 Okay, question number 34 in our line of energy. The question is we have toluene and it reacts with NBS. Okay, and promo succinamide, toluene is this. Right? When this reacts with NBS, what it does is, so what is the ally equation we have here? This is the ally equation, right? Double bond and then succinamide is this. So this is the ally equation. So from here we are going to add that form succinic acid and the product will be CH2. Okay, so it is phenyl bromamine. Option C will be the answer. Okay? 33. 34. 34. Phenyl bromamine. Okay. Whenever you see NBS and rho succinamide, it is ally substitution. So find out the ally carbon. They are going to remove confidylate. Okay? Understood. Reaction which alcoholic potassium cyanide is KCN. See now today we will finish this chapter and all the reabs discuss what is SN1 and what. So first reaction we are going to have, reaction which KCN, potassium cyanide. Alcoholic KCN, potassium cyanide. Right on alkyne N in alcoholic forms. So it is a simple reaction you see whenever you have alcoholic medium and when you heat this. So R combines with CN, you get RCN plus KS. This is the product you get. Okay? This is alkyne cyanide. Okay? One reaction you have to write. C2H5Cl combines with KCN, alcoholic medium on heating C2H5Cl bond N plus KCN. Okay? Type AC name of this will be ethane nitride. Nitride is the suffix we use for cyanolubes CN. Right? Ethane nitride. If we replace KCN with AGCN will we get an isosanide even in this case? There also we can get cyanide but we will get a precipitate of AGCN. There is no substitution where in this reaction there is no substitution. No it is like the placement or we can say exchange reaction we have. This is C2H5 plus CN minus K plus CN minus. So this attach with the CN minus and we will get KCN. Even if you take AGCN also there also forms AGCN or AGPR. Okay? So that will be the precipitate. Okay? This is the electrolyte. Now this KCN has you know the various of different different reactions this alkyne cyanide shows. Suppose if I take RCN right? This RCN under different different foundations gives you different different product. First of all one thing you have to memorize here. The cyanide whenever you oxidize cyanides you end up getting acid. CO is finally converted to acid. Okay? So first step will be you will get acetamide that is CO and H2 and then you will get acid. So whenever you have acetic hydrolysis or acetic hydrolysis. So this is what hydrolysis is in presence of an acid you end up getting carbazolic acid. Okay? So that is one of the preparation method of carbazolic acid also. That gets in the case we have seen of carbazolic acid chapter. Okay? First reaction in this you write down RCN will react with H2O and concentrated SCL. Right? The acid is concentrated SCL and H2O is the you know the aqueous medium we have concentrated SCL. This gives you amide which is RCN4 and H2O. This is amide concentrated CO and H2O amide. The first we will get amide and further when you again do the acetic hydrolysis of this H2O concentrated SCL you will get RC00H plus H2O. This is the final product we get. Means when you take excess of acid here in H2O finally you will get acid that is carbazolic acid. But not H2O it should be energy acid. It should be energy acid. It should be energy acid. It should be energy acid. So it should be H2O. That is also an oxidizing agent that is the peroxide that also we can take. Okay? This is so like you know of acetic hydrolysis of cyanide. When you do the reduction of this in presence of NaC2H5OH. Reduction of alkyl cyanide in presence of NaC2H5OH. You will get 4 moles of nascent hydrolysis here. And you get CH2NH2 which is 1 degree of reduction NaC2H5OH. You will get 1 degree amide. Suppose we have SNCl2 of HCl in the result of HCl. This gives you RCH double bond NH not HCl. This is the addition product you get first. And then when you use this again acetic hydrolysis of this again come. H plus H2O. You will get the H2O and the hydrolysis of RCHO plus NH2O. Second one if you reduce it you can use it as a preparation number alcohol. Which one? The one degree amide. One degree amide. From alcohol from one degree amide. How it is possible? If you add water and concentrate on HCl again then it is going to replace the H2O then OH2O. No alcohol preparation from amide there is one method. We cannot use HCl and H2O there. But I forgot the name of the reaction. We have a reaction called a test of amide there. Where we use nitrous acid HNO2. And that depends also on what degree of amide you have. We will do this in amide chapter. There is a test of amides. Like how we differentiate whether we have 1 degree or 2 degree or 3 degree amides. So there are different tests for that. Like we have nitrous acid test where we use HNO2. There is heensberg test where we have HgCl2. Heensberg is HnO2. And there are other methods also. We have one more method is there. That is I forgot the name for first part. And then it is mustard oil test. Something like that. Because 1 degree of amides when you do that reaction it gives a smell which is similar to panne. It is smell or mustard oil. That is why we call it mustard oil test. So what happens when you use your nitrous acid? Nitrous acid when you have 1 degree of amide then it converts into HNO2. So just you have to break this bond and you have to add to it. There is a discussion in amide chapter. It is not real. But here I am trying to make you understand. When we have cyanide. With cyanide we can form acid. We can form amide. And we can form carotidhyl also. And cyanide. How do we get cyanide? From alkali and you know. So from alkali this is the chain I wanted to make you understand. When I start this part. So these reactions also have to be memorized. There is no mechanism of these reactions. This particular reaction we call as the stiffness reaction in this one. S-T-E-P-H-E. This is the addition product we have. One more important point here. That is the addition product. It is not minus this product. So you see in our CNPF, our Singaporean seed. This is what the structure of this. Right cyanide. When we use NA-C2-H-5-O-H. This gives you complete reduction. Right. Means all the pythons will dissociate and convert it to sigma. That is why we have 4-H plus C-H. Two of this attach to the star one. Two attach over here. And two pythons raise to get R-C-H-2 connection. That is what the product is doing. Okay. What we have to analyze here. NA-C2-H-5-O-H gives you complete reduction. Complete reduction is what? Sigma. That is the complete reduction. But when we use SNC-H-2 with SC-H, you will have partial reduction. Why partial? Because out of two pythons, only one can convert it to sigma. One python is here only. Right. So here what we have. We are using two moles of nascent hydrolysis. One attach on the scarboard and other attach on this nitrogen. And another addition product with this. Okay. So this gives you only a stiffness reaction. We have what? We have partial reduction. Okay. Suppose the question is this. A compound on partial reduction with SNC-H-2-H-2 forms an intermediate addition product. And then upon acetic hydrolysis, it converts it to L-D-hyde. Right. Intermediate addition product. And then acetic hydrolysis converts it to L-D-hyde. What is that compound? That compound is cyanide. That is what the question could be. Right. So that is why these reactions are important. Okay. We get it from this and then at the end. This is the chain we have. You must keep this in mind. Okay. Copy down this. Okay. Next we have reaction with ammonia. A mixture of amines and quaternary ammonium salt. This reaction. Right. This reaction is known as quaternary ammonium salt. This reaction is known as analysis of alkydolene. Okay. So suppose we have C2H5Br. This is the alkydolene we have. And when this reacts with what? Ammonium. Right. NH3. So NH3 is suppose this. So whenever you have reaction of NH3, most of the reactions are the three hydrogens. And one by one these hydrogens take part in the reaction. So in the process of what happens? This H combines with this Br forms HBr. And the product will be C2H5LH2 plus HBr. Correct. What is this? What do you mean? No. With C2H5Br, we get the same molecule as we did. You can take different atoms. That's not an issue. Right. You can take different atoms. C2H4 and H2Br. H4? H4. Sorry. One H gets replaced with Br. Br. Okay. I'll write it. What is the part of the product? Tell me. What I said. This hydrogens is going to be attached. Right. So here we have two hydrogens. So one of these hydrogens will take this protein. And one of these hydrogens will take this protein. And one of these hydrogens will take this protein. And forms again. It's HBr. And this C2H5 tries to this nitrogen. Right. So the positive product will be C2H5 whole price. This is the mixture of mines. What do you mean? C2H5LH2. Okay. And that is for it. This is now C2H5LH2. And one more step if you go in this reaction. Which is again C2H5Br. Alcoholic medium. You heat this. You'll get ammonium salt. That is quaternary ammonium salt. Why it is quaternary? Quaternary because we have 4 degree hydrolysm here. C2H5, C2H5 plus. Okay. For nitrogen. And then you can write down like this. Br minus outside. Right. And plus Br minus. This is quaternary ammonium salt. This is how the reaction will get mixture of products. This method is not useful for the preparation of mines. Right. Because we are not getting that for the preparation of mines. You are getting mixture of mines. Correct. Mixture of mines we are getting. And it is very difficult to separate these mines from the solution. From the mixture. Right. So like you consider as a better method. When you get mixture of mines then again you have to separate all these mines from the mixture. And then again that will be not a good choice. Because again you have to put some extra tasks into that. Okay. So what happens? These methods however we get mixture of the mines are not useful or not used for the preparation of mines. That we will discuss again in our next chapter. Okay. Next one I will write down. Idoform. The other product we have is Idoform. How do we prepare Idoform? And a few properties. In this Idoform only one thing is important and on the basis of that we will ask the question. That is Idoform test. Okay. So which one only we have which forms or which shows positive Idoform test that you have to remember. The preparation part is not important. Okay. But Idoform test is the most important. You get questions from this in all the examples. Okay. Write down the heading first of all. Idoform. What is the formula of Idoform? CHI3. Idoform. It is prepared by when alcohol reacts with alkali solution like RKOH. When alcohol reacts like NaOH or KOH it forms Idoform. Reaction you see. CH3 CH2 OH plus I2 plus NaOH. It is Idoform. Alcohol reacts Iodain in presence. It is Iodain. I forgot to say that. You may mention over there. When alcohol reacts Iodain and NaOH like this. Iodain is similar. So the product will be CHI3 plus LCOONA plus you will get NaIU. This is the product. This reaction is not balanced reaction but the product you will use. This is Idoform. So in the CHD structures. The first panel. First panel. First panel, yeah? No.assen. CSP. We did also the CSP. Well, so the dynamics are central how do we get more and more? CSP, reaction I2CO3. It'll be CHI3 plus LCOONA. I don't need reactions are called Idoform reactions are called Idoform reactions Radius one Here in the reaction you can see where you come from Very great See this side And you can see where you come from We could just add a base and then this would precipitate out and then we can Produce amines from that When you add a base then No you won't react Because it is already a base See to the side when it's broken It's already a base And then we add a stronger base like NaO Then it will take out the HBR and precipitate And then we extract this That way you can produce amines Producing we can prepare HBR Like HBR from this Where where We can now You said that this isn't Preparation for producing amines It was a complex mixture So if you prevent it from falling on yourself At this stage it's like a preion The sodium hydroxide Then it take out the HBR and it can extract No no no It won't happen because any amount that you have here This will have more than Reaction of this particular molecule It forms a linear and a secret state That's because it's a base It's a base and you have to be using a base This won't react Okay Like I said Only one question they ask from here A bit of these compounds shows Idoform test or positive Idoform test Okay For that you have few things you have to memorize If you look if it is present in the molecule They show the Idoform test So that's what the point in looking at you Right on this point Right Idoform test It's shown by an alcohol This is from present C-H-O-H-C-S-A If this group is present It also shows positive Idoform test Okay Alcohol which has this loop And ketones And also why we hide ketones What is we hide ketones? See the whole point Means what if ketone is there And this group is present in ketone Then that particular point will show Idoform test You see here we have this group resist Okay With high alcohol is this Idoform