Okay, so we are now Organic Chemistry 2, I'm Dr. Sola, and we are now in Lecture 5, the second part of the lecture. So this is the end of our topic where we explore aromatic chemistry. And in this particular topic, before we were showing you about aromatic electrophilic substitution reactions in the series of four different ones where you have learned how to generate the electrophile, which you need to know, and how to predict and write down the substituent product. So in this one, we're going to explore two particular aromatic compounds, phenols and anilines. And then we're going to sort of try to rank the nucleophilicity of these particular substituted benzenes.
And again, you need to know that too. So we know what benzene is. We now understand. why it is an aromatic compound and why it is stable and how we can sort of essentially get it to react if we have a very good electrophile.
Let's move on from there and let's look at two types of benzene derivatives. And these are probably the two most important types. One of them is a phenol. And if you look at that, it's a hydroxybenzene.
This is like a benzene that's substituted with an OH group, which you're already thinking looks like an alcohol. It's called phenol in the old fashioned terms. Okay. And this one acts as an acid. So if it's acting as an acid, you're thinking it's going to be able to lose that proton there.
Okay. So it can be deprotonated. And the other class of compounds we're looking at that's very important is aniline.
Now, aniline is amino benzene. Okay. So it's a... benzene ring with an NH2, an amino acid. So an amino benzene aniline is the old-fashioned word.
And this acts as a base. So it means that this lone pair here can go and grab a proton and deprotonate something else. Okay, so this one acts as an acid.
This one acts as a base. As an acid, it will lose a proton. As a base, it will grab a proton.
So both of these compounds are more nucleophilic than benzene. If it's more nucleophilic, it means it's got more electron density in these compounds, which means it's going to be easier for them in the reactions we saw before to be able to push out a pair of electrons and attack an electrophile. It's more nucleophilic.
It's got more electron density to push out. So a little side note here, phenol has a very local origin story. So this was the first disinfectant that was used.
And it was fantastic because you could paint it on the body before you were going to do surgery. And it was done here. It was invented by Joseph Lister in Glasgow in 1864. So this was a great revolution when it came to doing surgeries because you can't really boil a patient to disinfect the patient, but you can paint on phenol onto them to disinfect them.
That's pretty cool, I think. And aniline is something I think is really exciting because this is used to make dyes. And this was originally isolated from a plant called indigo. So this is sort of one of the bases of the dye industry that just the synthetic dyes that took off around the world. So phenols, let's go to phenols.
Here is your benzene ring. There's an OH. told you it's going to it's going to react as an acid so it means we can lose that proton that's going to be more easily lost because it's acidic and if you think about it it's like an alcohol it's an OH group on an aromatic ring so you're going to have hydrogen bonding that happens with this but the fact that your OH group is bound directly to this aromatic ring means it's going to affect the character overall character of this ring. So it means that this is different to a normal alkyl alcohol, which we saw before.
It means that this is more acidic. If it's more acidic, it's easier to lose that proton there. And that's because we can be able to push electron density around the ring.
So remember when we're talking about alcohols, we said alcohols, the pKa is around 16 or 17. Like a hydrochloric acid, it's minus seven. So the lower the number, the more easy it is to lose a proton. And water is around 15.6.
And ethanol, like normal alkyl alcohols, is around 16. this case, phenol, the pKa is around 10. So it's much more acidic than a normal alkyl alcohol. It's much easier to lose that proton. So if you have phenol in a basic condition, so you've got this acidic compound and you've got a base around there, that base can come in and can pluck off that proton and form something called a phenoxide.
So when this is deprotonated, so if this is in basic condition, you're going to pull off that proton and form something called a phenoxide. So there's your phenol. It's an acid. And in a base, it's going to come in and pluck off that proton, push the electrons onto that oxygen, and we formed phenoxide.
The oxide, that minus charge there, phenoxide. And phenoxide, if you treat it with acid, So that's a conjugate base. If you treat it with acid, you'll reform your phenol.
Okay, so here we go. We treat it with acid and you form your phenol. So this will go back and forth. Why is this easier to remove the proton than like a normal ethanol proton? Well, let's show you what happens.
Once you form this phenoxide, there's that negative charge on there, you're sitting there on this aromatic ring and we can delocalize that negative charge throughout the ring. So that helps stabilize this negative charge on this compound. So if it's more stable, it means it's going to be easier to lose this. So how do we show that?
We're going to push a pair of electrons here to form a double bond here. But if we do that, we have to push this pair of electrons and we're going to just push it onto that ortho position. And now we have something here that's a formal negative charge in the ortho position.
But we can keep pushing these electrons around and we have to then push this one onto the para position. Okay so now we've got a formal negative charge para to where the oxygen is and let's form a double bond here and push this pair of electrons onto the ortho position. Okay and what if we push those back up here and this the electrons up to the oxygen. Look at that, we're back where we started from, but obviously the bonds are in different sides. So you can see we've got the more resonance structures you can draw, the more stability you have.
So this is very very well stabilized, this phenoxide, so that means we've got resonance stabilization, conjugative stabilization, you might have heard that word before. This ability to spread this negative charge on this compound all the way around in the ortho and the para positions like that, see here's the partial negative charge of the oxygen and partial negatives over the ortho and the para position, means it's much easier to pull off that proton. So that's why it's more acidic than normal alcohol. If we looked at the distribution of the negative charge mainly over the oxygen in the phenol, but there's some over the ortho and the para positions because we can push these electrons around anyway.
Once you form the phenoxide, we have a very similar, as we've just shown you here, with most of the negative charge sitting over the oxygen, but some in the ortho and the para position too. So you've got the stability of the compound. So for both of these species, phenol and phenoxide.
Look at the words phenoxide, negative charge there. The negative charge is mainly sitting over the oxygen, and that's not surprising because that's the electronegative element. It's also spread over the ortho and the para positions for both compounds because of that resonance stabilization.
And the take-home note here as well is both of these compounds, phenol and phenoxide, phenol and phenoxide. are more reactive than benzene on its own. So for an electrophilic substitution reaction, if you have a phenol or phenoxide, this is going to be more reactive with that electrophile than benzene on its own.
And that makes sense. There's more electron density in this ring, and that makes it easier to share electron density out to form a bond. And if you're going to compare the two phenoxide is even more reactive than phenol. But let's look at the other compound, aniline.
So aniline is an aryl amine, and it is actually less basic than an alkyl amine. So here's aniline, here's benzene with this NH2 group on there. And if you're just thinking about an alkyl amine, you might be thinking of, I don't know, triethylamine, like a nitrogen with three ethyl groups on it. Okay, so that's aniline, and that's a base. Okay, so.
Aniline isn't going to lose a proton. It's not going to be acidic. It's a base. It's going to grab onto a proton.
So it's going to be used to deprotonate something. Now, aniline is less basic than a normal alkyl amine because this nitrogen lone pair, this is electronegative, it's less electronegative than oxygen. this nitrogen lone pair that's sitting here is less available to be able to go out and attack a proton. And why is that? It's because we can push our electrons around this ring so that lone pair of electrons is delocalized into the ring, which means they're less available.
So it means that it's less basic than a normal alkyl amine. So as I said here, this nitrogen is less electronegative than an oxygen. So let's try to show you. Here's this lone pair on the nitrogen. We're going to push that into the ring here.
Let's form a double bond here. And we have to keep going and push that pair of electrons. And let's just push that onto the ortho position here. So look at this.
Now we have a formal negative charge here just because we push that pair of electrons here. And there's that formal positive charge here. And let's push that around onto the ring. Let's form a double bond here, which means this pair of electrons has to go onto the ortho position.
And if we do that, look at that. In the pair position, we have a negative charge up on the nitrogen. It's positive. And let's push that and form a double bond, which means this goes onto that ortho position. And we've got a resonance structure.
Okay, so that's showing you that that pair of electrons on the nitrogen in an aromatic amine, so this aniline, is actually... Okay. in resonance.
Okay, so it's not as available to push out and grab on a proton because it's involved in all of these resonance structures ongoing. So this is a resonance stabilization of aniline. Now, if you see aniline, and you think of an acid, you think, okay, well, this is going to go and attack a proton. Just be mindful.
Those four reactions we've showed you for the electrophilic aromatic substitutions of halogenation, Friedel-Crafts acylation, Friedel-Crafts alkylation, and nitration, those are all in acidic environments. So you cannot use aniline for those reactions. You can't use aniline to put on a chlorine or a bromine.
Because the minute aniline interacts with an acid, like an H+, any sort of acid, it's going to push its electrons out there onto the acid. and it's going to form this species here, which is protonated, right? That's called an anilinium cation. This is now completely, almost virtually, unreactive.
So the minutes in an acidic environment is going to form this, and it's not going to be able to do any other sort of electrophilic aromatic substitution reactions. So this is the conjugate acid of this base. Of course we can deprotonate this and reform the aniline base. Okay so the base can come in here, deprotonate, push those electrons back on there, and we have now formed aniline again.
So this is an example. This is this protonated anilinium cata and you can see the pKa is 4.5. It really is It's very happy to lose that proton, so it's quite acidic.
Whereas the alkyl protonated version of the anilinium cation, well, a cation, is pKa of 10.5, because it shows you this is much easier and wants to lose that extra proton than this would be. So what does this sort of lead us to? We want to think about... How nucleophilic, this list of aromatic compounds, which I'm about to show you, how do we rank them? Okay, so we want to go from most nucleophilic to least nucleophilic.
And as we said on the previous slide, when you're looking at phenoxide and phenol, phenoxide is the most nucleophilic because it's got this extra electron density that it can push into the ring. And that means that ring is primed to be able to push out of an electron pair to form a bond to a really good electrophile. After that is aniline. Aniline is very reactive.
So it's quite nucleophilic as we saw here, we can actually push electron density into the ring onto the ortho and the pair of positions. Okay. And after that is phenol.
Okay. Phenol also has that electron density in the ring, not as much as aniline phenoxide has the most. After that is benzene. So we're just.
ranking it in comparison to benzene. So benzene has less electron density to be able to push out as a nucleophile and these electrophilic aromatic substitution reactions. Phenol has more, so it's more reactive.
Aniline has even more. And then phenoxide has the most. But remember, if you have aniline and you try to use an acidic reaction, like the four reactions I told you before, the first thing that's going to happen is that aniline is going to attack an acid and form this anilinium cation, and that is essentially unreactive.
So if you have aniline and you use an acidic reaction, you're going to make it the least nucleophilic in the order of these compounds. Okay, so you need to remember this order and hopefully understand the reasons behind. So caution with aniline, avoid... acid, otherwise you deactivate it completely.
So the most nucleophilic compounds here have more electron density in the ring. You can push this electron density into the ring, push this into the ring, not quite as much. This one, not quite as much as those, but still reactive.
And we're just relative to benzene, which has nothing on the ring to add to or detract from it. And this one, yeah, it's not reactive at all. So you now have hopefully understood what aromatic compounds are, how to use electrophilic aromatic substitution on benzene to be able to add halogen, chlorine, acyl group, alkyl group onto it.
And you've learned the rate determining step. and the fast step was the deprotonation. Now you know how to make the electrophiles.
There's four different sets of electrophiles that you should be able to make. I know the reagents for that. You now have learned about phenols and phenoxides and now anilines, and you can also rank them in order of most nucleophilic to the least nucleophilic in terms of electrophilic aromatic substitution reactions.
Hopefully your brains haven't popped by now. Oh, and by the way, this is the end of lecture. I think this is the end of it. Just double check here. Yep, that's the end of lecture five and the end of topic three.