 Yeah. Shall we start? Yes. You there? Can you hear me? Hello? Yeah. Okay. So, like I said, this isomerism is done. Conformational isomerism is left. We'll take up that in the last. So we are going to start the next chapter in organic chemistry that is hydrocarbon. Okay. Hydrocarbon are the compounds, organic compounds, which are made up of, which are made up of hydrogen and, and carbon as the name suggests. Right. So we are going to study about the compounds which are made up of hydrogen and carbon in this chapter. Okay. Mainly, we have three types of compounds in this. That is, first of all, it is classified in two categories. Saturated hydrocarbon and unsaturated hydrocarbon, unsaturated hydrocarbon. Saturated hydrocarbons are alkanes. Saturated hydrocarbons are alkanes. Unsaturated ones are alkenes and alkynes. Alkenes, alkynes. Why these are unsaturated? Because on this, we can have addition reaction possible. Right. It shows addition reaction, addition reaction. Okay. The general formula of alkene we have, we know CN H2N plus 2. For alkene, it is CN H2N and for alkyne, it is CN H2N minus 2. If you talk about degree of unsaturation DOU, for alkene, it is 1. For alkene, degree of unsaturation is, oh sorry, for alkene, degree of unsaturation is 0. For alkene, it is 1 and for alkene, degree of unsaturation is, you know the formula of DOU? DOU equals to C plus 1 minus H plus X minus N divided by 2. You can find out for this, alkanes, alkenes and alkynes, you will get this as degree of unsaturation. Here in alkanes, all carbons are sp3 hybridized. Okay. In alkenes, at least one carbon, at least one carbon will be sp2 hybridized. Alkenes, at least one will be sp hybridized. Clear? Copy this down. Okay. So first we are going to do alkene here. So in organic chemistry, all the chapters have two different types of, you know, thing we need to understand here. That is preparation and properties, chemical properties mainly. All these are reactions only. We need to understand that what are the various preparation method of alkene we have and then what kind of reaction it shows. Okay. So all these are reactions you have to memorize. Okay. Like alkanes we can prepare by many different method. Okay. Heading you right down, preparation of alkene. The first method we have, the first method we have from aldehyde and ketone, from aldehyde and ketone. We know the aldehyde functional group is COH and ketone is RCOC, RCOR. Correct. So there are two very important reactions we have for the preparation of alkene from aldehyde and ketone. The first one we call it as clemention reduction. Name reaction very important. We call it as clemention reduction. Clemention reduction, it is generally possible in acetic medium. It is acetic medium reaction, acetic medium. Okay. The reagent for this reaction, reagent you must remember. The reagent is generally we use ZNHG, amalgamated zinc we call it as with concentrated HCl. This is the reaction we have amalgamated zinc with concentrated HCl. Okay. We can use zinc here or instead of zinc, we can also use amalgamated sodium NaHG. Mostly, mostly we use zinc only. This reagent you must keep in mind. Okay. So what happens in this reaction you see? Suppose we have a compound, a ketone, RC double bond O CH3. When this goes under the reaction with reagent, then this C double bond O, it converts into CH2. This is the product we get. Overall, the reaction that you have here in clemention reduction, you need to convert C double bond O with CH2. You need to replace this C double bond O with CH2 and this is the reaction we get. Okay. So you have to memorize this. Done. Okay. Next. The second reaction we have for this preparation method B is again the name reaction. It is Wolf-Kishner KISCHNER, Wolf-Kishner reduction. In Wolf-Kishner reduction, this reaction is possible in basic medium. Unlike first one, the medium here is basic. It is the basic medium reaction. We use strong base for this method like we can use potassium hydroxide KOH or we can also use potassium tertiary butoxide, tertiary butoxide. Like this we write. If you look at the structure of this one, tertiary butoxide is this, a CO-CH3, CH3, CH3 and K plus. This is potassium tertiary butoxide. The notation for this one is T stands for tertiary BUO minus K plus. This is the notation we have for potassium tertiary butoxide. So what happens? Suppose we have a ketone like phenylpropanone we are taking. So we have this. Okay. And when it is allowed to react with hydrazine NH2, NH2, then first it converts into hydrazone, which is this double bond NH2. And H2O goes out. This two hydrogen will take this oxygen goes out at this. We call it as phenylhydrazone. Then this phenylhydrazone is allowed to react with the strong base. We can take KOH and we are heating it. Okay. And then this double bond NNH2 gives us CH2 here and N2 gets eliminated. This is volficitinal reduction. So if you want to write down the product directly, this is aldehydo ketone. In the first step, we'll put hydrazine N2H4 and in the second step we'll put some base KOH and we heat it. And then directly it converts into this. This is the product we can directly write. So you should know how to write down the product directly. Okay. In the exam, this is the only thing which is required. Next is the second method of preparation we have from alkene. From alkene. We have different methods of preparation from alkene. First one you write down by catalytic hydrogenation. Catalytic hydrogenation means what? Addition of hydrogen in presence of a catalyst. What catalyst we can use? We can use nickel. We can use platinum. We can also use palladium for this. Like, suppose we have ethene, CH2 double bond, CH2 plus H2 Ni. CH2 CH2 both hydrogen will get attached from the same side. So this addition is, it is syn addition. Syn addition means both hydrogen get attached from the same side of the double bond. It's not like one is attaching from this and other one is attaching from this side. No. But it means that the both hydrogen is getting attached from the same side. Okay. So mainly when you have cis or trans alkene then it is important to understand. Okay. Like you see this example, what do you have to keep in mind in this or hydrogenation? Cis alkene. Cis alkene. On cis alkene we have what? Syn addition. Syn addition means addition of H2 from the same side. Syn addition of H2. And this gives mesocompound. It gives mesocompound as a final product. Look at this example. C double bond C. R. Suppose it is deuterium here. R and D. So obviously it is cis alkene. Both identical group present on the same side. Cis alkene plus H2 in presence of Ni, platinum or palladium. And the product would be, if you write down this fissure projection, since it is a mesocompound, so both H must be present this side. Mesocompound. If it is on the other side, it is not meso. This is the product we get. If you have trans alkene, trans alkene, obviously addition of H2 is syn only. Syn addition of H2. This gives resmic mixture. Resmic mixture. Same thing if you have here, this is the product we get. Resmic mixture. Like this. And the other one is also possible. The mirror image of this, which is, I'm sorry, HD, you have D over here. Mirror image of this. H, D, H, D, R, and R. This is the resmic mixture. We have just now, we had discussed what is the resmic mixture. This is stereochemistry of this reaction you must remember. Done? Yeah, like a few things that you have to memorize, but it's not like you have to mug it up simply. Mechanism, we can understand. We can discuss the mechanism, but yes, in the exam, if you want to do the answer, want to write down the answer quickly, you have to mug it up. Like how and what reactions, how the reaction proceeds. Point is, then you will be thinking like how many reactions we can mug up simply in the, in one chapter, you have so many reactions. Like this, we have so many chapters. So obviously, the entire one, entire reactions, the entire book you cannot memorize. But as we proceed in the organic chemistry, we'll discuss the mechanism also, and you will understand that the pattern of the reaction, how it is going on. So that will help you in keeping those reactions in your head. Like in this case, you can think of this kind of reaction may possible. Right. That's why I always say, organic chemistry, you have to revise at least twice before your exam. Okay. Once if you solve it, it won't help you. Okay. Physical chemistry probably once if you are done, it is done. Correct. So keep that in mind always that organic chemistry before your J mains exam, December, January, you have to finish it twice. Then only you will be able to do it, to do good into the exam. Otherwise, it will be confusing. And it won't, you know, like the thing is, you feel like, okay, B is the answer or C is the answer. But eventually, if you feel out of five, three or four questions, you will get it wrong if you're not very much sure in J kind of exam. Okay. So yes, answer for your question is, you have to keep this reaction in mind. We can discuss mechanism, but obviously mechanism you're not going to write in the exam. It's like, you know, in this condition, what reactions proceed and what product we get, how to write down the product that you should know. Mesocompound, which one? The first one you're talking about. No, it's not like that. Like, you know, anywhere you place, that's not a problem. Like, you will get the same thing. It's the same rearrangement actually. We'll have the mirrored image. If you place D here and here, then also R will be here. It's the mesocompound only. It's not about what atom we should place there. It's like we have a plane of symmetry there in between. And because of that only, it's mesocompound. Okay. So it is the first reaction of preparation of alkene from alkene. Now the second one we have by hydroboration by hydroboration reduction. So if the reaction is CH2 double bond, CH single bond, CH3. Hydroboration oxidation reaction. In that we are taking B2H6 diborane. And in the second step, we're using an acid, CS3COH. Okay. So in this case, the product here we get is CH2 single bond, CH2 single bond, CH single bond CH3. One hydrogen will get attached over here and one hydrogen will get attached over here. But what is important here and what you have to keep in mind that the source of this hydrogen, this hydrogen comes from B2H6. And this comes from the acid that we are using in the last step. That is this one. Means hydrogen from B2H6, it gets attached to the carbon atom, which has less number of hydrogen. Right. So it is kind of Markovnikov addition, we'll discuss that later what is Markovnikov addition. But the carbon atom, double bonded carbon atom, which has lesser number of hydrogen or more substituted will get hydrogen from B2H6. Okay. And acid gives hydrogen at the other carbon. How we can add H2 also directly. That's what we are trying to understand. Yes. So it's not like, suppose they are 10 methods. Okay. If you want to prepare, it's not like we have to prepare alkene. We need to understand what all methods we have by which we can prepare alkene. So if the question comes in the exam, why are you giving this reagent? Why don't you give it H2 directly? You can't say that. Okay. So you should know that what all methods we have by which we can prepare. Okay. So here sometimes what happens, they will give you the isotopes, like for example, suppose my same reactions, we have this molecule only. And we are using B2D6 here instead of B2H6. And the second step, we are using CH3COOH. And they'll give you all possible options. Like the options will be like this. They'll give you CH2, CHCH3, here H, here D. They can give you like this also CH2, CHCH3, here D, and here H. So you should know what is the source of this hydrogen and what is the source of this hydrogen here. D is the isotope you are using. Like hydrogen, we are using D here, deuterium. Correct. You can consider D as same as hydrogen as far as the reaction is concerned. Right. So this kind of question is reaction mechanism based now. So you should know this, that what is the source of this hydrogen? Is it the acid or the diborin? B2H6 is diborin. You should keep this in mind that the carbon atom, double bonded carbon atom, which is more substituted, their hydrogen from B2H6 gets attached. So since we are using here B2D6, so at this carbon, we have this D over there. Right. So we'll go with option B, not with option B. This kind of question sometimes there is. Clear. Okay. So from alkene, we have this two method. We can also prepare alkene by, like alkene by alkene, third one from alkene. It is very much similar to alkene method. Just we have to do the hydrogenation twice. Like for example, suppose we have RC triple bond CH. We have to prepare alkene. So both the pi bonds you need to break. So first step will do hydrogenation in presence of the catalyst, nickel, platinum, palladium. Then we'll have the syn adhesion. We'll get C double bond CH, H and H will get attached. Since we have to prepare alkene, now it is an alkene. So again, H2 with Ni. Then again, we'll get RC, HCH. H we already have here. RCH to CS3 will get as the final product. So basically we are doing hydrogenation twice here on alkene. Give me a minute. I'm coming. Yes, we have anti adhesion also possible. Okay. We'll discuss that. There are so many things we need to understand in organic chemistry. We'll just start it. So we'll discuss that. Okay. So I'm just trying to go through the portions first. When we start reaction mechanism, then you'll probably understand all these things. Okay. But yes, you should know all these things like the fear from where the hydrogen is coming in studio chemistry and all, wherever it is required, you have to keep that in mind. And if you see the portions of 12th grade, 70% of the syllabus is organic only. And it is something like this particular subject, organic chemistry, you cannot do on your own. Yeah. Although we're talking about like one of the doubt was, one of the doubt was like, is there any kind of addition apart from sin? So I was like, yes, we have anti addition also possible. So wherever it is, I will discuss that. Yeah, I was talking about like, yeah, I was talking about we have 70% of your 12th grade portion is organic chemistry only. So if you struggle in this particular subject, then your board exam will also suffer. And this subject is something that you cannot do on your own, right? Like I always say, notes are very important. Okay. NCRT won't help you at all in this. They probably helps you in getting marks in your school exam or board exam. But then also if you don't want to mug up things randomly, right, chapter wise, then you have to understand and you have to study a lot of things in order to understand those reactions which is given in your syllabus, right? So we'll do all those. Obviously time is a concern. Okay. Because we need to finish portions also on time. So we'll manage like that's how I did every year. Okay, so we'll do that. But yes, like I said, whatever we are doing, now you focus on this. What kind of addition reaction we have, what kind of substitution reaction we have, we have elimination reaction, right? Then we have elimination also different different types. There's so many things to understand. Okay, so we'll discuss once it comes. Okay. So next you write down, like I said, alkyne, say we can prepare this. Okay. And this reaction you will also get in alkyne chapter. Like it will be the chemical reaction of alkyne. Okay. So here you see in alkene, we have done the preparation of alkene from alkene. Now this reaction will also come in alkene chapter, but that will be the chemical reaction of alkene. Okay. So the once you finish this portion, organic chemistry, the next target is you have to build that bridge. Okay. Like you see, I'm going to do the next example here. Like we can prepare alkene from alkyne halide, right on the fourth step, fourth method from alkyne halide. Now we are doing this alkyne halide reaction here. This reaction, you will again see once you do this particular chapter that is alkyne halide, but this reaction will be there in chemical reactions of alkyne halide. Okay. So this you should know what are the common reactions we have. That is the second part. Once you revise, once the service is done, then second part you should be like this. Like how do we prepare, you know, alkene from alkyne halide, then alkyne halide to aldehyde, aldehyde to acid, acid to amines, many things. You have to build that bridge between the two chapter. Anyways, so we have two, three very important reaction here. First reaction in this, you write down from alkyne halide by reduction, by reduction. See, we can reduce alkyne halide and forms alkene. Like for example, you see, we have Rx alkyne halide. For reduction, we take hydrogen here, right. It gives Rh plus Hx. This is the reduction reaction. Now we can do this reduction by various different reagents. I will write down here what are the reagents we can use. The reagents we use here, we can take zinc copper couple, zinc copper couple we can use. We can also use alcohol, some alcohol we can use Rh with some metal like sodium we can use. We can also use LiAlH4. We will see this, it is a very strong reducing agent, LiAlH4, lithium aluminum hydride. LiAlH4, we can also use ethyl alcohol. Et stands for ethyl, ethyl alcohol with metals, Na, amalgamated zinc we can use in presence of water. Any one of this reagent we can place here and we'll get Rh with Hx. Next, write down the name reaction we have and it's a very important one. Not catalyst is the one which does not take part in the reaction. It just enhances the rate of the reaction. Either increase or decrease both ways possible. See reagent is, I'll tell you, suppose you have some metal, zinc, right. So actually what happens, the purpose of metal here, that it can provide free electron and this electron takes part in the reaction. Means if the reaction is going on, there must be the, you know, the, you know, that some driving force must be there, right. Like you cannot, A and B cannot react simply. How am I reacting? Right. We must have some condition that we must have some reagent. That reagent, you can say it initiates the reaction by providing electrons. Electrons can also say some solvent like suppose alcohol we are taking as a solvent. So this is the solvent actually. It's made the reagents and all those things are present, the reactant and then they are colliding. Some reactions are going on bond dissociation and bond breaking process, bond making and bond breaking process are going on and then we'll get the final product. Correct. So reagents are what? reagents actually initiates, it helps the reaction to proceed. Means A cannot give B directly. Sometimes what happens? Suppose you have a compound and you heat it at a very strong temperature at very high temperature and then the compound dissociate gives some product. So here we can't see, like we cannot see there is some reagent present. But the point is there must be some reacting condition. Then only the reaction proceeds. Like carbon when reacts with what oxygen it is not like it can react always at all condition and forms carbon dioxide. You have to provide that reacting condition. So you can understand the role of reagent is also something like that. If I tell you RX, carbon halogen bond, carbon halogen bond becomes weak in presence of free electron because it's the reaction. Suppose RX say, I have to make RH. So obviously you need to break this bond. Correct. Now to break this, you have to make this bond weak. Carbon halogen bond becomes weak in presence of free electron. That is the observation we have. Why it becomes weak? Because it is delta positive. This is delta negative. And here X as a leaving group, it goes out. So it is a part of mechanism actually. But yes, this is the role of reagent we have here. Okay. So yeah, the second reaction we were talking about from alkali to alkane is Wurz reaction. W-U-R-T-Z, Wurz reaction. Very important reaction. Okay. Definitely, you will get questions on this in your school exam also. Okay. It's a name reaction. Very important. Yeah, one second. I'll go back. So what happens in Wurz reaction, the reagent we are using is Na with dry ether. This is the reagent for Wurz reaction. Na with dry ether, sodium with dry ether. We usually take, preferably we take two molecules of alkohol, sorry, alkythylide. And this is allowed to react with Na. Okay. And then in presence of dry ether. So it gives R, R and NaX, R and NaX. So two molecules, two alkyl group combines here and forms symmetrical alkane. Okay. So basically, Wurz reaction, we have two mechanism possible. One is free radical mechanism. Free radical mechanism, you can understand the intermediates are free radical here. Or ionic mechanism, the possibilities. You can also have ionic mechanism. Ions basically forms. Okay. Free radical mechanism, we are not going to focus much here. We'll do that later. Free radical mechanism is something like, you know, in which like I said, the intermediate is free radical. And jahab every free radical form karova, the reaction in which the free radicals are involved, we do not have any control in that reaction. Right. So it is uncontrollable. We cannot control this reaction. It happens on its own. Right. We cannot control this reaction because free radicals are highly reactive. And the reaction proceeds on its own. Right. So what happens in this, suppose I have taken CS3Cl plus Na, two molecules of alkyl halide. So Cl CS3 with dry ether. So it gives CH3, CH3 and NaCl. So what property you have to keep in mind here that Wurz reaction may, we'll get higher alkanes. Higher and symmetrical alkanes we get. Okay. We get higher and symmetrical alkanes. So it is used to prepare symmetrical alkanes. We cannot prepare methane from this. Methane cannot be prepared because the carbon atom cannot be one because coupling would I write two methyl group are getting combined. Methane cannot be prepared. So actually what happens here? This bond dissociates and it forms radical. Like free radical forms. And this methyl radical, methyl radical club, it forms ethane. Okay. Now what happens? Preferably we take same alkyl halide, two molecules of same alkyl halide. Why? If you take different alkyl halide, it's not like the reaction is not possible. Suppose, for example, many one mole of CS3 Cl with Na2 Na and one mole of CS3 CH2 Cl. I'll go back once again. So here what happens? We are taking different alkyl halide. We require two more molecules of alkyl halide. But here we are not taking the same. We have one is methyl chloride. Other one is ethyl chloride. What would be the product? Could you tell me? Now tell me the product in this reaction. You have taken one mole of each. Okay. So yes, if you say propane, I won't say no, it's wrong. CH3, CH2, CS3. But it's not like only propane we get here. Along with propane, we get ethane. We get butane as well. CS3, CH2, CH2, CS3, butane. I said that the reaction, we do not have any control on this reaction. So what is the radical we get from this? We get methyl radical, CS3. And we will get ethyl radical, CS3, CH2. Or wherever we have radical, all these radicals have tendency to combine and get stabilized. We are taking one more. So we have so many such molecules. Any molecules we have. Means it's not like we have one single methyl radical. Along with this, like this, we have any radical present. Like this also, we have any radicals present. Okay. So it is possible that methyl radical combines with another methyl radical. It forms ethane. It is also possible that methyl radical combines with ethyl radical, forms propane. It is also possible that ethyl radical combines with ethyl radical itself and forms butane. So if they ask you, what is the total number of product in this reaction, your answer would be, we get three product here, three alkanes. Tell me, is it clear? Yes. Is this clear? Okay. Now, suppose we have two different ways, two different conditions you are assuming. Suppose you want to prepare, you want to prepare ethane. One method is what? You will take CS2, CS3, CL plus NA, we will get ethane, the previous reaction. And one more reaction is this. Which reaction is more efficient? The previous one or this one? Suppose if you want to prepare ethane, if you want to prepare ethane, then in this reaction, what happens along with ethane, you are getting two more alkanes whose boiling point is comparable. Then what happens? After this, you see alkanes, you need to separate the mixture here. You need to separate propane and then you need to separate ethane and butane. So you need to put some extra effort into it. Effort in the sense, these are the reactions, industry-based reaction. If you want to prepare ethane, then what all things you can do? So in the industry, cost is a major factor. You have to be efficient. You have to make more product in less expense. More amount you need to get so that you can use it and you can supply it into the market. So the point is, if you have two different ways to form ethane, we'll choose the more efficient one. Efficient means minimum input, maximum output. Are you getting me? Yes, correct. So what happens here? If you are going to use this method, means two different alkali, if you are taking, then we are getting a mixture of alkanes. For that, you have to put some more money. You need to make that arrangement, distillation and all. And then you have to separate ethane propane and butane. So why would we do this? If you have one simple way, it will take only methyl chloride. Directly, it will take only ethane. Simple, done. So that is why Bush's reaction, if you see, it is best for preparation of symmetrical alkanes. Symmetrical as in ethane, butane, if you have to make it make propane, then you need to get with this reaction. But you won't get only propane by this. Along with propane, you will get this one also, which is not cost effective. That's why, if the question is, which of this alkane can be best prepared with like in Bush's reaction, correct? So we will go by even number of carbon atoms if it is present, like ethane with 2, butane 4, hexane, heftane, like that. It's clear to all of you? Yes. Okay. Another reaction, which is very much similar to Bush's reaction, and the name of the reaction is, write down, Franklin reaction, third one. Third one from alkylolide C, Franklin reaction. What happens in Franklin reaction? Like I said, it is very much similar to Wohl's. In this case, we won't take Na, but we take Zinc. Rx plus Zn gives RR and Znx2. This is the reaction we have. So if you take Zinc, it is Franklin. Okay? Simple. Next, the preparation of alkane from Grignard reagent. Oh, one second. I missed one. The next one, you write down, number D. Wurge generally, excuse me, Wurge generally we use, we cannot compare two different reactions for the efficiency, but for symmetrical alkane, Wurge is preferred. And then it depends, like which one is easily available and less money you need to put in. Suppose Zinc is very difficult to get, or it's the cost of Zinc is more than sodium, then we'll go by sodium. Usually we say, Wurge reaction is the best method for the preparation of symmetrical alkanes. Okay. One more reaction we have, we call it as CoriHouse synthesis. C-O-R-E-Y. CoriHouse synthesis. In CoriHouse synthesis, we are taking again, all these are the preparation method of alkanes from alkyl halide. So we have alkyl halide and this is allowed to react with lithium Li in presence of ether first. So it forms Rli. Then this is allowed to react with copper halide, C-U-X, copper halide, then it converts into R2 C-U-L-I. Two molecules of this we take over here. R2 C-L-U-L-I, we call it lithium alkyl, dialkylcuprate. The name of this compound is lithium dialkyl, you have to memorize this one. Dialkylcuprate. Now this lithium dialkylcuprate is allowed to react with the another alkyl halide. R2 C-U-L-I is allowed to react with the another alkyl, like two molecules of that, R-X. You see, R- is not R here. Suppose it is methyl, then this is definitely not methyl. That is what it means. Okay, two different alkyl groups we are taking. When it reacts with two Ri R-X, it forms two R, R- plus Li-X plus C-U-X. This is the reaction. One very important thing here that you need to keep in mind that this R-X that you are using here, it should be primary for the best yield. Primary means one degree alkyl alkyl lead we should take for the best yield. That you will understand if you see the mechanism. We will see that also later. Okay, so we have three, four, like two, three methods more we have for the preparation of alkene and then we have chemical reactions. So most probably in next class, we will almost finish this chapter, means alkene, alkene definitely will finish. Alkyl chapter is not that big. Okay, you must finish the NCRT group 30 and 40 after finishing hydrocarbons. We will quickly go through it. Okay, fine. We will see you in the next class. Okay, we will continue with this preparation. Okay, okay, bye. Take care.