 Okay, there can we start now so we were talking about this question the city order right fourth option you see When this nitrogen loses its lone pair of electron then what happens? We'll have a double bond here This hydrogen will be as it is and here we have the positive charge right When it lose the electron to H plus or anything like that So because of this positive charge here the system becomes conjugated now and since it has 4 pi electron It is anti-aromatic Right, which is very unstable Right, hence its basicity nature nature will be leased in among all these four right Here you see I will discuss this second option in the last Here you see it can easily donate this lone pair of electron and here also it is possible but because of Minus I nature of this oxygen Right, if you compare first first and third the tendency for this molecule to lose Lone pair of electron will be high hence the basicity of first Will be maximum Right, then we'll have second Sorry third then we'll have second and in the last we'll have fourth because it forms anti-aromatic right Why third one is not? Why this second one is at the third position that you see You see this nitrogen It is actually there's a rule called Brett's rule. Have you heard about this? Brett's rule Yes, or no. Have you heard about this rule? Okay So right under this what is Brett's rule that sp2 hybridized carbon or atom sp2 hybridized atom atom at bridgehead position at bridgehead position is not Possible Right now, what is this bridgehead position? for example, you see this molecule right and This carbon use this carbon So at this position is the bridgehead position or this position is the bridgehead position at this position If you have any positive charge which makes this carbon atom sp2 hybridized this is structure. It becomes highly unstable then highly unstable Why it is highly unstable because you see because of this positive charge this carbon atom is sp2 hybridized And to make this a stable this two bond which is this one bond should be like this then only the Bond angle should be around 120 degree right this carbon means this bond should be this side this side. It should be like this But here it is like this you see one two and three so like this The bond angle is not possible and there will be a huge angle strain over here That's why it is highly unstable position. This we call it as breadth through Okay, like you see this molecule. It is not possible Even if you have this kind any kind of this kind of structure you see I'll give you some more example This kind of structure also this position is not possible However, this position is possible this positive charge here at this carbon. It is possible even if you have bond like this This kind of structure this positive charge is also possible here Right this two bond and positive here. This is also right. This is also possible. This is also possible Okay, at this point if you have positive charge it is possible But at bridgehead position it is not possible another application of this is what we cannot have sp2 hybridized carbon So it is not like always we should have positive charge over there Means in this example, you see if you write down double bond like this This is also not possible because again because of this double bond this carbon atom is sp2 hybridized Okay, this we call this breadth through what happens here because of this rule only when this loses this lone pair of electron it becomes positive charge here and The oh, I think this this is not the bridge bridgehead position. No. Yeah, actually, this is not bridgehead position Anyway, this you let it be this is this rule is not required for this question You let it be this one But here if you see if it loses this lone pair of electron, so we'll have positive charge here Right positive charge here, which is there in the plane perpendicular to this ring Right in the perpendicular to this thing. So obviously if you compare 1 2 and 3 Right this ring, which is perpendicular here is Comparatively less stable because of this conjugation here in this ring Right, so that is why this is lesser basic than these two compounds and this one will be the least basic Because it is anti aromatic Another way what you can say because of this resonance also because of this resonance also The nitrogen atom the basicity of nitrogen atom will be suppressed because if you compare these two This is also logic and apply this nitrogen is sp3 hybridized. I'm sorry. This nitrogen is sp3 hybridized This is also sp3 hybridized And this one is sp2 which is comparatively more electronegative That's why the basicity of this will be lesser than from one and third and this one is anti-aromatic. Did you get this? Breast rule is not applicable here. I got confused But this is also one point one, you know rule we have which you can keep in mind Understood this Got it next one question number 20. The answer is B Question number 20 Aditya Khushal is getting B What about others? Let me know the answer Okay, you see Living group ability we Defines with the electron withdrawing nature. Okay, so All these group With respect to oxygen since the options are given the more electron withdrawing group attached to oxygen Makes that particular group a better living group Correct so What did I say? more electronegative More electron withdrawing group attached to oxygen makes the group a better living group Okay, so if you see this is not electron withdrawing it is methyl So this one obviously it is the it is the what it is the least You know it has the least tendency to go out as a living group Second will be the least for sure and we have only one option Right second one will be the least we have only one option when oxygen attached to EWG electron withdrawing group Better electron withdrawing group easier will be the tendency for this molecule to come out as a living group Methyl group is a electron releasing tendency Second one will be the least this is the option and if you see CF3 is a electron withdrawing group and sulfur also attached with two oxygen Which is true in case of third and fourth molecule But then we have methyl in third and CF3 in the fourth so CF3 also withdraw electron So this will be the most or the best living group then third and then one and then two option D Okay This is the next question Got it. 21 is C intermediate during the addition of HCl to propene in the presence of peroxide Do we have anti-marconic off does it follow Anti-marconic of addition. Yes. Oh Wait, no Why amok is only for HBR. Yes. So first of all this Kharash effect or anti-marconic of addition or Anti-marconic of addition is only for HBR For HCl it is not possible Right and addition of HCl It also does not follow free radical mechanism free radical Mechanism ANC not possible But it follows what it follows Ionic mechanism ions forms Ionic mechanism and the reaction forms reaction follows with a with a stable Intermediate Carbonium ion or carbocation hence option B is correct Yes Free radical is for halogen. It is halogen halide Sorry, it is a hydro halide Got it. See this kind of mistake occurs in the exam Okay, and this happens because of the reason that you don't Read the question properly. You are hurrying your to give the answer. That's why it happens. Is it clear? next question number 22 option C is right because if minus I effect is less then It has very very less tendency to leave Or to release H plus ion furnace and furnace H plus ion, right? And that is only possible here in case of bromine and that too option C Lesser minus I lesser will be the tendency to release H plus ion and hence the smallest dissociation constant Okay, when when the molecule has less tendency to go into forward direction is dissociation constant will be Least next question 23 in this The bond length are same. What is the reason for this? What is the reason for this? Okay, one question one additional question. I'll give you on this you see This molecule and I am drawing the resonating structure of this. I'm drawing the resonating structure of this all of you Tell me the answer of this one So we'll get this positive charge and negative charge Now the question is the question is suppose. This is one two carbon and one two carbon Can you tell me the approximate value of bond order of P1 P2 carbon? This is the question. I am not asking the exact value. I want The approximate value 1.5. Okay, just a second this one you see do we have any difference in these two structure? suppose this is a first carbon second carbon third Ford One two three four five Again, what is the bond order here? Of C1 and C2 1.5 only is it what about others tell me the bond order that you know, you know the bond order and bond length are Propositional, but that also you can say Less than 1.5. Okay. Now you listen to me See this is that first of all these two are different structure both are what both are resonating structure of Benzene Both are resonating structure of Benzene Is it there? You know the case of equal resonance. Is it the case of equal resonance? Yes are no, this is the case of equal resonance. Yes or no, tell you Yes or no. Yes, right. Okay. It is the case of equal resonant Is this the case of equal resonance is this is the case of equal resonance? this one No, it is not equal resonance means what? When the pi electrons are equally distributed over the entire molecule which is what we have here. But since this part is not involved in resonance you see if you draw the resonance hybrid of this molecule your structure will be like this right it will be like this half only the half part is not involved in resonance. But when you draw the resonating structure of benzene ring the pi electrons are completely below the light and resonating structure is this so this is the when there is when there is uniform distribution or equal distribution of pi electrons are lone pair that is equal resonance and it is unequal resonance right. Now listen to me very carefully this is both it is the case of equal resonance so stability of both molecule will be equal stability will be same and when stability of the R s is same so both R s will contribute equally to the resonance hybrid right equal resonance means what I what I told you they will contribute equally to the resonance hybrid contribute equally to the resonance hybrid. But here it is not the case right here there it is not the case of equal resonance so one of the resonating structure may be what major contributors and other one may be minor contributors is it clear till here why I am saying it as major or minor is it clear since the condition is not for equal resonance it means the stability of these resonating structure will be different one will be more stable other one will be less stable correct the one which is more stable are major contributors right so we can say major contributors will have more stability fine this will contribute more into the real structure correct so in this case since the contribution is not same we cannot take the average value of bond order what did what did you do you have taken here one bond here two bond here one plus two divided by two resonating structure that's that's what you have done that part we can only do when the case is of equal resonance so in this case in this case bond order we can say it will be the number of sigma bond plus the number of pi bond divided by the number of sigma bond this is one formula right or we can also say the number of bonds number of bonds in resonance divided by number of RS number of RS okay now here you see both formula actually gives you the same value okay now number of bond in resonance means what since I am talking about C1 C2 carbon so we'll only count the number of bond between C1 and C2 so we'll have here what two bonds and here it is one bond two plus one there are three bonds so three divided by two RS we have the bond order will be 1.5 okay or number of sigma bond but you take care of one thing here we only take the bonds which is in resonance how many sigma bond we have here you see one two two sigma bond and one pi bond the two plus one divided by two again the answer will be 1.5 I can take one separate class for this if you want we can discuss all these things in so detail with so many examples I can give you okay but today since we are not discussing these things separately so I am just giving you a bit of idea okay if you see more number of questions more questions if you solve the things will be clear more right the point I am trying to make is what since we have equal resonance equal distribution of pi electrons so we can take average over here right but the same case will not have here we cannot take average over here and in this case we even cannot find out the exact value of bond order right but what we can say here that the bond order of C1 and C2 will lie in this range that is one bond here one two this is what we can say furthermore the bond order will have close to this value close to value one or two it depends on the stability of this or this rs the one which is more stable according to that we can say the bond order will have a value which is closer to two or closer to one got it is it clear now any doubt you can ask me guys speak up let me know if you have any right so in case of unequal resonance right in case of this case this case in case of unequal resonance I am just summing it up okay this is the last part if you can keep this one single line in mind you can understand in case of unequal resonance we cannot find out the exact value of bond order right but we can compare the bond order like in this case we can easily say that the bond order of C1 and C2 carbon will fall in this range right and in this also we can say it is closer to one because this rs is more stable than this because of two reasons since it is a neutral molecule more number of pi bonds no charge separation right so this molecule is more stable and it is the major contributor and that is why the bond order will have the value close to one and we can say it is in fact in between one to one point five right so in case of unequal resonance we cannot find out the exact value but and in that case we'll compare the bond order and like that only you will have the question they'll give you two questions two resonating structures and they'll ask which one will have more bond order or even they can ask you more bond length accordingly this because both will have inversely proportional relation correct exact value of bond order we can only find in case of equal resonance okay next class if you want I can give you many questions on to this right then you will solve you will understand this how to find out bond order and bond length even is it clear now like you see here in this case you should know to find out the stability of rs that's what I said that the rules that I've given you seven eight rule how to find out the relative stability of resonating structures that rule you must remember if you do not know how to compare the stability of these two you cannot say whether this bond order will be closer to one or closer to two so all these things are related I hope you understood this if you have any doubt you let me know we can discuss this I can give you more questions on to this for practice purpose into the next class okay okay next question can we discuss you know what is SIR effect SIP effect SIR SIT hysteric inhibition of resonance hysteric inhibition of protonation you know that okay so next class I will discuss few more things of GOC we'll solve few questions and few more things we'll discuss because in chemistry in these kind of chapters the more information you have the chances are better to get more marks okay so that is the thing don't solve all of you don't solve any tough questions okay basic understanding you should have that what you are doing actually okay anyways solve this question number 24 why answer will be eight why it is eight yes so again you see they are talking about energy of resonating structure okay that depends on the relative stability correct so again those rules you have to memorize okay more stability lesser will be the energy and we know neutral molecules are always more stable than the charged one okay so obviously option a is right okay there's no charge here so this one is the most stable structure separation of charge requires energy yes so you have to provide that energy obviously the energy will be will increase and the second thing what you can think just you find out the stability most stable compound will have the least in it right 25 what is the answer what is the error in first option a ram churn lead out are you getting it bond order comparison you understood ram churn you see this nitrogen is what here the nitrogen is bivalent only two bonds your answer is correct I just write it down it is bivalent and this carbon is pentavalent this is pentavalent and what yeah this carbon is also pentavalent right pentavalent so option b is correct one two four three yeah option b is correct okay next question we'll see guys you solve this just a second just okay what is the answer 27 one which is of incorrect pair we have to find right so okay first one we have resonance find second one second one is what second one is also a resonating structure right this is also a resonating structure rs not in equilibrium this is wrong right these two are in equilibrium what is this this is keto in all form right so this one is in equilibrium what is this option see this is right this is also right I think something is missing them I am also not getting it something is missing I'll just okay let it be this option I'll just cross check it okay I'll let it be this question means you are saying this is rotated like this it is a rotation about this you are telling I'll cross check this option you let it be now okay we'll see another question I'll cross check this you let it be this one okay this one you see tell me in the reaction this is ring expansion actually in the reaction shown below the six member ring is generated by something which bond yeah so what happens this bond this bond comes over here it is this h plus will come over here right then OH2 will go H2O will go out right H2O will go out this A will join over here 1 2 3 4 5 1 2 3 4 5 6 bond will have here this point will have this bond like this and then BR will attach here so this is the structure here right H2O will go out and then this A will come over here option A is right yeah correct this one propane in absence of peroxide the same question but there was the presence it is the absence of peroxide what will be the intermediate the intermediate again will be a carbonium ion carbocation answer A will be right okay addition of HI in presence or absence whatever the question is this would be the intermediate okay like we have seen in the previous example for the next one two questions for today 31 31 the answer most of you are saying A why it is A we are following most stick yeah charged species are more reactive and neutral molecule are more stable this is again see actually these are not resonating structure these are not resonating structure first of all A is a different molecule BCD are not the resonating structure of A why because since we know in in the resonating structures the charge balance must be there means if BCD are the RS of A since it is neutral all these molecules should be neutral overall but it is not the case we have positive positive charge here I think we can compare BC and D easily and overall since it is neutral so this should be more stable right overall it is neutral so this should be more stable but if you compare BC and D these are different structures no if you compare BC and D this will be the most stable because of hyper conjugation and resonance here because of hyper conjugation with this alpha hydrogen and hyper conjugation with this no resonance with this pi bond pi bond right and hyper conjugation over here here we have resonance with this resonance with this and I think this is the answer this we won't discuss simply the charge of the neutral species will be the more stable okay this will be the answer A right neutral species will be more stable okay second one what is the answer 32 the stable resonating form of vinyl methyl ketone vinyl methyl ketone vinyl methyl ketone is nothing but this CH2 double bond CH C double bond O CHC right so if it comes over here and this pi electron goes here so this will be CH2 positive charge single bond CH double bond CO minus CHC okay but we have to compare with all this obviously this one will be not stable because positive negative charge will see carbon atom at adjacent position this put this type of structure is highly unstable because they have very high tendency to form a double bond here so this is not stable at all right this is also not stable carbon oxygen bond has tendency to form double bond here and again the same reason will have here also right so this is the most stable why because you see one more thing can discuss here will have the positive negative charge as maximum possible distance right right negative charge is present on electronegative element right that's why option A is the right way after this A suppose order of stability if you have to write A will be the most stable after this A will have C right we have C and then B and D if you compare so B will be more stable because electron is displacing towards the carbonyl group and then we'll have D comparative if you see opposite charges must be closer then it is then it is not stable Sondarya you see this if it is closer right this has more tendency to form a double bond here charge separation at adjacent carbon is not at all stable for this see what happens you think like this you have this pi bond how do you separate this how do you dissociate this pi bond for that you have to provide energy and to maintain this charge separation this energy you have to continuously provide to this molecule and hence it will be at higher energy state which is not stable you decrease the energy a bit this will form this will convert into a double bond okay is it clear why C is more stable we'll have this positive see we are comparing the comparison of this and this will happen right let's discuss this we have one pi bond one pi bond right if you see the octet here right if you see the octet except this carbon all carbon has its octet and oxygen has its octet right here also you see except this carbon all carbon has its octet oxygen has its octet all the conditions are same for this two molecule the only difference is what the distance between the two charges so comparatively if you see depending on the distance of positive or negative charge we are assigning stability to A and C right so more the distance mode will be the stability because closer distance if you have the less distance if you have they have tendency to more tendency to form a double bond or to separate the charge like this you have to provide energy into it then only this kind of separation is possible yeah this one you are talking about this one that's what i'm also talking about i am comparing A and C right all the conditions are same we'll have one carbon with positive charge here also we have one carbon with positive charge yes unlike charges our largest distance is comparatively again unstable right but since it is a relative thing we are discussing no this is a relative thing just a second you convert this here oh like charges when it is closed it is unstable right oh yeah like charges close unstable so unlike charges close it should be stable right okay yeah i got confused we'll have unlike charges no positive like charges close is unstable so unlike close will be stable yeah correct correct correct my mistake my bad i'm sorry option c will be right option c will be right yes correct unlike charges are there got it got it i'm sorry my mistake anyway so we'll we'll discuss this right you see just one thing you tell me have you done this kind of questions where you have to find out heat of hydrogenation or heat of combustion because this kind of question they usually ask this kind of question they usually ask in the exam heat of hydrogenation heat of combustion thing right did you did you finish that kind of questions how to compare heat of hydrogenation with the stability of resonating structures and all this kind of questions have you done okay fine so what you will do your homework is what and that will be easier for all of us if you finish your homework you just finished your you know the relative stability of r s the various rule that you have i will discuss i will discuss that into the next class but for that you required this information how to this information of relative stability of resonating structure correct do you finish that relative stability of resonating structure on your own next class we'll discuss that heat of hydrogenation heat of combustion part and few examples again we'll see under bond order comparison and calculation fine so we wind up the class here only we'll see you in the next class finish your homework okay relative stability rules you must go through okay thanks all thanks all good morning