Reactions of Pyrrole, Furan, and Thiophene - Online Tutor, Practice Problems & Exam Prep

Reactions of Pyrrole, Furan, and Thiophene Concept 1

Hey, everyone. So in this video, we're going to talk about the reactions of our pyrrole, our furan, and our thiophene. Here, we're going to say pyrrole, furan, and thiophene undergo EES reactions just like benzene. Remember, EES is electrophilic aromatic substitution. Now, if we take a look, we're going to say we have here representing our heterocyclic compound in the form of pyrrole, furan, or thiophene. And we're going to say Z here could represent our nitrogen if it's pyrrole, it could represent oxygen if it's furan, or it could represent sulfur if it's thiophene.

In this reaction, we have this pi bond coming out and grabbing our electrophile. Our electrophile now connects to the ring, and we've used the pi bond to make this connection. This causes this carbon up here to become positive. All we do now is draw the other resonance forms, so this double bond can resonate here. When it resonates there, there's a double bond here now, and here becomes r positive.

Then this lone pair here can resonate here, giving us a double bond here, and a double bond here, and therefore making the Z, the heteroatom, positive. Lastly, we want to recreate our aromatic compound, so the carbon that gained the electrophile has to give up its hydrogen. So when hydrogen leaves, it leaves behind its electrons, so this bond falls here to make a double bond which kicks this double bond here, which makes this pi bond sit back on the heteroatom as a lone pair. We reestablished our aromatic ring, and now we have our electrophile. Remember, EES reaction works as an addition substitution type of reaction.

We add the electrophile and then later on we lose our hydrogen to reestablish aromaticity. Now, due to stabilization by the heteroatom of pyrrole, furan, and thiophene, they are more reactive than benzene. Because they're more reactive than benzene, they require milder conditions for EAS to work. And if we're talking about the reactivity order, we're talking about pyrrole being more reactive than furan, furan being more reactive than thiophene. Right?

So these are the key ideas you need to remember when it comes to EAS of these particular heterocyclic compounds.

Reactions of Pyrrole, Furan, and Thiophene Example 1

Everyone, in this example question, it says that furan can be acetylated using acetic anhydride and a Lewis acid catalyst in the form of boron trifluoride, while pyrrole can be acetylated directly without a catalyst. While pyrrole is more reactive than furan, why exactly is this? Alright. So pyrrole is more reactive because it is a basic aromatic.

Pedal cycles are typically more reactive. There's one thing I don't like in chemistry: the word "always." It tends to be a key giveaway because we know that exceptions can arise. We also know that we never discussed vouchicity being a reason for why one is more reactive than the other. This wouldn't work. Always being less electronegative and making the aromatic ring electron deficient decreases its reactivity.

So, oxygen is more electronegative than nitrogen. So, that's wrong too. The large atomic size of oxygen and poor orbital overlap with the aromatic pi system makes furan less reactive. Alright. So, remember when we're talking about our periodic trends, if we think of the periodic table as this really bad drawing that I made, just remember that as we move from left to right on the periodic table, your atomic size gets smaller.

We have nitrogen here somewhere and then we have oxygen to the right of it. Oxygen shouldn't be larger; it should be smaller. So by default, D will be our answer. Nitrogen is less electronegative than oxygen.

Yes, it takes up a positive charge more easily because it's less electronegative and better stabilizes the intermediate carbocation. That is true. So here, out of all our options, D makes the most sense.

So here, D will be our final answer.

Reactions of Pyrrole, Furan, and Thiophene Concept 2

Hey everyone. So in this video, we're going to talk about substitutions at carbons 2 versus carbons 3. Now here we're going to say that EAS reactions of Pyrrole, Furane, and Thiophene give major products by substitution at carbon 2. Now, if we were to take a look at this structure up to the right, remember Z is a placeholder for our heteroatoms that exist for Pyrrole, Furane, and Thiophene. For Pyrrole, it's a nitrogen, for furane, it's an oxygen.

And for thiophene, it's a sulfur. We say that the heteroatom gets location number 1, so that would mean that this would be carbon 2 or this could be carbon 2. And then this would be carbon 3 or this could be carbon 3. Now, here we're going to say substitution at carbon 3 produces minor products. Here, let's take a look at the different types of resonance structures that can form if we're doing substitution at Carbon 2 or Carbon 3 and why that shows that Carbon 2 is preferred.

Now, if we take a look at substitution at Carbon 2, we have this pi bond here which is going to come out and grab this electrophile. Remember, we've used this pi bond here to make this bond here causing this carbon here to become positive. Through resonance, we can form the other structures. So this can resonate here to make a double bond, causing now this carbon to become positive. This nitrogen can use its lone pairs to fall here to make a double bond.

And now we have nitrogen making 4 bonds, so it's positively charged. And remember, we want to reestablish our aromaticity so our hydrogen has to go. We're going to lose it. When it's lost, this bond forms our new pi bond. Falls here to make a double bond, kicking this here and moving this back as a lone pair of nitrogen.

So we have our structure here. We're going to say for carbon in position 3, again we have this pi bond here. This shoots out and grabs this electrophile. Now we're going to say, there's a hydrogen here. It's still there, and we've gained the electrophile.

Now it's this position becoming positive. We're going to say now that this nitrogen decides to resonate here to make a double bond. And now this nitrogen is positively charged. Right? And then next, we're going to say we want to reestablish our aromaticity.

So here, this bond breaks, falls here to make a double bond back. Gain this bond here to create our lone pair again, giving us this structure here. And we can see here that position 3 is more preferred because we have more resonant structures that happen, and we have more resonant structures that happen where the positive charge is on the less electronegative carbon which makes sense. In position 3, we have less resonant structures and one of them we have a positive on the more electronegative nitrogen. Right?

So here, this is just showing us that when it comes to EAS reactions of these particular 3 heterocyclic compounds, that position 2 is the favored position for EAS reactions to occur. Right? So keep that in mind when doing different types of reactions.

Reactions of Pyrrole, Furan, and Thiophene Example 2

Here in this example, it says, thiophene is highly reactive in comparison to benzene and undergoes bromination without a Lewis acid catalyst. Draw the major and minor products of the following bromination reaction. Now we know that, typically, if we saw bromination with benzene, it would be Br₂ over our Lewis acid catalyst, which is typically iron (III) bromide. But since thiophene is more reactive than benzene, we don't need to use that Lewis acid catalyst. We simply use acetic acid.

Now remember, when it comes to our heterocyclic compounds of pyrrole, furan, and thiophene, substitution of carbon 2 is preferred as the major product and substitution of carbon 3 would give us our minor product. So if we take a look here, remember, the heteroatom is at position 1, then we'll go 1, 2, 3. Here, we're going to draw our major product. And remember, in bromination, we're substituting a bromine, so Br goes here. It's in position 2 so this will be our major product.

And our minor product would have it go to position 3 or carbon 3 up here. So these will be our two answers. The first one on carbon 2 would be our major product and the product where it's on carbon 3 would be our minor product. Right? So, these will be our two final answers.

Furan undergoes nitration under milder conditions than benzene. Draw a mechanism for the formation of the major product when furan reacts with concentrated nitric acid.

Hey, everyone. So in this practice question, it says, urine undergoes nitration under milder conditions than benzene. Draw a mechanism for the formation of the major product when furan reacts with concentrated nitric acid. Alright. So here we have nitric acid and because it requires milder conditions, we have on the bottom this anhydride. Now, it says concentrated nitric acid which means that we're going to have 2 molecules of Nitric acid. One is going to help to protonate the other in order to form our electrophile in the form of the Nitronium ion. Alright. First of all, we know that this is going to favor Nitration of carbon 2. So let's just write down what the product will be at the end. This will be our product at the end. Now what's going to occur here is we're going to have one of these nitric acid molecules here. It's gonna get protonated. Just simplifying the second one here as this structure here. So this is gonna get protonated. Just created water, which is a great leaving group. So this can resonate here to make a double bond and kick out this water. This helps to make our Nitronium ion which will be our electrophile that the furan ring can attack. So here it goes. We're gonna then show furan attacking that positive nitrogen. So this pi bond breaks, attacks this nitrogen resonating one of these double bonds to either oxygen. So we're gonna get here. NO2 getting added to the furan ring, and there's an H+ here, and this is positive. Then we're gonna have the loss of hydrogen here. Which will help remake this double bond here and reform our aromatic ring. Now this loss of hydrogen or loss of H+ can be helped by you can see the water that left could play a role in coming in and deprotonating. This bond breaks and falls here to make a double bond giving us our final product here. So this is how we can see the nitration of furan in terms of the mechanistic steps. Right? So this would be our final answer.