 Good afternoon everyone. Hello Aditya. Good afternoon all. Who is going to write the exam on 8th of January? Anyone here? Sayuja, Vaishnavi, Mathili, Lalitha, Atmesh, Aditya, Ritvik, Sanjana, Shreyas, Vishal. Anyone of you writing exam on this 8th of January, first day? Aditya is there. Where is Amog, Naman and all? Okay, so we'll start now. See today we are going to solve some questions based on organic chemistry again. But it is not of one specific chapter. Okay? It is just a mixed question you have. You'll get mixed questions from all different chapters of organic chemistry. Okay? Then probably in the last we'll see some questions of polymer also. One or two different types of questions I have seen in polymer chapters and which has been asked in JEE exam. Okay? So those kind of questions also we'll discuss. Okay? One or two, just one or two different types of calculative problem in polymer chapter. Right? That we'll see. Okay? So we'll start with, let me check this. Yes, this one. Solve these questions. If you are not able to see, let me know. I'll just magnify it. Okay, Aditya. First one. What is the answer? Which of the following species is paramagnetic in nature? Okay? So all of you are saying B, free radical has one single electron, unpaired electron. That's why this is one of the intermediate we have free radical is paramagnetic in nature. That's very straightforward. Right? What is carbunium ion? What is the formula of this carbunium and carbene and nitrene we have already discussed? The carbene is this CH2, C with two double bond and one lone pair on it. Okay? This is carbene and this has six electrons. All are paired. So it is diamagnetic. See four electron, four bond pair of electron and two lone pair, one lone pair. Right? So answer is obviously free radical. Nitrene we have already discussed. Nitrene has suppose Rn with two lone pair and one vacant orbital. Right? We have discussed this thing in the Amine chapter. Okay? The formation of nitrenes and all various cases we have already discussed. Right? So this is nitrenes. Again, the number of electron is six. So it is all are paired. So it is diamagnetic. Right? What is carbunium ion? Tell me. What is the formula of carbunium ion? carbunium ion is pentavalent carbocation. Like suppose these H5 positive charge on it. This is carbunium ion. Okay? And this has again six electron. Right? Diamagnetic in nature. So if they ask you how many electrons are there? Answer will be six, six and six in carbunium ion, carbene and nitrene. Okay? Free radical has one unpaired electron. That's why it is paramagnetic. Question number two. Question number two. What is the answer? Question number two. Sirius is saying D. Simon is saying A. Pratik, Kondinia, B. Aditya Mishra, A. Let's pick A, D, A, A. So most of you are saying A and some of you are saying D. Okay, you see first of all the question is the following statements are wrong. Wrong statement we have to find out. You know Ozone is responsible for the greenhouse effect. Ozone is responsible for the greenhouse effect. So this is option A. You see directly it's saying it is not responsible for the greenhouse effect. So obviously option A is wrong. All other options are correct. Okay? Let me tell you the contribution of Ozone. It is about eight percent. Ozone is responsible for the greenhouse effect and its contribution is around eight percent, which is not that much important. For greenhouse contribution is around eight percent. You must remember this. Let me tell you again this is not one specific chapter we are doing. This thing contains the questions of all different chapters which is there in organic chemistry. Option A is correct. Ozone is responsible. Question number three, which one is the strongest nucleophile? Why D is the strongest nucleophile? Can we have B as the answer in this question? Question number three, can we have B as the answer? How many of you are confused with option B? B as in Bangalore? No one? Yes. B is not possible. Correct? See actually nucleophilicity or the nucleophilic order if you have to find out. So it depends on, if you write here, nucleophilic power. It is directly proportional to the plus i nature of the group attached to it. First of all the nucleophiles are negatively charged ion. So directly proportional to the plus i group. If you see this option D, we have cyanide group which is attached with CH2 minus. Correct? So this cyanide group has high electron with drawing power. Yes or no? Cyanide group has high electron with drawing power. That's why this will be a weak nucleophile because of the electron with drawing nature of cyanide group. Now if you consider option B and C, both has one methyl. CS3 CO minus, this has electron releasing nature and CS3 NH2, again the plus i nature of this will also enhance the or increase the electron density on this nitrogen atom NH2. Right? But if you consider the first option C2H5 SH, right? So here we have ethyl group which has the most electron releasing nature. Okay? That's why option A is correct here. Second point is what? Here we have sulfur but here we have oxygen and nitrogen. First thing is what? Methyl and methyl we have electron releasing group. But here we have ethyl. So here also we have higher electron releasing group. Correct? Second point is what? Sulfur nitrogen and oxygen if you compare. Sulfur is the least electronegative element in all these three. Right? That's why this one has the maximum nucleophilic power or strongest nucleophile. Is it clear? More electron releasing group, more will be the nucleophilicity or nucleophilic power. Find answer will be option A. Question number four. Fourth one is free radical substitution, right? D option is correct. You know the reaction is this. Chlorine molecule CLCl under light H nu forms chlorine radical. Chlorine radical and chlorine radical. Now this reacts with CH4. CH4 plus chlorine radical, it forms methyl radical, CH3 radical plus HCl. Right? Now this CH3 radical again combines with the chlorine molecule and forms CH3Cl plus Cl radical. Okay? And this continues till the chain terminates, right? This we call it as free radical substitution reaction. Option D is correct. Question number five. It's very simple, right? Here it is sp2 and here it is sp. So sp2 to sp. Correct? Question number six. Six you're getting A. Optical is not possible. What product you are getting? Write down the product and see optical isomerism is possible or not. Okay. You tell me what reaction is this? What is the name of this reaction? Question number six. CS3, CH, double bond, CH, CH, yeah. Final answer is correct. Yeah, it's aldol condensation. Why? Because there is alpha hydrogen present, right? So we know in aldol condensation we take two molecules of this, right? Acetyl dehyde is CS3, CHO. When this allows to react with dilute NaOH aqueous. First of all, we'll get aldol here, right? So aldol is what? CS3, CHOH, CH2, CHO. This is beta hydroxy aldehyde and beta hydroxy aldehyde is nothing but aldol. Aldehyde or ketone, we call it as aldol, right? However, in the question they haven't mentioned that we're heating this, right? So when you heat this, then this H and OH forms, H2O comes out and it forms CS3, CH, double bond CH, CHO, right? So this one shows geometrical isomerism, right? But this one since has one chiral carbon here, this will show optical as well. This will show optical isomerism. So the molecule as a whole, it shows optical as well as geometrical isomerism, option D is correct. So the reaction won't go towards the 100% completion. There will be a little bit of this product left, okay? So we get the mixture of these two, right? But you see the option A and D, so D will be more precise over here. If D option is not there, then obviously A is correct. So optical isomerism, this one will show optical because of one chiral carbon here. We know this fact, right? What is the condition for a molecule to show optical activity? What is the condition for a molecule to show optical activity? Condition for a molecule to show optical activity. Tell me, financial product won't show optical isomerism. One chiral carbon, chiral carbon, chiral carbon. So most of you are saying chiral carbon. You see the condition for optical activity is only one and that is chirality. Condition is this. When the molecule is chiral, it is said to be optically active. And what is chiral molecule? When it is not divided into equal half. If the molecule is not divided into two equal half, it is said to be chiral molecule, right? So when somebody asks you what is the condition for a molecule to show optical activity, your answer should be the molecule must be chiral, okay? And then you will define that what is chirality? When we say the molecule is chiral, okay? So the chirality is the property of the molecule by which it cannot be divided into it cannot be divided into two equal halves, okay? This is one part. Now, since this molecule contains one chiral carbon, right? So we are not discussing here about the chirality thing. We are saying what? The molecule contains only one chiral carbon and hence it is optically active, right? So presence or absence of chiral carbon is not the condition, okay? There are differences in these two. When you write down the answer in your board example, please take care of this thing. Chirality and chiral carbon has nothing to do with each other, right? Chirality is the property of the molecule. Chiral carbon is what? Chiral carbon is the sp3 hybridized carbon, which has four different atoms or group attached with it. Yeah, that is also correct. It should not have plane of symmetry, correct? So the point is presence or absence of chiral carbon has nothing to do with the optical nature of the molecule, right? The condition for optically active is what? Chirality, right? But it has been observed that if the molecule contains only one chiral carbon, then it is optically active, correct? However, presence or absence of chiral carbon is not the condition. I hope it is clear now. Next question. Question number seven. Tell me the answer. Question number seven. Question number seven is getting A, A is getting C, Sanjana A is getting C, okay? The correct answer will be option C. Why? Because this one is resonance stabilized. You see, this free radical is in resonance with this pi bond and then with this ring also, right? Maximum resonating structure will get over here. And that's why this one is most stable, right? So correct answer will be option C. Got it? The reason behind this is what? Resonance, right? Question number eight. Question number eight, what is answer B? Okay, most of you are getting B. This is the formation of what? Formation of Idoform, right? Which is CHI3. So obviously the Z or the final product will be option B, CHI3. First one is clear. We'll get X is what? X will be CH3, CH2, BR, alkyl halide. Now this alkyl halide hydrolysis we are doing, right? H2O. So we'll get Y and that Y will be what? An alkyl because HBR will go out. So we'll get CH3, CH2, OH. Now when this reacts with Na2CO3 with I2X's in it, you will get CHI3, which is nothing but Idoform. So this is just a reaction of formation of Idoform. Okay, it's a direct reaction. Option B is correct. See in these days, last one week you have, in these days you don't bother with the mechanism now, okay? Try to mug up the reactions, what product you are getting, in which reaction, what kind of, little bit of, you know, if you can relate with the mechanism if it is fine, but don't put so much of time in understanding the mechanism. That is the point I'm trying to make, okay? Question number nine. All of you are getting A. Why A is the answer? What is the condition of enantiomers? How did you do this? One pair of not ligand admins. These are not ligands here. One pair of molecule, you can say. Non-superimposable mirror image. Non-superimposable mirror image is fine. Another thing is what for non-superimposable mirror image? One molecule if it is R, other one must be S. If molecular formula is same and the configuration is this and this, R or S, then the molecule is said to be what? Enantiomers of each other, correct? This also we have discussed. And we know the, whatever the configuration of this, that is also you don't have to find out here, okay? If you compare here, we know what whenever we have odd number of exchange, one, three, five, seven like this, if these molecules have been exchanged, then the configuration changes from R to S or S to R. This kind of change is there. If even number of exchange is there, then there is no exchange. If R is the configuration and it would be R only. If it is S, then S only. There's no exchange in the configuration. We have discussed this already. Now you see with this A, the first molecule is option A. How do we get this? If you exchange the position of C2H5 and CS3 only. So one pair of exchange is there. That's why the configuration of this and this molecule will be different. Molecular formula is same. And hence this is the enantiomers of the given molecule. What is the relation of A, this molecule? What is the relation of these four? One, two, three, and four. One, two, three, and four. What is the relation? Why they are di-stereomers? Why they are di-stereomers? Sameer and Rithvik. See the molecular formula of all these molecules are same with same configuration. It means they are identical molecules. They are not mirror images, fine. But they have same molecular formula with same configuration. So they are identical molecules, same molecules only. You simple thing you try to understand like this. All molecular formula are same. Configuration is same. It means all the four molecule will have same behavior towards the PPL, plane polarized light. That's why they are identical. Di-stereomers is defined for different molecules. So option A is correct for this question. All these four molecules are identical. Next, tell me the answer. It's a direct reaction. 10th one is C. Mathali Ramchana is getting C. Suresh Sayuda is also getting C. Option C is correct. This is a direct reaction we have. Benzene mixed with air. So we have O2 present in the air. And catalyst we are using V2O5. Temperature is 775 Kelvin. In this you will get Malic Acid. Which is CH2COOH. Double bond CHCOOH. This two will not be there. So this one is Malic Acid. And since we are hitting this also, then condensation takes place. CH2O molecules goes out. Here OH and H will combine and comes out. And final product we will get here is CHC double bond OO. C double bond O. CH single bond C. Double bond C. This we call it as Malic anhydride. So option C is correct here. Malic anhydride. Cyclopentene. This is also very direct reaction we have. It's there in alkene chapter. Reaction of alkene in presence of alkaline KMN4. Two things we have to remember into this. Whenever you see the reaction of alkene with alkaline KMN4. So first of all it forms diol. Diol will get into this. That's what we have in all the options except this one. So when you have this information that forms diol. Option A you can eliminate easily. So one option we can remove. Now another thing is what the formation of diol. We must have two OH group must attach into this. And that's why the addition of this. You should know whether it is cis or trans or both. So alkaline KMN4 the addition of OH takes place from the same side. So what you have to keep in mind that two OH group will attach at the double bonded carbon atom from the same side. Means what? The cis addition takes place here. Cis addition of OH group. Two OH group from the same side. So we will get here cis 1 2 cyclopentane diol. Option C is correct. Must remember this thing alkaline KMN4 forms diol and cis addition is there. Both OH group attach from the same side. Question number 12. Question number 12. Do this one I am coming in a second. Psyme here is getting A. What about others? Some of you are getting A. Kondinia enzyme here is getting A. Ramcharan, Mathli and Lalita is getting B. So what is this reaction when Cs3 C triple bond Cs reacts with NaNs2. Again you see this reaction is again belongs to the hydrocarbon chapter alkaline. So when NaNs2 reacts with this alkaline, so the terminal carbon hydrogen will attach with NaNs2 and forms NaNs3. Like this you see. I will let you write down first. So the first reaction when we get A, in that reaction what happens here? This H will be taken up by this NaNs2 and forms NaNs3 and NaN will attach to this carbon. So the product A will have here will be this Th3C triple bond C Na plus NaNs3 will get. Now when this is secondly it reacts with C2H5Br. This is not A actually because we have second reagent also. C2H5Br then NaBr goes out and this C2H5 will attach to this carbon atom. So we will get CH3C triple bond C C2H5. Now what happens in the last step? What is the last reaction we have when we get B? A is this whatever I have written. What happens in the last step? What is the product B? When this molecule is allowed to react with H2, Lindler's catalyst, then what happens? Addition of hydrogen takes place and from alkyne will end up with alkyne double bond. But the point is whether this addition is in the cis manner or trans manner that's the question. And that's why some of you are getting A and some of you are getting B. There is the confusion actually you have probably. So whenever we have Lindler's catalyst, Lindler's catalyst always give the cis addition. So the product will be what? One triple bond will break and these two carbon atom will take two hydrogen atom. One here and one here and that will be on the cis manner from the same side and hence will get option B as the answer. Did you get it? You see why you are getting wrong answer some of you? That is because of you know the reaction actually that in this reaction hydrogen will attach and that idea you will have when you see the option also. That's why organic is little bit easy also when you see the questions in the exam. It is little bit easy but when you are studying it there are so many reactions which is very difficult to keep in mind means all at the same time. That's why at the same time it is a bit difficult also. The point is in this reaction where the confusion is that you got to know this thing that okay here the hydrogen will attach over here but how the hydrogen will attach either from the same side or from the different side there was the confusion. So Lindler's catalyst always gives cis addition or syn addition of hydrogen atom. So hydrogen will attach from the same side option B will get as the answer. Correct? All these are the smallest one information. This information you may have. You may write down in your notes or diary wherever you note down these points and try to keep this information in mind. These small things only helps you a lot in you know getting 4, 5, 10 marks in the exam when you write after one week okay. Anyways question number 13 next question. 13 what is the answer? This is also the direct reaction we have. What happens when alkene reacts with AlCl3 HBr? All these three reactions you see last three questions belongs to hydrocarbon family. One was from alkene other one from alkyne and this is alkyne. Have you ever heard alkyne can also give isomeric product with some reagent? Isomeric product have you seen these reactions? Okay Kondinia and methylation is getting B. B is correct. We are getting isomeric product into this. This reaction also you must remember. Alkyne whenever reacts with AlCl3 right or alkyne whenever passed over AlCl3 isomerization takes place. So we have N butane and when this reaction takes place we will get isobutane this product okay. Option B is correct over here I guess. Option B is correct yes. Isomerism takes place isomerization takes place and we will get option B. This is not multiple choice. It is not multiple choice question single correct. Major product we have to answer keep that in mind major product. Okay when it is single correct we have to answer the major product. 13th one is B question number 14. Okay just a second identify the reagent X in the following reaction. 14 again it is a direct reaction 14 all of you are getting C. Option C is correct here right what happens methyl ketone you see here this thing you keep in mind this is methyl ketone right. Methyl ketone are oxidized by sodium hypo halide hypo halide NaOX right we will get sodium salt into this and that is what we are getting here right. NaCl3 CSCl3 will go out chloroform will get and here will get ONA with sodium hypo halide and this has one carbon less than the carbon atom present in the molecule because one carbon atom is going out with chloroform right. That's why option C is correct. Question number 15 this one tell me question number 15. I am not getting your answers what is the answer question number 15 anyone there. Matthew is getting C okay so Matthew and Sayuzha is getting C and option C is correct. Okay question number 15 option C is correct you see actually correct answer is C over here. This CF3 when the trifluoromethyl right CF3 molecule when this molecule is attached at the benzene ring fluorine here all these three fluorine atom are electron withdrawing group right and the question is towards electrophilic substitution reaction. So electrophile required spot electrophile requires the charge density negative charge density to attack with right to attack on CF3 withdraw electron from the benzene ring and hence the ring deactivates towards the electrophilic attack on it right. When the electron is when the electron is like taken away by the CF3 molecule so obviously the electron density will decrease at this position right and hence the attack of electrophile will be difficult over there right. So the question is CF3 present on the benzene ring towards electrophilic substitution is correct. CF3 group activates the ring wrong because it is withdrawing electron so first of all it deactivates. CF3 group renders the ring basic that is also not possible because again it is withdrawing electron. CF3 group meta directing therefore deactivates the ring this one is correct. CF3 group is an ortho paradigm which is also wrong here right. So CF3 group is the meta directing group this information you should have it is a meta directing group right. Here we don't have any resonance thing it is eye effect right. The electronegative eye effect you are talking about it is a meta directing group and it deactivates the ring towards electrophilic substitution reaction. So option C is correct. Next question number 16 tell me the answer. Question number 16 under a hindrance condition tell me question number 16 what is the answer. 16th one some of you are getting D. Rithvik is getting D. Chai Meir, Sanjana, Mathleys, Ramcharan ok most of you are getting D and then some B and C also ok. You see in this question if you try to write down the steps here when an address condition is there. So first of all in this an address condition protonation takes place right and how this protonation takes place you see CH2 double bond CH then OCH3 H plus in presence of HBR. So what happens there are two possibilities here. One is if the electron goes on to this carbon atom then we have one product intermediate will get and other possibility is what if the electron goes on to this carbon atom and then we get the another product here. So in this case when the electron shift on to this carbon atom will get the blue one. So we'll write on here CH2 minus here right and with this H plus will attach on to this carbon atom. So we'll get CH2 H which is CH3 then we have CH positive charge on it and then we have OCH3. In the first in this case what happens this H plus will attack on to this carbon atom and we'll get CH2 positive charge on it CHH and then OCH3. So these are the two possible intermediate we have in these two reactions. Now the point is which one is more stable A and B according to that the product will get right will form the product ok. So now you see what happens here this positive charge is not involved in resonance and because of the I effect of this OCH3 group this carbocation is getting destabilized right but here what happens the lone pair on this oxygen atom this one you see the lone pair on this oxygen atom is in resonance with this positive charge or vacant P orbital carton over here. This is involved in resonance this carbon atom or the carbocation is comparatively more stable than this one ok. So this one is comparatively more stable carbocation here and BR minus will attach over here and the product will be CH3 CHPR OCH3. Answer will be option D most of you were getting this right option D is correct. Next one you see question number 17 tell me the answer. Question number 17 Khushal is getting C 2 promo 2 methyl butane. 17 we are getting C why because we are getting 3 degree carbocation over there and then bromine will attach right. So 2 promo 2 methyl butane is correct. Option C is correct. If you have doubt you can ask me otherwise I am moving on because all of you are getting C I guess ok. Here we will get 3 degree carbocation here and then the bromine will attach easily under that ok. That's a thing you can draw the structure you will understand. Next question number 18. Question number 18 18 you are getting C predominantly 2 butane because more substituted alkene is more stable right. 2 product will get here 1 butane and 2 butane. 2 butane is more substituted which is more stable product hence option C is correct here. Again this one also I am not solving ok. HBR will goes out and go out and then we will get question number 19. Reaction of 1 molecule of HBR with 1 molecule of 1,3,9 at 40 degree Celsius is predominantly. Addition of HBR it is. Tell me the answer. Question number 16. 16th one would we get option A if it was in the aqueous medium. Ok if you have H2O ok I will see this. If you have H2O present over there 3BR and temperature we will get CS3BR for sure. Then we will get some different product. Because you see then what happens the lone pair on the oxygen that will take the H plus from water first. And maybe we will end up with some alcohol maybe option 2 one of the product is correct. No it is not true because CS3CHO is there. So maybe we will see what happens when you have water present over there. That H plus from the water will taken up by the O CS3 molecule oxygen atom that you have over there. Means the lone pair of this oxygen atom this lone pair will take the H plus right. And then what happens here we have hydrogen postage onto this. Then either this bond and this bond will break right. So this bond will break and we will get vinyl carbocation here. CS3 OH will go out right. Vinyl carbocation CS3 OH will go out correct. And then we will get the other product that reacts with bromine and we will get the product. So we will not get CS3CHO in that case. We will get one of the product will be alcohol fine in aqueous medium. In one product we will have bromine and other one will have alcohol fine. Question number 19 what is the answer 19 you are getting A or D. Okay you see first of all in this reaction when we have 1, 3, V-dryne okay. There are two possible product here if I write down 1, 3, V-dryne is this CH2 double bond. CH single bond CH double bond CH2 1, 3, V-dryne. Addition of HBR gives you two possible product here. One is CS3 CHBR CH double bond CH2 and other one is CS3 CH double bond CH single bond CH2 BR. Okay this is 1, 4 addition product and this is 1, 2 addition product. Generally this product is very less stable okay. Right this product is very less stable because how this H plus and BR minus will attach over here when we will get negative sign here and positive sign here means this pi electron will shift here then only correct. But here what happens this reaction this is important okay you must keep this in mind and write down in your notes properly. This product is we call it as it is because of the kinetics of the reaction. We call it as kinetically favored product because of the kinetics of the reaction. Okay 1, 4 addition product it is because of the it is the thermodynamically favored product. Because of thermodynamics of the reaction because of the thermodynamics of the reaction. Two different product is possible here one is because the kinetics of the reaction and one is because of thermodynamics of the reaction. Here we will talk about the rate of the reaction here we will talk about the energy involved in the reaction because we are dealing with thermodynamics okay free energy change entropy and all we will discuss over here. Okay now this kind of product is possible when we have under vigorous condition we call it as. Vigorous condition means what when the temperature as it is given over here which is 40 degrees Celsius and all in this kind of when the temperature is more then thermodynamically favored product is possible. Right this kinetically favored product is possible when the temperature is very low around minus 80 degrees Celsius and less than that. Okay so both you must keep in mind. Okay according to the condition here it is 40 degrees Celsius right the major product in this case will be what one fourth. Addition product will get. Okay so that will be thermodynamically controlled or favored. So three bromobutene is not possible one two three. So we will have one bromobutene under thermodynamically controlled reaction. Okay option C is correct over here. Right so both condition you must remember according to the temperature. Right so both condition you must keep in mind very important one. When we have very less temperature then kinetically favored higher temperature than thermodynamically favored. Anyways when the temperature is high then the rate cannot be rate will be high right. So rate will not be the major factor here. Then we have to consider the energy involved in this reaction at higher temperature. That's why the thermodynamically favored product. Okay so C option is correct. Question number 20 all of you are getting C option C is correct. Question number 21 option D. Yeah for I do form taste the compound must contain CH3 CHOH group. This group must be there right. That's why option D is correct. Question number 22 tell me A to D addition of HCN. So all of you are getting option C. 22 option C is correct. All of you are getting C so I will not discuss in this. Question number 23. Acid catalyzed hydration of alkene. Is there any chances of getting primary alcohol? Is there any chances of getting primary alcohol? No right we have we'll get secondary and tertiary alcohol into this. In this what happens if you take H plus and H2. What is the product we'll get here? This bond pair will shift here. Right H plus will come over here and OH minus will go here secondary alcohol will get. There are no chances of getting primary alcohol. The secondary and tertiary will get tertiary will get by shifting of what? One true hydride or methyl shift according to the question we get. 23 option C is correct. Question number 24. Question number 24. Pent 3 into all into pent 3 into bond. Alcohol to ketone. Pyridinum chlorochromate affect double bond or not. Pyridinum chlorochromate affects double bond or not. It affects right. Chromic acid does not affect double bond. Pyridinum chlorochromate won't affect. Chromic acid what about chromic acid? Check this. Pyridinum chlorochromate won't affect. What about chromic acid? Yes Pyridinum won't affect 24. I think chromic acid also won't affect double bond. A and B I think both are correct in this. Double bond position will be as it is only alcohol has to convert into. Okay so PCC does not won't affect double bond. Chromic acid also won't affect double bond. If you see the structure here. CH3, CH2, CH double bond, CH3. Okay so this is the molecule we have here. And the product we get here is spent 3 in 2 bonds. So that will be CS3C double bond O, CH, double bond CH, single bond CH3. Okay so double bond won't affect it by Pyridinum and chromic acid. But we have here it is the 2 degree alcohol right. It is a 2 degree alcohol. I think because of secondary alcohol the answer should be B chromic acid we are taking. If it is primary then we can take any one of these A or B. Let me check. Lohetia is saying PCC not stable compared to chromic acid for secondary alcohol right. This is weak oxidizing agent right otherwise will oxidize to ester. That's what I said it's 2 degree alcohol that's why we are taking chromic acid not PCC. But Ramachan you are saying PCC is a weak oxidizing agent. Others is oxidized to ester. Now you see one thing first of all options C and D will affect double bond also. The first thing is that we have to preserve that double bond correct. And for that we have PCC and chromic acid correct. But PCC since it is a secondary alcohol we have so we cannot use PCC over here. That's why we are using chromic acid. Oh it's given in NCRT also. Fine so option B is correct right. Fine yes tell me. Option B is correct. Okay question number 20 24th you have done. Next one we will see. Just a second. Next one this one. Question number 28 what is your dot Lohetia? Someone mentioned I cause acetone effect on chromic acid. See chromic acid will take with aqueous acetone only. Acqueous acetone we are using as solvent over there in the mixture. So chromic acid with aqueous acetone that is the reagent we have. See that may be or may not be we cannot say that. But we usually take chromic acid with aqueous acetone only. The point is whether it will finally convert into ester or not that we can control the reaction. See the thing is first of all we have to choose the reagent which won't affect the double bond. Okay with that option D and C and D we can eliminate easily. Now we are left with A and B. So A and B we can control the reaction. If you want to like you know it's not impossible that we cannot form ketone from that chromic acid. We can if you provide the necessary condition or the required condition we have then we can end up the reaction with ketone. That's a different thing. But the point is the double bond is not affected by chromic acid over there that we have. That's why we are choosing option B over there. Understood? C and D we have eliminated directly. Since they are affecting double bond. It's another matter that whether we can control the reaction or not whether it is difficult to control the reaction or not. That's another thing. So that is not the question. Question number 28. Ethyl isocyanide on hydrolysis in acidic medium generates. Ammonium salt or it is B. Is it B? Isocyanide will oxidize into carboxylic acid with less than one with number of carbon atom less than lesser by one. Yes, correct. Option A whenever we have isocyanide that will oxidize into an acidic hydrolysis that will oxidize into carboxylic acid with one carbon less. So obviously we have ethyl isocyanide. Ethyl isocyanide is what? C2H5Nc. C2H5Nc. Plus we have acidic hydrolysis. So in this reaction first of all we will get carboxylic acid which is HCOOH and then we are getting amine C2H5Nh2. This will convert into acid and this will forms an amine over here. I have done this reaction in the class. I have given this reaction I guess. Isocyanide under acidic hydrolysis gives you carboxylic acid. So we have only two options possible here. A or D. B and C are not possible if you have disinformation at least. And always remember this carbon atom of isocyanide will form the carboxylic acid. This alkyl group will form the amines. Option A is correct here. Question number 29. All of you are getting D. Option D is correct. Option D is correct because I am not doing this one. Question number 30. Question number 30 based on resonance. Aritya is getting D. Mathili is also getting D. Apurva is getting D. What happened? What is the answer? What is the answer? What is the answer? What is the answer? Tell me. How many valence electron nitrogen may have? How many valence electron nitrogen may have? It cannot have more than 8 electrons. Am I right? Nitrogen cannot have more than... See all the resonating structure must follow the basic rule of bonding. Nitrogen can never have more than 8 electrons. You see in this structure, nitrogen has 10 electrons. And you don't have to observe this. If you go by these options one by one, obviously here it is written as 10 valence electron. If it has 10 valence electron, which violates the basic rule of bonding, that's why this structure is not acceptable as the resonating structure. So option C is correct here. Understood this. However, if you see the condition of resonance is possible here. This bond pair goes here and here we have the positive sign. The condition is true. But with all these structures, the basic rule of bonding must have to follow. That's why this C option is not correct. We cannot accept this as the resonating structure C. Next one, can you see this? It's fine or I have to enlarge this. Let me know. What happened? What is the answer? Question number 34. Vashnavi is getting A. Tell me the answer. Vashnavi is getting A. What about others? What happened? Okay, see. See, cyclobutyl bromide. The structure is this. Reacts with magnesium in dry ether. That's the reaction for the formation of Grignard reagent. And we get RMGX, which is where we are here. Magnesium with dry ether. This is the organometallic compound. Reacts with ethanol C2H5OH. So here we get what? CH3 and here we get OH simply. Because we have ethanol C2H5OH, correct? So how does this reaction take place? Suppose we have this reaction plus RMGX. How this reaction goes? Right? MGX will take this OH and R will attach with C2H5. Okay? And then we'll get H+. Okay? And we get here what? CH3OH here. Plus we'll get MGPROH. H2 also will have OH. It also will take H2. So this is the product B we have here, right? Compound reagent ethanol to give the alcohol B. So this is the alcohol we have, which is the answer option A. After this, if you want to write down the product, then what happens you see? Here we have CH3. Here we have OH, mild acidification. So we'll have H+. So with this H+, this H+, will attack on to this OH- and H2O will go out and we'll get a carbocation over here. So we'll have a carbocation here. The CH3 will be as it is and the carbocation. And then what happens? We have adding expansion into this. So this fourth member ring becomes five member, right? And one CH3 present on this ring and then H+. Also, suppose we have CH3 here, one H will be here. And here we have the positive sign, right? Now in this, what happens? This is 2TD carbocation. If this hydrogen comes over here, 1,2 hydride shift, then the carbocation we'll get here, that will be 3-degree carbocation, sorry, we'll get the product as this CH3. Here we'll have the positive sign, okay? Now this will take the Br- amount of HBr we are using. So Br- will come over here and the final product, which is not the question here, will be this Br here and CH3. This is the final product, okay? Any one of you, if you have any confusion in ring expansion, what you do in this case, just you number the carbon atom simply, that will be the easiest method, 4, 5 and 6. So what happens? Either this bond pair will come over here or this one, any one of those, both will give you the same result. Suppose this comes over here, right? The fourth and fifth will join and the first one will lose one electron from this. The first one will have the positive sign, okay? So this is the first carbon here, second, third, fourth, and this carbon is the fifth carbon, this carbon we have here. And this is the sixth one. Like this you see, first one will have the positive sign and this and the product will get here. OMGX will not change into OS. See what happens? When you have mild oxidation, the oxidation of water is too easy. And that too also what happens if water goes out, correct? If water, whatever you are saying, that will also take place a little bit. But this is the more stable reaction we have because ring expansion is taking place here. This step happens so quickly, okay? Because we are going towards the more stable carbocation, okay? So in this case, that is also possible whatever you are saying. In this case, since we are talking about the more stable product, so most of the H plus is taken up by this OH molecule over here, right? The final product will be this. However, the question is not this. We have to find out only B, option A is correct. Question number 35, you tell me. Thought in just a second, let me see. The organometallic compound reacts with ethanol to give an alcohol B, correct? After mild acidification, here we are using, no, see, the, this, here we have the condition, we have H plus H2O here. Then we will get this. Here we should write H plus H2O. Mild acidification we are using here. I did not write it over here. Then only we will get this, okay? This is fine. We are getting B. After a prolonged treatment of alcohol B with an equivalent amount of HBR. Now this HBR will give H plus. So after this we have this reaction. So this is understood. I thought, because we have done so many reactions on this in the class. Whenever we have reaction of RMGX, we are using this H plus over here. That is what mild acidification we are using. So we will not use this, OH will not go there. With mild acidification only, MG will get this OH and BR and then OH will attach to this. Correct. So this is the problem. Question number 35, you tell me. 35, Continia is getting C, Rithik is getting B. Vaishnavi tell me the answer. Remember you have asked the same question to me. A few days back. Opening of epoxide ring. See one more thing, I am not giving you a break today. I have to leave at 5.45. Okay, so we will continue the class till 5.45. We will not take any break. Yeah, it is option C is correct. Okay, you see this reaction is important also. This question they have asked in exams also, various other exams. Okay, and this type of question, opening of epoxide ring is very important in ether chapter. Okay, so yeah, C option is correct. Option C is correct. Just a second. Hello? No, no, no, I don't record. I don't record. Thank you. Yeah, so I will discuss the condition of this. Okay, but what happens in basic medium and what happens in acidic medium? Okay, first of all, you see, in this question that is given, this carbon atom is substituted, right? This carbon atom is substituted. And this is the more hindered site we have because this carbon contains only two hydrogen. This contains two methyl group. So this one, this site is more hindered than this site comparatively. This is one thing. Okay, if there is symmetry, then whatever bond you break, that won't make any difference. But here either you break this bond or this bond, there only you will get confused. You got confused maybe. So remember one thing. What I said, this carbon atom is more hindered. This one is less hindered. Now what happens in basic and acidic medium, you see, if the molecule I am taking this one, you must remember this. CS3 whole twice CCH2 with oxygen like this. When the reaction takes place in basic medium, remember, when the reaction takes place in basic medium CS3 OH, so O minus I am taking here, this will attack onto the less hindered carbon atom. Always in presence of base, less hindered carbon atom will attack, and the ring open up from the less hindered site. Since from this site, the rings gets open. So we are saying in basic medium, the ring opens up from the less hindered site. So here we'll get here, CH3 whole twice CO minus CH2 OCH3. Okay, which takes the S plus and we get CH3 whole twice COH CH2 OCH3. So this oxygen atom is here only, here. So this is B. Now when this reaction I'll write down again, CH3 whole twice CCH2 and we have a epoxide ring here. When this reaction takes place in acidic medium, H plus. Now in H plus what happens, the lone pair of this oxygen will take part in the reaction and this will attack onto this H plus and this H plus will attach over here, CH3C CH2, H and positive charge on it. Okay, now in this also there are two possibilities. Either this bond will break or this bond will break. Okay, since this is an acidic medium, so the carbon which is more hindered from where the ring, from that carbon, the ring will expand. Or what happens in another way we say, we have H2O also here. When this oxygen have the positive charge, this H2O will attack onto this carbon atom and this bond pair comes over here. So what you have to keep in mind, the ring opens up from the more hindered side. So what happens here you see, CH3 whole twice COH2 plus and then we have CH2OH. And then from this one, H plus comes out from this H2O to make this oxygen stable and we get the final product here at CH3 whole twice COHCH2OH. Option C is correct here. So in short, what you have to keep in mind in this kind of reaction, if the medium is basic, right, the ring opens up from the less hindered side, from this side. If the medium is acidic, then the ring opens up from the more hindered side. That is what you have to keep in mind, nothing else. So option C is correct. Next. Now these are the few questions of polymer. Just one by one you see, these are just the information based. Different monomers and all, just you revise. These chapters you must go through in these three, four days. You never know what question they are going to ask. One question if you can make it from this chapter, that will be good. Can you see this? Tell me the answer, one and two. The turbidity of a polymer solution measures scattering. It measures the scattered light. First one, option A is correct. Turbidity of a polymer gives you the idea of scattering of solution. Like scattering of light by solution. Okay, kind of, tindle kind of effect. Not exactly, but a little bit same. First one, option A is correct. Must remember that. Which of the following is not a semi-synthetic polymer? Cis polymer are naturally occurring. Natural polymers. Cis polymers are natural polymers. Those are not semi-synthetic. Question number two, option A is correct. Cis polymer are natural polymers. That is the key thing you must keep in mind. Next one you see. I have to fix this. Just a second. Question number three, number four, number five. Tell me first. Question number three, option A is correct. Because there is a condensation reaction between the monomers. H2 molecules generally, H2-H3 goes out and will get the polymer, will get the bond. Question number four, question number four, polymer obtained is polymerization is phenol formality height. C is correct. C is correct. Which is an example of thermosetting polymers? Thermosetting polymers, what is the example? Beckelite, right? Beckelite is an example of thermosetting. All these are given in NCRT, okay? See, one kind of question, just two, three questions more I'll give you then I'll show you one question, one different kind of question I'll show you in this. Which has been asked in JEE exam. A six. And then you do this one, eight. What is answer question number six? The following sets contains only addition polymers. Only addition polymers. So polythene we have, then PVC, acrylene. So terylene is not an example of addition polymer. This one is not. Option B is correct here. Which of the following is contains isoprene units. What is isoprene? Isoprene units are natural rubber. Isoprene is this one. I'll write down here CH2, double bond, CCH, double bond CH2. And here we have CS3. This is isoprene. All these things are there in your notes. You can check. Terylene, ninth one. What is the answer? Is a condensation polymer of ethylene, glycol and what? Ethylene, glycol and terfralic acid. So option D is correct here. Okay. Next you see this one you do. Previous your questions. Last nine question for today. First you solve all this question. I'll give you five, six minutes and then we'll discuss. Option number 50, B is correct. Uses also I have given the uses in the classes. Okay. This is there in your notes. What is 51? TICL3, we use for what? Is using Ziggler-Natta catalyst, right? Use as Ziggler-Natta catalyst. So A should be 2, right? And we are left with option B and C only. Tell me 51. TICL2 is used in Wacker process. In Wacker process, alkene is changed to aldehyde. Alkene is changed to aldehyde. So 51. I'll just write down. Question number 50, option B is correct. 51 is also option B. Which one is the condensation polymer? 52. 52 is A. Dacron. Dacron is the condensation polymer of ethylene, glycol and methyl terfitalate. Ethylene, glycol and methyl terfitalate. 53. Polyamide. Which is the following polymer is a polyamide. Nylon, correct? 53 is B, yeah, nylon. 54. 54 you let it be. We saw this in the last. 55 you tell me. 55. A polymer used for optical lenses. 55 D. Polymethyl metacrylate. See these questions, uses based question they ask. One is this catalyst thing. One is uses based question they are asking. You see this one also. What is the different monomer of one particular polymer, that kind of question and use based question they are asking. 55 is D. 56, cationic polymerization. 56. For cationic polymerization we required what? Electron deficient species. Right? For cationic polymerization we required electron deficient species. And which one is electron deficient here? A, L, C, L, 3. Yes, 56 is C. Beccalite is obtained from phenol by reaction with what? Tell me 57. 57 is formaldehyde, again H, C, H, O. Beccalite, phenol and formaldehyde. We have done this. The polymer containing strong intermolecular forces. Example hydrogen bonding. What kind of, in which we have this hydrogen bonding? 58. 58 is B, nylon 66. Okay. 54th, have you done this 54th question? 58 is B. 58 is B. Have you finished 54th one? Let me discuss that but let me, first you let me know. This type of question have you solved or not earlier? And you must have seen in most of the book they haven't solved this question. They have just given the answer. The answer is this. Yes. Okay, now you see. In polymer, usually one type of question they have asked. You have seen this. They have asked this in JMS question, 54th one. We'll see that question, but we'll see the concept first. When we talk about the average molecular weight, average molecular weight of polymer, then there are two types of average molecular weight generally we'll calculate. One is what? One is weight average. Copy down this, all of you. Weight average molecular weight, which has been asked in that question. And other one is number average molecular weight. Number average molecular weight. So first thing in 2013 in JMS, they have asked this question, which is the 54th one. We'll solve that question. They haven't asked this one. So maybe this concept they may ask this year. Or you never know, maybe they can ask this one also. But since they haven't asked this one, so this one is a bit important this year. So all of you copy it down properly. Now, little bit notation we'll use here. Mi, if you write, it is the molecular mass of each monomer. Monomer from which the polymer has been formed. Xi is the fraction of the polymer chain. What is the fraction of each monomer basically? Polymer chain. Now, suppose if I write down here, I'm taking some random data here. Mi, Xi. Suppose we have two different monomer, three different monomer, suppose we have, which is 14. I'll give you the exact question. I'll give you 14, 26, 38, 50, 62, 74, 86. These are the different, different monomers we have. Now, for this monomer, the fraction is given, which is 0.05. For this, it is 0.15. For this, it is 0.21. For this, it is 0.28. It is 62, 0.18, 0.10, 0.03. So the number average molecular weight, you have to only keep this formula in mind. Number average molecular weight will be the summation of Xi into Mi. Means these two, if you multiply, you have to find out this, Mi, Xi. Find out these values here, here, here. Like this, you find out the value and you add all this, you'll get the number average molecular mass. So this is the number average molecular mass. Now, the weight average molecular mass, weight average molecular mass is equals to summation of Wi Mi. Now, Wi is what? It is a weight average and weight average can be given by Mi Xi divided by Mn average. This Mn, okay, notation I forgot to write. Number average molecular weight is Mn bar. Weight average molecular mass is this. Mw stands for weight bar. So weight average molecular mass is this. Number average molecular mass is this. Wi is the average weight. Is the average weight for each monomer action. So you have to find out, in the next one, you have to find out Wi Mi. Find out these and you add these again, you'll get the weight average molecular mass. So in this question, let's come back to this question. In this question, in this one, if you calculate this, you will get 47.7 kg per mole. This value is given in kg per mole. Mn value it is. Right, and you solve this, the weight average molecular mass, you will get 53.9 kg per mole. Only you have to keep, yeah, it's like that only Simon. You have to keep this formula in mind. Remember Wi, you have to calculate by this formula. Right, average weight. Now in this, there are two more terms that is used to us. That is polydispersity. Polydispersity, Pd. It is calculated by the weight average molecular mass, Mw bar, divided by Mn bar. Number average molecular mass, which is nothing but 53.9 divided by 47.7, which is 1.1. Degree of polymerization. All of you have done this. They are giving you some extra data into this. They will not ask this much. Okay, but this thing, since I have to explain all these things, you never know what they are going to ask. I am discussing all this. You have to keep only this formula in mind. And with this, we'll find out polydispersity weight average molecular mass divided by number average molecular mass. Now, degree of polymerization, which is there in the question, degree of polymerization, there are two formula here. The first formula is number average molecular mass divided by M. This M is the mass of monomer, mass of monomer, or we also call it as repeating unit, right? If monomer is given. So in the previous question, I did not write here, but in this question, it is given that the mass of monomer is 56. 56, unit is the same kg per mole. So that value you can substitute here and you will get the answer. Now the another formula which is more important is, and this formula we use to solve this question. That is the molecular mass of the polymer of polymer divided by molecular mass of the monomer. Both are more or less same thing, but this one is more useful. Molecular mass of polymer divided by the molecular mass of monomer. Now we will go to that question. Is it clear till here? All of you have written the formulas, right? I request you to write down all this formula. If this question comes, you cannot do anything if you do not know. You will easily lose your four marks. That is why I am discussing this. Now you see the question. What question we have? This one. If a polythene sample contains two mono dispersed fractions, means we have monomer actually, two monomer actually, which is the ratio of 2 to 3, right? Degree of polymerization is given. DOP, degree of polymerization is given, which is 100 and according to the formula, what we can write molecular mass of the polymer, mass of polymer divided by the mass of the monomer. And what is the monomer here we have? The monomer is what? The monomer is polythene, whose mass is 28, right? Polythene. 28 is the mass of this. So 28. So mass of the polymer, one polymer, polymer one is 2800. According to this data. Another polymer you see, degree of polymerization is given 200. So mass of the polymer 2 or the second polymer that we have is nothing but 28 into 200, which is 5600, clear? Right? Now X1 for the first molecule is what? It is a fraction. So 2 divided by 5. Stop me if you don't understand this. X2 is what? 3 by 5. Now you see, we have to find out, I'll just, okay. Now we have to find out the weight average, see weight average molecular weight. So first of all, the weight average molecular weight, the formula is what? The formula is summation of wi mi, right? What is wi? It is the average weight, which is equals to mi xi divided by mn bar, number average molecular mass. So first of all, we'll calculate the mn bar, number average molecular mass. With the formula, you see it should be mx, m1 x1 plus m2 x2, right? m1 x1 is what? m1 x1 is 2800 into 2 by 5 plus 5600 into 3 by 5, right? When you solve this, you'll get 5600 into 4 by 5. I'm not calculating this, later on we'll calculate it. Mn average is this. Now we have to calculate this wi. What is w1? It would be m1 x1. m1 x1 is what? 2800 into 2 by 5 divided by mn bar is this, which is 5600 into 4 by 5 and when you solve this, you'll get 1 by 4, yes or no? Similarly, you calculate w2. 5600 into 3 by 5 divided by 5600 into 4 by 5 and when you solve this, you'll get 3 by 4, right? Now we'll calculate the weight average molecular weight, which is mw bar is w1 m1. w1 is 1 by 4 and what is m1? m1 is 2800 plus w2 is 3 by 4 m2 is 5600 5600. When you solve this, you'll get 49. Did you understand this question? Let me know. So answer will be option A. Is it clear? How to solve this kind of question? So number average molecular weight, they haven't asked. X i, M i, you have to calculate. Weight average molecular weight, I've given you. Degree of polymerization formula, I've given you. So two, three formulas are there. Only you have to do the calculation, that is it. Right? With that formula. I hope you can solve these kind of question now, yes? Okay. We'll wind up the class here only. Okay. Again, I'm telling you, don't read any new concept now. Okay, this is just one thing I saw it today, this particular question. I have given you this particular formulas, okay? But otherwise, don't try to understand some new concept. Some formula, if it is there, you can memorize, you can do some question onto that. Okay. Keep on revising things. Thank you. Take care. Goodbye.