 Okay, so we were talking about cyclic conjugated hydrocarbon which is nothing but anulines okay like for this example you see the name of this compound general name will write anuline and in the bracket we write a number which is equals to the number of carbon atom in the ring okay. So this number should be equals to the number of carbon atom in the ring okay. So this is an anti-aromatic compound because it has 4 pi electron so anti-aromatic we have already discussed this. Similarly benzene is also in a type of anuline and the name of this compound is anuline 6 right. So these compounds we already know it is aromatic compound 6 pi electron this compound name of this compound is anuline 8 because there are 8 carbon atom anuline 8 okay and number of pi electron if you see it has 8 pi electrons 8 pi electrons expected to be what expected to be anti-aromatic expected right. But since the carbon atom is more than 7 then the actual property is what it is non-aromatic in nature right non-planet because non-planet changes its plane and becomes non-aromatic okay. So this one is expected but not true okay. The structure of this if you see which is not actually important but I will just draw here like this it changes its plane okay and becomes non-aromatic okay. One very important example we have here in anuline which is this. Now in this structure what happens if you count the number of pi electrons the name of this compound is what it is anuline 10 carbon 1 2 3 4 5 6 10 anuline 10 okay the number of pi electrons if you see it has 10 pi electron right and hence expected to be aromatic but what actually happens here this hydrogen because this carbon is sp2 hybridized so we must have one hydrogen present this side and here also one hydrogen present this side. This distance is not that enough so that it can accommodate to hydrogen atom over here right. So because of this hydrogen-hydrogen repulsion one of the ring comes out of the plane okay one of the ring comes out of the plane and then because of this hindrance I'll write down because of this hindrance hindrance molecule becomes non-planet is non-planet to minimize the hindrance it changes the plane and becomes non-planet and when it is non-planet it cannot be aromatic but it is what non-aromatic in nature. So this kind of compound is non-aromatic if I write down this example similar kind of example we have is with this bond and this bond suppose of another atom is attached to the oxygen right these two means this oxygen is coming towards you it's it's this the whole ring is in one plane and this is coming above like it is at the upper surface of the plane coming out of the plane right so this bond if it is any atoms attached like this then this ring which is this ring is planar and the molecule is aromatic in that case but if any bond is any atom is not attached like this we have hydrogen hydrogen repulsion here and then the molecules becomes non-planet then it is non-aromatic these are few examples which you have to keep in mind okay normal logic if you apply you will get wrong answer okay that's why I'm telling you that these examples you must revise before going into the exam okay one last example we'll see the compound like this number of pi bonds if you count here one two three only one pi bond will count here four five and six so we have six pi bonds 12 pi electrons the 12 pi electrons means what it should be anti-aromatic but again number of carbon atom is more than seven anti-aromatic is not possible molecule is non-planar and it is non-aromatic this one this one nothing you see here what if you have hydrogen hydrogen like this okay if you have hydrogen hydrogen here present in the same plane of this ring then we have hydrogen hydrogen repulsion which because of that one of the molecule will come one of the one of the ring will come out of the plane actually okay one of the molecule will come out of the plane to minimize the repulsion okay and that is how one hydrogen come this side and another hydrogen will go away from you it's like that okay because of the repulsion here to minimize the repulsion now in this process the whole molecule as a whole this pi electrons are not in conjugation with these pi electrons and hence it becomes non-aromatic correct but this if one oxygen or any atom attached with to these two carbon here and this bond is coming above this plane right this oxygen is somewhere here on the top right the whole molecule is in one plane this whole molecule is in one plane and this this two carbon is attached with oxygen which is present above this plane here right in this case this molecule is planar and hence this aromatic understood so these are the examples you should keep in mind for aromaticity you must revise all these examples before going into the exam if any question they ask from this aromatic part you will be easily answerable to those questions okay you can easily answer those questions if you remember all these two three conditions in mind okay can we move on next so next you see we are going to discuss next is hyper conjugation we'll discuss hyper conjugation first because this application is also required there for hydrogenation and combustion so hyper conjugation will finish quickly and then we'll start hydrogenation in combination if the time permits today okay otherwise we'll finish this into the next class okay you see hyper conjugation we also call it as last class also I told you that it is sigma pi resonance because sigma converts into pi or we also call it as no bond resonance hyper conjugation is only possible in case of alkene alkyne alkyne or alkyne in case of carbocation and the last possibility we have in case of pre radical carbon ion does not show hyper conjugation okay hyper conjugative structure and you see if I draw this structure suppose we have CH2 CH3 CH never bond so for hyper conjugation the condition is what we must require sp3 hybridized alpha hydrogen right at alpha position we must have sp3 hybridized hydrogen present okay what happens because of this you see this sigma converts into pi and this pi converts into lone pair here the hyper conjugative structure of this will be CH2 this hydrogen will present here as H plus only double bond CH single bond CH2 lone pair negative charge right see since sigma converts into pi that's why we call it as sigma pi resonance and since H plus is present there without any bond hence we call it as no bond resonance okay now this structure you can draw with the other two hydrogen also right so the other two hydrogen with also you can draw the structure you'll get similar kind of structure and that structure I am not drawing okay but these three structures are we call it as hyper conjugative structure hyper conjugative structures right so if they ask you how many hyper conjugative structure we have you just have to count the number of alpha hydrogen right number of hyper conjugative structure is equals to the number of alpha hydrogen present over there number of if they ask you the total number of structure total structure possible that will be equals to the number of alpha hydrogen plus the given structure this structure this three plus this is the total number of structure possible correct other thing is what more number of alpha hydrogen more hyper conjugative structure you can draw and more will be the stability got it more will be the stability so the one factor will understand here that stability of alkene also affected by hyper conjugation stability of alkene and this stability is directly proportional to the number of alpha hydrogen present right number of alpha hydrogen present for example you see first question compare the stability of these alkene double bond this stability of these alkene a b c d which one is most stable tell me the first one number of alpha hydrogen here how many alpha hydrogen we have here three three six nine twelve alpha the number of alpha hydrogen here will be nine the number of alpha hydrogen here six number of alpha hydrogen here is three more number of alpha hydrogen more hyper conjugative structure you can draw and hence the stability will be this is the most stable compound and then this and then this right number of alpha hydrogen if you count here it will be what these are the alpha carbon one two three four alpha carbon with two hydrogen on each so we have eight and number of alpha hydrogen here will be one two right so one and two three right so number of alpha hydrogen is more here more will be the stability number of alpha hydrogen if you count here three five seven number of alpha hydrogen you count here that will be one so this one is more stable than the first one how many alpha hydrogen we have here and how many alpha hydrogen we have here where we have nine see here the number of alpha hydrogen is what three here and three here six this is the alpha carbon right number of alpha hydrogen two here stability will be this yes what about the stability of this molecule stability of these two why equal second one is more stable why because second one has more alpha hydrogen you see here what happens we have resonance possible conjugated system here we have hyper conjugation possible resonance is more dominant stability order is this so the order of stability follows this thing r h and then i if we have two molecules given in one we have resonance other one we have hyper conjugation then the resonance one is more stable okay similarly tell me the stability of this alkene oh cs3 then cs3 which one is more stable here here be all these things requires to solve the questions which is related to heat of hydrogenation heat of combustion because heat of hydrogenation and heat of combustion depends upon the stability of alkene so first of all you should know the stability of alkene and to stability of alkene to you know explain that you should know what is resonance what is hyper conjugation what is high high effect how it affects the stability of alkene okay how will compare the stability of alkene layer right so here we have lone pair present on this right so this lone pair is involved in resonance here right that is why this one and here we have three alpha hydrogen number of alpha hydrogen are three and here we have resonance correct and this is hyper conjugation okay so this one is more stable here because of because of resonance right so when you have this order or if you can identify the resonance and hyper conjugation you can find out the stability and with that we can compare the heat of hydrogenation and combustion also see in all these examples first four examples that I have given you here you see the first four examples that I have given here in all these examples the number of alpha hydrogens are different right number of alpha hydrogen are different but what happens if the number of alpha hydrogen are same in the given compound for example if I give you this example suppose we have a CH3 C double bond CH2 CH3 this is one compound and the another compound is this CH3 C double bond CH3 H H and the last one we have is CH3 CH double bond CH3 in these three molecules we have to assign the stability of the alkene so if you count the number of alpha hydrogen sp3 hybridized right so here it will be three and three six number of alpha hydrogen here if you see three here and three here six number of alpha hydrogen here it is six now so which one is more stable that's the question so now when we have alpha high number of alpha hydrogen same so what we do we'll try to you know write down the hyper conjugative structure for this if you write down the hyper conjugative structure that will be CH2 double bond C single bond CH2 right negative charge one CH3 will be as it is and we'll have H plus present here correct further this kind of you know five mode structure we can draw yes or no are you getting it let me know guys please five more structures will get similar kind of structure right if you draw the hyper conjugative structure here that will be CH2 H plus here double bond C single bond CH negative charge CH3 and H also we have right and similar kind of five more structure again we can draw okay similarly here also if you draw the hyper conjugative structure that will be H plus CH2 double bond CH single bond CH CH3 similar kind of again five more structure we can draw now what is the difference in these three hyper conjugative structure can you tell me the difference here what is the difference in these three hyper conjugative structure no no kushal hyper conjugation what is the difference here in these three structures tell me see this structure is it is one degree anion yes it is one degree anion right this is two degree anion and this is also two degree anion right and we know one degree anion is most stable so the five more structure that you draw here all this structure will give you one degree anion only and that is why this is suppose A it is B and this is C so one thing we are sure with that A will be the most stable alkene which is in all the hyper conjugative structure we are getting only one degree anion here you see in this two you can apply some hysteric hindrance factor right since both the methyl group are at maximum distance and both are two degree anion so because of this more distance the hindrance will be less and hence B will be more stable than C anyways trans isomer is more stable than the cis one so these two you can say easily this is trans this is cis sorry this is yeah trans and cis this one is more stable and since this one gives you one degree anion so this one is the most stable compound did you get this after this we can discuss heat of hydrogenation okay but we don't have time for that okay so we'll start heat of hydrogenation in the next class i'll take probably you know one hour more to complete this heat of hydrogenation company combination combustion and then we can start discussing problems right problems on this one more thing we'll discuss you know what is sir effect sip effect you know that hysteric inhibition of resonance how ortho substituted benzoic acid becomes more acidic than the para or meta substituted why it happens you know that okay so next in like next class in first one hour right we'll finish all these things okay hydrogenation combustion order how to define and then some acidic basic nature also what what all effects possible at different different places and how ortho substituted benzoic acid is more acidic than para or meta substituted it takes time okay so we'll discuss this into the next class so all of you must revise whatever we have done today okay aromatic compounds examples you just keep those examples in mind you won't get anything more than that yeah i will send you i will send you worksheet on that one pdf file i'll send you okay today only i'll send okay so we'll wind up the class here okay thank you yeah i'll send you the worksheet today i have one few set of questions which involves j e means question only that also i'll send you and there are a few basic questions also i'll send you because there are totally 30 40 questions will be there okay we'll send you that okay thank you guys i will see you in the next class take care bye bye