Michael Reaction

Concept: Concept: General Reaction

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Video 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: Concept: General Mechanism

4m
Video 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.

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

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

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Problem: Determine the product in the following conjugated addition reaction

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Michael Reaction Additional Practice Problems

Complete the following reaction supplying the missing product and showing correct regio- and stereochemistry where applicable. If a racemic or diastereomeric mixture forms show all stereoisomers; if no reaction takes place, write N.R.

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Complete the mechanism for the following Michael reaction. Be sure to show arrows to indicate movement of all electrons, write all lone pairs, all formal charges, and all the products for each step. Remember, I said all the products for each step. IF A NEW CHIRAL CENTER IS CREATED IN AN INTERMEDIATE OR THE PRODUCTS, MARK IT WITH AN ASTERISK AND LABEL AS "RACEMIC" IF RELEVANT. IN THE BOX BY EACH SET OF ARROWS, WRITE WHICH OF THE 4 MECHANISTIC ELEMENTS IS INDICATED IN EACH STEP OF YOUR MECHANISM (For example, "Add a proton").

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Predict the products for the following transformations. When necessary indicate the major product and relative stereochemistry.

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Draw the organic product(s) for the following reaction. Indicate stereochemistry where appropriate. Assume an aqueous workup, when necessary. A reasonable answer may be “ No Reaction.”

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α, β, γ, δ unsaturated carbonyl compounds frequently react as shown below. Draw a mechanism for this reaction and give a good reason for why a nucleophile reacts with an alkene at all.  

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Predict the product of the following reaction. 

 

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Provide a detailed mechanism for the following transformation.

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Predict the stepwise mechanism for the reaction shown below: 

 

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Predict the product(s) for the following reaction.

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Which of the following is the direct product of a Michael reaction (1,4 addition of an enolate to a enone)?

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Fill in the roles (abbreviated in parentheses) of the reactants in the boxes. 

Which one is the Michael donor ( D)? Which one is the Michael acceptor ( A)?

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Indicate the order in which the intermediates would appear during the conversion of   1 into 2. The asterisk (*) in the answers indicates the step where the quench is used. 

a) 1 → III – * → II → 2

b) 1 – * →  IV → II → 2

c) 1 → II – * → 2

d) 1→ II → I – * → 2

e) 1 → V → III – * → 2

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Predict the product of the following reaction:

 

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What is the final product of the following reaction?

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Which mechanism best describes the conversion of  A into B?

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