 Next reaction you write down, heliform reaction again very important, the reason we use for this reaction is X2 with in basic medium NaOH, in general we use I2, why we are calling it as heliform reaction because one of the product of this reaction is CHX3, CHX3, CHX3 where this X is what it can be chlorine, it can be bromine, it can be iodine, chlorine we do not use for this reaction, okay because carbon-chlorine bond is, bond is very high compared because the size is very less than chlorine, in comparison to the other three have chlorine bromine and iodine, so possible one of the product of this reaction is CH, CH, CH, VR3 or CH, I3, I3 we call it as iodopharm, this is chloropharm and bromopharm, right, you must have seen the reaction in which they ask the question like in which of these compounds shows iodopharm test, right, so iodopharm test is generally given by methyl ketone, right, there are other conditions also we will see that, but one thing is very important whenever the molecule contains C double bond O CS3, it gives iodopharm test, okay, even 2 degree alcohol will be oxidized, okay, 2 degree alcohol if you have under oxidation it may give ketone, right, it can give ketone, so 2 degree alcohol under oxidation, if it gives methyl ketone that also will show iodopharm test, so what are the conditions we have that we will see, first of all just to keep in mind methyl ketone if it is present it shows iodopharm test, okay, now the reaction is what in this, you see suppose we have R C double bond O CH3 and it is allowed to react with I2 with NaOH, I2 with NaOH, basic medium, sometimes they write simply OH-1, right, so you see in this reaction this OH- takes H plus from this carbon, acidic hydrogen is because this is electron with the drawing, so first step of the reaction is R C double bond O CH2H with OH-, this OH- takes this H plus and we get here what, carbon NaOH, okay, so the product of this reaction is what, R C double bond O CH2 negative plus H2, okay, now in the second step what happens, R C double bond O CH2 lone pair negative this attack onto this iodine molecule I2, okay and when it comes over here this lone pair of electron this I- goes out, okay, so product is what R C double bond O CH2 I plus I- which may come minus to this Na plus and forms NaI, okay, this is the first product of this reaction and generally for this reaction is I2 minus and NaOH we take in excess, okay, in general, if it is written it means it is an excess amount we have here, so suppose this reaction rate, suppose this rate rate is R1 we have, okay, for this reaction the rate is R1, in the second step what happens, now the molecule is this R C double bond O CHI hydrogen again this base takes this H plus from this molecule, right and it again forms what R C double bond O CH negative charge I plus H2, sorry, now next step what happens R C double bond O CH lone pair negative charge I it again attacks onto this iodine molecule and it converts into what same reaction actually, one more step we have R C double bond O CHI2 plus we will get I minus suppose the rate of this reaction is R2 and third step can you write down again this hydrogen goes off and we get C I3 over there and instead of CH I2 we will get C I3 yes or no, right, so all these hydrogens are acetic hydrogen that's why this reacts consecutively with this iodine in this region, so third step of the reaction is R C double bond O CHI and one hydrogen we have here and base again takes this H plus forms H2O and the product is R C double bond O CHI2 negative plus H2 and in the last step R C double bond O CHI2 negative plus I forms R C double bond O C I3 plus I minus, okay, copy down this, this step is this rate is R C. Now this C I3 makes this carbon carbon bond very weak, okay, this C I3 because it is electron withdrawing this bond becomes polar now, right, this bond is very weak now it can dissociate easily, right, so the next step what happens the OH minus again attacks on to the carbonyl carbon of that, okay, so next step of the reaction R C double bond O C I3 plus OH minus this OH minus attacks on to this carbon and this electron field goes on to this oxygen so it forms R C O minus OH C I3, okay, this lone pair comes over here since this bond is very weak the C I3 goes out as a leaving loop, okay, so what we get here R C double bond O OH plus C I3 minus yes or no, right, okay, now the medium is basic, so basic medium we won't get acid but what we get, we get carboxylate ion, right, so from this what happens protonation of this C I3 minus proton exchange because the medium is basic we won't get acid final product of this reaction will be R C double bond O O minus plus CH I3, Idoforms which is yellow in color, so this is Idoform reaction, so overall the product of the reaction is what with this we will get R CO minus and CH I3, this is what the overall reaction is, okay, now can you tell me R1, R2 and R3 which one, the rate is maximum for which reaction, R1, R2, R3 which one has the maximum rate of the reaction, why R1, why R3, it's with drawing electron, right, you see that's the right, R3 is the maximum rate, this is very fast reaction, R3, R2 and R1, why this is there because we have to take this H plus OH and this I is what electron with electron, right, so consecutive step the rate of the reaction increases reaction becomes faster because of the minus I effect of iodine, okay, removal of H plus is easy, right on heloform test is given by the molecule which has which has first one, which has first one methyl ketone group, methyl ketone group CH3 double bond of CS3, for example if I write down acetaldehyde CS3 C double bond OH, it gives Idoform reaction because we require what, we require acetic hydrogen, right, again you see this molecule CH3 C double bond OH CS3, again we have methyl ketone shows Idoform test, this molecule C double bond OH CS3 also gives Idoform test, okay, so wherever we have C double bond of CS3 ketone group present, methyl ketone group present, it shows Idoform test, all these molecules show Idoform test, second one write down, second case all alcohols which on oxidation gives methyl ketone group, all alcohol which on oxidation gives methyl ketone group, for example you see 2 degree alcohol, RCH OH CS3, under oxidation what it gives, RC double bond O CS3, this also gives Idoform test, okay, no 2 degree, 1 degree also forms sometimes, 1 degree may also form because this is 1 degree alcohol under oxidation, what it gives, aldehyde and what is the aldehyde formula here, CS3 C double bond OH, again you see methyl ketone group, okay, so 1 degree 2 degree alcohol may give this test, yes, because if you further take CS3 CS2 CS2 OH that won't give, because we have ethyl, then over there, okay, next point alkyl halide which forms alcohol on oxidation, alkyl halide which forms alcohol on oxidation which forms alcohol just to write down, may also show heloform reaction, because when it forms alcohol and then further it oxidize into aldehyde or ketone, one very important reaction we have in this, write down the product in this reaction, CS3 C double bond O CS2 C double bond O CS3 I2 NaOH, write down the reaction, it depends on if it's in Hs or not, maybe both okay actually, but then both will come, about to pass and you drink it yourself, I think they will put the ring, the ring with 3 colors, I do it, so there will be an oxidized concentration, maybe 4 or 3, the amounts of the acetic is the center one, acetic hydrogen is the center one, so OH- will take that one, not the side one, so central hydrogen will go, so you will get CH2 I2, active methylene go, so you will get CH2 I2, that one will go and I will now have extra hydrogen, that hydrogen can go again, also COHO, what is the product, first of all the reaction takes with a rich target, here or here, this one right, so the product if I write down and these two hydrogen get replaced, it will be CS3 C double bond O CS3, okay next what will happen, OH- will attack here okay, if this alkyl group is different, then what will happen, suppose here I am taking ethylene and this is ethylene, then attack where will it happen, ethylene will take a case of mode plus I, this will reduce the posture density here, so it will attack here, if it is the same, answer will be same, correct, so this OH- suppose attacks over here and this bond breaks right, so the product in this reaction will be CS3 C double bond O, CI2- plus CS3 C double bond O, okay, now this H plus will, because acid won't form, so this will rearrange itself, right, so this will be CS3 C double bond O, CI2- plus CS3 C double bond O, okay, but reaction here will not stop, okay, so finally after getting this molecule again reacts with what, OH- first which takes this hydrogen, okay and it forms CS3 C double bond O, CI2- plus H2O, okay and with I2 what it gives, CS3 C double bond O, CI3 with same mechanism we have discussed already, right and then what happens with OH- this will attack onto this carbon, this goes out as living proof, so it forms CS3 C double bond O, OH- plus CI3 minus, so it finally converts into CS3 C double bond O- plus CH, so it also continue with the CH3- COO, will the reaction continue? Yes, but actually because of this I2, this ideal is more acidic, so after the CH3 has been formed that will continue with those two CS3 C O minus also, right, after this the reaction rate is very slow because this becomes resonance stabilized, possible one-two step course of fire will happen, but this is resonance stabilized, this will reaction won't take this much, H plus our gas will come out, so we cross resonance will happen, okay, so usually after this ion that reaction won't take this, because the rate becomes very slow, understood this point, so overall the product of this reaction is what, we get very unusual product here, the product is CS3 C double bond O O minus, strike two molecules of this plus what we get, CHI, very important reaction, okay, molecule does not look like that it gives I2 form this, I2 form reaction, right, so I2 form is forming over here after this all these entire reactions, that's why this molecule also gives I2 form test, right on one note, ester and carboxylic acid, esters and carboxylic acid does not give I2 form test, does not give I2 form test, okay, so point you have to keep in mind is active with helen group if it is present, with helen group present, then this will also show I2 form test, okay, so this kind of reaction if you see, CS3, CHOH this molecule, CH2, CHOH and CS3, this is 2 degree alcohol, this will also show I2 form test because oxidation being ketone, C double bond, and then called the same reaction, so this molecule also shows I2 form test, if I talk about this molecule double bond O and double bond O, does this show I2 form test, it does, because it is something like this, active with helen group we have here, I2, I2 and then, because your product gives O, right, right on the product of this reaction, but this molecule shows I2 form test, that's how it will be first, chicken, of the whole here also, OH minus, did you understand this, this one is here, okay, so that H1 will be there, that H1 will be there, that bond will go and that will be O minus, this will be, so there will be CHI2 that bond will go, bond between the CHI2 and double bond, see I2 will be there, this bond won't be there and we have one negative charge here, okay, and then what will happen, it's deprotonation, it's a protonation, is it, I3 can't be done, so I2 only, and then it's protonated and then it will again go through the reaction and then do it, so it's deprotonation will happen, then what we get, double bond O and we get here CHI2 and this will be what, double bond O, O minus, again its reaction will go OH minus I2 with CHI3, CHI3 minus bond will be there, what is final product, final product I think we will get something like this, double bond O, O minus, yeah, O minus and double bond, is it something like this? Yes sir, this is CHI3, one carbon is gone with CHI3, so this carbon is not there, okay, this molecule also goes, so this active methyl in the example is very important for example, this question is very much asked in the example, which of these group shows idophon test, very common question, okay, whether it is J or any of the exam, very common question, write down the reaction, the product in this reaction, double bond O, C double bond O, tell me the final product, I2 with NaOH, I2 with NaOH and we are heating this also everywhere, okay, then only the CHI3 minus goes out, ready for Metski reaction we did, it's the name reaction we are almost done, today we will finish this, because we are doing the entire carbonyl compound, we are doing carboxylic acid reactions also, so aldehyde ketone is almost 98% is done, okay, just 2-3 reactions in the test, tollins, benedict and all we have, that will face today and I don't know, let's see how much we can finish carboxylic acid today, but by next class we definitely finish acid and then start amines also, 50% amines also will be finished by next class, when do you have exam? Next week, okay, what is the product here, this, here it will come, so I think it's, oh, why, that's all bond Methyl ketone, C double bond O, O minus and CHI3 will be the product, no, no, okay, but isn't the hydrogen between the two more acid than that side, yeah, but that reaction won't proceed after that, yeah, there'll be one, you get one I here and then the reaction won't proceed, yeah, and when this hydrogen gets replaced, the minus I effect of iodine, drag the reaction, see there are possibilities, you cannot nail it this, because this hydrogen obviously it is more acid than this hydrogen, so this hydrogen also take part in the reaction, but when they ask you this question, you do you have no other choice, you have to go with the iodoporm test, okay, so that's all, because again the reaction here also takes place, we get one iodine here and then the reaction won't proceed because there is no other driving force, but when this hydrogen replaced by this iodine, minus I effect of this will, you know, make the other two hydrogen more acidic and it comes out easily in express, okay, okay, so preparation we have done for this anti-hygidome, okay, we have finished on the new reactions, so what kind of am I in derivative by reaction protection of C double bond over diols reaction, diols reaction, diols reaction, diols reaction, the reaction with ammonia is derivative we finished and then we finished the diols, so we did two production, we did two production, and then you have the diol protection, non-reactive in this acidic medium, sorry, basic medium, okay, production of C double bond,