 Hello. Hello. Good afternoon. All are there. Can you please type in your name? Can we start now? Please let me know if you are there. Okay. So yesterday we have discussed about the preparation of alkene. Okay. So we have finished the various preparation methods of alkene. So today we are starting with the properties. Okay. So the heading you write down is the properties alkene. So again we will see physical and chemical properties. Again in chemical properties we have reactions. Okay. First of all you write down here the various physical properties we will see. So in this all the molecules in which the carbon number of carbon atom is from 1 to 3 from C1 to C3 carbon atom present in the molecule. All these molecules exist in the gaseous form. Right? So ethene, propene, all these molecules are belong to the gaseous form and at room temperature. So from C4 to C17, C4 to C17 the molecules exist in liquid state and then if the number of carbon atom is more than 18 or equal to 18 the molecule will be solid. All these states exist at room temperature. Okay. Room temperature. Next point you write down next one is except ethene, next property except ethene all molecules are all molecules are orderless. Okay. Ethene has pleasant order. Okay. Not that much important. Ethene has pleasant order. All other molecules are orderless. Okay. Next one you see the boiling point and melting point. The boiling point and melting point is higher than corresponding alkanes, higher than the corresponding alkanes. Right? This is the another property. Right? The pattern or the order or the you know the order of the boiling point and melting point it is same as for the alkanes we have and like alkanes we have discussed that the boiling point is directly proportional to the molar mass. So here also we have the same thing. Boiling point and melting point increases molecular mass increases. Okay. This is the same thing we have which we have done in alkanes also. Okay. Now since we have alkanes so these alkanes also shows geometrical isomerism means they have cis and trans isomers possible. Right? Cis and trans isomers possible. So according to that if you write for cis and trans isomers what we can write and this point is important here that if we talk about the boiling point. So the boiling point of cis isomer, boiling point of cis isomer is more than to that of trans isomer. Boiling point for cis isomer is more. Right? Melting point for cis isomer is less than to that of trans isomer. Why trans isomer will have more melting point because of symmetry and we can also say better packing here. We have better packing here. That's why the melting point of trans is more than to that of cis. Third point is what? Stability of cis isomer. Stability of cis isomer is less than to that of trans. Right? Trans will have more stability because of the less hindrance. Trans isomer will have less hindrance. That's why more stability. Okay? So cis, trans isomers, the stability will be more. Melting point will be more but boiling point will be less. Just a second. These are the few physical properties we have seen and in all these properties these three properties are very important. The last three properties. Okay? This you must keep in mind. You will definitely get question under this. Okay? Now we are going to see the chemical properties. Okay? Next one I write down chemical properties. See for alkene, since we have carbon-carbon double bond here, this is the formula representation we have of the smallest alkene. Okay? So since we have carbon-carbon double bond and we know pi bond is comparatively weaker bond, right? So this kind of pi bond if it is present, then most of the alkenes shows addition reaction. So what we can write, addition reaction, addition reaction are the most important reaction, important reaction for alkenes. Double bond breaks and addition takes place, right? So you see here, if I write down the general thing, C double bond, C we have, right? So what happens in this when the homolytic fission takes place? Homolysis of the pi bond. Then we get the free radical which is nothing but this, right? This is the free radical we have homolysis fission. Okay? In this if heterolysis takes place, heterolysis, then what happens? One carbon will have the positive sign and other one will have the negative sign, right? Unsymmetrical distribution of the pi bond we have here, pi electrons we have here, right? So this kind of distribution, unsymmetrical distribution, this is possible in polar solvents, possible generally in polar solvents, right? At low temperatures, right? So whatever this thing we have polar solvents and all, generally we observe this property, right? So most of the time it is true, but it is not the only thing, okay? So homolysis takes place or the symmetrical distribution takes place in presence of, in presence of light H nu free radical formation in non-polar solvent, non-polar solvent, high temperature, right? Non-polar solvent at high temperature, okay? This is the general thing which you should keep in mind. Now the next thing which is again very important we have and that is the reactivity order of alkene. Before going into the reaction, we'll first see one or two facts here. Reactivity order of alkene. And yesterday also we have discussed that more substituted alkene, more substituted alkene, alkene is more stable, less reactive, right? More stable and less reactive. So if I write down the reactivity order here, the reactivity order will be this. H2C double bond CH2. This is not substituted alkene, right? So this is what this is least stable and hence it is most reactive. So I'm writing down the reactivity order here, okay? This is the reactivity order. One hydrogen if it is replaced by an alkyl group, RCH double bond CH2. So this is substituted alkene, less reactive than this. Another one, if I write down RCH double bond CHR and this is almost equal reactive than this one. R2C double bond CH2. These two the reactivity order is almost same and this is more than R2C double bond CHR and this one is even more than CR2 double bond CR2. So this is the most substituted alkene we have. So this one is the last one is most substituted and hence this is most stable. Most stable means least reactive. So this is the reactivity order we have. You keep replacing the hydrogen atom the stability will increase but the reactivity will decrease and that is what the order we have. This one is important reactivity order. You must remember this, right? Now next you see the chemical reactions of alkene, right? Down next chemical reactions of alkene, okay? The first reaction is the first reaction is hydrogenation reaction, hydrogenation. Hydrogenation reaction we have discussed in preparation of alkene, discussed in preparation of alkene, okay? Like I said you will see the same kind of reaction in different different chapters, okay? This hydrogenation reaction it follows free radical mechanism, right? And the different reaction we have here and that we have discussed I'll just write down the name. The first one is subartier-sendrance reaction used for the preparation of alkene. Burts reduction also we use I have discussed this yesterday only. Burts reduction, right? Hydrogenation reaction is exothermic in nature, exothermic in nature and the stability is inversely proportional to the heat that is liberated in this heat of hydrogenation. This is also we have discussed, so we will not discuss this again. Heat of hydrogenation. So this is the reaction we have where the addition of hydrogen takes place. Like suppose if you have this compound eH2 double bond CH2 plus H2 in presence of any catalyst like nickel, platinum, palladium and all this hydrogen atom attached to this carbon atom and we'll get CH3, CH3. We have discussed this reaction in the preparation of alkene. So you can go through those reactions, the same reactions we have here. There is one more important reaction here we have and in that you write down cyclohexene, write down this point cyclohexene when heated with when heated with ethene oxidizes itself, oxidized itself to benzene forms ethene. So if I write down the reaction here cyclohexene is nothing but this plus when this reacts with H2C double bond CH2 so this gives you benzene to hydrogen atom comes out from this and those hydrogen atom attached to this carbon atom will get CH3 and CH3. So this is the reaction we have. So this is the hydrogenation reaction, the first reaction we have. So what we have discussed, check your notes preparation of alkene. Preparation of alkene, check your notes first. We discussed this subartier syndrome reaction. If not, then you write down in this, first of all the spelling is this A B A T I E R D E R E N S. Yes, correct Shreya. Subartier syndrome reaction is the hydrogenation of alkene hydrogenation of alkene in presence of a catalyst like nickel, platinum we can also use palladium here or we can also use rhodium. Okay. So it is the addition of hydrogen to alkene in presence of these catalysts. Correct. Next reaction you write down B that is halogenation reaction. Halogenation generally we take addition of bromine or chlorine. Addition of bromine and chlorine. And for this purpose we use inert solvent like carbon tetra chloride. For halogenation we use an inert solvent. The reaction takes place in an inert solvent and that is carbon tetra chloride C C L 4. When we have alkene right C double bond C so this alkene may be what may be cis or may be trans or we can say what may be symmetrical or unsymmetrical may be symmetrical unsymmetrical that is cis and trans. Okay. So in this the addition of this you know this halogen atom takes place like this. Suppose I am taking this bromine that is Br Br right. So in inert solvent what happens right in inert solvent there will be hydrolysis of pi bond. Okay. So what happens here one of the bromine will take the two electron will have the negative charge on it and other bromine will have the positive charge like this the electron transfers. Okay. Now if you have this carbon-carbon double bond so in this also when this pi electron shift over here so this negative charge will take this positive Br plus iron and we get C single bond C Br. We have bromine here and the two bond but here we have the positive sign now on this positive sign this bromine will attack over here and will get the product as T single bond C Br and Br right. This is the general reaction I have given you okay. But the addition of two bromine this addition of bromine atom may be from the same side that is sin addition or may be from the opposite side that is anti-addition both are possible. So when it will be sin, when it will be anti that we have to think okay and in both addition will get different different types of compounds here. Okay. So first of all you see you just write it down one word will use here that you did not understand now because it is there in isomerism you just write it down we will discuss this in isomerism okay. So if you have cis alkene cis alkene in cis alkene if sin addition takes place sin addition then it gives mesocompound M-E-S-O mesocompound if you have cis alkene and in this if anti-addition takes place then it forms resmic mixture have you heard this term reso and resmic meso and resmic if you have trans alkene and on trans alkene if sin addition takes place it gives resmic mixture resmic mixture trans alkene anti-addition mesocompound no I haven't explained this in an anti before it's there in again in isomerism okay. So it's all these things that I'm using now mesocompound resmic mixture sin addition anti-addition all these are there in stereo isomerism right that's why we haven't done this right but we'll do this for you to understand you see what is the meaning here here we have one side growing the product is this in this what we can say the both bromine atom is attaching from the same side the same side of that carbon carbon single bond so when same side addition is there this is sin addition but if the addition is like this carbon carbon the bromine we have this side and another one the bromine we have this side so both bromine atom is attaching from the opposite side one is from this side one is down the carbon carbon single bond another one is above right these are from the opposite side both bromine atom attached from the opposite side so this one is anti-addition this is what the meaning of sin and anti so this mesocompound and resmic mixture you let it be now you just write it down it's not there in both thing right so it is for J okay so when I discuss this in isomerism you will understand this what is the meso and what is resmic mixture okay just write it down in your notes we will discuss this later on it takes some time that's why I am not discussing this here right not useful for the board exam fine can we move on stereochemistry we will discuss after ionic equilibrium okay so you don't worry if you are not getting this okay we will connect all these things when we discuss stereochemistry don't worry with that can we move on tell me okay so next write down addition of halogen acids addition of halogen acids what is halogen halogen acid it is HX halogen halide okay so the order of reactivity of halogen acids are this write down the order of reactivity I is the most reactive and then HF order of reactivity is this you have to memorize this okay this reaction follows follows Markovnikov's rule what is Markovnikov's rule write down this first what is Markovnikov's rule tell me have you started this the negative part of the addendum will attach to the carbon with least number of hydrogen write down the definition of Markovnikov's rule write down Markovnikov's rule okay I am dictating addition of HX to an alkene in the addition of HX to hydrogen atom attached to the carbon atom the hydrogen atom attached to the carbon atom which has hydrogen atom attached to the double bonded carbon atom you write down then the hydrogen atom attached to the double bonded carbon atom which has which has less number of or lesser number of hydrogen atom which has lesser number of hydrogen atom so what did you write the hydrogen atom did you write hydrogen atom or the negative ion what did you write can you tell me hydrogen atom or addition of hydrogen atom or negative ion hydrogen atom hydrogen atom. Okay, okay, okay, then it will be something else. Okay, so what you just, you do this one correction here. Instead of hydrogen atom, you write down negative part of the reagent, negative part of the reagent. Okay, both way, you can actually define when you write hydrogen atom, then it will attach on the carburetor, which is more number of hydrogen. So both way you can write. So you just change this over and negative part of the reagent. And after this, all things are same. Instead of hydrogen atom, you just write down this negative part of the reagent. Okay, this one is important. Okay, so this reaction, addition of halogen acids, generally follows Markov-Nikov rule. Okay, but we have some exceptions into this that we'll discuss now only. Okay, but before that, you see some reactions over here. Okay, so here you see, first of all, if I take the example of unsymmetrical alkene. Okay, so here you see, with example, we'll try to understand for unsymmetrical alkene. For example, you see, if the molecule is CH3CH, double bond CH2, and the reaction is taking place with HI. Then the product will be, in this HI, we have what? H plus and I minus. So addition of H plus and I minus takes place. What is, here we have Markov-Nikov rule is what? Negative part of the reagent. So negative part is nothing but I minus. So this I minus will attach on the double bonded carbon atom, which has lesser number of hydrogen. So obviously the carbon atom should be this. So what happens here? The product will be CH3CHI single bond CH3. Right? So in unsymmetrical alkene, what happens? You will get only one product. Okay, this is the major product. If the another product, if you try to write down, that will be CH3CH2CH2I. Okay, so here what happens? Since the alkene is unsymmetrical, so we will not get this product. Okay, this product not formed. We have only one product here, which is nothing but this one. Okay, another way also we can define Markov-Nikov rule. This product is according to Markov-Nikov rule. Right? How this product forms? This pi electron shift over here and we get what? We get a carbocation here. That will be CH3CH positive charge into it single bond CH3. This negative charge will take this H plus. Right? So what we can say and how we get this product, this product we can get when this pi electron shift over here and we will get the carbocation here. So this carbocation is one degree, but this is two-degree carbocation. Okay? So Markov-Nikov rule we can define like this also. One thing is what? Like I have given you the negative part of the reagent attached to the carbon atom, which has lesser number of hydrogen. The another way is what? According to Markov-Nikov rule, the reaction proceeds with the formation of carbocation and that carbocation will be the stable carbocation. Stable means what? We have two possibilities here. Either we will get one-degree carbocation or we will get two-degree carbocation. So when the reaction follows Markov-Nikov rule, the carbocation that we get into this, that will be the stable carbocation, comparatively stable carbocation. So here we have one-degree and two-degree. Two-degree is more stable than one degree. So this reaction is followed up, followed by the formation of two-degree carbocation and this what happens? I minus will attach onto this and we get this product. This is Markov-Nikov rule. So Markov-Nikov rule we can actually define by two ways. One is by this that I have given you. Another one is reaction follows with the formation of more stable carbocation. Remember, we get only one product here. This product will not form at all. Understood? Now another thing you see, another one is what? In case of symmetrical alkene, what happens? So one example of symmetrical alkene you will see. In symmetrical alkene, we will get the mixture of product. That is the difference we have here. In case of symmetrical alkene, we will get the mixture of products. For example, CH3 CH2 CH double bond CH CH3. This is the reactant we have plus HI. And one more thing here, don't get confused with this symmetrical term that it should be symmetry over here. Here in this, we will again study geometrical isomerism also. Symmetrical means what the terminal carbon has same configuration here. Terminal carbon should have same configuration. Here you see it is sp2 and this is sp3, but here this one is sp3, this one is also sp3. So symmetrical alkene means what? The last carbon or the terminal carbon has the same configuration or the same degree of substitution also you can see. So in this case, what happens? In one of the product, I will attach here and in another product, I will attach on this carbon atom. So mixture of product will get here. One of the product will be CH3 CH2 CH2 CH3. Here we have the iodine plus another one will be CH3 CH2 CH2 CH3 and here the iodine will attach. So in case of symmetrical alkene, symmetrical alkene means what? The terminal carbon has same degree of substitution or same configuration. We will get the mixture of product. We will get 50% of this and 50% of this. So we will get two product into this. Isomers will get here, positional isomers. Understood? Now from this reaction, we can also define Marko-Nikov rule as write down the next definition of write down this. It is the electrophilic addition. It is the electrophilic addition, carbon-carbon double bond. A carbon-carbon double bond involves the intermediate formation involves the formation of, simply we will write down, formation of more stable intermediate carbocation, more stable intermediate carbocation. So we will get more stable carbocation, we will form more stable carbocation first and then the negative part of the reagent will attach. Order of stability of carbocation is this. Stability of carbocation is we have three degree maximum, then two degree and then one degree and then we have methyl carbocation, CH3 plus. Now one question I will give you and you try to get the product into this. Suppose the molecule is this CH3, CCH3, here we have CH3, single bond CH, double bond CH2 and this undergoes the addition reaction of HI. What is the product here? You just tell me one thing. I will just number this carbon atom. You have to tell me, suppose this is 1, 2, 3 and 4. At which carbon atom I minus will attach, that you tell me. At which carbon atom I minus will attach? Second carbon the I minus will attach. What about others? Tell me the answer. Now you see this reaction first. How the reaction proceeds by the shifting of pi electron and according to Markov-Nikov rule, the shifting of pi electron will be in such a way so that we will get what? We will get the more stable carbocation. So if the pi electron shift here will get positive charge here that is 2d d carbocation. If pi electron shift here will get the positive charge here that is 1 degree carbocation. So obviously 2d d carbocation is more stable. So shifting of electrons, pi electrons will be on the what? Will be on the first carbon. So in this case you see I will write down the reaction here CH3CH3CH3CH3 CH3 positive charge here. This will take this H plus so we will get what? We will get the CS3 here and we will get the positive sign here. So this is what 2 degree carbocation and like I said that the reaction proceeds by the formation of more stable carbocation. Now here we have one thing see if one of this methyl group shift onto this carbon atom then what happens? Suppose if I write here suppose one of these methyl group suppose this methyl group will take this bond pair of electron and attacks onto this carbon atom then what happens? So you see here CH3CH1 methyl shift here. So you will have CH3 and CH3. Now since this methyl has shifted over here and it takes this bond pair of electron so obviously this carbon will have the positive sign then. Okay so here we have here we have what? 2 degree carbocation but here we get 3 degree carbocation right and we know 3 degree carbocation is more stable than the 2 degree carbocation. So this kind of shifting we call it as 1, 2 methyl shift because the methyl is shifting and this shifting takes place on its own. Whenever the possibility of forming more stable carbocation we have by this kind of shifting we call it as 1, 2 methyl shift or if hydrogen is there then hydride shift okay. So first of all from 2 degree we will form this 3 degree carbocation and then the I- the negative part of the reagent that you have which is I- here which attach onto this carbon atom which is this. So this is the major product we have in this case. This kind of shifting is very common okay and you will get this kind of shifting in organic chemistry okay. So whenever I say first of all stable carbocation forms the stability of carbocation is this. So when if you are getting 2 degree carbocation and if there are chances of forming 3 degree carbocation by methyl or hydride shift then first we'll form the more stable carbocation and then the negative part of the reagent will attack. Understood? Yes or no? Tell me yeah. Now one more thing you see I'll just do some small changes into this question okay. Suppose the question is this CH3 CH3H CH double bond CH2 HI okay. Now in this question you see what happens. First of all this pi electron will shift here and that will take this H plus okay. So we'll get what CH3 CH3H we have here CH positive charge single bond CH3. So this is what this is 2 degree carbocation. Is there any chances of forming 3 degree carbocation or more stable carbocation in this? 3 degree tell me yes. How do we get that? See I'll just write down two different possibilities here either what happens if this one of this methyl group take this bond pair of electron and attack on to this carbon atom this is one possibilities and we get what CH3 CH3 CH3 CH3 and here we have positive charge. This carbon is again it is 2 degree carbocation and we do not have any difference in these two molecules right. But in this case suppose if this hydrogen shifts over here then what happens yellow shifting I'm writing the product with yellow only CH3 will be as it is see CH3 then we have CH2 CH3 and the positive charge here. So this carbocation is what it is 3 degree carbocation which is obviously more stable than this 2 degree carbocation. So now in this one this is the most stable carb possible carbocation we have here. Now in this the I- will attack and we'll get CH3 CH3 CH2 CH3. So this is the major product we get. So this kind of shifting we call it as what 1,2 hydride shift since hydrogen is shifting 1,2 right 1,2 hydride shift in the previous example you see since methyl is shifting you see here since methyl is shifting so we are calling it as 1,2 methyl shifts. So whatever shift according to that will slide down the name of that shifting is it clear now tell me it's not complicated it's easy the only thing is you are looking this you are you are having this first time you're getting this first time you're getting these reactions first and that's why it is a bit complicated you are feeling right the point is this kind of reaction is very common in organic chemistry only one thing you have to keep in mind whenever the addition of Hx is there will follow the formation of more stable carbocation okay. So 2 degree if there are any chances of forming 3 degree carbocation from 2 degree we'll get 3 degree carbocation first and there only the negative part of the reagent attacks okay so once you revise this you will understand okay now in this also see this this one is very important reaction that's why I'm taking too much time into this in this also there is one also one more thing that you write down next write down the heading anti-marconic off rule anti-marconic off rule we also call it as peroxide effect peroxide or kharash effect k-h-a-r-a-s-c-h both are same thing see first of all two points here you just have to memorize anti-marconic off rule or this peroxide and kharash effect is applicable only for addition of addition of hbr for hcl and hi it is not applicable understood peroxide effect or anti-marconic off rule is applicable only for the addition of hydrogen bromide it is not applicable for hcl and hi this you have to keep in mind there are reasons for this but that is not required okay so you just keep this in mind that addition of hbr follows anti-marconic off rule in presence of any peroxide so first of all you write down two three points into this write down the addition of hydrogen bromide hbr the addition of hydrogen bromide in presence of peroxide in presence of peroxide follows anti-marconic off rule follows anti-marconic off rule and this is not applicable this is not applicable for hcl and hi next one more point you write down okay akash stay alive okay one more point you write down after this anti-marconic off addition is caused anti-marconic off addition is caused not only by the presence of peroxide not only by the presence of peroxide but also also by by irradiation with light irradiation with light of a certain wavelength irradiation of with light by a certain wavelength so basically anti-marconic off addition takes place what is anti-marconic off addition it is just opposite to marconic off okay there the negative part of the reagent attached to the double bonded carbon atom which has less number of hydrogen here the negative part of the reagent will attach to the double bonded carbon atom which has more number of hydrogen okay that is what the anti-marconic off addition we have okay just reverse of that okay so this anti-marconic off addition takes place in presence of peroxide or in presence of light okay this is what the meaning of this here so now you see and this is applicable only for only for hbr not hcl or hi okay so if i write down this reaction suppose ch3 ch double bond ch2 in this if i add hbr in presence of any peroxide right so h plus since peroxide is present so this will get anti-marconic off product okay so h plus will attach to the double bonded carbon atom which has more number of hydrogen so here the product will be this ch3 ch3 sorry ch2 br and here we have ch2 right negative part of the reagent will attach the double bonded carbon atom which has more number of hydrogen this is anti-marconic off rule now in this only if i write only hbr there is no peroxide then the product will be what the negative part of the reagent will attach the carbon atom which has lesser number of hydrogen so we'll get this okay so this peroxide in presence of peroxide or light h new this anti-marconic off rule will be followed only for hbr not for hcl and hi is it clear now this peroxide effect parachute effect is very important okay you must remember this for both point of view also next one i write down addition of hypo hella acid hypo hella acid means what it can be hocl or hobr okay so write down into this write down this reaction is similar to this reaction is similar to the addition of halogen in presence of water halogen in presence of in presence of water okay so we can hypo hella acid is hocl or hobr but instead of hocl and hobr we can also write it down cl2 plus h2o for br2 we can write br2 plus h2o both will have the same thing in this when it reacts it forms hocl and hcl here we get hobr and hbr and then the reaction proceeds okay so both if you write cl2 plus h2o or hocl same thing right now you see what happens in this reaction if i write down the reaction here ch2 double bond ch2 plus h o minus and cl plus the product we get here is ch2 oh ch2 cl and here we have oh oh and cl will attach the reaction is the same as that we have done this one is not free radical this one is not free radical okay what happens here you see uh we will get first of all we'll get halonium ion into this okay halide ion halonium ion we call it as this ch2 double bond ch2 we'll take this cl plus first of all this pi electron will attack onto this cl plus and we'll get ch2 single bond ch2 and the chlorine will attach with this like this with with positive charge here on this chlorine atom okay now since we have water as a solvent so we have two nucleophile here right cl2 and h2o first of all it forms cl plus and cl minus so we'll have h2o as also a nucleophile since we have lone pair present here and cl minus also but since water we are taking as a solvent here so this is present in excess this h2o is present in excess so attack of h2o as a nucleophile will be easier here okay so what happens here this h2o will attack one of this carbon atom suppose on this one right and this bond will dissociate like this so we'll get ch2 cl and then we get ch2o h2 plus right oh h2 plus from this we get finally ch2 cl here and h plus will come out so we'll get ch2 oh plus h plus this is what the product we get here there's no free radical formation in this this is the mechanism that i have given just now understood in this what will be the product same reaction we have write on the product and tell me the name what is the iutc name of the product what is the answer we'll get here pi electron will shift here carbocation will get here on that the oh h minus will attack so the product here it will be ch2 br this is the product we get next write down addition of sulphuric acid yeah correct correct with it make sure i write down addition of sulphuric acid turn into this write on this reaction this reaction is is proton initiated initiated electrophilic electrophilic addition reaction initiated electrophilic addition reaction and this follows and this reaction follows follows marconi cough rule marconi cough rule in short i'll write down mr marconi cough rule okay so you have to memorize this that it follows marconi cough rule this is important again right so with the product changes if we add sulphonic acid acid instead of sulphuric acid just i will okay sulphonic acid if you add what is the formula of sulphonic acid r s double bond o double bond oh oh h i think the major product will be same into this other products will be different but major product i think it will be same okay so you see we are using here cold and concentrated sulphuric acid 80% concentrated cold and concentrated sulphuric acid you see this reaction ch3 ch double bond ch2 when reacts with h2so4 which is nothing but h plus and hso4 minus right so again this reaction is same as the reaction we have discussed pi electron shift here positive charge here will take the hso4 molecule right so the product here it will be what ch3 ch single bond ch3 and here we have hso4 okay this is alkyl hydrogen sulphate okay now when this alkyl hydrogen sulphate treated with water if you put water here h2o this h2o will give h plus to this hso4 and forms h2so4 again and OH will attach over here so we'll get ch3 ch ch3 and we'll get here OH so we'll get alcohol in this case you write down here when alkyl hydrogen sulphate this is alkyl hydrogen sulphate write down when alkyl hydrogen sulphate treated with water gives alcohol next you write down addition of oxygen next reaction we have write down into this alkene reacts with O2 in presence of silver catalyst alkene reacts with O2 in presence of silver catalyst in presence of silver catalyst at high temperature at high temperature and pressure form epoxide forms epoxide epoxides are nothing but cyclic ether epoxides are nothing but cyclic ether reaction you see our ch double bond ch2 plus when it reacts with oxygen in presence of silver catalyst it forms our ch single bond ch2 and over here okay write down one point here these epoxides these epoxides on hydrolysis hydrolysis gives diols we'll get diols we'll get diols on hydrolysis okay means what suppose we have this compound ch ch2 oxygen and oxygen when you do the hydrolysis okay acidic hydrolysis suppose I am using so this oxygen lone pair will take this h plus and when the bond dissociate we'll get one OH here and another OH here this is nothing but the diols we have that is how the reaction proceeds okay this kind of epoxide one more note you write down kind of this kind of oxidation also takes place with peroxy acids with peroxy acids what are peroxy acids we have CF3 CO3H we can also write C6H5 CO3H right or we can also write CH3 CO3 H one structure I'll write down CH3 CO3H is what it is this CH3 C double bond O O O and H so this is what the peroxy linkage we have peroxy linkage okay that's why these are peroxy acids for all these compounds the structure is this only okay so this kind of epoxide formation takes place in case of this peroxy acids also so that you have to memorize understood this so will the products change if you are in this we have already discussed can we move on next you write down addition of ozone