 Next we are going to write down from carboxylic acid right on from carboxylic acid R C double part O O H, first region we use Na O H with Ca O when you hit this. So this is from carboxylic acid. It forms R H, R H and CO2 molecules goes out. The name of this reaction is Sotoline decarboxylation. Sotoline decarboxylation. If you use this Na O H or Ca O H, anyone you can use electrolysis. You get R R L K. This reaction we got it as whole base electrolysis. So it is either 1 or K O H or K O H. Anyone you can use it. We can also reduce this with the help of red phosphorous and H I. It forms R C S 3. It is good reducing. Is that it? Which one is good reducing? Air phosphorous is good reducing. Li Al is good reducing. Na B is good reducing. Now what you have to observe here you see this molecule the alkene that you get. The number of carbon atom here is one less than the number of carbon atoms present. Because one carbon atom goes with CO2. So here we have one carbon less. Here we have equal number of carbon atoms in this method. Here we have the number of carbon atom is what? One carbon is less than this 2 into N number of carbon atom is here. This is the number of carbon atom. We hear with carbon because of the coxiality. So if you have N number of carbon atom here then here we have N minus 1 carbon atom. Here we have 2 into N minus 1 carbon atom and here we have N minus 1 carbon atom. Sometimes in the option you will get like this. You have to reduce the option you see the one carbon is less. It means we cannot use this method. We cannot use this method. Depending upon the product you can have the choice. If you have to reduce the length then we will go with this. Equal number of carbon atom present. Next write down the last method of preparation from alcohol. Last method of preparation. One important point. In this one you write down. In this one you write down. This reaction involves free radical mechanism. This reaction involves free radical mechanism. Next point the radical ion has property to release neutral molecule. And forms a new radical. Releases neutral molecule and forms a new radical. Basically if you take CS3 COOH. Suppose you take CS3 COOH. Acid thin NOH. What is the product we get here? Salt and water. Acid well based salt and water. So it forms CS3 COONA plus H2O. Now this will dissociate in the form of what? CS3 COO minus NA plus. Now the reaction takes place at anode and cathode. So this at anode the reaction is important. The reaction is we take two molecules of this ion. So it converts into two molecules of CS3 COO radical. And two electrons goes out. Radical and this electron goes out. Now like I said the free radical has the property to lose neutral molecule. And forms a new radical. From this what happens CO2 molecules goes out. If on two CS3 radical. And CO2 molecules goes out into this. We will get a new radical with high radical. Now this with high radical combines with each other. And forms and forms each other. So no can form new thing and make this? No you cannot form. But you can form other or number. Ethan even number we can form. No all. Propane we cannot form. Propane we cannot form. Got it. Up. So this is two guys. Now at cathode the reaction is right down. This is the reaction at cathode. Reaction at cathode is Na plus takes two electron and forms two sodium metal. Which reacts with water and forms NaOH plus H2. Now the question that they ask here. In electrolysis or colobae electrolysis. What gas evolves at anode? The answer is what? Carbon dioxide evolves at anode. And hydrogen gas evolves at cathode. She asked this question many times. Only even matter. So it is automatically. I think this has to work. Forms and everything. Yes. This should be there in the reaction. In the form of fire. Okay. Last one I will go over. Only one reaction we have. I will go over under reduction with red phosphorous and HRI. It is RH with equal number of carbon atoms. Write out the product in this reaction. Product in this reaction. Yes. How do you do C4S? C2S5. C2S5. C2S5. C2S5. C2S5. C2S5. C2S5. C2S5. RMVS plus H2. Except for C2S5. C2S5. C2S5. C2S5. C2S5. So that is C2S5 plus C2S5. No, C2S5 is non-cannot be that. Because both are minus one minus one minus one. C2S5 is a 4D Pidu H5. Oh! Okay, see. No, but in C2S5 this is C2S5. Yeah, again, if you want to understand this, How do I remember this? I can't remember The pattern you have to understand See, this is like a state This behaves as either Either way for nucleophiles Okay? So like nucleophiles, it will behave Well, it is carbonyl compound Okay, so here it is not carbonyl compound It is one What are the carbonyl compounds? The carbonyl compound is here Where the C-double bond is there L-d-high ketone L-d-high ketone Okay? So this here We do not have a right carbonyl compound L-d-high ketone So this behaves as a base here 2 plus and minus And the reaction of the base is Acid base reaction This will take this H Because it is an electronegative It has to be active hydrogen I told you Oxygen is hydrogen Hydrogen is hydrogen C-double bond is hydrogen This C2H5 will take this H plus forms C2 H6 Plus the other product is MGBR as it is And O-C2H5 Attach it This is acid base reaction Preparation of what? LK The one that we did is H2 This here H2 is up here This will take H2 less See, if you have a Carbonyl compound like this Or if here R1 Minus MG 2 plus X reaction Take it I don't know about this one Here this H is again acidate Oxygen is attached It is carbonyl acid Oxygen acid H plus is Q See, here what happens This bond is polar Right It is polar Here we have positive charge And this one is negative This R1 minus will attack On to this one Nebulophile now We are attacking Nebulophile here R minus What happens to oxygen? When it attacks over here So this will go up On to the oxygen Okay So what will happen R C O minus This two lone pair This R will be as it is And this R1 from this RMGS For this neutral show Here we will write MG 2 plus And X minus here is acidic Okay Now in the next step If you have H2O So this H will come here And OH will come here Not really what R, C, R, C, R1, OH Okay This is behaving as a nucleophile But here it is acid based reaction Why acid based? Because the nitrogen is active hydrogen Everywhere We have active hydrogen So RMGS with active hydrogen Always gives you LK Okay Okay RMGS with active hydrogen Always gives you LK What is the product here? So what is the LK? Active hydrogen Active hydrogen means Hydrogen where Hydrogen can go out as H plus Means Hydrogen must be attached With an electronegative element Okay Like oxygen Nitrogen This is also electronegative So if there is some Like delta class on the ocean This is also active hydrogen SP hybridized This is oxygen again This one is what? SP No SP 2 What will happen in this? But if hydrogen comes out from here It can It can It becomes resistive It becomes aromatic That is why it is H plus plus 2 Where the aromatic They are also active hydrogen So when it loses H plus It forms this Which is aromatic That is why this hydrogen is what? Acetic hydrogen Active hydrogen This will come over here Here you will get State of S6, State of S6 So MG MG beyond to You are over There is no question Okay Whatever is left here What is left here? LH2 LH2 will come over here Okay But for the next one For the next one This will happen MGBR C triple bond C C S3 Clear? Clear? Clear? Clear? Clear? So what do you have to memorize? RNGX When reacted Molecule Which contains active hydrogen It gives you LK Done What is in the cycle? What time is it? What time is it? Any hydrogen bonded to an active hydrogen All hydrogens are active But once it loses This makes it possible After this is not possible Next one I will write down Properties of alkane Finish all this All these the alkane will be C2 S6 We will get alkane with respect to the alkyl group Of ligand reagent Okay