 Good afternoon everyone. Can we start now? All of you please type in your name. Which subject you have tomorrow for the test? UT. Hello Shreya, Gargi. Okay, so both of you have English only. When is chemistry? Monday for KRM. Alkyne we haven't finished. Benzion reaction will do later, not in this chapter. Benzion reaction will do it later, not in this chapter. There are only few reactions given you can go through in NCRT. Alkyne have you finished? Alkyne's reaction? What have we done last class? Please let me know. Saturday internal. Which exam you have Gargi on Saturday? Quality quantitative analysis will do after equilibrium. Okay, so we'll fine. So Alkyne there are not much reactions, only few reactions are there in Alkyne's. We'll finish first Alkyne and then we'll start any equilibrium. Okay, we can finish this Alkyne chapter in one or some one or no one and a half hour or something. Okay. Okay, then we'll finish first Alkyne. Okay, then we'll see. You want me to start qualitative and quantitative after Alkyne's? Tell me. You want me to start qualitative or quantitative analysis? Correct. We'll finish Alkyne first then and then we'll start qualitative and quantitative analysis. Okay. When you have exam or HSR guys, when you have exam of this chemistry, let me know. Tell me now only we'll plan accordingly. Probably we'll have a class on Monday also. Okay. I tell me this coming Wednesday you have chemistry exam, right? And we'll finish first Alkyne quickly. So write down this Alkyne, the general formula of Alkyne is what? CNH2N-2. This is general formula of Alkyne that we already know. Okay. So like I said, we have only method of preparation and properties, correct? So we directly start the method of preparation of Alkyne's, method of preparation. First method you write down by dehydrohalogenation of by dehydrohalogenation of dihalytes. Okay. By dehydrohalogenation of dihalytes. Write down to this. The reagent we use for this purpose, there are two reagents generally we use. So the reagent we use, we can use. First of all, we can use alcoholic KOH with NaNH2, sodium NaNH2. Or we can also use NaNH2 in liquid NH3. Both reagent we can use for this purpose. Okay. You have to memorize these properties of reagents. So suppose we have dihalytes. What do you understand by dehydrohalogenation? Okay. Dehydrohalogenation means what? The elimination of Hx. Elimination of Hx, correct? So we have dihalytes, correct? Two halides are there. So if I write down the example here, suppose we have carbon-carbon single bond and we have Br. Here we have hydrogen. Here we have H. Here we have Br. And then here we have H. So in this, if I use alcoholic KOH with NaNH2, with NaNH2, then the product we get here is in this case, what since one moles of this we are using. So only one moles of this one molecule of HBr will come out. Okay. And we'll get carbon-carbon double bond. Here we have Br, H, H and B. H and H we have here, right? Now in this, again, one more molecules we use of the same reagent. Then we'll get alkyne, which is C triple bond. C here we get. Okay. So this reaction takes place in one by one step. In one step, one molecule goes out. Another step, another molecule goes out. Okay. In this what happens, if I use the another reagent, same types of reaction only we have NaNH2 with liquid NaNH3, two moles of this if you are using, then two NaBr goes out. Okay. And we get C triple bond, C over here. Alkynes will get like this. There's the carbon compound. Yeah, yeah, yeah, yeah. It has to be trans. It has to be trans. Okay. That's the thing. Now the important, you know, the application of this process is what? One note you write down after this. The important application is by this method, this is a conversion actually, by this method, we can convert, we can convert ethene into acetylene, right? For example, you see reaction. We have to convert ethene into acetylene. So ethene is H2C double bond CH2, right? If you do the bromination of this Br2 in an inert solvent, then we'll get what? We'll get Dibromo compounds, right? Which is nothing but C, CH2. Here we have bromine and here we have bromine. Now this is nothing but this compound we have. After this, when we use NaNH2 with liquid NaNH3, with liquid NaNH3, then according to this reaction, it converts into HC triple bond CH. This is what acetylene and this is ethene. The reaction is not in this chapter today. See, that's what I said in the beginning only. Here you see one chapter you will get in many different, what we say. See, we have done alkene and alkene, correct? If you want me to explain, I can explain, but it is not in this chapter. It is there in alcohol chapter mainly, right? So you must have seen we have used some reactions of ketone and aldehyde also, okay? So that is also there in, you know, ketone and aldehyde chapter, okay? If you want me to explain, I'll explain that. Let me finish this alkyne first. We will not take much time. Then we'll explain that dials elder reaction, okay, for alkenes. We'll also get alkyne into that, but it is not given in this alkyne chapter because we must know some different concepts, also some different reaction also to understand that, okay? So we'll see that also. First, let me finish this alkyne, okay? So this is what I'm discussing. This is the conversion of ethene to acetylene, okay? Next you write down, there is one more reaction we have when we take haloform, okay? With haloform also we can form alkyne. So write down the second point into this. When haloform heated with silver powder, when haloform heated with silver powder, then also we get alkynes. Then also we get alkynes or you can write down acetylene specifically if you want to write. Then also we will get alkynes or acetylene. You see the reaction here, we have haloform. Haloform is generally formalized CHX3, okay? And for this reaction, generally we use chloroform, C which is CHCl3 or idoform which is CHI3, okay? So and in this reaction we take two moles of any of this product, okay? Haloform is this CHX3. General name is haloform. When X is chlorine, it is chloroform. When X is iodine, it is idoform, okay? General formula of haloform is CHX3, haloform, okay? So we will take two moles of any of these compounds. Suppose we have CHI3 and this combines with Ag silver and we will take another molecules of this which is nothing but CHI3 again. When this is heated, when this mixture is heated, what happens? This combines with this silver and this iodine and it comes out. It is similar to Wood's reaction. Mechanism is not same but it has the same pattern. So and then it forms C triple bond CHH plus AgI, silver iodide. Six moles of AgI will get, here you'll put 6 Ag. This is the another method of preparation of acetylene, okay? Next method is we have laboratory preparation, laboratory method of preparation of alkyne. In this method you write down, in laboratory it is prepared by, it is prepared by the action of water and the action of water on calcium carbide. So calcium carbide is CaC2. When this combines with water that is H2O, when this is heated, then we'll get acetylene which is CH triple bond CH plus we'll get CaOH whole twice, calcium hydroxide, correct? We can also prepare this calcium carbide by this reaction when we have lime that is CaO combines with carbon charcoal at around 3000 degrees Celsius, very high temperature we have here. So it forms CaC2 plus carbon monoxide CO. This CaC2 we are using here in this reaction. Second reaction is not important. We can also prepare the next method you write down of acetylene by partial oxidation of methane. So write down methane, methane on partial oxidation gives acetylene with evolution of, evolution of carbon monoxide and hydrogen. So CH4 on oxidation with oxygen, partial oxidation we have, it converts into 2HC triple bond CH plus 2CO plus 10H2. There are many different products also we get, okay, but I'm just giving you the major product here on partial oxidation. Next you write down, all these reactions I'm giving you except this laboratory method is important. These reactions are not important, okay? Next write down magnesium carbide, next method magnesium carbide, formula is MG2C3, magnesium carbide on hydrolysis gifts, hydrolysis gifts propylene. So reaction you see MG2C3H2O, it gives CH3C triple bond CH plus hydroxide of magnesium, right? These are the few preparation method of alkyne we have seen, okay? Next you see the properties of alkyne, properties and in this first we are discussing physical properties, physical properties. First property we have here compounds from C2 to C4 carbon atom which contains 2 to 4 carbon atom are colorless gas, colorless gas. From C5 to C12 they exist in liquid form and if the carbon atom is more than equal to 13 then they are solids. Yes, we can also use electric arc with carbon and methane also, right? We can also use electric arc method is not important that's why I haven't given you, okay? Around 3000 degrees Celsius, around 3000 degrees Celsius, we can use electric arc on carbon and methane also if you want you can write down CH4. In this electric arc if you use 3000 degrees Celsius it converts into acetylene, this is also another method we have, plus hydrogen evolves. Electric arc we can also use for hydrogenation of carbon plus H2 with electric arc around, it is also around 2500 degrees Celsius. Here also we will get C triple bond CH plus CO plus H2, this only we get not CO and H2 will get you, okay? Yeah, this is what the method, electric arc, not important that's why I have skipped this but you can prepare. C13 onwards it is solid. Right on next, the melting point and boiling point, the melting point and boiling point of alkynes are slightly higher than the melting point and boiling point of alkynes are slightly higher than to that of alkene and alkene, alkene and alkene. So what we can write the boiling point or melting point is directly proportional to its molecular mass which is same for alkene and alkene, okay same thing we have here, okay? Next you see chemical properties, chemical properties. See chemical property the first one which is important here is the acidic character, acidic character. If you compare the acidic nature of alkene, alkene and alkene, which one will have maximum acidic character? Tell me the answer, maximum acidic character. What is the answer? Acidic nature, alkene due to SP, alkene is S, okay, alkene. Yeah, yeah, yeah, alkene is the most acidic one because you see alkene we have only single bond, here's your double bond and triple bond, single bond, double bond and triple bond, right? So due to hybridization here it is SP hybridized most electronegative, it is SP2 and this is SP3. So as percentage S character increases, acidity increases, right? Because of SP hybridization, the alkene is most acidic in nature, okay? Now the reactions, few reactions we'll see in alkene that is formation of, first reaction you write down, formation of sodium acetylides, sodium acetylides, okay? For this purpose the reagent that we use is NA with liquid NH3 or we can also use sodiumide. Sodamide is NA. Now what is sodium acetylides? The reaction you see we have acetylene CH triple bond CH when it reacts in presence of sodium liquid NH3 then this forms HC triple bond CNA plus half of H2. So this the sodium salt that we get here we got it as sodium acetylides. In acetylene one hydrogen is replaced by sodium, right? In this also when you use this reagent again NA with liquid NH3 then another hydrogen also will get replaced and we'll get this NHC triple bond CNA plus half of H2. This is disodium acetylides, okay? Disodium acetylides. One note you write down here, these sodium acetylides, these sodium acetylides are used for the preparation of higher alkynes, higher alkynes when it reacts with when it reacts with alkyl halide you see the reaction here. Alkyl halide with this react, right? So sodium acetylides is this NA here and when it reacts with alkyl lidarities XR then what happens this NANX forms NAX and R combines with carbon triple bond CH plus NAX. So this is what this is the higher alkynes we have, okay? Number of carbon atom increased over here, right? That is how this compound is used for the preparation of higher alkynes. Next one you see, next reaction that is formation of, formation of silver actually will get dye silver over here and the reagent we use here is and this is an important reagent not in this chapter but we'll see this reagent again in aldehyde ketone chapter that is AGNH3 whole twice OH. This reagent we call it as Tolens reagent, Tolens reagent, okay? This is an important reagent, okay? We'll see this again in aldehyde ketone chapter, okay? This is not important here but in aldehyde ketone we'll see it again. So when this combines with acetylene which is nothing but CH triple bond CH plus Tolens reagent, this gives all the hydrogen atom here get replaced by silver plus H2O plus NH4 NO3, 2 moles of this. This is silver acetylides. Now the important thing here it is what when in this silver acetylides if you mix dilute HNO3, dilute HNO3 since we have 2 silver here so we'll get 2 moles of HNO3 we are using, okay? Since we get, we are taking 2 moles we'll get 2 moles of AGNO3 plus H attached to this carbon atom and we'll get acetylene back. This is how we get acetylene from silver acetylides, okay? Reaction with dilute HNO3. If you get 1 mole of HNO3 you'll get AGC triple bond CH, only 1 silver will get displaced. Next reaction you see and this reaction we have already discussed. Write down addition of hydrogen, addition of hydrogen. So the reagent we use here and we have discussed this in the preparation of alkene if you remember. We can use H2O with various catalysts like nickel, platinum, palladium, second one. We can use Lindler's catalyst, Lindler's catalyst. So you don't have to memorize this again because this thing we have already discussed, right? In this what we use H2 with palladium in BSO4 or CSCO3. Third one we use with NA with liquid LH3 or we can also use LI LH4 which is also a reducing agent. Another agent we have here is LI with LH3 in ethanol, C2H5. We have already discussed this. In this reagent we'll get cis alkene. Since we are adding hydrogen to alkynes obviously the pi bond will dissociate and here we get what? Trans alkene. We have discussed this already. Reaction you see. We have HC triple bond CH plus hydrogen we are adding, right? 2H2. The catalyst we use nickel, platinum or palladium and the product here it will be what? Ethane C2S6. Now in this suppose we have this reagent RC triple bond CR and one reagent we are taking as Lindler's catalyst and other we are taking as NA in liquid NS3 suppose. So Lindler's catalyst gives you cis alkene. So the product here it will be RC double bond CR and hydrogen attached over here. This is H2 we are adding here and this will get trans alkene, right? So RC double bond CR H and H. You have to just memorize this. Where is this H2? This one. It is 2NH4NO3. 2NH4NO3. 2H2O plus 2NH4NO3. Can we move forward? There all of you? Yes. Next one you see. Addition of water. Addition of H2. No, I am not free right now. Can you call me tomorrow? Yeah. So addition of H2O here the reagent we use for addition of H2O is dilute H2SO4 we use HgSO4, silver sulphate. Okay, silver sulphate. Now here you see the reaction. Suppose we have alkyne this RC triple bond CH plus H2O. When this reacts in presence of dilute H2SO4, HgSO4 also we have. The product we get here is this RCH double bond CH2OH. But this compound here, here tautomerism is possible. Okay, tautomerism we haven't discussed. Okay, HgSO4. It is Hg, HgSO4. Okay, tautomerism when takes place then RC double bond O this alcohol converts into ketone. Okay, this is the tautomerism we have. What is tautomerism? We will discuss this later on. It is again there in isomerism. Okay. But one thing you must remember whenever you have this double bond with this carbon contains OH. So this is alcohol. Right. So it is all the suffix for this will be all. And this is ketone for suffix. We call it as own. Right. So this kind of tautomerism we call it as keto enol tautomerism keto enol tautomerism. What is enol? We have double bond and all. So enol. Okay, for double bond is en OH all. So it is ketone. So it is enol. And this is keto. So it is keto enol tautomerism. This kind of a product is actually spontaneous. How do we write down this kind of product? We have a double bond and then we have OH. So this hydrogen, you see here how this product forms, this hydrogen jumps onto this carbon atom and this double bond comes over here. So you'll get C double bond O single bond CHC. That's how we get. We will discuss this tautomerism in detail in isomerism. Okay, that will probably discuss in bridge course. Okay. So don't worry with that. It's not there in school exam. So don't worry with that. Right. So finally we'll get this how this reaction proceeds. You see one of these pi bond dissociates and this comes over onto this carbon atom. This acid gives H plus. Right. And this negative charge that comes over here, this negative charge will take this H plus ion. Okay. So what we get here you see we'll get RC double bond CH3. This one hydrogen and one hydrogen already we have here. It is CH2, not three. So this pi electron jumps over here. So we'll get positive charge on this carbon atom, two db carbocation. If the pi electron jumps over here positive charge will be here that is one degree not stable. Two degrees more stable. Right. Now for this two degree carbocation that you have here you have H2O molecule H O and H which has two lone pair. This behaves as a nucleophile attack onto this positive charge carbon atom and we'll get our C double bond CH2. And here we have OH H positive charge rate. Okay. Now since oxygen is an electronegative element positive charge on any electronegative element is highly unstable to make this stable deprotonation takes place. Means one of this hydrogen comes out as H plus ion leaving this electron pair behind. So what happens here? Suppose this hydrogen we have this will release its bond pair of electron to this oxygen and H plus comes out and we'll get RC double bond CHOH. And after this the atomism takes place and the product is this. Correct. One note you write down here. One note you write down in case of non-terminal in case of non-terminal unsymmetrical alkyne in case of non-terminal unsymmetrical alkyne the two isomeric ketones obtained to isomeric ketone obtains methyl ketone is the major product methyl ketone is the major product. So one step you see here what is the product we get here from this reactant. Here we have triple bonded carbon atom. So double bond attached with the triple bonded carbon atom and other carbon will get the hydrogen atom. Correct. That is how the product will write. Suppose the reaction is this CH3 CH2 C triple bond C single bond CH3. This is the alkyne we have H2O. We are using the same reagent dilute H2SO4 with HNO3 HDSO4 correct. So what happens one of the product will be what when the double bond O will attach over here and other carbon will take two hydrogen atom and other product will be double bond O attached over here and other carbon will get the two hydrogen atom. Two possible product if I write that will be CH3 CH2 C double bond O CH2 CH3 and the other one is CH3 CH2 CH2 C double bond O CH3. This is the two possible product we get. Okay mechanism is this only that we have discussed right. Now in this one this is methyl ketone. Why because one side we have methyl and ketone. Here we have ethyl and ethyl. So methyl ketone is this. So this one is the major product and this one is the minor product. These reactions you have to memorize. Next one addition of alcohol. You see the reaction here. The reagent we use for this reaction is COH we can use. We can also use KOH and we can also use HDSO4. You see this reaction we have acetylene CH triple bond CH when alcohol we are using is C2H5 OH. This mixture is heated in presence of HDSO4. Then the product we get here is this one of the bond will dissociate. This is C2H5 O minus and H plus. Right. So at one carbon will have H plus of CH2 double bond CHO C2H5. Okay. This we call it as ethyl and this position is vinyl position. Correct. So we call it as ethyl vinyl ether. ethyl vinyl ether. In this when you put H2O for the reaction you see then again this H2O will give H plus. Right. So this combines with this particular compound and finally we get CH3 CHO as the major product. Okay. The final product here we get depends on what you are taking here. Okay. From this H2O H plus will come on to this carbon atom. Right. You see how this reaction goes. The final product is this. You see the mechanism. What happens? H plus will come on to this oxygen. Later on in alcohol ether chapter you will see this kind of reaction is very common over there. CHO C2H5. This lone pair of this you know ether that you have that will take H plus from this and this H plus will attach here will have positive charge onto this oxygen. Now again the positive charge on oxygen is highly unstable. Okay. To make this compound stable what happens this oxygen will drag this bond pair of electron towards inside and this bond will dissociate. So we will get CH2 double bond CH positive charge on it plus we will get C2H5 OH. Sorry. C2H5 OH. Now in this only OH minus will attach here and we will get CH2 double bond CH and OH. Correct. Now in this also tautomerism takes place. If you remember the same thing tautomerism is enol and this is aldehyde. Right. So this from this what happens this H plus will come out and it will attach over here and this double bond comes over here. So we will get here. Understood this point. If you want to see this you see this H plus this H plus will jump onto this carbon atom and this double bond comes over here. So you get this. So this is the product we get when you take H2O over here. Okay. Now suppose after this we are taking one more alcohol which is C2H5OH. Then again it is minus N plus. So this H plus will attach onto this similar reaction H plus attached over here. Here this H plus attached will here. So we have CH3 one bond dissociate pi bond. CH will be as it is O C2H5 is already there in the first step. And in the second step this O CS2 C2H5 will attach over here and this is the product we get. So it depends on what is the second step what is the reagent you are taking here. Whether you have you are taking alcohol or you are taking water. One very important reaction of acetylene you just write it down. We will see the mechanism of this in carboxylic acid chapter. Okay. But sometimes they ask this question in this chapter also. That is why I am giving you this reaction here. Acetylene there are two reagents we have. One is suppose I am using acidified KMNO4 okay acidified KMNO4 and this I am using alkaline KMNO4 two different reagent I am using. Okay acidified KMNO4 breaks this triple bond completely. Okay this is an oxidizing agent. This will oxidize it gives three nascent oxygen with H2O. This will oxidize this completely. It breaks this triple bond completely and we get two moles of H C O OH here. Right alkaline KMNO4 does not break this triple bond but the product here it will be this double bond will dissociate and will have C O OH and C O OH. What you have to keep in mind when you have alkaline KMNO4 the number of carbon atom here and here it will be same. The only thing is the carbon atom attached with the multiple bond you attach O OH group O OH group to both the carbon atom. Right. So this is what the two different reagent you have you just keep this reaction in mind when you have the option you will understand what could be the product okay acidified KMNO4 and alkaline KMNO4. It is important number of carbon atom is same. Next the last next one write down isomerization write down on heating with on heating with alcoholic KOH on heating with alcoholic KOH or NAS2 in inert solvent in inert solvent the triple bond shifted towards triple bond shifted towards the center towards the center to form towards the center to form isomeric alkaline isomeric alkaline. So just you see this reaction here suppose we have CH3 CH2 single bond C triple bond CH when this is heated in presence of alcoholic KOH then this triple bond will shift towards the center the product here is CH3 right these two hydrogen also will shift here these two are positional isomers of each other. One note you write down here on heating with NAS2 in liquid NH3 on heating with NAS2 in liquid NH3 the triple bond the triple bond shifted towards shifted towards the end shifted towards the end means what if you have this compound only and in in this compound if it is heated with NAS2 in liquid NH3 then again we'll get isomeric product but here the triple bond will shift towards the end the product here it will be the same as this H2 also we're using right so these two reagents you must remember now the last reaction of this chapter we have is this that is hydroboration hydroboration oxidation reaction this reaction is very similar to the reaction we have done in alkynes alkene sorry so here in the case of alkyne we have simply triple bond present instead of double bond so this is suppose the compound we have for hydroboration oxidation reaction the reagent we use is BH3 in THF with H2O2 and OH- okay so in this first of all what happens is this compound when reacts with BH3 THF with H2O2 OH- right with H2O2 OH- so the product we get here is this CH3 CH2 CHO and behind we get here okay how this reaction proceeds that you see first of all this compound will take this reagent BH3 and THF and this forms CH3 CH double bond CH BH2 one of the hydrogen will jump over here right and this will get this double bond CH BH2 similarly this reaction goes twice one and two so after three like three attempt three after the third reaction in the first one one hydrogen comes over here second one this one of these two hydrogen will come over here and the third one again this hydrogen comes over here right the final product we get here is CH3 CH2 single bond CH whole thrice B after this this is derivative of hydride of boron correct after this we have this H2O2 and OH- this OH- will take this boron forms BOH whole thrice and the product we get here three molecules of CH3 CH double bond CH OH three molecules of this weekend because we have whole three over here now in this also plus we'll get this one BOH whole thrice which is not important in this only again tautomerism takes place this hydrogen comes over here and we'll get three molecules of CH3 CH2 CH OH in the same way correction let me make this should be CH double bond CH on this three is correct and we are taking here three moles of this we are taking so here also we'll get three moles of this three moles of that's fine this hydrogen right so what happens here we'll take three moles of this BS3 will give you one hydrogen over here so we'll get CH3 CH double bond CH right and the boron combines with this three moles we have already CH3 CH2 CH2 no it's not I have corrected it see see let me write down all these things you'll understand this see actually what happens okay I'll write it wait I'll just write it here see this reaction is this I have written there in one single step that's why you are getting the doubt but if I write down one step at once you will understand you have BH3 THF correct so one hydrogen comes over here we'll get CH3 CH double bond CHBH2 correct now in this again I am allowing this molecule to react with this again CH3 C triple bond CH okay so what happens here this we have already now this is reacting with this so again this one hydrogen will come over here we'll get again this part right so what we get here CH3 CH double bond CH whole twice BH because one again one hydrogen will comes over here here in this carbon again here also we use the same reagent so we'll get here CH3 C triple bond CH what happens then this hydrogen comes over here and we'll get CH3 CH double bond CH whole thrice whole thrice B is it clear now so keep on the we keep on using this reagent in each step that's why we have one two three moles that's why I've written there three moles of that so I have written that in one single step but the actual reaction is this understood this yes this is the last reaction of alkynes okay now one doubt Tridev has asked that is mechanism of dials elder reaction okay this reaction is actually there in reaction of benzene that is aromatic reaction okay but I'll just explain you this question a bit right on the last one here this alkyne chapter is finished okay now we are discussing this the last thing here after this we'll start qualitative analysis D IELS dials elder reaction it is the it is the addition of a conjugated diene addition of a conjugated dienophile dienophile this dienophile can be alkene or alkyne anything this reaction involves this reaction involves the pi electrons of both reagent that is pi electrons of conjugated diene and dienophile this reaction involves the pi electrons of both reagent in this reaction new bond forms which are energetically more stable than pi bonds, which are energetically more stable than pi bonds, okay? So you see the reaction here conjugated diene, conjugated diene are this, oh sorry, this is the conjugated diene suppose we have double bond here and double bond here, this is conjugated diene, okay? This combines with dienophile which can be an alkene also, right? Dienophile is this suppose, suppose we have an alkene here like this double bond, correct? So now when these two combines the product we get here is this, we will get cyclohexene here, double bond will be here. Mechanism if you see how this reaction proceeds you see this pi electron, pi electron here that will attack onto this carbon atom, right? Because it is you know resonance stabilized, this pi electron comes over here, right? And this pi electron will come here and further this negative charge since this pi electron is going onto this carbon atom, so this here, here we have slightly positive, here we have negative, finally this will attack onto this carbon atom, okay? So if I write down the number here you see, suppose it is 1, 2, 3, 4, this is suppose 5 and this is suppose 6. So when this pi electron goes here so 1 and 6 will join, there will be a double bond between 2 and 3. So it is 1, 2, 3, 4, 5 and 6. So there will be a double bond between 2 and 3, we have a double bond here, 1 and 6 joins, so 1 and 6 joins, 4 and 5 joins, 4 and 5 joins, so this is the product here. This reaction you see we can have this also, conjugated diene is important, okay? This is 2 double bonds diene and it is conjugated also, conjugated diene. Again in this one you see we have a conjugated diene plus we will have a double bond and suppose some atoms or groups are attached with here. So here the product will be what? We will get a cyclohex compound, right? With a double bond here everything will be same, only with this 6 carbon we have one more group attached, so 6 carbon we will have one more group here. If you have a triple bond then what happens? Conjugated diene plus we will have a triple bond, right? So what happens here? 1 pi bond will attack over here, this will attack over here, 2 bond will be left, so we will get cyclohex diene into this. This double bond will be 2 and 3 carbon atom and we have a double bond here also. This is the product we get. Is it clear? Did you understand this reaction? What happened? Are you there? Tell me. I am not getting any messages. Understood, Trudev? Okay, so this is it for hydrocarbon. Now can we start qualitative analysis? Okay, so you see in this qualitative and quantitative analysis we have to measure the quantity of various elements present in any organic sample. Okay, so there are various different methods we have. Okay, identification of radicals also there, correct? But those things are not there in this labors, J mains labors. Few things are there in J advance labors. Okay, that's why we don't discuss these things here in after J mains will discuss those things. Okay, those who was to prepare for J advance. Okay, but few methods are there which is useful for you know board also and then in J mains so that we'll discuss. Okay, so basically what is qualitative and quantitative analysis? Okay, qualitative analysis, generally we use this kind of these are the basically the laboratory method we use for the for a detection of various different types of element. Suppose I say carbon, hydrogen, you know nitrogen, all these elements halogen, what is the amount of these elements present in a given organic sample? Okay, for those estimation only we use these methods. Okay, so since we are going to suppose if I give you an organic compound like this and if I ask you find out the you know percentage of carbon and hydrogen present in this sample. Okay, so for that what we have to do? What are the reactions involved? Okay, what we'll do into the lab for that we'll get the desired result. Correct, so the first basically we're going to find out the amount of the element present in any organic sample. The first thing you write down here that is detection of carbon and hydrogen. Detection of carbon and hydrogen. Now like I said earlier also this thing is basically depends upon the concepts of mole, mole concept basically you have to use the unitary method for the calculation. Okay, the only thing is what you have to understand how do we do this? Okay, how many of you have done the practical on this? You have done this practical thing in the lab? So then you will understand it better. So they must have done the calculation also in the school, no? You didn't understand over there. Correct, you see. So first of all you see the sample that you have. I have given you a sample of you know organic compound and in this sample you have to find out the amount of carbon and hydrogen present. Right, then what you will do? First of all you have to understand or you have to find it out that in this sample whether this carbon and hydrogen is present or not. Correct, so for that what we will do. Okay, so the reaction involved here right the sample in this sample I will just write down the question first. In this sample you have to find out the composition of carbon and hydrogen. This is again, suppose you have to find out the percentage carbon over here. So now what we will do? This sample the reaction involved here you see this sample, we allow this sample to react with CuO, copper oxide. Right, CuSO4, copper oxide. Now the mixture that we get here, this mixture, this mixture is passes through, this is passes through, we allow this to pass through and hydrous and hydrous copper sulphate solution. Now this an hydrous copper sulphate solution is white in color. I am giving you this information, you have to memorize this. Right, an hydrous CuSO4 solution. So what happens here you see, if the carbon is present in this sample, okay, if any carbon is present into the sample then how the reaction takes place. This carbon combines with this oxide copper oxide and it gives CuO2 plus copper will eliminate here. Now this CuO2, now this CuO2, suppose you are getting this CuO2, some gas you are getting here and if you want to know that whether this gas is carbon dioxide or not then what we do this CuO2 we allow to pass through CaOH whole twice. Then what we get here, we get CaCO3 plus H2O. Now this CaCO3 that forms here, it forms, it turns milky, right, turns lime water milky, turns lime water milky, correct. So when this appears it means the gas was carbon dioxide and when the gas was carbon dioxide it means the carbon is present in the sample. This is how we detect the presence of carbon in the sample. Did you understand this, tell me. This is how we detect the presence of carbon in the sample. Now for hydrogen what happens you see. For hydrogen also, hydrogen, since we allow CuO2 to pass through it forms what? water and copper eliminates, right. Now when this is allowed to pass through CaSO4 solution which is white in color initially, the product we get here is hydrated, this is anhydrous, right. You see here anhydrous CaSO4, there is no water molecule. We will get hydrous copper sulphate solution.5H2O. Now this hydrous hydrated copper sulphate solution is the property of this, it is blue in color. So when the white turns blue it means the high anhydrous compound convert into hydrated compound, correct. That means the presence of hydrogen is there when the anhydrous compounds become hydrated in it means what by any means we are getting water over there and water is only possible when we have hydrogen, right. The presence of hydrogen we detect by the change in color of hydrated anhydrous copper sulphate solution, fine. These two reactions gives us the information that in this given sample we have carbon and hydrogen present, right. Now when the carbon and hydrogen is present then how do we find out the amount of these two elements in the sample? Can I move on next slide, tell me, correct. So you see here now few datas will be given in this kind of question I am assuming all the data here, okay. I am assuming all the data here and I will give you the final formula, okay. In the exam when it comes you have to use that formula and get the answer, correct. So I am using some I am taking some assumptions of amount I am assuming here. In the exam they will give you enough data so that you can find out, okay. Otherwise it won't be possible to get the amount of those elements, right. So what I am assuming you see the mass of organic compound the sample we have, the mass of the organic sample that is given to you, right in which we have to find out that amount that is suppose W gram, mass of the organic sample is W gram. The amount of CO2 that we obtain by the reaction that I also am assuming mass of CO2 suppose we are getting is Y gram, mass of H2O that we are getting here is X gram, correct. Now you see this reaction like I said before that you have to use the more concept thing over here, correct. So what is the reaction in which we get CO2 that reaction is this carbon plus copper oxide you will get CO2 plus 2 Cu here we have 2 Cu here. So from this reaction what we can write what is the molecular mass of CO2 44 gram what is the mass of carbon here 12 gram correct. So we obtained 44 gram of carbon from 12 gram of carbon, right. So what we can say see try to understand this logic okay you don't have to memorize this okay you don't have to memorize this or anything you don't have to memorize here, right. The only thing is what see according to this amount of carbon present in the sample according to this only we will get the amount of CO2 by this reaction yes or no tell me this any doubt so if the amount of carbon is more in the sample will get more amount of CO2 that's a simple thing correct. So what we can say 44 gram of CO2 we obtained from 12 gram of carbon, right. So Y gram of CO2 will obtain from how much gram of carbon that's what we have to write down, right. So you will apply here and we can write what 44 gram of CO2 contains 12 gram of carbon then Y gram of CO2 gives what 12 by 44 into Y gram of C. So what we can say in the given sample in the given sample the amount of carbon is what in the given sample the amount of carbon is found to be 12 divided by 44 into Y gram what is Y it is amount of CO2 obtained correct. Now since we have to find out the percentage carbon so percentage carbon will be what the mass of carbon which is 12 divided by 44 into Y this mass is present in what is the mass of the sample we have W gram correct this divided by W into 100 right. So the percentage carbon contained in the sample is 12 Y by 44 into 100 by W this is the amount of carbon percentage amount of carbon percentage carbon present in the given organic sample okay did you understand this yes W is the total weight of the organic sample that I have assumed did you understand this is it tough okay Agar's got it Shruti got it Rudev Gargi what happened Shreya did you understand this is it tough any doubt you have you can ask me okay now in this formula you see there are three unknowns there are three unknowns one is percentage carbon another thing is what amount of CO2 and then the mass of the sample so out of these three unknown two things will be given in the question okay the third unknown you have to find it out with this formula so you have to memorize this now we have to find out the percentage composition of percentage of hydrogen present correct so what we can write the another reaction the reaction for hydrogen and that we have is this 2H CuO gives H2O plus Cu right so what we can write here the molecular mass of this is 18 and this is 2 right similarly what we can write 18 gram of H2O contains what 2 gram of hydrogen so what is the mass of H2O we are assuming X X gram of H2O for that we required what 2 by 18 into X gram of hydrogen so percentage hydrogen is what X by 9 into 100 divided by W is it clear tell me can we move on next this is we have done the estimation of carbon and hydrogen next one you see that is estimation of halogen estimation of halogen so what we'll do here again we have to use some reactions to estimate this right so reaction that we use here is the reaction of silver nitrate AgNO3 with halogen X and this forms AgX some other product also will get but we are concerned with this only silver halide okay and this reaction takes place in karyus tube karyus tube not important for J or net exam but for board maybe you have to write it down okay karyus tube c a r i u s check the spelling probably this is the spelling we have okay now in this sample right this halogen we are getting from the sample that in which we have to estimate okay so in this sample this halogen we are it is coming from the sample organic sample that we have if in this sample any write down one note you write down here if any carbon hydrogen or sulfur any carbon hydrogen or sulfur present in the organic compound present in the organic compound it oxidizes to CO2, H2O and H2SO4 respectively H2SO4 respectively okay now again calculation we have to do mass of organic compound mass of the sample organic sample that is W gram mass of silver halide AgX it's suppose X gram now I also assume the atomic mass of halogen because it can be anything is supposed capital A gram atomic masses capital A so what we can say AgX AgX contains A gram instead of this AgX we can also write the molecular mass of this for silver it is 108 and for X I have assumed A so 108 plus A gram of silver halide contains A gram of halogen correct so what we are having X gram of AgX so X gram of AgX contains what A by 108 plus A into X gram of halogen percentage halogen will be what same thing A into X divided by 108 plus A into 100 divided by the mass of the organic sample percentage halogen it is X is this so this is the formula we have for the percentage composition of halogen present in the organic sample understood right this is what the estimation of halogen we have right but how do we detect the presence of halogen that's the second part of it okay so detection of halogen write down in this only you write down detection of halogen so here you see right down here the sodium extract or sodium itself sodium extract combines with combines with halogen to form sodium halide sodium halide okay so the reaction would be Na plus X fusion takes place we'll get NaX usually it is chloride chlorine bromine or iodine instead of X right now the presence of like X can be chlorine bromine or iodine so how do we know whether it is sodium chloride or bromide or iodide okay so for that you see the reaction here what we do in that Ag of NaX that you have in that NaX in sodium halide in sodium halide will add some amount of some amount of Ag Na3 correct so what happens what kind of reaction takes place that you see NaCl if it is NaCl right combines with Ag Na3 then it forms what AgCl plus Na Na3 if this AgCl is there then the precipitate we get here is white precipitate so this white precipitate confirms the presence of NaCl there right but this is also one thing we have this precipitate white precipitate it is soluble in in NH4 OH right first of all white precipitate and it is soluble in NH4 OH this confirms the presence of confirms the presence of chlorine in the sample correct second reaction you see this property you have to memorize white precipitate soluble in NH4 OH if it is bromine NaBr we will get plus Ag Na3 we get here what AgBr plus Na Na3 now the color of this is dull yellow precipitate dull yellow PPT correct dull yellow PPT this precipitate is also soluble in NH4 OH soluble in NH4 OH this indicates the presence of bromine presence of bromine just a second indicates the presence of bromine if it is iodine NaI plus Ag Na3 this also gives Agi plus Na Na3 now the property of this is the color of this is bright yellow the previous one is dull yellow this is bright yellow PPT but this thing is insoluble NH4 OH when it is insoluble this indicates the presence of iodine indicates the presence of iodine this is how we detect halogen first whether it is presence or not present or not and then if it is present we will find out the amount that is the estimation part okay estimation of estimation of halogen understood till here see you may have some theoretical question okay especially like these theoretical based question is it is important for need exam okay in J also they may ask I have seen few questions right now J immense is going on I have seen few questions they have asked very basic factual question they are asking okay if you you know like if I tell you they have asked one question yesterday what is the pH of rain water okay can you tell me what is the pH of rain water don't use google 0.6 this is the pH of rain water correct so they have asked this question in this study J mains exam okay so right now you know these days what is happening it's 5.6 these days you know what is happening because the students are scoring a lot in J mains okay they are scoring more than 90 percent even once in like last last year one of all of the student of jaipur I think I'm not sure he has scored 100 marks in J mains okay so the preparation level is too high now okay so if they ask some logical some good question there are few students who can solve those kind of questions also so now these days what I am seeing like what I'm like you know getting these few days that generate the exam has been started from last two three days what I what I have seen is that they have asked various factual based question like it is if you know it's fine if you do not know you cannot do anything yeah that's what so the point is so the point is three days like if you know then you know if you do not know you cannot do anything in these kind of questions so these data that I have given you here the color change soluble or insoluble these are actually factual based there is no logic in all these okay if you know fine if you do not know you cannot do anything correct so the pattern of the question that Jay is asking is is being changed these kind of questions generally they ask in need exam okay this factual based question they don't ask this kind of question in Jay usually but nowadays we are looking at these questions what we are observing that they are shifting little bit towards the factual based questions so when you are preparing for that you have to maintain the balance in these two logical thing and plus factual based question right after exam if you say so the question was factual is what can I do fine you cannot do anything but eventually you are not qualifying that's the worst part okay so anyways we'll just we'll you know discuss all these things sometimes like later on in the physical class but these factual data you must remember and then the numerical question will be asked that you can solve on the question on the formula that we get the percentage composition enough correct so fine we'll take a break here okay it's already 615 we'll start the class at 630 will it be fine 630 will resume the class okay so take a break now we'll resume the class at 630 we'll continue with nitrogen detection of nitrogen the last part of this take a break okay are you there can we start now you see the last part we have here halogen and carbon hydrogen we have discussed now the last one which is you know the most important one we have this one on this the question they have asked many times in both NEET and JEE exam okay and that is the detection of nitrogen detection and estimation so first we'll see detection of nitrogen this is a theoretical part in estimation again we have little bit of calculation of moles concept that we have done already same kind of calculation so how do we how do we detect the presence of nitrogen okay so the first thing here the there are two basically two tests we have okay they are basically two different tests or methods we have the first one is soda lime test soda lime test in this what happens the organic compound or the sample that is given in which you have to find out the no estimation of nitrogen you have to do so the organic compound or sample is allowed to react with soda lime soda lime and when you heat this if this compound contains nitrogen if this compound contains nitrogen then it will eliminate ammonia it will eliminate ammonia for example you see suppose the compound is CH3CONS2 as it amide reacts with NaOH CaO also will take that it forms CS3CONA plus NH3 okay one note you write down here one note you write down the organic compounds like nitro and diazo compounds organic compounds like nitro nitro means where we have NO2 group present and diazo diazo means N double bond N right organic compounds containing nitro and diazo does not liberate does not liberate ammonia ammonia on heating does not liberate ammonia okay so this is one of the exception we have okay on this also way we have one you know facts that i'll give you later on okay just write it down today we'll try to finish this portion okay today if not possible by the you know scheduled time i'll send you some question on whatsapp on the group i'll send you a few questions fine so this is the soda lime test we have the another test here another test here is lesson is method lesson is method now this method is useful on this they have asked question or twice in j e main exam 2013 and 2015 they have asked question on to this twice 13 and 15 j e main now in this you know what happens what we do into this one that will take sodium metal Na metal and this Na metal is heated so that it melts okay so here we have Na metal in melting is in in liquid instead actually not exactly liquid but it is in melting instead here we have not solid right now what happens in this we add the organic substance whatever the sample we have that we add into it will add organic compound organic compound means the sample okay now what we do this mixture all this mixture will add will heat strongly over here red heat will red heat strongly we heat this mixture over here right after heating after heating will cool down the mixture after heating the mixture is cooled and filtered and filtered this filtered liquid or mixture here this filtered liquid is known as liquid is known as sodium extract sodium extract or we also call it as lesson is extract sodium extract or lesson is extract which is usually alkaline in nature this filtered liquid that you get here this is usually alkaline in nature alkaline in nature if it is not then we add little bit of NaOH right on over here if it is not alkaline then we'll mix little bit of NaOH into it correct if it is not then we'll mix little bit of NaOH into it next you see what happens the Na extract that we get or lesson is extract in this we add FeSO4 and we heat this mixture again we heat this and then we add and then we add a few drop of FeCl3 FeCl3 and the resulting solution the resulting solution is acidified acidified with dilute HCl only what happens in all this blue color appears blue color appears which confirms which confirms the presence of nitrogen so all these you know practical thing this you don't have to memorize just you have to memorize the result don't miss this result out we'll get blue color that confirms the presence of nitrogen now what all reactions are involved in this that you see the reactions involved are suppose in the sample we have sodium carbon nitrogen this is present in the sample this actually these elements forms NaCl and when you add FeSO4 into it this is extract we'll get we add FeSO4 into it and then we get FeCl3 plus Na2SO4 and this also when we mixed four moles of NaCl which is already there in the sample then it finally converts into Na4 6 this is a coordination compound complex compound which finally reacts with FeCl3 and then it forms Fe4 FeCl6 plus 12 molecules of NaCl this compound we call it as ferric ferro cyanide cyanide and this is actually blue in color so this is finally forms which color is blue and that confirms the presence of nitrogen okay so you have to memorize this particular product ferric ferro cyanide blue color these are not important intermediate reactions are not important understood now we'll see the estimation of nitrogen how do we estimate this like what is the amount of nitrogen present here so there are two different methods for the estimation of nitrogen two methods we have the first one is Jeldahl's method and second one is Dumas method okay this method is also useful last year also in JEE they have asked one question on to this on Jeldahl method they have asked one question on to this last year also okay first of all this point you write down which has been asked again many times in a neat exam also this method this method is not useful is not useful for the compounds for the compounds containing containing nitrogen in the ring in the ring like pyridine cunolin cunolin compound containing compound containing nitro group NO2 and diso group which is nothing but n double bond n so this note is important okay you must keep this in mind that Jeldahl method is not useful for those compounds which contains nitrogen in the ring plus NO2 group if it is present or diso group if it is there okay now you see in this one I'll use the another slide all of you have copied this now here you see we have organic compound whose weight is known we are now we are trying to calculate the this is allowed to react with H2SO4 concentrated and when you heat this will get ammonium sulphate that is NH4 whole twice SO4 now into this we add NaOH and we get here Na2SO4 plus NH3 plus H2O now this NH3 reacts with the acid which is left here whatever acid left that is that will neutralize the ammonia over here and that it forms NH4 whole twice SO4 this is what this is unused acid here we are using acid whatever acid is left acid left is used to neutralize ammonia now here you see the organic compound the weight is known it is W gram H2SO4 will have some volume of it suppose V capital V centimeter cube and its normality will also given in the question V centimeter cube and N1 this NaOH which we use in excess and this is the standard solution standard solution right whose normality and volume will be known its normality is suppose N2 now you see few data I'll write down here volume of acid see here we are taking H2SO4 we can also take HCl also right instead of H2SO4 we can also take HCl here and then here also we'll have the HCl so the volume of acid taken is V centimeter cube the volume of acid left after this process is small V centimeter cube now this volume we have taken and this is the volume left so the volume used to neutralize NH3 is what neutralize ammonia that will be difference of these two V minus V centimeter cube now the unreacted acid that you have unreacted acid is estimated titration with with standard alkali standard alkali which is nothing but in this case we have N2 sorry NaOH okay so when you do this titration what we can write number of equivalents will be equal so for this acid will write N1 into V capital V is equals to N2 into a small V so from this relation we can calculate the volume of acid left right so when you have this V a smaller V if you have you can find out the volume used to neutralize NH3 that's the point that's the thing that you have to calculate first by titration we'll find out a small V and then we'll find out the volume which is used to neutralize NH3 that is V minus small V correct now you see the next what we'll do that thousand centimeter cube of N1 NH3 N1 is the normality one normal N1 NH3 contains 14 gram of nitrogen okay then V minus small V this centimeter of one normal NH3 contains 14 by 1000 into V minus small V N1 also will write so percentage of nitrogen will be what the mass of the nitrogen which is 14 into V minus small V N1 1000 into 100 divided by the mass of the organic sample this is nothing but W okay so this is the percentage of nitrogen present in this sample this is the formula previous slide you see is it clear it is based on titration next one you see this slide Akash understood Sridev okay so this is one method that Zeldal method last year also they have asked question on this on to this method in percentage of nitrogen okay so this is the one part now second method we have for the you know estimation of nitrogen is Dumas method this is the last one Dumas method in this method mark what happens right down the known mass mass of organic compound is heated with heated with dry cupric oxide dry cupric oxide so if any carbon is present in the sample that carbon reacts with CuO forms CO2 and Cu combines with CuO forms H2O and Cu nitrogen is there plus CuO forms N2 plus we'll get some oxides of nitrogen into this copper also eliminates we'll get few oxides of nitrogen also right which when you pass over Cu gauze it converts into nitrogen all these are gaseous mixture the point here it is what we are getting CO2 right we are getting N2 and if any other elements are there so we'll get oxides of that gas gas right vapor of H2O we are getting so all these are gaseous form we are getting gaseous mixture we are getting so next let me write down these gaseous mixture these gaseous mixture potassium hydroxide why we are taking COH because in potassium hydroxide nitrogen gas is not you know by potassium hydroxide nitrogen gas is not absorbed by the potassium COH this COH right down absorbed all gas this absorbed gas except that's why we are using this this nitrogen is you know start accumulating over that COH molecule that you have right it starts accumulating over there right so the volume of nitrogen produced volume of nitrogen produced is measured at room temperature room temperature suppose that volume is the centimeter cube right now how do we do the calculation we know what we know that 22,400 centimeter cube STP is it goes to what is the mass of this can you tell me this volume of N2 and STP will have what mass tell me what is the mass 22,400 centimeter cube of N2 will have what value of mass what how many moles we have in this volume 22,400 it's N23 there it's N2 right it's 28 gram 22,400 centimeter cube is equals to one more one mole of N2 gas is equals to 28 gram so 22,400 centimeter cube of N2 at STP is equals to 28 gram of N2 correct so what is the volume we are getting V centimeter cube so V centimeter cube of N2 is equals to 28 divided by 22,400 into V so percentage of nitrogen if you have to calculate that will be the mass of nitrogen 28 V divided by 22,400 into 100 divided by W well W this W is what mass of organic compound mass of organic compound right so this is the one question where we have the percentage of nitrogen if you have to find out in this one more type of question they ask sometime that I'll explain you now in all these question if you remember the formula these formula and plus the term that is written in the formula what is that particular term then you can do the questions because sometimes they ask you what is the volume of this gas at STP right at STP because you are doing this experiment at some temperature right that is the room temperature see we can apply that also but the point is you know we are doing this this is the practical organic chemistry today we are doing this observation the nitrogen gas will be there in the you know the tube or the burette that you have from there you can easily find out the volume mathematically you will calculate from PVS goes to NRT only but in lab how do you measure volume right with that tube and all burette and all you can easily see what volume you are taking see for per for percentage of N2 yeah PVS equals to NRT you can apply mathematically you have to use that one only I'm going to explain that part now PVS equals to NRT you can apply for that you should have the you know values volume RT number of moles you should have pressure should be given but in the lab you will get that you know data from there you can easily see and you can find out okay the volume is 10 ml or 20 ml that's why we are using this okay mathematically we'll use that only now coming back to see this we have to find out the because this experiment we are doing at some you know at some condition we have some value of room temperature and all so corresponding to this what is the volume of STP we'll get for these gas whatever the mass of this you'll get corresponding to this whatever what is the volume of this gas at STP that sometimes they ask you see the calculation here just last one now I'll write down the few assumption mass of organic compound is W mass of the organic compound is W gram volume of nitrogen collected is V centimeter cube atmospheric pressure atmospheric pressure is suppose we have P mmHg room temperature at which the you know experiment is taking place that is suppose T degree Celsius which is nothing but 273 plus T Kelvin sorry that is 273 plus T Kelvin aqua extension aqua extension because we'll have water also there is P dash what is aqua extension P dash mmHg aqua extension is the pressure exerted by water vapor right down here aqua extension is the pressure exerted by H2O vapor right so pressure of water vapor we have is this and this is at T degree Celsius whatever the temperature we have here at that temperature only weight like this okay so now as you see here the pressure of because you see the total pressure is P correct and out of this P dash is the aqua extension the pressure exerted by the all other components other than nitrogen all components other than nitrogen right so which is nothing but the water vapor we are assuming now so the pressure exerted by by dry nitrogen will be what it will be the total pressure P minus the aqua extension P dash mmHg so now you see there are two conditions we have one is the experimental condition at condition at which the experiment is taking place and other condition is what STP so we have to find out the volume at STP so what is the pressure of experimental condition is P minus P dash what is the volume we have here volume is nothing but V that is collected T1 is what 273 plus T Kelvin at STP what what is the pressure at STP 760 mmHg here also it is mmHg what is the volume here volume at STP is what we have to find it out temperature is 0 degree Celsius 273 Kelvin right now in this data you can easily apply P1 V1 by T1 is equals to P2 V2 by T2 from this you can easily find out V2 and this is the volume of nitrogen gas at STP you can convert the volume which is here calculated at the experimental condition to the volume at STP means what if the same experiment would have taken place at STP then what would be the volume this is the volume we have V2 centimeter cube did you understand this tell me okay so today we have done estimation of carbon hydrogen halogen nitrogen right so these are the formula we have we don't have anything into it okay radicals things are there in jay advance okay you don't have to worry for that correct so uh we'll you know we'll wind up the class here only few questions based on this i'll send you on the group okay i'll just see that i don't have a book right now i'll see those questions and send you on the group okay those few you know a little bit of question you can discuss out of all the discussion that we have done in practical organic chemistry the last two the estimation of nitrogen is the most important okay half sigma bond half sigma bond is okay just give me a minute what i said that the nitrogen is most important here right so what all things we have discussed in nitrogen that you must keep in mind okay they have asked question in jay also in neat also the nitrogen one is most important half sigma bond is the bond which has bond order half when half bond order is there it means we have suppose the for o2 o2 plus o2 minus o2 2 minus like this we have calculated the bond order correct to them so in that you see few for few molecules will get 1 1.5 2 2.5 like this okay so what is this 1.5 bond order and 2.5 bond order means this means what we have two complete bond and one half bond we have over there so half bond order means what when the bond half sigma bond means what when the bond order is half 1.5 bond order means we have one complete bond and one half sigma bond what is your doubt half sigma bond is this only you can you know send me the question then i can understand what exactly you are trying to ask you send me that question i'll see okay to them yeah i saw there was a question related to bond paramagnetic and diamagnetic behavior no o2 and all right the question was probably where the no the magnetic behavior changes from paramagnetic to diamagnetic or either way yeah yeah correct so that is what the half bond order whenever half sigma bond means what whenever the bond order you are getting half it means there is one half sigma bond one bond order means there is one complete bond one sigma bond two means two sigma bond like two bond we can have one sigma one pi means two complete bond will have if the bond order is an integer value it means two complete bond will have okay if it is an like you know fractional value point five like that it means there is the bond order is half over there that's that what it means okay is it clear now can we wind up the class here okay so next class we'll start with ionic equilibrium okay thank you