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Lecture Comments (9)

1 answer

Last reply by: Professor Starkey
Wed Jul 8, 2015 10:22 AM

Post by Neil Choudhry on July 8, 2015

Dr. Starkey,

In a Benzocaine synthesis Via Fischer Esterification where 250 mg of PABA, 3ml of absolute ethanol, and sulfuric acid is used, what exactly is driving the reaction forward, since is a reversible reaction? is it the acid catalyst, sulfuric acid, or the 3ml ethanol? I know excess ethanol can drive the reaction forward, but 3ml isn't excess?

2 answers

Last reply by: Jon Sorensen
Thu Jul 18, 2013 5:03 PM

Post by Jon Sorensen on July 18, 2013

Dr. Starkey,

I have a test coming up soon and as I was reviewing this information I have become confused with CTI's. In some CTI's an OR group is protonate in others it is not. When does a leaving group get protonated. Thanks for your help.


1 answer

Last reply by: Professor Starkey
Mon Feb 25, 2013 12:50 AM

Post by Sha'va Newsome on February 23, 2013

I don't understand how in the mechanism where LAH is reacted with the ester, how does -OCH3 become a leaving group if it isn't protonated?

1 answer

Last reply by: Professor Starkey
Thu Oct 27, 2011 11:08 AM

Post by Billy Jay on March 31, 2011

I'm confused. At 63:20 you mention that the Ester is less reactive than the Ketone (when explaining why adding NaBH4 wouldn't react with an Ester). Then when introducing the other Special Hydride Reagents (which stop at one reduction), you mention the reason behind this is because Acid Chloride is more reactive than ketones. In regards to stability, where exactly do Ketones and Aldehydes fit in? Are they less stable than Acid Chlorides and Anhydrides, but more stable than Esters (and the other Acid Derivatives that follow)?

Carboxylic Acid Derivatives

Rank the following compounds in order of increasing acidity:
  • CH3CH2-COOH is the least acidic because it has a pKa of 4.9
  • CF3-COOH is the most acidic because of the three electron-withdrawing F's. It has a pKa of 0.2
  • ICH2-COOH is stronger than CH3CH2-COOH but only has one electron-withdrawing group so it ranks lower than CF3-COOH
Draw the major product(s) formed from this reaction:
Draw the major product(s) formed from this reaction:
Devise a synthesis for the following reaction:
Draw the major product(s) formed from this reaction:
Draw the major product(s) formed from this reaction:
Draw the major product(s) formed from 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.


Carboxylic Acid Derivatives

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
    • Carboxylic Acid Derivatives
    • Preparation of Carboxylic Acid Derivatives
    • Preparation of Carboxylic Acid Derivatives
    • Preparation of Carboxylic Acid Derivatives
    • Overall: Order of Electrophilicity and Leaving Group
    • Carboxylic Acid Derivative Interconversion
    • Preparation of Acid Halides
    • Preparation of Anhydrides
    • Preparation of Anhydrides
    • Preparation of Anhydrides
    • Preparation of Esters
    • Preparation of Esters
    • Preparations of Esters
    • Preparation of Esters
    • Preparation of Esters
    • Preparation of Esters
    • Preparation of Amides
    • Preparations of Amides
    • Preparation of Amides
    • Preparation of Amides
    • Reactions of Carboxylic Acid Derivatives with Nucleophiles
    • Reactions of Carboxylic Acid Derivatives with Nucleophiles
    • Summary of Hydride Reductions
    • Hydride Reduction of Amides
    • Reaction of Carboxylic Acid Derivatives with Organometallics
    • Reaction of Carboxylic Acid Derivatives with Organometallics
    • Special Hydride Nu: Reagents
    • Addition of Organocuprates to Acid Chlorides
    • Intro 0:00
    • Carboxylic Acid Derivatives 0:05
      • Carboxylic Acid Derivatives
      • General Structure
    • Preparation of Carboxylic Acid Derivatives 1:19
      • Which Carbonyl is the Better E+?
      • Inductive Effects
      • Resonance
    • Preparation of Carboxylic Acid Derivatives 6:52
      • Which is Better E+, Ester or Acid Chloride?
      • Inductive Effects
      • Resonance
    • Preparation of Carboxylic Acid Derivatives 10:45
      • Which is Better E+, Carboxylic Acid or Anhydride?
      • Inductive Effects & Resonance
    • Overall: Order of Electrophilicity and Leaving Group 14:49
      • Order of Electrophilicity and Leaving Group
      • Example: Acid Chloride
      • Example: Carboxylate
    • Carboxylic Acid Derivative Interconversion 20:53
      • Carboxylic Acid Derivative Interconversion
    • Preparation of Acid Halides 24:31
      • Preparation of Acid Halides
    • Preparation of Anhydrides 25:45
      • A) Dehydration of Acids (For Symmetrical Anhydride)
    • Preparation of Anhydrides 27:29
      • Example: Dehydration of Acids
    • Preparation of Anhydrides 29:16
      • B) From an Acid Chloride (To Make Mixed Anhydride)
      • Mechanism
    • Preparation of Esters 31:53
      • A) From Acid Chloride or Anhydride
    • Preparation of Esters 33:48
      • B) From Carboxylic Acids (Fischer Esterification)
      • Mechanism
    • Preparations of Esters 41:38
      • Example: Predict the Product
    • Preparation of Esters 43:17
      • C) Transesterification
      • Mechanism
    • Preparation of Esters 47:58
      • D) SN2 with Carboxylate
      • Mechanism: Diazomethane
    • Preparation of Esters 51:01
      • Example: Transform
    • Preparation of Amides 52:27
      • A) From an Acid Cl or Anhydride
    • Preparations of Amides 54:47
      • B) Partial Hydrolysis of Nitriles
    • Preparation of Amides 56:11
      • Preparation of Amides: Find Alternate Path
    • Preparation of Amides 59:04
      • C) Can't be Easily Prepared from RCO₂H Directly
    • Reactions of Carboxylic Acid Derivatives with Nucleophiles 1:01:41
      • A) Hydride Nu: Review
      • A) Hydride Nu: Sodium Borohydride + Ester
    • Reactions of Carboxylic Acid Derivatives with Nucleophiles 1:03:57
      • Lithium Aluminum Hydride (LAH)
      • Mechanism
    • Summary of Hydride Reductions 1:07:09
      • Summary of Hydride Reductions 1
      • Summary of Hydride Reductions 2
    • Hydride Reduction of Amides 1:08:12
      • Hydride Reduction of Amides Mechanism
    • Reaction of Carboxylic Acid Derivatives with Organometallics 1:12:04
      • Review 1
      • Review 2
    • Reaction of Carboxylic Acid Derivatives with Organometallics 1:14:22
      • Example: Lactone
    • Special Hydride Nu: Reagents 1:16:34
      • Diisobutylaluminum Hydride
      • Example
      • Other Special Hydride
    • Addition of Organocuprates to Acid Chlorides 1:19:07
      • Addition of Organocuprates to Acid Chlorides

    Transcription: Carboxylic Acid Derivatives

    Hi; welcome back to

    Today, we are going to talk about carboxylic acid derivatives.0002

    Carboxylic acid derivatives take a wide variety of shapes and characteristics.0007

    Shown here is a list of such derivatives, and how they are related to carboxylic acids (and why they are called derivatives of carboxylic acids) is: instead of having an OH group attached to the carbonyl (like a carboxylic acid would), we have some other potential leaving groups.0013

    We could have a halide (usually chloride--this would be called the acid chloride); we could have an anhydride; so in this case, we have a halogen attached to the carbonyl.0029

    In the case of the anhydride, we have a carboxylic group attached to the carbonyl; the ester has an O-R group; the amide has a nitrogen group; and a nitrile looks a little different from all the others, but it too is a carboxylic acid derivative, and it has 3 bonds to a nitrogen.0042

    What most of these have in common, as I said, is having a leaving group attached here; and it is some group that has a lone pair of electrons--that is how I can identify something as a carboxylic acid derivative.0061

    We describe it as a heteroatom--something that is not carbon or hydrogen (so a halogen, oxygen, nitrogen, so on).0071

    If we want to learn how to prepare these carboxylic acid derivatives, we really need to have a good understanding of the differences between the various leaving groups.0080

    And so, one question we can ask is, "How do they compare, in terms of electrophilicity?"0087

    In exploring the features of a leaving group, let's ask which carbonyl is the better electrophile; let's compare the ester versus the amide--which of these is a better electrophile?0093

    To be a better electrophile means that you have more of a partial positive, right?--you are less electron-rich; that would make you a good electrophile.0103

    We could look at a few things: inductive effects and resonance effects.0114

    So, for the inductive effects, remember that oxygen is more electronegative than nitrogen.0117

    OK, so that means this oxygen is pulling electron density away from that carbonyl; what effect does that have on the carbonyl?0123

    Well, actually, we could state that, because it's more electronegative, that means oxygen is better at withdrawing electron density inductively (meaning through the bond).0130

    Oxygen is doing a better job pulling this electron density away than nitrogen is.0150

    OK, so as a result, the ester is more partial positive than the amide.0155

    So, just based on these inductive effects, we expect that the ester is the better electrophile.0163

    However, that is not the only influence that these leaving groups are having on the carbonyl; besides the inductive effects (the through-bond effects), we also have some resonance effects.0174

    That is because each of these leaving groups, remember, has a lone pair--at least one lone pair--of electrons.0185

    Any time you have a lone pair, we can have resonance; so the other thing that this oxygen is doing is: it is adding its electron density into the carbonyl.0190

    The oxygen can do that, and the nitrogen can do that.0199

    Let's take a look at those resonance forms, and see how effective that resonance donation is.0204

    OK, there is the resonance form for the ester, and here is the resonance form for the amide; we now have a double bond here and a positive charge on the nitrogen, and our carbonyl oxygen is now a single bond with an O-.0217

    OK, and what we want to ask is, "Which of these atoms (the nitrogen or the oxygen)--which of them is better at doing this resonance--at donating this electron density that way?"0231

    "And which of these resonance forms is a bigger contributor to the overall picture of the ester or the amide?"0240

    When I look at these two, I see that I have an O+ versus an N+; now, are each of those elements equally good at handling positive charge?0248

    They are not, because we know oxygen is more electronegative than nitrogen; so the nitrogen better handles a positive charge: that means this resonance form is a larger contributor to the overall picture of the amide.0258

    What we can say is: let's restate the fact that nitrogen is less electronegative than oxygen, so N+ is better than O+.0270

    In other words, nitrogen better handles the positive charge, because it is more electropositive than oxygen.0288

    OK, that means that nitrogen is better at donating its electron density (because that is what is happening in this resonance--the lone pair from nitrogen is going into the carbonyl--that is a donation; that is adding electron density).0295

    That means the amide carbonyl--what is the effect on the amide carbonyl?--it is more electron rich.0316

    Because the nitrogen is very good at sharing its electron density, the result is: this carbonyl is more electron rich than the ester.0326

    If we come over and we look at the ester situation, we could say that oxygen is more electronegative, and we could say it doesn't want to share its electron density.0334

    It doesn't want to share its lone pair: it's more electronegative--it is more likely to hold on to those, and it is less likely to share them, because that would put a positive charge on the oxygen.0354

    Therefore, the ester carbonyl is less electron rich: it has less of that electron donation, and therefore it is not as electron rich as the amide.0362

    Looking at these two arguments, what would you conclude about the electrophilicity--which is the better electrophile, the ester or the amide?0375

    The ester is the better electrophile: both of these point to the same conclusion--that the ester is the better electrophile.0383

    It turns out that that was the same conclusion we drew by the inductive effects; in this case, both the inductive effects and their resonance effects point to the same conclusion.0392

    Now, if we ever find that they point in different directions, which one do you think is going to win?0400

    I think the resonance effects are going to win, because those are usually more important in guiding the reactivity.0405

    But let's see: let's do another comparison.0410

    Let's compare, now, an ester versus an acid chloride--how about an oxygen, compared to a chlorine?0413

    What effect is it having on the reactivity of the carbonyl?0419

    Well, again, we could think about the inductive effects: both the oxygen and the chlorine pull electron density away from the carbon; they are both more electronegative than carbon (oxygen, even more so, because remember, oxygen is the second-most electronegative atom in the periodic table, so that is definitely a strong dipole).0423

    But let's jump right into the resonance effects, because that is where the most significant difference is going to lie.0440

    Once again, it is always going to be the same picture: we can take one of the lone pairs from our leaving group and share it with the carbonyl to draw a contributing resonance form.0446

    It will always put a positive charge on the leaving group and a negative charge on the carbonyl oxygen.0460

    And chlorine can do that, too--take one of its lone pairs and add it into the carbonyl.0465

    And then, just like we did before, let's look at these two resonance forms and see if we can predict which of them is having a bigger influence, having a bigger impact, on the overall structure--which is the bigger contributor?0474

    Now, we are looking at an O+ versus a Cl+; and this is a little more...looking at apples and oranges, because we are not looking at a direct periodic trend here.0489

    The oxygen and chlorine are not in the same row; they are not in the same column; OK.0498

    But there is something interesting about this chlorine: how often have you seen a chlorine involved in a double bond with carbon?0502

    That is a very rare structural element, and the reason for that is because the chlorine is so much bigger than the carbon that, when it tries to share that π bond with its large p orbital, it can't do so effectively.0510

    So, what we could say is that chlorine is too big for a good π bond.0526

    We could describe it as having poor orbital overlap; we are trying to overlap a 3p orbital with a 2p orbital, and there is not good exchange here.0536

    This π bond is a very weak bond; that means it is not a very significant contributor to the overall resonance hybrid.0552

    We could also describe that as saying that chlorine doesn't want to share, or is not capable of sharing, its electrons--its lone pairs.0560

    OK, and so oxygen now--compared to chlorine, oxygen is actually better at donating its electron density, because it's matched better with the carbon size.0574

    The oxygen is better at donating its electron density.0584

    What does that do to the ester carbonyl, compared to the acid chloride carbonyl?0594

    The ester carbonyl is now the one that is more electron rich, because it has more of this electron donation.0598

    The ester carbonyl is more electron rich.0604

    And so, who is the better electrophile in this case--the ester versus the acid chloride?0613

    Is being more electron rich a good thing for an electrophile?--just the opposite: that means the ester is a poor electrophile compared to the acid chloride, so both of these combine to tell us that the acid chloride is the better electrophile.0621

    Acid chloride is a better electrophile than an ester.0640

    Let's take a look at one more comparison--how about a carboxylic acid compared to an anhydride?0643

    An anhydride is kind of an interesting structure, because we have these two carbonyls; so now we are comparing an oxygen leaving group, versus an oxygen leaving group.0651

    We really have no difference, inductively, in terms of electronegativity.0659

    Let's, again, jump to the resonance, to see what those resonance forms look like: the carboxylic acid resonance form looks just like the ester resonance form (except we have a hydrogen here, instead of an R group).0665

    And then, for the anhydride, that oxygen can do the same thing.0681

    OK, what is the difference going to be, then?0690

    We have an O+ versus an O+; what is the difference?0694

    Well, are there any additional resonance forms that we can have?0698

    Sure; because we have two carbonyls here, we know that is going to be significant somehow; instead of having this lone pair come over to the left, to share with the carbonyl on the left, it can also move over to the right and share with the carbonyl on the right.0703

    So, let's look at that resonance form that we will get back to by shifting this π bond over on the other side.0715

    This carbonyl is intact, but now this carbonyl has the O-.0727

    And so, that is going to be the difference between the two: we have some extra resonance with the anhydride.0736

    OK, the big question, though, is, "How does that affect electrophilicity?"0742

    Remember, to be a better electrophile, you want to be more partial positive; we have a carbonyl that is a partial positive.0745

    Let's put it this way: this carboxylic acid has an oxygen that donates its electron density full-time to the carbonyl; there is nothing else for this oxygen to be doing, in terms of resonance, than to share with this carbonyl.0756

    We will say the acid oxygen donates to only one carbonyl: if we kind of look at it this way, perhaps you will see the difference.0771

    What is this oxygen doing?--it is spending its time delocalizing its lone pair to both carbonyls.0783

    The anhydride oxygen lone pair is shared between two carbonyls, and on to two oxygens.0791

    OK, so what does that do to being more electron-rich/more electron-deficient?--it looks to me like this carboxylic acid carbonyl is going to be more electron rich, because this has the full-time donation, where this carbonyl is not going to be as electron rich, because sometimes, some fraction of this lone pair is going to be shared with the carbonyl (but only half, because the other half is going to be shared with the other carbonyl).0809

    This does not have effectively as much electron donation.0835

    So, the ester carbonyl...I'm sorry, we're looking at carboxylic acid here--the carboxylic acid carbonyl is more electron rich.0838

    Who is the better electrophile?--the anhydride is the better electrophile.0852

    Don't think that it has something to do...because there are twice as many carbonyls; remember, just one carbonyl we are looking at is the electrophile (the other carbonyl is part of the leaving group).0863

    We are just comparing this one carbonyl to the carbonyl in the carboxylic acid; and what we are seeing is: we are not getting as much electron donation, because the presence of the other carbonyl draws some of that electron density away from the first carbonyl.0874

    So overall, we could continue doing these comparisons in pairs; but overall, we are going to be able to develop an entire scheme, an order of electrophilicity.0891

    It turns out that this order also parallels the order of leaving group ability.0902

    At the far end here, we have the acid chloride; let's take a look at the leaving group for each of these acid derivatives.0907

    The leaving group for an acid chloride would be Cl-, would be chloride.0915

    That is a pretty good leaving group: we have seen that before, in all sorts of substitution and elimination reactions--a halide would be a very good leaving group.0922

    The leaving group here is carboxylate (a carboxylate anion); the leaving group here--we are putting both of these kind of together--they are not hugely different in terms of their reactivity, so either an alkoxide or hydroxide as your leaving group...0929

    Both are similar in their reactivity--very similar.0951

    Here, we would have an NH2- leaving group; and what if we had a carboxylate itself as our electrophile, and we did a substitution--in other words, a nucleophile added into the carbonyl and kicked out this leaving group--what would the leaving group look like?0955

    I think it would be an O2-; that doesn't look like a very good leaving group.0971

    OK, so hopefully, you can see, as you compare these leaving groups, that we are getting poorer and poorer in terms of leaving group ability.0977

    So, let's take a look at the acid chloride and see some conclusions we can draw here.0986

    OK, the acid chloride at the far end here is the extreme: this has the best leaving group.0989

    What does it take to be a good leaving group--what defines a good leaving group?--it is something that is very, very a weak base; and so, those are some things we will look for.1003

    It's stable; it's a weak base; that is what determines leaving group ability; and like I said, a halide is a leaving group in all sorts of situations.1015

    We recognize it as a good leaving group.1024

    OK, it also has least amount of resonance donation by that leaving group: so remember, the chlorine was so large that it was not very effective at sharing its electron density with the carbonyl, so it did not have very much electron resonance donation.1027

    That makes it the most reactive; it doesn't have a lot of resonance, so that makes it less stable--it makes it more reactive.1047

    It also makes it the most electron-deficient carbonyl.1055

    All right, all carbonyls are good electrophiles; when you put a chlorine on there, that chlorine pulls electron density away inductively and doesn't add much back in by resonance.1063

    So, it's a very, very reactive, very electron-deficient carbonyl; and so, this is by far the best electrophile.1073

    This is great electrophile; an acid chloride is super reactive--we are going to see it react with any nucleophile you want to put in there; it is going to be happy to react with it.1081

    OK, that is slightly better than the anhydride: the anhydride is also a very good electrophile--it has a good leaving group; what makes this a good leaving group?1091

    Well, it has resonance stabilization, right?--that negative charge is delocalized over both oxygens; so that is a pretty good leaving group, too.1100

    OK, but that is a better leaving group than we would have on the ester or the carboxylic acid, because now alkoxide, hydroxide...these are strong bases; these are not particularly good leaving groups.1108

    And because this has more electron donation by the oxygen than the anhydride would, the carbonyl is also more electron rich and therefore less electrophilic.1119

    Remember, the nitrogen is really good at donating its electron density; so this is very electron rich.1132 N- is much less stable than an O-, so our leaving group is continuing to get worse and worse.1139

    And then, we get to this final end, where now, our species is negatively charged: that is an extremely poor electrophile, if you are negatively charged.1145

    Remember, we want to be electron deficient to be a good electrophile.1154

    Let's come all the way down to this last example, the carboxylate, and let's describe some things we see on this extreme.1157

    OK, this has the worst (meaning least stable) leaving group: least stable, strongest base--clearly, O2- would not be very happy leaving on its own.1168

    It also has the most resonance, and O- would be very effective in sharing its electron density by resonance.1184

    In fact, the second resonance form would be just as big a contributor as the first one, because it would be equivalent; so that is the best resonance you can have.1192

    That makes it the least reactive carbonyl--the least reactive carbonyl in terms of electrophilicity, because it is the richest (it is the most electron rich) carbonyl, and that makes it the worst electrophile.1200

    It is the worst electrophile; so really, you should be able to compare any of these carboxylic acid derivatives, like we just did for a few pairs; you should be able to compare them, and really, you should know this order.1223

    You should know this order of reactivity; you should understand it--this is not a random list that you should memorize; it is something that should make sense to you, because what we are going to be doing is: we are going to be doing these comparisons again and again.1237

    Because, the way you make a carboxylic acid derivative, typically, is: you start with another carboxylic acid derivative, and you just swap one group for another.1254

    OK, we call it a carboxylic acid derivative interconversion, and that is the way we are going to be synthesizing and preparing these various carboxylic acid derivatives.1262

    For example, if I were to take an acid chloride and react it with methoxide, I could look at the product where the chlorine and the methoxide have exchanged or done a substitution, where now I have an ester and the chloride leaving group.1270

    OK, now this mechanism is going to start with the acid chloride acting as my electrophile and the methoxide acting as my nucleophile; and it's going to add into the carbonyl; that is going to give me the following intermediate.1287

    It looks like a CTI; it's a charged tetrahedral intermediate, and when this collapses, this O- kicks back down; we have two possible leaving groups to kick out.1320

    Every time a CTI collapses, it's going to collapse to lose the better leaving group.1335

    It is critical that we know which is the better leaving group: who is the better leaving group here, methoxide or chloride?1343

    Chloride is the better leaving group, so this is going to come down and kick off the chloride and go back here.1350

    OK, and if we think about this reverse reaction happening, where the ester is my electrophile and the chloride is my nucleophile, this chloride can attack back into the carbonyl to go back to the CTI, but is it ever going to go back in this direction, once the chloride adds in?1358

    Is the O- going to kick out the methoxide?--no way.1374

    This reverse reaction is...this is a one-way street here; we are never going to be kicking out the poorer leaving group when we do our CTI; so that is how we can decide the equilibrium.1378

    The direction of the equilibrium is going to lie, as usual, in the direction that our leaving group is leaving.1391

    We can look at the better leaving group: this is the better leaving group.1398

    OK, the methoxide is the better nucleophile, and of these two electrophiles, comparing the acid to the acid chloride, which one is the better electrophile?1406

    Remember, the acid chloride was the best electrophile; so it is better than all the others, so it is the better electrophile and the ester is the poorer electrophile.1416

    All these things we can look at in a variety of ways: looking at the more reactive carbonyl, looking at the better leaving group...whichever way we look at it, we are going to get the same answer--we should get the same answer.1427

    In this case, it is the forward reaction that is going to be favored, where we displace the better leaving group with the better nucleophile.1436

    We go from a more reactive carbonyl to a less reactive carbonyl.1444

    OK, which means, since the acid chloride was the most reactive derivative with the best leaving group--that means we can start with the acid chloride and go anywhere.1448

    We can replace that with any other group that we want.1457

    So, the question, then, that I have for you is, "So, how do we make the acid chloride--how is it possible that we ever put a chlorine in there if it's such a good leaving group?"1459

    That is an excellent question; let's see how we do that.1468

    We'll call this the preparation of acid halides, and the chloride in particular, and what we are going to do is: we are going to take the...we are going to start with the carboxylic acid, and we are going to react with either thionyl chloride or phosphorus trichloride.1472

    These reagents should look familiar; these are the same reagents we used to take an alcohol and convert it to an alkyl chloride.1489

    And so, what these do is: they first react with the hydroxyl group here (so just a quick hint at what is going on--I'm not going to go through the complete mechanism).1498

    So, for example, if you had thionyl chloride, this oxygen can add in to the sulfinyl and kick out the chloride.1511

    What you do is: you make a great leaving group--you make a really great leaving group, one that is so great that even a chloride can displace it and kick it out.1520

    You can, in fact, make the acid chloride this way, using either SOCl2 or PCl3; if you ever need to make an acid chloride, you have to start with a carboxylic acid starting material.1530

    Now, how would I make an anhydride?--there are a couple of different ways we can make an anhydride.1546

    One possibility is by the dehydration of the carboxylic acid, and this would be a way to make symmetrical anhydride (so in other words, an anhydride where the two carboxyl groups on each side are identical).1550

    One way to get there is to start with the carboxylic acid, the parent carboxylic acid; so here, we have 1, 2, 3 carbon on each side of the anhydride, so we start with the 1, 2, 3 carbon propanoic acid.1565

    And, if we heat this mixture, we can lose water, so that is why we call it an anhydride: it is a carboxylic acid having had water removed.1578

    OK, and we need...this must be removed in order for this reaction to occur, because if you don't take this away, then what will happen is: the anhydride (very reactive anhydride) will react with water; we will get an addition-elimination; it will get a substitution to go right back.1590

    In order to push the equilibrium forward and move it forward, we need to remove this product to get the reaction going.1610

    What is happening is: we have two equivalents of our acid; so one is acting as the nucleophile, and one is acting as the electrophile.1618

    Just kind of the key...I'm not going to do a complete mechanism for this, but here is the bond that is forming: the OH group is adding in, and then eventually, you are going to be eliminating this leaving group and replacing that leaving group.1631

    That is one possible way to synthesize an anhydride.1645

    Let's see an example of that: this is succinic acid--it is an example of a dicarboxylic acid, and if you heat this up (especially in the presence of a drying agent--this is acetic anhydride), that is kind of an inexpensive anhydride that you can use that loves to react with water.1650

    So, you can put that in there, and it will act as a drying agent when it sees water or reacts with water to make acetic acid.1672

    It is going to make two equivalents of acetic acid; so this is just an example of a drying agent that can help get rid of the water and push the equilibrium forward.1680

    If we go to predict the product here, again, one of the carbonyls is going to be the electrophile; the other carbonyl is the leaving group part; that is going to be the nucleophile.1692

    There is the bond that we are forming; and we are going to do an addition-elimination.1705

    We are losing this OH, and we are going to lose this H as well; so that is how you can see that we are losing a molecule of water.1710

    What size ring would this form; well, it's a diacide, so when these two ends react together, it is going to form a ring; what size ring would that form?1719

    1, 2, 3, 4, 5: it is going to form a five-membered ring with an oxygen in there; it looks something like this.1726

    And so, this molecule is known as succinic anhydride, because it is the anhydride made from succinic acid; that is the way we name anhydride: we just take the parent carboxylic acid and replace that word "acid" with "anhydride."1738

    Now, if we wanted to make a mixed anhydride, a mixed anhydride has two different groups on either side of the oxygen.1758

    We couldn't do that by a dehydration, because we would have to throw two different acids in there, and we would have no way of knowing who is going to react with whom.1765

    OK, so instead, what we are going to do is: we are going to start with an acid chloride, and that is going to be our electrophile; we are going to add in a carboxylate nucleophile; and we have a sodium carboxylate, so any time you see sodium, that means you have sodium + and O-.1775

    So, we can drop out that sodium spectator ion, so I can see my nucleophile.1795

    I can see my acid-chloride electrophile: remember, really great electrophile; so let's see what the mechanism is going to look like to do this substitution reaction and get our product.1799

    We have our carboxylate; it is going to attack the carbonyl and break the π bond, so we'll have an O- up here; and what was the nucleophile we added?--it was an oxygen with a carbonyl and then a CH2 and a CH3.1812

    There is our addition; remember, overall, our reactions are going to be an addition and then an elimination.1833

    This elimination is the collapse of the CTI; this is the reaction that carboxylic acid derivatives undergo: addition of a nucleophile and elimination of a leaving group.1843

    So now, when I look at my CTI, my charged tetrahedral intermediate, and I decide which leaving group is going to be kicked out, which one wins?--because we have two possibilities.1857

    Remember, a leaving group is anything with a lone pair; so we have two possibilities.1866

    And, while this is a good leaving group, the chloride is better; this is our better leaving group.1870

    It kicks down and kicks off, and we go to our anhydride.1877

    It is possible that this reverse reaction can happen, but the favored reaction is right here, kicking off our better leaving group; so this is another reaction where we know we are going to push this reaction in the forward direction, simply because we are displacing a better leaving group than the one we are putting in there.1884

    That is why it is so important to recall that overall order of reactivity, and leaving group ability, so that you know which substitutions are allowed and which ones are not.1902

    OK, how do we get an ester?--an ester has an O-R group attached to a carbonyl; OK, there are many ways to make esters.1915

    One possibility is to start with a better leaving group attached to a carbonyl, and then introduce the O-R group.1924

    What better leaving groups are there?--well, we could look at our two most reactive acid derivatives, the acid chloride and the anhydride.1933

    The acid chloride--great leaving group; the anhydride--great leaving group; and so, in fact, in preparation of esters and amides and others, we will find that very often, anhydrides and acid chlorides are used interchangeably--they are both very, very good leaving groups.1943

    Let's see what will happen here: there is our leaving group; here is our nucleophile--the alcohol--now we know we are going to have to lose this proton at some point, so because we are losing an H and a Cl, that is why we need a base in here to do that deprotonation.1959

    We are going to get that salt; pyridine is a nice base that is used often in a case like this.1981

    And so, what does our product look like?--we will have our phenol with an oxygen, and now, attached to that oxygen will be the carbonyl.1986

    We did addition-elimination; we did an acyl substitution, addition-elimination.1996

    Whether I use acetyl chloride or acetic anhydride, I get the exact same product, don't I?--I still have the carbonyl with a methyl group; I added an acyl group in both of these cases, an acetyl group.2003

    OK, and for that reason this is described as an acylation reaction; you can say that you just acylated this alcohol with an acid chloride or with an anhydride.2015

    So, this is one very good way to prepare an ester.2024

    OK, another option is to start with a carboxylic acid and react with an alcohol; this is known as the Fischer esterification.2030

    The substitution reaction occurs in the presence of an acid, and what we are doing is: we are swapping the OH group for an O-R group.2038

    We use ethanol to add in, and ultimately, we kick out water.2050

    Now, here is a good question for this reaction, though: Which direction of the equilibrium do you think is going to be favored?2055

    Who is the better leaving group in this reaction?--because remember, the reverse reaction can happen, too--the reverse reaction is where water adds into the carbonyl and kicks out the O-R group.2064

    We saw that in the hydrolysis of ester reaction; OK, so we know this reverse reaction is possible; so how do we know which one is favored?2075

    Well, in fact, the leaving group--the ROH and the H2O have about the same leaving group ability.2085

    So, there is nothing naturally driving this equilibrium forward or backwards, based on the stability of the products; the stability of the starting materials and the stability of the products are all great.2096

    We are going to have to do something to push it: we need to drive off--we need to remove--a product, to push the equilibrium forward.2107

    OK, sometimes we can do something to react with the water, or move the water; one thing we could do, maybe, is distill--we can distill the ester off.2122

    This very often has a lower boiling point than our carboxylic acid and our alcohol, because these have OH groups; these are very, very polar starting materials; this is going to be less polar as the ester.2133

    So, sometimes the ester can be distilled off as it is formed.2147

    OK, but we need something--some strategy--to make the Fischer esterification work; or maybe you could have a huge excess, right?--if you have a huge excess of the alcohol, then that would be something that also can push it forward.2151

    OK, it needs to have some kind of acid catalyst; so sometimes, you just write HA; but it could be something like sulfuric acid, tosic acid, HCl...OK, any sort of acid in there is good.2164

    We wouldn't want to use H3O+ as our acid, though, because water is one of the products we are trying to get rid of, so you wouldn't want to be adding water into your reaction conditions--reaction mixture.2174

    OK, so what is the mechanism here?--again, it is addition-elimination for an acyl substitution; it turns out that it is the exact reverse of the hydrolysis mechanism we have already studied.2185

    Hydrolysis was when we took an ester in the presence of an acid, and reacted it with water, so the same number of steps that took us to the hydrolysis carboxylic acid and alcohol--the exact same steps, the exact same intermediates, are what we are going to be seeing today, and going in the other direction, doing the Fischer esterification.2195

    OK, so let's do the mechanism: where do you think we should start?2215

    I think we have acid-catalyzed conditions, acidic conditions, so I'm guessing the first step should be to protonate something.2218

    Now, there are a whole bunch of oxygens we can protonate: we can protonate here; we can protonate here; we can protonate here; OK, and those will all happen reversibly.2226

    We will protonate everywhere, but where is the place that will protonate that will actually move us in the forward direction?2234

    It is the carbonyl oxygen that, when we protonate there, we are going to get an intermediate that has some enhanced reactivity and therefore will start moving us toward the substitution product, OK?2240

    Because, when you protonate a carbonyl, you make a great electrophile; you turn that partial positive carbon that was already there into an even more partial positive carbon (even more electron-deficient), so it's a great electrophile.2257

    We will start by protonating the carbonyl; now, we have to find a nucleophile.2269

    Of course, our nucleophile is our alcohol--maybe even our solvent in this case.2274

    And so, it is going to attack the carbon and break the π bond, as usual.2283

    Now, this oxygen still has the ethyl group; it still has the hydrogen; so we have to put both of those on there.2293

    What else does it have?--it still has one lone pair, and it looks like a positive charge--just an oxygen with three bonds and a lone pair: 1, 2, 3, 4, 5; oxygen has 5; it wants 6, so we have an O+.2300

    OK, so we are getting there; we put in the ethoxy group; let's deprotonate; let's get rid of that positive charge.2314

    We can use A-; we started with HA, so we could just use A- to come back and grab that proton, because it is catalytic in acid, so every protonation step is going to be coupled with a deprotonation step.2321

    We are starting to move in the right direction: we know we have to add in this ethoxy group, and we just did that.2340

    How did we do it?--protonate, attack, deprotonate.2346

    Again and again, we see that pattern for acid-catalyzed mechanisms: protonate, attack, deprotonate.2351

    OK, so now, where do we go--how do we move forward--what else do we need to accomplish?2356

    Remember, it's a substitution reaction; so we added in the ethoxy--what do we need to take out?2361

    We need to remove this OH group and substitute that OH group; well, now we see either of these can be used--there is no distinguishing them anymore.2366

    But how are we going to make it eventually leave?2375

    We are in acidic conditions, so we need to protonate it to make it a good leaving group.2379

    Let's pick the one on the side here and protonate to make it a good leaving group; and how is it going to leave?2384

    How is that leaving group going to leave?2400

    It looks like I have a CTI, so with two arrows, I can collapse that CTI; that would be a great way to lose the leaving group--not just the leaving group leaving on its own, but having this oxygen lone pair push it out.2403

    OK, what is great about doing my mechanism that way is: look at where it brings us when we do that.2420

    When you lose a leaving group with assistance from that lone pair, it brings us right to an intermediate that looks this close to our product, right?--all we need is to get rid of that last proton.2429

    It is a very good habit to get into--using those two arrows.2439

    Notice, we just kicked off our water here, if you like to keep track of your leaving group; you can see that you just generated the other product that we are forming in this reaction.2443

    And then, for our last product, we can just bring back the A- and deprotonate, and we are done.2451

    We just made an ester; we did an acyl substitution reaction, addition-elimination; remember, it is acid-catalyzed, so what kind of charges do we expect for this mechanism?2460

    All our charges are neutral or they are positively charged; you can't use hydroxide as a leaving group; you can't use the methoxide as a nucleophile; and so on.2473

    OK, let's make sure we are using just strong acids to protonate and have no minus charges as we go through.2483

    It's about that Fischer esterification, a combination of a carboxylic acid and an alcohol.2492

    Let's see if we can predict a product: if I were to take this molecule that has a carboxylic acid and an alcohol in the same structure and treat it with some acid (remember, tosic acid is just like sulfuric acid--it's just organic soluble), and we throw that in with some heat, what reaction can happen here?2500

    Well, let's look for a nucleophile; let's look for an electrophile; I see my carbonyl--that is going to be an electrophile; this is going to be my leaving group that gets displaced, and where is my nucleophile?--this oxygen is going to be my nucleophile.2520

    So, the key step, the key bond that I am forming...without doing a complete mechanism, I can see the bond that I am forming, and we would be creating a ring in this situation.2538

    Now, what size ring would we form?--we want to make sure it's allowable before we do a cyclization.2548

    Let's count: oxygen is 1; 2; 3; 4; 5; 6 atoms--that looks like a perfect size ring to me.2554

    We have a 6-atom ring; one of those atoms is oxygen; so there are 1, 2, 3, 4, 5, 6; 6 was the carbonyl, and it was the carbonyl with the OH, but remember, one oxygen adds in and kicks out the other oxygen, and we end up doing a substitution.2564

    We lose this proton during the mechanism; we lose this OH during the mechanism; so water is our second product here--it must be lost in this Fischer esterification example.2583

    Now, another way to prepare esters is a reaction called a transesterification mechanism.2597

    Transesterification is when we go from one ester to another ester.2603

    Here I have an ethyl ester; if I took an ethyl ester and dissolved it in methanol, then I would get the methyl ester out and create some ethanol as a leaving group.2609

    Now, this might be a strategy you want to employ, because one ester is commercially available, but you want to have a different ester, so you want to replace that O-R group.2618

    But you know, where this really comes into play is, very often, as an unwanted side reaction.2631

    Let's say I was using this ethyl ester, and I just happened to use methanol as my solvent in my reaction without thinking of it; what is going to happen besides whatever reaction I wanted to have happen is some of this transesterification.2638

    So, any time you have a mismatch between the O-R group of your ester and the O-R group of your alcohol, we can get this swapping--we can get this substitution to take place.2652

    So, if I didn't want this transesterification to take place, what would be a better solvent for me to use with this ester starting material?2665

    I would use ethanol as my solvent, and then this transesterification reaction wouldn't be a problem: it would still happen (ethanol would still add in), but it would be kicking out ethoxide, and it would be an invisible reaction.2675

    OK, so we can't eliminate the substitution reaction; we can just put it in a nucleophile that matches the leaving group that is already there.2687

    OK, so once again, comparing our leaving groups: methanol and ethanol--you can't get more similar in terms of leaving group ability, so there is nothing that is naturally driving this forward or back.2694

    We need to use Le Chatelier's Principle, and either having a huge excess of one (like having it as a solvent), or we drive off the other in some way...we have to do something to push the equilibrium in one direction or the other.2704

    OK, the mechanism is exactly analogous to the Fischer esterification; we are going to protonate as our first step (I'll just abbreviate that as ethoxy).2717

    Let's go through this one more time, because it's great practice.2733

    We'll protonate as our first step; that makes super-electrophile here; we look around for a nucleophile; now we can have our nucleophile attack.2737

    So remember, acid-catalyzed: we do protonate, and attack, and then deprotonate; so here is our attack, and then we deprotonate.2748

    Now, you might think, "Well, wait: didn't we just have a CTI? Couldn't that CTI collapse?"2768

    Couldn't this CTI collapse?--it could, but who is the best leaving group in this CTI?--the leaving group is down here; it's the methanol.2775

    So, if this CTI collapsed, where would it bring us?--right back to where we started.2782

    OK, so yes, this is reversible; this is reversible; we can add in the methanol and kick it back out; OK, but again, if we have something pushing the equilibrium forward by Le Chatelier's Principle, we are going to keep moving forward instead.2787

    And now, we would protonate the ethoxy group, so that we could get that as our leaving group.2800

    Always keeping in mind where we are trying to go is how we can be guided in a mechanism.2810

    I didn't want this CTI, because it had methanol as my good leaving group; I want ethanol as my good leaving group.2819

    So, what I did was: I deprotonated down here, and I reprotonated up here; now, I am ready to do a collapse and kick out the leaving group I want to kick out.2825

    Here is my CTI, and here is my two-arrow collapse.2835

    I just lost a molecule of ethanol, and I recreated my carbonyl, and all I need to do is deprotonate, and I am at my final ester.2840

    This is called a transesterification: if you look closely at the mechanism on the previous slide, you will see that this is completely analogous to a Fischer esterification; but instead of having a carboxylic acid starting material, an OH here, so that you lose water, you have an O-R here, so that you lose an alcohol.2856

    OK, finally, another way that you can make an ester is to, instead of doing a substitution at the carbonyl (where you install an O-R group) start with an oxygen (like a carboxylate), and then you add an alkyl group to the oxygen.2880

    It is just doing a disconnection: the ester syntheses we have seen up until now have been forming this new bond by doing a substitution reaction at the carbonyl, and now we are taking a look at the possibility...well, another way you can make an ester is if we form this bond; that could also give an ester product.2895

    OK, so we could do that by an SN2 reaction; if this was our nucleophile and this was our electrophile, then we could have something like an SN2.2914

    If we had the carboxylate nucleophile, and we reacted it with an unhindered electrophile (like this primary alkyl halide electrophile), it looks like a pretty good SN2 reaction.2923

    Carboxylate is a good nucleophile; we expect it to do an SN2; and I just made an ester.2936

    OK, so obviously, this could only happen if you have an unhindered leaving group; so this is only good for building esters that have primary...or maybe methyl groups here...or possibly secondary, because this is a pretty good nucleophile.2946

    E2 doesn't compete too well, but it does have to allow the SN2 mechanism to take place here.2963

    Another interesting reagent that you can use to make an ester this way is called diazomethane; if you take a carboxylic acid and react with diazomethane...I have shown the structure here of diazomethane; the first thing that happens is: our carboxylic acid donates a proton to the diazomethane, which acts as a base.2969

    Diazomethane actually has a C-, so we can do a proton transfer here, and that does two things for us: it gives us a nucleophile now (now we just made the carboxylate, which is a good nucleophile); and it gives us a neutral carbon, bearing a great leaving group.2988

    What is the leaving group here if this were to get displaced?3013

    It would be nitrogen gas: so the products we get in this--just a back side attack--SN2 mechanism here: carboxylate now attacks the methyl that was just formed and kicks off nitrogen gas.3017

    We just formed a methyl ester.3035

    Diazomethane is a unique reagent that can only make a methyl ester, because it is a one-carbon electrophilic reagent that delivers that one carbon as a methyl group.3040

    That is an interesting reagent, a nice way you can make a methyl ester in particular, taking advantage of an SN2 mechanism.3052

    Let's see if we can do an example of a transform reaction.3062

    If we started with this alcohol, how could we turn it into this ester?3068

    So here, we see that we are forming this bond; that is one way we...actually, we can kind of look at it both ways: we could disconnect here or here, but let's take a look at this disconnection.3072

    OK, the oxygen was my nucleophile, and the carbonyl was my electrophile.3086

    If I had this O-, that would be a good nucleophile, and if I had this carbonyl, that would be a good electrophile; but the question is, what is attached to this carbonyl so that after the oxygen adds into it, I don't lose the carbonyl--I still have it?3097

    How can that happen?3113

    What I need is a leaving group--something like an acid chloride would be an ideal leaving group.3115

    If I had a good leaving group there, then this could do an addition-elimination, and I would retain my carbonyl.3121

    This looks like a straightforward acylation reaction: I just would use benzoyl chloride and acid chloride and some base (maybe like pyridine), and I would do addition-elimination, and the pyridine would be there to react with the HCl byproduct.3127

    OK, what if we wanted to make an amide?3149

    Now, remember that amide was the very last one of our carboxylic acid derivatives in terms of reactivity.3152

    It has the worst leaving group, so we can imagine that maybe we could start with any other leaving group there, and the nitrogen will always be good enough to kick it out and displace it.3159

    That is mostly true; let's see several examples.3169

    First of all, we can start with our acid chloride or anhydride: remember, those are the two best leaving groups, and that would be a great way to make an amide.3172

    So, for example, if I started with this amine, and I wanted to make this amide, I just added a 1, 2, 3-carbon carbonyl; so just like we did in the transform, if I used an acid chloride, I could do addition of the nitrogen and elimination of the chloride.3179

    Or, I could use the 3-carbon anhydride; so instead of a chlorine leaving group, I could have a carboxylate leaving group.3201

    Both of these would be great at doing the transformation (addition-elimination).3211

    OK, but remember, we always have a base that is thrown in here, because we have to remove that proton, and here we will lose a chloride; so we are forming HCl here.3217

    Or, if we used the anhydride, we would form the carboxylic acid as the leaving group; so we need some kind of base in here.3228

    And like I said, sometimes we use pyridine as a base; that is very common; OK, but because amines are bases, sometimes what we do is: we just add 2 equivalents of the amine.3237

    One of them will get acylated, and the other one can act as the base for the acid that is formed.3252

    Sometimes, if you see that there is an excess of the amine that is starting here, that is what it is being used for.3257

    Rather than use an amine, and then throw in a second amine to act as a base, or a pyridine, then we can just do two equivalents of the amine to act as the base, as well.3263

    OK, so one possible route to an amide--one of the best routes to forming an amide--is to start with the acid chloride or the anhydride and do a simple addition-elimination with an amine nucleophile.3275

    Another synthesis that is unique to amides is to start with a nitrile, a C-N triple bond, and do just a partial hydrolysis.3289

    We saw that one way to make an amide is to start with a nitrile.3299

    And so, one possible to do this transformation, for example, is to first install a cyano group: we would use sodium cyanide to do an SN2 here with that is a good way to introduce a cyano group to a carbon chain.3305

    And now, what we want to do is replace two of these C-N bonds with C-O bonds; that looks like hydrolysis.3326

    So, H3O+, mild, will do just a partial hydrolysis--will start hydrolyzing the nitrile; but if we don't add heat, then we can get it to stop at the amide product.3333

    This is another way that we can make an amide, thinking terms of a retrosynthesis, going back, doing a functional group interconversion back to a nitrile instead.3348

    What would be another way that we can do this transformation?3359

    We see we have 2 carbons here, and we are adding this one carbon that ultimately would be an amide functional group.3362

    Let's think of another possibility: well, let's go back and say, "What is another way that we can make an amide?"3372

    The first possibility we saw for making an amide was from an acid chloride; so what if I made this acid chloride instead--if I had this acid chloride, could I make the amide?3380

    Remember, this retrosynthesis is asking, "What starting materials do I need?"3392

    Well, if I had this acid chloride, I could just mix it with ammonia, and that would make the amide; so this would be a good possibility.3399

    Well, where do acid chlorides come from?3404

    Acid chlorides come from carboxylic acids: if I had this carboxylic acid, I could make the acid chloride and then go to the amide, right?3408

    So now, I'm looking at a synthesis where I am going from a bromide to a carboxylic acid; we have already seen the synthesis of carboxylic acids.3418

    In fact, we have seen the path where we do this disconnection; let's imagine breaking that carbon-carbon bond of the carboxylic acid and asking which of these carbons was my nucleophile and which one was my electrophile.3426

    I think the carbon with the carbonyl was my electrophile, which means the other one was a nucleophile; how do I make an ethyl group nucleophilic?3441

    The Grignard reagent is the way to do that: RMgX makes any random carbon group nucleophilic.3452

    And so, what electrophile do you react the Grignard with to make this carboxylic acid product?3462

    We need an electrophile with a carbon and two oxygens--I think that is carbon dioxide (yes, we saw the reaction of a Grignard with CO2 as a way to make carboxylic acid).3468

    This would be another path to doing this transformation: we could add in magnesium to make the Grignard reagent, where it just inserts itself into the carbon-bromine bond.3478

    And then, we can do carbon dioxide, followed by workup to install the extra carbon as a carboxylic acid.3491

    And then, we can convert that carboxylic acid to an acid chloride by a special reagent is needed for this; we use SOCl2 or PCl5...a chlorinating agent like thionyl chloride...3506

    And then, finally, how would I convert the acid chloride to the amide?3521

    Well, I could just use ammonia--I could use an excess of ammonia, because I know I need some extra in there to react with the HCl.3525

    OK, so this would be another path: many times, our synthesis problems or our transform problems will have lots of options, and we are seeing that there are a lot of ways to make these various carboxylic acid derivatives.3533

    OK, now one thing I want to point out, though, is: you might think, "Well, how about if I start with a carboxylic acid and I add in ammonia? Could I use that to make an amide?"3545

    Well, at first glance, it looks pretty good, because when we compare our leaving groups, we say, "Well, yes, the water is a better leaving group, and the amine is a better nucleophile, so I would think this would favor the products; OK."3557

    But the problem is: this reaction does not work.3569

    It does not work directly--it doesn't work easily; OK, if you really cook this and really add a lot of heat and did an industrial preparation, you could get this reaction to go.3572

    But in an ordinary laboratory setting, this reaction doesn't happen, because a different reaction happens.3581

    What is it about amines that are unique--what reactivity do they have, besides being a nucleophile?3587

    We just saw it on the previous slide: we saw that we use an extra excess of the amines so it can also act as a base.3594

    Amines are very good bases.3600

    The problem is: we are adding a base to a carboxylic acid; what reaction happens?3602

    We will have a proton transfer to make a carboxylate and ammonium.3611

    OK, and what is the problem there?--well, the problem there is: we now have a very, very poor electrophile (our carboxylic acid is now a carboxylate; remember, we said that was the worst electrophile we could have, because it's negatively charged).3624

    We have also lost our nucleophile, because we have...this is not a nucleophile.3641

    And so, like I said, if you had an excess amine, and you really heated it up, you could get this substitution reaction to take place; but under ordinary circumstances, there is no reaction that happens further here, so this would not be a way to make the amide.3649

    OK, instead, the way you would do this--how could you convert the carboxylic acid to the amide?--that is what we just saw.3667

    We saw: first, you convert this into a good leaving group by using SOCl2; first make the acid chloride.3674

    And now, you can throw in that amine and make the amide; so it's just going to take an extra step.3687

    The way we make an amide is from an acid chloride, rather than a carboxylic acid.3696

    OK, so we have just explored all of the various ways to make the different carboxylic acid derivatives (acid chloride, anhydride, ester, amide...) OK, now let's take a look at what kind of reactions they can undergo.3702

    They all have carbonyls, so they are going to be electrophiles, so what we are really going to be looking at is the reactions of the various nucleophiles.3718

    So, for example, your hydride nucleophile--let's review what a hydride does with a carbonyl such as an aldehyde or a ketone.3725

    If we had an aldehyde or a ketone, and reacted it with either sodium borohydride or lithium aluminum hydride--any of those sources of hydride (remember, we are putting quotes because it is really's bonded to the boron or the aluminum), that is going to attack the carbonyl and break the π bond.3732

    And then, after workup, or thanks to the protic solvent we have here, we would get an alcohol.3751

    We describe this as a reduction reaction, because we traded a C-O bond for a C-H bond.3758

    OK, when we try to do the same reaction with an ester--sodium borohydride with an ester--no reaction happens; there is no reaction.3762

    The sodium borohydride is not very reactive: we are going to be looking at sodium borohydride and lithium aluminum hydride separately.3773

    The sodium borohydride is not all that reactive: you can actually see that, because it coexists OK with a protic solvent; so you can tell it is not all that reactive.3783

    And remember that the ester is less reactive than the ketone.3793

    It is a poorer electrophile, compared to the ketone; why is that?3801

    Why is this carbonyl less reactive--what effect is this oxygen having on that carbonyl?3809

    Electron donation by resonance: OK, it adds an electron density that makes this carbonyl more electron rich, so it is going to be less willing to react with a nucleophile.3816

    OK, so here is one difference we can see from an ester to a ketone: reaction with something like sodium borohydride.3830

    If we try lithium aluminum hydride, however, that is more reactive, and so it will, in fact, react with an ester; it will react with an acid; it will react with every carboxylic acid derivative you can throw at it.3838

    It will react with nearly every functional group that you throw at it; OK, LAH is a great reducing agent.3848

    So, when you react it with an ester, you actually end up getting an alcohol product again, and it ends up adding two equivalents of the hydride.3854

    Let's see if we can do a mechanism for that: lithium aluminum hydride is a source of H-, so we can draw H- in quotes.3869

    And I'm pretty sure the mechanism is going to start as usual: nucleophile attacks the carbonyl.3879

    Let's see where that brings us: we now have a hydrogen added; we have an O- up here; any guess on what is going to happen here?3888

    As usual with our carboxylic acid derivatives, because it has a leaving group on that carbonyl, after nucleophile adds in, it is going to eliminate that leaving group.3897

    We are a CTI here, and that CTI is going to collapse to kick off a leaving group.3905

    We get a substitution, addition-elimination--an acyl substitution.3916

    OK, however, our reaction is not done yet, because we just made an aldehyde.3921

    Does an aldehyde react with LAH?--it does; in fact, the aldehyde is even more reactive than the ester that we started with--remember, this is an even better electrophile, so you can't stop here.3928

    This will not stop in these reaction conditions; they will continue to react with the hydride that is present, and a second equivalent will be added.3941

    OK, what now?--do we collapse?--do we continue?--do we get a third addition somehow?3956

    Is this a CTI?--is that a CTI (charged tetrahedral intermediate)?3962

    It is not: it is tetrahedral, and it has an O-, but what is it missing?--it is missing a leaving group: there is no leaving group.3970

    Remember, a charged tetrahedral intermediate has two groups with lone pairs--at least two groups with lone pairs.3979

    One group does the pushing out, and the other group with lone pairs is a good leaving group.3984

    In other words, this is an irreversible reaction; once you form a C-H bond, you are done--you are never going to break that bond.3989

    H- is not a good leaving group; methyl - is not a good leaving group; so you are done.3995

    This just waits around for step 2; we should have the same step 2, H3O+: so when you do a workup now, we are going to protonate this O-, and we get our alcohol product (which is what we predicted).3999

    We add two equivalents of hydride to the carbonyl: addition-elimination followed by a second addition.4019

    Let's summarize what we see with hydride: this is true not only for esters, but it is true for any of these carboxylic acid derivatives, so even the carboxylic acid would do this; the acid chloride; the anhydride (for sure, because they are even more reactive)...all of these will have addition-elimination of a leaving group, and then a second addition of hydride to get an alcohol product.4031

    OK, the only carboxylic acid derivatives that vary from this are the nitrogen-based ones (either the amide or the nitrile).4056

    OK, these will also add two equivalents of hydride, but the product you get is not the alcohol; you get the amine.4063

    You keep the nitrogen that was in your starting material; it doesn't leave.4071

    Remember how we talked about how poor a leaving group nitrogen is: it doesn't like having a negative charge, and this is evidence of how poor that leaving group is.4076

    It doesn't get kicked out.4085

    Let's take a look at this mechanism, though, to see how we can get to an amine from something like an amide.4086

    What happens in this mechanism first is a little unusual, and that is because our amide is slightly acidic, and we are reacting with hydride (which is a very strong base).4094

    And so, the first thing that happens is: we deprotonate the amide.4111

    Now, rather than just draw this as the anion, I'm going to add in this salt; remember, this was LiAlH4.4121

    OK, so the counterion we have here--we still have an aluminum with three hydrogens; that aluminum with three hydrogens and the lithium salt--this is all still going to be here, kind of like a counterion.4130

    This is a Lewis acid; it is going to be adding to this lone pair; it is going to be stabilizing it.4142

    OK, so it is not just a plain old nitrogen with a negative charge; it is forming a bond with this aluminum, and that is going to be stabilizing it.4148

    OK, so even though it is negatively charged, this is just how reactive hydride is: it will still get attacked by the nucleophile, now--by another equivalent of the hydride.4158

    OK, our nitrogen still has AlH3Li+, and now the oxygen has a negative charge, so let's draw its counterion.4175

    AlH3Li+...this is a very strange-looking CTI, isn't it?--it actually has two charges.4185

    OK, but it is still a charged tetrahedral intermediate; it looks pretty unstable, but it will form; the question is: when it collapses, who is the better leaving group?4196

    OK, they are both ugly--they are both looking pretty awful--but which one looks better?4208

    We have an oxygen that will have a negative charge, or a nitrogen that has a negative charge.4213

    Because oxygen is more electronegative, it better handles the negative charge; this is the better leaving group.4219

    What happens is: the hydride adds in, and the nitrogen--the existing nitrogen--kicks that oxygen out as a leaving group.4229

    That is how the oxygen gets lost.4240

    OK, so again, we saw before, rather than have an aldehyde with a carbonyl, we have this imine with a C-N double bond; but this is just like a carbonyl, in terms of its reactivity.4247

    And because I still have hydride around, we would expect that second addition of hydride to occur, and it does.4259

    OK, and again, this would be the salt here, but that is not going to be very relevant to us at this point, because we are just going to be working up at this point; there is nothing else for it to do.4272

    Again, this is not a CTI; it has no leaving groups; the only thing it is going to do is wait around for workup, so it can't be protonated.4281

    What product did we have?--we end up adding two equivalents of hydride, but we keep that nitrogen leaving group.4292

    The first hydride adds in, and then the CTI collapses to kick out the oxygen, and then the second equivalent of hydride adds in, as well.4300

    OK, the mechanism for the nitrile with the C-N triple bond is a little more straightforward, because you don't have that oxygen to get rid of; in that case, you just add hydride, and you add a second hydride with protonations in between.4310

    How about organometallic reagents--like a Grignard reagent: how would that react with a carboxylic acid derivative, such as an ester?4326

    Well, again, let's review what we know about ketones and aldehydes, and that carbonyl chemistry.4334

    If we used methylmagnesium bromide or methyllithium, that is a source of methyl -; again, we put that in quotes, because it is not ionic; it has that metal coordinated to it.4338

    But it is nice to draw it like this for our mechanism: of course, a great nucleophile; carbonyl is an electrophile; so, after our workup, we end up forming an alcohol.4349

    We get this new carbon-carbon bond, and with methylmagnesium bromide or methyllithium, we would add this new methyl group, and we would get an alcohol product out.4359

    OK, that is what we get for an aldehyde or a ketone: how does that differ when we use something like an ester?4371

    OK, well now, again, we will use methylmagnesium bromide, and it is going to add to the carbonyl and break the π bond (I'm sorry, this is not in equilibrium; it is a one-way street, because we are forming a new carbon-carbon bond).4379

    That makes an O-; we have an O ethyl; and we just added a methyl group.4396

    It looks very much like it did in the hydride example: the nucleophile adds in; it forms a CTI that is going to collapse to give a ketone; and can we stop at this point?4402

    We can't stop here: remember, the ketone is even more reactive than the ester; so we will assume we have an excess available of the Grignard reagent, so that we are going to be adding a second equivalent, just like we did for the hydride.4423

    We are going to add two equivalents of the Grignard's.4442

    We are going to get an alcohol product.4451

    So it's essentially the same exact mechanism as the hydride, same looking product, except instead of adding an H, you are going to add some kind of an R group (in this case, a methyl group).4452

    Let's try an example: it's always a little more interesting when we look at cyclic examples: this is called a lactone--a cyclic ester is called a lactone.4463

    What would happen to this product if we were to react it with an excess of phenyllithium?4473

    Well, phenyllithium means we have phenyl -; it's like we have phenyl -; let's see if we can predict this product without going through an entire, detailed mechanism.4476

    OK, I know that this is going to attack the carbon and break the π bond; that will give me an O- up here; what happens next?4488

    Then, that CTI, that O-, comes back down and kicks the leaving group out.4498

    What we are going to end up doing is: we are going to be breaking this carbon-oxygen bond here; let's see if we can draw that product--sometimes it's helpful just to redraw your ring, and then literally erase the bond that is broken.4505

    There is the bond that is broken; and we are going to have a carbonyl, because we did addition-elimination; we did a substitution; so we still have a carbonyl; we have introduced a phenyl; this we could draw as an O-, because it got kicked out as a leaving group.4518

    Is that our final product?4538

    If we have an excess of the phenyllithium around, we expect the product to repeat.4540

    OK, our phenyl - is going to kind of come back in again; so in our final product, we will have an alcohol with 2 new phenyl groups attached; and then, our O- upon workup is going to give us an alcohol.4546

    Let's just check to make sure we have...including the oxygen...1, 2, 3, 4, 5, 6; 1, 2, 3, 4, 5, 6; that is good--we still have 6 atoms there.4561

    A lot of times, when we are drawing our line drawings, and especially with cyclic systems, it is easy to gain a carbon or lose a carbon.4571

    OK, so we see our pattern: two equivalents of our Grignard has been added (our phenyllithium in this case), and we get an alcohol product.4579

    In this case, our leaving group is attached, so our leaving group is still there: we see the alcohol leaving group, in this case.4586

    Now, there a few special hydride reagents that I thought I would share some examples of with you, rather than just using lithium aluminum hydride, which is super reactive and reduces the maximum amount in each case.4595

    We can look at some other derivatives of that: for example, instead of just having 4 hydrides here, if you have 2 isobutyl groups attached to the aluminum (this is an isobutyl group: 1, 2, 3, 4 carbons--it kind of looks like an isopropyl: this would be an isopropyl--just those 3 carbons--then it has an extra CH2; that is called the isobutyl group), this is called diisobutyl aluminum hydride.4608

    This is called DIBAL for short; or sometimes, it's DIBAL-H to show that it has a hydride attached to it.4636

    And because of the presence of those isobutyl groups, it changes the reactivity; it makes it more selective and less reactive, and it turns out it only adds one equivalent of hydride.4645

    So, the H- equivalent is going to add, but when you form this intermediate, because it has the aluminum with the isobutyl groups on here, it turns out that this is stable and there is no collapse here.4655

    So, because there is no collapse, there is no opportunity for that second equivalent of the hydride to add in.4678

    This just waits until workup, and we trade those two C-O bonds from separate groups into C-O bonds to the same oxygen, and we get our carbonyl back.4685

    When you compare this product, you take an ester, and you can just do a partial reduction of that ester by doing a single addition-elimination, a single substitution reaction, and not having a second addition added in.4695

    You can go from an ester to an aldehyde with a special reagent like diisobutyl aluminum hydride.4711

    There are some other examples: you can put some tri(t-butoxy) groups on the aluminum to replace three of those hydrogens; so this one has that same sort of reactivity; sodium cyanoborohydride has some special reactivity with the presence of the cyano group.4721

    So really, there are dozens and dozens of reagents that you can use; we are really just looking at the tip of the iceberg--we kind of look at the most commonly used reagents when we are first introducing a topic.4735

    Another interesting example with carboxylic acid derivatives is the use of an organocuprate; an organocuprate has the formula R2CuLi.4747

    We have seen organocuprates adding (doing conjugate additions) to α and β unsaturated carbonyls, kind of like a Michael addition, a 1,4 addition.4760

    That is an interesting reactivity that cuprate has; it also has an interesting reactivity with an acid chloride.4770

    These have a different reactivity than a Grignard, but it is less reactive, and when we use it to react with an acid here, we have diphenyl cuprate: the phenyl 2- is going to add in, and then it's going to collapse down, and we will get our addition-elimination.4780

    But then, this product has no further reaction with diphenylcuprate: cuprates don't react with ketones.4802

    So, unlike the Grignard (where we had addition-elimination and then we added...again, with a Grignard, you are always going to add two equivalents, no matter what), it actually stops with the cuprate.4811

    Why does it react at all?--well, remember, the acid chloride is more reactive; that is a very, very high electrophile, very reactive, very electron deficient.4820

    The ketone is less reactive, because it no longer has that leaving group on there pulling electron density away.4829

    Any time you go to a product that is less reactive than your starting material, it might be a reaction that you can control and prohibit the further reaction going forward.4837

    OK, so that wraps it up for a look at carboxylic acid derivatives.4851

    "How do we prepare the various carboxylic acid derivatives, and what are some reactions that they can undergo?"4855

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