 Hello, good afternoon, guys. Please type in your name and let me know who are there. Hello, Suresh. Sangyana, hi. Good afternoon. OK, guys, to last class, we were discussing some problems of thermodynamics, like if you remember. So like I said, few questions were left in this particular chapter. So I'm just going to cover those questions first, and then we'll solve some questions of hydrocarbon with GOC, right? So GOC and hydrocarbon also will solve some questions in this session. But first, we'll finish a few questions of thermodynamics, which is left in the last class. So this is the first question that you can see on your screen. Try this and tell me the answer. Tell me the answer of this question. Assertion is spontaneous process is an irreversible process and may be divorced by some external agency. The reason is decrease in enthalpy is a contributory factor for spontaneity. Take an example of spontaneous process and then try to find out the answer of this question. Hello, Ramcharan. Take an example of spontaneous process. Then you'll probably be able to solve this question. What happens in a spontaneous process that you first try to understand with an example? Ramcharan is getting option A. Tell me, guys, answer. C, Suresh is getting B, Sanjana is getting D. OK. So some of you are getting D, B, and C. All the options probably I'm getting now. You see, in this question, the question is related to spontaneous process. So what is an spontaneous process, what's the fault? So you see, if you take an example of spontaneous process, we can take many examples. Suppose I am taking evaporation of water. Evaporation of water is an spontaneous process. However, boiling is not an spontaneous process. So what happens in evaporation of water? Water converts into, or we can say, liquid converts into vapor. Can we reverse this process? This process, can we reverse? Yes, we can reverse. How? We can easily reverse this process by applying pressure, applying pressure. So obviously, spontaneous process can be reversed by some external agency. So obviously, a certain A is right. Decrease in enthalpy is a contributory factor for spontaneity. Obviously, you see, if I take another example of this spontaneous process, let me take one more example. That is flow of water down the hill. So what happens in this process? Water is flowing from the hill to the bottom. So obviously, the potential energy is decreasing. So what we can say, this process is accompanied by decrease in energy. Decrease in energy. So we can reverse the process. And the energy also decreases here. That is what we have here, decrease in enthalpy. So obviously, of assertion and reason, both are correct. So A and R are true. A and R are true. So we can have the answer either A or B. Now, the other part of this, of these options are what? R is the correct explanation of A. So here you see what happens. This and this is not at all related. If decrease in energy takes place, this has nothing to do with the irreversibility nature of or irreversible nature of spontaneous process. Since both statements are true, assertion and reason both are true, but R is not the correct explanation of A. So this is what answer of this question will be option B. So simply what we can say spontaneous process is accompanied by decrease in energy and increase in randomness also. So the bottom line, which you have to keep in mind that spontaneous processes decrease in energy or enthalpy and increase in entropy, randomness. How we know this entropy increases because when liquid converts into vapor, right? So obviously this has more randomness. That's why we say, see what happens in this, what is this reason? Decrease in enthalpy is a contributory factor of spontaneity. See, spontaneous process we have discussed already, right? So any, see when decrease in enthalpy takes place, you can also understand this in a different way. I have explained this with the help of an example, but when decrease in enthalpy takes place, what happens? It means the process is going from high energy to low energy, high energy to low energy, decrease in enthalpy and low energy is what? It's more stable state. So if you try to understand the concept of any processes, they're either going towards the more stable state or more randomness, okay? So in the second state, decrease in enthalpy is the contributory factor of spontaneity, right? So fine, since decrease in enthalpy takes place, so the energy will be low over here and hence the stability will be more and that's why it is spontaneous in nature. Any process from high to low energy because of more stability, it will be spontaneous, right? So since spontaneity decrease in enthalpy takes place, so it is a contributory factor, correct? And obviously, R is not the correct explanation of A. Understood? But for delta G to be negative in enthalpy can increase or decrease, so that does not affect the spontaneity. It depends on temperature also. Ram sir, on temperature you cannot neglect here. See the only thing you have to, you don't know, when you consider delta G over here, right? Then you have to consider temperature also because it is temperature dependent term, right? Since we haven't discussed about temperature over here, so simply you can understand by two simple methods, two simple things you can think over here, right? Any process from high to low energy going towards more stability and when the process is towards stable state, right? The nature of this process will be what? It can be spontaneous, right? Most of the time it is spontaneous only. Second thing is what? If it's like, you see this example, this example, it is going towards higher energy state, liquid to vapor, you have to provide energy into it. It is an endothermic process, but since energy is increasing here, but at the same time, entropy is also increasing. So randomness is also increasing. So that's why we say the process always goes towards more stable state, which is the lower energy state or more randomness. Is it clear? If enthalpy increase or decrease is a factor over here. It depends on temperature and the nature of the process, whether it is exothermic or endothermic. Clear, understood all of you? Right, next question you see, solve this. Guys, this K is the equilibrium constant, okay? Now, first of all, you try to write down the relation of equilibrium constant, that is log of equilibrium constant log K, with temperature. Try to establish the relation first, then you can compare and get the term. Okay, I'll do this, you see. See for an isomerization reaction of A gives B, the temperature dependence of equilibrium constant is given by this, okay? The relation is again, we have to find out delta S, okay? So first of all, we know the value of this, the expression of this free energy with the reaction quotient. What is that? Delta G is equals to delta G naught plus 2.303 RTQ, right? Equilibrium, what happens? Delta G is equals to zero, you know this? So delta G naught is equals to minus 2.303 RT log Q, which we can also write minus RT log E base Q, or ln Q, okay? So this is the condition at equilibrium. Now, we know the relation of delta H and entropesis, entropy we have to find out. So we can write down again, delta G naught is equals to H naught minus T, delta S naught. Delta G naught is nothing but this, so that I'll substitute over here. It becomes RT ln Q is equals to delta G naught delta H naught minus T delta S naught. Oh, sorry, one thing I missed here. You see, equilibrium, right? Reaction quotient also becomes equilibrium constant K, right? That also I'll substitute here. So this Q is nothing but K here. This Q is nothing but K here. Here also it should be K, equilibrium constant, okay? So now ln K is equals to delta S naught divided by R minus delta H naught divided by R into one by T. So now you can easily compare this equation and this equation. You will get delta S naught divided by R is equals to four and delta H naught divided by R is equals to 2000. So then delta S naught will be what? Four R, so answer will be option A. Is it clear? Understood, guys. See, when I got the expression of ln K stress, okay? Why I have taken ln K, right? Because you see the question, the left hand side in the given equation, in the question, it is ln K itself, okay? So I'll write to write down the expression of ln K in terms of temperature, okay? So I'll try to write down the expression of this, similar expression I'll try to write down here. So when I get this ln K is equals to this, so this equation you see, this equation you can compare with the equation given in this equation, right? So you see this equation here, if this ln K is equals to four and here we have ln K is equals to delta S naught by R. Delta S naught by R is equals to this four here, right? That is what I have written here and we'll get the value of delta S naught. Understood stress? Okay, next question you see. This is one Carnot engine. Okay, 35th one, all of you are getting C. 533.3 Kelvin, yeah, it's the direct formula, okay? Efficiency of Carnot engine, I'll not solve this, I'll just write down the formula here. Efficiency eta is equals to T2 minus T1 divided by T2. So everything is given, 0.25, T2 you have to find out, T1 is given 400, you can substitute and get the answer, okay? So answer will be the temperature T2 is equals to T, which is nothing but 533.3 Kelvin, that formula we have written. Next question you see, question number 36. Vashnavi is getting A. So the formula here we use is delta G is equals to delta H minus T delta S. So delta G is zero at equilibrium. That is the only thing you have to know to solve this question. So T is equals to delta H divided by delta S. Both values are given in the question and once you solve this, you will get 400. Next question you see, what is the answer for this question? Tell me fast, I'll move on then. 37, what is the answer? 37 is V, right? Just now I have written the expression, so it's V only. Minus RT ln Kp you can write or you can write minus 2.303 RT log Kp, okay? So option B is correct. Next one you see, this one. All of you are getting C, okay? So the question is we have delta H and delta S value is given, okay? And we have to find out the temperature at which the reaction would not be spontaneous, okay? So for non-spontaneous reaction what we can write that delta G should be greater than zero. This is what the condition we have and we know the relation of delta GN, this delta H minus T delta S. So in this expression, if you apply the condition of non-spontaneous process, we will get delta H minus T delta S should be greater than zero, which is nothing but temperature is less than delta H by delta S. When you substitute all these value here, you will get the answer and the correct answer is option C only. So all of you have got the right answer. Next one you see, what about this match the column? So on the arrangement, stress and all of you are getting A. Any others, what is the answer? Drop your vaporization is always positive, decrease lowest entropy delta H of vapor by PB, correct? So obviously option A for option, for this A, this fourth one is correct, right? Because we know this is the formula we have. Deltation entropy when vaporization is there is equals to the enthalpy of vaporization divided by temperature at which the liquid boils. That is the boiling point of the liquid. And this value is always positive also, right? So A is nothing but two and four C, if you see all the option, okay? All of you are very sure that A must be four, right? So this, if you are sure with this, okay, when you see the option, you can easily eliminate option B and option C. So in match the column, you always try to use these tricks, okay? If you have one or two information with that, you can easily eliminate a few options, okay? Now we are left with or we are, no, if you're confused with two and third, then we can see the next option, K for spontaneous process. When the process is spontaneous, then equilibrium constant is what? It is always positive, right? You see, for B is two, B is two. So that is also not we have to look into that, okay? Option C you see. The slime solid, solid, crystalline solid state can have what? Crystalline solid state, right? It is the ordered arrangement. We have, you know, crystalline subsets are, will have the ordered arrangement, okay? So it always have the lowest entropy because the confusion is what? Whether it is three or one, right? So crystalline solid, solid, solid state decreases what? It does not make any sense, okay? So hence this C can be three also with the lowest entropy. So hence the answer will be option A, right? So all of you have got the right answer. I think, yes, all of you have got A only. Next question you see, solve this one. This question is a bit different. Okay, so resistance is getting D, the answer and D, all of you are getting D. Okay, so answer is 97%, that is D only, okay? So all of you are getting the right answer. Okay, you see this question in a fuel cell, the reaction is given. The standard enthalpy of formation of all these things are given. So first of all, we can find out the del G for the reaction, okay? So the reaction change in Gibbs free energy for the reaction will be what? Del G of formation of CO2 plus del G of formation of H2O, right? Since two moles are there, so into two. This minus del G of formation of CH3OH plus three by two del G of formation of O2. This is the expression we have. Now all the values are given, you have to substitute this value and you have to find out the del G for the reaction, okay? So once you solve this, the del G for this reaction, you will get 702, 702.6, okay? I have done this calculation, so I'm giving you the direct value over here, 702.6. Now the percentage efficiency of the fuel cell you have to find out. So percentage efficiency will be what? It is the energy that has been utilized, which is 702.6 divided by the enthalpy of the reaction which is given, 726 into 100, right? Then you solve this, you will get 97%, okay? Hence the answer will be option D, right? Next question you see, solve this one. Okay, we just is getting B. Vasanthi, Sanjana, B, Sandhya, B, okay? So all of you have got the right answer. See this is also formula-based question. Log KC we have to find out, del H and del S for the reaction is given, right? So we know the relation of delta G, not is equals to delta H naught minus T delta S naught. Since log KC we have to find out, so we'll substitute this del G in terms of KC, right? And the expression will be what? Minus 2.303 RT log of KC is equals to delta H naught minus T delta S naught, okay? So all these things, you see the value of delta H is given, delta S also given, temperature also given, RT, we know, so log KC we can find out directly. So when you solve this expression here, again you'll get log KC is equals to 10. So that's the answer option B. Well, you see the few questions on this relation, delta G is equals to this, this relation and this relation, there are, you know, there are many questions on these two relations, okay? So you must remember these two relations for the exam. Next question you see, this one. Tell me the value, set of thermodynamic parameters is given, what is the correct value? Liquid is converting into gas. Why it is D, option D, okay, delta S is positive fine because liquid is converting into gas. Why delta G is negative? Is it negative delta G? Okay, a veterinarian Sander is getting A. Why delta G is zero? Tell me, veterinarian Sander, yeah. And why it is negative? Tell me others. Tamsin, Shreya, Sanjana, Patnish, yeah, correct question. No, actually you see, this is what happens. I think all of you are sure with positive entropy, okay? That we are not discussing, right? Because liquid is converting into gas, so delta S must be positive. See what happens, you have water here, H2O, and you are heating this water at 100 degrees Celsius, and you'll get water liquid we have here, and here you'll get water, gas or vapor, whatever you say. 100 degrees Celsius, right? So now at this temperature, we know the liquid and gas both exist in equilibrium at 100 degrees Celsius, which is nothing but 373 Kelvin. Right? So at this temperature, both liquid and gas exist, so we have an equilibrium over here, so that's why delta G is zero at equilibrium, right? We are supplying heat into it, then only the temperature will increase. The process is not spontaneous then. See, this is the process, right? It is not a reversible process, it is not written here, but it is a process, we are heating this water, you'll get gas at 373. Now at this state, what happens? We'll have liquid also and gas also, so at that stage, what happens? That's the question is, understood? Ramchalan, no, for any kind of phase change, we cannot say something there. See, the process must be at equilibrium, then only we can say delta G is equals to zero, right? Phase change, this phase change takes place at higher temperature also. There will also be conversion of liquid into gas or any kind of conversion is there, but since temperature is not 100 degree Celsius, so that is not at equilibrium, right? So equilibrium thing, in case of water and vapor, water vapor and water liquid that you have, it always exists at 100 degree Celsius only, that's why delta G is zero. For any other liquid, if it's vapor and liquid will be at equilibrium at 200 degree Celsius, then at 200 degree Celsius only delta G will be zero, right? It depends on what temperature we have, otherwise we cannot say, right? Point is delta G is always zero at equilibrium, that is what you have to keep in mind. And since in case of water, we know the vapor and liquid both exist at 100 degree Celsius, it means we have an equilibrium over here, and that's why delta G is zero. I hope it is clear now. This one, all of you are getting D. Okay, so D is the right answer. Since the process is spontaneous, so reaction is going into forward direction, okay? And forward direction reaction is only possible. Forward reaction is possible just a second, I think. Okay, so forward reaction is possible when K equilibrium is greater than one, okay? Spontaneous, so delta G should be less than zero, negative, right? And we also know delta G is equals to minus NF E naught cell, right? For this to be negative, this E naught cell must be positive greater than zero, right? Hence option D is, right? Okay, so all of you have got this already. I'll move on to the next question. Okay, now this is, now we are discussing questions on GOC and, so this is the first question. You have to find out the acidic order of these. Ramchana is getting C, Sondriya, A, C, D, Sanjana is getting, okay? You see, this is experimental, this question, or the order is based on the experimental value, that is K or PKB or PKA value. Okay, so first of all you see, if we compare thiols alcohol, so since the alcohol general formula if I write, it is R, OH, general formula is R, SH. So generally what happens, thiols are, thiols are stronger acid, stronger acid than alcohols. First of all, this you must remember, thiols are stronger acid than alcohol because oxygen being more electronegative, it is very difficult to lose H plus over here, comparatively, with comparison to this RSH, okay? So thiols are the stronger acid than alcohol. So obviously if you see the option, the first is the most acidic compound we have here, right? If you compare the acidity of alcohol and water, okay? So in alcohol and water, this value you have to memorize, okay? The value of methanol, see generally what happens, I'll just write down this value also over here. Generally alcohol are, alcohols are weaker acid, weaker acid than water, weaker acid than water, but we have one exception here. And that exception is what? Exception is methanol. So what we can say, except methanol, all alcohols are weaker acid than water, okay? Except methanol, all alcohols are weaker acid than water, okay? So methanol, the value of the pKa value if I say, for methanol, the pKa value is 15.5. And on this basis only, we are saying this. That's why I said it is a bit experimental. For water, the pKa value is slightly higher than methanol, 15.7, you see? The difference is not that much, okay? And we know as pKa value increases, acidity decreases, correct? So that's why the order of acidity will be methanol is CH3OH is more acidic than water, than any other alcohol. So order will be, in the last, we have ethyl alcohol, etOH, and before this, we have thiol, which is nothing but CH3SH. So order will be first, and then second, and then third, and fourth. So first option is correct here. So this value or this order depends upon the pKa value. That's why I said it is a bit theoretical. Always remember thiols are more acidic than alcohol. Alcohol and water, if you are comparing, so methanol is more acidic than water, and all other alcohol are weaker acidic than water. Next question, nucleophilicity. Basically, you have to find out which one is a better nucleophile. One of you are getting A. What about first and second? First and second, if you compare, then what happens? Okay, you see. The answer for this question is option C. The correct answer is option C. First of all, we have discussed just now the acidic order we have discussed just now. And that is maximum for CH3SH, then we have CH3OH, after this we have H2O, and then we have ETOH, right? So if the acidity order is this, then the conjugate base of this will be CH3S-CH3O-O and then EtO-O. So more strong acid, its conjugate base is what? Weak but stable, right? For a strong acid, its conjugate base is weak and but stable, right? So if you write down the basicity order of all these, basicity order, it will be reverse like this. Basicity order will be this. Now you see the next thing here, there's one rule for nucleophilicity. And the rule is what I'll just write down here, that when nucleophilic center is different, you all write down this rule. When nucleophilic center is different, they belongs to the same group, they belongs to the same group, their nucleophilicity reverse of the basicity order. So here you see CH3S- and O-, okay? Nucleophilic center is different, right? Belongs to the same group. So their nucleophilicity is reverse of their basic order. So if I write down the correct order of nucleophilicity, that will be maximum for EtO-, then we have OH-, after that we'll have CH3S- and in the last we have CH3O-. Hence option A is correct over here, option one. Nucleophilicity order is same as basicity, okay? There's no difference into that, but one thing you have to keep in mind when the nucleophilic center is different and they belong to the same order then the order will be reversed. You can also compare these two like this. Oxygen with negative charge is more stable than sulfur with negative charge. Hence the tendency for this to attack as a nucleophile will be lesser than this, okay? Second thing is what? That since oxygen being more electronegative element, so it will attract the electron pair towards its side is strongly than SCS3S-, right? So here the attraction of this electron pair towards the oxygen will be more, here it will be less, so it can attack easily, so it is a better nucleophile. Understood? Correct? So all these points, all these points that I have given you here where you are making a mistake because you see all of you said first option. So fourth and third is fine. You get confused with these two, since you did not know this particular rule. So all these do you note it down so that you can revise it later on when the exam is there, okay? Like one or two or three, four days before the exam, okay? So these kind of rules you just keep in mind, okay? Next question, carbocation is least stable. SCS is getting C, SCS is getting C, all others are getting B, okay? See the option, okay? We have phenyl CH2, okay, fine. Here see here, first of all, this is vinyl carbocation, right? But the difference here, it is what? Double bond in B, we have a double bond here, I think. We have a double bond here, Atmesh. You see this carbon atom is SP2 hybridized. This carbon atom is SP, I'm talking about the carbon atom which has the positive charge, SP hybridized. This one is again SP and this is again SP2 hybridized. The question is, which carbon atom is least stable? That's the question, right? So we know more electronegative element on that, the negative charge is more stable, right? So if the question is NIN, then the answer would be anything else. That depends on some other factor that we won't discuss now. But the point is the difference in hybridization is there. And option B, the carbon atom is the most electronegative this one, since it is SP hybridized. So most electronegative element, positive charge on it, is the least stable, right? Hence answer will be option B. But we are getting B only, so I'll move on. Most stable carbocation. What is the answer? How it is D? Atmesh, why it is option C? All of you are getting different, different answer. Okay, Atmesh, why it is option C? Atmesh is getting A. Why A, Atmesh? Three degree carbocation, hyperconjugation. Hyperconjugation. Three degree carbocation are most stable. Where is three degree carbocation we have, Sanjana? SP3 hybridized. Where we have more hyperconjugating structure in C? How many alpha hydrogen we have there in option C? Where we have maximum number of alpha hydrogen? Yes. The maximum number of alpha hydrogen is present in A, correct? So in option A, we have maximum number of hyperconjugative structure possible. And that's why option A is the most stable carbocation due to hyperconjugation. Is it clear? So number of alpha hydrogen present here is five. Here number of alpha hydrogen is four. Number of alpha hydrogen, here it is zero. And the number of alpha hydrogen, here it is three plus one, four. Okay, so maximum number of alpha hydrogen present here, that's why maximum hyperconjugation and hence most stable. Understood? Next question you see, on which would give the most stable carbocation on dehydration? Okay, tell me. Most stable carbocation on dehydration. See, third option on dehydration, we are getting what, three degree carbocation? Yes, we'll get tertiary carbocation here, three degree carbocation on dehydration, one degree. And here we get again one degree. Tertiary carbocation is most stable, so hence option C, okay? Next. Following graph, the stability of different carbocation have been shown, potential energy you have to match. Okay, all of you are getting C. Why is it C? Why first one is the most stable? See the simple thing is the most stable carbocation will have the least potential energy, right? So graph C belongs to the most stable one and graph A belongs to the least stable one. Yeah, so the stability of carbocation, if you see in these three molecules, stability of carbocation that follows the order, first one is most stable, then we have second one and then we have the third one, least stable. Yeah, right. So stability order is this, hence the order of potential energy, more stable one, least potential energy, order will be reverse, right? So third one will have the least, then we have second one, we'll have the max, then we'll have second one and then we'll have the first one, okay? So the maximum potential energy for A, so third belongs to A, second belongs to B and first belongs to C. So one is C, second is B, third is A, so option C is correct, right? So all of you have got this. Next question you see, the crazy order of basic characters of the following. First answer, seventh one is A. Why is that A? The first thing, and from this only, you can write down the answer for this question. This molecule is what? Pyro, right? We have a double bond here and double bond here, right? So for this molecule, the lone pair is delocalized, right? This lone pair is involved in resonance. So since it is involved in resonance, so its basicity is minimum, minimum basicity. Why minimum? Because this is also, it can donate this lone pair also. This also can donate lone pair of electron and this also can donate because it is there in the P orbital which is perpendicular to the ring. It is not involved in resonance. This is Pyridine, right? So in Pyridine, the lone pair is not involved in resonance but in Pyro, the lone pair is involved in resonance. So hence if you compare all these four, this one should have the minimum basic character, okay? So second one is minimum which is only there in the first option. So this one is, right? Third one is C, why this third one is maximum? Because both these nitrogen can donate lone pair of electron. Yeah, that's what I'm discussing. Both these nitrogen atom can donate lone pair of electron. If you compare this, this lone pair is involved in resonance with this ring, right? And here also it is there, it is involved in resonance but whenever you draw the resonating structure here, one nitrogen atom will always have the lone pair on it. That's why this is slightly more basic than this. Understood? And how is one less because then four? You see this is Pyridine, right? This is NH2. Actually you see what happens if you compare these two. If you compare these two, this one will be more basic than this, okay? And similarly if you compare one and third also, third one is more basic than first because it simply has two nitrogen atom and both can donate lone pair of electron. Both has the capability of donating lone pair of electron. But if we compare one and fourth one, right? Here it is Enylene and here it is Pyridine, okay? So this is again, since this lone pair is again there in the P orbital perpendicular to this but this lone pair in Enylene, you see this and we have Pyridine. This is the lone pair. So this lone pair is involvement resonance, correct? And this is not involvement resonance. So if you consider only this, so first one should be more basic than fourth, okay? So first one should be more basic than fourth and second one is list. So you're not getting any option. Just once again, this is Enylene and this is Pyridine. It comes over here. If this is involvement resonance, so obviously the basicity of the first one, third one should be maximum and first one should be more than that of fourth one. And then we have the second one. I think this should be the order. These two we have to interchange, okay? Here it is involvement resonance and this is not involvement resonance. Why third one is maximum because we have two nitrogen atom which can donate lone pair of electron. Second one is minimum because it is involvement resonance, okay? We'll see the next question. This one you tell me, and the following order of increasing acetic strength. I have done this kind of question also, x, y and z. Tell me, which one will have, what is the order of acetic strength? So on the other hand, I mean it's getting A, automation versus it's getting B. Why it is B, automation versus, see, I just tell you. First of all, the carboxylic acid is the most acidic here, right? So x would be maximum because of the carboxylic acid. We'll have maximum x. Now both are amino acid here, right? So the effect of this, which is plus R effect, electron, sorry, minus R effect, electron withdrawing nature of this group. This group has electron withdrawing nature, right? Because of this group, the effect of electron withdrawing nature of this group is more over here than here, right? And hence, the removal of H plus ion from this part of this molecule is more easier, and hence the order should be xz, and then y, is clear? Because of electron withdrawing nature of this group, the effect of this, however, we cannot say here that it shows minus R or plus R effect. But since this is the bond with which the key C double bond O, OH is attached, right? So you understand like this, here we have NH3 plus NH3 plus. If you see, these two are electronegative elements, so all these bond pair of electron is shifted towards the oxygen atom. Carbon becomes slightly positive, and when this carbon becomes slightly positive, it will drag few electrons from this bond pair, right? And hence, the effect of this group is more over here than this, and hence, Z is more acidic than y. Clear? Next question. The increasing order of pkb value of the following, a, b, b, most of you are getting b. Okay. pkb value is inversely proportional to kb, and hence, it is inversely proportional to the basicity. So we have to find out the reverse order of basicity here. So which one is the most basic compound? CH, see, this carbon is 1, 2, sp hybridized. This is hydrogen, this is NH2 minus and CH3. So if I find out the conjugate acid of this, that will be CH4, NH3, H2, and this is acetylene. Right? So this is the, in all these, if you compare, this is the strongest base, right? This is the strongest base and least acidic. So it's conjugate base, least acidic. So it's conjugate base will be weak, but strong base. Weak but strong base. So if you compare this, this is the most acidic compound we have, correct? Since we have to find out the reverse of basicity order, basic order, then we're trying to find out which one is most acidic. That is nothing but the order of pkb. So whenever you have to find out the pkb order, you just find out the acidity order of this, acidic order. That will be the reverse of basicity, right? So which one is the most acidic compound in all this? Can you tell me? Tell me the acidity order. Try once, what is the acidity order of these compounds? No? Okay, you see here, acidic order you have to find out. That will be, obviously this NH2 minus will be more acidic than this first one. So what we can say, the third one, I'm comparing just one by one, third one should be more than one. Third one should be more than one. And fourth is also second and third if you compare. Second and third, right? Hydrogen atom is, Hydrogen is not that much electronegative, right? So this third, second will also have the more acidity than the first and the third one, right? And if you compare the fourth one and the second one, obviously this fourth one will be more. Okay, just a second. This is the order we have. We can surely say that the basicity of this this is more than is less than the NH2 minus. Basically the PKB of NH2 is more than to that of this. Means third one, basicity of NH2 is more than this. The third one is PKB of NH2 is less than the first one. Oh yeah, just a second. I've done the wrong thing here. Just a second. This one is not right. Okay. We can easily say that the basicity of NH2 minus is more than this. So PKB of NH2 is less than. Okay, so third one should be less than the first one. Third one should be less than the first one. And then the order will be reversed. I've given the reversed order. There's a first, third, second and fourth if you compare second and fourth, which one is more acidic, second and fourth, can you tell me which one is more acidic? Second and fourth, you tell me. Fourth is more acidic than second. Fourth is more acidic than second. Okay, so second and fourth. Yeah, right. So fourth is more acidic than second. Okay, now you see third one should be less than first and fourth one should be greater than two. So if you compare this fourth one greater than two, we are not getting any option here. Oh, there's something missing. He has given third and third here. Okay, I'll write down my answer. Okay, whatever I'm getting. The basicity of first one, PKB of first one is more than third. PKB of this is more than third. And second and fourth, second and fourth. So CH4, so this should be first, third. And then we have, I think second will have then, and then we have the fourth one. Is there any option? I think third option we are getting. Yeah, it's the third one, I guess. You're saying fourth one is more acidic than second. Okay, we have to find out the basicity, right, PKB. So basicity, reverse of basicity means acidic order we have to find out. So acidic, which one is most acidic? Okay, you tell me what is the hybridization of carbon here? It is SP3, right? What is the hybridization of nitrogen here? SP2, hybridization of this, you let it be, and this carbon is SP hybridized. So you tell me one thing. Okay, you tell me one thing, that the conjugate acid of this, and this is the conjugate base of that as, all the conjugate base, if you add H plus all into this, right? So which one is most stable here? This one is most stable, the first one, because this carbon is SP hybridized. Okay, so this one is most stable, means the conjugate base, this is the conjugate base which is most stable, then third and fourth, I'm comparing first, third and fourth. So this one is most stable, right? So acid gives, strong acid gives weak conjugate base. So this one is stable, but it is weak conjugate base. Means if it is weak, so this should be more acidic. So if you compare the basicity of first, third and fourth, the order should be, it is for fourth one it is maximum, then we have third and then we have first. Yeah, that's correct, now you got it. Fourth one is maximum, third and first. And H minus is obviously more acidic than these two, third one and the first one. So the answer will be, it is maximum for fourth, then one, and then third and then first. So answer is this one is correct, whatever I've done initially, this is not correct. Understood this, see what I have done, that this H minus, the acidity of H minus is more than to this, NH2 and this, the first one. Okay, and how we compare this first, third and the fourth one that we can do according to the, you know, the stability of these bases, conjugate bases. Okay, if you see, we know this fact that strong acid gives you a stable, but weak conjugate base, okay? So this one is, if you see the stability, this one is most stable. The stability order will be maximum for this, then this and then this, right? According to this, if you compare, you'll get the order of base acidity will be this and then you can include second one into this, you'll get answer as this. The fourth one is the right answer. Yeah, you're also getting the same thing. Okay, next question you see, this one resonating. I have discussed the stability of resonating structure. Tell me this one. There are seven rules if you remember. Tell me quick. 10th one is A. Reason, tell me the reason. Charge separation you have to check. See, the difference in these two is what? We have, okay, I'll write on the charge, maybe it is not visible. See, this is positive, this is negative, positive, negative, negative, positive, positive, negative. Charge separation you have to check. See, what you can say that whenever the like charges present adjacent to each other, like charges present adjacent to each other, then stability decreases, okay? So if you see here, all the molecule if you see, where we have charge present at adjacent atom, one, two, three, four, see continuous charges are present. Here also we have the same thing and compared to these two, here the charge separation is there, but here the charge separation is large, here it is less, means charge are placed very close to each other, right? So obviously if you compare all these four, these two must be less stable than these two, right? Charge separation we are considering. But in these two also you see, we have positive, negative charge present adjacent to each other and here we have positive, positive charge present adjacent to each other. Like charges will decrease the stability, hence the first one is the least stable compound, option A is correct. Got it? Yeah, next question you see. The correct order of stability of the following species. Eleventh one, which is getting D, D is right. Why you see? First of all, in this example you see, we have hyperconjugation as well as resonance. Since all these are positive charge, okay? And this is the lone pair of electron. So we have three plus three, six alpha hydrogen and resonance, okay? Here we have, we don't have oxygen present here, right? So we have only three alpha hydrogen and two alpha hydrogen here. But here we have three alpha hydrogen plus resonance. So first, in the first compound you see what all things we have. We have six alpha hydrogen and resonance possible, right? The second one, we have only three alpha hydrogen. Third one, we have three alpha hydrogen and resonance again possible. Fourth one, we have two alpha hydrogen. See, alpha hydrogen means what? We can have hyperconjugation possible. Alpha hydrogen means hyperconjugation possible. Number of alpha hydrogen, more will be the hyperconjugative structure, right? So in the first one, we have maximum hyperconjugation. Third one, we have resonance also there. Third one, we have resonance also there. But number of hyperconjugation structure is lesser than the first one. So first is obviously most stable. Then we have third one. Why third? Because with alpha hydrogen, we have resonance also. But in these two third and sorry, second and fourth, if you compare, second one will have more hyperconjugative structure possible. So we have two and then the last one is fourth. Hence option D is right. Correct. 23rd, which of the following molecules in pure form, unstable at room temperature? Why it is D? See, this compound, the third one, this is a conjugated system, contains only four pi electron, right? So it is anti-aromatic and hence it is unstable. We have four pi electron here also, but it is not a conjugated system. That's why this is not anti-aromatic. Okay, so answer will be option B. That's what four pi electron it has, but it is not a conjugated system. That's why it is not anti-aromatic. Got it? Next question. The hyperconjugative stability of tertiary butyl cation and two butane respectively are due to. See, you can do this question like this. Tertiary butyl cation means CH3, CH3 and CH3. This is tertiary butyl cation. So positive charge we have here. Now this positive charge means what? We have a vacant P orbital here, empty P orbital, right? Which form, which usually forms a sigma bond, but since it is empty now, so what we can say sigma P, empty. And there is only one option, empty sigma P. They have written two butane. I don't think it's two butane. It should be two butane, I guess. There is some mistake. Okay, let it be, I'll cross check this question, okay? This should be two butane, I guess. But I have to cross check, okay? So let it be this question now. But with this data, we can understand that in tertiary butyl cation, we have sigma P, empty orbital, right? P, empty orbital and empty P orbital we have there, which forms a sigma bond. So according to this information, we have only one option possible. That is option A. In these two, you see it is filled. Sigma, sigma star is not there, right? So according to this option, A should be there. But this is something missing over here. It should be two butane, I guess, but I'll cross check this question, okay? So let it be this question now. I'll give you another one. You do this one. Which is the following molecules in significant dipole moment? Yeah, we have done this in isomerism, right? That OH bond may be like this. Another possible is this OH, OH. So in this case, mu is equals to zero. In this case, mu does not equals to zero because of the orientation of this bond, right? So whether we have O or S, we have the same logic. For C and D, we'll definitely have dipole moment, so option B is correct. The logic is what here? For zero dipole moment, we must have linear molecule present here, like this cyanide group, okay? Monoatomic or linear molecule present, right? We have discussed this in isomerism, if you remember. Which is the following statement is wrong? Now it is alkane, B, upness is getting B, okay? So some of you are saying B and some of you are saying D. See, first of all, heat of combustion, two things you understand, okay? If for larger molecule, larger cycloalkane, I'll write down. For larger cycloalkane, the heat of combustion, the heat of combustion is more. So basically we say what? Heat of combustion per CS2 molecule increases, like this one, right? So this one is also right and this one is also right, right? If you see these two options, if you do not know anything in this question, you see this option A and C, both are same actually. So if this one is wrong, so this should also be wrong. And since it is the only one correct answer we have, so both cannot be wrong, right? So this is how you can eliminate option A and C. You understand my point? Both are actually the same thing, the logic is same to these two options. If this is wrong, this should also be wrong, but both cannot be wrong at the same time because the option has only one answer, right? So hence A and C you can eliminate, okay? Now, cycloalkane is a planetarity of that, if you see. In cycloalkane, except cyclopropane, this you must remember, except cyclopropane, cycloalkanes are nonplanned, are nonplanned. Generally, higher cycloalkanes are nonplanned, right? But cyclopropane is planner. Okay, this you must remember, this is again one information I'm giving you. Cyclopropane is planner and all other cycloalkanes are nonplanned. Generally, higher cycloalkanes are nonplanned only. So we cannot say cycloalkanes are planner. So this is the right answer for this question, right? It is not true for cyclopropane, which is given here also, you see, right? That's why if you remember, cyclohexane, the structure is this, oh, sorry. The structure is this, cyclohexane, chair form, right? And it is not a planner, structurally, okay? So generally, higher cycloalkanes are nonplanned, cyclopropane is planner in nature, okay? So option B is right, got it? This one, both alkane and alkene, tell me first. What is the answer? Cobreaction, yeah, it's correct. The correct answer is this, this cobreaction actually it is used for the preparation of alkane, alkene and alkyne, okay? It is used for the preparation of alkane, alkene, and alkyne. Williamson synthesis is used for the preparation of ether, if you remember. Wurz reaction, generally we get higher alkanes, symmetrical alkanes into it, okay? Sandmere reaction is used for the preparation of diso compounds, diso compounds. That's why the answer is cobreaction. It is the coal electrolysis actually, okay? It may give psilocyclic acid also depending on the reaction, reagent that you are taking, but there is a method, we call it as coal electrolysis. It is actually referring this reaction, coal electrolysis method. There we take the salt of sodium or potassium salt of acid, which gives alkane or alkene or alkyne. That one, psilocyclic acid is coal bisquit method probably, okay? It is coal electrolysis, right? Its answer will be option B. The structure of the compound with molecular formula, this, that has only two degree hydrogen atom. You have to draw all these structure, tell me. What is the answer? Wurz reaction, sandmere, it gives alkane, generally alkane, correct? And that alkane will be symmetrical alkane. See, actually, Wurz reaction, say in Wurz reaction, generally we take Rx plus two Na plus Xr. This alkyl group will generally take same alkyl group here, right? And that gives higher alkane, Rr plus two Nax. If you take R dash here, then we'll get mixture of alkane. R dash, R dash, we also get R, I mean, these three mixture of compounds will get, we'll not get alkene into that, right? So if you take different alkyl halide, you'll get mixture of alkanes. But if you take same alkyl halide, you'll get only one alkane, right? Higher alkane. Generally, Wurz reaction we use for the preparation of symmetrical alkanes, higher alkanes, okay? Yeah, if you have, if R has a double bond here, then you'll get alkene also over there, right? Depends on what you are taking. But generally, Wurz reaction, see, Wurz reaction, we never see this reaction for the preparation of alkene. We see this reaction for the preparation of alkanes always. So if it is not mentioned, we consume this as a preparation of alkane itself, not alkene. This R can be anything. You can take CH2, double bond CHCH3X, this you can take, right? In that case, you will get alkene, right? But that will be alkene only, not alkene, understood? This plus 2NA plus XCH3, these two are different, first of all. Then you may get mixture of alkene and alkene. But generally, Wurz reaction, we don't take different alkyl halide. That is why we are not considering this as an answer. If the question is multiple choice, more than one correct, then we'll take B and C both in that case, right? Tell me, what is the answer for 17th one? By looking at the option, you can say what is the answer and this is very basic. Yeah, 17th is B, because cyclohexane, all carbon atom is sp2 hybridized, this cyclohexane. All carbon atom is sp2 hybridized, contains only secondary hydrogen. So option B is correct. You can also draw the structure of other molecules. You can understand. Okay, if you want, you can draw, but I don't think it's required. Very basic thing. You can draw the structure and you can write down the answer. Next one is C29th. What is the answer? See, first of all, we have COOH. So you will get here the salt of this, COO minus COO minus and K plus salt of this, right? When you do the electrolysis of this, electrolysis here, right? Then oxidation takes place at anode and you'll get two different product here. One is this, we have a double bond and the other one is this, we have a double bond here, right? So these two are actually cis and trans isomers. So we'll get here geometrical isomers. K2CO3 goes out and then at anode, oxidation also takes place and forms double bond here and that will be the geometrical isomers. Is it clear? First of all, with this solution, acid, you'll get a salt of it and when the electrolysis takes place, K2CO3 goes out, okay? Oxidation takes place and we'll get a double bond and that will be the mixture of two isomers that is cis and trans we get. One more thing you think here, since it is only one option is correct here, when you look at this, you can easily eliminate option B. Enantiomer is not possible into this, right? Enantiomer is not possible, this you can eliminate easily. Now the positional isomer is nothing but the structural isomers also. So position comes under structural isomers. So if this is correct, this should also be correct, right? So you can also eliminate these two options accordingly, like this, if you think. Once you know if the answer has only one correct choice, then enantiomer you can opt in it because it does not form an enantiomer. However, it is a, what we can say, we are guessing it simply, but by looking at the molecule, you have this idea that it will never show any optical activity. So enantiomer is not possible into this and positional structure since both our position comes under structure only, so we can eliminate these two options also. Hence option A will be the more appropriate choice we have here. If you do not know this, then like this you can eliminate few options, okay? Next question you see, question number 30, which is the following reactant is suitable for the preparation of methane and ethane by using one step only, one step reaction you have to write down. Nineteenth one is C. What is the method we can use for this? Okay, C is correct. You can obtain alkene from this. So what are the possible method we have with this? Reparation of alkene from alkyne light. Wurz reaction we can use and we can also use reduction, reduction method in presence of zinc with acid. If you have RX and Wurz reaction gives you RR and here we'll get RH, right? Which is the following reactor is stable for the preparation of methane and ethane. Yeah, we'll not get methane from Wurz reaction, but we can get what? We can get this methane and ethane here from this reduction method. So Wurz reaction gives you higher alkene, right? So we'll not get methane over here, but here we'll get the Wurz thing, right? So in this reaction, okay, answer is C alkyne light. Highest boiling point, highest boiling point is expected for. This question is asked in JEE and octane, correct? Normal as branching increases, boiling point decreases, okay? As branching increases, boiling point decreases. All molecule you see, they have equal only eight carbon atom, all these molecule except this N-butane, right? So obviously as the molecule formula, mergers mass increases, boiling point increases, and when branching increases, boiling point decreases. With these two logic, option B is correct and octane, okay? Next, this one, okay? You see, first of all, this is an alkyne, okay? And cyclopropane, cyclopropane is highly unstable. This is highly unstable because of angle strain, high angle strain, right? So with H2 and platinum, this will completely, this will completely hydrogenate it. This will get the hydrogenation takes place here completely and it converts into octane, that is normal octane. Why octane? Because we have 3, 3, 3, 6 plus two carbon here, we have total eight carbon. So with H2 and Pt, complete hydrogenation takes place. Must remember this, complete hydrogenation with H2 and Pt. And this converts into N-octane, normal octane. So answer could be anything, B, C or D, anything between these two. Now here we have H2 with P2 catalyst, okay? This actually, hydrogenation takes place, but this will reduce this triple bond into double bond, right? Triple into double bond and we'll get cis form over here. Cis form will get over here, but complete hydrogenation does not take place here. Only one hydrogen, one pair of hydrogen atom will attach onto this carbon atom. So this triple bond converts into double bond and the answer will be this option D. So basically this P2 catalyst does not give complete hydrogenation, okay? And it converts triple bond into double bond. Option D is correct, okay? These two reactions you must remember, especially this P2 catalyst. Understood? Next question, this one. Do this is the last question for today. Easy one, try this. Oh yeah, yeah. Done. Is it C? Simon is getting C. Others tell me, what is the last step reaction Simon? That is from D to E. What is that reaction called? You see this, I'll lose this one. First of all, the first reaction is what? It is Wurz reaction, right? Naether. So the product we get here is this. The first step, it is B. Now, in this, if you do the chlorination of this one, two moles of this, chlorination with H nu, see this halogenation of alkene is always syn addition. Okay, must remember this. The two chlorine atom attached from the same side, Cl and Cl. Halogenation of alkene is always syn addition. Why it is syn? I'll just tell you one intermediate product here. The intermediate product, you'll get a cyclic halogenium ion here. Cyclic halogenium ion will be like this. Suppose we have X here, X, and this is like positive charge. This is a cyclic halogenium ion we get, okay? Okay, fine, fine. So it's not syn, it is trans addition actually, sorry. Because of this cyclic ion, what happens? The other halogenated ion or halide ion attacks from the opposite side, and this bond breaks like this, okay? And then halogen attached like this from this side and this side like this. So whatever it is, this is not a 3D representation. Just we have one thing here, that two chlorine atom attached at this position. You can draw this chlorine atom this side also. That would be actually better. This chlorine here and this chlorine here, because it is a trans addition, not anti addition, not syn addition, okay? So we'll get diachlorinated product here. Now in this what happens, next thing, when you are adding alkoholic k-wet, so two moles of alkoholic k-wet, we know it gives what? It gives alkene, right? So double bond we'll get here, double bond we'll get here. So we'll get this as the product here. Just 2KCl goes out. Now the last step we have what? That this reacts with ethene, CS2, double bond, CH2. Now this reaction is what? This reaction is, we call it as dials-elder reaction. Now this C double bond C, we can also write it as this. This is C double bond C. Now what happens here? This pi electron attacks onto this carbon. This pi electron goes here and this pi electron comes onto this carbon. And finally this carbon and this carbon joins. So we have a double bond here with two methyl group and one, two, three, four ring like this. So the product we get here is this. This is two methyl group and we have a double bond here. This is the answer we have option C. Last step reaction where we have this, let me draw this. You have a compound which has four carbon atom, conjugated compound, four carbon atom, plus a compound has two carbon atom, double bond. This reaction is dials-elder reaction. And in this, like I discussed over here, this pi electron jumps over here. This pi electron comes over here and this comes over here. And finally, since this is positive, this is negative, these two also joins. So we get a ring structure over here and the final compound is this. The name is dials-elder reaction. Did you understand this? Allogenation of alkene, you must go through. Allogenation of alkene. Alkene if you have, correct. Alkene if you have, that gives you trans addition of halogen. Okay, we'll discuss this later on in the next class. Claire understood. Okay, so we'll see you in the next class. This is almost done. I think three, four questions are left. We'll probably solve that also in the next class. And then we'll solve some other chapters, right? Okay, thank you all for joining. See you soon, bye.