 Next one is Michael addition. We also call it as 1,4 addition, 1,4 addition. Possible when one of the carbonyl compound is conjugate carbonyl compound, conjugate carbonyl compound. You see this reaction suppose we have R C double bond O C H 2 C double bond O plus C H 3 C H double bond C H C double bond O R or H ok double bond O ok. Now this R and this reaction takes place in presence of a base right. And solvent if you have to take, solvent will be conjugate acid of this B H ok. This R can be, this R can be it can be hydrogen also instead of this R you can take or you can also take O R. Esther also you can take for this reaction ok. Means if it is there aldehyde H C double bond O or O R C double bond O like this. Then we will have the same similar kind of reaction nothing change there right. Now you see this aldehyde is conjugated yes or no pi sigma pi right. This is conjugated aldehyde. So mechanism everything exactly same but in case of conjugation what happens this carbon won't take part in the reaction that is what you need to understand right. So what happens here you see first of all this carbon this C H 2 is active methylene group yes or no. Active methylene group you know what is active methylene group, methylene group is C H 2 ok, methylene group is C H 2. So when this C H 2 has two electron withdrawing group present. If you remember in toteminism we have done it right. Two methylene active in suppose this side also we have C H 3 and this side also we have C H 3. So if you see the acidity of this hydrogen this hydrogen and this hydrogen which one is more most acidic hydrogen the center one because both group is withdrawing electron. So for this carbon it is very easy to lose H plus correct. So this kind of group where C H 2 present between the two electron withdrawing group we call it as active methylene group C H 2 is already methylene right. It is active methylene group so if H plus has to come out in this molecule it will come from the middle carbon right. The active methylene group carbon because that hydrogen is most acidic hydrogen correct. So what happens here again you see the mechanism is same the only thing is what we need to generate the carbon ion which behaves as a nucleophile and that attacks on to the other molecule. What we are doing in the last reaction is that C N minus attack and then hydride shift and then we will get carbon ion and that carbon ion attacks on to the molecule right. So here we have to do the same thing. So what happens here this base takes H plus right. And where will this H plus come from which H plus is most acidic right. This H plus is most acidic right. So it takes H plus from this carbon and we will get a carbon ion here right and then this carbon ion attacks on which carbon. Other reaction if you see when there is no conjugation the carbon ion were attacking on to the carbonyl carbon. But since there is a conjugation so this will attack on to this carbon this pi shift over here and this electron shift over here. This pi is in resonance right. So this comes over here this goes here this goes here right. So this attack of this carbon ion takes place on the fourth carbon by fourth carbon because we are counting from this one. This is nothing related to our nomenclature one four the number is not related to nomenclature. It is just first position is this second is this third is this and fourth is this. And that is why we are calling it as one four addition. So Michael addition is just a name it is basically one comma four addition right. So when it attacks over here right this carbon also when you draw this electron here and here this positive charge resonance is stabilized right. That is why we will get more stable carbocation and hence it acts over here right. So basically you have to keep this in mind the reason is that stability only but when we have aldehyde or ketone is in conjugation right. And then the attack of this carbonyl carbon or sorry the carbon ion takes place on to the fourth carbon right. On to the fourth carbon because this is in resonance with this okay. So can you write down the mechanism first step what you do base takes h plus from this you will get lone pair negative charge here. And then second step it attacks on to this carbon this pi electron shift over here right on this two step first. I will write down this once we try this one step okay. So the first step what happens R C double bond O CH2 C double bond OR and base takes this hydrogen the product will be R C double bond O CH negative. The second step of the reaction the carbon ion attacks on this aldehyde conjugated aldehyde CH3 CH double bond CH C double bond OH. This attacks on to this carbon fourth carbon this comes over here and this pi electron shift on to this oxygen. So it forms R C double bond O CH C double bond OR and then we will get CH1 methyl group is here CH double bond CO minus okay. The next step from the solvent that is BH this oxygen gets protonates from the solvent in B minus it will be there in the solvent right. So it will be R C double bond O CH C double bond OR CH CH3 CH double bond CH OH. This is in all form yes or no in all okay in all in all form. So whenever we have in all form we will try to write keto form so must take care of this if you are getting this and in the option keto form is given. So try to write down keto form from in all form because keto is it is in general more stable right if there is no other factor like aromaticity right. So this after this the last step is the keto in all tautomerism okay. How do we write keto in all tautomerism? The H plus comes out from the first carbon first atom and attach on to the third so it is one three addition right triad we call it as. You have done this one triad triad and space total right it is triad this H comes out leaves this one pair of electron to this carbon. So if I write down this H is here this comes over here right this pi electron comes over here and it takes this H plus on this carbon. So product of this reaction will be R C double bond O CH C double bond OR CH CH3 CH2 C double bond O. So how do we write down the product because again mechanism you are not going to write in the exam okay. So karna gaya yaha saying we will take this H plus out okay and we add this here on to this carbon third carbon right. Everything will be as it is double bond or triad okay and this carbon attached with this carbon right. So one four addition if you get this compound one four addition karna directly if I write down this two will be as it is. Directly write down the product R C double bond O this is active with highly group CH and this attached with this thing will be same C double bond OR. And this attached with the fourth atom here so fourth atom is this which is carbon CH and this carbon we have this CH3. This CH3 will be here single bond CH single bond C double bond OH. Now this carbon the valence is not complete so write down simply here that is what the product we have okay. Not that important only thing you need to keep in mind if it is conjugation then one four addition over. Write down the product in this reaction try to write it on directly don't write mechanism okay. The reagent you see the reagent we have if you have OET present then the reagent will be the salt of this ET O minus NE plus. And the solvent of this is what ET OH okay. Other reagent we are using here is H plus H2 and in the last step we are heating this one. So one thing you take care here if it is OET then we will take the salt of this group only right OR if it is there then we take the salt with respect to that group. Next step H plus H2 also we are not using H plus H2. So what is the product we get the fourth carbon is this one two three four right and carbon I will get here under this carbon negative charge. So this carbon and this carbon will attach right. So the product is directly if I write down the product will have double bond O then CH 2 CH. CH CH CH CH see the whole bond O C double bond O OET and C double bond O OET is it connected yes殺 Balakar. Because this is the reaction that we did just now, one for addition, this is. And in that reaction, we are not using H plus H2. So, when you use H plus H2, this reaction is not different from it. This is the product of the first agent that we are using. If you use H plus H2 further, then this will, this H plus protonates this OAT and OAT. And finally, ETOH comes out and will get acid over it, CO1, CO2. See, this happens. You have to remember this. I will explain the mechanism here. Suppose you have this C double bond O, OR. Okay? H plus H2O, okay? So, this reaction, first of all, this OR gets protonates and it forms C double bond O, ORH positive charge. Okay? Now, this takes this bond pair of electron and it forms C double bond O, positive charge and ROH. I am just writing it down to make you understand this. Okay? The exact mechanism is not linked. Okay? And then on this positive charge, the H2O behaves as a nucleophile and it attacks on the electron. And it converts into C double bond O, OH2 plus. And the final step one happens to stabilize this positive charge. H plus comes out and it converts into C double bond O. The final product of this is what? Acetic hydrolysis of electron. It is an acid plus alcohol. Okay? Acetic hydrolysis of ester gives you acid and alcohol. So, when you do the acetic hydrolysis of this, you get what? All these things will be same and we get double bond O, CH2, CH, C double bond O, OH, C double bond O, OH plus what we get? Two molecules of ET, OH, acid and alcohol. Okay? Now, when you heat this, because we are heating this also. So, when you heat this and the carbon contains di-carboxylic acid, right? So, di-carboxylic acid, if you heat when it contains only one carbon in between, then CO2 molecules comes out. It is heating effect of compounds. Okay? If there is two carbon in between, you must have remembered this reaction. This is very important, okay? Because where it is, it is H2O-eliminate. Look at it. It is H2O-eliminate. It is CO2-eliminate. Right? So, what you have to keep in mind when the carbon, number of carbon atom is one in between the two carboxylic group, right? Suppose this is the reaction. Suppose I write this. Look at anhydride formation of C double bond O, OH, C double bond O, OH. Okay? Suppose this is the reaction. Okay? Now, when you heat this, in this case, what happens? H2O molecules goes out. So, you are getting what? Thalic anhydride. This one is thalic acid and we get thalic anhydride from this. Okay? So, what you have to keep in mind in case of di-carboxylic acid, if two carbon atom is present in between, the two carboxylic group, right? Two carboxylic acid group. Then H2O molecules goes out in the reaction, if you heat this. Okay? Or, if there is one carbon atom, then CO2 molecules goes out. Okay? It is very important. Only one carbon present, heat this CO2 molecules goes out. So, the final product of this reaction is what? CH3. When you heat this, any one of the CO2 molecules goes out and the product will be double bond O here, CH2, CH2, C double bond O. So, I can't load the CO2 molecule. Why is there no tissue? See, then for that, you need to understand the mechanism. Actually, what I have, which has a mechanism. I will tell you. See, when you heat this, because they are not using any reaction. So, one of the hydrogen atom, either this hydrogen or this hydrogen. Right? One of the hydrogen atom comes out, leaving this electron pair behind. Okay? So, after C double bond O double bond O. Or, this carbon will have what? Negative charge. Negative charge on it. Okay? Carbon has five bonds, so negative charge. Right? After this, this bond pair is like, it is there on this carbon and CO2 comes out from this reaction. What? So, this is not taking place with both the molecules. That's why the carbohydrate group on the same carbon is unstable. Slightly, you heat this, CO2 molecules goes out. But when we have motor carbons, that is stable. It won't eliminate CO2. Because, there is no reason for stability here. If this will come out, then CH3 will come here. Okay? And for that, you need very high heat. That is not possible. So, it won't happen. Okay? Also, you need very high heat. That is not possible. So, it won't happen. So, the point is, when you have two carbon atoms present between the two carboxylic acid group, on heating this, H2O molecules goes out and you end up getting an anhydride. Right? Only one carbon atom, then CO2 goes out and you will get acid. Okay? So, final product of this reaction is what? What is this? So, first step of this reaction is mygalation and then simply the acidic hydrolysis and heating. Right? So, these two things, one carbon and two carbon, is very important for this carbonyl compound chapter, especially carboxylic acid. Okay? Understood?