 What is the reaction? Write down, this reaction follows pre-radical mechanism and it produces symmetrical alkene. Next slide, it cannot prepare methane. The alkyl halide we use for this purpose, the alkyl halide we use for this purpose must not be secondary. So, it must not be tertiary. Tertiary alkali goes under elimination reaction. We will discuss that later. So, what you have to memorize here, the alkyl halide must not be tertiary. So, it can be secondary. Secondary is possible tertiary has higher intensity to go under elimination reaction. So, for secondary, each one will bond two more things. Like, if you write this, secondary is secondary. Secondary, you will have to clue. Even if we do not do a secondary also for this purpose. Priority is what? Priority, secondary, tertiary we do not know. Secondary, secondary, secondary. How the reaction goes, you see? We take two molecules of alkyl halide with Na which forms Rr and Ax. If you take two different alkyl halides, suppose you are taking Rx plus R dash x with the same region, then we get all kinds of possible alkene. We will get Rr, we will get Rr dash, we will get R dash R. Need JMA equation for today. How many products we get? The number of products is what? Three. One, two and three. I will come to that. The mechanism we will discuss in a real understanding. It is, you know, it is, you will not get one particular type of alkyl. That is why this reaction is not useful. That is why mostly we use two moles of one similar alkyl. And since we are, it is a coupling reaction to alkyl group is getting coupled here. That is why methane cannot be prepared. Methane cannot be prepared. Right? That is not possible. So that is the reason. Why this is happening? This thing you must take care of. We will get mixture of alkyl here. Three products possible. This reaction follows free radical mechanism. Now you see the mechanism of this. Sir, the first reaction is separate sodium halide from the Alkyl. No, no. This is different. Na is here. We will get this one. Salty. Alkyl is this. Now you see metal Na. Na is the metal, right? The metal contains Na in reaction that it gets free electron. Okay? Or like I said, alkyl light whenever you have, it presents a free electron, the bond becomes big. Polarize. So Na dissociates like what? It releases electron. Na plus plus one electron. Okay? How we get this Na x? Na plus and x minus. Okay? So we have to get x minus from this reaction. x minus, how we get this? We have Rx, okay? Rx or free electrons are the metals. Right? This free electron, this free electron will attack on to this halogen, right? And this electron is taken up by this alkyl brush and this is taken up by this. Because x has the minus side effect on R because you should be electron and alkyl R. It shouldn't be Na. See. See, actually, that is what. The mechanism is already explained, right? All these are the factual things. How the mechanism is going on, why carbon ion is not forming, why theoretical is forming, that is all explained. Because the condition that we are using, in that condition, electrolytic present is not possible. Okay? That is why homolysis is taking place. So and in case of homolysis only, free radical forms. And that's why the mechanism is what? Okay? So you have to memorize this. Logically, as soon as you go with them, you will get confused. You are in carbon ion or tuple sum. What condition makes this go on? See, obviously, when bonds dissociate things, what this requires some temperature. Okay? That temperature is not here. We are not using that temperature. That's why this homolysis is here. Okay? So it forms what you see. R dot, it forms plus x. Now this x minus N e plus combines forms Na x. Take it. In the next step, what happens? This R dot combines with another alkyl rope form. So, if I get this in a little bank, temperature, if you don't know, ketone is possible. But will the reaction sustain somehow that we cannot see? Because, you see, when you are increasing the temperature, sodium is losing electron. You never know like sodium can lose another electron on that temperature. Anything is possible. Theoretically, we cannot explain how it's going. So, what's the maximum temperature? It's not given, but around 250, around 250. Sir, what happens? Why do we have to write? Why do we have to write the words like that? It's not a G. It's not a G. No, in this retinase, it is not there. See, in case of dry ether, the free radical is reacting. If you take a polar molecule, then the reaction rate is very strong. See, there are many things. Yes, you can say it's a promoter. You can say catalyst also. See, you use a solvent for the reaction to present. And that solvent we use in which the rate of the reaction is sufficient. It's a moderate variable. It's not like very fast or very slow. So, with that purpose, we use either... So, how does dry ether react? Dry ether actually, in presence of dry ether, it's a solvent. In this solvent, the reaction is specific. So, basically, it makes the rate of the reaction moderate so that the product will get very fast. Polar solvent, we don't use in this process. That's the purpose of dry ether. Here, we are not using RMGX, whole of an happening. So, this is what we are getting. Now, you see this, the kind of ladle if it is different, we get radical of this and this also, right? And free radical is highly reactive. We do not have any control over these kind of reactions. So, R-radical may combine with R-radical. It may combine with R-radical. And R-radical also may combine with R-radical and R-radical. That's why we get all three protons. R-R-R-R-R, R-R-R-R-R. Okay, this question they have asked in the exam. How many products possible? Three products possible. Sir sir, the rx bond becomes more dark and with that incident, it becomes weaker. Yeah, it is easy to break. Sir then why do we use a non-polar solvent for the reaction? See, if you use H2O here, then this radical won't form, it probably gives you ROX alcohol. That's why we are using it. Sir, if you use it the same, I'll do a different reaction. You'll get something else. NaX swap in NaCl plus NaBr. You'll get all three products. But here what you'll get? You'll get two moles of NaX. If you have ClNBr here, then the product will be NaClNN. So similar, this kind of, we have two more reactions based on this. Right on the next name, I haven't written here, right on the next reaction. FITIC reaction, F-I-T-I-G. FITIC reaction. F-I-T-I-G. FITIC. We use aryl halide in this case. In this reaction, right down, in this reaction, aryl halide is treated with, what is aryl halide? Aryl halide is treated with sodium metal in presence of triacyl. It's just the name of a different reagent. Spray the benzene. Aryl halide means halogen on benzene. So suppose we have this benzene with Cl, same, exactly same reaction. Na, two moles of Na with Cl and benzene. Dry heat. Everything is same since we are using aryl halide here. The name of the reaction is FITIC reaction, F-I-T-I-G. The product is this. Another reaction it writes on. Wurz FITIC reaction. Wurz FITIC means what? One is aryl halide and other one is aryl halide. Same reagent. Is it forming alkene? It's not in the same thing but I've given you that. So it's basically the same thing. But it is not forming alkene. It's not the preparation of alkene. But since similar reaction is there, that's why I've given you that. And that's why I have not written it. The first one is FITIC, only FITIC. This is Wurz FITIC. So even over here it can form a lot of products. It can form like ethane and it can form all those things. Yes, yes, possible. Ethane plus two phenyl groups by phenyl. But the point is we are using aryl halide and the name is FITIC. One more reaction is this right here. Intramolecular Wurz reaction. Intramolecular Wurz. Intramolecular Wurz. What's the reaction? This is the certificate. You have to go and examine. See, you'll get a radical here. You'll get a radical here. But these two radicals are... So it has to be a... Anything. This is SPIDER. What is SPIDER? SPIDER. Yes, SPIDER. One product. Oh yeah, yeah, yeah. I'm going to buy it for this guy. I'm going to buy it for this guy. I'm going to buy it for this guy. Yeah, yeah. Oh, yeah. It's a... In the first one, can it form a polymer? So that's a base. So if you have an excess C, if you use excess of this one, then similarly this number of salt, LSE or whatever it is, from that to this. You won't get a reaction like this. This number of molecules, do you see if you get this N number of this? But it's not like on this, the... another molecule you combine over here. And then it will be this. We do not have any control over such reaction. But the major product will be this one. Some bicarbonate, you can say. What is the name of this one? Tell me. It's Bicyclo. Bicyclo. Bicyclo 1.1.0. Then... 1.1.0, butane. Which one is pyro? Four-carbons. Tell me, what is the product in the second half? See, one radical is here. One radical is here. We do not have that. What is the name of this one? It's not like on the first half. We do not have that. We do not have that. We have only one. We are not going to see that. We are going to see that. We are going to see that. No, it's a... You know, it's a... It's a radical. What do you have there? What do you have there? What do you have there? Spinal. So it's intra-molecular, within the molecule only it is happening, right? It's important. Now all the reactions you see, this reaction is franklin. By the way, NaCl or NABR form, major fraud will be visible. If you have 2Br and 2Cl, 2 moles of NaCl, 2 moles of NABR. You see, this reaction is at the same. Instead of Na, we are using zinc, that is it. So it's not hard at all. It's only zinc. You see, this reaction, the third step here, this step is SN2 mechanism. Look at the last one. Why do we have the Rx and Rn? What's the actual reaction of this? We use this to prepare anzymmetric molecule. So we have to use this Rx plus Zn plus Rnx.