 Hello, good afternoon. Can we start? Who are all there? Please let me know. Please type in your name. All of you, Shruti, Ritvik, Saimi here. Saimi, did you understand that question? You have asked one question. No, that has been asked in J.D. Main exam. You were saying the answer should be C and D, Ni2 plus and CO2 plus. The given answer is correct. Option D or fourth option is correct into that one. Did you understand that? Yeah, because Ni2 plus will never form low spin octahedral complex. That's why that is not the answer. Right? In CO2 plus, the electron may excite into the 4D orbital. That's the reason we have. Okay, so today we are going to solve some questions of hydrocarbon chapter. So we are probably going to solve 40 to 45 questions today in this session. Okay, all are from hydrocarbon. Right? So the first question you see it is there on the screen. Try this one and tell me the answer. Naman is getting B. Ritvik is getting B. Saimi is getting B. What about others? Tell me what should be the answer in this? First keep in mind for Wurz reaction. Generally, Wurz reaction is used for the preparation of symmetrical alkane. Symmetrical alkane. Right? It is the method for the preparation of symmetrical method. And that is only possible when you take same alkylide. Suppose we will take 2 moles of Rx, then 2 Na in presence of dry ethyl, then the product will be Rr. Right? But we take two different alkyl halide, R1x plus R2x with Na and dry ethyl. Right? All these regions are same. Then the possible product here it will be R1, R1 plus R2, R2 and R1, R2. These are the three possible product we will get into this. Okay? All possibility we will have over here. Right? So this is what happens when we have two different alkylides. So total three alkanes we will get. One, two and three. And this is what we have here. Since the question contains we have ethyl iodide and propyl iodide. So propane and propane combines, forms hexane, ethyl ethyl combines, forms butane. Right? So butane possible, hexane also possible. And when prop, ethyl propyl combines, forms pentane. Right? That is why propane is not possible in this case. The answer is option B. Right? This you must keep in mind they have asked one question this one. And how many products are possible in this reaction? In this reaction. Right? So always you remember different alkyl halide gives you three different alganes. So three product possible here. Right? But in this question the answer will be option B. Right? Next question you see. Solve this one. Try this. Alitya is getting B, Cymale, D, Shruti, D. First of all you see LiAlH4 is not possible because what it gives it converts acid into alcohol. Correct? So this is not possible option A. It is a reducing agent. Reduces acid into alcohol. Soda lime is generally used when we have salt of any acid. Like suppose if you have CS3-COOH, CS3-COO Na, salt of this. And when it is heated in presence of soda lime NaOH and we also use CAO over here because soda lime we have a mixture of these two. And when we heat this, decarboxylation takes place. Right? Na2-COO3 comes out into this. And the product we get here is CH4 plus Na2-COO3. Na2-COO3. Carbon dioxide is also. So soda lime is also used when we have salt of acid. Okay? Soda, soda, salt of acid. This is also not possible. Now rate phosphorus and HI we can use this reagent ZN with cong HCl. What is this reagent? And what is the method we have in which we use this reagent? Can you tell me the name of that process, that reaction? It is clemention reduction. Right? If you remember, clemention reduction, the reagent is this. And one more reagent we have that is NH2-NH2 hydrazine. What is this reagent? NH2-NH2 with C2H5O Na. This reagent we use in Wolff-Keschner reduction. Right? This is Wolff-Keschner and this one is clemention. Right? And clemention reduction, if you remember, it is used for the conversion of aldehyde or ketone. Aldehyde or ketone into alkane. And in short, what you can memorize, all these compounds C double bond O converts into methylene group that is CH2. Right? That is what the conversion we have. All other things will be same. But here we don't have aldehyde or ketone, but we have acid. Right? So for acid we use red phosphorus and concentrated HI. So answer will be option C. Red phosphorus and concentrated HI. One thing you must remember, excuse me, I'll tell you in this. If you have alcohol, right? And we are talking about this reagent, ROH, with red phosphorus and concentrated HI. It converts into RH. If you have RCHO, that is also we use here. Red phosphorus and HI if you are using, this again C double bond O converts into CH2. So this gives you RCH3. If you have ketone, then again the same thing. R2CO, same reagent. It converts into R2. The C double bond O converts into CH2. If you have RCOOH, then again it converts into RCH3 with same red phosphorus and HI. Okay? So all these compounds, alcohol, aldehyde, ketone or acid, converts into alkane where it is heated with red phosphorus and HI. Right? Answer will be option C. Is it clear? Which one? No, you see the product is given. No. You see acid is converting into, acid is converting into alkane. Right? Acid is converting into alkane. And this is the preparation of alkane you have. If you have revised this chapter, preparation of alkane, the topic will be like this. The preparation of alkane from the reduction of compounds containing oxygen as a functional group. There we have some other reagent also in HVG. You must revise that. We don't have only red phosphorus and HI present. I think there is I2 also we have there. Yeah. That's with iodine. Right? This is the preparation method of alkane from carboxylic acid or any other compounds containing oxygen, alcohol, aldehyde, ketone or acid. Okay? So you must remember this zinc concentrated HCl is Clemensian reduction and which is used for the conversion of aldehyde or ketone into alkane. Right? So this is the second question we have done. Now you see the third one. Tell me the answer. This is 6. It is written 6. Which I think it should be 2. 2 molecules of PhCH3. Yeah. It's toluene. Aditya, Naman, Amog, Arudev. Artevite is D. All of you are saying D. Why it is D? Explain this. What is the first product we get here? PhCH3 with CR2 H-New. What is the first product we get? Yeah. It becomes PhCH2Cl. What is the name of this compound? What is the name of the first product will be PhCH2Cl? Tell me the IUPSE nomenclature of this. Correct. Your answer is right. All of you are right. Tell me the name of this molecule. Is it benzyl chloride? What is the IUPSE? Yeah. Correct. What is the IUPSE name? Correct. Benzyl chloride. Any other name? Can you write down the other name of this one? Yes. Phenylchloromethane also you can write. Okay. That's what I wanted to know. Because when this functional group is attached with this molecule, so this becomes the substituents now. And for substituents of when Benzene ring presents, no. That's what. Benzene, we cannot use truthy. Because when Benzene is present as substituents, then we don't write Benzene in the name of this, but we write phenyl. Okay. So it is phenylchloromethane. Correct. Phenylchloromethane is right. Option D is correct. First of all, we'll get this. And then which reaction takes place? Two molecules of this will join together. And the product will be option D. Right? Question number four. Decreasing order of boiling points. Is it D? All of you are getting D? Right. Option D is right. See, two points. Boiling point is directly proportional, point is directly proportional to molecular mass. Or we can also say as the number of carbon atom increases, boiling point increases for a straight shield compound. And it is inversely proportional to branching. With branching, boiling point decreases. Correct. So first of all, we have pentane and butane. So obviously, butane will have more boiling point than pentane. So C is greater than A. Right? This option is not possible. C is greater than C, we have here. Now we have 2-2 dimethylpropane, which is nothing but the isomer of pentane only. What kind of relation between C and D? Can you tell me? What kind of relation we have between C and D? What kind of relation we have between C and D? Or you can say among C, B and D. Can you tell me? They are chain isomer. Right? Because the number of carbon atom in the parent chain is different. Right? So all these C, B and D are the chain isomers. Right? So in this chain isomer, we know that in isomers, we know as branching increases, boiling point decreases. So obviously, A will have the least. C is no end pentane straight chain compound maximum. Then we'll have B because there is only one branch. This is two branch. So C, B, D and then A option, D is correct. Okay? So they are chain isomers also. Next question. This is a factual question actually. So what are the carboxylation? Okay. Answer C option C is correct. Now one more question I may ask you in this one. That in all of these molecule, in which of this reaction or in which of this method will get higher alkane? Which of these methods will get higher alkane? Can you tell me that? Only Worc, only one. What is co-electrolytic method? Worc is sign is correct. You all know. But what is co-electrolytic methods? Aditya. See, first of all, the answer of this question is hodeline decarboxylation. And the reaction is this, which we have already seen just now. C is 3-C-O-O-N-A plus N-A-O-H. When you heat this, we'll get C-H4 and N2-C-O-3 forms. So let me think into this. Answer is option C. Worc's reaction will also give you higher alkane. Co-electrolytic method will also gives you higher alkane. Okay. I'll just write down. No, not D. It's A and B. See, what is this co-electrolytic methods? If you have salt of sodium or potassium. Suppose I'm taking sodium only. Salt of sodium and potassium. When you do the electrolysis of it, then we get R-R plus CO2 plus N-A-O-H plus H2. This reaction actually follows free radical mechanism. All these information, you must have free radical mechanism. And since we have electrolysis, so you must have some reaction at cathode. And we'll have some reaction at anode. So at cathode, the reaction is this. This metal you have N-A plus, accepts an electron and forms N-A radical. N-A radical. Okay. This N-A radical takes H2-O and forms N-A-O-H plus half of H2 hydrogen. Anode reaction is this. The anion we have R-C-O-O minus releases one electron because oxygen has one extra electron. That's why we have negative charge here. And forms R-C-O-O. This oxygen has one unfair electron and two electron will be as it is. So this is again a radical we have. And from this, we know radical has this property to eliminate a neutral molecule to form a new radical. So it eliminates CO2 to form alkyl radical. Now in this alkyl radical, coupling reaction takes place. So R radical combines and forms higher alkyl. So what is the point? Why I have written this question here? Because sometimes they ask this question, that code electrolytic method, what is the mechanism? Free radical mechanism. First of all, this question, you must keep in mind. And sometimes they ask this question also. At cathode and anode, which gas evolves? At cathode and anode. So at cathode, we can see hydrogen gas evolves. And at anode, carbon dioxide gas evolves. So these two products will get at anode. And these two products will get at cathode. So the kind of question they are asking nowadays, you never know what they are going to ask. So all basic, basic, small, small information you must have. Clear? Write down this equation and you must memorize these two, three points. Can you move on? Question number six. Okay, you haven't done this kind of question, right? We'll do this. So write down, see, first of all, this compound, which is N hexadecane, is nothing but cetane. This compound is cetane. Okay, so write down this one. The cetane number is defined as, where is defined as a percentage by volume, the percentage by volume of cetane, percentage by volume of cetane in a mixture of, in a mixture of cetane. I am reading, writing this N and all this. I don't know. This is also not there. In a mixture of cetane, this compound, alpha, methyl, naphthalene, for methyl, naphthalene. Okay? So cetane is nothing but N hexadecane, right? Now you see this question. The percentage of the letter is 10%, right? So this is given, the volume percent of the letter means this compound, alpha, methyl, naphthalene, the percentage of this is given, which is nothing but 10%, right? So what is the cetane number? It is defined as the percentage by volume of cetane in a mixture of this. So 10% by volume, methyl, naphthalene is present. So this cetane is present, what? 90% by volume. Answer is option B. Generally this kind of question they don't ask, but I have, you know, I took this question because the way they are asking the question, you cannot skip any smallest one information also. That's why I took this question. Understood. So cetane is N hexadecane, and cetane number is the percentage by volume of cetane in a mixture of cetane and alpha, methyl, naphthalene. Correct? Option B is right. Definition all of you write it down. Next question, highest volume point, B. I don't think I have to discuss this. Okay. Maximum molecular mass B and its state chain compound. Question number eight. Do this one. Question number eight. B is right. Next question. Okay, sir. Last? No. Rn, okay. No, no, no. Rn, okay. Question number is getting C. C, C, C. All of you are getting C. Tell me. Okay. And tell me in first option, there is two methyl pentane. How many monochlorinated product we get? In first option, how many monochlorinated product we get? Tell me the first one, second one, third one, and fourth one. For all these compounds, how many monochlorinated products possible? Six. Now you're getting different answers. Six. Four. Check your answer. Is it six? Five. Aditya and Vaishnava is getting five. Five is the right answer. You see, two methyl pentane. So pentane is this one, three, four, five. And here we have methyl. Right? So this position, we can have chlorine attached over here. One position is this. These two are equivalent position, right? So one, then two, then three, four. These two are equivalent position. The name of this, if chlorine attached here, the compound is what? One chloro two methyl pentane, right? If chlorine attached over here, then it will be two plus five, seven. And one plus two, three, four, five. So we'll count from this side. So one chloro four methyl pentane, right? So total five monochlorinated product possible in the first one. What about the second one? Two, two dimethyl butane. Butane is this. One, two, three, four. Here and here we have methyl. You see here, again, these three positions are equivalent, right? One, two, and three. Equivalent position. So the different product we get when this will attach. One chlorine will attach over here. And then one chlorine will attach over here. And then another chlorine attaches over here. So this three gives you three chlorinated product. One, two, and three. Okay? These positions are equivalent position. One, two, and three. So you'll get same product here. C is two already, right? N hexane if you have, then again we'll get three. N hexane also, but here also we'll get two. So here also we'll get three. Here we'll get two and here we'll get three, right? So see this kind of calculation is also important because in this only they can ask this question also at which one of these molecule gives the maximum number of monochlorinated product. Or they can also ask you minimum number of monochlorinated product, right? So that's why this calculation is also important, right? Correct answer for this question is option C. Next, question number nine. You must have done this kind of question. Neopentane, correct? Neopentane is this. You see all carbon positions are equivalent, right? One, two, three, and four. Equivalent carbon position we have, okay? That's why we'll get only one monochlorinated product here. Answer is option D. Just a second. Okay, you understood this, no? Next question you see. Aditya is getting A. Why A? Can you explain Aditya-Rithvik transposition? Yes, correct. Trans molecule we have minimum steric hand rinse and maximum stability, right? You see across how many stereocenters we have in these compounds? First option, how many stereocenters we have? How many stereocenters in the first one? Both double bond, right? This is the stereocenter and this is the stereocenter. Across this double bond and this double bond, GI is possible, geometrical isomerism, right? Why? Because both carbon atoms contain two different groups at it. I hope you remember all this because I have discussed this in the very first or second class of yours, right? Geometrical isomerism I have discussed all the three condition of GI, correct? The two stereocenter we have, this across this double bond if you see, the molecule is trans, right? Because higher priority on the opposite side. Trans or we can also say anti, right? Anti would be better. Here also you see it is also trans or anti, right? So both across both of the carbon atom, we have transposition or anti-position. So it is most stable, minimum in stereocenters and hence it is most stable, correct? Here you see these two are at cis position. This across this also we have cis position. Across this also we have, we have what? Across this we have trans, across this we have cis position, correct? Only these two stereocenters that you have at those stereocenters, at both the stereocenters we have transposition. That's why this is the most stable conformer, least steric hindrance, okay? Option A is correct in this, okay? So what we can write? Trans gives you least steric hindrance and hence most stable. That's why option A is correct, okay? Question number 11, next one. Two phenylpropene on acidic hydration. Write down the structure first of this molecule and try to get the product. Most of you are getting A. Two phenyl, two propanol. Okay, two phenylpropene is this. When you do the acidic hydration, which is nothing but the reaction with H plus H2O. So this reaction follows electrophilic, this reaction follows electrophilic substitution reaction. Or we can also write it down electrophilic reaction mechanism. And we know this electrophilic reaction mechanism follows by the formation of, formation of carbocation. And this carbocation should be, if possible, most stable, right? We'll try to form more stable carbocation and then we'll get the product, okay? So what happens in this reaction, you see, this double bond will shift over here, right? And we'll get the positive charge on this. This is a carbocation, right? Or what we can say, this double bond will take this H plus, actually. This H plus will come over here. Double bond shift here, negative charge will take this H plus. So we'll get CH3 over here. We'll have the carbocation over here. And this carbocation, like this, you can write down. I'm writing down all the steps here. CH positive charge, single bond CH3, single bond CH3. And in this, you have H2O. So H2O is this. Now this lone pair will attack onto this carbon atom. The product we get here is this positive CH3. And finally from this, since this oxygen has positive charge, this is stable, H plus comes out and we'll get the product as... So answer is 2 phenyl, 2 propanol. That's correct. So all of you are saying A only, that's correct. Those who have got B, you can understand from this mechanism. Is it clear? Next question. A is clear. Can you give me the answer? No. Can somebody give me the answer? No. Those who have got B connected, tomorrow morning they'll know. Okay. Name, we need to answer which is the problem. But the problem is this, you know, because your name was morning answer, you can give me the answer. Yeah. Is it B? Yes. Simon, you're getting C. It is not aqueous KOH. See, it is ethanolic potassium hydroxide. Okay, ethanolic. The purpose of this is what? The medium is alcoholic. That is it. So we have alcoholic KOH, right? And alcoholic KOH gives you alkene, not alcohol. If you take aqueous KOH, then you'll get alcohol, right? So answer in this will be propene. I'll write down here. Alcohol KOH gives you alkene, aqueous KOH gives you alcohol, right? So here we have alcoholic KOH, so we'll get propene with equal number of carbon atom. Correct. Option B is right. Is it C? Tell me others. What is the answer you are getting? What happened others? Tell me. What happened guys? Okay, you see. A gas decolorizes alkaline KMN of 4, right? So when this decoloration takes place, it means the gas must be unsaturated. The gas is an unsaturated compound. It is an unsaturated compound, which can be alkene or alkyne, correct? Unsaturated compounds may be alkyne or alkyne, okay? So this is the test of unsaturation if you remember. Double bond and triple bond, the presence will identify with the help of alkaline KMN of 4 and when decolorization takes place, correct? So obviously when you have, when you are sure with this unsaturation part over here, so you can easily eliminate two options into this, that is A and D, which are alkene, right? C and H2N plus 2. It is methane, all single bond and this is ethane. These two cannot be the answer. Now we are left with alkene and one is alkyne. Now you see next thing here, alkynes or terminal alkynes, okay? The next part we have got, terminal alkynes, which is nothing but the option C. It combines with sodium and sodium compounds and the silver compound like this here we have and forms sodium or silver salt, salt of sodium or like in this one what happens? The CH triple bond C minus AG plus will get here or if you have sodium, then we'll get NA plus over here. The salt of these elements will get, only terminal alkyne you must remember, not all the alkynes. You're talking about terminal alkynes now, okay? Must keep in mind. But this what happens, the alkene that you have, this not at all reacts with AGNO3. So find these two compounds can show decolorization with this KMNO4, but since C2H2 reacts with AGNO3 and forms the salt of silver, that's why this option is not true, but C2H4 is true, right? For example, you see this reaction, C2H4, which is nothing but C double bond CH2, H2 and this reacts with alkyne KMNO4, right? So we have KMNO4, alkyline medium. We'll get diol in this case, right? That is CH2, single bond CH2, OH and OH, ethylene diol, okay? But this does not react with AGNO3. With AGNO3, no reaction. But if you take alkyne over here, that will form salt of silver. That's why option C is not correct, B is the right one. All these are property reaction of alkyne and alkyline. You must, you probably did not, you did not revise those chapters. That's why you are missing all these, right? But no worry, you can memorize all this now also. You can keep in mind, take note of all this and that will help you in the coming exam. Next question you see. Anti-Marconic of addition of HBr. Easy one. Do it first. We have to finish till 13 before the break. C option is correct. So you should not have any doubt in this one. Anti-Marconic of addition, you know, that the negative part of the reagent will attach at the double bonded carbon atom, which has more number of hydrogen, correct? That is anti-Marconic of fluid. Anti-Marconic of addition is possible in case of HBr only. That is only in presence of light. We can also write H new, light or H new, or in presence of peroxide. This point to write down, okay? Anti-Marconic of is only valid for HBr. Addition of HCl, HI, HF does not follow, does not follow anti-Marconic of rule, even in presence of, even in presence of light or peroxide, right? So this is one important thing we have and you must keep in mind this thing we have asked many times in various examination, okay? The reason if you want to write why HCl, HBr, HI does not show anti-Marconic of addition even in presence of, you know, light and HBr. The reason is actually not important, but if you want you can write it down. I'll just give you. Write down this. See the light that we are using here, in presence of this light because of the wavelength of the light, wavelength of that light, the molecule HBr dissociates in its ion form, H plus and Br minus. And then the reaction proceeds, okay? It's not ions, actually it is radical. We should write down it as radical, not ions. Okay? So the homolysis happens over here we'll get hydrogen and bromide radical over here. But in case of HF, HCl, HBr, what happens? The bond strength order if you see it is maximum for HF, then we have HCl, then we have HBr, and then we have HI. Relative bond strength is this, right? So the molecule HF and HCl is, bond strength is too high so that it does not break in presence of light, right? In case of HI what happens, the bond strength is low of HI, right? But since iodine is a bigger molecule we have, iodine is a bigger atom, right? So being bigger in size, the I radical is not that much reactive. Just write it down. First of all, write down this one. The bond strength of HF and HCl is strong enough so that it won't dissociate in presence of light, correct? That's why this, too, does not show anti-marculation competition. HI, the bond strength is, you know, the bond strength is actually it is weaker than HBr. So it will dissociate in the forms of its radical, right? But since the size of iodine is big, right? Iodine has a bigger size so it is highly reactive in nature. So again the two iodine radicals combines and forms iodine molecule, I2, right? That is why HF, HCl, HBr does not show anti-marculation addition even in presence of light or peroxide, okay? Simply you have to memorize this. Now coming back to this question, anti-marculation addition of HBr is not observing. Anti-marculation addition, one more point to write down, right? You should not have any confusion. Anti-marculation addition, and this is only possible for HBr, which we don't need to write actually. Anti-marculation addition is possible, unsymmetrical alkene, unsymmetrical alkene. You see, all these alkenes are unsymmetrical, except the third one, but 2-in. That's why the answer is option C. Correct? Next question. All of you are getting C. Okay, first of all I'll write down the structure of this molecule, butene, right? So we'll have CH3, CH double bond CH2. Here we have CH3, CH3. It is 2-3 dimethyl butene. And when you do the ojournalysis of this, that is O3 with, and then we use what? ZL and H2O. There's a method to write down the ojournalysis product is what? The carbon atom which is attached with double bond, just break this double bond, add double bond O to this carbon and double bond O to this carbon. So what we'll get here, you see? CH3, CH, CH3, C double bond O, CH3H. This is the first product. Another product is this, HC double bond OH. So we'll get methanol and 1, 2, 3, 3 methanol, 2 butanol, right? So option C is correct. All of you have got the right answer. Yeah, option C is correct. 2 butanol. Next question. Let me answer. See, HOBR is nothing but HO minus NBR plus, right? Or this reagent, we can also write it as BR2 with NaOH. BR2 with, so now whether you have this molecule as the reagent or this as the reagent, both gives you the same product, okay? Because these two reacts and gives HOBR only, correct? So the addition of these to any alkene follows Marconi-Coff rule that you have to keep in mind. Marconi-Coff's rule, okay? So now you see the propane is CH3, CH double bond CH2 with HOBR, with negative part is this, positive part is this. So we know according to the Marconi-Coff rule, negative part of the reagent attacks on the double bonded carbon atom which has lesser number of hydrogen. OH minus will attack over here and BR will attach over here. The product here it will be CH3CH, CH2BR, OH. So one bromo, two propanol, right? Answer is option D, correct? Yeah, right, option D is correct. Next one, question number 17. Sorry. Actually right now I'm not talking about it by any means. Why does negative charge go to OH? Oxygen is more electronegative than bromine. The bond pair of electron will get attracted towards the OH group. Understood, Aditya? Since oxygen being more electronegative, so that will attract the bond pair of electron and that is how the bond dissociates. See what happens. Here you see we have bromine, right? BR, BR. So this will dissociates like one of the bromine will take this, here we have heteroetic cleavage. So we'll get BR plus and one of the bromine atom will have two electrons and negative charge on it. NAOH gives NA plus. So this NA plus will take this BR minus, forms NABR, right? And OH minus, this will attach with this so form HO or OH minus HOBR. That is how it happens, correct? Next one, what is the answer? Yeah, right. The answer is option B. Why this one is C? First of all, the intermediate form during the addition of HCl, propene prism of peroxide. First of all, we have peroxide here, but since HCl we have, so this won't be, won't follow by the anti-marconic origin will not be there, right? So propene is nothing but CH3, CH double bond CH2, right? So addition takes place according to Marconicoff rule, means negative part of this, which is Cl minus and H plus, this will attach at this carbon only. But what is the intermediate product? Since it is not the anti-marconic of addition, it is simply the ionic reaction we have. Means the reactant or intermediate product that you have, that will present in the form of ions, okay? We'll have heteroetic cleavage into this. So this bond pair will shift over here and we'll get positive charge on this carbon atom. So we'll get CH3, CH plus CH3, right? This H plus will come over here, okay? H plus. So this is what the intermediate product we have and finally on this Cl minus will attach, we'll get the final product. So intermediate product will be what? Option B. The ionic reaction possible in this, since HCl does not follow anti-marconic of addition, right? So free radical does not form in this reaction. Heterolytic cleavage takes place into this and then the reaction proceeds. Just keep this thing in mind that addition of HCl, in presence or absence of peroxide or even light also takes place for, we can write down like this, occurs by mechanism. Ions forms into this, okay? And we'll get option B as the product. Next question. Question number 18. What is the answer? See, the reactivity of alkyl halide, reactivity of alkyl halide is inversely proportional to carbon halogen bond strength. More will be the bond strength, less will be the reactivity, right? So we know bond strength as we go down the group, size increases and bond strength decreases, correct? So the bond strength order is what? For carbon and chlorine, it is maximum. Then chlorine and bromine. After this, we have carbon and iodine. This is the bond strength order. So reactivity will be reverse of this. So order of reactivity will be carbon-chlorine, then carbon-bromine, and then carbon-iodine. This is the reactivity of alkyl halide. And one more thing that you know that iodine is the best living group. That is why because you can dissociate this bond easily, right? When this bond dissociates, there is only reactivity or reaction proceeds. So that's why the reactivity is this. Option B is right. Next question. For this one, it is A-aditya, right? Okay. See, in this question, when you have this Cl2 and AlCl3, in this, only one chlorine atom will attach. Okay. So chlorination takes place. Monochlorinated product will get. Chlorination takes place with this reagent. If you have Cl2 with H-new, but again, this one, this one you see, it is actually the same as H-OCl. Right now we have discussed, okay? H-O minus and Cl plus. Okay. So in this, OH and Cl minus will attach. So this is also not true. This is also not true, right? Chlorination in presence of light gives you again one monochlorinated product into this. Okay. So this is also not true because we have to add chlorine at both double bonded carbon atom, and that is only possible by ionic reaction or ionic mechanism, and which is, can be possible with presence of chlorine only. Right? Chlorine atom, like I have already told you just now, chlorine and chlorine molecule, electrolytic cleavage takes place and forms Cl plus, plus Cl minus. Okay. And then first this molecule will attach with this double bonded carbon and later on, later on Cl minus will also attach and will get two chlorine atom attached with this carbon atom. Correct? And we get this problem. Right? Option A is correct in this case. Nineteenth D is not possible. I just know I have discussed this Sondaria, OH minus and Cl plus. One carbon will take OH, another carbon will take Cl. Correct? Next question you see. Pentane to pentane. Okay. What is the answer? We have to convert triple bond into, what is this reagent to H2SO4 and H2SO4? Okay. You see in this reaction, first of all, the simple one we have, if you have pentane, correct? So, the reaction is this CH3, CH2, CH2C triple bond CH. In this, the first reagent that we use or whenever you use this BH3 with H2O2 in basic media, which is NaOH or OH minus. Then what happens? Addition of H plus and OH minus takes place and that follows anti-marconic of addition. Right? So, on this carbon you see, if you have this reagent, you can go through the nodes we have already started this. Okay. This reagent gives the addition of H plus and OH minus. You'll get alcohol over here first. And this addition follows anti-marconic of addition. Right? So, you see these two carbon, anti-marconic of addition we have, the negative part will attach, negative part will attach at the carbon which has more number of hydrogen, correct? OH minus will attach over here and H plus will attach on this carbon atom. So, the product here it will be CH3CH2, CH2CH double bond CHOH. Okay? Now, here what happens, tautomerism takes place. So, this bond pair comes over here and this hydrogen will jump onto this carbon atom, alpha position. So, you'll get CH3CH2, CH2CH2CHO which is nothing but pentanol. Okay. Right? So, answer will not be HGSO4 and H2SO4. Okay. Most of you are, you know, saying option A. Okay. That you go through. You will understand the mechanism of this reagent. How this reagent reacts. Okay. So, I'll share this link on WhatsApp just now I'm sharing. Okay. You can go through it. If you have, still you have, if you have doubt you can ask me. Okay. Okay. I have since most of you, yeah, yeah, yeah, that's why, that's why I have shared the link you can go through. Okay. On your WhatsApp you will understand the mechanism of working of HGSO4 and H2SO4. Correct. You go through that, you know, there is a mechanism it is given over there. Okay. So, you go through that mechanism you will understand that. Correct. Just you go through once. Okay. If you have doubt then you can ask me. I'll explain that. But the whole mechanism is given in detail. We'll move on next. Okay. For this the answer will be option D. Like why not A? You will understand once you see the mechanism. Okay. That you go through once. Next question you see. Question number 21. Do this fast. This one is good. Tell me the answer. Tell me what is the answer? Okay. See, I'll solve this question. First of all, you see the hydrocarbon A on ozone releases give to isomeric forms of B and C. Right. And those isomeric forms will be what? What are those isomeric forms? Obviously that can be aldehyde or ketone. Right. Because we can convert these into each other easily. Right. We'll see that also. So first of all we'll get two isomeric forms B and C. B on oxidation gives D. Silver salt of D contains 59.6% of A. So fine. You see. The silver salt of D contains 59.6% of G. Okay. This is actually mass percentage. So it should write down over here the mass percent of it. It is understood actually here. Right. So the salt of silver that you have which is nothing but RCOO AG, the salt of silver. In this, if you, if the mass of this is 100 gram, if I take, then the mass of silver present into this is 59.6 gram. Mass of silver. Right. So the atomic mass of silver is 108. So for 108 gram of AG, what should be the mass of this salt we have? Right. So which is nothing but 100 divided by 59.6 into 108. Right. When you solve this, you will get 181 gram. The molecular mass of salt is 181 gram. This is the molecular mass of silver salt. So if I try to find out the molecular mass of acid of acid from which the silver salt has been formed, what is that? That would be the molecular mass of silver salt because we can subtract the molecular mass of, we can subtract the molecular mass of silver from it plus one. This silver you remove at H over here, you will get acid. Correct. So molecular mass of acid is this. When you solve this, you will get 182 minus 108 which would be around 74 I guess. Right. So it is 74 gram of the acid. Right. So acid must contain COOH first of all you see C double bond OOH and then here we have an alkyl group so that the molecular mass is 74. So by like when you by hidden trial what we can say the molecular mass of this part will be what? Carbon is 12. 16 into 232 plus 12 is 44 plus 1 is 45. Right. 74 minus 45 is what? Now in this 29 we can have 2 carbon plus 5 hydrogen. This gives you 29. So what we can say it is CH3 CH2. So the molecular mass or the molecular formula of acid will be CH2 CH3 COOH. This is the acid you have. I don't know whether you have solved this kind of question or not. This is nothing but gives D. Right. So this is nothing but D, the acid that we have. Right. Now since D on oxidation gives acid, D on oxidation gives D. Right. So oxidation, D gives you D. If D is acid, this must be aldehyde. And what is the formula of that we can write? CH3 CH2, CO, aldehyde. Aldehyde is are the functional isomers of what? Ketone. So another formula of ketone will be what? We have C double bond O because we have 3 carbon. Right. So C double bond O, CH3 CH3, acid, aldehyde and ketones are isomers of each other. This is C probably. Right. Now A on ogenolysis gives B and C. So these are the product we have. Or better if I write down like this. C double bond O, CH. Can you write down the formula of A with which on ogenolysis gives these two product? Did you understand this? A on ogenolysis gives these two product. Right. So what would be A? We'll add this carbon and this carbon by double bond. Correct. So the formula of A will be CH3, C, CH3, double bond, C, CH2, CH3 and hydrogen we have here. The answer will be what? I think option B. Correct. Is it clear? So a little bit of calculation we also had in this question. Moving on next. Let me know if you understood this. Question number 22. We are getting B. See, first of all cyclohexene is this. On ogenolysis, the product, here we have is, here we have CHO and CHO. Correct. Now this compound further treated with aqueous KOH. So if you remember aldehyde with alpha hydrogen reacts with aqueous KOH. This gives intramolecular aldol condensation reaction. Intramolecular aldol. Mechanism I have discussed in the class. Correct. Intramolecular aldol. So this base OH- will take the alpha hydrogen from here and will get a negative charge over here. CHO with a negative charge and this CHO will be here. Now this intramolecular means what? This negative charge will attack on to this carbonyl carbon. Right carbonyl carbon over here. So what happens here you see? 1, 2, 3, 4, 5. We will get a 5 membered ring. Sorry I should write down like this. Should be this. We have CH double bond O. So when this negative charge attacks over here, this bond comes over here. And here we have this carbon. We will have this hydrogen over here. So we will have OH and this carbon, this CHO group will be as it is here. Now in this what happens? Dehydration. H2O molecule goes out. Because we want aldol condensation. This is aldol reaction. Alpha hydroxy aldehyde or ketone is aldol. So now in this condensation takes place, H2O molecules goes out. Right. H2O molecules goes out from this carbon. And the product will be this. We have a double bond here and CH. So hence option A is correct. Understood. Next question. Tell me the product quickly. NO minus and CL plus. Finish this one then we will get a break. We will take a break then. So what is the answer in this? All of you are getting B. See the other molecule was HOBR. Right. Then we have taken HO minus BR plus. And OBR or NOCl. Right. We call it as nitrosyl chloride. Nitrosyl chloride. We also call it as Tilden's reagent. Tilden's reagent. And this reagent follows electrophilic addition reaction. And in this one it dissociates like this. We have NO plus and CL minus. Even NOBR also dissociates with the same fashion. NO plus and BR minus. Now in this the double bond will shift over here. We will get positive charge over here. CL minus will attach here and NO will attach here. We will get option A as the answer. Right. In case of HOBR it is fine. HO minus and then BR plus. Because we have hydrogen over here. Okay. So you must keep this difference in mind of HOBR and NOBR. Nitrosyl chloride that you have. Okay. We will take a break now. We will start in 15 minutes. Okay. We will start in 645. Fine. We will start at 645. Guys are you there? Can we start? What is the answer? You are getting B2 gaseous PNQ decolorized aqueous bromine. But only one of them gives a white precipitate with aqueous aminic acid solution. Okay. So first of all since it decolorized aqueous bromine. So the compound must be unsaturated. Right. So we cannot have any alkene possible here. A and C options are wrong directly. Ethane and propane. Right. Here you see we have here we have alkyne and alkyne. So butte 1 ion gives you what the silver salt of that. Right. Which is the white precipitate. And this you will get the another compound into this one. Correct. So the terminal alkyne that is butte 1 ion. And the other one is non-terminal alkyne. Right. So the white precipitate which aqueous aminic acid solution gives PNQ an angle to B. So the terminal alkyne gives you the white precipitate and other compounds will decolorize the bromine water. Right. So in this you see we have ethyne and propane. Correct. So both having both are actually terminal alkyne. Right. So both should give the white precipitate with this salt, which is not the condition we have because it says only one of them. So we should have only one terminal alkyne. And that is only possible in case of option B. Right. So option B is correct. Most of you are getting the same. Anyway. So we will move on. Question number 26. Tell me fast this one. What is the answer? 26 B. Okay. Option B is correct. You see in this question first of all this carbon is sp hybridized. This carbon is also sp3 sp hybridized and this one is sp3 hybridized. Okay. So we know the negative charge on more electronegative element is more stable. So sp is more stable than sp more electronegative than sp3. So obviously B and A is more stable than C. Correct. So B and A is more stable than C. So we can easily eliminate these two options. Right. So C is the minimum we have here. C is the minimum we have here. Now next one you see B and A if you compare CS3 has electron-relasing nature. And because of electron-relasing nature this will be little bit more, little bit less stable than this one because of electron-relasing nature it will increase the electron density and hence the electron, and hence the stability decreases. Hence option B is correct. B is most stable than A and then C. Correct. Question number 27. It's a direct question. What is the answer? Question number 27. Hot alkaline KMNO4. So oxidation takes place. Why it is C? Why A stress? See one-view time is nothing but this CH3, CH2, C triple bond CH. When hot alkaline KMNO4 if you are using so this forms acid, right? And the product will be with equal number of carbons. CS3, CH2, COOH plus the other product will be HCOOH. Other product will be HCOOH. So you'll get this product in one step, in single step, right? Sometimes what happens, these two also react but not completely and converts into CO2 and H2, right? So we'll have little bit of this also present and with this reaction we'll get CO2 and H2 also, right? So we can have any one of these as the answer, option C and option D, right? If I have to choose one in these two I'll go with option D, right? Because they are not talking about the final product it is not mentioned. It's just simple like one-view time on oxidation gives water, right? So this double bond will break, we'll get COOH here, COOH here, the product will be this. Sometimes in some book it is written also that these two also react and some part of it converts into carbon dioxide and water, right? That is how the reaction goes. But I'll go with option D in this one. Correct, next question. Tell me this one. The answer, B you are getting. Okay, the correct answer is B only. This is the hydration obviously H2OH plus, this is what? Hydrogenation H2PD. This is the chlorination reaction which is nothing but halogenation and D is nothing but the polymerization, three molecules of this polymerization takes place, right? So option B is correct, that question Next question you see question number 29. The hydrocarbon which reacts with sodium and liquid ammonia. It is also a direct question and we have discussed this also so far. Yes, option C is correct because terminal alkyne is required. We have discussed this today also many times that terminal alkyne forms salt with sodium and silver compound, right? So you'll get C plus, C minus Na plus over here, right? The hydrocarbon which reacts with sodium and liquid ammonia is nothing but terminal alkyne you have to keep in mind, right? So option C is correct. Okay, so in this one the product you will get here is CS3 CS2C, triple bond C minus Na plus like terminal alkyne only reacts with, sorry, forms salts with liquid sodium and liquid H3, correct? So option C is correct. Question number 30. Tell me the answer, question number 30. You see this RMGX, this is minus and this will have plus, right? And since it is a terminal alkyne, right? So it can easily protonate the RMGX over here because this hydrogen is what? This hydrogen is slightly acidic and why it is an acidic hydrogen? Because this carbon is SP hybridized. Okay? So since the terminal alkyne again we have here, so hydrogen is what? Acidic, so what we can say the terminal alkyne can easily protonate the Grignard reagent. So right down this point here terminal alkyne can easily protonate Grignard reagent. Protonation means what? Addition of H plus, right? So this H will come out and joins with this CS3 minus forms CH4 plus will get CH3C triple bond C MGX This is the product we get here. So option D is correct in this one, we will get methane. Always remember one thing Grignard reagents, from Grignard reagents if you are getting alkyne the alkyne will get with respect to the alkyl group Grignard reagent, okay? Option D is correct over here. Next one. Tell me this one I think you can do easily. Addition of HBR follows Markov-Nikov rule since there is no light or peroxide we have. Just you add BR minus an H plus Do we have any rearrangement in this? If yes then what kind of rearrangement you have done? You are getting option A. Option A is right. What kind of rearrangement we have done in this? Option A is correct. I will write down here. You see first of all, BR minus will attach at the second carbon. So we will have BR here. Double bond will get CH2 here. Right? Since we have one more molecule we have one more molecule. Again the shifting of pi electron will take place. Right? And this electron will shift here. And finally the bromine attached on the second carbon. So CS3, CBR, BR, CH3 with another HBR molecule. So it is 2, 2-dibromo-propane option A is correct. Similarly you can cross check other option. Yes, hydride shift is there. We have one hydrogen here. So we will have the hydride shift finally. So we will get this. Hydride shift will do to get the more stable carbocation. One thing here. H plus. This H plus will come. So we will get CH3, CH plus I think here we have CH2 plus. Like this you will get the product. Then hydride shift will be there. We will get 3-d carbocation and again HBR attached to this. Next question we will see. Which of the following has minimum boiling point? Tell me first. Minimum boiling point. It is minimum boiling point stress. Read the question properly. Minimum boiling point. So maximum branch will have the minimum boiling point. So it should be option D. Isobutene. With branching boiling point it decreases. Next question you see. 34. 34 and 35. Tell me these 2 questions. 34 and 35. 34 is A. What about 35? 34A is right. Both SSL and reason are correct. And this octatetrain is a non-planet molecule. Having 8 pi electron it is not an aromatic compound. Remember this one also. Non-planet octatetrain having 8 pi electron having 8 pi electrons and not an aromatic compound. So 34A is correct. What about 335th one? Nitration of benzene with nitric acid requires the use of concentrated sulfuric acid. It's fine. Because for nitration of benzene we need this the mixture of HNO3 and H2SO4. So for 34th one option A is right. Right on the answer at this. That is option A. So for nitration we will use this. Okay. And how this reaction proceeds? From this we will get NO2 plus plus HSO4 minus and then H3O plus. If you want to write down the balance equation is OH minus NO2 so OH minus and H plus from this we will get H2O then. H2O. So this is what this NO2 plus is nothing but the electrophile. We call it as nitronium ion. Now this ion only attached on the benzene ring and the reaction proceeds. Right? So that's why the first CUC first assertion we have the nitration of benzene with nitric acid requires the use of concentrated sulfuric acid. This concentrated sulfuric acid helps in the formation of nitronium ion. Right? It takes OH minus from this, forms H2O and we will get nitronium ion. So assertion is right. The mixture of concentrated sulfuric acid and concentrated acid produces electrophile NO2 plus which is also correct we have here. Right? Hence assertion and reason both are correct. Right? And R is obviously the correct explanation of A. So for this one also the answer is option A. Is it clear? Right. So we will wind up the class here only in this particular chapter we are left with 7-8 questions. Right? So I have to discuss one small concept into that one particular question you may get from that particular concept. Okay. So we will discuss that concept in next class. We will solve one or two questions on to that also. And then we will do some chapters of 11th class next class. 11th class we will do some chapters in the next online session. Correct? So probably the chapter will be states of matter. Okay? So if you can revise that would be better. Okay? Because we will directly solve the questions. Okay? So we will see you in the next class. Okay? First in the next class we will solve these 7-8 questions. One concept I will discuss and then we will move on to states of matter most probably. Okay? Thank you all for joining.