 Hello guys, good afternoon. Who are all there? Please type in your name. Hello, this is a question of how are you? Hello, Sainay, good afternoon. So guys, today in this session, this is not exactly a problem solving session, but we are going to see few basic concepts here. And that only we'll see the problems. Okay, so we'll first discuss a bit of theory and then we'll see some questions on that. Okay, like we have discussed last class that there are few problems that is based on heat of hydrogenation, heat of combustion, right? How to solve those problems, in which cases we'll have more heat of combustion or hydrogenation, how do we compare that? All those things we are going to discuss today, okay? So you see here, first of all, see all these things comes under resonance actually, resonance, hyper conjugation also you can say, right? So resonance, hyper conjugation, you should know the concept of hyper conjugation, resonance to understand or to compare heat of hydrogenation, heat of combustion, okay? But apart from these two applications, heat of hydrogenation and heat of combustion, we have other application of resonance also. And that is also important. And those things already you know a little bit, okay? But today in this session, we are going to discuss all those possible questions and concepts that you should know before going into the JEMN exam, which is coming in April, and any other, no, digital exams, engineering and computer exams, okay? So first of all, you see the resonance, just a second. So the first thing here, what all applications of resonance we have? Okay, so resonance, you all know what is resonance, what are the condition for resonance? So I'm not going into that, okay? I'm directly coming to the application part of resonance and we'll see how to solve the questions into this, okay? So first of all, you see the application of resonance, if I tell you, right? We can use resonance for the comparison of bond order, calculation, calculation and comparison, calculation and comparison, right? The next application we have, mesomeric effect. Mesomeric effect we have discussed last class, okay? So I'm not going to discuss this, we'll see problems on this directly, okay? Third one is aromaticity. Aromaticity, we all know what is aromaticity, but we'll see here some miscellaneous examples of compounds which shows aromaticity and non-aromaticity, anti-aromaticity, okay? So we'll see here some examples into that, okay? The next one we have stability of intermediate, stability of intermediate, and the last one here is heat of hydrogenation or heat of combustion. So today by the end of this chapter, you will be able to solve questions related to all these topics easily, okay? That is for sure, okay? And since, you know, in chemistry, especially in this, this is a part of GOC, right? So you should know information, you should have the information before solving the question, okay? So what all the information you require that will give you today, okay? So first of all, you see the resonance, and we know the resonance has, we can draw, what is resonance? Resonance is the phenomenon by which we can represent one particular molecule into one or more structural unit, okay? And these structural unit, because of resonance, I'll just, in short, I'll write down, because of resonance, we can represent a molecule into one, sorry, into two or more, or more different structural units. One molecule can be represent into two or more, more different structural units. Now these structural units are what? These structural units, we call it as resonating structure, structures, or we also call it as canonical structure, or we also call it as contributors. Why contributors? Because these structures contributes into the property of real molecule, correct? So now when I say contributors, there are two types of contributors here. Contributors or resonating structure are of two types, okay? The first one is equal contributing structure and the second one is unequal contributing structure, okay? Equal contributors, contributors, and here we have unequal, okay? So in short, if you want to write it down, you can write it on keywords, but keyword, you must write, okay? Because the main thing is to understand the problem, how to solve the problem. So we'll go to the problem very soon, okay? Now equal contributors, for example, you see, there are many examples of equal contributors. So for example, we have benzene ring. I'm not drawing the resonating structure here. This is an example of equal contributors. We have an example of carbonate ion. This is also an equal contributing structure, right? All these are equal contributing structure. Unequal contributors, again, we have many examples of unequal contributing structure. For example, this one, unequal contributors. CH2 negative charge, single bond C double bond O H, unequal contributors, okay? Now, the point that you have to understand, in case of equal contributors, the stability of all RS are same, right? Because all the RS will contribute equally into the real structure. So here we have the same stability, or stability is equal for all resonating structure. But here what happens in case of unequal contributors, stability of RS is different, RS is different. And that is why when I say that relative stability of resonating structure, that is only possible in case of unequal contributors. When I say relative stability, right? In this case, it is not possible because stability of all RS is same here. Here we have one RS is more stable and other one is less stable. So whenever I say relative stability of canonical structure or resonating structure, it means we are talking about unequal contribution. And since stability is different, so one is more stable, other one is less stable. The more stable contributors are what? Major contributors. And this one is minor. Understood? Major or minor contributors? Correct? So this is the basic theory that you should know. Is it any doubt you have here in this unequal contributors? Any doubt you let me know? Any doubt you let me know, guys? Now we are going to discuss the first application of resonance, which is bond order, okay? So you should have this idea. So now you see what is bond order actually? First part we are going to discuss here, bond order. Bond order is what? It is the number of effective bond between the two atoms. It is the number of effective bond between the present molecule, present in a molecule, okay? And we also know that the bond order is directly proportional to bond strength, okay? I don't think we required any discussion over here and inversely proportional to bond length. So if you have this information actually, you can also write down the order of bond strength. You can also write down the order of bond length, right? So they can ask you any one of these questions in the exam. They can ask you to compare bond length. They can ask you to compare bond strength. So for that we should know what is bond order and how to calculate bond order, right? So when I say bond order, we know this bond order has integral as well as fractional value also. Bond order may have integral as well as fractional value also. So this value of bond order can be integral. So the fractional value of bond order is only possible in case of resonance, in case of resonance. When resonance is not there, we'll have integral value of bond order. So this is possible when we have no resonance, right? Now, when I say resonance, there are two types of resonating structure. Again, we have already discussed and that is the first case, how to calculate bond order in case of equal contributors we are going to see, right? For example, for example, suppose if I ask you, now the question we are discussing here, suppose if I ask you, the first example I'm giving you here, tell me the bond order of this at C, double bond O, O minus, and the another resonating structure of this molecule is at C, double bond O, O minus. Okay? You have to tell me what is the bond order of this carbon oxygen bond, this bond and this bond, carbon oxygen bond, right? Bond order of carbon oxygen bond. What is a bond order? Can you tell me? Three by two. Why three by two? Because we are taking average of it, right? We are taking average of it. So the formula of bond order, and remember one thing, we are talking about the case of equal resonance, right? This should be what? The number of total bond, total bond in all RS, a resonating structure, divided by number of RS, which is here, this is the formula, right? So when you apply this formula here, so bond order will be what? I'll write down here. Bond order for this given molecule will be, will have two bond here, between carbon and oxygen only, we have to discuss two bond here and one bond here. Two plus one divided by two is equals to three by, the bond order, okay? Similarly, another example you see, what is the bond order in carbonate ions? See, carbon oxygen bond order in CO3 two minus. This is one structure, another structure is this, double bond O, single bond O minus. And again, the other structure is this, double bond O. What is the bond order of carbon oxygen here? Tell me, carbon, carbon bond, I'll write down bond order of sulfur oxygen bond. So here the bond order will be what? We have two bond present here, one and one, two plus one plus one divided by three, that is four by three. Similarly, here it is two plus one plus one, four by three. And what is the value here in the last one? Last one you tell me, which one is one by three, sign here, okay? So you see in this, the last one we have, this bond, we won't count here because this is not in resonance. Only bond, which is the part of resonance, that bond only we count, okay? Two, three plus two, five divided by three is the answer for this, okay? Now you see in all this structure, we are drawing the resonating structure, right? So it is difficult to draw the resonating structure in the exam all the time, okay? So this formula we are not going to use, but the formula we are going to use to calculate bond order is this, is equals to number of sigma bond plus pi bond divided by number of sigma bond. And only sigma pi we count, which is involved in resonance, okay? Which is involved in resonance. Sigma plus pi divided by sigma, this you remember. Now you see all these questions which we have done. Sigma plus pi plus sigma. You see this is the part which is involved in resonance. How many sigma we have? One plus one, two sigma, one pi. So two plus one divided by two, three by two. We are getting, here you see how many sigma we have? One, two and three because it is involved in resonance. So one, two, three sigma and one. So three plus one, four by three, right? Three sigma and one pi, four by three. Similarly here also. One, two, three sigma and two pi divided by, okay? Now, on the basis of this, if you use the second formula that I have written here, on this formula you don't have to draw the resonating structure, okay? For example, you see these examples. I'll write down a few questions here, okay? Double bond, double bond and negative charge. Bond order, you tell me, of carbon-carbon bond. C L double bond O, double bond O, double bond O, single bond O minus. Next one, P double bond O, O minus, O minus, O minus. The next one we have, benzene ring. Bond order, you tell me. C S three, C double bond O, O minus. Bond order of, what is the answer here? Last one you see. Now you think you, you see what? You don't have to draw the resonating structure of this. But what we have to count, since this is involved in all these molecule, so number of sigma bond is five, and we have two pi bond divided by sigma, seven by five. Here also, one, two, three, four sigma, three pi divided by four, so it is seven by four. Here you see, four sigma and one pi divided by four, so it is five by four. Okay, carbon-carbon bond order, here it is. We have six sigma bond, three pi bond divided by six, which is equals to nine by six, three pi. Here, we don't count this bond here because it is not involved in resonance, only this part is in resonance. So two sigma, one pi by two is equals to three pi. Got it, understood, right? Now you see, this is nothing but, like we are calculating the average bond order. You are taking the average value, right? If you see all these structure, it is nothing but the calculation of average, okay? Now, if you have unequal contributors, then how do we calculate and what do we do? So next point we have for unequal contributing structure, unequal contributors. So in case of unequal contributors, for example, suppose this is the, you know, example I am taking, CH2, negative charge, single bond C, double bond O, H, CH2, double bond C, single bond O minus, okay? Tell me the bond order of carbon oxygen bond, carbon oxygen. What is the bond order? You are saying three by two, correct? But in this case, this bond order should not be equals to three by, we cannot take the average here because the distribution of electron is not uniform, right? So we cannot take the average here. The first thing, what we'll do, I'll tell you. First thing is what? In case of unequal contributors, we cannot calculate the exact value of bond order, exact value we cannot calculate, but we can compare because distribution of electron is not equal, it is unequal contributors, so we cannot take the average, okay? And why we cannot calculate? Since its contribution is since it's, contribution is not equal. Now we cannot find out the exact value, but we can compare the value of bond order. For example, the example I have given you here, in this example, you see, if you ask me the bond order of carbon and oxygen here, so we cannot take the exact value, but I can surely say this, that this value will fall in this range one to two because here it is bond order is two and here bond order is one. So bond order will be in this range one to two. Whether it will be close to one or close to two, it depends on the stability of this structure and this structure. Since it is equal contributors, so one will be, since it is unequal contributors, so one will be major contributors, other one will be minor contributors. And how do we define that major and minor contributors? How do we define that? Which one is major contributor? Which one is minor contributor? Tell me, how do we define which one is major and minor contributor? Yeah, based on stability, right? So the one which is more stable is major, lesser stable is minor, correct? So, and for that stability, I have already given you the rules, relative stability of RS, okay? So that rule if you apply here, you see number of pi bonds are equal, right? Octet is also complete and we have a rule over there that negative charge on more electronegative element is more stable. So stability of B here and A, so B is more stable here, right? And in the more stable compound, the bond order is one here, right? So we can say the bond order of carbon option bond will be close to one, okay? So this example, this is not true, but bond order is close to one here or we can also say this bond order will fall in this range from one to one point five, right? Exact value, we cannot say, but the range we can say. Did you understand this? That is why those rules are important, the relative stability. See, what I said, first of all, did you understand this? Why the bond order will not be equal to three by two per week? Why it is not equal to three by two? Did you understand this? So when it is not equal to three by two, we cannot take the average. So bond order, see, these are what? These are the imaginary structure. The real structure is what here? The real structure will be this, CH2, C, single bond O, H. And the real structure will be this, the pi bond or the pi electron is delocalized from here to here. Here we have delta negative and delta negative like this. This is the exact structure. So bond order of carbon and oxygen, if you see, it is neither equal to one, not equal to two because it is not a complete bond. It is a half bond, right? So from this also you can understand, from this also you can understand that the bond order of carbon and oxygen will be in this range, one to two, correct? Whether it is closer to one or two, it depends on the stability of this and this structure. If this one is more stable and in the more stable structure, the carbon-oxygen bond has single bond, right? Hence we can say the bond order of carbon and oxygen will be closer to one, it won't be equal to or closer to two. That's what the point is. Did you understand this now? Correct. Now on this also you see this question. Tell me, now these are the questions only. You can get these questions and see it here. Suppose this is alpha and this is beta. Tell me or compare the bond order of alpha and beta. Alpha is carbon-chlorine bond in that chlorobenzene and beta is again carbon-chlorine bond in that chlorocyclohexate, order of alpha and beta. Alpha is greater than beta and why is it so? Answer is correct. Why alpha is greater than beta? Resonance, it's not resonance stabilized. The thing is because of resonance, reason is right, resonance. Because of resonance, here we have partial double bond characteristics, right? This carbon-chlorine will have partial double bond characteristics. That is why the bond order comparison if you do, it will be more for alpha and less for beta. If you write down bond energy or bond strength, the order will be same. Alpha is more than beta. Bond length, alpha is less than beta. All these three order or comparison you can do easily on the basis of this, okay? Bond order, bond length, bond energy, bond strength, okay? Any of these questions they may ask. All these depends on resonance and bond order. Now you see and this unequal contributors are more important basically. I'll write on few questions now. You solve this and tell me the answer. CH2 double bond CH, single bond OH, CH3, CH2, OH. Carbon oxygen bond, CH2 double bond CH, NH2. CH3, CH2, NH2. Carbon nitrogen, CH2 double bond, CH2. CH2 double bond CH, OH, carbon carbon bond. This is alpha and this is beta. Yes, so what do you have to compare? This carbon oxygen bond, this carbon oxygen bond. Carbon nitrogen, carbon nitrogen, beta. This is alpha, this is beta. Alpha, beta, this is alpha, beta, gamma. Alpha, beta, alpha, beta, gamma. Yeah, you write AB, that also easy to write. I can understand that also. Instead of alpha, beta, you write AB, AB like that. Alpha is A, beta is B. Okay, I'll discuss now. I think your answer is 50% wrong. 50% right, 50% wrong. First two, I think all of you have got the right answer for the first two questions. Because you see why alpha, the answer will be alpha is greater than beta. Why alpha is greater than beta? Because of resonance possible here. Right, lone pairs, sigma pi. So we have conjugation resonance possible here. Hence alpha is greater than beta. Similar condition we have here also. Resonance possible here. Resonance, I'll write down R. Here also resonance possible R. And because of that, again alpha is having more, alpha having more bond order than beta. Now in the third question, I can see some of you have got the wrong answer for this. See, we are comparing carbon-carbon bond, right? So because of resonance, what happens? You see, I know you must have got the, you must get confused with this bond. You may be calculating the bond order of this. You must understand one thing, the bond order we always calculate between the two atoms in a molecule, right? Since I have given you to compare the bond order of carbon-carbon bond, carbon-carbon bond. So because of resonance, what happens? You see here, this bond will jump over here, right? And the resonating structure of this will be CH2, negative charge, single bond CH, and double bond OH, positive charge, right? Now in this structure, you see, we have single bond here. It means the bond order of this beta, beta will fall in this range, will be greater than one and less than two, right? But the bond order of alpha here, it is what it is exactly two because there is no resonance here, integral value. Right? Hence alpha having more bond order than beta, right? Now in this case, you see, you are comparing alpha and beta here. This alpha value is what? Two, because there is no resonance, but beta is again in between one and two. Alpha is more than, correct? The value of alpha here, again, it is two. Value of beta here in between one and two, and the value of gamma here is one. Order will be alpha maximum, then beta and then gamma. In this molecule, what happens, you see? Because of resonance, this bond will become single, we'll double bond here and edge. Oh, sorry, one thing. We have negative charge here. This is negative charge, and this is positive, right? Again, we have single bond characteristics. So beta will be less than two greater than one. Answer will be this alpha is two. Here, what happens, you see? Alpha value is two, beta in between one and two, and gamma also in between one and two, right? But in beta, what happens, you'll see, let me draw the resonating structure here. We have double bond O, single bond O. This will come over here, and this will go here. So we get here double bond, oh, sorry, the order will be this will go here, and this will come here. So here we have positive charge, single bond, negative charge, okay? But if you have this structure, gamma, the first resonating structure will be this. This bond is here, and we can draw one more RS here, which is nothing but this single bond O minus, okay? So in this, we have one, I say beta is this bond, right? So in this, we have only one RS, which has single bond characteristics. In which we have, in this, we have two RS, which has single bond characteristics. That is why when we compare beta and gamma, the bond order of beta will be more, right? So order here will be alpha maximum, then we have beta, and then we have gamma. Which one, Purvik? That's what, it is on oxygen only. Yeah, correct. Did you understand this, how to compare the bond order? And when you know bond order, you can write down bond strength, bond length, anything. Yes, all of you understood. One last question we'll see here, and then we'll move on to the next application. Unequal contributors are more important here, right? So you must understand the comparison, how to do. Compare the bond strength or bond order here in this molecule. We have a benzene ring, nitrogen, CH3, CH3. You have to compare these two. This is alpha, carbon-nitrogen bond, and this is beta. Bond order, bond strength, bond length. Ramchandran, you there? Tell me the answer, Ramchandran. B greater than A. Why? Why B is more than A? A does not go and undergo resonance. Why, why is it so? Purvik, CH3 is attached to a place where there will be a negative charge. Not sure, I think, because carbon will have negative charge. CH3 resonance, negative charge. No, that's not the reason. Yeah, steric hindrance is the reason. See, steric hindrance is the right reason here. What happens with steric hindrance, you see? Since we have bulky group present here, because of this group, what happens? We have repulsion here in these groups, right? CH3, CH3, CH3, CH3, CH3. And to minimize this repulsion, this nitrogen changes its plane, right? And actual structure will be this. One will go, one molecule will go into the plane, and other one will come out of the plane. Now, the essential condition of resonance is what? The molecule must be planar. So whatever electron pair we have here, since it is not planar molecule, it has changed its plane, right? This lone pair is not involving resonance. So in this case, where we have this kind of group present, we have no resonance because of steric hindrance. And hence, the bond order alpha, we can say here it is equals to one, exactly one. Because a lone pair is not involving resonance. But here, the lone pair is involving resonance, right? So this bond order will be somewhere in between one and two, right? So that is why the bond order of beta is more than alpha. Bond energy of beta is more than alpha, and bond length of alpha is more than beta. This is the pre-order. Is it clear? Because of steric hindrance, resonance is not possible here. And when there is no resonance, bond order will have integral value, alpha is equals to one. Got it? Yes or no, tell me. Okay. Now, one more thing. If I ask you, what is the basic strength? Basic strength of first and second, A and B. Which one is more basic? Alpha or beta? One is compound A or B, A or B. A is more basic. Why? Because lone pair is available on the nitrogen. This lone pair is available here. That's why it is more basic. Yeah, correct, correct, Sandhya. So basic strength of A is more than that. Understood? Can we move on? Tell me. Last class we have discussed plus M minus M effect, right? I have given you plus M minus M group. Yes or no? Model group shows plus M effect minus M effect. Okay. Now I am writing down plus M minus one. I'm not going to discuss now, but we'll solve some questions onto this, okay? Stability and all we'll discuss in the last, today's session only. But first of all, you tell me, in these examples, which I'm going to write down now, which all effects are possible, right? Mesomeric, eye effect, which all effects are possible and which one is dominating. Suppose we have a benzene ring here and NH2 attached on it. On benzene ring, we have OCS3 attached on it. On benzene ring, we have NH3 plus attached on it. On benzene ring, we have CS3 attached on it. We have CHC11O, H attached on it. On benzene ring, we have chlorine attached on it. On benzene ring, we have NO2 attached on it. In this one A what all if it's possible here plus M minus M plus I minus I plus M minus I correct you see here we have plus M and minus I possible minus I because of plus M is electron releasing yes yes perfect plus means electron releasing whether it is M or I it is always electron releasing electron donating okay so plus M minus I why plus M because the first atom has lone pair on it and why minus I because nitrogen is more electric electronegative than this carbon okay but we know plus M dominates minus I mesomeric effect dominates inductive effect okay OCS 3 can I write plus M and minus I again but plus M again it is dominating why because we are lone pair on this oxygen to plus M now NH 3 plus what is the effect here NH 3 plus what is the condition for minus M Sondarya last class I have discussed the condition also I'm not getting this how it is minus M all of you are saying minus M how it is minus M what is the condition for minus M what is the condition for minus M tell me Ramcharan you there where is Vaishnavi yeah electron withdrawing is fine electron withdrawing is fine but how it is minus M what is the condition for minus M tell me last class I have discussed this tell me what is the condition for minus M did you write the notes in the last class when I was discussing this plus M minus M group if not then let me know I have to discuss many things today I'll discuss it again and move on double bond between first and second atom correct do we have double bond here or multiple bond do we have multiple bond here present in NS 3 tell me drawn withdrawing from the ring is fine it's right it withdraws electron but what effect we have here for a minus M effect this is the condition we have x double bond y or x triple bond y where the electronegativity of y should be more than x and another third condition is what we must have a vacant p orbital present on to this like CS 2 plus or bs 2 right but in this molecule when you write NS 3 plus right there is no vacant p orbital here there is no multiple bond also it won't show minus M is it clear yes or no but since nitrogen is more electronegative than oxygen right and one positive charge on it makes it more electron deficient so it only shows minus I effect did you understand this so if you do not have the understanding of this plus M and minus I it is very difficult for you to solve this kind of question first of all you should understand what group shows what effect and which effect is dominating if you have this understanding then only you can so all you can solve all those questions based on stability of intermediate acidity basicity but first of all you should understand this all of you understood this right this condition you try to apply okay think about it right now CS 3 attached here what group it is what effect it shows do we have plus impossible apply the condition of plus M possible or not tell me apply the condition of minus M possible or not tell me only plus I yes very good this shows only plus I right plus I'm not possible minus M is also not possible because of plus M we should have at least one lone pair onto this which is not there minus M not possible because we do not have any multiple bond present to plus I is possible and if you ask me any other effect it can show plus H also hyper conjugation but this I am not writing it down here right because hyper conjugation we haven't discussed we'll discuss it today also right but yes plus I possible here correct here what effect we have what effect we have here can I write down minus M and minus I where minus M is dominating minus M minus I multiple bond between first and second atom minus M minus I minus M dominating what about this chlorine yes plus sense plus S predominates here correct right but we haven't discussed hyper conjugation I am not telling you that okay but because of plus H and plus I your result will not change okay your result will not change both have the same effect or your result will not change but if somebody asked you what all effects possible here the answer will be plus I and plus H both right I'll discuss this hyper conjugation also after that you can you can you know combine all these and you can understand okay here what are the effects possible here chlorine has lone pair on it so it can show plus M since more electronegative and show minus I also but for chlorine we have discussed for halogen minus I is dominating over plus it okay hyper conjugation we have discussed last class fine then fine so in this case you see we have double bond and alpha hydrogen so hyper conjugation also possible you can see right and this is the dominating effect here okay but if you do not consider this hyper conjugation then only plus I right okay let it be this we'll discuss again minus I dominates plus or minus M in case of chlorine order will be this we'll have minus M and minus I possible minus M dominates minus M minus I possible minus M dominates is it clear there's no condition as such you see wherever we have resonance possible resonance possible hyper conjugation possible and inductive if it possible then the most dominance condition is resonance then hyper conjugation and then inductive okay okay Vaishnavi understood our edge and I this is the order we have see guys you give me some time okay you will understand all these things okay but let me finish this session today 3R session let me finish then if you have any doubt you ask me okay don't mix resonance hyper conjugation and all okay I am going in a in a way right I know what all things I have to discuss right if you want me to ask which one is dominating order will be this okay but don't worry how to apply hyper conjugation resonance and inductive effect for that will take the example for that okay and then I can you know tell you for sure that you will not have any doubt if you concentrate okay but don't worry with this hyper conjugation and I think I will discuss that also we'll take that examples for that so like this first of all you should understand how what all groups are you know there and they shows what kind of effect okay plus H minus H plus I minus I plus M minus M okay hyper conjugation little bit or we'll discuss today okay but again we'll discuss that now you see the next application like I told you aromaticity is the another application we have here so we'll discuss next aromaticity now I'll write on quickly there are three types of compound basically aromatic then non-aromatic and then anti-aromatic so what is the condition for aromatic compounds the molecule must be cyclic planet conjugated and it follows Huckel's rule which is 4n plus 2 pi electron right 4n plus 2 pi electron for example we'll take an example of benzene ring it has 6 pi electron and aromatic these 6 pi electron the molecular orbital that forms here like this the molecular orbital this 6 pi electrons are placed in this molecular orbital and all these electrons are paired actually you see one to who's rule 6 pi electron since all the electrons are paired that is why this molecule aromatic compounds are highly stable right now for anti-aromatic the first three conditions are same it must be cyclic it must be planar it must be conjugated and number of pi electron should be 4n pi electron this is the last like you know condition we have for example you see if I take this example cyclobutadiene it has 4 pi