 good morning all of you please type in your name okay good morning did you finish your assignment that I have sent you in the group did you finish the assignment right discuss that fine so what we do in the last half an hour because I am accessing now this YouTube thing so I cannot go to the WhatsApp group and check the no questions so what we do in the last half an hour you can send me your doubt your question number on the group I will discuss them okay there only fine last class we have discussed hyper conjugation right have you discussed aromaticity you examples of aromatic compounds have we discussed that aromaticity we have discussed or not so we have finished aromaticity and we have we have also discussed you means how many rationality let me know let me know guys quickly so just one example you tell me this one I think we have done this but just to make it sure I'm just giving you one example aromatic or what or even this one these two molecules A and B aromatic behavior you tell me okay yes correct so we have done it already right okay so move will move on okay so last class we have discussed hyper conjugation we have also seen that hyper conjugation is possible okay I'm moving forward now just five minutes quick revision of this and then we'll move on okay hyper conjugation is possible in unsaturated compound like alkene alkyne it is also possible in carbocation these three cases only hyper conjugation possible correct now if we talk about the stability of alkene this we have also discussed right and the condition what hyper conjugation is what we must have sp3 hybridized alpha hydrogen right and the stability of alkene is directly proportional to number of hyper conjugative structure which is also we can say the number of alpha hydrogen which is sp3 hybridized because it's the condition we have okay more number of alpha hydrogen more will be the hyper conjugative structure and more will be the stability right next thing is what the stability of alkene and the stability the stability of alkene in a given molecule there are possibilities that resonance is also possible I effect is also possible and hyper conjugation is also possible in a given compound right if there are different electronic effects possible in the molecule then the stability order will define according to resonance first then hyper conjugation and then I effect means the order we have to follow in this way next thing you have to keep in mind if the number of alpha hydrogen is different in all compounds then stability we can say is directly proportional to the number of alpha hydrogen we are considering only hyper conjugation here if hyper conjugation is only possible into this right but depending on this if I ask you one question here which alkene is most stable here tell me this is a CH2 double which one is most stable in all these change this in ABCD why is it so because you've seen all the molecule here the number of alpha hydrogen will be 12 number of alpha hydrogen is 9 here number of alpha hydrogen is 6 number of alpha hydrogen is 3 and here we have no alpha hydrogen right the more number of alpha hydrogen more will be the hyper conjugative structure and the stability will be more so a is maximum then B then C then D and then E and that is why you must have heard this or you must know this that more substituted alkene is more stable have you heard this the stability of alkene what we say more substituted alkene is more stable yes behind this is what because more substitution more will be the alpha hydrogen right you see this alkene is the most substituted among the all right that's why it is more stable right the reason is hyper conjugation only one more question we'll see tell me the stability of these two these are the questions only we are discussing in the form of concept we are discussing question only this one and which one is more stable a saw this question and discuss me this are there this okay the first one how many alpha hydrogen we have here 2 4 6 8 we have 8 alpha hydrogen and here we have 3 and 1 4 alpha hydrogen so a is more stable than B right number of alpha hydrogen here will be 3 5 and 7 number of alpha here in here is 3 so 7 since this one is more stable number of alpha hydrogen in this how many alpha hydrogen we have here and here tell me first compound how many alpha hydrogen we have 6 sp3 hybridized alpha hydrogen right so it is 6 and here it will be 2 right so this one is more stable in these two which one is more stable fifth one which is more stable question number 5 see in this what happens this compound is aromatic and this compound is anti aromatic and that's why this is more stable if you talk about this 6 1 which is more stable the point I'm trying to make here it is what the given molecule you never know that what all effects are possible you have to know analyze the compounds then like in all these compounds from 1 to 4 you see from 1 to 4 the you know there is hyper conjugation possible there's no resonance but when resonance comes into the picture then that will be the more stabilizing factor right so in the fifth example you see the first compound is aromatic and the second compound is anti aromatic we do not have hyper conjugation here but since the first one is anti in is aromatic it is more stable correct what about the sixth one which one is more stable for a stable why question of the reason tell me what is the reason for that correct correct so here we have resonance right so here we have resonance possible do we have hyper conjugation possible yes we have hyper conjugation also possible here because of alpha hydrogen present right here we have only hyper conjugation possible since resonance is there so this is the more stabilizing factor here first one is more stable than the second one okay just a second I have there in the center of there in the classroom yeah okay guys sorry got urgent ball actually right so here we have a resonance hyper conjugation this one is more stable again you see this pi bond here it is in conjugation right so hence this is more stable than the second one so here also we have resonance possible and hyper conjugation also there right but here we do not have resonance but hyper conjugation is there because of alpha hydrogen the resonance is the more stabilizing factor we have here right so from all these examples what you know what point I'm trying to make is what you have to find out the stability of alkene because once you have because if the question is related with heat of hydrogenation or combustion you have to deal with the stability of alkene that's why I have given you all these examples okay now the next thing we are going to start here is heat of hydrogenation so we'll start here heat of hydrogenation delta at hydrogenation like this now what is heat of hydrogenation it is the amount of heat released it is the amount of heat I'll write down in short you also write down it quickly amount of heat released one mole of alkene one mole of alkene goes under hydrogenation for example you see here to the next page I hope all of you have copied it down okay for example you see if you have this molecule suppose two molecules I'm taking here one is this that is butene and other one is this which is due to in this is supposed molecule A and this is molecule B right and when hydrogenation takes place it forms butane this also under hydrogenation gives butane okay since in both the reaction we are getting butane here so energy of product will be safe right and in these two can you tell me which one is more stable A or B which one is more stable A or B is more stable and why is that so why is that so what is the reason behind that more alpha hydrogen correct so this alkene is more substituted and number of alpha hydrogen here it is six and here the number of alpha hydrogen is two right so this one is more stable right so if I draw the energy profile of this reaction right energy profile so suppose this is the energy axis we have right and if I write down these two molecule A and B if I try to put the energy profile of A and B which one should be A which one should be B this should be A because B is more stable and less energy right it is more stable hence is it has less energy and A is less stable so we'll have more energy right so A will be here B will be here right but the product here we have that is suppose B the energy of that product will be same because in both the reaction we are getting butane only correct now here you see if I draw the energy profile for B it would be like this and for A also it would be like this right now the energy difference here you see it will be for B it will be this distance that is delta H B right and for A you see the energy difference will be this that is delta H A so if I talk about the magnitude of this energy which comes out in both the reaction the magnitude of A will be is more than the magnitude of B right plus I'll write down here delta H A because it is an energy releases in this reaction delta H B so magnitude of delta H A is more than to that of delta H B and which is nothing but this energy nothing but heat of hydrogenation heat of hydrogenation so heat of hydrogenation of B is less which is what which is more stable right this is more stable less energy right so from this example what we can conclude that more stable energy sorry more stable alkene gives sorry more stable alkene gives lesser heat of hydrogenation right so what we can write the the stability of alkene is inversely proportional to the heat of hydrogenation this is the first conclusion we can make here okay did you understand this stability of alkene you write down here stability of alkene one thing here you have to your the constraint we have since you see A and B both molecule has only one pi bond so this relation is only true when the molecule has molecule has equal number of pi bonds okay equal number of it will be true otherwise it is this we cannot say but what happens if the molecule has different number of pi bonds so for that we'll write here for the molecule which has strength number of pi bonds the heat of hydrogenation delta H hydrogenation is directly proportional to the number of pi bonds okay this is a two things you have to keep in mind to compare the heat of hydrogenation okay now one thing you see what happens here why see these these reactions are what these this process is this process is exothermic energy releases right exothermic process why it is exothermic you try to understand what happens in this reaction you see one pi and one sigma bond breaks and we'll get we'll get what two sigma bond because you see when this pi bond it means we have one hydrogen here attached with and one hydrogen here attached the pi bond breaks and we'll get in hydrogen here here also what happens we'll have an hydrogen here and we'll have an hydrogen here this hydrogen is nothing but this hydrogen and this hydrogen so what we can say you see one pi and one sigma bond breaks and we'll get two sigma one pi one sigma we'll get two sigma right so pi bond we know it is a stronger than the sigma bond so on the cost of one pi bond we are getting one sigma bond which is which has lesser energy right the difference in energy of this pi bond and one sigma bond will come out and that's why it is an exothermic process right so if you have more number of pi bonds right more sigma bond will form and more number of pi bonds will convert into sigma bond and that is why more energy releases in that process correct now these two things you have to memorize okay that stability of alkene when we have equal number of pi bonds inversely proportional to heat of hydrogenation when the number of pi bonds are different then heat of hydrogenation will be directly proportional to the number of pi bonds present is it clear can you solve some questions now on this last thing I'll tell you here when I say stability of alkene so to to you know to decide the stability of alkene right we'll use all those factors that we have discussed just now in those examples resonance hyper conjugation and all everything we'll see right okay so you see we'll are going to discuss your questions here you tell me the order of heat of hydrogenation right the question is the order of delta H hydrogenation the first question here is a and b then solve all those questions all the four questions then we'll start discussing it solve this I'll get some water done molecule which has different number of pi bonds right so more number of pi bonds more will be the heat of hydrogenation so order of delta H hydrogenation will be this right these two molecules has equal number of pi bonds number of pi bonds right but this alkene is more stable because of resonance and here we have hyper conjugation right the alkene which is more stable is this and we know the stability is inversely proportional to the heat of hydrogenation hence the order will be this you see that this alkene is stable because of resonance here only we have hyper conjugation so which one is more stable this one and the heat of hydrogenation order will be this now in this molecule you see we have we have different different number of pi bonds right so according to the logic you see the order should be what if it is a and c right we know more number of pi bonds more will be the stability more will be the heat of hydrogenation so order of heat of hydrogenation should be cb and then a but this order is not found to be correct this is not wrong this is actually this is wrong this is actually an exception here why it is an exception you see this compound is benzene which is highly stable because of aromaticity right so one factor is reducing the heat of hydrogenation since it is highly stable right the another factor which is number of pi bonds is increasing the heat of hydrogenation right so both factors are contradictory to each other and we know in chemistry when two factors are opposing each other then the result becomes experimental in that case okay and in this case the result is or the heat of hydrogenation is found to be in this order b has maximum then we have c and then we have a this is the order of heat of hydrogenation for this molecule this particular example you must remember you got it exceptions you must keep in mind okay the next thing here we have to discuss is heat of combustion co c heat of combustion now in this only we'll write down the similar way the heat of combustion is again inversely proportional to the stability of alkene stability of alkene and when I say alkene the stability of alkene for that the deciding factors are what will decide will decide the stability of alkene due to hyper conjugation due to resonance and due to aromaticity any of this factor can be there in the given question okay all these three factors you have to look after okay this is this condition is true when the two molecules has equal number of carbon atom this is true for the molecule has for equal number of carbon atoms number of carbon atoms but if the number of carbon atoms are different for different carbon atoms the heat of hydrogenation sorry heat of combustion h o c is directly proportional to the number of carbon atoms so these are the two factors then you have to keep in mind now in this we have third possibility also and that is if the molecule molecule has double bond different number of carbon atom this is not the special case it comes under this only molecule has double bond but different number of carbon atoms and we also decide the heat of combustion is directly proportional to the number of carbon atoms this factor is again dominating this is a three case possible this case the last one comes under this only but I have given here aromaticity is dominating okay I will discuss this also today see um for any carbocation if you have right then aromaticity always dominates resonance aromaticity always dominates resonance there is one factor which dominates aromaticity also and that is dancing resonance okay otherwise aromaticity is a dominating factor okay so we'll discuss that also first let me finish this thing quickly okay you understood these three points you have to keep in mind for heat of combustion now we'll see some questions on this of heat of combustion of these two molecules first one a b a b a b c b a b c b okay what is the answer guys the first one you see alpha hydrogen number of alpha hydrogen here in the first case will be what sorry number of alpha h is nine number of alpha h here it will be what we have three here and two here right so it will be five so this one is more stable and heat of combustion for this one is more inversely proportional to the stability right these two what is the answer what is the answer you have done for the second one h a is more you see in this this example is important okay and it's a good example also because when you look at this molecule it has more number of alpha hydrogen compared to this one right but the point is this molecule has more number of carbon atoms right since this molecule has more number of carbon atoms hence the heat of combustion of first will be more because it is directly proportional to the number of carbon atoms right so however hyper conjugation is there and more number of hyper conjugative structures are there but since it has more number of carbon atoms more will be the uh what we say uh heat of combustion now in the third one you see a b c d right so it is all the molecules has different number of carbon atoms and we know when the number of carbon atoms increases heat of combustion increases the fourth one you see fourth one all the molecules has equal number of carbon atoms right but this one is least stable least stable why it is least stable because this has maximum angle strain maximum angle strain so least stable so most will be the hyper conjugate sorry heat of combustion order so angle strain is also one of the factor right to to determine the stability one more example you see the fifth one of combustion order a b c d tell me the order what is the answer for this one c b a right more number of carbon atoms what we can say more number of c we have double bond right but the number of carbon atoms is also increased there are different number of carbon atoms the more number of carbon atoms more will be the heat of combustion order will be this let's say it is d c b a decreasing order it's not stability it's heat of combustion okay now this is these are the few examples we have discussed you will get questions like this only especially heat of hydrogenation is important so always consider the factor of stability of alkene we can have hyper conjugation we can have aromaticity right we can have resonance these three factors can stabilize alkene so whenever you have the questions identify the stability order according to these three factors whatever the possible factor we have right aromaticity is most stable than resonance and then hyper conjugation and inverse will be the heat of hydrogenation order right now like i said hyper conjugation is possible in case of alkene carbocation and pre radical right a few examples of carbocation also we see so we have hyper conjugation in carbocation so you see one example i'll write down ch2 ch2 positive charge here and hydrogen will have here right so again we'll draw the hyper conjugative structure of this this will come here ch2 h plus double bond ch2 this is the hyper conjugative structure and similarly we can draw two more hyper conjugative structure here because we have three alpha hydrogen present right number alpha hydrogen is three so we can draw three two more hyper conjugative structure right here also you see the stability of this carbocation right which is which is you know which is affected by i effect also but hyper conjugation also stabilizes carbocation right so what we can say that stability due to hyper conjugation if i write stability due to hyper conjugation is directly proportional to again the number of alpha hydrogen right number of hyper conjugative structure here it is what it is equals to the number of alpha hydrogen which condition should be considered for hoc stability what is hoc hype what is this hoc stability okay heat of combustion for heat of combustion stability or number of carbon atoms okay see heat of combustion is inversely proportional to stability right if you have equal number of carbon atoms right if you have equal number of carbon atoms then we consider hoc is inversely proportional to stability right but if the molecule has different number of carbon atoms suppose in the example third example you have if the molecule has different number of carbon atoms then more number of carbon atoms more will be this more will be the heat of combustion got it if the molecule has different number of carbon atoms then heat of combustion is directly proportional to the number of carbon atoms okay but for equal number of carbon atoms we'll check stability now you see if one example i'll write down here stability right opposite to number of alpha hydrogen because of hyper conjugation you see this example if you have this carbocation h3c will have so here you see since we have sp3 hybridized carbon atom also present alpha carbon atom also present so here we have hyper conjugation possible and plus i effect also possible the same thing we have here also hyper conjugation and plus i hyper conjugation and plus i right so which one is dominating factor here obviously hyper conjugation dominates over i effect but the stability with both the factor you can see that you'll get the same thing right since hyper conjugation we have discussed since the number of alpha hydrogen is nine over here number of alpha hydrogen is six over here number of alpha hydrogen is three over here so order of stability will be this sample you see right what all effect possible again we have plus i hyper conjugation plus i hyper conjugation plus i hyper conjugation sorry we have plus i hyper conjugation but which one is dominating in all these hyper conjugation is dominating and with hyper conjugation or plus i whatever you consider the stability will be maximum for this and then this because this molecule has seven alpha hydrogen this molecule has three alpha hydrogen and this molecule has one alpha hydrogen right but we compare the stability here for these two molecule which one is more stable we have ch2 plus ch plus ch3 which one is more stable just a second just a second we have ch2 and then ch2 plus this one comparison i'll give you tell me the answer third one which one is more stable fourth one b is more stable than a what about the third one third one you tell me a is more stable than b why a is more stable because this is resonance right and here we have hyper conjugation right so resonance is more stable fourth one you see we have alpha hydrogen two here here also we have alpha hydrogen two so we have equal hyper conjugative structure but at the same time here we have more plus i effect fourth one we have more plus i effect because we have profile group here here with high group and that is why the second one is more stable so all these things are relative you see if you see the hyper conjugative structure will have equal hyper conjugative structure but i effect is a is a you know deciding factor here because we have more plus i here lesser plus i here so overall the stability of this carbocation will be more got it similarly you see the same thing we have for free radical also but the only difference is in this carbocation we have complete transfer of electron but in case of free radical we'll have homolysis right one electron will go on to the one atom right so that's the thing so i'm not going to discuss that free radical part again right it's the same thing exactly same thing we have for free radical right now hyper conjugation like resonance and i effect also hyper conjugation also there are two of two types of hyper conjugation there are two types one is plus h and other one is minus h plus h are what these are electron releasing effect electron releasing group like we have plus i and plus r similarly minus h are electron withdrawing group which one the second one wait wait i'll see in the second one yeah this is the stability order right this is the stability order the heat of combustion will be diverse of this okay so plus h minus h we have discussed already so we are not going to discuss this again right but only thing minus h is possible in case of c cl3 okay that you must remember cf3 cbr3 and ci3 minus h is not possible in this three case okay ccl3 and then ch and you also know since we have discussed that because of electron releasing group erg or though our para position becomes what electron rich the ortho and para position becomes electron deficient the position is unaffected and this is true in case of what in case of resonance also right and the hyper conjugation plus h effect of cf3 is more than cd3 then cd3 this also i have discussed already so i'm not going to discuss okay now application of this you see we are going to discuss now next slide all right you see application of application of electronic effects right in this section what we are going to discuss we are going to discuss that stability with all electronic effects with respect to the all electronic effects okay so first of all you see if you're talking about carbocation this part is important because you're when you get the question you have to apply all electronic effects possible see the stability of carbocation affects by these factors actually maximum is this is the order i'm giving you okay in this order only will define the stability of carbocation the first and the and the most dominating factor we have here is dancing resonance i will discuss this first you write down dancing resonance and the second factor we have when this positive charge or carbocation is a part of aromaticity third factor we have plus m effect fourth we have plus h fifth we have plus i sixth we have hybridization hybridization seventh we have minus i minus h minus m and the last factor is anti-aromaticity meaning of this order is what the meaning of this order you see till from dancing resonance to this six the first six things are these are stabilizing factor this will stabilize the carbocation these are destabilizing factor destabilizing factor meaning of this is what suppose in the given molecules suppose core molecules are given right and in that four molecule we in one molecule we have aromaticity possible in other one plus m then plus h and plus i so the molecule which shows aromaticity is most stable then we'll have plus m then plus h and then plus this is the meaning of this understood we'll discuss few questions into this but first of all since we haven't discussed this dancing resonance we'll discuss that first see this dancing resonance is a special case you have to actually memorize this only one case it is there in case of or what we say this is a special case case where where sigma electrons taking part part in resonance sigma electrons are taking part in resonance remember one thing in resonance only pi and lone pairs are involved pi electrons and lone pairs are involved right but in this case sigma electrons are involved for example you see if you have this carbocation we have ch2 positive charge and we have a cyclopropyl group present if i draw the resonating structure for this one that will be this or this we have positive charge here positive charge here right now in this what happens this sigma electron any one of these two sigma electrons comes over here right and form this molecule if this comes over here we'll get this molecule correct why it happens because the cyclic cyclopropyl ring that you have it has very high angular strength angular strength and that is why this kind of resonating structure forms here okay so this is a special dancing resonance is actually a special case of cyclopropyl ring okay whenever we have cyclopropyl ring right so for example you see you must have seen this question the stability order of this molecule this is given in opitandon also they haven't given the reason here but the question they have mentioned the order of carbocation they have mentioned here you must have seen this cyclopropyl tricyclopropyl in benzene so here in this case what will be the order of stability which one is most stable which one is most stable no vaginal weights not the third one why i'll tell you when you have three cyclopropyl group then this kind of instruction you can draw even more here understood this sigma electrons also takes part into resonance this also and this also so here you can draw more resonating structures more similar resonating structure you can draw right and that is why the stability for this will be more than this and then this the thing is whatever you can do with one single ring right that you can do with the other ring also if it is present to the carbon got it so similar kind of resonating structure if suppose you replace this hydrogen atom with another cyclopropyl group here then another more structure you can draw correct hence this one is most stable than this and then this now the point is this order i have given you now suppose if i ask you that this is one molecule and along with this three i have given you one other carbocation like this and i ask you what is the stability order of a since in these three molecule involves dancing resonance right and we know dancing resonance is preferred over aromaticity right that is why this is the order of you know carbocation this is most stable than this and then this and then last year is it understood similarly if you have aromaticity and plus m then aromatic carbocation is more stable right plus m plus h plus m is more stable like this you have to compare so for that you should know what is the condition of plus m we have which all groups shows plus m effect where plus h is possible where plus i is possible you understood this on the basis of this so now the question that we are going to discuss now there we have application of all these factors which you know affects the stability of carbocation correct so now we are going to discuss the question here here we have discussed okay so first question i'm giving you tell me the stability of this carbocation we have c h 2 plus and here it is c n o c h 3 n r 2 c l and here we have c h 2 tell me these two first the first one is b c d a c b d a okay tell me one thing here first of all see these type of questions first of all you should know what is the electronic effects we have here because you see this part this part is common in all the compounds right the difference in stability will have because of the group attached here and the nature of these groups okay so first of all you tell me cyanide c n what effects it shows what effect it shows minus m correct so here we have so first of all you identify what effects are there minus m here can i write it is plus m n r 2 what about n r 2 plus m right what about chlorine only minus i however the chlorine has low unfair on it but it shows only minus i effective now minus m and minus i are destabilizing factor because both are electron withdrawing group right so the positive star density will increase and hence it is electron destabilizing sorry it is destabilizing factor plus m are electron releasing it is destabilizing factor so obviously b or c is more stable than a and d correct yes yes yes right personally minus i is dominating for halogen correct so b and c are more stable than d and sorry a and d a and d if you compare minus m dominates over minus i and this has this will decrease stability so obviously a will be a will be the least stable and then we'll have d correct now depending on the nature of n r 2 and o c s 3 which one has more plus m effect the stability we can decide then n r 2 has more plus m effect so stability stability will be b in the second and then c will be the most stable you understood this got it correct now similarly here you see c s 3 shows plus i n o 2 minus m c s 3 plus i right so plus i is what plus i is carbocation destabilizing factor right carbocation destabilizing factor and minus m is destabilizing factor right minus m is destabilizing factor and along with plus i here we can also have plus h possible hyper conjugation right alpha carbon sp3 hybridized right and which one is dominating here we know hyper conjugation dominates plus i so here if they ask you what is the dominating effect answer will be hyper conjugation right minus m so this b will be the least stable right plus i then c and the first will have a because of hyper conjugation yes i have given the order also for this and r 2 this this is see whenever you have two similar electronic effects okay plus m minus m whatever it is then you should know that which one dominates on other that you should know this comparison you should know i have given this order also and r 2 shows more plus m understood see whenever you have alpha sp3 hybridized alpha carbon present at ortho and para position will have hyper conjugation possible right so there will be the dominating factor is hyper conjugation now example you see tell me the stability order a b c d tell me the order guys done what is the answer t is maximum then c then a and then d question we have done it's it's simple only you see here what electronic effects we have here resonance what we have here again resonance this one is aromatic this one is also aromatic right which one dominates aromaticity dominates resonance right so obviously these two are more stable than these two right second thing here we can have more resonating structure because you have two phenyl group right so obviously this b is more stable than a because of more r s here third will be what both are aromatic compounds but here we'll have large conjugation more number of r s possible here more r s here right hence we'll have a b c and d will be the most stable this one you see one more example and then c then what is the answer both of you are getting exactly reverse why are you making these mistakes i don't know okay one thing you tell me first of all how to proceed this what effect we have here in a what effect possible a b c okay here what effect possible in a you tell me resonance correct hyper conjugation possible here because the carbon atom is sp2 hybridized no hyper conjugation here we have resonance again the both will have resonance if we compare the stability of this positive chart since we have large conjugation here obviously a is more stable than b because in a we have more conjugation correct resonance resonance is done here we have hyper conjugation possible right plus h sp3 hybridized alpha carbon atom hyper conjugation possible and we know hyper conjugation resonance dominates hyper conjugative structure order will be a b is it clear first of all you see what effects are possible and then you decide which effects effects dominates according to that you decide next question or next topic we'll discuss see we have discussed for carbocation similarly carbon and stability of carbon ion also follows this order the first one is aromaticity if the negative charge is a part of aromaticity second one is minus m resonance fourth one is minus h fifth one is minus i and then six one is hybridization these are stabilizing factor one two six these stabilizing factors are this we have plus i plus h right all these will stabilize because these are what electron withdrawing okay negative charge electron withdrawing will minimize the charge over there just one or two example we'll discuss here ch2 negative anode stability order first one and one more we'll discuss this stability order you see i will discuss this here we have minus m anode two minus m here also we have minus m so with minus m we cannot decide right because minus m is only associated with pi bonds it has nothing to do with the distance from this carbonyl but along with minus m it can also show what minus i effect here also it can also show minus i effect right so minus m is same at both ortho and para position but if you have minus i present here right in this case it is more effective here right because this closer to this carbon and a carbon ion right and that is why the stability of second one is more you see the dominating effect is minus m only okay you understand this dominating effect is minus m only but if you consider minus m you cannot decide the uh you know stability of carbon because both group shows minus m same group we have right and minus m is independent of distance it has only associated with the conjugation of pi electrons right that is same for ortho and para position so addition to addition to addition to this what we can think of we can think of i effect but ortho position we have minus i and para also we have minus i but since ortho is closer then minus i dominates over there and that will stabilize the carbonyl mode got it understood this one and discussing only some miscellaneous examples okay um you know good examples where you can make some mistake where there are possibility of you know doing some mistakes general thing if you understand the concept and various effects okay you can do general questions easily but these kind of questions you may make some mistake into the exam okay so this question you have understood question number one tell me second one and second one you see this is also very good example here see because of this group present methyl group present here we have steric repulsion here and here also we have a steric repulsion to minimize this steric repulsion what happens NO2 changes its plane right so one this group will go out into the plane and this one will come probably out of the plane like this okay dark this is a wedge line it's dash time this will change its plane and because of this change in plane minus m is not possible minus m effect is not possible here right so and here minus m effect is possible so this will stabilize mode order will be this if ortho positions are occupied then steric repulsion will come into got it okay some acidic order we'll discuss tell me the acidic order of this compound COOH here we have CT3, COOH, COOH, CH3, COOH, CH3, COOH, CH3 these three questions yeah I'll give you the break wait okay these three questions you do then we'll take a break okay I'll discuss this wait see first of all acidity little bit of idea you have right the acidic strength just give me 10 minutes we'll give you the break acidic strength but please concentrate acidic strength is directly proportional to the stability of of its conjugate base which is nothing but an anion right so acidic strength in other way you can say it is directly proportional to the stability of anion so that we have already discussed in case of carbon ion but here we don't have carbon ion here but anion will have right so if you form the conjugate base of this will get negative charge on this oxygen here and here now this stability will increase if you have electron with drawing group present correct electron with drawing group present so what we can say acidic strength is directly proportional to EWG inversely proportional to ERG is it clear till here yes or no guys so you see now again the same thing CT3 source this effect it shows plus H yes or no it shows plus I also but plus S is dominating CD3 also source plus H effect and CS3 also source plus H plus H effect are electron releasing effect right electron releasing effect because of electron releasing nature the acidic strength will decrease correct which one has maximum plus H effect which one shows maximum plus H effect tell me CT3, CD3 or CH3 maximum plus H tell me order I have given you already order of plus H effect of CT3, CD3 and CH3 which one shows maximum plus H effect CH3 methyl group it has maximum plus H nature means maximum electron releasing tendency hence the stability or sorry the acidity of this one will be least order will be is it clear guys tell me quickly we'll finish these two and then you can take a break now similarly here here we have plus H effect at CS3 what effect will have here at this position yes here we have only plus I because this is the meta position and we know and we know hyper conjugation and resonance does not affect meta position so we'll have only plus I present here both electron releasing nature plus H dominates over plus I right so order of stability will be this sorry acidity will be this right similarly CH3 here we have plus H NO2 minus M and here we'll have plus M OH right minus M increases the stability you see here directly proportional to EWG so it is EWG so this one is the most acidic first one plus M plus H electron releasing this will release more electron than this this acidity will be third and this will be second is it clear so what you have to do first of all you have to assign the effects electronic effects plus H plus M minus M or whatever it is right and then you have to see which one is electron releasing which one is electron with drawing according to that you can decide okay so I won't take much time here you we can take a break here have your lunch we'll start at 115 correct guys you there can we start now hello guys are you there hello can we start okay so you see we have discussed about this derivative of benzoic acid correct and if you see all the examples here we haven't taken ortho substituted benzoic acid right so now you'll see what happens if you have ortho substituted benzoic acid we have okay so there's a effect we call it as ortho effect you write down ortho effect what is ortho effect you see if you have we have C double bond O O minus and we have CH3 present here this is both groups are bulky here so what happens because of the steric hindrance present here this CO2 changes its plane and the structure will be like this the CS3 present as it is the CO2 molecule changes its plane and one of this atom is coming towards the observer O minus and other atom oxygen atom is going away from the observer like this okay because of this change in plane the CO2 minus or I'll write down like this CO2 minus ion is not in resonance with the ring within this molecule you see within this CO2 itself there is resonance possible right so it is not involved in resonance but CO2 minus can show resonance within itself and this is and forms I'll write down in forms equal RS equal RS because this is in resonance so we have equal RS present here now because of this equal RS the benzoate ion benzoate ion is more stable is more stable and hence acidic since this ion is more stable hence the whole molecule is more acidic so what we can say ortho substituted benzoic acid ortho substituted benzoic acid is more acidic more acidic than meta substituted this is ortho effect right since you see that resonance is hindered over here because CO2 minus ion changes its plane so we also call it as hysteric inhibition inhibition of resonance which is nothing but SIR so SIR effect or ortho effect are the same thing hysteric inhibition of resonance or ortho effect are the same thing right so this you must remember because of ortho effect the ortho benzoic acid ortho benzoic acid means ortho substituted is more acidic than para or meta benzoic acid got it ever you see ortho substituted benzoic acid it should be more acidic understood no it won't depend on the substituent whenever you have ortho substituted it is more acidic simple thing because COOH group is itself a large group here that won't depend on the substituent am i audible now am i audible okay you can hear me but you cannot see the screen right it's blank i don't some okay there are some you know some network issues though okay i'll send you the link again okay we'll send you the link again wait