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Lecture Comments (15)

1 answer

Last reply by: Professor Starkey
Tue Apr 21, 2015 1:01 AM

Post by sarit nahari on April 20, 2015

Hello, I am a medical student at Tel Aviv University. Because  your's  wonderful explanations, I'm starting to love organic chemistry.
Thank you very much.
Bono

1 answer

Last reply by: Professor Starkey
Wed Mar 11, 2015 1:03 AM

Post by Nicole Aquino on March 10, 2015

Thank you for the lectures. It helped me a lot. I'm a Medical laboratory student and i'm very glad to discover this page. :)

1 answer

Last reply by: Professor Starkey
Sun Dec 8, 2013 6:56 PM

Post by Annie An on December 8, 2013

What if i want to compare the water solubility of similar sized alcohols, aldehydes and carboxylic acids?
What kinds of things that i supposed to consider to compare them?

1 answer

Last reply by: Professor Starkey
Fri Jul 19, 2013 8:35 AM

Post by mateusz marciniak on July 18, 2013

hi professor i would like to thank you for your lecture series, it has helped me get through my summer orgo I course. i just wish you did some harder examples in each lecture series, some were difficult but others i thought a were a little too easy, nonetheless excellent lecture series

1 answer

Last reply by: Professor Starkey
Sun Apr 7, 2013 12:08 PM

Post by Edi William Yapi on April 6, 2013

Great Lecture !

1 answer

Last reply by: Professor Starkey
Sun Feb 17, 2013 5:23 PM

Post by natasha plantak on February 15, 2013

Your lectures are extremely helpful. Thank you so much!
~Natasha

1 answer

Last reply by: Professor Starkey
Thu Feb 2, 2012 11:01 AM

Post by Jason Jarduck on January 29, 2012

Hi Dr. Starkey

I really liked your lecture. I'm going to be looking forward to organic chemistry III as well. Very good information. Very clear.

Thank You

Jason

0 answers

Post by Billy Jay on March 28, 2011

Hmmm. Something's different...

Carboxylic Acids

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
Identify the starting compound in this reaction:
  • This is an oxidation of alkenes/alkynes reaction (Ozonolysis)
Devise a synthesis for this reaction:
  • Step 1:
  • Step 2:
  • Step 3:
Rank the following compounds in order of increasing acidity:
  • Cl is more electronegative than Br
Draw the mechanism leading to the product for this reaction:
  • Step 1:
  • Step 2:
  • Step 3:
  • Step 4:
  • Step 5:
  • Step 6:
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.

Answer

Carboxylic Acids

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
    • Review Reactions of Ketone/Aldehyde
    • Review Reactions of Ketone/Aldehyde
    • Carboxylic Acids and Their Derivatives
    • Ketone vs. Ester Reactivity
    • Carboxylic Acids and Their Derivatives
    • General Reactions of Acarboxylic Acid Derivatives
    • Physical Properties of Carboxylic Acids
    • Aciditiy of Carboxylic Acids, RCO₂H
    • Aciditiy of Carboxylic Acids, RCO₂H
    • Aciditiy of Carboxylic Acids, RCO₂H
    • Aciditiy of Carboxylic Acids, RCO₂H
    • Preparation of Carboxylic Acids, RCO₂H
    • Preparation of Carboxylic Acids, RCO₂H
    • Preparation of Carboxylic Acids, RCO₂H
    • Preparation of Carboxylic Acids, RCO₂H
    • Preparation of Carboxylic Acids, RCO₂H
    • Hydrolysis Mechanism
    • Hydrolysis Mechanism
    • Hydrolysis Mechanism
    • Applications of Carboxylic Acid Derivatives:
    • Ester Hydrolysis
    • Ester Hydrolysis Requires Acide or Base
    • Nitrile Hydrolysis
    • Nitrile Hydrolysis Mechanism
    • Use of Nitriles in Synthesis
    • Intro 0:00
    • Review Reactions of Ketone/Aldehyde 0:06
      • Carbonyl Reactivity
      • Nu: = Hydride (Reduction)
      • Nu: = Grignard
    • Review Reactions of Ketone/Aldehyde 2:53
      • Nu: = Alcohol
      • Nu: = Amine
    • Carboxylic Acids and Their Derivatives 4:37
      • Carboxylic Acids and Their Derivatives
    • Ketone vs. Ester Reactivity 6:33
      • Ketone Reactivity
      • Ester Reactivity
    • Carboxylic Acids and Their Derivatives 7:30
      • Acid Halide, Anhydride, Ester, Amide, and Nitrile
    • General Reactions of Acarboxylic Acid Derivatives 9:22
      • General Reactions of Acarboxylic Acid Derivatives
    • Physical Properties of Carboxylic Acids 12:16
      • Acetic Acid
      • Carboxylic Acids
    • Aciditiy of Carboxylic Acids, RCO₂H 17:45
      • Alcohol
      • Carboxylic Acid
    • Aciditiy of Carboxylic Acids, RCO₂H 21:31
      • Aciditiy of Carboxylic Acids, RCO₂H
    • Aciditiy of Carboxylic Acids, RCO₂H 24:48
      • Example: Which is the Stronger Acid?
    • Aciditiy of Carboxylic Acids, RCO₂H 30:06
      • Inductive Effects Decrease with Distance
    • Preparation of Carboxylic Acids, RCO₂H 31:55
      • A) By Oxidation
    • Preparation of Carboxylic Acids, RCO₂H 34:37
      • Oxidation of Alkenes/Alkynes - Ozonolysis
    • Preparation of Carboxylic Acids, RCO₂H 36:17
      • B) Preparation of RCO₂H from Organometallic Reagents
    • Preparation of Carboxylic Acids, RCO₂H 38:02
      • Example: Preparation of Carboxylic Acids
    • Preparation of Carboxylic Acids, RCO₂H 40:38
      • C) Preparation of RCO₂H by Hydrolysis of Carboxylic Acid Derivatives
    • Hydrolysis Mechanism 42:19
      • Hydrolysis Mechanism
      • Mechanism: Acyl Substitution (Addition/Elimination)
    • Hydrolysis Mechanism 47:27
      • Substitution Reaction
      • RO is Bad LG for SN1/SN2
      • RO is okay LG for Collapse of CTI
    • Hydrolysis Mechanism 50:07
      • Base-promoted Ester Hydrolysis (Saponification)
    • Applications of Carboxylic Acid Derivatives: 53:10
      • Saponification Reaction
    • Ester Hydrolysis 57:15
      • Acid-Catalyzed Mechanism
    • Ester Hydrolysis Requires Acide or Base 1:03:06
      • Ester Hydrolysis Requires Acide or Base
    • Nitrile Hydrolysis 1:05:22
      • Nitrile Hydrolysis
    • Nitrile Hydrolysis Mechanism 1:06:53
      • Nitrile Hydrolysis Mechanism
    • Use of Nitriles in Synthesis 1:12:39
      • Example: Nitirles in Synthesis

    Transcription: Carboxylic Acids

    Hello; welcome back to Educator.0000

    Today, we are going to be talking about carboxylic acids.0002

    Let's review what we know about carbonyls, that we have seen in aldehydes and ketones so far.0007

    A carbonyl is a C-O double bond, and what makes this functional group very special is that it has resonance, where we can draw a second Lewis structure.0013

    And any time we have resonance, that means that neither of these Lewis structures accurately reflects the actual structure of the carbonyl; it's some blend between these two.0024

    And so, each of them kind of describes the characteristics that the carbonyl has.0035

    So overall, every time that we see a carbonyl, we know that the carbonyl carbon has some significant partial positive character, and the carbonyl oxygen, some partial minus character.0040

    That means...the reaction as you have seen it so far is that the carbonyl carbon is electrophilic (in other words, nucleophiles add here), and the carbonyl oxygen, somewhat basic (in other words, we can protonate here).0052

    And, in fact, many, many reactions we have seen have started with...involved protonation of the carbonyl; and many, many reactions involve a nucleophilic attack on the carbonyl carbon.0071

    So, let's just review a few examples of these things we have seen already for aldehydes and ketones, because then we'll be able to compare that to how that differentiates (is different) from when we move into carboxylic acids and other function groups, known as carboxylic acid derivatives.0082

    For example, if we have a ketone, and we treat it with hydride (something like lithium aluminum hydride), that is a great nucleophile, we would expect that to add to the carbonyl, and then, after workup, we would protonate that O-, and we would get this alcohol product out.0097

    We describe that as a reduction reaction, because we have lost a C-O bond and traded it for a C-H bond and gone from a ketone to an alcohol.0119

    OK, a Grignard reaction, instead of an H-, gives us a C-; so this phenylmagnesium bromide would give us a source of phenol -; we could put that in quotes, just like we did we did for hydride (sorry, I didn't do that).0128

    We should put that in quotes, because this hydride is always coordinated with the aluminum; this carbon group is always coordinated with the magnesium; but it reacts just like that kind of nucleophile, and again, would attack the carbonyl.0142

    And then, after an aqueous workup, we would end up with an alcohol product.0154

    But, in this case, we would have a new carbon-carbon bond that we have also formed at the same time.0159

    This is just a few examples of nucleophiles that can add to carbonyls of aldehydes and ketones.0168

    Now, if we have heteroatom nucleophiles, like oxygen or nitrogen, our products look a little different; if our nucleophile is an alcohol, you can imagine that the same pattern we had for the hydride of the Grignard...our nucleophile, in this case, would a methoxy group; so we could add that in.0175

    But this alcohol product is no longer stable; having a single carbon with an OH and an OR group attached is not stable; so instead, this continues to react in the presence of acid catalyst, and we end up adding two equivalents of the alkoxy group, and we get out a product known as an acetal.0196

    Instead of having a carbonyl and two bonds to the same oxygen, we have two bonds to different oxygens; we have these methoxy groups.0220

    And, if we have a nitrogen nucleophile, again, you can imagine having this as an intermediate product--OK, but this, too, is unstable, having this alcohol, because this nitrogen with its lone pairs is going to come down and kick off that leaving group.0227

    At some point, we would probably protonate this, since we are acid-catalyzed, to make it a good leaving group.0246

    But ultimately, the products we get with an aldehyde or ketone and an amine are: we replace the C-O double bond with a C-N double bond, to give an imine product.0251

    So, these are all reactions that we have seen before for aldehydes and ketones, and we will see how those are going to be quite different from the next set of carbonyl-containing functional groups.0263

    We will see carboxylic acids and their derivatives.0275

    Now, a carboxylic acid is the functional group we have when we have a carbonyl; but instead of just carbons attached (like a ketone), or a carbon and a hydrogen (like an aldehyde), we have an OH group attached to that carbonyl.0278

    Now, that no longer is an alcohol functional group, or a ketone and aldehyde.0290

    This entire thing combines together to be described as a single functional group called a carboxylic acid.0294

    And if we want to think about its reactivity, we need to consider its resonance as well, and like every carbonyl, it has resonance that breaks the π bond and gives an O- and a C+.0303

    That is always true for any carbonyl; but by having this OH group, now, attached to the carbonyl carbon, we have additional resonance.0317

    That additional resonance comes about by this oxygen donating its lone pair of electrons and sharing it with that carbonyl (specifically, with that carbonyl carbon).0325

    We have a third resonance form that we can draw.0340

    How does that affect the reactivity of the carboxylic acid?0345

    Well, it turns out that this oxygen group, with its lone pair, donates electron density into the carbonyl.0349

    The OH group donates electron density, and that means that this carbonyl--the carboxylic acid--is more electron-rich than a ketone or an aldehyde.0358

    The effect of the oxygen is to add electron density into the carbonyl, making it more electron-rich.0369

    That makes the ketone or the aldehyde more partial positive and more electrophilic; so carboxylic acids and their derivatives (that we will see down the road), in general, are more electron-rich because of that electron donation.0375

    Certainly, if you have an oxygen group, you are like an OH.0389

    An example that we have seen of this before is when we used sodium borohydride as our hydride source, instead of lithium aluminum hydride.0395

    LAH is much more reactive; sodium borohydride is not as reactive--it's still reactive enough to react with a ketone and reduce it to the alcohol, but if we try to do that same reaction with an ester (the ester, remember, is less reactive, because this carbonyl is more electron-rich), we find that there is no reaction here.0403

    Sodium borohydride is...we say that NaBH4 is selective for aldehydes and ketones--it can reduce an aldehyde or ketone in the presence of an ester, because the ester is less reactive.0432

    Now, in this lesson, we are going to be talking about carboxylic acids, but in the next lesson, we are going to be talking about all of the related functional groups, called carboxylic acid derivatives.0452

    Let's introduce them at this point.0461

    If you have a halogen attached to a carbonyl, we call those acid halides; and usually, the x is chlorine; so it is usually the acid chloride that we are dealing with.0464

    If you have an oxygen surrounded by carbonyls on either side, that is known as an anhydride.0476

    If you have an OR group attached to carbonyl, it is called an ester.0482

    If you have a nitrogen attached to the carbonyl, we call that an amine.0487

    These all have the general structure of having attached the carbonyl--some kind of leaving group with a lone pair.0492

    So, this represents an atom with lone pairs; we describe it as a heteroatom, meaning "not carbon or hydrogen."0498

    If there is carbon or hydrogen, we describe it as an aldehyde or a ketone.0507

    But this has either a halide (we know halides, of course, are good leaving groups...and lone pairs); the leaving group on an anhydride is this oxygen with the carbonyl.0510

    So, for an anhydride, one of the carbonyls we consider as the electrophilic carbonyl; the rest of the group...the other carbonyl is part of the leaving group attached to that.0523

    The ester has an OR group attached as a leaving group; the amide has a nitrogen group attached as a leaving group; and then the nitrile doesn't really fit into the same pattern, but it, too, is a carboxylic acid derivative, because it has three bonds to a nitrogen, for example, instead of 2 to an oxygen and 1 to a nitrogen.0534

    It still has that same sort of pattern; we will look at nitriles a little later in the chapter.0555

    The reason that all of these functional groups are grouped together and considered as a unit is because they have the same general reactivity.0564

    OK, and a general reaction of a carbonyl with a leaving group attached is that, when it reacts with a nucleophile, the nucleophile attacks the carbonyl.0573

    And then, it might be tempting, now that you see something defined as a leaving group attached--it might be tempting to say, "Oh, well, that nucleophile just kicks out that leaving group, doesn't it?"0583

    Well, it can't; that would be an Sn2 mechanism: there are no Sn2's for carbonyls--carbonyls don't do that.0592

    But instead, the nucleophile attacks the carbonyl; and as always, it just breaks the π bond and puts those electrons up on the oxygen.0600

    OK, but when you look at the structure that is formed as a result, this structure is described as a charged tetrahedral intermediate (or a CTI for short).0610

    What defines a CTI is: you have a tetrahedral carbon (so in other words, an sp3 hybridized carbon with four groups attached)...so we have an sp3 hybridized carbon, and at least two groups have one or more lone pairs.0633

    So, for example, if we have an oxygen and an oxygen, or an oxygen and a chlorine, or a chlorine and a nitrogen--if we have at least two groups with lone pairs, we describe this as a CTI.0659

    What a CTI does is very special: it will collapse.0669

    When we see this pattern in an intermediate, we know that it is unstable, and it collapses.0674

    And what that means is that the leaving group will leave, but it leaves with the assistance of the other group with lone pairs.0679

    So, one group with a lone pair gets kicked out, and the other group with lone pairs helps push it out.0686

    And so, our leaving group leaves, and overall, what has happened is: we have had a substitution reaction take place (our leaving group has gone, and a nucleophile has taken its place).0695

    OK, but it is not a substitution mechanism we have seen before (for alkyl halides, for example); it is not an SN1 mechanism; it is not an SN2 mechanism; it is called an acyl substitution reaction.0706

    The mechanism can be described as addition-elimination.0718

    Our nucleophile adds into the carbonyl, and then the leaving group is eliminated by collapsing a CTI.0722

    We'll find it has both acid-catalyzed mechanisms and base-catalyzed mechanisms; we'll study both of those in today's lesson.0729

    Before we look at those reactions, though, let's think a little bit more about carboxylic acids as a functional group: what kind of behavior does it have?--what kind of physical properties?0739

    OK, so for example, let's take a look at this: this is acetic acid--this is a simple carboxylic acid: AcOH is how we could abbreviate this.0747

    The Ac group means we have a CH3 with a carbonyl; you will see that Ac group here in acetic acid, and it is common throughout all the carboxylic acid derivatives (acetyl chloride and acetic anhydride and acetic acid and so on).0756

    It is very good to be familiar with those common names.0775

    If you take a look at the physical properties, acetic acid is an interesting one because this is the acidic component that is in vinegar.0779

    So, the odor of acetic acid is very familiar to you; the bit that it has; the flavor that you have--that kind of burning sensation is this component in here.0790

    What does it have on here?--well, all carboxylic acids have the OH group; we know that the OH group is very polar, and any time you have an OH bond that is so polar that those groups can interact with other OH groups and can form hydrogen bonds, it's polar; it can form hydrogen bonds.0803

    It can form hydrogen bonds with itself; so, 2 molecules of acetic acid, in other words, can hydrogen bond to each other.0833

    That means it is going to have a high boiling point, and it can also form hydrogen bonds with water.0842

    That means it is going to have high water solubility.0852

    And, in fact, that is true: the boiling point for vinegar is 118 degrees Celsius, so even higher boiling than water--a very high boiling compound.0860

    It is miscible with water; so that means it is very highly water-soluble; and, in fact, vinegar is an aqueous solution, with acetic acid in there (a dilute aqueous solution).0873

    And so, you can't ever add enough acetic acid to water to have them separate out into two layers; it is always going to be a single solution, because it is miscible: it is soluble in all proportions.0886

    OK, and that is because it has this OH group; in addition, the second oxygen also has the carbonyl, which, of course, is also polar, and can do the same thing.0896

    It can hydrogen bond (form hydrogen bonds) with water; in other words, it can be a hydrogen bond acceptor.0905

    So, there are a lot of characteristics of carboxylic acid that make it very familiar with water and very much like water.0913

    OK, so when you see the RCO2H (that is how we abbreviate the carboxylic acid)--when you see that functional group, it is an extremely polar functional group.0922

    And remember, it has that resonance that we looked at on the first slide, too--on an earlier slide--that makes it even more polar with partial positive and partial negatives.0933

    OK, so it's an extremely polar functional group.0942

    Now, let's take a look at a different carboxylic acid; this one has, now, a longer carbon chain.0946

    It has two parts of the molecule: it has this carboxylic acid part, which is quite polar (can hydrogen bond and donate or can hydrogen bond and accept); so this part we would describe as being quite hydrophilic.0953

    But how about the rest of this carbon chain?0968

    Carbon-carbon bonds and carbon-hydrogen bonds are totally nonpolar.0971

    So that makes them hydrophobic, or you could even describe it as being greasy; this is something that would not like water at all.0977

    This actually has a pretty low solubility; this is insoluble in neutral water (this molecule).0994

    But if, instead of using neutral water, we were to use sodium hydroxide and water--basic water--what would happen is: the base would react with the carboxylic acid functional group (and we will look at this reaction next), and that would make an ionic compound.1001

    Now, that goes from being just a very, very polar functional group to an ionic functional group that is as extreme in polarity as you can get, when you have full charges.1024

    OK, so that makes it more soluble in water: ionic compounds love water.1035

    And so, what happens is: we can use base as a way to increase the solubility of a carboxylic acid by this chemical reaction it undergoes.1045

    The product of that acid-base reaction is an ionic, and therefore more water-soluble, carboxylate salt.1056

    Let's talk about the acidity of the carboxylic acid: it must be a significant physical property, if it is actually part of the functional group's name.1067

    If we go back to looking at an alcohol, an OH group that is just on an ordinary carbon change, that has a pKa somewhere around 18, and a carboxylic acid where that OH is attached to a carbonyl has a pKa somewhere around 5.1076

    That is 13 pKa units: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13; I don't even know what that number is--so many, many times more acidic by being a carboxylic acid OH instead of an alcohol OH.1089

    And so, we will talk about why that is in just a second; but right away, let's think about the choice of base.1109

    If you want it to deprotonate an alcohol, that is not a very easy thing to do; you have to use an extremely strong base.1114

    We use something like sodium hydride--a stronger base is needed, and furthermore, this is a base that is irreversible, because once you deprotonate, you would have the sodium salt here; and what is the other product you form?1121

    If you are using H- as your base, and it abstracts an H+, you end up with hydrogen gas.1143

    OK, so that was a way that we could make alkoxides.1150

    If we wanted to make an alkoxide, we would take an alcohol; we would treat it with sodium hydride--a very strong, irreversible acid-base reaction.1153

    Let's compare that to a carboxylic acid: most definitely, sodium hydride would effectively deprotonate a carboxylic acid, but we don't need a base that strong; we could use something like sodium hydroxide, and here a weaker base is OK.1162

    And, of course, we typically want to use the weakest base possible, because that is going to be cheaper; it is going to be easier to handle; and so on.1179

    OK, but that will give us the sodium salt; again, we get an O-, sodium +; this is called a carboxylate salt, when you have an O- attached to a carbonyl.1187

    Our other product here, when we use hydroxide, is water: that has a pKa on the order of 16, and so that wouldn't be strong enough to totally deprotonate an alcohol.1203

    But again, it very, very easily deprotonates a carboxylic acid; this reverse reaction is essentially nonexistent when there is such a big difference in the pKa's (16 compared to 5).1214

    OK, so one thing to keep in mind is that, any time you have a carboxylic acid exposed to basic conditions (things like hydroxide, alkoxide, something like that), you no longer have a carboxylic acid.1227

    You now have a carboxyl: we always need to keep that in mind--we can actually use that (like we just saw in the previous slide) for extraction purposes; that would be very handy.1242

    But we will also see some cases where we have to keep that in mind when we are doing our reaction mechanisms.1252

    OK, let's talk a little bit--since acidity is such a big component of a carboxylic acid's identity, let's talk a little bit more about these pKas.1258

    Why do we have this huge difference here?--can we explain that?--let's make sure we are clear on that.1267

    And, as usual, the way we are going to determine the difference, or explain that difference, is by looking at the conjugate bases.1274

    Let's look at alkoxide, versus a carboxylate, and see why it is so different getting to one versus the other.1282

    OK, if we have a carboxylate and an alkoxide, they both have a negative charge on oxygen, so they are both reasonable anions to have; oxygen doesn't mind having a negative charge.1293

    But one is so very different from the other, in terms of stability, and that is because this O- is attached to a carbonyl: that makes it an -ylic, and that makes it available for resonance.1303

    So, any time you have resonance delocalization, that is a good thing; any time you can delocalize the negative charge over two oxygens, that is a good thing.1320

    In fact, these two resonance forms are equivalent to each other; that is the best possible resonance stabilization you can have, because each of them is contributing 50/50 (they are contributing equally) to the overall hybrid.1328

    That is fantastic resonance stabilization.1339

    OK, so this has a...this is a resonance-stabilized conjugate base; we are really delocalizing the negative charge.1342

    That is a great thing: any time you can spread a negative charge out over multiple atoms, that is an excellent thing.1360

    That makes this less reactive.1365

    If you are more stable, that makes you less reactive; that makes him the weaker conjugate base: this is the less reactive, weaker conjugate base.1371

    The weaker conjugate base is the one that has the stronger parent acid.1384

    Is that what the pKa data told us?--yes, this was the pKa that was somewhere around 5--really low number--very, very good acid.1395

    OK, we compare that to the alkoxide: now again, oxygen doesn't mind having a negative charge, but it's all relative--we are just comparing this to a different O-; this one is resonance-stabilized; this one is not.1404

    So, this has no resonance; so there is nothing additional to stabilize that negative charge.1414

    There is way it can be delocalized over different atoms; OK, so if he is less stable (less stable--higher energy), that makes him more reactive; this is the more reactive, and therefore stronger, conjugate base.1421

    The stronger conjugate base has the weaker parent acid.1443

    And again, going back to that pKa, this is the pKa that was somewhere around 18.1452

    OK, so most definitely, you should have a good understanding of why it is carboxylic acids are such effective acids and what makes the carboxylate conjugate base such a stable species.1460

    OK, the take-home message that we have seen before and we are seeing again is that, if you have something that will stabilize the conjugate base, then that makes the acid a stronger parent acid.1472

    So always, what we are looking for is something to stabilize the conjugate base.1484

    Let's take a look at another example: let's compare acetic acid to trifluoroacetic acid; and the question is, "Which is a stronger acid?"1489

    Well, I don't see a difference in these two acids: they are both neutral; they are both stable; there is nothing unreasonable there.1499

    But, what we need to do is: we need to look at the conjugate bases, because those are charged species, and see if we can find a difference in their stability there.1504

    So, in other words, let's let each of these acids be an acid; let's deprotonate them; and of course, we are talking about the proton that is on the oxygen, that is always going to be the most acidic proton in a structure.1514

    Where we used to have an OH, now we have an O-; and the same thing over here--we used to have an OH; we have an O-.1528

    OK, now in this case, we are looking at a...they are both carboxylates (right?), so they both have the same resonance; so we need to find some other difference.1536

    The difference, of course, is that here we have hydrogens on the CH3; here we have fluorines, OK?--and what effect is that having?1548

    Well, what do you know about fluorine?--we know that fluorine pulls electron density toward itself; fluorine is electronegative--that means it pulls electron density toward itself.1558

    So, let's start by stating some facts here.1568

    We could say, "Fluorine is more electronegative than hydrogen," right?1571

    So then, we can say that fluorine withdraws electron density inductively--this is an inductive effect, meaning there is not a resonance form we could draw for this; we just use this arrow, showing that the electrons in these bonds are being pulled toward the fluorine--they are being pulled toward the fluorine, and therefore this carbon-carbon bond is being pulled, as well.1582

    OK, so those are the facts; here is the tricky part: is that a good thing, or is that a bad thing?1610

    Is that something that helps to stabilize this negative charge, or is it something that causes the negative charge to be destabilized?--that is the tricky part.1617

    OK, and let's think about what that negative charge means: a negative charge indicates to us that there is an excess of electron density; there are too many electrons around that oxygen.1626

    What are those fluorines doing?--they are pulling some of that electron density away.1636

    It kind of sounds like a good thing, doesn't it?--yes; it is delocalizing; it is helping moving and spreading it out through the molecule.1643

    OK, so this stabilizes the negative charge; that is a good thing.1649

    It delocalizes (right?): it puts some of the negative charge out on these fluorines.1665

    OK, so that means...let's call this one A, and B; so this is conjugate base A and conjugate base B; that is always a good idea, so that we can refer to things by name.1675

    So what we can say here is that conjugate base B is the more stable, and therefore less reactive, and therefore weaker, conjugate base.1690

    It is more stable, therefore less reactive, and it's the weaker conjugate base.1711

    Conjugate base B has the stronger parent acid.1715

    And that was our original question, right?--"Which is the stronger acid?"1728

    That was the...so B is the stronger acid.1731

    OK, something that stabilizes the conjugate base makes for a stronger acid; if we take a quick look at an energy diagram, and we compare the energies of A and B, they are relatively similar in energy; there is no big difference between A and B.1740

    But then, when we look at their conjugate bases, we see that conjugate base A is higher in energy than conjugate base B.1761

    The presence of those fluorines helps to stabilize it and lower the energy of that conjugate base; so B being deprotonated is an easier path than A being deprotonated; this is more favorable.1771

    Something that stabilizes a conjugate base makes the parent a stronger acid.1788

    We saw how resonance can do that, and here is an example of how inductive effects can do that.1795

    So, trifluoroacetic acid is a stronger acid than just ordinary acetic acid.1799

    Now, one more thing about inductive effects is that they decrease with distance.1806

    Because they are a through-bond effect, the more bonds it has to travel through, the less and less effect you see.1811

    For example, if you had a fluorine 10 carbons out from your carboxylic acid, the carboxylic acid wouldn't even know that the fluorine was there, and it would have no effect on the pKa, for example.1817

    Here is an example of some compounds that you can see that in: here we have carboxylic acid attached to benzene (this is known as benzoic acid).1827

    It is good to know some of these IPAC names, or a common name, for this carboxylic acid; we will see it a lot.1839

    Here we have some substituted carboxylic acids: here we have the fluorine in the para- position (this is called parafluorobenzoic acid); this is in the meta- position; this is in the ortho-position.1849

    OK, the benzoic acid itself has a pKa of 4.2, so what do you expect for these fluoro- substituted benzoic acids?--what do you think is going to happen to the pKa?1861

    We would expect these fluoro- substitutions to be more acidic; what does that do to the pKa?--it lowers the pKa.1873

    So, we expect them all to be lower than 4.2, and they are; but the one where this fluorine is really far away--very, very slightly lower: it's 4.1--just barely any difference at all.1880

    When it is meta-, it is 3.9, and when it is ortho-, it's 3.3; so almost 10 times more acidic, by having the fluorine here.1891

    This is quite far away from the O-; so it tells you that these inductive effects can be pretty powerful.1898

    They all have the same resonance; the carboxylate conjugate bases all have the same resonance; but it is these inductive effects that are affecting their pKas.1904

    Let's talk about the preparation of carboxylic acids: Where do they come from?--how can we get a carboxylic acid product?1916

    Well, because the carboxylic acid carbon is highly oxidized (it has 3 carbon-oxygen bonds), one way we can get there is by doing an oxidation reaction.1923

    We could start either with a primary alcohol or an aldehyde; both of these carbons already have a C-O bond, and they also have hydrogens that are capable of being lost to oxidation; so if we treat this with a very, very strong oxidizing agent, then we would expect that carbon to be fully oxidized to the carboxylic acid.1933

    All right, these are both two-carbon substrates, so we would get acetic acid as our product here.1956

    What are examples of strong oxidizing agents?--things like our Jones oxidation, chromic acid, sodium dichromate and H2SO4...KMnO4 permanganate and base is very good...1962

    OK, but notice: because we have base here, what is going to happen to the carboxylic acid product as it is formed?1974

    It will be deprotonated, and so, if you want the neutral product when you are all done, you have to do NH3-O+ workup; OK, that is just a little note on your basic reactions requiring acidic workup, if you want to get the neutral carboxylic acid product up.1982

    OK, the same thing for this oxidizing agent: this is known as Tollens reagent; when you use silver oxide, it also requires base, so we have to do an NH3-O+ workup in this case.1996

    This is an interesting reaction, because we use silver + to carry out the oxidation, and of course, the oxidizing agent itself gets reduced when it takes the electrons, and so it goes from silver + to silver 0.2007

    Well, silver 0 is silver metal, and so the Tollens test is a test for aldehydes.2021

    If you treat it with silver oxide, it will form silver metal, and the silver metal, if it is done just right, can plate itself out on the inside of the glass wall of the test tube, and you can get a mirror that is formed in this reaction.2028

    This is known as the silver mirror test for aldehydes, for RCHO, and the reaction involves oxidizing the aldehyde to a carboxylic acid.2044

    Of course, these qualitative tests are not nearly as widely used anymore, because we have spectroscopic techniques that can do a better job of quickly and easily analyzing our functional groups that are present.2059

    We can oxidize primary alcohols or aldehydes to get a carboxylic acid.2072

    We can also oxidize an alkene or an alkyne by doing an ozonolysis reaction; so the reaction of an alkyne with ozone totally breaks the carbon-carbon triple bond.2078

    All three carbon-carbon bonds go to carbon-oxygen bonds; so the product you get out is a carboxylic acid.2091

    We would get benzoic acid and acetic acid, in this case; so ozonolysis of an alkyne does that.2100

    Now, if we do ozonolysis of an alkene, if we do it with just an ordinary reductive workup, we would get out 2 aldehydes in this case--see how this double bond has a hydrogen at each position?2107

    We would get benzaldehyde and acid aldehyde, in this case; but we just learned that, if you had an aldehyde, you could take these, and you can oxidize them up to the carboxylic acid.2121

    So, you could use Jones oxidation or something like that, and you could get the carboxylic acid instead.2135

    OK, or we could go straight from here to the carboxylic acids, by doing ozonolysis, and instead of doing a reductive workup where you keep these hydrogens, you could do what is known as an oxidative workup, where instead of treating it with a reducing agent, you treat it with an oxidizing agent, like hydrogen peroxide, and then, all in that one pot, you go straight to the carboxylic acid.2142

    That is kind of handy: if you know you want the carboxylic product out, there is no need to reduce and then oxidize in a separate step.2167

    We can also use organometallic reagents to create carboxylic acids, using either organolithium or Grignard.2179

    If we treat those with CO2 (with carbon dioxide), we can form a carboxylic acid.2189

    Let's take a look at that mechanism: we have either an organolithium or the Grignard; we have an R- equivalent--that is a good nucleophile.2197

    And, if we take a look at the structure of carbon dioxide, we see a carbonyl here; so that looks like it would be a good electrophile.2208

    And so, we can have it attack the carbonyl and break the π bond; and let's take a look at what product that gives us.2218

    When we follow those electrons, look what we get.2229

    We get a carboxylate salt very close to being the carboxylic acid (right?)--after step 2, H3O+ will protonate our product, and we get a carboxylic acid out.2233

    That would be a great way of synthesizing a carboxylic acid; this is good for synthesis, because we just created a new carbon-carbon bond; we used a carbon nucleophile; we used a carbon electrophile.2249

    So, where the other ones, like the oxidations, use functional group in their conversions (oxidizing carbons have already existed), this is a way of introducing a new carbon and introducing that carbon as a carboxylic acid functional group.2259

    This is the new carbon-carbon bond that was formed.2273

    Let's see an example where this might come in handy: how about if we started with this alcohol, and we wanted to form this carboxylic acid?2283

    Now, when you compare your starting material and your product, you see: we had one; now we have two carbons; so this is a bond that we need to form, in this case.2289

    So, when we do our retro-synthesis, our retro-synthesis says, "What starting materials do I need?"2299

    Well, we recall that this is actually--this bond between a carboxylic acid carbonyl and the next carbon over--this is a bond we know how to form.2310

    We know how to make it, because we just saw that.2321

    Let's take a look at the two carbons involved in this reaction; we want these two carbons to come together; one of them must have been a nucleophile; one of them must have been an electrophile.2324

    The carbon is now a carboxylic acid; that carbonyl that was my electrophile--what did he look like before the nucleophile added?2335

    This was carbon dioxide: it was a carbonyl and another carbonyl.2345

    What nucleophile means--that means this methyl group was my nucleophile somehow; how did I make it a nucleophile?--I used a Grignard.2349

    The starting materials I need: I need methylmagnesium bromide, and I need carbon dioxide.2359

    If I had these two ingredients, I would be able to make this target molecule.2366

    That is a good plan: let's look back to see where we are.2371

    We are at methanol: we need methylmagnesium bromide; so let's think about the Grignard--how do you make a Grignard?2374

    You start with the alkyl halide, and you throw in some magnesium metal.2382

    So, what I will need to do is: I will need to convert this to methyl bromide (or methyl iodide if you want--any halide will do)--but we could use TBr3 to convert the alcohol to the bromide.2386

    Then, we can add in magnesium metal, which will insert itself there and give us the methyl-magnesium bromide.2398

    And then, we are going to take that Grignard, and we are going to react with carbon dioxide, but a Grignard reaction is never just "throw in the electrophile and you are done"; remember, we need to throw in the electrophile, and then we need to work it up.2406

    That is where this proton comes from; so this is always going to be a two-step procedure.2419

    Step 1 is carbon dioxide; Step 2 is H3O+; and work that up.2422

    It's very useful to use Grignard's and carbon dioxide in synthesis of carboxylic acids.2431

    Now, finally, another way that we can form a carboxylic acid is by starting with one of the carboxylic acid derivatives and converting it into a carboxylic acid.2440

    OK, we call this reaction hydrolysis; and if you take any carboxylic acid derivative (remember, most of them look like this, with a leaving group attached to a carbonyl, but remember, the nitrile was in this category too, and this reaction would work as well)--if you take these, and you react it with water plus some acid catalyst (this also works with base)--OK, when we do water and acid, we get a hydrolysis, and the product we get is a carboxylic acid.2451

    This is really what defines something as a carboxylic acid derivative: these are all compounds that, upon hydrolysis, give a carboxylic acid as a product.2480

    Now, you will either get the carboxylic acid, or if you use base here, what is going to happen in that carboxylic acid?--it will get deprotonated, so instead of getting the neutral acid, you would get the carboxylate salt.2491

    OK, but still, it's clearly just a proton away from being a carboxylic acid.2505

    What these all have in common, including the nitrile case, is: you are taking a functional group that has 3 bonds to heteratoms (nitrogen, oxygen, halogen, right?); with hydrolysis, you convert them to be 3 bonds to oxygen.2511

    OK, so hopefully then, even the nitrile--you can see we have gone from three C-N bonds to three C-O bonds; that is defined as being a hydrolysis.2529

    Let's look at an example of an ester: that is kind of a nice derivative to start with.2541

    The ester here has an O-R group attached to the carbonyl.2547

    Let's see that reaction with sodium hydroxide and water.2552

    Now, remember, I have a little H3O+ here at the end; we need that, because I want to look at the product where I get the neutral carboxylic acid.2557

    Now, this is another example of an acyl substitution: I hinted that that was going to be the sort of mechanism we will see for carboxylic acids and their derivatives.2566

    And overall, it is going to be a mechanism where we do addition and elimination--addition of our water nucleophile and elimination of our leaving group (our leaving group is right here--we are going to be losing methanol in this reaction).2574

    OK, and let's do our mechanism: we have the ester; we have hydroxide; the ester, as usual with our carbonyls, is going to be our electrophile; the hydroxide is going to be our nucleophile.2587

    What is going to happen?--well, the hydroxide is going to attack the carbon and break the π bond.2599

    So, as usual, we are going to get a nucleophilic addition to the carbonyl, which gives us an O-.2606

    Add in our lone pairs; so we add into the carbonyl, and then where does that bring us?2616

    Oh, that brings us, actually, to an intermediate we have seen--introduced in this lesson: we have a tetrahedral carbon, where at least two of the groups (in fact, three, in this case) have a lone pair, and one of them has a charge--we call this a charged tetrahedral intermediate.2624

    What happens when you have a charged tetrahedral intermediate?--it is going to collapse.2646

    This first step was our addition step; we had addition of our nucleophile into the carbonyl; and our second step here is our elimination step, where we eliminate our leaving group.2655

    What we are going to do here is: we are going to do two arrows to do this elimination; we call this collapse of the CTI, and it is where the O- re-forms the carbonyl and kicks that leaving group out.2668

    Our product is going to be our carboxylic acid.2689

    Now, we just kicked off OCH3-; we just lost that group; and we have our carboxylic acid product addition-elimination.2694

    OK, however, let me ask: are we done here--is this our final product?2705

    We just formed a carboxylic acid; what kind of reaction conditions do we have?--we have base: we have sodium hydroxide.2710

    This is not...we are not yet done here, because we have a carboxylic acid (let me redraw this down here)--we have an acid in the presence of base, and what I hinted is that every single time you see a carboxylic acid and you are around base, you will have a deprotonation.2718

    That is a very favorable reaction, so what is going to happen is: we are going to deprotonate...very much favored in the forward direction here...and we will get the carboxylate.2745

    The carboxylate salt is formed.2762

    That is our final product in this hydrolysis reaction; our base promoted--we don't call it catalyzed, because in this final step, we actually consume an equivalent of our base; we don't get it back anymore, because this reaction is so favored in the forward direction.2768

    OK, and then finally, now you can see why we need this step 2 H3O+: we can protonate our carboxylate and get our carboxylic acid as our final product.2788

    And, in fact, this deprotonation here drives the equilibrium forward, because (you can see) every step that we are doing is reversible; we get a hydroxide added in, and then it can kick back out.2805

    Forward-reverse, forward-reverse; every one of these steps is reversible, and they are in equilibrium; but this one last step, when you form the carboxylic acid--this deprotonation is essentially irreversible.2824

    That is what keeps the reaction moving forward, moving forward, moving forward, and affecting that overall hydrolysis.2834

    Now, let's think about this overall substitution reaction that we just did here.2842

    OK, in this case, our nucleophile was hydroxide that we were bringing in, and our leaving group was methoxide.2850

    Now, that is kind of interesting, because we have never seen an example where an alkoxide was a leaving group.2859

    That is because, in the mechanisms we have seen up until now, an alkoxide was not a permissible leaving group; so for example, if we wanted to an SN1 or an SN2, or E1 or E2, for that matter, but if we wanted to do a substitution mechanism that is unimolecular or bimolecular, something like this--if we thought of this hydroxide coming in and then just kicking out methoxide as a leaving group, that would never happen.2869

    That is an impossible substitution mechanism.2898

    We could only do that substitution if we have an excellent leaving group, like chloride, bromide, iodide, halide, or maybe a tosylate.2902

    OK, so we have never seen that as a leaving group; but it is an OK leaving group for collapse of a CTI; so if we have this mechanism, where we have an O- on the same carbon as that methoxide (remember, we have a tetrahedral carbon--that is our definition of a CTI: a tetrahedral carbon with 2 groups with lone pairs), then it is possible for that methoxide to kick out, because it is not just leaving on its own--it is being pushed out.2909

    So, we have this push-pull phenomenon going on; and using the two arrows helps you to see that it is not just methoxide dropping off on its own (kind of like an SN1 mechanism might be).2940

    It is only dropping off because you have this other group that is capable of pushing it out.2955

    That gives us a carbonyl, and that gives us our methoxide leaving group.2963

    OK, the driving force for this reaction--the reason that we can use methoxide in this case (because methoxide is pretty unstable--that is pretty hard to be on its own)--the reason it is OK is because we have formation of a carbonyl.2968

    What makes collapse of a CTI great--and again, you can see if you use those two arrows--is a pushing out of that leaving group; it gives a carbonyl--forms a carbonyl--in the process.2987

    And a carbonyl is a nice, strong, stable, resonance-stabilized functional group; so any time we can form a carbonyl, that is always going to be a good mechanism.2998

    Now, this base-promoted ester hydrolysis (that is the reaction we just studied) is called a saponification reaction; it is described as a saponification reaction because it literally can be used to make soap.3010

    Now, if you think historically--let's just stop for a moment and think about that word--historically, if you wanted to make soap (or in many parts of the world, where people make their own soap), the way you do it is: you take animal fat, and you treat it with ashes from a fire, and you mix it up with water, and you cook it--you boil it.3023

    The solution you get out is a soap solution.3043

    Now, a couple questions: "How is this working?" and also "Who in the world thought of this as a good idea to wash clothes?"3046

    The fact that you are going to take some bacon grease and throw it in some hot water, grab a handful out from the fire (some of the ashes), throw it in there, and cook it up--why would you ever think that would be a good idea to wash your clothes in?3055

    This has always kind of fascinated me.3065

    And the theory on how this evolved over time is that the women who were washing clothes would maybe have a bucket of water, or a little area of water (little puddle of water) that they were using; and they would maybe use a piece of wood to agitate their clothes as they were washing in the water.3068

    What they noticed was--after a while, after washing their clothes for a while, your washing solution would become more effective.3088

    It would be easier to wash clothes at the end of the week than it was at the beginning of the week.3101

    And so, what was happening was: the wood ash was coming from the wood that they were using to agitate, and the animal fats were coming from the clothes that we were washing.3106

    We are introducing more fats into the solution.3119

    So, that is when they kind of decided, "Hey, rather than wait for our soap solution to get good by the end of the week, or the end of the month, let's just add this stuff in the beginning"; and you would make out a soap solution.3122

    Now, as organic chemists, we could take a look at the structures of these components in these molecules to see what is going on.3134

    It turns out that animal fats are actually esters, and the wood ash is a source of sodium hydroxide.3140

    And so, this is the reaction that is taking place: we get a hydrolysis reaction to take place, and what is the product when you take an ester and you treat it with sodium hydroxide in water?3149

    Well, you do a hydrolysis, and you get the carboxylic acid as usual, but because it is in base, you don't get the carboxylic acid; you get RCO2-; you get the carboxylate salt.3162

    Now, how does a carboxylate salt act as a soap, and what do the structures of these molecules look like?3182

    Let's take a look at that.3188

    The saponification reaction takes a triglyceride (this is the structure of a triglyceride--you could call it a lipid or a fat or an oil); glycerol is the tri-ol, with three carbons, with an OH on each carbon; if you mix those with carboxylic acids, these are known as fatty acids, because they have really, really long carbon chains on them that make them fatty--make them hydrophobic.3192

    These acids can be either saturated, meaning they have as many hydrogens on them as possible--there are no double bonds; or they could be unsaturated.3218

    And of course, you have heard of monounsaturated fats, polyunsaturated fats, saturated fats--these all have to do with the structures of those carbon chains that are part of the triglyceride structure.3227

    We take this triglyceride--these are the animal fats that we are starting with; we treat this with sodium hydroxide and water; what we get is addition-elimination.3239

    We get a substitution reaction, where hydroxide comes in and kicks out this alcohol on each of these ester groups.3248

    We get back a molecule of this glycerol (that is your leaving group in each case), and then these carboxylic acid groups that are freed get deprotonated, and we get these carboxylate salts--again, these long carbon chains either have double bonds, maybe, or not.3255

    Now, if you take a look at this carboxylate salt, if you have a very, very long carbon chain, we would describe that as being hydrophobic.3274

    It is a nonpolar carbon chain, so that doesn't like water.3285

    And here, this carboxylate salt is highly polar; in fact, it is ionic; so you have this hydrophilic part.3290

    If you put those into water, what happens is: they congregate into something known as a micelle.3297

    They are going to arrange themselves in water into a sphere, where the outermost part of the sphere is coated with the hydrophilic heads (the ionic part), and all of these hydrophobic tails are buried inside of the sphere to minimize their contact with water.3303

    That is called a micelle; this is a very nice model of a micelle--of course, just a toy, but it works very nicely as a model.3321

    You can imagine each one of these little rubber strings as being one of those carboxylate units, and so the outside is ionic, and it's charged, and so there is a very positive interaction with water; the inside is all of our hydrophobic, greasy long carbon chains.3329

    Now, how does this work to clean things?3348

    Well, I have my soap solution with all of these micelles floating around in it; now if I take my clothing, and I agitate it (put it in a washing machine or scrub it on a scrub board or something like that), the grease and oils that are dirty on my clothes are going to make their way onto the inside of this micelle, where it is going to be very nicely dissolved by the interior of the micelle, which is hydrophobic and nonpolar.3350

    The grease works its way inside of this; we describe it as being emulsified when the grease kind of breaks up and gets trapped inside of these micelles.3379

    And now, when I wash my clothes with fresh water, these micelles get washed away, and our grease gets washed away at the same time.3387

    That is kind of how soaps work; now, you still could make soap using animal fats (and like I said, around the world, they still do that), but we can also have synthetic detergents that have this same general structure.3395

    It is not necessarily a sodium carboxylate salt here, because that has some problems making soap scum with hard water and so on.3409

    We can have different kinds of ionic (or very polar) head here, and various structures for the tails as well; and that is how we end up with a wide variety of soaps and detergents that all have very different cleaning powers.3419

    OK, back to the real world, then: that was the base-promoted ester hydrolysis mechanism; let's take a look at that same reaction, but an acid.3436

    We have an ester here, and now we are using H3O+.3445

    And in an acid-catalyzed mechanism, we can't have any strong bases around; so there are not going to be any negative charges, like we saw in the base-promoted one with hydroxide.3451

    What do you think our first step should be in an acid-promoted reaction--acid-catalyzed reaction?3462

    I think I should probably protonate something somewhere, because I have acid around; and that is where our carbonyl is going to come into play.3467

    I could just use HA to represent H3O+; our first step is going to be to protonate the carbonyl.3474

    Now, you could protonate down on this oxygen, and that certainly will happen; but remember, every protonation is reversible, so that is not something that is going to lead your products; so that is not something we are going to concern ourselves with right now.3484

    But let's protonate the carbonyl; and by doing so, we take a carbonyl (which is a good electrophile), and we add a positive charge to it; we make it even more electron-deficient.3496

    This is now a great electrophile--a super-electrophile.3507

    So, I'm going to look around for a nucleophile to add; what nucleophile is there?--of course, there is water; so H3O+ means I have H2O and some strong acid, HA.3513

    Sometimes you will see it drawn this way; sometimes you might see it drawn this way; but either way, you certainly have water around to be your nucleophile, and that water is going to attack the carbonyl and break the π bond.3524

    It gives me an OH up here, and this oxygen from water still has the two hydrogens on it, and what else?--how many lone pairs?3545

    It has just one lone pair, because the other lone pair is right here now; it is being shared as this covalent bond, so it has one lone pair left; and that looks like a 1, 2, 3, 4, 5 oxygen one...6...looks like an O+.3554

    I am going to protonate, and then attack, and then what do I do last?--I need to get rid of that positive charge.3568

    I am going to deprotonate.3578

    Protonate, attack, deprotonate: this pattern we are going to see again and again and again.3584

    I can use water to come back in, or I can use A-, like I had in that first step--something to deprotonate.3588

    What I have done so far is: I have done my addition of my nucleophile.3604

    Remember, I want to do a substitution; so I want to add in my nucleophile, and then what do I want to do?--I need to eliminate my leaving group; I need to get rid of my leaving group.3612

    Who is my leaving group going to be?--it is going to be this methoxy group.3622

    How do I make it a good leaving group?--an acid--how do I make it a good leaving group?--I need to protonate it.3627

    I bring my HA back in and protonate the methoxy group to be a good leaving group; now I have a good leaving group, and what else do I have?3636

    How would you describe this intermediate?3656

    It looks like I have a CTI; I have a charged tetrahedral intermediate; I have a tetrahedral carbon, two or more groups of lone pairs, one of them is charged; I am in the perfect situation to do collapse of a CTI.3662

    OK, remember: two arrows to do this collapse; so in other words, don't just lose the leaving group on its own; have one of these other groups.3677

    It actually doesn't matter which one you use; have one of these other groups push it out.3685

    That is the whole point of being a CTI and doing carbonyl mechanisms: you form the carbonyl as our leaving group leaves.3690

    OK, now, what leaving group did I have?--my methanol just got kicked out, so that is one of the products of my reaction.3698

    Notice: we need to have methanol as a leaving group--we can't have methoxide, in this case, because methoxide is a very strong base and is not compatible with acidic conditions.3705

    Remember: no O- (or CH3O-, in this case) in acid.3714

    I need to protonate first; that gives me a nice CTI; that gives me a good leaving group, and now I can collapse.3722

    OK, and now, when I do that with my two arrows, look how close I am; all I need to do is deprotonate up here, so I can use water to come back in, or I could use A- (I keep switching back and forth--sorry).3727

    I used HA down here, so now I could use A- to come in and deprotonate, and I'm done; I have done my hydrolysis of an ester to give a carboxylic acid.3739

    It is going to be common that acid-catalyzed mechanisms are going to be significantly longer than the base-catalyzed ones (or base-promoted ones); so the base ones are good to work on initially; get the hang of it; understand the addition-elimination and the use of CTIs.3754

    In those, we are working on negative charges; in acid, we are going to have positive charges; and the same general idea is going to happen.3769

    At some point, we are adding in our nucleophile; at some point, we are losing our leaving group; there are just a lot of other protonations and deprotonations mixed in to avoid our negative charges.3775

    Let me just point out: we have seen the acid-catalyzed; we have seen the base-promoted; I want to point out that one of those is required (some sort of catalysis is required--either acid or base conditions) in order for this hydrolysis to take place.3787

    OK, we cannot take a neutral ester (which is a weak electrophile) and neutral water (which is a weak nucleophile) and expect the two to come together.3801

    If we tried that...let's try it: just have neutral water attack a neutral ester; look at what we get.3815

    I now have an O- up here; and down here, I have an O+; OK, that is when you know you have made a mistake in a mechanism, because I can't have a strongly basic O-, ever, in the same sort of reaction conditions that I have a strongly acidic O+.3824

    There is no O- and O+ in the same mechanism: that is good to keep in mind, and that is good to help guide and help you pick up when you have maybe made a mistake.3845

    This is not going to happen; so instead, how does it work?3859

    Well, in basic additions, you increase the nucleophilicity of something; so how does this get adjusted in base?3862

    You don't have water attacking; you have a strong nucleophile, hydroxide, attacking.3872

    OK, then you have an O- that attacks; you have another O-; and so on.3880

    OK, how about an acid--how would I change this in acid to make the mechanism acceptable?3884

    Well, in acid, our very first step is protonating the carbonyl; so in acid, we don't have a neutral ester; we have a protonated ester.3890

    We have a protonated carbonyl, which is an excellent electrophile; so in base, you find negative charges and strong nucleophiles; in acid, you find positive charges and strong electrophiles.3902

    We are going to see this pattern coming in again and again and again; so this is an excellent time for you to get used to that.3916

    Now, let's just take a look at a nitrile example, because the nitrile looks a little different than the ester.3924

    All of the other carboxylic acid derivatives would have essentially the same mechanism as the ester, but the nitrile looks a little different; so let's see how we would handle that.3929

    Now, first of all, when we do a hydrolysis on an amide, there are two possible products you can get.3937

    If you do just a mild hydrolysis, we can replace two of the C-N bonds with C-O bonds, resulting in an amide product.3945

    We would describe this as being partial hydrolysis; and if we wanted to have more vigorous conditions (where we have, let's say, H3O+ and heat), we wouldn't stop at the amide; we would continue doing a hydrolysis of the amide.3956

    We know the amide is a carboxylic acid derivative, as well, right?--so it can also undergo hydrolysis; so we have addition-elimination, addition of water and elimination of the nitrogen leaving group.3973

    We would ultimately get a carboxylic acid product, OK?3983

    But this is kind of handy, because this is another way that we can create an amide (we will see down the line): by partial hydrolysis of a nitrile.3986

    Let's see if we can do this mechanism: this mechanism is just like the ester--very similar to the ester.3993

    We have already something on how to do this transformation; but let's see how we can go from a nitrile to an amide--that is a little less intuitive.4003

    We have acetonitrile (this nitrile); we have H3O+; our first step, I am thinking, is going to be to protonate, because we have an acid.4015

    Let's write that out: so protonate first; and where do we protonate?--well, the best course of action is to treat the nitrile, the C-N triple bond, just like it was a carbonyl.4033

    Do exactly what you would expect a carbonyl to do: I'm going to take a lone pair on nitrogen and protonate.4045

    Just like I would protonate a carbonyl, I can protonate a nitrile.4055

    And, just like protonation of the carbonyl made it a great electrophile, protonation of the amide also makes it a great electrophile, so we can look around for a nucleophile--we'll have water here, in this hydrolysis reaction, as our nucleophile.4061

    How are they going to react?--well, just like they would for a carbonyl: this oxygen is going to attack the carbon and break the π bond--totally analogous to a carbonyl.4077

    This oxygen still has two hydrogens on it, and a lone pair, so it's going to be an O+; and we now have a double bond between the carbon and the nitrogen, instead of a triple bond.4092

    OK, so what do we do?--we protonated, and then we attacked, and now we can deprotonate.4106

    The same pattern we have seen again and again for acid-catalyzed: protonate, and then attack, and now we can bring water back in as our base and deprotonate.4120

    OK, well, this doesn't seem like we are a lot closer, perhaps, but let me show where we are and think about where we are going.4137

    The initial product--the first intermediate product that we get in this hydrolysis is going to be the amide.4147

    Let me draw the amide (an upside-down amide); let's see if you can now think a little more clearly and identify what it is we have to do.4156

    This is a double bond with an OH here, and we are converting that to a carbonyl.4170

    OK, this structure is kind of like an -enol structure, which you may or may not have seen yet in a keto-enol tautomerization.4181

    It is a very similar mechanism; what we need to do is tautomerize--I'll just put this in parentheses, because you may not have seen that topic yet for ketones and aldehydes.4194

    But either way, when we compare these two structures, what we see is that here we have an NH, and now we have NH2; so one thing we have to do is--we have to protonate here.4207

    All right, at some point we have to protonate here, and down on the oxygen we have an OH, and it ends up being just an oxygen.4219

    So another thing we need to do is: we need to deprotonate here.4226

    It turns out that those are the exact two steps that we have to do for the remainder of this mechanism: we have to protonate and deprotonate, because we are in acidic conditions.4234

    Just like, up here, we protonated first, we are going to do the same thing here.4245

    We are going to protonate first, and then we are going to deprotonate second.4249

    How can I protonate this nitrogen?--well, you can think of using the lone pair, but our mechanism is going to be a little cleaner if we use the π bond.4254

    I can bring in another equivalent of H3O+ and protonate the π bond.4265

    Just like protonating the alkene, you break the π bond; you add a proton to one atom; and you get a positive charge on the other atom.4281

    And so, I protonate to get a carbocation; that was step 1.4291

    Step 2: I need to deprotonate--let's look at where we are and where we have to go--do you see how close we are?4295

    Let's bring water in as our base and grab that proton; and instead of having these electrons go and sit on the oxygen, it's going to be an O- and a C+.4304

    I'll go straight to the better resonance form, where those two electrons go to be a π bond; and we are done.4314

    We have formed an amide.4322

    OK, so we start out similar to what we would do for a carbonyl: protonate, attack, deprotonate; OK, but then, the second step is a conversion of one structure to another structure, and knowing where you are headed is really the key to getting this mechanism down correctly.4326

    OK, now the conversion of the amide to a carboxylic acid--we'll just say et cetera--that is going to be the same analogous one as the ester: we are going to do addition of the water and elimination of the nitrogen leaving group, in this case.4342

    Now, what is really cool about having a nitrile and knowing that a nitrile can be converted to either an amide or to a carboxylic acid is: we can use this--we can exploit this in synthesis, because cyanide is (there is a little lone pair here, too) a good nucleophile.4361

    There are strategies to get the nitrile functional group into a carbon chain; so let's see an example where that might come in handy.4379

    We have a 3-carbon chain here, and we want to convert it to this new compound; we still see our 3-carbon chain, but we see that there is this new carbon that is added on, and we need to synthesize that.4389

    OK, now let's think about our retrosynthesis: what starting materials do I need?4405

    There are actually a few different approaches we can have for this.4412

    One of them would be the nitrile that we just saw; so one approach is--we just saw that, where you now have a carboxylic acid, that could have come from a nitrile.4416

    In other words, if I had this nitrile, I could do a hydrolysis to get this carboxylic acid product.4430

    OK, so that would be a good strategy; now, we have to do this disconnection and ask ourselves, if we wanted these two carbons to come together and react, how could we do that?4437

    One of them must have been a nucleophile; one of them must have been an electrophile.4448

    Well, what we are remembering is that cyanide is a good nucleophile; so this certainly was my nucleophile, but that means this carbon was my electrophile.4452

    Somehow, I have to make that carbon electrophilic.4461

    Well, I'm starting with isopropyl chloride; I am starting with a leaving group in that position; it is already electrophilic.4464

    I think I have done enough planning here to think about my synthesis; so I can go ahead and do my transformation.4471

    I think my first step is to treat this alkyl halide with sodium cyanide to do an SN2 mechanism.4478

    That will replace the leaving group with the cyano group.4487

    And once I have the cyano group, now I can convert it to the carboxylic acid by hydrolysis; I want to trade those C-N bonds for C-O bonds; that is hydrolysis.4494

    H3O+...remember, some heat is required, because you want it to not stop at the amide; we want it to go all the way to the carboxylic acid: that is kind of a forcing reaction.4504

    OK, so nitriles are very handy this way, as a way to synthesize carboxylic acids.4514

    One last look at this problem: there is another strategy that we can think of; we have actually seen this as another way to make carboxylic acids--it would be exactly this disconnection, as well.4520

    What if we asked about these two carbons right away?--think about, instead of doing a functional group in a conversion right here back to a nitrile--what if we did a disconnection right away and asked about these two carbons?4533

    Which one could have been a good nucleophile?--which one was an electrophile?--what reaction have we seen that comes together and forms that carboxylic acid?4547

    How about if this carbon was my electrophile--what would that electrophile look like--a carbon with two oxygens?4559

    We could have carbon dioxide as an electrophile; what nucleophile would react with carbon dioxide to give this target molecule, then?4567

    How about if I had a Grignard as my nucleophile?--then that would react with CO2, and it would give this target molecule exactly.4579

    OK, so that is another approach that we can take; let's see how that would look.4589

    Well, we are starting with a chloride; I know, with a Grignard, we usually think magnesium bromide, but remember: a Grignard can be any halide, so we don't have to change that.4592

    We could just throw in our magnesium right away and make the magnesium chloride Grignard.4601

    Now, I have a good nucleophile; what did I want to do with that nucleophile?--I want it to react with the carbon dioxide.4610

    Remember, that is a two-step process: first, I could add in carbon dioxide; second, I could do an H3O+ workup to protonate; and so, this would give another strategy.4617

    In this lesson, we have been introduced to carboxylic acids, looked at their acidity and their other physical properties, and then we have also seen different ways to synthesize carboxylic acids.4627

    We looked at oxidation reactions; we looked at Grignard reactions to form them; and then, we looked at carboxylic acid derivative hydrolysis as another way to form carboxylic acids.4639

    We studied the mechanism of that hydrolysis reaction, and in the next lesson, we are going to look again at these carboxylic acid derivatives and think about "What reactions do they undergo?" and "How could we synthesize each of those functional groups?"4651

    I look forward to seeing you then; thanks for visiting Educator.com.4667