 3 and 4 are identical. Yes. 3 and 4, the relation we can say in third and fourth compound, they are identical in nature. Because if you look at from the top, you will see this. If you look at from the bottom, you will see this. Right? So both are the same molecule. View angle is different. Yeah. Okay. So next, we are going to start the last chapter that is hydrocarbon. Okay. Hydrocarbon, we have three portions in this. That is alkene, alkene. And the last one is alkyne. The first one is saturated hydrocarbon. Saturated hydrocarbon. All carbon atoms are sp3 hybridized and degree of unsaturation do you for this one would be zero for alkene. Okay. For alkene and alkyne, these are unsaturated hydrocarbon because they can show addition reaction. Okay. Alkyne carbon atom, at least one of the carbon atom would be sp2 hybridized, at least one degree of unsaturation would be one here. Alkyne, if you see, at least one carbon atom is sp hybridized, carbon must be there. And degree of unsaturation would be two. What is the formula of degree of unsaturation? Well, we have done this. Do you, we also call it as DBE equals to C plus one minus H plus X minus N divided by two. This is the formula we have. Okay. It represents the number of pi bond or ring present in the molecule. Okay. So in all these compounds, alkene, alkene, alkyne are in organic chemistry. We have only two parts. Here we have. Okay. No, you won't get any fractional value. Fractional value you will get when it is, when resonance is possible. Otherwise you won't get. Okay. I forgot to discuss that ascent. I'll discuss that one. So guys in organic chemistry, what happens in all the chapter you see, whether you are doing alkene, alkene, alkene, alcohol, alkyl, halide, aldehyde, ketone, carboxylic acid, amines, any, any chapter if you see, we have only two things. One is the preparation method and other one is the reactions, chemical reaction, that is chemical properties. Okay. So both contains reactions only. Preparation method also what, you know, what all method we have by which we can prepare one particular compound. In that also we have chemical reactions, chemical properties like one particular compound shows what kind of properties. Okay. That is also chemical reactions. So basically in organic chemistry, we only have chemical reactions, organic reactions. Okay. So you should know all these reactions. Now, obviously there are mechanisms, there are, you know, way by which we can understand how the reaction is proceeding and what are the various intermediates, whether the, you know, stabilizing factors we have. Okay. All these things we can discuss. So we will be discussing all these things in reaction mechanism next, there is a 12th grade. Okay. Here we'll have a brief discussion of all these reactions because actual mechanism if you try to understand, there are so many things before that you need to understand and that we'll discuss in reaction mechanism. Okay. So here, like, you know, we'll discuss wherever it is required here a little bit, but in detail we'll discuss over there. Because if you go in detail, it will take one month to finish more than one month in fact. Okay. So that's why I'm just taking up the important things plus, you know, the essential thing which is required for you to understand. Right. So all these reactions that we are going to do here, it will again come in reaction mechanism. Reaction mechanism basically is a backbone of all the organic chemistry, entire organic chemistry. Okay. So if you are comfortable in general organic chemistry and reaction mechanism, your 60% of your organic headache is gone. Okay. Because these two are the, you know, very important chapter we have in order to understand organic chemistry. If you do not get it, then you will feel like, okay, all the chapters you need to mug up simply, which is not possible at all. Right. Okay. So before going into that one doubt, I know one of you have asked that is what is SN2 reaction? Okay. So quickly it'll take five minutes. I'll just discuss then then move on to the alkane part we have here. So SN2 reaction is what it is a second order. That's why we have two over there. Second order nucleophilic substitution reaction. Nucleophilic substitution reaction. To understand this reaction, you should know what is a nucleophile. Nucleophilic means it comes from nucleophile. So you should know what is nucleophile and what is the substitution reaction we have. Right. So again, this is a part of reaction mechanism that we'll discuss later. Right. So what happens in this substitution of nucleophile with second order reaction? Order definitely we haven't done yet. It is there in a chapter called chemical kinetics in 12th grade. Okay. So just order you let it be not required. So it is basically the substitution of nucleophile. Nucleophile means what? It is nucleus lover. Nucleophile. Pile means love. Nucleophile stands for nucleus. So nucleophiles are nucleus lover. Okay. Nucleus lover. So what is the charge in the nucleus we have? Positive or negative or neutral? Positive. Right. So positive charge lover means what? The nucleophile must be must be having what charge? It must be negatively charged. Then only it'll attract it towards the nucleophile, which is positively charged. So nucleophiles are negatively charged or it can be neutral as well. But if it is neutral, then we must have the lone pair present on it. Means electron density must be there. Like for example, water. Water is also, what also behaves as a nucleophile because we have lone pair present on oxygen. Electron density is high. Cl minus is a nucleophile. OH minus nucleophile. There are many things we have. Let's not go into detail of all these. Okay. But nucleophile is this. So suppose we have, we have an alkyne halide. For example, I'm taking the simplest one, which is this. Cl, H, H, H, CH3. Correct. Now in this, what happens? I'll take another one, not this one. Okay. Which minus nucleophile? There are so many things. I can talk two hours on this. Okay. What is actually happening in this reaction? Okay. But let's keep it short. Okay. Because in detail, we'll discuss later. So what happens? You see basic understanding what you should have. The basic understanding is, do we have electronegativity difference from carbon and chlorine? Yes or no? Quickly. Yes. Which one is more electronegative? Chlorine, obviously. So chlorine will draw the electron towards its side. Delta negative on chlorine, delta positive on carbon. Because of this charge separation, this negative charged nucleophile, it has the natural tendency that it will attack onto this carbon atom. Right. It is, it happens on its own. It's very spontaneous. We are not doing anything into it. Right. So since because of this electronegative difference, we have this positive charge on this. So this nucleophile has tendency to attack. Right. And since carbon have no tendency to form five bonds, but there will be an intermediate. Look, carbon will be like this. And this OH is trying to make a bond with this carbon. Backside attack. I hope you understand a little bit of Hindi all of you. Correct. Yeah. Okay. So this is backside attack. Okay. OH minus backside attack we have. And this CL is what this CL is trying to leave this molecule. Okay. And that's why this CL we call it as leaving group. Okay. So over just trying to make a bond with this carbon and you're trying to push the chlorine out of the molecule like to jump out. Like, let me make the bond. You go out. Okay. So in between this, there will be a five, like this is transition state we get. Trigonal bipyramidal transition state, which is highly unstable. Why? Because carbon cannot show pentavalency. Right. So eventually what happens, eventually OH will push the chlorine outside, chlorine will take this electron pair will go out and it will make a bond with this carbon and we get a product which is this. Correct. This is SN2 reaction. Another thing, it forms for a very fraction of time. You cannot even imagine just before the dissociation of this bonded forms. Right. Dotted bond means dotted bond means that OH is trying to make a bond and CL is trying to leave the molecule. This is delta negative. This is delta negative. And this is delta positive. Finally, it goes out and we'll get a bond between OH and C. Yes. Since CL is a leaving group here, it has to go out. Then OH cannot attack from this side because in Dono Mayfield repulsion hindrance Hoganah. So best is what it always take the back door. It will try to get entered into the room and then push this chlorine out. Correct. So this is SN2 mechanism. It happens from, you know, it's a nucleophilic substitution second order reaction attacks from the backside. Okay. One step reaction and gets this product. So there's an inversion in configuration. Obviously it is not a, you know, optically active compound. If it is optically active compound, then we'll have some configuration of this molecule. Suppose it is R. Suppose it is not, but suppose if we have an optically active compound here, so that compound will have some, you know, some configuration R or S, I'm assuming R. And since if you have back, backward attack over here, this R will convert into S over here. This we call it as inversion of, of configuration. And this is a term given over here. And we also call it as Walden inversion. Walden is the name of the scientist who has given all these, you know, things that, and because of him only we are going to study all these things here. Walden inversion or umbrella. Have you seen that? In like hard storm, right? It happens. So that inversion, that same thing, you know, if you try to relate with this one, with this one, it's like umbrella inversion. No, you see. It was this like this umbrella, right? That's why it is umbrella inversion. If you could relate. Got it. Understood. End it out. So it's just over. Right? If you try to understand the actual reaction, we have a lot to discuss. Okay. Just give me one minute. Yeah. So this is the reaction we have. So we have just studied one upper layer of it. Okay. This may, we have a role of substrate, we have a role of reagent, we have a role of solvent, many things. Okay. And all those things, if you try to study, it will take, you know, two months for this properly, right? So we won't do that. School may not push the name for comparative exam. Important. Okay. All those concepts we need to learn that we'll do later. So I hope you understood this essence to Pooja, it's clear. Yes. Okay. Yeah. Okay. Okay. Okay. So we're talking about this alkene, right? So alkene preparation method you write down, what are the methods by which we can prepare alkene? Okay. So the first reaction we have that is decarboxylation of acid, heading right down only decarboxylation reaction, right down in this reaction, in this reaction, sodium salt of carboxylic acid, sodium salt of carboxylic acid is heated with soda lime, is heated with soda lime and forms alkene and forms alkene. So what is the reaction we have here? You see sodium salt of carboxylic acid. So we can directly take sodium salt. We can also prepare this. We can take carboxylic acid and base NaOH. Correct. When it reacts, so it is an acid, it is a base. So acid base reaction, H plus OH minus forms H2O. And we get a salt of this acid, which is RC double bond O, O minus Na plus. This is the sodium salt of carboxylic acid. Okay. So we'll take sodium salt of carboxylic acid and when it is allowed to react with soda lime. Soda lime is this, delu, Na. Soda lime is a mixture of NaOH and CaO, calcium oxide. This mixture NaOH and CaO is called soda lime. When we heat this, then it forms Na2CO3 and it forms Rh alkene. This is a preparation method of alkene from carboxylic acid. So what happens in this reaction? You see, this COO and Na and NaO. This two combines, forms Na2CO3 and it goes like this. Okay. You have to memorize these reactions. Okay. So carboxylic acid method of preparation is this. Second write down from aldehyde and ketone. From aldehyde and ketone. The reaction that we use here, we have two different methods by which we can prepare alkene from aldehyde and ketone. The first method we have, we call it as clemention reduction. Very important name reaction it is. Clemention reduction. Since we are preparing alkene from aldehyde and ketone, this reaction is there in aldehyde and ketone chapter as well. Okay. So mechanism we are not going to discuss here. Mechanism, you understand mechanism means what? Mechanism means the step by step description of a reaction, like how the reaction is proceeding and what is the ways or path we have by which we get the product. Okay. So here in detail of mechanism we are not going, that we'll do in grade 12, okay, aldehyde ketone chapter. But what happens in this? How to write down the product is very important and that is what you need to understand. Even if you know the mechanism also, obviously in exam, you're not going to draw the mechanism. Otherwise you would have solved, you could solve only one or two question only. Right. So you should know how to write down the product in a given condition. Okay. Mechanism is there that helps you to understand the reaction and to memorize the reaction. Okay. If you understand the mechanism, it will be easier for you to keep those reactions in your mind. Okay. But eventually it will be the given reaction, what should be the product. Okay. So in this what happens, we take an aldehyde. For example, suppose I'm taking RC double bond O, RC double bond O, I'm taking a ketone here, C double bond O, C, a ketone functional group, reagent you must know what reagent we use in claymuncheon reduction. We use amalgamated zinc, ZnHg, amalgamated zinc with concentrated HcA. This is what the reagent and when you get the question in the exam, it is the reagent only which will give you some clue that what reaction it is. Okay. So reagent you must remember. And what happens here? If you take here ZmHg and concentrated HCl, then what happens? How to write down the product? This C double bond O, instead of this, just you write down CH2. Simple. This is the product we get. So in this reaction, what happens overall? C double bond O converts into CH2. How? What is the step mechanism we have? That we'll discuss later, not now. It will take a lot of time. We have to, you know, finish this a bit early. But that you will understand once you see the mechanism. Okay. So for now you let it be. For now you let it be, you will understand. It's not like I, you know, there's any difficulty. I can also, I know, anyways, I can explain you the mechanism, how it happens. But that will, you will feel more complex. Like if you want, I can tell you this. See this, in this reaction, what happens? You see, first of all, we have an acid. Okay. So just you look at this one, one mechanism, you will, I think you will be convinced with this that what I'm telling you, suppose we have this molecule, correct? We have cong HCl. Cong HCl means concentrated HCl. It will give you H plus, right? Acids H plus will get it. H plus, what happens here? This lone pair here, oxygen, it will take this H plus, attack it on H plus. Electron and proton attack. So it forms this R, C, single bond OH and CH3 here. Why single bond OH? So double bond. This oxygen donated electron pair so oxygen pair will have the positive charge because it donates its electron. Now oxygen, you see, it is an electronegative element and positive charge on an electronegative element is highly unstable. It will drag the pi electron from carbon, it will draw. When it draws the electron, then what will happen? Carbon electron, oxygen will go out. Here CS3, carbon electron, oxygen will get the positive charge. So you will get this positive charge, okay? Now what will happen? We have metal. This is the acid game, right? This is done by acid. Now we have metal, zinc. Metal will lose the electron, always. It provides electron in the reaction, Zn2 plus plus 2 electron. Now what will this electron do? It will attack this carbon. Right? So we have this reaction in the next step, R, C, OH, CH3, positive charge and here though electron, it will sit on this carbon atom, positive negative attraction. Then we will have this COH, CH3, right? And here it will be 2 lone pair negative charge, okay? Now this, again acid will give H plus, COH, CH3, right? Again this OH will get protonated, 2 lone pair on oxygen, attack on H plus. So it forms R, C, O, H, H, positive charge, H, CH3 and then it will take the electron pair, goes out as H2O. You see where oxygen goes? Oxygen goes with H2O. So we will get R, CH, positive charge, CH3 and then again Zn say you will get 2 electron from Zn, okay? This will attack on to this. Once again stress, I'll go back. Let me finish this. R, CH, negative charge, CH3 and then acid say H plus, negative charge will take this H plus. So it becomes R, CH2, CH3. And that is why I was saying, the mechanism just let it be now, okay? So Anurag, where did this oxygen go? Correct, yes. So that is not required actually, okay? You should know how to write down the product and all these things happen. I would say, mechanism will have maintained a distance now because little bit you should know about organic chemistry more. This was a GOC you need to go through the GOC again. Plus reaction mechanism, once we do, they will discuss the entire thing. After that, you will be comfortable with all this, okay? That's why I'm refraining from this mechanism. I'll go back once I can stress. Yeah, purpose of metal in the reaction mostly, I won't say always because organic chemistry, there's nothing certain, okay? So we cannot generalize any statement. We always have some exceptions for every statement, okay? Mostly. That's why, you know, in general we say wherever metal we have, that will provide free electrons into the reaction, okay? Mostly we use for alkyl halides also because alkyl halides, carbon halogen bonds becomes weaker in presence of free electron, okay? What kind of reaction is this? It's a name reaction basically. We cannot say it is a substitution or elimination. It's a name reaction. It involves, you know, just bond dissociation and bond formation process, that is it. It's not like, you know, all the reactions will come under one particular name, that is substitution or elimination or something like that. Not always good, okay? Yes, I'll go back stress one second. So I hope you understood this, that mechanism, obviously it will take time and it won't help you much at this stage, okay? So did you copy this mechanism? Yes, so like this. You just have to keep in mind that clay mentioned reduction converts aldehyde or ketone into alkene. C double bond O may, you just need to replace with CH2. Finish. So for example, if you have aldehyde, for example, you see this, Rc double bond OH, aldehyde. CaO is, you know, it is actually when Na2-CO3 forms, there is CO2 evolves in that process. And CO absorbs that CO2. So it helps in the elimination of carbon dioxide from the reaction. We also say in the soda lime decarboxylation reaction, CO2 evolves and this CO traps that carbon dioxide and it helps in the removal of carbon dioxide, okay? RcOH, ZNHG, conch HCl, if you produce here. C double bond O, what should we write? What is the product we get here? Anybody? What should be the product? C double bond O, this may convert to CH2. One H we have already, so RcH3. No, it's only RcH3 because number of carbon atom is not changing. Number of carbon atom is not changing. Yes, Rc. Yes. So this is the clay mentioned reduction happens in acidic medium. Another reaction in this from aldehyde ketone, we call it as Wolf-Kishner reduction. Wolf-Kishner. And don't ask me, where did Wolf-Kishner come from? Maybe a scientist might have written in the book anywhere. Right? Wolf-Kishner reduction. What happens in Wolf-Kishner reduction? Any aldehyde or ketone you take, for example, I am taking this one, pH stands for phenyl benzene ring, C double bond O, CH3. And this is allowed to react with NH2, NH2, hydrazine to react with NH2, NH2. So Wolf-Kishner reduction is a reaction in base, in presence of a base hydrazine. Clay mentioned is for acid. Okay, this is for base hydrazine. Okay? So what happens here? This two hydrogen will take this oxygen will go out as H2O and pH C double bond N, NH2, CS3. We'll get this. First of all, we'll get this. And then this compound is allowed to react with KoH, hydroxide. So C double bond N, N2 will come out and it converts into pH CH2, CH3, plus N2 will go out. So Anurag, where did this oxygen go? Yes, it became water. Because yes, this two hydrogen takes this oxygen, goes out as H2O, and this nitrogen, the double bond gets attached to the carbon atom. So what are the reforms? KoH reacts with this. Okay. And it eliminates N2 and hydrogen will get attached to this. We'll get CH2, CS3. No, no, no, not any longer. We have that mechanism, lone pair of nitrogen will take part in the reaction. But that is not required. Okay, don't try to, you know, get into mechanism of all the reaction. It won't help you. I didn't get you on drug. Resonance is a conjugation, and then we'll show resonance. If there is conjugation, then it will show resonance. Okay, see, let's not make it complex. What simple reaction. If you have aldehyde and ketone, we have two methods to prepare alkene. Wolf-Krishna reduction is also the same thing. You see, eventually, what happened here? C double bond O converted to CH2. This is what it happens, right? Same thing happens over here. Eventually, the result is same. C double bond O, this converts into CH2. Isn't it? That's what happened. But the mechanism is different, right? The reagent is different. The first reaction, Clemensian was taking place in a basic, in an acidic medium. Here, the medium is basic. The reaction is happening in the basic presence. That is the difference we have here. Clear? This is what you need to memorize. Yes, yes, you have to memorize this. You have to memorize this. Organic chemistry, take my word. You have to finish at least twice you have to do before your exam. Not always. Here, we can have anything. Are also possible. Okay? At least two times. Organic chemistry, at least twice you need to finish. If you do it honestly and properly, okay? Third time, if you do it, that would be the best for the preparation. Okay? So, understand this. Okay? This is from like carboxylic acid we finish, aldehyde ketone we finish. Now, you see third reaction from alkene. How do you prepare alkene from alkene? Alkene from alkene by catalytic hydrogenation. What do you mean by hydrogenation? Addition of hydrogen. Right? Hydrogenation means addition of hydrogen in presence of a catalyst. Okay? Catalyst. So, what will happen? We have the reaction supposed CH2 double bond CH2 reacts with H2 in presence of a catalyst and catalyst we can take any one Ni, platinum, palladium, anything. Anyone. So, it converts into CH3 CH3. So, what happens here? First, this hydrogen get adsorbed at the surface of catalyst. Supposedly, catalyst surface. So, H2 will get adsorbed on this surface. I am not saying adsorbed. I am saying adsorbed. AD, right? Adsorbed. And when it gets adsorbed, then it will get attracted towards this carbon atom and slowly this bond forms and this pi bond breaks and you will get CH3 CH3. The important thing in mechanism. But yes, you have to keep this in mind. If both hydrogen gets attached from the same side, it's not like one is from the bottom and another one is from the top. No, both from the same side, either from the bottom or from the top. And hence, this addition, we call it as syn addition. Syn addition, same side's addition order. Then, next, B. The second method from alkene is write down. It is hydroboration reduction of alkene. Hydroboration reduction. Reagent. What reagent we use? We use BH3 in the first step. And then we use, in the second step, we use an acid CS3 COOH. BS3 and acid we use. We can also use dimer of this BS3 that is B2S6 also we can use. So reaction is what? You see the CS3 CH double bond CH2 when it's allowed to react with in the first step with BH3 with CS3 COOH. Then it forms an alkene, which is CS3CHCH2 one etch here and another etch here. What you should know that this hydrogen is coming from the source of this hydrogen is BH3 and source of this hydrogen is CS3 COOH. No, here we do not say syn or anti. That's not possible here because the double bond breaks. It forms a single bond and again first. Yeah, that will make a difference. Okay, because sometimes they won't give you BH3, but they will give you BD3. BD3, if they give you deuterium, isotope of hydrogen, then D you cannot place at this carbon. It will be at the carbon, which is more substituted. Source of these hydrogen, all these are parts of mechanism. Source of these hydrogen you must know. If they give you BH3 with CS3 COOH, then you should know that CS3 COOH is where in the molecule. So what you have to memorize that on more substituted carbon, hydrogen comes from BH3 and the other carbon it is coming from the acid that is CS3 COOH. Clear? Butte 2 in is symmetrical compound. So whatever you make, it's fine. In that case, there is no difference. But when we have unsymmetrical, then we have to be a bit careful that the etch is coming from there in case of unsymmetrical. Yes, run up. See, butte 2 in is symmetrical. No. See, I'll tell you. Wait, just a second. I'll write down. I know it's a bit difficult to digest for all of you at this stage because all of a sudden the entire thing has been changed a lot. You will have a lot of doubt in your head, like what is going on and how the reaction is proceeding. Like Anurag asked, where does this oxygen goes? This kind of thing, this is a very natural thing that you are having now. But obviously, if you spend some time with organic chemistry, you have to be patient and spend some time with this particular subject. Slowly, it will come, it will place online. See, if you have this, so whether it is, like if it is, it didn't get to you, okay. So here what happens? We have CS3, CH, single bond CH, CS3 and hydrogen here. Since it is symmetrical, so it does not make any difference. Okay, so this is hydroboration reduction reaction. Next method we have from alkyl halide. Right down from alkyl halide. One second. Next, from alkyl halide. Right down alkyl halide. The first method we have for the preparation in this, we call it as Wurz reaction. W, U, R, T, Z, very important reaction, Wurz reaction. In this reaction, we take two molecules of alkyl halide. Like for example, Rx, it is allowed to react with Na and Rx. Two molecules we are taking here. In presence of dry ether, this is a solvent we have in which the reaction is taking place. So when this reacts, this X combines with Na and forms NaX sodium halide and R combines with R plus 2 NaX. This is the product we get. So we are getting higher alkenes here. For example, you see if you take two molecules of CS3Cl and 2 Na in presence of dry ether, you will get CS3, CS3 ethane and 2 NaCl. This is the product we get. This reaction involves free radical mechanism. Free radical mechanism. I am coming to that point of this second. Free radical mechanism. And we are getting symmetrical alkenes here. Drawback is methane cannot be prepared. Why we cannot prepare methane? Methane cannot be prepared because two alkyl groups are clubbing. So at least two carbon atoms must be there. Yes, at least two carbon atoms must be there. Hence, methane we cannot prepare by this method. What happens if you take two different alkyl groups? For example, Rx and 2 Na, and 1 Na, R dash, different alkyl groups. Then in this case, we will get mixture of alkenes. You will get another product as R dash, R dash and another one as R, R dash. So in the exam, they ask this question like how many products are possible. So if you take different alkyl, you will have all three possible combinations. Three products will be there. Three different alkenes you will get. Okay? You must have doubt. Ask me the doubt. No, it doesn't. Dry either doesn't. Yeah. Okay. And this is not, you know, useful. We don't take this alkyl, this kind of reaction. Burj reaction may generally, we take same alkyl highlight. Up in the industry, in the book, whatever we have given that we actually do in industry. Okay. I won't go into detail of that by ether we are using in the solvent. For any reaction, we require, you know, mobility of molecules so that they can collide and the reaction proceeds. And for mobility, we require solvent. And hence we have this ether over here. Ether is polar, a proteic solvent. Okay. I don't want to go into detail of all those who are cutting about them. Just let it be. Just to keep that in mind, if we have a inverse reaction, dry ether is the solvent we are taking. Role of solvent, let's not discuss now. We'll do that later. Okay. I'll try to understand this that. No, fine. Water obviously we cannot use. Question is why ether we are using. So ether, why we are using this, let it be. Why we have dry ether because obviously if you have water present, then water may also react in this because water has lone pair on oxygen. It's a polar solvent. Right. So it is reactive. It may react and forms alcohol instead of alkene. So that is the reason of dry ether we are using because we have to be a water free solvent. Otherwise water may also react in the, no reaction. That's why we are taking dry. Why ether? Because ether is polar, a proteic. You asked me last week what is polar or proteic solvent. Then I have to go in a different direction. So let's not go over there. Just to keep that in mind for now, that would reaction metal is sodium solvent is dry ether gives you higher alkane or our combines gives you alkane here. Now, why we are not preferring two different types of what name would reaction is it? Dry ether. It is written over here. Dry ether means water free. So you see which reaction polar, proteic and polar, a proteic. I said dry ether is polar, a proteic solvent. Water is polar, proteic solvent. So just solvent part, you just let it be now. Now you want to know what is polar, proteic or polar or proteic. Okay, one more thing. Polar, proteic solvents are those in which electronegative atom or hydrogen bond is like alcohol. Polar, proteic. Okay, ether, you see, there is no oxygen and hydrogen bond. It is polar, a proteic. DMSO, polar, a proteic solvent. DMF, polar, a proteic solvent. Okay. I don't want to go in all these, not required. Abhi chahiye nahi ye. Okay, so let's focus on reaction and how do we get the product in this. Okay, esko baat ne karengi. Water is polar, a proteic solvent. Right. Alcohol is polar, a proteic solvent. Acid is polar, a proteic solvent. Okay, phenol, polar, a proteic. Okay. So the question I was just discussing this one that Woods reaction is used to prepare only symmetrical alkane. Because if you take different alkyl halide, then you will get mixture of product here. RRV bando rahe, RDS, RDS, VORB bando rahe. Means, agar suppose meh CS3Cl le raho, with 2NA and then C2H5Cl, then kya ban jayega? Ethane banega? Butane banega? And butane? And eth-propane banega? CS3, C2H5 banega? Suppose we have to prepare ethane. Industry mujhe sip ethane banana hai. Okay. Ethane banana hai. We have two methods. Ek tu ye kar sakta hai. Dusala ye bhi kar sakta hai. Which one is more useful? Could you answer? Common sense. Which one is more useful? First one, because isme there is no by-product, right? Ethane banana tha ethane bane gaya. But yaha pe we are getting mixture, right? Bhi ye karne ke baal, what we need to do? We need to separate the alkane now. Ab ethane kualak karo, butane kualak karo, propane kualak karo. Uske liye you have to put some, you know, some other processes. Uske liye you have to, you required manpower, you required electricity, you required, you know, money for that. Means you have to have the setup in which you can separate this mixture, correct? So, that's why industry mujhe karate hain jo ki minimum input and maximum output, correct? So, that's why Woods reaction is usually used for preparation of symmetrical alkane, okay? And we use identical halides in this, different halides we don't prefer in this reaction. First, dusra kya ho kya? Mene kaha, mechanism is free radical, okay? So, when I take, when I liye one mole of this, one mole of this, so one mole me do, there are so many molecules, n-a molecules. So, we'll get n-a methyl radical here, n-a methyl radical bane gaha. Kase baal ne karenge mechanism? n-a ethyl radical bane gaha, okay? Ethyl radical bane gaha. Because we have n-a and n-a of this, so it is possible that methyl radical combines with methyl radical, to ethane manjaya gaha. Ethyl radical combines with ethyl radical, butane manjaya gaha, methyl radical combines with ethyl radical, propane manjaya gaha. So, all three possibilities we have, correct? And we cannot control this reaction because the reaction involves free radical, and free radical reaction is not under our control, that the intermediate free radical is highly reactive, and we cannot control this reaction. It happens on its own. If you take only one molecule, if you take only one molecule, then we have ethyl, single molecule if you're taking. So, we prefer Wurz reaction to get symmetrical alkene, okay? Same alkyl halide we'll take for this purpose, clear? Understood? One note you write down here. Tertiary, write down tertiary alkyl halide is not preferred for this reaction. Tertiary alkyl halide is not preferred for this reaction. Tertiary alkyl halide is not preferred. Okay, a similar kind of reaction, which we call Franklin reaction. B, Franklin reaction is also similar to Wurz reaction. The only difference is instead of zinc, sorry, instead of Na, we are taking zinc here. Metal is different, name is different, type of reaction is same. Okay? Third reaction, write down in this, Cori house synthesis, Cori house synthesis, Rx reaction with Li ether. As it is, you have to memorize this Li ether, then we have R Li, which is then allowed to react with Cu x converts into R2 Cu Li, di alkyl cuprate, lithium di alkyl cuprate. The name of this compound is lithium di alkyl cuprate. And then this lithium di alkyl cuprate R2 Cu Li reacts with R dash x, another alkyl halide we are taking here. It was Rx initially, and it is R dash x. Okay? And the condition here, this alkyl halide should be one degree alkyl halide for better yield. R R dash plus Li x plus Cu x, this is the product we get. Only one key point we have here about this. The alkyl halide that we are using in the second step should be one degree. This method is used for, write down, this method is used for, used for the preparation of symmetrical as well as unsymmetrical alkane. This method is used for the preparation of symmetrical as well as unsymmetrical alkane. Done? No, it's not like that. No, it's different. The mechanism is different. Here, we do not have free radical mechanism. In this, we have ionic mechanism, not free radical. Okay? Free radical mechanism, we cannot control. Okay, I'll tell you what happens here. See, first of all, Li is again a metal. So, this Li, you know, first group can release one electron. So, two Li we are taking, so it forms two Li plus plus two electron. And we know in presence of electron, the alkyl and halogen bond becomes weak. So, these are two electron. Because it's delta positive, this is delta negative. And when this happens, this X minus will go out. So, we'll get R minus and X minus. Then this Li plus, we'll take this X plus, forms Li X. X minus with Li plus, it forms Li X, lithium halide. R minus, because we have two moles of Li plus, one is used over here. Another one is this, forms R Li, right, R Li. This R Li is allowed to react with Cu X. So, we are taking two molecules of this, copper halide, it forms R2 Cu Li with Li X. So, we are using here copper halide. We are not using only copper. So, Cu X minus will go out and from R Li plus, Li plus will take this X minus, forms Li X. And this Cu, it bonds between this alkyl group and lithium and forms a lithium dialkyl cupidate. So, basically Cu X provides X minus by releasing electrons in the reaction. And then the last step, we have R2 Cu Li reacts with R dash X. Okay, done. Yes, what happened? Okay, next slide done from Grignard reagent. Have you heard the name from Grignard reagent? Same halogen also we can take, we can take different halogen also. Okay, Grignard reagent is RMGX. It is a complex compound, sigma bonded complex compound in which we have R minus, X minus and MG2 plus. We also write it as R minus MGX plus, R minus MGX plus. Now, this Grignard reagent behaves in two different ways. Okay, first of all, how do you prepare this? Alkyl halide if you take Rx and this if you are allowed to react with the metal MG in presence of ether if you heat, then it forms RMGX. Mechanism is, we will do it later. But like this we prepare Grignard reagent, RMGX. Now, we have R minus and MGX. It has two types of reactions. One is nucleophilic substitution, other one is acid base reaction. Like whenever R minus MGX reacts with a compound containing acidic hydrogen or active hydrogen, it behaves as a base and takes hydrogen from this. So, it forms Rh and MG over HX. This will become Rh, MG over HX. Correct? What happened? R minus is H. Whenever hydrogen is attached to an electronegative element, it is active hydrogen. We can also think of this with alcohol, Rh. If you think of CS3COH, Rh means alkene would be same with respect to the alkyl group of Grignard reagent. Other component if you see with this reaction, the second reaction would be Rh, MG, H, it forms O, C double bond O, CH3 and this X would be as it is. For this one, the another one would be Rh say H is gone, MG, O, R, XC. But this product is not important, important alkyl. So, whenever we have active hydrogen, means hydrogen attached with an electronegative element like oxygen, nitrogen, then we have acid-based reaction with Grignard reagent forms alkene. Next, write down from red phosphorus and HI. Red phosphorus and HI is a powerful reducing agent. It's a property you have to memorize. It's a powerful reducing agent. So, what happens? Whether you take alcohol, RCH2, OH, or you take aldehyde, RC double bond OH, or you take ketone, RCOR, R you take acid, RCOH. So, whatever compound you take, all these compounds reacts with red phosphorus HI and converts into alkene with equal number of red phosphorus HI converts into alkene. Red phosphorus HI converts into alkene. And that's why there's a saying in Hindi for this. Red phosphorus HI, takraega, alkene banjaega. So, reducing agent, powerful reducing agent reduces alcohol, aldehyde, ketone, acid into alkene. Mechanism is not there. You have to mug this up. Always keep this in mind that the number of carbon atom on the reactant side must be equals to the number of carbon atom on the proud side. Done? Okay. We have last reaction, preparation method of alkene. This we call it as Colbe's electrolysis method. In this reaction, sodium or potassium salt of an acid, sodium or potassium salt of an acid on electrolysis gives higher alkene. Sodium or potassium salt of an acid on electrolysis gives higher alkene. So, we'll take two molecules of this salt, RC double bond O, ONA. Sodium salt, two molecules of it. When it goes under electrolysis, it forms higher alkene. RR banega, CO2 niklega, two molecules, NaOH banega, and H2O banega. Gaseous form, gaseous form. Okay? This is the reaction we have. You have to memorize here that CO2 gas evolves at anode. Anode pis CO2 evolveoga, and H2 gas evolves at cathode. Cathode pis H2 evolveoga. Okay? Alkene physical property, if you see, C12, C4 carbon atom, means the molecule in which number of carbon atom is 124, it is exist in gaseous state. Gaseous state. Okay? Gaseous state may exist. Sometimes they ask this question, how many gaseous product we get in this reaction? Let's say, for example, if you have a compound say CH3 C double bond O, ONA. What is the product we get? Could you tell me? The product here, ethane, CS3 CS3, and CO2, H2, and NaOH. If the question is, how many gaseous product we get in this reaction? How many gaseous product we get in this reaction? Answer? Three. Very good. Because, only two carbon atom gaseous state. So, one, two, three gaseous will get, neat me a question. Number of gaseous state one. Okay? We get three gaseous here. Only if the carbon atom is 124. Okay? Anode pis CO2, cathode pis H2. So, this is it for the preparation of alkene. Okay? Next class, most probably we'll finish this chapter. Okay? We'll take two more classes, max to max, to finish, you know, because a little bit of P-block also we'll discuss, group 13, group 14. Yeah, tell me. Pranav, okay? Yes, Pranav, tell me. Then what? Tell me, you know, related to what? Achha, propane and HBR. Okay. So, propane and HBR we'll do next class. No. You will get Marko-Nikov addition, right? And like it go like this, CH3, C triple bond CH. So, HBR will be added. So, it will form CH3, C double bond CH2, BR. And Fitts addition will HBR form CH3, C, BR, BR, CH3. Gem dihalides you will get. Oh, sorry, here will be BR. Gem dihalides you will get. HBR addition is Marko-Nikov rule without any peroxide. Right? You know what is Marko-Nikov rule? So, we'll discuss this next class. That's what you are directly doing this, you know, the reaction. What is the logic behind this? You are not getting. We'll do this next class. I'll tell you one small thing here. Look, CH2 double bond CH2. So, HBR say H plus BR minus we'll get. And then this pi bond will take this H plus. So, one of the carbon atom will get positive charge and CH3. And then this PR will attack onto this carbon atom. If you have triple bond, the same thing happens twice. Okay? And we'll get more stable carbocation in that. Okay? So, we'll get more stable carbocation and their BR minus will get attached. Okay? I'm doing this in a hurry because there are a lot of things you have to understand. So, next class we'll discuss this in detail. Okay? Yeah. Okay. Take care. Bye-bye. You have exam, right? KPI? Yes. So, all the best. Just give your best and leave the rest. Okay? Don't think about the result. Just try to give your best in the exam. Okay? The result is always the outcome that is not in your hand. Right? What you can do, that only you should focus on. Okay? All the best, guys. Take care. Bye.