 Hello guys, good evening. All right, so today we are going to start a new chapter that is hydrocarbon, right? Isomerism we've finished. Only one conformation is left. We'll do that in the last, OK? So, OK. So what we are going to do today, hydrocarbon, see? So if you look at this chapter, hydrocarbon, we have three different portions in this, OK? Three different portions. The first one is alkene. Then we have alkene. And the last one is alkali. Organic chemistry in all the chapters, two things we have to study, OK? One is the preparation method of that particular compound. And the other one is the properties. Properties could be physical or chemical properties. If you look at the chemical properties, again there we have chemical reactions. So mostly we have chemical reactions in organic chemistry, OK? We can have chemical reactions for the preparation of alkene, alkene, alkyne, right? We can have chemical reactions which reflects the chemical properties of these compounds, OK? That is what the entire chapter we have. Whether you are doing hydrocarbon or you are doing alkylalite or you are doing alcohol phenol ether, aldehyde ketone, anything. It includes methods of preparation and then chemical properties, which we have only reactions. If you look at this compound, alkene, OK? It is the saturated hydrocarbon. It is a saturated hydrocarbon. Saturated hydrocarbon means it does not show, does not show, addition reaction. Addition reaction. Later on we'll see what is this addition reaction and why we call it a saturated hydrocarbon, OK? All the carbon atoms, all carbons are sp3 hybridized, OK? There's no unsaturation in this. Hence, the degree of unsaturation is 0 for this molecule, OK? This is all about alkene. We'll see the preparation methods and chemical properties. But general idea you must have. Alkene and alkyne comes under unsaturated hydrocarbon. I'll light on this for you. It is unsaturated hydrocarbon. Unsaturated hydrocarbon is the one which shows addition reaction. Or you can also understand which contains double or triple bond. That is unsaturated hydrocarbon, OK? So alkene is an unsaturated hydrocarbon, right? At least one of the carbon atom is sp2 hybridized. At least one, we can have more than one also, right? Degree of unsaturation for alkene is equals to 1. Similarly for alkene, at least one of the carbon atom is sp hybridized. And degree of unsaturation for alkene is 2. What is the formula for degree of unsaturation? Anybody? Degree of unsaturation, we also call it as double bond equivalent, DBE. And this is equals to C plus 1 minus H plus X minus N divided by C. When double bond forms, right? So the bond strength of double bond is more than to that of single bond. However, we do not compare the stability of alkene and alkene. We say both are stable. And that's why they exist, OK? But if you want to compare this alkene and alkene in general with a comparable molecular mass, in general what happens? The double bond has more strength. So the stability of a double bond is more than to that of a single bond, OK? But this is not relevant, actually. This kind of question, they won't ask you that alkene and alkene, which one is more stable, OK? So this is the basic understanding you must have for hydrocarbon. As the name itself suggests, it is a compound made up of carbon and hydrogen. Well, that is what hydrocarbon? Hydrocarbons are the compounds made up of carbon and hydrogen, OK, hydrocarbon. Alcohol, alkene, halide are the derivative of hydrocarbons, OK? Organic chemistry is all about the study of hydrocarbon and its derivative, OK? So one by one, we'll discuss all the three portions here. First, we'll see alkene, the various preparation method of alkene, then alkene and then alkene, OK? We have many reactions in this, like I said. For all the reactions, we are not going to discuss the mechanism, OK? Because that is not important. And mechanism, whatever it is required, we'll discuss. But in detail, we'll discuss in a chapter called Reaction Mechanism that will start in 12th standard, OK? So heading you write down, all of you. Heading is alkene. The general formula of alkene is, we all know, Cn H2n plus 2, OK? And hence, the DOU is 0. You can calculate from this. Yeah, one second. I'll just, yeah, I know there is some construction going on. It will be there. Actually, I tell you the reason. In here, in my apartment, actually, what happens? 2 to 4 is a break time, right? 2 to 4 is a break time. So they cannot do any cutting or chipping during this time, 2 to 4. And luckily, we have class at 4 only. So once the time goes to 4 o'clock, they start all this cutting and chipping. So it will develop, therefore, like I would say, like 30, 40 minutes, not more than that. So you won't have any disturbance after that, OK? So yeah, correct. So I think it's better now. I'm using the headphone now. Probably it's better. Can you hear me, all of you? Properly, yeah, OK. So alkane, we were talking. Alkane is a general formula, CN, H2N plus 2, right? All the carbon atoms are sp3 hybridized. What are the different methods we have by which we can prepare alkane, OK? So write down the heading, methods of preparation. Methods of preparation. Like you see, there are different, different methods. In NCRT, only few methods are given. If you look at the portion for competitive exam, we have to study a lot. And the reactions of alkane and the reaction of alkane, alkane, alkane, it will be there in all the chapters, right? Like it's not like you are done with this chapter, so it's done. Organic chemistry, all the chapters are linked. You study chapter A, you have to have application of things that you have studied here A in chapter B. Then it is required in C. All the chapters are interlinked. You must have the information of the previous chapter when you're going to the next chapter. So once you finish all this chapter, then the second time, once you want to revise, you have to build the bridge between the two chapter. Like from here, how do we go to this? From B, how do we go to C? So that's why I always say, when you finish organic twice before your exam, then only you will be in a better position. Once if you're solving, it won't help you much. Okay, you will only get 50% of it. You finish organic chemistry only one. I'm talking about the competitive exam, okay? So you should set your target like this only that before December, coming December, you need to finish it twice. That's why in 12th grade, I prefer to finish organic chemistry first so that you can revise and you can finish the second time also on your own. So that's how we need to go in 12th standard. I'll let you know all these things, what is required. So methods of preparation, we can prepare alkene from different, different compounds. We can prepare from aldehyde ketone. We can prepare from acid. We can prepare from alkyl halide. So there are various methods we have. So what method we have which can convert an aldehyde or ketone into alkene? Okay, what method we have which can convert an alkyl halide into an alkene? That is what we need to understand, right? So the reaction that we are going to do here for aldehyde and ketone, here it is there in the preparation of alkene. But when you do the chapter called aldehyde ketone, this reaction will also be there but that will be in the chemical properties of aldehyde and ketone. Okay, so you will get common reactions in this chapter. That's why the revision, second time revision is very important for this one. Anyways, the first method of preparation you write down, the first method of preparation we have, preparation of, preparation from, from aldehyde and ketone. Aldehyde and ketone. Two different methods we have by which we can prepare alkene from aldehyde and ketone just a second. Okay, like I said, two different methods. The first one is, the first one is you write down, Clemensian reduction. This is a name reaction, right? Clemensian reduction. You have to memorize this, how it happens. We can discuss the mechanism but mechanism won't help you much. Okay, you should know these reactions, how it forms products. So what happens in this reaction? First of all, you should know that what reagent we are using for this reaction. You have to memorize the reagent because in the reaction, in the test or exam, it is the reagent only which gives you the idea of what reaction it is actually. So in Clemensian reduction, we use amalgamated zinc. Okay, we use amalgamated zinc, Zn with Hg we are using. This is amalgamated zinc, Zn Hg with concentrated Hcm. This is the reagent we have for Clemensian reduction. We can also use amalgamated sodium in instead of zinc. Na Hg also we can use. But rarely you will see this present here. Mostly we have Zn Hg only. Okay, but this is also fine. We have the same reaction. So mostly we use zinc with Hg. Okay, so what happens in this reaction? Suppose I am taking a ketone. Ketone is RC double bond O CH3. And it is allowed to react with the reagent called Zn Hg and concentrated Hcm. So in the exam or test, you will understand the reaction by this reagent only. You will see, okay, Zn Hg cong Hcm. It is the reagent for Clemensian reduction. So it is a Clemensian reduction here. What happens in this reaction? In this reaction, the product that you get here is our CH2 CH3. This is the product you get. What you have to memorize? You have to memorize that whenever you have a carbonyl carbon, carbonyl group like this, C double bond O reacts with Zn Hg cong Hcm, the same reagent here. Then this C double bond O, instead of this, you need to replace CH2, that is it. C double bond O into CH2, okay? C double bond O, you have to replace this with CH2, correct? This is the Clemensian reduction we have. What is this? It is an acidic medium reaction because we are using acid here. This is what you have to keep in mind. It is an acidic medium reaction. In presence of acid, the reaction is taking place. We have one more method by which we can prepare alkene from aldehyde and ketone. And the name of the reaction is, we call it as Clemensian reduction, sorry, Wolf-Keschner reduction, number two. It is under aldehyde and ketone, only number two. It is Wolf-Keschner reduction. Both reactions are important. Okay, you should know these reactions, okay? As far as the product is concerned in both reactions, it is same, same product you get, okay? But the reagent is different, mechanism is different, okay? So what reagent we are using here in Wolf-Keschner? The reagent is, first of all, we use NH2, NH2, that is hydrazine. It is hydrazine, it's a base, basically. And in the second step, we use a base, like KOH. In presence of KOH, we heat in the second step. So this is the base we are using in the second step. So this reaction, unlike the first one, it is taking place in presence of a base. First one is an acid, okay? So what happens here? We have a reaction, suppose I'll write down this reaction. We have a phenyl group. We can have anything over here, C double bond O, CS3. Suppose this molecule we have. And this you are allowing to react with NH2, NH2 first, hydrazine. And then in the second step, we are getting, we're taking a base, KOH and we are heating it. We can also take here, NaOH is a base, right? That's also possible. Again, the same thing, C double bond O will convert into CH2, that is it. We have CH2, CS2. This is the product we get here. Obviously, Clemensha and Vulkrishna, as far as the product is concerned, is the same thing. Here also, C double bond O you have to replace with, you have to replace with CH2. And same thing we do in the first reaction also, but yes, the reagent is different. The mechanism, part of the reaction is different. Understood? This reaction actually takes place in two steps, Vulkrishna, okay? However, if you remember this, at C double bond O you need to replace with CS2, that is fine, you can do the questions from this. But how this reaction proceeds, in between what happens, this first convert into, like it happens this way, I'll just show you. See, you have this molecule, I'm just taking a simplest example, RC double bond O and CS3, suppose I'm taking. Okay. So when this is allowed to react with hydrazine, hydrazine is this NNH2 and this nitrogen has two hydrogen. So what happens, just to memorize the reaction, how to write down the product I'm telling you. In first step what happens, this O goes out with this two H, forms H2O. So the product would be RC double bond, H2O goes out, it is NNH2 and CS3. This we call it as hydrazine. And in the second step, this hydrazine is heated with a base, KOH. Then this gives you RCH2 CS3, like this it happens. But yes, this intermediate step is not very much important. You should know C double bond O should convert into CS2, right? It discuss the mechanism, how it happens, what is the process we have, but not required. It's not required, so I'm not discussing. Okay. Could you tell me the product we get in this reaction? Suppose we have cyclohexanone, cyclohexanone with NNH2, NNH2, hydrazine, okay. We can take here NNH2, NNH2. In the second step, any base we can use, one of the base I've given you, KOH, but we can also use tertiary butoxide ion, PBUO minus K plus, right? And solvent we can use here DMSO. See, I'm just telling you what base we can use here. If you don't want to use this, directly KOH you can use. But you never know what reagent is given in the reaction, in the question, right? You should know all the various possibilities. Tertiary butoxide ion is also a, and I'll tell you what, wait, one second, Gayatri. What is tertiary butoxide ion? It is also a base. What is the structure of this? Structure of this would be tertiary butoxide. Tertiary butoxide, how would it be? CS3, CS3, CS3, this is tertiary group. And butoxide, because we have four carbon, O minus tertiary butoxide ion. K plus, it is a salt, so K plus, NE plus you can write. So tertiary butoxide ion, the structure is this. We write it this way, T stands for tertiary, B-U-O butoxide ion, like this. DMSO is a solvent, it's polar aprotic solvent, which is polar aprotic, I don't get you, Prakul. Where is this acidic and basic things are coming? I'm not discussing acidic and basic thing here. What is exactly your doubt, tell me. I'm not discussing acidic, basic thing here. What is your doubt, tell me. I'm not getting you. Yeah, one second, whatever I will scroll. One second, guys, I haven't finished it. One second, I was getting a doubt, that's why I stopped. You know, the strengths are different. How do they use the same product? We are getting only one product here. You're talking about this base and this base, right? See, when the bases are different, they can affect the rate of the reaction. If you have very strong base, right? In that particular case, maybe the reaction rate would be very fast or slow, depends upon the mechanism, right? You have to understand the mechanism for this. But the product would be same. The nature of product would be same because both are bases, so purpose of base is same over here. Very strong base or weak base can affect the rate of the reaction accordingly, like the mechanism, whether it either it will increase the rate of the reaction or decrease. But if you talk about the product, that will be same. We can say we can achieve the product or get the product in a lesser time, sooner or later, we can say. But base is base, so the behavior of base in the reaction would be same, whether you are taking very strong base or weak base. I'll just stall the left. Give me just one second, okay? So this is, I was talking about this base and this is a solvent we have. We call it as dimethyl sulfoxide. Dimethyl sulfoxide. We will discuss this in reaction mechanism, like I said in detail we'll discuss over there. Okay, dimethyl sulfoxide, the structure is this, CH3S double bond O, CH3. It is polar up-protic solvent. We have two types of solvent, polar-protic and polar-up-protic. This one is polar-up-protic solvent. You must have heard about SN1 and SN2 reaction, isn't it? SN1, SN2 reaction affected by the presence of solvent also. Like just for your information, if you have heard, polar-up-protic solvent favors SN2 reaction. What is SN1 and SN2? We'll discuss later in reaction mechanism, but yes, solvent also has an effect over here. What is polar-up-protic solvent? It is a solvent in which hydrogen is not associated with any electronegative element. That is polar-up-protic solvent. You see, this hydrogen is associated with carbon. We don't consider electronegativity difference between carbon and hydrogen. If you talk about alcohol, ROH, hydrogen associated with oxygen, water, hydrogen associated with oxygen, electronegative element, acid, hydrogen associated with oxygen, electronegative element, right? All these are polar-protic solvents. What? Polar-protic solvent, right? And in polar-protic solvent, SN1 reaction favors. This solvent favors SN1 reaction. We have reasons for this also. Why this favors SN1? Why this favors SN2? But yes, that is not the point that we need to discuss now. We'll just let it be. We'll discuss it later, right? Understood polar-protic and polar-up-protic solvent? Polar-protic solvent are those solvents in which hydrogen is associated with an electronegative element like oxygen, nitrogen, et cetera. In polar-up-protic, we don't have hydrogen and electronegative element bond. Like we have DMSO, DMF dimethylformamide is another example. Ether is another example of polar-up-protic solvent. All these things here, right? So we are using this base. So basically in Wohl-Kirchner, you have to keep this in mind. In second state, we are using a base over here. So what would be the product? Product would be same, C double bond O. We'll replace C double bond O with CH2. So this is the answer. Product would be same. We can achieve this product in a lesser time or more time depending upon the rate of the reaction, but product would be same. It won't change. Yes, no doubt. Ether is polar-up-protic. Ether is this, obviously, ROR. There's no oxygen-hydrogen bond. So it is polar-up-protic, clear? Yeah, I'll just scroll left one second. Okay, then copy this if you haven't. Which question? The one I discussed in the last, Shradha, this one. Yeah, this is the answer, you see. Cycloalkane. We didn't know. C double bond O, we just finished, convert this into CH2. That is it. Okay, understood? No, it's not. It's written here 2H, but this carbon is there only. This carbon with 2H. So C double bond O, and here we have CH2. That is what we are doing over there. Our same thing, that also we can understand. And you just remove oxygen and attach to hydrogen there. Same thing. Okay, so this is the two methods by which we can prepare alkene from aldehyde and ketone, clemention and volucleicinate. Clemention in acidic medium, acid is present over there, and volucleicinate is basic medium, base is present over there, okay? Now, the second one you write down, second method of preparation, from alkene. What all methods we have by which we can prepare alkene from alkene? What is alkene? Alkene is C double bond C, right? If you want to convert this into an alkene, so you have to break the double bond, pi bond, sorry, like this, okay? You have to dissociate or break this pi bond. So for that, what we can do, so that the pi bond we can break. So one simple method is what? We can use hydrogenation. Attach hydrogen here and remove this pi bond. So first method by alkene we have, yes. First method by alkene we have method A you write down under this. From alkene, the first method by catalytic hydrogenation. By catalytic hydrogenation. Terms also you try to understand. Catalytic hydrogenation. First of all, what do you mean by hydrogenation? Hydrogenation is just the addition of hydrogen, right? Catalytic hydrogenation means in presence of a catalyst, right? So basically it is the addition of hydrogen in presence of a catalyst. What catalyst we can use? We can use nickel, we can use platinum, we can use palladium, anyone. Anyone of these, you know, catalyst we can use for the, you know, what we say, hydrogenation reaction. So hydrogenation is addition of hydrogen, of hydrogen. Now you look at this reaction. Suppose we have an alkene. I'm taking the simplest one, ethene, CH2 double bond CH2. Hydrogenation, so plus H2. Anyone catalyst we can use, I'm using nickel, platinum, palladium, anyone you can use. It converts into CH2, single bond CH2 and the two hydrogen get attached to this carbon atom, forms an alkene, correct? Addition reaction, it is. Now, since it shows addition reaction, that's why alkenes are unsaturated compound. Alkene make a say over addition reaction, you see? Not possible, no. You need to break the bond, right? It was substitution over addition, right? Since we have only single bond here, hence it does not show addition reaction. That's why it is saturated compound, alkenes, okay? So what happens in this? You must memorize this particular property. This is a syn addition reaction. Syn addition means both hydrogen get attached from the same side, either from the bottom or from the top. It's not like one hydrogen is attaching from here and other one attaching from this side, either from the top or from the bottom, okay? How it happens? The purpose of catalyst here, that hydrogen molecule get adsorbed at the surface of the catalyst. Suppose it is nickel catalyst. This is the surface we have. So this hydrogen atom get adsorbed at the surface. Adsorbed, I'm saying. I'm not saying adsorbed. It is adsorbed. Adsorption is a surface phenomenon. It will happen at the outer surface only. Like you see, if you drop one, if you put one drop of ink on a white chalk piece, right? Suppose you have white chalk piece, just one drop you place over here, right? And if you break this chalk piece, suppose this is a one drop of ink we have, blue ink, right? If you break this, you will see, inside this chalk piece, there is nothing. The ink particle won't be there. It is just at the outer surface only. Yes? Yes or no? This ink particle presents at the outer surface only, correct? So this phenomenon is adsorption. It's called, we call it as adsorption. And what happens if you just place this chalk piece in the bottle of the ink, right? And keep it for two, three, four, five minutes. Then this ink particle will get absorbed in the core of this chalk piece also. So that is the process of absorption, okay? When the particles goes into the core of the substance, okay? That is absorption. But when it is at the outer surface only, it is not present in the core of this substance. It is adsorption, correct? So absorption, what happens in absorption? First we'll have adsorption. And then the particles goes into the core of the substance. That we call it as absorption. So here what happens? Hydrogen molecule or atom gets adsorbed at the surface of nickel, right? This bond is, you know, is about to break here. This bond is about to break hydrogen resin bond. And then it gets attracted towards this carbon atom like this. And slowly what happens? This pi bond breaks, these two bond forms, and this gets free from the surface of the nickel, okay? So this is what the process feels. And hence it is a sin addition. Since it gets adsorbed at the surface, so it is not possible that one hydrogen is coming from the bottom and other one is coming from the top. Either from the top or from the bottom. This is the only possibility there. Hence it is sin addition. Sin addition means both hydrogen get attached from the same side of the carbon atom. Same side of the double bond, okay? Now the next method we have from alkene is by hydroboration reduction reaction. B, right down. By hydroboration reduction reaction. Hydroboration reduction reaction. What happens in this? We have the reagent in this. In the first step we use BH3. BH3 or B2H6 also we can use. Anyone. And in the second step, we use an acid, weak acid, Cs3, COOH we can use, okay? What happens in this reaction you see? Suppose the reaction is CH2 double bond, CH single bond CH3. And it is allowed to react with BH3 first and CS3 COOH in the second step, okay? So the product we get here is CH2, single bond CH, single bond CH3. This hydrogen here and this hydrogen here. This two hydrogen will get attached. Important point here is this hydrogen, the source of this hydrogen is BH3, coming from BH3 and source of this hydrogen is the acid that we are using, okay? Why this is important, the source here? Because sometimes what happens, they will give here like this. BD3 deuterium with CS3 COOH, like this they'll give. You should know that, see in this one we do not say it is synone anti because the mechanism is different, okay? It happens in a different way. Means both carbon atom, it does not attach with the hydrogen at the same time. First, BH3 reacts and then in the last step CS3 COOH reacts. So here synone anti, we do not talk. Yes, yes, B2S6 is that only, okay? Three centered two electron bond, okay? So here we do not talk about synone anti addition. When you see the mechanism, you'll understand why it is not synone anti. So in this case, what happens? BD3 will give D here, CS3 COOH will give H. So you see, the carbon atom, you consider this one, the carbon atom which has the lesser number of hydrogen, which means more substituted carbon atom will get H from BH3, isn't it, okay? Beachvala, H from BH3, lesser substituted one gets H from the acid. So BD3 will give D, this D will attach at this carbon or this carbon, what we can write. It should be CH2, CH, CH3, and D, yoga. And we know CS3 COOH will give H and this H will attach to the carbon atom which has more number of hydrogen, which is this. So here we have the H. So they leave all possible options. One of the options will be this, another one will be D here, H here, another one will be both D, like this little frame. Understood? This is hydroboration oxidation reaction. Okay, next slide down from alkyne. Alkyne, we know alkyne in which the carbon atom has triple bond, the two pi bond you need to break and forms C single bond C, like this. This is what the reaction should be. Okay, so what we can do in this method? We can also do the hydrogenation of it, right? So in hydrogenation what happens? Catalytic hydrogenation, H2NI, first it gives you alkyne because one pi bond will break and two hydrogen will get attached, like we discussed in the case of alkyne. Again, you will do the same thing, H2 with NI. We'll have C double bond C, C single bond C, H, H. This is the answer we have. Basically, we are doing hydrogenation twice over here. Next reaction, write down from alkyne halide. Very important reaction we have. We call it as Wurz reaction, W-U-R-T-C. That also we can do for cool. There are many methods for the preparation. Okay, we can do that also. Okay, what happens in Wurz reaction? Very important reaction, this one is, we take two molecules of alkyne halide in this. Suppose two Rx, alkyne that we are taking and is allowed to react with sodium, two molecules of sodium in presence of dry ether. Dry ether is the solvent here. Dry ether means there's no water present here. So what happens? This molecule gives RR, higher alkanes plus two NAX. This is the reaction we have, higher alkanes we get. Let's take an example here. Suppose we have CS3Cl plus two NA, and since we have two molecules of this, so another one is Cl, CS3, dry ether. So in this reaction, this NA takes this two chlorine, two NA takes these two chlorine, forms two NaCl and goes out and the product would be CS3 combines with CS3 and two NACl. Okay, so what is the key points we have here? The key points are, first of all, we get higher alkanes, higher alkanes. Can we prepare methane from this method? No, we cannot prepare because the number of carbon atom for the molecule it forms will be minimum two, two or more than two, it's possible. So this is a drawback of this reaction that methane cannot be prepared. Methane cannot be prepared. Okay, another one is used to prepare, used to prepare symmetrical alkane. Ethane can be prepared. Ethane can be prepared. Butane can be prepared. Hexane can be prepared. Symmetrical alkane, okay. Two mechanism possible. Just write down the name, we'll discuss this later mechanism. Reaction mechanism we'll discuss. Two mechanism possible. One is hierarchical mechanism in which hierarchical forms. Other one is ionic mechanism, ionic mechanism in which ion forms. Product would be same. We can have two different path for this, hierarchical or ionic mechanism. Okay, one very important reaction you see. Usually what happens in virtual reaction, we take symmetrical, sorry, same alkyl halide. Means suppose you are taking CS3Cl, then we'll prefer to take the same molecules, like two molecules of the same compound, like CS3Cl, CS3Cl, right? Usually the reaction goes like this only. We prefer to take same alkyl halide. Yeah, one second. Same alkyl halide we prefer. What is the reason behind this? We'll discuss now. Copy this. Done, Dave. Suppose we take two different alkyl halides. It's not like the reaction is not possible with two different alkyl halides. That also we can do. But what is the drawback in that reaction? That I'll show you just a second. Yeah. So what happens here you see, if you take two different alkyl halide, for example, I am taking CS3Cl, methyl chloride, plus 2NA, and another molecule is ethyl chloride, CS3CH2Cl, in presence of dry ether. What is the product we get here? Could you answer? Two NACL obviously forms and along with NACL, the product would be propane. That is propane, right? Only propane, anything else? Anything else could we get? See if they ask you that how many products possible in this reaction? The answer would be propane plus will get ethane, CS3, CS3, plus will get butane as well. CS3, CS2, CS2, CS3. So we'll get the mixture, yes right Madhavati, very good. We'll get the mixture of alkene over here. Why mixture we get? That try to understand here. See it is a free radical mechanism. So we'll get CS3 radical here and we'll get ethyl radical here, which is CS3, CS2 radical. All these are present in the reaction mixture. Now free radical mechanism is what? It is not in our control. It means we cannot control this reaction. Last one we have no butane, you see. I'll tell once again, just let me explain, okay? What I said? Free radical forms and we do not have any control on this reaction. Means wherever we have free radical mechanism, we do not have any control on that reaction. Free radicals are highly reactive. They combines with another radical and forms the compound. So what happens? It is possible in this reaction that methyl radical combines with methyl radical itself. Since we cannot control it. One second, Meghana, one second. One possibility is what? When methyl combines with methyl radical, so it forms ethyne, combination of methyl with methyl. If ethyl combines with ethyl radical, it forms butane, CS3, CS2, CS2, CS3. And when methane combines with ethyl radical, sorry, methyl combines with ethyl radical, it forms propane. That is CS3, CS2, CS2. And we do not have any control on this. That's why all three possible products we can get in this reaction. Tell me now, what doubt you have, okay? It's very simple. If you take only CS3, CL, correct? So CS3 will combine with CS3, gives you ethyne. Similarly, if you take different one, then why not the CS3 combines with CS3 itself or ethyl radical itself or methyl and ethyl radical itself? All possible combinations we can have. No, percentage combination, percentage comparison we do not count here. We do not calculate. We just say we get the mixture of alkene and the number of product in this reaction would be three. Percentage composition we cannot count here. That they won't ask, don't worry. No, no, no. We can't say that. We must have some way, we can understand based on the stability of radical and all, but that is not required to calculate and all. No, no major product in this. Like I said, we get mixture of product, not major or minor, okay? Now you see the problem is what in this reaction I'll tell you. No, if you take X of CS3, CL, then also you will get ethyne as the product because CS3, CL will convert into methyl radical, right? And we cannot think of that two methyl radical combines will get ethyne, will get ethyl radical. That's not possible. All the radical combines and the chain terminates over there, will get ethyne simply. It's not like two radical combines and forms a new radical, that's not possible. And then that radical combines with the other one, that's not possible. The bigger chain we won't get, if you take X of CS3, CL, the product only would be ethyne, that is it. Yes, correct. Now I was talking about here that what is a drawback in this reaction? Now, suppose if you want to prepare ethyne in industry, suppose if you want to prepare ethyne, we can do two different reactions here. One is we can take two molecules of CS3, CL, the first reaction that we did, plus two and A in presence of dry ether, this forms CS3, CS3, right? One method is this. And another one is done in the top like this one. You also, we are getting an ethyne. Which one is a better way? Better method to prepare ethyne. First one or the second one? Second one. Second one is the better one, because in this we do not get the mixture of product. The composition of ethyne will be maximum here because only ethyne is present. Here what you need to do? Once you do this reaction, you'll get mixture and then you have to separate the mixture in this. And for that again, you have required some manpower, you require electricity. You have to put some, you have to make the arrangement for separation of alkene. So for that, you have to put money into the industry, right? And that's why this is not the preferred way to prepare alkene. Got it? So in industry, what we do, the always the objective is what? Minimum input and maximum output. So if we have a way by which we can prepare only ethyne, where we don't need to separate it, means we don't need to put extra money into this for separation, then obviously this one is a better method we have because we are getting only one product that is what we want. But here, some unexpected product also forming. We don't want this ethyne and butane, right? Hence, this method is not the preferred method we have. So in Woods reaction, what happens? We always prepare symmetrical alkanes and we prefer to take same alkyl halide for the reaction. Understood? Correct? Understood? Okay. Now, similar reaction we have, name and the reagent is different. Second reaction, right down from alkyl halide, we call it as Franklin reaction. Franklin's reaction. In Franklin reaction, what happens? Again, we take two molecules of alkyl halide, two Rx. Two Rx is allowed to react with zinc. The metal is zinc over here we are using. So it forms Rr and ZnCl2, Znx2. This is Franklin method, okay? Not then important just keep that in mind. Another method we have from alkyl halide, yes, see the reaction path is different, mechanism is different from Woods reaction. But if you see only the reaction in one step, it looks very similar to Woods, only the reagent is different, metal is different. So more or less it is same. You can memorize in order to write down the product, how to write down the product. But yes, reaction mechanism is different, a different path we have in this reaction for the preparation of alkyl. See, name of the reaction is Corey house synthesis. So what happens in this reaction you see? First of all we will take alkyl halide Rx and it is allowed to react with lithium. Lithium is a metal in presence of ether. So it forms Rli, lithium alkyl here. This is further allowed to react with Cux. Copper halide. It forms R2 Cu Li. This we call it as lithium dialkyl cuprate, lithium dialkyl cuprate, okay. So what happens in this you see? Next method, next step, this lithium dialkyl cuprate is allowed to react with R-x, alkyl halide, a different alkyl halide. We can take same also, different also we can take, okay. But we take two molecules of this. And in this one, a product would be two R R- plus Lix plus Cux. What you have to keep in mind that this alkyl halide that you're using here is to ask an exam, this alkyl halide that we're using here, it should be one degree. You know what is one degree, two degree, three degree halide? One degree, two degree halide, you know? Yes, if the carbon containing halogen, CH2X attached with only one alkyl group, it is one degree, right? If the carbon which has halogen, it attached to two alkyl group, it is two degree, secondary. If it is attached with one more alkyl group, then this starts to be three degree. So here we are taking one degree halide here, right? Preferably for better yield. It's not like with two degree, the reaction is not possible, but the yield won't be great over there, okay? Because it's an elimination also possible over there. Anyways, point here is, we can take same alkyl group here or different as well we can take. This method is used, write down. It is, it is, or this way I don't, this method is better than Wood's reaction. This method is better than Wood's reaction and can be used and can be used to prepare and can be used to prepare symmetrical as well as, unsymmetrical, symmetrical as well as unsymmetrical alkanes, symmetrical as well as unsymmetrical alkanes. Remember, symmetrical alkanes is the one in which we have even number of carbon atom, unsymmetrical alkanes, which has odd number of carbon atom. Okay? So in this reaction, in this method, Corey-House synthesis, we can prepare both symmetrical as well as unsymmetrical alkanes, okay? It depends what alkyl halide we are taking here in this step, okay? If it is same, symmetrical, different and unsymmetrical, okay? Next reaction, next method of preparation from Grignard reagent. Have you heard Grignard reagent? Grignard reagent is RMGX, this is Grignard reagent. Sigma bonded, sigma bonded organometallic compound, sigma bonded organometallic compound. Organometallic compounds are those compounds in which we have carbon and metal bond present. No, it is different, alkyl halide we are done. It is the different methods now from Grignard reagent we are preparing, okay? Alkyl halide is done, three methods we have done. So it is sigma bonded organometallic compounds. Organometallic compounds are those compounds in which we have carbon and metal bond, carbon metal bond present, okay? We'll study this organometallic compound little bit in coordination compound in 12th grade, okay? Three types of organometallic compound we have. One is sigma bonded, other one is pi bonded and other one is sigma and pi bonded, okay? So we have three types of organometallic compound. Anyways, RMGX, in this if you look at the charge, alkyl group has minus one charge, halosin has minus one charge and magnesium has plus two charge on it. So overall, Grignard reagent we write this way, R minus and MGX plus. Like this we write Grignard reagent, okay? How do you prepare Grignard reagent? You take any alkyl halide, Rx. Rx, you allow this to react with magnesium in presence of ether. It gives RMGX, preparation method of Grignard reagent. Alkyl halide with magnesium and then ether, it gives RMGX, correct? Now, this Grignard reagent can show two types of reaction here. One in which it behaves as acid base reaction and another one in which it shows substitution tendency, substitution tendency. Substitution tendency, it shows in case of carbonyl group, carbon, in case of carbonyl compound, okay? Substitution in case of carbonyl compound, correct? So carbonyl compound, if you are taking, so R minus behaves attack on the carbonyl carbon and so substitution reaction. We are not concerned with substitution now. Acid-based reaction we need to see. When it shows acid-based reaction, when it reacts with, write down, when reacts with compound containing active hydrogen. Active hydrogen means what? Hydrogen attached to electronegative element. Hydrogen with electronegative element. Call it as active hydrogen. For example, you see. If you take RMGX, okay? R minus MGX plus and you allow this to react with water, H2O. So H2O may, if you see, it is the hydrogen attached with an electronegative element, oxygen, this hydrogen. This R minus will take this H, okay? So what are we writing? RH, okay? And whatever is left after removing H, that part will get attached with magnesium. So you would have MG, OH, X. Alkene. So we got alkene with respect to the alkyl group of Grignard reagent, understood? So suppose if you have a compound, say NH3, then also the alkene would be same, RH will be the same. This will be attached to this and you will get RH. Other compound would be MGX. So it will be attached to NH2. If you have acid, CS3, CO, OH. Again, see, this hydrogen is active hydrogen. Again, it forms RH. And with this, this will get attached. O, C double bond O, CS3, right? So whatever is left, that will attach with magnesium and we get alkene here, clear? S3PO3, you have to look at the structure. S3PO3's structure is like this. So one and two. Next method of preparation from red phosphorus and HAI. This is the reagent we have, red phosphorus and HAI. It is a powerful reducing agent. It is red phosphorus HAI. It is a powerful reducing agent. And reduces, reduces aldehyde, ketone, reduces aldehyde, ketone, alcohol, carboxylic acid, reduces aldehyde, ketone, alcohol, carboxylic acid, reduces aldehyde, ketone, alcohol, carboxylic acid into alganes. Into alganes, okay? So all the compounds here converts into alkene when it is allowed to react with red phosphorus and HAI. First example you see, RCH2OH. It's an alcohol, red phosphorus HAI. Gives RCH3. Okay. So you must take care of that the number of carbon atom here equals to the number of carbon atom here on the product side. That would be same. There's no change in the carbon atom. Alcohol, we are done. If you are taking an aldehyde, red phosphorus HAI, again it will give RCH3. If you have a ketone, RCH2R. If you have an acid, red phosphorus HAI, RCH3. So we have equal number of carbon atom on the product side also. Okay. Two more methods we have for the preparation of alkene. Right down the next one by Colby's electrolysis. Right down in this reaction, in this reaction, sodium and potassium salt of an acid. Yeah, once again I'll go back. Okay. Right down in Colby's electrolysis. In this reaction, sodium and potassium salt, sodium or potassium salt of an acid, sodium or potassium salt of an acid, goes under electrolysis, goes under electrolysis and forms, goes under electrolysis and forms higher alkanes. I'll repeat once again, Shradha, goes under electrolysis and forms higher alkanes. I'm repeating. I said in this reaction, sodium or potassium salt of an acid, sodium or potassium salt of an acid, goes under, goes under electrolysis and forms and forms higher alkanes. Okay. What happens in this reaction? Overall, the reaction goes like this. We'll take two molecules of the salt of an acid. For example, R-C-O-O-N-A, sodium salt of an acid. When it goes under electrolysis, it forms R-R, it forms CO2, it forms H2 and NaOH. Okay. This is the reaction we have. H2 is a gas. Obviously, carbon dioxide also is a gas. The reaction of anode, if you see, the reaction of anode will take two molecules of R-C double bond O-O-N-A. It forms ions, R-C double bond O, O-NA+, and then in this one, free radical forms here. How you see this? This O- that we have over here, this electron pair, it comes over here and this sigma bond breaks and goes onto the alkyl group, two electrons. Two electrons on two alkyl group. So one alkyl group will take one. So we'll get two radical here, two alkyl radical and plus CO2. And then these two alkyl radical, R radical, R radical, combines and forms R-R. So higher alkanes we get. Point is that this carbon dioxide evolves at anode, carbon dioxide. The reaction of cathode, if you see here, we have two NA+, plus two electron in water. It gives two NAOH, plus H2. So hydrogen gas evolves at cathode. So all these questions they ask at anode what gas evolves at cathode what gas evolves. One more thing you must take care of here that in alkane, if you have carbon number from one to four, means first four member from methane to butane, first four member of alkane, methane to butane, their physical state is a gas, exists in gaseous form. So sometimes they ask you this question, how many gaseous product we get in this reaction? Suppose the reaction is this. I am assuming a reaction here, CH3, C double bond O, ONA. It goes under electrolysis. Number of gaseous product. Could you tell me here? It is not the gases getting collected on anode. Gas evolves at anode. At anodic reaction. Okay. Gas evolves in the anodic compartment you can say. So the product here would be, we have CS3, CS3, ethane, carbon dioxide would get evolved, hydrogen will get evolved, and NaOH will form. So CO2 and H2 is the gaseous product we have here. CO2 and H2. But since alkane contains only two carbon atoms, this is also the physical state is gas only. So in this case, the answer would be three. Three gaseous product we get here. Okay. One note you write down. We also get ester as a byproduct in this reaction. Okay. Last method of preparation we have. Soda lime decarboxylation. Last method write down. Soda lime decarboxylation. So write down. Carboxylic acid. Carboxylic acid goes under. Goes under. Decarboxylation reaction. Carboxylic acid goes under. Decarboxylation reaction. Carboxylic acid. Goes under. Decarboxylation reaction. In presence of soda lime. In presence of soda lime. And gives alkane. Okay. In presence of soda lime. And gives alkane. So reagent here. For this reaction. The reagent is soda lime. Soda lime is a mixture of NaOH. And CaO calcium oxide. Yes. I said. Carboxylic acid. Goes under. Decarboxylation reaction. Carboxylic acid. Goes under. Decarboxylation reaction. In presence of soda lime. And forms alkane. Okay. Carboxylic acid goes under. Decarboxylation reaction. In presence of soda lime. And forms alkane. Okay. So what happens in this you see. Suppose we have an acid. I'm taking RCOOH. Carboxylic acid. RCOOH. We just need to heat this acid. With soda lime. Soda lime is NaOH. And CaO. What happens. CaO2 will get eliminated. RH forms. And carbon dioxide gas evolves in this reaction. How it happens. This CO2. Comes out. This CO2 comes out. RH will get combined. And we'll get an alkane. Okay. So when you heat an acid with soda lime. Right. Soda lime. We call it as decarboxylation reaction. This CaO. Traps carbon dioxide. And helps the elimination of. Carbon dioxide in this reaction. Got it. Okay. If you. Try to see the mechanism in this reaction. I'll just few steps I'll explain. How the mechanism proceeds. It is a base. Right. So base will give. OH minus. This OH minus will take this H plus. It is an acid. H plus will release. It takes up by OH minus forms what. H2O and carboxylate iron. Carboxylate iron. Then what happens. This comes over here. And the sigma bond breaks. Forms are minus. So it is the anion forms are minus. And CO2 eliminates. Right. So intermediate is a car. Anion here is. Okay. So intermediate is a car. Once this are minus forms. In the next step, what happens. Our minus. Will take active hydrogen. From water, like it was happening in case of. Reagent. And forms. This is how the reaction proceeds. Intermediate is a car. Anion. More stable car. Faster will be the reaction rate. Okay. These are the various preparation methods of. Alcane we have. Next write down properties of. Alcane. We'll see physical properties. Physical properties. First point. Write down. Like I already told you. C1 to C4. First four carbon atom. The physical state. Is a gas. C5 to. C17. Till 17th carbon atom. It is liquid. Only one exception we have in this. And the exception is. Neopentane. Except. Neopentane. What is the IUPAC name of neopentane. IUPAC name. Two comma two dimethyl propane. Yes. Two comma two dimethyl propane. There are two mythologies. A second carbon atom. So this exists in gaseous form actually. Right. Gaseous forms. Otherwise all compounds are in liquid state. If you have carbon atom. More than equal to C17. More than C17. Not equal to. Or if I write down this as. C18. Then it is. Wakes like solid. Not that important. First point you must remember. C1 to C4. Write down alkanes are. Non-polar molecules. Second point write down. These are non-polar molecules. Why it is non-polar? Because all carbon atoms are. SP3 hybridized. There is no electronegativity difference. Hence no polarity. Third point. Melting point and boiling point. Melting point and boiling point. Is directly proportional to. Molecular mass. More molecular mass. More will be the melting and boiling point. Okay. Next point to write down. Among isomeric. Molecules. Among isomeric molecules. Among isomeric molecules. The melting point. And boiling point. The melting point. And boiling point. Decreases with increase in branch. Among isomeric molecules. The melting point and boiling point. Decreases with. Increase in the number of branches. Right. So basically. As branching increases. Melting point and boiling point decreases. Because surface area decreases with branching. Okay. Hence the answer is. One more note you write down in this point. One note write down. It is observed that. It is observed that. Even number of carbon atoms. It is observed that. Even number of carbon atoms. Has more. Has more melting point. Then odd number of carbon atoms. Even number of carbon atoms. Has more melting point. Then odd number of carbon atoms. And this is true for comparable. Mass. Means we cannot compare. See for. And this is true for comparable mass. Means we cannot compare. See five and see 10. Means the molecule which has five carbon atom. And the molecule which has. 10 carbon atom we cannot compare. In that case the molecule mass will be the factor. But suppose we have C5 or C6. It is C5 and C6. Suppose we have. Like this. So in this case. This molecule. And this molecule. The melting point. Would be. Would be more than this one. It has more melting point. The reason is. You see that. The terminal carbon atom. Is on the opposite side. So we can have better packing in this. Yes. Butane is slightly more than. Painting the melting point. Because of packing. That is a reason in trans isomer. Transka. Melting point is more than. To that of cis. Because in trans may we have opposite. And that's identical molecules present. So we can have better packing over there. Less hindrance better packing. Which gives more melting point. Melting point is associated with the. Packing in the crystal. Better packing. Strong bonding. More will be the melting point. Okay. So we have slightly more melting point than. Painting. Exen will have slightly more melting point than. Painting. So one carbon. Like more or less we can compare. But we cannot compare C6 and C13. In that case, obviously we'll go with melting point. Sorry. The molecular mass. Okay. So this you must observe must keep in mind. So there are a few physical properties we have for. Next write down chemical properties. Like I've already told you. In chemical property also we have chemical reactions of. Alkane. So first we are going to see halogenation. Very important reaction. For Kevin by point of view also. Okay. Halogenation means what. Reaction of alkane with halogen halogen. Okay. Write down. Alkane reacts with. Reacts with. Bromine. Or chlorine. We are to our seal to. In presence of sunlight. Sunlight UV light. UV light. Or in dark. Or in dark. But if the reaction is taking place in dark. The temperature must be high. Right. Either in place of light or in dark. At higher temperature. At higher temperature. Around. 250 to. 400 degree Celsius. This temperature you don't have to memorize. 250 to 400 degree Celsius. Alkane reacts with bromine. Chlorine in presence of sunlight. UV light or in dark at higher temperature. And forms. Forms alkyl halide. Alkyl halide. Direct chlorination is difficult because of strong. Floating floating bond. Okay. Direct iodination is also difficult because the reaction is reversible. Backward reaction favors. I'll tell you what. To write in chlorination and iodination. But mostly. The halogenation is true with. Bromine and chlorine. So what happens in this reaction you see. Suppose I have taken CH4. Methane. When it is allowed to react with CL2. Chlorine in presence of light. H nu. Then H and CL combines. It is a free radical mechanism. Forms CS3, CL. And HCl. Right. This is what it happens. Mechanism is free radical mechanism. Write down. Free radical mechanism. Okay. If you have only one molecule of this. If CL2 is not in excess. This is the product we get. If this is in. Limited quantity. But if this CL2 you are taking in excess. For example. CS3 H. CL2 excess. H nu. Then the reaction won't stop. At this point. At CL will definitely form. This is the first step. But the reaction won't stop here. Okay. It continues. The reaction continues. With another CL2 molecule. Again one more. H will get replaced. CS2 CL2 with HCl. Further. This again reacts with CL2. Because it is in excess. So CSCl3. Further. This also reacts with CL2. Forms CCl4 plus HCl. Here the reaction will stop. So it continues to. Carbon tetrachloride. Eventually we get. This molecule is called. Chloroform. Chloroform. CHCl3. This is possible with. Free radical mechanism. Sorry. Excess of chlorine. Mechanism is free radical. Direct is not possible. Fluid with fluorine. Direct fluorination is not possible. I will come to that. Direct fluorination is not possible. Okay. This is active. Then we have chlorine. Direct fluorination is difficult. Because a strong fluorine-fluorine bond. If you want to get alkyl fluoride. You can use another substance. For alkyl fluoride. If you want to prepare. We use some fluorinating agent here. Which can give. F minus ion. For alkyl fluoride. See this. We have many different fluorinating agent. C2H5. We have two C2H5. Br. With. Hgf2. Hgf2 is the fluorinating agent. It forms two C2H5. F. And Hgbr2. One note you write down here. Direct fluorination. Fluidination. Is. Explosive. Direct fluorination is explosive. And can be achieved by. And can be achieved by. By the action of. By the action of. Inorganic fluorides. Inorganic fluorides. On Bromo or Ido. Bromide or iodide. This is the inorganic fluorides. On Bromide gives you this. You see this order. From here we can also conclude one thing. Right. Down. Bromination is. Bromination is slower than. Chlorination. Bromination is slower than. Chlorination. And usually carry out. And usually carry out. At higher. Temperature. And usually carry out. At higher. Temperature. Okay. Bromination is slower than. Chlorination. Right. Hence to get the. Substantial amount of product. Bromination usually carry out. At higher temperature than. So that will have a. Significant rate of the reaction. And we get the product sooner. Right. Up next point to write down. Direct iodination of. Alcane is difficult. Since it is. Since it is. Reversible in nature. The reaction is reversible. All the cows you must know. I would a nation is reversible. And floor in a nation is explosive. Okay. That's why we use some inorganic flow rights. SGF to. Direct. I would a nation of alkane is difficult. Since it is reversible in nature. So because of this reversible nature, it has backward tendency. Means what happens if. This CH4 suppose. Is allowed to react with I2. Correct. Reaction is. Reversible. So it forms CH3 I. And. But again it reacts. And it has high tendency to go backward. So we won't get the. Product here. Significant amount of product. Significant amount of product. So what we need to do. Hence the. Iodination. Hence Iodination carried out. All these informations are important. Carried out. In presence of. In presence of. An oxidizing agent. An oxidizing agent. So what is the. Use of oxidizing agent you see. Suppose we have this reaction. I2 it is already written I won't write it. We have this reaction CH4 I2 gives this. Okay. So if you keep on removing this. I continuously. Then we'll force this reaction to go in. Forward direction and we get the product here. Right. So for this. We can use. We can use. Nitric acid or Iodic acid. The oxidizing agent that we're using here is. Nitric acid. Nitric acid. HNO3. Or we can also use. Iodic acid for this purpose. And the formula of Iodic acid is. Hi O3. This we can use for the oxidation. So what happens this. Hi. When reacts with Iodic acid. Hi O3. It converts into I2. And H2O. Correct. It transforms this oxidizing agent oxidizes. Hi into I2. And hence the reaction goes in forward direction. That is what the purpose we have. Okay. Three, four more reactions we have its direct reaction. We have no mechanism. You have to mark this up. The reagent and how to write down the product here. Next slide down. Nitration. Second reaction. Nitration. For nitration. The reagent we have nitric acid. Everywhere for nitrogen will take. HNO3 only mostly. HNO3. At around 400 degrees Celsius. What happens in this. You see, suppose you have an alkan. CS3. CH2. CH3. Heated with HNO3. So you will get all possible products here. Means. You can get CS3. CS2. CS2 NO2. You can get CS3 CH. CH3 NO2. At the second carbon. And it is also possible. Because the we are hitting at high temperature. It is also possible that carbon carbon bonds also break. CS3. CS2 NO2. And CS3 NO2. All possibilities we have. Mechanism is again three radical. No need to understand or memorize the mechanism. Because you should know this. HNO3 will get all possible. Nitrate. Next slide down. Sulfonation. See for sulfonation. The rate of sulfonation first of all you see. Rate is. Maximum for three degree. Then two degree carbon. And then one degree carbon. That is the rate of the reaction. That's why preferably we take. Tertiary. Carbon here. Like this. Three degree carbon we have. Reagent for sulfonation is sulfuric acid. H2SO4. This gives. R3. C. SO3H will get attached. And H2O goes out. This H and OH combines. Because you see. H2SO4 is what? S double bond O. S double bond O. OH. OH double bond O. So this H and OH combines. You see here. This H and OH combines. H2O forms. And carbon will get attached with SO3H. Which is this. Yeah. So. Next slide down. Can we do with one molecule? One molecule is fine. One molecule is not required. One molecule we can take. Okay. Yeah. Sulfonation is done. Next slide down. Isomerization. Fourth one. Isomerization. Reagent for isomerization. That you have to memorize here is AlCl3. You just need to heat the compound. In presence of AlCl3. Around 400 degrees Celsius. Chain isomers here. CS3, CH2. CH2, CS3. When it is heated with AlCl3. It forms. 2 methyl propane. This is how we get the isomers here. Just we need to heat this. Okay. No need to run the mechanism. Not important. It is not given also. Just need to memorize how to write down the product. Chain isomers you need to write down. Suppose you have heptane. Okay. Few examples you have to keep in mind. CS3, CH2. 5 times. And CS3. So it is heptane. When you heat this with AlCl3. You will get CS3. CS3, CS3. Iso octane you will get. This is the product we get. Iso octane. Sorry. Iso heptane it is not octane. Even in octane also you will get the same thing. Okay. We have two more reactions in this last two reaction. Second last one is combustion. Combustion is. When you heat the compound in presence of oxygen. Okay. Atmospheric oxygen. Yeah, one second. Done. Yeah. In combustion what happens any hydrocarbon. If you heat with axes of oxygen. That reaction is combustion reaction. Okay. Hydrocarbon under combustion. It always eliminates carbon dioxide. And what. For example. Suppose we have a hydrocarbon CX. HY. Heated with O2. Excess. Excess of oxygen. It forms carbon dioxide. And water H2O. Always. Any hydrocarbon. Any hydrocarbon. X if you balance this X. This is why it is Y by two. And this would be X plus Y by four. The balance reaction. Okay. Always exothermic. This reaction is. Delta H. Is less than zero for this reaction. You know, and I. The heat always releases in this reaction. Always exothermic. Last reaction is aromatization. Very important, but sometimes they ask this also. The reagent to get an aromatic compound. In the reaction is. Very important. This one is. And that you will get this. Just you need to know the reagent is CR2O3. With L2O3. This is the reagent we have. And we heat in presence of this region, the molecule. Okay. So suppose you take N hexane. So you take normal hexane. And you heat this with. In presence of this reagent. Then it hexane. So it gives you benzene death. This is the reaction. If you take N heptane. Heptane means we have seven carbon atom. With same reagent. Then obviously the benzene will form. We always have benzene in this. But we have seven carbon atom. So one CS3 will get attached to this benzene. So you take normal hexane. And you heat this with in presence of this reagent. So one CS3 will get attached to this benzene ring. We call it as toluene. The name of this compound. One CS3 on benzene. We call it as toluene. Similarly, if you have octane. And octane. Then again a benzene ring. With. Two CS3. At two different position like this. And we have one more possibilities. Where we have the benzene ring. And on which we have CH2. CH3 group attached. This we call it as orthozylene. The molecule is. Orthozylene. And this one is simple. We call it as ethyl benzene. Copy this down. N means normal hexane. Normal hexane means we don't have any branch. This one. This is normal hexane. No branch. That is in C isomerism. What happens when hydrogen atom will shift. Okay. Like you think of this one. You have to memorize that you will get only those reactions only. Okay. Because we don't have mechanism. But if you really want to understand you can think of like this. Suppose we have CS3. CHH. Sigma bond here. CH2 CS3. So when you heat this. Then carbon. Carbon hydrogen bond breaks. And carbon carbon bond also breaks. All. Like whatever the possibility we have that will happens. Because we have alkanes only two possibilities we have. Either we can break carbon carbon bond. Or we can break carbon hydrogen bond. So when it breaks, so it is possible. That this methyl will get attached to this carbon. And this hydrogen will come to this. Carbon. Hence we get this one. CS3 CH. CS3 CS3. So you need to write down. Chain isomerism. You need to write down chain isomercy. With that you can understand. Similarly, if you have heptane. So heptane is this one. Two. Three. Four. Five. Six. And seven. In this we have many possibilities. Means we can break this carbon carbon bond. Okay. And the hydrogen will come to that carbon. We can also think of any carbon atom present in between over here. The but the one which is more stable. That only will write. And that would be. When you get the branch. Isomer here, which is this CS3. CS3. CH CS3. This is the most stable. Other products also possible, but we'll write this one. Only one to reactions. We have that you have to memorize. There are many possibilities. You'll get many competitions here. Okay. But only one or two compounds. We have that you need to keep in mind. Understood this. This is it for alkane chapters. Okay. We are done all the reactions of alkane. Now we'll take a break. And after the break, we'll start with alkene. Okay. Preparation of alkene. And then chemical reactions of alkene. Correct. Fine. So we'll resume at 6 30. Okay. Take a break now.