Get your favorite dishes delivered home or online. Okay. Hello everyone. Aryan is there. Please type in your name. Okay. Aryan Admesh. Okay. So, see today we are going to revise few concepts of GOC and then mainly we'll focus on questions. Okay. So, the question that has been asked, that only we'll discuss because there are so many questions in all these chapters. Okay. So, only important questions and those questions which has been asked in the previous exam, we'll do this revision along like around those questions only. Okay. So, first of all, few questions I am going to give you. Okay. These questions are based on like nomenclature because this nomenclature, general organic chemistry, combinedly they give one question on this. Okay. So, few like miscellaneous questions whose rules are actually like the rules for those kind of questions is a bit different. Okay. So, those kinds of questions we'll see so that we can revise those rules also. Right. So, first of all, I'll just write down the few compounds here. You have to tell me the name of these compounds. Okay. So, we are discussing today general organic chemistry. So, first of all, you see this question and tell me what is the name of this question? IOPAC name. We have a benzene ring and on this we have CHO NO2 and here OCH3 attached. What is the IOPAC name of this question number one? Question number two is this CH2, CH, CH2. Here we have CN, CN and CN. Third question, all these questions has been asked. Okay. What is the IOPAC name of this compound? The name of this compound, OH, CN, NBR. Fifth one, two methyl group attached. Sixth one is this, seventh. Here we have chlorine and here we have BR. Eighth one, CH3, CH3. And the last one is this, this one. Okay. The first one is, finish all of you first. Done. Okay. See, the answer for the first question will be, you have said three methoxy 2-nitrobenzaldehyde. Okay. 2-nitro, 4-methoxybenzaldehyde. Okay. So, benzaldehyde is correct because order will follow here, the priority order. So, this carbon will have the first position. It's one, two, three and four. So, at fourth position we have methoxy group and at second position we have nitro group. Okay. So, Atmesh has written 2-nitro, 4-methoxy, which is wrong because we have to follow the alphabetical sequence. Right. So, we should write, so we should write what? We should write methoxy first and then nitro. Right. So, the name of this compound will be 4-methoxy nitrobenzaldehyde. 4-methoxy 2-nitrobenzaldehyde. Okay. Second one you see, we have three functional group present here. Okay. Three same functional group or carbon containing functional group. So, whenever we have more than two carbon containing functional group directly attached to the unbranched alkane, then we do not consider this carbon or functional group into the parent chain. That is the rule we have. Okay. So, parent chain will be this one, two and three. Right. So, three carbon and cyno group attached to it because these are not in the parent chain now. So, we cannot take the suffix of this, which is nitrile, but we'll take the prefix of this here. So, we'll write what? 1, 2, 3 tricynopropane. So, name of this will be what? 3-cyno-propane. Okay. So, the rule for this is what? When more than two carbon containing functional group directly attached to the unbranched alkane. Right. Then we do not consider the carbon of the functional group into the parent chain. Okay. That's the rule we have. So, this one you see, I'll just give you one more question on the basis of this rule. Suppose we have CH2, CHCH2, and here we have COOH and COOH. Tell me the name of this compound and this compound. Here we have COOH. Here we have COOH and here we have COOH. These two, what is the name for these two? Quickly we'll finish this, then we'll start some GOC concepts we'll discuss in this. What about others? Propane 1, 3 di-oic acid. Which one it is? Propane 1, 3. See, Atmesha and Lalita, what is the basic mistake you are making here? You are writing down di-oic acid, right? And di-oic acid, if you write, it means you are considering the carbon atom of functional group into the parent chain. That's why you are using a suffix of COOH. Understood? You are using what here, Atmesha and Lalita. You are using the suffix of these functional group, which is oic acid. And you are writing what? Di-oic acid. It means these two, you are considering into the parent chain. And if you are considering this into the parent chain, it means the number of carbons should be what? One, two, three, four and five. Right? So it cannot be propane. Is it clear, both of you? Right, Aryan, correct. Okay, so first of all, you see, these are the very basic problem we have, which being a JAS parent, you should know these rules, okay? You'll get one question from this based on these rules. We are only considering a miscellaneous rule. You must remember all these after this class, okay, don't forget this.  Here, this compound you see. What is the rule I said that if rule you write down for this kind of compound, I guess you haven't done this kind of question. If more than two, more than two, carbon-containing functional group, carbon-containing functional group are directly attached to the unbranched alkane, unbranched alkane, then we do not consider the carbon atom of functional group into the parent chain. So this is the rule we have, right? So now you see the first compound, we have only two group directly attached. The important thing is what it is directly attached. It should be directly attached, okay? So we have only two groups. So we'll consider these carbon atom into the parent chain. So number of carbon in the parent chain reward, one, two, three, four, five. And if you remember, we have carboxylic acid, cyanide, amide, okay, in all these functional group, we always start numbing from the carbon atom of the functional group, right? Generally, we do numbing like that only. We always give first position to the carbon atom of the functional group, right? Functional group of carboxylic acid. So one, two, three, four, five. Five carbon we have, so the name should be what? Five carbon, so pentane, one, comma, five, di-oic acid. Pentane acid should be small a here, to take care of that small a, right? One, pentane, one, comma, five, di-oic acid. If you consider this compound, according to this rule, we have three functional group, directly attached to the unbranched alkane. We will not consider this carbon into the parent chain, right? So now in this, the numbering will be what? Either you can start from this side or this side. So both side, the position of this is same, that is three only, so that won't matter wherever you are starting the numbering. Suppose I'll start the numbering from here, one, two, three, this is four and this is five. Right, so the name of this will be what? Pentane, pentane, one, comma, one, comma, three, comma, five, tri-carboxylic acid, okay? The name should be pentane, one, comma, three, comma, five, tri-carboxylic acid. It is the prefix of this, okay? C-O-H will write carboxylic acid. Understood this, now I'm coming back to this. What is the name of this compound? Tell me the third one, bicyclo-210 pentane. What about others? Okay, see there are two carbon atom common into this, these two carbons, so this is a bicyclo compound, okay? And in bicyclo compound, we always start numbering from this bridge carbon. These two carbons are bridge carbon, the common carbon atom, right? This is a bridge carbon. So we always start, the rule is what for bicyclo compound, we always start numbering from the bridge carbon, then we go towards the larger ring, okay? Numbering will start from this bridge carbon, so we can either go this smaller ring or this larger ring. So the rule is what? We will start numbering from this carbon atom and we'll go towards the larger ring, two, three, four, and five, okay? This is the numbering we have. Now since this is a bicyclo compound, so we'll write down the name, simply it starts with bicyclo, open bracket, number of carbon atom, here it is five, so we'll write pentane. Now in this open bracket, we'll write down some number and that is the number of carbon atom present in the two ring, right? Except this common carbon atom, which we also call a bridge carbon, right? So this one, you see this four-membered ring, these two are the carbon atom common here, bridge carbon, so we'll not consider these two. We have two more carbon here, so we'll write two dot, the numbers must be separated by dot, comma, we cannot use, otherwise it will be wrong, okay? One carbon we have here, one dot, and there is no carbon with which this bridge carbon is connected, right? This will be zero, pentane, this is the name of this. For bicyclo compounds, we always write down number here in descending order, descending order, right? Another one you see, one more example I'll give you for here, for a spiro compounds, okay? Spiro compounds you see, in the spiro compounds, the spiro compounds in which the two ring are connected with a quaternary carbon, like the simple example is this. Spiro compounds, two rings are connected with a quaternary carbon, okay? In this, the name we'll write here is a spiro, then open bracket, then the number of carbon atom, according to that we'll write down. Now here the thing is what, if you do the numbering of this, we will start numbering from the smaller ring, okay? And we don't start numbering from this common carbon atom, okay? So we start numbering from here, one, two, then it is three, four, and five, like this we'll do. So a spiro, it is pentane, and here we'll write down the number of carbon atom present here, that is two in one ring, two in another ring, this carbon we don't count, but here in this case, we'll write down the number in ascending order, ascending order, like if I write down this compound, the another example for this, this one. This is also a bicyclo compound, okay? In this we'll start the numbering from the smaller ring first, so this one, one, two, right? We will not start from here, but in bicyclo, we are starting from the common carbon atom. So one, two, three, now here you can write down fourth to this carbon also, or this carbon also, depending on the substituents. Suppose if I write down one substituents here, then you have this, you know, foundation that you have to go four, five, six, and seven. Because if you go like this, then the position of substituents will be four, five, and six, sixth position. If you go by this way, then four and fifth position. Lowest Locant's rule will follow, right? So since there is no substituents present here, we can write down fourth carbon at any of these carbon, so four, five, six, and seven. Total seven carbon we have into this. The name will be Spyro, Open Bracket, Heptane. And here the name will write, and the number will write in ascending order. So first we'll write these two carbon. So for two dot, here we have, except this we have four carbon, right? Two dot four, Heptane. Is it clear? It's not hexane. You see the number of carbon atom is five only. The first one, RN. Yes or no, tell me first. Okay, so I hope you understood this. This one you see. So bicycle two, one, zero, pentane. You must remember one thing, the number, the sum of these number that you get here, two plus one, that is three, you must have two carbon more than this number here in this parent chip, in bicycle. But in Spyro, we have only one carbon mode. Yeah, actually the number of atoms in the ring, generally we have only carbon as atom in the ring. That's why I'm saying carbon, okay? But actually it is the, how many member are present in the ring that we have to count? Count number of carbon into this, number of atom present in the ring, not bond, okay? Now, this one, what is the name of this one? Tell me, let me tell you, the highest priority of this nitride. This is the first carbon, okay? Priority order you already have there in the book. This is the first carbon. Now you tell me the name of this compound. Tell me the name. Two bromo five hydroxy benzo cyanide. Others six bromo three hydroxy benzene cyanide, okay? You see, first of all, we will start numbering from this carbon atom since this is the highest priority here, cyanide. Now, numbering we can go to this side also and this side also, right? So one, two, three, four, five, or one, two, three, four, five. You see in this one, the sum of locations is same, whether you go to this direction or this direction. The sum of locations are same. That is two plus five, seven. Two plus five, seven, right? So when the sum of locations are same, so we'll give priority to the alphabetical, according to the alphabetical sequence, right? So alphabetically, this is hydroxy and this is bromo. So bromo comes first. So we'll do numbering like this. One, two, three, four, five and six. So two bromo, five hydroxy. The name will be two bromo, five hydroxy, benjonitrile, not cyanide. Because this is now, in the functional group we have, right? So it is there in the ring. So for that we'll write benjonitrile, not benjoncynide. Cynide only we use when this is the, not the primary functional group, not the main functional group. No, actually, I'm not giving the preference bromine over OH because then we have multi or polyfunctional group present more than two. Then one functional group will be the primary functional group. Others will be the substituents. Now if they are substituents, whether they are functional group or not, we will consider the rule of substituents and that lowest set of logents and all, right? When we have considered this as the primary functional group, these two become substituents now. Now according to substituents we'll follow. We don't see again the priority of this and this, bromine hydroxy. Understood? Fifth one you tell me, fifth one. One, three, three, trimethane cyclohexene. What is the name of the seventh one? Seventh one you tell me first. Three, bromo, one, clodocyclohexene. Okay, fine. So this is correct. The rule is what? Whenever the ring contains double bond, right? Whenever the ring contains double bond, then we always give first and second position to the double bonded carbon atom, always, right? Whenever the ring contains double bond, then we always give first and second position to the double bonded carbon atom. So here you see, here you can write on one and two or one and two, whatever, because there is no substituents. Since here we have substituents present, so out of these two carbon atom, we have to give first and second position. So this should be first and this should be second. Then it will be third. So one comma three comma three trimethane cyclohexene. The name is correct, okay? So this one you have done. This one again, ring contains double bond. So we'll start numbering from the substituted double bond, one and two and three. So that three bromo, one clodo-cyclohexene, okay? Similarly, when you do this one, what is the numbering in this, you tell me? Eight one. What is the name of this one? What is the name of the eight one? Three five dimethyl-cyclohexene, okay? So if we start numbering from here, so one, two, three, four, five, six. So four and six position, 10. One, two, three, four, five. So lowest sum of locations will follow. One, two, three, four and five. So three comma five dimethyl-cyclohexene, correct? So this one is also correct. Now what is the name of this one? What is the name of sixth and ninth one? Four ethyl trimethyl-cyclohexene. One, two, three, four ethyl trimethyl-cyclohexene, correct? Alphabetical order will write down ethyl first and then methyl. This one is correct. Sixth one, yeah, it's correct. Sixth one, you tell me. Cyclo-propyl-cyclobutane, yes, it's correct. Primary functional group will be this because it contains four carbon, here only three carbon we have. So we have cyclo-propyl present onto this. Cyclo-propyl-cyclobutane. Cyclo-propyl-cyclobutane. Now just two, three more questions we'll see. What is the name of this one? CS3, C-O-N-H-B-R, these four compounds. And we have this. Now what is the name of this first one? What is the functional group we have in this? What is the functional group? Okay, see. First of all, the functional group in this is amide. The structure of this is what you see. CS3, C-Double bond O, N-H2 if you write down, then it is amide. Okay, C-O-N-H2 is amide group. But here we have N-H-N-B-R because one hydrogen of this nitrogen is replaced by B-R here. So again in amide also, this will start coming from this carbon that was a functional group, one and two. Right, C-O-N-H2 is amide. When write down C-O-N-H-B-R, it means one of the hydrogen of nitrogen is replaced by bromine, right? So in this compound, whatever atom or group or substituents attached with nitrogen, the name will write like this N, capital N stands for nitrogen. Since B-R we have here, so we'll write N-domo. This means that this bromine is attached with nitrogen. And we have started from this carbonate one and two. So ethane, and this amide group we have, ethanamide. This is the name of this compound. N-bromo ethanamide. Now the name of this compound, the second one is what? Ethyl cyclopropane, correct. This one is methyl ethyl. So it is ethyl cyclopropane. What about this one? One ethyl cyclopropane is also correct. See, cyclopropyl ethane we cannot write, Atmesh. It is ethyl cyclopropane. Why the rule is what? If the ring contains any carbon chain, right? If the ring has any carbon chain attached with it, so either you can take a ring as the primary chain or this carbon chain as the primary chain, correct? Now the point is what? Whenever the ring contains the carbon atom, that is more than or equal to the number of carbon atom present in the side chain, right? Whenever the ring has more or equal number of carbon atom present in this chain that you have, then we always consider ring as the parent chain. So if the ring is the parent chain, so this is substituents now. Substituents is ethyl cyclopropane. So here also you see the number of carbon atom here and in the ring is same. So the parent chain will be this ring. Okay, so this will be propyl cyclopropane, correct? Propyl cyclopropane. Now in this case what happens? Since the carbon chain has more number of carbon atom, so this becomes now the primary chain and this becomes substituents. Yes, yes, yes, yes, submission. So this is the primary chain and this is the substituents now. So we'll write what is the name of this one? Tell me. Cyclopropyl 4-carbon we had, so butane. Okay, cyclopropyl butane is correct. Now what is the name of this compound? See for this kind of compound, when the carbon containing functional group attached directly to the ring, and this is another case, right? When carbon containing functional group attached directly to the ring, then we use the following suffix. If COH is attached directly to the ring, then we write carboxylic acid. For COH we'll write carboxylic acid. If CHO is there directly present to the ring, then it is carbidehyde. If CN is directly attached to the ring, then it is carbonitrile. This suffix we use. So we write here what? Cyclopropyl butane, carboxylic acid. This is cyclopropyl butane, carbidehyde. This is cyclopropyl butane, carbonitrile. Clear? Last question. This is, you don't have to write down the name for this. This is question based on ether. Once they have asked these questions, I'll just give you one specific question here. All these are oxygen. All these are oxygen and these are carbon, right? One, two, like this, they have carbon here. This is the structure of crown ether. General name is crown ether. So we write down the name of this compound. The gen, or the standard form we have here is X. This is the general name of this compound. X crown, Y. The question is what is the value of X and what is the value of Y? Once they have asked this question also, X and Y. You know this? Okay, see, X, you write down. X represents, in this compound, like the structure is given here. So X represents the number of carbon atom plus number of oxygen atom in this. And Y represents only the number of oxygen atom. So if you have this information, you can do this question easily. Now you count the number of carbon and oxygen atom here. We have two, four, six, eight, 10, 12 carbon and six oxygen. So value of X will be 18 and value of Y will be six. That understood? Just this two definition you have to write down. Whatever the first thing here, that represents the number of carbon plus oxygen and this represents the number of oxygen, that is it. So we already know what is I effect and all, plus I and minus I. Okay, there are very less plus I group, but minus I group, there are very various number of different, different minus I group we have. And depending on these groups, we can easily understand the stability of carbocation, carbon ion, acidic nature, basic nature and all, right? So one thing is what you can simply memorize that minus I group should be this, this, this, okay? Electron with drawing group, minus I is electron with drawing group, sigma electron with drawing group, okay? But if you memorize those order, it is fine, but you can easily compare and by common sense, you can say that which of these group will show more minus I or less minus I. Like suppose, few examples I'll give you here, fluorine, chlorine, bromine, iodine, if I ask you, which one of these will have maximum minus I effect, okay? Your answer will be straightforward that fluorine will have maximum minus I effect since it is more electronegative, right? So basically, this electronegativity difference you have to compare, electronegativity order you have to find it out. So this one is very straightforward. Suppose on the basis of this only, if I give you another example, suppose one thing here, I'll write down OH and SH you have to compare, OH and SH, which one will have more minus I? Which one will have more minus I? OH or SH, NH2, pH2. Next one we have NH3 plus NH2, yeah, OH is right. Why OH? Because you see with this oxygen, this group is attached to the chain, right? So obviously this oxygen is more electronegative than SH, right? Tendency to withdraw electron for OH will be more than to that of SH, right? Now when you compare NH2 and pH2, nitrogen and phosphorus, which one is more electronegative? Obviously nitrogen, answer will be this, right? In this compound, we have same nitrogen and nitrogen, but this nitrogen has positive charge on it, right? So this has more tendency to withdraw electron, right? So order of minus I will be this, okay? Similarly, you see if you compare this one, OH2, OH, H positive charge on it, one lone pair. And if you compare OH, which one will have more OH2 plus or OH, obviously OH2 plus, because oxygen here will have positive charge on it, more will have a tendency to withdraw electron, right? Next one you see. If I ask you to compare this F, OH and NH2, what is the answer in this? F, OH and NH2. In this, if you compare fluorine, oxygen and nitrogen, right? So fluorine will have maximum electronegativity, maximum minus I, then we'll have oxygen and then we'll have NH2. So in this, all these examples you see, we can easily compare the electronegativity of the atom and we can say which one is, which one will have more minus I or less minus I, okay? Next one you see, if I ask you to compare SH and chlorine, SH and chlorine, chlorine here is more electronegative. Size is also one of the factor we have here, right? But since both belong to third group, third period, so size will not have the major factor here, electronegativity of fluorine is more, more minus I. Some more example, if I give you C triple bond N, like this kind of question, CH double bond NH and CH2, NH2. What is the order of minus I? Like this one you see, C double bond OH, C double bond O, CL. Here you see, I'll just, I will explain this one directly, okay? Here you see in this question, all these are carbon-carbon atom. The difference is what? This carbon is sp hybridized, this carbon is sp2 and this is sp hybridized. So now we know this sp hybridized carbon is most electronegative, right? So order of minus I will be this more electronegative, more electronegative, more electron withdrawing nature, percentage factor, fine, that also you can say. These two if you compare, see chlorine will have tendency to withdraw electron, it is electron withdrawing, more electronegative, right? So when this withdraw electron C double bond O, C double bond O is same both the compound. Since this has more tendency to withdraw electron than hydrogen, so order of minus I will be this, okay? Another one you see, S double bond O, OH, double bond O and S double bond O, OH. Which one will have more minus I nature in these two? This one will have more minus I because it has one extra double bond O group which has electron withdrawing tendency. All things are same in both the compounds because of this double bond O, the minus I nature of this is more, right? So like this you can compare easily compounds, the minus I of compound and then based on this, the various application we can do that also we'll see. Few exceptions we have into this. The minus I nature of NR3, NR3 plus, the minus I nature of this is more than to that of NH3 plus. Group R electron releasing, so that will decrease the positive charge density here. So ideally this should have less electron withdrawing nature comparatively because you have hydrogen here. This positive charge electron, positive charge density, here it is more. So it has more tendency to be electron withdrawing, right? But this is found to be more plus I characteristics, sorry, minus I characteristics. This is an exception we have but we can understand with the help of percentage S and P character, but that you let it be. Just you keep this in mind, this exception. The same logic here also if you see, if you compare OR and OH, R has electron releasing nature, right? So here the electron density is comparatively more than this OH group. So it has more tendency to release electron rather than accept electron, right? So but ideally should be this, but actually the minus I effect is this. All these order are minus I effect. You don't get confused with this. Minus I effect, this is also minus I effect. It is not because of hindrance. It is because of here we have the P orbital of this alkyl group, the carbon that is present here and P orbital of this will take part in the bonding. So P orbital of nitrogen is involved in bonding. So there is more S orbital available here, right? So we say the percentage S character of nitrogen available percentage of hydrogen is more here than this and that's why it has more electron withdrawing tendency. Here you see hydrogen has S orbital. So SS overlap, SS, SS overlap like that, okay? So the S orbital of nitrogen is involved in bonding here. So percentage S character of this is lesser than this. That's why the plus minus I effect is this. But I would suggest you to memorize these two exceptions. Next one you see the number of plus I group is relatively very less, okay? This order sometimes it is important. If you write down the isotopes of hydrogen that is CT3, then CD3, then CH3, then the plus I nature of CT3 is maximum. This is plus I nature I'm talking about, okay? Bond length order if you have to write it down it is reverse CT. This is the bond length order. Because of bond length only, the tendency to release electron is always this. We'll see some questions on this. The stability order of carbocation. The stability order of carbocation you have to find it out. First one, obviously we know three degree is more stable than two degree than one degree. So those kind of question I'm not giving you, okay? But these are also simple question. Like you see this one, oxygen we have here. Here we have the positive sign. And one more question is this. Here we have the positive sign. These two, the stability order of this carbocation. CH2 positive CH3. Here we have NF3 plus CH2 positive, two greater than one. Why two greater than one? Because this oxygen has electron withdrawing nature. This will withdraw electrons. So here the positive charge density increases, stability decreases, order will be this, okay? This NF3 plus has maximum minus i nature. And if you compare this carbon and this N, which has a positive charge, obviously the electron withdrawing tendency of NF3 plus will be far more than CH3. CH3 will have electron releasing tendency. So this will stabilize this positive charge here comparatively. So order of stability will be this. Okay? One more question you see. We have this ring. Here we have CH2 plus CH3, CD3. And the last one we have CD3, CH2 plus. CH3 is more stable. Why CH3? See the electron releasing nature of CD3 is maximum than CH3 and CD3. The order at this now I've given you. CD3 maximum, then CD3, then CH3. Electron releasing tendency plus i. So order of stability should be this. Okay, what is the last one? Stability order greater than one. The answer will be this. Because this carbon is what? This carbon is sp2 hybridized. And this carbon is sp3 hybridized. So it is more electronegative and it has more electron withdrawing nature, right? This positive density increases over here, charge density and here it is lesser. So stability of second is more than that of first. Okay? So like here we have carbocation. We have done it for carbocation. The same order we have for free radical also, right? Free radical and carbocation has same kind of nature. Okay, so like here the plus i group increases stability. So for free radical also plus i group increases stability and minus i group decreases stability. So that is the same thing. If you have carbon ion, then the stability will be reversed. I hope these two, you know already. If yes, then I will move forward. Otherwise I'll give you some question on this. The order will be reversed and for carbocation free radical are same kind of nature, same concept we have. Yes or no? Okay, so I'm just skipping these two. Free radical and carbon ion. Now you see the next one, the acetyl basic nature or acidity, suppose we'll discuss first, acidic nature. So any acid you have, suppose HK, now if you have to compare the stability of this or to know whether this acid is strong or weak or with respect to some other molecule if you have to compare the stability of this or nature of this acid whether it is strong or weak, then what we do, we just try it, we just form the conjugate base of this acid which is nothing but A minus is the conjugate base of this acid, okay? What we'll do to find out conjugate base, we'll just remove H plus from the molecule. So A minus plus H plus. So this is nothing but the conjugate base of this acid. Right, so it is conjugate acid base pair, right? Now if this conjugate base is stable, right, then the molecule is acidic. Why it is acidic? Because in organic chemistry, we only, we mainly will define acidic nature of the compound by the tendency of this compound to lose H plus ion generally into water, right? So we just remove H plus and then we'll compare the stability of its conjugate base, okay? Now you see here, if I write down two compounds that is first one is formic acid H and then acidic acid, CS3, C double bond OOH. If you have to compare the stability or the acidic nature of these two compounds, first of all what we do, we just remove H plus from this and we compare the, we form the conjugate base first. So conjugate base of the first compound will be at COO O minus and for this it will be CS3 C double bond OO minus, right? Now this is the conjugate base of this molecule, this is the conjugate base of this molecule. Now we compare the stability of these two conjugate base, okay? So here you see because of the plus ion nature, the stability of this one plus the nature of CS3, the stability of this one is less. So it is less stable and this one is more stable. So stability order is this and since the conjugate base is more stable, it means the compound of this or the acid of this conjugate base is more acidic, right? So that's why the formic acid at COOH is more acidic than the acidic acid, CS3 COOH. So what we can see here from this, the acidic strength is directly proportional to what? Minus I group because if this group present here which is electron withdrawing, so that will stabilize this negative charge and hence the acidic nature will increase, correct? So acidic nature directly proportional to plus I and inversely proportional to minus I group. Sorry, directly proportional to minus I and inversely proportional to plus I, reverse. Now if I give you one simple question onto this, that is suppose we have CH2F C double bond OOH, CH2Cl C double bond OOH, CH2Br C double bond OOH, right? So obviously all these chlorine, chlorine, bromine are electron withdrawing, right? Electron withdrawing. So electron withdrawing group is what? Minus I, all these are minus I. So minus I group we know increases the acidity of this compound, any compound, so mode will be the minus I, mode is the minus I, mode will be the acidic nature of this, okay? So acidic nature will be maximum for this and minimum for bromine. Clear? Yes or no? Now, you see the question that asked in the exam generally has two, three different nature, okay? So now we are doing the acidic strength comparison, acidity strength comparison. See, first of all, since to understand the acidic strength, we have to compare the stability of what? The stability of its conjugate base, correct? In ionic equilibrium you must have studied that a strong acid will have weak conjugate base but stable, right? Strong acid, if you remember ionic equilibrium, strong acid gives weak but stable conjugate base. Stable conjugate base. If conjugate base is not stable, then the acid will not be strong because it has no tendency to release H plus ion. Conjugate base forms when it releases H plus ion. If after releasing H plus ion, the base is not getting a stable, then it won't release H plus ion, okay? That's the point. So weak means it is a weak base, stable means stable, it is a stable, right? So these two things are different, okay? If you have weak acid, the weak acid, its conjugate base is what? Unstable but strong. Strong but unstable conjugate base, correct? So first of all, in acidic strength, we have to form the conjugate base of this and then we have to check the stability. Now the stability of conjugate base, so I'll just write down here, acidic strength, we have to check the stability of conjugate base, right? So in these three possibilities may be there by which you can compare the stability of conjugate base, okay? The first one is electronegativity or first we'll see suppose size. Size factor will also affect the stability of conjugate base, electronegativity factor, right? And one more type we have that is not a factor like this, but when we have the hydrogen attached with the same kind of atom, okay? This third one I'll write down later on, just let it be, okay? Now, if you consider the size factor first, okay? With size, how the conjugate base stability will change, so you see here, size and electronegativity we can discuss here together. So first of all, if I ask you, the acidic strength of this compound, Hf, HCl, HBr, Hi, this is the first one, I'm discussing these two together, right? So that you can differentiate easily what concept we are applying there, okay? First compound I'm taking this and second compound, suppose I'm taking H2S, that is hydrogen sulfide and HCl. Both these molecules, you have to compare the acidic strength. Tell me, what is the order of the first one? What is the order? Tell me, are you there? What is the order? I'm not getting any response. Okay, now I got, wait. Okay, so you're saying HCl is more acidic than H2S. What about the first one? Okay, let it be. Now you see here, we have to first form the conjugate base of this and that we can form by removing H plus.  I is more, okay. So now you see, there are, if you check the stability of this compound, right? So any NIN, the negative charge, there are two, you know, there are two factor we have here. The first is what? Negative charge, electronegative atom, electronegative atom is more stable. Okay, and the second one is what? That uniform distribution of charge, uniform distribution of charge, whatever charge it is, positive or negative, leads towards stability. More uniform distribution, more will be the stability, right? So now you see here, if you compare the, if you like compare the stability of these conjugate base by the first factor with this one, negative charge, more electronegative atom is more stable. So F minus is more stable. This gives you what? The stability should be this, F minus, then Cl minus, then Br minus and then I minus. And if you compare this one, if you check this factor, then it should be I minus, Br minus, Cl minus and F minus, right? Now you see there are two contradictory fact we have. These two are contradicting each other. So the order of acid will be this, you are correct, you are right, order of acid will be this. It means what? This factor is dominating here, right? Another thing you see here, right? So this negative charge on more electronegative atom, we will not consider in this case, okay? Now you see this question. Its conjugate base is Hs minus and its conjugate base in Cl minus here will go by the first factor by this one, by this factor will go into this. So according to this factor, the stability of Cl minus is more. So the order of acidity will be this and this is also correct, this is also correct. The point is here we are considering this size factor uniform distribution of charge means size more will be the size, more will be the uniform distribution of charge and this we are doing by electronegativity factor. Understood this? Till now you understood, now you'll ask me sir when we'll apply this charge factor and when we'll apply size factor and when we'll apply this electronegativity factor, okay? Now so there are actually two types of question. The first type when the size varies and the second type when we have electronegativity varies, okay? So in this one you see the size is getting increased as we go down the group you see, the size is getting increased. Here you see the fluorine is in 2p orbital, this has 3p, this has 4p and 5p, right? So down the group we are going for fluorine, chlorine, bromine, iodine, one extra shell is adding in all these element. That's why here size factor is dominating and we'll consider size factor to determine the acidity of this compound. But here you see both sulfur and chlorine belongs to third period, right? This has 3p orbital, this also has 3p orbital. The same orbital we have here, so here the size factor is not dominating but electronegativity factor is dominating. So we consider the stability of conjugate base according to what? According to electronegativity factor? Here we consider according to size factor. Is it clear? Your answer is correct, but did you understand the logic when we apply size factor and then we apply electronegativity factor? Now on the basis of this I'll give you some question, okay? Just you tell me the answer. If you keep all these facts and logic in mind you can easily solve this kind of question. In a second you can solve. You see we have OH and then in the ring we have SH. So obviously you see oxygen and sulfur. This is 2p and this is what? 3p. So size factor is dominating here. So which one will have the? See this one. See here, which factor is dominating? That depends on what molecule you have. Here you see, here the size factor is dominating because we have fluorine, fluorine, bromine, iodine. Here we have electronegativity is dominating because this S and Cl belongs to the same period, third period, 3p and 3p. So here the size won't change that much. You see in the period table we have oxygen, sulfur and then we have fluorine and chlorine, right? So in these two molecules and these two molecules the size difference is not that much. But if you compare these two molecules we have size difference. We have size difference. So when you are going from top to bottom we are considering size left to right we are considering electronegativity. Yes. Now you see these two oxygen and sulfur the order should be this. Similarly if I write down suppose alcohol or OH and RSH, what should be the order? This sulfur down the group we are going, oxygen and sulfur. If I write down H2O and H2S, what will be the order? Sulfur is more electronegativity, sorry, more size so more uniform distribution of charge, more stability of the conjugate base, right? So these kind of compound where we have difference in size the first type we have type one. First type of question that they give you that is when we have variation in size variation in size we consider when you are going from top to bottom, right? And in this case what we can say the acidity of this molecule is directly proportional to the size of an ion. Because larger will be the size more uniform distribution of charge and more will be the stability. Now type two we have what? Type two we have where we have the difference in electronegativity, electronegativity. And we don't have difference in that much in size. Size difference is not that much like we are comparing what sulfur and chlorine not much difference in size in that case, right? Electronegativity factor is dominating and acidity is directly proportional to electronegativity. Oh, you have briefed, actually I have class now I forgot to inform you since I got some problem in my phone. So I could not access any contact so I am using this number in a different phone I don't have any contact now. So I could not inform you actually I have class till 7.30 and there were some because the school is holiday, not today in the school. Right, so that's why. So you have reached in the center. Okay, so what do you do now? No, you can solve some questions, you know that. Okay, and not tomorrow because tomorrow I have to go to Rajinagar. Okay, Wednesday I'll be there. Okay, sorry I could not inform you because I got some problem in my phone, right? So, okay, we'll see that out on Wednesday. Okay, I'll let you know. But you solve some questions, you have come so you don't go home simply, okay? Solve some questions at this half an hour then you can go. Okay, yeah, bye. Okay, so difference in electronegativity. So acidity is directly proportional to the electronegativity we have. Okay, so like all these examples we have done for size factor. Now for electronegativity one example I have given you for this one you see. R C triple bond CH then R C double bond CH2 then R CH2 CH3 obviously you see this carbon atom is sp hybridized this carbon atom is sp2 and this is sp3. So more electronegative atom is this so order will be this, right? So this kind of question you may get difference in size and electronegativity part. Now the third type that we have in this is type three that you write down when hydrogen atom, when hydrogen atom is attached with is attached with same kind of atom. So like first example we'll see into this that is CS3, CS2 OH, this is first one. Second one is this CS3 CH OH CS3 CS3 and then CH3 COH CS3 and CS3. First, second, and third. Which one is more acidic? Tell me the answer, one, two, and three. One is greater than two, greater than three. Okay, so now why is it so because of the plus i nature of this group, right? So actually in this type, when the hydrogen atom is attached to same kind of atom then different electronic effects comes into the picture like i effect, r effect and all. So according to that we'll decide, okay? So here you see in this case when we have i effect so more i effect less will be the acidity. Acidity order will be this one is maximum then two and then three, okay? So you see, so what we can general thing what we can write here in this case as this is your write down for type three, acidity is directly proportional to minus i effect and inversely proportional to plus i effect. Yes, now you'll see here just one or two example more we'll see. What is the acidic order of this and three? In all this compound you see the oxygen atom is attached with oxygen, same atom. All these are type three order of these two. One, two and three quickly. Three to one for the first one for second one, two, three. So it is three to one because it is three plus i group, right? Here we have only two and here we have only one. So primary, secondary and tertiary. So order will be this, okay? Acidic order here because of the electron withdrawing nature of chlorine, right? The effect of this will be maximum at first position because see it has nothing. Suppose if I give you a benzene ring also here, this has nothing to do with the acidic nature of this compound because i effect depends on what the distance we have. So only we have to see the difference. It has nothing to do with the resonance or pi electrons here, right? So obviously the order will be closest here. So first, second and then third, right? Now one type of question, the last two questions I'll give you into this. One type of question they ask like this, you see. Hydrogen, hydrogen. Here we have double bond O, double bond O and double bond O, hydrogen and hydrogen. Suppose this is alpha hydrogen. This is beta and this is gamma. I'm just giving some symbol to this hydrogen, okay? It has nothing to do with alpha position, beta position and all, right? So which hydrogen is most acidic here? What is the order of alpha, beta and gamma? So order of acidity will be what? Alpha, since we have two electron withdrawing group here. So any of these two hydrogen that is the tendency to release these two hydrogen will be maximum here. So alpha will be most acidic. Then we have beta and then we have gamma, right? So this is the order. This kind of question also they ask depending on this, or you know, if I have very various electronic question. One more question on this type I'll give you. This is the last one for this. This is alpha, this is beta, this is gamma and this is delta. Tell me the order, alpha, beta, gamma and delta. Okay, see here, since this is carboxylic acid. So obviously this alpha will be most acidic here since the hydrogen attached with the carboxylic group, carboxylic ion, right? So alpha will be maximum, right? If you compare this beta, so this carbon is what? This carbon is two degree and this carbon is also two degree. So the tendency of this beta and gamma will be equal because both attached to the two degree carbon atom. Here you see this is three degree. This carbon is three degree. This carbon is three degree, right? So this two will have equal strength, right? And the least will be this. So what we can write alpha is equals to greater than beta, almost equals to gamma greater than delta. This should be the order, okay? But theoretically we say this, that beta and gamma will have equal strength, but this depends on what group attached here and what group attached here, right? So if you compare this carboxylic acid as a stronger electron with drawing nature than this OH, okay? So obviously that will also, whatever group we have here and here, that will also affect the releasing of this beta and gamma hydrogen. But that we cannot say theoretically, it is a practical thing, right? But theoretically this kind of question they ask the answer will be this only. We cannot differentiate these two here now. But suppose that this position in this question only if I make some changes and what is that change? Suppose this, I'll just write it down here. This is the ring and this is same. We have COOH, it is alpha. And here suppose if I write OH like this, here we have OH and here this is delta, this is gamma and this is beta and this is alpha. This is alpha. So here obviously the alpha is maximum. This is now one degree. This is two degree and this is three degree. So now we can differentiate it easily, right? So alpha, then we have beta, then we have gamma and then we have delta. This is the order. Understood this? Yes or no? Now one more change you can do into this, this question. Suppose if I add here, fluorine. So with this, the order is this for the first question. Okay, don't get confused. Order is this is the first question. And this only if I add fluorine here because of the minus sign nature of this and this group is closer to this hydroxyl group. So then beta will be more acidic than gamma. So in that case the order will be this alpha maximum, then we have beta, then we have gamma and then we have delta. Is it clear? So with this fluorine, with this question the answer will be this. If fluorine you remove the answer will be this and for this one the answer is this. Understood this? Okay, so now you can understand this that nothing is absolute in this kind of question. One group you remove, one group you add, change your position, the order may be different, completely different. Okay, so you have to understand the concept and these kind of question is always relative. Okay, whatever options are given in that with that options only you have to compare and then you have to find it out that which one is more stable, more acidic or basic whatever the question is. Okay, so then nothing is absolute in this kind of questions, one group you change things will change. Suppose if I add here CS3, right? If I add here CS3, then what happens? This is electron, suppose with CS3 what will be the order then? If I substitute here CS3 tell me the order now then instead of floating if you have CS3 then what is the answer? ACBD, right? If you add CS3 here so because of plus a nature of CS3 this hydrogen will be lesser acidic than this hydrogen. Then we have alpha, gamma, beta and delta, right? So like this they can change the question. Again, if you add this floating here then that won't change the answer. The answer will be this only because the effect of floating at this position on this side and this side will be same, right? So this you must keep in mind don't try to memorize this kind of question. Our last question which is important and I think I have discussed this in carboxylic acid also. If I ask you the stability or the acidic nature of this compound what is the answer? C double bond OOH, then we have benzoic acid COOH and last we have acetic acid CS3COOH. What is the order? One, three, one, three, two. Okay, you see here the answer is this. One is maximum, then we have this and then we have this, this is the answer. The reason is what in this compound we have equal resonance. Equal resonance, I'm talking about the conjugate base of this compound. The conjugate base of this compound will have equal resonance and that's why that conjugate base is most stable. Order will be this is maximum, then we have, see one note you write down into this first. This you must memorize, okay, this is important also. Right now the aromatic carboxylic acid and carboxylic acid is more acidic, is more acidic than aliphatic carboxylic acid except for mick acid. Aromatic carboxylic acid is more acidic than aliphatic carbon, excuse me, aliphatic carboxylic acid except for mick acid. So, formic acid, must remember one thing, formic acid which is SCOH is the only aliphatic acid, aliphatic carboxylic acid, which is more acidic than the aromatic carboxylic acid. All other aliphatic carboxylic acid is less stable, less acidic, sorry, all other aliphatic carboxylic acid, except formic acid, is less acidic than aromatic carboxylic acid. Clear? The reason is what? Equal resonance. So, this one you must memorize, this is important, okay? Now, according to this only, you can compare the basic strength, right, tends to lose lone pair of electron, okay, any component basic strength you have to compare, compare acidic strength and then reverse the order, okay? So, that also you can do, and we'll do again, we'll do next thing we'll do is resonance here, resonance you write down, and in this we are not going to draw what is, we are not going to do what is resonating structure, how do we draw resonating structure and all that you have already done and I think you know all these things, how to draw the resonating structures and all. So, I'm not going to give you all those things again, okay? What we are going to do, we are going to compare again the relative stability of resonating structure, okay? So, write down the heading here, the relative stability of RS, relative stability of RS. So, we are comparing resonating structure of the same compound, stability of the resonating structure of the same compound, right? So, there are seven, eight rules we have in this and those rules we have to memorize, we have to apply those rules only to compare the relative stability. Must remember one thing, we are talking about the stability of RS of same compound, okay? So, if one compound we know already for one compound, we can draw two, three, four, five resonating structure, possible, right? Suppose any compound suppose, I don't want to draw the resonating structure again because that will know, you already know and that will waste our time. The point is what suppose any compound that is C is any compound we have and with this C we can suppose draw a resonating structure like this, okay? So, this three structures combines together and that will give the resonance hybrid which is the what, real structure of the molecule, true structure, real structure and these are what? These are imaginary structure, right? Imaginary structure and hypothetical structure, not true, okay? The actual molecule will not exist in any of these three resonating structure form but that will exist in what? Resonance hybrid, all these things you know, yes or no, tell me fast. Yes, are you there? Okay. We know this resonance hybrid is the real structure and resonating structure are the imaginary structure. So, these resonating structures have different stability also and the most stable one will contribute maximum into the resonance hybrid because we know all these resonating structures will contribute into the properties of the resonance hybrid, right? The properties of this real structure will be somewhat similar to C, A and B and collectively they will have the property of the entire molecule. Some property this molecule will get from A, B and C means whatever the resonating structure we have, each resonating structure will contribute into the property of the resonance hybrid. The resonating structure which contributes maximum into the properties of resonance hybrid is the major contributors or major contributor, right? Or the resonating structure which contributes minimum that is minor contributor, okay? So we can have one as major contributor and other as minor contributor, right? So the one which is major contributor is most stable also, right? So, sometimes why we need this relative stability of RS because of this contributing factor of this molecule, right? The most stable resonating structure will contribute maximum into the resonance hybrid and that's why we have to know this at which resonating structure is more stable or less stable. That's why we are doing this, the relative stability of resonating structure. So like I said, few rules we have, seven, eight rules we have and that we have to keep in mind to compare the relative stability of resonating structure. So rule one you write down first of all, quickly you write down rule one, more pi bond, more stability. In short, you write down, okay? I just want to finish this one today. More pi bond means more stability. Example, you see just one or two examples we'll see in each rule so that you can understand. Suppose the first one is this double bond here in double bond O and this resonance is possible if you see. This double bond, this pi bond will go here and this pi bond comes onto this. So we'll get this as the resonating structure. We have double bond here, O minus and here we have positive sign. So in these two A and B, if you are comparing, the one which has more number of pi bond or we also call it as more covalent bond, more stability, right? So more number of pi bond, more stability, so stability of A is more than to that of B. So difference in the number of pi bonds if you have, then you can directly say more pi bond, more stability, okay? If you compare this one, one more example we'll see. Oxygen, we have lone pair on this oxygen and then this pi bond and the resonating structure if we draw here, it will be double bond here. This will come here and this, right? So here we have, okay, we have positive charge here also. So here also we have the double bond. So here we have, sorry, just a second. So here we have oxygen, positive charge and this is the structure. So this is, if it is A, if it is B, so obviously B is more stable than A because it has two pi bonds onto this. Is it clear? Yes. Now the second rule is sometimes what happens, rule two you write down, sometimes what happens in some of the resonating structure, the number of pi bonds are equal, right? So if the number of pi bonds are equal, equal pi bond, equal pi bond, then stability of the resonating structure or simply stability is inversely proportional to the charge separation, to the charge separation. Means what? In the one molecule if there is no charge and in another molecule if you have some charge separation, so that stability will be less. For example, you see, if I take phenol, right? This is OH and when you draw the resonating structure of this, the lone pair of oxygen will take part into resonance and we'll get double bond OH, positive charge, right? And we get this negative charge here. This lone pair comes over here and this pi electron comes over here, right? So in this molecule you see this A and B, the number of pi bonds are equal, but here we don't have any separation of charge, here we have charge separation. So obviously when charge separation is there, stability will be less. So stability of A is more than that of B. Is it clear? Another one you see, if I take aniline, tell me the stability of these two. NH2, lone pair on it. Double bond NH2, positive and then we have double bond, double bond negative. What is the stability of this A and B? Which one is more stable? Obviously, one is more stable than two. Why? Because here we do not have any charge separation, okay? So when the number of pi bonds are equal, then we consider what? Charge separation. Rule 3 you write down, okay? Now in this also, if pi bonds are equal and same charge separation is there. Same charge separation. Then obviously we cannot do the stability check from rule 1 and 2, right? Because these are not different. Then in this case what happens, what we say? That stability is directly proportional to the complete octet, complete octet of number of atoms. Means if you have more number of atom with complete octet, then more will be the stability, okay? See first example, BH2, CH double bond CL with three lone pair on it, one negative charge. If you draw the resonating structure of it, that will be pi bond comes over here, right? And that will be BH2, so BH2 negative double bond CH single bond CL with three lone pair on it. Okay, so now in this one if you see, carbon and chlorine has same kind of bond. You see, eight electrons for chlorine, eight electrons for carbon. For boron if you see, there are two hydrogens, two, two, four and four electrons here. So here we have eight electrons for boron, but for boron here you see it has only six electron.  So obviously the stability of B is more than A. Is it clear? In this example you see positive charge on this carbon and then if I draw the resonating structure, the structure is this double bond positive charge here and this lone pair comes here. A and B, which one is more stable? A will be more stable. See oxygen octet is complete here, carbon octet is not complete, right? You just check this carbon which has the positive charge here. This does not have complete octet, right? But here you see this carbon has complete octet along with this, what we say oxygen also because you see here this lone pair comes into bond pair. So it is associated with oxygen here also. This oxygen has already has one lone pair and one lone pair becomes bond pair. So octet of oxygen is complete here, you see 2, 4, 6, 8, but along with this carbon oxygen is also complete here, right, 2, 4, 6 and hydrogen here, 8. So obviously B is more stable than A. Understood? This says negative charge sometimes the first three rule, we cannot judge from the first three rule, then we use this rule 4, okay? The negative charge on more electronegative element, more electronegative atom is more stable, more stable and positive charge, charge is left here, and positive charge on less electronegative element or more electropositive element on more electropositive element is more stable. But the first three rule fails, then we can judge the stability with this rule also, okay? For example you see the simple one, we have CH3 C double bond O, CH2 with one lone pair negative charge, resonating structure if you draw this pi electron goes here, CH3 C O minus double bond CH2 plus, we don't have plus here because carbon has 8 electron only, you know, so no extra charge on it, okay? Now if you apply rule number one, which is pi, number of pi bond, same pi bond, what is rule number two? Can you apply rule number two into this? What is rule number two? Charge separation, right? So can you apply charge separation into this? That also we cannot apply because only one negative charge. Rule number three is what? Stability, right? Now octet, now octet of all atoms are complete now, here you see, carbon has, okay, if you compare this, octet rule chalo will try, 8 electrons, oxygen also has 8 electrons here, this carbon 8 electrons and this carbon has 2, 4, 6, 8 electrons, so all atoms has complete octet. So like you see, these three rules we cannot apply here, then what we say that negative charge on more electronegative atom is more stable, here the negative charge on carbon, here the negative charge on oxygen, so order will be B is more stable than A. Simple one you see, 2 minus, then we have double bond O, negative charge here. Number of pi bonds equal octet complete, like charge separation, 1-1 negative charge, we cannot do this by three rules, first three rules, so this rule will apply, negative charge on more electronegative atom, A is more stable than B. One more example, see all these are questions only, in GOC whatever examples we do, those are questions itself, okay, so we are discussing question only, suppose we will write on this CH2 negative C double bond O, NH2 negative, the first resonating structure of this will be this negative charge comes over here, this pi electron goes here, so we will get CH2 double bond C single bond O minus and NH2 minus, next is structure what we can draw from this only, this negative charge comes over here, this pi electron goes here, so CH2 negative double bond CO minus double bond NH2, sorry one thing, this two is not there, this two is not there, this two is not there, okay, A, B and C, tell me the stability order of these three, BAC, BCA, BCA is correct, why it is correct you see, in B, so first of all electronegative order element, the order will be what, oxygen is most electronegative, then nitrogen and then carbon, so if we can according to this compound I am writing down this, okay electronegativity order, if negative charge is there, so we will try to put negative charge on oxygen atom, that will give more stable product, now if there are more negative charge we have then first we will give it on oxygen, then nitrogen and then carbon, oxygen and nitrogen if they have negative charge it is the most stable structure we have, right, which is option B, sorry, the structure B, oxygen and negative, then what we can have, in the C you see oxygen and carbon we have and A we have carbon and nitrogen, so obviously since oxygen is most electronegative, so negative charge on oxygen gives you always more stable product, so B and then we have C and then we have A, this is the order, then rule number five, rule number five, stability will be more, stability will be more when opposite charge are closer, when opposite charge are closer and like charges, when stability will be more, when opposite charge are closer and like charges are far apart, more the distance between the like charges more will be the stability, less the distance between the opposite charges more will be the stability, okay, so now you see this one, first example, I am taking the example of phenol, OH, so the resonating structure of this, one of the resonating structure is this one, double bond OH positive charge and this pi electron will jump here, negative charge, double bond and double bond, another structure if you draw in the possible resonating structure is negative charge here, double bond, double bond, double bond OH positive charge here, A, B and C, now you tell me the order here, tell me the order I will just come, one second, okay, what is the answer? A, C, B, tell me the answer, A, C, B, okay, you see here, like suppose you get this question in the exam, okay, don't like try to use rule 5 directly here, like suppose you are getting this question in the exam, then all these rules you can apply, right, so first of all what you will think, you will think number of pi bonds, so equal number of pi bonds, so you cannot do that from rule 1, rule 2 is what, when there is charge separation, stability will be less, right, rule 2 is what, charge separation, stability will be less, so this rule if you apply A, B and C, do we have charge separation in this? No, so A is the most stable one because B also we have charge separation, C also we have charge separation, right, we will not have any charge separation, so obviously stability of A is maximum, now with this particular like conclusion you can at least one or two, one option you can eliminate if you get this question in the exam, now coming back to this B and C, this B and C we have to compare, right, now this rule 5 is what, more the charge separation between the opposite charge, less will be the stability, correct, like charges if you have positive, positive charge, so because if positive charge are close enough, then we will have repulsion over there, that will decrease the stability, that's why we are saying that same charge or like charges must be far apart and opposite charges must be closer to each other, right, you see in the B option this two charge are closer than the option C, so obviously B is more stable than C, understood, so this the conclusion that A we are getting maximum stability, it is from rule 2, right, from rule 2 charge separation, B and C we are concluding from rule 5, yes, just a second, hello, so A, B, C is the order, one more example we'll do, here we have negative and this charge will be as it is, now from this the another structure we can draw, that will be double bond here, here and here negative charge, A, B and C, A, B, C, is it right, tell me others, A, C, B, how it is A, C, B, R, M, opposite charges must be close to each other, okay, A has no separation of charge, so obviously A will be stable, yeah correct Lalita, yeah, so first of all the first we can apply rule 2, so we are going to rule 2, A is most stable, no separation of charge, B and C we have separation of charge, so we cannot do this with rule 2, so for this we can apply rule 5, here we have distances small, here we have large distance, opposite charge we have, so then we have B, smaller distance then larger distance C, is it clear, R, N, yes right, first we'll apply rule 2 charge separation and then rule 5, distance between the charges, okay, now another rule you write down rule 6, see in this rule actually I'll just make you understand with this, this with some example directly, okay, we simply see here the eye effect and all various electronic effect, like you see this one, first example if I give you the molecule is this, here we have positive charge double bond methyl and when you draw the resonating structure of this, this pi electron will come here, so we'll get this double bond here, here and here we have the positive charge, right, now the point is what, this positive charge, the two positive charge here we have equal number of pi bonds, right, we have equal charge also octet also if you can consider that you'll get the same thing in octet also, so here what we do, we just check the possibility of this carbocation, plus i, plus i and plus i, not this one because that carbon is sp2, right, not this one, so we have two plus i here and here you see this is one degree carbocation, only one plus i, so obviously this is more stable than the second one, we just only check the eye effect and all in this, second example you see, suppose we have this one, the molecule is double bond CH2, positive charge here and then we have CH2 plus, this is not the double bond, okay, take care of this, this is single bond only, not double bond, which one is more stable here or B, which one, A is more than B, okay, obviously A is more than B, correct, why because this is carbocation, it is two degree and this is one degree, so we know two degree carbocation is more stable, what happened at miss, oh for second one you are saying, for second one, what is the answer, second one, what is the answer for the second one means the last question, A is more stable, why, this is for the last one, right, third one, third question, see this is this is carbonion, right, so it is two degree carbonion, this is three degree carbonion and we know two degree carbonion is more stable than three degree, this is carbocation and this is carbonion, negative charge, understood, okay, now rule seven, right on this, more benzenoid form, benzenoid form is more stable, right, if you have more benzenoid form then that stability will be more, okay, now what is this benzenoid form, suppose if I write down the example of naphthalene, naphthalene, the structure is this, another one if I draw, the structure will be this, this is naphthalene, okay, now in this two you see, how many benzenoid form you see here in this, in this structure, how many benzenoid form you see, see when this, this side just let it be this, this, suppose this ring is not there, then this is one benzen ring, yes, and now you see since this double bond you can write down this side also, that won't make any difference in the molecule, this double bond can be this side also, can be this side also, so with this double bond if you consider here, this is also one benzen ring we have, right, this is one benzen ring separately you see, and this is the another benzen ring, so in this naphthalene we have two benzenoid form, right, we have two benzenoid form because you can consider this pi bond with this ring also and this ring also, so one benzen ring is this, another one is this, two benzenoid form, now here you see, here we don't have any pi bond present here, right, so here we have only one benzenoid form, benzenoid form, so obviously when we have two benzenoid form, so first one is more stable than the second one, is it clear, more benzenoid form, more stable, understood, another one the same kind of example you see, anthracene, the structure of anthracene is this, we have three ring here, one, two and three, with one double bond here, double bond, double bond, oh sorry just a second, one and this also you see, one, two and three, so one, two, three then this and this, this is anthracene, okay and when you draw the resonating structure of this you will get one, two, three, four, five, six, seven, so in this the number of benzenoid form is what, the number of benzenoid form here and here, can you tell me, here we have two benzenoid form, one and two and here we have only one, right, one benzenoid form, so obviously the first one is more stable, right, so this is the rule we have for number of benzenoid form, the name of this rule we call it as Fry's rule, F-R-I-E-S, Fry's rule is this, number of benzenoid form, name is not that important, the last rule we have here to comparison the stability and you must remember, you must take care of this thing that we are comparing the stability of resonating structure of the same molecule, we haven't taken now the two different molecules, right, but we can use these for two different molecules also, we are comparing the resonating structure of this, stability of resonating structure, now rule eight we have and this is the special case when we have aromaticity, it is simply aromaticity, so when aromaticity is there, when the compound is aromatic, then that will overrule every other thing, right, aromatic compounds are the more, most stable compound and the stability order if you see here, it is first of all it is aromatic, this order they have asked in need exam many times, since directly this thing aromatic and then non-aromatic and the last one we have anti-aromatic, right, so this order they have asked many times in need exam, this exactly this order, okay, now you see the example here, suppose the first compound is this CH2, you draw the resonating structure CH2, positive charge here, here we have negative charge when this pi electron comes here and then the another structure it is B and for here it is CH2 negative and here it is positive, maybe this pi electron may jump here also and here also, so if it jumps here you will get this, if it jumps here you will get this, now for the aromatic compound Huckel rule should be followed, right, 4n plus 2 pi electrons, right, where n can be any number 0, 1, 2 and so on, right, anti-aromatic and it must have 4n pi electrons, see this Huckel's rule and this one it is just a reference we have, there are many other compounds which does not follow Huckel's rule but they are aromatic in nature, there are compounds which follow Huckel's rule but then also anti-aromatic but then also it is not aromatic, so still the research is going on on this, okay, a few compounds are there, I will give you in another class what do compounds are there for aromatic things, okay, the point is for aromaticity the molecule must be planted, right and in all those molecules where we have complete delocalization of electron, uniform distribution of electron, right, then the compound is said to be aromatic, like you see if you take this one, this example, here we have negative sign and this, right, in this structure you will draw the resonating structure of this, you will get 5 resonating structure, 5 Rs you will get here and in each of the Rs you will see or in all these 5 Rs if you see the negative charge will present on all the carbon atoms, here, here, here and here, if you see all the resonating structure you should draw, right, so like when we have this type of structure where the charge is distributed equally among the atoms present in the ring then the compound is aromatic, that is also one way to find out aromatic compounds, right, so uniform distribution of charge leads towards stability because uniform distribution is possible in the case of aromatic compound and for aromaticity the molecule must be planted, right, so there are few compounds we have which you have to keep in mind, okay, like here you see in this one, when you calculate the number of pi electrons, we calculate lone pair into that also because lone pair has the tendency to form pi bond, okay, so in this compound if you calculate the number of pi electrons, number of pi electrons here it will be what, 2 plus 2 4, 4 plus 2, 6 pi electrons, right, so for n is equals to 1 it follows Huckel's rule, n is equals to 1 it is 6 pi electrons, so this follows Huckel's rule, H-U-C-K-L-E-S Huckel's rule and hence it is aromatic, right, follows Huckel's rule and planted also, the molecule is planted also, that's why it is aromatic compound, in this one you see it has only 2 plus 2 4 pi electrons, so for n is equals to 1 it is true, hence this one is anti-aromatic AA and this one is non-aromatic compound, I will discuss this aromaticity also in the other class, right, basically few examples we have for aromatic and non-aromatic compounds, those you have to keep in mind because it's still, there is no one logic with which you can say the compound is aromatic or non-aromatic or anti-aromatic, like I said the molecule must be planted and you should know which all the molecules are non-planned, so in there are a few examples we have 5, 6, 10 examples that I'll give you in the next class, okay, so here you see since this compound is aromatic so order of stability will be first of all we have the aromatic, then we have anti-aromatic, sorry non-aromatic and then we have anti-aromatic C, this is C, right, so order is B, A and C, is it clear, like I said in Toyota that when aromaticity is there then this will overrule every other possibilities, because of aromaticity this is most stable, aromaticity is a special case, in the beginning only I said that when aromatic, when aromaticity is there then that will overrule all other possibilities of stability, right, that's the reason, understood, now one last thing we see this aromaticity only, one last thing we'll discuss, one, just one example we'll see here, like I said that there are compounds which follows Huckle's rule but not aromatic, one example you must keep in mind and the example is this, oh sorry, in this one is not, now in this one you see this molecule, how many pi bonds we have in this molecule, number of pi bonds, sir my class, so number of pi bonds are 5, so number of pi bonds are 5, so how many pi electrons we have here, number of pi electrons are 10, right, so if you see Huckle's rule 4n plus 2 pi electrons, for any value of any integral value of n, can we get this 10 electrons or what value of n will get 10 electrons, n is equals to 2, right, n is equals to 2 if you substitute here you'll get 10 pi electrons, okay, so according to Huckle's rule you see there are 10 pi electrons, so molecule must be what, aromatic, right, molecule must be aromatic because it follows Huckle's rule but this molecule is not aromatic because the molecule is not planar here, what happens, this bond we are here, we have hydrogen and hydrogen, so this distance is not that much large so that it can accommodate to hydrogen here, right, so because of this repulsion here, one of this molecule to minimize this repulsion that will change its plane, so this molecule will come up and the other ring will come down, so the two ring are in two different plane, right, they are not in the same plane, the two ring are in two different plane, so when the molecule is non-planar, so complete delocalization of electron is not possible, uniform delocalization of electron is not possible and hence the molecule is not aromatic, you understood this, yes, did you get this, the point is the two ring is not in the same plane, why, because of this repulsion here, because of this repulsion, the molecule two ring are not in the same plane, one ring will be up and other will be slightly down, right, to minimize this repulsion and when the ring is not planar the uniform, uniform distribution of or delocalization of electron is not possible and hence the molecule is not aromatic, so like you see, it follows Huckel's tool, if you look at this molecule you can say okay it follows Huckel's tool aromatic, no, there are many exceptions into this, one of the exception is this, you should know this kind of molecules which are planar and non-planar so that you can do this kind of questions, anyways in the next class I will give you some examples of compounds which follow Huckel's tool and does not follow Huckel's tool aromaticity and non-automatic compounds, okay, so those things we will discuss in the next class, so these are the eight rules we have with these rules you can compare the stability of resonating structure and on this they ask question in exams, okay, so you must memorize those rules, okay, which is not that tough also, only thing you have to keep in mind is what the rules that we have, what all rules we will find out, we will use to find out the stability, understood, okay, so this is it for today, okay, we will see other class, we will see some more concepts, okay and you saw some question on this, okay, any book if you have try to solve some question on this, okay, okay, bye bye, take care