 Yeah, just a second. So the next one is, is alkyne. The general formula of alkyne is Cn H2n minus 2. Okay. And the basic difference, like the one of the most very important difference we have in the properties of alkene, alkene and alkyne is alkyne is acidic in nature. Why it is acidic? Because it has sp hybridized carbon atom and the hydrogen attached to this carbon, like for example, you see, we have CH3 C triple bond CH. If it reacts with a base, suppose NaOH, then we have here acid base reaction. This hydrogen, since the carbon is sp hybridized here. This carbon is sp hybridized. This hydrogen is acidic in nature because of highly electronegative carbon atom here. And this OH minus and H plus combines forms H2O and we get a salt over here, which is the sodium salt CS3 C triple bond C minus Na plus. This is the salt we get. So one very basic difference in properties of alkyne, alkyne and alkyne is among the three, alkyne is the one which is acidic in nature. That is very important property we have. We have different methods by which we can prepare alkyne. Like a few examples we have already done. Similar way, one more step you need to go, you will get the final product. Like for example, you see right on the heading method of preparation. Method of preparation. Method of preparation, suppose we have, first we are having here from gem dihalides. We know what is gem dihalides, two halogen atom present on the same carbon atom. Look at this reaction. The reaction is RCH2 CXH. This molecule, when it is allowed to react with alkoholic KOH. Remember the preparation of alkyne. We have done the same reaction, only once we have done with mono halides. So we had mono halides, we have done it once, HX goes out and we will get the pi bond. When we have alkoholic KOH, first we get alkyne RCH double bond CXH and then again in the next step, we are using a different base that is NaNH2. Again HX comes out, it forms RCH triple bond, RCC triple bond CH. This is what we get, the final product. Okay. Now here you see one very important thing. In the next step, we cannot use alkoholic KOH. NaNH2 soda amide, we call it as soda amide. It is a stronger base than alkoholic KOH. Stronger base than alkoholic KOH. Okay. So alkoholic KOH is not strong enough, so that it can remove hydrogen from an sp2 hybridized atom, carbon atom. The role of base is what? It takes OH H plus from this. Since it is sp2 hybridized carbon atom, so it is difficult to remove hydrogen for this base, alkoholic KOH. Hence, in the second step, we require a stronger base and hence we use NaNH2, forms alkene, alkyne like this. Similar way we have alkoholic KOH, acid-base reaction and then base takes hydrogen from the beta carbon and the reaction process. Same reaction we have. So I am not going to drill up all this. We can also prepare alkyne from visceral dihalytes. Exactly same reagent and same reaction we have, I will write down here, from visceral dihalytes. Visceral dihalytes, you see we have carbon, carbon attached with hydrogen here and two halogen atom, alkoholic KOH, HX goes out. It forms first an alkenyl halide, which is this double bond carbon with X. And then the next step, we are using a stronger base, NaNH2, forms an alkyne. So all these information you should have that NaNH2 is a stronger base than KOH. This you have to memorize. Okay, the next one is tetrahalogen. Tetrahalogen means if you have tetrahaloalkanes, from that also we can prepare. Third one, write down D-halogenation reaction or simply D-halogenation from tetrahaloalkanes. R, C, C, X, X, reagent we are using, zinc dust. So when you heat this with zinc dust at around 300 degrees Celsius, temperature you don't have to memorize. First this halogen goes out, X2 will get a double bond, X and X. Further you heat this with the same reagent, temperature you need to increase a bit. And then this X2 also goes out, forms an alkyne. This is the product we get. We can also have Colbe's synthesis for the preparation of alkyne. We take here Hc double bond C, Hc double bond O, O minus K plus. Same thing we have, just the double bond we'll have over here, so that we can get one more pi bond and we'll get an alkyne here. It goes under electrolysis, it goes under electrolysis and it forms Hc triple bond CH plus two molecules of CO2, H2 and KoH. This is what we get. A very important reaction we have, fifth one write down, is laboratory preparation. How do we prepare acetylene in laboratory? In lab, acetylene is prepared by, in lab acetylene is prepared by the hydrolysis of calcium carbide. Very important reaction, yeah I'll go back. Here also we have seen same thing, up to four carbon atom, it is gas only. Yeah, so in lab write down, in lab acetylene is prepared by the hydrolysis of calcium carbide, the hydrolysis of calcium carbide. Calcium carbide is this, Ca2 plus, C triple bond, C negative charge, negative charge. This is calcium carbide, the structure is this, we write, but the molecule is obviously CaC2. Where it goes under hydrolysis, 2H2O, it forms acetylene, CH triple bond, CH and slagged lime, which is CaOH whole twice. Important reaction, in J also they have asked this. Okay, one more method of preparation we have, this is first method, one more method we have, in which we take chloroform, chloroform the formula is CHCl3. If you have CHI3, it is Idoform. General name is haloform. Okay, halogen, so haloform. So chloroform, when it reacts with silver, 6 AG, it forms AGCl, precipitate and we get acetylene, which is this. These two reactions we have, both reactions are important. Now, one last method we have, in which we can prepare higher alkenes from the smaller one, means one alkene can be converted into another alkene, sorry not alkene alkene. Okay, so right on preparation of higher alkenes, higher alkenes. Higher alkenes can be prepared by the help of Grignard reagent. Grignard reagent is RMGX. I have discussed Grignard reagent before. Yes or no? A little bit of information about it. Have I discussed? Yes, Grignard reagent. So we have RMGX class here. Right, so what happens in this? This Grignard reagent is allowed to react with any smaller alkene. Suppose we have HC triple bond CH. So we know this hydrogen is acidic, so Grignard reagent with active hydrogen gives acid-base reaction. So it forms an alkene RH and HC triple bond C minus MGX plus we get it. Okay, we are not concerned with this alkene. If you want to get higher alkyne, then again here you add alkyl halide, RX. So this RX, MGX takes this X minus, forms MGX2 and R will get attached to the terminal carbon, which is this. So this is the higher alkyne we get from the smaller one. Okay, so properties, you see physical properties, we have similar only like melting point, boiling point, you know, it increases with mass and decreases with branching. So that's common. We are not going to do that. We'll see some chemical properties, chemical reactions here. Chemical reactions of alkynes are also very much similar to alkene. Okay, you can easily correlate and memorize this. First reaction we have hydrogenation. Hydrogenation, we all know it is an addition of hydrogen in presence of a catalyst. Correct. So suppose we have an alkyne, our C triple bond CH reacts with H2 in presence of a catalyst Ni high temperature. It forms an alkene. We know this reaction, both hydrogen get attached on the same side. Again, we do the hydrogenation of this one. Nickel high temperature, you will get an alkene RCH2 CS3. Very simple reactions. I guess you don't have any doubt in this. Okay, we can use any catalyst here. We can use nickel, we can use platinum, we can use palladium, many catalysts we can use. But when you use nickel catalyst, nickel catalyst, the reaction is called, this hydrogenation reaction is called, subartier centerance reaction. Subartier centerance reaction. So name reaction and keep this in mind. Okay, second one you see, hydrogenation. This reaction also we have discussed. I'll just show you. Suppose we have an alkyne, CH triple bond CH, hydrogenation means addition of hydrogen in presence of various catalyst. So we can use H2 with PDBSO4 and we heat this. Or we can also use H2 with NA liquid NH3. H2 with NA plus NH3 liquid. It is NA-NH2 only burst reduction that we have. So when you have H2 PDBSO4, it gives you cis alkene. Right, here cis is not possible, but we'll get this. Or if you write down this CH3 CH3 triple bond CCS3 on addition of water, sorry, H2 you'll get cis alkene, which is this. H CH3 CS3, cis you will get. We know this gives you cis reaction. This gives you trans. In the first one though, we'll get this only. We'll get CH2 double bond CH2. But if you have the second one, in the second one you will get trans, C double bond C, CS3 CS3 H and H. We have discussed this reaction again in alkene chapter. We'll get this one second. Just a second guys. There is one more reaction in this we have where we use B2 H6 di borane. Right down. Third one, reduction of alkene or simply you write down reduction with di borane. What is the formula of di borane? Anyone? Formula of di borane is B2 H6. Okay, it is a dimer of BH3 borane, right? B2 H6. So alkyne reacts with B2 H6 and the reaction follows with acidic hydrolysis. C triple bond C, R dash or I'll take R only. First it is allowed to react with B2 H6 and in the second step we'll get here H plus H2. In this one what you have to memorize the product would be a cis alkene. So all these reactions you see we are getting alkene only because of the reduction triple bond converts into a double bond since hydrogen is getting added and cis alkene we get. Okay, I think all of you can you see the screen? Is it frozen? Yes, all of you can see properly. What else you can rejoin then? Okay, next write down the reaction with Hx hydrogen halide. Remember the reaction of Hx with alkene. What rule we follow over there? Hx with alkene. What rule we follow? Marconi Coff rule, right? So here also the reaction follows Marconi Coff rule. The reaction follows here Marconi Coff rules. You know what is Marconi Coff rule? The negative part of the reagent. Okay, so first we get suppose we have a reaction Rc triple bond CH with Hx. So negative part of the reagent will get attached to the carbon atom which has lesser number of hydrogen. We'll get CH2. Further it reacts with Hx. It forms gem dihalides where both halogen atom get attached to the same carbon atom and we'll get this Marconi Coff rule in both steps you need to follow. Okay, carbon with X will get a negativism. No, it's not like that. See what happens if you have this halide over here, like it's not like there's a carbocation intermediate forming over here. Right, both H and X will get attached at the same time. Okay, okay, so that's fine, not a problem, yeah. Yeah, so I was talking about this. Yes, this reaction you see it's not like the carbocation forms over here. Right, both H and X will get attached to the carbon atom simultaneously at the same time both will get attached. Okay, if carbocation forms then you can say the positive charge over here because of minus I effect of halogen, this won't be stable hence it won't form. But here the carbocation is not forming there. Okay, if you look at the mechanism here, right, this reaction takes place in presence of mercurinium and I'll tell you since you've asked this a question I'll just explain I'll tell you the mechanism of this one. You write down here Hg2 plus. Okay, this reaction takes place in presence of Hg2 plus iron. Okay, how the reaction proceeds you see in the first step you have this alkyne Rc triple bond cr with Hg2 plus. Okay, then this pi bond attached with this Hg2 plus and we'll get a cyclic mercurinium iron in between which is this C double bond cr and this Hg2 plus 2 plus here forms a cyclic ring like this cyclic mercurinium iron we call it as then what happens from Hx from Hx we get X minus and X minus will attack to one of the carbon atom. Right, when this X minus attached to one of the carbon atom. Okay, this sigma bond the ring that we have here it breaks and this goes on to this mercury so it forms Rc Hg plus because it takes electrons one positive charge will get reduced double bond Cx here we have and this side we have R we get this now the H plus that you have over here this H plus will get attached to this and Hg plus will come out correct so we'll get here R I'll write down this so we'll get here Rc double bond CxR and H will come over plus Hg2 plus will go out then the same thing you know we have to repeat again here same thing we need to repeat again this Hg2 plus forms a cyclic mercurinium iron here so we'll have this Rch this pi bond is about to break Cx and R we get this here again we have Hx so X minus from this it will attack on this carbon atom but this carbon atom right so this X minus attacks on the same carbon and this goes up so the point is here your concern is the positive charge that forms on this carbon should will be un-stabilized because of the electron don't electron withdrawing nature of this halogen which is not forming here it forms a ring we don't have an intermediate carbocation here that's why we have this understood right again you have to proceed the same thing here from here we'll get the answer okay so mechanism is not you don't need to know the mechanism you just need to know that we'll get a gem dihalides in this reaction okay this mechanism anyways is not that important yes so uh next reaction we have addition of Hcn hydrogen cyanide hydrogen cyanide Rc triple bond C H when reacts with Hcn the reagent we are taking here CuCl temperature around 700 770 degrees Celsius and it forms C H2 or I'll write down here R C H H double bond C H single bond C N okay so basically what happens like HCl H plus comes over here and Cn minus comes over here okay we'll get the product here H plus and Cn minus we'll get attached which one are you which one are you previous which reaction you are talking about what show this one you see this one I was talking about marconi cough rule in this just to write down the product you can write down product easily once you follow marconi cough rule this two carbon atom we have the one which has hydrogen atom and this one we won't have any hydrogen atom so X minus will get attached to the carbon atom which has no hydrogen here that is what the product will write mechanism is different right marconi cough will be generally followed write down the product directly in one step yes so we'll get this few more reactions are there we'll finish this up we don't have much to do next write down I think the last two reaction we have write down oxidation oxidation of alkyne in NCRT they haven't given the reaction much okay we are doing like we're not following NCRT at least over here right it's not you know for comparative NCRT organic chemistry is not that good okay write down acetic or alkaline KmnO4 acetic or alkaline acetic or alkaline KmnO4 breaks alkyne into two parts into two parts and forms and forms acids carboxylic acid with example you'll understand suppose we have RC triple bond CH oxidation we are doing reagent is acetic or alkaline KmnO4 this triple bond just you need to break write down COOH with this carbon COOH with this carbon so product would be RC double bond OOH and HC double bond OOH especially for these kind of reactions oxidation you just know how to write down the product mechanism they won't ask you ever okay mechanism is not at all required in comparative exam also you just need to know how to write down the product suppose we have a H over here HC triple bond CH goes under oxidation reaction it gives you a formic acid two molecules of formic acid okay but we have one change over here one exception in fact what happens here two reaction I'll write down this gives you this I am taking here acidic KmnO4 but with this reaction if you are taking alkaline KmnO4 alkaline KmnO4 it forms oxalic acid this one is oxalic acid so this is an exception we have for exception we can't do anything this is an exception so acidic it gives formic acid with alkaline it gives oxalic acid and this reaction is an exception next reaction write down ojonalysis in ojonalysis what happens similar kind of reaction we have we have alkyne suppose we have R-C triple bond CR with O3 it gives ozonide first you see this reaction you'll find it is a similar reaction we have but this sigma bond would be there there the sigma bond breaks here we have three bonds so one will be there as this this one it forms first okay and then if you do the hydrolysis of this one H2O hydrolysis it gives you R-C double bond O C double bond O R this is the product we get dye ketone it forms this reaction is there in aldehyde ketone chapter also one additional point in this is what if this dye ketone goes under oxidation with hydrogen peroxide it converts into carboxylic acid R-COOH plus R-COOH this is not that important if you want you can write it okay obviously you will feel like there are so many reactions how do we memorize it so let me tell you in one go you won't be able to memorize all this okay with practice you'll give some time then slowly you will get to know that okay this could be the product and organic chemistry you always remember options are always you know half of the information you always get the idea that what could be the possible product in the given reaction okay slowly with when you give time a bit more then you will understand and it then it then it will be easy for you to memorize also because there are so many things that we need to know I would say once you you know understand the concept of GOC properly plus the reaction mechanism then it will be very easy for you to memorize all these things okay but the problem is in school slavers and all there's nothing called reaction mechanism they won't teach much because not in the slavers but for comparative exam you need to understand those how carbocation forms what are the process of rearrangement we have free radical mechanism how free radical mechanism how free radical forms what are the mechanism we have there's so many things so that we do actually in the class but this year in every year in the event that's very difficult to do that's why I always take up initially when we start the grade 12 you know classes so that we'll discuss over there I am sure when we finished reaction mechanism you will be able to understand most of the reactions easily and obviously I am not saying you don't have to memorize things there also you have to memorize but you can have a balance between the two where to mug up where to understand what concept we have how to relate concepts in order to memorize the reactions those things helps you a lot one last reaction in this we have the acidic nature of alkyne write down the last one acidic nature of alkyne in the beginning only I discussed that alkyne's has acidic hydrogen because of sp hybridized carbon atom and hence it shows acidic properties so reaction with metal you see reaction with metal I'm taking sodium here write down alkyne reacts with metal alkyne reacts with metal and and evolve hydrogen alkyne reacts with metal and evolves hydrogen gas because of its acidic property so suppose we have a reaction to our single bond C triple bond CH with metal okay so metal provides electron and in presence of that electron this hydrogen can go out easily and it forms to our C triple bond C minus which takes any plus so we'll have a sodium salt of it and hydrogen gets released in this reaction this is acid base reaction you can see it forms salt and hydrogen gas removes okay clear so this is it for hydrocarbon chapter okay many reactions we had done in this the mechanism part of few reactions like hydrolysis dehydration okay that we'll discuss in detail in reaction mechanism you will have a you know better idea of that but here the chapter hydrocarbon is finished next class yes we have oxidative ozone analysis I will discuss that in reaction mechanism reductive ozone analysis we have oxidative ozone releases you won't get aldehyde in that case you will get acid I'll just give you one example here suppose you have RC triple bond CH over here and CH2 for example this if you go with O3ZN H plus H2O when zinc is present it is reductive ozone analysis and reductive ozone analysis gives you aldehyde like this okay but when you have this reaction RC double bond CHR with O3 and then in the second step we are not using zinc when zinc you do not use then it is oxidative ozone analysis so we have the mechanism of this also we'll discuss in reaction mechanism and in this case when you have oxidative ozone analysis you won't get aldehyde but you get acid like you see this bond you break everything will be same you'll get ketone plus the another molecule you will get aldehyde first but this won't be the final product this won't be the final for further it oxidize into acid so only one thing you have to keep in mind in oxidative ozone analysis carboxylic acid you will get you won't get aldehyde ketone if forms fine you'll get ketone only in reductive ozone analysis zinc is present then aldehyde can form here and dehyde won't form we'll discuss this later on okay anyway so this is the end of hydrocarbon chapter and another thing is next class we'll do conformational a bit plus we also do p-block okay whatever possible we can finish in one class because I think probably next class will be the last class we have for this session okay so we'll finish these two p-block I would suggest you come prepare let's just go through a like just give it a read in the book we'll just discuss the important topics and I tell you what to memorize and what to do in p-block okay that will finish next class okay guys thank you take care bye bye inorganic you must follow ncrt if you have time for nc after ncrt then you can go through the another books okay ncrt you must do yeah okay take care bye bye