 one more reaction here. CS3, CS2 same reagent I am taking. C, CS3, CS3 and BRG. Base is, okay, these are the solvent we asked, sort of this, so K minus K plus also we can write NA plus, okay. So, since it is a bulky base, right, this bulky base forms what product? CS3, CH double bond, C, CS3, CS3 and another product is CS3, CH2, C double bond, CS2. So, this is a major product, Hoffman is the major product. The reason of this is what? Because this is again a bulky base, large base. So, this O minus has more tendency to take hydrogen from this carbon, because this methyl group also represents this group to attack on to this hydrogen. So, it said what is the difference this carbon, this carbon we have. Here we have three hydrogen and here we have one methyl group. This O minus when he tries to attack or takes this hydrogen atom, this methyl will produce some repulsion to this group. But here we have the hydrogen only, C, H2 and H. So, hydrogen repulsion will be less. That's why it has more tendency to take hydrogen from this carbon. And hence, this product is dominating major problem, okay. So, because of bulkiness of this group, the hindrance of methyl group and this will be more here. Hence, the extraction of H plus from this carbon is difficult. Hence, we get Hoffman as the major product. Here we have hydrogen, which is negligible. But here we have methyl repulsion. So, this repulsion of methyl is more than hydrogen. That's why this product. So, you have to memorize this. Large base Hoffman is the major product. Next point to write down. So, what? Just a minute, sir. So, this kind of show is here. No. Then, next write down, note on this point, dehydrohalogenation of alkyl fluoride. Dehydrohalogenation of alkyl fluoride follows less substituted product. Dehydrohalogenation of alkyl fluoride follows less substituted product. So, here the less substituted means Hoffman only is the major product. Why? Suppose, we have a reaction CH3 CH2 CS3. Suppose, this is the alkyl fluoride we have. Now, when the reaction takes place with a base called CS3O minus and when the base is CS3O minus, what is the solvent? CS3O L. This is the solvent we use. Sir, we use fluoride and alkyl. Yeah. Base. Alkyl chloride. Chloride also you can use. Now, we use bromine as our product. Means generally these reactions are true for bromine and fluorine, alkyl bromide and alkyl fluoride. But in case of fluorine, sorry, alkyl bromide and chloride. But in case of fluoride, the product is different. So, the first example that we took for alkyl bromide, same thing will happen for chloride also. Okay, no change. Chloride and bromide is same. What is chloride? No, it is chloride. Chloride we have. Chloride. No, no, chloride. I said chloride, right. Yeah, it is a chloride. Yeah, it is a chloride. CS3 CHF CS2 CS3. Chloride. Okay. So, in this case the major product is CH2 double bond CHCH2 CS3 and the minor one is CS3 CH double bond CH CS3. This is major. This is minor. The reason for this is since carbon chlorine, fluorine bond is strong, stronger than carbon chlorine or bromine. So, what happens this CH2 hydrogen, this carbon here, this is the intermediate stage. One more intermediate is wrong. For this first two intermediate stages we use. Okay, now this what happens, chlorine, poor living tendency in comparison of living. Yes, iodine is the best, imine is the best halide and living growth. Okay, this has the least living tendency out of the poor hydrogen atom and more electronegative this one, least living tendency. So, it actually what it does, it delayed the extraction of H plus from the beta carbon. Okay, so it is not like this base will take this H plus directly and goes off. Right. So, we will have an intermediate stage form into this one like this delta negative charge forms over here during the base which extracts H plus from the beta carbon. So, like this and there is the two possibility we have where this delta negative charge forms on the carbon atom intermediate. Right. So, this is a characteristics of carbon ion. Right. Carbon ion characteristics developed, carbon ion nature because of this poor living tendency of F minus. This is the carbon ion and it develops here and here also. Fine, this one and this one if you compare, this is one degree carbon ion and this one is two degree carbon ion. Which one is more stable? First one is more stable. That is why this is more stable and hence this one is the major problem. The reason is it forms an intermediate carbon ion. Okay, it is not the completely developed carbon ion. We call it as non-classical carbon ion. Okay, partial negative charge, non-classical carbon ion. Delta negative, delta negative. Delta negative? See, actually this was the negative already. It tries to take this H plus. Right. When this H plus is going away, this carbon has this two electrons. Okay. So, charge is developing under this and it is not like the charge is developing. It goes off and this charge comes over here then F minus goes off. Okay, done. Next point. There is possibility of forming conjugated alkene. If there is possibility of forming conjugated alkene then we always form, then we always form conjugated alkene as the major product. So, in respect of everything else? Yes. Because of conjugation, the alkene becomes more stable. That is why. Okay, suppose this is an example. We have C H 2 double bond, C H single bond, C H 2 C H chlorine, C H 2 C H 3 with C 2 H 5 O minus and C 2 H 5 O H. Tell me the major product here. So, the double bond will come at the third one. C H 2 double bond, C H single bond, C H 2 double bond. These two are the beta carbon, right? Yes. So, they will come on the other side. Double bond here and here. This double bond is in conjugation with this double bond. So, the product is the major product. C H 2 double bond, C H single bond, C H double bond, C H single bond, C H 2 double bond. This is the major product. Okay. So, these are the reactions you must remember. Right, we will discuss E 1 mechanism also. So, so that if there are two conjugations possible, we will always go with extended conjugation. Yes. Cross extended is always more stable than cross. E 1 mechanism, you see, suppose we have an alkaline line, C H 3, C C H 3, and V R. You see, that is the alkaline line. So, it prevents elimination reactions, that is the alkaline line. So, when this reaction takes place at a certain temperature, this V R minus goes out, right, and forms a carbocation. Like I said, E 1 reaction is what? Two step reactions. First step, we will get a carbocation, positive charge, and V R minus goes out. Okay. Now the next step, since we always use this, do this reaction in a polar solvent, and suppose we are having water here, H 2 O. So, water here behaves as a base and it takes H plus from this carbon. This first carbocation forms, then this will expect the H plus. This bond pair comes over here and forms a double bond. So, product of this reaction is C H 3 C C H 3 double bond. The mechanism is first step carbocation forms a reversible reaction, and then the polar solvent takes H plus from the alpha carbon, that's right down next action. We use this extractor of H plus is there. So, we use solvent which we behave as a base and solvent. That's why we see amine we have. Nothing, H plus has been made into H 3 O plus. That's H plus combines it, we are going to have it forms H B R. Okay, next reaction. The hydration of alcohol. The reaction of alcohol. The hydration of alcohol right on the hydration of alcohol takes place in presence of an acid and generally we use what concentrate S2SO4 any acid which can produce proton H plus and hence this reaction is known as acid catalyzed dehydration and hence this reaction is known as acid catalyzed dehydration see one thing I forgot to tell you the first reaction we have done E2 elimination E2 elimination we also call it as a 1,2 elimination same thing 1,2 elimination and since it extracts H plus from the beta carbon we also call it as beta elimination here both are all are same thing beta elimination 1,2 elimination or beta elimination right on dehydration of alcohol right on all of them means what E2 elimination or beta elimination E1 generally it is 2 step reaction so in the second step the carcogatine forms and then we take alpha hydrogen from there so with respect to carcogatine the carbon which uses hydrogen is the alpha carbon that's why we are not calling them as beta elimination right on dehydration of alcohol this reaction this reaction forms and intermediate forms and carcogatine intermediate forms a carcogatine intermediate right this reaction forms a carcogatine intermediate and then carcogatine intermediate and then from the alpha carbon H plus goes out to form an alkene carcogatine intermediate and then from alpha carbon H plus goes out suppose the reaction is CH3 CH2 this is the reaction we have and we are using concentrated H2 SO4 okay acid base H plus now what is the use of this acid H plus the use of acid is that it protonates this oxygen this is H plus and this oxygen takes H plus from this acid okay so protonation of hydroxyl group takes place and it forms CH3 CHOH2 positive charge CH3 okay now the advantage of this is what OH minus we know it is a poor living room but H2O is a good living room right so this cannot go out as OH minus but when it protonates this H2O goes out as a neutral molecule and it forms a carcogatine that is CH positive charge CH3 plus what is the last step last step is what from any one of this carbon atoms suppose here C2 and H this takes this H plus leaving this bond pair behind and it forms CH3 CH double bond CH2 plus see actually when you know the structure of S2 SO4 as double bond OH double bond OH OH and OH these are acidic hydrogen so this goes out we have OH minus H SO3 H is left so this is the conjugate base we have this OH minus takes this H plus forms H2SO4 and so it's not simple then which one will it attack right see in this reaction the carcogatine is forming as an intermediate and in all those reactions where carcogatine is forming we always try to get the most possible stable carcogatine by rearranging okay so in this rearrangement of carcogatine is possible in order to get the most stable carcogatine rearrangement of carcogatine is true for all those reactions where carcogatine is forming as an intermediate