 The structure that you see here is of para-nitrophenol which is basically an NO2 group getting added to the para position. The making of para-nitrophenol is one of the intermediate steps in the production of paracetamol which is usually prescribed by the doctors for fevers. So how do we make this para-nitrophenol? There are some clues in the name itself. We are going to be taking phenols and then nitrating them. So let's see how it's done. But before we go to the nitration, let's just look closely at phenol. So if you look at this oxygen, this oxygen has lone pairs of electrons. And what it's going to do is it's going to share these electrons with the ring because of which the overall electron density increases. But there's also something here. Because there are double bonds, you can actually break these bonds and check the resonance that happens here. And when we do that, there's something that we notice. So like when we break this double bond, we get a negative charge here and this double bond forms. Then when we break this double bond, the negative charge is here, there's a positive charge here and the bond is formed here. And later when we think of breaking this double bond, we have the negative charge here, the positive charge comes back here and then this bond is formed. So basically if you follow the negative charge, what you'll notice is that the electron density is higher at the ortho and the para positions. So what this means is apart from the OH group activating the ring, the electron density is much higher at the ortho and the para positions. So this is already giving us some clues about the products of the nitration reaction. Because the electron density is higher at these fixed positions, when phenol will undergo an electrophilic aromatic substitution reaction, that is basically a reaction with an E plus, which is an electrophile, it will get added to these positions. Like that could also indicate that probably we'll not get a meta product because the electron density is not higher at the meta position. And since it is because of the OH group that the electron density is higher at the ortho and the para positions, the OH group is called an ortho-paradirecting group. So now we have some clues about the reaction. So let's see how the actual reaction takes place. Now because we wanted a nitration, let's start with a commonly used nitrating agent that is nitric acid or HNO3 with sulphuric acid or H2SO4. Now you know the story about this phenol. It has lone pairs which is going to share with the ring because of which the electron density will be higher at the ortho and the para positions. And you'll remember from the resonance structures that when the double bond breaks, you'll have a negative charge at these positions. So let's pick the para position for now. So if you break this double bond, we'll have a negative charge here. And if you look at this nitrogen, because these oxygens are pulling away electrons, this nitrogen will have an overall positive charge and there's a negative charge here. So you know what's going to happen. Because there's a negative charge here, because there's a positive charge here, this phenol will go and get attached to this nitrogen. But there's also a hydrogen here. So when this double bond broke, the phenol lost its aromaticity and then it got attached to the nitrogen. So to regain its aromaticity, it's going to throw off this hydrogen and this N is going to get rid of this OH, giving out water here and this NO2 will be added to the para position. But the thing is, the electron density was higher at the ortho positions as well. So what we see is the NO2 will not just get added at the para position, it will also get added at the two ortho positions. So instead of the para-nitrophenol that we wanted, what we got was 2, 4, 6 trinitrophenol which is commonly called picric acid. So let's break down this reaction. Now initially we wanted only para-nitrophenol, but what we see is the NO2 got added to the ortho positions as well. And this was sort of expected because we know that the OH group is an ortho-paradirecting group. So the electron density was higher at the ortho positions as well, which is why they got attached to both of these positions. And another way to think about that is, HNO3 is a strong oxidizing agent and this can be a bit confusing, so it's not related to the addition of oxygen here. That's not happening. What's happening is if you look closely at this step, there is a hydrogen here and when this double bond is broken, this ring is attached to both the nitrogen and the hydrogen. And then because of the loss of aromaticity, this structure is unstable. So the hydrogen is kicked off, this OH is kicked off forming water and the NO2 is added. So this removal of hydrogen is also oxidation. And because HNO3 is a strong oxidizing agent, it has gone ahead and removed hydrogen from all three positions. So that's another way of looking at it. But in any case, we did not get the product that we wanted, which is para-nitrophenol. So what do we do? Now one way to think about it is that if we can reduce the number of NO2 groups available, it will probably get added at fewer positions. And how can we do that? So what if we took some of the HNO3 and added it to water? So effectively, what if we dilute the HNO3? And what happens in dilution is we are not changing the reagent. We are only making fewer of the NO2 groups available in the solution. And because there are fewer number of NO2 groups in the solution, maybe it will get added only to a few positions and not all of them. Let's see if this strategy works. So this time we are performing the same reaction, but with dilute HNO3. And like we discussed before, the reagent is not changing, the mechanism is not changing. We are only making a fewer number of NO2 groups available in the solution because the rest is water in the dilute solution. And we are hoping that because there are fewer NO2 groups available, it will not get added to all of the positions. And the way this reaction is going to proceed is in the same way as before. We have this oxygen which has lone pairs and it is going to contribute these to the ring, thereby activating the ring. And we have these double bonds which can be broken. And from resonance you know that the negative charge will be on the ortho and the para positions. So if we break this double bond, we have a negative charge here and because the oxygen atoms are pulling electrons, we have a positive charge on the nitrogen atom. So there's a negative charge here, a positive charge here and you know what's going to happen. This nitrogen is going to be attached to this ring and there's also a hydrogen here though. So as soon as the nitrogen is attached, the ring loses its aromaticity and it becomes more unstable. So to get back its aromaticity, it will kick out this hydrogen and the nitrogen now has 5 bonds, 1, 2, 3, 4, 5. So it's also going to throw out this OH, they'll form water and you have the NO2 which will get added to the para position. But this reaction is different, we're using Diute HNO3. So we're going to get different products and you can already see I've made two of these so there are two products. We're going to get para-nitrophenol which is what we wanted but we're also going to get ortho-nitrophenol. So we're going to get two products. One is the para-nitrophenol and we also have the ortho-nitrophenol. So one thing to think about here is that we've got a para product and an ortho product. Why didn't we get a product where, which is a 2-4 product in which on one phenol we have NO2 added at both the ortho and the para position? We didn't get that. Can we think of why we didn't get it? One way to think about this is that when we use diluted HNO3, we reduce the number of NO2 groups that were available in the solution. And if there were a fewer number of NO2 groups from dilute HNO3, the number of phenols was much more. And you know that every phenol was activated by this OH group and the electron density was much higher at these three positions. So this activated phenol with a higher electron density is looking for a positively charged electrophile, which is this nitrogen. So as soon as the very few NO2 groups are seen, it will quickly go and grab hold of them, which is why it will become very difficult for one phenol itself to get hold of two NO2 groups. And this is an intuitive and less formal way to think about it. The more formal way is of course in terms of concentration. And by controlling how much of concentrated HNO3 we're adding to water to create this dilute solution, we're controlling for its concentration, which is of course the factor which is controlling the number of available NO2 groups. So there's that. But because there are two products, how do we separate both of these products? So let's see how we can do that. So before we think of how we can separate these products, I've just written both of them down. This is the ortho-nitrophenol and this is the para-nitrophenol. You can see some oxygens here, there is a hydrogen here. So there is something that you could expect. Some sort of interaction between them. And you probably already know what it is, which is, yep, hydrogen bonding. So there is some hydrogen bonding happening here. So first look, let's look at the ortho product. So what you can see here is, there is this hydrogen here, there is this oxygen here. There will be some sort of hydrogen bonding here in between these two. And which is called intramolecular hydrogen bonding. Let me just write intra here, which is within the molecule itself. So that's one type of hydrogen bonding. And in the second case, which is the para, there will be intermolecular hydrogen bonding. But how would it actually happen here? So it's not very clear. So let's do one thing. Let's keep another molecule besides this one, but I've just flipped it around, right? So now you can see something here. Again, just like this, there is a hydrogen here, there is an oxygen here. There is going to be hydrogen bonding like this. So just by looking at these two structures and the nature of hydrogen bonding in both of them, there is something that I can say about this one that is going to be different from this. What is it? So because both of these molecules are interlinked in this manner, we can think that this will be slightly more stabler or it'll be less volatile, it'll be more linked. And how is this going to be reflected in the properties? So one way it is going to be reflected is that this one will have a higher boiling point because it is less volatile and more stable because of intermolecular hydrogen bonding. Oh, I forgot to write it down, just this is intermolecular hydrogen bonding. So the point here is if these two will have different boiling points, now I want you to think about it, if you were in charge of separating them, if you were given both of these and you were told that the boiling point of the para product is higher and the ortho product is lower and you had to come up with some way of separating them. What would have that been? Give it a thought. Maybe if you want you can pause the video and think about it and then we'll continue. Okay, so you probably guessed it, you're going to use heat. So since both of them boil at different temperatures, whatever boils first, you can separately collect it and that'll be your separation. So the process that we use to do the separation is called steam distillation. And that's how we can separate the ortho and the para products. Now let's go back to our reaction. So now we have separated the ortho and the para products. But if you look at how much of the product was formed, you'll notice that more of the ortho product was produced in this reaction compared to the para product. Or you could say that the ortho product was the major product of the reaction and the para product was the minor product of the reaction. So then the question is, why did we get the ortho product as the major product? And some clue for this can be found in the hydrogen bonding that we discussed before. So we can think about it in this way. The ortho product formed more, which means it was easier to form. So the hydrogen bonding in some way helped the formation of the ortho product. And you can see that here. So let's say if we had the Enoto group approaching at the ortho position here, we have this nitrogen, we have this double bond O here. And for now I'm going to write this as O negative. So you can see because of hydrogen bonding, this hydrogen is sort of pulling the oxygen atom and keeping it in place, which is why the sort of nitrogen aligns here and gets attached and we get more of the ortho product. So there's one more way to think about it. Now because out of the three positions with high electron density, two of them are ortho positions. We have twice as many as ortho positions than there are para positions, which is why it is twice as likely that the ortho product is formed. So this is another way in which we can think that why the ortho product is the major product. But when we set out our reaction, we wanted to get the para nitro phenol as the major product. So what we need to think is, is there some way in which we can tweak this experiment such that we get the para product as the major product. And based on what we saw just now, if we can somehow prevent this NO2 group from getting added at the ortho position, if we can somehow restrict its movement, we can then force it to form the para product as the major product. So let's see how we can do that. So as we discussed before, we want to block this ortho position. And what I'm going to do is I'm going to perform the same reaction, but this time with HNO2 instead of HNO3. And you'll see why in a moment. So first let's see how this reaction takes place. It's going to be very similar to the reaction with HNO3. You have these two oxygen atoms which are pulling electrons away from the nitrogen because of which the nitrogen develops an overall positive charge. And because this phenol is activated by the OH group, the lone pair and the resonance, when this double bond breaks, there will be a negative charge here. And this will get attached to the nitrogen. But there's also a hydrogen here. And as soon as the nitrogen is attached, the aromaticity of the ring is lost, which is why immediately this H is kicked out and the nitrogen will also kick out this OH, leaving out water. And this NO will get added here, which is a para-nitrosophenone. But there's a question to be asked here. The OH group was ortho-paradirecting. So we should have also got ortho products here, right? But why didn't we get ortho products? To understand why this happened, let's look closely at the structures of NO and NO2. So if I draw out NO here, what you can see is this is a planar structure where the nitrogen and the oxygen are in the same plane. And at the same time, if I look at NO2, both of these oxygens are sort of coming out of the plane. So now think about the NO getting added here. So if I just draw one NO here, if this NO wants to approach this position, the nitrogen will get added here. But what will happen is this oxygen will be in plane with this oxygen from the phenol. So there will be some sort of resistance here. And again, this resistance will not be there in case of NO2 because what's happening in case of NO2 is that both of these oxygens are away from the plane. Like one is above and one is below. So the resistance from this oxygen of the phenol, which is in this plane of nitrogen will be very less in case of NO2. So the NO finds it very difficult to get attached to the ortho position. But at the para position, there is no such hindrance, which is why you get the para product as the major product. So there's one more thing to note here. You may remember that in the case of NO2, there was hydrogen bonding because the hydrogen from the OH and this oxygen, which is in a plane different from that of the nitrogen, would form the hydrogen bond. But in case of the NO, the oxygen is in a different plane from that of this hydrogen. So here the hydrogen bonding is also very less. And because of both these factors, that is the hindrance because of this repulsion between the two oxygens and the lack of hydrogen bonding is what results in the para product being the major product. That is why when we reacted this phenol with HNO2, we got para nitrosofenol as our major product. So now all that is left is to make this NO into an NO2. And the easiest way to do that is to simply oxidize this. So let's react this with dilute HNO3. Now we know how HNO3 is a strong oxidizing agent. So what it's going to do is it's going to convert this NO into an NO2. And there we have our para nitrofenol, which is the major product also in this case. So to sum up everything, we started our nitration of phenols with concentrated HNO3 and got picric acid, but we wanted para nitrofenol. So we tried diluting the reagent, which gave us some amount of para nitrofenol, but more of the ortho product, which is why then we had to rethink the process and we got the phenol to react with HNO2, fixing the NO at the para position and then oxidizing it to get para nitrofenol as our major product.