 The hydrohalogenation of alkyl halide goes under dehydrohalogenation reaction minus of Hx. Dehydrohalogenation is removal of a hydrogen halide. Alkyl halide goes under dehydrohalogenation reaction in presence of a base. This reaction follows E1, E1 and E2 elimination, E1 and E2 elimination reaction. See this E1, E2, SN1, SN2 like I said all these reactions will discuss in detail in reaction mechanism. Okay, here I will give you a bit of idea how this reaction works. So, first of all this 2 and this 1 stands for order of the reaction, it is second order elimination reaction, this is first order elimination reaction, okay. E2 reaction is possible with 1 degree and 2 degree alkyl halide, okay 1 degree and 2 degree Rx alkyl halide, if you have then E2 reaction possible and even reaction generally possible in tertiary alkyl halide in polar solvent. This is polar solvent, generally E2 reaction takes place in non-polar solvent, okay. So, if that was in a non-polar thing then E1 wouldn't happen. No, E1 reaction takes place in polar solvent. So, if I don't have a tertiary thing in a non-polar solvent. We won't take, we won't take. Like SN1, SN1 in E1 are similar reaction like similar in terms of the number of steps. In SN1 also will have an intermediate carbocation, 2 step reaction, carbocation forms and nucleophile attacks. Here what happens, again carbocation forms and then H plus eliminates from this, that's why it is elimination reaction, okay. So, it is similar to SN1 in terms of the number of steps involved or the intermediate forms, correct. So, it is also 2 step reaction. SN1 is substitution of nucleophile and this one is elimination. So, product is completely different. This is 1 step reaction. Okay, these 2, 3 things we must remember now and see the reaction here. Suppose we have a compound alkyl and I say carbon, carbon halogen and here we have hydrogen present, okay. Now for this reaction we use a base. The base can be small base also we can use like C2H5O minus, okay. We can also use some bulky base. In presence of bulky base the product is different. For small base the product is different, okay. We will see what product we get. What in general the steps of the reaction is, this base extracts H plus from the beta carbon. Okay, what is beta carbon here? The one which has hydrogen, right. The halogen is attached with the carbon atom. That carbon is alpha carbon, right. So, this carbon is alpha and this carbon is beta. So this base B minus extracts H plus from the beta carbon, right. And this bond pair of electron it comes over here and it comes down here to make a pi bond. This halogen atom goes out as leaving look as X minus, okay. So overall what happens? The product of the reaction is we have C double bond. C plus halide ion goes out and the conjugate asset of this base. Is this bulky or small? It's a general step of the reaction. You will see what is the bulky base and what is the small base. Okay. But in general, no it is a Edo reaction. It's Edo reaction. It's one step reaction. In general the reaction takes place like this one. The base takes H plus ion from the beta carbon and this bond pair forms pi bond here and then halogen atom goes out. Okay. Now right down. Base used for this reaction. So why is it called Edo? Because it is second order elimination reaction. The order of this reaction based on this and this. The concentration of base and the reagent. And calculate. That's why it is second order. Okay. Now firstly what is order? Order of putting in chemical kinetics. Okay. Just you understand the second order elimination reaction. Order will discuss in chemical kinetics. Okay. Right down. Base used for this reaction. The first point to write down. If we use a small base like. If we use a small base like ethoxide ion. Or hydroxide ion. C2H5O minus. If we use a small base like ethoxide ion C2H5O minus. Or hydroxide ion OH minus. Then the product. Then the major product is. The set chip product. S-A-Y-T-Z-E-F. C2H5O minus and OH minus. Hydroxide ion, ethoxide ion. If we use a small base like ethoxide ion or hydroxide ion. Then the major product of the reaction will be according to the set chip rule. Or we can also say it as set chip product. Okay. Set chip is S-A-Y-T-Z-E-F. Set chip product, the major product. Or we also call it as Z-Safe. Both are the same thing. Basically we will get more substituted elements. This means we will get more substituted alky. More substituted alky. Okay. The major product. S-A-Y-T-Z-E-F. S-A-Y-T-Z-E-F. Very first you write down. We will discuss with example. Okay. Next write down. If we use a bulky base. If we use a bulky base. Tertiary buttoxide ion. Tertiary buttoxide. Tertiary buttoxide we write like this. T stands for tertiary. V-U-O-minus. Tertiary buttoxide ion. What's the structure of that? C-H-3. Both sides C-O-minus. C-O-minus and C-H-3 group attachment. Tertiary buttoxide ion. Triethylamine. All these are large base. Sir what a large one? A bulky means a big size. Okay. That's it. Vulki. Okay. Triethylamine. Triethylamine. Etc. Then the major product of the reaction is Hoffman product. Then the major product of the reaction is Hoffman product. Okay what is Hoffman product? less substituted algae, region also we will see. But you see the reaction. Okay, suppose the alkynolide is this CS3, CH2, CBR, CS3 and CS3. And in the reaction we are using a base say C2H5O-. This is the base we are using. They can also give you C2H5O and Na like this, salt of this. Same thing here, Na plus and this minus. So this also behaves the same thing. Now one thing is here, this reaction takes place in the solvent. Right, so whatever base you are taking here, the conjugate acid of this is building the solvent. So what is the conjugate acid of this? H5OH. Just we add H plus into this. Okay, suppose you are taking OH minus as a base. Then the solvent is what? H2. This is the condition we have. We always take the conjugate acid at the solvent. Right, so this is the agent for this reaction. Now the product name, can you tell me the product? Product will be CS3, CH double bond, CH3, CS3. We also get CS3, CH2, C double bond, CH2, CS3. This is the two product you get. Okay, what is this product? This is sets of product, more substituted alkene. Right, and this is Hoffman. How this reaction proceeds? See this is the base we have. This base takes H plus ion from the beta carbon. What is the beta carbon here? This one is alpha, this one is beta and this one is also beta. Even this one is also beta. However these two positions are same position. Right, if we take H plus from here or here we get the same product. Okay, so the possibility is what? That this base, this base C2H5O minus attacks onto this hydrogen, takes H plus. This bond pair comes over here and this BR goes out as a leaving to BR minus. Product is setger product beginning. This is path A supposedly. So setger we get with path A. Now if this base falls, so this is the base, when it attacks onto this hydrogen, this is also the possibility to break. This is path B. So with this the product is what? Hoffman product The point is this base has tendency to extract this hydrogen and this hydrogen. The major product depends upon the stability of alkene. We know more substituted alkene is more stable. That's why this is the major product, setger product. Okay, so what I said that this base takes H plus from the beta carbon. It always attacks onto the beta carbon. Right, this H plus is taken up by this base and this bond pair of electron, it comes over here to form a pi bond. Right, when it comes over here, so this carbon has to lose one bond, so that it completes tetravalency. Right, so this BR minus, we know halide ions are good leaving group. Right, this BR takes this bond pair of electron and goes out as BR minus leaving group and we'll get this alkene. Okay. So does this work with fluorine also? Fluorine generally we don't use because the bond of bond strength of carbon fluorine bond is very strong. Okay, in that case we'll get a different order. Have we discussed the language? Okay, but controlling the case means the things will be different. Generally, Hoffman product is favorable there. Sir, how big does the base have to be for us to call it now? That you don't have to think about. You just have to take this C2S5 over its smaller base, over H minus ion is a smaller base. In fact, CS3O minus is also a smaller base. When you have a large base, it will be some branch will be there, tertiary group attached over. This is try ethyl amine. Right, try ethyl amine kya ho gya and ke sa three ethyl group attached. So, when you look at this group, it looks like bulky group. Right. This is a parameter of this. Yes sir, is ke baad bhiya consider the bulkiness of that group. But when you do some questions, when you solve some questions, you understand which one is bulky, which one is not bulky. Like the example that I have, you know, mostly you'll get this tertiary oxide. T, B, U, O minus. The same thing will be written in the question also. Okay. Or suppose some tertiary alcohol is there. C, CS3, CS3, CS3, and OH. That's just bulky. Tertiary groups will be considered as a bulky base. So, gaya baad hiya. Okay. This is sedgeff stable Hoffman, the minor product of this. Sometimes again, I'm telling you, across this double bond, GI is possible if these two groups are different. Suppose I have C2H5 here. Okay. But this becomes C2H5 and CS3. It means across this, geometrical isomerism is impossible. So, it's a piece of trans. Two products will be possible there. Right. Both will be equally found. Yes, equally found. This will be measured, but in this, we consider two products. Not four products. Okay. So, if they ask you number of products. Okay. So, whenever alkene is forming, always consider this geometrical isomer as per condition. Okay. Next one at hand down, makes points. Why do we take C2H5OH? How does that happen? See, any reaction takes place in the solvent. The molecules are moving, right? Yes. And in this way, only this ion will attack onto this at least. So, why it doesn't mean C2H5OH at all? Generally, see, you can use another solvent also. But the point is, we require a proper range of the reaction. So, even if you say in physical chemistry, you have a concept called electrode potential. So, electrode potential may be electrode go. There's a metal rod. Suppose it's zinc metal rod. Right. So, this will dip into a solution, which has the same ion present in it. So, condition is what? If you have zinc electrode, this will always dip into ZNS4. Then only you'll get half cell. Okay. The point is, there are many conditions. You can take another solvent also there. But in that case, the reaction is not that physical. It's difficult to proceed with the reaction. So, maybe if you use some other solvent, so the mobility of this ion is affected in that solvent. Okay. And it is, you are getting this from this only. This is not getting reacted into the, getting consumed in the reaction. So, when this H plus comes out, you see with this C2H5OH minus, O minus will get C2H5OH and BR minus goes out. So, this is not getting consumed in the reaction. If it is different, then you'll get some different product. The rate of the reaction, many things will be there. Let's affect it. And since, and that is why we are not using different solvent. Another way what you can say, that this phase, you are getting from this solvent only. This will dissociate like what? C2H5OH minus. And that C2H5OH minus will attack. So, for any, even if you see, if you take another, I will discuss that example also. Solvent, if you take water, water, why we cannot take water here? Because if you take water, water in water, oxygen also has to go on there, right? So, water also behaves as a nucleophile. So, there is possibility that this lone pair also attacks on to this hydrogen. And it is also the possible, right? To minimize that fact also, we are using the solvent of this one. You can ask if it is fair. You do it for OH minus and H2O is the solvent. Ah, that's funny. He said what? Why we cannot use another solvent here? Suppose we use water, then let me also attack on to this. That's why we are not using it. Preferably what we use, that it is the given, you know, condition. Preferably is what? Base and its conjugate acid as a solvent. That is the ideal condition for this kind of thing. Even in organic combustion, this thing is true in all the chapter.