 Bromination of phenols is a classic example of electrophilic substitution reactions. And it is also one of the common intermediate steps in a lot of synthesis reactions. So let's check out how bromination happens in the case of phenols. But before we go to the bromination, let's look closely at this bromine molecule. So as you know bromine has 7 valence electrons. This bromine will also have valence electrons and between them they share this pair of electrons and that is how the bond is formed. So for the bromination reaction to happen what we want is we want to break this bond. And when bromination is performed we usually don't add bromine directly to whatever we want to react it with. Typically what we do is bromine is a liquid so we'll add a small amount of bromine to a solvent. And then we use this solution for our reaction. So in this case for bromination we are going to be using bromine water which is basically we'll add a small amount of bromine to water. And we're going to use that solution for our bromination. Now the thing to note here is what exactly is happening because of the water. One of these bromines will get an electron and the other one will get a positive charge. So let's assume that this bromine will get the electron. Even when it has not fully got the electron, let's say that it is pulling the electrons more towards itself. So it will have a slightly negative charge. And because the electron is away from this bromine, this one will have a slightly positive charge. But in the solution this bromine is not alone, there's water all around it. So what happens because of the water? Now water is a polar molecule which is to say that because the electronegativity difference of oxygen and hydrogen is high, oxygen will tend to pull the electrons towards itself which is why it will have a overall negative charge and the hydrogens will have a slightly positive charge. And this is also why we call water a polar solvent. So this bromine is surrounded by water and look at what's happening because of the bromine. So this bromine has a slightly negative charge and there are two hydrogens here which have a slightly positive charge. So this water is essentially pulling this bromine away on this side because this bromine is positively charged. This oxygen is pulling this bromine away and you can see what this water is doing to the bromine. It is helping the split of this bromine and that essentially is the role of the solvent here. So because water is a polar solvent, it has both positive and slightly negative charges on different atoms because of which it is easily able to pull this slightly positively charged bromine and this slightly negatively charged bromine apart and that helps with the reaction. So now with this background, let's see how the bromination takes place. So to start the reaction of phenol with bromine using bromine water, we already know something about the phenol. This oxygen has lone pairs of electrons that it is going to add to the ring thereby activating this ring and we know that when we break these double bonds and draw the resonance structures, the electron density is higher at the ortho and the para positions which is why the OH group is called ortho-paradirecting group and if we come to the bromine here we know that this bromine has been dissolved in water and the water is pulling apart this bond because of which we have this bromine which is say positively charged and this bromine which is negatively charged and this happens when water has helped in breaking this bond. So now what happens? If we break this double bond, we will get a negative charge here and this will get attached to the bromine but what has happened is when we broke this bond, the aromaticity of this ring was lost and we also know that there is a hydrogen here. So what is going to happen is this hydrogen is quickly going to be kicked off, the double bond is going to be formed again, restoring the aromaticity and this H will float around and find the bromine with a negative charge and it is going to form HBr. So what we will see is that this bromine has got added to the para position but we know that the OH group is ortho-paradirecting. So we will not just get the bromine here. The bromine will also get added to both of the ortho positions and what we get is 2, 4, 6 tri-bromophenol and this reaction is fun to watch in the lab because you can actually see the change in colors. So bromine water has a reddish-yellowish color like rust and as soon as you add the bromine water to the phenol and the reaction is complete the liquid turns colorless and there is a white precipitate at the bottom and this white precipitate is our product here which is 2, 4, 6 tri-bromophenol. So now the question is what if we want bromine only at the ortho position or only at the para position? What changes can we make to get these products? So let's see how we can do that. So earlier reaction you remember what was the role of the polar solvent that is water. It was helping break these BR-BR bonds easily because it was pulling both of these bromines apart. Now if we want only mono-bromophenols and we want them to be added only let's say at the ortho position or only at the para position, one way to do that would be if we don't allow these bromines to break apart easily only few of them will be available in the solution and the phenol is reactive. So it will want to attach with the bromines that are available but only if very few bromines are available maybe we will not get multiple substitutions on the same phenol. So let's try this. So what we are going to do is we are going to change the solvent we are going to use CS2 or carbon disulfide as the solvent and if you look at the electronegativity values of carbon and sulphur they are quite close so CS2 is a non-polar molecule and thus it's a non-polar solvent as well but it will not easily pull apart both of these bromines and effectively there will be very few bromines available for the reaction and also let's just do one thing let's perform this reaction at a low temperature so with these two conditions let's see how the reaction proceeds now we know about phenol that the oxygen has lone pairs that it is going to share with the ring and we can see by drawing the resonance structures that the electron density is higher at the ortho and the para positions and this time the solvent is CS2 which is non-polar so it will not really help in pulling apart these two bromines but even then for a few of them that break let's say this one has a positive charge and this one has a negative charge if we break this double bond we get a negative charge here there's a negative charge, there's a positive charge this immediately bonds with the bromine here but as soon as this bromine is now attached this ring has lost its aromaticity so there is a hydrogen here and it will easily kick off this hydrogen and regain its aromaticity and this H plus will find a negatively charged bromine and form HBr so this is similar to what we saw in the previous reaction but this time we will get different products and because I've drawn two you already know that there will be two products so what are the two products? one is when you get the bromine attached to the para position and one is when the bromine will be attached to the ortho position so we have two products para-bromo phenol and ortho-bromo phenol and one more thing to note here is that the para product is the major product and the ortho product is the minor product two things happened here first because we used CS2 which is a non-polar solvent and we performed the reaction at a low temperature this bromine-bromine bond was not broken up easily and we had very few bromines available for the reaction and on the other side there were too many phenols and all of them are electron rich and as soon as they see this bromine electrophile they quickly go and grab it which is why you see the addition of bromine at either the para position or the ortho position and because the bromines are few in number it is very difficult to get two bromine additions on one phenol itself which is why we don't see a product which is 2-4 bromofenol so that's one thing now another question that we can ask is why is it that the para product was the major product here and to understand that think of the minor product now if you look at the intermediate steps there was something which prevented the addition of bromine here and made it difficult and that difficulty was not there at the para position now if you look closely bromine has a lot of valence electrons and oxygen also has electrons so when the bromine will approach at the ortho position there is going to be some repulsion between both of these which makes the bromine difficult to attach at the ortho position and no such difficulty is present at the para position so the bromine will easily get added at the para position which is not possible in the ortho case and if you remember from the earlier reaction it's not like the addition at ortho is impossible earlier we got a tri-bromoproduct in which the bromine got added to both the ortho and the para position so then the idea is that it's a combination of both of these factors we had very few bromines available for reaction and it was more difficult to get the bromine attached to the ortho position which is why we got a para major product so what I want you to take away here is that the solvent had a very important role to play here in the bromination reaction in the earlier reaction we saw how water being a polar solvent made it very easy to break this bromine bond and a lot of bromines were available which is why we got a 246 tri-bromophenol product which is the white precipitate and as soon as we changed the reaction conditions we introduced a non-polar solvent and we performed the reaction at a lower temperature only one bromine got attached to the ortho or the para positions and even there because of this difficulty in attaching the bromine at the ortho position because of the repulsion we got a para product which is the major product and the ortho product was the minor product