Show how the following products might be synthesized from suit...

Question

Show how the following products might be synthesized from suitable Michael donors and acceptors.

(b)

Accepted Answer

Hello, everyone. On this video. I want to show how the given compound can be prepared using proper micro donor and micro acceptor string materials. So my condition is a newco addition of a micro donor which is usually an active methylene on a micro acceptor which is usually a alpha beta unsaturated carbon neal. And this is a carbon carbon bond formation reaction. So the mechanism of the reaction involves the pronation of active methylene by a base, followed by a 14 edition of the generated Newco file to an alpha beta unsaturated compound. The hydrolysis of the generated intermediate leads to the product. So in order to remove the group from the product, a deep carboxyl can be performed. So I'll go ahead and draw out the general mechanism. So we can get an idea of the micro addition. So starting off with my Michael donor, we'll have an R one group, we just call it as we have several R groups. Then we have a key to here, then our next Q to over to the right. And another a group will label as R two. Of course, we have this sort of central proton. So coming in with our base, we'll just use, oh We use any base here. Of course, we know that we have three lone pairs on our hydroxide. So this base is gonna go ahead and deron this proton here. And now what happens is that the bond between the carbon hydrogen will go ahead and break forming a lump pair on the central carbon. So now we see that the R one group remains intact as well as our ketone there, R two group ketone all remains intact. And now we have a lump pair on the central carbon forming a negative one formal charge because this is fair and stable. We're gonna want to do that. The lump pair forms a pi bond between the carbon and carbon. And then this pi bond here will go ahead and break forming a lump pair on the oxygen. So what's gonna happen there sort of like president in structures again, R one group remains intact. R two group remains intact. We formed a new pi bond and then now we have a single bond between the carbon and oxygen. Oxygen gains a new lone pair now bears a negative one formal charge. Now the fun part comes in when we add in the micro acceptor. So go ahead and draw general micro acceptor. So we have maybe an R three group of some sort. If is there, you mean another R groups R four and the ketone group there. So what's gonna happen here is that, of course, we said that carbon Neals really like to form. So the loan pair from our auction will go ahead and come down making a pie bond to relieve the central carbon. Right over here is carbon Neo carbon. This pi bond will go ahead and attack our micro acceptor. Then this pie bo will go ahead and move over to the right closer to the carbon neo carbon. Then this pie bond will go ahead and break and form a long pair here. Let's go ahead and write out actually that we have this one being our Michael donor. And then this right here is my Michael a suter. All right. So I'm moving on with our reaction. What we're gonna have now is a big product. Again, our one group remains all intact. We're gonna go ahead and push down our original molecule. So our two group goes there. Now we're adding in the Michael receptor. So our three group comes down here, then we have our four group, new pi bond, which is strong on the bottom. Actually, then we have this oxygen here getting a new long pair bearing a formal one, formal negative, one, formal charge. All right. So what's gonna happen? Then again, the carbon Neil has to reform the pi or the lump pairs will go ahead and come down to form a pi bond. Then this new pie bond or double bond will go ahead and break and form a lump pair on this carbon. So what we're gonna get now, let's see here. So again, one, R one, the ketone remains intact, R two and ketone remains intact. Then the new edition here R four remains intact. R three remains intact. But this time now on this carbon connecting the R three group, we have a lump pair bearing a negative one from one charge. Then what's gonna happen is that we have our water molecule. So just floating around, what's gonna happen is that the carbon bearing the negative one from a charge, its lom pair will go ahead and itself and of course, forming lone pairs on our oxygen, so we can have our negative charges in certain basic conditions. So now if that happens, what's gonna happen is that again, our original molecule, the micro donor remains un intact. So we have our one group, we have R two group with our keto remaining intact. Then we have our R for R four group, the bound oxygen and R three group. Now this carbon will have a, will not have no charge because we have sort of broken or took away this long pair I formed and Perin our carbon. So now I'll just have the correct amount of bands of electrons there. So the mag condition product contains a 15 diar part in the given problem. We have an alpha beta on saturate nitro compound. We will have a 15 relationship between the carbon neal and the nitro, the newly formed carbon carbon bond is the C two C three bond. The bond that was broken during the reaction is the C three C four pi bond. So to identify these starting material, which is our micro donor micro acceptor, we disconnect the C two and C three bond. So take a look at this over here, we can go ahead and number those starting off with the central sort of ketone that I think is. So we have 123, four, we can call the five. And of course, they would just kind of draw it out this nitrogen. How I'll have those two oxygens if we just sort of ignore this oxygen or 02 there. So what's gonna look like is that we have one oxygen there, double bound oxygen there. This single bonded oxygen will of course have three lump pairs and negative one formal charge. Then this nitrogen will bear a positive one formal charge. All right. Again, Mary said that we're disconnecting the C two C three bond. So we're disconnecting right over here to adding my retro synthesis areas, the michael donor, it's gonna be where we have the left side of the molecule, the six memory ring with the two ketone groups. And then now we have a coo et again, this is my Michael donor and from my, my, you have a double bond there and then nitrogen still having that single bond and oxygen with three lump pairs bearing a negative charge and then double bound oxygen making the nitrogen bearing a plus one formal charge. And again, this is going to be of course, my Michael a acceptor. So what happens then is that my Michael donor and Michael Septer will go ahead and react with each other. So we have a negative charge here. So I'll put together an form, see starting off with my Michael donor, we have the coo et so we kind of have like that art group going on then connecting my micro receptor still bearing that negative one formal charge on that single bond oxygen. Now, this step bo and oxygen will now have a single bond as well bearing a negative one formal charge and this nitrogen will have again this plus one formal charge. What's gonna happen there, there's two different parts of the reaction. First one being hydrolysis followed by a dec carboxyl. If we do that, we'll get our final product. Again, the six member bring gear remains intact along with the two key to grips the coo et will go ahead and leave still having that single bonded oxygen with a negative one formal charge. Our double bound oxygen will go ahead and reform still having the nitrogen bearing a formal one or a plus one formal charge. So all of this imper is going to be my proposed mechanism from our proposed micro donor and micro acceptor going to our final product here, these coming together to form this bigger molecule here. And then with hydrolysis and deep carboxyl, this is how we form our product.

Show how the following products might be synthesized from suitable Michael donors and acceptors.
(b)

Key Concepts

Michael Addition

Michael Donors and Acceptors

Synthesis Strategies in Organic Chemistry

Watch next