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Lecture Comments (5)

1 answer

Last reply by: Professor Starkey
Sun Sep 13, 2015 10:17 AM

Post by Jinhai Zhang on September 11, 2015

Dear Prof. Starke:
For Reduction of -NO2 to NH2on the benzene ring, you mentioned that we can use H2, Pd to reduce it  
In my textbook, they also mentioned that we can use Sn, HCl and OH-. how does this work? thank you very much

1 answer

Last reply by: Professor Starkey
Wed Oct 8, 2014 1:22 AM

Post by Brandon West on October 8, 2014

Dr. Starkey,

Do you have any info on Birch reductions?

0 answers

Post by Jason Jarduck on February 26, 2012


Great Lecture!!!

Thank You

Jason Jarduck

Aromatic Compounds: Reactions, Part 2

Devise a synthesis for this reaction using diazonium salt:
  • Step 1:
  • Step 2:
  • Step 3:
  • Step 4:
  • Step 5:
Devise a synthesis for this reaction using diazonium salt:
  • Step 1:
  • Step 2:
  • Step 3:
  • Step 4:
  • Step 5:
Draw the major products for these reactions:
Draw the major products for these reactions:
  • Benzylic carbon can't be quaternary:
Draw the major product(s) for the following reaction:
  • This is a nucleophilic aromatic substitution reaction.
Draw the product(s) for the following reaction:
  • This is a nucleophilic aromatic substitution reaction.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.


Aromatic Compounds: Reactions, Part 2

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Reagents for Electrophilic Aromatic Substitution 0:07
    • Reagents for Electrophilic Aromatic Substitution
  • Preparation of Diazonium Salt 2:12
    • Preparation of Diazonium Salt
  • Reagents for Sandmeyer Reactions 4:14
    • Reagents for Sandmeyer Reactions
  • Apply Diazonium Salt in Synthesis 6:20
    • Example: Transform
  • Apply Diazonium Salt in Synthesis 9:14
    • Example: Synthesize Following Target Molecule from Benzene or Toluene
  • Apply Diazonium Salt in Synthesis 14:56
    • Example: Transform
  • Reactions of Aromatic Substituents 21:56
    • A) Reduction Reactions
  • Reactions of Aromatic Substituents 23:24
    • B) Oxidations of Arenes
    • Benzylic [ox] Even Breaks C-C Bonds!
    • Benzylic Carbon Can't Be Quaternary
  • Reactions of Aromatic Substituents 26:21
    • Example
  • Review of Benzoic Acid Synthesis 27:34
    • Via Hydrolysis
    • Via Grignard
  • Reactions of Aromatic Substituents 29:15
    • C) Benzylic Halogenation
    • Radical Stabilities
    • N-bromosuccinimide (NBS)
  • Reactions of Aromatic Substituents 33:08
    • D) Benzylic Substitutions
  • Reactions of Aromatic Side Chains 37:08
    • Example: Transform
  • Nucleophilic Aromatic Substitution 43:13
    • Nucleophilic Aromatic Substitution
  • Nucleophilic Aromatic Substitution 47:08
    • Example
    • Mechanism
  • Nucleophilic Aromatic Substitution 50:43
    • Example
  • Nucleophilic Substitution: Benzyne Mechanism 52:46
    • Nucleophilic Substitution: Benzyne Mechanism
  • Nucleophilic Substitution: Benzyne Mechanism 57:31
    • Example: Predict Product

Transcription: Aromatic Compounds: Reactions, Part 2

Hi; welcome back to

Next, we are going to talk about some more aromatic reactions that can happen.0003

So far, the reaction we have seen is one called electrophilic aromatic substitution, and it looks something like this: we start with benzene (or some substituted benzene), and one of the hydrogens on that ring ends up getting replaced by an electrophile.0008

We end up attaching an electrophile onto the aromatic ring: we call that electrophilic aromatic substitutions, and we have seen a wide variety of electrophiles that can be introduced to an aromatic system this way.0028

If you use Br2, FeBr3 or Cl2, FeCl3, that would be your way of forming Br+ or Cl+; so we can brominate or chlorinate a benzene ring this way.0040

HNO3, H2SO4 combine to form NO2+: we could do a nitration.0053

SO3, H2SO4 is a way of adding a sulfinyl group, an SO3H group, a sulfonic group.0060

And all of these various methods are ways of generating carbocations; if you have a carbocation, you could do a reaction called a Friedel-Crafts alkylation reaction, in which you add an alkyl group.0069

We form a carbocation by protonating an alcohol with a Brönsted acid or a Lewis acid and causing water to leave, or using a Lewis acid to remove a leaving group or protonate an alkene, and so on.0091

And, if we were to take an acid chloride and aluminum trihalide, that can pluck off the chlorine and make an acyl cation, called an acylium ion, and that reaction is called a Friedel-Crafts acylation.0103

We can add an acyl group or an alkyl group or a variety of these other substituents.0116

We can use combinations of these electrophilic aromatic substitution reactions to make all sorts of interesting benzene rings and disubstituted benzene rings and so on.0121

But there are a few other reactions we can do: one such synthetic strategy is using something known as a diazonium salt; it has an N2+ group here, and that is shown right here.0130

The way you make an N2+ group is: you start with a nitro group (and that is interesting, because we know how to put on a nitro group--that is good to know); we reduce the nitro group to an aniline derivative.0145

We take the nitro and we reduce it to an aniline: we could do that with either catalytic hydrogenation or some kind of metal reduction (like iron or zinc or tin--some acid and metal).0160

OK, so we could reduce it to NH2, and then what we do is: we react the aniline with sodium nitrate and HA; that is going to...our NO2- and our H+ are going to combine to form nitrous acid.0172

HONO: that is not an abbreviation for the acid--that is the actual structure, HONO: it is HNO2.0191

This is a very unstable acid: it can only be formed in situ; so this is formed in situ (meaning in the reaction conditions from these reagents).0198

What happens to an NH2 group when you treat it with nitrous acid is: it gets converted to an N2+ group.0208

I'm not going to get into that mechanism of that transformation, but we just need to know the reaction conditions, the reagents, in order to prepare this diazonium salt: it's two steps from the nitro group.0216

What we have just done is: we have introduced a great leaving group onto the benzene ring.0231

If this were to leave, it would be leaving as a molecule of nitrogen gas: you can't get much more stable than nitrogen gas, so by putting a great leaving group here, it gives us the opportunity of doing some substitution reactions, where we can replace that leaving group with a variety of nucleophiles.0238

Some of those reactions are called Sandmeyer reactions; we call it a Sandmeyer reaction when we use a copper salt, especially.0256

So for example, if we had a diazonium salt, and we treated it with copper bromide or copper chloride, we could replace an N2 group with either a bromine or a chlorine.0262

Or, copper cyanide--we could put in a cyano group; so that is pretty cool: we are seeing, now, how you can make an aromatic nitrile.0273

OK, there are a variety of other nucleophiles: if you use potassium iodide, you can make the iodide, and HBr4 puts in the fluoride, so this is nice, because...we have already seen how to brominate and chlorinate, but now we can see...with this method, we can have access to all 4 halogens and ways to put them on the benzene ring.0282

If we react the diazonium salt with water, we could make a phenol; we could install an OH group; and finally, we could reduce it by using H3PO2.0303

Notice, it is H3PO2: H3PO4 is a strong acid; H3PO2 is a reducing agent.0313

It is missing those oxygens, so you know it is going to want to get oxidized; that makes it a good reducing agent.0323

That is a way that you can convert the diazonium salt to just have a hydrogen here.0329

Now, the mechanism for this reaction is not entirely understood; maybe it's an SN1 type mechanism, where the leaving group leaves and then the nucleophile comes in, or it could be more concerted.0334

So, we don't have to worry about the mechanism here; it is just a synthetic tool; we are going to use the reagents in a variety of ways to make all sorts of interesting products.0347

Ultimately, what we are saying is: we can replace the NO2 group with a wide variety of groups; we can replace it with an NH2, simply by reduction, and then we can convert it to any one of the halides (Cn, OH, or hydrogen) via the diazonium salt.0356

This gives us all sorts of great synthetic possibilities.0378

Let's try a few transformations: if I started with benzene, and I wanted to make meta bromochlorobenzene, we can think back to our electrophilic aromatic substitution, and say, "Well, I already know how to put on a bromine and a chlorine."0382

But what is the directing power?--once you put on one group, how does it direct the next group coming in?0396

Each of these halogens, because they have lone pairs--that makes them ortho/para directors, so it would be impossible to do this transformation using electrophilic aromatic substitution alone, because as soon as you put on one of the halogens, the second one would come in para to that.0404

So, it can't be done with electrophilic aromatic substitution alone; we need somehow to get a meta director in there to get that meta group.0421

Luckily, either of these can be put in via the diazonium salt; and where does the diazonium salt come from?--it comes from a nitro group.0431

The very first thing we are going to do here, actually, is nitrate the benzene ring.0440

It doesn't matter where we put it; let's put it at the top here.0445

What are the reaction conditions to do a nitration? HNO3, H2SO4.0448

We can nitrate the ring--and how does a nitro group direct?--because it is an electron withdrawing group, it is a meta director, so now we see how we can get that meta relationship going.0456

Now, if I chlorinate--do an electrophilic aromatic substitution to add in a chlorine--it will come in meta.0468

To chlorinate, I use Cl2, FeCl3, and that chlorine is going to go into the meta position.0475

So now, what I need to do is convert this NO2 group into a bromine (replace it with a bromine); what I have to do first is convert it to the diazonium salt (make it a good leaving group), so that is a two-step process.0486

The first step: we could use catalytic hydrogenation (that is kind of my favorite reducing agent, since I know that so well); that will convert the NO2 to an NH2, and then we need to treat this with nitrous acid.0498

Now, I know I need nitrous acid, HONO, but that is not a reagent; you can't go to the stock room and say, "I would like some nitrous acid"--it doesn't exist in a bottle (it's not stable).0513

OK, so the reagents we use to generate nitrous acid are NaNO2 and some acid--any acid, usually something like HCl.0523

We use NaNO2, HCl to make nitrous acid; that converts the NH2 group to an N2+ group; and now we are ready to replace it with any nucleophile we want.0533

In this case, we want a bromide; so this is where we use our copper bromide to do that substitution.0545

OK, let's try another one: Synthesize the following target molecule from either benzene or toluene (those are your possible starting materials).0556

Well, there is no methyl on the ring, so we won't need to start with toluene in this case; but let's think about doing a retrosynthesis.0563

A retrosynthesis asks, "What starting materials do I need--what could I make that would be able to form this target molecule?"0572

Well, I see that I have an ester down here; and the way I make an ester is: if I disconnect that bond, I go back to the alcohol at this position; if I had this alcohol, I would be able to convert it to the ester.0579

OK, so let's treat this as our new target molecule.0593

We can put in a cyano group via a diazonium salt; we can put in an OH via a diazonium salt; the question is, "Which do we put in first so that we get the proper para relationship?"0597

If we put in the cyano first, the cyano group is an electron withdrawing group: that makes it a meta director.0611

The OH with its lone pairs is an ortho/para director; it is a very strong electron donating group.0619

The first group I need to add is the OH, which means the first group I can take back out is the cyano.0628

In other words, what I want to do is first make phenol from benzene; and then I can work on putting in the cyano group, and then I can convert the OH to an ester.0635

So this is going to be the direction of my reaction: I'm going to start with benzene, but it is very good to do this retrosynthesis, this planning, first, so that you know what you have in store.0648

Our synthesis, then, starts with benzene; and we want to make phenol--we want to put in an OH group.0657

The only way we could do that is via a diazonium salt: we have to prepare the diazonium salt.0666

Starting from benzene, that is going to be three steps: first we have to put in the nitro group--let's draw the nitro group down at the bottom, because that is where we want our OH to be.0671

And how do we nitrate benzene? HNO3, H2SO4: you are going to be using these reagents so often, in doing these problems, that it is really going to become second nature to you--some of these reagents.0682

So now, we have nitrated, and now it's two steps to convert it to N2+: first we reduce the H2 palladium, and then we add nitrous acid (NaNO2, HCl--remember, these combine in situ to form HONO--to form nitrous acid).0694

That converts our NH2 to N2+, an excellent leaving group; and now, we are ready to displace it with whatever leaving group we want.0717

In this case, we want to make the phenol, so we just add H2O, and we have made phenol.0730

OK, now we want to put in a cyano group: how do we put in a cyano group?0741

We already have our ortho/para director, so we know that whatever group we add next is going to come into the para position.0746

To add a group just to a plain old benzene ring, there needs to be some kind of electrophilic aromatic substitution; we need to add a good electrophile, and the electrophile we are going to add that could be, ultimately, converted to a nitrile, is the nitro group.0752

We are going to effectively repeat this process again in order to put in the cyano; so you don't have to draw an arrow and draw those structures after every single step--let's do it over one arrow.0766

Let's see if we can imagine the steps we need to take to go all the way to having the cyano here.0779

What do we have to do first to make the N2+ group?0788

First, we have to nitrate: HNO3, H2SO4--so first, we have to introduce the nitrogen.0792

Then, we have to reduce it: catalytic hydrogenation--remember, you could use tin, is not the only reagent you could have.0802

Tin, HCl...something like that will reduce it, as well.0810

That makes the NH2, and now, NaNO2, HCl makes nitrous acid, which converts the NH2 to N2+; now we have a great leaving group.0815

And step 4: how do we get the cyano in there?--we are going to use copper cyanide--this is one of the Sandmeyer reactions.0825

Overall, starting from benzene, it is a 4-step process to put in a nitro, convert it to a good leaving group, and replace it with a nucleophile.0834

So now, we have inserted our cyano group--excellent.0843

We are close, but we are not quite at our target molecule: our target molecule doesn't have an alcohol down here--it has an ester.0847

Let's think about where an ester comes from: I see that acetyl group, so I know I need this; the reagent is going to have something with that in it.0857

What do I need attached to this carbonyl?--I want to install an oxygen there, so I need some kind of leaving group for that oxygen to displace.0868

I want a substitution: I want an addition-elimination: I need to substitute, so an acid chloride would be perfect for that.0876

One way to make an ester is to take an alcohol and react with an acid chloride; put that in here with some base like pyridine, and we would have our ester; we would have our target molecule.0883

OK, let's see one more example: how about if I wanted to make this target molecule?0897

Now, I have been given a starting material, but one other thing that I wanted to note about this particular product is: you might say, "Well, I need to add a nitro, and I need to add a methyl, and I need them meta to each other, so I would want to add the nitro first, and then do my Friedel-Crafts alkylation."0902

But remember that a Friedel-Crafts alkylation won't go if you have an electron withdrawing group on there; so even though this looks like it might be a simple target molecule, it is not.0924

In fact, it is not something we could do by traditional electrophilic aromatic substitution.0938

OK, and either way, we are given this as our starting material--we already have the methyl on here; we have to figure out how to move the nitro.0943

Now, that is weird: have we ever seen a group pick up and move over on a benzene ring?--we have not.0951

That is not what is happening here: instead, we used to have a nitro up here--now what do we have?--we have a hydrogen.0957

We simply have to replace that nitro with a hydrogen; we used to have a hydrogen here; now we have a nitro.0966

These are going to be two different substitution reactions that we need to figure out.0975

We have to figure out how to do that, and we have to figure out which one to do first.0980

OK, would it be possible to just nitrate right now, and get the nitro group in at this position?0986

So, in other words, if I just did HNO3, H2SO4, would that put the nitro group where I want it to go?0993

Remember, this is a meta director; it's an electron withdrawing group; the alkyl group is an ortho/para director; so where would a nitro group go?1002

It would go meta to the nitro and ortho to the methyl, since the para position is blocked; it would get this product out.1014

It wouldn't put the nitro in the right place, and furthermore, I now have 2 nitro groups: how would I distinguish between them, if I wanted to do something with one and not the other?1022

This is a dead end--this is a bad idea.1030

OK, so instead, what we are going to do: we need to put this nitro group in here; we have to change this from being a meta director to something else.1033

But remember: how do we do a substitution reaction here?--we need to convert this NO2 to an NH2 on its way to being a diazonium salt; so that is going to do it for us, isn't it?1043

If I had an NH2 here now, how do we reduce an NO2 to an NH2?--we do catalytic hydrogenation, or let's do tin, HCl--we could do one of these metal-promoted reductions.1056

OK, now I have an ortho/para director, and another ortho/para director; who is going to win?1075

If we had an electrophile coming in, the nitrogen wants the electrophile coming in here; the methyl wants the electrophile coming in here; who would win--who is the stronger director--who is the stronger activating group?1085

Well, anything with a lone pair is going to win out: this is very good at donating electron density, so this is a strong ortho/para director, so it wins.1101

OK, the problem is the problem: if I did HNO3, H2SO4, I do not get the product that I want; I get this product instead.1111

It doesn't go ortho to the NH2; it goes meta to the NH2; I'm sorry, I wanted to do nitro.1124

Now, what is going on there--what happened?--I thought I planned this so beautifully; why did it not go to the right position?1130

Remember something special about nitrating an aniline derivative?1138

We have these strongly acidic reaction conditions; this nitrogen is going to be protonated in the conditions, and it is no longer going to be an ortho/para director.1143

OK, so this is not the product we want, so how do we get around it?--we need a protective group.1153

We need to put on a protecting group; so before we can do our nitration, we need to protect the aniline.1163

Typically, we will do that by making an amide: we will put on an acyl group.1172

OK, we just saw in the last synthesis problem how to put on an acyl group, so we could use acetyl chloride and base, and a little pyridine; that makes the amide.1179

Now, we have, again, still a very strong director--stronger than the alkyl group; so this will direct ortho/para, and since the para is blocked, when we do our nitration now (HNO3, H2SO4), we expect that nitro group to come in right here.1191

OK, so this is starting to look like our product: we have the methyl and the nitro where we want them to be.1222

Now, we have to get rid of this nitrogen group and replace it with a hydrogen.1228

Now again, we can do that via the diazonium salt; we just need to turn this into an N2+ group, so how do we do that?1233

Well, first we have to get back to our aniline; we can remove our protective group now; NaOH, hydrolysis to cleave that amide group off.1239

And how do we turn this into a good leaving group?--we react it with nitrous acid; that is NaNO2, HCl; remember, that makes HONO; that makes nitrous acid, HONO.1256

There is N2+; and finally, we want to replace that N2+ with a nucleophile; in this case, we want a hydrogen nucleophile.1272

We get that by using H3PO2 as our reducing agent, as our nucleophilic source of H, and we finish our transformation.1284

Quite a lengthy transformation--quite a lengthy synthesis; it requires a lot of planning; it requires a lot of know-how--not just what groups we know how to add, but how the existing groups direct and what are some side reactions we have to avoid, and so on.1296

These transform problems can get to be quite involved.1311

OK, let's now talk about, not adding groups to the benzene ring itself, but let's take a look at the substituents attached to benzene and look at the reactions that they can have.1317

For example, how about reduction reactions?--we have already seen some reduction reactions.1329

For example, if you have a nitro group on a benzene ring, we have seen that you can reduce it to an amino group; you can do that with catalytic hydrogenation, tin-HCl.1333

OK, so that is a reduction reaction; we also saw that if you had a carbonyl--if you had an acyl group on a benzene ring or some other aromatic ring--you could reduce that very easily, all the way down to a CH2 (a complete reduction).1342

That is because this carbonyl is benzylic; we saw this when we got this by the Friedel-Crafts acylation--we saw that we can reduce it all the way down to an alkyl group.1357

OK, we could use catalytic hydrogenation, only because this is benzylic.1367

It is because of that benzene ring that makes this ketone possible to completely reduce.1374

But then, you could also have a zinc amalgam, an HCl, the Clemmensen reduction, the Wolff-Kishner reduction...those worked for any ketone, so this isn't really a reduction reaction that is special to aromatic side chains; it is just something I put in there because we have seen it recently, in the context of the Friedel-Crafts acylation.1383

OK, how about oxidation reactions?--it turns out that, just like the benzylic position is easily reduced, the benzylic position is also easily oxidized; so benzylic, remember, means that you are next to the benzene ring.1405

And so, this methyl group--if we were to treat it with an oxidizing agent--a strong oxidizing agent--we would expect to oxidize this, all the way up, as much as we possibly can; and that would mean going to a benzoic acid.1418

That would make benzoic acid this way.1436

A very unusual oxidation: we have never seen an oxidation before of an alkane--usually, our oxidations either start with a functional group (like a carbon-carbon double bond or a triple bond), or we already have an oxygen in there (like an alcohol can be oxidized).1438

This is actually an alkane, but it is not just an alkane; it's an alkyl group attached to a benzene ring.1453

We call this arynes when you have both an alkyl group and an aromatic group; that combination--we call them arynes.1458

This is an oxidation of arynes, because those benzylic carbons are easily oxidized.1472

What strong oxidizing agents work here?--something like chromic acid, sodium dichromate, H2SO4--we call that the Jones oxidation.1478

OK, or KMnO4, NaOH...this is permanganate, also a very strong oxidizing agent, because these are basic conditions; we need some H3O+ at the end to protonate our carboxylic acid product.1487

OK, but strong oxidizing agents will oxidize the benzylic carbon.1501

This reaction is even more interesting, because this even breaks carbon-carbon single bonds: we have never seen that before in a reaction, other than ozonolysis, when we had a carbon-carbon double bond (both bonds were cleaved).1506

Here, we have a plain old carbon-carbon single bond, and when we treat it with a strong oxidizing agent (like sodium dichromate, an acid...), here is our benzylic carbon that is going to be oxidized, and it is going to even...not just trade in C-H bonds, which we are used to seeing for an oxidation--it even trades in its carbon-carbon bonds for oxygen.1519

And so, once again, the product we get out is benzoic acid.1540

The whole rest of the carbon chain gets chewed up and oxidized, and we are left with a benzylic carbon that is oxidized as much as possible, meaning a carboxylic acid.1545

OK, so this happens to any alkyl group that you have attached to a benzene ring, except in the situation where our benzylic carbon is quaternary (meaning it has all 4 single bonds to carbons).1555

In that case, when we try to do an oxidation, there is no reaction.1572

OK, with that exception, every other benzylic carbon will be converted to a carboxylic acid.1575

Let's see an example.1581

Here we have KMnO4, sodium hydroxide, heat: definitely a strong oxidizing agent...oxidation conditions.1584

And so, what we need to do is: we need to identify any benzylic carbons we have and then oxidize those.1592

Where do we have benzylic carbons?--this looks like a benzylic carbon; the carbonyl is next to this benzene ring; this is a benzylic carbon--it is next to this benzene ring; but these two carbons are directly attached--there is no carbon that is next to a benzene ring here.1598

So, these are the two carbons I am going to be oxidizing; I am either going to be breaking that carbon-carbon bond as part of my oxidation reaction...and so my product--if I just erase that bond, it has come back to: these are the two carbons that I said were going to be oxidized.1613

What do they get oxidized to?--carboxylic acid, carboxylic acid; so I'm going to get a dicarboxylic acid here (let's see how I can draw it in here without getting too crowded).1633

Every benzylic carbon gets oxidized to a carboxylic acid, unless it has four carbons attached to it, making it quaternary.1644

OK, now oxidation is a way to make benzoic acid; we have seen some other reactions that can be used to make benzoic acid--let's review those while we are here.1656

One way we can get a carboxylic acid is by hydrolysis; so if we had something like a nitrile, we could trade those three C-N bonds to C-O bonds by H3O+ reaction with water and heat.1665

That could be base-promoted or acid-catalyzed; both of those are possible.1679

We just saw that we can introduce a cyano group via a diazonium salt; so we know how to put the cyano onto a benzene ring, and we also want to remember that cyano groups can get converted to carboxylic acid.1684

That would make a benzoic acid derivative.1697

OK, we have also seen Grignard reactions to make carboxylic acids; so we know how to chlorinate or brominate a benzene ring.1700

Once you have a halogen on a benzene ring, we can make a Grignard, and then a Grignard can get converted to a carboxylic acid; so let's think about that reagent.1709

A Grignard is a nucleophile; we need some kind of an electrophile that has a carbon and two oxygens--what electrophile has a carbon and two oxygens?1718

Maybe CO2?--that is exactly what it is.1729

A Grignard plus CO2, followed by workup, will give carboxylic acid; so in this case, we would also get benzoic acid; so if your substrate is not suitable for doing oxidation reactions, then maybe you have hydrolysis conditions or organometallic conditions.1732

They are all quite different, so sometimes, one synthesis is going to be better suited than another.1749

Now, back to our reactions of our aromatic side chains: we saw reduction reactions were easy; we saw oxidation reactions; we can also do halogenations at the benzylic position, doing a free radical halogenation.1758

So, if I were to take this aryne and react it with Br2 and hν...OK, that will form a radical; the Br2 can cleave homolytically to form two equivalents of bromine atom.1772

And then, that bromine atom is going to find some alkyl group from which it can abstract a proton.1793

So, that is not something that is going to react with the benzene ring, but it will react with this carbon chain.1800

Which hydrogen is it going to take?--it is going to take the one that results in the most stable radical intermediate.1805

If we go to this benzylic position, that is going to give the most stable benzylic radical; so what we do is: we do a hydrogen atom abstraction; we break this bond; one electron pairs up with the bromine; the other electron stays behind.1814

Let's see if we can fit it a mechanism here.1831

We are going to get this benzylic radical; this benzylic radical is resonance-stabilized, so that is the best one we can have.1833

Let me draw that up here, so I have some room for my mechanism.1841

This is resonance stabilized; we can move that radical around the ring; we can move it over here; we can move it down here; so the benzylic position is a great place to have a radical, because you can have this resonance stabilization.1844

OK, once I have this radical here, how do we get a bromine there?--well, now this reacts with Br2 again and does another atom abstraction.1864

This time it abstracts a bromine atom, breaks the bond, plucks off a bromine, and this regenerates Br.; so remember, our radicals are chain reactions; it gives as our product a bromine in the benzylic position.1871

OK, so remember, halogens are a good leaving group; so it is easy to introduce a leaving group in the benzylic position by doing a free radical halogenation.1891

We just had a little review of that; so, in this first step, we could also note that we just kicked off a molecule of HBr; HBr is another product in this reaction, but we get the free radical halogenation.1902

OK, a little review also: this is good for benzylic; it would be good for any allylic carbon or a tertiary carbon; that hydrogen would be good to replace, because that would also be a very stable radical.1915

OK, those are great places to do a bromination; those would all be better than secondary, and better than primary, and better than methyl.1928

We always look at the alkyl group and try and find the most substituted one to give the most stable radical intermediate that we can have.1935

Sometimes, we use Br2; but remember, Br2 can also undergo electrophilic additions with the ring itself.1944

If you had some Lewis acid around, that could happen, or if you have a double bond; if you have an alkene; bromine would definitely react with it; so sometimes, instead of using Br2, we use this reagent, N-bromosuccinimide.1952

N-bromosuccinimide, which is known as NBS for short, is another good source of Br.--is another good source of radical bromine.1966

And so, you could just imagine--just like we had Br2 cleaving and being a source of bromine radical, the NBS can also be a source of bromine radical.1973

So, sometimes you might see NBS as part of your reaction conditions for a radical halogenation.1982

OK, and if we did have a leaving group in the benzylic position, we also know that substitution reactions are very favorable at the benzylic position.1990

OK, both SN1 and SN2 mechanisms are favorable when the leaving group is on a benzylic position.2000

Let's take a look at this reaction on the left here: if I took this bromide and reacted with sodium hydroxide and water, I know we have hydroxide here; that looks like a very strong nucleophile.2008

This would love to do the SN2, and so that is the back side attack; that is the one-step substitution mechanism.2023

It would come in from behind the leaving group--the leaving group is a wedge, so the hydroxide has to come in from behind.2031

So, my product here--what would you expect for the stereochemistry?--how would we describe that back side attack's stereochemistry?--we call it inversion of stereochemistry, so you would expect that OH back here.2037

OK, now remember that hydroxide is also a strong base, which means it can do E2; but in this case, because it's benzylic, the SN2 is favored.2050

Even though it is secondary, we are going to get majority SN2, because that transition state--there is a p orbital in the transition state of the SN2; that sees the p orbital right next door on the benzene ring; and that lowers the energy of the transition state and makes the SN2 more favorable, so there are not the steric concerns that we usually have that lead it to E2.2062

We have the opposite: we have something that stabilizes the transition state.2083

OK, so just a reminder that SN2s are awesome on benzylic carbons; SN1s are also great; SN1 is the step-wise substitution mechanism--that is where a leaving group just leaves on its own.2087

That happens when we have something like H2O, which is a weak nucleophile; a weak nucleophile means we can't do a back side attack--instead, we are just going to do a unimolecular substitution.2101

Our leaving group just leaves on its own; and what is great about that reaction--about this SN1--is: because we have a stable carbocation because of resonance (again, we can kind of draw what that resonance looks like), if we use one of our π bonds from our phenol group (we can draw this so you can see it a little easier; let's draw a benzene ring)...if we show this π bond moving over, then we can show the resonance stabilization.2114

But of course, benzylic is one of our very stable carbocations, which means it would be a good SN1.2157

Let's think about the stereochemistry of this product: we are going to have water come and attack the carbocation and then do a deprotonation; so ultimately, we are going to add an OH group, just like we did over here.2166

But what about the stereochemistry--is the water going to attack the top face of that carbocation, or the bottom face?2178

Well, remember, this is actually a planar carbocation; it's planar, so it's statistically just as likely to attack from either face; so we are going to get both products, where the OH is a wedge, and the OH is a dash--whether you attack from the top or the bottom.2185

Remember, we are going to get a racemate out, a racemic mixture, when we do an SN1 mechanism through a carbocation.2204

OK, so for obvious reasons, because of the resonance, we know that SN1s would be good at the benzylic position, but less obvious is the fact that SN2s are also very favorable.2213

Benzylic leaving groups are really hot--very reactive.2224

OK, so let's try a transform where we go from toluene to this carboxylic acid, and we have just seen a reaction (haven't we?) that does oxidation.2230

So, what if I did a Jones oxidation--if I did sodium dichromate, Na2Cr2O7, H2SO4?2243

Would that oxidation work in this case?--is something wrong here?2251

Yes, here we have one carbon; this oxidation always gives benzoic acid as a product.2258

This is not benzoic acid: there is an extra carbon here.2263

This carbon is right here; this is a new carbon-carbon bond--there is an extra carbon in our product, so it is more than merely an oxidation process; we have to do something to introduce a new carbon.2267

Let's think, instead, about this carboxylic acid product, and ask, "How did we get there--what is a reasonable disconnection--what are some ways we can make a carboxylic acid?"2281

We can't do oxidation, in this case, of the benzylic position, so what other reactions have we seen?2294

We just reviewed a few today.2301

One is: we could have done it by hydrolysis; instead of having a carboxylic acid here, we could have had a nitrile.2306

We could have had a cyano group; OK, and the cyano group we could put in by forming that bond: let's think about these two carbons that we want to react.2313

One of them must have been a nucleophile (and we are continuing our retrosynthesis here).2324

One of them must have been a nucleophile, and one of them must have been your electrophile: the cyano, as cyanide, would be a great nucleophile; this was my nucleophile, which means the other carbon was my electrophile.2329

If I had, for example, a leaving group here (like a bromine and sodium cyanide), I think that would do a nice SN2 to make the cyano, and then we can convert it to the carboxylic acid.2344

OK, that is good--that is one option; what is another option?2357

We have seen this synthesis of a carboxylic acid before, where we do that disconnection; and we do this disconnection right away.2360

This is more like we did a functional group in a conversion first, and then we did a disconnection; what if we did this disconnection right away and thought about one of those carbons being a nucleophile and one of them being an electrophile?2370

The carbonyl was my electrophile, which means this carbon was my nucleophile; how do we make a benzylic carbon a nucleophile?2384

How do we make any carbon a nucleophile?2395

How about using a Grignard reagent?--if I had a Grignard here as my nucleophile, what electrophile could it react with to give a carboxylic acid product?--a carbon with two oxygens?2398

Oh, we are back to CO2 as our electrophile.2411

So, as usual, we are going to do a retrosynthesis, working backwards to a reasonable nucleophile and electrophile that we know can come together and give a reaction and continue a transformation.2415

OK, but either of these syntheses requires that we have a leaving group in the benzylic position; we don't have that--we have toluene.2428

The first thing we have to do is convert toluene into benzyl bromide or benzyl chloride.2437

How do we go from a methyl group to a bromine--could you use something like PBr3?2445

Where have we seen PBr3?--that was if you had an OH here that replaces...that substitutes an OH for a leaving group; that is no good.2452

How about Br2, FeBr3?2461

That is another bromination reaction; where would that brominate?2467

That is electrophilic aromatic substitution; that makes Br+, which would put the bromide in the ring down here para to the electron donating methyl group; that is not going to work, either.2472

It is very important to keep all of our reagents straight in our minds, so that we know which one to bring out.2482

What do we need here?--we need to replace a hydrogen on an alkane with a bromine; that looks like free radical halogenation.2490

OK, so instead, we are going to use bromine and light energy, or heat energy, hν; we need to generate a radical...or NBS, hν; we need to form a radical, and in this case, there is only one place you can halogenate, and that is on the methyl group.2497

We can now make benzyl bromide, so if we did this first path, we would treat this, then, with sodium cyanide to make the nitrile.2516

And then, we need to convert that C-N triple bond to three C-O bonds; when you are trading a nitrogen for an oxygen, that looks like hydrolysis.2530

So we do something like H3O plus heat, or base and water and heat, and with an acid workup to protonate.2542

OK, so we could do hydrolysis there; so that would be one good way to make our target molecule; what is another way?2550

Instead of treating it with sodium cyanide, we can convert this benzyl bromide into a Grignard; we can turn it into benzyl magnesium bromide by adding in magnesium.2558

We already knew that was a possibility, because we could react this with carbon dioxide; we would do this as a two-step procedure--carbon dioxide followed by H3O+; and that would be another way to make our target molecule by a different disconnection.2571

There are all sorts of interesting paths, now, to make a carboxylic acid target molecule.2586

OK, let's talk about another mechanism that we can have for aromatic rings: we talked about electrophilic aromatic substitution--there is another reaction called a nucleophilic aromatic substitution.2595

Let's take a look at an example of that: we are going to have a starting material that looks something like this: we are going to have a leaving group attached to a benzene ring, and we are going to need some electron withdrawing groups, appropriately placed.2606

They are going to be in the ortho positions or the para position to that leaving group; and we will see why those are important soon.2625

But when we have this kind of intermediate, and we react with a nucleophile, what happens is a two-step mechanism.2633

First, that nucleophile attacks where the leaving group is; and it doesn't kick off the leaving group.2640

That would be...what mechanism would that be, if you attacked the carbon and kicked off the leaving group all in one step?...that would be an SN2 mechanism; we can't have an SN2 mechanism on an sp2 hybridized carbon.2649

So instead, the nucleophile attacks and breaks the π bond.2659

So we have: our leaving group is still here, and our nucleophile is now added, and we now have a lone pair, which puts a negative charge on this carbon.2668

OK, so we have a carbanion intermediate; now, hopefully, you can see why we need those electron withdrawing groups: they are here to stabilize this anion.2684

If you have a nitro group right here, you can have resonance with that carbanion.2694

Or, we can move the carbanion down here, and it can have resonance with this nitro group.2700

OK, so we will look at this more closely on the next slide; but this is why those are critical.2703

OK, so step 1 is: we have the nucleophile; step 2--we eject the leaving group.2710

What we have is a lone pair of electrons; the next carbon over--we have a leaving group; this is kind of like collapse of a CTI.2715

The lone pair comes down; the leaving group gets kicked off.2721

Even though that first step is difficult, because you lose your aromaticity, the second step is great, because you gain your aromaticity back; and so, once again, we are observing a substitution mechanism where our benzene ring stays intact.2728

That is going to be the only kind of reaction that an aromatic ring is going to want to undergo.2741

OK, so it is a substitution reaction involving a nucleophile; so that is why SNAr, a nucleophilic aromatic substitution.2747

OK, in this case, the ring is acting as an electrophile; a nucleophile is attacking it, so the ring acts as an electrophile.2755

If we have electron withdrawing groups on that ring, pulling electron density out of that ring, that is going to make the ring a better electrophile--more electron deficient and ready to react with a nucleophile.2762

We say that the ring is activated toward SNAr by electron withdrawing groups.2774

Now, that is the exact opposite of what we saw when the ring was trying to react with electrophiles, in the electrophilic aromatic substitution.2781

We said that electron withdrawing groups would deactivate the ring.2788

OK, but in this case, because we are trying to react with a nucleophile, electron withdrawing groups activate the ring.2792

OK, and the leaving groups that we are going to see--we will see the halides, as we normally do, but what is very interesting is that fluoride is the fastest leaving group, in this case.2798

Usually, in all of the other substitution reactions we have seen, we have never seen fluoride, ever, as a leaving group; but it is a good leaving group for this mechanism, because it actually helps to activate the ring as an electron withdrawing group, itself.2810

OK, and the iodide is the slowest.2821

Let's see an example: OK, what if we had this fluoronitrobenzene substrate, and we react it with NaOH?2827

OK, NaOH is our nucleophile; do we have a leaving group?--I know it is not what we are used to, but when we see a fluoride on a benzene ring, especially with those nitro groups, we need to recognize it as a leaving group.2838

What is our product going to look like?--wherever we used to have our leaving group, we now have our nucleophile.2850

It is a substitution, replacement of a leaving group with a nucleophile; so we know exactly where it is going to go every time.2859

There is no stereochemistry to worry about, because this is all planar; so in terms of predicting the product, it is a very straightforward reaction to predict...well, as long as you are recognizing that it is a nucleophilic substitution.2867

OK, let's see if we can do a mechanism for this.2881

It is going to be a two-step mechanism: we are going to start with our nucleophile attacking the carbon bearing the leaving group (because that is where it is going to end up bonded), and we break this π bond to get a carbanion intermediate.2885

OK, and here is the C-, and let me draw out the nitro group, just as a reminder of what the nitro group looks like, and a reminder on how this ends up being a good thing for the negative charge.2900

When you have a negative charge α to a nitro, we can have resonance that delocalizes the negative charge onto an oxygen; that is always good resonance, if you can delocalize it onto an oxygen.2919

Any other resonance forms?--yes, you can actually use this nitro down here, as well; so we can have (let's see if we can go all the way from there)...we can give back this nitro group, and then move the extra electrons into that nitro group.2937

I'm kind of skipping some other intermediate resonance forms in between, just trying to make a point here.2954

But the important thing is: when you have nitro groups in the ortho and para positions, you will be able to use them to help you localize the anion intermediate.2965

And so, they are critical: without those electron withdrawing groups, the reaction doesn't go; you can't just form a carbanion that is not resonance-stabilized.2977

OK, so when we are all done, that is our first step; this is all just resonance, so these are not adding extra steps; it is just redrawing the same intermediate lots of different ways.2985

Our second step (so step 1 was "add the nucleophile"), step 2, is: eject our leaving group.2995

Our fluoride needs to get kicked out, but something is going to help kick it out; let's see if we can go from this resonance form straight to the product--where is our extra electron density up here?3005

Now it's all the way down in this oxygen; so we can start with this oxygen; we can bring this back down to re-form my nitro group; we could put this π bond back in the ring; and that will shift this π bond over, and that is what kicks our fluoride out.3016

We should be able to go from any resonance form all the way to our product.3030

There is our nucleophilic aromatic substitution.3039

Let's see if we can predict a product: we have aromatic starting materials--how do we know that we are going to have a nucleophilic aromatic substitution, and not an electrophilic aromatic substitution?3045

All right, it is kind of easy, maybe, to start getting these all jumbled up; OK, well, in an electrophilic aromatic substitution, what do you need?3059

You need some kind of really hot electrophile, like a carbocation--like a Br+--like an NO2+, right?3068

If you look at these reaction conditions, there is no electrophile; there are none of those electrophilic reaction conditions or reagents, so that means there is going to be no electrophilic aromatic substitution.3078

That reaction can't happen, so what other reaction can happen?3094

Do we have a leaving group on the benzene ring?--we do.3097

Here is a tosylate--that is a good leaving group.3101

Do we have a nucleophile that can displace that leaving group?--we do: here is our nucleophile.3104

Do we have an electron withdrawing group that would help stabilize the carbanion intermediate?--we do; it's right here.3112

This is the ortho/para; in this case, we just have one, but we have an ortho/para electron withdrawing group.3119

OK, those are the ingredients that are necessary for a nucleophilic aromatic substitution to take place.3124

OK, so what does our product look like?--we are going to replace our leaving group (which is the tosylate down here) with our nucleophile.3130

It's a nitrogen; it's an aniline; so a nitrogen with a benzene ring--how many bonds does nitrogen like to have when it's neutral?--just 3 bonds; so I know, at some point in this mechanism, I'm also going to have to deprotonate here.3141

But this would be our neutral product...nucleophilic aromatic substitution, where I replace the leaving group with a nucleophile.3154

This was my nucleophile.3162

OK, and one last mechanism to take a look at for aromatic compounds: let's take a look at the following reaction.3167

If you react this substrate with NaNH2 and ammonia and heat, you get two substitution products out.3174

You get one where this nucleophile...OK, this is a nucleophile, so that part I see; we do have a leaving group; so I can see how we are getting some kind of nucleophilic substitution.3185

I have one product where the nucleophile directly replaces where the leaving group used to be, but I also get a product where my nucleophile is one carbon over.3197

Now, let's think about it: now, this is the experimental observation; these products were isolated.3207

So then, it is up to the chemists to decide, "Well, what mechanism could we have to account for those products?"3213

Could it be an SN2 mechanism--could it be an SN2?--well, first of all, that would be impossible to get this product, because the nucleophile didn't come in where the leaving group was; but also, it can't be SN2, because our leaving group is on an sp2 hybridized carbon.3219

SN2 back side other words, can the nucleophile just attack and kick off the leaving group--back side attack--concerted mechanism?3235

It has to be a tetrahedral carbon in order for the nucleophile to come in do that; so no, it cannot be SN2.3243

How about SNAr?--we just looked at a nucleophilic aromatic substitution; could we have that two-step process, where the nucleophile adds in and then kicks off the leaving group?3249

Well now, again, that could form this product, but that would be impossible to form the second product; but furthermore, what do you have to have in order for the nucleophile to attack the ring in that fashion?3260

You have to have some kind of an electron withdrawing group: is a methyl group an electron withdrawing group?3272

It is not--it is an electron donating group; so it can't be SNAr, because there is no electron withdrawing group; so we can't even use SNAr to make this first product.3277

There is actually a new mechanism, a different mechanism, to make this product, and it works as follows.3286

The NaNH2 is not just acting as a nucleophile; it is acting as a strong base.3293

It is deprotonating the benzene ring and causing an elimination to take place.3302

I am having an E2 elimination; I am having a single-step elimination, which forms a triple bond in my benzene ring.3310

A triple bond--this is clearly a highly strained intermediate; what geometry does a triple bond want--what hybridization do you have?3321

It is sp hybridized; it wants to be linear; so clearly, you can't have a triple bond in a 6-membered ring without having a huge amount of ring strain.3329

OK, this is known as benzyne, because you have a triple bond in a benzene molecule--a highly unstable intermediate.3337

But this kind of intermediate would help explain the products we get, because what happens next is: we have: the NH2- can (I'm sorry, I forgot about that lone pair) add in now as a nucleophile.3349

It gives me back my benzene ring; so it takes care of that strain; but it still leaves this anion--very highly unstable anion, but it can form here.3370

And then, we can protonate it using the ammonia that is around.3378

And that is how we get one of our products; that is how we get the substitution product where the nucleophile comes in where the leaving group was.3389

OK, but because the benzyne has 2 carbons with the triple bond that can be electrophilic, the nucleophile coming in doesn't have to go where the leaving group used to be; it can also come down here, at the other carbon, and that is how we get the mixed regiochemistry.3399

I have this flipped over--sorry--from the way I drew the product; but we can still, hopefully, see it's the same thing.3419

Protonation of this with ammonia leads to our second product.3425

Benzyne is a very interesting intermediate--a highly reactive reaction intermediate--but it can be formed and will give these products when an aryl halide is treated with a strong base (like NaNH2).3434

Let's see an example of that: let's see if we can predict the product, here, of a benzyne mechanism.3452

Again, we have a strong base; we have some heat; we have vigorous reaction conditions here; we have a leaving group on the benzene ring; so this will be a benzyne condition.3458

Where can it deprotonate--where can the hydroxide deprotonate?3470

There are no hydrogens over here, so it has to come to this side (let's go ahead and show the mechanism, since it is a one-step mechanism to form the benzyne); so the benzyne intermediate we are going to get looks like this.3474

And now...that can't ever be our final product; now, a nucleophile is going to attack it to give a product, to lead to our final product, and where can that nucleophile attack?--it can attack at either side of the benzyne--either one of those carbons.3492

What products would it lead to?3506

We are going to observe: for benzyne mechanisms, we are going to get the nucleophile coming in exactly where the leaving group used to be, and we are going to get the product next door, where the OH is one carbon over--next door to where the leaving group used to be.3511

So, if you ever see this kind of product mixture or product distribution, that is a very good clue that this mechanism involved benzyne, because that is the only way to get both of those carbons involved in the mechanism.3530

That wraps it up for all of our aromatic reactions.3542

I hope to see you again soon at; thank you.3546