 Ok, one more point right now. Ok, one more point right now. Ok, one more point right now. Ok, one more point right now. Ok, one more point right now. Ok, one more point right now. What is H2O2? H2O2 is a solvent we are using. Sausband actually we need a medium for a reaction 2%. Sausband in Delhi like there will be a mixture over there. In that a reaction is taken. No, HPR condition according to Antimodiconic of his imprinted paraffin antimodiconic. Here for reaction it requires H2O2. It is a name reaction right. So reagent everything is taken. What is the product we will get in this reaction? Done. Yes, H2O2 is the addition, the situation is the same. The first reaction the product is what? It is oxymarculation demarculation right. It follows Marconi Coff rule addition of O-H and H right. O-H is the product. Get it right on the mechanism. First one, right. This one the answer is what? Addition of again O-H and H. Antimodiconic. Here we will have 3 more samples. Yes, yes. So it is found that if you use W2-Lambor, Is that okay with Lambor? What about Lambor? If wouldn't Lambor like O-H could go anywhere. Anywhere. If I use excess Lambor, it will get rid of O-H. What I say is that O-H and O-H are on one side. Because anti-addition is too much for O-H. On the other side. So the lower one on the left of O-H. No sir, I am saying if there was a second double bottom, Then on the left side the lower one. This side you are saying? No. This one? Yes sir. This double bottom is there? No sir, I am saying if there was a double bottom, There on the left side. On top there? On top there. No sir. I am saying if there was a double bottom, Then would I get a quills? Exactly sir. You understood the mechanism right. Here we have cyclics remembering And then this S2 attack from the backside. Right and then this ring can open up with this, We will get to know. Let's write down oxidation reaction. That we always do in carbocation formation. Then only we will know rearrangement of carbocation. Okay, write down next oxidation reaction. Write down an alkene can be oxidized. Not all the esters. Write down. An alkene can be oxidized to cis 1,2 glycol. On reaction with potassium permanganate. On oxidation with potassium permanganate. In a cold basic solution. And an alkene can be oxidized to cis 1,2 glycols. Either by, what did you write? By reaction with, Km in a cold basic solution. And by osmium tetra oxide OsO4. Osmium tetra oxide. Both reagents I can use. Both can be used. Osmium is in the geometry. Osmium tetra oxide formalize OsO4. We also call it as osmic acid OsO4. The reaction you see. CS3CH. Double bond CS3. KmNO4. Temperature is very, it is not including the temperature. Less than 5 degree Celsius. Otherwise diols can be reduced into alcohol or ketone. It is possible by elimination. Less than 5 degree Celsius. So the product of this reaction is CS3CH CS3. And we will have the cis addition of this. Yes you have to memorize this. Mechanism, little bit I will discuss. But it is not the entire mechanism. CS3CH. Even this same product will get, when you have osmium tetra oxide with H2O2. We will get cis addition of 4H. Diols reaction, again you will get some H2O2 formation. If you simply heat this in acidic medium. Then H2O goes out and after that even osmium tetra oxide. This goes out. And we have a carbocation here. So this carbocation and this is here. If you have any nucleophile that will attack on to this carbocation. If you heat this, you must really get some nucleophiles. So H2O formation. What is cis? What is cis? Cis means the same side of the reaction. Same also you get. Cis means the same. Cis means the same. Okay, possible. We will get the same product. KmO4 converts it to mnO4- K plus mnO4- This OH- will have excess of it. It will attack one by one on cis to carbocation because this bond is colder and eventually it goes out. mnO4- So how can mn form so many bonds? mn all in all. This means that last your p and your s. You can also form so many bonds. mn can also form so many. So what's the thing of mn? What's the atomic number of mn? What is the atomic number of mn? What do you say? 25. Did you say it's well? 25. After mn is what we have? 26. After mn is we have iron. 35. T-blog element. It has expanded operative. Okay, what? So you said there will be problems. There will be big problems. There will be big problems. What will be the shape of mn? Tell me. It's still there. It's still there. So all these four oxygen atoms inside the condom are different. So this is the stuff. Basically you have to memorize this. Diol forms and both over it on the same side. Another write down, next point. Let me not write. When you heat this in acidic medium and you have supposed some new mn in Cm. Acidic medium with this water molecule goes out and has a positive charge here. This will attack over here. And this is the formation of helo-hybrid. High blade, we'll see that later. I mean you're going to write it. No, no, no. It will be OH. And this is Cm. H plus comes over here. H2 goes out. Positive charge. And this is a nucleophile. Where does that pass out? I'm telling you that's what. When you heat this in acidic medium. You have to provide for permission of reaction. Next write down. Second point. Next reaction is under the oxidation part. Let me write down. No, oxidation. We can use different, different reagents. First clue is this. Next write down. Oxidation with hot alkaline. Oxidation with hot alkaline. Permanent. Acidic permanent. Acidic permanent. Acidic dichromate. Acidic dichromate. Or potassium per iodate. Na, i, O4. Or potassium per iodate. Na. Okay. Potassium, sodium both we can do. Any metal you can do. Potassium per iodate is Ki, O4. Sodium per iodate is Na, O4. Potassium per iodate. In presence of catalytic amount of Km, O4. Catalytic amount of Km, O4. What is catalytic amount of Km, O4? Katm is Km, O4 is behaving as a catalyst. Like the... It is present in a small amount there. The effect won't be there in the reaction part. Okay. Catalytic amount of Km, O4. Forms, aldehyde. Ketones. Carboxylic acid. Or carbonyl dioxide. Aldehyde, ketones, carbonyl dioxide. Or carbonyl dioxide. This reaction you have to memorize. Mechanism is not there. 2-3 points we have into this. Okay. According to that we write down the product. Okay. Aldehyde can form, ketone can form, acid can form or CO2 also. Okay. Terminal CH2 group. CO2. Terminal CH2 part. See if you have... Terminal CH2 is this. Double bond CH2. Can you write it down? Terminal... Methyl... Terminal CH2 of alkene is completely oxidized to CO2 and CH2. Terminal CH2 of alkene is completely oxidized to CO2 and CH2. This is first one. Next point, second point we write down. If an alkyl group is attached to Alkyl group is attached to the double bonded carbon atom double bonded carbon atom then it is oxidized to aldehyde first and it is oxidized to aldehyde first which further oxidized into carboxylic acid. See you always remember this. When you do the oxidation of 1 degree alcohol you will get aldehyde always. Further if you oxidize, aldehyde converts into acid always. 1 degree alcohol to aldehyde to acid. 2 degree alcohol to ketone which may go under bare villager oxidation because it involves ester. But 1 degree alcohol to aldehyde and then acid always. If you reduce this, then it will dark. Acid to aldehyde, aldehyde to 1 degree acid. Ok. So right on next point, third one. A double bonded carbon atom having 2 alkyl group is oxidized to ketone. A double bonded carbon atom having 2 alkyl group is oxidized to you see what suppose you have an alkene say CH3, CH2 CH2 double bonded CH2 points are given you. We can write the product of this reaction if it goes under oxidation. Suppose the reaction is very commonly used reagent is KMNO4 OH-CH CHs Now the product of this CH is a terminal double bond right? This converts into what? Terminal CO2 So one of the product is CO2 and this carbon dioxide is this carbon atom. Ok. Now this will also oxidize right? This will oxidize into Altehyde with equal number of carbon atoms CH3, CH2, CH3 this is Altehyde first it forms Altehyde and further since it is a strong oxidizing agent so we cannot stop the reaction till here right? So it further it oxidize into acid which is CH3, CH2 CH2 This is a water So if you want it to stop over here then how do you We have to use some weaker mild oxidizing agent like bromine water ok. But in this reaction it is not possible see if the oxidizing agent is strong then this will convert into this If it stays here so you have to make this weak so that it won't convert this into this Altehyde wherever the double bond is present if non-terminate double bond is there maybe it will reach me right? Then first you will get Altehyde and then you will get you will get what? Acids ok? So this part will go under oxidation and this part will go under oxidation since this is terminal so this gives you CO2 oxidation of this gives you Altehyde first with equal number of carbon atom and then acid with equal number of carbon atom Now suppose if I take this example CS3, CS double bond, CS CS3 and it will be the product in this one same Altehyde same Altehyde first and then carbon solid gases can I write down the product is two molecules of CS3 CO4H or CHO and then CS2COH ok I will write down that CS3, CHO which further oxidize into CS3, CO4 always remember there is a KMLO4 is a very strong oxidizing agent you would not get Altehyde as a product in this what is the product in this reaction see this carbon has two alkyl group right? so you will have here ketone forms ketone so this is here and this converts into so if terminal double bond is not there then we always get acid or ketone in my own form in this reaction second reaction in this one no see it is the product it forms and then it converts into so final product is this one in the mixture you will not have antihyde you will have acid so this is here so keep on making here here you will have acid in this CS3, CHO and further that will oxidize into this that is it antihyde is not forming here all these converts into this yes all the double bond will break it is everywhere you have to break and write then you will get acid into this double bond is to break and write down the carbosolic acid ok if you have to form antihyde then we use that Na, NaO4 or KiO4 write down with potassium pariodate or sodium pariodate antihyde is the major product or alkene gets oxidized into an antihyde KiO4 or NaO4 is a weaker oxidizing agent and KMNO4 so it cannot oxidize antihyde further into acid tell me the product in this reaction CS3, CH double bond CH2 and Ki4 product in this reaction no it is not done it is not CH2 it is CO2 we get CO2 and antihyde product for the first one is and for this one it is CS3 so all these 2-3 rules we must remember