 Hello guys, good afternoon. Good afternoon, Ritik. Who all are there? Please type in your name. Only two of you, Ritik and Visist. Where are the others? Okay, hello Saimid. Have you started preparation for board exam? Okay, Visist, your exam is already going on, right? Okay. Okay, so last class, we have discussed you, you know, board questions. So today also, we are going to discuss a few questions that is, you know, board level questions only. You see, the first question like the one marker, what is the answer? IOPSC name. Two phenyl ethanol. Others tell me the answer. Two phenyl ethanol. Okay. Again, you see Saimid here. You have written this two phenyl ethanol, but there should not be any space. Okay. So when you write down the name, the answer will be two phenyl ethanol without any space here. You should not write any space between this phenyl and ethanol. Okay. That you must take care of, otherwise you'll lose your mark. This question you tell me? We'll see some reasoning question also today. No, it should be small only. What is the answer answer? Tell me the name. What is the IOPSC name of the structure that you have written? What is IOPSC name? I am not getting any response, guys. Done. What happened? Tell me. Okay, I'll write it down. You see. Write the structure of an isomer of the compound, this which is most reactive towards SN1 reaction. Okay. So first of all, we'll try to, you know, find out the degree of unsaturation. Okay. However, in this, this is a board level exam. The board level question only one mark is there. You don't have to, you know, find out, show this DOU, degree of unsaturation there in the answer sheet. Just you calculate it for your own sake. Like for this one, you see four plus one, the formula we have H plus X minus N divided by two. So the degree of unsaturation is zero. Means there is no double bond or triple bond or ring present here. It is an alkyl halide. So since SN1 reaction we have, and for this, you know, the alkyl halide must be what? Tertiary. Tertiary alkyl halide. So here you see the structure would be CH3C, CH3, BR, CH3. Why it is tertiary? Because, you know, for SN1 reaction, it follows by the formation of carbocation and tertiary carbocation is more stable. So this will be the possible structure. Okay. Understood. Next question. 10 minutes one. PBNO3 on heating gives a brown gas with no dimelization. Okay. Answer is correct. It's NO2. And the reaction will be what? This reaction you have to write down. The reaction will be PBNO3. Heat this, it forms PBO plus NO2 and O2. This is not a balanced reaction. So this NO2 undergo dimelization and forms N2O4. So the answer for this question will be what? NO2 is the gas. Right. This reaction you have to write down. If you memorize last class also that we have discussed in one more question, they have asked one question based on this reaction, this kind of reaction. There we have P4 when reacts with some compound. If you remember HNO3 or something. Right. So one reaction based question they ask here in one marks. What happened? Why you are messaging? Just let me check it. Okay. I saw some messages now. Okay. I could not. The message was not, you know, reflecting over here. Just now you saw your messages on your WhatsApp. Anyways, anyways, that's why I was waiting. Yeah, correct, correct. So this is the answer for this question. Now, the next question you see, this is again you see, this is sixth number question. So I'm not discussing all the questions, right? Two important questions which is there. I'm just discussing those. This is two marks question. Do this one. What is the unit of K and order and molecularity? Order is zero. Molecularity is two. Yeah, right. Zero order is correct. Molecularity two is correct. Unit K unit you already know. Second inverse mole per liter one minus N where N is the order of the reaction. Okay. So this you see, like I said, order is a experimental quantity. Okay. So you cannot, you know, find it theoretically or mathematically. It is an experimental quantity. Right. So one thing you just keep in mind, I tell you that any photochemical reaction, right, always proceeds with a chain mechanism. Right. And there is no any rate determining the step in photochemical reaction. Right. Like in this case, as 2 plus CL 2 reacts from HCL in presence of light. So this is again, photochemical reaction. So all these photochemical reaction proceeds with a chain mechanism. And in chain mechanism, there is no any rate of reaction determining the step. Okay. Hence the rate of the reaction is constant. Right. Rate of the reaction will be constant. Hence order will be zero. Molecularity is what? One molecule of H2. One molecule of CL2. So molecularity is again one plus one two. Right. And then the formula of K we already have, which is this formula is liter inverse, sorry, second inverse. Here we have second inverse mole per liter one minus N where N is the order of the reaction. One thing I have mistaken here, this should not be second, this should be time actually. Right. Time would be the correct one. This can be anything second or minute anything. So when order is zero, the unit will be mole per liter time inverse. This is the unit of K we have here. Right. So order is zero. Why zero? Since it is a photochemical reaction. Molecularity is two. This is a very common question. The difference of order and molecularity for both point of view. The difference between order and molecularity is a very common question for both point of view. Okay. Next question. This one. Right chemical equation involved in following reactions. Hoffman bromide degradation carbide amine reaction. What is Hoffman bromide reaction and carbide amine reaction? What happens in this reaction? You can write it down. What is Hoffman bromide degradation reaction in Hoffman? Okay. All of you have done. You see in Hoffman bromide reaction, amide reacts with bromine right in aqueous or ethanolic solution. Right. So Hoffman bromide gives you, this gives you primary amine. Primary amine. Amine. And in this what happens the amide, which is R C double bond O NH2. Amide is allowed to react with bromine in presence of NaOH. And the product we get here is R NH2, which is one degree of amine plus Na2 CO3 NaBr and then H2O. This is the product we get. So we will get primary amines in this reaction. Okay. Hoffman bromide. Carbide amine what happens in this? Little bit of theory of this also we write down that what happens in Hoffman bromide that you write down first two, three lines and then you write down the reaction. Similar thing you do with carbide amine reaction also. In this what happens, aliphatic or aromatic primary amines with chloroform. Right. So we can have this R NH2. This should be also primary, but this can be aliphatic or aromatic both. Primary amine. Can be what? Aliphatic or aromatic. When this primary amine is allowed to react with chloroform C HCl3 in alkoholic KOH solution. It forms carbide amine RNC isocyanide. It is basically carbide amine is nothing but isocyanide which is 3 KCl plus 3 H2O. Yeah. So this primary amine can be aromatic or can be aliphatic both. So this you must remember. So whenever you have any name reaction like this, first of all you write down the statement of that reaction that what happens in that reaction. Okay. Then you write down the reaction like this. That is the best way. Okay. Okay. Next question you see. This is again the logical question. The next one I'm giving you. Do this one. Gas A more soluble in water than gas B at same temperature because the following two gases are higher value of KH and Y. Okay. So see actually what happens here. The Henry's law according to Henry's law if you discuss. What is Henry's law? We have two, three expressions here and we can write Henry's law is supposed to work. The concentration of gas dissolve. Right. So concentration I'll suppose I'll write C only concentration of gas dissolves is directly proportional to what pressure of the gas. Right. So when you remove this proportionality sign C is equals to KH into the pressure of that gas. Right. Where K H is what? K H is the Henry's law constant. Okay. Now what happens? The question is which one of these two gases have higher value of K H. Okay. So what happens if K H increases when this K H increases solubility should also increase which is nothing but the concentration. Right. If the K H is higher, it means the concentration will high. And then we can say the solubility will be high is high with higher K H value solubility will be high with higher K H value. Correct. So you see here it is given the question of following two gases have higher value of K H. So the gases will have higher value of K H will be more dissolve. Right. So since A is more soluble in water, so we can say the K H value of gas A is more than to that of K H value of gas B. Right. This is the answer of the first question. Second one is again, you know all this maximum boiling as you throw shows gas B should have higher K H. Just a second. Gas A is more soluble in water at the same temperature. More soluble means more will be the concentration its partial pressure is more solubility decreases with increase in K H value. More solubility lessens the partial pressure. Is it? See all the gases dissolve according to its own partial pressure. Yeah. So higher will be the pressure more will be the dissolution of gas partial pressure I'm talking about. Higher will be the partial pressure. Easier is the dissolution of gas. Just a second. If these was to K H X into right. Okay. Okay. Just you this question. I'll just come to this question again in five, 10 minutes. I'll give you another question. I'll think on it. Okay. Let me think on it once you solve this question next one. You do this one. I'll just come to that question again. Question number eight will do it again. Just you solve this question first. Well, a coordination compound is mixed with a GNU three three moles of this right the formula. Done. Okay. So there are three moles of agCl precipitated. Right. So three chlorine atom must be outside the complex square bracket. Right. So the formula of this will be co NS three whole six. And cl three will be outside. Okay. And the name of this compound is. It is hexa mine. Cobalt oxidation is says is status three. So hexa mine cobalt three chloride. Yeah, correct. The answer is correct. Okay. Okay. Next one you see question number 10. Then we'll see some five marks question. This one. Then we are F three. The hybridization will be seven into four. That is 28 28 by eight is three. So three bond pair and two lone pair for BRF three for BRF three. We have three bond pair and two lone pair. So steric number is five hybridization is SP three D. Right. And what is the structure. The shape will be what T shape. If you write down XZF four. In this, the number of lone pair and bond pair is what eight 36 by eight. So we have four bond pair. And two lone pair. Right. So the hybridization will be six is a steric number hybridization will be SP three D two. Right. The geometry will be what and shape will be what the square planet. Here the shape will be T shape and geometry is what diagonal bipyramidal diagonal bipyramidal. Okay. So this is the first part. Now you see the second question. The second part of what happens when as to gases passed through an echo solution of Fe three plus salt. Okay. So what happens this Fe three plus salt in that the sulfur dioxide act act as a reducing agent. Right. And it reduces Fe three plus two Fe two plus right. So this S O two here S O two act as a reducing agent. And it reduces Fe three plus two Fe two plus. Most of you have written the same thing. The color of Fe three plus is red, which is not required here, but I'm giving now, but color of Fe two plus is green. This is a color change. Right. And now what happens the last part X E F four reacts with S B F five. Okay. The product that we get here is X E F three and S B F six. This is the product we get. Right. So all these Zenon Zenon fluoride that we have act with Lewis acid. Okay. So Zenon fluoride act with Lewis acid and forms complexes. So this is the reaction we get here. Got it. Yeah. Right. We get the complex. Next one right down the product in these reactions. What is the product we get here? The product is tertiary butyl iodide. The first one will get tertiary butyl iodide. And then we'll get methanol. Okay. So all of you are getting the same product. What about the second one? Then the second one. Is it butene to butene? The first answer is correct. And what happens this lone pair I'll write on here only. This lone pair of oxygen takes this H plus and we get C H three C C H three C H three O C H three. Here we have hydrogen and positive sign. Further what happens here? This bond pair comes over here. We'll get tertiary carbocation which is C H three whole thrice C positive charge plus C H three O H. Now this will take the I minus and we'll get C H three whole thrice C I tertiary butyl iodide. Similarly, when you have alcohol that is two degree alcohol, right, which is heated is in presence of copper as a catalyst. This converts into ketone, two degree alcohol. The product here it will be C H three C H two C double bond O C H three. This is two butanone. Okay. And in the last one, the product will be if you have phenol that is C six H five O H H five. When it is heated with chloroform with an O H and then the second reagent is some acidic acid we are putting in. So this converts into hydroxy benzoic acid. This is the product we get right in presence of acid that aldehyde again oxidize into acid. Actually, you can write down the both product first to get aldehyde which finally converts into acid if it is a complete oxidation we have. Rithik you'll write butane two own or two butanone. Anyone of these two you can write but two own is not correct because when you write but two own, it does not, you know, give the information that where do we have any double bond present or not. Or do we have all sigma bonds there with acid it becomes salicylic acid. Do this conversion. Next question. Do this conversion. Total benzene to biphenyl. Yes, the first one is fitting reaction. What reagent we use here? Can you tell me? It's similar to Wood's reaction propane to Ido propane. How can we convert this propane into alcohol like propanol? Can you convert propane into propanol first and for that what reaction we use into all antimarconic of addition of what? HBR you are using but it is Ido propane. Can you see it is not bromopropane it is Ido propane. Hydration gives you what kind of product? Antimarconic of or marconic of? Okay, then you are using Finkelstein correct, okay. Yes, so with marconic of will you get one Ido propane then? Iodine will be there at first carbon with marconic of? Try that, write it down, try to see. Okay, so to make this Iodine to add this Iodine at the first carbon, right, we need antimarconic of addition. Okay, have you heard about oxym sorry hydroboration oxidation reaction? Hydroboration oxidation, DH3 with H2O2 and all? Yes, so when we do the hydroboration oxidation reaction if you remember this reaction follows antimarconic of addition and addition of OH- and H plus takes place, correct. So when you do the hydroboration oxidation of propane you will end up getting propanol, correct. Yes, you will get antimarconic of addition and OH will attach at the terminal carbon there and will get propanol. When you get propanol then you can allow this to react with HIE, you will get H2O out and Iodine will attach at that carbon. Did you understand this? So first step we can use is hydroboration oxidation and then we will use the reaction of alcohol with HI, clear? Do you want me to write down this reaction? So propane is CH3, CH double bond CH2. So hydroboration oxidation reaction, BH3 with H2O2 gives you antimarconic of addition of H plus and OH minus, we will get CH2OH, right, antimarconic of addition. Now in this if you put HI then H will come over here, H2O will go out and this Iodine will attach. The final product will be CH3, CH2, CH2I, Ido propane. What about the third one? Two bromobutane, two butane, done. Okay, one more thing you see in the 11th C question. Just now Ritik has sent me one image. Okay, in NCRT it is given salicylic aldehyde, right. So the thing is salicylic aldehyde forms when you are assuming partial oxidation, right. So if you write down salicylic aldehyde, it means you are assuming partial oxidation when you write salicylic acid, it forms when you have complete oxidation. We know aldehyde on oxidation it gives acid, right, that's the only thing. So if you write aldehyde also it's not wrong, acid also it's not wrong. Only thing is assuming complete or partial oxidation, right. But in NCRT it is given salicylic aldehyde, so better you write aldehyde only, okay, that's the thing. Yes, the next question we have 12th, third one, bromobutane to butane, right. So we use alcoholic KOH it converts into two butane very straight forward direct question we have. Okay, next you see here, the next question is question number 13 I guess we will see. Yeah, this one. What is the structural difference between starch and cellulose? What type of linkage present in nucleic acid peptide bond? Okay, correct, correct, right. A few things you have you should add into your answer here. I'll dictate you just write it down. A few points to write down in this. Write down starch is branched homopolysaccharides of D glucose. Starch is the branch of or we can say starch is branched homopolysaccharides of D glucose and contains and contains two types of glucose polymer and contains two types of glucose polymer. That is amylase and amylopectin that is amylase and amylopectin. These are three marks question. Okay, two types of glucose polymer and they are amylase and amylopectin. Now what is amylase? It is a long unbranched chain of amylase. Write down. It is a long unbranched chain of D glucose residues connected by alpha one comma four linkage. Next line you write down. The branched polymer of glucose, the branched polymer of glucose, where in alpha one comma four glycosidic linkage where in alpha one comma four glycosidic linkage joins successive glucose with branched point having alpha one comma six linkage. Okay, cellulose just one point to write down. Like you said it is correct that it is linear unbranched polymer of glucose. So cellulose is the linear unbranched polymer of glucose, unbranched polymer of glucose and you add this line, glucose residues linked together by beta one comma four linkage. Okay, second one you have done. Next one give an example of each fibrous protein and globular protein. Fibrous protein is silk, right? Globular protein is egg albumin. Next question you see. Globular protein is egg albumin and fibrous is protein. Next you see question number 19, we'll see. Question number 19. Solve this question. For the first or third world decomposition reaction of data given here. Calculate log constant. Yes, yes. Log constant means rate constant. Then or trying, yes, yes, yes. If they have even, right, they have written log only it means log base 10. Otherwise they'll mention E over there. Minus two, which is got minus two. Tell me the answer. Yes, calculate the log constant it is given. You have to find out log K. Yeah, log K. What is the value of K you are getting wishes? Tell me just a second. Are you done? What is the value of K wishes you got? Okay, I'll just do this now. See the reaction here is C2H5Cl C2H4 HCl. The initial pressure is given 0.30. And then it is 0.30 minus X. X and X. Right, so when you add all this the total pressure Pt is equals to what? 0.30 minus X. Plus this X plus X gives you two X. 0.30 plus X is equals to 0.50 from this data. So X is equals to 0.20 when you solve this, right? Now since it is given that first order third-worldly competition it is. So we can write down K is equals to 2.303 divided by T log of A by A minus X. A is the initial value. Now when T is 300 what is A minus X? A is this, X is this we have calculated. So when you put this you get 2.303 by 300 log of A is 0.30 and this minus 0.20 is 0.10. So that will be log 3 itself. So it should be through 0.303 into log 3 value is given 0.4771 divided by 300. What is this value? Tell me. What is this value? Tell me. Is it the same 0.301 by 300? Why? It is 0.30, total pressure. So when it decomposes pressure will decrease. X is not the degree of dissociation. It's the X pressure which reduces. Tell me this value 2.303 into 0.477 by 300. What is this value? Total pressure is increasing. Yeah, correct Ramchar. Total pressure initial it is 0.30 because initially what we have? We have only this present initially. But when it when it starts decomposing we are getting two different products C2H4 and HCl. Right, so because of that pressure may increase after sometime when the reaction proceeds. Total pressure 0.59 into 10 to the power minus 3. Okay, so K value you got and then you can find out log K also. Okay, you can further solve this 1.59. So I'll assume this as 1.60 into 10 to the power minus 3. Right, and 1.60 we can write 1 plus 2 into 0.30 which further we can write 1 plus 2 into log 2. Right, so that will be 1 is log 10 plus log 4. That will be log, okay, so next what we can write. I just trying to solve this. This term into 10 to the power minus 3 into log 40 is equals to K. So log K is, I don't think we can solve like this. Can you solve this further? This will be log of 10 to the power minus 3 into log of 40. Sorry, log of whether we write log A plus log B log of log 40. Anyway, so this data must be given here. What is the log value of this you are getting? I'll just do this log of 1.6 you put here. That will be log 16 minus log 10. So minus 3 log 16 is 4 square 2 log 4. So 2 into 0.60 minus 1. So minus 4 plus 1.20. That will be minus 3.8. This is what I'm getting. Minus 3.8 we're getting. If they ask K value, answer will be this log K. You just put log and solve this minus 3.8 approximately you'll get. Okay, so I'm just getting minus 3.3 close enough. Whatever I do not know. You only told me that the value is this. If it is this, the answer will be this. Otherwise you solve this. You will get the value here of K and then you find out log K. It just need the calculation. Nothing else. We'll move forward. One more question we'll discuss on here. The numerical question. Question number this one, 22. Solve this one. Again you see the pattern is again same. In 3 marks question 2 numericals are there. Question number 19 and 22. 23rd is the abstract question and 24 to 26. Again you will get some reactions, series reactions, numericals, name reactions. These kind of questions you will get for the 5 marks. Numerical questions, series reactions, ABCD product, you have to find out into that. And then name reactions you will get. Question number 22. Solve this. Assume universe complete ionization. Boiling point of solution. Boiling point of solution. What is the value of delta TV you are getting? Delta TV is 0.346. Vaishnavi. 0.173 fresh. 0.346. So what is the boiling point then? What is delta TV? Then how you are getting 373 or something? Ramcharan, check your calculation all of you. Yeah, correct. Vent of actor is 2 because MGSE 4 will dissociate as MG 2 plus as a 42 minus. I value is 2. Yeah, 100.346 degree Celsius you got. So delta TV is 346 only, right? 0.346, sorry. Delta TV is 0.346 Kelvin or degree Celsius, right? So you have to find out the boiling point. Boiling point is what? Delta TV is equals to TV dash minus TV where TV is the boiling point of original solvent. 100 degree Celsius in case of water. Okay, so that will convert into Kelvin. So that will be 373. So answer will be 373.346. This is what you are getting. 373.346. Yeah, that's right. 373.173 is wrong. And don't write Ramcharan these solution questions. You don't write the approximate answer. Write down the exact answer that you are getting. Like in this case, it is not 373. It is 373.346 Kelvin, right? When you write 373 only, it means that delta TV you are not considering at all. It is 100 and then it will have 373. That is simple. Okay, so write down the exact value. Don't round it off. All of you have done this. Do I need to solve this? Correct. I'll move on. Okay, now you solve this one. This is electrochemistry. So you will get two numerical questions here. Either from any of these. See, you are getting two numerical questions from physical chemistry. Okay, so the chapter would be solution, electrochemistry, solid state or chemical characteristics. So from these four chapters, you will get two numerical questions of five marks here in the last portion. Solve this question number 25. So one data is missing in this question. I'll just write it down. The E0, these data you see for the second question, B. E0 for A2 plus 2A. It is given minus 2.37. And that is for B2 plus 2B. It is given minus 0.14. These two values are also given. Again, the answer and you are rounding it off. Don't round it off. Why A is better? Tell me the exact value. Don't round it off all of you. Why A is better? Tell me. See, the first answer is correct. This will be 1.99 approximate. You have to find out E cell, right? E0 cell. Formalize E cell is equals to E0 cell minus 0.0591 by N log of product by reactant. Okay. So with that, you are getting 1.99, which is correct. In option B, in the question B, you need to find out which using the following E0 values of A and B predict which is better coating the surface of ion, whether the potential of ion is given. Okay. So here we have to understand for electroplating, we need a cathode couple which gives positive value of E0 cell. Right. Which gives positive value of E0 cell. Okay. Positive value means what? The potential or reduction potential of A or B. Right. It should be lesser than, lesser than the value of ion. Okay. The given value of ion. That is what we have to find out. Okay. Magnitude we are talking about. See for ion it is 0. Minus 0.44 it is given. Right. For ion, it is minus 0.44. For A it is minus 2.37. So if we find out by this formula, which we have E0 cell is equals to, is equals to E0 of A2 plus 2A conversion minus E0 of Fe2 plus 2Fe. Which is nothing but minus 2.37 minus, minus of 0.44. The value we get here is minus 1.93, 97. 93. Correct. 93 volt we get here. Okay. Now similarly if you find this is for with A. Now if you find out with B E0 cell with B. So for B it is 0.14 plus 0.44. You see this value is 0.30 which is a positive value and that is why we require. So B will be the, is the better coating, is the better for coating on the surface of ion. Okay. See what happens if you take this B and this ion. E0 of the cell is positive and what it means when E0 of the cell is positive, the reduction potential you see Fe2 plus 2Fe and B2 plus 2B both are reduction potential. So reduction potential of this metal whatever it is B is lesser than to that of ion. Right. So when you couple this ion with this B, ion will go under reduction and this will go under oxidation. Right. So that is what it requires for the, you know, for better coating, so for better coating that is what it's required. Hence the answer will be B. You must, one thing you must keep in mind that E0 cell for better coating must be positive. That is what you have to keep in mind. See both has negative potential only. Point is we have to compare A with ion and then B with ion. Then what happens? Both you can use as a coating for the coating of the surface of ion, but which one of these two is better, which gives you positive E0 cell value. Okay. One more question we'll see series reaction I want to to give you this question and then we'll take a break after this question number 24. Yeah. Okay. We solve the first part here. This or part we'll discuss after the break. Solve this part first. This part. What does this H2PDBSO4 does? What does this S2PDBSO4 do? Tell me. Yeah. Yeah, correct. So first one is CS3CHO. Then what happens? What happened with NH2OH? What is this CS3CHNOH? This is oxym formation, right? Formation of oxym. C double bond NOH is oxym. Yes. Yes. Correct. We have done this in the class also. Mechanism we also have seen that we have a rearrangement in this. Right. So first product will be CS3CHO will get aldehyde with equal number of carbon atom. And then the reaction with NH2OH gives you the formation of oxym, which is nothing but C double bond NOH. And H2O will go out into this. Second reaction you tell me. RMGX reacts with CO2 in presence of acetic hydrolysis. Correct. Then with PCL5, COOCl or COCl, C2H5, Cym-L. How did you get C2H5, Cym-L? It should be CS3 COCl. Correct. Okay. Next one you do. Distinguish between all these. C6H5COCH3, C6H5CHO. Which one is tolerance test? This can be distinguished by hydroform test also. Right. Methyl ketone, ketone gives positive hydroform test. C6H5COCH3, that is acetofinone and it is the methyl ketone also. The compound which contains CH3CO group shows positive hydroform test. Right. Which gives yellow precipitate of hydroform with sodium hypoiodide. Right. So what happens? The C6H5COCH3, which is acetofinone gives positive hydroform test but benzaldehyde does not give this test. Right. So hydroform test we can use. Next one is CS3COH, acetic acid and formic acid. What happens here? You see, the formic acid easily decolorize pink color of KMNO4. Write down this. Formic acid easily decolorize pink color of KMNO4 while acetic acid does not again perform this test. Right. Formic acid decolorize pink color of KMNO4. Acetic acid will not do the same. Right. Third one is the boiling point. What is the order of boiling point? Acid, alcohol, aldehyde. Correct. Acid will have the highest boiling, highest boiling point, maximum boiling point and then we have alcohol and then we have aldehyde. Vaishnavi and all of you, you take care of one thing. Last class also I told you this thing. What is the question you see? Arrange the following in increasing order of their boiling point but you have written it in decreasing order. Understood. All of you, Vaishnavi, Saimee, Rithvik, Mrs. Strahm, Ramcharan. If the question given in increasing order, you have to write down in increasing order. It is the decreasing order you have written. Correct. Do take care of all these things. Okay. You know all the things but if you write down like this, you won't get marks, unit and all, everything. Okay. Cool. So we'll take a break. We'll start in 15 minutes. Fine. Right. Can we take a break now about the distinguish between C.V. and M.N. I think you can use bromine water. Yeah. You can use bromine water. Okay. We'll discuss this after the break. We'll discuss this. Correct. Yeah. Bromine water I think you can use. Okay. We'll take a break now. We'll start the class at 5.55. Okay, guys. Can we start? Are you there? Okay. Now solve the second part. Wolf-Krishna reduction, you know. What happened in Wolf-Krishna reduction? C double bond O converts into what? Okay. Tell me the reagent. What reagent we use in Wolf-Krishna reduction? Yes. So C double bond O, hydrazine, NH2, NH2 we use. And C double bond O converts into CH2. Right. So carbonyl compound converts into alkane. So any of these reactions you can write down. Okay. So any carbonyl compound if I write, R C double bond O. R. And when this is heated with hydrazine, NH2, NH2, C2H5O1 or KOH also you can write down. When you heat this, it converts into R CH2R. So basically in this reaction, C double bond O carbonyl group converts into CH2. And other things will be same. This is Wolf-Krishna reduction. Okay. Now arrange the following in increasing order of their reactivity towards nucleophilic addition reaction. Which one is maximum? Suppose I'll write down this as first, then second, and then third. Take care of this thing. Increasing order we have to write. Okay. Increasing order. What is order? See this order depends on two factor. Right. If you have the carbonyl group, C double bond O. So we know this bond, carbon oxygen bond is polar. This is slightly negative charge and this will have slightly positive charge. Where the nucleophile attacks. I didn't get any response stress. I can see your message. I gave my response or you said, but I didn't get any response. Anyways, I'm doing this. So the attack of nucleophile depends on two factor here. One is the steric crowding here. One is this positive charge density. More charge density, more will the attack of nucleophile. More crowding, less will be the attack of nucleophile. Correct. So you see the three compounds we have here. CH3, C double bond O, CH3. Then we have CH3, C double bond OH. And the last one we have PHC double bond O, CH3. In all these example you see, because of the plus high nature of these two group, this positive charge density will be less over here in comparison to this. Right. So if it is this one is two, this one is three, and this one is one. So obviously, second one is more reactive than the third one. Second one is more reactive than the third one. Now here what happens, we have plus high nature of this. Phenyl has electron withdrawing nature. But at the same time what happens, this produces steric crowding also because of large size. So because of steric crowding, the attack of nucleophile will be lesser than these two, two and three. So the order will be this. First, third and two will be maximum. Okay. Tell me the next one. Why carboxylic acid does not give reactions of carbonyl group. See in this carbonyl group we have either C double bond O or C double bond OR or H we can have here. Right. But in carboxylic acid we have RC double bond O OH. So this lone pair is involving resonance in carboxylic acid. Right. So the carbon is not that much electron deficient here. If this bond will come over here, this lone pair will come here and we'll get again a double bond here. So this carbon atom or carbonyl carbon is not that much electron deficient as it is there in aldehyde or ketone. That's why the reaction of carbonyl compound is not given by carboxylic acid generally. Last question you tell me. Write the product in the following reaction. A is propanol B is acetone. Yeah, it's correct. Right. It's a direct reaction to functional isomer. Okay. You can easily find out degree of unsaturation and then you will have the idea that we'll have either one double bond or one ring. Right. And then the direct reaction with NaOH and I2. Okay. Next question we'll see. An element crystallizes in a BCC lattice with cell edge of this density is this. How many atoms are present? Done. What is the answer? Tell me.