 See, for this chapter, like I told you already, preparation methods are not important. All the name reactions are important. Right on next, art-eastert reaction. Right on, by this reaction, an acid converts into another acid with one more carbon atom. Right on, in this reaction, acid chloride, treated with disomethane, disomethane, and forms, and forms di-Azo-ketone, and forms di-Azo-ketone, which goes under rearrangement, which goes under rearrangement, under rearrangement, and forms, and forms ketene, which reacts with, which reacts with silver oxide in presence of H2O, and gives higher acid, and gives higher acid. Okay, reaction you see here, we have to use acid derivative generally. R-C-O-O-H, if it is allowed to react with SOCl2, it forms acyl chloride, R-C-O-Cl, acyl chloride. Now this acyl chloride is further allowed to react with two moles of disomethane. This is disomethane, right, two hydro replaced by N2, disomethane we have here. This H, and this Cl forms HCl, and it gives CH double bond, C double bond O, CH N2. This is di-Azo-ketone, right, this is di-Azo-ketone. When you heat this, and it goes under rearrangement, and it forms R-CH double bond, C double bond O, and this is ketene. This ketene with silver or silver oxide, we can also take with H2O. It converts into R-CH2-CO-O-H, this is the acid we get here. Overall reaction if you see, R-CO-O-H converts into what? R-CH2-CO-O-H. So one with hyaline group you have to insert between the alkyl group and CO-O-H. The mechanism you see after this, di-Azo-ketone is this. R-C double bond O, CH double bond N, double bond N, positive charge on this nitrogen and negative charge. This is di-Azo-ketone, this one. When you heat this, so in this reaction also, with this ketene, N2 also goes out. Nitrogen gas evolves into this. This electron comes over here, right, and this bond pair of electrons shift onto this carbon atom so that this bond breaks and N2 goes out in the reaction mixture. So it forms R-C double bond O, CH with one lone pair, negative charge on it, plus we will get N2 gas. This is actually a carbene intermediate, this one, carbene intermediate, right. Now in this, whenever this carbene intermediate forms adjacent to the carbonyl group and this is true for nitrates also, that we will see in Amine's chapter, okay. Whenever this Amine's or Nitrate's or carbene's intermediate forms adjacent to this carbonyl group, there's a rearrangement takes place, okay. So R plus rearrange itself here, rearrangement takes place and this rearrangement we call it as Wolff rearrangement. This comes over here, this electron pair goes here and it forms R-CH double bond C, double bond, right, which on reaction with H2O with AG converts into R-CH2CO, right. So basically H plus comes over here and this takes OH minus, R-CH2CO, okay. One very important point here is this H2O I'll write down this way. If you take this H2O, then you'll get acid here. So this H is coming from this H2O, right. If you take R-OH here, then the product will be R-CH2CO-O-R. So this is coming from this solvent that you are taking in the last, that will get acid. This one write down properly, right. So we are getting acid which has one carbon more here, R-CH2CO, one CH2 we have to introduce. If you are not taking H2O in the last step, you won't get acid. Alcohol if you take, you'll get acid, okay, R-CH2CO-O-R. This is in preparation we have only few methods. There are few important names of this compound we have which we should know. CH2N and COOH, okay, right. Now when N is equals to zero, the molecular formula is what? And this we call it as oxalic acid. This name you must remember. N is equals to zero. The molecule is oxalic acid. Similarly, N is equals to one. It is malonic acid. N is equals to two. We call it as succinic acid. N is equals to three. N is equals to four, which is adipic acid. N is equals to five. We call it as pymelic acid. So how do you memorize this? You need to give this in mind. This three structure of the compound you must remember. What is this compound? It is compound PXCH double bond CH, COOH, pylolic. It is not pylolic. What are the reactions? Synamic acid. Synamic acid, okay. PX program is CS3, then this one is protonic acid. Lactic acid, what is the formula? CS3, COOH, H, COOH. It is lactic acid. These two are GI of each other, right? Cis and trans. Right, the cis one is malic acid. And this one is pylomeric acid. So this is the structure of the compound. This is a carbazoleic acid. Reactions? Reactions earlier. How was the preparation? How was the preparation? Reaction last class. There are some name reactions we have. That is Huffman, Bromind, Curtis. That we'll do in Amines. Because for that the intermediates are nitrenes. So first we'll see what is nitrenes. And then we'll see the reactions of Amines. Okay? There are a few more name reactions we have in this chapter. That we'll discuss in Amines chapter. First we'll see intermediate nitrenes. And then we'll discuss that. What is? Etard reaction. Etard reaction. Yeah, we have that reaction also. Are we finished? When? This afternoon. I haven't done it. Write down. It's that again name reaction we have. Write down in this reaction, in this reaction the methyl group, CRO2CRO, in this reaction the methyl group, which is attached to an aromatic ring, generally it is benzene ring. It's aromatic compound reaction we have. Attached to aromatic ring, converts into, converts into aldehyde functional group, converts into aldehyde functional group, when it is heated in presence of, when it is heated in presence of an oxidizing agent, which is, which is generally a, which is generally chromyl chloride, CRO2Cl2. Solvent we can use, inert solvent like CCl4 we can use, chloroform also we can use for this purpose. So the reaction is, suppose we have a ring with Cs3 attached here, and when it is heated with CRO2Cl2, chromyl chloride, CCl4 is a solvent, forms C double bond O and H. This is the reaction we have. We will get some oxides of chromium also into this, and then as of HCl also we will get into this reaction. This is also a direct name reaction we have, reactions of aromatic compounds, aromatic compounds. Mechanism is not there, see actually, CRO2Cl2 and CRO2HCl2 vice with C H, that is the intermediate form in AR and H plus. How do we get this? We have CRO2Cl2, and this is an oxidizing agent, right. So react with H2, so the O2 is react with H2, so from there H2 is from CRO, from CRO, CRO2Cl2 in brackets, CRO2HCl2 and then twice. Like this. So it actually reacts with CRO2Cl2. Look at this, is it? This is the intermediate form. And then, and then, CRO after react with H2. Which means single reaction? So there is no reaction. So we can assume that here, OCRO HCl2, which I am going to try to correct. With H3O plus, H2O, C H O twice. What is this? It looks like a hydrogen. It looks like a hydrogen. It looks like a hydrogen. It looks like a hydrogen. It looks like a hydrogen. It looks like a hydrogen. So this is it, and then its reaction will be H2. So double bond, and then we have an OH. There is this H. Oh, you have OH OH. I don't know what it is. Can you just say, burn stuff? This H plus, will go on to this oxygen. Then you will get more H. We have ring here. Or this molecule is nothing but this. So with H plus, this comes over here. And this bond breaks. So we will get what? So I have to need two H. OH will come on C carbon. And we have one more molecule. We have two OH. So we have this, and the same thing we will have here. CR double bond OH, H, Cl and Cl. So finally this goes out when this bond breaks. And we get this, COH, OH and H. And then preheat the microscope. H2 will come out. H2 will come out. Mechanism? I don't see it. How do we get this? How do we get this? After this did you understand? This two molecule we have. So with H plus H2O, this oxygen, this oxygen gets protonated. Positive charge here. Then this bond goes here. This two molecules goes out. Take this OH minus from this solvent. Correct? And when we get this OH OH here, two molecules goes out. We will get aldehyde here. That's what the product we have. CHO. So after this, fine. But how do we get this? So CR will have a double bond with both the OHs. We are saying this double bond OH, double bond OH. So one double bond will break and attach the OH. So we have Cl and Cl. So we have two molecules. We will get single bond. And then again, another molecule will have this. And the other OH will attach to the C? Yes. And then this will attach to the C. That's the important mechanism. Okay, but the real mechanism is not important. This methyl group basically converts into CHO. We are using bromide chloride, oxidizing agent in an inert solvent, CCl4, carbon tetragloride, carbon disulfide also we can use here. We use HPSS2 gives you benzyl dehyde, okay. CHO. Okay, so next class we will start our minds. Thank you.