 Bromination reactions are a type of electrophilic substitution reactions and it's one of the most commonly performed experiments in the lab. So let's see how bromination of aniline happens. So we already know something about aniline. It has a lone pair here that it is going to share with the ring and because of resonance we know that when we shift this pi bond the electron density increases at the ortho and the para positions. Or in other words we know that this NH2 group is ortho-paradirecting and the next important thing to note here is about this bromine. So we know that in this reaction we use bromine water and what is a polar solvent? So it helps easily break this bromine-bromine bond and what we get is a bromine with a positive charge and this bromine ends up with a negative charge when this bond breaks. Because we know that NH2 is ortho-paradirecting and this bromine is our electrophile we find that the bromine gets added to both of the ortho positions and the para position and we have a product which is 246 tribromoaniline. So this reaction proceeded as we expected and the bromine got added to both of the ortho and the para positions. But what if we want only one bromine addition? The idea here is that this NH2 is activating the ring. It is sharing its lone pair making this highly reactive and also we use bromine water. So water being a polar solvent is easily helping break this bromine bond. The question is can we somehow modify this process so that we have only fewer bromines available? Let's see if trying a different solvent helps. So what we're trying differently is we are using a non-polar solvent this time which is this CS2 or carbon disulfide and what it's going to do is this is a non-polar solvent and what the solvent does is it helps break this bromine bond. The more the polarity more the dissociation because the polarity helps pull both of these bromines apart. But in the case of CS2 because it is non-polar it will not easily break these bonds and so we'll have less of the electrophile which is this bromine with a positive charge available for the reaction. And also we're going to perform this reaction at a low temperature. So what we're trying to do is we're trying to make very few bromines available in the reaction and since the electrophiles will be few in number maybe only one of them gets added instead of all three that we got before. This is what we're expecting. But then if we go ahead with the reaction what we'll see is the bromine again gets added to all three positions. There is the two ortho and the para position and again we have this 246 tribromo and lean but why did this happen? So the culprit here is this lone pair and again as we saw before the lone pair is sharing its electrons with the ring and resonance happens so the bi bonds shift and the electron density increases at these positions. But the thing is nitrogen is not a very electronegative atom like in oxygen. So what happens is the effect of this sharing of lone pair is much higher because the nitrogen is not keeping any electrons in some sense. It's just pushing them towards the ring and highly activating it because of which this ring becomes very reactive. So instead of just picking up one bromine it goes and grabs whatever bromines are available in the solution and although we know that because of using a non-polar solvent we have very few bromines available because this ring is very reactive multiple substitutions happen on the same ring and we get this tribromo product. So that means we need to rethink our approach. Somehow if we are able to engage this lone pair then maybe we can think of getting only mono bromine substitutions. Let's see if there's a way we can do that. So as we discussed before what we want to do is we want to somehow block this nitrogen and make it sort of less reactive so that it does not share the lone pair fully with the ring and the idea is we need to somehow engage this lone pair and one way to do that would be to react it with this acetic anhydride. So what is going to happen in this reaction is this carbon oxygen bond will break and this group will attach itself to this nitrogen. So the product of this reaction will have a carbon cure with a double bond O and the CH3 which is basically this group which will take off this hydrogen and get added at this nitrogen. So now if you look at this lone pair apart from being in resonance with the ring it also has the option of forming resonance structures with this double bond. So what we have done by initiating this reaction is we have engaged this lone pair in resonance at this step. So you can think that the lone pair is now contributing to a lesser extent to the ring and therefore the activation is way lesser. So now what if we carry out the bromination after this step? So let's see what happens. So if you perform the bromination now what we see is again the electrophile is bromine which is in the form of this bromine with a positive charge and blocking this group here also made it difficult for the electrophile to approach at this position because this is a bulky group and it sort of blocks the way here. So instead of attaching at the ortho position the bromine will go and attach itself at the para position and now what is left is to restore this by breaking this bond and getting back NH2 here. So how we can do that is by simply reaction with an acid or base. So we have H plus or OH minus which attaches with this group and you have the NH2 back here and so by this process of blocking we can get a monobromo substitution. So another way to think about this is that because this lone pair was engaged in resonance with this oxygen as well this entire group here activated this ring to a far lesser extent than the NH2 or another way to think about this is that this group is less ortho-paradirecting than NH2. Apart from this we will also get an ortho product but it will be a very minor product because of the steric hindrance from this group. The key idea here was we started with bromination and we got tribromoproduct and even after changing the solvent this lone pair on the nitrogen made this ring very reactive. So the approach that we took was to block this nitrogen off by using this group which made the lone pair be in resonance with this oxygen here as well and because we engage this lone pair and this group is a bulky group it became very difficult for the bromine to get attached to the ortho positions which is why we got the para product as a major product. So that was the story about bromination. All of this happened because of this lone pair. There is one more example of electrophilic substitution reaction where this lone pair causes problems. Let's see what that is. So we saw before how this lone pair on the nitrogen gave us unexpected results. Let's see one more example of that which is Friedlkraft reactions. So you may remember from before that Friedlkraft reactions are electrophilic aromatic substitution reactions. So the Friedlkraft's alkylation looked something like this where let's say if you had an aromatic ring reacting with this CH3Cl in the presence of a Lewis acid which is ALCl3. This CH3 would get substituted onto the ring giving us this product and here the ALCl3 would be a catalyst. So what happens when this aniline will undergo Friedlkraft's alkylation? So the setup is the same. Let's say we have CH3Cl in the presence of the catalyst which is ALCl3 and we know something about ALCl3. This is a Lewis acid so it accepts electrons and if you look closely at this nitrogen it has a lone pair which it wants to give off and so this will act as a Lewis base. So what will happen in this reaction is immediately you have the ALCl3 getting attached to this nitrogen the nitrogen then develops a positive charge the aluminium gets a negative charge and it forms a complex and if you look at this complex this nitrogen is now pulling electrons from the ring so the reaction does not proceed any further after this step. So instead of acting as a catalyst which is what we know from Friedlkraft's alkylation this ALCl3 forms a complex with this lone pair and a similar situation happens in Friedlkraft's acylation reactions as well. In that case also a complex is formed in a similar way and the reaction does not proceed any further. So the most important idea to remember about this is that this complex formation happened because of this lone pair which is why aniline does not show Friedlkraft's alkylation or acylation reactions.