 Good afternoon sir. Good afternoon. Let others join and then we'll start. Okay sure sir. Sir can you hear me properly today? Yes fine. Okay sir. Okay can we start now? Yes sir. Yes sir. Okay so what have we done last class? Oxidation reduction. So what were the last class and we did? Tolenz reagent. Tolenz. Yes sir. Tolenz reagent is no. Yes sir. Yes sir. Okay. Sir you wrote, you told us to write Felling's test heading. So okay one question you see and then we'll start with Felling's solution. Just a second. Suppose we have this molecule and there is OH here and OET here. This reacts with two molecules of CS3, MgCl, H plus H2O. Tell me the product. One second sir. Sir I think I'm done. What is the answer? How many products you got? I got two. Okay what is the name of the product? As a triangle plus CH3. Triangle and cyclopropane. Yes sir. And? MgOH here. No. What is the other one? Like any answer? Simple no. See. First of all we have OH present right. This is active hydrogen. So with CS3, MgCl it gives acid base reaction and it forms what? Plus we get the conjugate base of this acid which is O minus and OET. This is one product but we are using two moles of it. So if this lone pair comes over here this goes out as a living group and it forms what? Double bond O plus OET minus. This is ketone and on this the reaction of CS3, MgCl takes place with H plus H2O and it converts into OH CS3. Tell me any doubt in this? One second sir. Clear sir. No doubt? No. So the next one, the next test we have that is felling solution test. Felling solution test. We use two solution for this test and both we call it as the mixture of both is means the combination of both is the felling solution test. Fellings A and fellings B. Yeah. Fellings A, fellings B you can say or felling 1, felling 2 you can say anyone. So it is felling 1 that we take as aqueous CuSO4 solution. This is felling 1 and the felling 2 is sodium potassium tartarate. And the combination of both we use in two steps of these two. Both we call it as the felling solution. Okay. This is felling solution. Okay. Okay. What is sodium potassium tartarate? Sodium potassium tartarate is this. We have OH, OH, HH, COONA and COO. This is sodium potassium tartarate. When you look at this molecule sodium potassium tartarate could we say that this is the salt of weak acid and strong base. Is it? Yes, sir. Why weak acid? You have to COOH groups and organic compounds. COOH, COOH here and when it reacts with NaOH and KOH it gives you these two. Correct? Salt of weak acid and strong base. In Ionic equilibrium we will discuss this. We have discussed this already, right? And its aqueous solution is what? On hydrolysis if you do on hydrolysis, on hydrolysis the solution will be acidic or basic. The solution is basic. Why basic? Strong base. Because there is strong base. Whatever is strong is there and that would be the nature of the solution. If you take a strong acid the solution will be on hydrolysis the solution will be acidic. Strong base, the solution will be basic. Correct? So in the second step we have basic solution. That is what the conclusion of all this discussion is. So now what happens here that you see? Basic solution we have and we have Cu2 plus present over here. Right? So with Cu2 plus in the basic medium, Cu2 plus in basic medium we have OH minus, CuOH minus and it forms what? It forms CuOH which on heating converts into CuO plus 2H2. Now this CuO is the oxidizing agent we have. Aldehyde when allowed to react with this, it reacts with the oxidizing agent CuO and this converts into Cu2O. It gets reduced and this oxidizes aldehyde into an acid that is C double bond OH. The conformation of this reaction we get by the color of this which is brick red color or red color. Right? The Cu2O that form that is also very, you know, it is an oxidizing agent. It can behave as an oxidizing agent. So if any aldehyde is left in this reaction, that also reacts with Cu2O and converts into CuO. So this is the test of aldehyde. Okay? Failing solution test. Yes sir. So in the next step one more reaction we write down here. If any aldehyde is left, R C double bond OH reacts with Cu2O. It gives copper and acid. So the red or brick red color that you get that confirms the presence of aldehyde. This is the test of aldehyde. Right? All aldehyde gives this particular test. Ketone does not give. Write down one note over here. All aldehyde gives this test. All aldehyde gives this test. And with an exception of benzaldehyde. With an exception of benzaldehyde. So benzaldehyde does not show failing solution test but it shows tolerance reagent test. Okay? Important in this one. Why sir? I said it is an exception. Okay sir. So if there is any reason, then it is not an exception. Okay. Okay sir. Okay. So it is an exception. One second sir. One second. All aldehyde. Gives this test except benzaldehyde. Yes sir. Okay. So this is important that benzaldehyde shows tolerance test but not benedict solution test. Okay. Ketone does not give this test. Write down. Ketone does not give this test. What about this one? Sir this also gives a links test. Why? Because of active hydrogen. Active hydrogen is the reason. Hemiacetyl gives this test. O-R-O-H-N-R. Hemiacetyl. Ketone does not give this test. What about this one? This one down? No it won't give. Sir neither sir. It will neither form a Hemiacetyl structure or is it an aldehyde? This also gives this test. Okay. One hydroxy to ketone. If it is present, it gives this test because it goes in the tautomerism and forms aldehyde in basic medium. And that is the reason of fructose also which gives this test. Remember in biomolecules. Yes sir. D. I won't be yelling at this point. No. That fructose gives ferring solution test. Benedict this one. Tolerance reagent test. Why? Because we have one hydroxy to ketone. It goes in the tautomerism and one of the form it has aldehyde as a functional group present. That's why the mixture shows this particular test. Even the same thing happens over here. See the medium is basic. The medium of this reaction is basic. OH-. This OH- takes this active hydrogen H+. Right? So you said active hydrogen but not exactly correct. Only active hydrogen won't give this test. You have this, correct? Now when this act here to form a pi bond, then this goes out as a leaving group. And eventually we get what? We get an aldehyde. And that's why this shows this test. So if they give you a simple molecule aldehyde, you can say it shows toluzol ferring. But the problem is they won't give you the simple one. Right? So please you must take care of. Whenever you have one hydroxy to ketone, it shows toluzol as well as ferring solution test. This time. Okay? If you have any astral structure, this also shows ferring and toluzol reagent test. Yes, sir. Okay. Third one to write down. It is Benedict solution. All these are weak oxidizing. Okay. Benedict solution test. What is Benedict solution test? It is again the similar kind of test, exactly similar. We take aqueous CuSO4 with sodium citrate for this test. Sodium citrate is what? It is this one. Only the molecule is different here. But this serves the same purpose here. You know, the previous one, sodium potassium citrate. What kind of salt was that? Mixed salt. What is mixed salt? Basic salt. Basic salt, right? Solution of salt of weak acid in strong base. And this is also the same thing you see. Weak acid in strong base. Yes or no? Also goes on hydrolysis and makes the solution basic. The OH- that you get here. From this solution, we get OH- and then we have the same thing. Reaction of OH- with Cu2 plus like we did in the last one. Do you want me to write down? Or it's fine? Yes, sir, please. I won't write. Okay, sir. From here to here, it is single. Mayut, do I need to write down this? No, sir. There also the solution is basic. Same thing here, basic solution. Until here, the way by which the solution is getting basic, that way is different. We are using different compounds over there. Right. But the purpose is same. That also gives OH- this also gives OH- which reacts with Cu2 plus and gives this product. Even you have- you don't have to write down this again. Just after that you write down. Then it is same as filling solution test, finish. Nothing much. Okay, sir. One second, sir, please. Just after this you write down. Same as filling solution test. Finish. Yes, sir. So these three tests that we have done. The medium of these tests is what? Basic. So all these are basic medium tests for aldehyde and ketone. Correct. Now the next one is- you write down. The next one is acidic medium test. What all tests we have which takes place in acidic medium. These tests are not important. Okay, tollens, fillings, benedict are very, very important. Okay. So those things are far more important than this acidic medium test. Okay. Acidic medium test we'll see but it is not that important. The first one is HgCl2 test. HgCl2 test. Okay. The reaction you see we have an aldehyde. RC double bond OH reacts with HgCl2 with H2O and this converts into RC double bond OH plus Hg2Cl2 plus 2HCl. This the color of this is white. Further if any aldehyde is left that also reacts with Hg2Cl2 in presence of water and converts into the color of this is grayish black. Gray black. This is the change in color that gives this particular test. Like I said this is not at all important. Okay. But just you need to keep this in mind. Only we have two tests into this one. At HgCl2 and one more we have that we call it as CHIPS reagent test. The second one. CHIPS reagent is rosa aniline. The reagent is rosa aniline hydrochloride. There is no reaction just you need to know what change happens into this one. Okay. Rosa aniline hydrochloride. This the color of this is or magenta. The color of this is pink or magenta. When it reacts with SO2 this converts into this will get oxidized and we get oxidized form of it. Okay. Which is colorless actually. Which is colorless. Okay. Now in this solution if you place aldehyde here in the solution if you place aldehyde this converts into this converts into acid. The aldehyde get oxidized converts into acid and we get the CHIPS reagent back onto this. What we say on reaction with this CHIPS reagent restores its color and it again confirms and it again forms pink or magenta color over here. So this color restores the reaction the CHIPS reagent restores its color that confirms the presence of aldehyde. And we also say that aldehyde restores the color of CHIPS reagent. The color of CHIPS reagent. Ketone does not do so. Okay. And that is the test. Only the color change you have to keep in mind. So if I ask you glucose or fructose which one gets this test glucose or fructose? Glucose. Glucose. Fructose won't get this test because the medium is acidic. Right. And fructose won't convert into acid. Sir. Done. Once I say glucose gets this test. What exactly is rhodaniline? Your voice is not coming to me. What exactly is rhodaniline hydrochloride? The structure I will show you. Okay sir. The structure I will draw. I don't have space here. I will draw it in the next page. But it is difficult to memorize. Okay. Just know how the color changes take place. What color changes are there? Rosa aniline hydrochloride is the structure. Yeah. So this one is rhodaniline hydrochloride. Okay. This structure. The name you should keep in mind that stiff reagent is Rosa aniline hydrochloride. Yes. Okay. These two tests we have which we perform in acidic medium. Okay. And it is not important at all. Okay. Since it is there in the slivers we have done. Okay. Now the next reaction is we have seen oxidation of antihyde. Okay. The next we have to discuss is oxidation of ketone. Oxidation of ketone. I have already told you that it is done in dusty condition. It is not easy to oxidize ketone since it does not contain hydrogen on the carbonyl carbon. Right. And this reaction is also not much important. Okay. When I say not much important. Not much important. I'm not asking you to leave this. Okay. If you revise you should be more focused on, you know, on antihyde part, not ketone part. That is what I mean. Okay. Now in this one, we use a very strong oxidizing right on the oxidation of ketone is not easy. The oxidation of ketone is not easy. And it is done. And it is done in drastic condition. In drastic condition with the help of with the help of acidified KMNO4 with the help of acidified KMNO4 and acidified K2CR2O7. So with the help of acidified KMNO4 and acidified K2CR2O7. Oh, this time it's acidic KMNO4. Yes. Always we used to use alkaline. This is different. Okay. So this is, we have a very strong, both the reagents are very, very strong oxidizing. Okay, sir. And that's what we have studied that for ketone we require a strong oxidizing agent. Mild won't oxidize it. Right. Biomolecules also we have discussed we use bromine water for to differentiate antihyde and ketone. Right. Yes, sir. Fine. So how do we write down the oxidized product of ketone. Okay. Oxidized product of ketone. Here we do not have the mechanism like I said, the mechanism of oxidation and, you know, reduction reaction. There are a few reactions whose oxidation mechanism is given in the books like if you see pitocytes and this thing. But eventually if you understand the mechanism, then also you have to memorize how to write down the product. Okay, so the mechanism part here won't affect much, whether you get it or not. Okay, so we are not going to do the mechanism here. I'll just understand to make you, I'll just make you understand how to write down the product in this. Okay. Few things you have to keep in mind. Suppose you have a ketone like this. So in acidic medium, acidic reaction of ketone later on in other chapters also we see when we write down the reaction of it, we always draw the enol form of it. Okay. And the enol form of this molecule is this. No doubt in all form. And then when you do the oxidation of it, this reaction is exactly similar to the ojournalysis reaction. Okay, oxidation ojournalysis reaction, you have to just break this double bond like we do in ojournalysis. So what we get here, we get OH here and double bond O plus we get O double bond C. We have two hydrogen here, but the medium is acidic. So this also will get oxidized OH and OH and further it dissociates into CO2 and H2O. This is the product we get. Usually we get this. This one is forming over here since we have two OHs present on the same line. Just draw enol form, break that double bond like ojournalysis and attach oxygen. And you won't get aldehyde into this. Okay. You always get acid. This one you see. Could you write down the product in this one? Double bond O. In this one what happens? Sir, could you explain how do you get that diol? Which one? That is alcohol and further oxidation now, sir. Like the yellow stone. This one now? Yes, sir. You break this bond. OH is as it is. Here we get double bond O. Yes, sir. This carbon will have double bond O this side. This one. And the two hydrogen here, that will also oxidize into acid. Okay, so one will get the normal carboxylic acid product and then another thing, the same carbon with two OH groups like that. If the carbon atom contains H, you just need to convert into OH. Okay, sir. Like we did that one, oxidative ojournalysis. It is exactly similar to oxidative ojournalysis. Okay, okay, sir. Means one line symbol you keep in mind. You draw in on form, more stable in on form you draw and write down the product of oxidative ojournalysis that is. Okay, yes, sir. So if you remember in oxidative ojournalysis, don't, sorry, anti-hype. Yeah, yes, sir. Yes. That's one. Okay. Okay, so in this one, what we do will draw the more stable in all four and that would be this OH here and double bond here. Now after this, if you do oxidation, so we'll write down the oxidative ojournalysis. You just break this bond and that would be CS3 C double bond OH. The CS3, the next one is selenium dioxide. So in this one, how do we write down the product you see suppose we have CS3 C double bond O CS3. So if you hate this, then the product would be CS3 C double bond O C double bond OH that is it. I'll show you some more examples pH C double bond O CS3 with the reagent, you will get pH C double bond O C double bond O H. Okay. If you have a six member ring with a double bond O present here, then the product here again a six member ring with two double bond O present at the adjacent position. This is what we do. Like, you know, when you have carbonyl compound ketone, then with SEO to selenium dioxide, you have to write down one more carbonyl group. Addison to the carbonyl group present, that is it. Number of carbon atom bond change. Here you see CS3 CO. Instead of this CS3 will write down C double bond OH. Same thing we have here, same thing. Okay. So all these reactions you have to memorize how to write down the product. Finish. Yes, sir. Okay. Now in the next one, suppose if you have alkene, write down one note. Alkene on oxidation with SEO to alkene on oxidation with SEO to the hydroxy group, OH group, the hydroxy group, OH group gets attached to the allylic carbon gets attached to the allylic carbon. So what you need to do here, you just need to do this for alkene ketone. I have told you what you need to do for alkene. Suppose we have CH3 CH double bond CH2 this with SEO to if you're reacting, then this carbon is the allylic carbon. So here what happens, we get OH CH2 OH single bond CH double bond. CH2, that is it. This is the product you get. Okay. But in a given compound, we can have more than one allylic carbon possible. Right. More than one allylic carbon is also possible. Okay. So you should know the rate of reaction for different, different allylic carbon one degree, two degree and three degree. Write down the order of rate of reaction, the order of rate of reaction for allylic carbon, the order of rate of reaction for allylic carbon. And this order is if you have two degree allylic carbon, then it's the rate of reaction is maximum, then one degree and then three degree, two degree, then one degree and then three. Means suppose if the allylic carbon is one degree and three degree, then OH group get attached at the primary carbon one because the rate of the reaction is more. If it is two degree and one degree, then OH group get attached at the secondary carbon that is two degree. Right. This reaction SEO2 is important for example. Okay. You just need to memorize two things over here. Okay. One is for ketone will have an carbonate group adjacent to the keto group, which is already present. And if you have a key, then an allylic position hydroxy group get attached at allylic position. An allylic position could be more than one in a given reaction. For that we have ordered two degrees maximum, then one degree and then three degree. Okay. Based on this information, this information you write down the product in these reactions with SEO2, with SEO2, with SEO. Both carbon allylic carbon are two degree. So anywhere you attach, this won't make any difference. OH is the answer. This one is one degree, but this one is two degree. Two degree, the rate is more. So the product of this would be this OH. This one, the reaction place takes place at secondary carbon. OH is done. Yes. Next one is bear villager oxidation reaction bear villager oxidation. This is used for mainly for ketone, but it can oxidize obviously aldehyde also. So it is used for the oxidation of aldehyde or ketone. Right. So what happens in this? Suppose you have a ketone, R1, CO, R2. And in this, we take a peroxy acid for this purpose, peroxy acid. And this converts into this product. So ketone is this and it converts into ester in this reaction. We use peroxy acid for this purpose, peroxy acid. Any peroxy acid we can use for this purpose. Got it? So just what we need to do, we need to insert an oxygen between the alkyl group and the carbonyl carbon. But which alkyl group is the question? Correct. So we'll discuss the mechanism for this first and then we'll talk about it. Okay. Did you write this? Yes, sir. Okay. Now you see the mechanism. Ketone. Okay. This is a peroxy acid, peroxy acid. And this peroxy acid, this lone pair attacks onto the carbon atom. And this pi electron moves onto the oxygen. So what we get here, R1 CO minus R2 OO C double bond O, R H and positive charge. Right. Now because of this, no, H, this is not stable. So from this H plus comes out and we get this C double bond. O. We get this. Now here we have peroxy linkage. So this is not stable. Right. And one of the oxygen takes the bond pair of electron and goes out and it converts into R1 CO minus two O positive charge plus O. Now after this, what happens, one of the alkyl group here, one of the alkyl group, suppose this one, which has more migratory aptitude, this will migrate onto this oxygen. Means it takes this electron and migrate onto this oxygen. Okay. So this comes over here. This lone pair comes over here and this migrates onto this. So the product of this reaction is R1 O C double bond O R2. This is the product we get. The one which has more migratory aptitude. Okay. What is that exactly? That one get migrate onto the oxygen atom. So migratory aptitude of R1 is greater than R2, I am assuming. Migratory aptitude. What? Why exactly does this migration take place? Your voice is not coming. No sir. Very low. Come again. No sir. Better. So why exactly does this migration take place? Why migration happens? Yes sir. Okay. I will discuss this later. Okay sir. See, first of all, this lone pair comes over here. No. Right. And this positive charge on this oxygen. Oxygen is an electronegative element. So positive charge is not stable at all. So what it does, it attracts the sigma electron towards its side. And if the attraction is great, then the sigma electrons breaks. The sigma bond breaks and one of the alkyl group moves towards the oxygen atom. Oh, positive charge. Right. So migration happens because of high electronegativity of oxygen one. Plus there's a positive charge on oxygen. Right. And that's why it attracts the sigma electron, any one of these two sigma electron. And hence the migration takes place. Okay sir. Okay. So migratory aptitude you have to memorize. I'll just write down here. If you look at the migratory aptitude. And again, I, like I told you, mechanism is not important here. You just need to know migratory aptitude, nothing else. So migratory aptitude of hydrogen is maximum. Then we have three degree, two degree. Then we have phenyl one degree and then alkyl. This is the migratory aptitude. Right. So the one which, which is a better migrator between that and the carbon, you attach one oxygen. That is the answer for that question. Nothing else. So suppose if I have this question, CH3 C double bond, Oh, pH with CF3 COOH, the answer for this one, C double bond. Oh, oh, right. Per auxia signal. Don't size does not matter. It electron releasing power. More will be the migratory aptitude. Okay. In this too, which one is a better migrator? Sorry. Finale is a better migrator. The one which is a better migrator. You just need to introduce oxygen there. And that is it. Nothing, no mechanism. This one in this one, which one is a better migrator? That's the group. So the product here is O C double bond. Oh, CS3. Okay. In this one, both are symmetrical. So we'll have this, which one is a better migrator? Here it is two degree. And here it is one degree. Right. Two degrees of better migrator. Four member ring convert into five member ring. And here we are. Sir, I didn't get the last one. Why? I know. Last one is a problem. Last one. This carbon atom is what? This is two degree, right? And this is one degree. Two degree is a better migrator than one degree. Hence this carbon and this carbon, you have to introduce one oxygen. So four member ring convert into five member. Yes, sir. Next. Oxidation of diol. Okay. Oxidation of diol we use. H I O four. I would ask it. So suppose here also. How do we write on the product? CH2 CH2 OH. And when you allow this to react with H I O four. It is a viscinal diol. H I O four. You just need to break this bond, the carbon carbon bond. Attach OH on both carbon. Okay. So the product is CH2 OH. OH. Plus we get OH. CH2. Again, on the same carbon atom to OH is not stable. So H2O comes out and we'll get at CHO. H2O comes out and we'll get. H CH. Okay. So here also mechanism is not there. Just you need to break the carbon carbon bond, carbon carbon bond, which contains OH group. Those carbon carbon bond, you have to break and attach. OH on those carbon. Another example you see. Right on the product in this. What happens? We'll break this bond. And attach OH. Again, this is not stable. The two H2O comes out. And we'll be end up getting this molecule. What is the name of this compound? Hexane diol. Hexane diol. One, two, three, four, five, six. One, two, three, four, five, six. Hexane diol. Okay. Write down the answer in this one. Write down the product. I'll just come and take water. Okay. What is the answer in the first one? All the carbon carbon bond. You have to break. Attach. And you know the fact that on one carbon. Two OH group is not stable. Right. Then eliminate water. You will get the final product. Okay. And this is what this oxidation reaction we are doing. Right. Alcohol. Right. See. We know alcohol on oxidation gets aldehyde, which further oxidize into acid. Right. So here also when you do the oxidation of diol, you will get either aldehyde or a ketone here. Right. So you see all these bond will break this bond. This bond will break. And each of these carbon atom contains one one OH group. So. We'll get two molecules of this. H2. See OH. OH. And the middle carbon. Which has one hydrogen, but it contains three OH group. Yes or no. Now after this, what happens from here? H2O eliminates. And we get two molecules of. H2C double bond. Oh. That is formaldehyde. Right. And here also. H2O goes out and we get formic acid. So this is the product of this reaction. Correct. Yes, sir. Okay. Okay. In this one, what happens in the second one? So the first and second carbon bond breaks. See this bond breaks. And. This bond breaks. And. Here this bond and this bond won't break because. At a decent position, OH is not present. This side and this side of it. So this bond and this bond won't break. And we get. And. One people. All. And from this. Oh. Oh. Oh. Oh. So now. So oxidation with. For takes place only at the adjacent carbon. Right here it is not possible. No oxidation for this one. It will be as it is present over here. This present as this. Try this one. No. On complete oxidation. What we get here. Complete oxidation. Hmm. Oh, you can hear me. Now yes. Okay. Previous two times. No. Okay. I was telling products that I could. I don't think you could hear me that time. I didn't get. Okay, sir. Now it's fine. No, it's not fine. I'm not speaking now, sir. I mean now I'm speaking but. When I told I was not speaking before that I was not speaking. I got it. Okay. You see all these bonds break here. All these. Correct. So you look at this carbon atom. This one we have to watch this one to watch this one to watch this one to watch. So this four carbon atom gives you. The similar product. Okay. So we get four molecules of. H. See. And the carbon atom will have one OH already present on it. And two OH will get because of dissociation of two bonds. Yeah. Yes. One second, sir. No one second. No, sir. Okay. See the carbon atom, the number of bonds you are breaking equal number of OH group you attach on the carbon. That is. Now you think we'll get four more. Okay. Now you think we'll get four molecules of this. And we get one molecule of this one. One molecule of this. This three. Tell me is it right. Okay. Yes, sir. Yes, sir. That's what I was thinking what happened to the special things at the end. No. This what happens. H2O comes out. Formic acid. Four balls of this. H. C double bond. OH. One more already present. So we get total how many. I shall write down four only because this reaction I have written. Plus H C. OH already is there. This one. Plus this one gives you a formaldehyde, which is this. Yes. Hence the answer for this question is. We get five molecules of formic acid. And one molecule of formaldehyde. Done, sir. Okay. So with this we have done with. The various oxidation reaction. Okay. Next we are starting with. Reduction reaction. The various reducing agent be used for this purpose are. The first one. Is. A very strong reducing agent. That is ally acid. A very strong reducing agent. That is ally. The name is lithium. Lithium aluminum hydride. Lithium. Aluminium hydride. This is ally. It is. A coordination compound. A coordination compound. And it is represented as. Plus. Minus. Okay. The two components are different from the simple compounds. The part which is written the square bracket. This part we call it as. Coordination is fear. And the part which is written outside the square bracket. We call it as. Ionization is fear. Okay. This is coordination is fear. This ionization is fear. Okay. The coordination compounds differs with. Or the normal compounds in this way. That when you dissolve this into any solvent. Then the part which is written in the square bracket. This part the coordination is fear. This we also call it as a complex part of the molecule. And hence it is a complex compound or coordination. Why complex part because it won't lose its identity in the solution. Lose its identity means what? Aluminium hydrogen bond does not break into the solution. But if you take NaCl right. Na plus Cl minus both get hydrolyzed completely. And in solution then loses their identity. Hence this kind of compounds are normal compounds. The compounds which. When you dissolve in water. Does not lose its identity completely. And does not break down into the smaller ions. Are called complex compound. Basic difference is this. This part won't lose its identity in the solution. When we start coordination compound. We'll talk about it in detail. So here we just, you know, the basic idea you must have. I LH4 is the. Coordination compound complex compound. And it's IOPSC name is lithium. It's IOPSC name is lithium. Tetra hydride. Hydride. Aluminate. Aluminate. This number is the oxidation state of the matter. Okay. To write down the IOPSC nomenclature of this compounds. Again, we have certain rules for this. Those rules will discuss in that particular. Like we have a IOPSC nomenclature of normal organic. For this also we have the same one. Okay. Now. This is the reagent we have that is the reducing. Strong reducing. Now, little bit will understand about the reduction reaction. The reduction is the reaction in which. There is an addition of hydrogen. Addition of hydrogen is reduction. Hydrogen is an electropositive element. Okay. So removal of. Electro negative element is. Of production. That is oxygen. Second way we can release this. Third way. Acceptance of electron. Right. Acceptance of electron is again reduction. When you accept electron oxidation state decreases. Decrease in oxidation state is. All these four different way. We can understand a reduction reaction. Now what happens with aldehyde. When we aldehyde and ketone in this reduction reaction. Did you copy this? Yes. Okay. So. Heading you right down and the height keto. And the height keto. We use R1 C double bond. Oh. R2. R1 C double bond. Oh. R2. With LILH4. And H plus H2O. LILH4. And H plus H2O. The product we get here is R1 R2. You let it be as it is. With carbonyl carbon you attach. H and. OH. So what is happening here. He toned into. Two degree alcohol. Obviously reduction. Two degree alcohol gives ketone on oxidation. So ketone on reduction gets two degree alcohol. Right. Mechanism you want. Mechanism you want to know. Yes. Yes sir. After. 10 seconds. So mechanism here is important. That's why we'll discuss mechanism. But those kind of reaction that I all. And. There are. All those reactions mechanism not at all. Okay. Okay. Yes. So. First of all. What happens? First of all, this LIH4. Like I said. It is a complex part and it dissociates as. plus and AlH4 minus. AlH4 minus, the structure that you see, it is like this. Aluminium has four hydrogen attached with it and a negative charge present on aluminum. So in the reaction what happens, the whatever you take, aldehyde or ketone, for example, for this one you have taken this ketone. From this, the AlH4, aluminum hydride that you have, this hydrogen takes this electron pair, means whatever the temperature we are using for this, that is enough to break this bond. The aluminum hydrogen bond is not that strong. Okay, it is inic and it is not that strong bond. Okay, so we can dissociate this bond very easily and H minus comes out. So H minus comes out, that is the point here. So basically, this dissociates into AlH3, which is a neutral molecule stable and hydride ion H minus comes out. And this H minus attacks onto this carbonyl carbon. Okay, and the reaction proceeds. So what happened? H minus comes out and this H minus attacks onto this carbonyl carbon, pi electrons shift onto this oxygen and we get here R1 CO minus. And for to neutralize this, Li plus R2 is here and hydrogen is attached. Okay, next what happens plus what we get? Plus we get AlH3. Next what happens with H plus H2O? Solvent. So we get R1 CO H, R2 H plus we get LiO H. Sir, like, I'm not asking you to oppose anything, but I generally like mechanisms and all, but how exactly will it be helpful and competitive example? How exactly? Like, I like mechanisms and all, but how exactly will it be helpful and competitive Okay, now the like the question that you asked, the answer for that is, could you tell me if I write down this product from where this hydrogen is coming? Okay. Okay, what? Tell me. Because Li AlH4 is a coordination compound and that gives AlH3 and H minus, which is very important. So this H is coming from what? Coming from Li AlH4. If they give you Li AlD4 here, you will get D over here, not hydrogen. Oh, okay, that's things. Option may be here. But the mechanism why it is important and why I'm discussing this here, because in question you will get sometimes here D plus D2O or here Li AlH4. Oh, okay, yes sir. Li AlD4 or here H plus H2O. Okay. You see, first of all, the important thing is that this hydrogen here, it is coming from Li AlH4 or this hydrogen is coming from the solvent. This hydrogen is coming from the solvent and this is coming from Li AlH4. So you know this fact already, the ketone 2 degree alcohol on oxidation gives ketone. So obviously, our reduction here, like ketone is going under reduction. So obviously, the ketone converts into the secondary alcohol, correct? Now to get some secondary alcohol R1, R2, you don't have to do anything. You have to add here hydrogen and here hydrogen on carbon and oxygen. But you need to take care of that this hydrogen is coming from which what is the source of this hydrogen and what is the source of hydrogen coming over here on this oxygen? Yes, Li AlH4 and solvent, done? Yes sir, done. So what you need to keep in mind? Wherever you see this Li AlH4, right? We need to see this H minus R a has a reaction. Simple. Here Li AlH4 de ka hydride and it will give and then reaction takes place. Nothing much you need to do. okay fine now you look at this question so once again sir i saw you moving your cursor so i thought you'd be ready to ask us hello sir yeah when you get just a second i am in a call okay unless we try okay sorry sorry sorry sorry okay thank you yeah yeah tell me when that i got a call from one of the parents sorry i have to pick tell me what happened did you finish it yes i finished it okay so this li alh 4 and x plus s2 so obviously we have formaldehyde so we need to add h on the carbon and on the oxygen so the answer would be what answer would be cs3 OH right this one is an aldehyde obviously it gives one degree alcohol and that would be cs3 ch2 OH this is ketone it gives secondary alcohol cs3 ch OH cs3 because here we have everywhere we have a hydrogen so we don't have to think anything over here but now here we have d plus d2 and h less we know the hydrogen that attached with the carbon atom that comes from the reducing agent li alh 4 so here we have this hydrogen but here we have od instead of OH correct in this case what happens here we have OH here from the solvent and instead of hydrogen we have d is this clear yes this is what the product of this reaction okay this one you try tell me the final product so is it an alkene alkene name square double bond ch yes sir what is the answer answer tell me sir i got an alkene the same thing okay here the product we get is this now this is a with acid dehydration protonates h2o comes out and will get a carbocation but then we have ring expansion possible right with ring expansion you'll get a five-membered ring with positive charge present on this and then h plus comes out from the adjacent carbon and the answer here would be an alkene cyclo alkene sir you are muted yes i got your point ma'am just give me some time i'll arrange your call back to you on this because you know i'm not related with nine and ten standard i'm not taking classes into that so obviously like regarding the admission is concerned i'll let you know you know you can take the admission that's not an issue yeah hello hello sir i don't know why i'm getting these calls third call today from the parents anyways so what is the answer in this one sir first step you get a secondary alcohol sorry primary alcohol see i have discussed all these things if you do not pay attention that if carbocation is forming then there must be some rearrangement if it is possible you have to consider that what you have done you're just taking this h plus out and make a double bond over here yes right so take care of all these things that's why when you revise genoc on a regular basis you'll keep this thing in mind that what could happen okay anyways so what happens in this one final product tell me is there any ring is there any rearrangement carbocation one second sir hydrangea yeah one comma two hydrangea from primary to secondary carbocation so first of all i think a i'll write down directly we'll get it as this with carbon ch3 and here we get ch2 OH correct yes and then with acid we'll get a carbocation here which is this c ch3 and i'll write down positive charge here with ch3 after rearrangement yes or no yes okay after this is it possible like anything anything possible here to get more stable carbocation now i'm very sure you forgot whatever i have discussed in goc no say now it can go further to the ring what can go further to the ring so what is the condition for any expansion uh you need to increase stability like we know that uh three carbon ring and condition for ring expansion the carbon atom attached to the ring sorry the carbon atom adjacent to the ring adjacent to the ring right the positive charge present on the carbon atom just adds into the ring so here we have the condition of ring expansion if you remember i have discussed when the positive charge present on the ring this is the condition of ring contraction remember that yes okay so here what happens the carbon atom here it is the ring expansions one two three four five so we'll get a five member ring after the expansion and at first carbon we have two methyl group present so this is the two methyl group and we have a positive charge on the second carbon which is this now you tell me after this what happens is this the most stable carbocation uh no sir when you have a three-degree carbocation from a two-degree you can go to a three-degree carbocation how do we do that hydride shift where is the hydrogen on methyl shift so we have a methyl shift here for the correct yes sir yes you are the methyl shift you see this adjacent adjacent methyl methyl yeah that's what i meant and then h plus comes on could you tell me how many structure we get over here what is the most stable one here and the double bond is between those double bond in between the methyl groups methyl groups uh left side double bond constituted alkene this is the major product we get but a double bond another product possible where we have double bond over here and here minor products yes these are minor products the most stable one is this the major product right you see so like this they frame the question okay if you miss one thing you will get minus one or zero simple yes right more substituted alkene more stable so we'll get that okay so it is you don't have to memorize this you just need to know that when carbocation forms we always form more stable carbocation by various different methods ring expansion hydride shift methyl shift okay so there is no other rule for this you have to aware of it that this is happening this kind of reaction is possible and hence this is the final product okay no doubt in this the point is li lh4 the carbon atom takes hydrogen from li lh4 and oxygen takes hydrogen from from what from the solvent that is exclusive tell me the number of product possible in this reaction number of product possible in this reaction li lh4 h plus h2 number of product you tell me uh one second sir so two i got two isomer seven you get the alcohol there in the hydrogen yes yes sir okay so alcohol dash wedge for different configuration alcohol and hydrogen uh that also yes so tell me how it is too so once alcohol comes front uh and the alcohol and c h3 will be in the same plane next time hydrogen and c h3 will be on the same plane only one yes sir oh one second one second how many how many how many how many tell me what will happen you'll get a wrong response two two two carol carbon two carol carbon then how many uh what is the answer then two to the power two then this four four then why did you say two that time process uh nothing nothing nothing okay so the correct answer for this one is sir but try to understand see see i have drawn this uh that will do some configuration difference what configuration this make your product amala tell me how many parts you get into this one same reaction do we have any difference in this too we difference yes yes how many products we'll get here one sir poh h and this method right near the guy near the area there yes sir how many carol carbon in this two sir two this one and this one yes sir how many uh structures the different products possible here sir four four this one could be r and this one could be r so one possibility is both r r another possibility is both s s another possibility is one is r other one is s first one is s next one is r this is the four possibility here yes or no go yes yes sir correct all of you understood this yes sir when you have this diagram it means it is the position of methyl is fixed it's very logical and simple when you draw this it means isca configuration is fixed now r s whatever it is but it is fixed we cannot say isca rb yoga sb yoga like we are doing over here because the methyl is fixed it is coming out of the plane towards the observer right so for this carbon here also we have two carol carbon in this molecule we have two carol carbon this kind of question i have discussed already before but since it is the uh you know veg dash form over here so methyl the configuration of this carbon is fixed because the methyl is fixed here we cannot assume this that this methyl is going into the plane so whether it is r or s suppose i am assuming this as r i am assuming this i don't know whether it is r or s i am assuming this so this would be r only but this can be what this can be r or s both so answer is what answer we can have r r or we can have this one can have s and this one is r only we cannot change this borrow did you get it yes sir right so when we draw this structure or if you draw this one also will be two in this reaction because this is fixed here we we cannot assume this to go out of the plane and into the plane like here we are doing so this is the difference in these two questions got it amla understood this so this you must keep in mind there is a difference of this structure and this structure here we cannot change the configuration of it only one configuration r or s but this configuration could be r and could be s both so these two possible combinations am i clear yes sir okay next one you see derivative of acid acid and it's derivative how many li alh4 the reaction of li alh4 we are we are looking at but there's no difference between the first and the third one no vidyata there's no difference as far as the number of product is concerned the only difference is the view angle is different the first one you are looking from the top and the first one if you look at from the bottom you will get the third molecule did you get it vidyata yes so as far as the product is concerned the number of product there is no difference in the view the view angle is different for the two okay next you write down derivative of acid where is adweta any any information no no sir sir she told she texted you why she isn't here oh i did not see okay you see this compound r c double bond o l living group okay l is the living group li alh4 so i'm not going to write down this again and again how h minus we get from this okay can i write this directly from this will get hydride ion h minus no issue right yes sir this will attack on to this carbon the spy electron shipped on to this oxygen what do we get we get r c o minus h here and this h comes from li alh4 so the product is what r c o h and aldehyde and aldehyde we know if you use excess of li alh4 this further reduce into what li alh4 this further reduce into clear no doubt so what we are getting r c double bond o l l is the living group here could be cl b r i o c s 3 o ph o c o cl 3 anything right r c h 2 o h we are getting write down the product in these reactions with you c h 3 c double bond o cl li alh4 h plus h 2 c s 3 c h 2 c double bond o o c s 3 same thing r c o o h same product uh yes sir the first one can i write down the product directly the product will be c h 2 o h first we'll get aldehyde with same carbon atom and then aldehyde reduce into one degree alh4 this is what we'll get c s 3 c h 2 or just a second hydrogen so we'll get c h 2 o h plus we'll get c s 3 o h from the living group this also we get this converts into r c double bond o h which further reduce into r c h 2 o h remember this is possible when li alh4 is in excess right then only it happens this one is a bit different uh you know reaction i'll show you how acid it is exactly same mechanism we have but one thing you will get maybe you'll get confused and that's why i'm discussing this r c o o h with li alh4 li alh4 which gives h minus iron h minus iron attacks onto the carbon and it forms o minus h and o which further what happens this lone pair comes over here and this o h minus goes out as a living group r c double bond o h plus o h minus now you see till now we have discussed that o h minus is not a good living group correct yeah alcohol always goes under protonation this converts into h 2 o plus and then h 2 o goes out this is what we were discussing till now yes sir but here it is happening and that's why i'm discussing this mechanism our if you don't see this then the mechanism is same like we're getting the product similarly like we're getting another reaction but if you look at the mechanism you may get confused that how this o h minus is leaving out correct so why it is happening here o h minus is a poor living group but why it is happening because we don't have any other choice actually either you see either o h o h minus will go out or h minus will go out that is what possibly and if you compare these two then o h minus is more stable than h minus because the negative charge present here on oxygen and electronegative value so if you have to compare o h minus and h minus o h minus is better than h minus in terms of the living nature and that is why it is happening here otherwise it is a poor living group yes sir okay that is what it happens so after this again what happens this goes under reduction with li alh 4 and converts into alcohol that is r c h 2 o h okay similarly we can talk about esters also see this with li alh 4 could you tell me the product here this is a living group o c o c s 3 tell me the product c h 2 o h and c h 2 c so we get here r c double bond o h first of all from this part and the living group forms c h 3 c double bond o o h acid acid further reduced into aldehyde c s 3 c double bond o at this mechanism if you see and this further converts into what and alcohol that is c s 3 c s 2 o h any doubt in this this also get reduced into r c s 2 o h no sir no doubt okay so this is the reaction of li alh 4 okay you see this uh suppose i'll take one example like this alh 4 major product tell me that yes sir what is the answer cyclo benzyl cyclohexylenethanol why cyclohexyl so the positive charge in the final answer or the intermediate answer final answer is this final answer is correct no i can't hear you when cut c here what we get amla we'll get alcohol here yes or no yes sir that alcohol is this c h 2 c h 2 o h okay and this is again an acidic hydrogen that is clear no yes fine when cut you are not audible your voice is not coming okay so anyways we'll take a break now yeah we'll take a break now we'll resume the session at 650 okay guys yes take a break hello yes sir yes sir yes okay so next reaction write down of li alh 4 reaction of amide rco n h 2 with li alh 4 and h plus h 2 okay the product here we get is rc n h 2 and 2 hydrogen means from amide to amine that is primary amine one degree okay i would suggest you to memorize the reaction like this one rco n h 2 converts into rc h 2 n h 2 okay mechanism we'll discuss you copy down this so in the first step rco n h 2 from li alh 4 it forms hydride ion right this comes from li alh 4 and this attacks on to the carbon atom this goes up and the product is r c o minus h and n h 2 okay to neutralize this we write here li plus okay further the lone pair on nitrogen forms a pi bond and this goes out here as a living group and the here we get r c h double bond n h h here okay this molecule further we have a positive charge on to this nitrogen because it comes back here to neutralize the positive charge this h plus comes out and this is taken up by this li o minus or we can say o li minus takes this hydrogen from here and it forms r c h double bond n h plus li o h this molecule is amine okay now in this again li alh 4 gives hydride ion and this again attacks on to the carbon atom and it forms r c h 2 and this electron pair shift over here forms n h negative charge on it with in the last step with h plus h 2 o converts into r c h 2 n h 2 r c h 2 n h 2 this is the mechanism right also once again i this hydrogen these two hydrogen comes from li alh 4 only this you have to memorize and this hydrogen comes from solvent that is what are you sir the initially from n h plus h the o minus li abstracts that hydrogen and goes away as li o h okay the so then it becomes r c h and h not plus the plus also is removed where the white steps are middle bro h plus goes with this no so electron pair is taken up by this nitrogen oh yeah okay just one second sir now is my voice audible right so i would suggest mechanism you don't focus much just you see amide is converting into amines and carbon atom the both hydrogen is coming from li alh 4 uh one second sir done yes sir okay one more point here that you need to take care of that the derivative of amide if it is there if this nitrogen atom does not contain any hydrogen then the reaction is not possible suppose here we have two alkyl group attach then this cannot take the alkyl group from this nitrogen correct so after this the reaction is not possible so for this reaction obviously we have amide we can have derivative of amide but again the nitrogen must have at least one hydrogen on it so for this reaction at least one hydrogen the functional group which contains nitrogen all reacts with li alh 4 and converts into amines this is the general thing you have you can keep that in mind any functional group you can take amide you can take cyanide you can take emi you can take nitro all these converts into amine one degree of mine so what so could i repeat those again the functional group that contains nitrogen atom okay one second sir one second functional group the functional group which contains nitrogen atom so as in which one rco in this case any functional like amide ho gya nitrogen atom is there rc okay rcn ho gya right rch double bond nh emine functional group rn o2 right so functional i'll i'll give you the example right on the first this the functional group which contains nitrogen on reaction with li alh 4 on reaction with li alh 4 gives primary amine primary amines okay for example you see this converts into rch2 nh2 rch2 nh2 here again both hydrogen here is coming from solvent from solvent and this hydrogen is coming from li alh 4 okay cyanide converts into primary amine if you have emine this reaction just we have seen before with the same reagent nh it converts into rch2 nh2 if you take rn o2 it converts into rn h2 if you take auxine again it gives one degree of amine so these are the reactions we have for example tell me the product in this one li alh 4 h2o sir hex again ch2 and h2 this two reaction tell me the product and so the the source of the hydrogen is going to remain the same in all the cases right okay sir just a minute sir okay so all these reactions are given in amine's chapter okay last of the organic chemistry but reduction reaction we are doing it here only so we are not going to do any oxidation reduction reaction after this we are covering all the chapters okay sir done here what happens rcn converts into what rch2 nh2 correct so it is a six-member converts into ch2 nh2 here also you see i said nitrogen must have at least one hydrogen so this converts into cs3 ch2n this would be as it is and one hydrogen remember this hydrogen is coming from solvent here it is not like the nitrogen retains this hydrogen over here no it is coming from the solvent if you take d2o here d plus d2o you will get n with d this hydrogen on carbon it is coming from li alh 4 yes sir next reaction you see epoxide this reaction is exactly same like we did in case of brignard reagent the only difference is here we have h minus not r minus and that h minus comes from what comes from yes that is a reducing agent li alh 4 li alh 4 gives h minus and this h minus like alkyl group there it attacks onto this carbon atom this comes up here and this gives you hydrogen then ch2 ch2o minus anti-position right o minus and then with h plus h2o it converts into what it converts into h ch2 ch2o h remember one thing that this is this one and this one present at anti-position present at anti-position this comes from this comes from comes from li alh 4 this hydrogen this is again this hydrogen comes from solvent here clear exactly same reaction we have same pattern of reaction we have with rmgx if you see only r will be replaced the only difference is what there we have r minus that comes from rmgx reagent you are changing right so obviously the attacking particle will also be different but the same pattern we have over here there's no any doubt as in two type attack if you have un-symmetrical epoxide then this h minus will attack onto the onto the less hindered side for example could you tell me what is the product we get in this reaction cs3 ch ch2 oxygen same reagent tell me the product is it primary alcohol yes it is propane to all yes what is the source of this hydrogen water this hydrogen that one li alh4 and this OH is water always if you have this one this one li alh4 with nh4 cl nh4 cl also gives you the same product okay nh4 cl also gives you the same product the product of this reaction is right this h minus comes from the top onto this comes from the top this comes over here this won't attack onto this carbon because it is crowded correct so this cs3 cs3 will be up here because when this goes when this bond pair comes over here oxygen will be down so OH is here and hydrogen from the li alh4 will be here and this hydrogen here as it is now we were using h plus h2o here correct here it is nh4 cl but then also the product won't change here so what does this do this it's nh4 cl okay you see what hydrogen nh4 cl is a salt of what again weak base and and strong acid strong acid so it's it's uh it goes on hydrolysis and it gives what it gives acidic solution correct because acidic is strong acidic solution means in the solution we have h plus and hence the medium is acidic hence if you write h plus h2o or nh4 cl it gives you the same product correct understood yes okay so this is it for li alh4 sorry reaction of aldehyde ketone we have discussed we have discussed epoxide cyanide amide we have discussed we have discussed imine hydrox this one oxene many compounds we have discussed on the next one we have to discuss with li alh4 that is the reduction of alkene and alkyne with li alh4 okay li alh4 with alkene and alkyne suppose we have a reaction rch double bond ch single bond ch3 see this reaction with li alh4 it does not give any reaction no reaction here why no reaction because this too this gives what this gives hydride and both sides directly hindered and this here also we have five electron cloud right this electron cloud repel this hydride ion if it is an electrophile it may attack onto the carbon atom which is not the case here yes that's why the reaction is not possible here with alkene and alkyne but in this case you see this one x versus 2 in this case when you take here al cl3 then the reaction possible that is rcs2 cs3 you will get the product okay no no no that's right winker that's not possible so that's right and not possible no sn2 with aryl vinyl is not possible why are you discussing this no sir just told us a same thing no sir same principle aryl and vinyl compound halide does not go under sn2 reaction yes sir that's what i want for aryl sn1 sn2 both very difficult because neither the carbocation forms stable nor the bond dissociation is possible over there for essential okay sir yes okay sir understood thank you like you do similar principle no sir hence i told it nothing else sir okay fine so this is the reaction possible with al cl3 al cl3 is what al cl3 is a louis acid is the louis acid so any louis acid may give this kind of product so look at the mechanism here this pyelectron it takes here in this vacant orbital of aluminium and it forms rch ch delta positive delta negative okay now after this what happens since this charge develops here so this h minus this attacks onto the carbon atom here ch h ch2 al cl3 this happens this bond breaks and comes over here okay now next what happens from this here we have the positive charge on carbon atom just a second so next what happens this bond pair is taken up by this carbon rch ch ch2 negative charge on this and cl al cl3 goes out why this happens because the electronegativity of carbon is more than the electronegativity of aluminium so it takes the electron pair negative charge on it now the next step with h plus h2o it gets protonates and it forms this hydrogen this hydrogen this comes from this comes from solvent this comes from li lh4 sir one second sir i didn't could understand this done i know sir could you one second sir i'm getting confused here why one second sir i don't know just a minute sir like i'll go through this so sir the aluminium will form complex because it's in louis acid it'll accept it right and then h minus is generated as usual because of li al h4 okay and then del minus and del plus develops on aluminium and carbon respectively so after that i don't understand nothing then see here why the dexmo is not taking place because of pi electron now this electron is involved with aluminium okay carbon gets some positive charge here and there this h minus can attack on this okay now after this epoxide this ring will open and then it goes up oh okay okay sir got it so basically you just have to break the double bond and attach hydrogen on the both carbon yes sir but which hydrogen came from where that's also important right done sir okay now the next one is similar reaction we have with alkyne rc triple bond c cs3 same thing happens twice actually so it goes with first it forms an alkene and then further the reaction takes place and it converts into an alkene so eventually you'll get alkene only in this one so same mechanism lo sir same mechanism lo sir peroxide li al h4 h plus it gives two molecules of OH okay heterolysis happens here one of the hydrogen takes oxygen takes h plus and similarly hydrogen attached over here we'll get two molecules of alcohol right double bond if it is present in conjugation this this is the product we get double bond get reduced when it is in conjugation like this here aldehyde will reduce into alcohol fine but this double bond will get reduced when it is in conjugation if it is not in conjugation for example cs3 ch double bond ch co double bond will be as it is double bond won't get reduced here since it is not in conjugation these are some specific reactions that you have to memorize done yes sir another reducing agent we have li al h4 we have done the another one is nabh4 this sodium sodium borohydride sbh we also write it this is also a complex compound complex compound like li al h4 the name of this is iopec name sodium tetrahydrido tetrahydrido borohyd iopec name of this okay the structure of this if you see like i said it is similar to that of li al h4 okay now boron attached with four hydrogen and one negative charge on it okay now this boron hydrogen bond is less ionic lesser ionic than lesser ionic than aluminium hydrogen at four minus so this one is this one is lesser ionic right and it is what lesser ionic means it has more covalent character more covalent character correct now more covalent character means what the bond is stronger we have a stronger bond over here stronger bond means you it is difficult to dissociate this bond and produce h minus ion okay h minus ion is difficult to produce right hence hence this bond is what this nabh4 is a weaker reducing agent weaker than li al h4 got it okay weaker reducing agent than li al h4 since the bond is strong here one more point here this nabh4 can't reduce can't reduce ester can't reduce ester this is important okay nabh4 can't reduce ester because it is lesser uh you know uh weaker reducing agent than li al h4 now based on this two li al h4 and nabh4 sometimes they ask you to compare the rate of reaction so compare compare the rate of this reaction with respect to with respect to li al h4 and with respect to nabh4 compare the rate done no no that's not correct pancut see first of all how do we compare the rate of this okay by comparing leaving groups no we compare the positive charge density on the carbon this one then only h minus will attack on this the first step is what the first step is the attack of hydride ion from these reagents correct so if this positive charge density is more then more will be the tendency to attack on this now we compare the positive charge density here this is what minus i or what it shows minus i this shows yes sir there's no effect here and this shows plus i or plus h both we can say okay now because of plus m the positive density here decreases and goes to the minimum value that's why the rate of b would be minimum for sure minus i this electric positive charge density increases and hence the rate of a would be maximum here between c and d if you compare here we have plus h and there is no effect here so c is more and then we have t is it clear yes sir okay now nabh4 nabh4 can't reduce ester so that is last only we have no only we have to eliminate this is not possible for b so it would be d then c and then a not valid for me so it's a positive charge on carbon atom facilitates faster reaction increases rate of reaction because hydride ion can attack easily easily yes okay sir so nabh4 can reduce only nabh4 can reduce aldehyde with the reagent nabh4 x plus s2 it converts into one degree alcohol ketone gives you two degree alcohol if you have acyl chloride it first converts into aldehyde then it converts into alcohol so this is for nabh4 correct next we have to start with catalytic hydrogenation you want to start it today just one second i was writing that actually this is again just write down catalytic hydrogenation okay so we'll start from this next class first you copy down this done sir done yeah yes sir okay now one reaction we'll see it is not oxidation reduction reaction just one small concept i would like to know discuss here okay that is the concept of neighboring group participation so this one you see this we call it as a type of nucleophilic substitution reaction okay in short we treat it as snnj okay nucleophilic substitution reaction where the neighboring group participates into this okay we also call it as this thing we also call it as ankymeric assistance ankymeric assistance what is this reaction you see suppose we have a compound h cl h sulphur and phenyl here ph sulphur is known there on this now in this what happens suppose we have a nucleophile OH minus here for neighboring group participation the group here and here you see this is a leaving group leaving group is this so the attacking particle present in the same molecule and the leaving group is present on the at the anti position of this attacking part okay so what happens here you see this group that is attached over here it must have lone pair present on it right lone pair present on it so this lone pair it is like intramolecular it attacks onto this carbon and this leaving groups goes out you see the neighboring group takes part in that reaction that's why it is neighboring group participation so we'll get a this reaction you see sulphur here bonded with this ph positive charge onto this sulphur atom and hydrogen will be here and here also we'll have the hydrogen correct just a second yes always anti this should always present at the anti position of this leaving group okay and the two hydrogen will be here rearrange itself like this okay this attack this attack is first of all it is SN2 type okay so there will be a change in configuration here first so whatever the configuration we have here here we have the opposite configuration means r becomes s over here in the next step what happens the nucleophile of the reaction OH minus OH minus attacks here on the same carbon atom and it pushed this electron pair onto this sulphur atom it goes back onto its original position right so what happens the same type of attack happens here and OH from the bottom H on the top then here again H on the bottom SPH on the top sulphur and phenyl on the top even this attack is also SN2 type so again this configuration changes over here so whatever the configuration we have here here we retain the configuration finally retains configuration so this is SN2 type attack two-step SN2 type neighboring group participation SNNGP we have right so the no the property of this reaction is what mechanism is true SN2 but we have two steps of SN2 so configuration we retain the configuration of the molecule r is r only in between it becomes s but then again it converts into r okay when it takes place we have SPH present like this lone pair must be here and leaving group present on the opposite side okay the second property of this reaction is what this reaction is extremely fast reaction okay the rate of reaction of this r o r of SNNGP is thousand times faster okay thousand times faster than faster than SN1 or SN2 reaction happens so fast right now the group that shows SNNGP reaction r I have given you this sulphur if it is sulphur present then it is possible if you know the group that shows sulphur hoga so SNGP possible agar iodine is there SNGP possible bromine is there SNGP possible lone pair we need lone pair even for phenyl also it is possible because its pi electron react P H pi electron we take part in the reaction okay so this is SNNGP important it is not okay not much important but yes rate like you know on the basis of rate they can ask you some question plus configuration basis question they may ask one second sir yes i return this down okay now on this what kind of questions forms in this take or though question you must see suppose we have a phenyl ring present here and we have CH2 CH2 suppose this carbon 14 we have just to differentiate the two carbon atoms C 14 here and here we have CL okay this is the benzene ring okay OH- is the nucleophile so you see this is here and this is present here so here what happens we have ankymeric assistance okay that is SNNGP this pi electron shift onto this carbon atom and then this attacks onto the C 14 so that this leaving group goes out right this is SNNGP SN2 the first step of the reaction I'll write down the overall mechanism is SNNGP but the first step it is SN2 configuration changes okay so what happens here it forms like this this CH2 will be here and this C 14 H2 is here and this bond will be positive charge okay next what happens the attack of OH- again SN2 okay so this attack this will attack onto this carbon atom and this goes up forms a double bond here so this will be a benzene ring everything is as it is right down the product this car ring attached with this carbon atom and this 14 H2 with OH here this is a product we get so could you please explain the yellow step again sir see it attacks onto this you will get a three-member ring in this so okay three-member ring and this may attack onto this this goes up and we'll get the structure oh okay here the possibility of OH- is equal to attack onto this carbon also because both are equivalent yes other product for this is this one CH2 if this attacks over here it's the second so this attacks onto this carbon if see i'll change the color here if this attacks if OH- attacks on this carbon then what happens this bond goes over here then what happens you see this carbon will be two and it is 14 right this attacks the ring and we'll get CH2 OHU understood this oh yeah yes sir so can you repeat it just a second ah tell me what i said this OH- has equal probability to attack on this carbon and this carbon yes fine yes okay so when this attacks onto this the white line you see this attacks over here this bonds goes over here and the ring attached with this carbon with C14 carbon and OH this is the product clear yes sir now but when this will attack onto this carbon atom this bond goes over here now the ring attached with C14 and this carbon will have OH yes now they may ask you the relation between the product here and the relation is what they are positional isomers positional isomers so this kind of question they may answer oh yeah yes sir one more question you see the last one for this suppose you have a compound like this this is D this is H the a nucleophile in the molecule and leaving group much must present on the opposite side right yeah sir in anti positions yeah in this one the reaction is possible or not and here we have chlorine with OH- the reaction possible in this in this also the reaction possible because it is a single bond no carbon caramel single bond so this will rotate itself in opposite direction at the anti position and then the same thing will have over here right so if you have an open bond like this the acyclic structure then whether it is at the same position means sin position or anti position the reaction is possible and the product in this one would be SCH3 CS3H configuration though won't change because it retains the configuration and OH here so this why because the rotation across this bond is possible right and then it will go at the anti position but when you have ring of structure this is not anti position that's what I said it is not anti but they can rotate itself across carbon carbon single bond and then go to the anti position and then reaction possible but this kind of rotation is hindered in this structure when you have when you have cyclic structure like this or in this one when you have cyclic structure like this then the rotation is hindered so in this case it must be at the anti position but when you have acyclic compound this one the last one acyclic compound then whether it is present over here or here it gives us the reaction because the rotation carbon carbon bond possible no the bulkiness won't be there but reaction is possible it can rotate itself we can have conformational instruments of this no sir in the previous case what hindrance in the previous case is because of bulkiness no sir where we have hindrance you rotation cyclic structure the rotation is not possible no no this is though a nucleophile no molecule yeah this are anti position phenyl and chlorine if it is not also it's fine because it is carbon carbon single bond oh right okay sir understood yes sir this is SNNGP nucleophilic substitution neighboring group participation okay so this is it so next class we'll start with what catalytic hydrogenation okay next class we'll finish this on learn is the assignment has been uploaded you can solve those questions okay yes sir okay thank you thank you so much thank you sir thank you