen if you replace this oxygen with SH. This is also plus m. You won't get this anywhere but oxygen and sulphur you can do. So what is R? R2 is the alkyl group. R2 is this. That's why I wanted to do nomenclature first. R is this. Both alkyl groups are present. All this hydrogen you replace by this alkyl group. Okay? See, one thing you just try to understand here. This group means what these groups when attached to the ring, it donates electron to the ring. Okay? Now if you compare the plus m nature of O CS3 and O C double bond O CS3. Which one has more plus m nature? Why O CS3? Answer is correct. See, here first of all both will show plus m. Any doubt on this? Because it can show, that one can show it. But here what happens, this lone pair is also involved in resonance with this oxygen. High sigma. Pi. Lone pair is also a pi. This pi electron will withdraw this electron towards the side. Electron releasing tendency is less for this. In comparison to this. See, nitrogen has lone pair on it. And this lone pair is in conjugation with this pi bond? Yeah, so what we can do? We can put this lone pair into bond pair like this. And this pi electron shipped onto this target. So what, eventually what is happening? This NH2 group is giving electron to the ring. That's why it is plus m group. Later. Sir, is that O CS3? Can't it also be O R? Yeah, but we can accept it. Because O R is also... O R, in general, you can write O R, not O R. O R in his right hand. This is very important. You have to identify first the nature of the group. Then you can find out the other properties. So for plus m, the first atom must have at least one lone pair on it. The next one you write down. NH, this is one example also. NH, which was there? These are plus m group. The group which shows plus m effect. Now can you tell me if you thought there's a plus m effect of this one and this one. Which one has more plus m effect? NH2 will have more. NH2 will have more. The logic is same. Okay, right now thank you. What about halogens? X? X has lone pair. X has three lone pair, right? Does this halogen show plus m effect? Yes. Condition of plus m is there? Yeah. So when this halogen attach to the range, any chlorine whatever it is attached to the ring. Since it has lone pair, so it can show resonance like this. So it has plus m nature. Right? It has plus m nature. Plus m nature it has, but its effective effect is this. Minus i. Usually what happens? Resonating effect dominates inductive effect. Means plus m always dominates plus i or minus i. Means effect of resonance is more than that of inductive effect. Means suppose if you have two molecules are there, both has carbocation. One carbocation is getting stabilized by plus m effect and other carbocation is getting stabilized by plus i effect. The one which shows plus m effect is more stable. Right? Because plus m dominates plus i. Basically in general we can say mesomeric effect or resonance dominates i effect. So usually what happens? You see when oxygen attach here, for this example you see nitrogen and oxygen bond we have, right? Nitrogen is more electronegative. Yes. So this nitrogen shows minus i also. If I ask you what all effects possible here, you can answer minus i and plus m. Both effects possible. But which one is dominating according to that we will get the answer. But we know mesomeric effect dominates i effect plus m of NH2 dominates minus i nature. Only case of halogen. Yes, only case of halogen. Only case of halogen its minus i effect dominates plus m nature. You must remember this is very important. Any doubt in this? Right? So condition for plus m is the first atom must have at least one lone pair. Halogen has three lone pair but its minus i dominates. Okay? Not plus m. Kind of exception. Yes. Usually plus m dominates minus i or plus i. Okay? This example you see. Carbocation stability. Which one is more stable? That one. That one. That one how to ask? Second one. Second one. First one or second one? Second one. Second one. My second one? Five of them. See in this what effect we have? What effect? Electronic effect? That series of plus i. Plus i. This carbocation will have plus sign yes or no? Yes. CS3 is giving electron CS3 CS3 and C2H5 is this side right it is alkyne group no alkyne group has plus sign nature it is CH2 CS3 both are plus sign effect alkyne group are you are not the first class I will share the video I forgot to share also okay so these are alkyne group and alkyne group are electron releasing plus sign nature so here we have only plus sign effect here what we have these two CS3 shows plus sign but because of this pi bond this positive charge is in conjugation plus sign right this shows plus sign effect so effective the plus sign is effective over here means we can say the second carbocation is getting stabilized through resonance mesomeric effect plus sign effect and plus sign dominates plus sign that is why the second one is more effective order is this. So can't we also just see it has more time for less? No that is only possible for different decision RS of same volley is that different volley? Because it is pi sigma vacant orbital first type of conjugation first type of conjugation plus sign you can say it has also resonance so resonance that is the statement better write resonance not plus sign resonance 0 resonance don't write plus sign resonance right Or can we just say 1 pi bond 0 pi bond No 1 pi bond 0 pi bond is only applicable for RS of same molecular ion that 8 rules have given you know it is you can compare RS of same molecule these are two different molecules what I am telling you we have two carbocation we have to compare the stability so first method eye effect second way resonance dominance eye effect second one more stable that is it so when you finish resonance hyper conjugation and all you will get the you know questions in which eye effect hyper conjugation resonance all are taking place according to the dominating effect we will find out the answer even in this one I did not discuss plus H effect hyper conjugation here plus I and plus H is also possible hyper conjugation right and hyper conjugation dominance eye effect resonance dominance hyper conjugation so order is RHI decreasing order okay so you have to first identify what all effects are there and then you can compare okay so we do an exam first we will understand okay but you have to focus okay so write down next minus M group just reverse of this okay minus M group you write down it is electron with drawing group okay electron with drawing groups are those groups actually in which between first and second atom will have multiple bonds like suppose X is attached to the ring okay X is I will suggest you to memorize this general condition because there are many examples it is difficult to memorize the nature of the group plus M minus M what is that so general thing you must remember the first atom which attached with the ring pH and the second atom why suppose okay we must have multiple bonds between the two it can be a double bond it can be a triple bond and if it is there then why it is minus M because you see this example if you have this ring right and one group say CHO is attached CHO is this suitable bond OH right so since this double bond is present pi sigma pi conjugation we have correct so this oxygen takes electron pair and this withdraws electron from the ring and that's why it is minus M so this conjugation if it is not there then it cannot withdraw electron from the ring since carbon does not have lone pair so it cannot donate the electron and this pi sigma pi conjugation we have so it can this pi goes onto this oxygen then this carbon takes pi bond from the ring so eventually it is withdrawing electron so the condition is reverse or minus M exactly minus M is electron withdrawing from the ring plus M is electron releasing so minus M is the condition between first and second atom we must have double bond or if the atom which is attached with the ring has vacant orbital then again we have pi sigma vacant orbital possible so this is also the type of conjugation here if this three condition is there then the box means vacant orbital this box means vacant orbital okay vacant orbital example you write down NO2 you see NO2 the structure is this N double bond O double bond we can also write this L d height C double bond OH okay we can also write C double bond O R C double bond O OH okay BH2 if BH2 is attached with the ring it also shows minus M effect minus M effect because boron has vacant T orbital on it boron is vacant pure so BH2 is phenol right phenol this right phenol it is phenol it is phenol oh it is benzene with bond this is phenol one open bond here this group can be anything can be anything BH is this phenol we can have cyanide C triple bond N all these group shows minus M effect electron withdrawing nature got it then I write down some examples here you have to tell me first what all effects are possible and which one is dominating okay for example you see you have benzene ring NH2 group is attached O CS3 what do you want to say why why you are asking that what all effects are possible it's okay what do we get by finding out which plus M effect or minus M effect only stability you know see when you when you get that question no so you should know what effect is taking place then only you can compare right it is not required here suppose if I ask you what effects possible here what is the answer plus M and minus M minus I because nitrogen is more electronegative than this carbon so if I ask you what all effects possible you can answer plus M and minus I but the nature of the compound is according to plus M because plus M is dominating yeah so when you find out the stability or anything so you have to think according to plus M effect not minus M okay so in this one one more okay NH C level bond over okay here what all effects are there plus M and minus I plus M dominates here what we have plus M minus I why minus I we have because carbon oxygen will have electronegative effect right and that is a permanent effect so that bond pair of electron is always shifted towards the more electronegative okay what about this plus M dominates minus I C level bond over minus M and only minus M because this carbon is sp2 this carbon is also sp2 there is no electronegative only minus M even here also this sp and this sp2 so minus M and minus I it is sp it is sp2 sp is more electronegative so minus M minus I but minus M dominates minus I here it is plus M minus I plus M dominates so plus M or mesomeric effect always dominates inductive effect except in the case of what except in the case of halogen so if you have halogen attached on this halogen has three lone pairs it can show plus M effect it can also show minus I effect but minus I of halogen dominates plus M right in case of halogen inductive effect dominates okay only halogen okay so this is the next class we will start with aromaticity aromaticity will be my class is here