 Okay, can we start? The next reaction we have here is the addition of ozone and we also call it as ozone releases. Addition of ozone and we also call it as ozone releases. Okay, this is an important reaction. So right down into this, addition of ozone takes place, addition of ozone takes place in presence of an inert solvent, presence of an inert solvent. Inert solvent example we have CCl4 chloroform from CHCl3 or we can also have CH2Cl2, dichloromethane. In this reaction, write down in this reaction first, first ozonide forms, which further decomposes, further decomposes reduction with, reduction with ZnH2O or we can also use Zn with CH3COOH or we can also use H2PD etc. But most commonly for reduction purpose we use this one, Zn with H2O, whatever reaction you will see, mostly we will use this only, Zn and H2O. So you see this reaction, in this reaction the product will be, will be aldehyde ketone. Any one of these product will get an aldehyde or ketone. So you see the reaction, first of all suppose we have the reaction if I write carbon-carbon double bond, alkene reacts with ozone O3, in presence of suppose dichloromethane CH2Cl2, temperature we use around 200 Kelvin, which is again not required to memorize. The intermediate product which is the ozonide, we call it as, we get here is, this three oxygen will attach with this carbon atom, like this, this we call it as ozonides. When the reduction takes place with Zn and H2O, so what you can do here, you see the bond dissociates, this bond comes over here, bond pair of electron comes over here, this bond pair of electron goes here and this also dissociates. So actually what happens, here you see, one of the bond is this which dissociates, this carbon will go with this oxygen or the carbon will go with one of this oxygen and this oxygen is taken up by this Z and zinc oxide will go out in this reaction. So one oxygen with this zinc and two carbon will take one one oxygen each. So the product here it will be C double bond O as one of the product plus O, C as the another product. When this bond pair comes over here, we will get C double bond O, this bond dissociate goes over there here, we will get C double bond O and this dissociates which takes this Z and forms Z and O. So we will get here carbonyl compound, you see C double bond O, we have carbonyl compound. Now if you see both these are, suppose it is R and R, so you will get what here, ketone, if it is H and H, you will get aldehyde. That is why I said that in this reaction you will get either aldehyde or ketone, any one of this product you will get. Depending on what is the molecular formula of the alkene we have. Okay, understood this? This is how the product we have to write it down. So since you have to write down the product, so here if you do the original list of this in one step, how do you write down the product? You will do the original list of it and then you use Z and with H2O. So what is the product you do? What is the product you will write? You just break this double bond here and add double bond O to this carbon atom, double bond O to this carbon atom. So the product will be C double bond O plus C double bond O like this. This is the product you get. So whenever you have alkene and you are doing original list of it, just you break the double bond and you add oxygen with both the double bonded carbon atom. See first of all, you see here, how do you write down the product C double bond C like this? And what we are doing? We are doing original list of this and in the second step we use Z and with H2O. So first of all, how do we write down the product? If this is the molecule we have, you have to write down the original list product. Just you break this double bond and with this carbon you add double bond O. All other bonds will be as it is. Another double bonded carbon, you add double bond O. All other bonds will be as it is. This is how we write down the product. But how the reaction goes here with this oxygen, three oxygen here, this will break these two bonds and the both carbon atom is attached by the oxygen like this. It forms a ring here. So when you heat with Zn and H2O here, one of the oxygen is going with, like it combines with Zn and forms zinc oxide and then this bond pair comes over here, this bond will dissociate. So you see here when bond pair comes over here, so you will get C double bond O. This bond is not there anymore because this bond pair has been shifted over here. So you will get C double bond O and this bond is also break and it goes with this oxygen which combines with zinc, forms ZnO. So you will get C double bond O, C double bond O and ZnO also goes out. This is how the product we get here. So this is what since we have C double bond O and C double bond O, so this is what the carbonyl compound. If these two are R and R, so you will get ketone. If these two are H and H, so you will get what? You will get aldehyde. So in this reaction what we see, what we get, you get either aldehyde, either aldehyde or you will get ketone or we can get mixture of aldehyde and ketone. So how do we get mixture of aldehyde and ketone? In this example you see if I write down here R, R, C double bond, R, H, then how do we get here? We get ketone this one and this one will be R, CO, H. So this is aldehyde and this is ketone. So we can get any of these three products. If all are R only, so we will get only ketone. If all are H only, so we will get only aldehyde. So depending on the alkene, you can get any of these three products possible into this. Did you understand this? Tell me quickly, we will finish this alkene today. Now one more type of question that they ask here that they will give you the product. They will give you like alkene undergoes ogenolysis reaction and the product is this. Suppose they will write down the product. Here you see alkene undergoes ogenolysis reaction and one of the product is this. The product we get here is CH3 C double bond OH and the other product will be H C double bond OH. So with an alkene when ogenolysis takes place O3, you will get a mixture of product and the first product is this and second product is this. You need to find out the formula of this alkene. What is the formula of this alkene? Molecular formula of this alkene. This kind of question that they ask. So to solve this kind of question, what do you have to do? You just you see the product that we get in the previous example. See if this is the product we have, right? Now this is given in your question and you have to find out what is the formula of this alkene, means this formula you have to find out. How do we get this formula? We remove this O and this O and we attach these two carbon atoms. Yes or no? We will remove the oxygen, double bonded oxygen and both carbon atom will attach and we will get the alkene, the formula of alkene, right? So that is what you have to do here. You see here in this, the product is these two. So we can also write down this molecule as this CH3 C double bond OH plus O double bond CH and H. So what we will do? We will remove this oxygen atom, right? We will eliminate this oxygen atom and we will join these two carbon atom and we will get the alkene. So the formula of alkene will be what? C double bond C, this carbon has two hydrogen and this carbon has CH3 H. So this is what? Propene, right? So formula of alkene is what? The alkene is nothing but propene. So this kind of question also there. What do you need to do? Just remove that double bond O and join those carbon atom, okay? So you'll get propene. So this is, again one important reaction we have. Now you see oxidation reaction of alkene. Next write down oxidation reaction, oxidation of alkenes. So oxidation of alkene is possible, write down the oxidation of alkene is possible with different, different reagents, right? Oxidation of alkenes is possible with different, different reagent, okay? The first reagent that we use for this purpose or the reagent we have, we call it as Bayer's reagent, B-A-E, Y-E-R-S, Bayer's reagent, okay? So this Bayer's reagent is, it is important, it is alkaline KMNO4, alkaline KMNO4. This is pink, that also you write down. It is of pink color, alkaline KMNO4. Write down, in this reaction, sorry, this reaction is also known as hydroxylation reaction. This is also known as hydroxylation reaction. Write down, in this reaction, alkene converts into diols, alkene converts into diols and the pink color disappears and the pink color disappears, which is the test of unsaturation, right? Which is the test of unsaturation, okay? Test of unsaturation means what? If I give you one compound and if you have to find out whether there is double bond present or not, then we use this reagent, Bayer's reagent, okay? If the pink color of the Bayer's reagent disappears when the reaction takes place, it means unsaturation is present in the molecule, okay? That is what the test of unsaturation we have, okay? So when you have this compound, suppose CH2, double bond CH2 and we are using alkaline KMNO4. So alkaline we have H2O also and we have some nascent oxygen since oxidation is taking place. The reagent we have alkaline KMNO4. So in this, what happens? The double bond or the pi bond dissociates and will get two OH group attached with the double bonded carbon atom. Addition of OH, this is important, write down, addition of OH follows OH group, follows addition means the OH group will attach onto the double bonded carbon atom from the same side, okay? Now the few more examples of this you see, suppose you have this cyclic compounds, okay? Double bond here we have, here we have CH3 and here also we have CH3. When this is allowed to react with alkaline KMNO4, syn addition of OH takes place, okay? Means both OH group will attach from the same side. Suppose we have OH here and here and the other side we have CH3 and CH3. Syn addition is important, from the same side the OH will attach. One more point here you write down, if we have cis alkene gives meso product, again this meso and resmic mixture we will discuss in stereo isomerism, just you note it down. Cis alkene gives meso product, if you have trans alkene it gives resmic mixture. You just write down this thing, okay? Don't think on it, okay? We will discuss this later on, don't worry with it, okay? Just write down your notes, correct? So there are various other reagents, one reagent I have given you, Bayer's reagent. Apart from this we have other reagents also that I will write down and you have to keep in mind, like in this case when you have Bayer's reagent, syn addition is there, that is what is important here. OH group will attach from the same side, this is what you have to keep in mind. Similarly, if you have osmium oxide which is OSO4, this is with H2O2 if you use, this is also an oxidizing agent, okay? This reagent if you use, again the OH belong will only attach little carbon atom. Here also the addition will be syn. Another reagent is hydrogen peroxide H2O2 with HCOOH. This will also give the addition of OH group, but this addition will be anti addition. The two OH group will attach from the opposite side. If you have peroxy acid, peroxy acid, which is HCO3H, peroxy acid, this also gives you anti addition, right? So in all these reagent, oxidizing reagent 1, 2, 3 and 4, first two gives you syn addition and the last two gives anti addition. In all these four, this Bayer's reagent is the most important one, okay? Another reaction we write down, that is isomerization, isomerization. Write down, write down an alkene, an alkene can be converted, converted into other by hitting at high temperature, preferably at low temperature, hitting at high temperature or preferably at low temperature in presence of, in presence of preferably at low temperature, in presence of a catalyst, a catalyst like anhydrous aluminium chloride, anhydrous AlCl3, okay? So two possibilities we may have into this. We have seen this in case of alkene also. What is the doubt Shreya? Other means other alkene and alkene can be converted into other, means other alkene, okay? Different alkene, different product, different compound actually, okay? So suppose we are taking pentene or pent vanine, which is nothing but C-H3, C-H2, C-H2, C-H double bond, C-H2. When you heat this in presence of AlCl3, so one possibility is what? This double bond will shift, there is no mechanism for this. You have to memorize this reagent, C-H single bond C-H3, pent 2 in, this is right? One pentene, pent or two pentene. So these are the isomers of each other. Another possibility is what? Shifting of double bond we have here. Another possibility is what? Shifting of or migration of methyl group, right? Migration of methyl group. So for this if I write down C-H3, C-H2, C-H double bond, C-H2. When you heat this at with AlCl3, then C-H3 group will migrate and will get isobutene in this. C double bond, C-H2. So this is the isomers we have. We get isomers like this. Understood? Next reaction you write down. Next reaction, oxymercuration, demercuration reaction. Oxymercuration and demercuration reaction. In which one will have the mixture of product? The previous one. Here, see actually this alkene, see this one, if you have substituted one methyl group onto this, okay? That one and this one if you compare, okay? So this one is obviously more substituted. So the major product will be the most stable product. See, whenever we have double bond in the terminal, on the terminal carbon, then this double bond will migrate to form more substituted alkene. That's the case, okay? Here also you see, here you see if this double bond comes over here, then this carbon and this carbon, these two carbon you consider, this carbon and this carbon, both carbon are substituted here. One of the methyl group will migrate over here. But in that case, this carbon will not be the substituted carbon. Yeah, the major product will be the product which is most stable. And we know in case of alkene, more substituted alkene is more stable actually. That is how we get, okay? So you don't have to worry with this. There are only two possibilities, okay? There are only two possibilities. One is what shifting of double bond and other one is what shifting of methyl group. But in both cases, in both cases, the objective is to form the more substituted alkene, more substituted alkene. And if you are confused with that, why not the CS3 will migrate over here? So you don't worry. You will have option also into these kind of questions. So they will not give you all possible product because this is not the multiple choice question, okay? So only two possibilities are there. One, shifting of double bond and other one, the shifting of methyl group. In both case, we have to have more substituted alkene, right? So according to the option, you will choose the answer. Is it clear? Coming back to the next one, oxymercuration and demarcuration reaction, okay? So you see in this one, there are two step reaction. In this, the mercuric acetate ion will attach first and that's why we are calling it as oxymercuration. And in the second step, it gets de-attached, right? That's why we are calling it as demarcuration reaction, okay? So the only thing you have to memorize here is the reagent, okay? Write down it first into this. This reaction takes place, this reaction takes place mercuric acetate, mercuric acetate in presence of H2O THF solution. THF is tetrahydrofuran, Jeff solution. Followed by, now the second step reaction is what? This is followed by, this is followed by the reduction with, reduction with ABH4 NaOH. It produces the hydration of alkene, hydration of alkene according to Markovnikov rules, Markovnikov rules, okay? Try to see the reaction in this. We have this CH3 whole thrice C, CH double bond CH2. And when this reacts with mercuric acetate is this CH3 C double bond O. This we use in THF tetrahydrofuran with H2O. Now the product here it is what? This double bond will break and this HGOO CH3 will attach over here. So the product we'll get here is this CH3 whole thrice C single bond CHCH2 O C CH3 HG CH3 HG. This is oxymer curation step. Now when this is allowed to react with NaBH4 with NaOH you will end up with CH3 whole thrice CCH CHCH3 and you'll get OH here. See NaBH4 is a reducing agent. It produce hydrogen here. So this hydrogen will attach onto this carbon will get CH3 and with this NaOH will get OH over here and will get OH here. So what happens here you see the reagent you have to remember in the first step the reagent is this like I'll write down here you see you don't have to memorize or you don't have to write down this step, okay? You may be confused with this or that how do we get this but this you don't have to bother with. Only you have to keep in mind is what there are two reagent which is CH3 COOH whole thrice HG and with this we are using tetrahydrofuran in H2O and in the second step the reaction or the reagent we have is NaBH4 with NaOH, right? So with this if the reagent is this we are using with this reagent. So the H plus, see the negative ion is what happens here you see. Here the H plus will attach. So here the OH minus will attach and here the H plus will attach. So obviously in all these reaction the addition of H plus and OH minus takes place and what it follows? It follows Marconi-Koff rule, right? Which says what the negative part will attach on the double bonded carbon atom which has lesser number of hydrogen. So these two are the double bonded carbon atom and the negative part is OH minus. So obviously OH minus will attach on this carbon and H plus will attach here and you can directly write down the product here is CS3 whole thrice CCHOH single bond CS3. See we can discuss the mechanism of this also but I am not discussing the mechanism now it's there in alcohol chapter. So we'll discuss it there only. Understood? One important thing here we have it is what? See we are using here H2O. This is important you must take care of. We are using here H2O. So this OH minus this H2O will give OH minus the nucleophile. Nucleophile we are getting from here and the H plus we are getting from this reagent NABH4. So here what happens in some of the reaction? Instead of this water we can have some alcohol also. So that reaction will write it down but first you see the point is whatever the molecule you have here according to that only will get the product. We are adding OH minus here but always it is not necessary that OH minus will be attached over here. It depends what we are taking here. If you are taking H2O then OH minus will attach. Okay Tridev we'll send you this. Send me this on WhatsApp better. Tridev send me this question on WhatsApp. Okay I'll send you a solution. Anyway we'll finish this. The point here is what? Next point you write down continue with this. Write down if we use alcohol if we use alcohol instead of water instead of H2O as nucleophile as nucleophile then the reaction is known as then the reaction is known as coxy alcoxy mercury reaction or demercuration reaction and it forms ether. Now you see this reaction. We have CH3 double bond CH2 and the reaction the reagent we are using here is first reagent we have HGCH3 COO whole twice with THF and here we are not using water but suppose here we are using alcohol CH3OH and the second reagent is same that is NEBH4 with any base any OH right. So if you have water here then H2O will give what OH minus if you have CH3OH so the nucleophile for this alcohol is what it is CH3O minus. So instead of OH minus this will attach everything will be same Marconi Koff rule and all. So here you see the product in one step if you write CH3CH H plus will attach here right and negative part of the reagent is what OH CH3 minus so we will get OH CH3 here understood this. Suppose one more reaction I will write down the last one is this we have this ring double bond and CH3 here. In this the reagent we are using is HGCH3 COO whole twice THF C2H5OH and the second reagent we have is NEBH4 with OH minus. What is the product we get again what happens Marconi Koff rule the nucleophile is what here it is C2H5O minus so that will attach according to Marconi Koff addition. So here we will have OH C2H5 and this will be CH3 as it is so this is the product we get understood this tell me did you get this okay so this is it for alkene okay next class will finish alkene okay one more class probably will take okay alkene is not that much big okay we'll finish that so we'll finish alkene in the next class fine we'll wind up the class here only see you soon correct revise all these reactions that we have done okay it is a bit confusing if you do not revise you will forget okay so don't forget to revise solve some questions go through ncrt okay thank you all bye