Hello, good afternoon. Guys, please type your name in the comment box. Today, we are going to discuss some questions of alkyd halide from organic chemistry. The first question we are going to discuss today is this one. Tell me the answer of this question. Tell me the answer. What is the main product of the reaction of 2-butene and chloroform? Okay, see I'll do this question. Okay, so the question is 2-butene, which is CH3, CH double bond CH, CH3, when reacts with chloroform. Right? So this reaction, it simply gives the addition of H plus and CCl3 minus as a nucleophile here. NaOH also, this reaction takes place in presence of NaO, which that is also it is understood. It actually forms CCl3 as a reagent. So here in this reaction, the product will be CH3, CH, CCl3 will attach over here, then CH2, CH3. Well, the first reaction is this. Now, when the base that you have here NaOH, and we actually take here three moles of NaOH, because we have three chlorine here. We'll take three moles of NaOH that will form three NaCl. So we'll get CH3, CH, NaCl forms and we'll get C, COH, coltris here. And this will be as it is, CH2, CH3. Right? Now, there is one thing which is missing over here. We should heat this also. Right? This will heat also, generally in some, most of the time, we'll take this heat. But whenever this carbon contains three OH groups, this is a bit unstable. And hence, what happens, one of the H2O molecules comes out from here at some temperature. So this H2O will come out from any of these three hydroxy group, OH, H, OH, H, or like that. So H2O goes out and the product here will be CH3, CH, CH2, CH3. And here we'll have COOH. This is the final product we get. Okay. So answer here really what? 2, 1, 2, 3, 4. Okay. 2 methyl, 1, 2, 2 methyl, deutenoic acid option B. Alkene, whenever reacts with chloroforms, it is an addition reaction always. H plus and CCl3 minus will get this product with any OH, any Cl goes out and OH will attach over here. And this structure is highly unstable. One H2O molecule goes out and we'll get this methylmethic acid. Next question, you see. Repression of alkene halide in laboratory is less preferred by, why it is D? Why it is D? Because halogenation of alkene, this follows what mechanism? A halogenation of, I'll write down here. Halogenation of alkene, this follows free radical mechanism, free radical mechanism, which is very difficult to control. Right. And that's why we get mono substituted product, di tri tetra. Okay. So that is why this direct halogenation of alkene is the least preferred method for the preparation of alkali light because of free radical mechanism will have mono di tri tetra substituted alkene product you get. Okay. So hence all of you have done this. Option D is correct. Third one, you see. Tamtrin, podric, cyanide is getting here. Why A is correct? Why A is the answer? Tell me the reason. Okay. AGCN is covalent. The carbons of the nitrogen attacks. AGCN is covalent. So N develops slight negative charge. The carbon in the carbon. So N attacks N, not the carbon atom. N attacks N the compound, not the carbon atom. Correct. Okay. You see this AGCN that you have. AGCN, it shows covalent nature. Or we can also say the greater covalency of silver and carbon bond. Right. So here what happens, N is the donor atom in case of AGCN. Right. In KCN what happens, CN minus exists as a free ion. Right. In KCN I'm talking about, CN minus exists as a free ion and better donor carbon forms the bond. Right. So that's the thing over here. So here you see the reason behind this is first of all, I'll write down the greater covalency of, covalency of silver and carbon bond. In KCN, CN minus exists in, exists as free ion. Exists as free ion. So that is why it happens. So the question is iodine from iodine ethane can be substituted by cyanide group in a number of ways. Which is the following is true statement. Okay. So with AGCN find ethyl isocyanide that you have while with KCN, ETCN is found as a major product. Okay. So that is what we have. This is getting replaced by the cyanide group and ETCN forms over here. So here you see option B that you have AGCN, ETCN, ethyl cyanide and with KCN ETNC form. This is not correct ETNC we don't get here. With either KCN or AGCN. Okay. Point here is what AGCN is true because of greater covalency of silver and carbon bond. But KCN it is not true. Right. It is true when this KCN reacts with ethyl isocyanide and then ethyl cyanide forms. Okay. Like you said, this is given directly in NCRT also. So this thing you have to memorize. Option A is correct. All of you have done this. You must have created somewhere in NCRT. Okay. So this kind of question actually you have to memorize. Okay. There is if you haven't memorized it, you cannot do this kind of question. Okay. You'll get confused in the example. Option A is correct. Greater covalency of silver and cyanide and in KCN CN minus exit. So reaction will be sober. Next question you see question number fourth. Hydrocarbon has molecular mass 7.2. It gives a single monochloride with two dichlorides on photochlorination. The two chlorides are tell me the probable formula. What is photochlorination? Chlorination in presence of light sunlight. Correct. Now tell me the formula here. What is given the question is 7 point. Oh, you see the options. When you see the option, I don't think 7.2 is correct. It should be 72. Do this correction. The molecular mass here is 72. Then this is the easy one. I think you can do this. There's nothing to memorize here. Okay. So answer that you're getting is C. Okay. So first of all, you see the question says a hydrocarbon has molecular formula 72. It gives a single monochloride and two chlorides on photochlorination. Right. Two chlorides are correct. So you see first of all, we assume the hydrocarbon molecular formula is CNH2N plus 2. Now what is the molecular mass of this? 12N plus N plus 2 into 1 and this should be equals to 32. So 14N minus 270. So N is equals to 5. So we are getting the number of carbon atoms are 5. So the molecular formula of the hydrocarbon will be C5H12. So it is a pentane. Right. Now the question is it gives a single monochloride and two dichlorides on photochlorination. Right. The single monochloride is possible only when it has neo-eskeleton. Right. Which is CCH3, CH3, CH3 and CH3. In this, all these coefficients are equivalent. This gives you monochlorinated product. Okay. So I think according to the option you go CL2, CL2. This is the one of the option we have. I think option C is correct here. Other structure, this gives you monochlorinated product. Other possible structure is what? If I write down N pentane, in this you tell me how many product possible, chlorinated product in this structure? Tell me this one. In this structure, how many chlorinated product we get? Three. Correct. And how it is three? One, two and three. What if I write down this structure, this one? Again one position is this, other one is this. Any other structure possible? Option you see here, these are the options given. CH, CL2. Okay, fine. Actually I was thinking something else. Anyways, anyways, right. This is one dichloride. Another dichloride is what? CH3, CH3. Correct, correct, correct. This is actually not required. These are not required here. This is not required. These two things are not required here for this question. I was thinking something else. Anyways, you see with this value, we have got the value of N. So the formula is pentane. So when we draw this structure of pentane, we'll get monochlorinated product. And in this one, if you have to find out the dichlorides, so again, there are two possibilities. Both chlorine atom will attach on the same carbon. So we'll get this one. When both chlorine atom will attach on the different carbon here and here. So we'll get this one. So this is what two dichlorides we get. Hence the possible answer will be option C. In this kind of question, you have to go with through the options. Then you will have a better idea. Correct. The question is about dichlorides. Means in one molecule only two chlorine atoms should be there. Yes, can we move on to the next question? This one. One mole of hydrocarbon A reacts with one mole of bromine to give a dibromo compound, which is this. A on treatment with cold dilute alkaline solution forms a compound. This or ozone is correct. Do this one. This is the city's reaction. This one is this compound. This compound is C5 H12 O2. And here it is C6 H10 Br2 C6. Sime here is getting A option A. Okay. So some of you in Sanjana are getting A only. So first of all, you see the information that is given here. One mole of hydrocarbon A reacts with one mole of bromine gives a dibromo compound. So what could be the possible compound? What is A here? General name you tell me. Is it alkene, alkene, alkyne, what it is? Alkyl halide, what it is? When A reacts with one mole of halogen, that is bromine, forms a dibromo compound. What is A? You see, first of all, I'll just give you a slight information of this alkene. See, if you have C double bond C and this one reacts with bromine, in this anti-addition of bromine atom takes place. Anti-addition why? Because this bromine converts into Br plus and Br minus. Okay. And this and this pi electron attacks onto this Br plus. So what happens here? We'll get C single bond C and here we'll get one cyclic halogenated compound. This we call it as cyclic bromonium ion. Bromonium ion. Now because this cyclic compound here, the Br minus which attacks onto this, this Br minus which attacks, it always attacks from the back side. It cannot attack from this side because here is a cyclic compound. And when it attacks over here, this bond will shift onto this. So the product we get here is this. So addition of halogen always takes place from the opposite side on alkene. This is the first thing which is not actually required here in this question, but this information I'll give you. Right? So what happens when hydrocarbon A reacts with one mole of an bromine gives diabromo compound is C6H10Br2. Right? Which on treatment with KMNO4 solution forms a compound this, which is on ozone analysis, like this A on ozone analysis gives equimolar quantities of propanol and ethanol. So first of all, I'll go from the bottom of the from the last like the compound that we are getting here. Right? So we'll start this question from reverse. Right? So propanol is what propanol is nothing but this C double bond O CH3 CH3 and we get what? Ethanol. Ethanol is CH3. Again, there is a mistake here. This should not be ethanol. This is ethanol. Am I right? See, ozone analysis of alkene gives you either aldehyde or ketone. It won't give alcohol. So this should be ethanol. So ethanol will be this O double bond C CH3. Correct? Anyone of you have doubt in this ethanol and ethanol? Right? Addition, ozone analysis of alkene always gives you either aldehyde or ketone. So we cannot get alcohol here. It should be ethanol. So this is ethanol and this is propanol. So can we write down the possible structure of alkene here from these two ozone analysis product? And that structure will be what? This will take out and this carbon and carbon joins to the double bond. So C double bond CH CH3. Here we have CH3 and CH3. This is the possible structure of alkene. Hence the compound will be what? Option A. You don't have to do anything with this KMLO4 and all. Understood? Right? Now, for this, you know, for this, for learning purpose, we also do this the second part also. When this is allowed to react with KMNO4, KMNO4, which is alkaline KMNO4. And what is this reagent? This is Bayer's reagent, right? Bayer's reagent. And this gives diol and both OH from the same side it attaches, right? So the product here it will be CH3C CH3 OH single bond COH H CH3. This is the product we get, okay? So hence we have done the answer. Answer is option A only. But when this reacts with Bayer's reagent will get diol. Both OH will attach from the same side. We'll move on to the next question. How will C6 be formed when there are only five carbon atom? Where is C6? Where is C6 it is given? Oh, in the question, first sentence, right? Then let me cross check the question. Maybe I think it should be C5 only. Let me cross check the question first, just a second. Yeah, no, it's C5 only. So it's a mistake here. This is not C6. It is C5. Okay, that's a mistake. This is not six. It is five only. Okay, so all everywhere we have C5, not C6, right? So it's a mistake, a printing mistake. It is it cannot be C6. Okay, next question number six. Tell me the answer. Why are you getting A? You see it reacts with NaOH in the last. If suppose COH is there also, which reacts with Na, H2O will go out and we'll get the salt of its CONA. So you see, toluene reacts with excess of Cl2, CH3 reacts with Cl2 and it forms, first of all, we get CH2 Cl and HCl forms. But since this is present in excess, right, so we end up getting Cl3 over here, right, which reacts with three moles of NaOH, we'll get three NaCl out. And again, this carbon atom will get COH whole thrice. Again, this is the same thing that we have already done. COH whole thrice finally loses H2O molecule and we get benzoic acid, COOH, right? It is allowed to react with NaOH. Again, H2O goes out and we get sodium salt of benzoic acid. Answer will be option B. Option B is correct. Next question. Ramjanan is getting A. See, NaOH is given in the reactions I made in the last step. When H2O goes out, it is given when it finally reacts with NaOH. You see the previous one, it is given sunlight to the product which on hydrolysis followed by the reaction with NaOH. Hydrolysis followed by the reaction with NaOH. So this is it. Hydrolysis and followed by the reaction with NaOH. All of you are getting different answer in this, A, B, D. What happened, guys? Tell me. Give me one answer.  So three are saying A, two are B, one is D. Hans Degel reaction. Okay. Double bond O. When this reacts with CH3MGBR, this is minus and this is plus. So this negative charge will attack onto this carbon atom and this pi electron goes here, which takes this MGBR. So the product here we get is CH3OMGBR, which on acidic hydrolysis, H plus H2O, it is given. CH3MGBR, H3O plus. It forms what? OH will get here. This H plus will come over here. So here the product will be CH3 and we'll get here OH. Now in this compound, you heat with H plus. So this H plus will come over here. We'll get what? OH2 plus. And finally, H2O will goes out. So we'll write down OH2O minus H2O. And here we'll have the positive charge with one CH3. Just a second. We have H plus and H2O. We have H plus in heating. Now what happens here, this H2O goes out and from this also at alpha position, at any of these alpha positions, here we have C, here we have two hydrogen, two hydrogen and three hydrogen. So from alpha position, H plus also goes out and we'll get a double bond. So we have two possible product here. If you write down this one, this is one possible product and the other possible product is this CH2. So obviously this one is the more stable one. Hydrogen comes out from the carbon which has lesser number of hydrogen. So that's how we get an alkene over here. Now this is the most stable compound here, comparatively more stable compound. When this reacts with HBR and H2O2, this gives what? Addition of HBR in presence of peroxide gives you anti-Marconic of addition, anti-Marconic of rule. So according to that, H and BR minus will attach and BR minus will attach over here and we'll get CH3. Hence, option A is correct. If you take only HBR, then this BR will attach over here. This BR will attach over here in case of only HBR. Since peroxide is there, so anti-Marconic of rule is followed and we'll get BR over here. And we know this anti-Marconic of addition, which we also call it as, here also there's one thing that is peroxide effect or carousel effect, right? That is only possible in case of HBR. That also you'll keep in mind. Okay. Hence, answer option A is correct over here. Question number eight, Sondaria is getting C, Ramsundan is getting C. Okay, all of you are getting C. Option C is correct. It is just the chlorination reaction in presence of HCl3, right? And we always get ortho-parasubstituted product here. See in alkythylide, all the reaction with that is stratocraft alkylation, acylation, sulfonation, halogenation, nitration, all these reactions always gives ortho-parasubstituted product. Okay. And that is why here also we'll get the mixture of ortho-paraproduct. Hence, ortho-paracluoro toluene is the answer. Understood Purvik? C is correct. It is ortho-paradirecting group. It is ortho-paradirecting group. That's why we'll get ortho-parasubstituted product, right? Ortho-paradirecting group. That's why we'll get ortho-parasubstituted product. Okay. This we call it as, see the reaction of this, which is the, you know, actually it is a preparation method of aryl halide. Okay. This reaction actually follows electrophilic substitution of arenas. Okay. Electrophilic substitution, substitution. Where we'll get ortho-parasubstituted product, which is, takes place, if I write down the exact reaction here, the toluene CH3 plus X2. Now this X can be chlorine bromine or iodine. This is when heated with ion in dark or hydrous FeCl3, anhydrous FeCl3. Then we'll get ortho-parasubstituted product and the product here it will be, this CH3 will be as it is, X will have here or this X may attach at para position, CH3 will be here as it is. This is the two product possible here. Okay. Next question you see. Do this one. It is again a factual question. Then what is the answer? Okay. Some of you are saying D and most of you are saying D. Why is it D? Any one of you? Why both assertion and reason are incorrect? Phenyl chloride is more preferred, better reactions than obviously reason should be also be incorrect. Okay. Actually you see this phosphorous chloride that you have, especially this phosphorous trichloride is slightly basic in nature and phosphorous pentachloride has complex structure. Okay. The structure is not simple and that is why and we always study the, you know, the preparation of alkyl chloride is always better or it is the better way always to use thionyl chloride as a Cl2 for the preparation of alkyl chloride. Okay.  That is also a fact. That's why it is a factual question. Okay. Yeah. Right. Correct. So, so that's why we always preferred thionyl chloride for the preparation of alkyl halide. Okay. Over phosphorous trichloride or pentachloride. Correct. So the answer in this question is option D, assertion and reason both are incorrect. Okay. Option D is correct. Next you see question number 10. Phosphorous chloride gives pure alkyl halide. Right. Phosphorous halide if you use, it also forms POCl3, POX3. I have done this reaction in the class. 10th one, all of you are getting A. It's correct. Both are correct. Assertion reason. Yeah. I think it's Wood's reaction. So tertiary, deutyl bromide gives you 2-2-tetra methyl. Correct. Assertion is correct and Wood's reaction alkyl halide with sodium, dry sodium and dry ether to give hydrocarbon containing double the number of, yeah, correct. Option A is correct. I don't think we need to discuss this. Okay. Next question. Question number 11. Highest melting point. 11th, all of you are getting D. Why it is D? D answer is correct. Para dichlorobenzene. Answer is right. What is the reason behind this? Packing efficiency. You can say para dichlorobenzene has symmetrical structure and hence better packing. You see, chlorine, chlorine. With these three, orthometa and para, you cannot differentiate on the basis of the molecular mass, right? But since here we have symmetry, better packing, and hence the highest melting point, right? Paraporation more stable or you can say better packing, symmetrical structure, hence highest melting point, whatever you can say like this, any of this reason. Option D is correct. Question number 12. All of you are getting C and we know as branching increases, boiling point decreases. More branching, lesser will be the boiling point. Okay. So third one will have minimum boiling point and second one will have maximum, so option C is correct. So boiling point, inversely proportional to branching, right? Next question you see. 13. We are following compounds increasing order in their density. 13 is A. All of you are getting A. Reason behind this? Why it is A? Because the molecular mass, right? More mass, more density, correct. So fourth one will be maximum, then third, then two and then one option A is correct. You see, this is very easy since they have given two halogen, two halogen one and all like that. Okay. But if you have iodide, bromide, chloride and fluoride also present, then only for alkyl halide, the density follows this order. I minus for iodide it is maximum. Again the same thing of mass. Then we have bromide, then we have chloride and in the last we have fluoride. This is also the order of density. Question number 14. Isopropyl fluoride plus A on heating gives two ethoxy propane. The compound A is what? Find out this. Answer, you can check it with options. 14, again all of you are getting D. C2H5ONA, right? So I will try to write down the reaction here. Isopropyl chloride. Isopropyl chloride is this and here also we have CH3. Isopropyl chloride, if it reacts with, first of all, the possibilities either B or D because it contains NA. So I will first check, it's very simple one. NA, C2H5. This forms NaCl, it goes out and the product here it will be CH3, CHO, C2H5 and here we have CH3. Right? So this is ethoxy propane. Hence option B is correct. Ethoxy group is attached at second compound. So two ethoxy propane. This is what the name. So I am here and Mashravi is getting C. Sondarya resist. Okay, again you are getting C and alkaline hydrolysis of a tertiary halide by aqueous alkali. If concentrated of alkali is, if the concentration of alkali is doubled, the reaction rate. Okay, what is the mechanism? It follows tertiary halide with aqueous alkali. What will be the mechanism? Since you see the halide is tertiary, so what we can say that tertiary halide follows SN1 mechanism. SN1 mechanism and we know SN1 mechanism is independent of the concentration. Right? It is independent of the concentration of solution. Right? So whatever concentration you take, its reaction rate will not change. It will be same. Right? If this halide is primary halide, in case of primary halide, the reaction follows SN2 mechanism. And when SN2 mechanism is there, it depends on the reaction and hence the rate also gets double. If the question was this, with primary halide, rate also becomes pulse. Option C is correct. Next you see question number 16. 16 is Sondarya and Rithvik is getting A. Sanjana is also getting A. Which halide will be least reactive with respect to hydrolysis? And why is it so? Why benign fluoride is least reactive? Why benign fluoride is least reactive? The positive charge lies on more electronegative carbon atom. Is the screen is visible continuously? Or you are getting some connection problem. Is it there? Okay, you see you are I think, so you see benign fluoride is this CH2 double bond CCl. Now here what happens, the lone pair of this chlorine is involved in resonance. It is resonance stabilized, delocalized. We get the structure as CH2, CH double bond CCl, positive and negative. Okay? So since this lone pair is involved in resonance, this bond has partially double bond characteristics and hence it is least reactive. Option A is correct. Okay? Next question you see question number 17. This one, the product of which of the following reaction on dehydration gives 17 is A. All of you are getting A. See 1,2 dibromethane. When you use KCl, KBr forms and you will get Cn attached with the two carbons. Dioxymate, which on hydrolysis gives acid. Done. What happened? Option A is correct. This reaction we have also done in the class. 1,2 dibromethane. 1,2 dibromethane. Correct. So the molecule is this CH2Br and CH2Br. When it reacts with KCl, two moles of this actually we are taking. So you get CH2, Cn and CH2, Cn, which on acidic hydrolysis, H3O+, which on acidic hydrolysis converts into acid with same number of carbon atom. COH, CH2, CH, which on dehydration gives the compound, which is CO, CH2. This is the compound we get. Okay? This compound is, if you remember, it is succinic acid. And hence, option A is correct. Next question. Which of the following compounds will give resmic mixture on nucleophilic substitution by OH-? What is the answer? Ca, C. Which of the following compounds will give resmic mixture? Okay. Resmic mixture. Rethic C, Corbic C, Simonis C, Mrs. Trost C. Sanderah is getting D, Sanderah is getting A. Okay. So for this question, I don't know why you have, most of you have got wrong answer in this question. The correct one is A. And that's why option A is correct. Okay. Now you see why. Nucleophilic substitution reaction. So if the molecule is this CH3, CH, ET, BR, then the carbocation we get here is CH3, CH, ET, N+, right? This carbocation, you see, I'll write down other compounds as well. CH3, C, CH3, ET. And the carbocation here it is CH3, C, CH3, ET. And the last one we have is CH3, CH, C2H5, CH2, BR. And this gives you CH3, CH, CH2 positive charge, C2H5, right? Now the question is we have to form a resmic mixture. So what is resmic mixture? It is an equimolar mixture of the two DNL form of an optically active of a compound basically. When I say DNL, it means it is the optically active compound. So for optically active compound, what is the condition we have when a compound said to be optically active? What is the reason for this? When a compound is optically active? How do we check the compound is optically active or not? For an optically active compound, the compound must be what? The compound must be chiral and this chiral has nothing to do with chiral center, right? You have to check the chirality of that molecule and with chirality, this chiral is associated. It has nothing to do with to do with chiral center, right? So if there is no plane of symmetry, then the molecule is optically active or no symmetry, either it is plane of symmetry or center of symmetry. And the other thing is what? If the molecule contains only one chiral center or chiral carbon, then it is optically active, right? So there are two ways. If any compound you have to check optically activity or not, if the compound contains only one chiral carbon, only one chiral carbon, if it is there, fine, it is optically active or there should be no POS or COS in the molecule. This is the two things you have to check. However, let me repeat this again, that presence or absence of chiral carbon has nothing to do with the optically active nature of that particular compound. The necessary condition for a compound to show optically active behavior is the compound must be chiral, okay? And chiral molecule are those molecules which shows chirality properties. Chirality is the property of a molecule by which it cannot be divided into two equal half, correct? So now when you see all these molecules, a nucleophile you have OH minus. So this molecule is what? It is sp2 hybridized carbon atom, right? Which is trigonal planet, which is same for all these carbon atom, but this is not chiral carbon. So no chiral carbon. Here also this carbon contains two hydrogen atoms. So this is not the chiral carbon. This is not the chiral carbon because two methyl group are present. This carbocation you see contains what methyl group? One ethyl group and one hydrogen. Hence, this is a chiral carbon. And since it is planar, so OH can attach from the top or from the bottom, both side. So one side it gives D, other side it gives L. Hence, this one gives optically active compound, right? Which is nothing but the mixture of D and N, which is the rest of the mixture. Is it clear? Right? This question was very logical. If you have this, you know, knowledge of optically active compound, rest of the mixture and all, you can easily do this. Okay, so don't make these kind of mistakes. Something if you do not know some factual thing that I can understand that you do not know, you cannot put any logic into this. But these kind of question is pretty logical. I have taken the class of this optically active thing or not. I think the first class I have taken was this only. How to determine optically active nature of a compound? Okay, revise that also, right? Next question you see. Question number 19. The order of reactivity of following alkyne light for SN2. Tell me. 19th one is D. Okay, right. RI, RBR, RCL, RF. Next one. Question number 20. Think about it. It is also logical. Purvik is getting D, Sander is getting A. Now next, tell me D. Okay, Purvik is getting C. This is getting B. Why B is correct? Why the order B is correct? Why B is correct? The order B is given here. Or randomly you said, third carbocation will be more stable. You have to alternate double bond. And two is more stable than one because resonance is possible. Okay, it looks more likely than one. Correct. Okay, now you see here.  Living group will go out. We'll get a carbocation here. We'll get a carbocation here. And we'll get a carbocation here. Now SN1 reaction we have, correct. It follows by the formation of carbocation. Right? Intermediate is carbocation. Right? More stable carbocation, more stable carbocation would be the reactivity, will be the reactivity. Okay, so basically we have to find out, correct. No, now Vaishnavi has got the right answer. We have to find out the stable carbocation. This carbocation you see along with the lone pair of this oxygen atom, it has 6 pi electron, which is aromatic. Right? Hence, this carbocation is the most stable. So second is the most reactive molecule. Okay? So obviously we can have the answer between C and D, A and B are not possible. Now when you see this carbocation, right? So this carbocation is here and the ring is what? Anti-aromatic because it has 4 pi electron. So it is anti-aromatic least reactive. Least anti-aromatic, I'll write it on AA here. Least reactive. Right? If you see this one, this is allylic carbocation, which is resonance stabilized. Right? So the most stable, most stable carbocation is this one. Okay? Hence, second is most, then allylic carbocation again is stable. Then fourth one, right? Because the third one is anti-aromatic least stable. Right? So order will be 2, 1, 4 and 3. Right? When you have ring like of structure, you must consider this aromatic factor. It is not like you do not know this. The only thing you didn't take care of that aromaticity is also one kind of, you know, factor we may have. So in organic chemistry, you must take care of all these things. Planetarity, hindrance, aromaticity. Okay? Whenever stability things comes into the picture, all these factors you have to cross check, especially when you have cyclic compounds like this. Aromaticity you must think of. Right? Next question you see. Question number 21. 21. SN2 reaction proceeds with the inversion of configuration. SN2 reaction proceeds with the stereochemical inversion. SN2 reaction follows second auto-kinetics, which is incorrect statement. Okay? It's 21. The answer is you're getting A. Why A is wrong? Why A is wrong? SN1 reaction is followed by the formation of carbocation. Right? Yes. It proceeds with resimization. So you'll get the same configuration as well as the opposite configuration also. Right? But SN2 reaction we have already discussed this. There's a term called Walden inversion. Right? So Walden inversion takes place because backside attack is possible over there. That's why in SN2 reaction we always have inversion. But that is not true in case of SN1 since it follows by resimization. Okay? Hence option A is incorrect. Question number 22. Bromination of alkene. SN1 reaction obviously it is resimization. Elimination of HX from alkylhyde. It is said Jeff. Right? So bromination of alkene is what? It is vicinal di bromide. Just now I have discussed bromination of alkene follows by the formation of a cyclic bromonium ion. Right? Or a halonium ion, cyclic halonium ion. Anti addition is there. So we'll get this dihylide over there. Right? And alkylidine halides gives you Zem dihylide. Okay? So obviously option A is right here. Next question. Question number 23. Done? A drying. Okay. Do this. D. Cymate is getting D. Okay. Tell me what happens if we use alkali k-ov-H and then NaNH2. First carbon. In the first option bromine is attached over here. This is bromine. And here bromine is attached in the second option. Let me check then. One, two, three. Yeah, I think. Again it is, okay, I'll just do this. Again it is a printing mistake. So this Br8 is attached over here. One, two, three and four. Correct? I mean. Now I'll do this. You check. See. See, first of all, if you find out the degree of unsaturation of EoU of the given compound A, which is 5 plus 1 minus h is 9, x is 1, n is 0. So h plus x minus 1, h plus x minus 1 by 2. So this will be equals to 1. So degree of unsaturation is 1. It means in this compound A, either we have one double bond and one ring present. And that is what we have to find out. The given option contains one double bond. So for this also, DOU is 1. And for this also, DOU is 1. So any one of these or both can be the answer. Now you see, suppose if I take the first compound, with this I'll try to find out the given condition of the question. The first one compound that is option A is what? CH3, CH2, CH double bond, CBR, CH3. Now when you do the hydrogenation of this compound, hydrogenation, we will get C5H11Br. For this, if you find out the DOU, that is 0. So it is an alkyl chloride simply. So that compound is CH3, CH2, CH2, CBR, CH3. You see, there also we have one hydrogenation. This is the compound we get on hydrogenation. Now the next condition is what? When reaction of A with alkoholic KOH first and then NaNH2. So this reaction with alkoholic KOH and then NaNH2, it gives you alkyne. And the product here it will be CH3, CH2, C triple bond, CCH3. This is the product we get. On alkoholic AOH with NaH2, it gives alkyne that we have already done. You must have done this in the alkene chapter. If not, then you can go through. Next thing is what? Lindler catalyst. What is Lindler's catalyst? Lindler catalyst is palladium BASO4. Now this is very important because this does partial reduction. Partial reduction. Now partial reduction means what? If you have triple bond, this converts into double bond. So the product here we get is C double bond, CH3H and then ethyl group we have here, which is C2H5. Lindler catalyst that is H2PD BASO4 gives partial reduction. Instead of this, if you use H2 with nickel, platinum, this gives complete reduction and alkyne converts into alkene if you use this reagent. This gives complete reduction. Alkyne converts into alkene, but if you have alkyne, Lindler's catalyst, first it converts into alkene. Again, you use H2PD BASO4, then you will get alkene. So it is a two-step process. So this again here, H2PD gives you cis isomer, which is important here. This also allows to react with Na with liquid NH3.  C2H5, H, if you have done this fine, otherwise you can memorize this now also. Na liquid NH3, H2PD BASO4, both will partially reduce the triple bond. This gives trans, right? And this one gives cis. This is the important thing you have to remember. It is a factual again. So obviously these two are isomers of each other, right? Now you see, produces C with evaluation of NH3, Lindler catalyst to gives D on reaction with Na with liquid NH3 produces E. So this is E and this is D and both D and E are isomeric, isomers of each other. So obviously option A is correct. It satisfies the condition option A is correct. Similarly, if you take this option also, so first of all, when you do the hydrogenation of this one, this converts into alkene like this. We'll have a BR over here. Alcoholic KOH and NH2 you use, again, you'll get a triple bond here. You will get alkene, which again reacts with these two reagents, gives you cis and trans isomers. So this also gives you the same isomeric product, same kind of isomeric product. Hence for this question, the answer will be both option A and B. Is it clear? Okay, so we'll take a break now. We'll start at 7. Will it be fine? So we'll start at 7. We have half an hour more. We'll finish probably 10 more questions on this than this chapter will be finished. Right? So take a break. We'll start at 7 o'clock. Okay, can we resume the class? Guys are there. So next question you see, question number 24th, which shows complete stereochemical inversion during SN2 reaction is? Yes. 24th one, the answer is A because one degree alkyl halide thrown towards SN2 reaction. Right? And that is why option A is correct because all of you have got this. Next one, question number 25, the decrease order of the rate of above reaction with nucleophile from A to D. See the reactivity of nucleophile depends upon two factor. It's charge density and the hindrance, steady hindrance. 25th one, you're getting B. Okay. See, first of all, if you compare CH3O- and OH-, which one is better nucleophile? Obviously CS3O- because of IE effect of CS3. More charge density on this oxygen atom and more better nucleophile it is. Right? Now, so obviously this D and here it is ACO-, so that would be CH3, C double bond OO-. Okay. And we have phenyl O- that is pH O-. So this group has electron with drawing tendency. This group also has electron with drawing tendency. So all these four, if you compare, so which is CS3O- is the best nucleophile we have. So D is maximum. Right? D is maximum. And then we have OH-, which is nothing but C. Then A and B if you compare. Right? A and B. So here what happens because of this, see that both has electron with drawing nature, but phenyl has larger size here. Well, just a second, phenyl has larger size CS3 if you compare pH O- and CS3COO-. So this has electron with drawing nature. This also has electron with drawing nature. And this one has electron with drawing group of size of. So here you see the phenyl O- we have. So this is nothing but, okay, this is nothing but this one and O-. So if you compare the nucleophilicity of this one and this one, right? So this oxygen and this oxygen, it is attached to the phenyl group, which is electron with drawing and this is also electron with drawing. Right? So electron with drawing nature of this and this if you compare. Which one has more electron with drawing nature? C double bond O or this phenyl group. Electron with drawing nature. See, here we have oxygen and here we have carbon-carbon, that too also we do not have double bond. We have partial double bond characteristics here. That's why you see the phenyl, since it has more electron with drawing characteristics, so B is more than that of A. Right? Hence, I think you are saying B only, right? So DCB with drawing tendency, it means less nucleophile, weak nucleophile. Okay, so option B is correct. Right? Yeah. Right? So all of you have got the same thing, B only. Right? So this is the thing, right? Like charge density and size factor also sometimes comes into picture. Next you see question number 26. Portal-suntaining, these two are halide, those are original levels. They were labeled A and B for testing. A and B were separately taken in a test tube and boils with NaOVET solution. The end solution on H2O was made acidic with dilute with H1O3 and some solution was added. The following statement is true. Tell me. See, there is one line missing in this question and the line was is this. The end solution in each tube was made acidic with dilute H1O3 and then some H1O3 solution was added. Right till here. So after this point added, this line you must write, after this added point we have this Substance B give a yellow precipitate. This is given. Substance B or bottle B that you have gives a yellow precipitate. This is also one of the information given. One line is missing here. Okay, after this added, Substance B gives a yellow precipitate. Then which one of the following statement is true for the experiment? In this question, you see what happens. We have phenyl iodide first, iodo-benzene. And the second one is this CH2i. So this bond is strong. Here what we can say, the carbon and iodine bond is strong. So this bond dissociation of this CNi is not possible here. But here this CH2 and i bond, this bond is comparatively weaker and it can dissociate and when you put AGNO3 into it, it forms precipitate of AGI which is yellow in color. So the yellow color solution that we are getting or yellow precipitate that we are getting, it depends on the bond strength of carbon and iodine. So carbon and iodine here it is weaker, here it is strong and we know already that phenyl carbocation is not stable. So this bond won't dissociate. This will dissociate and gives yellow precipitate. So B gives yellow, it means B is what? This compound is B and this compound is A. So A was C6H5. So this is correct. A is the right answer. 27. Tell me first, final product is given. Yeah, option B is correct. We will take alcoholic NaOH and we will get alkene over here. With alcoholic NaOH, we will get alkene here and then the Markovnikov addition of HBr in presence of an acid gives you B. So first we will get alkene and then addition of HBr followed by Markovnikov gives you the product, this which is given in the question. Question number 28. 29. What is the answer? The most reactive halide towards SN1 reaction and butyl bromide, not possible because it is fun degree. Secondary butyl tertiary, oh yeah, it's very simple one, tertiary butyl bromide. Tertiary carbocation, most stable. That's why we are taking option C. Isopropyl fluoride undergoes hydrolysis by, it's 2-degree alkyl halide. See, allyl carbocation is stable through resonance, where shall we? Allyl carbocation is stable through resonance, but SN1 reaction you see, it is, we are talking about SN1 reaction, right? So we'll take the stability of carbocation, right? So in resonance, what happens? We don't have actual, it's like the electron is delocalizing continuously, right? Electron is moving from one point to another point, right? In resonance, but in tertiary or secondary carbocation, what happens? There is no movement of electron continuously, but the positive charge over there and there is electron releasing group which stabilize that positive charge, okay? So because of inductive effect only, the substitution reaction is favorable over there. Why nucleophile attack it there? Because of positive charge. But in resonance, this charge is delocalizing continuously, right? Electron is delocalizing. That's why allyl we are not considering. Yeah, question number 30. I see isopropyl chloride is 2-degree alkyl halide. So mechanism can be anything, SN1 and SN2. So option C is correct here, right? Option C is correct here. Depending on solvent, we can have any mechanism possible, right? Generally, we say what in polar solvent, SN1 mechanism is favorable, okay? But tertiary or sorry, secondary alkyl halide can follow any one of these mechanism, SN1 or SN2. Yeah, it's 2-degree carbocation. Question number 31. Marine atom in chlorobenzene is ortho and paradigmatic because, tell me, resonance effect predominates over inductive effect. Inductive effect predominates over resonance. Option A, this question you let it be. I'll discuss this next class. This question I'll discuss. I have studied one thing I can recall now that in chlorobenzene, what happens, this chlorobenzene, this electron is in 3P orbital, right? And this electron here, the carbon atom, here we have 2P orbital, right? So that's why that's why the resonance effect is not, you know, dominates over here. Here we have more minus IE effect because of 3P and 2P orbital, right? I have studied this one somewhere I can recall, okay? But the answer is also given option A. So I'll just cross check this, okay? What is the, you know, if it is ortho-paradirectly, then it is only possible when this electron pair comes over here, this comes over here, this comes over here, then only it is ortho-paradirectly. So we don't have any doubt with this when it is ortho-paradirectly. But the point is with the option that it says resonance effect predominates over the inductive effect. Here I have some doubt whether it predominates or not because I have studied this, I know 2P and 3P orbital, the resonance effect is not that much effective, okay? Did you understand my point? So answer for this question is what is resonance only? If this lone pair of electron is involved in resonance, then we'll get negative charge at ortho-paraposition and hence it is ortho-paradirectly, that is fine. But my question is, it says resonance effect predominates over inductive effect. This I will cross check again because I have studied this because of this 2P and 3P orbital here, resonance is not that much effective, okay? So we'll discuss this next class, we'll let it be this now, correct? But the answer if someone asks you that chlorine in chlorobane gene is ortho-paradirectly because of resonance. Whether it predominates or not, I will let you know. Understood? Understood? Can you move on to the next question? Yeah, so we'll discuss this next class, this thing, okay? 2P, 3P and all. So we have done this, no? Same question. This one, 2-2 dichlorobutane on boiling with aqueous KOH gifts, tell me, easy one, 2-butanone, 2-2 dichlorobutane, it is CH3, COH, OH, CH2, oh sorry dichlorobutane, it is no? CH3, dichlorobutane. When this reacts with aqueous KOH and we take 2 moles of KOH here, so we'll get here OH and OH, so I'll write down the product CS3C, OH, OH, CS2, CS3. And when the carbon contains more than one OH group, more than one OH group is present in the same carbon, then dehydration very easily possible. From this H2O molecule goes out and we'll get CS3C double bond O, CS2, CS3. This is 2-butanone, option B. Next question, less reactive than alkyl halide towards nucleophile. What is the answer to this one? Tell me fast. What is the answer? It is because of resonance, right? Where we have carbonium ion in this? Yeah, we do not have carbonium ion. Answer is option A. Due to resonance only, it is less reactive towards nucleophile. No, because of resonance, we'll have negative charge at ortho-n paraposition. When you draw the resonating structure, you will have negative charge at ortho-n paraposition. So that's why, since we have negative charge present, that's why nucleophile will not attack. Nucleophile will attack at the point where we have negative charge deficiency. Understood? Next question, which reaction is incorrect? The last two questions, tell me. Why A is wrong? Poor Vick. See, ammonia is a strong base. In case of, remember this point, it is very important. In case of secondary or tertiary alkyl halide, it favors elimination, elimination reaction over substitution. So elimination reaction is favorable in case of secondary and tertiary alkyl halide. Okay? So first option you see, it is a tertiary alkyl halide. Substitution not possible. Hence the reaction is wrong. Answer is option A. If you see option B, it is tertiary and this is the elimination product. Okay? Elimination reaction. Here also you see secondary alkyl halide elimination product. Here you see it is the primary alkyl halide. So substitution product it is. This is substitution. This is also substitution, which is not possible in case of tertiary alkyl halide. Hence answer is A. Right? So always remember this point. Ammonia is a strong base with secondary and tertiary alkyl halide gives elimination reaction. Right? Last question. Okay, we do not have any question. So we are done for the day. Okay. Next class, we will again see some other chapter. Some more questions we will do. Okay. Any doubt you have, you can ask me, otherwise we will wind up the class. Any doubt? Tell me. Okay. So we will see you in the next class. Okay. Thank you all.


Hello and welcome to the session. In this session we discussed the following question which says, state the cardinal number of the following sets. We have first a set A equal to set of the letters in the word umbrella. Then next we have a set B equal to the set containing the elements X such that X plus 1 equal to 0 where X belongs to N that is the set of natural numbers. Before we move on to the solution let's first define the cardinal number of the set. We have the number of distinct members of a set is called the cardinal number of the set. So a cardinal number of a set say A is denoted by N of A. This is the key idea that we use in this question. Let's move on to the solution now. First we have a set A which is the set of the letters in the word umbrella. So the set A would be equal to the set containing the elements U, M, B, R, E, L and A. Since the repetition of the elements are not allowed in the set so we are writing the letter L only once. Though it is repeated two times in the word umbrella. Let us now count the number of distinct elements in the set A. It's 1, 2, 3, 4, 5, 6, 7. Since according to the definition we have that the cardinal number of a set is the number of distinct members of a set. So we can say that the cardinal number of the set A which is denoted by this is equal to 7. This is the answer for the first part. Now we have the second part in which we have a set B that is equal to the set containing the element X such that X plus 1 is equal to 0. Where this X belongs to the set of natural numbers N. Now since X belongs to N this means that X would take the values from 1, 2, 3 and so on. But we know that minus 1 plus 1 is equal to 0 and this minus 1 does not belong to N. So this means that there is no number or no natural number which when added to 1 gives us 0. Thus the set B is equal to 5 that is it is an empty set. Also the cardinal number of the empty set is 0. Thus cardinal number of the set B is equal to 0 since set B is an empty set. So this is the answer for second part of the question. So this completes the session. Hope you have understood the solution of this question.


Good evening, I'm Mulanideh Adyemi, and this is the News Now. Abdulrahman Cheneh, the leader of the Nigerian junta, says the military are ready to explore diplomatic dialogue. Cheneh spoke on Saturday during a meeting with the Nigeria's intervention team, comprised of Islamic scholars. The scholars were led by Balilaw, the national chairman of Jamal II, Zalatul Bidal Sunar, and met with Cheneh for several hours in Niger's capital, Nani. A delegation from the Nigerian junta was received in Kanukri by Gennan military authorities who asked for rainfall support to face the challenges ahead, at a time when the military regime in Nani is under threat of a staffrican intervention. The Gennan head of state, Connell Mamadi Damboya, whose regime also emerged from a coup d'etat in September 2021 held tax in Kanukri on Saturday with a delegation led by General Moussa Balil, the Canadian television reported on Saturday evening. In a statement at the end of July, Kanakri had expressed its disagreement with the sanctions recommended by the economy community of a staffrican state, including military intervention. To get to hit the notification button so you get notified about fresh news updates.