Ch. 22 - Condensation ChemistryWorksheetSee all chapters
All Chapters
Ch. 1 - A Review of General Chemistry
Ch. 2 - Molecular Representations
Ch. 3 - Acids and Bases
Ch. 4 - Alkanes and Cycloalkanes
Ch. 5 - Chirality
Ch. 6 - Thermodynamics and Kinetics
Ch. 7 - Substitution Reactions
Ch. 8 - Elimination Reactions
Ch. 9 - Alkenes and Alkynes
Ch. 10 - Addition Reactions
Ch. 11 - Radical Reactions
Ch. 12 - Alcohols, Ethers, Epoxides and Thiols
Ch. 13 - Alcohols and Carbonyl Compounds
Ch. 14 - Synthetic Techniques
Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect
Ch. 16 - Conjugated Systems
Ch. 17 - Aromaticity
Ch. 18 - Reactions of Aromatics: EAS and Beyond
Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition
Ch. 20 - Carboxylic Acid Derivatives: NAS
Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon
Ch. 22 - Condensation Chemistry
Ch. 23 - Amines
Ch. 24 - Carbohydrates
Ch. 25 - Phenols
Ch. 26 - Amino Acids, Peptides, and Proteins

The Michael Reaction is a 1,4-conjugate addition of an enone with an enolate and always forms 1,5-dicarbonyls.

Concept #1: General Reaction

Transcript

In this video, we're going to focus on a specific form of conjugate addition to an enone called the Michael reaction. The Michael reaction is a 1,4-conjugate addition. Remember that when I say 1,4-conjugate addition, I'm talking about an enone and I’m talking about adding right there, of an enone with an enolate. Remember that if specifically, if your nucleophile is an enolate, that is called a Michael reaction if that is your nucleophile.
Basically, if you think about it, this is kind of like an aldol times two because your first aldol reaction created the enone. Now, I'm doing a second aldol reaction. I like to think about Michael reaction as an aldol times two. What's cool about these is that they're always going to form the same thing. They’re always going to form 1,5-dicarbonyls. Let me show you how to draw the products first and then I’ll show you the whole mechanism. The products are really easy. In this situation, because we’re doing a conjugate addition, my rules of where to line up your enolate and your electrophile for aldol go out the window. I don't want that anymore. What I prefer you do is to draw the enone at the bottom exactly the way it is but without its double bond and then draw a single bond coming off of the conjugate position, so position 4. Then just draw that attached to your enolate. That's how you draw the product. It's literally enone, enolate, new bond, and then obviously I pointed this out earlier, no pi bond. Don’t draw a double bond there. What I’d like to do is regardless of whatever the nucleophile is, I like to just draw it right on top. But we have to go into the mechanism.
In this next video, I’m going to show you the full mechanism for the Michael reaction. This is also going to help you guys understand why that pi bond isn't there because that can be a little confusing why is that pi bond gone. We're going to talk about it. That's coming up in the next video.

Concept #2: General Mechanism

Transcript

Alright guys, let's bring the molecule down and we're going to draw the four mechanism, okay? Oops, it got right, cool. Okay. So, I got this molecule and I've got my needling, okay? So, guys what's going to happen is that. Remember that I said I have two electrophilic regions, I have this electrophile here, we're used to seeing that but this is also an electrophile now, it doesn't look like it because it's a double bond but it is because it can resonate that positive charge can resonate. So, in my first step I actually get this kind of strange arrow of the negative attacking this carbon, okay?

Now, if I make that bond I have to break a bond. So, I'm going to break this bond and make a double bond here, I'm going to break this bond and make a lone pair there. So, what I'm going to get is something that looks like this, O negative, double bond, single bond, that single bond has a new bond that's attached to my old enolate, okay? So, is that making sense so far? I have a negative charge and I have a double bond, okay? Now, what's going to happen is that this molecule gets protonated because notice that we have an enolate, right? So, there must be some kind of conjugate that took off that H. So, let's just say that it had been hydroxide. So, we'll use water to protonate, okay? So, now we're going to protonate, wow it's terrible arrow, we're going to protonate and that's going to give us a molecule that you should know something about, let's draw it first. Alright guys. So, I have a molecule, that's similar to my 1,5 dicarbonyl but what do we call this species down here? what is that? guys, do you recognize that that's a vinyl alcohol or more specifically this is an enol tautomer, right? That's an enol tautomer, right? So, now this is in a basic environment. So, through tautomerization that double bond that enol tautomer is going to switch to the keto form. Remember, that the keto form is pretty much always the most stable unless it's a beta dicarbonyl. So, through tautomerization I'm going to regenerate the keto form and I'm going to get this, okay? if you want more information on the tautomerization mechanism definitely feel free to go through the tautomerization videos that I have and I explained everything there but at this point of the course you just need tools to use tautomerization you know what to draw every time, okay guys? So, that shows you guys the full mechanism for micro reaction which also applies to any conjugate addition guys, just so you know this mechanism, this is the same mechanism for any conjugate addition, okay? So, now you know what that looks like and you know how to make a 1,5 dicarbonyl. So, let's move on to the next video.

Practice: Determine the product in the following conjugated addition reaction. 

Practice: Determine the product in the following conjugated addition reaction. 

Practice: Determine the product in the following conjugated addition reaction