 Good evening students, am I audible? Can you hear me on? Okay, so yes. So last class I think we finished alkene, right? Okay, so I think there's still more, there are some people to join. We'll wait for like couple of moments and then we'll start. Yeah, so I think we are done with alkene. Colby electrolysis is the last one, yes. All right, the action of alkene we didn't do. All right, we did only preparation, right? Fine, we'll continue with that. All right, do you get any information like when is the last class for this session? Which date? So write down properties of alkene, properties of alkene. First one, write down physical properties, physical properties. Physical property you see the first one, the physical state in physical state, what happens? The first four carbon atom, C1 to C4. The first four carbon atom exists in gaseous state, gaseous state, carbon number if it is C5 till C17. The state will be liquid and if the carbon atom is 18 or onwards, C18 and above, it is solid, solid like wax, wax like solid we have, okay? The second one we have an exception here, that exception is neopaintain, okay? In this neopaintain, if you see, the physical state of neopaintain is gas. So this is an exception we have. Neopaintain, the formula I guess you all know, it is CH3, CH3, CH3, CH3, CH3, CH3, CH3. And CH3, could you tell me the IOPACI name of this molecule? It's 2-2 dimethylpropane, yeah? Next slide down, melting point and boiling point, melting point and boiling point. Melting point and boiling point is directly proportional to molecular mass, molecular mass, right? If you have isomeric alkene, see among isomeric alkene, among isomeric alkene, melting point and boiling point decreases with increase in branch. If the branching increases, then melting point, boiling point decreases, valid for isomeric alkene. One more term we have in this, one more point we have into this. If you talk about melting point, melting point for even number of carbon atom, for molecule has even number of carbon atom, even number of carbon atom is more in comparison to odd number of carbon atom. This is valid for comparable molecular mass, comparable means if you talk about or if you compare the melting point of butane and pentane, then butane will have slightly more melting point because it has even number of carbon atom, right? You cannot think of butane and nonane, in that case, molecular mass dominates. For next higher molecule, we can compare like this because actually what happens in melting point with better packing, melting point will be more. Like you see, if you have butane, so it is like this. Butane is this, pentane if you see, one, two, three, four, five. Pentane is this, right? So here in butane, we have better packing. Hence butane, the melting point is more because both methyl group is on opposite side. No, it is kind of trans, not trans we cannot say, but when we have trans molecule, same identical group on the opposite side, then we can have better packing in this. That's why trans we always have more melting point than the cis one. Similar kind of thing we have here. Two methyl group, terminal methyl group are on the opposite side. So slightly more packing we have here, better packing we have here. Hence more, what we say, hence more we have melting point, okay? If you compare pentane and hexane, see, two, three, four, five, six. Hexane again you see both methyl on the opposite side, hence hexane the melting point is more than to that of pentane. But when you compare this two, then butane more melting point than pentane, okay? Because of better packing. Next slide down, chemical properties. Okay, we have done just one more slide before this. You can see that later on the channel, okay? And you can watch the video also in the share. First reaction we write down halogenation, halogenation. Write down, halogens reacts with alkene. I'll write down here. Halogens and mainly we'll talk about chlorine and bromine here, chlorine and bromine. Halogens mainly chlorine and bromine reacts with alkene, reacts with alkene in presence of sunlight, sunlight, UV light you can take, sunlight, UV light and in dark at high temperature and in dark at high temperature forms a mixture of substituted product. One point to write down, this reaction involves free radical mechanism. All these free radical mechanism we'll discuss later in reaction mechanism chapter, okay? So look at this reaction, what happens in this? Suppose we have methane, CH4. Methane reacts with chlorine in presence of sunlight, H-nume, H-nume in sunlight. Then what happens? This H combines with CL, forms at CL and CS3-Cl, this is the product we get. If you have excess of chlorine, then this chlorine continues to react with this molecule. Further it converts into CH2-Cl2, which further reacts with CL2-H-nume, CH-Cl3, which further converts into C-Cl4. C-Cl4, here the reaction stops. That's why we say we get a mixture of substituted product. All the hydrogen atom one by one, it's substituted by chlorine. Done, copied, H-nume is yeah. It is the energy of one photon, H-nume, light. That means light, yeah. So in the form of light, we are providing energy. That's why the carbon-hydrogen bond breaks, chlorine-chlorine bond breaks, and then the hydrogen-chlorine forms and CS3-chlorine forms. Done, copied. So remember, the reaction is free radical mechanism. And you keep that in mind that we do not have any control in this kind of reaction. Free radical reaction, we do not have any control. So whatever the products are possible, all the product forms. We cannot control this reaction. We cannot say only CS3-Cl forms. We can do this, but we need to provide only one molecule of Cl2 in the reaction then. If you have excess of Cl2, all these products will get and hence we say we get a mixture of product, okay? So this we can do with chlorine and bromine. We can find out the chlorinated product or brominated product, okay? Iodide we cannot get from this. Fluoride we cannot get from this, okay? So how do we get fluoride in this reaction? Suppose you need to form alkyl fluoride, what we need to do, right? Down, direct fluorination, all of you copy this, direct fluorination is not possible. Direct fluorination is not possible. Direct fluorination is not possible. Hence in order to obtain alkyl fluoride, right? Direct fluorination is not possible. Hence in order to obtain alkyl fluoride, we use following reaction. And the reaction is we have two C2H5Br which reacts with HgF2, reacts with HgF2 and forms two C2H5F plus HgBr2, plus HgBr2, okay? So this is the halogen exchange reaction. From one halide, we are getting alkyl fluoride. Direct fluorine, if you add in alkene, the reaction is very explosive. We won't prefer that, okay? If you want to prepare iodine, iodide, then what we need to do, right? Down, iodination is reversible, is reversible carried out in presence of, in presence of a strong oxidizing agent, strong oxidizing agent. So when you have alkene, suppose CH4 with I2, this reaction is reversible. Then it forms CH3I and HI. But this reaction, for this reaction, you have to use an oxidizing agent like iodic acid, we can use HIO3 or HNO3 also we can use. It is iodic acid, oxidizing agent. So what is the use of this? The moment HI forms, the oxidizing agent oxidizes HI into I2. So this reaction, it continues, it continuously removes HI by this oxidation reaction. Hence the reaction is forced to go in forward direction. Otherwise, because of reversible nature, it has an tendency to go in backward also, right? To eliminate this possibility to have a backward reaction, we are eliminating HI by the use of this oxidizing agent, which converts this into I2. And the hence reaction goes in forward direction always and we'll get iodide, no, not required. Only this one, oxidizing agent, that's fine, okay? Right, next slide down. The second reaction is nitration, right down, alkene reacts with nitric acid, alkene reacts with nitric acid and forms, and forms nitroalkane and forms nitroalkane. This reaction is called nitration. So in this reaction, what happens? All possibility will have here. Suppose we have a reaction CS3, CH2, CS3, and the reaction is carried out in presence of HNO3, nitric acid, nitrating agent is this, at around 400 degrees Celsius. You don't have to memorize the temperature here, not required. At all possible position, you remove H and attach NO2. So we can have one of the possible product is CS3, CH2, CH2, NO2. Another one is CS3, CH2, CH3, and at the middle carbon, we have NO2. If you attach NO2 here, it is nothing but this product, but one more product is possible when you break the carbon-carbon bond, CS3, NO2, and CS3, CH2, NO2. This is the possible product we get because we are having this reaction temperature very high, 400 degrees Celsius, and this is good enough to break the carbon-carbon bond also. That's why carbon-carbon bond also breaks and it gives all possible nitroalkanes here. Not very important. Third reaction I write down. Third one is isomerization. See, the reaction of alkane are very direct reaction. You just need to know the reagent and write down the product. Okay, mechanism is not important here. We won't discuss that, not required actually. Isomerization is what? Lower alkanes, see what happens? Write down butane or higher alkanes. Butane or higher alkanes when heated with, when heated with AlCl3 at around 400 degrees Celsius, at around 400 degrees Celsius, converts into, converts into stable isomers, isomers by rearrangement reaction. So what happens in this, you see, we have CH3, CH2, CH3, this when heated with AlCl3 at around 400 degrees Celsius, then this hydrogen comes over here and this methyl shift over here. Generally you will get chain isomers in this. I know it is difficult to understand how to write down the product. That's why I'm telling you. You just remember chain isomers you need to write down and just memorize the reagent here. When you get this question with option, you can easily identify that what could be the possible product. Actually, when you heat this, then what happens? This hydrogen jumps over this carbon and then this carbon-carbon bond breaks and this carbon gets attached to this carbon. You'll get this. Next slide down, combustion. Two more reactions we have. Fourth one, write down combustion reaction. Combustion, we all know. It is a reaction of with oxygen. Alkene when reacts with any hydrocarbon, for example, reacts with oxygen, goes under combustion reaction and forms CO2 and H2O, always. Hydrocarbon on combustion reaction gives you carbon dioxide and water always. You can balance this reaction. X, C, Y is here. So we have Y by two. So it is X plus Y by four. The balanced reaction is. This reaction is always exothermic. Energy releases always in combustion reaction. So always exothermic delta H is less than zero for combustion reaction. Next one we have, the last one, that is aromatization. This one also, you just need to know the reagent. So what happens in this? Suppose you have n-hexane, n-hexane. N-hexane, when heated with the reagent, say, CR2O3 with AL2O3, CR2O3, AL2O3, if you heat this. CR2O3, AL2O3, if you heat this. This reagent means aromatization you need to consider. Six carbon atom here, with six carbon atom, the aromatic compound that we have is benzene. This is what we get. If you take n-heptane, I hope you all know what is aromatic compound with the same reagent. Obviously the aromatic compound with seven carbon atom, it is benzene only, you get benzene. But since we have five, seven carbon atom here, so along with benzene, we'll have one CS3 present on this. Okay, the name of this compound, the common name is toluene. If one methyl group present at benzene, it is called toluene. If you have n-octane, eight carbon atom, with the same reagent, so we have two possibility here. One is, obviously we'll have the benzene thing. One possibility is you have one CS3 on the top, one more CS3 over here. Another possibility is, we'll have one ethyl group present on this ring, ethyl benzene. This two possibility here. Benzene thing we always get in this reaction, always. No, 10 carbon aromatic ring is not possible. It may form, it's not like it's not there, it may form, but depends again. If it is aromatic, then it may form. We can think of para, we can have isomers of this. No, for aromatic compound, we must have at least benzene ring. It's not like lower carbon atom aromatic ring is not possible, we have other things for that. Like this molecule you see, it is an aromatic compound. But this kind of ions we won't get in this reaction, so we are not talking about this. Even three carbon atom with this, it is also an aromatic compound. So ions we are not considering. You don't worry much here. That is fine, you can get the isomer of this CS3, maybe here, maybe here, that's fine. In this reaction, don't think much, we just have, you know, this three reaction. You always get the derivative of benzene in the reaction. Other things you won't get here, okay? So this is it for alkene, okay? Like I said, we don't have much reactions in alkene. In alkene, alkene, we have more number of reactions in comparison to alkene. So next slide down, we are going to start the next second part of this chapter, that is alkene. Heading all of you right down, alkene. You know, alkene, the general formula, we have CNH2N, degree of unsaturation DOU is one, okay? One carbon atom, at least one carbon atom is as P2 hybridized, okay? What are the different methods of preparation of alkene? Methods of preparation. Methods of preparation, we can prepare alkene from halides, okay? Alkyl halide we can take, we can take mono halides, we can take dihalides also. We can also prepare alkene from alkyne. So first reaction you write down, from alkyne, from alkyne. What happens in alkyne, you see? Suppose you have a molecule, RC triple bond CH, this is an alkyne. On hydrogenation H2, we use a catalyst here, right? That is PDBSO4, palladium BSO4, right? It forms our C double bond C, H, H and H. Or do one thing, okay, you'll get like this. The two hydrogen will get attached to these two carbon atom and one pi bond breaks, okay? You look at this reaction. If you have RC triple bond CR dash, another alkyne, reacts with H2 in presence of PDBSO4, the catalyst we are using. Here you'll get cis-alkene. Cis is this, RC double bond CR dash and both double bonded carbon atom has hydrogen attached on the same side, okay? So this is cis-alkene we get. But you see here, if the same molecule, if the same molecule reacts with H2, in presence of any liquid NH3, the catalyst we are using. Sodium with liquid NH3, the catalyst we are using. This reaction gives trans-alkene, RC double bond C, R dash H on the opposite side. This is trans, okay? So you have to memorize this reaction. It's important one. Now what you have to keep in mind here that this reaction, H2 with PDBSO4, this we call it as a catalyst called Lindler's catalyst. Lindler's catalyst it is. This reaction here, it is called Burge reduction. B-I-R-C-H, Burge reduction. So Burge reduction gives you anti-addition, trans-product you get, Lindler's catalyst gives you cis-addition, cis-product you get. Burge reduction and Lindler's catalyst. Lindler's catalyst gives you cis-alkene, Burge reduction gives you trans-alkene. Now, the reaction we have right down from alkyl halides. Second method of preparation, alkyl halides. Right down alkyl halides goes under elimination reaction. I'll write down it. Alkyl halides goes under elimination reaction in presence of a base, in presence of a base and forms alkene and forms alkene. So the reaction here is, suppose we have C-H3, C-H2-X. This compound is heated with alkoholic KOH, for example. Alkoholic KOH. KOH is a base, alcohol is a solvent we are using here. Okay, so it forms C-H2 double bond C-H2. So what happens here? This HNX combines and forms this compound, alkene. Okay, copy down this. Yeah, so see here, okay, KOH, does KOH do here. See, actually what happens, you see, we have this KOH in presence of an alcohol. It is a part of mechanism actually. Suppose I'm using an alcohol here, C2H5OH. Okay. It reacts with KOH, reagent is this whole. This mixture is the reagent, alcohol with KOH. So what happens, you see, KOH is a base, it is an acid. So it forms water, first of all, and the salt. That is C2H5O- and K+, now this behaves as a base and takes hydrogen from beta carbon. You see, this carbon, which is attached with the alpha carbon, like halogen, which is attached to the carbon atom, that carbon atom is alpha carbon, and a distant one is the beta carbon. So what happens here? The base that we have, which is C2H5O- is the base, it takes this hydrogen from the beta carbon. This sigma forms here by an X goes out as a living group, you'll get this. So along with this, what you will get, you'll get C2H5OH, the alcohol, plus this X is taken by K, it forms KX. This is the total reaction we have. You can take KOH, you can also take another alcohol, any OH also, you can take another base, any OH also, you can take. Another example in this only you see, CH3CH2, I'll write down CH2 as this, CHClCH3. Now in this one, two different product possible, because it is unsymmetrical one, KOH with alcohol, two different product possible here. When this two combines, you'll get what, a pi bond here. So one possibility is CS3CH2, CH double bond CH2, this is one. Another possibility is when CL takes left wall hydrogen, this one, this is also possible, because two beta carbon we have, this one, this one, two beta carbon. Another product possible is CS3CH, double bond CH3CH3. Now in this one, one product is major, other product is minor, okay? So which alkene is the more stable one, first one or the second one? Second one, that is because of hyper conjugation, right? More alpha hydrogen, yes. So this one is major product, the one which is more stable is the major product, the one which is lesser stable is the minor product we have. So when we have the possibility, like these two products to write down, we'll write down both the product. We can also conclude one thing, you see the two beta carbon we have? This one has three hydrogen, this one has two hydrogen, or we can say this beta carbon is more substituted because we have one methyl group present over here. So in order to get the major product, what you need to keep in mind that to get the major product, we take hydrogen from the beta carbon which is more substituted, or we can say we take hydrogen from the beta carbon which has the lesser number of hydrogen present on it, okay? So we have two hydrogen here, we have three hydrogen here. So from this we'll take out, we'll get this. This rule, we call it as SETJF rule, S-A-Y-T-Z-E-W-F, SETJF rule, right? The product according to this rule we get is called SETJF product. What is SETJF rule right down here? Hydrogen comes out from the beta carbon, hydrogen comes out from the beta carbon, which is more substituted, okay? SETJF rule, hydrogen comes out from the beta carbon, which is more substituted, okay? This is the monohalide reaction. Another one you write down, third one from dihalides to halogen atom. Dihalides also we have of two types. As you see, R, C-H, X and X, this is one type of dihalides. Another one is R-C-H, C-H-2, X here, X here. Basic difference in the two molecule, we have both are dihalides only. As a beta carbon is the carbon atom, which is adjacent to the alpha carbon. Like you see, this carbon atom is attached to chlorine, attached to the functional group. This is the alpha carbon, and the carbon which is attached to the alpha carbon is called beta carbon. So this one and this one, okay? Right, you see this two structure that we have both are dihalides only. When the two halogen atom attached to the same carbon atom, we call it as gem dihalides, gem dihalides. When two halogen atom present at the adjacent carbon atom, we call it as visceral dihalides. We'll have many reactions for this in organic chemistry. Must remember the name, visceral dihalides, okay? So write down gem dihalides, gem dihalides when heated with Na, zem dihalides when heated with Na in presence of ether, in presence of ether forms higher alkene. Presence of ether forms higher alkene, okay? Again, I'm repeating. Zem dihalides when heated with Na in presence of ether forms higher alkene. See this reaction? Suppose we have RCHRCHX2 heated with two Na in fact, and we'll take two molecules of this. So another molecule is this. It is very much similar to Woods reaction actually in presence of ether. So what happens in this? Two X and this two X takes the two sodium molecule and forms two NaX plus this carbon and this carbon will get attached with a double bond. Oh, just a second, one correction here. It's not Na, it's zinc actually. Here also ZNCl2. Correct this? Two ZNX2, heated in presence of zinc, right? So similar reaction we have like we had in Woods reaction. What is the product we get into this one? Two CS3 CHCl2 with zinc. It forms ZNCl2 and the two molecules of this combines the double bond CS3 CH, double bond CHCH3. Yes, it is one of the gem halides. We can also form with viscinal. I'll tell you what to do. Similar reaction we have from viscinal dihalides, you see. It is heated with zinc dust for better surface area. We are taking finely divided zinc, zinc dust heated at 300 degrees Celsius. It forms RCH double bond CHR and ZNX2. This is the product you get. Okay, so it only forms alkene when we have adjacent halogen atom present. If it is not at adjacent position then what happens you see? Suppose we have a compound CH2, CH2, CH2, X and X. It is not at the adjacent position, the two halogen we have. Can be different also, that's not a problem. Can be same, can be different. Our group can be same, can be different, okay? Now when this if you heat with zinc dust then you won't get an alkene but you'll get a cyclopropane here. This bond breaks, this bond breaks and this two carbon atom joints you'll get cyclopropane. It forms actually radical and then the two radical compounds. Next I done preparation method from alcohol. I done alcohol at 160 degrees Celsius, forms alkene by dehydration. Dehydration is removal of water. So water eliminates out and forms alkene. So what happens in this reaction you see? I'll take the simplest example. We have CH3, CH2, OH. So when heated with H2SO4 at 160 degrees Celsius it converts into an alkene CH2 double bond CH2. How it happens you see? Acid gives H plus and this lone pair on this oxygen takes this H plus, right? So it forms CH3, CH2, OH, H and positive charge over this. This we call it as protonation. Addition of H plus on alcohol. It is protonation of alcohol. Then you see what happens very important step. This oxygen is electronegative and positive charge on oxygen is not stable at all. Oxygen wants to get stabilized. So what it does? It takes this electron pair in which one electron of itself and one electron of this carbon. It takes both electron pair and goes out as H2O. So H2O eliminates, goes out and forms CH3, CH2 and positive charge on this. And then when you heat this from the adjacent carbon H plus comes out, leaving its electron pair behind and it forms CH2, CH2 plus H plus. And this H plus taken up by HSO4 minus and forms H2SO4 again, right? So acid is not getting consumed. The concentration of acid won't change in this reaction. First step, it gets consumed, H plus. And the last step, the H plus gets released, which maintains the concentration of an acid. So basically acid is not taking part in the reaction. It won't get consumed. So we call it as acid catalyzed reaction, behaving as a catalyst, is not taking part. Acid catalyzed reaction, understood? Now the key points of this reaction is right down the first point in this. Right down, it is an acid catalyzed reaction. It is an acid catalyzed reaction. Second point, involves carbocation as intermediate. Involves carbocation as intermediate. Rearrangement of carbocation possible. Rearrangement of carbocation possible. Rearrangement means what? First point was it is an acid catalyzed reaction. Acid is behaving as a catalyst. So it is an acid catalyzed reaction. Suppose you have this molecule. We have CH3, CH, CH3, CH, CH3OH. When it is heated with H2SO4 concentrated, what happens in this reaction? What is the product we get? Anyone try this? Okay, so the product here you can write is this. CH3, CH CH3, CH double bond CH2, yes? Or you can also write CH3, C CH3, double bond CH CH3. Double bond CH CH3, yes? Let's see how to do this reaction. First of all, what happens? I'm just discussing the mechanism here. Remember in detail, we'll discuss this thing again in reaction mechanism, but how it happens you see. H plus comes from the acid. Then this lone pair takes the H plus, protonation of alcohol. So what we get? We get CH3, C CH3, HCH, CH3OH2, positive charge on it. So like I said, oxygen posture is not stable. So what it does, it takes this bond pair of electron and goes out as H2O. H2O plus CH3, CH CH3, CH CH3, and positive charge on this, right? And then you can eliminate one hydrogen from here or one hydrogen from here. You'll get this two product, right? But this is not the exact mechanism here. What happens further in this? There will be rearrangement of carbocation. Whenever carbocation forms, we always try to get the more stable carbocation. If that is the right data mining step. So all these things, like I said, I'm just giving you the most important part over here. Detail I'm not going. That we'll discuss in reaction mechanism. So here this carbocation forms. What are the factor here which stabilizes the carbocation, could you tell me? Can we say hyperconjugation? Yes. How many alpha hydrogen we have here? Number of alpha hydrogen? It is four, yes. Three, this one, and one this one. Alpha carbon, the hydrogen attached with the alpha carbon is called alpha hydrogen. So CH. One hydrogen is this, another hydrogen is this. Three plus one, four. Okay. Now imagine this. Imagine this, this hydrogen will take this electron pair and rearrange itself onto this carbon atom. Then what happens? Hydrogen takes both electron here. One of carbon, one of hydrogen itself. Since carbon is losing one electron, then we'll have a positive charge on this carbon atom. A new carbocation we get. Can you tell me the alpha hydrogen in this carbocation? Eight. So which one is more stable? Obviously the second one is more stable. Right? Second one is more stable. Since second carbocation is more stable, then this carbocation automatically converts into this one. In the reaction, it will always happen. We cannot do, but the reaction never give you the final product without this step. It happens always. So in all reactions, wherever carbocation is forming as a rate, in the rated mining step, we always have this arrangement possible. There are various factor which stabilizes carbocation. This is one of the factors. We call it as one to hydride shift. It is one comma two hydride shift. Why hydride shift? Because hydride ion is shifting, hence one hydride shift. So because of this one comma two hydride shift, we are getting more stable carbocation. Further, if it is possible, again we'll do. We'll follow this step till we'll get the most stable carbocation. If you do not get more stable carbocation, again, you can think of shifting of this hydrogen here. You can do that. And then you will check the stability of the new carbocation forms. If it is more stable than this, then that step always will also happen. Will also occur. If not, then it won't occur, right? So this is the most stable carbocation we get. Now from this, either from this carbon, this hydrogen comes out or this hydrogen comes out. So finally, the product would be this only, this or this. There's no change in the product. One second, one second right here. Finally, the answer would be this, but the path is this. Okay, you cannot form this carbocation and eliminate hydrogen from here, and here you'll get this. After this, this will happen coincidentally in both the cases, whether you do this or not, you're getting the same product here coincidentally. But this step always happens, okay? Just a second, Aitya, I'll come back to your doubt. Suppose you'll have here methyl group present. For example, I'm taking here suppose methyl present. Then this we call it as one two methyl shift. Suppose you have a phenyl present. Here we have phenyl present. Then this is one comma two phenyl shift. The purpose of this kind of shifting is to get more stable carbocation in the next step, which is this one, right? So in detail, we'll talk about it in reaction mechanism, but this is enough for now. Yes, now what is the doubt we have? So but how can hydrogen take the electron from carbon? See, actually what happens, it doesn't happen this way. I have this, we just understand it this way. Actually, this positive charge attracts this sigma electron towards this side. And this attraction is so, what we can say is so great that finally hydrogen comes out and attached to this carbon atom. So whenever we have a decent positive charge or vacant orbital like this, then this electron pair attracted towards this and hence this shifting happens. All these things happens in the solution, okay? In solution, this happens. I understand it is a bit difficult to digest that how it happens, like we whenever, like whatever we want, we break the bond this way. But organic chemistry is like this only. You should take this as, okay, this is what is happening in the solution. This is your learning, okay? Obviously in the solution, what is happening that we cannot understand theoretically. You have to do that experiment in the lab. You have to check what product we are getting. If the product we get this, then the possibility is what? Then the possibility is that the reaction may proceeds via these steps. How do you get to know that? I can talk about it one more hour if you want, okay? This kind of shifting we got to know about this shifting because when we draw the graph of this reaction, we get two different intermediates here, like this we get, two different intermediates we get, one is this, other one is this. So this intermediate means what? We are getting two intermediate over here. That's why this kind of shifting we can conclude, okay? Maybe the shifting is happening because carbocation is getting more stability. So there are so many research behind this kind of conclusion that we draw for a reaction, okay? So actual thing if you want to understand, it is a lot of thing that we need to discuss. This graph is not at all required, but how do we get to know the two different intermediates we have? Graphs gives us this information. This particular point that you have, it gives you the number of intermediates. This is reactant, this is product. So if you have two intermediate here, it means this kind of shifting is possible. So there are so many things I understand. It is a bit difficult to understand this particular chapter. It's not about it is tough, but this is the first chapter in which we are doing this kind of reaction. So slowly once you practice some questions, you read out theory books on organic chemistry, you will be understanding this, like what it is happening and how it is happening. So organic in the first case, you should take this as the story, like, okay, this is what it is going on. It is more like a story if you're starting it first time. That's why once if you finish, you'll only get 50% of it, 60, 70% of it, okay? Twice you need to finish organic chemistry before your exam. If you really want to do well in the exam, because second time if you do, then you will be start connecting with it. Okay, this is why this is happening. Okay, I've studied this point over there also. There also it happens this way. Okay, so always keep this in mind in the book, whatever reaction is given, all reactions are done in the lab. It is proved that this reaction gives you this. So first we get the product and then we have the explanation of it. Then we have the mechanism of it. Then we have the steps involved in it, right? So we say, okay, this is the possible way by which the given product is forming. So it is actually other way. Okay, so what we can understand here that whenever carbocation forms in the rate determining step, I'm using this term rate determining step. We haven't discussed it. Don't bother about it, just ignore. We'll discuss this later, okay? So whenever carbocation forms, we'll try to get more stable carbocation. First by rearrangement. If it happens, fine. Otherwise we'll write down the product simply, okay? So this is the answer we get. All of you understood this. In this too, obviously this one is a major product because more stable alkene will write down. And this one is the minor product. Any doubt in major and minor? Any doubt in major minor product? Product one cannot be right. It is product one won't form, yes, correct. Product one won't form, so I was about to do it. It is not possible in this. That's right. Product one won't form. What is the another product we get? The another product we get here is this. One is so this hydrogen will come out. And other one, the minor one would be this. Hydrogen from this carbon comes out, right? So it is CH2 double bond CCH3, single bond CH2, CH3. This is the minor product we get. Yes, major product would be same. This we call it as dehydration of alcohol. Why? Because water is getting eliminated. Dehydration is taking place. So alcohol forms alkene through dehydration reaction, okay? Next write down the next reaction we have. Colbe's synthesis. Write down aqua solution of aqua solution of sodium or potassium succinate, right? Sodium or, sorry, potassium or sodium succinate. S-U-D-C-I-N-A-T-E succinate is electrolyzed and forms ethene at anode. And forms ethene at anode. Similar reaction that we did in the alkene preparation. Reaction you see? We have CH2, CH2, C double bond OOH. So if you write down this C double bond OOH, it is succinic acid, okay? It is succinic acid. But we have to take the salt of it, okay? Salt would be what? We can take sodium or potassium salt of it, minus K plus, minus K plus. It is what I, is potassium succinate. Forms from succinic acid. When it goes under electrolysis, electrolysis, we have the similar mechanism like we did in alkene preparation. CO2, CO2 forms will get two molecules of CO2. CH2, double bond CH2, two molecules of CO2, okay? And we get H2 at cathode, 2KOH at cathode. So this is at cathode. This is at anode. It is ethane. Yes, it is a gas. C2 to C4 carbon atoms are a gas in alkene also. Done? Clear guys? Copy? Now the last method of preparation we have, we call it as Wittig reaction. Write down, it is the method of preparation of alkene. It is the method of preparation of alkene. Method of preparation of alkene from carbonyl compound. Carbonyl compound means what? Aldehyde or ketone. From carbonyl compound, aldehyde or ketone. So for this reaction, we use a reagent called Wittig reagent. Wittig reagent is what? Wittig reagent is triphenyl phosphorane. It's a direct reaction. Just you need to memorize the phosphorane. Memorize the reagent in this. Triphenyl phosphorane is this. We have P double bond, C. And with this phosphorous, we have three phenyl group attach. pH, pH, pH. And with this C, we can have two alkyl group, same or different. We can have two hydrogen atom. We can have two hydrogen atom. We can have two hydrogen atom. Or we can also have one alkyl group or one hydrogen atom. Any, all three possibilities we have, RH, right? Presence of alkyl group or hydrogen at this carbon atom won't affect the product. Means product would be an alkene only, right? We can have different, different name for that. But product would be an alkene, right? This RH won't affect the major product. Major product will be alkene only. So what happens in this? Suppose I'm taking the simplest one here. For this reaction I'm just assuming C single bond, single bond, H and HCH2. This is vitig reagent. Very simple reaction, obviously mechanism we are not having in this. So what, because this reaction, again it will come in carbonyl compound chapter. There we can discuss. So we have pH3P, triphenyl group, pH3P, double bond CH2. This reacts with an aldehyde I'm taking. So I'm taking H, C double bond OR. And we are heating it, okay? We are heating it. Now what we need to do, we just focus on this double bond and this double bond. This oxygen will be replaced by whatever group present here with phosphorus. So oxygen would be here and CH2 would be here. Product would be pH3P double bond O plus we get RCH as it is double bond CH2. So this is an alkene we get. I missed one word over here. It is, since we have taken CH2 over here, so it is methylene also. Methylene triphenylphosphoryl. Yes, exchange of position. Oxygen will take the position of the group attached with this with phosphorus and this group will come over here, simple. We can discuss the mechanism, it's just an arrangement we have, okay? But that is not required again. Directly you can memorize this. Mechanism based question also you won't get in this reaction. If you know how to write down, it's fine. Nothing much is required. This is it for the preparation of alkene, okay? Now we'll do properties of alkene, physical properties. First one, C2 to C4. If the carbon atom from 2 to 4, the physical state is a gas, okay? If the carbon atom is C5 to C17, C5 to C17, it is a liquid and C18 onwards, alkenes are solid, okay? Among isomeric alkenes, boiling point decreases with, decreases with branching. As branching increases, surface area actually decreases, right? With decrease of endowalls force of attraction, hence boiling point decreases. So it is true everywhere actually, wherever the branching is there, among isomeric compounds, boiling point is less. More branching, less of the boiling point. We have already discussed it. Like melting point of cis isomer is more than, is lesser than, is lesser than to that of the trans isomer. The reason we know better packing, we have also discussed this in isomerism chapter, okay? Better packing. Next write down chemical properties. First one we have reaction with hydrogen. Reaction with hydrogen. This reaction we have already done in the preparation of alkene from alkene. I'll write down here, you will understand this, what exactly the reaction is. Suppose we have an alkene RCH double bond CHR. Alkyl groups, same or different, anything can take. With hydrogen in presence of a catalyst and I. It gives RCH single bond CHR and both hydrogen will get attached from the same side, alkene it forms. Remember this reaction we have done in preparation of alkene. Stereochemistry of this reaction you must remember. What do you need to keep in mind? If you have cis alkene, cis alkene, you are doing hydrogenation of cis alkene. It gives you mesocompound. This you have to memorize very important information. Mesocompound, okay? That means what? Suppose we have C double bond C. We have alkyl group attached to this carbon and we have deuterium. When you allow this to react with H2 in presence of nickel, it forms a mesocompound and we know what is mesocompound. We have a plane of symmetry, internal compensation. So both D and H are present on the same side. CS3, CS3, remember mesocompound? So once you know this information that it forms mesocompound, you will draw this structure only. So that will have a plane of symmetry within the molecule, isn't it? Yes or no, guys? On the other hand, if you have trans alkene, for example, you see this, if you have trans alkene on hydrogenation, it gives you resmic mixture. It gives you resmic mixture, okay? Look at this reaction here. Trans alkene is this. Suppose I am taking this example, plus H2 with Ni. You know this information, it forms resmic mixture? How do we draw resmic mixture? You see two compounds it will form. One D is this side, D, H, H opposite side, CS3, CS3, plus the other one would be the mirrored image of this one. Yes or no? H, D, H, D, CS3, CS3. This information that we did just now, mesocompound and resmic mixture, we call it as stereochemistry of this reaction. Achai, by mistake, I have written it. No problem, that's not a problem. You can also place this as CS3. Then also it will be a mesocompound. You can say this could not be a product of this one if I write down R here. But if you look at this compound individually, not about this reaction, look at this compound. It is a mesocompound, you see. There's a plane of symmetry here, internal compensation. Yes? Achai, you are asking this question. If first one it won't be meso, if R is not same as R, then we can say correct, definitely right. If both this alkyl group are not same, then obviously we don't have the mirrored image, then it's fine, it's not a mesocompound in that case. Yes, correct. Next you see another reaction, right down halogenation reaction. Right down, it is an electrophilic addition reaction. It is an electrophilic addition reaction. Okay, so in this one what happens, you see, we have a reaction RCH, double bond CH, same or different, plus X2 in presence of a polar, a protic solvent, CCL4 we are taking, carbon tetrachloride, it is a polar, a protic solvent. Polar, a protic solvent if you take, it won't ionize this halogen molecule. Since we have a bond like this, this bond won't break easily. Delta positive, delta negative. So what happens in this? The pi bond breaks and the two halogen atom gets attached to this carbon atom. So it forms basically visceral dihalytes, visceral dihalytes, correct. Next slide down, the third one, reaction with HX, hydrogen halide. You have an alkene, C double bond C, reacts with HX and forms alkyl halide. C single bond C, X and H. Pi bond breaks, H plus and X minus will get attached. Okay, so in this one, what happens? You see, suppose you have an unsymmetrical alkene, I'm taking this example, CS3CH double bond CH2. If this is allowed to react with HBR, then this BR and this H, right? You should know that where this BR should attach and where this H should attach. The product in this reaction would be this. So unsymmetrical alkene, you have to be careful, like whether BR will attach on this carbon or this carbon. How to decide? We'll discuss that. If it is symmetrical alkene, then wherever you attach BR or H, it won't make any difference. We'll discuss how to understand this, which will attack where, but this reaction you see, two, three points you'll write down, involves carbocation formation, this reaction, involves carbocation formation. And when carbocation forms, again, there will be rearrangement. Okay, reactivity order of HI is maximum. Then we have HBR, then we have HCL and we have HF. This is the reactivity order. Okay, so you see here, the same reaction I'll write down, the reaction is CH3CH double bond CH2, reacts with HBR. So we have H plus and BR minus. So this pi bond shift onto this carbon so that we'll get a positive charge here and this positive charge carbocation is more stable than this one. It comes over here and then it takes this H plus. So what we get here you see, we get here CH3CH single bond CH2H from HBR and we'll get a positive charge over here. Other way if you think, then it happens this, it forms this CH3CH hydrogen will get attached over here, CH2 positive charge. Which carbocation is more stable? First one or the second one? First one, right? Because of more alpha hydrogen, first one. Hence this carbocation won't form at all, okay? In the first step, we always write down the more stable carbocation, correct? Hence this will form. Now in the next step, what happens? The BR minus will get attached to this carbocation. If there is no rearrangement possible, directly it will get attached to this. If there are possibility of getting more stable carbocation by rearrangement, we'll do that first and then BR will get attached, okay? This product, we can also write down according to a rule called Markovnikov rule, Markovnikov's rule. Statement, all of you write down in this. Statement, write down. According to this rule, according to this rule, the negative part of the reagent, according to this rule, the negative part of the reagent will get attached to, will get attached to the double bonded carbon atom, the double bonded carbon atom, which has lesser number of hydrogen. The double bonded carbon atom, which has the lesser number of hydrogen. With this rule, directly you can write down the answer. See, you know this rule? Then what you need to do? You have this reagent, negative part of the reagent is BR minus, H plus and BR minus has to get attached. BR minus is the negative part of the reagent. This rule says what? The negative part of the reagent will get attached to the carbon atom, double bonded carbon atom, which has lesser number of hydrogen. Only one hydrogen we have here. Hence, this is the product we get. Correct? We have one more rule that is anti-marconic off rule, AMR. It is exactly opposite of marconic off rule. I'm just telling you, this won't write this down, just keep that in mind. Anti-marconic off rule, okay? It is applicable in presence of, this you can write down, anti-marconic off rule is applicable in presence of peroxide, of peroxide. So it is exactly opposite. Here, the negative part of the reagent applied, like attached on the carbon atom, which is the lesser number of hydrogen. In anti-marconic off rule, it is Ulta. Negative part of the reagent will attach to the carbon atom, which has more number of hydrogen, okay? In presence of peroxide, it happens. It is applicable, applicable only for HBR, memorize this very important one, applicable for HBR only. This, we also call it as peroxide effect or Kharash effect, K-H-A-R-A-S-C-H, Kharash effect. This is the name of the scientist actually, based on his name only effect is given. Remember always, it is applicable for HBR only. This AMR rule, it is based on free radical mechanism, I'm not going to detail off mechanism over here. This two, three information you can keep in mind, you can solve all questions from this, right? Again, I'm repeating this point, that anti-marconic off rule is not applicable for HCL or HI. It is only applicable for HBR, for HBR only. Based on these two, three information, you should see this. Suppose we have this molecule, CS3-CH, double bond CH2, and this is allowed to react with, we have HCL, we have HBR, we have HBR in presence of peroxide, we have HCL in presence of peroxide. The product if you see, very simple one, it's a marconic off addition you can think of here. The negative part of the reagent will attach to the carbon atom which has lesser number of hydrogen. So on the middle carbon will add chlorine, CS3-CH. Same thing on the middle carbon will add bromine. Since it is peroxide here, so it follows anti-marconic off rule, Kharash effect. So in that case, it is Ulta. So the carbon atom which has more number of hydrogen will take bromine and here we'll have hydrogen. In this case, it is HCL and we know in case of HCL, AMR rule is not applicable. The product here is same as the product we get in the first reaction. This is the product we have in all four. Marconic off rule is for any alkyne halide, okay? Anti-marconic off rule or peroxide effect is valid only for HBR. Then yeah, next you see oxidation of alkyne. For oxidation, we use different, different oxidizing agent. Like the first one you see, we are using acetic KMNO4, acetic KMNO4 or hot KMNO4. Both serves the same purpose. They are strong oxidizing agent. How to write down the product is very important, okay? Suppose we have an alkyne, R, C double bond, C, RH. Acetic KMNO4 we are taking for oxidation purpose. So this double bond you need to break and you need to write down acid across this double bond. So, C, this carbon, double bond we break and write down this carbon with this carbon will write down acid. Plus, we also get the same thing, HO, C double bond O, right? So alkyne will get oxidized into acid, double bond we need to break and write down the carboxylic acid. Same thing suppose if you have this molecule, RCH, double bond CH and H we have, acetic KMNO4. This we call it as terminal alkene. Terminal alkene. Why terminal alkene? Because the last carbon contains double bond. Here you see, last carbon contains double bond. Here we have R, so obviously this is not the last carbon. So when we have terminal alkene, then the terminal carbon gives you carbon dioxide, CO2 and H2O. And this side you will get acid, R, C double bond OOH. So you'll break this from here, you'll get carbon dioxide and water and this part you'll get acid. Very simple one, you can understand this easily, how to write down the product, okay? Like I'll write down this as an example, CS3CH, double bond CH, CS3. You're doing the oxidation of it with the same reagent. So what happens, this bond breaks, COOH, COOH here. You'll get two molecules of CS3COOH, acetic acid, okay? So remember always, if you have terminal alkynes, terminal alkene, sorry, terminal alkynes gives you carbon dioxide, one side, another side, we have acid. So acetic KMNO4 gives you carboxylic acid. So if you have hydrogen connected to this, if the R group is not there, then it would be this only, HCH double bond CHH. In that case, acetic medium, you'll get two molecules of CO2 and two molecules of H2O. Another oxidizing agent we have, alkyline KMNO4. Alkyline KMNO4 and one more we have and osmium tetroxide, OSO4, osmium tetroxide. You see, in this osmium tetroxide, the oxidation state of osmium is plus eight, which is the maximum oxidation state of any element we have in the prior table. Plus eight is the maximum oxidation state of any element. Many a times I've asked this question, osmium source shows the maximum oxidation state. Both gives you similar kind of product in the reaction, okay? It gives similar kind of product in the reaction. First we'll see KMNO4 and then we'll see osmium tetroxide. You see, KMNO4, if you are taking cis alkene, write down the way I am writing it down, okay, exactly the same way you write down. RC double bond CR over here will have HCH. KMNO4, you see, I'll just show you one thing over here, like this, KMNO4 also dissociates as K plus and MNO4 minus. MNO4 minus, the structure, if you see, it is MN double bond O, single bond O minus, double bond O, double bond O. It is like this, MNO4 minus, okay? So what happens here from this KMNO4, we'll get MNO4 minus like this. O minus double bond O and double bond O. So this O minus attack to this carbon, this pi bond shift onto this carbon, once again, yeah. I'll write down O minus here, O minus here and here we have a double bond O like this. So what happens, this pi will shift here and this pi will shift here, then this will attack onto this. This pi will shift here and this pi will get attached to this. So you'll get a cyclic structure here, which is this. C single bond C, single bond C, single bond C, then we have MN here, MN double bond O, single bond O minus and here we have R, R, H and H will be as it is. This happens in between, intermediate is this and then this reaction is allowed to react with O H minus, basic medium with H2O. Then finally, this converts into a diol, O H, O H on the same side, H here, H here and R. Do you see the previous one? Acidic KMNO4 gives you acid, but here we are getting diol. So this is what you see. The product that you get over here, it is a mesocompound, isn't it? It is a mesocompound we get. So if you have sysalkine, sysalkine gives you a mesocompound in this reaction with alkaline KMNO4. See, it is not a tough thing. Only thing we have so many oxidizing agent, okay? Acidic gives you something else, alkaline gives you something else. You should know this that alkaline gives you diol. Sys gives you mesocompound with alkaline KMNO4. Similarly, if you take trans, trans gives you a restrict mixture in this reaction. I'll write down the reaction in the next one. Directly I'll write down H double bond CRH with reagent would be what? Alkaline KMNO4, alkaline with OH minus H2O. Gives you mesocompound. So mesocompound means what? Both OH on the opposite side, H, H, R, R. Rest of the mixture we need to write down and the mirror image of this would be OH, H, H, OH. R, R. So if it is trans, trans gives you resmic mixture. This is important here. Like cis gives you meso, trans gives you resmic mixture. Right, so alkaline KMNO4 we have, hence we have base over there. Osmium tetra oxide if you take. Osmium tetra oxide also gives you the same thing. But only one change, one difference we have over there. I'll just write down here. If you have cis alkene plus OSO4, osmium tetra oxide. Only difference is what? Like we have alkaline medium. Here we have in the second step basic medium. Gives you cis we have, gives you meso. If you have trans with OSO4, osmium tetra oxide, acidic medium gives you resmic mixture. Exactly same, both reactions gives you the same product. The difference is the first one we have alkaline medium. Here we'll have the acidic medium. So one last reaction we have in this, we'll finish this and then we'll take a break. Done, okay. Last reaction we have in this, write down. Ojonolysis, important reaction. All these reactions we'll have again in reaction mechanism. They will study in a different manner of all this. So in Ojonolysis what happens? We have carbon-carbon double bond alkene allowed to react with O3, ozone first in the first step. And in the second step, we use Zn with H plus H2O. Zinc with H plus H2O. Zinc with H plus H2O we are using. Very simple one. Easily you can write down the product here. What you need to do, just you break this double bond, add oxygen to this, oxygen to this. So the product would be this COO double bond C. This is the product we get. In between what happens with ozone, first of all, ozonide forms. Ozonide is something like this. Ozonide is this. Then what happens? This bond, when it is allowed to react with zinc, then zinc will take this oxygen, forms ZnO. This bond comes over here, makes a pi. This bond comes over here, makes a pi. And this bond is taken up by this oxygen and zinc forms ZnO. Hence, we'll get along with this, we'll get this. Double bond O plus zinc oxide we get. When you take zinc, it becomes reductive ojonolysis actually. Reductive ojonolysis. Okay, alkene converts into aldehyde or ketone. It gives you, it gives aldehyde, ketone or mixture of both, aldehyde or ketone. Depends upon what alkene we have, what molecule you have taken. Suppose if you take this Hc, double bond C, R, R and R. Right, it's O3. Then we have ZnH plus H2. Regent is fine, what we need to do? Take the double bond, attach oxygen with the two carbon atom. So it will be R, C double bond OH plus R, C double bond OR, aldehyde and ketone. If this hydrogen you replace by alkyl group, you'll get only ketone. If all these, if this alkyl group you replace by hydrogen, you'll get aldehyde here again, formaldehyde, okay? All alkyl group replaced by hydrogen, only aldehyde we get. That's why we have all possibilities. We can get only aldehyde, we can get only ketone or we can get a mixture of both here, understood? Sometimes they also ask you the Ulta question. Ulta question means what? They'll give you an alkene or ojournalysis gives you this product. What is the structure of alkene? So what you need to do? You need to just remove this oxygen and attach the two carbon atom with the double bond, that is it. So Ulta also they ask sometimes, okay? So you see here, some questions on this, you see? Write down the product of ojournalysis in this reaction, CH2 double bond, CH single bond, CH double bond, CH2. So one more thing you must take care of when you have multiple double bond, all the double bond you need to break, all the double bond means what? You break this double bond, attach double bond O with this, double bond O with this, double bond O with this, double bond O with this. Hence the product would be H, C double bond OH, this is the one product. Another one would be H, C double bond O, the middle one, C double bond OH, the another one would be, again, one molecule of this only, two molecules of this, one molecules of this. Is it clear? Yes, tell me. Understood this. Suppose we have a molecule this, double bond, double bond with ojournalysis reaction. In this, what we need to do? You break this double bond, break this double bond, attach oxygen with all these carbon atoms. So the product would be this, double bond, double bond O, O, and another product would be again the same, double bond, double bond O. Any doubt in this? Yes? Well, last example, write down the product in ojournalysis of benzene. Try this. Yes, what is the product we get? All the double bonds you need to break, right? You need to break this double bond, you need to break this double bond, you need to break this double bond. You see, this is C, C with double bond O, double bond O. So we'll get three molecules of H, C double bond O, C double bond OH. Isn't it? Yes? Any doubt in this? Yes, could you respond? Any doubt in this? Last one? Yes, today I've gone a bit fast, I know, because we need to finish this today. We don't have that much time, that's why I've gone a bit fast. Yeah, which one? This one, this one. The last one. See, you look at this one, this way, just a second, I'll just draw the bigger one. Suppose this is the benzene we have, correct? Okay, ojournalysis we are doing. We know we need to break the double bond, so you need to break this, you need to break this, you need to break this. All bonds need to break. So if I just write down here only, probably this you'll understand. Suppose if you break this double bond this way, you also break this double bond, and this double bond you break, correct? And all these carbon atoms with double bond O we need to attach, right? Like this. So one molecule is this. One molecule is this, isn't it? C double bond O, C double bond O, H and H. Like this, we have three molecules, two and three, and that is what I have written here. Is it clear? Clear? So everywhere the double bond, you need to just break the double bond and attach oxygen over there, okay? If you have to go opposite Ulta, then remove the oxygen, attach the carbon atom with the double bond. Clear? No doubt. So this is it for alkene. We'll take a break now. We'll resume at 635 and we'll start alkyne, okay? Take a break.