 In this reaction alkyne reacts with alkyne reacts with alkyne reacts with mercury acetate in presence of water in presence of water to form hydroxymercureal compound is that hydroxymercureal compound in bracket right on oxymercureation this first step is oxymercureation which on reduction which on reduction with NaBH4 which on reduction with NaBH4 gives NaBH4 is sodium borohydride good reducing NaBH4 is sodium borohydride which on reduction with NaBH4 gives alkyne in bracket right on demercureation ok it is demercureation ok alcohol the reaction is suppose we have CH3 CH double bond CH2 on hinted with HG OAC fold wise this is the first step reagent HG OAC fold wise in H2O in the second step we use NaBH4 alkyne radio ok the product of this reaction is alcohol right so the product here is CH3 CH2 OH acetic acid cct ac acidic acid it C is this Cns3 C double bond ok this slide S scenario is this ув The entire thing if I write down here it is HgO, OcCs3, C double bond, O with HgO. So what is the problem directly after the first step? I will write down, mechanism will discuss 2-3 points to try to understand the mechanism and discuss, okay. We will get a cyclic ring into this, intermediate cyclic ring, okay, no rearrangement here. You see overall what is happening in this reaction, across this double bond H and O H get attached, right. So the first step you will get an epoxy mercurinium ion, the intermediate stage, right. And that on reduction gives this alcohol, okay. I will discuss the mechanism also. One note you write down here first. In this reaction, in this reaction, carbocation does not form, hence no rearrangement. Carbocation does not form, hence no rearrangement, okay. Next point. This reaction follows, this reaction follows anti-addition of, anti-addition of H plus and O H minus. H plus and O H minus. Next point. The first step is oxymerculation, oxymerculation, oxymerculation in which mercurinium ion forms, mercurinium, mercurinium ion forms, which is a three member HG plus, I will write down the reaction, mercurinium ion forms, which is a three membered cyclic ring, three membered cyclic ring, okay. So the first step of the reaction, if I take this example, CS3, C double bond, CH2, CH3. This on reaction with HG, OAC, O2, and H2. Or first step of this. So what happens, this pi electron is taken up by this metal and we get a mercurinium ion here, CS3, C, CH2, HG, delta positive OAC. Here we have CS3, intermediate. This we call it as mercurinium, cyclic mercurinium ion, which on one reaction with water, which is present here on reaction with water. This lone pair attacks onto this carbon, this is also delta positive because this loses pi electron, okay. This lone pair attacks onto this carbon, always remember here this attack is on more substituted carbon and not this one here, right. So it forms CS3, C, CS3, OH2 is positive, CH2, XG, OAC. This attacks over here, this lone pair is taken up by this silica. Now from this what happens? The removal of proton H plus forms OH here. So the product is CS3, C, CS3, OH, CH2, HG, OAC. This loses electron, right. And because of this bond also mercurinium ion, mercurinium also one has slightly positive parts. Sir, if it loses bond shouldn't it be a proper plus rather than just a delta plus? No, no, no. It's not like, it's a, tendency to form a three-membered, it is not completely developed. Okay. Because it is very less stable, three-membered ring. Angle strain is very high. Yes. So it won't form there. Okay. Sir, where does that OAC hold, where does one more? It forms a complex, we'll have a mercury compound there. It forms a complex that we are not writing it up. Here also it forms. We'll get a silver metal and in the final reaction you'll get this also. Acid. Acid. Okay. This is here. HG is here only, right. In the next step it will come out. De-mercuration. Since mercury is getting attached with this carbon atom, so this step is oxy-mercuration. Right. Oxy-mercuring acid is attached here. It is oxy-mercuration. Now the next step when you add NAVH4 in basic media, that NAVH4 provides what? Hydride ion that will replace this, forms metal of this and acetic acid. So after this, the second step you can directly write down. CS3, TCS3, OH, CS2, HG, OAC with NAVH4, OH- it gives CS3, CCS3, OH, CS2, HG plus acetic acid. So what are we trying to achieve now by doing this? So what's the point? So how do you drop that acid? How do you drop that acid? Or does it come out itself? This one. This one? Yeah. This is positive charge, right? Yeah. It is, it is not stable. So she said being electrocuted. Leave and we get a carbocation. No carbocation does not form in this reaction. So to stabilize. So in the dehydration of alcohol, we remove the H2O plus. Yes, yes, yes. For that we have, we heat that. Elimination always requires some temperature for that. You have to heat this and it will come out. Sir, the second one, we said we put it in, like in the prime of OH-, whether that means that we put it in water. We are getting emotional. Water, if you put it in water, then it will attack the lone pair. See, for elderly ones, for any reaction, the path mechanism is different. You cannot say sir, why can't we put water here? If you put water, then has its lone pair present water molecule? That it will attack onto this. You will probably get tired or exhausted. We do not know what will be the product in that case. Okay. That's why we are using this. Actually, this Na, why we are using this base here? Because this will take this OH-, forms NaOH, right? And then this boron pH board releases hydride ion, which attacks onto this and the hydrogen attacks with this. Silver, sorry, mercury material form and AC OH also form. So for all the reactions, you cannot change solvent, right? You do not know what is the behavior of that solvent under a given set of conditions. Okay. See, all these things, mechanism are not important. What is important that we have to keep in mind is the attack of this, see this H2O is behaving as a nucleophile here, right? H2O is a nucleophile. OH is not a nucleophile. Eventually, OH gets attached with this carbon atom when H2O attack onto this. But nucleophile is not OH, nucleophile is H2O. Nucleophiles are the molecules or species which attacks onto the positive charge, the carbon atom. So this is behaving as a nucleophile, right? It is a nucleophile here. Okay, it is a nucleophile. So you always remember the nucleophile attacks on the more substituted carbon, okay? This is one very important point you have to remember, right? Okay, overall what is happening? The addition of OH and H plus. According to what rule? You see, according to Markovnikov rule, right? So this reaction follows Markovnikov rule. Addition of OH and H takes place here. You're right on this. Follows Markovnikov rule. Addition of OH and H. So how are OH and H easy on different sides? Yes. How are they on different sides? There's an attack over here. This is all over here. No, no. This is what this will be. This is what I have written. This is why I am not doing this on this side. Because this attack is anti-antibiotic. So why is that? Because if this attacks on this side, right? Then this bond is difficult to break. And there will have the hindrance of this HG bond. There's an entire group with a bulky group. This will produce hindrance to this attack of this water, which is a nucleophysium. That's why it's an attack from the backside. Anti-antibiotic. Okay. Sir, if you find an attack with the most functional carbons, it doesn't mean that an electrophile attack will lead to an over-touch. No, it's not like that this year. Actually, this part of the time is stable because of this tool. So, since this is stable, so it is stable. So the positive charge here forms very easily in comparison to this. That's why this will attack on the side where it's a more positive attack. Understood? Yes. So it is Markovnikov rule. Addition of OH- and F- plus on the double bonded carbon atom. First step till here is oxymarculation. Next step is removal of the compound of mercury. That is demarculation reaction. Okay? Markovnikov reaction. Demarculation reaction. Okay? Now, one thing very important here, sometimes in the question what they do, they don't write here or give here H2O. Listen to the gap for this one, right? Because you see from where this OH is coming from which molecule? From H2O. So if they replace this H2O with CS3OH, then the product will not be an alcohol, but it is an ether in that case. Okay? Why? Because the nucleophile here is what? CS3OH. Here CS3OH will attack on to this and then H-plus goes out, so we'll have 4 CS3 will attack on to this carbon atom. Got it? Okay? So if you have CS3OH here, the name of the reaction is oxymarculation. No, not oxymarculation. It is alkoxymarculation demarculation reaction. Got it? Write down the next reaction. Same thing. Instead of H2O write down this, hitting you write down alkoxymarculation demarculation. What? It's not antidex. Antidex. Antidex. Back from the opposite side. It is alkoxymarculation and demarculation reaction. Sir, I'm still confused about how this is going to work out. See, this pi electron, you can combine this with Hg. But it's forming a bond. Right? This will be like a bond. This thing, this was the 2 electron. Distributed between the 2 atoms. Now it is distributed in 3 atoms. Okay. Yeah. So what? It forms a complex. Actually it is like bromine halogenation. Yeah, halogenation. Similar to that. Bromonium ion forms over there. Halonium ion like this one. Similarly here also. Sir, but here the C isn't getting formed. Here also we'll have carbon dioxide. Same thing here also. Okay. Sir, why doesn't that carbon halogen? This carbon also halogen. But this one is more stable because of these 2 with halogen. That's why I'm showing you. So that it will attack over here. Okay. Okay. So we'll write down this reaction with HgOac, CsGOH. The name of the reaction is alkoxymarculation, demarculation. And the product in this reaction is an ether, not alcohol. Okay. Done.