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Lecture Comments (15)

1 answer

Last reply by: Professor Starkey
Sat Feb 20, 2016 10:23 AM

Post by Tatiana Betancur Giraldo on February 18, 2016

Dr. Starkey---  Is there a possibility to do a lecture with formation of benzynes with organometallics? I currently stuying aromatic compounds in organic chemistry II, but the professor in my class has talked about X and L-ligands, and I cannot find those topics to study for my test.



1 answer

Last reply by: Professor Starkey
Sun Jan 25, 2015 12:16 AM

Post by Samuel Tindell on January 23, 2015

Would NaBH4 and LiAlH4 also work to reduce the carbonyl in the friedel-crafts acylation example at 78:42?

1 answer

Last reply by: Professor Starkey
Sat Jul 19, 2014 10:36 PM

Post by Anhtuan Tran on July 19, 2014

Hi Dr. Starkey,
I just have a couple of questions.On the slide of Friedel-Crafts Alkylation Drawbacks part A (can over-react, you used ethanol with sulfuric acid to react with benzene. Since alcohol was in the acidic environment, the ethanol must be protonated. And then water was leaving to form the carbocation, but the carbocation that was formed was the primary carbocation and isn't it a terribly unstable one? If it is unstable, how can it be favored to form and how can it become an electrophile to attach to the benzene ring?
I read in my book and it said "Methyl and ethyl halides do not form carbocations when treated with AlCl3, but do alkylate benzene under Friedel-Crafts conditions." Can you explain to me what is Friedel-Crafts conditions?
Thank you so much.
By the way, your lecture is very great, organized and detailed. I really enjoy it. Thank you!

1 answer

Last reply by: Professor Starkey
Mon Feb 10, 2014 10:14 PM

Post by Jude Nawlo on February 10, 2014

Hi Dr. Starkey!

When I was considering dienes and dienophiles for Diels-Alder reactions, -NH2, -OH, etc. counted as electron withdrawing groups. Why do OH groups count as electron-donating groups in aromatic reactions (I associate its higher electronegativity with electron withdrawing characteristics, and was wondering if I was doing something wrong with this analysis).

These lectures are the best. Thank you so much for all of your help! :)

1 answer

Last reply by: Professor Starkey
Tue Mar 27, 2012 11:21 PM

Post by Lukasz Skora on March 27, 2012

Hi Dr. Starkey,

These lectures are terrific and this website is phenomenal. This for me is a tremendous leap forward!

Thank you so much!

Lukasz Skora

1 answer

Last reply by: Professor Starkey
Tue Feb 21, 2012 9:42 PM

Post by Jason Jarduck on February 19, 2012

Hi Dr. Starkey,

Another great lecture. I really enjoy your lectures and I'm interested in taking Orgnic chemistry 3, biochemistry 1 and 2. Keep up the good work!!

Thank You

Jason Jarduck

1 answer

Last reply by: Professor Starkey
Thu Oct 27, 2011 10:59 AM

Post by Jamie Spritzer on September 12, 2011

in 39:00 what makes FeBr3 (or FeCl3) a Lewis acid?
why does it grab onto the Br from Br2? What makes
FeBr4 a great leaving group?

0 answers

Post by alexandra ortega on March 8, 2011

why cant you just protonate the alcohol using an acid making it a good leaving group? does it have to do with the lone pairs on the oxygen?

Aromatic Compounds: Reactions, Part 1

Rank the following common substituents on benezen rings in order of increasing activation:
-X [X = F, Cl, Br, I], -NH2, -NHCOR, -OR
-X < -NHCOR < -OR < -NH2
Draw all the resonance structures for the following compound:
Draw the product for the following reaction:
  • Electron Withdrawing makes the compound react more slowly
  • Meta director
Draw the product for the following reaction:
  • Both the OH and CH3 groups are ortho, para directors
  • OH is the stronger activator so substitution occurs ortho or para to it
Draw the product for the following reaction:
  • This is a Friedel-Crafts Alkylation reaction with rearrangement
  • Step 1:
  • Step 2:
  • Step 3: Carbocation rearrangement
  • Step 4:
  • Step 5:
Devise a synthesis for the following 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 1

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.

  1. Intro
    • Reactions of Benzene
    • Electrophilic Aromatic Substitution
    • Electrophilic Aromatic Substitution on Substituted Benzenes
    • Effects of Electron-Donating Groups (EDG)
    • Regioselectivity: EDG is o/p Director
    • Effects of Electron-Withdrawing Groups (EWG)
    • Regioselectivity: EWG is a Meta Director
    • Effects of Halogens on EAS
    • Summary of Substituent Effects on EAS
    • Directing Power of Substituents
    • Electrophiles for Electrophilic Aromatic Substitution
    • Electrophiles for Electrophilic Aromatic Substitution
    • Electrophiles for Electrophilic Aromatic Substitution
    • Electrophiles for Electrophilic Aromatic Substitution
    • Electrophiles for Electrophilic Aromatic Substitution
    • Electrophilic Aromatic Substitution: Nitration
    • Nitration of Aniline
    • Nitration of Aniline
    • Electrophilic Aromatic Substitution: Sulfonation
    • Electrophilic Aromatic Substitution: Friedel-Crafts Alkylation
    • Friedel-Crafts Alkylation Drawbacks
    • Friedel-Crafts Alkylation Drawbacks
    • Friedel-Crafts Alkylation Drawbacks
    • Synthesis with Electrophilic Aromatic Substitution
    • Synthesis with Electrophilic Aromatic Substitution
    • Intro 0:00
    • Reactions of Benzene 0:07
      • N/R as Alkenes
      • Substitution Reactions
    • Electrophilic Aromatic Substitution 1:24
      • Electrophilic Aromatic Substitution
      • Mechanism Step 1: Addition of Electrophile
      • Mechanism Step 2: Loss of H+
    • Electrophilic Aromatic Substitution on Substituted Benzenes 5:21
      • Electron Donating Group
      • Electron Withdrawing Group
      • Halogen
    • Effects of Electron-Donating Groups (EDG) 10:23
      • Effects of Electron-Donating Groups (EDG)
      • What Effect Does EDG (OH) Have?
      • Reactivity
      • Regioselectivity
    • Regioselectivity: EDG is o/p Director 14:57
      • Prove It! Add E+ and Look at Possible Intermediates
      • Is OH Good or Bad?
    • Effects of Electron-Withdrawing Groups (EWG) 20:20
      • What Effect Does EWG Have?
      • Reactivity
      • Regioselectivity
    • Regioselectivity: EWG is a Meta Director 23:23
      • Prove It! Add E+ and Look at Competing Intermediates
      • Carbocation: Good or Bad?
    • Effects of Halogens on EAS 28:33
      • Inductive Withdrawal of e- Density vs. Resonance Donation
    • Summary of Substituent Effects on EAS 32:33
      • Electron Donating Group
      • Electron Withdrawing Group
    • Directing Power of Substituents 34:35
      • Directing Power of Substituents
      • Example
    • Electrophiles for Electrophilic Aromatic Substitution 38:43
      • Reaction: Halogenation
    • Electrophiles for Electrophilic Aromatic Substitution 40:27
      • Reaction: Nitration
    • Electrophiles for Electrophilic Aromatic Substitution 41:45
      • Reaction: Sulfonation
    • Electrophiles for Electrophilic Aromatic Substitution 43:19
      • Reaction: Friedel-Crafts Alkylation
    • Electrophiles for Electrophilic Aromatic Substitution 45:43
      • Reaction: Friedel-Crafts Acylation
    • Electrophilic Aromatic Substitution: Nitration 46:52
      • Electrophilic Aromatic Substitution: Nitration
      • Mechanism
    • Nitration of Aniline 52:40
      • Nitration of Aniline Part 1
      • Nitration of Aniline Part 2: Why?
    • Nitration of Aniline 56:10
      • Workaround: Protect Amino Group as an Amide
    • Electrophilic Aromatic Substitution: Sulfonation 58:16
      • Electrophilic Aromatic Substitution: Sulfonation
      • Example: Transform
    • Electrophilic Aromatic Substitution: Friedel-Crafts Alkylation 1:02:24
      • Electrophilic Aromatic Substitution: Friedel-Crafts Alkylation
      • Example & Mechanism
    • Friedel-Crafts Alkylation Drawbacks 1:05:48
      • A) Can Over-React (Dialkylation)
    • Friedel-Crafts Alkylation Drawbacks 1:08:21
      • B) Carbocation Can Rearrange
      • Mechanism
    • Friedel-Crafts Alkylation Drawbacks 1:13:35
      • Want n-Propyl? Use Friedel-Crafts Acylation
      • Reducing Agents
    • Synthesis with Electrophilic Aromatic Substitution 1:18:45
      • Example: Transform
    • Synthesis with Electrophilic Aromatic Substitution 1:20:59
      • Example: Transform

    Transcription: Aromatic Compounds: Reactions, Part 1

    Hi; welcome back to

    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 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 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 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, 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) 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' 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 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