Organic Chemistry Aromatic Compounds: Reactions, Part 1

Hi; welcome back to Educator.com.0000

Next, we are going to talk about some of the reactions that aromatic compounds like benzene can undergo.0002

Now remember: benzene and other aromatic compounds have a very special stabilization called aromatic resonance.0010

For that reason, we are not going to have ordinary reactions like an alkene might have; so for an alkene, we saw that you could do an addition reaction where you break the π bond (to react with HBr, for example); you add a hydrogen and a bromine across the π bond.0016

Well, this reaction does not happen--this is not formed; and the reason this is an unfavorable reaction is: by losing one of those π bonds, you would lose aromaticity.0033

That would be quite unfavorable; so instead, the reactions that benzene undergoes (and certain materials like benzene)--we have substitution reactions occurring instead.0047

We have a benzene ring; we react it with bromine, and where we used to have a hydrogen, we now end up with a bromine.0060

This is a favorable reaction, because it is still aromatic.0068

That is going to be typical of the reactions that we see in this lesson: we are going to see substitution reactions, and at the end of that reaction, we are still going to have our benzene ring intact.0073

The most significant reaction that we are going to spend most of our time on is this one, called the electrophilic aromatic substitution.0086

It is called that because we are reacting the benzene ring with an electrophile; we are just going to call it an E⁺ for now, and then eventually we will see what electrophiles we can add in this fashion--what the actual structure of that positively charged species is.0092

But what happens is: we take a benzene ring; we react it with E⁺; and we end up with an electrophile attached onto the ring.0106

So again, it is a substitution reaction, because we substitute an E⁺ for an H⁺.0115

We remove a proton, and in its place we have an electrophile.0124

Now, this is a 2-step mechanism, and the very first step is addition of the electrophile; now, we just said how unreactive benzene is, and how much it wants to have this aromatic stabilization.0128

So, this has to be a very, very special reactive electrophile; we are going to talk about what reaction conditions are needed to initiate this reaction.0143

OK, but if we do have an appropriately reactive electrophile, what can happen is: one of the π bonds in benzene can act as a nucleophile, and the electrophile can add to it.0152

Now, if I break that bond and react an electrophile, much like if I protonated a π bond, I am going to get a positive charge on the ring--I'll get a carbocation intermediate.0165

Now, this carbocation is allylic, so it will always have resonance stabilization, and in fact, it's still allylic--I'll be able to use both of these π bonds, and I will always have these 3 resonance forms for this carbocation intermediate.0176

Now, this is a very difficult step--this is a slow step: what is the problem with this step?0199

Not only are we forming a carbocation, which we know is a high-energy intermediate, but we are starting at such a stable, stable low-energy state, because it is aromatic, that it is an even further reach; so it's a very difficult step, because we lose aromaticity.0205

This is difficult to happen, but if that electrophile is reactive enough, it is going to happen.0221

OK, and then, another thing we can note is that, in the intermediate that we have, the positive charge is always going to be either ortho or para to the carbon that added the electrophile.0227

So, here is our electrophile: the positive charge is either in the ortho position or the para position or the other ortho position; and again, that pattern will always be true, so that is going to help us be able to draw this mechanism and know we are doing the right thing.0241

OK, that is our first step in the mechanism: the second step in the mechanism is deprotonate.0255

Remember, we are doing a substitution; so first, we add the E⁺, and then we remove the H⁺.0259

And so, what we are going to do is: we are going to take a base (any base will do--let's just use :B), and we are going to go after the proton that is on the same carbon as the electrophile, because remember, that is where we are doing the substitution.0264

We are going to grab that proton; these two electrons are going to go fill in that vacancy of the carbocation; and let's take a look at the product we get.0279

We get our product: we get our benzene ring with the electrophile attached to it.0291

This, now--the first step was a rate-determining step: this step is super easy--that is why you could use any base at all: super easy; really fast; what is so great about this second step?0295

Well, not only do we fill this octet and get rid of the carbocation, but at the same time, we gain our aromaticity--we regain that.0305

Any step that gains aromaticity is going to be a really downhill race and a very fast step.0312

Now, when these reactions get interesting is when, instead of just using benzene, we use substituted benzenes and do electrophilic aromatic substitution.0322

If we already had a group on the benzene ring, and we had an incoming electrophile, there are three possible places for that group to come.0333

It can either come to one of the adjacent sites, the ortho positions, or it can come to the 1,3, and that would be the meta position, or it could come in opposite to that group and be a para position.0345

The regiochemistry, the place where that electrophile goes, depends on the type of substituent, this type of group, that is already on the ring.0357

We are going to classify those groups into three different categories: we can have electron-donating groups, electron-withdrawing groups, or halogens.0366

Let's discuss each of those and their behaviors, and then we will see many, many examples of those after.0374

OK, so these electron-donating groups--we describe them as being activated toward electrophilic aromatic substitution; and this is versus benzene.0382

So, compared to benzene, by putting an electron donating group on there, we make it more reactive.0392

The regioselectivity for this type of group is described as being an ortho/para director.0400

Activated means more reactive than benzene toward an electrophile: this is just for electrophilic aromatic substitution--this one reaction.0409

We describe it as being an ortho/para director because an electrophile comes in, and there is already an electron donating group on the ring; that electrophile gets directed, gets pointed, toward either one of the ortho positions or the para positions.0422

Those will be our major products in this kind of a reaction.0434

OK, and some examples of a group that would be described as an electron-donating group: any kind of amino group, or an OH, or an O-R, or if you have a nitrogen with a carbonyl attached (that would still be an electron donating group), or an oxygen with a carbonyl attached.0439

So, if we had an amide or an ester group, those would also be electron-donating; or just a plain old R group--having an R group...an alkyl group of some kind, is also an electron donating group.0461

OK, and one by one, we are going to be looking at each of these in detail: so right now we are just looking at an overview, and then we are going to be explaining each of these different components down the road.0484

OK, an electron withdrawing group is something that is deactivating toward the ring: we describe this as being deactivated, meaning it is less reactive, now, compared to benzene, toward an electrophile.0493

And we are going to get the opposite regioselectivity from an electron donating group: these are known as meta directors.0504

In this case, the incoming electrophile is going to go to this position, one of the meta positions; of course, it doesn't matter which side you go to, because this is a planar molecule, so that would be the same compound, whether you went to the right side or the left side.0510

And some examples of electron withdrawing groups: well, we have actually used that term before (EWG), and so all of those groups we have called EWGs before are still EWGs (electron withdrawing groups)--things like carbonyls, a ketone, aldehyde, ester...something like that...0524

A nitro, a cyano...those you have seen as electron withdrawing groups; an SO₃H, a sulfonyl group--we will see those in this lesson; you will see where that comes from.0538

If you had a positively-charged group, like an ammonium group, that would be electron-withdrawing.0552

OK, so we will see these in certain cases, and when we do, we will recognize them as EWGs, electron withdrawing groups.0557

OK, and the halogens are in their own category, because they don't fit perfectly with either of the first two we have discussed.0564

It turns out that these halogens are deactivating, or you could say that this ring is deactivated.0571

So, in that respect, it is like an electron withdrawing group; but it is an ortho/para director, so it has that in common with an electron donating group.0582

We will see how it has both of these personalities and where that comes from.0590

OK, so we can point out here that it would go to either the ortho position or the para position, if you had a halogen already there.0596

And the halogens, of course--we will just look at chlorine, bromine, and iodine; the fluorine is kind of in its own category, so we won't be doing reactions with that as a substituent.0605

OK, so one by one, let's take a look at these and see if we can understand all of these different details.0617

The first thing is the electron donating group: an example of an electron donating group, we said, was an OH.0625

So, if you took phenol, and you did an electrophilic aromatic substitution, we describe this as an electron donating group (which we just said on the last slide means it is an ortho/para director), and that is observed because, when we brominate the ring (this is one of the electrophiles, we will see, that you can add to an aromatic ring), the bromine comes into either para or ortho to the existing substituent.0629

We get a majority of para and less ortho, but we don't get any meta product out; so that is what...this is an example of a reaction with an ortho/para director.0658

OK, and this is pretty typical: ortho is usually minor, compared to the para; usually, even though it's an ortho/para director, usually para is the major product.0669

That is simply because, even though there are two ortho positions at which you could react, bringing in the electrophile next to an existing group automatically increases some sterics and slows down that reaction.0682

OK, so it is the sterics that slow it down and typically make that the minor product.0694

OK, so let's ask why that is: Why is it that an incoming electrophile is going to prefer to go to the para position or the ortho position?0701

What we want to ask is: what effect does an electron donating group have (like an OH group)?0710

Well, we call it electron donating because there must be some way that it adds electrons to the ring; and the way it does that is: it takes one of its lone pairs (which is allylic--these are benzylic lone pairs), and any time we have a lone pair next to a π bond, that lone pair can come in, and the π bond can move over, and we can draw a resonance form.0715

We can draw another Lewis structure for this molecule, for phenol, and so that means that this Lewis structure contributes to the overall picture.0737

Is that the only Lewis structure we can draw?--well, we will find that, when it comes to resonance in dealing with aromatic compounds, they are always going to come in threes, because we have the three π bonds that we are dealing with initially.0747

So, sure, this is still allylic, so we can move the negative charge down to the bottom carbon; and it is still allylic, so we can move it to the other position, as well.0758

So, in fact, there are four resonance forms in total for phenol, and when you look at that resonance, then, and knowing that this resonance exists, what are some conclusions we can draw?0779

How does this affect the reactivity of the molecule?0792

Well, the electron donating group, you see, makes the ring...is it more electron rich or less electron rich?--well, you can see there is actually negative charge character in that ring because of that electron donation, so clearly, it makes it more electron rich with that electron donating group.0797

Since it is reacting with an electrophile, that means the ring is acting as a nucleophile; does being more electron rich make it a good nucleophile?0816

It does; so this ring is a better nucleophile than benzene is: that is why we describe it as being activated toward electrophilic aromatic substitution.0824

It is activated; the OH is an activating group; OK, so by making it more electron rich, that makes it better able to react with an electrophile.0839

OK, how about the regioselectivity?--we said it was an ortho/para director; why is it an ortho/para director?0849

Well, if we just look at this resonance, what does the overall structure look like?0854

Well, we know that this oxygen has some partial positive character, and the ring has some partial negative character; where does it have partial negative character?--in the ortho position and the para position and the ortho position.0860

We have partial negative; so take a look at that: I'm just predicting, just thinking about that resonance, that if I were an electrophile...where would I want to go?0873

I would want to go to one of the ortho or para positions; so that does help explain, or at least predict, that regiochemistry.0881

OK, but just this prediction is not enough to really prove why, in fact, this is an ortho/para director; so let's take a look at a process by which we can prove that.0888

OK, the way we prove it is: we add the electrophile in each of the positions (the ortho and the meta and the para), and we compare the possible intermediates.0898

OK, we look at the different pathways it can take, and we decide which one is better.0909

So, if we added the electrophile in the ortho position, that would break this π bond (as shown in this Lewis structure), which means the electrophile goes here, and we would have to put a carbocation here.0913

Remember, that first step always gives a carbocation intermediate.0924

If we added to the meta carbon instead, that would be still breaking the same π bond, but our positive charge would go here.0928

And if we put the electrophile on the para position, that would break this π bond, and the carbocation would go here.0934

These are our three competing intermediates.0941

Now, we know each of these is resonance stabilized, because they are all allylic charges; so let's take a look at the additional resonance forms we can have.0944

We can move the positive charge over here, and then we can move it up to that top carbon.0955

OK, the meta addition intermediate also has two additional resonance forms we can draw; it moves the carbocation here.0968

Notice our pattern: every time we move the π bond, the carbocation jumps from the first to the third atom.0979

That is what every allylic carbocation does when we show resonance.0987

OK, notice that the carbocation can never be at the carbon bearing the electrophile, because that has two groups on it already; so it is going to be on alternating carbons (let's draw this down here).0992

We can move it to the top, and then we can move it over to the bottom right.1011

OK, good practice: so here are three competing intermediates; which one wins--which one is better?1020

Is there any difference between the three intermediates that makes one a better path or a worse path?1026

Well, it looks like, when you look very carefully, the difference we have is: in the ortho and in the para intermediates, we have a unique carbocation resonance form, where the positive charge is on the same carbon as the OH.1034

That is what sets apart the ortho and para intermediates, compared to the meta intermediate.1053

OK, so we found the difference, but here is the key question: is that a good thing, or is that a bad thing?1058

Is the OH a good thing for the positive charge?1065

Well, we know oxygen is electronegative, so you might think, "Oh, that might not be good for a carbocation," but remember, we call the OH an electron donating group.1069

Why do we call it an electron donating group?--because it has lone pairs that it can donate, and if you had a vacancy right there, look what can happen: we can have an additional resonance form to delocalize that charge...additional resonance.1079

How about for the para-can we do that?--sure; instead of moving this π bond up, we could move this; move that lone pair down; and the same thing--we get extra resonance.1106

OK, so that is a good thing: that is what makes the ortho and para paths preferable.1125

OK, so is it good or bad?--it's good: OK, we can say that the OH group stabilizes the positive charge by resonance, by offering an extra resonance form and delocalizing that charge out, over an additional atom.1132

OK, so that means the ortho and para intermediates, and therefore their transition states, are going to be lower energy and favored.1155

OK, if you bring down the energy of your intermediate, you bring down the energy of your transition state: so that means that the ortho/para reactions are going to be faster...these are faster reactions.1171

And therefore, they are going to be the major products.1186

OK, so any time we are talking about the rate of a reaction and which one is faster, that means we are discussing kinetics; this is about kinetics.1190

It is not that we are saying the ortho product and the para product look better or are more stable; it is not about product stability (that would be a thermodynamic argument); instead, it's a kinetics argument, just saying, "Hey, the one with the better intermediate is the one that wins."1200

The lower-energy intermediate is the faster reaction, and that is what leads to the major product.1214

OK, let's see what we can about electron withdrawing groups.1221

A carbonyl is an example of an electron withdrawing group; let's ask the same question--what effect does it have on the benzene ring: why is it called an electron withdrawing group?1225

There must be some way that it pulls electron density out of the ring: well, it does that as shown here.1236

The π bond could be taken out of the ring to put a negative charge up on oxygen.1245

That gives me a positive charge here, and those two π bonds are still there; of course, that means we have additional resonance forms to now delocalize that positive charge.1259

It can move to the bottom carbon, and it can move to the top right carbon; so this is the resonance picture of a carbonyl substituted benzene ring.1272

OK, what does that do to the reactivity of the molecule?1289

What did the electron withdrawing group do? Did it remove or add electron density by resonance? It removes electron density by resonance; that is why it's called an electron withdrawing group.1294

The ring is now more electron deficient as a result.1306

What kind of charges do you see in the ring?--you see positive charges: it has lost electron density because of that resonance withdrawal.1311

It is more electron deficient: so is that good for a nucleophile to be electron poor? That is not what we want for a nucleophile, so the ring is a poor nucleophile, and that is why we describe it as being deactivated toward electrophilic aromatic substitution.1319

Electron withdrawing groups are deactivating groups, because they make the ring a poor nucleophile.1338

OK, how about the regioselectivity?1344

If it did react with an electrophile, where would it go?--well, again, let's consider this resonance: overall, what does my structure look like?1348

I know this oxygen has some partial negative character, and I have partial positive character at the ortho and para positions.1356

So, if I were a positively charged electrophile, where would I go in this ring?1365

Well, I know I certainly wouldn't want to go to where there is already partial positive character: in other words, an electrophile would be repelled by those ortho and para positions.1370

Where would it go?--the only other options: it would go to the meta positions.1380

We are predicting that it is going to be a meta director, and that is in fact what is observed: we can tell that just by looking at the resonance of our starting material.1386

But again, let's go through the process of proving it beyond a shadow of a doubt, so we can see why, in fact, meta is the best place for the electrophile to go.1395

We are going to do the same thing: let's add our electrophile to the ortho position, meta position, and para position; I have left out the π bonds here so we can fill them in and get a little more practice with that.1404

If I added it here, then this would be the π bond that it broke; the other two π bonds are still here, and so our carbocation would be in this position.1416

If I added at meta, that would involve that same π bond; so these two are still there, and there is our carbocation.1424

If I added at para, I would have to break this π bond to add at para; so these two π bonds are still there, and my carbocation is now here.1432

Make sure you draw the correct initial intermediate, and then your additional resonance forms will be correct; if you draw this incorrectly, then you have no chance of drawing the remainders.1442

The electrophile, though, is going to be adjacent to the carbocation.1456

And then, we are always going to have additional resonance forms that we can draw, additional Lewis structures: we can go here; we can move this down; always, these three resonance forms--at least these three resonance forms.1459

OK, and as you get a little better, and you get some more practice, you will find that there are maybe some shortcuts you can take; let me share with you one of those.1480

So maybe when you are doing homework or trying to sketch something out on your own, this is something that can save you a little time.1487

But, of course, on an exam, or any time you want to do a complete explanation, you would certainly want to draw out all of the relevant resonance forms.1493

OK, but without drawing all of the resonance forms here, I can imagine where the positive charge is going to move in the additional resonance forms.1500

It is right here: where would the positive charge be in the next resonance form?1510

Remember, every time we do resonance, it jumps from the first to the third; it alternates carbons; so I know that there is another resonance form that will put the positive charge here.1516

I'm going to just put in parentheses that there is a positive charge character that can go there, and a third resonance form would alternate down here; I know I could put the positive charge down here.1524

This is kind of a shorthand way that you can represent--you can kind of acknowledge--the three resonance forms that would be possible.1535

Draw one complete resonance form, and then you could use the parentheses to indicate what the others would look like.1545

And the same down here: let's do that same shorthand.1550

The positive charge starts here, but I can also move it up here in parentheses, and I can also move it down here.1555

OK, so once again, let's look at those three intermediates and see which one is better.1562

Once again, we are going to always find the same incident, where the positive charge ends up at the same carbon as the substituent, where it's ortho and para, but it is never on that carbon when it's meta.1570

We found the difference--we found the unique resonance form--but the question then is: Is that a good thing, or is that a bad thing?1583

Now, let's take a look at a carbonyl: is a carbonyl group good or bad for a positive charge?1593

Well, what do you know about a carbonyl?--it's electron withdrawing, right?--it's electron withdrawing, because it has this resonance; every carbonyl has this resonance.1601

Now, that resonance doesn't help delocalize the positive charge, but what does it point out?--it points out the fact that there is some positive character.1611

Every carbonyl carbon has some positive character, because it has this resonance that it's capable of.1622

Is it a good thing to have two partial positive charges adjacent to one another?--no, that would be a bad thing.1629

What we have now is destabilized by adjacent partial positive.1638

OK, so is it good or bad for a carbocation?--it is bad for a carbocation.1654

What we could say here is that the electron withdrawing group destabilizes the C⁺, because it has our adjacent positive charges.1660

Which one is favored? In this case, the meta one is favored, not because there is anything great about the meta; it is just not as bad as the ortho or para.1676

Meta is favored--again, the same argument we said: the lower energy intermediate means the lower energy transition state, and therefore a faster reaction.1685

It is favored, and it is the major product.1696

The two things we observed for the electron withdrawing groups: they are going to deactivate the ring and make it less reactive; and they are going to be meta directors, because that is the best place to put the electrophile.1700

OK, so let's take a look at halogens: halogens don't really fit perfectly with either electron withdrawing groups or electron donating groups, so what effect does something like a bromine have on the benzene ring?1715

Well, in this case, it kind of has two things going on: it has inductive withdrawal of electron density, because bromine is more electronegative than carbon; so we know we have this, which kind of makes it like and electron withdrawing group.1728

It pulls electron density, and this is just inductively; so this carbon-bromine bond--these electrons in the σ bond--are being shared unequally, moving towards the bromine.1742

OK, but we also have going on some resonance donation; so because the bromine has lone pairs that are allylic, that are benzylic, it does have the possibility of doing resonance, and it can do resonance, and it does do resonance.1752

That makes it Br⁺ and a C^- and so on; any other resonance forms...we could say "etc."; we could kind of think about where that negative charge would go...1773

It could also be placed here, in the para position, and here, in the other ortho position.1785

OK, so it looks like, "Well, hey, it has resonance; why is this not just an ordinary electron donating group?"1791

The problem is: with the halogens (chlorine, bromine, iodine), you are getting bigger and bigger and bigger; OK, so it has resonance, but this is poorer resonance, because we have a weak interaction between our 2p orbital on the carbon and the bigger 3p orbital on the bromine.1798

OK, so because that interaction is weak, this π bond is weak, this is not a strong electron donation, like we would have with an oxygen or a nitrogen.1821

OK, so it is poorer resonance, but it is enough to direct ortho/para, right?1830

We have a negative charge character on these ortho and para carbons; if you look at our overall picture here, we have a partial positive character on the bromine--a little bit, but still there, and partial negative on the ortho and para positions.1843

So, that is where our electrophile is going to be attracted.1859

You can prove it, just like we did; I'm not going to go through that step, but you could try it yourself: add it (ortho, meta, and para), and what you will find is that the bromine will stabilize an adjacent positive charge; it is a good thing.1864

It will have resonance donation for that carbocation, making the ortho and para intermediates lower energy and a better path to take.1887

It is, in fact, an ortho/para director; but, because the resonance isn't so strong, and because it has this inductive withdrawal, that makes the ring electron poor; the ring is electron poor, and therefore, a poor nucleophile.1896

When you are looking at the rate of the reaction, it is going to be slower than just plain benzene without a halogen on there.1917

We describe this as being deactivated toward electrophilic aromatic substitution.1924

OK, so you can see that it has both some characteristics that it shares with the electron withdrawing groups (because of inductive effects) and some characteristics that it shares with electron donating groups (because of that resonance effect).1938

Let's summarize what we have seen: the electron donating groups (something like an oxygen, nitrogen, or even just an alkyl group)--what they all have in common--what makes them electron donating, and therefore ortho/para directors, is the fact that they can all stabilize an adjacent positive charge.1954

The way that the oxygen does it is: it has lone pairs that it can add in, right?--it has that resonance, so that activates the ortho and para positions and stabilizes that positive charge.1975

The nitrogen has a lone pair, so the same thing--it's easy to spot an electron donating group: anything with a lone pair.1989

And then an alkyl group: now, an alkyl group doesn't have a lone pair, but remember, inductively, these are electron donating groups; so it would be favorable to have a carbocation in this position, because it would be a tertiary carbocation; we know that that is more stable, because of hyperconjugation.1995

OK, so all of these are good for an adjacent positive charge; that is what makes them ortho/para directors.2010

Our electron withdrawing groups are things like carbonyl, cyano, nitro--these are things that have π bonds to a heteroatom (like oxygen or nitrogen), so these pull electron density out of the benzene ring.2017

That is what makes them deactivated--that is what makes them less reactive; you could do that for cyano; you could do that for the nitro and for this positively charged nitrogen.2033

It can't have resonance delocalization, but it can, again, inductively pull electron density away, and that is what makes an electron withdrawing group.2042

All of these would destabilize an adjacent positive charge; it would be a bad thing to have a positive charge next to an N⁺ or a carbon with positive character.2052

OK, so this is a summary: we need to be able to identify any substituent as either electron donating or electron withdrawing, so we know where to put an incoming electrophile.2063

Now, what if we had more than one substituent on the ring, and they were competing on directing the incoming electrophile--who wins in that competition?2077

In order to determine that, we need to know something about the directing power of the substituents.2085

OK, the NH₂ group and the OH group, the O-R...these are all described as strong activators; they have a lot of resonance donation--that makes them very influential on the benzene ring.2090

And so, we will see them winning battles, if they have a competition.2108

OK, when we have a carbonyl attached, we describe these as weak activators; OK, and remember, we actually said the halogens can be described as deactivators, so sometimes you see the order of the alkyl group and the halogens swapped here.2114

OK, but what is important to realize is: these are all not as good as having a nitrogen or oxygen donating electron density, and none of these are as good as just a plain old amine or an oxygen.2132

Why would an ester not be as good at donating electrons as an alcohol?2145

What is the difference here--what happens when I put the carbonyl on an oxygen; how would that affect the donation into the ring?2152

Well, of course, by putting a carbonyl, you now off another place for those electrons to delocalize, and so it is going to be spread out; it is not going to be able to donate as much into the benzene ring as if you just had an OH or an NH, right?2158

The same thing with the amide: the amide is not as effective as a donator, because it is sharing its time, now, with the carbonyl.2172

OK, and then, finally, all of these are going to win out: you can see a big jump here, when you get to the electron withdrawing group; this is deactivating, and so it loses every competition; it is never going to beat out another group, because this isn't telling the electrophile where to go; it is just kind of telling it where not to go.2181

Electron withdrawing groups always lose.2199

Let's see an example: if we had an electrophile come into this disubstituted benzene ring, where would it go?2202

Well, we have a methyl group here; OK, let's think about that methyl group: how does it direct?2210

It is an electron donating group, which means it directs ortho/para, so that means it wants the electrophile to come in ortho to itself, so it can come here, and this ortho position is already blocked; or it can come para to itself.2215

The methyl group wants the electrophile to go in one of those two positions.2233

OK, but we have a methoxy group here with its lone pairs: that definitely makes an electron donating group, and even a stronger electron donating group, because it's more activating.2239

This, too, is an ortho/para director, but it has a louder voice; so where does it want the electrophile to go?--it wants it to come in ortho to itself, or para to itself.2253

Here we have a competition, because both groups can't be satisfied; one of them has to give in, and it is the group with the lone pairs that is going to win the battle.2266

Our major products are going to be one where the electrophile adds an ortho to the methoxy (we are just going to call it an E for now, because we don't really know what groups we are adding), and para to the methoxy.2276

The methyl group is not going to have a chance to tell the electrophile where to go, because it is not as strong of an activator as the methoxy group is.2294

OK, so these are my two products: which one do you think might be major?2303

Well, this one has some steric hindrance, so that is probably the minor product.2306

Typically, the ortho one...or any product that has steric hindrance...is typically formed more slowly, and therefore is a minor product compared to the other one.2313

Finally, let's take a look at some electrophiles; we have just been calling it E⁺ for now--what electrophiles can we add to benzene and other aromatic compounds?2325

OK, let's go through a variety of them, just, again, as an overview; and then, one by one, we will see some examples of those.2336

The first one is halogenation: you can add a halogen, like a bromine or a chlorine, to a benzene ring.2342

These are our reaction conditions: either bromine and iron tribromide or chlorine, iron trichloride, and more.2350

OK, this is just a very, very small sampling of brominating conditions or chlorinating conditions; in fact, there are dozens of ones you can use, so if you see slight derivations of these, it's not a big deal.2357

Typically, what you have is bromine and some kind of loose acid or chlorine and some kind of loose acid.2369

What these conditions generate (and that is the important part) is: they generate Br⁺ or Cl⁺; so that is the positively charged species that is going to end up attacking the benzene ring.2375

Where do these electrophiles come from?--well, what the iron does is: it grabs onto one of the bromines from Br₂, and that makes this a very good leaving group.2386

You can imagine this just leaving to give the Br⁺, or maybe it kind of hangs on until the benzene ring displaces it; but the Lewis acid goes in, grabs onto the leaving group, and helps pull it off, and that is what leaves behind this highly electrophilic reactant species, Br⁺.2401

Or, Cl⁺ would be the same sort of mechanism, with fluorines in place of bromines here.2420

We call that reaction halogenation.2426

Another thing we can do is a nitration reaction; in nitration, we add the NO₂ group, and the conditions we have for nitration are sulfuric acid, combined with nitric acid, combined with sulfuric acid.2429

When you mix these two, you end up forming the NO₂⁺ group; there is the structure for it.2442

And the way this NO₂⁺ is generated is: what happens when you mix nitric acid and sulfuric acid?--well, believe it or not, the sulfuric acid actually protonates the nitric acid.2450

Here is the structure of protonated nitric acid: that makes a very good leaving group right here; it looks like it's water as a leaving group.2462

And so, what can happen is: this O^- can form a π bond and kick that leaving group out, kind of like collapse of a CTI here, except it is not tetrahedral--but the same idea.2473

You have one group pushing it out as the other group leaves, so this loses water, and you form NO₂⁺.2483

NO₂⁺ is the electrophile that undergoes the electrophilic aromatic substitution, so we get a nitro group.2490

This is called a nitro group that we can add on in this method.2498

We can also do a sulfonation: that adds the sulfinyl group, SO₃H; we do that using H₂SO₄ with some SO₃ in there; this is called fuming sulfuric acid (that sounds like fun, doesn't it?).2506

But remember, these reaction conditions all have to be really very harsh and very strong electrophiles; that is the only way we can get benzene to react.2519

So, most of these electrophilic aromatic substitution reaction conditions are extremely vigorous.2527

OK, in this case, SO₃...here is the structure of SO₃ that is your electrophile; and where does SO₃ come from?2534

Well, again, if you manage to protonate H₂SO₄, if you add an extra proton to H₂SO₄, you make that good leaving group again.2540

That leaving group could leave; you can have it leave with assistance; and then deprotonation can get to the SO₃; but this is kind of the key step in generating the SO₃, but that is often part of the ingredients, as well.2550

Now, what is cool about this reaction is: this is great if you want to add a sulfinyl group, like if you wanted to make tosyl chloride or something, this would be a way that you get that sulfinyl group onto the aromatic ring.2567

OK, but also what is very interesting about it: this reaction is reversible.2577

These are the reaction conditions to put on the SO₃H group; if you were to heat that substituted benzene, you could lose the SO₃H group; so we will see some examples where we can take advantage of that reactivity and that reversibility.2581

Now, there is an important class of reactions called the Friedel-Crafts reactions; these are important because they involve carbon electrophiles.2601

If you have carbon electrophiles reacting with benzene, you are forming new carbon-carbon bonds; so Friedel-Crafts is really important in the synthesis of carbon chains containing aromatic rings.2607

One of them is called the Friedel-Crafts alkylation reaction; that is where we add some kind of alkyl group.2619

And the way we add that alkyl group is as a carbocation.2625

There are many, many ways we can form carbocations: some we have seen before--for example, down here, if we just take an alkene in the presence of some acid (HF is a good acid to use in this case), what happens is: that alkene gets protonated, and as a result, you form a carbocation.2630

Well, a carbocation, a very strong reactive electrophile, would react with a benzene ring.2648

OK, some other ways we have seen forming carbocations: if you take an alcohol and treat it with a strong acid, the acid will protonate the alcohol to make a good leaving group, and that leaving group could leave.2655

OK, so these are ways we have seen in the past; some methods that are employed typically for electrophilic aromatic substitution involve...instead of using a strong Bronsted acid, we use a strong Lewis acid (like boron or iron here).2667

We could use boron with the alcohol, and Lewis acids are kind of like big protons; so just like a proton would add to this oxygen and make it a good leaving group, a boron or aluminum could add to this oxygen and make it a good leaving group.2683

This is now a good leaving group; it would be stabilized when it leaves; so it can just leave--that would give the carbocation.2699

OK, we could also have a leaving group, a halogen, on a carbon chain; we know that those can leave on their own to give carbocations away.2705

To initiate that leaving, or really force them to leave, is to add in (kind of like we did with the bromine and chlorine) iron tribromide, iron trichloride, something like that.2713

That can go in and grab that leaving group and pluck it off even faster.2722

OK, so again, this makes an even better leaving group, and so the leaving group can just leave and form a carbocation.2727

Lots of different methods to form a carbocation...any time we have a carbocation reacting with a benzene ring, we call it a Friedel-Crafts alkylation.2735

And then, finally, there is a reaction called a Friedel-Crafts acylation; that means we are adding an acyl group (that is a carbonyl group like this).2744

The way we add a carbonyl group is: we start with a carbonyl with a leaving group (like an acid chloride or an anhydride--here is our leaving group; here is our leaving group; OK), and we want that leaving group to leave.2754

It is on a carbonyl, so we have never seen that reaction before; but if you add a Lewis acid, like aluminum trichloride, what happens is: that adds to the leaving group and pulls it off.2768

That is, again, a way of plucking off a leaving group; in fact, this can leave with the assistance of that lone pair to give this resonance form; this is called the acylium ion.2783

When you have the acylium ion, this would react with a benzene ring to install a carbonyl group.2794

We call it the Friedel-Crafts acylation.2801

OK, so there is an overview of the various electrophiles we can add; let's look at them one by one and see some examples and some experimental details--some mechanisms and so on.2804

Let's take a look at that nitration: if we were to predict the product here...OK, our reaction conditions are HNO₃, H₂SO₄.2814

We are going to have a lot of reagents in this lesson--a really good time to pull out some flashcards to keep them all sorted out on who does what, because when we go to predict a product, we have two questions we have to ask.2823

The first question is: What are we adding?2835

So, how do these reagents combine to form an electrophile; what is that electrophile?2837

Well, these are the nitration conditions; so these will add an NO₂ group.2843

Now, the second question is: Where are we adding it (in other words, the regiochemistry)?2850

If it's just benzene, you just add it anywhere, and you have nitrobenzene; but this is chlorobenzene we are starting with, so we already have a group on the ring; we need to decide what kind of director that is, to know where the nitro group is going to go.2857

OK, so what do you think?--this is a halogen; I know halogens have lone pairs--that is our clue for what kind of director it is.2872

Anything with lone pairs is going to activate that ortho, and therefore the para position; it is an ortho/para director.2878

Halogens are ortho/para directors; so that means my nitro group is going to come in either ortho (it doesn't matter whether you draw it on the right or left--it is the same compound; there is no stereochemistry to worry about with aromatic reactions, because everything is planar--everything is flat)...so we can add at ortho, and we can add at para.2888

Our major product is probably going to be the para substituted, because there is a little less steric hindrance there (para is major--sorry).2911

We get some ortho, and we probably will get para as our major product.2922

OK, so we should be able to do a complete mechanism for this.2927

Here are our ingredients: we have chlorobenzene, nitric acid, and sulfuric acid; let's look at our mechanism.2932

Now, we saw before that the electrophilic aromatic substitution was a 2-step mechanism; that is assuming you have your hot, reactive electrophile ready to go.2937

If you don't...in this case, we don't: we need to make that first; that will always be the initial step(s) of the electrophilic aromatic substitution.2948

Our first thing we are going to do is: we are going to take nitric acid, which has an O^- and an N⁺ (nitric acid is always a strange-looking Lewis structure)--we will take nitric acid, and we are going to protonate it with sulfuric acid.2957

Sulfuric acid is a very strong acid--so strong that we can actually protonate nitric acid; I'm going to protonate it on this oxygen, because when I do that, it's going to make a great leaving group, right?--it makes water.2982

When I protonate over here, it forms a water leaving group; so this O^- can kick down and kick it off.3002

I kick out water as a leaving group, and I make this structure, which has a positive charge on nitrogen; here is my NO₂⁺.3013

So, nitric acid and sulfuric acid combine to make NO₂⁺.3022

Now, what does that NO₂⁺ do?--that NO₂⁺ reacts with benzene, so now we are ready to do our substitution reaction.3028

We have chlorobenzene; we want to add it to the para position--let's show the mechanism to get to the para product, which means I'm going to have to use this π bond down here that is attached to the para carbon.3038

It attacks the nitrogen, and then what do you think happens?3048

I think I'm going to break one of the π bonds; very good.3052

These two π bonds are still here; we have a new group added here; we have a nitrogen, single bond, O^-, double bond, O; end with an N⁺.3060

There, we have just made a nitro group: we made an NO₂ group.3072

OK, and what happens to the benzene ring?--we broke this π bond, so there is a positive charge here.3075

Now, we know this has resonance stabilization; I could put the positive charge here; I could put the positive charge here; that is how we would explain why it preferred to add at the para position and not at the meta position.3079

OK, but we don't need to draw out those resonance forms every time we do a mechanism; we can just use this Lewis structure and then continue on with our next step.3091

What is our second step?--remember, it's a substitution: the first step is: add the electrophile (add the E⁺).3100

The second step is: remove the H⁺.3107

So we have to deprotonate as our second step; where do we deprotonate?3110

A very, very common mistake is to grab the proton from the carbocation: that is not going to help us, and that is not the proton we need to lose.3117

The proton we need to lose is the one that was originally in the para position, that is no longer going to be there in the product.3124

OK, so all we can do...we can just use A^-, or we could use the conjugate acid here; we could just say :B...anything you want to use there--anything at all--will be strong enough to deprotonate and give us back our aromatic ring.3131

There is our complete mechanism: however many steps it takes to make the electrophile, and then two steps to do the substitution part.3148

This is nitration--addition of NO₃.3157

OK, let's take a look at an example of this reaction with aniline as our starting material.3161

Aniline is a benzene ring with a nitrogen group and an H₂ group.3165

So, what are we adding, and where are we adding it?3171

HNO₃, H₂SO₄ is a way to nitrate; it's going to add the NO₂ group.3175

Where do we expect it to add?--well, we need to look at the NH₂ group and think about how that directs an incoming electrophile.3181

Let's draw in that lone pair, because I know those are relevant.3189

It looks like an ortho/para director: it looks like an electron donating group, and that makes it an ortho/para director.3194

Remember, we have this action going on: that resonance tells us everything--that is all we need to know.3200

OK, so what do we expect as our major product?--we expect the nitro group to get added to the para position as the major product.3206

That is an excellent guess: however, in this case, that doesn't happen.3215

The major product is, in fact, the meta product.3220

What is going on here--how does that happen?--after everything we have just learned, we would not expect this.3231

OK, so when we ask "Why?", what we need to think about is: look more closely at our reaction conditions--how would you describe these reaction conditions?...nitric acid, sulfuric acid--extremely acidic; definitely acidic.3238

I know I have plenty of acids around: well, an amine, like aniline, is a base.3252

Do you know what is going to happen here?--an acid-base reaction--a proton transfer.3258

Any time an acid-base (or proton transfer) reaction can take place, it will: it is the fastest reaction mechanism we have.3262

So, what is going to happen is: we are going to protonate that amine; so in acidic conditions, we do not have aniline; we have protonated aniline.3269

This is what is going to react with the NO₂⁺.3283

Now, what kind of group do we have up here?--we have a positive charged nitrogen; there is no longer the lone pair to donate; so, in fact, we just turn this into an electron withdrawing group.3288

How do electron withdrawing groups (EWGs) direct?--they are meta directors.3301

So now, when the nitro group adds in, it is going to add in the meta position.3309

We are assuming a neutralization step here, so that we get out the neutral aniline there; I probably should have that as part of your conditions, so that we know, after workup, we would get our neutral product back.3316

But it was in the reaction, when we had our electrophile, that it was the protonated form; and that would change it to be a meta director.3330

OK, so this is something to keep in mind for this reaction in particular, because nitration is always strongly acidic; aniline is always a good base; and so, this is something that we would expect to have happen.3340

Now, what if I wanted to make this product--what if I really wanted to make para nitroaniline--how could I do that?3354

Is there any way I can prevent this protonation from occurring and changing the behavior of my substituent?3361

Well, there is, in fact: OK, the way we handle this is: we are going to protect our amino group--we are going to add a protective group to kind of shield it from the acidic conditions.3369

What we are going to do is: we are going to make an amide; so we are going to convert the amine to an amide.3382

What kind of reaction conditions would we need--what reagent would cause that conversion?3387

I know I need this acetyl group of a carbonyl and a CH₃; what has to be on this carbonyl?3394

It looks like we want the nitrogen to do an addition-elimination: we want to still have this carbonyl.3405

We need a leaving group; so if we use an acid chloride and some base to react with the HCl that is formed, we can get a substitution, and we would get the amide product out.3410

OK, that is how we make an amide.3423

Now, think about this amide: this nitrogen still has a lone pair, so it is still an ortho/para director, but because it has this resonance with the carbonyl, that no longer makes it basic: it is no longer basic because of that resonance.3425

That lone pair is delocalized in resonance, so it wouldn't want to get protonated.3449

So now, when I do my nitration reaction, it comes in: this is still an electron donating group, so it is still an ortho/para director, and it comes in the para positions we want.3454

And then, as usual, when we put on a protective group, we need, at some point, to be able to take it off.3464

The way you take off an amide group is base and water.3469

It undergoes hydrolysis: what mechanism is that?--again, addition-elimination.3474

We are going to cleave this bond, add in hydroxide, kick out the leaving group; and we get back our amine.3478

This is kind of classic protective group strategies, where you hide your functional group, do a reaction, and then regenerate your functional group when you need to.3486

OK, let's take a look at sulfonation: when we have this fuming sulfuric acid, we can add the SO₃H group; so that is what the SO₃H group looks like.3498

Now again, we have seen that kind of pattern when we have used tosyl chloride or tosic acid; this is very similar to that.3514

Tosic means it has the tolyl group, so there is a methyl group here in the para position.3522

OK, but either way, this is how you could introduce that functional group onto a benzene ring.3527

But what makes it especially interesting as a reversible is: if you add heat, if you heat a sulfonate, you can cause that group to come back out and get resubstituted back with a proton.3533

OK, we are not going to look at the mechanism for that; it is just simply the reverse mechanism.3545

But instead, what we are going to look at is an application of this that we can use in synthesis.3550

And because we can add this group on and then take it off, it kind of looks like and kind of sounds like it can act like a protective group.3555

That is exactly how it can be used: we can use it as a blocking group.3561

Let's look at this example: let's say I had phenol, and I wanted to make ortho chlorophenol.3565

How could I do that?--could I just...the way we put in a chlorine group is: we add chlorine and FeCl₃; Cl₂ and FeCl₃ generates Cl⁺; so would that work?3572

Well, the oxygen is an electron donating group, which makes it an ortho/para director; but would ortho be your major product?3585

It would not be your major product; it would give para as the major product.3598

OK, and even if it wasn't the major product, you still would definitely be getting para product, and so you would have a mixture that you would have to separate.3605

That would not be a very good synthesis, if you are getting your desired target molecule as a minor product or one of several products (or one of two products).3612

Instead, what we are going to do is: we are going to treat this with SO₃, H₂SO₄, and instead, sulfonate at the para position--add the SO₃H group at the para position.3620

And what is cool about that is: now, if we chlorinate (Cl₂, FeCl₃), what happens?3635

This oxygen is an ortho/para director, which means he is directing to one of (let's put it over here, because that is how we have it drawn in the product) the ortho positions, but the para position is now blocked.3643

I can't have a substitution at that position, because there is no proton there.3664

So, it can only go to the ortho position; but we have added a second group here, so we need to think about its reactivity--how does it direct?3668

If you had this substituent on the benzene ring, would you expect that to be an electron donating group or an electron withdrawing group?3676

It has a π bond, so I would expect that it would be something that could pull electron density out of the ring; it, in fact, is an electron withdrawing group, so it's a meta director.3684

It is a meta director...so where does the sulfinyl group want the chlorine to go?3695

It wants it to go meta to the sulfinyl, which, in fact, is the same place that the OH group is telling it to go.3703

So, in this case, we don't have a competition between the two groups; both of them are pointing to the same position, saying, "That is where we want the electrophile to go."3709

So again, that is great for a synthesis, because now we have one major product that we are expecting.3716

And now, all we have to do is heat this (usually with some acid); we heat it, and we lose our SO₃H group, our blocking group, and we get the desired product.3724

So, because of this reversibility, it kind of makes it a unique electrophile and has some interesting applications, like we just saw.3734

OK, let's talk about the Friedel-Crafts alkylation: this was the reaction where we generated the carbocation one way or another; and it is that carbocation that ends up adding to the benzene ring to make an alkyl benzene.3745

Like I mentioned, this is a very good reaction to know about, because it is a way of forming new carbon-carbon bonds.3762

That is huge--hugely significant when we are talking about synthesis, or talking about creating new organic molecules.3771

OK, one thing to point out about the Friedel-Crafts akylation and acylation that we have seen before is: none of these Friedel-Crafts reactions work if there is an electron withdrawing group present.3777

It just has to do with the reactivity of the electrophile: remember, an electron withdrawing group is deactivating.3788

We know it already slows down the reaction with an electrophile; but it turns out that it slows it down so much that the Friedel-Crafts doesn't go at all.3801

You might see that on occasion--no reaction here and there--and that would be why; so we want to avoid those Friedel-Crafts when we are doing our planning.3809

OK, so let's see an example: we have benzene; we are mixing with cyclohexene and some acid.3818

What reaction do you think we are going to have happen here?3825

Well, let's start with the mechanism, and see if that helps us with our product.3827

We are going to combine these two; and what happens when you have an alkene in the presence of an acid?3832

That alkene is going to get protonated, right?3841

We look around for something to protonate with the acid; and we can protonate it to give a carbocation.3843

Right away, I see what my electrophile is; so I know what group I'm going to add to the benzene.3853

Let's draw our product, then: it doesn't matter where we attach it to this benzene, because the benzene has no substituent on it already: we are just going to be adding a 6-membered ring to that benzene--that is our electrophile.3858

We add an alkyl group--any Friedel-Crafts alkylation (in this case, a cyclohexyl group).3871

OK, so we made our electrophile: now, once we have our electrophile, we can do our substitution reaction.3878

We'll bring in benzene; it doesn't matter how you draw benzene; it doesn't matter which π bond you pick here--any π bond will do.3885

It is that π bond that is going to attack the electrophile: if I use that π bond, I can draw it down here.3894

OK, that is step 1, remember: add the electrophile; that is how we do our mechanism for the electrophilic aromatic substitution.3907

Step 2 is what?--lose a proton: deprotonate.3915

Think about where you need to deprotonate before you do this next step: where is the hydrogen?--right here.3920

It was whatever carbon we added the electrophile to; that is where we are going to deprotonate; and again, just use anything at all to deprotonate; and you are done.3927

The Friedel-Crafts alkylation was three steps, because it only took one step to form the carbocation intermediate.3938

Now, let's talk about some details about the Friedel-Crafts alkylation, OK?--because there are a few drawbacks with this method.3945

OK, one problem is that it can overreact: we can get a dialkylation, or even a trialkylation, rather than just a monoalkylation.3952

OK, so here is an example: let's say we had this alkene, and we mixed it with this alcohol and an acid: now, what happens when you combine an alcohol and an acid?3960

That is going to protonate the alcohol; let's just do a little bit of a mechanism here for practice.3971

I would expect that to give us a protonated alcohol; now we have an excellent leaving group, so what can happen to that leaving group?--it can just leave.3982

We can lose water, and we can form this carbocation.3992

So remember, one of our questions we have to ask is: what are we adding?3996

When you have ethanol and sulfuric acid, you are going to be adding the ethyl group.4001

Here it is: we added the ethyl group, but the problem is: how does an alkyl group affect the aromatic ring?4006

How does it act?--as an electron donating group?--an electron withdrawing group?4016

Remember, alkyl groups, inductively, are electron donating; so the problem here is that this is now more reactive than your starting material--than the benzene that you started with.4020

When you have this carbocation around...remember, we don't just have one molecule of these things--we have millions of them; and when you have this carbocation around, and it goes to react with something, it is going to prefer to react with the ethyl benzene, rather than just the benzene.4040

What product would it get?4055

This is an electron donating group, so how does it direct?--it is an ortho/para director, so I could also get some diethyl benzene--some para diethyl benzene--as another product.4056

And, in fact, I could get trialkylations and so on, because this is now even more electron rich.4071

Of course, now it has some steric hindrance, so the future reactions are not quite as fast; but certainly, you can have this dialkylation happening.4075

If we have a huge excess of your benzene, then you can suppress this a little bit; but it's always a battle that we are facing, and that is going to be true any time the product you are forming is more reactive than the starting material--this is always going to be something that you are fighting.4084

OK, another problem is that we are involved with a carbocation--the mechanism...we know that carbocations can rearrange, so if you have a very stable carbocation, this is a great mechanism.4102

If you are trying to put in a carbocation that is not stable, it is going to be a very difficult mechanism--it is not going to work, as a matter of fact.4112

Let's see another example: if we have this propyl bromide, n-propyl bromide, reacting with a Lewis acid, that is going to pluck off the leaving group and give us a carbocation.4118

But, rather than give us the 1, 2, 3-carbon chain adding right here...why would it add to the para position?--let's check that regiochemistry really quickly.4135

Why would it add to the para position here?--because this has lone pairs; this methoxy group has lone pairs; that makes an electron donating group, and therefore an ortho/para director.4143

Everyone is an ortho/para director, except for electron withdrawing groups.4155

He is an ortho/para director, but instead of getting the n-propyl group as we might predict, that is not formed; instead, we get isopropyl--which is great, if you want to form isopropyl, but not if you want to form the other one.4160

So, let's take a look at this mechanism: let's propose a mechanism to get to this isopropyl, and then we will talk about, "Well, OK, coming back to this problem--what if I really wanted to make n-propyl?--what if I wanted to make this compound--how could I do it?--because I obviously can't do it using Friedel-Crafts alkylation."4173

OK, so let's look at a complete mechanism.4189

Aluminum trichloride--like BF₃, this is a Lewis acid; it is a Lewis acid because it has a vacancy--this aluminum does not have a filled octet; there is no lone pair here.4192

It just has 3 bonds, and so it is very, very reactive, because it has no filled octet.4203

It is seeking a source of electrons: any time it sees something like a halogen, something with electrons, that is going to add to it.4209

Let's just go to a line drawing here: and it is going to bond to the halogen.4219

That bromine now has 1, 2, 3, 4, 5, 6 electrons; we know halogens want 7, so it is missing an electron.4225

That is a Br⁺; and this aluminum wants only 3, so that is why it was neutral over here, but now it has 1, 2, 3, 4; this is now negatively charged.4235

OK, so the Lewis acid is kind of like a giant proton; instead of protonating the bromine, we are going to put the aluminum trichloride on the bromine; that turns this into a great leaving group--a super leaving group, even better than bromine was on its own.4246

OK...which means it leaves, but when it leaves, it gives a primary carbocation: is that a good carbocation?--that is an awful carbocation--very unstable.4262

And, if there is any way that it can arrange to a more stable one, it will do that; and, in fact, right next door, it has a secondary carbon, so the carbocation can get over there by taking one of these hydrogens from the middle carbon and shifting it over.4275

That is going to be a very favorable rearrangement; we call that a hydride shift, because the hydrogen with its two electrons (it is like an H^-) picks up and moves over to the next atom.4291

And, as a result, now this carbon is missing a bond; so that, now, is where the carbocation is.4304

We have moved to a secondary carbocation, which is more stable.4310

Always, always, always, if we have a carbocation, and it has a way of rearranging, it's going to do that.4316

And so, we really need to keep that in mind with this mechanism.4320

OK, let's finish up our mechanism: how do we go from the isopropyl carbocation to the product?--well, now we do our two-step substitution mechanism.4324

We use methoxybenzene anisole here, and this is going to act as our nucleophile; the carbocation, of course, is a great electrophile; so in the electrophilic aromatic substitution, our benzene ring acts as a nucleophile.4335

We will use this para carbon, so that we attach the isopropyl group to the para position.4348

That leaves these two π bonds intact and gives us something missing here; it gives us a carbocation in this position.4357

That was what puts the carbon group on there; how do we finish this off?--we need to get our aromaticity back; we need to deprotonate.4368

We added the electrophile, and now we lose a proton.4376

It is right here that we want to lose that proton; we could just use an A^- here, or a B--anything at all can come and grab that proton, and we are done.4382

Always be aware, in a Friedel-Crafts alkylation, that if a carbocation can rearrange, it will rearrange.4395

Let's come back to this n-propyl: what are some strategies we can do to install an n-propyl group when we know we can't do it through a carbocation, because a carbocation would most definitely rearrange?4403

Well, in that case, one strategy we can have is: instead of using a Friedel-Crafts alkylation, we can use a Friedel-Crafts acylation.4417

So, instead of using just propyl chloride, we are going to use this acid chloride and a Lewis acid; and what happens there, again, is: just like we saw before, the aluminum is going to add on (let's just show that mechanism really quickly) to the chlorine leaving group, and it plucks it off.4424

Lewis acids grab onto leaving groups--halide leaving groups--and pluck them off.4451

What we can show is: we can show it actually getting kicked out with the help of this carbonyl, and that forms a resonance-stabilized acylium ion.4455

OK, now what is nice about an acylium ion is: there is no chance for a rearrangement, because it is already resonance-stabilized--if you just had the leaving group leave on its own, you would get this resonance form.4473

It is this carbocation, or you could draw it as the O⁺; it is already resonance-stabilized, so it has no other better place to go.4491

There is no chance of rearrangement, and it only adds once--because remember, another problem we had with the Friedel-Crafts alkylation was that it could add multiple times; you could get a dialkylation...4498

In this case, it only adds once; what is the difference here that makes that true?4510

Why is it, when you add an acyl group, the reaction is done?4516

Well, let's think about the effect that this acyl group would have on the benzene ring: is it an electron donating group or an electron withdrawing group?4521

When it has the carbonyl, this is an electron withdrawing group, and so that deactivates the ring and makes it less reactive.4530

So remember, the electron withdrawing group is deactivating.4539

So now, our product is les reactive than our starting material; so it is very easy for it to add just a single time.4545

In fact, remember: when we said once there is an electron withdrawing group on the ring, the Friedel-Crafts reaction is so slow that we say it is no reaction.4550

So, we are pretty much guaranteed it is only going to add once here, and with no rearrangement.4558

OK, the other cool thing about the Friedel-Crafts acylation is: after you put this carbonyl group on here, it can be reduced: the carbonyl can be reduced.4565

And, if we had an appropriate reducing agent, it is possible to take that carbonyl and convert it all the way to a CH₂ to completely reduce it.4577

And so, look what we get: we ended up adding a 3-carbon chain in a linear fashion, an n-propyl group that we couldn't do when we tried to do it with the Friedel-Crafts alkylation.4586

So, this strategy is a great way of adding straight carbon chains to a benzene ring.4597

OK, let's talk about these reducing agents, though: what can we do to effect this reduction, if we wanted to go all the way from a carbonyl down to a methylene?4606

So, it's a complete reduction--not just to an alcohol, but all the way to a methylene.4616

Well, there are a few things we can do here: we can do either tin or metal amalgam with HCl; this is our Clemmensen reduction.4619

We have seen this before--it is called the Clemmensen reduction: a metal-promoted reduction reaction.4637

Or, we could use hydrazine and KOH and heat; this is called the Wolff-Kishner reduction.4643

We have actually seen these reductions before for general ketones: so these two methods are good for all ketones.4651

They will take a ketone and completely reduce it down to a methylene (and we have seen that before).4660

OK, however, in this case, because it is benzylic--because the carbonyl is benzylic, the benzylic position has some special reactivity, because it is next to these p orbitals.4665

It turns out that catalytic hydrogenation, even, will completely reduce the carbonyl to an alkane.4677

This is OK for reduction of benzylic carbonyl only; OK, if you try to do an ordinary carbonyl with catalytic hydrogenation, it wouldn't work: remember, catalytic hydrogenation is usually something we use for alkenes, alkynes, carbon-carbon π bonds...4685

So also notice: it is not reacting with the benzene ring; so benzene is not an ordinary alkene--it is not going to be reduced under ordinary catalytic hydrogenation conditions; but a benzylic carbonyl would.4711

OK, let's look at a few examples of synthesis problems that utilize the electrophilic aromatic substitution.4726

Here is a transform: so what conditions would you use to convert the given starting material into the desired product?4732

OK, we see that we have a 1, 2, 3, 4-carbon chain here; those 4-carbon chains are still here; so this looks like a new carbon-carbon bond that is being formed.4740

And this is a bond we just learned how to make: we could attach a carbonyl to a benzene ring by doing a Friedel-Crafts acylation.4750

What are the two ingredients we need to do a Friedel-Crafts acylation?4765

For this part (the carbonyl group), we need a really good leaving group here (either the acid chloride or the anhydride--either one is fine), plus AlCl₃; that is what we need to generate that great leaving group, so that we get the acylium ion.4769

OK, and what did the benzene ring look like before it did the Friedel-Crafts acylation?4788

What was on this carbon of the benzene ring?4795

Is there a leaving group?--no, it's just the hydrogen; all you need is a hydrogen on an aromatic ring, in order to do a Friedel-Crafts acylation.4799

So, it is just like we erased that group and all we need is plain old benzene; so I just want to make sure we are pointing that out...on what your starting materials are, not just being able to draw what your products are, for any electrophilic aromatic substitution.4806

OK, so what does our synthesis look like?--we start with the carboxylic acid; we need the acid chloride; so what reagents do we use for that?4822

Well, we could use thionyl chloride, SOCl₂, to make the chlorine here for the acid chloride.4830

And then, what do we do with the acid chloride?--well, we just mix it with benzene and some aluminum trichloride, and we get our Friedel-Crafts acylation.4838

That is just going to be a 2-step synthesis, but being able to do the retrosynthesis and recognizing that this is a bond that is formed in a Friedel-Crafts acylation is the key to this problem.4847

OK, other problems with electrophilic aromatic substitution involve synthesizing multi-substituted rings, aromatic rings; and these get interesting, because what we have to decide is not only "How do we add these groups--what reaction conditions?", but "In what order do we add the groups?"4860

So, the key here that we have to consider is "Which do we add first?"; we know how to add a nitro--we have seen that reaction; we know how to add a bromine--we have seen that reaction; which one do we add first?4879

We can either add in the nitro first, and then the bromine (there are only two choices, right?--so let's look at both of them), or we could add the bromine first (let's draw it down here, just so it matches the orientation of our product), and then nitrate.4891

Which is the proper one to choose, or are they both going to work?4913

Well, the key here is: remember, once you add one group onto the benzene ring, that group is going to influence the regiochemistry of the second group, on where it goes.4918

How would you describe the relationship of these two substituents?4927

They are meta to each other; so how do we get the meta?--we need to add the meta director first.4932

Which one is a meta director?--the nitro group is an electron withdrawing group; that is a meta director.4944

The halogen is an ortho/para director, because it has these lone pairs that make it donate electron density by resonance.4951

We can't add the bromine first, because if we added the bromine, then the nitro would add para to the bromine, and that would not be our target molecule.4960

OK, so keep that in mind when we are doing our planning: so this is the better plan--this is the one that is going to work; we have to nitrate first, and then we have to brominate.4973

So now, it's a matter of reviewing those flash cards and thinking about those different reagents.4982

What are the reagents that we need to do each of these transformations--what is it that adds a nitro group, an NO₂ group?4986

We need HNO₃, H₂SO₄.4993

Nitric acid, sulfuric acid: those are our nitration conditions; and then how do we add a bromine?4998

We are going to use Br₂, and then some strong Lewis acid.5006

Something like FeBr₃ is what you typically use, but you could also use AlCl₃ or FeCl₃; there are a lot of different variations here that you might see, and so do recognize that there is more than one answer when it comes to the reagents, but it is the strategy that will only have one solution.5010

OK, so this wraps it up for our first part of aromatic reactions; we are going to see some reactions other than the electrophilic aromatic substitution in the next lesson.5031

I hope to see you soon; thanks.5042