 So the most important thing that I want you to know about coal-based reaction is that we're going to be starting with a phenol and our end product is going to be salicylic acid which is basically a COOH getting added to the ortho position and this reaction is important for many reasons. One of which is that salicylic acid is used to make aspirin which is usually prescribed by the doctors when you have minor pains or fevers or things like that. So that's why this reaction is important. So I've told you what the end product is but I haven't told you what the reagents are or how we get from here to here. So let's see that. So I've broken down the reaction into two steps. The first step is reaction with NaOH and the second step is reaction with CO2 and the addition of an H+. And you can probably see what's happening in these reactions. Like in the first step NaOH is a base and phenol is an acid. So this is an acid base reaction and in the second step there's a CO2 here. So maybe this is what is going to be a part of the COOH here. So let's see how this happens. But before we go to the reaction there's something to know about phenol. So as you know this oxygen here has lone pairs that it is going to share with the ring because of which the electron density in the ring increases and because of resonance the electron density is higher at the ortho and the para positions. So let's see how the resonance happens. Now if we break this double bond we know that there'll be a negative charge here and a positive charge here but there's also an oxygen atom here. So what's going to happen is with the lone pair on oxygen a double bond is going to form here and there's going to be a positive charge on this oxygen atom. Next what we can do is we can break this double bond because of which there will be a positive charge here and a negative charge here and you can see that the negative charge here from before and this positive charge forms a double bond here. And lastly if we break this double bond we can see that there will be a positive charge here and a negative charge here. So this positive charge along with this negative charge will form a double bond here. So these are the resonance structures of phenol. Now if you follow the negative charge here across all the three of these structures you'll notice how the electron density is higher at these ortho positions and this para position which is why we say that the OH group is an ortho para directing group. So that's one thing about phenol. So we saw how the phenol is ortho para directing. Now let's think about what happens if this hydrogen is taken off. So if we take off the hydrogen what we get is this phenoxide anion. Now the idea here is that once we have taken off the hydrogen and there is a formal negative charge on this oxygen atom it makes it very easy for the resonance to happen because here in the case of phenol it was the lone pairs which were contributing to the resonance. So you always had a positive charge on this oxygen but now with the hydrogen gone this negative charge will make it way more easier for the resonance to happen. Effectively what is happening is the phenoxide anion will be more reactive or we could say that it is more ortho para directing because this negative charge is easily transferred to these ortho and para positions. So now with this information that the phenoxide anion is more ortho para directing or more reactive you can probably guess what the first step of the reaction is going to be. So I told you before that the first step is reaction with NaOH. Let me just write it down here and you can already guess what's going to happen. So because we want a phenoxide anion what this base is going to do is going to take off the hydrogen from here. So what we're going to have is we'll have a negative charge here and there will be one Na plus here and this is the first step of the reaction. So now at the end of the first step we have got this phenoxide anion which has a negative charge and it is more reactive than the phenol that we started with and because we have this negative charge that is distributed to the ortho and para positions this is now ready to attack in the next step. Let's see how that happens. So the second step started with reaction with CO2. So I've just written down CO2 here. It's written in a slightly different manner but essentially what we're doing is we are initiating the reaction with CO2 here and this is the phenoxide anion from the earlier step which is more reactive. So think of what's happening here. Now CO2 is a weak electrophile but we do have this oxygen pulling electrons away from the carbon and the carbon having a slightly positive charge that's developed here and also there is this interaction between this sodium and this oxygen. So maybe it's just helping it align a little bit and getting it ready for the reaction and then what happens is because it's a phenoxide anion let's say we break this bond and we get a negative charge here and a double bond has formed here and because the bond is broken so it will immediately go and get attached to this carbon here and now what happens is this CO2 is here and there's also a hydrogen here and something drastic happened in the step. If you notice the aromaticity of the ring has broken so to quickly get back to the stability of the aromaticity what happens is this etch is thrown out and this bond is completed so then what happens we see that the etch has got added here the CO2 is here and the double bond is formed restoring the aromaticity of the ring and of course there'll be a negative charge here. So now it looks like we are almost there there is some small thing missing so let's do one thing let's provide it some etch plus so if we go ahead and give it some etch plus it'll just go and attach itself here and we have our COOH here which is added to the ortho position so that completes our reaction we started with a phenoxide ion which was more reactive we got a weak electrophile for the next step the CO2 was here and then we got COH finally when we added a little bit of etch plus but the question is you may remember from before that the OH group was ortho-paradirecting so the electron density had to be higher at both the ortho and the para positions which means we should have got a para product as well so where did it go well it is there we will get some para product as well but what we'll find is the ortho product is the major product let me just write it down here so this is the major product and the para product is a minor product let's try to understand why this happened so usually when you have a major product which is ortho or which is para the reason for this is that there is something in the intermediate steps where the formation is being stabilized and here if you remember I told you that this sodium has the some sort of pulling effect for the oxygen which is sort of setting it in place and that makes it slightly more stable which is why we see more of the ortho product now sometimes it is hydrogen bonding which does this or any such sort of interaction but basically that is one reason why we got the ortho major so now to bring together everything we started our coal base reaction with phenol and the first step was to make it more reactive so that it can react even with a weak electrophile like CO2 and for that we used a base which is NaOH and what we saw was initially we had this O negative Na plus forming and this was our phenoxide anion which we used for a second step our second step was reaction with CO2 and two things happened because this is more reactive and because this sodium was sort of directing the oxygen from CO2 it got aligned towards the ortho position and the electrophile quickly bonded with the carbon and there was a hydrogen here which got kicked out and we got a double bond O and O negative here and finally when we added the H plus this negative charge was replaced by a hydrogen and that is how we went from phenol to cellosilic acid by reacting first with NaOH and then by CO2 and then throwing in an H plus from an acid