🔥 TG PGECET Pharma Chemistry-I Super 100 Series | Vijaya Prasthanam 2.0 | Most Expected Questions

Ajouté : 2026-06-04

[00:00:41]Hello dear students. Good afternoon everyone.

[00:00:44]Okay, is my screen visible to everyone?

[00:00:48]My voice audible clearly?

[00:01:01]Hey, so welcome everyone to superund series. So today we're going to start with our superund series for TGPGCT as you have your exam very soon. So today we will be solving some questions top superund questions from pharmarmacist subject. So starting with our first question then I want participation from everyone. So everyone need to participate in all the questions. Yes. So starting with our first question we have here in our screen that says geometrical isomearism is due to Okay. So the options are the restricted rotation about a double bond the presence of an keto group the presence of co group or the presence of an asymmetric carbon. So geometrical isomeism. I guess everyone is aware of that. Yes. Very good. Yes, that is actually the restrict restricted rotation about the double bond. Very good. Yes. Yes. Everyone is right. A is the correct option for the geometrical isomeism. Yeah. Okay. So, A1.

[00:02:09]So, this is the explanation. You can go through that. Okay. So, basically the single bond it has a uh uh we can say without any hindrance we have rotation.

[00:02:19]But in geometrical isomearism we have a hindered rotation across a double bond. Okay. So next coming to the question arsenic guess in limit test for arsenic is produced by the reaction between. Okay. Everyone know about the limit test of arsenic.

[00:02:38]Yes we use gazit apparatus for that.

[00:02:42]Everyone is aware of that. I guess everyone has seen that apparatus.

[00:02:46]It looks somewhat like uh this kind.

[00:02:50]Yes. And we basically have a mercury chloride paper here. And the color the yellow stain there the intensity of that yellow stain uh tells us about that what is basically uh we compare that stain the intensity of that stain with the standard and with the help of that we check on the limit of the arsenic. Okay.

[00:03:14]Uh so if we talk about the reaction there so that mercury chloride paper it uh basically reacts. So what happens there we have the arsenic arson gas it along with what it along with hydrogen gas it reacts and the reaction gives the yellow stain. Okay with the help of that we get yellow stain. So basically the arsenic arcinius acid and hydrogen along the together they form arson gas that is responsible for what? Responsible for the yellow stain that we get in the limit test of the arsenic. Okay. So the answer is the D1. The arsenic arcinous acid and the hydrogen together they form the stain that we get. Okay. Now coming to the next question. Yes, the answer is D.

[00:04:10]Uh, which my moty is present at C7 position of benzoizopins? Yes, benzoizopins the hypnotics and sedatives everyone know about that. So, electron attracting substituents, phenile ring, CH2 or alky substituents. So, I guess the benzoizopen rings everyone remembers this is the benzene ring. Then we have a a nitrogen two nitrogens basically double bond here. The ring somewhat looks like this only. We have a phenile ring here and in this benzoizopin at seventh position we have a group attached. The group like Cl2 okay so these groups are basically what they are electron withdrawing group.

[00:04:57]Yes, electron withdrawing group or we can say electron attracting substituents. Okay, so B they basically result in increasing the activity of our benzoizopins. Okay. So A1 yes the A1 is very right. Yes the A1 is the correct option here. Okay. So electron attracting substituents are needed in the benzoizopins. Okay. So this is the structure of benzodizopen.

[00:05:24]Here you can see and here you can see the X and we prefer all the electron withdrawing groups here.

[00:05:33]Next coming to the question in HLC the analytical performance improves when okay so HLC high performance liquid chromatography so the analytical performance is improved when okay so particle diameter is increased particle diameter is reduced coarser particle is paired with shorter columns or low temperature is used yes Anyone who want to answer this one the HLC analytical performance what happens to that when when when it get improved. So basically what happens now this improvement in the particle performance is due to the reduction in the particle size. Why?

[00:06:26]caused by reducing the particle size it increases the surface area. Yes, if the particle size is reduced then the surface area is increased and due to the increase in the surface area the theoretical plate number also increases and due to that increase in the number of theoretical plate it improves the separation. Okay, it improves what? It improves the separation also and it also improves the resolution. So that's why we prefer particle diameter to be reduced whenever we have to increase the HLC analytical performance. Okay. Yes.

[00:07:12]Yes. Many Kumar is right here. Okay.

[00:07:14]Again yes the B option is the correct one. Very good money. So you can go through the explanation part. Coming to the next question we have here. What is the value of n in huckle's rule? Okay, what is huckle's rule? Write in the comment section. I want to know the huckle's rule. If anyone remember the formula when a compound have nine pair of electrons okay so they have given nine pair nine pair of what electrons they have given and they have asked that what is the actually the value of huckle's rule there. So what is Huckle rule formula? Anyone who's going to tell me? So 4 n + 2 pi electrons that is huckle rule I guess. Yes. So here they have given nine pair of electrons. So nine pair nine pair means we have total 18 electrons. Okay. They are talking about how many? 18 electrons. So if we place the value of pi here 18 so 4 n = 18 - 2 that is 16. So n = 16 by 4 that is 4 4s are 16. So a1 the a is the correct one. Yes. Okay. So the answer is four. Okay. The nine pair means 18 electron. So we can solve it with the help of 4n + 2 pi electron formula.

[00:09:00]Okay. So it's a simple one. I guess everyone got it. Yes. So that is our huckle rule. Okay. For the aromatic compounds we have that. Yes. Very good.

[00:09:13]Okay. I uh you want my volume to be incre increased a bit. Okay. Is my voice not clearly uh audible? Arush, now is it okay or should I increase it more?

[00:09:30]Okay. Yes. Very nice.

[00:09:34]Yes, the formula is 4n + 2 pi electron.

[00:09:37]That is very okay. Okay. Is my voice now audible or should I increase it a little more? Please confirm so that we can continue with the session then.

[00:09:49]Okay, I am getting the answer. Okay.

[00:09:51]Okay. HCl and NaOH titration is an example of Okay, that's a simple one.

[00:10:02]Yes, the answer is NaOH is also a strong base and acid. HCl is also an strong acid. So means it is an example of an strong acid and strong base titration.

[00:10:18]Okay. So that is again a simple one.

[00:10:20]Yes. So A is the correct answer again.

[00:10:23]Okay. Very nice. You are correct here.

[00:10:27]Okay. Continuing as per Bronstead Lori concept acid is defined as such a simple question. I guess everyone knows the answer here. The Bronstead Lori concept.

[00:10:39]Yes. based on what the donation of the protons here. Okay. So, electron pair acceptor any substance or molecule that can donate proton any substance or molecule then that can accept proton or electron pair donor. So the Bronstead Lori concept is basically based on what it is based on. So they are specifically asking about acid. So what is acid? acid is a compound that can uh easily donate the proton. More easily it donates the proton better the acid. Okay. So I guess B1 goes right. Yes A you are very right.

[00:11:20]The B1 is correct. Okay. Yes Arush the B1 is the correct one according to the Bronstead Lori concept. Okay. So there are different theories for acid base.

[00:11:31]Arinius Lori Lewis. Okay. We can go through that. We all know about that.

[00:11:37]Yes. Then moving to the next question we have here. Angular fused polyuclear aromatic hydrocarbon is read the question carefully. Angular fused polyuclear bifphenile napylene anthrain or finanthrine. Okay we have a lot of polyuclear aromatic hydrocarbons here.

[00:12:03]Okay. Napylene is also one of them. Bif phenile, enthras and phenthrine all all four are polyuclear fused angular one aromatic ones. But out of all these four which one is angular? So napylene, bifphenile and enthrasin they are linear. Here you can see this is napylene. It is linear one. Enthrasin again linear.

[00:12:29]Yes. If we talk about finanthine we have it like this. somewhat.

[00:12:36]Yes, we have an angular thing here. So the finanthine is the one that is angular. Here also you can see the classification. So we have linear and angular and finanththerine is an angular one. Yes. Dd1 is the correct answer.

[00:12:50]Yes. Okay. So manikumar yes very good.

[00:12:54]Okay. So coming to the next one we have here. Okay. Replacement of oxygen at C2 of barbituric acid by a sulfur. Okay, we are replacing oxygen. Okay, the oxygen that is present in barbic acid nucleus we are replacing it with sulfur. Then what will happen? Whether it will increase the lipid solubility, decrease lipid solubility, show no change in the lipid solubility or none of this. So basically if we replace the oxygen with sulfur. Now sulfur results in what?

[00:13:33]Sulfur results in increasing the lipopilic nature. So as the lipopilic nature is increased so lipid solubility is also increased. Yes.

[00:13:45]So yes. So the add uh so the changing or substituting oxygen with sulfur results in the increasing lipid solubility. Yes.

[00:13:54]Yes. Very nice. Okay. So the answer is the A1 like thopental is an example for that. So basically we have replaced this oxygen out here with the sulfur. Okay.

[00:14:07]This oxygen replaced with sulfur that is thopentil and it increase the lip lipid solubility. Okay. Then next which of the following are the types of plane bending?

[00:14:21]In plane bending they are talking about in plane. The question is from IR spectroscopy. The movement the vibrations in IR whether it is twisting or wagging in scissoring or rocking.

[00:14:34]Scissoring or twisting or none of them.

[00:14:37]Okay. So starting with our first option twisting and wagging. Twisting and wagging they are in out of the plane bending. They do not have in plane bending. If we talk about scissoring scissoring is what? like a scissor. Like if we say this is the thing. So one is moving here. So one will move here. It will be in plane and it will be like scissor like the scissors move. Okay. So that is the scissoring. Okay. So that is in plane. If we talk about rocking what is rocking? Rocking is the opposite of scissoring we can say. So basically what happens here both in same direction here then they will come back together also.

[00:15:20]Okay. So the this is rocking means one here, one here that is scissoring.

[00:15:24]Rocking is both on the same side. Okay.

[00:15:27]So they are both in plane bending. Okay.

[00:15:30]Next is scissoring and twisting. Now twisting is out plane. So the answer for this is the B1. Okay. Scissoring and rocking that are the inplane ones. Okay.

[00:15:42]So this is in plane bending. The scissoring and rocking. Okay. I guess everyone knows that the IR spectroscopy we are aware with that. Okay.

[00:15:53]Coming to the next question. The sodium rose bengal I131 is used in. Okay. So that is a radioactive one. Radioactive substance. The iodine radioactive one. Yes. Study of potassium ion exchange, plasma volume determination, brain scanning or liver function determination.

[00:16:18]I guess that's a simple one. The I that is a radioactive compound. So what happens? We basically this radioactive agents we use them for the determination of the functions functions like liver function determination. So they act as a marker there. Okay. They act as a tracer marker there. So basically for the diseases like for the evaluation of the liver diseases if we talk about cerosis if we talk about hepatitis so for that we use this rose Bengal okay so the D1 the liver function test that is the correct answer here okay the liver function test we use this rose ben sodium rose Bengal I31 okay next question we come to that What reagent is commonly used in reduction?

[00:17:13]Birch reduction. Yes. What is birch reduction? It is basically a partial reduction reaction of what? Of aromatic rings. reduction. It is a very common type of name reaction. I guess everyone is aware of that. And what we use there? Yes, you're right. Very right. Yes. What we use in birch reduction? in the reagent NaOH Na in liquid NH3 peladium and carbon H2SO4 or we can say sulfuric acid. Yes, Arush you are also right. We use what? We use sodium in liquid NH3 there. Okay. What we use? We use sodium in liquid NH3 in birch reduction. It is basically the reduction of the it is not actually fully reduction. It is partial reduction of aromatic rings like benzene. Here you can see we haven't done the full reduction only partial reduction is there. So that is bridge reduction here.

[00:18:18]Okay.

[00:18:20]Next coming to now you keep these questions in your mind. Okay. They are very much uh uh possible that these type of questions may come in your exams.

[00:18:30]Okay. Okay. So uh watch them clearly.

[00:18:34]Okay. and understand them clearly so that you don't miss them when they come in your exams. Next, which of the following HLC detector is not a soluble property detector? Solute, sorry, not soluble solute property detector. So, which is not a solute property detector?

[00:18:53]Is it UV visible detector? It is based on solute. Now, the solute particle, the UV is based on that. Yes.

[00:19:01]Then photo diode array detector. So then fluorosense detector and refractive index detector. So photo diode array fluosense detector they are also based on the solute property detector only.

[00:19:15]But if we talk about the refractive index detector that is the one that we do not have anything related to the solute property detector. Okay. So the D1 is the correct one here. Yes. Yes.

[00:19:27]Arouch. Yes. Siman you are right. Okay.

[00:19:32]Okay. So coming to the next question then. Okay. You can go through the explanation part. I'm not going to that next one. The chemical name of caffeine.

[00:19:44]Caffeine. Everyone is aware of that.

[00:19:46]Basically a CNS stimulant we talk about that is a very good uh good example of CNS stimulant. Now if we talk about the chemical name what is the nucleus that we have in caffeine? Anyone remember the nucleus? We have a purine nucleus or we can also say we have what? We have zenthin nucleus. Yes. What is the basically the structure of caffeine? Let me draw it for you.

[00:20:14]So this is basically the structure of caffeine. We have yes double bond O here. Double bond O here. And then we have CH3 at three different positions.

[00:20:29]So this and here and here we have CH3.

[00:20:34]If we start the numbering. So from here the numbering starts 1 2 3 4 5 6. This is the seventh position. 8 and 9. Okay.

[00:20:44]So at first at first at uh sixth and at at first position at fourth fifth ninth position and at seventh position.

[00:20:59]Basically at uh these position we have what? Sorry the CH3 group is not here.

[00:21:04]Sorry I have made it wrong. Let me draw draw it again for you so that it can be more clear for you.

[00:21:12]So like this we have the structure. Here we I have double bond. Here we have double bond.

[00:21:21]Here we have a CH3 group. Here also we have a CH3 group. And then here we have a CH3 group. Here and here we have a double bond. So this is the structure.

[00:21:31]Numbering starts from let me do it from a different color. 1 2 3 4 5 sorry 4 6 7 8 9. So at first position at third position and at seventh position we have a methile group. Okay. And the ring present here it is called as a zenthin ring. Okay. So that is basically the structure of caffeine. So if we look onto the options, the A option goes with our structure. Yes. So the zenthine ring we have here. So A1 is the correct one.

[00:22:09]Okay.

[00:22:10]Next question. Scor synthesis can be used to synthesize scorsis.

[00:22:17]Analine, napylin, furon or finanthrine.

[00:22:21]All four are very widely used uh heteroscyclic rings. Not all are heterrocyclic but most are aromatic rings. we can say. Okay. So, score synthesis are basically useful for the synthesis of what? Yes. Anyone want to answer?

[00:22:40]Okay. So, the score synthesis is basically used for our finan.

[00:22:46]Okay. Let me show you the reaction. So, score synthesis that is specially for finanththerine preparation we use it.

[00:22:54]Okay. So, the D1 is the correct answer here. Okay, coming to the next question. Brag's equation is what is Bragg's equation?

[00:23:07]So, basically Brag's equation is anyone remember what is Bragg's equation and why we use Brag equation anyone? Do you know why we use Brag equation and what is actually a Brag equation? So brag equation is this n lambda 2d sin theta that is our bra equation and basically for the uh we can say in the crystal when we have angle of incidence and defraction for that basically brag equation is used okay yes uh okay for the PDF thing I guess many of you are asking for PDF I'll talk to the team okay so for that PDF thing I can't send you directly Okay, let me talk to them right now. You focus on questions here as this video is there in YouTube. So you can look onto the YouTube also. Now you can just take it in 2x. Okay. You can try to solve them.

[00:24:05]You can try to solve them with me. Yes.

[00:24:08]So uh right now focus on the questions here. Okay. Don't say PDF, PDF and all.

[00:24:13]Right now solve it. Then we will look on to all the queries you have related to PDFs.

[00:24:20]Which impurity is tested using Nestler's reagent? Okay, look onto the question carefully. Nestler's reagent, they're talking about not the cylinder, they're talking about the reagent. Okay, arsenic ammonia chloride or sulfate. First question, what is Nestler's reagent?

[00:24:43]Nestler's reagent is potassium mercuric iodide. Okay, Nestler's reagent is potassium mercuric iodide and we use Nestler's reagent for the impurity testing of ammonia. Okay. Yes, the ammonia we use B1. Okay. Not Cai B.

[00:25:06]Okay. So, potassium mercuric iodide that is the reagent, the Nestler's reagent we have. Okay. Coming to the next question.

[00:25:16]Arrange the following electromagnetic waves based on order of energy. Okay, it's an interesting question. I guess everyone know about the energy order.

[00:25:28]Which energy has the highest one? We also have a trick for this. Kaga.

[00:25:34]Yes. Kaga X UV. Remember?

[00:25:39]Yes. The the gazit apparatus is for arsenic. No, we are not having anything.

[00:25:44]Huh? Yes. Yes. Yes. Victory wives that is the gazget apparatus for arsenic we use but here we are talking about not the apparatus but the nestler's reagent.

[00:25:55]Okay.

[00:25:57]So arrange the following electromagnetic. So which one has the highest electromagnetic wave? Cosmic ray that has the highest one. But here in options we can't see cosmic. Then after cosmic we have gamma rays. So gamma rays are highest after that. So gamma rays are highest. Then ka x is for x-ray.

[00:26:19]Then we have x-ray there. Okay. Then uv first is ultraviolet. Then visible. So UV then we have microwave. Okay. So the highest order of energy electromagnetic wave is for cosmic then gamma. So the B1 I guess here goes with our answer. Yes.

[00:26:41]The gamma has the highest X-ray, then UV, then microwave. Okay, B1. Yes. Is that clear? B1 is the most favorable or most best answer I can see here. So the B1 is the correct here. Next question we have in our screen. Which among the following is a butyrofenone derivative?

[00:27:04]The question from the classification of drugs. What is butyrofenone? There is an antiscychotic drugs we have under antiscychotic class. Yes.

[00:27:16]So if we talk about its classification it it is in derivative of what?

[00:27:21]Procycloperasin heloperidol fluenazin or oxy oxiputin. Yes. Which one? I guess butyoproofenone comes under under what?

[00:27:34]Under heloparidol.

[00:27:36]Yes. So under heloparidol we have butyrofenon. Yes very good victory wives. Yes Akshai you are also right. So hello pararidol is the correct one here. Yes the others are also right.

[00:27:51]Very nice. Yes the B1. Next question.

[00:27:55]The emitted radiation in fluosense will be okay. The emitted radiation in fluosense according to you which one?

[00:28:05]The shorter, the longer, both short or long or none of this? Yes. The emitted radiation in fluosense. Can anyone tell me which one? They are longer, shorter or none. Yes.

[00:28:22]So they are basically the longer ones.

[00:28:25]No, not shorter. Longer. Okay. In fluosense the emitted radiations are longer.

[00:28:32]Yes, you can go through the explanation part. I guess everyone is very much aware of the fluosense one fluosense part. So the A1 the longer ones. Next question. What is the primary product obtained from Stephen's reaction?

[00:28:48]Stephen's reaction. Anyone aware of that? Stephen's reaction. What is that?

[00:28:55]Anyone heard of that? Stephen's reaction. So Stephen's reaction is also called as Stephen's eldihide reaction.

[00:29:05]Yes, Stephen's reaction is also called as Stephen's alihide reaction. And what happens there in this reaction? Liihide we basically we convert it to nitral group. The CN is attached to that. Okay.

[00:29:21]So nitral group it converted to that. So basically nitral group the nitral group the CN group that is basically it gets converted into an alihide. So it is also called as tifen alihide reaction. A1 is the correct one. Here you can see alkyle nitril in the presence of sn2 it is amino chloride that further gives us the formation of an alihide. Yes. Very good.

[00:29:52]Okay. So the alihide the A1 is the correct one here. Very nice. Next we are coming to the next question. Which one of the following is used as a systemic alkalizer?

[00:30:05]Alkalizer to increase the pH what we use systematically. Okay. In the systemic in our system in our body the alkalizer that we use sodium chloride, sodium bicarbonate, sodium sulfate or sodium acetate. Naturally it is also present. I guess this one is very simple one. Is it sodium chloride? NaOH is NaCl sorry sodium bicarbonate sodium sulfate. Yes very nice. That is sodium bicarbonate.

[00:30:36]Very very nice. Okay. So B1 the sodium bicarbonate is the correct one here.

[00:30:42]Okay. Yes everyone is right here. The sodium bicarbonate we can go through this explanation part. Yes, it's not very difficult. Just the theory has been given. Next, we come to our question.

[00:30:56]Following is an example of hydrophilic mobile phase. Hydrophilic.

[00:31:03]Yes, hydrophilic that is water loving. So they are saying all are water loving except mobile phase that all are water loving except isopropenol, ammonia and water. Okay. So there is water. So how we can say that it is not water loving? Okay. Then isopropanol and water. Then we have isopropanol and ammonia. Then we have dimethile, ether and cylohexane.

[00:31:33]They are organic compounds. Yes. The D1 they are organic compounds. So they are Yes. They are hydrophobic ones. They are hydrophobic. Mobile phase example. Not the Yes. Yes. Yes. You all are very right. Yes. So the D1 is the correct one here. Okay. So they are hydrophobic, hydrophilic. They are of two type. You can go through the examples we have here. Yes. So next we come to our next question. Very right.

[00:32:05]Okay. Product obtained by furon and benzene dials elder aduct. Okay. So in Dil's elder adduct basically we combine two one compound with Din. So one should be D in other should be what? Okay. So basically they are saying whenever we do Dil's adduct reaction between furan and benzene then what product will we get?

[00:32:36]Bifenile napylene enthrasin or phenanthine.

[00:32:41]Let me show you the reaction for this.

[00:32:44]If anyone is Yes, you all are very smart. You know every answer. Okay. Yes, the answer is actually napylene only.

[00:32:52]Okay. So whenever we combine basically die in with our the reaction we have the of the aromatic ring. So basically benzene we have here and here we have furon they together with diar reaction they forms a napylene ring. Yes, the napylene is the correct answer. The B1 ion exchange capacity expressed in the terms. Okay, that is a unit. I guess everyone is aware of that.

[00:33:23]Some what?

[00:33:26]Not very confusing things. It's a simple one. Yes. No. Not in microgram. No.

[00:33:33]Ion exchange capacity. ion exchange capacity mill equivalent per gram. Okay, not microgram. The C1 is the correct here.

[00:33:44]Okay, you can go through the explanation part also. Okay, the ion exchange raisin you see. Yes, the C1 is the correct one. Okay, theopile and theob broine. I guess very familiar names you must have heard. Theopile and theob broine. Yes, like they are the derivatives of caffeine only. They also have what? They also have a purine nucleus are structurally similar and have the system. Okay. The question is only that quinolin, napkine, phinoin or zenthine. Which one? Which one? Yes.

[00:34:17]Theopilin and theob broine. Which one?

[00:34:20]It's a very simple one. The same ring that is present in zenthine.

[00:34:25]Oh, sorry. The same ring that is present in caffeine. Yes, I already said the answer. Yes, the D1 zenthine is the correct one. Okay, they all are derivatives.

[00:34:37]So, zenthine and zenthine ring that is present in theopilin and theogloin. The same ring was present in the caffeine also if you remember. Yes.

[00:34:50]Next question. Okay. Effective dose of radiation is tiss in tissue is measured with effective dose of radiation.

[00:35:06]Yes. In tissue is measured with rowing, rbe or radiation per unit. Which one?

[00:35:17]Yes. Again a question that we need to know. No, not D1.

[00:35:26]27D.

[00:35:29]No, don't get confused. It is Rio engine 1. Okay. The dose equivalent we check in not the radiation per unit. Okay. Yes.

[00:35:46]Yes. Victory Wives says said it right.

[00:35:50]Okay. Yes. The D1 is the the B1 is the correct one here. Okay. Next question.

[00:35:58]Sulonation of napylene gives thermodynamic products. Now read the question carefully. Here it is talking about the thermodynamic product. Okay.

[00:36:09]We have to look onto this particular thing. Thermodynamic product. Sulonation of napylene. Now as sulonation is happening so there will be attachment of sulfonic acid only. Okay.

[00:36:24]Now thermodynamic product. Thermodynamic means this whenever we do the sulonation of napylene it is done in two different temperatures. Okay. It can be some one is little lower and other one about 60° we can say and one is about 180°C. So 180° C can be called as thermodynamic product. And if we talk about the thermodynamic product we get two sulfonic acid. At second position we get sulfonic acid. And if we talk about the uh 60° if we do the sulonation of napylene we get one sulfonic acid. Okay not d no that is wrong. the B1 napylene to sulfonic acid that is the thermodynamic product of the napylene.

[00:37:14]Okay. So basically done around 180° centigrade you can take that temperature. Yes the major and minor products are possible here and that is based on the temperature. When there is high temperature the major product is the napylene 2 sulfonic acid and whenever there is lower temperature the major product is napylene 1 sulfonic acid. Remember this they have clearly written thermodynamic product. So that's why the B1 is the correct answer here.

[00:37:47]Okay.

[00:37:49]Next. Napylene undergoes nitration with HNO3/ H2SO4 at 60° Celsius to give mainly.

[00:38:02]Again they have given the temperature they are undergoing the nitration as the nitration is going on that is very obvious that we will add a nitro group there. Yes. So what will be the answer for that? Whatever. Yes. The at 60° centigrade they are asking.

[00:38:24]Yes. Yes. Yes. The 60° centigrade. Which one? Napthylene nitration. Yes. The A1 is the correct one. No Arush. Not B A1.

[00:38:34]Okay. The A1. The one nitronapylene that we will get. Okay. At the lower temperature at 60° C. This will be the product the A1. Okay. Oswald's dilution law is applicable to now what is Oswald's dilution law and why it is applicable to what is it applicable to weak electrolyte strong electrolyte nonelerolytes or all the electrolytes yes yes which one I'm waiting for your answers The Osworld's dilution law is applicable for the Yes. The A1 that is weak electrolyte.

[00:39:30]Yes, very good. A1. Yeah, that is right.

[00:39:34]Weak electrolyte. The Oswell dilution law is especially applicable for weak electrolytes. that states the degree of dissociation of weak electrolytes increases with the dilution of the solution. Okay. So weak electrolyte is it specially talks about that.

[00:39:58]Next question. Now the question is little bit uh we can say calculation is here. Okay. So it is an interesting question. Very easy calculation we have.

[00:40:08]I'll I'll tell you. Okay. The molar absorptivity of a compound is 2500.

[00:40:15]Okay. So let's write it. Molar absorptivity like means E they have given 2500.

[00:40:23]Okay. What is the molar absorptivity?

[00:40:25]2500.

[00:40:27]Then for 1% transmittance means percentage transmittance they have given is 1%.

[00:40:38]Okay.

[00:40:39]Calculate concentration. C we have to calculate and they have given length as 1 cm.

[00:40:50]Okay. So this is the given data we have.

[00:40:52]Now we have to calculate. Now how we will calculate? First thing 1% transmittance is equals to we can say 1 by 100. That means percentage transmittance is 0.01.

[00:41:10]Is that clear to everyone? This is simple. Yes. Next we come to now we have percentage transmittance. Now there is one more formula for absorbance. Okay.

[00:41:20]Now first thing I have to tell absorbance is A is equals to epsilon into C into L. That is the simple formula for our absorbance. Now for this absorbance we do we have E value here we have L value here we have to find out the concentration and we do not have an absorbent so we have to find out an absorbance how can we find out an absorbance absorbance is equals to minus log t where t is the transmittance we have the value of transmittance we can put in that so a is equ= to minus log of 0.01.

[00:42:05]Yes. Or we can say a is equ= to minus log of 10 ^ -2. We can write it like this. Or we can say a is equ= to - -2 that is equals to a = 2. So we get an absorbance here. Yes. Yes. Yes. Very very good. Yes. The victory wipe says the formula. Yes. You are very right. So absorbance we got a now we can put that in formula 2 is equal to 2500 into c into 1.

[00:42:43]Yes. So then we get c is equ= to 2 upon 2500 that is the thing. Then if we do do it like this. So like we get about 0 eight. Yes. 28 16 0.008.

[00:43:13]So that is the answer. 0.008 molar. This one we will get. Okay. You can calculate that with the help of calculator. Yes.

[00:43:21]Victory wives your answer is very correct. Yes. So the answer is the B1.

[00:43:26]Okay.

[00:43:27]So B1 is the correct answer. Got it everyone? Yes. This is the solution. You can go through that. Okay. So coming to the next one. Which approach involves modifying a known drug to improve its property while keeping the same therapeutic effect? Okay. Modifying known drug we are modifying the drug and we are not improving its property but we are not changing its therapeutic effect.

[00:43:56]So what is that effect called? Selective optimization of side effects denovo design random screening or high throughput screening. High throughput screening no not at all. Random screening screening is nothing related to our therapeutic effect. So no denovo design. Dean over design is like we are doing it from start and we are changing the therapeutic effect may be changed there. Okay. So D over design that is not no not again no a big no. Next is selective optimization of side activities. So yes the without changing the therapeutic effect we are improving its properties. So that is selective optimization of side activities only the sa we can say sosa. Okay. So yeah so sa victory wipes yes you are right aka right no A is not right he's saying B so A1 in universal indicator a pH of 7 is shown with the pH in universal indicator pH of 7 is so shown by which color yes the pH7 the neutral pH is shown by yellow green blue or pink that's a simple one So the yes the neutral pH is shown by the green color.

[00:45:20]Yes. The green color shows the neutral pH. Okay. The seven. Yes. Yes. Yes. The seven. Castor oil is dash type of cathartic. Okay. Castor oil is cathartic. Yes. You Yes. Castor oil is cathartic. And they are asking which type of cathartic whether it is bulk forming whether it is stumulant whether it is lubricant or whether it is stool softener. Which one? So castor oil is an stumulant cathartic. The B1. Yes. Very nice. The B1. The stimulant cathartic.

[00:46:03]That is the answer. Okay. If you can go through the explanation part. Coming to the next question. From the Hook's law, the value of K represents the Hook's law. Everyone remember what is Hook's law. We have studied that in what? In our instrumental analysis, we study Hook's law. So the K represent plank's constant, wavelength constant, bond strength or the force constant. Yes. The derivation if you remember you may remember the for the K value also. So the K is the force constant. Yes. Yes.

[00:46:44]Yes. Braha you are very right. The force constant. Okay. The K you can go through the vibrational frequency derivation here. So the force constant is the correct answer. Okay. Next question.

[00:47:00]Which of the following statement regarding arhinous acid is correct?

[00:47:04]Okay. Arinous acid. Such a simple question we have here in the screen now.

[00:47:09]So answer should come from everyone all the participants. Dissociates into O when dissolved in water. Dissociates into H+ when dissolved in water. Donates electron pair or accepts proton. Arinous theory. Such a simple theory. Yes. The first theory of acid and base. According to that, what is acid? Acid is whenever we add a substance into an water, it will dissociate into H ions. Yes, the B1 very very good. Yes, everyone is right here. The B1 is the acid. It goes with arenous theory. Okay. But he was not able to explain for all the acids because all the acids don't dissolve in water and forms H+. So the theory was uh not very uh we can say acceptable very much and other theories came after that.

[00:48:10]But Arinius said this which of the following statement about beer Lambert law is incorrect. I guess beer Lambert law the UV spectroscopy. Everyone has studied beard Lambert law. So absorbance is directly proportional to concentration. Molar absorptivity changes with concentration. Absorbance is adep uh is additive for multicomponent system or the law holds well at row concentration.

[00:48:39]Which one? Which one is incorrect about beer limb's law? Yes. Which one?

[00:48:48]Yes. I'm waiting for your answers. I guess easy one. The beard law. Everyone remembers that absorbance is directly proportional to concentration. Yes, absorbance is directly proportional to concentration. How can it be incorrect?

[00:49:03]Don't you know the formula? A is equals to epsilon into C into L. So A is actually related to concentration. Now here you can see. So that is not wrong.

[00:49:16]B1 you see now it says molar absorptivity changes with concentration.

[00:49:22]that they are not related. No molar absorptivity will not change with concentration. The B1 okay then next theropin is an aster of is an aster of what?

[00:49:43]I guess atropen everyone knows about is the combination it is basically a reimmic mixture. Yes, atropen is a reimmic mixture and it is a reimmic mixture of hyiosyamine that also everyone remembers and that is formed by the aster formation by the asterification reaction between tropin and tropic acid. Okay. So the answer the entropen is an aster of basically an asterification reaction happens there between tropen and tropic acid. The B1 is the correct one here.

[00:50:27]Okay. Not the mandelic acid. The yes the tropen and the tropic acid that is correct. The B1 remember this at tropen.

[00:50:37]Yes. Simple simple questions they are but they are very important.

[00:50:42]A spectroscopic instrument that uses a filter to select the wavelength is called spectroscopic instrument uses a filter to select the wavelength. Yes.

[00:50:56]Which one? Spectrometer, spectrograph, photometer or inferometer?

[00:51:03]Which one? The filter is used by the photometer.

[00:51:08]Okay. The C1 is the correct here. It uses filter to select the wavelength instead of the dispersion device.

[00:51:17]Next question. Alkyhide are converted into alkanes. Very very simple. The name reaction very very familiar name reaction we have here. Yes, photometer is right. Coming come to the next question. Now the alkyhelides are converted into alkanes by what? Birch grigard reagent. Sabeture senders reaction or the woods reaction. Which one? The alkyle helide is converted.

[00:51:48]Yes, the woods reaction. Very very good.

[00:51:51]We use sodium metal here along with the dry ether to convert the alkyle helide into our alkan. Yes. Okay. Next. As a result of dash in concentration of buffer components, the buffer capacity becomes okay they're asking concentration and buffer component and buffer capacity what is the effect. So whenever we what what we do with the concentration so that so that our buffer component is and our buffer component what we do with the concentration of that that our buffer capacity increases.

[00:52:33]So basically what we do whenever we increase the concentration of our buffer components. Okay. Whenever we increase the components that will result in making of the good more better buffer.

[00:52:45]No. So that means the buffer capacity will become more. So whenever we increase yes we increase the buffer concentration components. Okay. So that will result in increasing the buffer capacity also. So A1 is the correct one here. Yes. It's very very nice. Very good. You all are right. So this is the thing. Next in NMR spectrum of a substance the environment of absorbing protons with respect to the environment of neighboring protons tells us about NMR nuclear magnetic resonance. I guess everyone know about that. The NMR spectrum, position of signal, intensity of different signals, number of signals or splitting of signals, the substance, the environment of absorbing protons with respect to the environment of neighboring protons. What does they tell the neighboring protons?

[00:53:44]Yes, very good victory wives, you are very right. That is the splitting of the signals. That is the correct answer.

[00:53:53]Very very good. Okay. The D1 splitting of signals. We see that in NMR, right?

[00:53:59]The singlet, double, triplet, we get quadrate. Yes. Sexit sometimes. Septit sometimes. Okay. That we get. Yes. Spin spin coupling is there. That is also the true thing. But according to the options we have here is we have is the splitting of signals we have. Yes.

[00:54:20]Which among the following do not have an active metabolite? Okay, simple one.

[00:54:25]Just read the options, you will get the answer. Morphine, neostigment, dasipam, digtoxin. Following does not have an active metabolite. Which one? Yes, simple simple simple one. Morphine, neostigin, dioipam or the digtoxin.

[00:54:48]The neo stigman. Yes, the answer is the B1. Very very nice. Okay.

[00:55:01]The Gunda boy. Okay. He has uh written his name as Gunda boy. Okay. You are right. Akshai is also right. We have Sahita. Yeah, you all are right. Okay.

[00:55:13]Praha is also right. Victory Vibes is also right. That is neostigman. Very good everyone. Okay. So morphine, dioipam, digtoxin they all have a active metabolite but not the neostigment.

[00:55:27]Then okay what for? What thing the neostigin is mostly used? The major the major use of neostigment two different things we majorly use neostigment. What are they? I want to know in comment section if you remember. Maybe they ask the question from there. So I want to know that also if you will not I'll tell you but I want to know whoever want whoever knows that can answer me.

[00:55:50]Absorption in the IR region is due to the changes in which energy in molecule.

[00:55:56]Absorption in IR region is due to change in what? Vibration rotational vibrational electronic rotation and translational or electronic and vibrational.

[00:56:09]Yes, the vibrational and rotational these are the two energies the two energies that change in the vi region.

[00:56:20]Okay, so based on that we show our result. Okay, no one has answered me the neostigment one. Yes, the use of neostigment. Don't anyone know about that?

[00:56:37]Okay. So coming to the next question.

[00:56:41]Yes. Victory wipes. Yeah. Right. The treatment of mastthenia gravis the drug of choice. And one more thing the cobra bite. Okay. Whenever there is a cobra abide in that particular situation also neoigman is the preferred drug. Okay. So two things you remember myia gravis and cobra bite for neostigman. Okay, that the question also comes from this part but it was not mentioned here that's why I asked okay yes it is an anticolon estray that is also right what which type which type it is water soluble and reversible also okay what happens to the solubility of a gas in liquid as temperature increases solubility of gas whenever we increase the temperature will the gas in the liquid it will increase its solubility decrease its solubility it will remain same or it depends on gas and liquid involved.

[00:57:48]Yes, gas is a little bit different. Why?

[00:57:52]Because whenever we increase the temperature it will be more it will have more kinetic energy and it will get more freedom to move here and there and it will not dissolve. it will not be soluble. So it decreases the solubility the temperature. Okay.

[00:58:13]So decreases with increase in the temperature due to the increase in the kinetic energy.

[00:58:20]Sorry. Yes. The next question. Which list below gives only spin active nuclei? Okay. Spin active nuclei. NMR.

[00:58:30]Yes. NMR1. Spin active nuclei. List 1 H 2 H 12 2 carbon 1 H 12 C 19F 1 H1 13 C 19F 2 H 12 C1 19F Which one? Yes. Yes.

[00:58:48]Yes. Spin active nuclei. Simple one I guess.

[00:58:53]Yes. Yes. The C1. Yes. Very nice. The C1 is correct. Yes or no?

[00:59:03]So they all are what the spin active nuclei they will be active in the NMR you can go through that. Yes. So D C1 we go with that.

[00:59:17]Next question.

[00:59:19]What will be the correct reactivity order for electrofile in the following compound?

[00:59:27]Electrofile.

[00:59:29]Yes.

[00:59:30]the reactivity order for electrofile. So basically electron uh we can say loving okay electron loving whatever if we talk about that. So if you look on so here we have nitrogen here we have oxygen we have a nitrogen. So here the nitrogen and oxygen they both are responsible for the their lone pairs are respons well the pyrol ring has the best electrofile nature. Then then we talk about the oxygen. Why? Because oxygen is much more electrogative as compared to the nitrogen. So that's why nitrogen is more electrophilic. Then we have oxygen and in last we have the pyidin ring. Okay.

[01:00:15]So the option A is the correct one here.

[01:00:18]Yes, you are right. The A1 is the correct one.

[01:00:22]Next when two substances have same absorbance that wavelength is called same absorbance. Same absorbance is called as what? It's a simple one. Same absorbance.

[01:00:38]Isobestic point. Yes, the same absorbance is the isobestic point. The C1. Okay. Next question. A metabolite of spironolone.

[01:00:52]Spironolone.

[01:00:53]Metabolite. It is metabolite. It is metabolized to what? Yes, you are very right. The spirolone it is metabolized to our caninone. Okay, it is metabolized to our caninone. Okay, the C1 is the correct one here.

[01:01:16]Coming to the next question. A molecule leak used for introducing the sample, determining the amount of sample introduced, metering the sample to ionization chamber or pumping action.

[01:01:30]For which one we use molecule leak?

[01:01:33]Molecule leak is used for what?

[01:01:37]For metering the sample to ionization chamber. Yes, that is the correct one.

[01:01:42]The C1. Yes. Yes. Yes. The C1 is the correct one here. For metering of the sample to the ionization sample, we use a molecular leak there. Okay.

[01:01:56]Yes. The next question here in the screen we have. Yes. Everyone is not participating. Everyone need to pass it.

[01:02:03]Participate. Okay. The difference between the electrode potential of two electrodes when no current is drawn through the cell is called. The difference between the electrode potentials. Yes. Electro cell potential, cell emf potential difference or cell voltage. Which one?

[01:02:26]Yes. Waiting for your answers.

[01:02:30]Uh-huh. Yes. The answer is R B1. The cell emf.

[01:02:37]That is the difference between the electrode potential of two electrodes when no current is drawn. No current.

[01:02:44]Okay. So that is the cell emf. Okay. EMF whenever there is no current.

[01:02:54]Next question. Which of the following is known as white white withdraw? White withdrawal. Which one? What is called as white withdraw? Also known as white withdraw. Zinc sulfate. Ferrus sulfate.

[01:03:06]Copper sulfate, magnesium sulfate. Which one? Yes. Yes. Which one? Which of the following is known as white withdraw?

[01:03:16]I guess that's an easy one. Zinc sulfate.

[01:03:20]Yes. A1. Very nice. Okay. Yeah. No. Da.

[01:03:25]The white withdraw is a.

[01:03:29]Okay. The ring sulfate we have as white withdraw. Emic. It is used as that.

[01:03:39]The repeat uh the repeatability of a measurement opt obtained from multiple sampling of homogeneous sample is repeatability of a measurement obtained from a multiple sampling of homogeneous sample. Accuracy, precision, ruggedness or robustness. Yes. Which one?

[01:04:06]The measurement obtained.

[01:04:10]Yes. Accuracy, precision, ruggedness or robustness. Which one? Yes. Very, very nice. The answer is the B one. The precision.

[01:04:19]Okay. The precision is the right one.

[01:04:21]The B option.

[01:04:26]The an uh the ethnic acid is derivative of. Next question. The ethronic acid is then derivative of acetic acid phenol chlorenol or phoxy acidic acid.

[01:04:39]Anthroenic acid. Yes. Ethrrenic acid is an derivative of which what? So this is an example.

[01:04:50]Okay. So yes it is an example of what?

[01:04:53]It is an example of yes the D1 the the phoxy acetic acid. Okay. The phoxy acidic acid it is in derivative of that.

[01:05:05]This is the structure we have of that.

[01:05:07]Okay. So the D1.

[01:05:11]Yes. Very nice. You all are correct here. Next question we have. Which of the following is correct? TLC. Everyone remember TLC? It's a simple question. RF value they are asking the retardation factor. They are asking here the distance traveled by solvent by distance traveled by solute. Distance traveled by traveled by solute by distance traveled by solvent. Distance traveled by solution. Distance traveled by solute or distance traveled by solution or distance traveled by solute. Yes. Which one? So the retardation factor is basically Yes. The B1. Yes. Very very nice. The B1 is correct. Yes. Ranja.

[01:05:55]Yes. Praa you are very right. The answer is B. The distance traveled by solute upon distance traveled by a solvent.

[01:06:03]That is the correct answer. Okay. The B1 we can go through this. Yes, this is the TLC plate we have. This is the sample we put on here. This is the distance traveled. This is the solvent front. So this upon this is called as our RF factor. Okay.

[01:06:23]Quil's method is used for the measurement of cautil's method. You must have heard about that in your inorganic chemistry thing. Yes. Depression of freezing point, elevation of boiling point, lowering of vapor pressure, osmotic pressure.

[01:06:41]Yes. Yes, you are right. Victory wives, you are very right. The B1, the elevation of the boiling point is method. Okay. The B1 morphine show absorbance of ultraviolet light at wavelength. Morphine's wavelength the absorbance wavelength of morphine is the okay that is a learning question I guess so I think some of you maybe someone knew that yes the morphine one morphine has no it's not C no victory wise no it's not C it's a the 286 nanometer not 76 okay so 286 now you Remember that.

[01:07:29]So the A1 the 286 nanometer that is the correct for the morphine match the incorrect pair. Which one is incorrect?

[01:07:42]Then Robinson Gabriel synthesis for oxazole.

[01:07:46]Then we have girop synthesis for pyodin.

[01:07:50]Then we have bisclar nepilaxim synthesis for iso for the quinolin. correct one here incorrect pair okay you can go through the explanation part here that's a simple one next in the NMR spectroscopy which energy is absorbed simple such a simple question nothing can be much more simpler than this question okay if you know NMR if you have studied NMR you can't get confused here cosmic vibrational radio cosmic energy no vibrational is for IR rotational Again for IR radio frequency yes for NMR C1. Yes you are right. The C1 is the correct.

[01:08:36]Next chloro group present in which position of chlorioide? Chlorioide that is a type of what? Chlorioite is what?

[01:08:44]It is a diuretic. We all know. Yes.

[01:08:47]Chlorioide. Thioides are the diuretics.

[01:08:50]They act on the distal convulated tibule. That also we know. So what at what position we have in the structure we have a chloro group that chloro groupoup is important for its activity and that chloro group is present at the sixth position yes you are very right ranjana okay ranjina I'm talking ringa okay sorry ringana okay so that is the chlorioazide the chloro group I don't know the spelling whether I'm wrong or right okay for the pronunciations of your So sorry for that.

[01:09:26]Calculate the absorbance maximum for the compound given. Okay, the compound is little bit uh disturbed. Okay, let it be. Let's move on to the next one. Okay, so sorry for that because the structure is not clear. So we can't solve that. Doesn't matter. Let's come to the next question.

[01:09:46]Okay, Rochelle salt is Rochelle salt.

[01:09:51]Yes, it's a very simple one. Roshchelle salt. That's also the name for potassium sodium tartrate, silver nitrate, sodium s of thioulfate and potassium chloride.

[01:10:04]Felling solution B that also uses Roshchelle salt with NaOH.

[01:10:11]So what is Rochelle salt? Roshchelle salt is Yes. Very nice. The A1 the potassium sodium tart rate that is our Rochelle salt absorption in IR region is due to the change in which energy in a molecule. We have already done this question.

[01:10:32]No no no not A. A1 is Yes. Now you are right. A1 is copper sulfate. Yes you are right in that. Yeah. B1 is Rochelle salt with NaOH.

[01:10:42]Yes. The absorption in IR. We have done this so many times. I guess vibrational and rotational three four times we have done this in today's session only the pheninoiain derivative used for Parkinson disease okay all our pheninoiasin derivatives in which one is the Parkinson disease we use which one it can cross the bloodb brain barrier so we can easily use it Parkinson disease and it is exoproioin ethoproathioin sorry okay so ethoproioin is the pipino derivative that we that we can use in our Parkinson disease. Okay.

[01:11:26]The relationship in osmotic pressure can be expressed as the osmotic pressure we can express it as what? PV NRT P1 N1 P1 TF KFM or TB KBM. Yes. Which one? I guess it's a simple one. PB PVNRT. Yes, that is the equation for osmotic pressure. The R here is the ideal gas constant and we have the absolute temperature 2. Yes. So that pi here is an osmotic pressure only. So the PV equals to NRT that is the equation. Yes.

[01:12:07]65bs here. Right. Next question.

[01:12:11]The morphin with ali group in the morphin with ali group in which one where we get morphen with ali group present. Which drug?

[01:12:26]Which drug has it? Yes. Can anyone tell me which one? Codin, the pathin they all are almost similar structure. But if we talk about the Nellor Nelorin, Nellorphin we have here you can see Nellorphin has the what? It has the morphin. Okay. So Nellorphine is the drug here that has a that has a morphin. Okay.

[01:13:02]Next question. What is the correct order of acidity? Okay, don't go to the question. It's a very simple question.

[01:13:08]Okay, very simple.

[01:13:14]The acidity, the order of acidity from weakest to the strongest acid. Yes, they have given four and we have to tell which is the correct order. So I guess if you see this is alcohol, this is an caroxilic acid, this is again alcohol and this is O. Yes. So here you can see we only have an acid here caroxilic acid. So I guess second one will be the strongest acid out of all these pro four. Then phenol also is somewhat acidic. Yes phenol is also somewhat acidic. So third one we will take it as second one. Okay. Then two are left. Now this is alcohol. This is carbonal. This is carbon with alcohol. Okay. So I guess this one is more acidic as compared to this. Okay. First because first we have this compound then we have O. Okay. So more acidic is this alcohol directly. So then comes the first one. Then comes the fourth one. Okay. So I guess third one goes with our thing. Yeah. Second then third then first then fourth. Yes. The C1 it goes with our answer. Yes. The C1 the ranjana again is right. Yes. So C1 is the correct one.

[01:14:32]Coming to the next. Oxa zipam is a metabolite of simple simple question.

[01:14:37]Oxazipam dazipam teazipam a and b none of these. Simple question. Oxazipam is a derivative of both a metabolite. The oxazipam is a metabolite of diazipam and teaspipam. Yes. The c1 is the correct one here. Yes. The C1. Yes. Okay.

[01:14:59]Now coming to the next question.

[01:15:02]Vender wall equation is relationship between pressure, volume and temperature with pressure volume temperature and what the vendor wall equation. Yes. The what the amount of the real gas. Yes. Okay. The A1 R we also considered that now. So basically the real guess the amount of real gas that also we see in the vendor wall equation. Okay.

[01:15:41]Yes. 69th A that is correct. When acetone reacts with Grigard reagent followed by hydraysis it gives. Simple question. Okay. Uh you will only tell me this question. Grignard reagent we are reacting it with acetone. Which one is the correct for this? The acetone is reacting with grigard reagent. Simple question.

[01:16:05]It will give we will add grigard reagent adds and CH3 also. Okay. Remember that.

[01:16:12]No no no no no not primary.

[01:16:15]Yes. Yes. Victory wives is right. The tertiary alcohol we will get. You can see the reaction here. We add a CH3 group extra here we add we get get to add that and after hydraysis what we have we have an O also to tertiary alcohol is formed so the C1 the tertiary alcohol is the correct one yes the C which of the following determinant is a in a buffer solution by Henderson hasselbeck equation which one is determined by the buffer equation by Henderson hassleback equation. Simple question. Very simple question.

[01:16:56]Henderson Hasselbach equation in buffer solution we also call it as buffer equation. Remember buffer equation Henderson hassleback equation. What we find in that? We find out the pH value.

[01:17:11]Yes. The 71. Yes. Yes. Yes. The B1 is the correct. The pH value is determined with that.

[01:17:19]Next question.

[01:17:21]Controlled alkyation of ketone via an enamin intermediate is named as controlled alkyation of ketone via an enamin intermediate is named as what?

[01:17:37]It is named as yes. Who's going to answer me this one?

[01:17:43]It's a simple one.

[01:17:45]It's just a name reaction.

[01:17:49]It is Stokes reaction.

[01:17:53]Okay. It is Stoke reaction. It is an alkalation of ketone via interminent intermediate. Okay. So that is Stoked reaction. So this is ketone and here also you can see basically the Stokes reaction. Okay. Next question. Not manic stokes. Okay.

[01:18:14]The temperature of the system remains constant in Very easy. Okay, that is very easy. Same constant temperature. What we will call it? Yes. What we will call it? Yes, that is isoothermal.

[01:18:31]Yes. The thermal. Iso means same thermal means temperature. So that is the constant temperature we have here.

[01:18:39]Isothermal process.

[01:18:42]The B1. Yes, you are right.

[01:18:51]Which of the following compound is an example of aromatic ether?

[01:18:56]Nothing can be simpler than this.

[01:18:58]Aromatic ether. They are directly asking the uh name of the ring. Ethileacetate dimethile ether any sole or diile ether.

[01:19:08]Okay. The options are very simple.

[01:19:12]The ethile acetate, dimethile ether, any soul or diile ether.

[01:19:20]Okay. The any soul.

[01:19:24]What is any soul? Yes, the C1 is the correct one. Any soul is basically the this this is called as any soul. So this is an aromatic ether only. The C1 is the correct answer. Any soul. Okay. C1 any soul. We also call it as mythoxyenzine C1. Okay, remember any soul is an aromatic ether.

[01:19:48]Which parameter is used to represent electronic effect in 3D QSR? The question from drug discovery. Yes. The electronic effect it will be shown by it will be shown by R. Yes. Yes. Yes. It will be shown by electrostatic field. Very nice. Yes. The electrostatic field that will be our 3D QSR electronic effect. 75 B option. Yes. Yes. Arouch.

[01:20:23]Yes. Yes. Yes. Very nice. Nice. Okay.

[01:20:25]The B1 we can go through the explanation part again. Law of osmatic pressure is explained by Okay. That is also simple one. We have done this. We did the uh formula for this a few questions before only we did the law of osmetic pressure pi if you remember. Yes. So it was it was Yes. It was remember don't you remember that right now only we didn't with the vent rule.

[01:20:59]Yes or no?

[01:21:02]So the D1 which of the major product of the following react what is the major product of the following reaction? Major product yes the enoline it reacts with the NaO2 Na N2 and HCl at 0° centigrade. The major product we will get will be we will get the benzene dasonium chloride. Yes, we have a dazonium here.

[01:21:34]So benzene dasonium chloride we will get at a position. Yes, that benzene dasonium chloride will further undergo cuc undergo with the reaction in the in the presence of that. So cuc will result in the addition of an CN group. CN group also called as sino group or we can also call it nitral. So we will have a benzene with a nitral. So we have a benzo nitral. Yes. So the answer is the A1.

[01:22:04]Yes. The A is the correct answer. Okay.

[01:22:08]So this is the reaction.

[01:22:12]Addition reaction are the reaction of what can be much simpler than this addition where we can add something.

[01:22:24]Okay, if we want to add something we should have a vacancy for the addition.

[01:22:29]Now if we will not have any space then how we will add? So the vacancy we have where when we have some bonds there. So in single bond is it possible that we can add something like it's a sing single bond okay like uh let's say not C we can say this this this okay we have singles all so where we will add everything is already settled so for addition we need some double bond we remove this double bond and we add something so that is the addition reaction ction.

[01:23:09]Okay. So, double bonds and triple bonds are important. So, A and C D1 that is the correct. Yes, you are you all are right.

[01:23:18]Okay. So, D1 is the correct here. Next.

[01:23:21]Which reaction mechanism not allows rearrangement of substrate?

[01:23:26]Rearrangement of substrates. It substrate it is not allowed in E1 E2. Both E1 E2 neither E1 E2. Which one? the rearrangement. So the rearrangement is possible not possible not allow the rearrangement is our which one the B1 yes very good Arouch E2 okay E2 is the reaction that do not undergo the rearrangement okay how are aromatic e ethers typically synthesized easy and synthesized the aromatic ethers. Mostly the aromatic ethers are synthesized from I guess that's a very easy one.

[01:24:27]Yes. Yes. It's very obvious. It's also written the Williamson ether synthesis.

[01:24:32]So we form aromatic ethers from the Williamson ether synthesis only. Okay.

[01:24:38]So the uh sorry the C the C1 is the correct one here. Oxidation reaction of finanththerine produces we are oxidizing finanththerine and what we will get?

[01:24:53]Yes. Finanththerine oxidation. What we will get with the help of finanthrine oxidation?

[01:24:59]Finanthrine oxide 23 finanthraquinon 278 dioxopenanthrine or dphinic acid which one the oxidation of the oxidation of our finanthine what it will do.

[01:25:24]This is the reaction oxidation of finanthine. Whenever we oxidize it the final product we will get the dphinic acid. The D1 yes you are right the D1 is the correct one. Okay the diphinic acid the most common type of reaction in aromatic compounds. Okay that one that one is right. Yeah. The most common type of reaction in aromatic compound elimination, addition, electrofphilic substitution reaction or the rearrangement reaction? Yes. Which one?

[01:26:01]The aromatic compound?

[01:26:04]No, no, no, not rearrangement. How can it how can the aromatic compound rearrangement reaction is the most common one? We are not asking that is it possible or not. We are asking that which one is most common. It will undergo mostly parametric compounds.

[01:26:21]Electrphilic substitution reaction. Yes, the C1 is correct one. Yes. Yes. Yes.

[01:26:26]That one is correct. Yes, Ranja. Okay.

[01:26:29]So, the electrphilic substitution is the most common one.

[01:26:36]Next question. Electrphilic substitution reaction on thazole at position off.

[01:26:43]Okay. electrofphilic substitution on thiezole.

[01:26:50]Thiozole means one sulfur and one nitrogen.

[01:26:54]That is the ring. And what it will undergo the electrofilic substitution at what position?

[01:27:05]So basically thazole undergo the electrphilic substitution reaction at fifth and fourth position that are the most common places where the electrofilic substitution reaction takes place. Okay.

[01:27:25]What is the major product of the following reaction? The major product of this following reaction this is the mythoxy. Okay. Okay, the mythoxy group right now I told you the uh and the name of this particular product I want to know the name also okay you will tell me the name in the comment section the major product of the following reaction HBr they are saying HBr yes what we will get we will get what the CH3 will be gone the CH3 will be gone with Br3 Here we can get and we will get what? O the O. And what is this ring called? It is called as a phenol ring. Phenol ring.

[01:28:16]Okay. The major product will be the phenol. Not the D1. No, the benzile bromide. No, no, no. That is wrong. We will get the phenol.

[01:28:27]This type of CH3 Br.

[01:28:31]That's a simple reaction with HBr. Okay.

[01:28:37]A1 that is the correct one. Identify the correct statement regarding geometrical isomeism. Again the geometrical isomeism we started with geometrical isomeism.

[01:28:48]Again we have the question from there.

[01:28:50]Cis and trans isomer are in animer.

[01:28:52]Okay. Cis and trans isomers are destr.

[01:28:56]Geometrical isomeism occur in cyclic compounds.

[01:29:00]rotational rotation around C double bond C bond is free leading to a rapid inter conversion.

[01:29:08]D option I guess is that is absolutely wrong. Thus the carbon double bond carbon is not a free rotation. It is the most hindered rotation we can say.

[01:29:19]That's why it is undergoing the geometrical isomeism. Geometrical isomeism occur only in cyclic compounds.

[01:29:28]This is not a cyclic compound but it can undergo geometrical isomeism. So again C is also wrong. Okay. Here we can A B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B A We can write anything. Okay, so that is also wrong. D and C are wrong. A cis and trans isomers are enhances.

[01:29:50]They are nons super imposible. I don't think so. the geometrical isomers they are the mirror images of each other. No.

[01:29:59]So di isomer di is stereomr are the best uh option correct option we can say. So the what the diomer they are the different they are the thing. So basically cis and trans isomer they have different spatial arrangement. Okay there is the arrangement difference and they are not mirror images. Okay. So they those are dither.

[01:30:26]The B1 is the correct. Next arrange the compound in decreasing order of reactivity. The nucleophilic addition reaction. Yes. The re the arrange these compounds in decreasing order of reactivity for the nucleophilic addition reaction. Yes. Which one? Acid chloride, alihide, ketone. What will be the correct order of rearrangement?

[01:30:54]Decreasing order of reactivity.

[01:31:00]Sorry, I have shown you the answer.

[01:31:03]Okay, which one?

[01:31:06]So I guess which one? The acid chloride that will result in the best reactivity.

[01:31:13]Then we have alihide. Why alihide?

[01:31:16]Because it has one hydrogen there. Okay.

[01:31:20]So due to that reactivity is much more possible better ketone do not have a hydrogen very difficult to for the reaction. Okay. So decreasing order of reactivity than aster. Okay. So 1 2 3 4 that is the answer. The A1 is the correct option. Okay.

[01:31:40]Any soul is formed when phenol is treated with any soul. How can we form an any soul?

[01:31:49]How? With the treatment of what? First of all, what is any soul? Remember what is any soul?

[01:31:59]Yes, the answer for that.

[01:32:03]So basically what happens any is the ether. Okay, the anol. So whenever the phenol it interact with CH3I okay and NaOH they can undergo the reaction that will have OC3.

[01:32:21]Yes. This same reaction can be taken place in this side also.

[01:32:28]Yes. So that can also be the thing.

[01:32:32]Remember we have the similar kind of reaction. What is that reaction called?

[01:32:37]Can you tell me if we I take it in a reverse uh way. Okay. What will be this reaction called?

[01:32:45]So that is the any soul. So we need NaOH and CH3 CI that is the things we need with the reaction for the formation of our Nissol from the phenol.

[01:32:59]Which of the following is a nucleophile?

[01:33:01]Nucleus file. Nucleophile. Nucleus loving. What we can use as a nucleophile? BF3, Alcl H2O or H+ ion.

[01:33:10]Which one? Yes. Which one?

[01:33:14]Easy one, I guess. Who's going to answer me? This one.

[01:33:20]The nucleophile.

[01:33:22]Yes, the answer here is the C1. The H2O.

[01:33:28]That is a nucleophile we have. Okay.

[01:33:33]So H2 we use in the nucleophile the SN1 SN2 reaction if you know the nucleophile H2 only we use there also now. So nucleophile the H2O which of the following is not a method for separating enanchans.

[01:33:52]Which of the following is not a method for separating enanchans?

[01:33:56]Simple one I guess such a simple question. Chyal chromatography crystallization with chyal resolving agent distillation and enzyatic resolution options are so so so clear that we can't be wrong in this question distillation I guess how can with the help of distillation we can separate in numerous see what yes distillation the absolute configuration of the kyal center is determined by which rule what is the rule We determine the absolute configuration of a chyal center. Yes.

[01:34:33]Which rule?

[01:34:35]Uh-huh. Rule. Which one?

[01:34:44]Yes. We have the CIP rule. The can in old pre-log priority rule. We have for the determination of absolute configuration on the basis of Molecular weight we determine the the mass we see the weight we see. Yes. Remember the CIP rule based on the ranking of the we give the priority based on that. Yes. So that is our CIP rule. Okay. Based on the atomic number we do that.

[01:35:21]Next question. The IUPAC name of tartric acid. I guess for that you should know the structure of tartric acid.

[01:35:32]Do you know what is tartric acid? This is tartric acid. Yes. And here in tartric acid you can see you have two caroxilic acid and you have two hydroxile group. So if you see in the options so two t two caroxilic acid two caroxilic acid as are at first and fourth position and we have the hydroxy group and hydroxy group are present at second and third position. Here you can see second and third position we have hydroxy. So the option A is right. Yes.

[01:36:06]Yes, that is the correct one. The A option. Okay.

[01:36:11]Diile amin and methile amin expresses which type of isome? Diethile amine and the methile amine. Okay. So dethylamine. Which one are you telling me? Diileyamine. Let me write here for you. Diithylamine and methileamine.

[01:36:33]CH3 and H2.

[01:36:37]So which isomer? position functional metamor or none of this. I guess this is a very simple one. Yes. The isomer the what isomers are diyamine and methileamine.

[01:36:49]Which one? Yes.

[01:36:52]Yes. None of these. They are not isomer.

[01:36:55]They are totally different things. They can't be similar. They show no isomeism with each other. So none D1 none of these alihide of our first oxidation product of simple question. What is oxidized to form the alihide? Which alcohol? Primary, secondary, tertiary or monohydric alcohol? I guess this is also simple one. Yes, the alihide first oxidation product is the Yes. Which one?

[01:37:30]The primary alcohol, the primary alcohol only undergo the oxidation reaction and it results in the alihide. Okay, A1 that is the correct ketoarism is observed in.

[01:37:45]So what was the condition for ketoenol toarism?

[01:37:50]First condition for ketoinol toarism is that it should have an alpha hydrogen.

[01:37:58]If we look alpha hydrogen in this we have alpha hydrogen yes we have it in this we have alpha hydrogen no we don't in third do we have it again no alpha hydrogen here yes we have it okay so first and four C1 that goes with our answer yes the C1 they can show the toarism one and four Yes, those are the ketool form that can show the toarism.

[01:38:33]Now the next question n butane exist in how many number of confirmational isomeism?

[01:38:41]The nb butane it exist in how many number of confirmational isomeism?

[01:38:47]Yes.

[01:38:49]Do you remember the formula the confirmational isomearism for butane?

[01:39:00]Yes. Yes. Yes. Who's going to answer me?

[01:39:02]This one goes one and one goes two 1 and2 goes 2 and2 goes 2 and1 goes. Which one?

[01:39:14]So basically we have six but out of these six we have similar kind of also okay so we prefer this entire one the gauze one and the gauze one okay the others are somewhat similar only so we don't consider them so total are six but out of these three are all total different okay so basically three forms we have two go and one empty okay so the First one the one N and two B that is the correct one for the Nutane.

[01:39:51]This reaction mechanism is favored by substrates low temperature polar solvent and concentration of nucleophile. Yes.

[01:40:01]Which one? The E1, E2, S1, S2. First thing is either it is elimination or the substitution.

[01:40:10]substitution reaction or substitution nucleophilic reaction or elimination reaction. So basically the we can say we are using what we are using the nucleophile.

[01:40:23]So it will be nucleophilic reaction only. So substitution nucleophilic. So A and C we can remove them from list.

[01:40:30]Okay. Next we have the SN1 and SN2. So low temperature polar solvent polar protic solvent and the concentration of nucleophile and favored by substrate. So that is our yes that is our which one the which one the reaction. Yes. Okay. Next, which with respect to the confirmers of ethane, which of the following statement is true? With the confirmers of ethane, which of the following statement is true? Bond angle remains same but bond length changes. They are talking about confirmers. Remember that they are talking about conformers. Okay? Bond angle changes with bond length remain same. Bond angle and bond length changes both both bond angle and both bond length remains same. Which one is for conformers? Conformers for ethane. Yes.

[01:41:40]So in conformers what happens? There is no change in bond angles or length.

[01:41:46]Conformers do not result in the changes in the length or the angle.

[01:41:52]Okay. So basically the answer for this particular question is the D1 because confirmers they do not result in the change in changes in its what the angle and the length they remain same whenever we talk about the confirmer. So D1 is the correct one here. Okay. Now coming to our last question for today's session. We have the 100th one. Yes, the everyone need to answer this question as we have reached till here. The last question, what reagent is typically used to facilitate benzoilation in the Scottton warming reaction? Sodium hydroxide, sodium chloride, sodium bicarbonate or sodium cyanide. Which one facilitates the benzilation in the Scotenb reaction?

[01:42:47]Yes, benzilation is facilitated by what?

[01:42:54]Which one?

[01:42:56]The Yes. The A1 that is the sodium hydroxide. Yes. Okay. Sodium hydroxide is the one that undergoes the benzoilation the facilitation of the benzoilation reaction under goes with that. Okay.

[01:43:16]So I guess the questions to you were clear to everyone till here and the series was helpful also I guess. Yes to everyone I guess everyone understood the concept also. If you have any problem you can question uh you can put your question in comment section. We will try to solve your problems. Okay. And I want to wish you all the best for your exam.

[01:43:42]I I know you will do your best and you will definitely definitely qualify your exam. Okay. So all the best to everyone and thank you everyone for joining the session. Okay. So I'll end it here for now. Bye-bye everyone.