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Lecture Comments (23)

1 answer

Last reply by: Professor Starkey
Tue May 1, 2018 12:05 AM

Post by Carol Moghnieh on April 28 at 11:01:26 PM

Hello Dr. Starkey,

In Claisen condensation, we did a final deprotonation that was the driving force for claisen, but after workup its the exact same molecule before having deprotatonated it. I'm not too sure what was necessary with that step, can you please clarify it for me


2 answers

Last reply by: Brijae Chavarria
Mon Mar 9, 2015 12:59 PM

Post by JaeYoung Sim on April 1, 2014

I download all power point and print out to study but why the slide order dose not match with lecture slide order? I am just so confused. How can do arrange these slide order?

2 answers

Last reply by: Some one
Wed Apr 24, 2013 4:17 AM

Post by Some one on April 23, 2013

Hi professor, we mentioned that it usually takes strong nucleophiles to attack the carbonyl carbon, such as LAH and Grignard reagents. What makes the enolate such a strong nucleophile in the aldol reaction that it would disrupt the resonance stabilized carbonyl. Thank you so much.

1 answer

Last reply by: Professor Starkey
Mon Apr 22, 2013 6:55 PM

Post by Some one on April 22, 2013

Hello professor, why is C-alkylation preferred over O-alkylation on an enolate?

1 answer

Last reply by: Professor Starkey
Sun Mar 31, 2013 12:35 PM

Post by Marrbell Martey on March 30, 2013

What would the product be if aldehyde is used in Dibenzalacetone formation

1 answer

Last reply by: Professor Starkey
Sun Apr 29, 2012 9:38 AM

Post by Rachel Paquette on April 27, 2012

at 41:00 you are explaining the mechanism for the aldol condensation, and you say that the alpha hydrogen will be deprotonated because that is the theme but if you have OH wouldn't you be able to attack the carbonyl carbon and then move the electrons in the double bond move up to the oxygen? how do you know whether you are deprotonating the alpha hydrogen or attacking the carbonyl carbon

0 answers

Post by Misael Nieto on March 28, 2012

I love the detail in this lectures. Thankyou

1 answer

Last reply by: Professor Starkey
Mon Jan 23, 2012 11:29 PM

Post by Jason Jarduck on January 22, 2012

Hi Dr. Starkey,

I like your lecture alot!! Today I watched the complete lecture to find the answer to a question.

Thank you

Jason Jarduck

0 answers

Post by Jamie Spritzer on August 12, 2011

in 15:49, where is the counter-ion Na+ going to be? If there are 2 resonance forms, is it with one O- or the other one, depending on the resonance form?

1 answer

Last reply by: Professor Starkey
Thu Oct 27, 2011 11:02 AM

Post by Jamie Spritzer on August 10, 2011

In 65:22 the base LDA is used to deprotonate, but in the second step at the end, there is an acidic workup with H3o+. How can you have both acid and base in the same mechanism?

1 answer

Last reply by: Professor Starkey
Thu Oct 27, 2011 11:05 AM

Post by Jamie Spritzer on August 7, 2011

Is the acidity of the alpha proton in a carboxylic acid about the same as for an ester?

0 answers

Post by Senghuot Lim on July 17, 2011

i bow down to the god of Ochem

Enols and Enolates, Part 1

Draw the tautomerization mechanism for this reaction:
  • Resonance-stabilized cation:
Draw the stepwise mechanism for this reaction:
Draw the product for this reaction:
Draw the product for this reaction:
Draw the product for this reaction:
Draw the product for this 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.


Enols and Enolates, 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
    • Enols and Enolates
    • Keto-Enol Tautomerization Mechanism
    • Keto-Enol Tautomerization Mechanism
    • Formation of Enolates
    • Formation of Other Carbanions
    • Formation of an Enolate: Choice of Base
    • Formation of an Enolate: Choice of Base
    • Formation of an Enolate: Choice of Base
    • Other Acidic 'α' Protons
    • Other Acidic 'α' Protons
    • How are Enolates Used
    • Alkylation of Enolates
    • α-Halogenation
    • α-Halogenation
    • Aldol Condensation
    • Aldol Mechanism
    • Aldol Mechanism
    • Aldol Summary
    • Acid-Catalyzed Aldol Mechanism
    • Acid-Catalyzed Aldol Mechanism
    • Crossed/Mixed Aldol
    • Crossed/Mixed Aldol
    • Crossed/Mixed Aldol Retrosynthesis
    • Claisen Condensation
    • Claisen Condensation
    • Claisen Condensation
    • Intro 0:00
    • Enols and Enolates 0:09
      • The Carbonyl
      • Keto-Enol Tautomerization
    • Keto-Enol Tautomerization Mechanism 2:28
      • Tautomerization Mechanism (2 Steps)
    • Keto-Enol Tautomerization Mechanism 5:15
      • Reverse Reaction
      • Mechanism
    • Formation of Enolates 7:27
      • Why is a Ketone's α H's Acidic?
    • Formation of Other Carbanions 10:05
      • Alkyne
      • Alkane and Alkene
    • Formation of an Enolate: Choice of Base 11:27
      • Example: Choice of Base
    • Formation of an Enolate: Choice of Base 13:56
      • Deprotonate, Stronger Base, and Lithium Diisopropyl Amide (LDA)
    • Formation of an Enolate: Choice of Base 15:48
      • Weaker Base & 'Active' Methylenes
      • Why Use NaOEt instead of NaOH?
    • Other Acidic 'α' Protons 20:30
      • Other Acidic 'α' Protons
      • Why is an Ester Less Acidic than a Ketone?
    • Other Acidic 'α' Protons 25:19
      • Other Acidic 'α' Protons Continue
    • How are Enolates Used 25:54
      • Enolates
      • Possible Electrophiles
    • Alkylation of Enolates 27:56
      • Alkylation of Enolates
      • Resonance Form
    • α-Halogenation 32:17
      • α-Halogenation
      • Iodoform Test for Methyl Ketones
    • α-Halogenation 35:55
      • Acid-Catalyzed
      • Mechanism: 1st Make Enol (2 Steps)
      • Whate Other Eloctrophiles ?
    • Aldol Condensation 39:38
      • Aldol Condensation
    • Aldol Mechanism 41:26
      • Aldol Mechanism: In Base, Deprotonate First
    • Aldol Mechanism 45:28
      • Mechanism for Loss of H₂O
      • Collapse of CTI and β-elimination Mechanism
      • Loss of H₂0 is not E2!
    • Aldol Summary 49:53
      • Aldol Summary
      • Base-Catalyzed Mechanism
      • Acid-Catalyzed Mechansim
    • Acid-Catalyzed Aldol Mechanism 54:01
      • First Step: Make Enol
    • Acid-Catalyzed Aldol Mechanism 56:54
      • Loss of H₂0 (β elimination)
    • Crossed/Mixed Aldol 1:00:55
      • Crossed/Mixed Aldol & Compound with α H's
      • Ketone vs. Aldehyde
      • Crossed/Mixed Aldol & Compound with α H's Continue
    • Crossed/Mixed Aldol 1:05:21
      • Mixed Aldol: control Using LDA
    • Crossed/Mixed Aldol Retrosynthesis 1:08:53
      • Example: Predic Aldol Starting Material (Aldol Retrosyntheiss)
    • Claisen Condensation 1:12:54
      • Claisen Condensation (Aldol on Esters)
    • Claisen Condensation 1:19:52
      • Example 1: Claisen Condensation
    • Claisen Condensation 1:22:48
      • Example 2: Claisen Condensation

    Transcription: Enols and Enolates, Part 1

    Welcome back to

    Today, we are going to talk about enols and enolates, which are components of aldehydes and ketones.0003

    So far, what we have seen for aldehydes and ketones involved the reactivity of the carbonyl, the C-O double bond; and the most significant part that we have seen so far is the polarity of the C-O double bond due to the resonance, which makes this carbon partially positive and makes it an electrophile.0010

    In other words, nucleophiles add here; and that is certainly a significant reactivity of aldehydes and ketones--of any carbonyl.0030

    What we are going to shift toward in this lesson is looking at the alpha carbon (that is the carbon right next to the carbonyl): and consider the protons that are attached to that carbon.0041

    It turns out that these protons, the alpha protons, are acidic: and what does it mean to be an acid?--it means you donate a proton.0052

    In other words, they can be deprotonated.0063

    In this lesson, we are going to explore the deprotonation of the α carbon and the consequences of that deprotonation.0069

    One other component that is of interest to us in this unit is the idea that any carbonyl-containing compound (like a ketone or an aldehyde) is in equilibrium with another form called an enol.0077

    Now, this structure is called an enol because it has an alkene (a carbon-carbon double bond) and an alcohol (OH), and they are on the same carbon.0093

    The interconversion between a ketone and its enol form is known as a tautomerization, and as you can see, the equilibrium is favored in the reverse direction here: the ketone is more stable.0104

    Having that carbonyl there is preferable, so that is the form that it prefers to be, but what is important to recognize is that there is always a small amount present here of the enol--a small amount present at equilibrium.0117

    So, in other words, any time you have a ketone or an aldehyde or any carbonyl-containing compound, you also have some of the enol form around; that is going to be important to us when developing some of the mechanisms, down the road.0136

    Let's look, first, at this tautomerization mechanism: how do we go from a ketone to an enol?0149

    First, let's draw the enol form, because that is going to be helpful for us to get from one place to the other.0155

    Let's think about where we are headed; and when we compare these two structures, we see that there are two things that need to be accomplished, and that is why we have a two-step mechanism here.0165

    It looks like we have an oxygen here, and now it's an OH; so one of my steps is: I need to protonate here at this oxygen--I need to add a proton at some point.0175

    And here we have a CH3, and now it's a CH2, so I need to deprotonate down here.0187

    And, of course, the π bond has moved as well; that is going to be part of our mechanism, and could be seen as resonance, too.0195

    The only question we have is: those are our two steps--what step are we going to do first?0203

    Well, typically, this tautomerization--the presence of enol is going to be something that we find in acidic conditions, so in acidic conditions, let's protonate first; so that is going to be our step 1 (protonate), and step 2 is going to be deprotonate.0207

    So, I can protonate; let's add in our lone pairs on this oxygen: step 1 will be "protonate the carbonyl oxygen."0220

    That is going to give an oxygen with three bonds and a lone pair: 1, 2, 3, 4, 5; that is a positively charged oxygen.0239

    OK, and my second step, then, is going to be to deprotonate this α carbon.0248

    It turns out we are deprotonating right away; so instead of making a CH3, let's make this a CH2, so that we can see one of the hydrogens involved.0255

    What we want to do is deprotonate that hydrogen; we can use the A- that we formed in this first step to be a base; and can you see how the mechanism might work to get us to our enol structure?0263

    Typically, when you do a deprotonation, the electrons just go and sit on this atom; but instead of moving them here, we can move them over to be a π bond, which would push this π bond up onto the oxygen; and now, we end up with the best resonance form here, where we have no formal charges.0280

    Any time we need to do a tautomerization, it is going to be a two-step mechanism; and we should practice this both forward and backwards until we get pretty good at it, so that it will be automatic when it comes to part of a larger mechanism.0298

    Let's take a look at that reverse mechanism.0312

    We have an enol, and we want to go back to the ketone.0316

    This is now the more favorable mechanism that we will see, and what do we have to do?0321

    Well, again, it is still going to be two steps: the reverse mechanism (we already have it written down here) is going to be the same number of steps and involve the same intermediates as the forward mechanism.0328

    And so, what do we need to accomplish?--well, we have an OH that now needs to go back to an oxygen; so one thing I need to do is: I need to deprotonate.0339

    It really helps to think about where you are headed and what you have to accomplish before you dive into a mechanism.0350

    And my CH2 becomes a CH3, so I am going to protonate in this position.0355

    Those are going to be my two steps: and what do you think we should do first?0364

    Well, I am thinking again, because we are in acidic conditions, we should protonate first; so that should always be our first step in our tautomerization.0367

    Let's redraw this: I need to protonate at that CH2; so I can bring an acid over here, HA, and how could I protonate this carbon?0375

    Well, I can use this π bond as my base; I can grab that proton, and that way, I can add a proton to that position.0388

    What happens to this carbon now: it only has three bonds, and so I have a positive charge.0401

    Just like any time I protonate a π bond, like we saw for alkenes, you get a carbocation as your product.0405

    Our second step (that was step 1: to protonate): step 2 is going to be to deprotonate, so I can bring my A- in here to deprotonate this oxygen.0412

    Once again, rather than deprotonate and give an O- next to a C+, my mechanism would be much more efficient if I grabbed that proton and took these two electrons and filled them in as a π bond between the carbon and the oxygen.0423

    There we go--our two steps, and we have gone from our enol to our ketone.0442

    Now, I mentioned at the beginning that we are going to be able to deprotonate α carbons, and let's think about why a ketone's α protons would be acidic.0449

    As usual, any time we want to determine something about the acidity of a compound, we should take a look at the conjugate base, and see how stable that is.0459

    Here is an α proton; the pKa for such a proton is on the order of about 20, and if we imagine deprotonating it (treating it with some strong base that can come in and remove it), we would get this carbanion.0466

    Now, normally a carbanion is not a very happy charge, because carbon doesn't like having a negative charge; but there is something about this conjugate base that makes it reasonably stable.0483

    Well, that is because the lone pair is allylic; it is next to a π bond; in fact, it is next to a carbonyl, which means it has resonance; I can delocalize this charge.0494

    Any time I have an allylic lone pair, I can bring the π bond in, and bring this π bond up, and I can draw a second resonance form.0506

    Any time I could draw a second resonance form, that is always a good thing; and that explains why this is a reasonable place to deprotonate, because the conjugate base is resonance-stabilized.0516

    It is a resonance-stabilized carbanion; so it is a carbanion, but it is stabilized by resonance.0533

    And we can not only delocalize that charge, but we could delocalize that charge to the more electronegative oxygen; so this is an excellent resonance form.0543

    Now, there is a name for this resonance-stabilized carbanion: it is called an enolate.0550

    We just saw what an enol looks like: an enol has a double bond with an OH attached to that, right on that double bond; an enolate is the charged version of that structure.0555

    We have a double bond with an O- attached to that same carbon.0566

    And so, as you could see, the title of this lesson is "Enols and Enolates"; so those are going to be the two species that we are going to be interested in both forming and reacting.0569

    It turns out that we are going to be using both the enol and the enolate; if we have acid-catalyzed mechanisms, that is when we will be using the enol version of this compound, and in a base-catalyzed mechanism is when we are going to be using the enolate version.0586

    So, we will see many examples of these in the following slides.0601

    Now, before we progress, though, I want to talk a little bit about deprotonating a carbon; that is not a very easy thing to do.0607

    We are going to see repeatedly, throughout this unit, that we can do it for α carbons; let's think about other types of CH's.0614

    OK, if you have an alkyne, that has a reasonable pKa, as well--somewhere around 25; that means...that is a low enough number that, if you do treat it with a strong base, it is possible to deprotonate here and form this carbanion.0623

    Remember, this is sp hybridized, and that is what made this possible; and this is an OK carbanion to make.0637

    We have made these; we have used these in synthesis; and this is something that you should recognize as a staple carbanion.0647

    OK, however, if you tried to do that same thing with an alkane or an alkene--if you try to take an sp3 hybridized carbon or an sp2 hybridized carbon and try and do the same thing and react with a base, you would get a completely unstable carbanion here and a completely unstable carbanion here.0654

    We cannot deprotonate alkanes and alkenes; OK, it is possible to deprotonate alkynes, and the focus of this unit, of course, is deprotonating α carbons, such as aldehydes and ketones.0674

    Now, if we are going to be protonating, that means we need a base; let's think about what bases are possible.0689

    If you had to name a base, you might say something like sodium hydroxide; we know that is a good base, so let's try that.0693

    Let's react this with sodium hydroxide and think about what product we would get.0699

    OK, I could use hydroxide here as my base; it can grab this proton; and rather than have the electrons sit right on carbon, most often, when we deprotonate an enolate, we are going to go right to that better resonance form.0707

    We are going to take these electrons and form a π bond and move those electrons all the way up to be on that oxygen.0723

    That is the preferred resonance form--that is the better resonance form.0729

    We can go ahead and draw the enolate that way.0733

    OK, we can get that enolate; and what is the other product here?0737

    If I used hydroxide as my base, I just formed water.0741

    Now, water has a pKa of about 16, so when we compare these two acids, the acid in the forward direction has a pKa of 20; the acid in the reverse direction has a pKa of 16.0746

    Which is the stronger acid?0759

    Water is the stronger acid; and how does the equilibrium lie in a proton transfer reaction?--it always lies in the direction of the weaker acid-base pair.0761

    So, this reaction can happen, but just to a small extent; it is actually the reverse reaction that is favored in this case.0772

    So, if I were to use a base like sodium hydroxide, what would happen is: I would deprotonate just a small amount of this ketone; I would make a small amount of the enolate; a small amount of enolate will be present at equilibrium.0780

    But, for the most part, the ketone would still be a ketone; it would be the neutral form.0802

    This is exactly the situation we want in certain reactions; so in those cases, we use sodium hydroxide.0808

    We also want to recognize--let's just make that note here--we have a large amount of the neutral ketone or aldehyde present at equilibrium.0817

    So, with a weak base like hydroxide, we have mostly ketone with a tiny little bit of enolate around.0830

    If we wanted to completely deprotonate the ketone or aldehyde to make it an enolate, 100%, we are going to need a much stronger base.0837

    One of the bases that has been developed for this, that is used very widely, is this one; this is called lithium diisopropylamide (that means we have an N-).0846

    Lithium diisopropylamide is very conveniently abbreviated as LDA; and this is a great base, because it is very strong base, and it is very bulky (having those two isopropyl groups introduces a lot of steric hindrance).0858

    That makes it a non-nucleophilic base.0876

    So, in other words, LDA is just a base; all it does is find a proton to take, and that is its only purpose.0882

    Where do we have acidic protons?--of course, the α protons are acidic, so we can deprotonate here.0890

    Again, we could show that mechanism; the N- can come and be the base, and we can make this enolate.0897

    This would be the lithium enolate; it would be the lithium cation here, because it's lithium diisopropylamide.0907

    We most often don't have the counterion there, but of course, it's there every time we have a negative charge.0914

    OK, but the other product here, now, is going to be diisopropylamine: diisopropylamine has a pKa of around 40--this is a very, very weak acid.0920

    That means this reverse reaction of having an enolate try and deprotonate diisopropylamine is very, very small: essentially nonexistent.0930

    We can think of this as a one-way street--as a way to completely deprotonate a ketone or an aldehyde and convert it to its enolate form.0939

    Now, we don't always have to use LDA to do a complete deprotonation; there are cases, like this one, known as an active methylene; an active methylene is when you have this situation, where this carbon is between two carbonyls.0951

    That makes it even more acidic: the pKa...let's compare it to this α carbon (this α carbon has just one carbonyl, so it is similar to an ordinary ketone): this has a pKa somewhere around 20.0971

    OK, but when you put one next to 2 carbonyls, you have a pKa somewhere around 11.0986

    So, it is orders of magnitude more acidic.0992

    When you think about deprotonating anywhere else, neither of these is at all acidic, being next to an oxygen or just being a plain old alkyl; remember, we are never going to deprotonate an ordinary alkyl hydrogen, so these are not acidic at all.0997

    It is only the ones that are going to be α to carbonyls that we are going to be considering deprotonating in this unit.1016

    Sodium methoxide is a good base; so we could use that as our base, and if we did that, we would (let's just draw a line drawing here) deprotonate this completely.1021

    Again, the reverse reaction is minimal; we can completely deprotonate this α proton, and that is because this enolate not only has the carbanion (and we could delocalize it into this carbon on this oxygen on the left--the carbonyl on the left), but this could also have resonance to the other carbonyl, and that is what makes this conjugate base so very stable, and therefore makes the parent acid so acidic.1038

    OK, my other product here--if I used ethoxide, it gives me ethanol; that has a pKa of somewhere around 18.1070

    So again, comparing a pKa of 11 to a pKa of 18, that is a huge difference, and that strongly favors the forward direction and negates the reverse direction.1079

    This is called a stabilized enolate: when we have something that is next to two carbonyls that resulting enolate has extra resonance, and so we describe it as a stabilized enolate.1091

    In this case, you could use LDA; but as usual, we always want to use a reagent that is the lowest reactivity, most stable; that makes it easier to handle and less expensive.1108

    And so, sodium ethoxide would work just as well here, so we would use that.1118

    Just a quick question: we talked about how sodium hydroxide is something that can be used to form a small amount of enolate: why not use sodium hydroxide in this case?1122

    That certainly is on the order of the same basicity as ethoxide, so it would also do a good job of deprotonating this.1133

    But is there another reaction that can happen if I use sodium hydroxide as a base instead of sodium ethoxide?1141

    Well, take a look here: we have an ester in this starting material; so what would happen if you took this ester?--there is another reaction that can happen, besides just the acid-base reaction.1150

    We could also have hydroxide add into the carbonyl, and then come back down and kick off the ethoxide; so we could also get hydrolysis substitution--an acyl substitution--here, if we used hydroxide.1163

    Notice: because this is an ethyl ester, what base did we choose to use? we used the ethoxide.1177

    OK, so the key here is: for esters, when we go to choose a base and we want to deprotonate α to an ester carbonyl, we are going to use the matching base.1186

    So, if we have an O-R ester, we are going to use the alkoxide that matches that; so if it's a methyl ester, we use methoxide here; we had an ethyl ester--we used ethoxide; and that is to avoid the substitution reaction that can happen with the carbonyl.1202

    If you have ethoxide, ethoxide can add into this carbonyl, but what is it going to be kicking out?1219

    It is going to be replacing ethoxide; so that reaction becomes invisible, and no longer a side product.1223

    Now, I just started to introduce how you could have esters be deprotonated at the α carbon; and that is true--we are focusing most of our reaction on aldehydes and ketones initially.1231

    But you can have any carbonyl; so here we have a proton that is α to a carbonyl that has a pKa somewhere around 20; we can deprotonate here.1243

    Here we have, again, an α proton that is next to a carbonyl, but that carbonyl happens to be an ester; that is still acidic--not quite as acidic as the ketone, but it is still acidic, and we still can deprotonate here.1253

    OK, but carbonyls are not the only groups that impart acidity to the carbon adjacent to it.1266

    OK, the nitro group, the NO2 group, also makes this carbon acidic--in fact, significantly more acidic than having just a carbonyl.1273

    And the cyano group is another group that makes α carbons acidic, kind of like the same as an ester might.1285

    OK, and let's think about...let's draw those conjugate bases to see if it makes sense to us; this should make sense on why each of these hydrogens is acidic.1294

    If we have a ketone, and we put an anion here, why would that be an appropriate place to have an anion?--well, because it is next to the carbonyl, and it can have resonance.1307

    OK, how about if I had an ester?--I'm sorry...I have an ester...I have a methoxy group over here, instead of just a methyl group.1322

    Well, we could still have resonance here, and this is still a resonance-stabilized enolate, so you can make the enolate of an ester, just like you can for a ketone.1332

    OK, but how about the nitro or the cyano?--well, let's take a look at the structure of the nitro group.1343

    Let's imagine deprotonating; so we just lost a proton here, and we now have this anion.1348

    When you draw out the nitro group, it looks something like this; and how would that be a stable carbanion?1356

    Well, this can have resonance just like a carbonyl can; we have an N-O double bond instead of a C-O double bond; but just like the carbonyl could, we can use that to delocalize the negative charge and move it to an oxygen atom.1370

    Of course, having a negative charge on an oxygen atom means this is a very good resonance form.1385

    OK, plus you have this positively charged nitrogen pulling electron density and stabilizing inductively; that is why this is even more stable than an enolate; it makes the α proton so much more acidic.1390

    How about a nitrile?--let's imagine this anion and draw out the nitrile: a nitrile, of course, has a C-N triple bond.1401

    Well, again, we have an allylic negative charge and an allylic lone pair, so the lone pair can come in, and the π bond can move over.1411

    We can have resonance; it puts the negative charge on nitrogen; that is not as good as oxygen, but it is certainly better than carbon, and it is because of this resonance stabilization that it makes it a reasonably acidic proton and something we can deprotonate with a strong base like LDA.1420

    These are enolate-like species; when we say "enolate," we mean a carbonyl of some kind--either a ketone or an aldehyde or an ester; but these are enolate-like species, as well.1437

    Just a quick question: there is a big difference here--5 orders of magnitude less acidic for an ester, compared to a ketone: why do you think that is less acidic--what is there about this conjugate base that is better than this conjugate base?1451

    Or why would a ketone be less willing to donate a proton?1467

    It must have something to do with this O-R group: well, remember: what effect does this oxygen have to the carbonyl next to it?1471

    It is an electron-donating effect to that carbonyl.1479

    So, an ester carbonyl is more electron rich; so that means it does not want to have this negative charge--this negative charge, now, has to compete with the oxygen's electron donation for this carbonyl, so there is less resonance stabilization that can go on to delocalize that negative charge.1485

    OK, so it is because the ester carbonyl is more electron rich that that makes it a weaker acid.1509

    Just to summarize: the carbonyl, the nitro, the cyano...each of those groups are described as electron-withdrawing groups (or EWG for short); and what they all have in common is: if you put a negative charge right next door to them, they would be able to stabilize that negative charge by resonance.1519

    That puts these three groups together in a certain category; we are going to see lots of examples, actually, where the EWG comes into play, and imparts a particular reactivity to the groups attached to it.1539

    So, now that we have taken a look at how to make enolates with a mechanism, what kinds of bases we can use, what types of α protons are going to be acidic...what would we do with an enolate--how are they used?1555

    Well, they are carbanions; they are examples of carbons with a negative charge, which makes them good nucleophiles: anything that is electron rich is going to be a good nucleophile.1567

    And so, we are going to react it with a variety of electrophiles.1578

    Electrophiles are things that are electron poor: things like an alkyl halide--that is an electrophile we have seen: we know that this halogen (chloride, bromide, iodide...) pulls electron density from this carbon, making it partially positive.1582

    So, we can have nucleophiles attack here, doing something like an SN2 mechanism, for example; and so, an enolate would be an example of that.1600

    OK, or even just Br2 or a halide like that--it doesn't have a carbon bearing a leaving group, but it has a halogen bearing a leaving group.1613

    For that same reason, you could have an SN2 occurring with that; we will see reactions of enolates with something like Br2.1624

    Now, a carbonyl has a partial positive/partial minus; so we have seen carbonyl reactivities as electrophiles, and we will see something like a ketone, aldehyde, ester...OK, in general, any carbonyl would be a good electrophile; and we will see enolates react with those.1635

    And finally, an epoxide would be a good electrophile, as well; remember, the carbons of the epoxide are partially positive, because the oxygen pulls electron density.1657

    There is a lot of ring strain involved in that: it's a three-membered ring, and so those also readily react with nucleophiles.1667

    Let's look at some of these examples.1674

    The first one, reaction of an enolate with an alkyl halide, is described as an alkylation reaction, and let's take a look at this reaction sequence.1677

    If we were to take cyclohexane and first treat it with LDA (remember, that is lithium diisopropylamide), every time we see LDA, we know that we have a base, a strong, bulky base.1685

    And what do bases do?--they deprotonate.1699

    So, we look at cyclohexane and ask, "Where can we deprotonate?"1703

    And of course, it is that α proton that is going to be acidic; so step 1 is going to deprotonate the α carbon to give us a carbanion.1706

    OK, and what I'm suggesting is that this carbanion, now, is going to be a very, very good nucleophile.1719

    In step 2, when I react with methyl iodide (which is a good electrophile), what reaction can happen here?1726

    I think the nucleophile is going to attack the carbon and kick off the leaving group; that looks like a back side attack--it looks like an SN2 mechanism.1736

    That is exactly what happens, and I have now installed a methyl group at the α carbon; so this is called an alkylation or an α alkylation reaction.1744

    It is this two-step process: we deprotonate, and then we treat it with an alkyl halide.1759

    Now, because it is an SN2, remember, that back side attack requires minimum steric hindrance, so we have to have an unhindered alkyl halide; a methyl group would be great; primary would be fine; but as soon as we get to secondary or tertiary, then E2 is going to be preferred, because the enolate is still a pretty strong base.1764

    Now, as shown here, the enolate is a nucleophile at carbon: in other words, C alkylation--putting the alkyl group on the carbon--is preferred, rather than O alkylation.1783

    OK, and this is important to point out, because remember: the enolate has two resonance forms.1794

    It doesn't matter which resonance form you draw; you should still be getting the same product.1800

    OK, so let's take a look at the mechanism using the better resonance form, the preferred resonance form.1804

    What we'll do is: we will have our base, our LDA, come in and deprotonate; but instead of putting the negative charge on this carbon, the more appropriate place to put that negative charge is on the oxygen.1810

    I am going to draw the enolate like this.1825

    Now, that is the better resonance form, because it has the negative charge on the more electronegative oxygen; in other words, this better represents the actual hybrid of the enolate structure.1831

    However, it is very tempting now, when you look at this, to think, "When I bring this together with my electrophile, it's very tempting to use the negative charge from oxygen to go and attack."1841

    That would be described as O alkylation, where the alkyl group ends up on the oxygen; that does not happen.1854

    That does not happen; it is the α carbon; it is the α carbon that is the nucleophile in an enolate, no matter how you draw it; it is the α carbon.1860

    So, how do I get this carbon to react?--well, this is what enolates do: the enolate is here: the enolate is a nucleophile; and I'm going to start with the lone pair up on the oxygen, but instead of having that attack the electrophile, it's going to come down and re-form the carbonyl, and that is going to push these electrons out.1871

    We are going to form this bond here between the α carbon and the methyl group.1889

    These two electrons are going to come and attack the carbon, and that is what kicks off the leaving group; so it is the α carbon that is nucleophilic and does the SN2.1894

    So now, when we follow the arrows around, I see that I am back--my oxygen has two lone pairs; I have my carbonyl back; and attached to that α carbon, I have my alkyl group.1905

    OK, this is typically how we are going to be drawing our enolates for the rest of the chapter, and how we can use them.1917

    This is not wrong; this is still an acceptable drawing of an enolate; it is just not as good a representation as this one.1924

    We want to get used to using this mechanism.1933

    What else can we do with the α carbon?--we can also halogenate at the α carbon; that reaction is described as the α halogenation.1940

    Here is an example: if I take this ketone and treat it with chlorine and base with water, I can install a chlorine in the α position.1947

    Now, that mechanism is analogous to the one we just saw for α alkylation; we deprotonate the α carbon, and then we attack the Cl2 to put in the chlorine.1959

    OK, and we can look at that reaction; but this reaction is not that useful, because this product that we formed--if you take a look at that α carbon, it turns out that this is now more acidic than your starting α carbon.1969

    And why is that?--well, we just added a chlorine here; we know chlorine pulls electron density; or any halogen is electronegative and pulls electron density here; so that enhances the acidity here and makes it easier to deprotonate here.1987

    So now, I am going to be forming some of this enolate, and I'm going to end up adding a second chlorine; so we could have multiple halogenations.2000

    This would not be a very good way to make this 2-chloro, (1, 2, 3, 4, 5) 3-pentanone, because it would be hard to stop in this case.2010

    Any time your product is more reactive than your starting material, it is difficult to control those reactions.2021

    OK, in fact, it is so reliable that there is a test called the iodoform test that counts on this overreaction.2027

    It is when you have a methyl ketone (so I have a CH3 here, and attached to a carbonyl, that is called a methyl ketone)--if you treat a methyl ketone with iodine and sodium hydroxide, what happens is: you end up replacing all three of those protons with iodines by successive α halogenations.2035

    And that turns this group into somewhat of a leaving group; now, it's very unusual to have a carbon leaving group, but because that carbon has three iodines on it, all pulling electron density away, it is a reasonable leaving group.2058

    And so, hydroxide can add into the carbonyl now and do a substitution reaction.2073

    We can get a CTI, and again, even though we don't normally think of this carbon as a leaving group, it can, in fact, get kicked out; and the product we are going to get--we will get CI3-, which gets protonated to iodoform.2081

    We will get the carboxylic acid product out here when we are done; and this ends up being a precipitate; this is a yellow precipitate, and this is the test.2098

    The iodoform test is when you treat a sample with iodine and sodium hydroxide; if a precipitate forms, then that is a chemical test, a qualitative test, to indicate the presence of a methyl group attached to a carbonyl.2110

    So, these kinds of tests are outdated, now that we have such powerful spectroscopic techniques, like NMR, where you can really see a methyl ketone a lot less ambiguously.2124

    But this still has some historical note, and you will often see this reaction still, in textbooks.2135

    With a base-promoted reaction, we expect to have multiple halogenations.2141

    If we want to do a single halogenation, that is possible; but what we are going to do is: we are going to use acid-promoted reaction conditions instead.2147

    In this case, if we used Br2, but with acetic acid now instead of a base (because here we see a carboxylic acid, so we recognize that these are acidic conditions), it is possible to get a single halogen in here with good yield.2157

    OK, but because our conditions now are acid-catalyzed, rather than base-catalyzed, we can't use the enolate; we are going to use the enol instead.2174

    The first two steps are going to be making the enol (remember, we already looked at that mechanism).2183

    It might help to think about what the enol structure looks like; the enol has (instead of a carbonyl) an OH, and then it also has the double bond between what used to be the carbonyl carbon and the α carbon; so that is the enol.2193

    What are the two steps to get there?2207

    Well, first we need to protonate; we will just abbreviate our acetic acid as HA; so step 1 is protonate.2209

    And then, step 2--I need to deprotonate.2223

    This is our tautomerization mechanism; it's just a series of protonations and deprotonations; I'm going to take this α carbon; I'll use the A- that I just had, since it's catalytic in acid, because I use the acid in the first step, and I get it back in the second step.2230

    I form the π bond and move the π bond; so 2 steps--we should always be able to form an enol.2245

    An enol, now, is a nucleophile, just like an enolate.2252

    It is not as easy to see, because we don't have a negative charge; but it is nucleophilic, and it is nucleophilic at the α carbon, just like the enolate was.2255

    So, if we bring in Br2 as our electrophile, notice, it has that leaving group attached to the bromine.2267

    How are we going to get these two together?--well, we are going to react the enol, just like we did the enolate.2274

    What we are going to do is: we are going to start up on this oxygen, this lone pair; we are going to form the carbonyl, and that is going to kick the π bond out, right?2281

    We are forming this bond between the α carbon and the bromine and kicking off our leaving group.2289

    That installs a bromine at the α carbon; this gives me an oxygen; I still have the hydrogen on here; what else do I have?2297

    I have the π bond, and I have just one lone pair left, and a positive charge.2306

    So, I am very close to my product, right?2313

    What do I have to do to get to my product?--I simply have to deprotonate the carbonyl; we can bring A- back in, or we could use the Br- that we just formed; and we can deprotonate.2315

    So, as we will see is common in these reactions, the acid-catalyzed reaction is going to be more steps than the base one.2334

    And we will be using an enol as our nucleophile, instead of an enolate.2344

    But this would be a way that we make...we effectively can do an α halogenation on a ketone or an aldehyde.2350

    We have seen how to add a halide to the α carbon; we have seen how to add an alkyl group to the α carbon; what other electrophiles are there?2359

    Well, the carbonyl, of course, is a big electrophile, and we will study that reaction next; we are going to be spending quite a bit of time, actually, on that reaction.2368

    It is known as the aldol condensation, and it is the reaction of two equivalents of a carbonyl-containing compound, an aldehyde or a ketone.2379

    We'll take a look at an example where it uses base, but it can also be acid-catalyzed; we will see both mechanisms here.2392

    The product looks like this: now, we use two equivalents of the aldehyde, and here is one of them, and then here is the other one.2399

    It is like we had our...if we drew the second acid aldehyde molecule down here, you can see, here is the reaction that is happening.2412

    We are forming a bond between this α carbon and this carbonyl carbon.2422

    OK, we will take a look at that mechanism next.2427

    Now, this product has both an aldehyde component and an alcohol component; so this is described as an aldol product, because it has an aldehyde and has an alcohol.2429

    This is actually one of the few cases in organic chemistry where this is not named after Professor Aldol; this is not a named reaction that is capitalized; it is simply described as an aldol because that is what the product might look like.2442

    OK, and we will also see that, if you heat an aldol product, you can cause it to lose water; so this aldol product will react further upon addition of heat.2456

    It readily loses water to form this carbon-carbon double bond; so we'll look at both of those mechanisms, "How do we get to this aldol product?" and then "How does that aldol product lose water?"2474

    OK, let's look at the base-catalyzed mechanism first: and when we are in base, we are going to deprotonate first--let's ask where would be a good place to deprotonate.2487

    Certainly, it's going to be the α carbon, because that is the theme for this unit; and so, this is not an α carbon; this is not an α proton; this is attached directly to the carbonyl.2499

    That is not at all acidic; but here is my α carbon; here is an α proton; let's change that to a CH2, so we can see one of those hydrogens.2511

    And, absolutely, that is going to be our first step: to deprotonate that α carbon to form an enolate.2518

    Now, every one of these steps is reversible, so we will use our equilibrium arrows here to go from one to the other.2525

    OK, so my first step is to make an enolate, and what kind of reactivity do we expect for an enolate?2535

    Well, it is a nucleophile; so we need to look around for an electrophile.2541

    OK, and this is where it becomes important the choice of base that we use.2548

    We only used hydroxide; we didn't use LDA in this case--we just used hydroxide; hydroxide is a weaker base, so it has taken some molecules of this aldehyde starting material and made the enolate.2552

    But, by and large, most of this aldehyde is still the aldehyde; so the aldehyde is still present.2567

    Therefore, it can serve as our electrophile; the carbonyl can be our electrophile.2575

    So, we let our enolate do its thing: form the π bond; form the carbonyl; kick the π bond out; and attack the carbonyl; and then, what happens when a nucleophile attacks the carbonyl?2582

    We break the π bond, and we move those electrons up.2594

    Let's redraw our enolate part here: we have our carbonyl back; we have a new bond from this carbon to this carbon; this carbon is this carbon; now what do I have on this carbon?2598

    I have an H and a C; I have a hydrogen and a methyl; and I have a single bond O-.2616

    That is a nucleophile attack on an electrophile.2630

    We're not quite done yet: we have an O-; how can we get rid of an O-?2634

    All we need to do is protonate; so we can bring...we just use water; we formed water in this first step, and hydroxide acted as a base; so we could use water as our acid.2637

    We can protonate the O-, and we have our aldol product; here is our aldol product.2652

    Notice, it has a new carbon-carbon bond, because we had a carbon nucleophile attacking a carbon electrophile; so here is the new carbon-carbon bond in the aldol product.2662

    That is going to be important to us, to identify when we try and do synthesis problems.2675

    OK, and overall then in base, it is going to be a three-step mechanism: first, we deprotonate, and then we had attack of the enolate onto the carbonyl, and then we protonate.2681

    Deprotonate, attack, protonate: we are going to see that pattern again and again and again for base-catalyzed mechanisms.2699

    OK, so a summary then: an aldol--what does it mean to be an aldol reaction?2707

    It forms a new carbon-carbon bond between the α carbon of one carbonyl compound (in this case, a ketone or an aldehyde) and a carbonyl carbon of another.2711

    Between the α carbon and the carbonyl carbon, we describe that reaction as the aldol reaction.2722

    Now, we said that this aldol product can lose water; so let's draw the aldol product we have so far.2730

    We said that this, now, can undergo loss of water, and that mechanism is also a two-step mechanism; it is described as a β elimination.2740

    OK, and it starts--the first step is going to be to deprotonate; so I'm going to take that hydroxide again, and I'm going to deprotonate the α carbon, so let's make that a C-H.2753

    We are going to deprotonate the α carbon and make an enolate.2768

    And now, typically, after we have made an enolate, that enolate that we have used thus far on mechanisms--that enolate has always formed the carbonyl and kicked this π bond out to some external electrophile.2779

    But, in this case, what we have is a leaving group that is in the β position.2797

    Remember, the α position is the first carbon next to the carbonyl; the β position is the next one over; and hydroxide can act as a leaving group.2805

    So, when you have a leaving group in the β position, that enolate will re-form the carbonyl, shift this π bond down, and kick the leaving group off.2812

    We are going to deprotonate as our first step; and step 2--we are going to inject the β leaving group.2823

    And, when we do those three arrows, the product we get is the aldol product that we are expecting: the carbon-carbon double bond in between the α and the β carbon.2831

    This is our mechanism: this step here kind of looks like a collapse of a CTI, a little bit.2846

    Remember, collapse of a CTI was when we had an O- kicked down, and on the same carbon, had an OH; it would kick off the OH.2858

    This is an extended system, but it is still the O- kicking down and forcing the OH off.2864

    OK, now think about it: we just lost hydroxide here; that is not a typical leaving group, but we have seen it before for collapse of a CTI; so that is acceptable.2872

    And now, we are seeing that, because this mechanism is similar, it is also acceptable for a β elimination mechanism.2883

    You can lose hydroxide as a leaving group; OK, and remember, the reason for both of these--the reason that it is OK--is that your driving force is your formation of the carbonyl.2888

    In both of those mechanisms, what is pushing the hydroxide out is an O-, which is forming a carbonyl at that same time, so because I am forming this carbonyl, it is OK to lose hydroxide as a leaving group.2907

    OK, just a quick note of caution: this loss of water--this mechanism is not an E2 mechanism.2919

    OK, it is very tempting here (I have redrawn it)--if I wanted to go from this aldol product and eliminate water, what is very tempting is that I just use my base to deprotonate.2928

    And, instead of using it to form an enolate, like I need to, I think, "Well, let's just form this π bond and kick out the leaving group; wouldn't that be a fast, easy mechanism--shouldn't that be better?"2940

    OK, that would be an E2, when we have a single-step elimination mechanism; that is described as E2.2951

    OK, but that does not happen: there is no way we can do an E2, because hydroxide is not an acceptable leaving group for such a mechanism: we have to have a traditional leaving group like bromide, chloride, iodide, tosylate, something like that, in order to enable this.2958

    The only reason this elimination can take place is because of this carbonyl, because you have this resonance-stabilized enolate intermediate, and so it is critical and must be used as part of your mechanism.2975

    OK, so just a reminder not to do a single-step elimination here of water, but to do a two-step β elimination.2985

    OK, so let's summarize the aldol: an aldol is what we have when we have an aldehyde or ketone; we are going to treat it with either base or acid; the ketone (in this case) is going to serve both as the nucleophile (the α carbon is doing the nucleophile), and it is going to serve as the electrophile (the carbonyl of the second molecule is going to be the electrophile, so there is the bond we are forming).2996

    We form a new carbon-carbon bond between the α carbon in one and the carbonyl carbon in the other.3020

    OK, so the product we get could be described as a β hydroxy ketone, right?--not in the α position, but in the β position, where it's going to have this hydroxy group.3027

    And then, if we heat this, we can lose water, and so now we form a double bond between the α carbon and the β carbon; so this structure is described as an α β unsaturated ketone, because we have a point of unsaturation; we have a double bond between the α carbon and the β carbon.3039

    So, when we do an aldol, we can kind of choose which kind of product we want to draw: it's either the β hydroxy ketone, or sometimes it's an α β unsaturated ketone; very often, the trigger on whether or not you are going to lose water will be the addition of heat.3061

    OK, sometimes this is spontaneous, because this dehydration is a relatively easy one.3076

    OK, let's think about why that is: typically, when we dehydrate an alcohol, we need strong conditions: sulfuric acid, phosphoric acid, heat--it's very hard to dehydrate an alcohol.3085

    This one happens very, very easily: you don't need a strong acid; you could even do it in base; and you just need to warm it up a little.3095

    In many cases, it is very, very facile.3103

    OK, but think about your final product: you are not just forming an ordinary carbonyl; it is not just an ordinary alcohol; it is a β-hydroxy is β to a carbonyl, and so this resulting double bond is now going to be conjugated with the initial carbonyls.3106

    This is a conjugated system, and that conjugation--when you have a double bond right next to another double bond, that means you have resonance, and it is a very stable structure.3128

    So, that is why aldol products are very likely to lose water and go to an α-β unsaturated carbonyl.3144

    OK, so so far, we took a look at the base-catalyzed mechanism; we will look at the acid next, but let's draw some comparisons.3155

    In the base-catalyzed mechanism, everything was either...all the species were either neutral, or they had a negative charge; the nucleophile we used was the enolate; the electrophile we used was the ketone carbonyl.3162

    OK, now how is that going to change for an acid-catalyzed mechanism?3182

    Now, an acid-catalyzed mechanism--you can't have a strongly basic species, like an enolate.3185

    You need to have either neutral compounds or positive charges: you are not going to have any O- in an acid.3193

    So, instead of the enolate, what species could we use that would still be nucleophilic?3198

    How about if we protonated that to make it an OH; what would that be called?3204

    That is the enol, and that is going to be the nucleophilic species in the acid-catalyzed.3209

    We also can't have a neutral carbonyl, because if you attack a neutral carbonyl, that forms an O-; you can't have an O- in an acid-catalyzed mechanism, so what we are going to do is: we are also going to protonate the carbonyl, and it will be a protonated carbonyl that gets attacked.3214

    So, it turns out that this mechanism is going to be a little longer, in order to accommodate these species; but let's take a look at that.3232

    We said we had to make an enol; those will be our first two steps of the mechanism, then: we will start with our...we are doing acetone, in this case; and we have some acid present.3244

    So, to make an enol, first we protonate; and then the second step to make the enol is deprotonate; so we take this α proton, A-, and we have an enol.3256

    That has to be our first two steps: to make the enol.3281

    Now, I have my nucleophile: what electrophile is it going to react with?3288

    Well, we just saw on the previous slide: it is not going to react with a neutral carbonyl; it needs to react with a protonated carbonyl.3292

    We just showed the mechanism on how to form the protonated carbonyl, so we can use that again; but this is going to be our electrophile.3300

    What is the bond that we are going to be forming?--it's going to be between this α carbon and this carbonyl carbon, so our mechanism to get there is: our enol starts at the lone pair on the oxygen, forms the carbonyl, and that is what kicks the π bond out.3308

    It kicks those electrons out; those attack the carbonyl of the electrophile and break the π bond; that is the key step in any aldol--when the α carbon attacks the carbonyl carbon.3327

    Let's see where this brings us: let's redraw the enol.3341

    It now has a carbonyl and still has the proton up here; and we just formed the new bond between the α carbon in one and the carbonyl carbon in the other; what does this carbon have on it?3346

    It still has the two methyl groups, and it has a single bond OH.3358

    What do we have to do to finish this up?--it looks like we just have to deprotonate; so we'll bring A- back in here, and we are done.3368

    There is our aldehyde; so once we made our enol, then we protonated the carbonyl; we attacked the carbonyl; and we deprotonated the carbonyl.3385

    OK, but it is going to be a little bit longer mechanism to get where we want to go.3395

    Notice, overall there are no O- charges in the acid; everything is positively charged or neutral--that is going to be consistent with any acid-catalyzed mechanism.3397

    Now, let's think about losing water from that aldol product to form the alpha, beta unsaturated ketone; we describe that mechanism as a beta elimination mechanism; we saw, in base, it was just two steps.3416

    The first step was: we deprotonated to make the enolate.3428

    And then, the second step was: that enolate kicked down, kicked down, and kicked out the leaving group; so eject the beta leaving group.3432

    Now, how do we have to adjust that--who is the beta leaving group?3439

    It was hydroxide in the base-catalyzed.3442

    So, how are we going to adjust this for the acid-catalyzed version?--we can't have an enolate, so instead, we are going to make the enol.3446

    OK, and what do you think about hydroxide--do you think that would be an acceptable leaving group in acidic conditions?3455

    No way--much too strong of a base; so what would be a leaving group that would be OK to have around in an acidic solution?3460

    We would have to lose water; OK, so with that in mind, let's think about our mechanism: let's give it a try.3468

    We are going to start with our aldol product; we want to lose water from this, so the first thing we need to do is: we need to convert this to the enol.3475

    How do we make the enol?--two-step tautomerization mechanism--2 steps.3489

    First, we protonate; that tautomerization practice is really going to pay off when you get to these more complex mechanisms, so that hopefully this part will be somewhat automatic: protonate at the carbonyl, and then deprotonate at the α carbon.3495

    That is the first thing we have to accomplish: we have to make the enol.3520

    OK, and now, we need to get rid of our beta leaving group, but it is not a good leaving group right now.3527

    If this kicked down and kicked off, the leaving group would be hydroxide leaving.3532

    So, how do we make it a better leaving group?--we need to protonate, and so now it is going to look like water when it leaves; it looks like water right now, as it's attached.3535

    This is a good leaving group in acid; so when we said that hydroxide is an OK leaving group for collapse of a CTI for beta elimination, that is true, but that assumes you are looking at a base-catalyzed mechanism, because you have a hydroxide.3554

    In this case, we need water as our leaving group; so now, we are ready to do our beta elimination and eject that beta leaving group.3569

    The same idea as with the enolate, though: we are going to start up here in the lone pair on this oxygen and form the carbonyl; that is what pushes this π bond down; that is what kicks our leaving group off.3577

    All right, we just lost water.3590

    And now, we track our product; see where we are; this, now, is an O+ up here, so we are this close again; we just have one proton to go.3594

    All we need to do is deprotonate: we can bring A- back in.3606

    A-...of course, that is the only negative charge we can have in an acid-catalyzed mechanism: A- represents some very stable conjugate base of a strong acid (like sulfuric acid conjugate base, let's say).3611

    OK, so that would be an OK thing to have in acidic conditions.3624

    A final deprotonation, and we have done our elimination.3631

    So again, it was just two steps in base and several steps in acid; but practice with the base--get used to that; understand the logic, and be able to talk yourself through the mechanism, and know where to go, so that you can make those adjustments and make it work for acid.3634

    So again, it's all positive charges or neutral species.3648

    Now, in all the cases we have looked at so far, aldol has always been a self aldol, meaning one molecule of a ketone has reacted with the exact same structure of another molecule of a ketone.3657

    OK, when we have two different ketones or aldehydes coming together, this is something that can be reasonable, if only one of the compounds has α protons.3670

    So, for example, if we bring it together--acetone and benzaldehyde with base (sodium hydroxide and water), we think about who could be the nucleophile and who could be the electrophile.3681

    Well, we know we have a base here; so we can deprotonate, and we are going to look to our α carbons to deprotonate.3693

    Well, acetone certainly has an α carbon; so this one--when we deprotonate it, that would be our nucleophile.3703

    We have one nucleophile here, but what about benzaldehyde?--there are no α protons on the benzene ring, and this hydrogen is not alpha, so this has no α hydrogens; so that means this cannot be a nucleophile in any way.3708

    It has to be an electrophile.3725

    We only have one nucleophile; how about looking for electrophiles?3728

    Well, both of these structures have carbonyls, so you might think they both could be the electrophile in the reaction.3732

    OK, however, we have a ketone and an aldehyde, and the aldehyde is the better electrophile.3738

    Now, let's just take a quick look at why that is: OK, well, remember: alkyl groups donate electron density; so when we look at a ketone, we know that these carbons are going to be adding electron density to the carbonyl.3750

    That makes the ketone more electron rich.3761

    OK, is that good for an electrophile?--that is not good for an electrophile.3765

    OK, this is more electron rich; so this is a poorer electrophile, and the aldehyde is the better electrophile, because this only has one alkyl group donating; this has a larger partial positive.3770

    So, comparing an aldehyde and a ketone, an aldehyde is going to win, in terms of electrophilicity and reactivity.3784

    And so, comparing acetone and benzaldehyde, benzaldehyde is a better electrophile.3791

    We have one electrophile; we have one nucleophile; which means we can expect to get one major aldol product--and what would that look like?3798

    Well, we could start by drawing our acetone: we know we are going to form a new bond between...this α carbon is our nucleophile, and where is it going to attack?3809

    It is going to attack this benzaldehyde carbonyl; so this used to be a CH3; it is now a CH2, because we deprotonated there, and that is how we have room for this bond.3818

    And then, when we are all done, what do we have on this carbon?--we have a phenyl, we have a hydrogen, and we have an OH.3830

    It would be an O-, but we had protonated at some point, so we get this β-hydroxy ketone; this would be our major product.3837

    However, this probably would lose water spontaneously; we wouldn't really have to warm this up, probably; just having it at room temperature would be enough for this to lose water.3845

    In this case, it is very likely that you would get this as your major product.3860

    OK, and let's think about this particular example: why is it, in this case, that it would be very difficult to isolate this alcohol--very difficult to not have it undergo this dehydration step, this β elimination step?3868

    Well, that is because we have...this double bond is not only in conjugation with the carbonyl, like it always is in an aldol; but we also have this benzene ring right here, so this is conjugated with the benzene ring π bonds, and that makes this π system an extended, conjugated π system--very, very stable.3881

    So, in the case with benzaldehyde, it's very common to seemingly spontaneously lose water and go straight to the alpha, beta unsaturated carbonyl, even though you haven't seen the addition of heat explicitly.3905

    Now, what if I wanted to do a mixed aldol between two ketones or two aldehydes--two groups that have similar reactivities?3923

    Well, we could exert control there; we wouldn't just want to mix those two together in the presence of base, because then, you can get all sorts of possible products out.3931

    You have two possible nucleophiles; you have two possible electrophiles; so in a mixed aldol like that, you might have up to four possible products.3939

    But the way you could exert control is: instead of just mixing everything together, you could do it step-wise by using something like LDA.3947

    Remember, LDA is our strong base; and so, if we started out with just a single ketone, and treated it with LDA, we would expect that ketone to be deprotonated.3954

    Now, this is an interesting case, because we have two different α protons in this case; how would we know which one is the major site of deprotonation?3967

    Well, remember: lithium diisopropylamide had those two isopropyl groups; it is a very bulky base; LDA prefers to attack the less hindered α carbon (and therefore, α proton).3978

    So, it has to do with sterics; it has to do with kinetics; it goes to the less hindered side.3995

    So this is going to deprotonate to give the following enolate: this is a we have just one nucleophile; we have this enolate being formed.4001

    Remember, the reason we use LDA--we use a full equivalent of LDA so this gets completely converted to this enolate.4021

    OK, and now, after the enolate is formed, now we add in a second carbonyl; we have just one electrophile, so there is really just one major product that can occur.4028

    What is the bond that we are going to form?--the electrophile has reacted with the carbonyl carbon; the enolate is reactive at the α carbon; and so, we expect the enolate to attack the carbonyl.4039

    And so, what is our product going to look like after workup?4055

    We are going to form a new bond to the α carbon, like we would in any aldol; this carbon is part of the 6-membered ring; and then, on this carbon, we also have a single bond O-, which then gets protonated upon workup.4059

    Now, we would do a separate workup: when our reaction is all done, we get this β-hydroxy carbonyl.4076

    Now, I want to point out just some stuff to you, some common mistakes: it is very common to draw as the product of this, this product; and what would be the problem with drawing this product?4084

    You have attached the carbonyl carbon to the carbonyl carbon; something is missing here.4098

    OK, you have an α-hydroxy carbonyl; that is not the product you should be getting--it should be a β-hydroxy carbonyl.4105

    So, especially when it comes to line drawings, it is very easy to lose carbons when you are drawing your zigs and zags; so make sure you explicitly show your α carbon that was your nucleophile.4111

    There is a new bond to your β carbon, which was your electrophile (that carbonyl carbon).4121

    And that way, you won't make mistakes of accidentally losing carbons when you are dealing with line drawings.4126

    Let's see if we can take one of these problems and do it as a retrosynthesis: if we are given this starting material...this target molecule, rather...and asked to synthesize it, how could we do that?4136

    Can we predict what reagents we need for the aldol to give this target molecule?4149

    Remember, a retrosynthetic arrow asks, "What starting materials do I need?"4156

    "How could I create this target molecule?"4164

    Well, it looks like an aldol product, even if it doesn't give us that clue; it looks like an aldol product, because between the α and the β carbon, we have a double bond.4167

    This is an alpha, beta unsaturated carbonyl (ketone, in this case).4176

    And that is the type of functional group pattern we would get if we did an aldol reaction.4185

    So, it is possible to just kind of disconnect the bonds that we know that are being formed; but I think maybe an easier step backward would be to say, "OK, well, remember: I got this double bond because I lost water from the β-hydroxy ketone."4192

    Maybe you can kind of add water back in; let's abbreviate this as just a phenyl group; let's add water back in, so that we add the hydrogen, and we add OH.4208

    We already had hydrogens here, so we can leave those in if we want, so we don't get too confused.4221

    Let's add water back in to go back to the β-hydroxy ketone, because this is also a pattern that we could get from the aldol reaction, but it's a little easier to grab onto this one, because we have some more interesting functional groups.4225

    OK, the key disconnection we make: remember, what is an aldol?--it is the reaction of an α carbon to a carbonyl carbon.4243

    This is the key disconnection that we make in an aldol; it is between the α and the β carbons.4251

    This carbon--remember, one of these carbons was a nucleophile; one of them was an electrophile; that is how they were able to come together to make this product.4256

    The α carbon, of course, was the nucleophile; this α carbon was the nucleophile--in other words, that was the enolate in the mechanism.4267

    And that means that this carbon was my electrophile; what electrophile could I have that, after it gets attacked, we end up with an alcohol at that carbon?4280

    It was a carbonyl; if I had a carbonyl here, and an enolate attacked it, it would break the π bond, and I would get an alcohol there.4292

    That is the logic we can use: when we do this disconnection, then, we end up with just a phenyl and a 1-carbon aldehyde (it looks like benzaldehyde, again, is involved in this).4302

    That would be our electrophile, and what would our nucleophile be?--it would be a phenyl group with a carbonyl and then a CH3.4313

    So, if I had acetophenone and benzaldehyde, I could actually just mix these two, because this is similar to the mixed aldol we just looked at--because this is a ketone and this is an aldehyde.4328

    I could just mix these two together in the presence of base (sodium hydroxide and water, or methoxide and methanol--something like that), and a little heat to afford the dehydration.4341

    These actually would come together to form this single product as our target molecule, what we expect to get.4352

    This is our major aldol; we wouldn't have to bother first treating this with LDA and then adding the benzaldehyde separately.4359

    OK, but this would be a good process by which we could come up with the starting materials required for an aldol synthesis.4366

    Next, let's take a look at a reaction called the Claisen condensation: this is very much related to the aldol condensation; in fact, it is an aldol reaction; but instead of using a ketone or an aldehyde, we are going to be using an ester instead.4376

    We call that a Claisen.4388

    Let's look at an example of it and see where it takes us.4390

    Let's say we have this ester and we react it with ethoxide, sodium ethoxide, and ethanol; this is a base.4393

    Any time you see hydroxide, know it is a base.4401

    And then, after that, we are going to do some mild workup; let's see where it brings us, rather than looking at the product first.4404

    What do you think the first step will be of this mechanism?4410

    We have a base, so let's deprotonate something; where do we deprotonate?--α carbon.4415

    α carbon, α carbon, α carbon: that is the whole theme of this lesson.4420

    So, I am going to look: and remember, these carbons are not attached to the carbonyl; those are not α carbons, but this one is; so I could use ethoxide as my base, and I could deprotonate the α carbon.4425

    This is done reversibly, because ethoxide is not LDA--it doesn't do so in just one direction.4439

    But I will get out some of this enolate; let's just abbreviate this--this is an ethyl group here, so we can abbreviate, OEt, and save us a little time drawing it.4448

    OK, so deprotonate to form an enolate: that is a good idea; and what kind of reactivity does an enolate have?4461

    We know that is a nucleophile; so let's look around for potential electrophiles.4468

    Well, remember: because we use this weaker base, ethoxide, we still have most of this ethyl acetate, still as the ester form; so that is still around.4474

    We can bring in another molecule of the ethyl acetate as our electrophile; so see how it is just like an aldol?4486

    We have an enolate of a carbonyl-containing compound attacking a non-deprotonated copy of itself; so our enolate does its thing; re-form the carbonyl; break the π bond.4493

    And then, what happens when you attack a carbonyl?--always, always, always the same: break the π bond and move it up on oxygen.4507

    Let's draw this product over here; let's draw our enolate again--it now has a carbonyl.4516

    Single bond to the α carbon; new carbon-carbon bond here; very good, and then we have...what is on this carbon?...we have a CH3; we have an O-; and we have an O ethoxy--we have an O ethyl.4526

    OK, now if this were an aldol reaction, if we were using aldehydes and ketones, our reaction would be done right now...essentially--almost done.4547

    We deprotonate; we attacked; and then, all we would do is reprotonate; we would have our β-hydroxy ketone product.4554

    OK, however, when you are dealing with an ester, it takes a different path, because what happens when a nucleophile attacks an ester?4562

    You now have an O- on the same carbon as a leaving group; we call this a CTI (a charged tetrahedral intermediate).4571

    So, this is not a stable intermediate: this is not just sit-around-and wait-for-something-to-come-and-protonate-it.4581

    What happens to a CTI?--we use two arrows, and we collapse that CTI.4590

    The lone pair comes down; the leaving group gets kicked off.4597

    And so, our product ends up being this: we kick off ethoxide.4604

    All right, what happens with esters when they get attacked by nucleophiles?--we get addition-elimination; we get a substitution at that carbonyl.4615

    So, taking the place of this ethoxy group is going to be the enolate group.4621

    OK, and it looks like our reaction is done here, because we are at a nice, stable, neutral, happy product; and this is the ultimate product that we will be isolating.4630

    However, in the reaction conditions, this product is not stable, because our reaction conditions are basic.4640

    And what do you see about this product that makes it want to react with base?4648

    We have an active methylene here; we have an active methylene that has two EWGs, two electron withdrawing groups; it has an ester and a ketone attached to it; that makes it very acidic, and because we are in basic conditions, we are going to deprotonate this.4653

    Now, you can draw it, if you like.4680

    We could deprotonate the α carbon.4684

    OK, to make this stabilized enolate, of course, this has lots of resonance; so this is very, very stable; and in fact, this last step--this final deprotonation--is the driving force for the Claisen.4692

    If we did not have this last step--if, for some reason, you didn't have a hydrogen here, and you couldn't deprotonate--the Claisen actually wouldn't go; the retro-Claisen would happen instead.4708

    It is because this final step is essentially irreversible--this is such an acidic proton--that we get this as our final product.4722

    OK, so that is what happens after step 1; and now you can see, perhaps, why we need this step 2: we need to have an acidic workup to reprotonate this enolate that has formed as the final product, and that is how we can get the product that we kind of imagined we were going to get.4732

    We can go back to this neutral intermediate that we had for a fleeting instant, and we come back to our actual Claisen product.4753

    So, how do we describe the Claisen product?--it is described as a β-keto ester.4762

    We have added a new group; we have added a new bond to our α carbon; and the group that we have added is another carbonyl, because we had addition-elimination at that ester group.4770

    We had a substitution there; so that regenerates the carbonyl in the β position, so we get a β-keto ester when we do a Claisen condensation.4781

    Let's take a look at an example and see if we can predict the product here.4794

    OK, we have, in this case...we have a methyl ester; so that is why we are using sodium methoxide and methanol.4798

    Remember, any time we want to have a base around an ester, we want to use that matching base.4804

    And so, what do you think would happen here?--well, I think this base is going to deprotonate our α carbon.4812

    Where do we have an α proton?--right here.4818

    So, I think we are going to have this anion as our nucleophile (I'm just seeing if we can do a little work here, without going through a complete mechanism; let's see if we could come up with the product).4822

    What is that going to react with?--well, there are no other ingredients in here, so this is going to just be a self Claisen condensation; we are going to have another molecule of our ester.4834

    We can draw the carbonyl down here, if you would like, so it's a little easier for it to attack.4848

    I'll bring a second structure in here, and that is our electrophile; our key bond, then--just like the aldol: the α carbon of one to the carbonyl carbon of the next.4856

    But this is going to attack the carbon and break the carbonyl π bond; but then, what is going to happen next?4868

    That O-, in the following step, is going to re-form the carbonyl and kick out the ester.4875

    OK, so it is not an SN2 mechanism; you don't want to just bring the arrow in and show it kicking out--that implies a back side attack; but even if you want to do some of this work in your head, you can imagine it going up into the carbonyl and then coming back down out of the carbonyl and kicking off the leaving group.4881

    OK, and what it does--affect the substitution.4897

    And, after our aqueous workup, our mild workup, we will be able to get to our neutral products; so, can we draw our product, do you think?4902

    Well, let's redraw this enolate part: CH3, CH2, CH; and then our ester group; and so, what we did was: we had to take off one of those α protons so we would have this room here.4909

    This is our new carbon-carbon bond--just a little reminder--just like we have in the aldol.4925

    And then, what is that going to be attached to?--that is going to be attached to this carbonyl, which has a propyl group attached to that CH2, CH2, CH3.4931

    And does that kind of look like the product we would expect?--we get, not α, but β we have a β-keto ester; it is exactly the kind of product we get from a Claisen condensation.4947

    α carbon of one; carbonyl carbon of the other; so if you can see its relationship to the analogous aldol, that is really going to help you work through the Claisen when you need to.4958

    Let's take a look at one last example: what if we had this diester, and it's a diethyl ester, and we treat it with sodium ethoxide, and we just saw what was going to happen?4970

    What product can you get?4983

    OK, well, any time you are reacting something with base, that means we are going to lose a proton; we are going to deprotonate somewhere, and we are going to lose what kind of proton?--we are going to lose an α proton.4984

    We are going to deprotonate the α carbon.4998

    So, this is a symmetrical molecule: it doesn't matter which side we deprotonate; let's deprotonate over here.5001

    We could do that; OK, there is our α carbon--that is our nucleophile; and now, rather than having a second molecule that we can attack, because this has two functional groups, let's see if possibly we can get an intramolecular reaction to take place.5010

    Do we have any electrophile, internally, that is the appropriate distance away?5025

    So, let's check: if this is our first carbon, this is 1, 2, 3, 4, 5 atoms away from a carbonyl which is electrophilic.5031

    Would forming a 5-membered ring be a favorable ring to form?--absolutely: 5- and 6-membered rings are the ones that we are always going to be looking for.5044

    So, what we could do is: we could draw a nice 5-membered ring: 1, 2, 3, 4, 5, and number it: 1, 2, 3, 4, 5, and then just add in our groups on what is missing.5054

    On carbon 1, what do we have?--attached to that is where we have our ethyl ester, because carbon 1 was the α carbon, so it should be α to a carbonyl.5066

    And then, we had a CH2, a CH2, CH2, and then carbon 5 has what?--it has the new bond to carbon 1, of course.5078

    Here we can show it as the bond that we are imagining forming; so it has the new bond to carbon 1, and what else does carbon 5 have?5086

    It has an O ethyl, and it has an O-.5099

    That is what we would get as one of our intermediates, after we deprotonated and attacked.5107

    All right, we can imagine this attacking and breaking the π bond up; and where do we go from here?--well, this looks like a charged tetrahedral intermediate, and so we are going to have this collapse with our two arrows.5113

    Our O- comes down and kicks off our leaving group, and we end up forming a ring that still has that same pattern of a beta-keto ester that we expect for a Claisen.5127

    An intramolecular Claisen condensation has its own name: it is called a Dieckmann reaction; but is the same exact sort of reaction.5142

    Aldol reactions can happen intramolecularly, as well, to form rings, and a Dieckmann is when we have a diester cyclizing to do an aldol-type condensation.5151

    This wraps it up for part 1 of our enols and enolates discussion, and we will continue the discussion, looking at some different types of species, different types of electrophiles, that enols and enolates can react with, next.5163

    Thanks for visiting