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Lecture Comments (5)

2 answers

Last reply by: Francesco Frigo
Mon Jul 7, 2014 9:55 AM

Post by Francesco Frigo on July 4, 2014

Hello dr. Starkey, I have one question regarding the last step in th synthesis of the epoxyketone. At minute 136:50 you use the MCPBA as an oxyating agent to transform the ketone in an epoxyde. But wouldn't this reaction cause an attack to the carbonyl too, generating a trasposition known as the Bayer Villiger reaction? that is why I used a different approach involving the haloidrine, a cloroidrine actually. I trasformed the ketone using cl2 and water in a cloroidrine (H20 adds to the most substituted Carbon, Cl- adds to the less substitute C) and then, with the use of a base, the newly generated alcol does a Sn2 to the clorine closing the epoxyde ring. But maybe the MCPBA approach works in this particular situation, that's why I'm asking here. Thank you in advance!

1 answer

Last reply by: Professor Starkey
Fri Jun 27, 2014 11:05 PM

Post by Luke Memet on June 26, 2014

Thank you for your outstanding work Dr. Starkey, this is tremendously helpful!

Organic Synthesis Strategies

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
    • Organic Synthesis Strategies
    • Example of a RetroSynthesis
    • Synthesis of Alcohols by Functional Group Interconversion (FGI)
    • Alcohols by Reduction
    • Alcohols by Hydration of Alkenes
    • Alcohols by Substitution
    • Synthesis of Alcohols by Forming a New C-C Bond
    • Other Alcohol Disconnections
    • Synthesis of Alkyl Halides
    • Synthesis of Alkyl Halides by Free Radical Halogenation
    • Synthesis of Alkyl Halides by Substitution
    • Synthesis of Alkyl Halides by Addition
    • Example: Synthesis of Alkyl Halide
    • Synthesis of Ethers
    • Example: Synthesis of an Ether
    • Synthesis of Amines
    • Gabriel Synthesis of Amines
    • Amines by SN2 with Azide Nu:
    • Amines by SN2 with Cyanide Nu:
    • Amines by Reduction of Amides
    • Reductive Amination of Ketones/Aldehydes
    • Example : Synthesis of an Amine
    • Synthesis of Alkenes
    • Synthesis of Alkenes by Elimination
    • Synthesis of Alkenes by Reduction
    • Synthesis of Alkenes by Wittig Reaction
    • Example: Synthesis of an Alkene
    • Synthesis of Alkynes
    • Synthesis of Alkynes by Elimination (FGI)
    • Synthesis of Alkynes by Alkylation
    • Example: Synthesis of an Alkyne
    • Synthesis of Alkanes
    • Synthesis of Aldehydes & Ketones
    • Synthesis of Aldehydes & Ketones by α-Alkylation
    • Acetoacetate Ester Synthesis of Ketones
    • Example: Synthesis of a Ketone
    • Synthesis of Carboxylic Acids
    • Example: Synthesis of a Carboxylic Acid
    • Malonic Ester Synthesis of Carboxylic Acid
    • Example: Synthesis of a Carboxylic Acid
    • Synthesis of Carboxylic Acid Derivatives
    • Alternate Ester Synthesis
    • Practice: Synthesis of an Alkyl Chloride
    • Patterns of Functional Groups in Target Molecules
    • Patterns of Functional Groups in Target Molecules
    • Patterns of Functional Groups in Target Molecules
    • 2-Group Target Molecule Summary
    • Example: Synthesis of Epoxy Ketone
    • Example: Synthesis of a Diketone
    • Intro 0:00
    • Organic Synthesis Strategies 0:15
      • Goal
      • Strategy
    • Example of a RetroSynthesis 1:30
      • Finding Starting Materials for Target Molecule
      • Synthesis Using Starting Materials
    • Synthesis of Alcohols by Functional Group Interconversion (FGI) 6:00
      • Synthesis of Alcohols by Functional Group Interconversion Overview
    • Alcohols by Reduction 7:43
      • Ketone to Alcohols
      • Aldehyde to Alcohols
      • Carboxylic Acid Derivative to Alcohols
    • Alcohols by Hydration of Alkenes 9:28
      • Hydration of Alkenes Using H₃O⁺
      • Oxymercuration-Demercuration
      • Hydroboration Oxidation
    • Alcohols by Substitution 11:42
      • Primary Alkyl Halide to Alcohols Using NaOH
      • Secondary Alkyl Halide to Alcohols Using Sodium Acetate
      • Tertiary Alkyl Halide to Alcohols Using H₂O
    • Synthesis of Alcohols by Forming a New C-C Bond 15:47
      • Recall: Alcohol & RMgBr
      • Retrosynthesis
    • Other Alcohol Disconnections 19:46
      • Synthesis Using PhMGgBr: Example 2
    • Synthesis of Alkyl Halides 26:06
      • Synthesis of Alkyl Halides Overview
    • Synthesis of Alkyl Halides by Free Radical Halogenation 27:04
      • Synthesis of Alkyl Halides by Free Radical Halogenation
    • Synthesis of Alkyl Halides by Substitution 29:06
      • Alcohol to Alkyl Halides Using HBr or HCl
      • Alcohol to Alkyl Halides Using SOCl₂
      • Alcohol to Alkyl Halides Using PBr₃ and Using P, I₂
    • Synthesis of Alkyl Halides by Addition 32:02
      • Alkene to Alkyl Halides Using HBr
      • Alkene to Alkyl Halides Using HBr & ROOR (Peroxides)
    • Example: Synthesis of Alkyl Halide 34:18
      • Example: Synthesis of Alkyl Halide
    • Synthesis of Ethers 39:25
      • Synthesis of Ethers
    • Example: Synthesis of an Ether 41:12
      • Synthesize TBME (t-butyl methyl ether) from Alcohol Starting Materials
    • Synthesis of Amines 46:05
      • Synthesis of Amines
    • Gabriel Synthesis of Amines 47:57
      • Gabriel Synthesis of Amines
    • Amines by SN2 with Azide Nu: 49:50
      • Amines by SN2 with Azide Nu:
    • Amines by SN2 with Cyanide Nu: 50:31
      • Amines by SN2 with Cyanide Nu:
    • Amines by Reduction of Amides 51:30
      • Amines by Reduction of Amides
    • Reductive Amination of Ketones/Aldehydes 52:42
      • Reductive Amination of Ketones/Aldehydes
    • Example : Synthesis of an Amine 53:47
      • Example 1: Synthesis of an Amine
      • Example 2: Synthesis of an Amine
    • Synthesis of Alkenes 58:20
      • Synthesis of Alkenes Overview
    • Synthesis of Alkenes by Elimination 59:04
      • Synthesis of Alkenes by Elimination Using NaOH & Heat
      • Synthesis of Alkenes by Elimination Using H₂SO₄ & Heat
    • Synthesis of Alkenes by Reduction 1:02:05
      • Alkyne to Cis Alkene
      • Alkyne to Trans Alkene
    • Synthesis of Alkenes by Wittig Reaction 1:03:46
      • Synthesis of Alkenes by Wittig Reaction
      • Retrosynthesis of an Alkene
    • Example: Synthesis of an Alkene 1:06:57
      • Example: Synthesis of an Alkene
      • Making a Wittig Reagent
    • Synthesis of Alkynes 1:13:09
      • Synthesis of Alkynes
    • Synthesis of Alkynes by Elimination (FGI) 1:13:42
      • First Step: Bromination of Alkene
      • Second Step: KOH Heat
    • Synthesis of Alkynes by Alkylation 1:15:02
      • Synthesis of Alkynes by Alkylation
      • Retrosynthesis of an Alkyne
    • Example: Synthesis of an Alkyne 1:17:40
      • Example: Synthesis of an Alkyne
    • Synthesis of Alkanes 1:20:52
      • Synthesis of Alkanes
    • Synthesis of Aldehydes & Ketones 1:21:38
      • Oxidation of Alcohol Using PCC or Swern
      • Oxidation of Alkene Using 1) O₃, 2)Zn
      • Reduction of Acid Chloride & Nitrile Using DiBAL-H
      • Hydration of Alkynes
      • Synthesis of Ketones by Acyl Substitution
      • Reaction with R'₂CuLi
      • Reaction with R'MgBr
    • Synthesis of Aldehydes & Ketones by α-Alkylation 1:28:00
      • Synthesis of Aldehydes & Ketones by α-Alkylation
      • Retrosynthesis of a Ketone
    • Acetoacetate Ester Synthesis of Ketones 1:31:05
      • Acetoacetate Ester Synthesis of Ketones: Step 1
      • Acetoacetate Ester Synthesis of Ketones: Step 2
      • Acetoacetate Ester Synthesis of Ketones: Step 3
    • Example: Synthesis of a Ketone 1:34:11
      • Example: Synthesis of a Ketone
    • Synthesis of Carboxylic Acids 1:37:15
      • Synthesis of Carboxylic Acids
    • Example: Synthesis of a Carboxylic Acid 1:37:59
      • Example: Synthesis of a Carboxylic Acid (Option 1)
      • Example: Synthesis of a Carboxylic Acid (Option 2)
    • Malonic Ester Synthesis of Carboxylic Acid 1:42:34
      • Malonic Ester Synthesis of Carboxylic Acid: Step 1
      • Malonic Ester Synthesis of Carboxylic Acid: Step 2
      • Malonic Ester Synthesis of Carboxylic Acid: Step 3
    • Example: Synthesis of a Carboxylic Acid 1:44:53
      • Example: Synthesis of a Carboxylic Acid
    • Synthesis of Carboxylic Acid Derivatives 1:48:05
      • Synthesis of Carboxylic Acid Derivatives
    • Alternate Ester Synthesis 1:48:58
      • Using Fischer Esterification
      • Using SN2 Reaction
      • Using Diazomethane
      • Using 1) LDA, 2) R'-X
    • Practice: Synthesis of an Alkyl Chloride 1:53:11
      • Practice: Synthesis of an Alkyl Chloride
    • Patterns of Functional Groups in Target Molecules 1:59:53
      • Recall: Aldol Reaction
      • β-hydroxy Ketone Target Molecule
      • α,β-unsaturated Ketone Target Molecule
    • Patterns of Functional Groups in Target Molecules 2:03:15
      • Recall: Michael Reaction
      • Retrosynthesis: 1,5-dicarbonyl Target Molecule
    • Patterns of Functional Groups in Target Molecules 2:06:38
      • Recall: Claisen Condensation
      • Retrosynthesis: β-ketoester Target Molecule
    • 2-Group Target Molecule Summary 2:09:03
      • 2-Group Target Molecule Summary
    • Example: Synthesis of Epoxy Ketone 2:11:19
      • Synthesize the Following Target Molecule from Cyclohexanone: Part 1 - Retrosynthesis
      • Synthesize the Following Target Molecule from Cyclohexanone: Part 2 - Synthesis
    • Example: Synthesis of a Diketone 2:16:57
      • Synthesis of a Diketone: Step 1 - Retrosynthesis
      • Synthesis of a Diketone: Step 2 - Synthesis

    Transcription: Organic Synthesis Strategies

    Hi, welcome back to Educator.0000

    We are going to talk, today, about organic-synthetic strategies; now that we know a lot about organic transformations that we are capable of, how do we apply those to building specific target molecules?0002

    The goal of a synthesis is the synthesis of a target molecule (so if you ever see the letters t.m., that is what it's referring to; that is the goal--the organic compound we are trying to prepare).0016

    And the way that we typically approach a problem like this is: we start by looking at the target molecule and dissecting it--thinking about what disconnections we can make: how can we break it down?0028

    And that process is described as performing a retrosynthesis.0039

    The goal is to use commercially available starting materials; so we want to keep working backwards to get simpler and simpler starting materials, until finally we have something that we know we can acquire commercially (and hopefully, inexpensively).0044

    And what we are going to do is: we are going to use known functional group interconversions.0056

    So, we know about reductions that are possible, and oxidations, eliminations, substitutions...the kinds of reactions we have seen; so we are going to use reactions that we are familiar with.0060

    And, in order to do that, when we do our retrosynthesis and our disconnections, what we are going to be looking for are recognizable nucleophiles and electrophiles.0072

    Nucleophiles are the electron donors; electrophiles are the electron acceptors; and we are going to be looking for ones that we have seen before, that are familiar to us.0082

    So, for example, if this was my target molecule (I wanted to make this nitrile), and I wanted to plan, what is a way that I could synthesize this compound?0091

    I am going to perform a retrosynthesis: now, we use this special arrow--this double-lined arrow (it's called a retrosynthetic arrow); and it asks the question, "What starting materials do I need?"0100

    So, in other words, we are asking, "What could I have started with?"--and using a reaction that we are familiar with in order to create, synthesize, this carbon structure.0112

    OK, and what we are going to be looking for, in this case, is a good disconnection: in other words, what bond was likely to have been formed in the synthesis reaction?0124

    And it is probably going to be this carbon-carbon bond; it's the most likely one that we can start with, because we have this functional group here--the cyano group--and that is a functional group that we can add in.0135

    So, if we look at these two carbons that are involved (we are going to be forming a new carbon-carbon bond to form this product), we have to recognize that, as the starting materials, one of those carbons must have been a nucleophile; one of those carbons was the electrophile.0146

    That is the only way those two are going to come together and form a bond.0161

    So, if we remember the cyano group--we have seen the cyano group as cyanide (Cn-); so we could say that this carbon was the nucleophile, which means, somehow, this carbon was the electrophile.0164

    One thing we can do is: we can simply break that bond; we are just going to erase that bond, and at each position, we are going to put a charge that is associated with being a nucleophile or electrophile.0178

    To be a nucleophile (the electron donor), I have a negative charge; and to be an electrophile, one thing you could do is put a positive charge there.0192

    Now, this first possible step that we are doing to create what is known as synthons.0202

    Now, these are imaginary structures that have the perfect reactivity: in other words, if I had this carbocation and this cyanide nucleophile, clearly they would come together and form the bond to make our target molecule.0208

    And, in some cases, the synthons are recognizable and reasonable reagents we can use; so in other words, cyanide is a stable anion; we could simply purchase sodium cyanide and use that as a nucleophile.0223

    But this carbocation is an impossible reagent to use; you can't go to the stockroom and ask for a carbocation, even if it's a stable carbocation.0239

    Carbocations are just fleeting intermediates, not stable reagents.0246

    So, we need to ask, "Well, what is the compound--the stable organic molecule--that has the reactivity of this carbocation?"0253

    We have a 4-carbon chain; and off this end carbon, we want it to be electrophilic; we want it to be electron deficient.0263

    Well, a lot of times, what we can do--the approach we are going to take is: we can add a leaving group.0269

    And so, for example, if I put on a bromine or a chlorine or something like that, or an iodide, now we have an alkyl halide...a recognizable, stable electrophile; it has a leaving group, and that is something that would be attractive, and could react with the cyanide; and what reaction do you think would happen here?0275

    Let's look at our synthesis, now that we have planned it...that is called the retrosynthesis; this is all the imaginary planning we are doing.0297

    Just like, if you are planning a road trip from New York to LA, you don't just hop in the car and go (you first look it up; you plan your trip, and then you can take the trip), what we have done is: we have planned our synthesis with a retrosynthesis.0303

    And now, we can perform the synthesis: we said we need butylbromide, and we are going to treat this with sodium cyanide, and what reaction is going to happen there?0318

    Sodium cyanide, of course, is a source of Cn-, and it looks like we have perfect conditions for a back side attack, SN2.0327

    Attack the carbon; kick off the leaving group; and voilá, we have our target molecule.0336

    OK, so what we are going to do is: we are going to be analyzing various target molecules, based on the functional groups present; and we are going to be looking for reasonable reactions, reasonable disconnections, reasonable interconversions we can do that will give that target molecule as a product.0342

    We are going to study one functional group at a time.0358

    So, for example, if we wanted to synthesize an alcohol, what we want to ask is, "What reactions have we seen that give alcohols as products?"0361

    OK, one way we can do that--many reactions we have seen are called functional group interconversions, or FGI's for short--in other words, not changing the carbon chain at all, just swapping out one functional group for another.0371

    So, for example, we have seen that, if we want to make an alcohol (so if this is our target molecule), we can think of several different retrosyntheses.0384

    Remember, this retrosynthesis arrow asks, "What starting materials do I need?"0394

    We are imagining--"What reaction have I seen--what could I have started with that could be converted into an alcohol?"0400

    For example, we have seen that if you have a carbonyl, that could be converted to an alcohol; that looks like a reduction reaction, because I started with two C-O bonds, and now I have only one C-O bond.0406

    And what kind of reagent would do that?0417

    If I had this ketone, and I wanted to convert it to the alcohol, the reagents I would need would be something like maybe lithium aluminum hydride, followed by workup.0419

    A hydride source would convert a carbonyl to an alcohol; so that is one way we could make an alcohol.0433

    Another way we can make an alcohol is to start with an alkene.0438

    If we have an alkene, and we added water across the π bond (so hydration of the alkene), that would also give an alcohol product.0441

    And there are many different hydration conditions we have for that.0448

    Or maybe we could do a substitution: in other words, if we have a leaving group already on the carbon chain, then we can replace that leaving group with a hydroxyl group.0452

    Let's look at these one by one.0463

    If we have any kind of carbonyl--a ketone or an aldehyde or any one of the carboxylic acid derivatives (well, not all of the carboxylic acid derivatives, but if we have an ester or an acid or an anhydride or an acid chloride--most of the carboxylic acid derivatives), if we treat it with lithium aluminum hydride, that is a source of H-, and that H-, of course, is going to attack the carbonyl and break the π bond.0465

    After workup, we are going to end up with an alcohol.0491

    Let's look at the ketone case; that would give us a secondary alcohol, because this carbon, bearing the OH, has 2 carbon groups on it.0495

    If we started with an aldehyde, that would give us a primary alcohol, because there is only one carbon group attached to the carbon.0506

    And, if we start with a carboxylic acid derivative, remember what happens to the carboxylic acid derivative: when hydride attacks, it breaks the π bond, but rather than stay at that O-, we get a charged tetrahedral intermediate.0516

    The CTI that collapses kicks back down and kicks off the leaving group; carboxylic acid derivatives have leaving groups.0529

    So, we do an addition-elimination reaction; and then, hydride can add a second time.0537

    What we end up with is the addition of two equivalents of hydride added.0543

    And we end up with a primary alcohol again.0552

    OK, so just a review of some of the various reactions we have seen that lead to alcohol products; we could start with a more oxidized form of carbon (ketone, aldehyde, carboxylic acid), and then reduce it.0557

    Now, if we wanted to hydrate an alkene, we have seen several different methods for that.0570

    One is simple addition of H3O+; so that is going to break the π bond; it's going to add an H and an OH (the components of water) across the π bond; and we learned about the regiochemistry of that.0574

    Remember Markovnikov's Rule?--Markovnikov's Rule says that the hydrogen goes to the carbon with more hydrogens; and so, the product we would expect would be the more substituted alcohol; so we call that Markovnikov addition.0585

    Because, when we do our synthesis, we have to make sure that the synthesis is occurring with the proper regiochemistry to put the functional groups exactly where we want them.0604

    An alkene has two carbons that are reactive; we want the alcohol to be in only one of those positions.0611

    Now, this mechanism involves a carbocation; and remember, carbocations can rearrange, and we can have some other issues going on.0617

    These H3O+ conditions are not suitable for all alkenes (hydration of alkenes), so another one we could do--this is called oxymercuration reduction, or oxymercuration demercuration.0629

    And this synthesis--this two-step procedure--is just a more controlled way to do the same transformation--a Markovnikov addition of water across the π bond.0642

    Synthetically, that is a little better, more reliable technique to use, rather than just acid and water.0655

    And then, you remember this one: this is called hydroboration oxidation.0663

    Hydroboration oxidation--what was special about this reaction, this 2-step sequence, is that it added water, but it did it in anti-Markovnikov orientation.0666

    So, we add the hydrogen to the carbon with the less hydrogens, and the OH to the other end.0677

    This is very nice, synthetically, because now, we have tools to control our regiochemistry and add the alcohol to one carbon or the other, and get a few different target molecules as a result.0685

    OK, what if we wanted to do substitution?--well, that means we need a good leaving group on a carbon (like a bromine).0703

    And, simply using sodium hydroxide (that is a very good nucleophile, and so, we would expect, in this case, that it would do a back side attack, SN2), in fact, we could get this primary alcohol.0711

    If we start with a primary alkyl halide, meaning it's on a primary carbon, this is very little steric hindrance; this is a great SN2.0726

    This is a great way to make a primary alcohol.0734

    But, if we wanted to try the same reaction on a secondary alcohol, now we have more sterics; and remember, back side attack is very sensitive to sterics; and hydroxide is not only a good nucleophile, but it's also a strong base.0738

    The reaction that competes and actually wins is going to be the elimination reaction instead: E2 is major.0754

    That is going to be a problem if we are trying to synthesize the alcohol: we can't use hydroxide and do an SN2 reaction.0764

    So, what can we do instead?0772

    Well, we have some tools that we can employ to get to a desired product by using a slightly different route (a little detour, if you will).0773

    Here is what we are going to do: instead of using hydroxide as our nucleophile, we are going to use sodium acetate.0786

    OK, now sodium acetate--still a good nucleophile--would love to do an SN2, and the difference here (let's draw the product)--now, the difference here is that sodium acetate is a weaker base than hydroxide.0792

    Can you see why it would be a weaker base?0810

    What does this O- have to stabilize it and make it less reactive, that this O- doesn't?0812

    It has resonance; this is resonance-stabilized.0819

    OK, so that means it's less reactive, so it's a weaker base, so the E2 does not compete as much in this case, and we get a good yield of the SN2 product.0827

    OK, now how does that help us?--well, if we wanted to put an OH in this position, we have succeeded in replacing the bromine with an oxygen; and now, if we treat it with hydroxide and water, we could do hydrolysis of this ester.0836

    We just made an ester, and we can convert that ester to the alcohol by clipping off this carbonyl group.0851

    Now, how is that working?--well, hydroxide, again, can come in as a nucleophile (kind of like we saw hydride doing with LAH); it adds into the carbonyl, but then that CTI comes back down and collapses.0859

    And we are going to get, as a final product, our alcohol.0871

    We describe sodium acetate as a synthetic equivalent: we are going to use that word quite a bit--a synthetic equivalent of hydroxide, because it ultimately gives us the same product we would have gotten if we had used hydroxide as our nucleophile.0875

    And so, that is why we call it an equivalent.0895

    But, it enables us to do a reaction that hydroxide itself would not be capable of doing.0898

    OK, so these are all using a strong nucleophile--hydroxide or acetate; if we have a tertiary alkyl halide, we can't do an SN2 mechanism on it, because it's too hindered, but we could use a weak nucleophile (like water) and do SN1.0904

    In other words, a leaving group can just leave to form a carbocation; then water can attack, and so, ultimately, we can get an alcohol substitution taking place.0922

    But, this is going to be a different mechanism; so if you have an alcohol on a highly-substituted carbon, then one possibility would be maybe to start with an alkyl halide in that position and use an SN1 mechanism and a weak nucleophile.0931

    Now, another way to synthesize an alcohol is by forming a new carbon-carbon bond; so the reactions we have just looked at are functional group interconversions (FGI).0949

    So, in other words, this is not by functional group interconversion.0960

    FGI means our carbon chain is already fixed, and all we are doing is replacing a halide with an OH, or replacing a carbonyl with OH, or replacing a double bond with an OH, right?--that is changing one functional group to another.0965

    But when we are talking about organic synthesis, what we are really interested in is synthesizing new organic compounds with new carbon chains.0977

    So, we have to think about reactions that form new carbon-carbon bonds: those are really going to be the most interesting to us.0985

    And so, how can we do that--what reaction have we seen that forms new carbon-carbon bonds that give alcohols as products?0992

    Well, we have seen the reaction of using a Grignard. A Grignard is an equivalent of an R-; that is a very good nucleophile; and so, that would attack the carbonyl.0999

    And, if we followed it up with step 2, doing an aqueous workup (H3O+), we could protonate (let's put on our lone pairs) that O-, and that would give us an alcohol product.1015

    This reaction results in an alcohol, and it also, at the same time, forms a new carbon-carbon bond.1033

    Really, this is an ideal synthetic technique to use, if we want to create alcohols.1042

    What does the retrosynthesis look like?--remembering that that reaction exists, now, if we look at an alcohol target molecule, rather than just trading the alcohol for some other functional group, we could think about making an actual disconnection, breaking a carbon-carbon bond.1049

    What we are doing is: we are disconnecting the bond that is going to be formed in the forward direction.1067

    I have this R group shown here; of course, any of these carbon groups can be disconnected, but let's look at this R group.1071

    Now again, if we are disconnecting this bond, we look at these two carbons, and we say, "In order for those two to come together, one of them started out as a nucleophile; one of them started out as an electrophile."1081

    Who is who--which is which?1091

    Well, the carbon that now has the single bond, OH--what did that look like as a starting material?1094

    That used to be the carbonyl, the C, double bond, O; so this carbon was the electrophile as a carbonyl, which means this carbon was the nucleophile.1100

    If it's going to react with an electrophile, it must have been nucleophilic.1113

    How do you make an R group, a random carbon chain, nucleophilic?1116

    You attach a metal to it and use an organometallic reagent.1120

    So, our again, if you want to look just at the synthons, we can do that; this is a good example, where we can look at the synthons.1125

    Let's just break the bond, move the electrons to the nucleophile, and now we have an R- and a C+; these are our imaginary synthons.1134

    And so, we say, "OK, well, what reagents do we have that actually have this reactivity?"1146

    OK, we can kind of think of this lone pair kind of coming back down and filling in that vacancy, and then it brings us to a recognizable starting material, a ketone; that can be our electrophile.1152

    And an R---the way we make an R- substrate is: we use a Grignard.1167

    So now, we have come back; our goal, as usual, is to come back to recognizable, simple starting materials, with the desired reactivity, and they would be able to come together to form the target molecule.1175

    Let's look at a few other examples of slightly different alcohol disconnections, still using Grignard reagents.1188

    Let's use phenylmagnesium bromide as one of our starting materials in our synthesis.1195

    But let's look at this target molecule and think about this disconnection: what is unique about this one is: the alcohol carbon has two phenyl groups--has two identical groups attached to the alcohol carbon.1202

    We have a special disconnection we can do for that: we can actually disconnect it twice, all at once; now let's think about this synthon.1214

    We need two positive charges here, and two nucleophiles here, so again, here is our phenylmagnesium bromide; we have that settled; but what electrophile could this be--what electrophile do we have that would actually add two equivalents of the Grignard to it?1226

    Well, we can think about getting rid of one of those positive charges by making it a carbonyl, like we did before; but then, we still have another positive charge--we still have another electrophilic site that the Grignard will attack.1248

    Well, how about our method of adding a leaving group?--and that is exactly what we are going to do here.1262

    The leaving group that we will typically add to a carbonyl--you could add a chloride, but a more common one is to make the ester.1267

    Remember, an ester adds two equivalents of a Grignard reagent, just like it added two equivalent of hydride (as we saw before).1277

    So, we could do a very rapid synthesis of this by adding both of those phenyl groups in, by starting with the 3-carbon ester.1288

    Our synthesis is going to start with the ester; and very often, we use the ethyl ester.1297

    You could use any ester, but the ethyl ester turns out to be pretty commonly available commercially; so, if you have to pick an ester, that is a good one to pick.1302

    If we react this with excess phenylmagnesium bromide, let's just review that mechanism really quickly.1313

    The phenyl - is going to add in first; that results in a CTI, a charged tetrahedral intermediate, which is going to kick out the leaving group.1324

    And it is going to result in a ketone; but we can't stop at the ketone, because ketones also react with Grignards.1344

    So, another phenyl - if we have excess here, another phenyl - is going to come in, and after workup, we are going to get step 2, H3O+; so that is our complete reaction.1350

    We are going to first add in the excess phenylmagnesium bromide; and then (step 2) we are going to do our aqueous workup; and we are going to have our target molecule.1364

    I just wanted to review a little bit of our mechanism, because it may have been a while since you saw that reaction.1375

    If we are going to use it in our synthesis, I want you to remember what is going on here.1379

    OK, how about this next disconnection?--we want to use phenylmagnesium bromide, which means this is the new carbon that we are forming.1385

    But this is a different disconnection than what we have seen before, because it is not breaking the carbon bond at the same carbon as the alcohol; it's the next carbon over.1394

    So, let's go backwards to our synthons, because I have given you some guidance here, saying we have to use the phenyl Grignard; so I know we're going to use phenyl minus as our nucleophile; and then, let's look at this synthon.1403

    We need this carbon to be electrophilic--not the same carbon as the carbonyl; so hopefully, you see that this is not going to be the same as a carbonyl, because that would give the phenyl group and the OH group on the same carbon.1417

    So that is not going to work; so what other electrophile have we seen that has reactivity where the nucleophile adds to one carbon, and the OH ends up at the other carbon?1429

    Can you see what it might be?1441

    Let's think about these lone pairs: what if we closed the ring here--what would that give us?1442

    That would give us an epoxide; epoxides are great electrophiles--they react with Grignards, and they do exactly what we want to, in this case.1447

    You attack one carbon, and it ends up adding to the...1457

    You attack one carbon; the OH ends up on the next carbon over.1464

    We're going to start with ethylene oxide, in this case; we are going to add (step 1) phenylmagnesium bromide; so here is our synthesis.1467

    Remember, after doing our planning, now we actually have to go and execute the synthesis: this is when we would be going to the stockroom and getting our reagents.1474

    We would be looking up the literature to get precise experimental procedures.1481

    This lecture is not focusing on the experimental procedures and what solvents to use and how long the reaction goes and how you are going to isolate the products...1486

    OK, what we are talking about here are the strategies that we use to plan a synthesis.1495

    How do we dissect a target molecule and come up with a reasonable plan?1500

    So, the synthesis, then, would start with ethylene oxide; we would add phenylmagnesium bromide, and again, reviewing our mechanism, the phenyl is going to attack at this carbon, break open the ring (so this is an SN2 mechanism), and the phenyl adds up.1504

    So, if we have this carbon (1, 2), it attacks this carbon (1, 2), and the O- ends up on the next carbon over.1519

    Does that look familiar?--that looks like we are getting to our target molecule.1528

    Step 2: H3O+: as usual with our organometallic, we need an aqueous workup.1533

    We can protonate, and we end up with our target molecule.1541

    A very reliable way to make an alcohol is a Grignard plus a ketone, but that is not the only possibility: you could have a Grignard plus an ester (and that would add two equivalents), or you could have a Grignard plus an epoxide (and that would shift the position of the OH group).1546

    Let's think about alkyl halides: what if we have our target molecule with a halogen in it--a bromine or a chlorine or an iodine?1568

    What are some reactions we have seen for that?1577

    One possibility: if the alkyl halide was our target molecule, one possibility would be to start from an alkane, so we could do a free radical halogenation of an alkane.1579

    We can maybe start with an OH, an alcohol, and do a substitution.1590

    Or, we could start with an alkene and do an addition reaction.1596

    Let's look at each of these three possible starting materials, to do functional group...again, these are all functional group interconversions.1601

    This one doesn't really have a functional group to start with; it's un-functionalized--you are adding a functional group; but the others are functional group interconversions.1610

    In other words, all of these have the carbon chain intact; we are just adding a halogen to it somehow.1618

    Free radical halogenation looks like this: we start with an alkane; we have no functional group whatsoever; and we add in bromine and hν, or maybe we use NBS as a source of the bromine radical.1625

    But what bromine does, when it sees light energy, is: it cleaves homolytically; so this bond breaks, and one electron goes to each bromine, and it forms two equivalents of Br..1637

    We get a bromine radical; and so this is your cue that we have a radical reaction; and what is the radical is going to do is: it is going to pluck off one of the hydrogens of this alkane, and it's going to remove whichever hydrogen leads to the most stable radical.1651

    Remember, radicals have the same stability as carbocations, so the more substituted, the more stable.1668

    A tertiary is the most substituted, over secondary, over primary, over methyl.1674

    It is going to preferentially abstract this secondary hydrogen (I can't stop doing mechanism here, so we'll just look at the mechanism and do a quick little review), and we'll get this secondary radical.1678

    And then, that is going to react with the Br2, so we do a hydrogen atom abstraction first, and now we do a bromine atom abstraction.1695

    It's a quick mechanism--just two steps--and that is going to give us the bromide.1705

    Tertiary is the best; if you have an allylic or benzylic hydrogen you could remove, that would be great, because the resulting radical would be resonance-stabilized.1711

    These are very easy carbons to functionalize with a bromine.1721

    Now remember, chlorine and hν is not as selective; so unless it's a very symmetrical molecule that only has one possible halogenation product, you are going to get a mixture of products.1725

    So, really, this method is only reliable for bromination--to put in a bromine.1739

    OK, what if we wanted to do substitution?--well, we could start with an alcohol and have an OH here, and replace it with a halogen.1748

    Now, one way that we learned to do that is to use an acid (HCl, HBr, or HCl and some zinc catalyst, because it's a little less reactive).1754

    And what does that do?--it has two parts: it has a strong acid, so it protonates the OH; remember, an OH is not a good leaving group--hydroxide is not a good leaving group--so if we want to do a substitution, we have to take that into consideration.1767

    So, by using a strong acid, it will protonate the OH to turn it into water (which is a very good leaving group).1785

    And then, it also provides the halogen to do the actual substitution.1791

    This is a good way to make a halide; however, these acidic conditions that we have in this case--acidic conditions favor SN1 mechanism (carbocations).1796

    And now, this is good if you want to make a tertiary; it's very good for tertiary alkyl halides, but for secondary we might have the possibility of doing rearrangements, and primary would be really bad--it would be a lot slower, because we would have to do an SN2; it's not as good at that.1812

    OK, so, yes, this works, but it has limited applications; much better will be these alternative reagents, like thionyl chloride, phosphorus tribromide, or phosphorus and iodine.1834

    What these do is: these reagents first react with the OH to make it a good leaving group, and then they deliver the halogen to replace that good leaving group.1845

    So, these are going to be more reliable reactions to use.1853

    Thionyl chloride is a great reagent to install the chlorine.1857

    PBr3 gives us a bromide; and then, phosphorus and iodine gives the iodide.1865

    Starting with an alcohol, we can go to any of the halides quite readily.1869

    And now, again, remember: the goal here is to talk about strategy, not specific reagents and specific reaction conditions.1872

    I am listing thionyl chloride here, because that is what you usually find in most organic textbooks as the reagent to convert an alcohol to an alkyl halide.1881

    In reality, there are dozens and dozens of reagents that will do this transformation--and reaction conditions.1890

    Now again, thionyl chloride--you might not have access to that, and so you might need to find an alternate chlorinating agent; but the key here is: we are reviewing the fact that one way to put a chlorine in a structure is to start with an alcohol in that position.1897

    And we just learned how to make alcohols, so this is a great way to extend that, now, to convert it to alkyl halides.1916

    OK, another possibility is to do an addition reaction; if we start with an alkene, and we add HBr, it is going to break the π bond; it is going to add a hydrogen and a bromine; and just like we saw with hydration, this starts by protonation of the π bond, so we get Markovnikov addition in the hydrogen.1923

    The proton goes to the carbon with more hydrogens to give the more stable carbocation; and so, once again, we are going to get Markovnikov regiochemistry--Markovnikov addition of HBr.1940

    OK, now there is an alternate mechanism that happens when we add in peroxides; we usually just list R, O, O, R to represent some generic peroxide, so again, the one you select can be a variety of different peroxides.1956

    But now, what peroxides do is: just like the Br2, this is a very, very weak bond--very easily cleaves homolytically to form radicals; and so, this is now going to do a different mechanism, a radical mechanism, where Br. adds to the alkene, rather than a proton.1972

    And so, what we are going to get is the opposite regiochemistry; we are going to get anti-Markovnikov, where the bromine adds to the end carbon.1991

    This is pretty handy, again, to give us the complementary mechanisms, where we can add the bromine to either carbon, depending on what our specific reaction conditions are.2002

    OK, so this, now--there are two ways to add something with anti-Markovnikov regiochemistry across a double bond: we either use hydration or oxidation to make an alcohol, or you use HBr and peroxides to make a bromide.2018

    OK, now again, a good note here is: this mechanism does not work--we can't use HCl--we can't just apply it to some other reaction we have never seen before.2033

    The chlorine radicals have quite different energies than the bromine radicals, and the bonds that we're breaking and forming, and so this reaction does not exist--does not give good yields of our anti-Markovnikov product.2042

    This is restricted to making bromides only.2054

    OK, so let's take a look at a sample reaction and see if we can apply what we have learned so far.2059

    And, rather than doing it as a synthesis, I'm giving the problem to you as a transform.2065

    And so, a transform simply means: provide the reagents necessary to convert, to transform, the given starting material into the desired product.2071

    OK, but this is our target molecule; so the way we should approach this is to start with that target molecule and work backwards--do a retrosynthesis--and ask ourselves, "What starting materials do I need? What reaction have I seen that would give this product?"2081

    We look, and we see that we have an alkyl halide; and then, we think: well, what reactions have we seen that make alkyl halides?2100

    OK, and now, we have a lot of possibilities; but in this case, because we were given the starting material--we have to start with this chlorobenzene--that kind of restricts what possibilities we might be starting with here.2107

    And so, one that looks like a reasonable option is starting with an alcohol in that position by doing a substitution.2122

    So, instead of having the chlorine, if I had an alcohol here, I know I can convert that alcohol to the chloride.2129

    OK, now, what is nice about having an alcohol is: we learned about lots of different disconnections for alcohols, including ones that form new carbon-carbon bonds.2136

    So now, I can go to my disconnection, because I know that that is a carbon-carbon bond that needs to be formed during this synthesis.2144

    Now, I can do this disconnection; let's look at these two carbons involved in the disconnection and ask ourselves (we're doing our retrosynthesis here and coming back to our starting materials--we have to ask ourselves), "One of those carbons started out as a nucleophile; one of them started out as an electrophile; that is the only way they are going to form a bond; so who is who?"2152

    The carbon that now has the single bond, OH--that used to be the electrophile; this was the electrophile, because it was a carbonyl--which means, in order to react with it, this was a nucleophile.2172

    How do I make a phenyl ring nucleophilic?--any random carbon chain--how do I make it nucleophilic?--I use a Grignard.2186

    So, what I need to do this synthesis (here is my planning)--I need an MgBr attached to the benzene carbon, and I need a one-carbon aldehyde (a 1-carbon carbonyl, which turns out to be just formaldehyde--just one carbon).2197

    If I had this nucleophile and I had this electrophile, would the reaction work?2215

    What is cool about a's challenging to imagine the reaction and to come up with the starting materials, but what is great about it is: after you have proposed a possible combination of nucleophile and electrophile, you could check your work!2220

    You could say, "If I had phenylmagnesium bromide, and I added it to formaldehyde, would this be the product I would get?"2233

    It would be; every reaction we are going to study is going to be a simple reaction.2240

    Once we do that difficult work, we can check our work and know that we have the right answer.2244

    OK, so how do I make a Grignard reagent--how do I make MgBr?2250

    Well, I start with a bromine in this position, and I insert magnesium.2255

    OK, so this looks like the plan now: this looks kind of like what we have here.2262

    But I picked bromine (because that is usually how I see a Grignard reagent, MgBr), but instead of a bromine, I have a chlorine.2266

    Now, is that a problem--do I have to somehow switch this to a bromine first, to make the Grignard, or can I make a Grignard from a chloride?2272

    I can make a Grignard from any halide; so, instead of the bromide, let's use the chloride, because that is the starting material given.2282

    OK, MgBr is the most common one, but you can have MgCl or MgI just as easily.2290

    OK, so now, I see that I have finally worked backwards to the given starting material, and I am ready to do my synthesis.2296

    What does my synthesis look like?2303

    Now that I have mapped out my path, now I know that the first thing I have to do is add in magnesium to make the Grignard.2305

    And then, after I have the Grignard, now I am going to combine it with formaldehyde, and I'm going to say step 1 is formaldehyde, because step 2 has to be H3O+ workup.2316

    That adds the one carbon and gives us an alcohol product; and how do we go from an alcohol to the chloride?--SOCl2 is going to be the ideal reagent.2333

    HCl would work fine here, because this is a benzylic carbon--very good for SN1 and SN2; that would work OK here, as well.2348

    So again, we are going to have a lot of different options when it comes to reagents; there is very often going to be more than one possible answer to a synthesis problem.2356

    Let's move on to ethers: an ether has an oxygen with two carbon groups on either side, so the disconnection we are going to make is between the oxygen and either one of the carbon groups.2368

    OK, and so, if I do that disconnection, and I look at these two atoms, one of them has to be a nucleophile; one of them has to be an electrophile.2381

    Well, oxygen, we know, readily handles a negative charge; so this was my nucleophile; this was my electrophile.2388

    And I can just go back to the synthons and think about where that leaves me.2399

    This is our O-; have we ever seen that as a reagent?--certainly; that is a very stable anion.2406

    So, we could just an alkoxide--something like a sodium salt--that we could either make, or we could buy.2414

    And the R+--we have seen that synthon before; if we want to make an alkyl group electrophilic, what we want to do is add a leaving group to it.2421

    So, if we had an alkyl halide...if we have an alkoxide plus an alkyl halide, it gives an ether.2432

    And what does that mechanism look like?2445

    Alkoxide plus an alkyl halide--what mechanism do you expect to have happen here?2450

    It looks like good old SN2--a good nucleophile replacing a leaving group--back side attack.2456

    And that is going to form that carbon-oxygen bond and synthesize an ether.2464

    That is going to be the typical disconnection we will see for an ether.2469

    Let's see an example of that: here we have TBME (this is called...a common name for this is tert-butyl methyl ether, TBME); this is an additive that is used in gasoline.2473

    And what if we wanted to synthesize that from alcohol starting materials?2486

    So, instead of having this as a transform problem, I am simply restricting the various starting materials you can use to be only alcohols as your source of carbons.2490

    And that is a really common instruction to have, because alcohols are readily available and you can do a lot with them.2501

    So, let's look at this: well, any ether...well, an ether that has two different R groups here actually has two possible disconnections you can make.2509

    And so, let's look at both of those and see which one might be better; let's look at this as disconnection A and this as disconnection B.2518

    Disconnection A--remember, we are always going back to an alkoxide and an alkyl halide; so, in disconnection A, we could put a bromine, chlorine...your choice; we could have this alkyl halide and this alkoxide.2524

    And what does disconnection B look like?2542

    Now, with disconnection B, the oxygen is with the t-butyl group, and the bromine is on the other carbon; so, we're going back to an alkoxide and an alkyl halide, a nucleophile and an electrophile.2545

    Always go back to a nucleophile and an electrophile.2559

    OK, so both of those, theoretically, would both form this target molecule; but now, in a practical sense, let's think about running this reaction: would both of these reactions work equally well, or is one a better choice than the other?2563

    What was the mechanism that we had between this nucleophile and electrophile?2578

    It's an SN2 mechanism; so what do we know about SN2?--back side attack; something that slows it down is steric hindrance.2583

    And so, where do we not want that steric hindrance?--we don't want the steric hindrance on the carbon bearing the leaving group.2595

    How would you describe this carbon?--this is a tertiary alkyl halide, so would an SN2 happen here in this case?2602

    Definitely not: there is no way an SN2 is happening on a tertiary carbon; in fact, it is not that there is no reaction; a different reaction happens, because the alkoxide, remember, just like hydroxide, is a strong base.2611

    What happens when a strong base sees a tertiary alkyl halide?--we get an elimination reaction.2625

    In other words, it's going to attack--not do the back side attack on the carbon, but it's going to attack one of these β hydrogens, form a π bond, and kick off the leaving group.2631

    We're going to get an E2 mechanism exclusively.2640

    OK, so when we're planning an ether synthesis, we want to look at both carbon groups and decide who is going to be the less hindered alkyl halide for the better SN2.2644

    Down here, we have a methyl halide; that is the least hindered one possible; this is a great SN2.2653

    And so, that would give us our target molecule.2660

    OK, so that is our plan: we did our retrosynthesis; now we want to do our synthesis.2664

    But remember, we have to start from alcohol starting materials; so I can't just start with t-butoxide; I can't start with bromomethane.2668

    So, we are going to have to make these; so how do you think you would make bromomethane (let's do that one first) if we had to start with an alcohol?2676

    What if I just started with methanol?--I could convert that to bromomethane.2684

    And I could do that by simply adding PBr3.2691

    Now again, if I wanted to make the chloride, I could use SOCl2, thionyl chloride--your choice.2694

    That is how I could make the alkyl halide; now, how could I make this alkoxide?2700

    Well, I would start with the alcohol (tert-butanol--remember, we had to start with an alcohol), and I want to convert that to tert-butoxide (t-butoxide), so it looks like we are doing a deprotonation.2705

    We could use a nice, strong base--like sodium hydride is great for making alkoxides, because remember, the hydrogen gas bubbles off here, so it's irreversible.2718

    Or, we could use sodium metal to do a redox reaction; in fact, what we probably would use is potassium metal, because this is not so reactive.2729

    Remember, potassium t-butoxide is a really common reagent we see.2739

    Sodium hydride is fine, but in reality, we probably would be using potassium metal here to synthesize this anion.2744

    But, once we have the t-butoxide, we can put in the bromomethane (let's show our mechanism here, just for fun--just as a reminder of what is going on here).2750

    And in fact, we have made our target molecule.2761

    OK, how about an amine?--what if we had an amine?--that means there is a nitrogen in our structure.2767

    Well, clearly, if we think about what we just did for the ether, again, we would do a disconnection at the heteroatom; that is a reasonable place to do a disconnection.2772

    We could come up with an alkyl halide and an amine; this would be a great electrophile; this would be a great nucleophile--amines are excellent nucleophiles.2784

    OK, but that is actually a bit of a problem here, because nitrogen is such a good nucleophile that it's difficult for it to do just a single SN2 mechanism.2793

    We can have one example of an SN2, and then a neutralization; but then, it can continue to attack and attack, because when we have...we would make methylamine.2802

    OK, and this, now, is actually--because the carbon is an electron donating group, this is actually a better nucleophile than ammonia was.2822

    This is now going to compete for the methyl iodide that is still present, and so it can attack again and again and again.2832

    And in fact, if you have excess methyl iodide, the nitrogen is going to keep attacking until it has no more lone pairs left, and we are going to get this quaternary...this is called a quaternary ammonium salt.2840

    That is how reactive amines are.2854

    OK, so unless you want to make a quaternary ammonium salt, we need some strategy to control the substitution so that only a single substitution takes place.2856

    And so, what we're going to do is: again, we are going to come back to these synthetic equivalents: instead of using NH3 as our nucleophile, we are going to use something else.2866

    What can we use?--one possibility is this one: it's called the Gabriel synthesis; this is called phthalimide.2876

    And if we treat phthalimide with a strong base (we could just use sodium hydroxide here), it's going to deprotonate the NH to give this N-, and this is now a good nucleophile.2887

    This phthalimide itself is a poor nucleophile; it has a lone pair on that nitrogen, but why do you think this would be a poor nucleophile compared to ammonia or some other amine--what else can this lone pair do besides be a nucleophile?2903

    It can have resonance with both these carbonyls, right?--it's resonance-stabilized, and so this is now quite unreactive.2919

    But if we deprotonate it, and we have a negative charge, now we have something that is, in fact, nucleophilic.2930

    So, if we react it with an alkyl halide, we can get an SN2.2935

    But after that single SN2, it's no longer reactive, because we are back to a neutral structure.2941

    OK, so what did this just do?--it just replaced a leaving group with a nitrogen, and now, if we use hydroxide and water, we could do hydrolysis.2948

    This is a carboxylic acid derivative, and so, hydroxide could add into the carbonyl and kick out the leaving group, and add into this carbonyl and kick out the leaving group; we are going to end up cleaving both of these groups and giving us just a carbon chain with an NH2 group.2957

    Phthalimide is described as a synthetic equivalent of ammonia, because it's a way of replacing a halogen with an NH2 group.2977

    We could do the same sort of reaction using azide as the nucleophile.2991

    Azide (N3-) is a very good nucleophile, and so that can do an SN2.2995

    And then, we can reduce the N3 group; this is a reduction.3003

    It gives off nitrogen gas, and it keeps one of the nitrogens and reduces it down to an NH2.3009

    So, same idea: azide is a synthetic equivalent for ammonia, and it lets us do the reaction, a single substitution of our leaving group, without having multiple substitutions taking place, or multiple alkylations.3016

    Another possibility is to use the cyanide ion: now, cyanide, we know is a good nucleophile--we have seen that before--and so that can do an SN2.3033

    But we didn't just add a nitrogen in this case; we added both a carbon and a nitrogen, so this would be a synthesis that adds both an extra carbon and a nitrogen.3045

    Now again, this could be reduced, either by catalytic hydrogenation (so just like we can reduce a triple bond down to a single bond) or--because it's a carboxylic acid derivative--we could use LAH reduction for this.3056

    In either case, it's going to effectively reduce both of those π bonds like it's adding hydrogens to the carbon and the nitrogen; and we get an amine in this case, as well.3069

    This is another effective synthesis of amines, and it adds an extra carbon, as well.3082

    We can also do functional group interconversions using an amide; so how do we make an amide?3092

    Well, we start with an acid chloride, and we add an ammonia (oops, I added an extra zigzag here).3098

    This is how we would make an amide; and the reaction of an amide with, remember, earlier we looked at how lithium aluminum hydride reduction of most other carboxylic acid derivatives gives the alcohol product.3111

    Amides and nitriles are the exception to that, because this is such a poor leaving group, it never leaves; and we end up reducing the carbonyl completely down instead.3125

    We still get our two equivalents of hydride being added, but it's going to be giving an amine product instead of an alcohol.3136

    This is another possibility: we can think about...all we are looking at is: if we wanted to make this amine, how can we do it?3146

    One of the possibilities is to start with an amide, or a carboxylic acid that can be converted to an amide, and then reduce it down to an amine.3153

    Another possibility is to start with a ketone or an aldehyde and do a process called a reductive amination.3163

    We take an amine, and you add hydrogen and nickel, or you could use a special reducing agent like sodium cyanoborohydride.3172

    OK, neither of these reducing agents will attack the carbonyl; but, when you mix it with an amine, what happens to the carbonyl?--it gets converted to an imine, and then that imine, in situ, gets reduced by either the hydrogen gas that is present or the sodium cyanoborohydride.3182

    This is called a one-pot reaction, where we mix all three of these ingredients together--first the imine and the carbonyl react, and then the reduction takes place; and in the end, we get out the amine product.3212

    Let's take a look at an amine and think about some different ways that maybe we can synthesize; again, there are going to be a lot of possibilities, but let's consider a few of them.3228

    What if I had to synthesize this target molecule?--I have a secondary amine, meaning it has two carbon groups; and so, what are some ways that we can make it?3236

    Well, maybe we could do a functional group interconversion, convert it to...3246

    Again, the reverse of a functional group interconversion...well, one way, we just saw, is: we can make it from an amide; so what if I added a carbonyl on either one of these sides?3252

    If I had this amide, then I could reduce the amide to the amine by LAH, lithium aluminum hydride.3267

    OK, that is good; and how do I make an amide?--well, that is this disconnection...we'll actually talk about carboxylic acid derivatives in just a bit, but if I started with this carboxylic acid, I could make the acid chloride and make the amine that way.3275

    I would need this amine; if I had this amine and this carboxylic acid, I could combine them to form the amide, and then I could reduce it to make my target molecule.3289

    That is one possibility.3301

    Another possibility we just saw was, remember, the reductive amination; if I made this imine--if I had this imine (even just in situ--not even that it's isolated, but) that would be something that could form this target molecule.3303

    And the way you make an imine is from an aldehyde or a ketone and an amine.3322

    This is another possible way to make the target molecule: to start with this propynal and mix it with this amine with the sodium cyanoborohydride or nickel catalyst with hydrogen, and that would make the target molecule, as well.3331

    OK, so these are a few options that we have, but both of them still have a pretty complex amine starting material; so let's think about how to make that amine starting material.3347

    Let's say this is not commercially available, or we had instructions that you can only have so many carbons in your starting materials, or so on.3358

    You will usually get some guidance, so you know, "How far do I have to go back with my retrosynthesis? What starting materials are suitable?"--those would be instructions that come with a given problem.3367

    OK, but let's think about how to make this: now, we want to do that disconnection and add in that nitrogen group, but remember, we can't use ammonia itself with very good yield; so let's try something like the Gabriel synthesis, where instead of...3376

    We could just do a functional group interconversion; we can say, "Well, that nitrogen maybe came from the phthalimide group," and now we could do our disconnection; so we could just do a functional group interconversion, and now we could do our disconnection to do the Gabriel, which means I would need the phthalimide nucleophile, and what do I need at this position?3392

    This one was my nucleophile, and what do I need at this carbon?--it must have been an electrophile, and so I would need this alkyl halide.3415

    I just attach a leaving group--chlorine, bromine, your choice.3427

    OK, or what if I wanted to do it by the cyanide option--what if this came from a cyano group?3432

    Now, that would mean that these two--the carbon and the nitrogen here--were a cyano; and then we have a (1, 2, 3)...we have this isopropyl group attached.3441

    This could be another possible starting material that could be converted to the target molecule; so where would this come from--what is our disconnection now?3453

    Our disconnection would be at the cyano group; and again, this cyano group was my nucleophile, so this one was my electrophile; so in this case, I would need sodium cyanide, and I would have the isopropyl chloride or bromide--as my nucleophile and my electrophile.3462

    A lot of different options for making amines--most of them require using some kind of synthetic equivalent for the nitrogen, to initially get that nitrogen onto the carbon chain.3484

    What if we wanted to make an alkene--install a carbon-carbon double bond?3501

    Well, one possibility is to do an elimination reaction--so I could have an alkyl halide here and do an E2.3507

    I could also do an E1 elimination; the most common one we have seen for that is dehydration; so if we start with an alcohol, you can dehydrate it to make an alkene.3515

    You could start with a triple bond and do a partial reduction to the double bond; or you could do a Wittig reaction--a Wittig reaction would start with a ketone or an aldehyde--so a lot of different starting materials that can be converted into alkenes (give alkene products after reaction).3526

    Let's look at elimination reactions.3545

    If we have a good leaving group here (alkyl halide), and we use hydroxide as a strong base, what is going to happen is: it's going to attack one of these β hydrogens; and remember, it's going to go for the more substituted one.3547

    We call that Zaitsev's Rule: the more stable alkene is formed (that would be the more substituted one).3566

    And so, that would give us a double bond between these two carbons.3574

    OK, so we would also get the trans alkene, as shown; that would be the more stable alkene.3580

    We call that anti elimination; so if we have stereocenters involved, we have to make sure that the leaving group and the β hydrogen are anti to each other, 180 degrees: so those are some of the details that we had for the E2 elimination.3585

    OK, another possibility is to start with an alcohol and do an E1 elimination; OK, this is an E1; our alcohol will get protonated, and then it can leave.3598

    OK, but this mechanism is not a 1-step mechanism like the E2; the E2 is very reliable, because it is a 1-step mechanism with known stereochemistry, regiochemistry.3616

    Those are the kinds of reactions we want for synthesis: we want to do reactions that are going to give us high yields of a single product.3625

    That is either a target molecule or something we can carry on toward a target molecule.3633

    But any reaction that involves a carbocation, now, we have the possibility of rearranging; this is a secondary carbocation, but it can rearrange to be another secondary carbocation by a 1,2 hydride shift.3638

    And then, once we have this carbocation, now we have something closer to this substituted carbon, so this can eliminate and give a more stable alkene.3657

    So, most certainly, especially because we have heat, we have equilibrium, every step is those cases, we always go for our thermodynamic product, the most stable product.3670

    And so, we most certainly will expect rearrangement; and in this case, we would do that to get the trisubstituted alkene, which is the most stable.3680

    Still following Zaitsev's Rule, but knowing that we can have rearrangements, means that we are really going to want to get out whatever stable alkene we can get.3690

    Where, for the E2, we only have a choice between these two hydrogens; there is no way that the double bond can make its way over here, so we would get a pretty good yield of a single product.3700

    Here, we would probably get a pretty good yield of a single product; but if you wanted your alkene to be in the other position, this would not be a good synthesis of that.3708

    Dehydration is a useful strategy when we want to make a very stable alkene (meaning highly substituted).3716

    Now, we can also get to an alkene by reduction; so this is another functional group interconversion, because we are not changing our carbon structure, but if we start with an alkyne (a triple bond) and we only partially reduce and get rid of one of the π bonds, we would be left with an alkene.3727

    So, we had two methods for doing this: one is using catalytic hydrogenation and Lindlar's catalyst.3741

    Lindlar's catalyst is a poison that we add in; it is described as a poison catalyst, and it stops the reduction after a single addition of hydrogen.3747

    And it is special because, remember, catalytic hydrogenation does syn addition, and so this is going to give us the cis alkene.3756

    It adds just one equivalent of hydrogen, and it does so from the same face, and so, that is a great way to make a cis alkene with good stereoselectivity.3767

    If we want to make the trans alkene, we will use dissolving metal reduction; here we use sodium metal or lithium metal, so that has one electron; it's a 1-electron donor; it's a good reducing agent.3777

    And this has a very different mechanism through a radical anion, and as a result, we end up adding the hydrogens; the hydrogens that come from the ammonia end up being the proton source.3788

    We end up adding the hydrogens trans to each other, and we get the trans alkene.3799

    This strategy is very useful if our target molecule contains a cis alkene or a trans alkene; there is really no better way to ensure the proper stereochemistry than to start with an alkyne in that position.3805

    That would be a very good retrosynthesis, in those cases.3822

    OK, and then we also have an alkene synthesis that is a true synthesis--a formation of a carbon-carbon bond.3827

    And so, again, that is why this is a really important reaction for synthesis, used widely throughout organic synthesis for making complex molecules.3835

    That is called the Wittig reaction; the Wittig reaction, recall, uses a Wittig reagent.3843

    This is called a Wittig reagent, and we can draw it two ways: we can either draw it as shown, with a double bond between the phosphorus and the carbon (but this is actually a resonance-stabilized carbanion); you could draw it with a negative charge on the carbon.3851

    And actually, this is the resonance form that makes a little more sense when we are thinking about the mechanism--"how does this react?"--because now we see we have a carbon nucleophile.3867

    A Wittig reagent is actually a carbon nucleophile, and we could imagine that carbon attacking, and that is the key carbon-carbon bond that is forming.3876

    OK, the rest of the mechanism I won't get into now, because it's a little complicated, and it tends to lead students astray, away from the right answer, because the product is quite simple to predict.3884

    We simply look at the carbon group of the Wittig reagent, and we replace the oxygen of the carbonyl with that carbon group; so where I used to have a ketone, I now have an alkene.3897

    OK, so this is a great way to make alkenes; and you have perfect selectivity, because you know, exactly where the carbonyl was, that is where your double bond is going to be.3912

    OK, there are also conditions you can have that can affect the stereochemistry; so it is possible to maybe get the cis alkene preferentially, or the trans alkene; I'm not going to get into that now.3923

    But let's take a look at an example where that might be used: so let's say this is our target molecule.3935

    Now, we can do this by functional group interconversion, but let's say we wanted to do a synthesis of this, forming new carbon-carbon bonds.3940

    So, what we would do for a retro-Wittig is: we would completely break apart the carbon-carbon double bond, because both of those bonds are formed in the Wittig reaction.3947

    And now, we look at these two carbons, and one of them was the Wittig reagent, and the other one was a carbonyl; and actually, it's your choice on who is who, because we end up with the same functional group on either side--just an alkene.3956

    We can either decide that the cyclohexane is the part, and we make cyclohexanone (and in that case, it would be Ph3P and then a 2-carbon Wittig reagent), or we could make the Wittig reagent from the cyclohexane moiety, so that would be a double bond, phosphorus...Ph3.3975

    And it would then be the 2-carbon aldehyde.4003

    Wittig reagents react with aldehydes and ketones.4009

    Let's try a synthesis of an alkene: here we are doing it as a transform example, so we know exactly what bonds have been formed and changes have taken place.4018

    A good strategy is to number--1, 2, 3, 4: we have 4 carbons to start.4028

    And this methyl--this ethyl group is still an ethyl group, so that is still 1 and 2; but this is 3 and 4, so that tells us that this carbon-carbon bond is new in the product, and it points to the disconnection that is required.4034

    That looks exactly like the disconnection that we just saw for the Wittig, and so it's our choice: we can either say: OK, we had this as the aldehyde, and then we reacted with a 1-carbon Wittig reagent; or, we could say we wanted this Wittig reagent and the 1-carbon aldehyde (formaldehyde).4053

    Your choice--those are both reasonable places to go.4086

    Let's look at both of those, because I want to show you how to make a Wittig reagent if you need to: that is good review.4090

    And so, let's just work backwards a little bit more, to see where we get this.4097

    So, the question here is: we need this aldehyde--how do you make an aldehyde?--what reactions have we seen that give aldehydes as products?4103

    We haven't really talked about those yet; but one possibility would be a functional group interconversion back to an alcohol; if I had an alcohol, I could oxidize it to the aldehyde, so that is a possibility.4114

    And the way we make a Wittig reagent is: we start with a halogen in the position that the phosphorus is going to be.4128

    So, what we need to make is this alcohol or this bromide; and then we can work our way to our target molecule.4137

    So again, remember: the way we are doing a synthesis is: we are starting with the product; we are doing a retrosynthesis--we are doing a planning.4145

    Now, we look back, and we say, "OK, how did we get there?"--so how do I make this alcohol?4152

    Can I convert an alkene to an alcohol?--I can; that is the hydration reaction.4156

    Notice the regiochemistry: do we need Markovnikov or anti-Markovnikov regiochemistry here?4162

    It looks like the OH is adding to the carbon with more hydrogens, so this is anti-Markovnikov; so this is when we kind of work with that toolbox of reactions that we have--that stack of flashcards that we worked so hard with--and recall what are the reaction conditions--what are the reagents--needed to do anti-Markovnikov hydration.4169

    That was a 2-step process: hydroboration (BH3-THF or B2H6...hydroboration), oxidation (H2O2, NaOH).4188

    Hydroboration-oxidation gives me the alcohol I need; and then, how do I convert an alcohol to an aldehyde?--I need some kind of an oxidizing agent (PCC is a great choice here).4201

    And then, how did I convert the aldehyde to the alkene?--I just used a Wittig reagent.4216

    So, if the instructions don't say otherwise, you can use a Wittig reagent as a commercially available reagent, just like you could use a Grignard reagent.4223

    OK, but let's look at how to make them, if you ever have to make a Wittig reagent.4231

    And that would be the second path; so the second path turns this alkene into the Wittig reagent first, by making it a bromide; and now, we have to break the π bond, add HBr across the π bond, and once again we need to do it with anti-Markovnikov regiochemistry.4236

    We don't just use HBr; we use HBr with peroxides--adding those peroxides gives us the anti-Markovnikov addition.4255

    OK, now how do we turn this into the Wittig reagent?4264

    The Wittig reagent is a 2-step procedure: first we add triphenylphosphine (phosphorus is right underneath nitrogen in the periodic table, so just like nitrogen is a great nucleophile, phosphorus is an even better nucleophile, because it's bigger--more polarizable).4269

    This is a great nucleophile--loves to do the SN2; so that is our first step--that is what puts the phosphorus group on here, and it makes a phosphonium group, a P+; now it has 4 bonds.4285

    And then, the second group, we want the negative charge: remember, it's a P+ and a C-, so the second reaction is...that is a butyllithium...we add in a very strong base.4302

    Butyllithium is often used in this case, and what that is going to do is: it is going to deprotonate the carbon next to the phosphorus.4315

    And you could either draw it as a C-, or...let's bring it right in to do the resonance form that we showed in our retrosynthesis; remember, there are two resonance forms for the Wittig reagent; you could draw it either way.4326

    And that is how you make a Wittig; so it's two steps from the alkyl halide: triphenylphosphine, followed by butyllithium.4340

    OK, once we have this Wittig reagent, what do we want to do with it?--we want to react it with formaldehyde, and that would give us our target molecule.4347

    OK, so if you look at a target molecule like this, with an alkene at the end of a carbon chain--very unstable--it's impossible to do this by dehydration, and it might be problematic to do by elimination as well.4359

    Maybe we could reduce an alkyne; we will talk about making alkynes; or a Wittig reaction is a great way to install an alkene at a maybe unstable position, especially at the end of a carbon chain.4377

    What if we wanted to make an alkyne?4390

    We have two double bonds, so it's possible that we insert both of those by doing a double elimination.4394

    If I had two bromines, and these both had hydrogens, I could lose HBr twice and make the triple bond.4400

    OK, or I could start with a terminal alkyne that has a hydrogen here, and replace that hydrogen with an alkyl group; so we call that an alkylation reaction.4409

    Let's take a look at both of these.4419

    The synthesis of an alkyne by elimination is simply a functional group interconversion, because our carbon chain is already intact; we are just changing the leaving groups into π bonds.4423

    How do you get the halogens--the leaving groups--in there?--well, you can maybe start with an alkene; you can brominate the alkene.4436

    Remember, this does a trans bromination, anti bromination.4443

    This is plus the enantiomer, so we would add in two bromines (the stereochemistry doesn't matter much here) anti to one another; and then, if we do KOH heat, we could do one elimination, one E2 elimination, to give an alkene.4451

    And then, we can--with heat--with more vigorous conditions, because this is a more difficult one, because we have a vinyl leaving group...but it is possible to do a second elimination to form a triple bond.4479

    OK, so it is possible to do a double E2 if you have two leaving groups.4495

    OK, another way to make an alkyne is to start with another alkyne--to start with a terminal alkyne (we call it a terminal alkyne when it's at the end of a carbon chain, so that has a hydrogen at one end).4503

    If you treat this with a strong base (sodium amide is a very strong base), it can deprotonate it.4516

    So, terminal alkynes are acidic (they have a pKa somewhere around 25), so they can be deprotonated, and that is because this resulting carbanion is reasonably stable, because it is an sp hybridized carbon.4522

    Normally, we don't deprotonate carbons; but we can for alkynes.4539

    And so, just like cyanide, this is a reasonable anion; it's a good nucleophile; and so, if we react it with an electrophile such as an alkyl halide, we would expect a good old SN2 back side attack, and we do an alkylation reaction.4543

    We can convert a terminal alkyne to an internal alkyne, and see, we formed a new carbon-carbon bond: so this is another great reaction to have in our toolbox for creating new carbon chains, extending our carbon chains.4561

    What does the retrosynthesis of an alkyne look like?--if we have an alkyne as the target molecule, we can look at either end of that alkyne and do a disconnection there.4578

    We consider the two carbons that we are bringing together: one of them was a nucleophile; one was the electrophile; and where is our nucleophile going to be?4587

    The triple bond is a position, we know, that can handle the negative charge; so this was my nucleophile; the other carbon was my electrophile.4595

    If we want to go to the synthons, we can do that (in other words, just break the bond and make a + and -).4605

    And have we seen this anion before?--we have; we can make that by starting with the alkyne and treating it with a strong base (NaNH2 is the base we usually use in this case); it's common to see NH3 as a solvent here.4614

    So, you might find that there; that is OK.4631

    And then, how do we make an alkyl chain, a carbon chain, electrophilic?4635

    We can't use a carbocation; we can't ask in the stockroom for some ethyl carbocation--that wouldn't work.4641

    What we do is: we add on a leaving group.4646

    We make an alkyl halide as our electrophile.4651

    It's a great retrosynthesis of an alkyne.4657

    OK, so let's try another transform: we are starting with an alkyne in this case; we have 3 carbons (1, 2, 3); and so, where are those carbons here?4661

    It looks like this is our methyl, and our alkyne has been converted to an alkene; and so, we know that this must be a new carbon-carbon bond.4671

    There are two things we have to do here: we have to make a new carbon-carbon bond, and we have to make an alkene.4680

    What is the stereochemistry of that alkene?--we have to make a trans alkene.4687

    Those are transformations we have done before; we need to think about what order we want to do them in.4693

    OK, because we have the alkyne already, this is the position at which we can deprotonate and add on the new carbons; so that is the retrosynthesis we are going to do--say, "OK, this is going to come from the alkyne."4702

    We are going to take the alkyne and turn it into a trans alkene, and now that you have an alkyne, now you can do the disconnection to make this anion and this electrophile--this nucleophile and this electrophile.4720

    OK, so the first thing we are going to want to do is alkylate: so NaNH2; NH3 is our solvent; that will deprotonate, and now we want to add in a...this is another 2 carbons we are adding; we need another 2-carbon chain with a leaving group in there, so we can do an SN2.4739

    And then, how do we convert?--now we have our carbon chain in place; and how do we convert a triple bond into a trans alkene?4766

    Remember, hydrogenation adds syn, so if you want to add trans, we need dissolving metal reduction; so that is Na/NH3, dissolving metal reduction.4777

    Take a look at these reaction conditions--very similar to reactions with alkynes.4787

    Remember, NaNH2 means we have NH2-, so that is a strong base; Na on its own means it's a sodium atom; that is a reducing agent.4790

    NH3 is just our solvent.4803

    OK, just a reminder: if you tried to do this in the reverse order--if you did Na/NH3 first to reduce--one problem is: there is no way I can control my trans stereochemistry; but the other problem is: now that I have an alkene, I can no longer deprotonate.4807

    OK, so it's only the alkyne that has the acidic proton; so we have to take advantage of that functional group and use it in order to make the new carbon-carbon bond.4834

    Then, we can worry about whatever functional group interconversion we're doing and transform the alkyne to something else.4843

    OK, what if we wanted to make an alkane?--what if our target molecule had no functional groups on it?4852

    Well, again, we think backwards, and we ask ourselves, "What reactions have we seen that give alkanes as final products?"4858

    And most of them are reductions of some kind; we can either have an alkene or an alkyne and reduce the π bonds, or we could maybe have a leaving group here, and if we replaced with a hydrogen, a hydride, that would be a substitution (also a reduction).4867

    But another possibility is: we can use an organometallic reagent; if we use the organocuprate, organocuprates react with halides to do a coupling reaction.4882

    So, that is another possible way of forming a new carbon-carbon bond as you are making an alkane target molecule.4891

    How about aldehydes and ketones?4904

    Aldehydes and ketones have two carbon-oxygen bonds, so one way that you can get there is to start with one carbon-oxygen bond (in other words, an alcohol) and oxidize it.4909

    PCC or Swern conditions are kind of mild oxidizing agents, selective oxidizing agents; they only partially oxidize.4919

    If we have a primary alcohol, that is our biggest concern, because a primary alcohol has two hydrogens you can replace.4928

    So, if you want to replace just one of those, we need PCC or Swern, and that will partially oxidize it to the aldehyde.4935

    If you start with a secondary alcohol, it's going to oxidize it to the ketone; so we're replacing one C-O bond with two C-O bonds.4948

    We can make aldehydes or ketones that way; we can do ozonolysis of an alkene--if we start with an alkene, ozonolysis completely breaks the carbon-carbon double bond and replaces each of those with carbon-oxygen double bonds.4959

    If we do a reductive workup, like zinc or dimethyl sulfide, that is called a reductive workup, and that keeps...4979

    Notice, there is a hydrogen on this alkene; that hydrogen stays intact; and so we can get a mixture of ketones or aldehydes, depending on what our initial alkene looked like.4989

    Those would be all oxidation reactions, starting from alcohols or alkenes.5000

    We could also do reduction reactions; so if we started with three heteroatom bonds, we can reduce it down to two heteroatom bonds with a special reagent called diisobutylaluminum hydride.5005

    We know LAH has aluminum with just four hydrogens on it; diisobutyl has isobutyl groups (oops, that is just a single bond here); so an isobutyl group looks like an isopropyl with an extra carbon.5017

    When you add these isobutyl groups onto the aluminum, it changes its reactivity and makes it more selective; and what you could do is: you can achieve partial reductions.5036

    So, you could add in the hydride to an acid chloride, and that can kick back down and kick off the chloride, and then stop.5045

    The intermediate, actually, that you have is stable with the diisobutyl group, so it doesn't go further than that.5056

    Same thing with an ester: you could do partial reduction of the ester, and a nitrile, followed by hydrolysis, would do the same thing: it only adds one hydride in, and then the C-N double bond can be replaced with a C-O double bond.5063

    This is a unique way, a special way, to make aldehydes only; we could start with one of these carboxylic acid derivatives and do a special reduction of it.5084

    OK, we could also start with an alkyne and add water across the alkyne; that is going to add H and OH across the triple bond.5096

    Now again, we could either do it with acid, or we could use a borane reagent to do anti-Markovnikov.5107

    We have a special borane reagent, disiamylborane; it's a little bulkier and improves the regiochemistry a little better to add to the terminal hydrogen.5119

    But here we get...if we add an H and an OH across an alkyne, it adds just one equivalent, because we end up with an enol intermediate.5131

    We know enols are not stable; they will tautomerize to give, instead, a ketone product--or an aldehyde, if we started with a terminal alkyne and did anti-Markovnikov.5141

    This is simply a functional group interconversion, changing a triple bond to a ketone or an aldehyde; it's not an oxidation or a reduction, because we added both an H and an OH, so it's just...a hydration, we call it--just a hydration.5155

    Now, another possibility is to make a ketone by acyl substitution; so just like we did the DIBAL, adding only one equivalent of hydride, there are certain conditions where we can add just one equivalent of a carbon group.5174

    The organocuprate will do that with an acid chloride, and again, the organocuprate will add in, and then come back down and replace that chloride, but we can control the conditions with this reactivity so that it adds just a single time, and the cuprate will not react with this ketone; it will only react with the more reactive acid chloride.5187

    Remember, acid chlorides are incredibly reactive electrophiles, and so the cuprate will react with that, but then it will stop once it gets to the ketone point, so we can get just a single substitution there.5209

    Remember, if we used a Grignard reagent, the Grignard would continue reacting with the ketone, and we would end up adding two equivalents of the Grignard.5222

    And similarly, when we use a nitrile and a Grignard reagent ("and a Grignard reagent"--in this case, we can use a Grignard), it is going to add just once (I have a prime here) to give this kind of an intermediate, just like it would attack a carbonyl.5234

    And then, the H3O+ workup would do a hydrolysis that would replace the C-N double bond with a C-O double bond.5256

    That would be another way of making a ketone.5267

    These are some special synthetic tools that we have to form new carbon-carbon bonds, starting again with carboxylic acid derivatives.5269

    Now, one great way of making an aldehyde or a ketone is to start with an aldehyde or a ketone and add a carbon group to the α position; we call that α alkylation.5281

    And that takes advantage of the fact that the α carbon is acidic; the α protons have, again, a pKa of around 25, and so they can be deprotonated.5291

    The base we usually use for that is LDA, and in a case where we have more than one type of α proton to remove, LDA is governed by kinetics.5306

    We do it at -78 degrees (I probably should add that in there); we do it in cold conditions--it's governed by kinetics, and so it's governed by sterics; it simply attacks the most available proton and removes that one.5318

    So, it is not going to go after the more sterically crowded one; it is going to deprotonate here, and it is going to give us an enolate intermediate.5331

    An enolate is a great nucleophile; and so we are going to be using that in synthesis, and that is going to give a variety of ketone products out.5347

    If we react it with an alkyl halide as our electrophile, remember, it's the α carbon of the enolate that is reactive.5355

    Remember that we have a resonance form here; we could show that really quickly, because we can really draw it either way; you can show the resonance form with the C-.5362

    Usually we draw out the better resonance form with the O-, because the negative charge is on the more electronegative atom.5371

    But either way you draw it, it has the same reactivity; it's the α carbon that reacts.5376

    So, this is the bond that we are forming between the α carbon and the alkyl halide carbon, and the way we do that mechanism is: our O- reforms the carbonyl and kicks the π bond out, and that is what does the SN2.5380

    This is back to an SN2.5395

    We just formed a new carbon-carbon bond.5400

    This is a really great way of synthesizing new ketones and aldehydes.5404

    What would the retrosynthesis of a ketone look like?--one possibility would be to identify that α carbon and do a disconnection from the α carbon.5410

    And when we are looking for who is the nucleophile and who is the electrophile, always the α carbon is going to be the nucleophile, because that is where we can have an enolate.5423

    That means this carbon (whatever group we have here) was the electrophile, and in this case, if we just want a plain old carbon chain, it would start as an alkyl halide; so our retrosynthesis goes back to this enolate and this carbocation...5434

    I'm sorry, the carbocation is the synthon, which is the alkyl halide, as the electrophile.5453

    OK, now sometimes that reaction doesn't work so well; enolates can be quite reactive, pretty strong bases, and so instead of using...this was the enolate that we just said we needed on the previous screen...we can use a different one: this is called acetoacetate ester, or ethyl acetoacetate.5464

    And we can use that as a synthetic equivalent for this less stable enolate.5489

    OK, and here is what it looks like: what we could do is: we could deprotonate with a weaker base now; we could use a weaker base, because this α proton is now α to two carbonyls; it's more acidic, so we don't need LDA.5495

    So already, our synthesis is simpler, because LDA is very difficult to work with (you have to prepare it fresh and use it fresh--very difficult to handle).5510

    We could use a much weaker base, and we can deprotonate and then alkylate very easily; these would probably be much higher-yielding steps.5520

    And then, if we react this with H3O+, we have an ester here, so we could do hydrolysis of that ester.5533

    It turned it into the carboxylic acid, so how did that happen?--we have addition-elimination (addition of water and elimination of the ethanol leaving group).5543

    OK, and now, what we have is a β carbonyl carboxylic acid.5551

    We have a carboxylic acid, and not in the α position, but in the β position, we have another carbonyl; so we could call this a β-keto acid, for example.5561

    Any time we have this kind of structure, when you heat it, it is unstable; it decomposes, and it evolves carbon dioxide.5570

    We call this reaction a decarboxylation: this structure can be decarboxylated.5578

    It ends up just clipping off this CO2 molecule--I'm not going to go through the mechanism for that, but it's just a cyclic transition state, where we tip this OH up here, and we get a cyclic transition state, and then tautomerization of the enol that results.5584

    And we end up forming...we get back our 3-carbon chain that we wanted, and we would make the same target molecule we made on the previous screen, but we would do a much higher yield.5603

    This is described as the acetoacetate ester synthesis, and we'll find that any time we want to make the enolate of acetone (in other words, just the 3-carbon ketone--if we want to make that enolate), instead, we are going to use ethyl acetoacetate.5614

    This is commercially available, readily available, and even though it's a longer synthesis, it gives a much higher yield of our products: we can make acetone derivatives here.5632

    It's acetone with another carbon chain--with another carbon group attached.5646

    Let's take a look at an example here: what if I wanted to synthesize this ketone?5653

    One good place for a disconnection would be looking to the α carbon; and so, let's do that.5659

    That would be a reasonable carbon bond to break.5667

    We look at our two carbons here and identify who was the nucleophile and who is the electrophile; of course, the α carbon was the nucleophile, so this one was the electrophile.5673

    We work back, and we could just go to the synthons; we have this synthon, C-, and this synthon, C+.5688

    OK, and the C+, the carbocation--all we need to do is put a leaving group here.5697

    OK, but again, we have a secondary leaving group; this enolate is a pretty strong base--we would probably have some E2 if we tried to bring those together.5705

    And so, instead of just using this unstable enolate, we are going to add an ester group here to help stabilize that negative charge.5714

    We describe this as a stabilized enolate.5728

    It is a weaker base; it's a better nucleophile; it is going to give us a better yield.5731

    So, here is the acetone enolate; and any time we want to use that, we are going to use ethyl acetoacetate anion, instead.5736

    OK, so our synthesis is going to be starting with ethyl acetoacetate, and step 1 is going to be: we need to deprotonate.5746

    So, we could just use sodium ethoxide here; if we have the ethyl ester, we are going to use the ethyl base so that there is no substitution that can occur here.5757

    And after we deprotonate, now we can add in the bromocyclopentane.5767

    And so, hopefully, you see how we have moved toward our product: we have this extra ester group on here; we added that to help stabilize our intermediate anion; but, apart from that, you can hopefully see the basic carbon framework of our target molecule coming together.5778

    OK, so the only thing that is left, then, is: we have to remove our extra activating group, this ester group; and the way we do that is: we have to first hydrolyze it, so it's the carboxylic acid that we'll decarboxylate, not the ester.5800

    What we do is H3O+ and heat; so we're going to hydrolyze, and then we're going to decarboxylate; we can do this all kind of in one pot, and that is going to remove CO2 and leave us with our target molecule--ethyl acetoacetate ester synthesis.5816

    What if we wanted to make a carboxylic acid?5838

    A carboxylic acid--a highly oxidized carbon: this carbon has 3 C-O bonds.5840

    One way that we can do that is by some kind of oxidation reaction; we could start with an alcohol; we could start with an aldehyde; those already have C-O bonds to start, so we can increase those numbers of C-O bonds by oxidation.5846

    We can also do hydrolysis: a nitrile is a carboxylic acid derivative--it can be easily converted to the carboxylic acid itself, simply by hydrolysis.5861

    Or, we could do an organometallic reagent, a carbon-carbon bond-forming reaction, using a Grignard and CO2.5871

    Let's look at each of those.5878

    Let's look at an example that has some of those--excuse me.5881

    Here is a transform that we want to do, and not only have we formed a carboxylic acid, but we have also added a new carbon.5884

    OK, so that tells us that we are going to have to do that disconnection.5895

    This is not just a functional group interconversion; this is not just going to be an oxidation.5898

    It is going to be a formation of a new carbon-carbon bond.5903

    The two ways that we saw that we can make a carboxylic acid: one, we said, was using a Grignard reagent; so, if I had this Grignard and CO2, then that would form this bond and make a carboxylic acid.5907

    That is one option; another option that we just saw on the previous page is: we could have a nitrile first, and the nitrile could be converted to a carboxylic acid.5924

    Well, that is useful, because now a nitrile has a good disconnection: we saw this in the very beginning of the lecture; that is a reasonable disconnection, because when we look for our nucleophile and our electrophile, we know that cyanide is a very good nucleophile; this was our nucleophile; this was our electrophile.5937

    And so, we could start with a bromide and sodium cyanide.5958

    Both of these mechanisms--in this case, both of them are very good; in some cases, our reactions might not be stable to Grignard conditions.5965

    Maybe imagine if you had a multifunctional target molecule: most target molecules have lots of functional groups; they are not simple structures like these; and so, sometimes you can't do a Grignard reaction, because that would react with some other carbonyl, or something else in the rest of the molecule.5975

    In that case, the cyanide approach might be better.5992

    OK, but this requires an SN2 mechanism; and so, if this carbon is not something you can do an SN2 on, then a Grignard would be a better approach.5998

    These are complementary, in the sense that, in some cases, one approach would be better than another.6006

    In this case, because it's a secondary halide, both of them work fine, so if we wanted to do the Grignard, we would add a magnesium to make the Grignard, and then we add in CO2.6012

    Remember the structure of CO2?--I just want a reminder here of what that mechanism looks like.6029

    Now, we have a carbonyl and a single bond, O-; so a Grignard plus CO2 would give a carboxylic acid.6036

    Of course, we need step 2, H3O+, to protonate that O-; so that would be one way to synthesize this carboxylic acid.6043

    Another option would be to add in sodium cyanide and do an SN2.6051

    And the key here is recognizing that nitriles are, in fact, carboxylic acid derivatives, and as such, they can be hydrolyzed to get back to the carboxylic acid.6059

    So, hydrolysis simply means reaction with water; so H3O+, and probably some heat (because nitriles aren't particularly reactive, and it first hydrolyzes the amide, which is also pretty unreactive, and so we would probably need all the heat).6070

    In fact, the base-promoted is probably a better method--probably more commonly used--because you have the driving force that, as the carboxylic acid is formed, it will get deprotonated in the reaction conditions, and then, that kind of pushes the equilibrium in the forward direction.6091

    We could use NaOH in water, and probably some heat too, because it's maybe not so reactive; but then, if we wanted to get the neutral product, we would have to do H3O+ workup at the end to get to the neutral carboxylic acid.6110

    This is kind of like saponification; remember, it gives a carboxylate salt here.6128

    It's just a little easier to write H3O+; but in reality, in the lab, usually we do base-promoted hydrolysis instead of acid.6132

    OK, this is not an oxidation; we don't need an oxidizing agent; we are trading C-N bonds for C-O bonds, so this carbon is not being oxidized; it is just being replaced with water--hydrolyzed.6141

    Now, there is another way to do carboxylic acid synthesis: this is called the malonic ester synthesis, and it's taking advantage of alkylating the α carbon once again, by using an enolate.6155

    And this is analogous to this ethyl acetoacetate synthesis, except now we are using a diester.6168

    Instead of a methyl group here, we have a diester.6175

    So, what would happen to this if we treated it with sodium methoxide?--we would deprotonate very easily to give an extra resonance-stabilized enolate, a stabilized enolate with two electron withdrawing groups.6177

    And then, reaction with an alkyl halide would be an SN2.6195

    Deprotonate, and then an SN2; so we can do an α alkylation quite easily with this active methylene--with this very reactive methylene.6201

    And now, we could get rid of one of these carboxylic acid groups; first, we're going to hydrolyze to make the diacid (of course, it would hydrolyze both esters, so we would get the dicarboxylic acid here).6215

    And now, again, we have a β carbonyl acid; we have a carboxylic acid, and in the β position, we have another carbonyl.6236

    This, upon heating, will decarboxylate.6247

    It doesn't matter which one you remove; it would give the same product; but our final product looks like this, and what we have made is an acetic acid derivative.6251

    All right, this 2-carbon group ends up being retained, and we functionalized--we added in another carbon--we alkylated the α position; and the resulting functional group is a carboxylic acid.6268

    This is another great way to make carboxylic acids: to start with diethyl malonate.6284

    Let's look at an example--see if we can work through this one.6294

    Here is another example where we're doing a transform; so we're telling you what disconnection to make, because we are starting with a phenyl, plus one extra carbon; here is phenyl plus one extra carbon.6299

    We know that that carbon-carbon bond has to be formed in the reaction.6310

    These are the two carbons that are reacting: one is the nucleophile, and one is the electrophile; and what do you see that is special here?6317

    I see that carbon is α to a carbonyl; so, as soon as I identify that we have an α carbon, I know this was my nucleophile, and I know this one was my electrophile.6326

    OK, very convenient--because we already have a leaving group on here; it's already starting out electrophilic.6339

    OK, so let's do our retrosynthesis, asking what starting materials we need.6345

    We can start with the synthons; that is a good habit to get into (until you get more experience, and you can jump right to the reagents).6352

    Our electrophile has a positive charge; our nucleophile has a negative charge; the positive charge--we could just add a leaving group and make an alkyl halide.6362

    Like I said, we already had that, so that is great.6370

    But how do we make this anion?--we can't make this anion, because this is a very acidic proton; so this is an impossible anion.6373

    And even if we protected this as the ester, it would be a pretty unstable anion; a better way to stabilize it would be to have a second ester group, as well; so here is where we use our diethyl malonate.6383

    Again, it's commercially available, even though it has these two functional groups: commercially available, great to use--diethyl malonate.6398

    And we would get that just by deprotonating...this is the enolate of diethyl malonate; we would get that by just deprotonating the diester.6407

    All we need to do here is to react this with the anion of diethyl malonate; so we would deprotonate that first, and then combine it with the benzobromide; that would do an SN2.6415

    We just added these extra carbons...oops, I didn't add my extra carbon; excuse me.6434

    Be careful that we number our carbons; here is the benzylic carbon--sometimes it's good to add little dots or numbers here.6441

    This is the new carbon we have added, and that is the carbon...a really common mistake that I just made there, when you lose a always double-check your work after you draw your structure.6447

    Here is the original carbon; here is the new carbon that I have added; there is the α carbon, and that has the two ester groups attached.6458

    And then, how do I get rid of that extra ester group that I added in there?--I can just hydrolyze it and convert it to the diacid, and then heat, and that will do our decarboxylation.6466

    Hydrolysis with heat does decarboxylation.6481

    What if I want to make a carboxylic acid derivative?6487

    We just learned how to make carboxylic acids; those, in turn, can be converted to any of the derivatives by conversion to the acid chloride first, and then it can be converted to an ester or an amide.6489

    The way we go from a carboxylic acid to an acid chloride is: we use thionyl chloride, SOCl2, the same thing we used to turn an alcohol into an alkyl chloride.6505

    And then, all we need to do is add in the appropriate nucleophile (the alcohol or the amine) to do the substitution with the acid chloride.6517

    We put in a very good leaving group--the best leaving group there is, the acid chloride--and then you put in any nucleophile, and it's going to do addition-elimination and replace that chloride.6527

    Some other options that we have for esters, though--because those are a really common functional group, so there are a lot of different ways we can make it--we could do the Fischer esterification.6540

    The Fischer esterification takes a carboxylic acid and mixes it with an alcohol, and then we do addition-elimination, and it replaces the OH group with an O-R group, and what else is formed here?6548

    We lose this proton and this OH, so we are forming water in this reaction.6564

    Well, if you look at the reverse reaction, the reverse reaction is also possible: the ester can react with water to do hydrolysis to go back to the acid and the alcohol.6570

    So, the Fischer esterification--both of these mechanisms are acid-catalyzed, and so the Fischer esterification is in equilibrium; so the only way we can get it to work is if, somehow, we can remove the water as it is formed--or remove the ester as it is formed (sometimes that can be done, because it has a lower boiling point).6579

    Remove a product to push the equilibrium forward.6602

    Fischer esterification works, but we need to do something to help it go toward the ester.6610

    We can also do an SN2 reaction, and actually, we saw this earlier, when we looked at a way to make alcohols.6619

    We could start with the carboxylate--a very good nucleophile--and react with an alkyl halide; and so, that is a way of forming an ester; but now, we formed this bond, where usually, with the ester, we are doing the disconnection at the carbonyl.6628

    So, instead of doing an acyl substitution, you can add an alkyl group onto the oxygen; so that is a possibility.6649

    And then, there is a very unique reagent called diazomethane; this is if we want to make a methyl ester specifically--we can use this special reagent, diazomethane.6656

    Diazomethane has an interesting structure: so when we draw it out, we have...this middle nitrogen has a positive charge; this carbon has a negative charge (it's a zwitterion).6668

    And so, what happens with diazomethane--when you mix it with the carboxylic acid, first this acts as a good base because I have a C- here, and we know carboxylic acids are quite acidic--it acts as a base and deprotonates.6682

    And now, I have a carboxylate nucleophile, and I have a methyl group with an outstanding leaving group on it: what would that leaving group be, if it were to get kicked off?6698

    It would be nitrogen gas; you don't get much more stable than that, so what happens is an SN2 to kick out nitrogen gas, and we get a methyl ester.6710

    It is kind of could also use iodomethane, but diazomethane is a special reagent that both deprotonates and installs the methyl group.6725

    Now, another possibility is to start with an ester and just functionalize, alkylate, at the α position--so just like we did for ketones or aldehydes, you could do the same thing; you could deprotonate with LDA and then do SN2 with alkyl halide.6735

    So, just like you can make an enolate from a ketone or an aldehyde, you can also make an enolate of an ester; and that also would react with an alkyl halide to do an SN2.6754

    That is another possible disconnection we can make for an ester target molecule: we can focus on that α carbon, just like we did, and assume that the ester functional group is already in place.6777

    Let's try synthesis of an alkyl chloride, using some of the techniques that we have seen.6794

    There are a lot of different possibilities that we have here.6801

    First of all, how do we make chlorines--how do we install a chlorine into a carbon chain?6806

    We could maybe start with an alkene and add HCl; but one possibility is to start with an alcohol; and what is nice about starting with an alcohol is: there are a lot of different ways to make alcohols.6815

    So, for example, we could disconnect from the alcohol carbon; so we could disconnect here--let's call that disconnection A.6834

    Or, we could disconnect here; or we could even disconnect one carbon over, like this; and so, let's think about where those would take us.6845

    Disconnection A--we need a nucleophile and an electrophile; so that looks like...maybe if we had a Grignard, methylmagnesium bromide.6857

    And this would be a good place to pause the lecture, actually, if you want to try and work out these disconnections on your own--see if you can figure it out on your own!6869

    We could do this Grignard reaction (that would be disconnection A), or disconnection B would break it up into this aldehyde and this Grignard.6891

    We have a 4-carbon Grignard and a 2-carbon aldehyde.6900

    Now, I didn't talk about this earlier, but one of the things that is ideal when we're looking for a good disconnection is: we want to try and simplify our molecule as much as possible.6904

    So, any time you can...if you have a 10-carbon chain, if you can break it somewhere in the middle to get two 5-carbon chains, or a 4- and 6-carbon chain, that would be a very good retrosynthesis, because we are getting much more simple, smaller molecules.6917

    If you have a 10-carbon chain, and your retrosynthesis just breaks off one carbon, well, now you still have a 9-carbon chain that you have to make.6931

    And so, that would typically not be as good.6937

    So, when you look at these, this approach might be a little better, because the parts are more nearly equal.6939

    In fact, C is even better in that regard, because C takes the 6-carbon chain and breaks it up into two 3-carbon chains.6947

    But C is an interesting disconnection because it is removed, now, from the alcohol carbon; so how do we make this carbon electrophilic?--what if we have an epoxide?6955

    If we have an epoxide, it would be attacked at this carbon by a Grignard; and this is a really nice synthesis, because now we have two 3-carbon starting materials.6967

    But, in reality, all of these are commercially available, simple molecules; so none of them would be bad.6982

    OK, so that is one possibility (to disconnect an alcohol); another possibility is to take our alcohol and say, "Well, where could that alcohol have come from?"6988

    Let's do a functional group interconversion on it and go back to a ketone and say, "Well, if I made this ketone, then I could reduce that to the alcohol, and then convert it to the alkyl halide."7000

    That is nice, because ketones also have several different disconnections we can use.7014

    For example, now we can focus on that α carbon.7022

    We could focus on that α carbon and do a disconnection.7027

    And we can either disconnect it directly, or we can remember that we need to add an additional electron withdrawing group; if I added an extra ester group on, now this would be a really great disconnection, because my anion, my enolate, would be extra resonance-stabilized.7032

    So, this goes back to ethyl acetoacetate as a great nucleophile...and then the alkyl halide to do my SN2.7050

    You can either do the disconnection here and say, "Oh, I have this anion; I need to (let's just make those two equal, rather than an arrow) convert that to ethyl acetoacetate," or you can kind of add the activating group in first, before you do the disconnection.7064

    But either way, instead of using the enolate of acetone, it would be better to use ethyl acetoacetate instead.7084

    OK, I think there is maybe another FGI; how can I make this alcohol, besides from a ketone?7097

    I could imagine starting with an alkene; if I had this alkene, would I be able to convert it to this alcohol?7107

    Now, remember, I wouldn't want any rearrangement, so I could do oxymercuration reduction; that would be a way of doing Markovnikov addition.7121

    How do I make an alkene by synthesis?--well, this looks like a good application of a Wittig reaction.7127

    I could have this Wittig reagent and this aldehyde, and maybe this aldehyde is commercially available, or maybe I can make that by doing a disconnection, etc.7139

    So really, once you start thinking about the different transformations that we are capable of, and the common nucleophiles we have seen and the common electrophiles we have seen--once you kind of build up that library, and you take this systematic approach to doing disconnections of your target molecules and looking for the right bonds to break and the right interconversions you can do, then you start to see a wide variety of synthetic possibilities jumping out at you.7153

    But it takes a lot of practice; so don't be discouraged if you find it challenging, because I think synthesis is one of the most challenging things that you do in studying organic chemistry.7183

    The next thing to consider is: what if we have, not just a target molecule with a single functional group--but what if we have patterns of functional groups?7194

    There are certain patterns of functional groups that are characteristic of certain reactions.7202

    OK, so for example, if you recall the aldol reaction, the aldol reaction takes two equivalents of a ketone or an aldehyde (and let's draw a second molecule of one right here); and what happens is: it serves both as the nucleophile and the electrophile.7207

    The α carbon is the nucleophile; the carbonyl carbon of the other is an electrophile; and so, that is the bond you form.7226

    And so, when the enolate attacks, we end up with an alcohol here.7236

    There are our two carbons that were formed...the carbon-carbon bond that was formed; so this is called the aldol reaction.7241

    And remember, an aldol product, especially when we heat it, will lose water to form a double bond between the α and the β carbons.7248

    OK, again, we still see our two units of acetone molecule that we started with, either in this form or this form; this is a self-aldol condensation, meaning that the same molecule has been both the nucleophile and the electrophile.7257

    Knowing that this is the general pattern we see in the products of an aldol reaction, that can help us in doing a retrosynthesis when we see those patterns.7273

    So, if our target molecule looks like this--if it's a β-hydroxy ketone (so here is a ketone, and not in the α position, but in the β position), we have an OH; that tells us that this product could have come from an aldol reaction.7282

    The disconnection we make is between the α carbon and the β carbon; we look at these two carbons, and we ask ourselves, "Who is the nucleophile? Who is the electrophile?"7297

    And, because we see this is an α carbon, we see this; we know this was the nucleophile as the α carbon; so this was the electrophile.7310

    And if it now has a single bond OH, it must have been a carbonyl.7320

    We can work backwards in our retrosynthesis--we can work backwards to the enolate and the electrophile--the carbonyl...the α carbon and the carbonyl.7327

    OK, and it's a little more difficult, but it's the same disconnection: if you see, instead, that you have an alpha, beta unsaturated ketone (that means we have a double bond between the alpha and beta carbons), it is the same disconnection.7341

    It's the same disconnection: we're breaking through this carbon-carbon bond.7353

    If you want, you can kind of do it in steps: you can say, "Well, let's just do a functional group interconversion and add the water back in," because you know that you got to this double bond by dehydrating.7357

    And so, you can add it back in to get to the β-hydroxy ketone--because I think that's a little easier to break apart and see who the nucleophile and who the electrophile are.7371

    It's kind of a big jump to go all the way from here; so if you want, you could do it stepwise; but still, you are going to come back to the same starting materials: you are going to come back to carbonyl compounds--aldehydes or ketones.7384

    Now, another reaction we studied of enolates is called the Michael reaction; the Michael reaction has an if we have epoxide as our base, for example, we could deprotonate here; we are going to have a stabilized enolate (that is what is needed for a Michael reaction), so it usually has two electron withdrawing groups on here.7397

    And when we react with an alpha, beta unsaturated ketone, this β carbon is partially positive, so it's electrophilic; and that is the carbon that is going to get attacked.7416

    That is the bond that we are forming; that is the bond that we formed; and we end up protonating the α carbon as part of our mechanism.7428

    We have a nucleophile to add down here, and a proton to add, to the second carbon of the double bond.7441

    Our retrosynthesis, notice the pattern that was formed here: every time we do a Michael reaction, we get a 1,5 dicarbonyl (1, 2, 3, 4, 5).7448

    What does our disconnection look like?--well, once again, we are going to focus on that α carbon, and we are going to disconnect from the α carbon.7459

    The α carbon was our nucleophile, so the other carbon was our electrophile; and when we do our disconnection (let's just look at the synthons)...OK, now I recognize this synthon--this is just the enolate.7466

    This is an enolate, so I know I can deprotonate a ketone to get that; so that is simply this plus base.7486

    OK, but how about this one--how do we make a β carbon electrophilic?7494

    Now, you might think, "Well, I don't know; we have always put on a leaving group, so can't I just put on a leaving group and start with this starting material?"7498

    Well, you have to be careful in using starting materials that have multiple functional groups that you haven't seen before; we can't just randomly pull functional groups together and assume A) that it's commercially available, or B) that it's stable.7506

    In fact, this molecule is not stable; it has a leaving group in the β position--not in the α position, but in the β position.7518

    And compounds with β leaving groups are unstable--very easily subject to elimination; so in fact, this molecule would very readily lose HBr.7527

    So, this is not a stable molecule; it spontaneously loses HBr to give the following structure--to give the alpha, beta unsaturated...because this is a very stable alkene that you form, because it is conjugated with the carbonyl.7537

    But guess what--this alkene has the exact reactivity that we wanted, because we wanted this carbon to be electrophilic, and because of resonance (right?--this is resonance-stabilized); let's just review the reactivity of an alpha, beta unsaturated ketone.7552

    That β carbon is electrophilic; it has partial positive character; so all we need to do is add a double bond between the α and β carbons.7573

    We don't add a leaving group; we add a double bond instead.7582

    That is the disconnection of the Michael reaction; we go back to an enolate and an enone--and alkene, an alpha, beta unsaturated ketone.7585

    OK, and then a third reaction that we studied is the Claisen condensation; the Claisen condensation is just like an aldol, except now we have two esters; so we have one ester reacting with a second ester (let's draw them together here).7599

    And so, once again, it's going to be the α carbon of one as the nucleophile, the carbonyl carbon of the other as the electrophile.7616

    But what happens when a nucleophile adds into an ester?7622

    It breaks the π bond, and then the O- kicks off and ejects that leaving group; so because the ester has a leaving group, we get a substitution taking place at that carbonyl.7627

    The enolate adds in and kicks out the ethoxy leaving group, and so we end up with a β-keto ester; we end up with a carbonyl in the β position.7638

    This is called the Claisen condensation; so the retrosynthesis of this pattern, a β-keto ester--we still look at the carbon in between the two carbonyls; we look at that α carbon, and we disconnect one of these carbons--we disconnect, on either side...7651

    The α carbon was the nucleophile, and so the other carbon was my electrophile.7672

    Let's go back to our synthons and see if we can translate those synthons: what do those look like?7679

    This one is simply an ester plus base, plus sodium ethoxide; if we had the ethyl ester, we would use the ethoxide base as a good base.7689

    We could make that enolate; how would we make this electrophile?7701

    What are we going to put in this position to make it electrophilic?7705

    Now we attach a leaving group.7709

    And the leaving group we are going to use is, again, just an ethyl ester; we don't need to use the chloride--the ester is much more stable and much more commercially available, easier to work with; so that is the best leaving group to use.7712

    You can see, in this case, that it's just a self-condensation; both the nucleophile and the electrophile are ethyl acetate, and so that is really the best disconnection we could ever make: to break the molecule exactly in half, to two identical starting materials.7726

    The summary of the two group reactions that we have seen looks something like this: these are all the different patterns we can have, and in every single one of these, where we are going to focus in on is the α carbon.7744

    OK, we are always going to disconnect at the α carbon, because we know the α carbon has the characteristic reactivity; it's a good enolate.7760

    We are going to make this enolate, in each case, which we can draw as the C-, just to make it easy--but, of course, the better resonance form, the way we usually draw it, is as the O-.7769

    OK, so that is our nucleophile in each case; and all we are varying to get these different patterns is the electrophile.7787

    Who is the electrophile that is going to give us an alcohol?--the electrophile, in this case, is just a ketone, so we call this reaction the aldol reaction.7796

    What did we have in this place to now get the carbon-carbon double bond?--well, this is also the aldol reaction.7808

    If we had a carbonyl, we could make the alcohol; and then, if it lost water, we would get the double bond.7814

    This is also the carbonyl; these are both the aldol condensation.7820

    If we retain our carbonyl after the enolate adds in, how do we keep the carbonyl?--by putting a leaving group on it, so the enolate adds in and then kicks out that leaving group; so here is where we are going to use an ester.7829

    This is the Claisen disconnection, a 1,3 diketone; this would be a mixed Claisen instead of the self-Claisen we saw before, but the key is having a β-keto structure.7842

    And now, we have a 4-carbon (1, 2, 3, 4-carbon); it's going to be this β carbon that we want to be electrophilic, rather than the carbonyl carbon.7856

    And the way we make the β carbon electrophilic is: we add a double bond.7868

    We make it an alpha, beta unsaturated carbonyl.7873

    Let's look at a few sample syntheses to wrap this up.7880

    These are a little more complicated target molecules; and I have instructions, for these last two problems, that we have to start with cyclohexanone.7885

    So, it kind of gives you some guidance as to where we are going to be disconnecting the target molecule, and how we are going to break it down into commercially available starting materials.7893

    Let's look at this to start: OK, let's look at our functional groups--we have a ketone, and we have an epoxide.7904

    So, let's look at that epoxide first: what reaction have we seen that forms an epoxide?7912

    That is what we have to imagine: remember, this is asking: what starting materials do I need?--what functional group was there initially?7920

    We have seen epoxidation reactions: what functional group do you start with to get an epoxide?7927

    You start with an alkene--alkenes can be epoxidized.7934

    So, let's do that disconnection first--that functional group interconversion, saying an epoxide could have come from an alkene; and by doing that, we now have a pattern of functional groups.7939

    We no longer just have a ketone and an alkene that we are going to handle separately; we are going to look at them together and recognize them as an alpha, beta unsaturated ketone.7950

    The alpha, beta unsaturated ketone is a product we can get from an aldol reaction.7964

    Recognizing this as an aldol product is going to lead us to the proper disconnection that we want to make.7970

    OK, and that disconnection is going to be from the α carbon; again, if you want to do it in smaller steps, we can do that.7977

    We can first do a functional group interconversion, knowing that this alpha, beta unsaturated ketone came from the β-hydroxy ketone.7984

    This is probably well worth your time, when you are beginning, to go through this initial step so that you more accurately get the correct starting materials.7997

    OK, so now, you see the disconnection we are making; now we are ready to do our disconnection.8009

    Here are our two reactive carbons; one of them was a nucleophile; one of them was an electrophile.8014

    I see the α carbon, so I know where my nucleophile is; this carbon now has an OH, so it was my electrophile, and what did it look like?8022

    If it now has a single bond OH, it used to be a carbonyl.8031

    So now, I can get to my two starting materials; I need this enolate, and I need this electrophile; here is my nucleophile, and here is my electrophile.8037

    There is my plan: remember, this is the most important part--to plan things out before you jump into your synthesis, so you know where you are going.8051

    OK, my synthesis, then, starts here: I start with cyclohexanone, and how do I make the enolate of cyclohexanone?--I use LDA for that.8060

    Let's draw it the old-fashioned way; and then, I'm going to mix this with formaldehyde, and I'm adding my electrophile; so I'm doing a crossed aldol condensation with two different reagents, and so I can do it stepwise like this, with LDA, and then by adding this in.8079

    And then, I could just do my workup and heat to ensure that I'm going to dehydrate in the end; and it is going to give me my aldol product.8098

    Now, in this case, because formaldehyde is such a great electrophile, and because this has no α carbons, I can actually do this all in one pot; I don't need to do the stepwise synthesis using LDA.8113

    I could just combine cyclohexanone and formaldehyde, and some weak base like sodium hydroxide, and heat it--because when sodium hydroxide goes to deprotonate, the only place it could deprotonate is the α carbon of the cyclohexanone, and the formaldehyde is a better electrophile than cyclohexanone (an aldehyde is a much better electrophile than a ketone).8130

    So, we essentially just have one nucleophile and one electrophile; so this is the only aldol that can take place--so we really don't need to do all of these crazy steps.8159

    We could just throw these two together with some heat, and we get our crossed aldol condensation to take place.8166

    OK, and how do we get to our target molecule?--we need to epoxidize this; so again, thinking back to that giant list of reagents we have in our head, it was some peroxy acid...that is what we need (like mCPBA--mCPBA is a great reagent for that); some peroxy acid--you could use H2O2..any peroxide will do this.8173

    But that is how we can make the epoxide, which was our final target molecule.8198

    So sometimes, we have our patterns that we look for, and then it gets manipulated a little further; so sometimes we have to work backwards a little bit before we can get to that place where it's OK to make a disconnection.8203

    OK, let's try this one: again, we want to work from cyclohexanone; so we have two new carbon-carbon bonds that we are forming, and we might want to think about which of those we want to add in first or second.8218

    And remember, if you have two carbonyls, then that would be a really stable enolate; so let's take out that methyl group first, because I think (let's also abbreviate my phenyl ring) see what I'm saying?--if I have this α carbon, this would be a very easy α carbon to deprotonate and alkylate.8236

    Let's do that part last; let's add in this carbonyl group first, so that it will give us an active methylene; that is going to be a better synthesis than trying to go the other way.8260

    Now, we do our disconnection to get back to cyclohexanone; and let's look at our two carbons and decide who is our nucleophile and who is our electrophile.8275

    This is our α carbon, so this was my nucleophile--my enolate; this is my carbonyl carbon; this was my electrophile, but it is still a carbonyl, so what does that tell you?8290

    It was an ester, so this is a mixed Claisen condensation.8305

    So, I need this enolate, and I need this ester.8312

    If I have those two--if this was my nucleophile and this was my electrophile, that would, in fact, give us our target molecule.8321

    OK, so really, in a lot of these cases, dealing with enolates and difunctional compounds, it's the retrosynthesis that is kind of the challenging part.8331

    The synthesis itself is usually quite straightforward, because now we just bring the reagents together that we decided we needed.8340

    OK, so how did we make this enolate?--again, we could do it stepwise, if we want; we could use LDA to deprotonate (and I could just draw it this way so we could see it both ways).8350

    And then, I can add in my ester, and that is going to do addition-elimination; we would actually need a workup here, because in these basic conditions, this very acidic proton would get deprotonated, so to neutralize it, we need a workup.8364

    And then, now, to alkylate--again, LDA...we don't even need LDA here; we could just use sodium hydroxide, because it's so much more acidic--and then methyl iodide to do the SN2.8384

    And there is our target molecule.8405

    So, lots more problems for you to practice; find a good textbook with lots of examples in there to work on, because really, the way you get good at synthesis is "practice, practice, practice."8407

    But hopefully, soon you will find that you are getting the hang of it, and you are ready to tackle even the most challenging problems.8417

    Thanks a lot--I hope to see you soon.8423