 One of the common electrophilic substitution reactions we study is that of Nitration and the reagent used for nitration is usually this HNO3 or nitric acid in presence of H2SO4 or sulphuric acid Which gives us an electrophile which is an NO2 with a positive charge And this electrophile is going to be substituted on our reactant which is aniline But where will the substitution happen and what will be the final product and how does this happen in this video? We're going to discuss all of that. But first let's look at aniline a little bit closely So we know something about this nitrogen It has a lone pair of electrons and what it's going to do is it's going to share these with the ring because of which it is Going to activate the ring and there are also double bonds. So there's going to be resonance So let's just quickly draw the resonance structures and see where the electron density increases So here once the lone pair shares its electrons We'll have a double bond here and this pi bond will shift to the ortho position Which is why we have a negative charge here. Then next we will have this pi bond shifting here So there will be a negative charge here and similarly we have this pi bond because of which we have a negative charge here So if you follow the negative charge, you'll notice that the electron density is higher at the ortho and the para positions Which means that this NH2 because of this lone pair is activating the ring and NH2 is a ortho-paradirecting group So now based on this information itself, we know that when we have an electrophile in the reaction Something like this E+, it will probably attach at the ortho and the para positions So let's see how the reaction happens So here we have our nitrating mixture which will give us a NO2 electrophile And from before we know that the NH2 group will activate the ring and the electron density is higher at the ortho and the para positions So we know that the products that form will have a NO2 at these positions So first we have NO2 attaching to the para position and then we also have this NO2 getting attached at the ortho position But if we check how much of the product is formed, we find that the para product was around 51% And therefore this is the major product but the ortho product is only 2% Because we know that the NO2 is a large group and there will be repulsion when NO2 approaches the ortho position But still this is a really small number and also there's the question that what is the remaining 47% And looking for this is when we realize that there is one more product Which is we have this NO2 getting added at the meta position as well And surprisingly this gives us a very high yield which is the remaining 47% But this was not expected We thought we will only find the ortho and the para product And the electron density was not high at the meta position So then how did the NO2 get attached here? To understand this let's look a little bit closely at the reaction conditions Maybe one of our assumptions was wrong So if we look at the reagent, we can see that the reaction is actually happening in an acidic medium So there will be some H+, which is generated And this aniline has a basic character So maybe something is happening here that is not making this NH2 ortho-paradirecting Let's see what's happening We saw before how there is H+, from the acidic medium And now in the solution this H+, will react with this lone pair on the nitrogen of the aniline And form this anilineum ion And so the formation of this ion is deciding which products are formed So to understand that, let's go back to one of our initial steps Where we had drawn the resonance structures So we had started with this aniline And we said that there is this lone pair here Which is going to activate this ring And because of which we had these resonance structures So we saw that the electron density was higher at the ortho and the para positions Because of which we were expecting ortho and para products But now as we saw, because of the H+, this anilineum ion is formed So now like this case, there is no negative charge here Which is moving across the ring Or in other words, there is no increased electron density at the ortho and the para positions Which partly explains why we did not get the products as expected Because we started with this condition Where we thought that the lone pair will be in resonance But now there is no lone pair So then the question is, what is determining the products here? And for that, we need to closely look at this nitrogen So we know that this nitrogen has a positive charge So it will want to pull electrons from this ring So it is now deactivating the ring But what is happening because of this pull of electrons? So if we think of this structure where there is this group X Which is pulling electrons So if it is pulling electrons from here Now I have not drawn the hydrogens here So you can think that there are hydrogens here But for now we are going to ignore them And what I want to focus on is If we have this X Which is a group that is pulling away electrons So it will first try to pull out the electrons from the carbon that it is attached to Which is the closest to it Which is this carbon Number one Now if it was a very strong electron withdrawing group It will also try to pull from this carbon But the point is This second carbon is further away from this first carbon So the effect of the pull will be lesser here And this effect This electron withdrawing effect decreases with distance So the further we go from the electron withdrawing group The lesser is the electron pulling effect So now if we imagine this same chain here You can see that this nitrogen will pull electrons Most from this position Then the second carbon This third and this fourth And because of this we know something If we look at the electron density The electron density will be least at this position Because this nitrogen is closer And it's pulling away electrons And the electron density will increase as we go from one to four So if you look at these three positions There is two, three or four Or in other words If we look at ortho, meta and para positions Electron density will be highest at the para position Then at the meta position And then at the ortho position And we know that in our reaction The electrophile Which is in the form of an E plus Will get attached to the ring At places where the electron density is higher And we saw how the electron density was higher At the para position Which is why we get a parametri product Then next it is at the meta position Which is why we get a significant meta product And it is least at the ortho position Because of two things First at this position The electron density is very low Because this nitrogen is pulling away electrons And also we can see that As this NO2 group approaches this position There will be hindrance here at the ortho position So because of both of these reasons The amount of ortho product form is very less Which was around the two percent that we saw So the formation of this anilinear mine And its electron pulling effect Explains why we got a major para product And a significant meta product And very little ortho product But what if we don't want the meta product So the formation of this ion Happened because of the H plus reacting with this lone pair What if somehow we can block this lone pair And engage it elsewhere So that it does not form this ion And by that can we prevent the formation of a meta product Let's see if we can do that So as we discussed before What we want to do is We want to block this NH2 off So that it does not destabilize the ring And we then let the nitrogen happen So I've just written it down in a different way But this is basically NH2 And what we're going to do is We're going to react it with acetic anhydride Which looks something like this Another way of writing it down is in this form But this is what it looks like So when we initiate the reaction What happens is This acetic anhydride will break at this point And we will have this group attach itself at this nitrogen So we can write it as this C double bond O and this CH3 So now a couple of things have happened at this stage First you can see that there is this double bond O here Which means now this lone pair Is going to be involved in resonance at this point And the other thing to notice that Because this is a huge group We have effectively blocked off the ortho position And in a sense what we have done By blocking this group Is we have made this lone pair on the nitrogen Less reactive So now what we can do is We can initiate the nitrogen at this step And let's say we have this electrophile Coming out from our nitrogen mixture Of HNO3 and H2SO4 This NO2 will now get attached at the para position And again because this is a huge group It will be very difficult for the NO2 To get attached at the ortho position So now what's left is to convert this back into the NH2 And for that what we do is We can simply react it with H plus or OH minus In either case what will happen is This particular part will go out Because of which we will get back the original NH2 here And there we have our para product Which in this case will also be the major product And we got this by blocking this NH2 So again why this works is By adding this group here We are forcing the lone pair to go in resonance with this oxygen here And because the lone pair is engaged now with this oxygen It activates this ring to a far lesser extent Than the NH2 And the other thing is We saw how NH2 was ortho-paradirecting When we looked at the resonance structures So although the lone pair is getting shared with this oxygen This group on the whole is also ortho-paradirecting Because it's not like the lone pair Is not going to be shared with the ring Is just being shared much lesser Because of which the ortho-paradirecting effect is lesser in this case So the point here is It's not like we're going to not get an ortho product When we did the nitration here It's going to be very minor Because of the hindrance because of this group And there is no hindrance at the para position Which is why we got this product as a major product