EAS Reactions of Pyridine - Online Tutor, Practice Problems & Exam Prep

EAS Reactions of Pyridine Concept 1

Hey, everyone. So in this video, we're going to take a look at electrophilic substitution reactions of pyridine. Now here we're going to say unlike the five-membered aromatic heterocycles, pyridine is significantly less reactive than benzene. And in fact, we're going to say that the ring itself is electron deficient. Now remember, nitrogen is more electronegative than carbon, so the electron density is pulled towards the nitrogen atom.

So, it's going towards it. This makes the ring more electron deficient and less likely to undergo electrophilic substitution reactions. Now, nitrogen also gets a positive charge under electrophilic substitution reactions. Here, the nitrogen itself, since this lone pair is not part of the pi system, can be protonated giving us NH⁺ here, and it's positive. Or if it's in the presence of a Lewis acid catalyst, it could actually interact with it.

This would help minimize the potential for Friedel-Crafts reactions. So we'd have our nitrogen forming a linkage with aluminum chloride. Again, aluminum, nitrogen is making four bonds with positive. Now we can say that activating groups can help pyridine undergo electrophilic substitution reactions more effectively. So, again, because of its nature of the heteroatom, this particular heterocycle is less reactive towards electrophilic substitution reactions.

EAS Reactions of Pyridine Example 1

Here, with this example, it states that unlike pyrrole, pyridine undergoes nitration under extreme conditions. But why? Pyridine tends to form radicals under acidic conditions making the nitration reaction difficult. Here, we never talked about radical formation when it comes to EAS reactions being less effective with this heterocycle, so that would not work. In the pyridine ring, carbon 3 attacks the proton instead of the nitronium ion.

This was also never really mentioned. Under nitration conditions, the nitrogen atom in the pyridine ring is protonated, which makes carbocation formation difficult. Now here, we talked about one of the issues with our nitrogen heteroatom is that it can interact with H+ ions. So, yes, the nitrogen atom can be protonated. If it is already positive, it makes it less likely to form a carbocation somewhere else on the ring.

So, this makes sense. And then pyridine undergoes nitration in multiple steps requiring a lot of energy. Again, we never talked about the energetics involved with this. We discussed the electron deficiency. We talked about the nature of the lone pair on the nitrogen.

This would not work. So, out of all my options, it is option C that is the correct answer.

EAS Reactions of Pyridine Concept 2

In this video, we're going to take a look at substitutions at C₂, C₃, and C₄. Now, S_N reactions of pyridine yield products by substitution at C₃. Here, we're going to say that substitutions at C₂ and C₄ usually do not take place. How can we be sure of this? We're going to do resonance and see what happens.

Now, we know that in all these reactions, our pyridine is going to attack our electrophile initially and will create a carbocation intermediate. In the first reaction, the positive charge will be here, and will resonate our double bonds to move this positive charge around. Moves here. Now in this reaction, moving it around this way gives us a positive nitrogen, which is not ideal. You wouldn't want the positive charge to be on the more electronegative nitrogen.

But luckily, remember, we're going to lose a hydrogen here, an H⁺ in order to reestablish our aromaticity. So we're going to say that this bond here breaks to fall here to make a double bond, this double bond, and we've reestablished our aromaticity. When it's at C₃, we're going to say this comes in, grabs it, creating a positive charge here. We have a double bond here, and here still. This can resonate over here, in order to make a double bond and get rid of that positive, but then it comes over on this side.

And then this can move down here, giving us a positive charge up here. And through the loss of our bond here, our H⁺, we reestablish our aromaticity. So we've reestablished our aromatic ring, and we did not create that unstable positive nitrogen. So that's why substitution at C₃ is preferred. Now if we look at C₄, again, we're going to use this π bond, and we're going to add our electrophile.

We're going to add it up here, So that means that this is positive. And we're going to say what can happen is this could move right here to make a pause to get rid of that positive, giving us a positive nitrogen again, which is not ideal. This could resonate down here, making this positive. And then again, remember, we lose H⁺ to reform our aromatic ring right here. So although all of these could help make the aromatic ring once again, C₃ is preferred because we never create that unstable positive nitrogen.

Here, we rather have the positive charge on the less electronegative carbon. That only happens in the C₃ position.

EAS Reactions of Pyridine Example 2

Here it says in this example, why does the pyridine ring not undergo nitration at c₄? Well, when we worked out all the resonance structures of our pyridine, we saw that the c₃ position was the most preferred. That's because we never created the unstable positive nitrogen intermediate. All the positive charges were distributed on the carbon atoms, which are less electronegative. So, if we take a look here, the one that most fits in with this description, being the furthest from the nitrogen, no.

The reaction intermediate has an electron-deficient nitrogen with a positive charge. Yes. This is the answer. The pyridine ring is more easily nitrated at the nitrogen atom instead of carbon 4? No.

The c₄ position gets protonated under the acidic conditions of nitration. Again, no. It has to do with the formation of that very unstable positive nitrogen intermediate. That is not ideal for positions c₂ or c₄. So that's why we do nitration at c₃ instead.

EAS Reactions of Pyridine Concept 3

This video, we're going to take a look at the directing effects on substituted pyridines. Now similar to 5-membered heterocycles, ortho, meta, and para positions are assigned through the carbon framework. Rule 1 says that the directing effects are the same as electrophilic substitution on benzene rings. And for polysubstituted rings, the most activating group takes precedence. Remember, we have our meta directors here: the nitro group, our positive nitrogen, sulphonic acid, nitrile, and carbonyl.

If you're not one of these groups, then you're an ortho para director and therefore activating in nature. You would have more of a say in where the next group goes over these meta directors. And the rule here is that when it comes to pyridine, the C3 substitution is always preferred. If we take a look here, we have our pyridine with OCH₃ attached, and we're trying to do bromination. Now remember, pyridine, because of the heteroatom in the form of nitrogen, is less reactive towards electrophilic aromatic substitution (EAS) reactions.

So, higher temperatures, harsher conditions will be required to do our basic types of EAS reactions. Notice this temperature of 100 degrees Celsius. Here, the OCH₃ group, our methoxy group, is an ortho para director. Nitrogen, which is part of our heterocycle, prefers the C3 position. So we're actually going to say C3 is preferred and this is ortho para.

So OCH₃ wants it to go ortho or para, but the para position already has nitrogen. And then the nitrogen wants C3, which is this or this. So, we could put a bromine on either one of these. So, I'm going to put it right here. Now the next one, we have a pyridine with a carboxylic acid.

Now we're trying to do nitration here, but notice the temperature is much higher than we had above. Remember, this is an ortho para director. They are activating in nature. They help to spur EAS reactions to happen more readily. This is a meta director.

It's deactivating in nature. So pyridine already has a hard time doing EAS reactions, and I'm slapping on their meta director which makes it even harder for an EAS reaction to occur. So how do I counteract these two negatives? I increase the temperature by a lot. That's why it goes up to 300 degrees Celsius.

Nitrogen again, C3 is preferred. So the carboxylic acid, it wants to meta to itself. So where it's located, we can see it as 1 for itself and then meta to that would be, this position right here. The nitrogen wants the C3 position, which would be right here. So they both would want the NO₂ group to go there.

So under nitration, we have the NO₂ being attached here. So remember, when it comes to directing effects, we have to take these into account, the activating groups as well as the preference of the heteroatom nitrogen, wanting the C3 position if it can get it, if it's open.

EAS Reactions of Pyridine Example 3

Hey, everyone. So here it says predict the product of the following reaction. So here we have our pyridine that's substituted. It has a methyl group, which is ortho para directing, and an NH₂ group. Now the nitrogen has lone pairs on it, which makes it even more activating than the methyl group.

It wants it to go ortho para as well. The ortho position already has a methyl group, so here's the para position right here. Nitrogen wants the C₃ position, so 3 C₃ to itself would be here as well. So both it and the NH₂ group wanted to go in that position. So that's where we're going to get bromination occurring.

So here, we're going to have our methyl group still here. We have our NH₂ group here, and we're going to add our bromine group right here. So this would be our final answer.

Draw the structures of products A, B, and C in the following reaction sequence. Select final product.

• A.
• B.
• C.
• D.

Pyrrolopyridines are a group of fused heterocycles where a pyrrole ring is fused with a pyridine ring. Nitration of the following pyrrolopyridine produces the 7-nitro product while the 5-nitro product is not seen. Draw resonance structures to show why the 5-nitro product is not formed. (Hint: Draw 5 resonance structures).

Everyone, in this practice question, it states that pyrrolopyridines are a group of fused heterocycles where a pyrrole ring is fused with a pyridine ring. Nitration of the following pyrrolopyridine produces the 7-nitro product, while the 5-nitro product is not observed. We are tasked to draw resonance structures to show why the 5-nitro product is not formed. Draw five resonance structures. Here, we identify the position 5 on the ring, and we will show all the different types of resonance structures that can be created, explaining why the 5-nitro products are not favored. We'll start by drawing our ring structure. In our discussion, when the nitro group attempts to attach to position 5, there's initially an H there, and then we attach an NO2 group. The nitration at this position initially gives a positive charge. Continuing the resonance, when we move the electrons around, we end up with several possible structures: The first results in a resonance structure where the positive charge is transferred but remains within the ring structure unfavorably. The next resonating shift moves the positive charge closer to a nitrogen atom, creating an unstable positive charge on a more electronegative nitrogen, which is unfavorable. Following that, additional shifts attempt to delocalize the positive charge, but each time, the charge ends up on nitrogen or requires nitrogen to form an unfavorable fourth bond, making it positively charged and violating the octet rule. By the fourth and fifth resonance structures, we illustrate that every attempt to stabilize the charge through resonance results in the formation of a less stable, electron-deficient nitrogen. We conclude that the configuration of the nitro group at position 5 leads to a series of resonance structures that all stabilize poorly due to the presence of a positively charged nitrogen atom. Therefore, these structures explain why the 5-nitro product is not observed, favoring instead the 7-nitro product where such destabilization does not occur.