 Hello guys. Good evening. Good evening sir. Yeah. Good evening sir. So guys, are you solving this one? Center module for organic chemistry. Are you solving those modules? So for alcohols, I just started. Yes, same idea. Yeah. Louder. Yes sir, I started alcohols. Yeah, actually the thing is for module 4 and module 3, the solution is missing. I'm trying to get it. We have given the book for scans so that we can circulate the PDF copies of it. Okay. So now I see this in those PDF, a few chapters are missing. It is not there. Okay. So some of you have asked me already for the solution. So I'm trying to get it. Okay. Maybe if you know any hard copies then the office, then I can, you know, they can give you the scan copy of it. Okay. So I'm on it. I'm trying to get it. Okay. Since the office is closed nowadays. So that's why I'm getting some difficulty into this. So we can use MS. Yeah. So whenever I get it, I'll just share with you. Okay. Most probably I'll get it soon. Okay. So I think last class we were discussing this reduction of reaction. Correct. And we had finished reduction. The last reaction we did. Okay. So we'll start with open up, open our oxidation and. invasion. Correct. Yes, sir. Yeah. Okay. So let's today we'll start with a, there are go many more reactions. You have to do in this. So we'll start today with. Oxidation of Aldehyde and ketone. Okay. Oxidation of aldehyde and ketone. And then we have to do dials and then reduction reaction. So right down the heading, oxidation of aldehyde and ketone. So we know aldehydes are good reducing agent, right? Can be easily oxidized, easily oxidized to carboxylic acid, carboxylic acid. And hence, aldehydes are good reducing agent. Right, these are good reducing agent. Okay, as compared to ketone. Ketone, the oxidation is not that easy. We have already discussed it in bio molecules, right? So relatively, oxidation of ketone is difficult. Right, ketone is difficult to oxidize. Difficult to oxidize because there is no hydrogen attached with the carbonyl carbon. Difficult to oxidize. Ketone get oxidized under very drastic condition, very strong, you know, oxidizing agent required for this, like acidified K2CR207, KMNO4, all these things required. Okay, so what are the different reducing agent that can oxidize, sorry, oxidizing agent that can oxidize aldehyde and get reduced. Okay, so mainly we have three different reducing agent that we use and these are the test of aldehydes and ketones also. The first one we write down, tolerance reagent test. Tolance reagent test. Like I told you this, that we'll discuss this in oxidation reduction chapter when we were doing bio molecules, right? Tolance reagent test. So first of all, what is tolerance reagent? Tolance reagent is the ammonical solution of silver nitrate, NH4OH plus AGNO3. This is tolerance reagent, ammonical solution of silver nitrate. This question also they ask sometimes in need exam, not in JEE, but need to have asked this question many times, what is tolerance reagent? Okay, now we have this mixture, so what happens with this, you see? AGNO3 reacts with NH4OH and it converts into AGOH plus NH4NO3, ammonium nitrate. AGOH, when you heat this two molecules of this, when you heat it converts into silver oxide plus H2O elevators. Now this silver oxide, it is a mild reducing agent, mild oxidizing agent. Mild oxidizing agent and this oxidizes aldehyde into acid. Not an issue, just now we have started, okay, this is the first slide we are discussing, we are having, right? We are discussing oxidation of aldehyde and ketone, correct? It is a mild oxidizing agent. So when you take an aldehyde, so this aldehyde goes under oxidation in presence of tolerance reagent, which gives you, which ultimately gives you AG2O, so it converts into RCOOH and AG2O silver oxide get reduced into silver. Now suppose the tube you have in which the reaction is taking place, this silver atom that we have, suppose this tube, here the reaction is taking place in this tube. So we have here aldehyde and all these reactions taking place here. So this silver, which comes out here, it get deposit, deposit onto this tube, on the surface of this tube. This you can say this is the deposition of silver on its surface. And this appears like a mirror here. This deposition of silver, this is silver AG and this appears as a mirror when you look at it from outside. And hence this test we also call it as silver mirror test, right? Tolerance reagent test or silver mirrored test. So very important, okay, for all kind of exams, very, very important. Finished? Okay, now the most important question that they ask here, that which of these compounds gives tolerance reagent test or positive tolerance test? Okay, so all aldehyde gives, sorry? Sir, could you move to the previous slide? Done, sir. Done? Okay, so this is a very common question that they ask, which of these compounds shows positive tolerance test? Okay, for example, you see HC double bond OH formaldehyde, CS3C double bond OH acetyldehyde, then we have benzyldehyde acetone, then we have CS3COH, ORH. What is this structure? Could you tell me? So that's a hemiacetyl. Hemiacetyl, very good. It's a hemiacetyl structure. Okay, so like I said, all aldehyde gives positive tolerance test. This one shows, this one shows, this one shows, this won't show. This one shows, this is also hemiacetyl structure, OR and OH. OH, OR, hemiacetyl. So all these three are hemiacetyl structure, right? So the tolerance reagent, what is it again? What is it? Yeah, what's the compound again, sir? It is the mixture of, what we say, it is a ammonical solution of silver nitrate, that is NH4NO3, NH4OH plus AGNO3. Ammonical solution of silver nitrate. Okay, sir. What is this structure? This is acetyl structure. Simple, hemiacetyl may what we have, 1OH and 1OR. When we have 2OR, then it is acetyl. And acetyl structure won't show tolerance reagent test. Okay, you see, this is very, very important question. They'll give you 4, 5 options or suppose 8, 10 compounds they'll give you in integer type and they'll ask you which one of this shows positive tolerance test. Okay, just a second guys, just a second. So this is acetyl. Acetyl won't show tolerance reagent test. Okay, hemiacetyl shows. That's why you see in biomolecules we have discussed fructose has ketone as a primary functional group, but it shows tolerance reagent test. See, actually if you look at the structure, the basic medium mechanism, I'll just discuss this. They just give me 2 minutes. I'll discuss it. Okay, so first of all, you should know what all compound shows positive tolerance test. Aldehyde shows very simple. They won't ask you this. Usually how they frame questions. Suppose in need and all they can give 4 compounds, which of these shows tolerance test or does not show tolerance test. Or advance in integer type, they can give you 8 to 10 compounds. How many of these molecules structure shows positive tolerance test? Like this they can ask. So simple aldehyde if you have, all will answer this question. Aldehyde shows positive tolerance test. What is important here? You need to keep in mind hemiacetyl shows as still does not show. Why? We will discuss it just a second. Now, one more type of compound also shows positive tolerance test. And that is, suppose we have this one. CH2OH, we have here OH. And in the second carbon, we have C double bond O. This kind of compound means at, if the compound is one hydroxy to keto, one hydroxy, one hydroxy and two keto compound shows positive tolerance test. This is also you need to memorize. Now, it is not like I am doing the nomenclature of it, not like that. Just to name this compound, just to have an idea that what kind of compound it is, hydroxy and keto group, if present at the adjacent carbon, then it shows positive tolerance test. That's why we are calling it as one hydroxy to keto. Now, the question is why does it show the tolerance test? Okay. So if you look at the reaction, the reaction is what? The reaction is basic medium reaction because we are taking NH4OH over there. Right? So this is basic medium oxidation reaction. This is important one that you need to keep in mind. Okay. If you look at this, because of this, it is the basic medium reaction we have, basic medium. Now, in basic medium, what happens, you see, if you have hemiacetyl structure. So from here, what happens? The base OH- takes this H plus acidic hydrogen and it converts into its conjugate base, which is this, plus H2O. Further, what happens? When this comes back to form a pi bond here, this goes out as a living group. Hence, it converts into an aldehyde. And when we have aldehyde formation possible, then it shows positive tolerance test. So actually what happens when you put this molecule into the failing solution, medium is basic, this converts into aldehyde and hence the solution shows positive tolerance test. But it is not possible with this acetyl structure because we do not have acidic hydrogen here. Did you get it? Yes, sir. Yes. So we do not have acidic hydrogen. So hence, acetyl won't show positive tolerance test. Okay. Why this one will show? Because you can draw the tautomer of it in basic medium. You can draw the tautomerism of the tautomers of it. Tautomers, if you draw between this 1, 1, 2, and 3. So what we get here? We get CH single bond OH. And from the third carbon, this oxygen comes over here on the first carbon. We have a double bond here, OH and then R. Further, we can draw one more tautomeric structure of it between 1, 2, and 3. Sorry. We'll go this way. 1, 2, and 3. Third say, okay. So from the third, this hydrogen comes over on the first carbon and this pi bond shift over here. So eventually this converts into what you see? It converts into an aldehyde. That's why 1, 2 hydroxy group also shows positive tolerance test. Did you get it? So, sir, if we had a heavy acetyl structure instead of 4, if we had some other leaving group, like IE, will that also show tolerance test? Yeah, possible. If you have good living group here attached, then it may convert into aldehyde and it shows. So this is an idea I'm giving you. Usually they ask questions for hemiacetyl enol. But obviously the purpose of this OR is what? That it behaves as a living group when this lone pair forms a pi bond here. Correct? So whenever you have a living group present here, a good living group, okay, then it shows tolerance test. Here it is keto form. C double bond O is more stable. Keto form is generally more stable than enol. No doubt in this. Correct? So to sum up this, what we can say, tolerance reagent test or tolerance test is given by all aldehydes plus hemiacetyl plus 1, 2 hydroxy, 1 hydroxy 2-ketone group. 1 hydroxy 2-ketone means at adjacent carbon, we have double bond O and OH present. Correct? Next. So the tolerance reagent test, it is a basic medium reaction. Okay? Second one, we have failing solution test. Failing solution test. So failing solution test is what? We take, in this case, we take two solutions and we call it as failing one, failing two or solution one, solution two, failing A, failing B like that. Okay? So suppose we are failing one, this solution is aqueous CuSO4. And failing two is, failing two is sodium potassium, sodium potassium tartrate. Sodium potassium tartrate is this, OH, OH, COOK, COONAH, this is sodium potassium tartrate. Okay? The combination of these two is failing solution. This plus this is failing solution. Okay? So first of all, this one, the sodium potassium tartrate that we have, what is it a salt? Right? Is this a salt? What kind of salt it is? Basic. The salt of? Basic salt. The salt of? Weak acid. Weak acid and a strong base. Correct? And since we have a strong base, its solution is what? Its solution is? Basic. Since it is basic salt. And it is basic solution, basic salt. So it gives what? OH minus in the solution. So you can take here any basic salt. Usually it is a name reaction failing solution test. So we take sodium potassium tartrate. Okay? The solution is basic gives OH minus. Now this OH minus reacts with Cu2 plus which is there in the first solution, failing solution one. Cu plus two plus two OH minus, it converts into CuOH whole twice hydroxide. And when you heat this, it converts into CuO, CuO plus H2O. This is an oxidizing agent. It gets reduced into Cu plus and oxidizes aldehyde into acid. Okay? So the reaction is RC double bond OH plus CuO, the oxidizing agent. It converts this aldehyde into acid and CuO converts into Cu2O. Right? So Cu2 also capable of oxidizing any aldehyde left. This Cu2O again serves as an oxidizing agent. Converts aldehyde into acid and copper metal. So this is a reaction. Again, this reaction is basic medium oxidation reaction, basic medium. Here also the compound that shows, the compound that shows a failing solution test are aldehydes. Generally ketone does not show this test. We get a brick red color of it. The copper oxide or this we get, we have a brick red color of it. That confirms the, in some book they have written simply red only, but brick red or red we can say, not that important. Okay? So aldehyde shows this test again, failing solution test. Okay? So HCHO, HCHO gives this test. Okay? CS3, CHO gives this test. Bengealdehyde is an exception. It does not give failing solution test. It does not give failing solution test. It is an exception. Okay? Tallins reagent, Tallins test is given a Bengealdehyde, but failing solution is not given a Bengealdehyde. I must remember this. It is an exception. If you take a ketone, ketone also does not show this test. Right? Hemiacetyl, Hemiacetyl shows this test. Right? 1-2 hydroxy ketone also shows this test. So everything is similar. The only difference here is that Bengealdehyde does not show this test. No, we can say aromatic high does not give this test. Okay? One more. See, this one is also basic medium test. One more test we have, which is exactly similar to failing solution test. And the name of this one is Benedict solution test. Benedict solution test. Okay? Like I said, it is exactly similar to failing solution. The only difference is we have CuSO4 aqueous here. Right? Plus we have sodium citrate here. Sodium citrate. The failing, failing 2 is different over here. There we have sodium potassium tartrate. It is sodium citrate. Okay? So you have support same here. So for the previous thing, Hemiacetyl did not give it an acetyl. Hemiacetyl gave an acetyl did not correct it?. Yes. Okay, sir. CH3, COO na. Here we have CH3, COO na. And here we have OH. Okay? This is sodium citrate. Now, when you look at this compound, sodium citrate, is it a basic solution? It is the salt of what? weak acid and a strong base again, NaOH, NaOH, NaOH, strong base. This is also a basic salt, gives OH-ion in the solution. And once you get OH-ion, again Cu2 plus reacts with OH-. After this, all those reactions are same, like we had in failing solution. So these three tests we have given by aldehyde. Okay, it is a basic medium test. Okay, the first two failing and tolerance are the most important one. Okay, now the next we are going to see some acidic medium test. Only two we have into this one, not much important this one. Acidic medium test. Acidic? Yeah. Sir, could you go back to that Benedict one? I did not write down the entire reaction. You get OH- after this it is same as we had in failing solution. Yes, sir. Thank you. Okay, so we have only two, you know, tests here in acidic medium and it is not important. Okay, acidic medium test. So the first one we write down, we have HgCl2 test, HgCl2 test. What happens in this? We take aldehyde with HgCl2 two molecules of it in presence of water. It converts into an acid plus will get white color of Hg2Cl2. HgCl2 can further react with aldehyde. Just we have this reaction, nothing much into this one. H2O, it converts into acid again, plus we get two Hg and two Hg. This is what the reaction we have, acidic medium reaction. The color of Hg that we get here is gray-black or gray-black also we can say. Copy this down. Next write down the second test we have and in this one we do not have a reaction also. Just you need to know one property. Sir, can you go back to the previous slide? So one second. Yeah. Yes, possible, Vikas. How does the HgCl2 test oxidize benzaldehyde? That's possible, it can oxidize. Aldehyde, it can oxidize. If there is any exception, we will discuss that. We have exception only one. That is for this one, felling solution test which could not oxidize benzaldehyde. Otherwise it is true for aldehydes or aldehydes. So even benedict can't do benzaldehyde, right? Yes, benedict also won't. That's what I said. Okay. Felling and benedicts are similar reaction. They cannot oxidize benzaldehyde into benzoic acid. That is the only difference change we have, exception we have actually. Yes, sir. Done. Okay. Now another one here, another one is shifts reagent test. Shift reagent, you just need to know what happens into this. Reaction is also not required in this one. Okay. So what is a shift reagent? Shift reagent is it is rosa enylene, rosa enylene hydrochloride. This is a shift reagent, rosa enylene hydrochloride. The color of this is pink or magenta, pink or magenta color, rosa enylene hydrochloride. When this is allowed to react with sulfur dioxide in the first step, this get oxidized. So we'll write down here, oxidized form of this we get oxidized form of it, which is colorless. So first of all the pink or magenta color, so to decolorize the rosa enylene hydrochloride or a shift reagent. Now the test of aldehyde is what? In this colorless solution, if you add aldehyde, then this aldehyde restores the color of the shift reagent. So what we get? This aldehyde get oxidized into acid here and this will get reduced into rosa enylene hydrochloride. That is a shift reagent. So we get back a shift reagent when we add aldehyde into it and this color is again pink or magenta. So what we say, the test of aldehyde is what? That it restores the color of shift reagent in the reaction. And that is the test of aldehyde shift reagent test. Nothing much you need to know into this one. Only one thing which is also not at all important, but the structure of rosa enylene hydrochloride, I am drawing it here. But whenever they ask, they'll write down the name only. Overall, the reaction is not that important. Sir, add it out. Tell me. Sir, why does glucose not give shift reagent test? Glucose. Glucose, yeah. Yeah. Sir, I read that it doesn't give shifts or reagent test. It does not give, right? Yeah. Why? Just a second. Okay. So glucose has hydroxy as well as aldehyde group also present, correct? Medium is, it is an acidic medium reaction because you have hydrogen chloride, HCM. Acidic medium reaction, what happens? The hydroxy group reacts with this and it goes under dehydration, right? The lone pair on oxygen atom in aldehyde, right? And the hydroxy group, the lone pair you see both has tendency to react. Okay. But the lone pair on oxygen in aldehyde group and this one if you see, so hydroxy group has poet linkage. So when it gets protonates, then H2O easily eliminates. When oxygen protonates, that instead of getting a carbocation when the pi bond shifts, so stability is not there that much. Okay. And when here we have the, you know, carbocation forms over there, that is more stable than the other one. That's why usually the aldehyde group of glucose does not go under shift region test. The medium is what? The medium is acidic here, hence it is. Okay. Hydroxy group takes part in the reaction. Aldehyde group does not convert into acid into that. Okay. So this is Rosa Aniline hydrochloride. Three benzene ring we have here, not three, two benzene ring, double bond NH we have here, dot HCM. This is also benzene ring. This is the Rosa Aniline hydrochloride. Obviously they won't give you this structure. They'll write down the name on it. Okay. Usually what they ask question this, like one, this question, glucose question they ask in biomolecules in general. In that chapter, you may, you may be, you get those reactions. But in aldehyde and ketone, what they ask that which of these molecules restores the color of shift reagent like that. They frame the question. So you need to take aldehyde over there, not any other compound ketone or acid. Okay. This is the only question that they ask. They won't even ask you, they may ask you the name, what is shift reagent, but they won't even ask you the structure of Rosa Aniline hydrochloride. Okay. So over here, even hemiacetyl won't give it, right? No, hemiacetyl we have only in basic medium, not in acetic medium. Okay. Then only if basic medium is there, then only that acetic heroin goes out top and that converts into aldehyde. Yes, sir. Okay. So these are the five best of aldehyde we have discussed. Three in basic medium, tolerance reagent, felling source and benedict solution. And two in acetic medium, agency halitone, it's a free interest. Okay. Now write down oxidation of keto. Oxidation of ketone. Also we have two, three actually methods here. Right. Again, the two methods, how to write down the product. We have to see the mechanism is not there. Okay. So write down the oxidation of ketone, but the last one that is bare villager oxidation there will discuss the mechanism. Like I said, ketone is difficult to oxidize. The oxidation takes place in drastic condition of ketone. Okay. So if you have any ketone, we can use the oxidizing agent as hot acidified KMNO4 or K2CR2O7. This we can use. Okay. Oxidizing agent of ketone. What we need to do here, if you have any ketone, I'm just trying to tell you how to write down the product in this. So we always write down, we always write down the enol form of this ketone. That is OH here and double form. Enol form of this. In the sum of the reaction of ketone also you see, we always draw the enol form first and then we'll write down the reaction. Those reactions we'll see in that particular chapter. Okay. So how to write down the product. First you write down enol form of it and then you write down the product according to oxidative or generalysis. So this reaction is similar to, this reaction is similar to oxidative or generalysis. What we get in oxidative or generalysis? What is the difference we have in oxidative or generalysis? So there's no zinc. So no zinc. So you won't get aldehyde but we get acid. Yes sir. Correct. So what you need to memorize here just from ketone, you write down the enol form. I'm not discussing the mechanism that won't help you. Okay ketone, you write down the enol and then enol from enol you write down the oxidative or generalysis product. So could you tell me the oxidative or generalysis of this? What you will get? Ethinoic acid plus methanoic acid. Ethinoic and methanoic. Yes sir. So how do we write down, see the reagent is this only. The reagent is H plus K2CR2O7 or KMNO4. The reagent is this only. But the product will write down similar to the oxidative or generalysis. So this bond you have to break and you have to add double bond O and double bond O here. So here we have double bond O and OH. And here what we get? O double bond C has two hydrogen but since it is oxidative, so we'll get here OH not H. So further this one is not stable because we have two OH present on the same carbon atom. H2O molecules eliminate and it converts into carbon dioxide. So what is the compound called? The second one. This one? Yes sir. We can say this is a, there's like one general name is not there but we can say it is a diol, we can say diol form of, what we can say methanone. We call it dihydroxy methanol. That's the way you can, you can say like that because we have two hydroxy group present. So it is methanone first of all. So dihydroxy methanone, you can say. But common name in general we can say it is a derivative of ketone or diol simply we can say. The common name. IUPAC name will be one carbon double bond O. So it is methanone, two OH present. So it is dihydroxy methanone. Okay. So two OH on this carbon is not stable. H2O eliminates forms CO2. The product will be an acid and carbon dioxide. Okay. So all the molecules, you'll write down like this one. First you write down the enol form of it and then oxidative or generalizes. Write down the product in this one. Means the question will be like this. I'll show you the question. The question will be like this or which they won't give you the reversible sign here. This goes under H plus KMNO4. This is the question actually. Now what we'll do to write down the product, we'll write down the enolic form of this, more stable enol. That will be this OH here and double bond here. The other one will be the less stable minor product. Okay. Now after this we'll do the oxidative or generalizes and hence this converts into, right, this bond will break and add double bond or double bond. So what we get here, CH3C double bond O, this carbon. This side we have H, so OH plus CS3COOH. So two molecules of this we get oxidative or generalizes. We won't get aldehyde in this one. Write down the product in this two. Okay. So first of all we'll write down the enol form of it. The enol form would be this and then we have oxidative or generalizes. So that would be one, two, three, four, five carbon, correct. Double bond OOH, double bond OOH. This is the product we get, right? This one we have four. So we have one, two, three, and four double bond OOH. One, two, three, four, five OOH. So we have double bond OOH, okay? One, two, three, four. One, two, three, four. Di-Ioic-Ioic acid we get here. Okay. So just we'll write down the enol form and then oxidative or generalizes we'll write down the product. Now the second one we have selenium dioxide, the second compound, SEO2. This is also very simple one. You just need to know the reagent and what it does in the reaction. So suppose we have the oxidation of this compound you need to do with SEO2. So what you have to do here at the adjacent carbon of C-double bond OOH, just a second, you pick up. Yes, those oxidizing agents which can oxidize ketone, they are strong oxidizing agents because oxidation of ketone is not easy, okay? So in this case, what do you have to do? Just adjacent carbon to the carbonyl carbon you have to add one double bond OOH like this. Hydrogen, that is it. Another example you see, you'll get it. pH C-double bond OCH3 with SEO2. So obviously on this phenyl group we cannot add. So it will be pH C-double bond OOH. This carbon will have one more double bond OOH and one hydrogen attached to it, okay? One more reaction. Suppose we have double bond OOH like this, cyclohexanone. The product of this would be on any adjacent carbon just add double bond OOH like this. This is the product of this. This is also used to form or to attach hydroxy group on the allylic carbon. Copy down this first. I'll go to the next page and then I'll write down. This one, the allylic carbon, this is a bit important. Only one point you have to keep in mind. Like the last class we have discussed, MNO2 can oxidize only allylic alcohol, right? Similarly, here what happens at allylic position, OOH group you have to attach. That is it. Suppose we have double bond CS3CH, double bond CS2 alkene. If you have ketone, then you have to add double bond O on the adjacent carbon. That is it. Done. But if you have alkene, then on the allylic carbon, we'll have a hydroxy group. This is the product of this reaction. I'll go back. These two reactions you have to memorize. For all these oxidation and reduction reaction, the mechanism is not given. If you find out this in some good books, some higher standard books if you study, it is quite complex also. If you look at the mechanism, then also you're not going to write down the mechanism in the exam and get the answer. They won't ask that question because there's no possibility here. Just need to add OOH over here. It's not like some D2O or H2O we are using so that it will change the product, nothing. These kind of reactions, no, you need to memorize it this way only. At allylic position, OOH you have to add. What is important here, I'll tell you. The important is what? The rate of this reaction, write down, the ROR rate of this reaction, it follows the solid. 2-degree carbon has highest rate, then 1-degree and then 3-degree. This is the rate of the allylic carbon we have, the reaction at allylic carbon. Means if you have 2-degree allylic carbon, then OOH group present at 2-degree allylic carbon, then 1-degree and then 3-degree. For example, suppose we have cyclohexene. Cyclohexene, the reaction with SeO2. Obviously this is allylic, this is allylic, both are 2-degree, correct? Both are 2-degree, so anywhere you add OOH, this is the answer for this question. But if you have a compound say like this, with SeO2, this carbon is what? This carbon is allylic carbon, this carbon is allylic carbon, but this one is 2-degree and this one is 1-degree. So OOH group get attached at secondary carbon, that is this one, done, finished. One more example you see, this is 1-degree, this is 2-degree and this is quaternary, there is no hydrogen present here. The reaction takes place obviously at 2-degree. So we get OOH here, this is the answer. So I would suggest this two reaction for SeO2, you must remember, mechanism is not at all required for this reaction. No, not like that here. First of all we won't take this SeO2 in excess. Still we have one more hydrogen present here. So we won't take that maximum of excess of SeO2 and all these positions are oxidation takes place. We generally don't consider that. In a molecule at only one point and only one carbon, the oxidation takes place. And you will have option also, so you can go through by option, but usually it is not possible, they won't ask you that. Okay, now the last one, the oxidation of ketone is, we have bare villager oxidation. It is again a name reaction and this one is important also. Write down the third one, bare villager oxidation reaction. Done. Okay, what is bare villager oxidation? Bare villager oxidation reaction is the reaction of aldehyde or ketone. It is given for ketone first, but since it can oxidize ketone, it can also oxidize aldehyde. So aldehyde and ketone oxidize from this particular reaction. Right, so if you have a ketone, suppose R1 C double bond O R2 and we use a peroxy acid for this purpose. Peroxy acid we use which converts this into R1 C double bond O O R2. That is it, the product we get. So it is ketone and on this oxidation ketone converts into esters. That is what the bare villager oxidation we have. So ketone into ester, yes. Sir any peroxy acid works here? Sorry? Does any peroxy acid works here? Any peroxy acid you can take. We will show you some examples. Sorry? Normaldehyde won't work right? Formaldehyde, no, that is also possible. It converts into formic acid then. Okay. It converts into formic acid, that is possible. I will discuss the mechanism, I will discuss. But first of all, again for this one also mechanism you are not going to write in the example. So you should know how to write down the product. What we will do, carbonyl carbon and one of the alkyl group will attach or insert one oxygen atom between the carbonyl group and one of the alkyl group. Which alkyl group is the question? Where do you have to insert? This side or this side? This alkyl group or this alkyl group? So the alkyl group, suppose I have inserted over here, right? It means this alkyl group is the better migrator. The alkyl group which is a better migrator, you have to insert oxygen between that alkyl group and the carbonyl group, clear? That is what we need to know. Now to do this question, what you need to know, just first of all we know better electron donors are better migrators. Second thing, the migratory aptitude which we have already discussed, the migratory aptitude you should know for this particular question, this particular reaction. We will discuss the mechanism here. But these two things you know you can do this. You don't require mechanism for this. Migratory aptitude you know. We may have both side hydrogen we may have present. So the migratory aptitude of hydrogen is maximum. Then we have 3 degree. Then we have 2 degree. Then we have phenyl. Okay, then we have 1 degree. And then we have methyl. So this is the migratory aptitude in general. We write down this. Means what happens? Could you tell me the product in this one? Suppose if I take this one CS3, C revolve bond OH. Which suppose one peroxy acid I am taking. The CF3, COOH. What we get here? Ethnic acid. CS3 C revolve bond OOH. That is it. Hydrogen and methyl. Which one is the better migrator? Hydrogen. So we will insert oxygen between this carbonyl group and this hydrogen. This is the answer. Okay. If you have this CF3, COOH. What is the answer we get? Which one is the better migrator? This one is 2 degree. This one is 1 degree. 2 degree is the better migrator than 1 degree. So we will insert oxygen here. And it converts into a 5 membered, 6 membered ring. Double bond O. Any doubt in this? Is it fine? Tell me. Understood? No doubt. Is there like a acid which is like most commonly used in this kind of thing. Like peroxy acid. CF3, COOH we use most commonly. We can also use MCPBA. Metacluid per benzoic acid. Okay. These are the most commonly used. But how does CF3, COOH have peroxy linkage? I missed this. Okay. Peroxy. Okay. Any peroxy acid you can take. Fine. So this is how we will write down the product. Just you need to know this one. That is it. Okay. Migratory aptitude you need to apply and insert one oxygen. Now what is the mechanism of this reaction? See the mechanism. Okay. So what happens? We have suppose a ketone, RC double bond O, R2. And a peroxy acid. Suppose we have HO, O, C double bond O and R. This is a peroxy acid. So what happens? Season which attached to the hydrogen in peroxy acid. This oxygen. Not this one. This oxygen, it donates its lone pair of electron, lone pair of electron onto this carbon atom. And this pi electron shift onto this oxygen. And why this happens? Because once it donates, oxygen get positive charge. Then H plus, this H plus comes out to stabilize the positive charge on the carbon atoms. So if I write down this in one step, we'll get this R1, CO minus R2, O, O, C double bond O, R. Did you understand this? H plus, clear? Yes. End out? No. Okay. Now what happens in this? Since we have peroxy linkage, this bond is not stable. We have electron, electron repulsion here. This per sigma bond takes up by, is taking up by this particular oxygen. Okay. It goes out creating a vacant orbital onto this oxygen. So R1, CO minus single bond O, two lone pair, and a vacant orbital positive charge onto this oxygen. This R2 will be as it is. Okay. Now, this takes this H plus forms RC over. This goes out as a leaving group, LG. Plus RC double bond O, O minus goes out as a leaving group. Now here what happens, the one R1 or R2. It is an irreversible reaction. The one which is the better migrator R1 or R2. Suppose R2 is a better migrator. It takes this sigma electron pair and this rearrange itself onto this oxygen because it has one vacant orbital and we can take this to electron under this. In this way, what happens? We get R1, CO minus O, R2. And this carbon has positive charge to neutralize this. This sigma, this lone pair forms a pi bond here. So R1, C double bond O, O, R2. Again, I'm telling you this mechanism is fine, but you have to write down the product direct. You don't have to think of mechanism over here. Then. Okay. Write down the product in this reaction. CS3C double bond O pH with, suppose, with, suppose MCPBA. Meta clodo per benzoic acid. Nothing. You can, you know, keep it with you. I'll just write it down. You just cross check. So which one is a better migrator? CS3. Ph is a better migrator, right? So we'll add O between this C double bond O and pH. The product is CS3C double bond O, O pH. Your three degrees better migrator. So the product is O C double bond O CS3. Hydrogen is a better migrator. OH. This converts into a six member ring, including oxygen present in the ring. Okay. So this is the oxidation of ketone. Only three. Yeah. So isn't the second one quaternary? This one. No, this carbon is this. No, we are talking about, we are not talking about with respect to this. This carbon attached with one, two, three. Hence it is tertiary here. Tertiary group attached to this carbon primary group, this methyl group attached to this car. That is what it is. Okay. So three different reactions we have seen. One is the reaction of reaction of what SEO to the first one is H plus KMN O four H plus K to CR two or seven acidified KMN O four K to CR two or seven. Second one is selenium dioxide SEO to and third one is bare villager oxidation. Bare villager. We need to apply migratory aptitude SEO to we have two types of reaction. One is for ketone. One is for alcohol. Sorry. In case of alkene, we'll get O H on the ally carbon. And when we have ketone, then adjacent carbon will have a double bond. Oh, that is it. Okay. Three reactions, three oxidation of oxidation reactions of ketone. Now the other one, write down the oxidation of diol. Okay. Oxidation of diol. Only few reactions. Just one, you know, the reagent we use for this purpose. Again, mechanism is not required here. That is H by O four. What do you need to do? Suppose you have any diol. For example, CS to O H CS to O H with H I O four. If you heat this. Okay. Then what happens here, the carbon carbon bond dissociates. This bond dissociates and you have to attach O H with this carbon and O H with this carbon. Means that adjacent position, if you have O H or in that case, suppose functional group like and the height present, then also it is possible. Addison carbon contains O H on reaction with H I O four. This carbon carbon bond dissociates and we add O H group on each carbon. Plus the same thing we get from the lower carbon. That is this. Which further converts into H C H O plus H two O. And this also converts into H C H O plus H two. This is the product in this. Just you need to dissociate to break the carbon carbon bond. Right. And product you need to write down. Similarly, you see what is the product we get in this reaction. I'll write down the next place 123456. So we have 123456. We'll get here double bond O and H double bond O and H. Plus we'll get two molecules of water. So H I O four is a good oxidizing agent. Right. So does aldehyde remain aldehyde or does it become acid? No, it will be aldehyde only here. It won't convert this aldehyde into acid. The final product in case of aldehyde will be in case of diol will be an aldehyde. If it is forming an acid, I'll show you some more example. If you are getting an acid directly, not by the oxidation of aldehyde, then the product will be an acid. Means aldehyde won't further oxidize into acid. Okay. Yes, sir. So it is a good oxidizing agent, but not good enough to oxidize aldehyde into acid. This one you see. Write down the product in this. Again, you see this group is not at the adjacent position. Right. So this won't oxidize. The first one, what happens. Three molecules of HCHO. This two will break, right? Yeah, that's what you and I thought. H2COHOH first plus H and we get three hydroxy group attached to this carbon. And the last one is also this. H2COHOH. I'm sorry. Correct. Now from this, what are any minutes? It converts into HCHO. Water eliminates. It converts into HCHO. Water eliminates. It converts into HCOH formic acid. So two molecules of formaldehyde and one molecule of acid. How does it become three hydroxy groups in the second structure? Is it too sadistic? No, no, no. One OH is already present. Okay. This bond breaks one OH here. One OH here. One OH here, one OH here. Okay. The middle one, two sigma bond breaks and hence two OH on the carbon. For a given carbon atom, the number of sigma bonds you are breaking equal number of hydroxy group you have to attach. Methanol, Richard, I guess. Not methanol. I'm in the methanol. Okay. So now in this, what happens? This one, you have to break this carbon-carbon bond. So we'll get this one forms what? If I can write down this, it is HCHO, one molecule. Yes or no. And this one won't get oxidized. So from this to this carbon, it will be as it is. So we have CH2OH. The lower carbon is this. Then the middle one is this. And this one has two OH present. This side OH, this side OH converts into C double bond OH. This is the product we get. Plus number of water molecules. If you count, it will be two H2O. Okay. So this is it for oxidation reaction. We have done alkene, alkyne, alcohol, then aldehyde, ketone, diion. Okay. All these things we have done oxidation reduction. Next reaction, you write down reduction reaction. Okay. So like oxidation reaction, we use various oxidizing agent. So for reduction reaction also, we use various reducing agent. Various reducing agent. Correct. So what all these reducing agents we have? The first one is LiAlH4. It is lithium aluminum hydride. Lithium aluminum hydride. Okay. This LiAlH4, it is a coordination compound. Coordination compound will study coordination chemistry. That is there in inorganic chemistry will do that chapter also. And we'll discuss this coordination compound in detail into that chapter. Okay. It is a coordination compound, which we also call it as complex compound, complex compound. Why complex compound? Because it has a complex part present into it. Because if you look at the structure of this, the bonding here, we have Li plus and we have AlH4 minus. This is the actual structure we have. Okay. So coordination compound is like this only. For example, you see K4 FeCN6. It is also a coordination compound, complex compound. So you always write down one part in a square bracket and one part outside the square bracket. Okay. So this part, which is written in the square bracket, we call it as coordination sphere. Coordination sphere and which is written outside the square bracket. We call it as ionization sphere. It is not related to this, you know, reduction reaction, but just I'm giving you a bit of idea of coordination compound. Okay. Like normal compound, these coordination compound won't lose its identity in the solution completely. Like suppose NaCl if you add. So it dissociates completely in water. Na plus equals Cl minus equals. This also will dissociate, but this part loses its identity completely. But the part which is written in the square bracket, this is the complex part. And since it does not lose its identity into the solution, hence we are calling it as complex. This is one aspect in which these kind of compounds differs from the normal compound and we call it as complex compound. It does not lose its identity into the solution. Here also you see to dissolve this in water, K plus will lose, but this one, this square bracket, this ion won't lose its identity into the solution. Right. This is the basic difference between a normal and the coordination compound. Okay. So one more difference we have, like in this, we have two types of valencies, primary valency and secondary valency. So all these things in detail will study in the chapter called coordination compound. A very conceptual chapter, the only conceptual chapter of inorganic chemistry where you can have, you can apply concept and you can get the answer. Sure, you will get one questions from that in your J exam. Okay. Any exam you'll get questions from that. If you are going to write neat exam, then you will get two, three questions at least from that particular chapter. And this one part of this coordination compound is the isomerism also. So the isomerism that we have done already. And this chapter, they mix that two chapters and forms one question of isomerism. So there are different, different, different types of questions you will get from this chapter. Most important one, the only chapter, conceptual chapter, which you can understand. And it is coding also. If you understand it properly, you'll definitely get, you know, correct answer into this. There are a few things that you need to memorize here also. Few things, like, you know, 10, 15% you need to memorize. But then, you know, the other things you can understand it. So LiAlH4 is a complex compound. Correct. Now you see what happens. This part, you look at the structure of this part. AlH4 minus. It is a reducing agent. So obviously it gives H minus hydride ion to the solution, right? So AlH4 minus is this. Al with aluminium, we have four hydrogen attached. Negative. And around this, we have Li plus present. This you can write in a square bracket. Aluminium hydrogen bond here, right? It is, it has more ionic character and less covalent character here. Its ionic character is more. Aluminium hydrogen bond. Its ionic character is more. More in the sense that if you look at the next reducing agent, right? NaBH4. So there the ionic character is less. So I am letting out this ionic character is more in the context of NaBH4. That you must mention over there. Ionic character is more in comparison to NaBH4. So because of this, what happens since it has ionic character. So this Li AlH4, this AlH4 minus Li plus, there is no role for this. It just neutralize the ion over there. Right? The major thing is this Li AlH4 minus ionic bond is, you know, weaker bond we have here. So it easily converts into AlH3 and releases hydride ions. H minus. I am sorry I have written AlH4 here. It is AlH3. It forms AlH3. So Aluminium also becomes neutral here, you see. And releases hydride ions. So because of this hydride ion since it releases, so it is what? It shows the reducing character. Any doubt here? No. NaBH4 we will discuss after this. First we will see the reaction of Li AlH4. Again here also we have for certain reactions, we have mechanism. Certain reactions I will give you where the mechanism is not required. So again I will suggest you understand the reaction that we only, how to write down the product. Okay? Now NaBH4 you see, like I said, here we have more ionic character. NaBH4, we have more less ionic character, little bit more covalent character in comparison to Li AlH4. So BH4 ions that you have over there, the BH bond is stronger than the AlH bond. Hence for BH4- it is comparatively difficult to lose hydride ion. That's why NaBH4 is a weaker reducing agent in comparison to Li AlH4. Did you understand this? Don't write this NaBH4 part right now. We will discuss NaBH4 separately and then you can write it down. So NaBH4 is a weaker reducing agent than Li AlH4 because of a stronger boron and hydrogen bond, relatively. Correct? This is the one point you must remember. Now we know alcohol, primary alcohol on oxidation gives aldehyde. Correct? So aldehyde on reduction, what it should give? Alcohol. Alcohol 1 degree, so it is exactly Ulta. Correct? 1 degree alcohol on oxidation gives aldehyde. Aldehyde on reduction gives alcohol. 2 degree alcohol on oxidation gives ketone. So ketone on reduction gives 2 degree alcohol. Got it? So we will see this basic thing you must remember. Right? Everything is based on this information only. Alcohol to aldehyde, aldehyde to alcohol and ketone as well. Correct? So we will see the reaction first of all and then we will discuss here the mechanism also. Okay? This reaction, the mechanism is important. Wherever it is important, required, I will tell you to understand the mechanism. Okay? Don't worry for that. Okay? Suppose we have ketone, aldehyde ketone, the reduction of aldehyde ketone we are discussing. So whenever you see from now onwards, whenever you see, what did I write? Whenever you see LiAlH4 or NaBH4, right? You should see H- not LiAlH4 because the reaction is because of H- that is coming out from LiAlH4-. Agreed? No doubt. So suppose we have ketone like this and what did I use? The first step I use LiAlH4 and in the second step I use H plus H2. So acidic hydrolysis in the second step. We always use it. Acidic hydrolysis in the second step. Okay? So what this reducing agent gives, could you tell me? 2 degrees alcohol. This gives H minus. H minus, yes. Correct? So from here you get H minus. H minus you see how easily you can understand this. RNGX reaction you remember, right? Nucleophilic addition. So what happens from RNGX? We get R minus and that R minus attacks on what? Attacks on carbon. We don't have R minus but we have what? We have H minus. Any doubt? No? Tell me come on. No doubt, right? So this H minus what happens? This H minus attacks onto this carbonyl carbon and this will push the electron on this oxygen. Exactly similar to the reaction that we had in the case of RNGX. Got it? So these here, the product that we get R1 CO minus H R2. This is the first step we have. The second step we are using acidic hydrolysis H plus H2O. So this O minus converts into what? Converts into OH R2 and H here. You see ketone converts into 2 degree alcohol. So the H will go out as H plus right? Sorry? So the H should go out as H plus right? This one, no, no, not this one. Sorry, sorry. This H plus simply attacks onto this. Got it? Understood the mechanism, right? So what is important here? Why mechanism is important in this reaction I am telling you? Could you tell me the product in this reaction? The reaction is this. Suppose I am taking, tell me the product here. Could you tell me the formula here we get? So CH3 CDH. We get here CH3 COH D H. What happened if you take here D plus D2O? So you should know the hydrogen that attached with this carbon. What is the source of this hydrogen? Could you tell me the source of this hydrogen? LIH4. LIH4, right? So this hydrogen which attached with this carbon, the source of this is it comes from the reducing agent LIH4. And this comes from the solvent that we are using. That is H2O. No doubt. That's why in this type of reaction, the mechanism is important. Where, you know, when this thing change LID4 or suppose LIALH4 D plus D2O, then you should know the carbon, the carbonyl carbon, which is gaining hydrogen, one on carbon and one on oxygen. So from where these two hydrogens are coming onto this. Only this thing is important here. After this, we have same thing. If you have ketone, then two degree alcohol, aldehyde, then one degree. Any doubt in this? Right? Reduction reaction, we know it is the addition of hydrogen. So hydrogen is getting attached. Reduction is taking place. No doubt. Okay. One more thing here, it is important that you already know that this carbon becomes what? This carbon becomes chiral carbon now. So it has two form, R and S. That always you have. Okay. That kind of question. So I'm giving you some questions. Write down the product in this. LIALH4 H plus H2O. CS3 C double bond O CS3. Same reagent. LIALH4 H plus H2O. ALH4 D plus D2O. So here the product is CS3 OH. Here the product is CS3 CH CS3 OH. Here the product is we get OD here and H. Finished. Write down the product in this one. This one you have done. A is what? No doubt in this. After this, what happens? Rehydration reaction. SN1 mechanism, right? So H plus comes over here. H2O goes out. So we'll get a carbocation there. That is CS2. CS2 positive charge plus H2O. Next, what happens? Hydride shift. Hydride shift with hydride shift. We'll get this one. CS3 positive charge. Then what happens? One, two, three, four, five. So we'll get a five membered ring. And the positive charge will be at the second carbon, which is this suppose. So first we'll have one methyl group. And here we have one positive charge. Then what happened next? Hydride shift. Hydride shift. This converts into this. And a positive charge here. After this, what happens? After this, we'll get an alkene. H plus comes out from the adjacent carbon, means the alpha carbon. So the possible product is one of the product is this. Another product is this. Any other product possible? Second one is more stable. Second one is more stable. Any other product possible here? No, right? Because this double bond and this double bond is the same molecule. So these two products possible. This one is the major product. So in the first one, we'll get RNS, right? In the second one, we'll get. Not in the first product, like in the question one. This one? Or this one? Yeah, that one. In this one, you'll get RNS, yeah. If they ask you the number of products including studio isomers, RNS, two products, we'll get here. So this funda that you know, wherever it is applicable, we'll apply that. Okay. In this case, you see, for this molecule, we have geometrical isomerism possible here. This shows GI, right? Across this double bond. So cis and trans are easy isomers possible here. ENG. So number of products, if they ask you in this one, you'll get two plus one minor. That is three product will get. Got it? Yes, sir. So tell me the number of product possible in this molecule. This is methyl. And this is allowed to react with LiLH4 with H plus H2. Two product. Okay. Okay. Just a second. I'll tell you. Yes. Two products is correct here. The product we get here in this reaction is this OH over here. One methyl group is already present here, right? So this is what this is a carol carbon, right? And this is also a carol carbon. We have two carol carbon here. But what happens here? This, this carbon has two possible configuration that is RNS, but this one won't have two possible configuration because this is a veg dash representation, right? So configuration of this is fixed. This carbon, the configuration is fixed because it is given this veg representation. It can be R or S. Anyone you can consider. You cannot consider R and S both for this particular carbon. It can be R or S. If you have this molecule, suppose double bond O and one methyl group present with the same reaction, then the product would be this OH and methyl. This methyl, you can consider here two possible configuration of this because this is also has two carol carbon. So if this one is R, this one is R, if this one is S, this one is S, this one is R, this one is S. And if this one is R, this one is S. So in this case, we have four possible combinations that we have R, R, RS, then we have SR, SR and SS. These are the four possible combination we have for this particular compound because this is not fixed. This is not fixed. When it is fixed, so it can be either R or S. Suppose I am assuming, I am assuming R this one, assuming the configuration of this is R. So what we can write? This one could be R and the first one could be S. This one could be R or S. Second one is always R, so R, R, SR combination. Two possible combination answer is the number of products are two in this reaction. Understood? Can I move on? Next, write down derivative of acid, aldehyde ketone we have discussed with LI-LH4. Now the derivative of acid, RC double bond OL with LI-LH4. So I have already told you whenever you see LI-LH4, you should consider this H-. And this H-attacks onto this carbon atom. It forms RC O-H. And since this is the living group L, so when this lone pair converts into pi bond here, the living groups goes out and it converts into RC double bond OH plus L-. But we have aldehyde, so this further get reduced into RC H2 OH. So with LI-LH4, it is understood that we are using H plus H2. There's no doubt about it. H plus H2O in the second step, it is understood that we'll use it. Clear? No doubt. Living group can be what? C-L-B-R-I-O-R-O-P-H, etc. All these can be the living group. What is the product we get here first? First, this converts into aldehyde and then it converts into CS3-CH2-OH. Okay, reagent I'm not writing it down, it is understood now. What reagent we are using? If you take ester, this converts into CS3-CH2 C double bond OH and CS3O minus takes H plus from the solvent converts into CS3-OH. Further it gets reduced into CS3-CH2-CH2-OH. Is there any way to stop it after the first step, just keep it as an aldehyde? With LI-LH4, it's not possible. It will eventually gives you alcohol. But we have some catalyst, some reagents. With that, we can do this. Okay, we'll discuss that later. We can use poisons, the catalyst, so that the activity or the reactivity of the catalyst reduces. And in that case, it is possible. But with LI-LH4, we eventually get alcohol. So with lignar reagents. So we had the same reaction, right? There, you said only if it's excess, it goes. But that is not reducing a reaction now. We are discussing about reduction reaction. So in reduction reaction with LI-LH4, you won't get aldehyde. With RNG, is it possible? But that reaction is not reduction reaction. Yes. We can use one weaker, you know, this thing, reducing agent. That is dibal-AH, diisobutyl aluminium hydride. Okay. So in that case, we'll discuss. We'll discuss that also. But here with LI-LH4, it is not possible. So with dibal-H, it will stop at aldehyde. If you don't take it in excess. Yeah. That actually, because of the hindrance, it's reactivity is less. So that gives aldehyde in that reaction. Okay. So we'll discuss dibal-AH also later on, not now. But with LI-LH4, you won't get aldehyde as the final product. Okay. Now you see one thing here. Till now we have discussed that OH- is a very poor living group, right? And that's why alcohol goes under protonation reaction. And then H2O goes out living group and as living group and forms carbocation. That is what we have discussed so far. Right. But here what happens in case of LI-LH4 H plus H2O, OH- also goes out as a living group. But it does not mean it is a good living group. It is not a good living group. But we don't have any choice here. Right. So what happens in this reaction you see. First of all it gets reduced into acid reduced into aldehyde and then aldehyde reduced into alcohol. So this is the reaction we have overall. Okay. What happens with LI-LH4, the H- it comes here, attacks onto this carbon atom, this goes up and it forms our CO-HOH here. Okay. Now when this lone pair comes back to form a pi bond here then we have only two choice. Either H- goes out or OH- goes out. So OH- is the negative charge on oxygen in case of OH- is more stable than the negative charge on hydrogen like H- right. That's why OH- relatively is a better living group goes out and forms and OH- and this is taken up by LI plus of LI-LH4 forms LI-OH. After this you know the mechanism RCHO converts into RCH2H. Right on the negative charge on oxygen. The negative charge on oxygen is more stable than the negative charge on hydrogen atom. The negative charge on hydrogen atom. Hence OH- goes out as a living group. Hence OH- goes out as a living group. Correct. Acids also converts into aldehyde and then aldehyde into alcohol with LI-LH4. Yeah. Just a second. We'll finish one more reaction just last reaction we'll finish and then we'll take a break. Just a second. The last one before the break you see for anhydride. Anhydride is RC double bond O. OC double bond OCH3. So this is also a good living group. This is also a good living group with LI-LH4 H plus H2O. It gives you RCHO first of all plus CS3 C double bond O OH. This again reduce into CS3CHO reduce into CS3CH2 OH. This reduce into RCH2. Any doubt in this? Sir how do we know which side leaves over here? That's what I wanted to ask. Which one is the living group? It doesn't matter right? No it does matter. The CS3CHO is the living group. Suppose if it is unsymmetrical then if this H minus attacks onto this you will get CS3CHO. And if this attacks onto this you'll get suppose ETCOCH3. So if it is unsymmetrical so you have to see where this H minus attacks. Suppose you are thinking that it doesn't matter here. But it does matter when you take LI-LH4. In that case... You have a smaller group right? Because positive channels still exist. Why smaller group? Sir because a bigger group will donate electrons and that positive channel... Yes correct. The point is what? It is nucleophilic addition type. No it is nucleophilic addition. One very important and interesting comparison we have on this only. We'll discuss that later not now. You see it is nucleophilic addition. So it depends upon the positive charge density on this carbon atom. Agreed? Right. You can tell me sir why not steric hindrance you are considering. Suppose we have ET present here and CS3. I would say H minus will attack onto this. So ET and CS3 are different groups. Static hindrance we won't compare here. Why? Because suppose it is... First of all it is nucleophilic addition. Nucleophilic addition involves what? The positive charge density on this carbon atom first of all. And if suppose size is also big then hydrogen the attacking particle the size is very small. Are you getting me? That's why the hindrance won't affect much. The hindrance or the crowding around this carbonyl carbon won't affect much this reaction. That's why we consider the positive charge density on this carbonyl carbon. Did you get it? So if it's unsymmetrical then what will happen? Like where will the HR... I'll take one example. So what you need to check here? You need to check the one with the group. The group which shows lesser I effect plus I effect. In that case only the positive charge density on this carbon will be more. I'll take an example and I'll try to make you understand this. Suppose we have CH3, CH2, C double bond O. O, C double bond O. And then I'll take your CS3. Li, L, D4, H plus H2O. So this D minus attacks on the carbon atom which has more positive charge density. Which has more positive charge density. So here we have plus I effect of CS3. Here we have plus I effect of ethyl group. Which one has more plus I effect? Obviously ethyl because larger group shows more plus I effect. This will donate more electron and hence this will reduce the positive charge density on the carbon atom more. And hence this D minus attacks on this carbon. Any doubt? This will attack on the one which shows lesser I effect. Lesser group you can say or basically lesser plus I effect. Yes. Lesser plus I effect, correct? Anyone is thinking about hyper conjugation here? Hyper conjugation will give you the same answer, right? Anyway like CH3 has more alpha hydrogen. Right. That's why what we say that smaller group takes the H minus of or D minus of Li, L, D4. So could you tell me the product? If I had CH2F on the left side then it will go there. Right. Then you can see comparison will be done. There will be something like suppose if you write down CS3 and here CS2F then here it will attack. Means the question you will get so that you can compare. Means contradiction won't be there. Like it's not like two factors are opposing. In that case it is difficult to understand or to decide where this H minus or D minus will attack. Suppose if you are talking about this one, if you take this one, CH2F C double bond O, O C double bond O CS3. So in that case obviously this D minus will attack on this carbon because because we have a minus I effect increases the positive density over here. Clear? No doubt. Tell me the product in this reaction. The product we get here is CH3. So propanol and C double bond OB plus ET C double bond OH. Correct. Now this converts into CH3, CH, OH, D. Suppose here I am using LIALH4 H plus H2. Then we'll get H here. If you use LIALD4 then it is again D over here. Correct. Same thing that we have already done. LIALH4 ET C double bond O and H. Then it converts into ET CH2 OH. No doubt in this. So it's in the question. They give like LIALD4 only in the first step. And then after that they don't give anything to the assume LIALD4. In that case it is understood that you have to assume LIALD4 only. And anyway you will have options like that. But suppose here if I do not write anything. Okay. Then it is understood that this is what we are using in the next step. Generally they give this kind of compound because you know when they give you this you see this carbon becomes what? Caramel carbon. Caramel carbon. And then we can have two more products here possible. So when they ask the number of products including studio isomers they can change this D and H. Got it? Yes. One more just last example of this. Tell me the product in this reaction. The question is this H plus H2. First carbon and second carbon. Where is H minus attacks? So the parallel. First one or second one? The first one. Because this group shows. Minus minus N. And this group shows because it is that metapogy. Metapogy. Here the posture density will be more. So the product would be what? Ending C-H-O-H. Yes. This is the product. No doubt. Fine. So we'll take a break now after the break we'll see the reactions of amide. Okay. We'll resume the session at 635. Take a break. Yes sir. Yeah. Okay. So we have discussed and I tried. The next one is. Amide. So in case of amide what happens? Different types of reaction it gives. One thing you need to. Keep in mind is what? The functional group in which nitrogen is present. Right. In which nitrogen is present. This gives primary amine in the reaction. Okay. Primary amide. So this. On reaction with LiAlH4. H plus H2O converts into this. Double bond O you need to replace by two hydrogen here. Where this both hydrogen comes from LiAlH4. This one. Both hydrogen comes from LiAlH4. Now for this reaction the condition is what? The condition is. This is again. We have finished two parties. The condition is. The nitrogen. At least contains. At least contains. One hydrogen. More than one is fine. But if there is no hydrogen present here. Then the reaction is not possible. Okay. So this is the overall reaction. Whenever you have nitrogen containing functional group. You will get one degree amine as the product. You won't get alcohol. This one simple thing you must remember. We'll discuss mechanism also. But yes. Mechanism. From mechanism you get to know that this high two hydrogen comes from LiAlH4. Which is the only important. Okay. How this reaction proceeds. You see. The first step of this reaction is. RC double bond. H2. With H minus and this hydride and comes from. LiAlH4. This attacks onto this carbon and this goes up on to this oxygen. So we have our. See. O minus. H. And then NH2. Loan pair on this nitrogen. To neutralize this O minus here we can write here Li plus. Next step. Li plus. Next step, what happens? This lone pair forms a pi bond here and this takes this electron pair and moves out as Li O minus. Okay, so these forms are C double bond N H H and H here, which positive charge on this nitrogen atom plus Li O minus we get. Now this Li O minus takes H plus from this, leaves this bond pair of electron behind and it forms emine. RCH double bond N H will get emine plus Li OH. This hydrogen is the hydrogen of nitrogen atom, this one. So what happens if this hydrogen is not present here, two hydrogen is not present here, at least one is not present here, then this Li O minus won't take this hydrogen and the further reaction is not possible, emine won't form in the reaction. This is emine. Okay, after this, again H minus from Li LH4, this behaves as a nucleophile attacks onto the carbon atom and this electron jumps over here. You see this H is coming from Li LH4, this H is coming from Li LH4. So the carbon atom which takes two hydrogen here, both hydrogen comes from Li LH4 and this we have single bond N H negative charge. So in the last step, what happens with the solvent H plus H2O? This protonates the nitrogen atom here and this convert into RCH HNH2. So important point here that you have to memorize that this hydrogen comes from solvent H2O and the other two hydrogen, this one and this one, it comes from Li L, got it. So nitrogen containing functional group gives you one degree emine with Li LH4 on reduction. Similarly, you see if you have cyanide RC triple bond N with Li LH4 H plus H2O. With cyanide, we again get one degree amine that is CH2NH2. Both hydrogen on carbon comes from Li LH4, these two hydrogen comes from solvent. If you have emine, this reaction we have already seen the mechanism of this. CH double bond NH, this is emine with the same reagent, it converts into RCH2NH2, again one degree amine. Nitro group RNO2, RNH2 unit, one degree amine. Oxene, RCH double bond and one lone pair on this nitrogen atom and OH here. Oxene converts into RCH2NH2. Okay, so just you need to know this reaction. Wherever we have nitrogen containing functional group, this gives you amine in the reaction. Write down the product in this to reaction. So could you go back to the previous slide for a second? So the product in the first one is what? RCH2NH2? So in this mechanism, can there be anything expansion or anything like that? No, no, nothing. You don't have carbocation. Okay. Remember this thing here? This H is coming from the solvent, right? So if I just change this so that if you revise it, you will get it. Suppose if you use your D plus D2O with Li LH4, then you will get here ND not NH. Okay, next you see epoxide. Reaction on epoxide is exactly similar to the RMGX reaction we have. Okay, so we have suppose CH2, CH2, just two example we'll see. CH2, CH2, Li LH4, H plus H2O. The product here we get is CH2, CH2, OH. One H here. This hydrogen, this hydrogen, this comes from Li LH4 and this one comes from the solvent that is H2. How does it happen? You see this H minus, this attacks on the carbon atom and the ring open up. This O minus takes H plus from water gives you this. So this is the terminal carbon here, the last one which takes this hydrogen from Li LH4. So we have this product. What is the product? So could you go back to the previous slide? I have question. Just a second. The product in this one with the same reagent, like in that second thing, RCHNH, if we took Li ALD4, then would it be RCHD NH2 or RCD2 NH2? No, where see this carbon, here the carbon that two hydrogen is coming from or hydrogen is coming from Li ALH4 only. So if you take here Li ALD4, then it will be RCH2D NH2. So I do not understand what you are telling again. I will write down it. See on the carbon the hydrogen is coming from Li ALH4, correct? Yes. So if you have this one, RCH double bond NH, the reaction with Li ALD4 H plus H2O, then the product would be RCHD NH2. Okay. So does both the hydrogens come from Li ALH4? So can it be CD2 and LKS? No, we have already CH, no. Even has already formed after this we have Li ALH4. But if you use here Li ALH4 with H plus H, sorry, H plus D plus D2O. In that case it will be RCH2, then NH already we have and D coming from this one. So in the first example, then it will either be like CD2 or NH2 depending on the. The first one you are talking about? Yeah, RCN. This one? No, sir, the same slide RCN. If you take a Li ALD4 then it will be. Yeah, the two hydrogen is from this one. This two hydrogen is from solvent and this two is from Li ALH4. So basically you see what happens. From here you will get immune first and then this immune converts into amines. So this is the intermediate part you'll get in this reaction. So it is two step reaction from one from this to this and second from this to this. So finally you'll get from this to this. Okay. You want me to write down the mechanism of it? Okay, see just there's two minutes. See, we have RC triple bond N. So with Li ALH4 you will get hydrodial H minus. So this H minus attacks onto this carbon atom, pi electron shift over here. RCH double bond N and negative charge on it. So this will get protonate by the solvent, RCHNH. Again, what happens? The same thing. H minus from the Li ALH4 attacks onto this. This comes over here. So RCH2 single bond NH negative charge. Last step could you tell me? So you see what happens from where this H minus and this H minus are coming. It is coming from Li ALH4 only. But nitrogen is getting protonate by the solvent. You see by the solvent. What is the product in this reaction? This one? It's a propanol. Propane 2 on, correct? Yeah. So it would be CH3CHOHCH2H. The source of this hydrogen and the source of this hydrogen. This one from Li ALH4 and this one from the solvent. So this H minus from Li ALH4 like R minus in RMGX, this will attack on the less hindered side. Okay. And then this comes over. Okay. Now the next one, write down the reduction of alkene and alkyne. So the reduction of alkene and alkyne. See if I write down this to reaction. So could you go to the last page? Last page is this one, right? No. Yes. So this one. So why does it attack on the second one again? It's the same reason as RMGX? Yes. Yes. Less hindered side. Okay. Both way you can think. Less hindered plus the plus side of this reduce the positive density here. Attack will be less. Tendency will be less. Epoxide may be always considered a hindrance. The site of attack must not be sterically hindered. Okay. Done. Yes. Okay. So alkene and alkyne. You look at these two reactions. R CH double bond CH CH3 with Li ALH4. I'll take one alkene with Li ALH4. So this also, this is no reaction. This also gives no reaction. In both of the reaction, the reaction, the product won't form. The reaction is not possible at all. Okay. Even for alcohol also, we have the same thing with Li ALH4. It won't give any reaction with alcohol. Cannot reduce alcohol. Why it does not happen here? Because this Li ALH4 gives H minus. Right. So this H minus is the nucleophilic end. Right. And here also we have electron cloud because of pi bond, pi bond, lone pair, electron cloud. Right. So this won't attack onto these two carbon atom because of the electron cloud only already present around the carbon atom over here. Right. So when H minus comes over here, this pi electron, this pi electron, this lone pair repels this hydride ion and hence the reduction of alkene and kind alcohol is not possible. Okay. But if you consider this reaction, RCH double bond CH2. This on reaction with Li ALH4 ALCl3 with H plus H2O. I forgot to tell you one thing over here. I see this H plus H2O that we are using here. It is an acidic solution, H plus H2O that we are using. So we can also use, instead of H plus H2O, suppose NH4Cl if it is written in the second step. This also serves the same purpose here. Why? Because NH4Cl, it is first of all, it is acidic solution, acidic hydrolysis. NH4Cl in water, it is also an acidic solution. Why? Because it is the salt of what? Strong acid HCl and weak base NH4OH. Are you getting me? So the hydrolysis of this or this salt on hydrolysis gives acidic solution. Yes or no? Acidic solution means H plus water already present over there. So if you write Li ALH4, then NH4Cl in water or simply NH4Cl also they write sometimes or H plus H2O. Both serves the same purpose. You will get the same product into this. Got it? Yes. This we must take care of. Sometimes they will give you NH4Cl also in the next step. So it is a salt of strong acid and weak base. So serves the same purpose. On hydrolysis, it gives acidic solution. Important this one also. Okay. With ALCl3, the reaction here, the product we get here is RCH2, RCH2CH3 alkane. Okay. And in this alkane, this hydrogen on the middle carbon, this comes from Li ALH4 and this comes from the solvent. Now, first of all, why does it happen or how does it happen? Why in presence of ALCl3? The reaction is possible. So ALCl3 is Lewis acid, first of all. The structure that you see for ALCl3, it has three sigma bond and one vacant orbital. So because of this vacant orbital only, it behaves as a Lewis acid. Okay. So what happens here? This RCH double bond CH2, double bond CH2, it donates this pi electron into this vacant orbital of this aluminium and it forms a cyclic three-membered intermediate ring. Okay. CHCH2 and here aluminium attached with chlorine, chlorine and chlorine. Right. Here we have delta positive and delta negative. Now, in this case, what happens? Now, the advantage of this is ALCl3 that it creates a non-classical carbocation here. We always consider secondary, more stable one, not this one. Okay. Non-classical carbocation it creates and hence further, H minus from Li ALH4 can attack onto this carbon. And when this attacks over here, this comes onto this aluminium. So here after this, it is type of epoxide. But remember one thing in epoxide, this attacks on the less hindered carbon atom. But here, since we have non-classical carbocation, so two degrees we are considering here. So this will attack on this two degree, not on the terminal carbon. This bond comes over here, the ring opens up and this converts into, and this converts into RCH, single bond H. What is the previous page? Yes. Yes. Let me just finish this. ALCl3 and negative charge on this aluminium. Further, what happens? ALCl3 negative charge here. Carbon atom, what happens here? This carbon atom drags this bond pair of electron towards the site. And the reason for this is the electronegativity of carbon is more than to that of aluminium. That is why it happens. And this becomes neutral. And we get here RCHHCH2 lone pair negative charge on it, like we had in the case of Amines NH2. In the last step with H plus H2O, protonation takes place and this gets protonates converts into RCHCH2 H, H where this hydrogen comes from on the secondary carbon comes from LILH4 and this hydrogen comes from solvent that is water. I'll come back again to this page. Let me go back first and copy this down. So in presence of Lewis acid, it is possible, but when you do not have Lewis acid then that carbocation, non-clustical carbocation won't form and the pi electron cloud repels the hydride ion that comes from the reducing agent and hence the reaction is not possible. So is the reaction important? I mean like the mechanism? No, mechanism is not important. You just need to see why we are discussing mechanism. The whole purpose is this only, from where this hydrogen and this hydrogen is coming. So if you can keep these two in mind that this is coming from solvent, this coming from the reducing agent, this is it. Otherwise there is no role of mechanism. Yes. So what about a symmetrical alkenes in that? How do we figure out which one takes which? We have to see the stability of carbocation. The basic thing is that if it is symmetrical then I think both way you will get the same product. But overall if this is more stable then this will attack on stability of carbocation. However it is not a classical carbocation. It is partially developed, but then also stability will consider and will attack H-1. Yes. Okay. One more thing, AlCl3 you see, it is not getting consumed in the reaction. Okay. Plus we get here AlCl3 also. So it gets released. It is behaving as a what? Catalyst in the reaction. No, it is the reaction of alkene. Any alkene goes under this reaction, this kind of reaction. Yes. Yes. Any alkene you can use. For alkyne also the whole process is same. Only the reaction takes place twice. Means first we will get alkene and then alkene converts into alkene. Same thing we have. So we have suppose RC triple bond CCH3 with the same reagent LiAlH4 AlCl3 H plus H2O. First it converts into RCH double bond CH single bond CH3 and then in the next step with the same reagent we get RCH2 CH2 CH3. Okay. So these are the reactions of LiAlH4. We covered almost all the reaction of LiAlH4. Now the next reducing agent we have that is NaBH4. NaBH4 like LiAlH4 it is also a complex compound. Write down the second one. NaBH4 sodium borohydride. SbH also we call it as in short that is sodium borohydride. Okay. The structure is same. We have Na plus and bH4 minus. I have already told you that this bond the bH bond strength is more than the bond strength of LiH because it is less inic. Less inic. More bond strength. Less inic more bond strength means NaBH4 the bond it is difficult to dissociate the bond and hence NaBH4 is a write down is a weaker reducing agent than LiAlH4. First of all write down it cannot reduce NaBH4 cannot reduce ester cannot reduce ester. It can reduce NaBH4 can reduce aldehyde ketone acid halide acid halide. Now it is also nucleophilic addition type reaction. The nucleophile is H minus again that attacks and the reaction. Can you go back to the previous yes. Done. Thank you. Okay. So if you compare the rate of the reaction write down the question. Compare the rate of reaction with respect to with respect to the first one LiAlH4 and the second one NaBH4. For the compound we have R C double bond OCl A, R C double bond OOR, B, R C double bond OH, C, R C double bond OCH3D. Done. Yes. What is the order with respect to LiAlH4? The order with respect to LiAlH4 is see it is nucleophilic addition type. So we have to check the positive charge density here on this carbon and on this carbon. Cl shows minus i, this shows plus m, this shows no effect. Here we have plus i. Tell me where we have the maximum positive charge density. In case of A minus i effect, hence the rate of the reaction, it is maximum for A. Then we have C, then we have D and then we have B. According to the positive charge density on the carbonyl carbon. If you talk about NaBH4, then we know NaBH4 can't reduce ester. So we just have to eliminate the second compound. Order is this. Can't we also do it just by seeing which is the best leaving route for LiAlH4? No, it's not. The rate of the reaction depends upon the rate determining step. Correct. If you remember the carbocation, the reactions in which the carbocation forms will compare the stability of carbocation there to compare the rate of the reaction. Yes. So why do we consider the stability of carbocation there? Because the formation of carbocation is the slowest step. That is the rate determining step. So rate, we always compare with the rate determining step. So here in this case, the nucleophilic addition reaction, the rate determining step is the attack of H minus ion that you should know. And this one is experimental that you cannot find out logical by looking at the reaction simply. Okay. So like you have remembered that carbocation formation is the rate determining step for example, like addition of HBr or dehydration reaction of alcohol. Here also the nucleophilic addition reaction, the rate determining step or the slowest step is the attack of nucleophile that is H minus or R minus in case of RNG. Clear? Yes, sir. Right. So this is also very simple, similar to LiAlH4. So if you want to write down the product of this, RC double bond OH, NABH4H plus H2, it converts into RCH2OH. If you take ketone, this converts into secondary alcohol, RCHOH. All those things are same from where H are coming on this carbon and oxygen atom. Okay. So I am not discussing all those things again and again. It is exactly same. Asyl chloride also first converts into aldehyde, then converts into alcohol. If they ask you this question that rate you need to compare with respect to the reagent, LiBH4, LiAlH4 or NABH4. Obviously the answer will be LiAlH4 because it is a stronger reducing agent, forms H minus faster and hence can react also with a faster rate. Okay. Now, the next, the third reagent we have, important, it is catalytic hydrogenation, a reduction through catalytic hydrogenation. Okay. So this is used for unsaturated compounds, unsaturated compounds. Like for example, we can talk about alkene, we can talk about alkene, unsaturated, we can talk about cyanide, we can talk about ketone. For catalytic hydrogenation, suppose you are considering aldehyde or ketone, right? So I would request, I would suggest to you what, that this C double bond O you consider as C double bond C only, because the pattern of the reaction is exactly same. Suppose you are considering cyanide, consider this as alkyne only. The reaction, the way the reaction takes place here and here, similarly it takes place here and here, nothing much. Okay. How the reaction takes place, we'll see this one. Okay. Alkyne you see first of all, alkene. So if you look at this reaction, C double bond C and catalytic hydrogenation H2Ni, this converts into alkene by syn addition of hydrogen atom. It is syn addition of hydrogen atom, exothermic or endothermic reaction. So you break a pi bond to get to sigma. So sigma is stronger bond, no more energy releases. One pi converts into two sigma, so it is exothermic. Bond formation is an exothermic process, energy releases and two stronger bonds are forming, right? If you consider HH bond, if it breaks, then also one sigma and one pi converts into two sigma, right? So stronger, more stronger bond forms over here, more strong bond forms over here. Hence the energy comes out exothermic. Why it is syn addition? First of all, the metal you hear you see, suppose this is the surface of metal. So hydrogen first of all, adsorb, get adsorb on the surface of metal. Suppose this is the metal surface we have. Okay. And hydrogen is here, it is present over here adsorb at the surface of metal. It is adsorption, right? When it comes closer to this alkene, then it has tendency to make a sigma bond with this hydrogen like this. So it since it is adsorbed to the surface, so this can come either from this side or from this side. It is not possible that one hydrogen is coming from this side and other hydrogen is coming from this side. And hence it is syn hydrogenation, syn addition of hydrogen, correct? So hydrogenation of alkene, it is syn, not anti addition. Exothermic reaction energy comes out. If you talk about the rate of the reaction, see the more stable alkene, the lesser will be the tendency to go under this reaction. So rate is inversely proportional to the stability of alkene, right? And stability of alkene is directly proportional to crowding. It is because of what effect? It is because of hyper conjugation. So more branch at the double bonded carbon atom, more will be the stability. So this is the configuration, sorry, adulation of rate we have. One second. One note if you write down here. Alkyne has lesser surface area than alkene. Alkyne has lesser surface area than alkene. And hence it has more reactivity towards, it has more reactivity towards hydrogenation reaction. Basically, alkyne is more reactive. This you need to keep in mind because both carbon atoms are sp hybridized. So they are concentrated in a small reason relatively. And hence the rate of the reaction is more over there. Can you repeat it again? Alkyne has lesser surface area than alkene. And hence the rate of the reaction of alkyne, rate of hydrogenation reaction you write down, rate of hydrogenation reaction of alkyne is more than to that of alkene. Okay, then this is very important and you will get different, different types of questions on this. Okay, suppose first of all, you need to compare the rate of these alkenes. Could you tell me the rate at CH2 double bond CH2? CH3 CH double bond CH2. And then we have CH3 CH double bond CH CH3. Order of rate? So increasing order, right? For the first one. Okay. So I'm asking you have to put it in increasing order, right? Any order, it is in decreasing. That's not it. Okay, we know as crowding increases, the rate of the reaction decreases because rate and crowding is inversely proportional. One thing. Second thing, alkene, if it is more stable, it won't undergo. It has less tendency to go under the hydrogenation reaction. So order of reaction is this. This alkene is more stable. This is more reactive. Alkyne and alkene, if you have, you can go with alkyne always, go with alkyne always. This one is more stable because of conjugation. So order is this rate of the reaction. Now the addition is sin. So if you say, if you take cis alkene, for example, CH3, C double bond, C CH3, H and H, D2 Ni, for example, D2 Ni. So cis alkene goes under sin addition, gives erythro compound, right? Cis, sin, erythro. Erythro means both the D present on the same side. This is erythro. Optically inactive due to internal compensation, mesocompound it is, right? If you take trans alkene, D2 Ni, trans on cis addition, gives 3O, TST, okay? 3O means both D will write here like this. Addition is sin only, but will write this way and will have a mirror image of this also. So here we get enantiomers. Rescuing mixture, these two are enantiomers of each other. Optically inactive due to external compensation. Sir, in the previous slide, could you explain the third question? I didn't get that. This one? Sir, the rate ones, yeah, can't we say triple bond would be harder to break? No, no, no. See, it's not like the bond dissociation thing we are considering. We are considering the surface area of this. Lesser surface area means this pi electron is concentrated in a small region. Hence, it can take the hydrogen atom easily towards it. That's why the reason that we say alkyne has lesser surface area than alkene. The pi electrons are concentrated in a smaller region relatively. Hence, can take the hydrogen atom easily on this. Like, can attach the hydrogen atom to it easily. That's why the reduction, the rate of the reduction is easier for alkyne and then alkene. More rate of the reduction here. Okay, sir. Thank you. So reason, sometimes they ask you this theoretical question also in the exam. The reason is lesser surface area that you must try. All other things comes because of that surface area only. Okay, so CSE and TST. Right, cis, syn, erythro, trans, syn, trio, erythro is meso, optically inactive, trio is optically active. Overall, it is optically inactive. Rescuing mixture in enantiomers. What is the product we get in this reaction? One equivalent H2NI. See, this alkene, this pi sigma pi is in conjugation. More stable, less will be the rate of the reaction. But this pi is comparatively less stable. So reaction takes place at this carbon-carbon double bond. Since we have only one equivalent, so we have to select like this, where the reaction is taking place. These two double bond will be as it is. And the two hydrogen attached like this syn addition on these two carbon. Is it fine? It's the same property, right? Yes. Rate of the reaction of alkene is inversely proportional to the stability. The stability of this pi sigma pi is more because of conjugation. This one is not in conjugation. Less stable, more will be the rate of the reaction. Correct? Sir, we can say about crowding also here, right? Crowding also you can consider. Yeah, obviously. Here we have less crowding, so more rate of the reaction. Same thing. Yes. Okay. Now, if you have this compound, the application of this is huge actually, this one, you know, hydrogenation reaction. You look at different, different types of compound in the question. H2Ni, again, one equivalent I'm using. What is the product we get here? H2Ni, one equivalent. H2Ni, one equivalent. So you see in this one, the product would be this because the compound becomes aromatic now. If you have butadiene like this, okay, then it gives 1,4 addition product, TCP. Hydrogen here, hydrogen here on the same side. It is TCP thermodynamically controlled product, 1,4 addition. Okay. First and fourth position, the addition takes place. If you have alkene and alkyne both present in a compound, then we know the rate of the reaction of, reaction of what? Alkyne is more than to that of alkene. So the first reaction takes place at alkene. You'll get a diene. Further, if you have, if you add one more equivalent of H2Ni, then it converts into TCP that is 1,4 addition product and the product of the reaction would be this. So why is this product formed like this TCP product? 1,4. Yes. It goes under resonance actually and hence this point and this is the most stable form we'll get when you draw the resonating structure. So that takes part in the reaction. Okay. Got it? Yes sir. Okay. Now, suppose we have, we take this reaction H2Ni, excess delia, so what we get here, we get first alkene, right? And what kind of alkene? Cis alkene, both hydrogen attached from the same side. Further, it again goes under reduction reaction and it converts into, converts into an alkene. This is the product we get. So finally, we are getting alkene into this. If you have, if you have NF of H2Ni, if you take in this reaction. So from alkene, we are getting alkene. Suppose if you want to obtain alkene, suppose if you want to stop the reaction over here, further, if you don't want to go, then what we need to do for this, correct? Are you getting my point? We want alkene from alkene, not alkene. Then what we need to do, we need to reduce the activity of this catalyst Ni because if it is active, then it continues with the reaction and eventually we'll get an alkene here, right? So when we need to reduce the reactivity, you can say or activity is the better word we should use over here. Catalyst, the activity of catalyst we need to reduce, okay, which is also the active mass. Active mass is nothing but, you know, we take it as concentration, but there is, you know, a difference in these two terms. Okay, we'll discuss this in chemical kinetics. What is the active mass of a catalyst? Suppose the surface of the growth may or mutate like that. Okay, sorry. Just I saw the message, you know, that's fine. Hello? Yes, sir. Yes, sir. Can you hear me now? Yes, sir. Yes, sir. I don't know, I lost my connection somehow. Anyways, so we were talking about this, okay, so we were talking about if you want to get an alkene from alkyne, right? Then what we should do. So I was talking, what sir, your voice is breaking, at least for me. Even for me. Yes, sir, same for me. So hello. Hello. So, yeah, so I was talking about this active surface area, correct? So you have suppose another molecule. Obviously, this is one reactant, another reactant or catalyst we have. So on this also we have some active surface area present, right? So when the collagen takes place, between the active surface area of the molecule, then there will be, no, the possibility of product will be there. I cannot say that the product will form in that case. That is one of the criteria. This we call it as orientation barrier of the reactant molecule. Okay, orientation barrier means when the molecules, when they are colliding with each other, they should collide in a proper orientation so that the collagen takes place between the active surface area of the molecule. So if it happens, then this condition is satisfied. Orientation barrier is, you know, is fulfilled over there. Plus, there is one more condition that we call it as energy barrier. Okay, if these two conditions are satisfied, then only the reactant converts into product. Okay, so we are concerned, here we are concerned with this area, this active surface area of the catalyst here. If you reduce the active surface area, obviously the chances for this, you know, a proper orientation would be less and hence that will reduce the possibility of forming of product and hence the rate of the reaction also reduces. Reduces both things. The possibility also plus rate of the reaction also. The point is, if you reduce the surface area of this catalyst Ni, then the rate of the reaction reduces. Right, its activity reduces. Correct? Since alkyne is more reactive than alkene, then this alkyne will manage to react with this hydrogen and convert into alkene. But we know alkene is lesser reactive than alkene. Plus, the catalyst is also not that active because we have reduced the activity of catalyst. So it is also not that active. This also lesser reactive than alkene, alkyne, sorry alkene, hence this conversion is not possible in alkene. Right? So to sum up this, what we can say, if you want to obtain alkene from alkyne, not an alkene, then we have to reduce the activity of catalyst. And for that purpose and for that reason, we poison the catalyst. Okay? We call it as the poisoning of catalyst. So when you do this poisoning, that will reduce the activity of catalyst, the active surface area. But alkyne is enough reactive. It will, even in that case also it reacts with this, converts into alkene. But further the activity is not, the reactivity is not that great for an alkene and further reaction is hindered in that case, restricted. Right? So this, you know, the alkene we can obtain from alkyne by poisoning the catalyst, by reducing the active surface area of the catalyst, the activity of the catalyst. Okay? So how do we do this? Suppose we have a reaction R, yes, R C triple bond C R2. Right? Two reactions we have here. H2 with Ni. And this Ni is poisoned with BASO4. This is the poisoning of catalyst to reduce the surface area. H2 with Ni. H2 with Ni. And it is poison. Sorry. CaCO3. Okay. Poison with CaCO3. So this, in this way, you will get alkene as the final product. You won't get alkene in this reaction. So the product would be a cis-alkene, R1C double bond C R2 H and cis-alkene. Okay? This, you know, reagent H2PD with BASO4, we call it as Rosenman catalyst. You must have heard about it. Rosenman's reaction. Yes? Yes, sir. This is Rosenman catalyst. This also we use the question, Param, you were asking that time. This reagent also we use to convert acid halide into aldehyde. It won't reduce further aldehyde into alcohol. Got it? Yes, sir. So what is the exact difference between this and Lindler's catalyst? Lindler's catalyst, we are using different reagent, that is it. Nothing much. Both give cis-alkene into this. Both reduce the C. The purpose of this is what? To reduce the activity, no. So we have different molecules, which we use to reduce the activity of this. So when we use CaCO3, it is Lindler's catalyst. When we use this, it is Rosenman's catalyst. That is it. Nothing much. Okay. You see this one with CaCO3, it is Lindler's catalyst. So different reagent we are using here, CaCO3 and BASO4. The purpose is saying that in both case, we'll get cis-alkene, not an alkene. Correct? One note you write down here. If you want to obtain alkene from alkene, if you want to obtain alkene from alkene, we need to reduce the activity of catalyst. We need to reduce the activity of catalyst, comma, write down. The poisoning of catalyst, write down the poisoning of catalyst, reduces the surface area of the catalyst, the active surface area, write down, reduces the active surface area of catalyst and hence the rate of the reaction. With this poison catalyst, write down, with this poison catalyst, the alkenes are not much reactive. With this poison catalyst, alkenes are not much reactive so that they can react and convert into alkene. Alkenes are not much reactive so that they can react and convert into alkene. Hence, eventually, we get cis-alkene as the product. Cis-alkene as the product. Got it? So, H2NI-BASO4 or H2PD-BASO4 also we can take. H2NI-CACO3, these are Rosenmann and Lindler's catalyst respectively that it gives cis-alkene. Now, this reaction you see, suppose we have an alkyne, CS3-C triple bond, CCH3, the reaction of this with H2PD-BASO4 gives A and when this A reacts with Br2-CCL4 gives B. What is A and B? The first A is what? A would be a cis-alkene. So, C double bond, CCH3 and both H attached on this side. Now, on this reaction with Br2-CCL4, what it gives? Cis-on-sin addition gives what? Ilythro compound. So, we will get here Ilythro compound. So, the product would be Br2-CCL4 is anti-addition, right? So, what we get here? This gives you two products. Both we are on the opposite side, right? H, CS3, CS3 and we'll get a mirror image of it. So, this is what this is. Resinic mixture, enantiomers, all those things you already know we have discussed many times, okay? This is the product we get here. So, like this they can, you know, relate the reactions and ask the questions like this. So, we have one more reaction into this that is Burge reduction, okay? We won't discuss it today. Burge reduction. So, just these three reactions you keep in mind, all these three together. The first two, Rosenman-Linders gives you cis-alkene, but Burge gives you trans-alkene in this, okay? This gives you trans-alkene. We'll see this reaction next class and there are some more reactions left that will finish next class most probably this chapter and we'll start with the alcohol, right? The few, I think 90% of the reactions of alcohol we have done. So, those reactions which are left will start next class only, okay? One more thing, assignment on Lernist on this chapter it's already uploaded, okay? You can do that. You can do that and one more PDF I'll share with you on the group for oxidation reduction. You must try that also, correct? Nothing, like not much critica I think one and a half hour max to max, not more than that. So, after this chapter can we start with coordination, confirms and then we'll go to alcohols? Coordination Mehul will do, but because we have done it, so this thing is fresh, right? So, we can finish alcohol quickly then. So, let me finish that alcohol chapter and then we'll do coordination, okay? Okay, sir. Okay, thank you. Take care. Bye. Thank you, sir. Thank you, sir.