 The name is important, the name of CS3, C level point O, C O O H. If the formula is given, you can say this shows I to formula because we have C O C L to represent this. But sometimes they give you name also. This name and the formula you must remember. Which is nothing but R, C H, R and this one, right? To L kind of attach with the star, which attaches with the nitrogen. Which is too busy, I must know that case, okay? You know the formula, lactic acid is this and C O, this is lactic acid, okay? So you will see it's C H over CS3 group is present. This is nothing but this. C H over CS3 group is present, lactic acid also shows positive I to formula. Correct? Is this an optical reactive compound or something? Yes? Why is it optical reactive? Lactic acid is two I to formula of this. One is R, the other one is S and both shows I to formula. Okay? Next one you write down carbide amine reaction. Carbide amine reaction. Means it is a reaction with C H I3 where we get carbide amine. Okay? So there's two reactions we have. And you see this will start this isomerism. Like the total isomerism is left because that will do and then we do the commercial isomerism. Okay? So you write down in this reaction, in this reaction when one degree of amine, when one degree of amine reacts with hydroform, reacts with hydroform in presence of alcoholic KOH. In presence of alcoholic KOH it forms isocyanide, it forms isocyanide. And this reaction is carbide amine reaction. So in carbide amine reaction we use this. Did I say one degree of amine? You write down anything, not amine. Anything. Or you can also write in this. C6 H5 and this is also the same. Anything. When reacts with C H I3 in presence of alcoholic KOH. Can you write down the product in this? What is the product we get? Isocyanide. Isocyanide is what? Nc. Rnc. Rcn is cyanide. And when R is recorded as isocyanide. Cn- is an amine-tinted ligand. Amine-tinted ligand we are discussing. Got an issue in the Hong Kong chapter. Okay. The point is this. Carbon and this nitrogen both has an Nc to attach with another KOH. When carbon attached with this. So wherever you have one molecule. In which more than one atom has an Nc to attach with another KOH. Like in this case we have carbon and nitrogen. So when we have more than one carbon atom present. Which can attach with the amylated ligand. So Cn- ligand is nothing but this donates a bone pair. To form a bond between the two molecules. Right. That is called as ligand. So here since carbon and nitrogen both has an Nc to attach. Okay. So whenever Cn attached with R it is known as isocyanide. Okay. So this is a carbide amine reaction. Which we also call isocyanide test. So in this reaction isocyanide forms. Okay. And the isocyanide will be. Vengine ring Nc. Plus acl. Plus. And this we call it as phenide isocyanide. So wouldn't it be Ki? What? Ki isocyanide. Okay, phenide isocyanide. Formation of phenide isocyanide. Yes. In hell you will have headache and you will not feel well. So isocyanide is endless. It's not that good. Okay. And we have a very bad reaction. Something like, you know, Rotten fish type. Rotten fish type. Rotten fish. Rotten fish. It's very. Rotten fish. Rotten fish. You want to take it actually. Rotten fish. Rotten fish. Rotten fish. Oil. That kind of smell. All this rotten. This is potato. That is also very good. Yes Rotten potatoes and smell is given by the side. So are good? No, it's not. But they are not different.agen.っぽ height So, there is also some dried fishes there, right? They used to keep that fish in sunlight and then dry. Which in Bengali origin, we call it Sukhapa right now, right? So, that is very ridiculous. You cannot make that. Heated in presence of hydroform, heated in pressurized and H.I. It forms phosphorus and H.I. It forms CH2-I2. This is one important reaction. CH-I3, red phosphorus, H.I. And it forms CH2-I2. But when this is heated in presence of silver powder, it will end up getting acidity. Silver powder will end up getting AGI and acidity. You can make two more of this here. So, this one is more important. This one also you can keep in mind. And this is I sign it. So, only these two three reactions for hydroform make up. Like I said, you will have this hydroform test. Every year they ask this question. You must remember those conditions. What are the groups Mahadev shows hydroform. And that is the most important part. And after that these two reactions. What do you like, AGI powder? It's power, right? Lagoon. Okay. Next you will have carbon tetrachloride. CCL4. Carbon tetrachloride. CCL4. How do you prepare carbon tetrachloride? Write down. This way I am already done, right? What is that? Correlation of methane. Exes of chlorine. The first method of preparation you write down. When this CH4 combines with CH2, excess has to be written down. You write down all those steps here, theoretical mechanism and all. And it forms CCL4 plus HCA. Carbon tetrachloride. Okay. There are two points written to this. First we already know. It follows free radical mechanism. It follows free radical mechanism. And next point. The formation of methyl radical. The formation of methyl radical. In chain propagation step. The formation of methyl radical. In chain propagation step. Is the rate determining step. This is important. Sometimes they ask you this question. This is the rate determining step. Okay. The coordination reaction. Formation? Methyl radical in chain propagation step. Is the rate determining step. This is one point. Another method is what? We can use carbon disulphide with CL2. We get here CL4 plus S2Cl2. CCl4 S2Cl2. Carbon tetrachloride. And S2Cl2 we call it as sulfur monochloride. It's sulfur monochloride. Sulfur monochloride is basically the same. Because we have two molecules of SCl. Sulfur monochloride is a monomer basically here. We use the catalyst here for this reaction. And then catalyst reaction is not important. The catalyst can be iron or we can take I2 also. Or we can take NCl3. One of these catalysts we can use as well. S2Cl2 can be combined with S2Cl2. It forms carbon tetrachloride. It forms carbon tetrachloride. So whatever the carbon disulphide is left here. That also has a reason to combine with S2Cl2. It is the presence of any catalyst of course. CLCl3. It forms CCl4 plus sulfur monochloride. So whatever combines with this product. And again converts into carbon tetrachloride. Next slide down. Reaction of CCl4 not carbon tetrachloride. Reaction of CCl4 carbon tetrachloride. And only one reaction we have here. Which we call it as CCl4 carbon tetrachloride. Important as well as important. And forms. And forms. Reaction you will see what reaction we have. Penal. O H plus CCl4. C O H here. Plus NaCl and S2. What did you say? Salis Ali Kashe. Salis Ali. Salis Ali. Salis Ali. Salis Ali. Salis Ali. What happens most? Very happy. What did you say? Salis Ali Kashe. Salis Ali. Whatever you say. This is the product. Right. What happened Krishna? Nothing. So this is the product we get. C O H at powerful collision attack. Right. If you try to understand the mechanism. You can understand. This actually we take 4 moles of NCl. Right. These two might react. CCl4 plus 4 NaCl forms. Let the 4 moles. And then 2 moles of H2O will come out. 2 moles of H2O plus we get carbon dioxide. C O H gives its boundary to this carbon. And this O H gives its boundary to this. So we have C double bond, C double bond O. And 2 moles of H2O. Right. And we get carbon dioxide. Now this benzene or phenol that you have. O H plus C double bond O double bond O. This pilot attack on this carbon dioxide. The carbon on this carbon dioxide. Because 2 double bond O we have. So this carbon becomes what? Because of eye effect of oxygen. It is draw electron. This carbon has positive. A partial positive channel. Right. This partial positive channel is have. Attack by electrons of benzene. Right. And this electron will jump over. So here what we get? It has tendency to gain. Right. So what happens for that? From this carbon it loses this H plus I. Or we can also say the proton exchange takes place. Between here and here. So what happens? This hydrogen is here again. Because that is a more stable state. Right. So the product proton exchange. We write it down here. Proton exchange. And the product will be O H. From this is one of the method of this. Because H, C, L, D. When it is flow of water. Right on this. It is prepared to distillation. Distillation you know what is that? It is prepared by distilling. Distillation. It cannot have what acid told. With leaching product. What acid told with leaching product. What is leaching product? C, L, C, L, D. Right. Can you tell me the oxidation state of. What is the oxidation state of each protein atoms? Tell me. Oxidation state of each protein atoms. Right. I don't know. Oxidation. I don't know. Oxidation. So, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 40, 40, 40, 40, 41, 41, 41, 42, 43, 44, 44, 44, 45, 46, 47, 47, 47, 47, 48, 49, 49, 49, 49, 49, 49, 49, 45, 50, 2, 3, 3, 4, 5, 6, 7, 8, die n y to the minimum and keep playing. If I play the game like this, I give myself an average of 5 patterns of like, maybe a few more turns and I hear something like that, react with the u-ring here and it forms CH3, CH0s, clarity height plus, plus, you get, this is the same, you get clarity height. Now in this clarity height, react with u-ring, CH0 plus, this product is, flow rate is flow rate is flow, and the height is flow rate, the CH0 is flow rate. Now this is allowing to react with this straight line, the first product we get. What we take here, we take two moles of this and two moles of this, C8 over H, over H, one more of straight line and then two more of this and this CTL3 combines this edge and this CTL3.