 right down next, next reaction we have T-Tinco reaction, T-I-S-C-H, all these reactions have similar kind of mechanism, Aldon-Kanesar-O, T-Tinco-Reformance scheme, Fulcane, whatever we did have similar kind of mechanism. So, in this reaction what happens will take two molecules of aldehyde, two molecules of aldehyde, R-C, I take two molecules of aldehyde, R-C-O-H and it reacts with aluminium ethoxide, aluminium ethoxide, E-L-O-E-T both rights and we hit this right. The product forms in this reaction is an ester, okay? So, the product will be R-C-O-C-H-2-R, R-C-O-C-H-2-R which is an ester. See how this reaction takes place, this reagent aluminium ethoxide, aluminium ethoxide, it behaves as a Lewis acid, okay because it has one vacant p orbital, so it can accept a pair of electron, it's similar to AL-H-3-B-H-3 like that when aluminium moron has three bonds, it has one vacant p orbital, okay? So, it can accept one pair of electron, so behaves as Lewis acid and hence what happens in this, this lone pair, this lone pair as tendency to like it can donate its lone pair to this aluminium here, right? So, the first step of the reaction is R-C, for this reaction the mechanism is a bit different, R-C-O-C, right? Oxygen will have positive charge because it zonates electron and it attached with AL-O-E-T both rights and negative charge on this aluminium because it accepts electron, okay? Now, to stabilise this positive charge, this pi electron is taken up by this oxygen and we'll get carbon positive charge here on this carbon, right? So, next step is R-C-H positive charge O-AL negative charge O-E-T both rights, okay? Now, this positive charge by the other molecules of aldehyde, this lone pair again attacks on this positive charge of this carbon, right? So, next step of the reaction is R-C double bond O-H plus this molecule is R-C positive charge H-O-AL-O-E-T both rights, okay? So, this lone pair attacks on to this carbon and this oxygen carbon bond we have. So, the product here R-C-H double bond O and this O attached with this carbon, R-C-H-O-AL-O-E-T both rights, aluminium will have negative charge that won't change, okay? Oxygen will have positive charge here, right? Now, again this bond pair is taken up by this oxygen. So, it converts into R-C-H-O-C-R-H-O-AL-O-E-T both rights and we'll have positive charge on this carbon, right? Now, there's a hydride strip here. This hydrogen takes this layer of electron and rearrange itself on to this carbon and it converts into R-C-H-2 O-C-R positive charge O-AL-O-E-T both rights. In the last step, this AL-O-E-T whole price goes out in the living room. The bond pair shift over here and the product here we get is R-C-H-2 O-C-11 bond O. This reaction is T-shell co-reaction converts aldehyde into ester, okay? reagent is AL-O-E-T aluminium with oxide. Done? See, this is the mechanism of this reaction but you don't have to do all these things in the exam, okay? How to write down the product that you see? Whatever the aldehyde is given, two molecules of this aldehyde, you write down one ox, it's like to write down the product, you can consider this as canizaro reaction, right? What happens in canizaro reaction? It is disproportionation reaction, oxidation and reduction, right? So, one molecule of this, if it gets oxidized, what it forms? Acid, R-C-double bond O, OH. So, one molecule, you write down oxidation product. Other molecules, you write down reduction product, R-C-H-2 OH, okay? So, write down the product, what do you have to do? One oxidation and one reduction. And then, and the heating, what happens? H2O goes out, right? So, this OH N-H forms H2O and the product is R-C-double bond O, O-C-H-2, R, that is what the product is getting. So, this is the method you have to use to write down the product in this reaction, oxidation and reduction, okay? Mechanism is a bit different from the other reaction but you don't need to, you know, write down the mechanism in the exam, okay? Regent, you must take care of. What is the reagent? Aluminum ethoxide for this reaction tissue for reaction, understood? So, whatever the reactant is given, aldehyde, write down the oxidation and reduce product, reduction product and then remove water molecule, okay? Write down the product in this reaction. We have two molecules of same reagent. What is the product for this reaction? One molecule will get oxidized, so it forms R-C-double bond for OH and other one can reduce O-H-CH2, right? And then, H2O molecules goes out, so the product will be, okay? Here again, what is the product? CH3, CH2, is this correct? Yes. 5 carbon both sides will have equal number of carbon, okay? This is the intramolecular reaction, okay? So, this will get oxidized, this will get reduced, okay? So, the product here is what? C-double bond O-OH, CH2, C-double bond O... No, sorry, CH2-OH. CH2-OH, so I will write down this way, CH2-OH. And then, when you heat this, H2O molecules goes out, right? This H takes this OH, goes out and this oxygen attached with this carbon. So, we have 1, 2, 6, 1, 2 plus 4, 6 and 7th carbon. So, we will get a 7th member ring, right? 7th member ring and oxygen is present in the ring, okay? So, it is, this is the ring we have, 7th member and O attached with the oxygen, carbon which has double bond O. This is first, sorry, this is first and this is 7th carbon and the second carbon has double bond O, which is this. So, 1, 2, 3, 4, 5, 6, okay? This is the carbon composite. Next reaction I will write down, benzoin condensation. Have you finished preparation of this early identity test? First class, preparation of benzoin condensation. Benzoin condensation is again a reaction of mildehyde and generally, we take benzolehyde, P H C double bond O H, 2 molecules of this, okay? The reagent we use for this purpose is KCN, okay? KCN is the reagent we use. What is the product we get here? This reaction has a single mechanism, that is the same mechanism. First, we will get carbon ion and that carbon ion attack onto the other molecule of this. How do we get carbon ion? This nucleophile is what? What is that? I think reagent here, Cn minus, right? This Cn minus will attack onto this carbon because carbon is partial positive charge, carbon oxygen bond 4 and this pi electron goes on to this oxygen, right? In the first step of the reaction, what happens? P H, CO minus H and Cn, right? After this, there is intramolecular acid-based reaction, okay? So this hydrogen shift onto this oxygen. Losing H plus gains H plus, right? So intramolecular acid-based reaction. So P H, CO H and here we have Cn and this carbon will have negative charge, right? Yes or no? Because H plus goes out to this bond field will be here on this carbon. Now, this behaves as a nucleophile for the other molecule, okay? So other molecule is what? This is the first step of the reaction is this and then the second step, we have P H, C double bond OH plus P H, CO H, Cn and this attacks onto this carbon and this pi electron goes out to this carbon, okay? So it forms P H, CO minus H, CO H, Cn and this P H, okay? Cn minus is a leaving group here. There is no leaving group present onto this carbon. If it is there, then this bond pair, this bond pair of electron comes over here and one of the leaving group goes out, right? But there is no leaving group here, right? So what happens in this reaction? Again, there is a hydride shift, this H plus comes over here, protons shift basically and we get P H, CO H, H, CO minus Cn and P H, okay? Now, when this lone pair of electron forms one pair, this Cn minus goes out as a leaving group and that is why this KCn is a catalyst here. Concentration of Cn minus does not change, okay? So product of this reaction is what? P H, C H, OH, C double bond OH. So KCn behaves as a catalyst here. It will take place in a basic way, right? We need some nucleophyte. So Cn minus is behaving as a base here, you can see what I'm saying. That's why it attacks because here we have two attacks. So suppose you put some medium-mass mode OH minus on us, then KOH should be taken up and then that will yield Cn. Yes, correct. So when you change the reagent, you will get some other product. Mechanism will be same but that reaction is not benzoin condensation. It's a name reaction. So reagent and everything is fixed for this. Okay, name reaction, you cannot change the reagent. If you take some other molecule like KOH you are saying, so same kind of reaction will be there. Mechanism will be same but the name of the reaction is not benzoin condensation. Okay, next write down, finish this one. Remember they won't give you the name of the reaction. Benzoin condensation they won't say. They won't say parking reaction. They won't say canizaro, I'll do it. You need to identify this. Okay, for Vatsky they won't mention the name. So for that what you need to do, you need to memorize the reagent for each name reactions, right? So that is important. Benzoin condensation, the reagent is KCn.