Now, the last reaction which we discussed is oxidation by MnO2. In oxidation by MnO2, in only oxidizers which all compound, tell me, it only oxidizes which all compound, allylic and benzylic, very good, right? Next thing, okay, next is oxidation of side chain of aromatic compound, third point, I think this is the third point, oxidation of side chain of aromatic compound, oxidation of side chain, oxidation of side chain of aromatic compound, how to oxidize the side chain of aromatic compound? Now, what to do in this? Here you have a benzene ring, let's say, and if you use KmnO4, tell me students what will happen? Any idea? Alkaline KmnO4 on heating, do you think what will happen? Tell me, yes, we are going to start a new chapter after finishing this chapter, we know, yes students what will be the product, any idea, KmnO4, benzene, okay, no reaction, okay, benzene can't do anything, but if there is a benzene and there is a carbon, students you won't see all this thing anywhere in textbooks, so please, please make, please pay attention here, if you see benzene ring and on this carbon, if you have either hydrogen, deuterium, pretium, triple bond, double bond, oxygen, any of these, and if you use KmnO4 in basic medium, all things will be oxidized to carboxylic acid or I want to say benzoic acid, everything will be oxidized to benzoic acid, if it has a hydrogen, deuterium, tetrium, double bond, triple bond, oxygen, anything, example, example, let's say this is a benzene ring and benzoic acid, this is called as toluene, if you use KmnO4 in basic medium, what will happen students, quickly tell me, quickly tell me what will happen, toluene with KmnO4, I am waiting for your answer, very good Aaryan, okay, benzoic acid, fine, so students, not just this, try to understand, I am going to put so many reactions now, so here, if you have this type of compound also, CCH3, double bond, CHN, any N number and CS3, that will also give you benzoic acid, if you have something like this, benzene with C triple bond C and CH3, that will also give benzoic acid, if you have something like, if you will have something like this, CH2, CH2, in place of that you have a CL or NH2 or OH, anything, you will get benzoic acid, if you have something like C, C double bond O, CH2, N, CH3, this will also give benzoic acid, all these things will give you benzoic acid, okay, then sir, yes, but that will also give the same product, okay, that will oxidize and give the CO, that COH means, that will also give the benzoic acid in shock, no, I am telling you that is a major product, no, major product will be the benzoic acid, rest this group will be broken down and they will be further oxidized, that is the different part, we will deal them separately, okay Arjit, but the major product with the benzene ring will be benzoic acid, example, but there is one case, let's say you have this compound, what will happen with this, what will happen with this, what will happen with this, no reaction, because there is no hydrogen attached to this carbon, so it can't oxidize, wherever there is a hydrogen, there are the possibility or double bond or anything like that, but carbon-carbon single bond is present, it won't oxidize, carbon-carbon double bond, it may oxidize to give you benzoic acid, but carbon single bond no reaction, okay, there are some more questions which can be asked, let's say this is a benzene ring and again this is a benzene ring, if it oxidized with KMNO4 in basic medium, what will be the product, what will be the product, so here the point very good, kingshock, correct answer, students you need to think there is a carbon present with a double bond, there is a carbon present with double bond, we won't touch this benzene ring, otherwise its aromaticity will break, we will break this compound like this and the product obtained will be dye carboxylic acid, is it clear, is it clear, okay now students oxidation means removal of electron, okay now a very interesting question which I am going to ask, think use your brain and tell me what will be the product, you need to use your brain and tell me the product, don't simply write sir this is the product, there is some concept involved here, I am giving you a hint oxidation is removal of electron, oxidation is removal of electron, students now I will give you, I will tell you what will happen, this is a NO2, NO2 is a strong minus M group, NH2 is a strong plus M group, plus M group will increase the density on this, minus M group will decrease the density on this, if I use KMNO4 oxidizing agent, oxidizing agent means remove the electron, now tell me from which ring removal will be done, if I want money from someone, this means the other person should have money so that I can take, which ring is having electron so that we can oxidize it, which ring is having electron, ring with NH2 or ring with NO2, ring with NH2 is having electron because NH2 is a plus M group, so it is having the electron density, so this will get oxidized, NO2 will remain as it is, NO2 will remain as it is and this will give you the di carboxylic acid on this, is it clear students, yes or no, yes or no, next, next is oxidation by V2O5, oxidation by V2O5, vanadium pentoxide, now students here this will oxidize your, you can say that benzene ring also, if you have a benzene ring to use V2O5, yes you need to increase the temperature to 500 degrees Celsius, what will happen, it will oxidize that and break this bond, which bond it will break, I will tell you one minute, I will tell you which bond it will break, this is the, bonds are like this, this bond will break and this bond will break, so the product will be COOH, please remember these are very specific question and due to very high temperature, due to very high temperature water will be removed because di carboxylic acids are there, so heating effect of di carboxylic acid, I discussed something about this also in class, water will be removed due to very high temperature, water will be removed and the product will be C double bond O, O, C double bond O and dash, malic anhydride will form, this is called as malic acid, this is malic anhydride, malic anhydride, okay students this type of question is also asked, okay let me give you one question, if there are two benzene rings like this, tell me what will be the product, if there are two benzene ring and V2O5 on heating, yes what is element about carbon in final product, about carbon is oxygen, yes V2O5 again both ring will not break, only one ring will break and product will be malic anhydride, means this anhydride will form, is it clear students, yes or no, okay now next is, so this V2O5 you need to remember, okay students there is one more reaction which is generally asked, just remember this reaction, if there is a naphthalene, okay naphthalene is there and you use K2CR207, K2CR207 a strong oxidizing agent in acidic medium or CRO3 in acidic medium, the product will be one benzene ring will remain as it is, one benzene ring will remain as it is, one benzene remain as it is, the other benzene will have O double bond O, double bond O, this will get oxidized like this, okay these are the benzylic positions where you get the oxygen, okay don't think much about this, this will be the product, okay now students the next one is itard reaction, what is the itard reaction, write the heading itard reaction in which one of that we have already discussed, so write the heading itard reaction and very important reaction, itard reaction, in itard reaction you take toluene, you take toluene, toluene with KMNO4 in basic medium and heating gives you carboxylic acid I have already discussed, okay that I have already discussed, so what other reactions are there, so this we have already discussed, this is here if you use CRO2Cl2 also, itard reaction the reagent is CRO2Cl2, this will give you benzene with benzene dehyde, this is called as itard reaction, toluene with CRO2Cl2 will give this, okay CRO2Cl2 will give this product, okay next will be sometimes, so students here you can say there is one more reaction which you write, which you can write here, if you use that in first step CR3CO, CR3CO whole twice oxygen, if this type of compound given to you in presence of water then also benzene dehyde will form, then also benzene dehyde will form, done student any doubt in this, okay so the CRO2Cl2 sometimes it return it is in presence of carbon disulfide or something like that, but the important thing is CRO2Cl2 must be there, okay so this is itard reaction, sometimes it is written in presence of CS2 carbon disulfide, so that it gives you itard reaction and these three reactions of toluene are important, okay fine, so next is fine, bromide chloride, this is bromide chloride, okay whatever fine, so okay here you can say that this is another reagent, no not specific name even to this, but this reaction for whenever there is also this is also a chromium anhydride formed by chromium acid, okay chromic if you have a acid of chromium if anhydride formed by that that also give the benzene dehyde, okay anhydride is taken, next is next is oxidation by HIO4, fourth one oxidation oxidation by HIO4 by HIO4, okay here you can write here you can write oxidation of HIO4, moles of HIO4 require, moles of HIO4 require, moles of HIO4 require, moles of HIO4 require is equals to is equals to number of bonds broken, number of bonds broken down, so moles of HIO4 here stoichiometry importance is there, okay example if you have this compound if you have this compound like it reacts with di-carbonate or it reacts with viscinal diol, so number of bond which will going to break that will decide the moles of HIO4, even it will react with C double bond O, COH, alpha hydroxycarbonate, even it will react with alpha hydroxycarbonate, okay so here what will be the reaction we will see first that so example if you have this compound let's say what will happen let's say you have CH3, C double bond O, CH2, COH, CH, COH, CS3, so students what you need to do when HIO4 is used here only one mole is needed why because we have one the system CHOH, CHOH, only one mole of HIO4 needed to break this bond and when you break this bond just add OH and OH on both side just add OH and OH on both side or see when you add here OH and here OH what you are going to get what you are going to get CH3, C double bond O, CH2, CH, OH and OH in next step one carbon to OH group unstable water will be removed minus H2O, CS3, C double bond O, CH2, CH2, C double bond OH, you will end up getting an LD high okay this is the one product and one product will form from this end also one product will form from this end also that will be the another product will be I will draw the another product also another product will be also LD high and that will be C double bond OH and CH3 so another product is this how to know which bond will break whenever because I told you HIO4 box for dye you know that will C double bond O, 1, 2 dye carbon in vissinal dye all like that here you have 1, 2 dye carbon in and you see in the reactant 1, 2 we have vissinal dye all in this case so only that bond will break vissinal dye all bond will break and give you give you with HIO4 this reaction will be used in in glucose bio molecule chapter also there I won't tell again see this is the mechanism or this is the process I will directly say okay we have already discussed this end that place so CS so the product will be one product is this another product is this so I will put them in a box one product is this and another one is this another one is this okay one student in one student has asked me over whatsapp lights are in ZN I by mistake in my notes I have written when ZN, HG and HCL used with conjugation uh C double bond O is in conjugation with double uh with double bond so both will get reduced who asked that question both bond will be reduced someone asked I don't remember again in my last class when I explained the ZN, HZ and HCL when double bond is conjugation with C double bond O then both will get reduced I told this thing but in the product I have kept C double bond O please change that thing make it a complete alkene okay fine there is one question one of your friend has asked this question okay let me ask let me give you this question if you have CS3 C double bond O C OH CS3 could you tell me the product students right by yourself HIO4 only one more needed because one double bond is there one one this type of system is there alpha hydroxy carbonate students product done break this bond break this bond CS3 C double bond O HIO4 taken right so what will happen here you have when you have C double bond O so one will be the carboxylic acid right and how to make the product here you have if I add OH here if I add OH here so it will be a carboxylic acid right and if I add OH here also what I'm going to get CS3 OH and OH so the product will be not CS3 sorry CS3 carbon one minute another will be this two carbon with OH and one CS3 so product will be you can remove water from this here you have one hydrogen so another product will be CS3 O H students is it clear yes or no just break the bond add OH break the bond add OH one product is this one product is this okay fine next part is oxidation of aldehyde ketone I think you must have learned this thing in school also if they have started aldehyde and ketone chapter so next part is oxidation of aldehyde and ketone oxidation of aldehydes and ketones aldehydes and ketones student there is very very important reaction A is tolerance test tolerance reagent or tolerance test it is also used to test the aldehyde and the tolerance reagent is what is the tolerance reagent tolerance reagent is ammonium hydroxide along with AG NO3 it is also called as ammonical silver nitrate ammonical silver nitrate okay the above further ethanol will not get oxidized I told you this is this oxidizing HIO4 only work when there is on next carbon on two carbons there are O present then only it will work is it clear whether like only three cases where I explained one two di carbonyl visceral diol and alpha hydroxycarbonate okay I'm coming to this tolerance reagent is this fine what happens students there are two reactions when you use alkyne with tolerance reagent there will be very interesting reaction NH4 OH AG NO3 students what will be the product can anyone tell me can anyone tell me the product it's not difficult to predict okay I will help you first you try then I will help you no no very good inshore correct answer inshore write the product so students product form from the tolerance reagent will be product form from the tolerance okay here the product is acid base reaction this is the acidic hydrogen it will this will be removed and since you have an AG plus ion so C triple bond C minus AG plus so you can write it like this also it will form up over and bond so AG this is white ppt white ppt will form okay if someone asks her what is the tolerance reagent how it is prepared so students first and how it is used for aldehyde and ketone if this is the question so I want to tell you first we will write like what happens when you use how it is prepared tolerance reagent is prepared by this reaction if you want to prepare tolerance reagent you need to have AG NO3 plus ammonical silver 9th rate the reaction will go and form AG OH plus NH4 NO3 double decomposition reaction in next step in next step your AG OH will react with your AG OH will react again with ammonium hydroxide and silver forms a very very stable complex in coordination compound I will teach this thing silver form of very complex very stable complex with ammonia this reaction is used to extract silver also from its soul because silver forms a very stable complex with ammonia and OH minus will form along with water so this is called as tolerance reagent this is called as tolerance reagent this is called as tolerance reagent when I'm writing NH4 OH and AG NO3 this means the same compound is this because when you mix them this compound exists in complex formation complex form now now I'm telling you how it will act like an oxidizing agent if you use any aldehyde RCHO and you use tolerance reagent along with that if you use tolerance reagent along with that tolerance reagent along with that OH minus okay RCHO OH minus here you have a tolerance reagent and the product will be RCOO minus because aldehyde is there oxidizes to carboxylic acid so but since ammonia is present you will get ammonium salt RCO minus and NH4 plus because carboxylic acid with base gives the salt along with that AG will form because aldehyde is oxidized something has to reduce the silver right now is in plus one oxidation state this get reduced to silver zero and this is giving you the mirror this is called as silver mirror this thing deposit you know over the inner wall of test tube along with that ammonia liberated and water will form so this AG will get deposited on inner wall of test tube this is called as silver mirror test okay ketone does not give this test ketone does not give does not give this test this test okay now who aromatic and both aromatic and aliphatic both aromatic and aliphatic aldehyde gives this test aliphatic aldehyde both of them give this test aromatic and aliphatic aldehyde gives this test means both will become RCOO minus is it clear students yes or no yes or no next reaction sir is this the only reaction for aldehyde and ketone oxidation no again a very important reaction is called as felling's test felling's test in felling's test this is these two are very important for your board exam also first you need to write the felling's reagent what is the felling's reagent felling's reagent is formed by two substances a plus b okay what is this a plus b don't write this in exam felling is formed by two substance a and b you need to write a and b also what is that a is aqueous Cu SO4 very good to be a is aqueous Cu SO4 and your b part is alkaline solution of sodium potassium tartarate alkaline solution of sodium potassium tartarate tartarate this is a salt sodium potassium tartarate okay so this is also called as ruchelle salt what R O C H E double L E ruchelle salt now these two things are mixed and we call that thing as a felling reagent okay so what happens student's general reaction is ld hide and the hide plus felling solution gives the reddish brown ppt reddish brown ppt oxidation happens and give the why i'm why i'm emphasizing on in these two tests more because they are the identification test of ld-headed ketone so ld-hide can be oxidized to something and you get the reddish brown ppt what actually happens i will write the equation if you have r C H O then copper ion will react why copper ions are because we have taken felling reagent aqueous Cu SO4 okay so Cu SO4 will give you copper ion plus O H minus ion five times when you balance it later in later i think you in class 11 to learn balancing also how to balance the ionic equation so i'm just directly writing a ionic balance ionic equation ld hide will oxidize to acid ld hide will oxidize to acid and your copper 2 plus will get reduced to Cu 2 O this is called as 2 plus oxide this is called as 2 plus oxide and it is a red ppt red precipitate okay or sometimes it is written like reddish brown so instead of writing red right reddish brown ppt reddish reddish brown ppt okay reddish brown ppt now sometimes it is written red ppt sometimes it is written reddish brown but i think we will we will assume that okay red ppt is also fine so Cu 2 O will form and free water so our copper will get reduced ld hide will get oxidized so this is again a one reaction which is very very important but there is one issue in this reaction what issue not give test for this reaction not give test for for aromatic ld hide this does not give test for aromatic ld hide okay this does not give test for aromatic ld hide so definitely these two are not going to because we are discussing the oxidation of ld hide so these two reactions won't give the test of ketone is it clear now felling solution is over so any doubt in this students any doubt please remember that in this in when we have a and b part we mix them in equal amount we mix them in equal amount felling solution a and falling solution b this is called as felling solution a felling solution b mix them in equal amount then you will get this compound is it clear students yes or no if someone asks her what is the salt of sodium and potassium iterate how it looks like it looks like this c o o minus h o minus h o minus h o c o o minus and they both these both these both oxygen are going to attach with offer two plus this is this is what felling solution is it is used okay that is why we have written directly copper two plus and this thing forms due to the sodium and potassium territory because we have a territory acid this is a territory acid if i put h and h everywhere this is called as territory acid okay so that will give you the this test students next text test is okay shifts test third point next is shifts test shifts test this will be given by l d hide only r c h only not elephant not aromatic l d hide okay r c h o here you have r c h only here you have shifts base shifts base what is that para rosanne rosanne lee rosanne lee hydrochloride hydrochloride this is the shifts base most of the time it will be given shifts base in the reaction it is pink in color when it is pink in color now when it is passed through so to it become color less it become color less and this is called as shifts reagent this is called as shift shift reagent now if aliphatic l d hide is taken now if you are aliphatic l d hide i have written here r c h only then here you are aliphatic l d hide taken aliphatic l d hide taken you will get the r c o o h that will oxidize to acid that will oxidize to acid and the pink color re-stool and the pink color re-stool okay or pink color re-stool so the point is whenever color less shifts reagent first you make the shifts reagent using the shifts base and when it is used with the aliphatic l d hide that will thus this color will start appearing again and this this give you that this tells you the presence of l d hide in your compound is it clear students any doubt yes or no next is benedict test next is fourth one fourth one is benedict text test benedict text test what is this benedict test it is a aqueous copper sulfate you take plus sodium citrate sodium citrate sodium citrate students aliphatic l d hide only r c h o only for aliphatic l d hide only for aliphatic l d hide it is difficult to oxidize the aromatic l d hide okay again here you have r c h o you take benedict reagent to take benedict reagent benedict reagent and the product will be r c o minus plus sodium citrate you take aqueous copper sulfate you take this will oxidize l d hide will get oxidized and you will get c u because copper 2 plus you have taken same thing as what you got in the felling solution c u 2 o will form red color will tell you red color will tell you okay there is a presence of an l d hide both felling solution and benedict test are similar the only thing is the sodium salt are different there you have a sodium and potassium salt of tartarate tartaric acid here you have a sodium salt of citric acid that is the only difference is it clear students yes or no okay students there is one more reaction which we will discuss later also but you can write here oxidation of ketone how to oxidize ketone methyl ketone so right oxidation of ketones in bracket right methyl ketone okay generally generally it is very difficult to oxidize ketone okay we have discussed the oxidation of l d hide that is over now oxidation of ketone generally it is very difficult to oxidize the ketone but methyl ketone can be oxidized okay sir how we are going to oxidize the methyl ketone so here you write iodo form reaction by iodo form iodo form reaction now this iodo form reaction is very very important for your good exam point of view for all the competitive exam what is methyl ketone first we need to understand what is methyl ketone methyl ketone is something like c h 3 c double bond o dash this is called as methyl ketone okay students now meet all the methyl ketones will give this test and what happens during the reaction we will see so if methyl ketone is taken the same reaction will come in carbonyl compound also there i won't discuss that in detail because i'm taking in the this section in oxidation section so when there i will directly say refer this page now what happens in methyl ketone oxidation you write reagent use this sodium sodium hypo iodide reagent use this reagent is sodium hypo iodide sodium hypo iodide okay sodium hypo iodide now here what we have gives carboxylic acid having one carbon less than gives carboxylic acid gives carboxylic acid with one less carbon with one less carbon carboxylic acid will form with one less carbon what is the general reaction there will be a carbonyl compound in presence of carbonyl compound you can be having NaOi sodium hypo iodide you have this or sometimes it will be returned most of the time it will be written like this NaOH and i2 based with i2 later i will explain this reaction also in v blog chapter not now NaOH and i2 will give you RCOO if this thing will get oxidized O Na will form and CH i3 this is very very important compound called as iodo form iodo form okay this is called as iodo form now this iodo form is yellow ppt yellow ppt this tells the presence of ketone so ketone can be oxidized but only only methyl ketone okay so question can be asked question can be asked or what is the reaction for iodo form production iodo form reaction you can write that NaOH when i2 is used it will give you NaI plus NaOi plus H2O this reaction is generally taught in p blog chapter there i will again emphasize on this reaction and i will tell for all the halogens reaction for all the halogens but as of now you can just write this any doubt students any doubt in this okay so one question if i ask see if question is question is what will be the product what will be the product of this reaction quickly write the product will it give iodo form test will it give iodo form test question is this will it give iodo form test methyl ketone is needed methyl ketone will it give iodo form test okay fine student it will not give iodo form test okay it won't give iodo form test fine no reaction why there is no methyl ketone to have iodo form test there will be methyl ketone sir if someone is thinking sir could you please tell us the mechanism what will happen so the mechanism is let's say this is the methyl ketone CS3COOR this methyl ketone when reacts with Na NaOi I have written so NaOi can be written as OI minus OI minus is a is a nucleophile as well as base since I told you methyl ketone is needed it will take the acidic hydrogen it will take this hydrogen once it will attack on this hydrogen once it will attack on this hydrogen you will get CH2- you will get CH2-C double bond O R in next step since you have taken I2 also since you have taken I2 this negative will attack on this vacant orbital of I and this I minus will be removed this I minus will be removed product will be highly reversible reaction no doubt so product will be RC double bond O CH2 I CH2 I in next step in next step what will happen then see the same thing will repeat again and again again there is acidic hydrogen again there is acidic hydrogen same thing will repeat and the product form will be RC COO CI3 CI3 do you remember students this has do you remember what is this few days back we discussed something about this this product this this substrate which type of reaction this substrate gives few days back we discussed RC double bond O CI3 which type of reaction this will give if there is OH minus ion any idea forgot it is a strong minus I group it is a impure carbonyl impure carbonyl reaction S impure carbonyl reaction S characteristic reaction of impure carbonyl a characteristic reaction of impure carbonyl CS3 COCl impure carbonyl which type of reaction it gives very good Archit impure carbonyl characteristic reaction pure carbonyl gives nucleophilic addition impure carbonyl give very good thing show SN2 TH mechanism SN2 TH this will go this will go CI3 will be removed product will be RC OOH along with CBR3 minus or CI3 minus and this is a colorless solid this is a colorless solid fine then okay we are not colorless solid here it is here no solid CI3 if it is a BR3 I will tell you later it is a yellow solid okay once it takes the hydrogen what happens in next step this is acid in next step it will take hydrogen from water you can say there is a water taken so in next step it will it will form CI3 it is a yellow solid so this is the mechanism of iodopharm reaction but so that is why I do you use methyl ketone is it clear this is a yellow solid okay so students some people ask are is it only possible for iodopharm is it not does it not giving chloroform or bromopharm yes it will give chloroform bromopharm also CsCl3 is chloroform colorless liquid colorless liquid that is why we don't use it if it is CI3 it is colorless solid CsBR3 it is colorless solid CsBR3 is colorless solid but we are interested in some color that is why we use iodopharm we use iodine because CI3 is yellow solid that is why we prefer iodopharm reaction okay rate is same for all the three which is the okay which is the RTS step student RTS step is this why I said rate is same for all these because RTS step is this rate determining step is this first step because this is very less ascending hydrogen which it has been removed so our generation of nucleophile generation of C minus is a RTS step is it clear students is it clear any doubt okay sir which all groups now few groups few other groups also give this test few other groups also give this test student first tell me did you understand or not did you understand or not iodopharm reaction yes or no yes okay fine C which all groups can give your methyl ketone Cs3 COR this is the most important but even this also gives this test COHR sir why it gives test because can anyone tell me why this reagent also gives two degree alcohol also give iodopharm test can anyone tell me why why okay fine students very good answer it will oxidize first it will oxidize first by NAOI and it will convert into methyl ketone only so even this will give the positive test this will give the positive test okay fine anything else CS3 COOH this also gives positive test because methyl sir this is anti-hyde how this will give test yes this will give test because there is a methyl group present next to C double bond O next to C double bond O methyl group present this will also give test though it's anti-hyde no worries okay will this give the test will this give the test just check will this give the test here you have taken two degree but here you have taken one degree only okay so students there is a methyl group present here this thing here you have see even this will oxidize first to this state even this will also first oxidize to this state so this will also give the iodopharm test means don't think directly carbocation will form this will oxidize to this state first it will also give iodopharm test but impure carbonyl does not give iodopharm test like this G with a lone pair this is called as impure carbonyl this is called as impure carbonyl this won't give iodopharm test does not give iodopharm test is it pure students or helo form this is also called as helo form any doubt yes or no so students there is one interesting part here there is one interesting part here right one thing okay there is one more thing if you have this compound CS3 C double bond O CH2 C double bond O CS3 will this give iodopharm test will this give iodopharm test student think first think about the mechanism and then write think this will give iodopharm test think about mechanism think about mechanism and tell students does not give iodopharm test sir what sir why okay why students i told you in mechanism base is attacking on this hydrogen base is attacking on this hydrogen do you think there is one good option for base to attack in this case do you think there is very one good place to attack for base active methylene group is this hydrogen more acidic is this hydrogen more acidic if i'm talking about hydrogen number a and hydrogen number b which is more acidic which is more acidic thing i taught you the active methylene group thing which hydrogen is more acidic a or b a is more acidic hydrogen why can anyone tell me why why a is more acidic hydrogen why not b why a is more acidic hydrogen think okay sir i don't know so if you don't know then draw if you don't know then draw the two states where the hydrogen is removed if you don't know and draw and check tell me which carbonyne is more stable tell me which carbonyne is more stable if this is compound one and this is compound two which is more stable there is a resonance between two oxygen there is resonance with two oxygen it has resonance just with one oxygen this negative has resonance just with one oxygen so it is more stable yes yes is it clear students yes or no so if it is more stable this hydrogen is more acidic so attack won't happen on this so what condition is there carbon should have more acidic hydrogen in methyl ketone that methyl the methyl of ketone should have more acidic hydrogen then only heloform reaction will take place the methyl of ketone methyl of ketone should have more acidic hydrogen then only you will get heloform reaction otherwise it does not hear heloform reaction okay so methyl group of what you can write as a note methyl group should have more acidic hydrogen methyl group should have more acidic hydrogen then only you will get heloform test then only heloform test is given okay otherwise attack will happen on this your methyl group won't participate also any doubt students just last line related to the heloform rate of i2 is equals to br2 is equals to cl2 no change because they are not involved in rbs step any doubt fine so students we have done so many questions in this now let us move to the next part of the same thing oxidation of alkanes last part then we will start aromatic compound chapter last part of this chapter oxidation of alkanes so it's not needed to cover this because alkanes chapter you learned in last year but still i'm covering so that those who missed that part can first one is complete oxidation first one is complete first of all i told you that alkanes are very very less reactive so complete oxidation means your cn h2n plus 2 reacts with oxygen okay i will write a balance chemical equation those who don't know this is the balance chemical equation o2 gives n times of c o2 plus n minus n plus 1 times of h2o so this is a balanced chemical equation okay any doubt 2n n n plus 1 right balance chemical equation plus energy this is one oxidation now this is a complete oxidation so what is incomplete oxidation then there is something called as incomplete oxidation also that you know that carbon will form or carbon monoxide will form so that is the incomplete oxidation i will write that that's a methane gas reacts with oxygen gas in a limited amount in a limited amount gives carbon solid gives carbon solid black incomplete oxidation plus 2h2o liquid these are the two oxidations of reactions of alkanes which you all know but someone says sir i want chemical you know you want sir controlled oxidation if you want to have a controlled oxidation there is a third reaction right controlled oxidation less important but still this is the oxidation technique used for alkanes here you have a regulated supply of oxygen regulated supply of oxygen so you have c h h h m o2 o3 m o2 o3 when you heat it you have only controlled oxidation 2 OH will be converted into 2 hydrogen will be converted into 2 OH and you will stop you will just get aldehyde don't think about acid you will get aldehyde so this is the controlled oxidation just very specific example molybdenum oxide is used i am moving to the next reaction chemical oxidation okay now the last one last reaction sir can k m n o4 also oxidize if someone think so yes k m n o4 also oxidize but right fourth one chemical oxidation how it will oxidize we will see if you have possible only with possible only with those alkene having three degree hydrogen possible only with those alkene those alkene having three degree hydrogen three degree hydrogen can be oxidized by alkaline k m n o4 please remember this part if you have this type of alkene three degree hydrogen if you use alkaline k m n o4 also it will oxidize it to alcohol it will oxidize it to okay yes and there is one reaction one more reaction but okay fine that's it these are the general reaction you can write one more reaction that is methane reacts with oxygen in presence of copper at 577 kelvin you will get methanol this is also a preparation of methanol with the help of methane using copper with this this chapter is over okay if there is some reaction which are left that will be covered in some other chapters i think mostly all reaction i have already taken today i will be sharing two worksheets one for reduction one for oxidation please don't forget to complete and students in last class i shared the worksheet for substitution i extended its time also but still i can't see many submission substitution reaction i gave you long back but still you people have not submitted your answer this is very very disappointing you need to submit your homework otherwise there is no point so how many of you have done the homework but not submitted for substitution tell me last time i gave you the homework for substitution reactions how many of you have done but not submitted tell me amritha who else only one okay fine so students please submit it after the i will extend the date if needed but you need to complete the homework and those who have submitted i have the list so in ptc i will tell all those things those who have submitted and those who have not submitted next is let us start a new chapter okay students see we have done almost uh not but many of the reaction we have already covered most of the reaction we have covered in the reaction mechanism okay and in whatever mechanisms we have discussed now this is the aromatic compound sir why you are taking this chapter because if i don't take this chapter every chapter is incomplete why because if you see in ncrt there is a alcohol and a phenol so phenol itself a very different compound it is not showing reaction like alcohol so which type of reaction it shows we will see in this aromatic compound chapter so and it is very important for your je point of view or any competitive exam point of view okay you won't say see any chapter like aromatic compound in your in your school textbook but yes you must find any you must find phenol you must find benzene something like that so those all will come under this section so here it is very very important to understand this chapter okay students after today's class i will be sharing one one pdf with you okay that pdf will have a aromatic compound related things so please go through that pdf also that as a very we can say that important topics and so many examples also so that the pdf you can use okay now students first i will start with the aromatic compound here it is a cyclic unsaturated compound or two types cyclic unsaturated compounds are two types cyclic unsaturated means which are a cycle and double bond or triple bond like that so they are of two type one is the or instead of saying two type they will have a three type one is aromatic one is anti aromatic anti aromatic and one is non aromatic non aromatic okay students first tell me one thing last year did you learn how to check whether something is aromatic something is anti aromatic and something is non aromatic yes or no did you learn that Huckle's rule yes yes yes so you have learned that so i won't be discussing that in detail okay so Huckle's rule you all have learned so for anti aromatic the only difference is it has it has okay what are the primary conditions they should be cyclic they should be planar cyclic planar okay fine cyclic compound planar or ring atoms all atoms should be or atoms should be either sp2 or atoms should have either sp2 or sp hybridization sp3 is not allowed otherwise the ring will not be planar complete cyclic delocalization complete cyclic delocalization localization okay those who are new for them i am going to tell this again and complex cyclic delocalization what else what else fourth one is should follow 4n plus rule should follow should follow 4n plus 2 rule 4n plus 2 pi electron pi electron rule rule of Huckle 4n plus 2 pi electron rule of Huckle this is highly stable this is a highly stable and sixth one is burns with black smoking flame burns with burns with black smoky flame okay now aroma that is have a fragrance smell okay so that is having aroma and the last one is diamagnetic in nature they all are diamagnetic students did you learn this last year all these all these seven eight points for aromatic compound yes or no these should be the necessary condition for any aromatic compound necessary condition or you can say yes these are the characteristics they are highly stable burns with necessary conditions are these which one these four cyclic planar complete cyclic follow 4n plus 2 rule okay the next is anti aromatic in anti aromatic first rule second rule third rule are same are same but in fourth rule it differ it has 4n pi electron 4n pi electron where n is equals to 1 2 3 4 whatever highly unstable this is highly unstable highly unstable and the last one is paramagnetic in nature they are paramagnetic aromatic compounds are anti aromatic and aromatic compounds are diamagnetic non-aromatic compounds are non-planar first of all the primary condition they are non-planar because of any reason because of any reason not just sp3 or whatever because of or because of any reason if it is a non-planar we will say it is a non-aromatic compound okay it shows property of shows property of alkenes and alkynes it won't show property of aromatic compound is it clear students let us do some examples these are the conditions i think you must have written all these things let us do some questions if you have this compound students please tell me either it's aromatic or not is this compound aromatic tell me first is this compound aromatic tell me whether they are aromatic anti aromatic non-aromatic this is compound one this is compound two and this is compound three okay fine done students few more questions i want to give you so fourth one is okay this one and fifth one is fine therefore to many questions okay fourth one fifth one will be so students this is fifth one minute one two three four five six seven and eight done anyone who has done all the right questions i ate member drink with this okay quickly tell me answer of this the power yes okay fine students i think you have already done just a quick recap this is anti aromatic this is anti aromatic four and plus two five electrons okay this is what is the answer for second question what is the answer for second question tell me here you have double bound second question okay you have done tell me fast second question okay second question is non-aromatic first of all you can't say it as aromatic because there is not some complete cyclic delocalization first of all there is not complete cyclic delocalization because here delocalization breaks double single double single single don't pair how you can say it's a complete cyclic delocalization and second why non-aromatic because this carbon is sp3 this carbon is sp3 okay hope you understood this yes next one next one third one so this compound is actually the second one is non-cleaner third compound third compound this means you people have not learned it properly last time you have not learned it properly just see the lone pair delocalization lone pair single double single complete cyclic delocalization cyclic delocalization means double single double will repeat so complete cyclic delocalization number of electron pair two electron two electron two electron six electron aromatic compound six electrons are involved in delocalization two from nitrogen two from purple four from purple no yes and lone pair not participate in why why it won't participate it will participate no no no no no okay fine always remember this is very less basic this compound is very less basic in nature because this is a highly this is aromatic compound no in furan the in furan also furan also oxygen is delocalized always remember here the here you have nitrogen present the lone pair is in delocalization with the double bond wherever possible wherever possible okay fine if you are comparing it with pdd that is a different thing in pdd nitrogen oxygen nitrogen lone pair is not in delocalization why because there is already a complete cyclic delocalization there this lone pair is not in delocalization not in delocalization not in delocalization but not in this case here lone pair will be in delocalization so it is aromatic compound two two two six electrons is it clear to all of you here here it is also aromatic here it is also aromatic but these two electrons are not delocalized so we will count only these other six is it clear yes or no okay next fourth fourth one complete cyclic delocalization yes fourth one aromatic anti aromatic non aromatic what Gayatri correct answer the wrong answer okay one minute one minute one yes there you have 48 plus 2 pi electron yes yes yes yes sorry Vavav correct Vavav correct Gayatri wrong and rest everyone is correct suti check again this compound is aromatic two electrons are in conjugation only two electrons but complete cyclic delocalization aromatic any doubt do this first one the rings are attached with a single bond we will count the individual rings we will count the electrons of individual rings when they are attached with a single bond so here you have six electron here you have six electron aromatic and all our planar everything is there what about seventh one yes Deepa correct pittig reaction forms by tonight seventh one seventh one Ruchita correct answer Archit correct answer okay so here what will be the answer so here we will say two one one total six electrons six electron in delocalization complete cyclic delocalization aromatic tell me answer for this so okay fine i'm asking for okay eighth one what is the answer of eighth one eighth one students eighth group okay fine i will tell you eighth one will become like this oxygen is more electronegative electron will be pulled by this and it will get positive okay sorry now see i forgot to put one dash there now tell me now tell me whether it's aromatic anti aromatic or non-aromatic okay fine answer is aromatic aromatic yes it's not second is different they are the resonating structures understood it will shift like that when electron will shift like that we will say it's a aromatic compound it forms aromatic compound so whenever these type of missions are given you will count the rings you will count the number of electrons in a ring then for the second one it is benzene second one it is benzene no no no second one is not in delocalization at all because this carbon has nothing it's not complete delocalization above you have just one pair on one atom okay yes students ninth one whenever rings are attached together we will add a total number of double bonds when the rings are attached we will add a total number of double bonds what will be the answer fine so this will be the aromatic why because total total electrons are 10 electrons are in delocalization 4 and plus 2 what i have left the two questions question number 10th i have left and question number six so students there is an exception in question number 10th and 6 but i move to next slide i have not discussed question number 10th and question number 6 it seems like question number six there's in the question number six it seems like it is aromatic compound but actually it is non-aromatic why what happens in question number six what happens in question number six this is double bond this is double bond here there are hydrogens present here they are here you have hydrogens which are creating a repulsion these two hydrogens creates a repulsion you know it is called a transangular strain there is some angular strain created by these hydrogens this repulsion is there so what happens they make this ring non-planar they make this thing non-planar and once it become non-planar even if you have even you have complete delocalization or whatever the compound is non-aromatic any doubt any doubt now the eight member thing this is question number six for question number 10 the eight member thing generally exists as this the eight member thing generally exists like this okay this is what form this is a tub shape this eight member ring exists in a tub shape okay there is a double bond i will draw double bond like this there is a double bond like this so no delocalization because for that ring should be planar okay this is a non-planar non-planar so no cyclic delocalization tub shape non-planar tub shape so these exceptions you need to remember and here you have a pure double bond here you have a pure double bond pure double bond means it is not in delocalization so it is non-aromatic is it clear students non-aromatic is it clear okay now if there are one student has asked okay now see if this is sir in the fifth question if one ring is aromatic yes yes i'm coming to that i'm coming to that great page someone has asked this question wait now what will be the answer for this and coming to your question what will be the answer for this if there are two rings are like this aromatic non-aromatic anti-aromatic sure fine students whenever the rings are connected by the two double bonds when our ring is connected with double bond we will shift double bond some to some of the atom like we have done in that question we have done in that question when ring next atom means when atom of the ring is attached with double bond we shift the double bond and check whether it's aromatic or not here also this is attached with double bond i will shift to the atom so you are going to get something like this you are going to get something like this negative and positive now tell me what about the two rings you can separately check what about the two rings so if you see what about this ring aromatic or anti-aromatic two electron aromatic what about this ring are you not able to imagine this are you not able to imagine this is the situation one ring is like this one ring is like this one is like this one is like this there is a positive there is a negative double bond this is a aromatic because to an electron complete cyclic delocalization two electrons aromatic but here it is anti-aromatic so whenever such situation arises the overall compound is non-aromatic we don't say that aromatic not anti-aromatic we will directly say the compound is non-aromatic is it clear whenever you have this situation both compounds are aromatic then only we will say aromatic compounds both rings should be aromatic example example one more question one more question if you have this thing a very interesting question if you have double bond double like this yes tell me past whether the compound is aromatic or anti-aromatic or non-aromatic you need to shift electrons such that such that you get an aromatic compound try to get aromatic compound shift electron in that way students I am un-sharing my screen for once one minute I will show you something very interesting yes fine students can you see my screen can you see my screen yes or no okay fine so here if you have this type of situation the question which I have given is this I am going to share this PDF with you there are lot of examples given in aromatic and anti-aromatic everything can't be covered in class so I will be sharing this PDF with you but the question which I have shared just now is this one minute the question which I have shared just now is is this okay there are two possibilities there are two possibilities one possibility is the electron will be shifted like this then both ring will be aromatic aromatic one possibility is this that one become anti-aromatic other become anti-aromatic but we will ignore the second option because we are getting one option where both are aromatic so the overall system is aromatic but here in the previous mission overall system is non-aromatic because you are not getting anything where you are getting both aromatic rings so please underline please write this line in such type of system it can never be anti-aromatic and such type of system is aromatic then both ring must be aromatic after displacement of double bond okay so students there are lot of hundreds of example given in this in this PDF which I will be sharing after the class and you can go through this properly okay fine so this is all about the aromaticity of any compound here I want to just tell you one more thing like azulins what are azulins you can see here okay in a monocyclic compounds there are so many see this are azulins are like this one minute yes so four pi electrons these examples I have already taken in class these examples are over now what are the azulins these type of rings okay non-cume non-pi bond in monocyclic you can see that there are so many pi bonds in a monocyclic compounds so here you have is isomer of naphthalene so if you have this isomer you shift the electron and you check there are the resonating structures okay so here if you have this type of compound then what will happen all those things you can see here okay like one question is asked generally asked like generally whatever anulin here you see there is this is also a compound 12 pi this is anti aromatic this is anti aromatic you count the number of double bond one two three four five six seven eight anti aromatic this is also anti aromatic this compound is also anti aromatic why so here two to four to six to eight to ten to twelve why it is anti aromatic any idea yes because what happens students this can anyone tell me why this carbon why this compound is anti aromatic this one this one count number of double bond two four two four six eight ten twelve and fourteen so students what happens we don't count this double bond at the center because it is not present in delocalization your double bonded center is not present in delocalization actually electron will have a look delocalization in periphery of the ring what is the periphery of the ring this part electron will remain in delocalization in this periphery of the ring like one minute this part it will remain in delocalization on periphery of the ring so this electrons is not in delocalization that makes this compound anti aromatic is it clear yes or no yes or no okay so find one more question and the last question related to this topic so if i take if i take this example if i take this example right best i think it's clear but if you'd like one to take one more example i will take it here right this is the last thing if you have this would you comment whether it's aromatic or anti aromatic or non aromatic or whatever one two three four five six seven eight tell me fast aromatic or non aromatic think fast and tell me very good syncing so correct answer synch and correct answer fine lucida correct answer students i told you electron will delocalize on the periphery of the periphery of the ring electron will delocalize on the periphery of the ring your this double bond will not be included in delocalization so that this compound is aromatic is it clear i hope we have taken sir then naphthalene even though as a double bond is in between no that is different thing in naphthalene if you have this one two okay fine that is a different thing here this is a part of aromatic ring okay here in naphthalene we don't say because this is a part of aromatic ring okay but as the ring increases the number of ring increases then this thing we say this is non aromatic or means then it is not included because there is a complete delocalization and extended delocalization fine now next is formation reaction in which aromatic compounds are formed so write the heading formation of aromatic compound formation of aromatic compound one minute students okay formation of aromatic compound what all formation involved for aromatic compound students can you please tell me what will be the product if this is a compound given to you if this is a compound br with aqueous agn o3 quickly write the product i will tell you here you have this seven eight the same ring here but if you see this forms a aromatic compound if br is removed if br minus is removed there will be a positive charge and this forms an aromatic compound plus an o3 minus and that is how ag br is formed tpt so this reaction is how this is how the aromatic compounds are formed okay because you always remember students sometimes this is a highly stable compound that is why br is removed this is a highly stable compound that is why br is removed okay now if you have this equal this diagram or this reaction tell me what will be the product aqueous agn o3 what will be the product name i will write seven bromo cyclo epta one three five trying one three five trying yes yes students what could be the product the correct answer very good no reaction because if you form the product because if you remove the br minus a compound which is going to form is something like this and this is anti aromatic compound very very unstable so there is no reaction this compound won't form this reaction doesn't happen is it clear students is it clear what will be the product this question we have already done what will be the product quickly tell me think this reaction we have already discussed okay i'll help you little bit very good deeper correct answer this reaction we discussed in reaction of metal with active hydrogens with active hydrogens done reaction of metal with active hydrogens done find students with this hydrogen become become highly active because once it will go once it will go the compound will be an aromatic compound so this hydrogen prefers to leave with sodium and the compound which is going to promise aromatic compound is it clear so this is how the aromaticity aromaticity leads some reaction aromaticity of compound if you have this type of reaction what will happen there are bf3 now bf3 will say okay i will i can accept the halogen bf3 is electro deficient so f minus will go and form this compound when f minus will go the compound form will look like this compound form will look like this positive and positive this is a aromatic compound you can count number of electrons 2 to 4 to 6 6 electrons complete cyclic delocalization planar molecule planar molecule all the four conditions are satisfied here okay planar complete cyclic delocalization at 4n plus 2 and cyclic compound so plus bf4 minus 2 molecules of bf4 minus 1 is it clear students yes or no even if you are whole way electro this is reaction also even if you are whole way electro this is reaction if you remember that also goes in forward direction because of the formation of this is also whole way electrolysis this is also one way to prepare a benzene ring preparation of aromatic compound okay fine there is one more reaction which we have already discussed in some of the chapter in the in last part of substitution reaction we have discussed this reaction also if you have this type of compound br br single bond single bond if you have this type of compound with zinc and heating effect what will happen students remember can you imagine what will happen here yes br will leave zinc will give its electron what will happen br will leave the compound zinc will give its electron to this atom the select so br will leave the component zinc will give its electron can i directly write like this can i directly write like this that what will happen i can directly write it like this instead of drawing the zinc and all instead of drawing this i can directly write br will give its bond to this this bond will call like this and this br will leave and you will get a very interesting molecule this is also a preparation technique of benzene anaphthalene this this this is it clear this is also a reaction when i told you when there is a conjugation attack take place such that we will see like in which case the compound become more stable now we are coming to the chemical properties chemical properties of aromatic compound chemical properties of aromatic compound what all chemical properties we have in our saliva in this so first you are trying to understand what is aromaticity what is aromaticity students it is the property due to which compound becomes compound is the highly stable this is the property of aromatic compound due to which compounds are highly stable so here you can write one thing that the property due to which cyclic and saturated compounds are thermodynamically more stable property by which by which or due to which cyclic unsaturated cyclic unsaturated compounds are thermodynamically more stable compounds are thermodynamically thermodynamically more stable more stable than than expected one than expected one whatever you expect it is more stable than that compound okay so here you can say the property due to which highly unsaturated compound behave like a saturated compound due to this reason they behave like saturated compound due to this they behave like behave like saturated compound and saturated compounds have some characteristic property that is substitution reaction so they also give substitution reaction so here write gives substitution reaction they also give substitution reaction because this is a property of saturated compound but if these compounds are giving this this thing this means they are behaving like a saturated one example example you see that if you have a hello alkane if you have let's say cs3 br plus nucleophile cs3 br plus nucleophile it gives you cs3 and u substitution reaction but unsaturated compounds give addition reaction unsaturated compound give addition reaction so so here the benzene also if you see benzene in presence of electrophile benzene in electrophile also give the substitution reaction and you get a benzene with electrophile plus h minus plus h plus okay this also gives substitution reaction now students there are two ways to tell the substitution reaction there are two ways to tell nucleophilic substitution there are two ways even the substitution is divided in two types i have already discussed those two types one is electrophilic and one is nucleophilic substitution one is nucleophilic aromatic substitution nucleophilic aromatic substitution okay and this is electrophilic i will directly write eas electrophilic aromatic substitution okay now as i already told you students this part that nucleophilic aromatic substitution is not favorable why why nucleophilic aromatic substitution is not favorable because if you have some this compound if you have a normal benzene ring and nucleophile will attack then nucleophile has to leave h minus with nucleophile nucleophile has to form a new plus h minus now the problem is this h minus is highly unstable so nucleophilic aromatic substitution is not favorable not so favorable either it form h minus or if there is a some some other okay area of hydrogen right if there is some other nucleophile that will have a double bond c this is one case otherwise if you have this compound if you have this compound cl and nucleophile why this reaction is not favorable why this reaction is not favorable tell me by this reaction is also not favorable any idea of chlorine no you can say that sir this will form this compound a nucleophile plus cl minus why this reaction is not favorable any idea there is chlorine chlorine if you see if i draw it like this chlorine double bond double bond why this reaction is not favorable the reason is it is very difficult to remove sear from the benzene ring because it has a double bond characteristic so this bond is having a double bond characteristic double bond characteristics that is why difficult to break difficult to break that is why both of these reaction here h minus is unstable here h minus is unstable so in any case nucleophilic substitutions are not favorable these are not favorable so the only reactions the only reactions which we are going to discuss is is electrophilic aromatic substitution this we are going to discuss there is a benzene ring there is a electrophile that will give you the substitution of electrophile plus h plus this we are going to discuss is it clear students now so we have discussed the reactions of nucleophilic substitution already we have covered that far reactions of nucleophilic aromatic substitution remember when anode 2 is there or a strong strong electron withdrawing group is present remember yes or no if a strong electron withdrawing group is present then the reaction will electron withdrawing group is present at ortho and para position then only reaction will proceed now let us move to the next reaction if there is ns2 okay and there is a electrophile what will be the product students quickly tell me what will be the product electrophile in presence of ns2 this you have learned last year yes plus m group and electrophile no idea students please answer how many of those who know the answer please write your answer what will happen when electrophile is taken to this where it will attack okay if I ask you the simple mission like this if there is a this this you learned last year if electrophile is taken what will be the product toluene toluene with electrophile yes opposition so students this you know that you learnt it last year that okay so I will generalize this thing I will generalize this thing whenever there is a benzene ring whenever there is benzene ring and g is present it will be now there are two cases it will be ortho-paradirecting it will be ortho-paradirecting like what all groups f plus m plus i groups then it is ortho-paradirecting if g is plus m plus i if g is plus m plus i then it is ortho-paradirecting and if g is g is meta-directing g will be meta-directing g will be meta-directing if it is minus m minus i yes or no the reason is quite obvious the reason is obvious why why it is like this students if it is like this n h2 if this is n h2 what is going to take place yes there is a n h2 you put electrophile electrophile you put what will happen if you see this electron will shift here okay try to understand your electrophile will wants to attack either on when the due to the delocalization what will happen due to delocalization what will happen this negative will appear here or negative will appear here or negative will appear here so electrophile wants to attack on these or only these positions example of many people explain it like this also if it will attack on para position let's let's have a attack on para position then i will explain see if this electrophile attacks on para position so electron will shift here and it will attack on para position so what will be the product the product will be our n h2 electrophile electrophile positive charge double bond n h2 double bond this is the case in next step it will delocalize this is called as aranium ion this is called as aranium ion so it will again delocalize and the negative will come on this carbon n h2 positive lone pair and double bond are like this and edge and electron now students your positive is getting complete octet students do you remember i told you this is a highly stable situation this is highly stable situation when the positive is getting a complete octet by the lone pair so this is octet completion octet completion step that is why it is highly stable and that is why attack is favored on para position so what will happen lone pair will be donated and the the product which you are going to get is something like this and edge 2 plus n h2 plus this this electrophile hydrogen a very very stable highly stable intermediate highly stable intermediate this is why this is the reason why electrophile attacks on para position the same thing is applicable when you draw when you write when you draw the attack on meta ortho position also when you draw same thing is applicable for ortho position same is applicable for ortho position applicable for ortho position but if you do the same thing for meta position you won't get this ion at all you won't get this ion where the octet of all atoms are complete you won't get positive charge on this atom so that octet completion can take place that is why these are ortho pair directing plus m are always ortho pair directing so i students i don't think that minus m needs to be explained is it needed or you can do it by yourself because in minus m if attack will happen on or para position negative will come on this place and if negative will come you know because it's a minus m it will be very very unstable situation like by by minus m effects are meta directing because if they are let's say here also be considered that it is a group n o 2 group is there let's say minus m and if it electrophiles attack here if electrophiles attack here so what you're going to get electrophiles attack on this attack on this so you will get something like this n o 2 positive then it will undergo delocalization it will undergo delocalization you will get something like this n o 2 positive n o 2 positive double bond double bond like this and electrophiles very very unstable very unstable situation why because positive is next to minus m group it is already withdrawing the electron and carbon got the positive charge next to the n o 2 very very unstable situation because n o 2 is a minus m group and carbon attached to it is getting positive charge so very very unstable situation that is why whenever there is a minus m group we always have we always write the attack will take place on meta position not in ortho and para is it clear is it clear students yes or no any doubt any doubt with this sign so now what is about the reactivity react benzene compound always favor the electrophilic substitution reaction at and if there is a plus m group it always favors the you can say that ortho and para position so question whenever you see this type of question g group with electrophile the product will be g and electrophile para position g and electrophile ortho position g with elect g with electrophile ortho position these will be the situation where g is plus m and plus i okay now so students let's say you have these type of reactions now one thing is very clear that these are the let's say you have this compound let's say you have this compound benzene with benzene another benzene another benzene tell me where the electrophile will attack any idea if electrophile is taken where it will attack right whenever electrophile is taken in this ring it will attack here right it will attack here this is called as this is called as kcp product kcp kinetically controlled product and whenever it is like this whenever it will attack on the next position then it is thermodynamically controlled product i will tell you why it is kcp and tcp tcp thermodynamically controlled product okay tcp why it is like this we will come to that fine so this is the electrophile how the electrophile will attack now next is fine so students if you see the question is like if you have this type of situation benzene ring there is second benzene ring and there is third benzene ring then where we'll attack then where your electrophile will attack any idea any idea where the electrophile will attack okay it will attack on it will attack on these two positions which two positions these two positions electrophile will attack on these two positions so the product will be this electrophile okay so similarly there are many other cases let's say this is one reaction this is all because of the stability of uranium ion i told you everything in the organic chemistry of aromatic compounds are decided by stability of uranium ion so which is the rhanium ion if you forgot the rhanium ion this is the rhanium ion rhanium ion is this this ion which is forming after attack after electrophilic attack the ion which is forming is called as rhanium ion we see its stability okay students please write this thing later otherwise you will forget this is the rhanium ion rhanium ion this we will see the stability of this ion and based on that we will decide where the attack will happen so let's say here whenever these things these rings are there we won't bring the aromaticity of these two rings but there are two positions all are symmetrical so the attack will happen on these positions electrophile will attack on these positions so the product will be benzene ring benzene ring and one more benzene ring and your product will be something like this okay the tcp and tcp comes like that in that compound really not everywhere okay if more than one group are present in aromatic ring now students okay so how to know because when the electrophile attack on that place the rhanium ion which is going to form is most stable this is the way to check whether attack will happen on which position okay you need to draw the rhanium ion and you will see that its stability okay so here if attack will happen on this ring okay see these are all the this dash position represents the benzylic carbon benzylic carbon everywhere wherever I put the dash these are all the benzylic carbon so attack will take place on benzylic carbon here also I will put the benzylic carbon this is also benzylic carbon benzylic carbon whenever attack happens on this the compound which you are going to get if you okay see one student has asked her why how to decide I am going to take just one example not more I am just going to take one example let's say attack happens on this position let's say attack happens on this position okay so so your positive will form somewhere here okay the above see here this is going to form right or wrong when attack will happen on benzylic carbon and let's say if attack will not happen on benzylic carbon then this is the second product this is the second possibility one minute yes so let me see positive electrophile is coming so here you are going to get the yes so positive will form here in one case but if attack will happen there positive will form in this position right positive will form here or you can say you can say if the attack will happen on meta position electrophile will come here on meta position and positive will form here like this okay done is it clear there are two possibilities only no third possibility I told you benzylic position is favored I said this is favored yes or no okay first tell me did you understand this part what I explained I'm taking this first example I'm taking this first example someone has asked sir why attack will happen on these positions why not on these positions why you are writing electrophile on these okay this is a tcp product tcp product is fine but generally we don't write tcp product tcp means when you increase the temperature then only product will come at this place tcp means on increasing temperature on increasing temperature we will get this product acp means thermo kinetically controlled product means if you don't increase temperature this will be the product only attack won't happen at this position so question is why why it is like this so reason is students tell me if attack will happen on this position there will be how many resonating structure you can draw there are so many resonating structures how this this positive is in extended delocalization with whole benzene brain right right so see there is a delocalization when we draw the delocalization so yes yes I'm taking break this is the this is one resonance structure this is one resonating structure this is one resonating structure positive is present here negative is present here there is one more resonating structure there is one more resonating structure right understand when this negative will this double bond will delocalize electron will come here double bond will come here positive will come three possible resonance structure yes or no if I keep one aromatic ring separate one aromatic ring separate three possible resonance structure now do for this there is only one possible resonance structure because this positive and this double bond is not in delocalization at all so resonance will take place between these these only is it clear only one resonating structure your this positive won't won't have resonance with double bond is it clear so less resonating structure okay I am I am I am keeping the rest aromaticity of one compound intact kinshok I am not breaking the aromaticity of one one benzene ring then I am comparing the stability of aranium ion so you can see that aranium ion is more stable in in this type of attack when attack takes place at benzylic position is it clear to all of you so attack will always take place on benzylic carbon yes wherever I didn't receive your reply any doubt okay so benzylic positions are always so now students we will have a break of 15 minutes then we will continue at 620 we will start at 635 okay students am I audible so whenever there is an attack whenever the benzene ring will attack on electrophile we will prefer the electrophile to attack on ortho-benzylic position okay students there are some other points also which are important whenever you see any reaction taking place for benzene related compound you need to take care of those things let's say here you have okay fine first question if you have something like this CH2 double bond CH2 is it a activating group is it ortho-pera directing or not just tell me ortho-pera directing or not it's matter directing okay okay what about this there is n double bond O there is a lone pair also and code double bond O is it ortho-pera directing or not and there is one more SO S double bond O O H ortho-pera directing or not fast okay fine students tell me whether these are ortho-pera directing or not or activating or deactivating quickly tell me ortho-pera directing or not activating or deactivating so fine these are see whenever there is a lone pair present on an atom ortho-pera directing means as I told you that lone pair will stabilize the radium ion whenever you have this chance that lone pair will stabilize the radium ion they are all ortho-pera directing but here since overall effect is minus M effect they will create a minus M effect so they are deactivating is it clear is it clear because their effect is minus M effect so they are deactivating ortho-pera directing but they are deactivating by ortho-pera directing because they will stabilize the radium ion ortho-pera directing and deactivating now let's say you have this situation let's say you have this situation let's say you have this one chlorine chlorine bromine iodine quickly tell me this you have done last year past whether they are ortho-pera directing or not activating or deactivating activating or deactivating whether they activates the ring towards the electrophilic aromatic substitution or deactivates the ring ortho-pera directing is because of the lone pair but they are deactivating deactivating why deactivating because their minus i culminates because they have a minus i effect is it clear it is they have a minus i effect so students in general in general what you can write in general if there is a g g plus m plus i that is g must not be halogen g must not be halogen in this case ortho-pera directing ortho-pera directing as well as activating as well as activating but it is having a more reactivity than benzene ring your benzene will have a more reactivity than the halogen x or anything which is plus m because of that it is ortho-pera directing minus i because of the due to the minus i it deactivates the ring deactivating plus m is because of ortho-pera directing but their minus i culminates like it's a halogen like in halogen they are ortho-pera directing no doubt but the point is they are deactivating due to their minus i effect and the last one is why where it has no lone pair last one is why this is matter directing matter directing and deactivating deactivating example all minus m and minus i groups all minus m and minus i this is a reactivity towards electrophilic substitution reaction this is the reactivity towards electrophilic aromatic substitution yes minus i okay in case of halogens it does not dominate in case of halogen it does not dominate is it clear in case of halogen it does not dominate minus i is more than minus m example now so one Christian can be asked like this quickly tell me what will be answer for this if there is a CH3 there is CH3Cl CH2Cl and there is CH2 and there is CH2Cl2 sorry CHCl2 students which all are ortho-pera directing and activating and which all are ortho-pera directing deactivating quickly tell me and the last one is this CCL3 answer this question ortho-pera directing deactivating or deactivating like that it decreases left to right very good fine what about ortho-pera directing thing which among them will direct the electrophile towards ortho and para position okay last part is wrong last part is wrong students there is three hyper conjugation two hyper conjugation one hyper conjugation and here you have a strong minus i 0 hyper conjugation no hyper conjugation so due to hyper conjugation it is ortho-pera directing ortho-pera directing ortho-pera directing meta directing say it is more activating so it is activating I will represent it like big A activating will represent by big A small a more small a it is actually deactivating deactivating is it clear students yes or no because CCL3 there is no hyper conjugation just negative hyper conjugation means definitely it is meta directing right so there is no hyper conjugation due to hydrogen this means it is not ortho-pera directing right yes students there you can write the TCP and KCP which I told TCP I already told you it takes place at high temperature TCP product is obtained at high temperature okay fine next is so normal reactions which we discussed when when two groups are present then when two groups are present when two groups are present here you have let's say you need to tell me where the electrophile will attack here it is most activating or activating group is present like this most activating or Fine. Most deactivating now. Most deactivating and deactivating group is present like this. Where the electrophile will attack? Within a benzene ring. Most deactivating and deactivating. And the last one. Most deactivating and deactivating. Tell me if electrophile wants to attack. Where it will attack on these benzene rings? When two groups are present. M A represent most activating. Most activating. Like you have learnt about the M effect. If NS2 is present it has a more M effect. Most activating let's say NS2. And less activating is let's say. Activating is let's say OH. It is less activating than NS2 group. If these two are present then attack will happen on which carbon atom out of these four atoms. Students once you finish it say done. One para to one. Ok fine. Students whenever this type of situation arises we will decide by activating group. We won't decide by deactivating group. So here in first case most activating is this and this is activating group. Where the electrophile will attack? Where the electrophile will attack? Decided by most activating group. Ok more activating group will decide. So it has either para ortho position or para position. So tell me where attack will happen? Where attack will happen either on this position or on this position? Yes or no? Any doubt? More activating this or this? Like this. This is activating. There are three possibilities. But tell me which is the best possibility? Here which is the best possibility? This is ortho of one para of other. This is ortho of one and para of other. Will attack happen here? Will attack happen here? Ortho of both? Will attack happen here ortho of both? Ok try to understand. According to this activating group its ortho position is this and its para position is this. According to this activating group this is ortho position to it and this is para position to it. So these two positions definitely there will be attack will happen. But will attack happen on this position? Will attack happen on this position? No. Why? Because it is hindered. So not possible. Similarly we can say for every atom. Most activating so ortho and para position these two and these two places attack will happen. Though it is a meta for other group. Though it is a meta for other activating group. But most activating will decide. Most activating will decide. In these two cases when there is activating and deactivating. Activating will decide. So activating will decide means either attack will take on this position or on this position. No other position. Again activating and deactivating. Activating will decide. But when this activating will decide either ortho position or para position. Ortho and para. But major one will be when it is like this major one will be which one? This is ortho, para. Okay fine. When there is a A and D attack can happen on these two positions. But not happen on this position because it is ortho of other. Ortho of one and ortho of other as well. Now students which one will be the major one? Here you can say this is a deactivating group. Deactivating group deactivates the position of ortho and para. So this is ortho, this is para. But according to this, this is ortho, this is para. So fine. Here you need to see the molecule, whether there is an intense or not. What about this? Attack will happen on ortho or para. Activating group ortho position. Hope you understood this. Most of deactivating and deactivating. Most deactivating will decide. Most deactivating will decide. So where the attack will happen? Tell me. Whenever most deactivating, so meta position of this. Meta position of this. Either on this place or on this place. Any attack can happen on any place. Is it clear? Okay. Now in these two places also, where are the more chances to have attack? Where are the more chances to have attack? Tell me. Out of these two. Okay, this is preferred one. So I will put up this is preferred one. Here attack will happen because it is para to other. This deactivating group deactivates its para position. Okay. This deactivating group deactivates its para position or ortho position. But here the ortho position will have a little bit hindrance. Ortho position have a little bit hindrance. So that is why attack will happen on this place. More. What about this? Most deactivating deactivating. Most deactivating will decide. So attack will happen on this. Meta to most deactivating. Okay. No other position. No other position. Here also this and this. Most deactivating will decide. These two position will be the equivalent position where attack can take place. Is it clear students? Yes or no? Any doubt in this? Fine. So students, this is about the activating and deactivating. Now, last year you have learned all those reactions which I am not going to cover is halogenation. Students, halogenation, you know that general reaction of halogenation reaction. Benzene ring reacts with anhydrous, anhydrous ClCl3, YIPS, chlorine and HCl. So I am not going to discuss that again. The mechanism and all. You all know this thing. Halogenation. I am not going to discuss. Okay. One question. Like why iodination is reversible? So Hio3 and all these are formed that is the iodination is reversible. Because those things you need to remember. Second is sulfonation. I am not going to discuss which you have already learned last year. Sulfonation. Not going to discuss. Okay. What is the electrophile in sulfonation? Just tell me this thing. Electrophile in sulfonation. Anyone? SO3 is the electrophile. Not H2SO4. Electrophile is SO3. Okay. Fine. So next is. You can take oleum and all. Okay. So fine. Next point is. Okay. If you want me, I will directly write the equation in sulfonation. What happens? There is a benzene ring concentrated H2SO4 and you will get. You think you will get sulfonic acid. That's it. SO3H. Fine. Next is de-sulfonation I will discuss. I think you have not discussed the de-sulfonation part. What is de-sulfonation? So students, I am skipping some part because you have learned last year. Skipping means I am not telling you the mechanism and all. So de-sulfonation means if there is a SO3H group and if you use dilute H2SO4, dilute HCl or steam, what will happen? This SO3H is a very good living group. This will depart and you will get a benzene ring. This is point as de-sulfonation. Okay. But if you use concentrated H2SO4, what will be the product? You tell me. If it is a concentrated H2SO4, what will be the product? Any idea? Where it will attach? Very good archit. Insocorrect answer because SO3H is a deactivating group. SO3H is a deactivating group. Students, how to decide activating and deactivating? I think someday I told you loan pair is present means it is active. Loan pair is present means it is a plus M and double bond is there it means minus M. So based on that you will decide auto and para directing. Here you don't have any loan pair. Sulfur has a double bond. So it is a meta directing. SO3H has this structure. There is no loan pair. SO3H. This structure. There is no loan pair. So it is meta directing. SO3H will bond. Is it clear students? Yes or no? Okay. Do one question. Do one question. This question is asked. Like we have this bromobenzene. We have toluene. We have toluene. But I want only bromobenzene. I want bromine substituted only at para position or let's say only at ortho position. Only ortho. Only product. Is it possible to get that? Is it possible to get that using this above equation? Whatever I explained in Sulfonation part. Think and write the answer. Ruchita no. Think. Think. Not possible. I don't think bonding is not possible in that. Double bond thing. Okay. Whenever there is a double bond. If this is attached to a benzene ring. How to check whether the group is ortho para directing or meta directing. If this sulfur atom has a loan pair. It is ortho para directing. If it is having just double bond. Just double bond. Then it is meta directing. Meta directing for double bond. Ortho para directing for loan pair. If it has a loan pair and double bond. Then it will be ortho para directing. But deactivating group. Remember I took the example of Anno. Anno has a double bond and loan pair both. Student just a minute. I will go to the previous slide. Those who don't have clarity on this. Ortho para directing and this thing. I told you this part. See here you have HSO3. They are ortho para directing. Why? Because they have a loan pair. But they are also bonded with double bond. That is why they are deactivating. If they don't have double bond we could have said that they are activating as well as ortho para directing. Loan pair tells that ortho para directing because they will stabilize the positive which will come here. During the Arrhenium ion formation. They will stabilize this positive charge. By donating their loan pair. That is why they are called as ortho para directing. Whenever there is a loan pair. But whether they are activating or not. It will be decided by whether they are bonded with the electronegative atom or not. If they are bonded with oxygen and double bond. It is a strong minus i group. Strong minus i. So minus i means it will be deactivating. I hope it is clear. Now student this question. A very easy question I have asked. Now how to use this equation. The above equation. As I told you de-sulphonation is used. So how to get this? Actually first it is reacting with H2SO4. H2SO4. SO3 is an electrophile. It will attack on ortho and para position. It will attack on ortho and para position. But since para position is major. So I will write SO3 which will attack here. Because it is a very big group. Very big group. Sulfonation gives a major product at para position. Because it is a very big group. Now once the SO3 will reach on this position. What you can carry out? What you can carry out after this? Any idea? Yes tell me. Allogenation. Very good. So when you carry out the allogenation at this place. What you are going to get? You can reuse BR2 oblique FE. Allogenation. Where the bromine will attack? Activating group. De-activating group. Activating and deactivating. Who will decide? Activating will decide. Activating and deactivating group. Both are present. Activating will decide. Attack will happen on these two positions. Bromine will attack here. CS3. BR. SO3H. SO3H. In next step what should we do? In next step what should we do? Tell me fast. What should we do so that we get this? De-sulfonation. Heat. Steam. Getting in presence of steam. So when you heat it is a de-sulfonation. De-sulfonation. Is it clear? So this reaction is used to block the para position. Sulfonation is many times it is carried out to block the para position. Why para-sulfon? It is only product because SO3 is a very big group. When it attacks here it is very very unstable. It will go to the para position fast. Because it is very big group ortho creates a hindrance. What else? Sir, FCS3. Ortho para directing. Why didn't we just brominate? Ortho para directing if you brominate you will get the product at ortho and para. I'd ask you only product I need para. I don't need. I need only ortho product. King shock. That was my question. Okay, fine. So this is a normal. Okay, we will do some other sulfonation reactions also. Students which will come in other compounds. Let's say you have this reaction. You have this reaction. Okay, first thing. If you have a CS3, a very interesting question I'm going to ask. CS3 and D2O and D2SO4. D2O and D2SO4. What will be the product? If you have OH phenol reaction with D2O and D2SO4. What will be the product? Students dilute D2SO4. D2O with D2SO4 tells you that dilute D2SO4. Dilute sulfuric acid. Dilute sulfuric acid as an electrophile. Dilute H2SO4 as an electrophile. H plus. Dilute D2SO4 as an electrophile. D plus. The product will be... D plus is an electrophile. What will be the product? D, D and D. Yes or no? Okay, next is... Here if you have a phenol. Okay, leave the second question. Later I will take the second question. First let me explain the simple sulconation. If you have this with concentrated H2SO4. Tell me what will be the product? Here you have 15 to 25 degree Celsius and in next case concentrated H2SO4 at 125 degree Celsius. Students this question will come in your phenol chapter. Done? Fine. Here you have concentrated H2SO4. Phenol SO3H. SO3 will be electrophile. Product will be... Because it is an ortho-paradirecting. Because it is ortho-paradirecting. At ortho attack will take place. Which SO3H at low temperature. At low temperature why? Can anyone tell me why? Sir you only told that SO3H won't attack in ortho position. But why you are writing like this? If someone is wondering. So tell me why I have written SO3H at this position? Very good Weber. Correct answer. Why I have written SO3H at this position? Kingshook correct answer. Because this SO3H forms. This SO3H forms. Hydrogen bonding. This SO3H forms. Hydrogen bonding with. Hydrogen bonding with. Intramolecular hydrogen bonding. What? That is why it is stable. Intramolecular H bonding. That is why it is stable. But if you have concentrated H2SO4 at high temperature this bonding will break. This intramolecular hydrogen bonding will break. Your OH will form and SO3H will come at the para position. Is it clear students? Yes or no? Any doubt? Because at high temperature. Bond will break. Now students a very interesting question I have given you. You need to think. You need to think like. If you have normal this compound. With BR2 and FE. BR2 and FE. What will be the product? Very interesting question. Which will clear all your doubts. About the. Plus M minus M. Ortho-paradirecting. Whatever things. Done. This question will come in your nitrogen containing compound. Time students. I will tell you. Students whenever you get confused here. BR plus will be the electrophile. Now where it will attack. So students oxygen has a lone pair. So it will be. Ortho-paradirecting. Activating also. And carbon is bonded with oxygen. It is having a minus M. I told you to detect the plus M and minus M. Plus M will be having a lone pair. Minus M will have a double bond. So this is a deactivating ring. This ring get deactivated. This is a activated ring. And activated ring. Ortho and para will be the product. But since there is a huge hindrance around the ortho position. Attack will take place on para. And product will be this. Product will be C double bond. This. This. This. This. This. This. This. This. This. This. This. So there are 2 , 3 concepts. Very good guy. Three. There are 2, 3 concepts you need to see. Para will be favored because it has a less hindrance. So on Para, BR will form major product. No doubt ortho will be also there, but that will not be the major. Okay, next reaction, sulfonation of amines, very, very important. Always remember, whenever you see aniline and phenol, they will have some exception. They won't show reaction as usual we get for toluene. They will show some different reaction. Whenever aniline and phenol comes in electrophilic substitution reaction, fine. Fine students, done. This is a base. Don't think that electrophilic aromatic substitution will take place. If you have written it electrophilic aromatic substitution, then you are wrong because I told you acid base reactions are fastest. So the product will be NH3 plus SO4, HSO4 minus. How many of you have written this? Anyone? Anyone who has written this? Oh, very good. HSO4 minus acid base reaction. In next step, if you heat, if you heat at 200 degree Celsius, if you heat at 200 degree Celsius, then you might get the product which you have expected. Then you get your product which you have expected that NH2 and SO3 will come at this place. Is it clear students? Any doubt in this? So next is, this is the reaction. Okay, one student asked this. Sir, what happens in desulphination? Which type of mechanism? Students, the desulphination mechanism, I will tell you. Mechanism of desulphination is called as Ipso attack. Ipso attack. What is Ipso attack? There is SO3H and H plus reacts in presence of water. Because you have taken H plus in presence of water. What happens? This electron, Ipso attack means attack on pre-occupied position. Attack on pre-occupied position is called as Ipso attack. So it will take the new electrophile that is H plus. But this atom only will attack, not its ortho or para anything. This is called as Ipso attack. So this SO3H will be present here. Hydrogen will come. Hydrogen will come like this and there will be positive charge. After some time, since this is called as Sigma complex or Arrhenium complex, Arrhenium ion. Arrhenium ion is also called as Sigma complex. I can write here Sigma complex. Now what will happen? Since SO3H is a good, SO3 is a good leaving group. SO3 is a good leaving group. It will say, okay, I am going. You stay here. I am going means I will lose. I will give my electron to this. So what will happen here? You will get SO3 benzene ring. You will get a benzene ring along with SO3H plus. In next step, it will lose H plus. In next plus, it will lose H plus. You will get SO3 glass. So this is called as electrophilic Ipso attack. Electrophilic Ipso attack. Okay, right here. This is a type of, this is an electrophilic Ipso attack. Or you can write here Ipso attack. One minute Ipso attack. Okay, sir. Can there be any nucleophilic Ipso attack? Yes, there is something called as nucleophilic Ipso attack. Right next thing. And these are all under the sulphonation topic only. Next is nucleophilic Ipso attack. What is that? In nucleophilic Ipso attack you have let's say nucleophilic. Same thing, SO3H will be there. SO3H will be there. Nucleophile will come and attack. So one minute, one, one, one, one. Here you have nucleophile. Let's say you use any OH. If you use any OH, your nucleophile can also attack. Very specific reaction. Don't apply it everywhere. What happens? What happens? First any OH is taken. So acid-base reaction will take place. Acid-base reaction will take place so it will form SO3-. First is acid-base reaction. Okay. Once acid-base reaction takes place after that what will happen? Here your any OH is present. And this OH-, this OH- will attack on this position. And when it will attack on this position instead of, okay fine. What will happen? This electron will shift on this position. In next step, instead of getting positive charge, here you are getting negative charge. Because it is nucleophilic Ipso attack. Attack is taking place on that atom only. Where the living group is present. SO3-. Like this. SO3-. There is OH- also present. OH- also present. This, this. In next step. In next step because it is a strong Electro-negative atom and a good living group. This will go like this. This will go like this. And the product obtained will be this. You will get Phenon. Oh my God. Students this is one of the technique to prepare Phenon. You will learn in NCRT but no one will tell you that this is Ipso attack or whatever attack. But I think now you can easily make it out that Phenol is prepared by nucleophilic Ipso attack. From SO3-. So if someone say write the product, students can you write the product quickly tell me what will be the product if KCN is used.  Just write the final product and tells are done. Once you write, say done. Fine. How many of you have done this product? How many of you have written this? Very good. Very good. Very easy one. Right. Next is now students next reaction is Nitration. You tell me student what is Nitration mixture? Nitration mixture is Yes. Yes. Nitration mixture is we have already learned this last year. But I am taking all these reactions for Phenol and Emily. Nitration. For Nitration the reactants are first the general reaction. This is general reaction. Concentrated HNO3. Concentrated HNO3 and H2SO4. Concentrated HNO3 and H2SO4. Concentrated H2SO4. Nitration mixture and you will get a nitro benzene. This you learnt last year. Okay. So here if someone wants to know how Electrofile is generated. First step is always Electrofile generation. So first step is generation of Electrofile. Generation of Electrofile. Okay. So this is generated. But still I will tell you those who are who don't know this. Because Concentrated H2SO4 is a strong acid. It behaves like acid. It will take and your nitric acid will behave like base. Nitric acid will behave like base and acid base reaction will take place. You will get HHO NO2. Acid base reaction will take place. You will get this positive here. In next step. Oxygen is getting positive. Oxygen will say okay. I am leaving as a water. And NO2 plus will be the Electrofile. If anyone wants to know how Electrofile is generated. This is the process. NO2 plus is the Electrofile. Okay. Now what all other reagents can be used? Reagents can be anything else also. Reagent can be only Concentrated HNO3. Reagent can be this is B part. B reagent. C reagent is HClO4 and HNO3. Sometimes this is also written in textbooks. D HNO2 in presence of HNO3. Very rarely these are written. Most of the time you will see the A part only. But these are also the reagent. Sometimes it is written. Dileute HNO3 also. Dileute HNO3. But Dileute HNO3 work for phenol. Or phenol nitration mixture is Dileute HNO3 also. Because phenol is highly activated ring. You don't need concentrated HNO3. So these are the few things. Which are given as a. This thing. Sometimes you know very interesting thing. If someone wants to. Sometimes it is also given. HNO2F plus BF3. So HNO2F plus BF3 is nothing but. It is HNO2 plus. And BF4 minus. So it is also giving you the HNO2 plus. HNO2 plus electrophile. So overall we want HNO2 plus electrophile. That is why this whole. Thing is going on. Okay. Let us start doing the questions in nitration thing. Students normal nitration is this. If you have benzene ring. Nitration can okay fine tell me students. If you use concentrated HNO3. And conch H2SO4. No doubt the nitration will give you this product. No doubt. But. But this reaction takes place at 60 degree Celsius. But if you carry out this reaction at 90 degree Celsius. Can you tell me what will be the product. And if you carry out this reaction at 120 degree Celsius. What will be the product. Temperature students. Think okay fine. I will help you a little bit. This thing will further react. This thing will react with HNO2. At high temperatures. The product formed in first reaction will further react. Tell me it will direct HNO2 to which position. This compound A will direct HNO2 to HNO2 to which position. Very good Ruchita correct answer. Gayatri correct answer. King Shoth wrong answer. Archit correct answer. HNO2 is a deactivating group. HNO2 is a deactivating group. Ring become highly deactivated. But the temperature will help. For further reaction. High temperature will help for further reaction. And the deactivating group is always. Wrong answer. Because HNO2 is a deactivating group. It will. Attack will happen on meta position. The best position for electrophile. When any deactivating group is present. Because electrophile. Does not want to go on those positions. Which are highly deactivated. Ortho and para positions are highly deactivated. Due to this HNO2. But in meta position. There is only minus side effect of HNO2 is acting. That is why your electrophile attacks on meta position. Sthoti. Sthoti and King Shoth. Did you understand. Yes or no. Not Sthoti. Fine. High temperature. Tell me. Now it become a highly deactivated ring. But still if you increase the temperature then. Your this compound will further react. Both are same. Now think deactivating and deactivating. Both are same. Very good Ruchita correct answer. Archit correct answer. Gayatri correct answer. Fine. Students one more meta position will be occupied by HNO2. Is it clear. Let us have a good. Let us have one question. If you have this reaction. Toluene. If toluene is reacting with. Concentrated HNO3. Plus concentrated H2SO4. That A compound is formed. Then it is reacting with alkaline KMNO4. Alkaline KMNO4. Gives B compound. Which will react with. NaOH and CaO. NaOH and CaO. NaOH and CaO. Gives B compound C. Predict A, B, C. Predict A, B and C. Done. Fine. So students answer our first question is. Concentrated HNO3 you have taken. Okay. So here the product will be. You need to do one by one. Step. Concentrated HNO3. So can I say. Nitration mixture. CS3 is taken. So what is going to form. What is going to form. Okay. Can I say how many of you have written this. CS3. HNO2. HNO2. And HNO2. How many of you have written this. Try nitro toluene. In next step. Alkaline KMNO4. I think there is a methyl group. That will oxidize to carboxylic acid. Today in today's class only we discuss. HNO2. HNO2. HNO2. That is my Alkaline KMNO4. In next step. HNOH and CaO. Which reagent is this. Which reagent is this. Soda lime. Remember soda lime. Soda lime. What is the purpose of soda lime. Decarboxylation. Decarboxylation not E2. Decarboxylation. COH group will be removed. Yes. No, no, no. Mechanism of soda lime is EI mechanism. Remember. EI. Last thing which we discussed. Not E2. Intermolecular elimination reaction. Intermolecular elimination reaction. Intermolecular elimination reaction. Intermolecular. So this is the answer. How many of you got all the three answer right ABC. Just write me. Fine. Okay. So intermolecular elimination soda lime. COH will be removed. Students what will be the product of next two reactions. These are the last questions of today's class. Okay. As I told you whenever aniline and phenol will come. Reaction won't be as simple as you have expected. Reaction won't be as simple as you have expected. So I will write for first. You have expected. Okay, sir. Aniline will come. Aniline will come everywhere. But the point is. The point is. The product of the first reaction is. NH2. Para position. 51%. Para position is 51%. Okay. Aniline will come here also. Some product. And aniline will come. Some ortho product also. It is 2%. Your metaphors. Percentage will increase. Meta will be 41%. Okay, sir. Why sir. If someone is thinking why it happens. Students. Why not acid based reaction. Yes. Actually first only acid based reaction going to take place. First acid based reaction takes place. So you will get NH3+. Okay. Student first acid based reaction will take place so that you get NH3+. Any doubt. Because this is base this is acid. But what happens after that this become a deactivating group. This become deactivating group. And once it become the earlier it was activating group. And as a lone pair. So once it become a deactivating group as well as meta directing. Because there is no lone pair left with it. It become meta directing. That is why your meta product is huge. Is it clear students. Yes or no. Sir but why not leaving group. Where is the leaving group idea. But why not leaving group. Where is leaving group idea. But why not leaving group. Where is leaving group. How NS3 will leave. No, no, no, no. Double bond characteristics. Double bond characteristics. How it will leave. I told you substitution reaction is not possible in benzene. Very few examples are there. Ipso attack is one of the substitution reaction of nucleophile. Otherwise everything is electrophilic substitution. And electrophilic substitution takes place at the end of substitution. Why you are going to have a Ipso attack here. Ipso is just for SO3H. Is it clear. So these are the exceptions. Whenever aniline phenol will come they will show some exceptions. Okay fine. Student what is answer of 2. This is the last question. I will write. The one product because phenol is a highly activating group. The one product is this. 2, 4, 6 tri nitro phenol. No doubt but it is a minor. This is minor. This is also called as picric acid. Picric acid. Picric acid is a minor product but the major product is this. Oxidation is the major product. You know the phenol will get oxidized. No one has thought that concentrated HNO3 act as oxidizing agent also. No one has thought that okay fine. Very good question you have asked. Very good question you have asked that sir why even para product is formed. Why even para product is formed because your amine is not that strong base that reaction goes always in forward direction. It is having equilibrium with NS3 also. Like this is a equilibrium reaction. Don't think everything becomes NS4 plus. Because it is not very strong base. So most of the part is ammonia only. Is it clear? Reaction is mostly toward backward direction. So students these are the compound which are going to form. Okay please remember all these things. These are very important. Okay now students I am telling you very important thing. Now after few minutes the assignment allotted to you for reduction and oxidation reactions. Please complete your assignment without fail and I will put comment after checking your assignment before next class. Don't forget to submit your assignment. That's all for today. Thank you bye have a nice day. If you have any doubt you can stay back. Okay students thank you.