Propose a mechanism for the following reaction.

Question

Accepted Answer

Hello, everyone. In this video, I want to suggest a plausible mechanism for this reaction right over here. So in Robinson and relation is a reaction used to synthesize conjugated cycle hexagons. For this given reaction here, the star materials is a cycle hex non derivative on the right or left and a metal vinyl ketone right over here. So while the product is going to be a conjugated cycle hex unknown right over here. So by definition, then then this given reaction is indeed a Robinson annual ation. Let's go ahead and propose a possible mechanism. So of course, we're basic conditions. Let's go ahead and draw our alpha carbon in. So that's going to be this proton right over here. Of course, we have our hydroxy coming in as a base. Let's draw that in as well. Hydroxy has three lone pairs bearing a negative one formal charge on the oxygen. So this lone pair is gonna go ahead and print itself grabbing this proton on a molecule, the bond between the hydrogen and carbon will go ahead and break forming a new pi bond. This carbon carbon wants to relieve itself. So we're gonna take one of the pi bonds from the carbon E um making it a long pair on our oxygen. So what's gonna happen? Then we'll move downwards here. So we can see the left side six member ring is not touched. So draw it in with alternating pi bonds. Alright. So we have formed a new pipe bond right over here. And then our carbon Neil kind of functional group now becomes a single bond with three lone pairs on an auction, of course, bearing a negative one formal charge. So now we basically have president's going on. So what's going to happen is that theme pipeline between the carbon and oxygen wants to reform? So that's exactly what's going to happen is that one lone pair will form another pipe on there. This pipeline will now become a lone pair on this carbon right over here. So following our arrow pushing, we see again on the left side of the molecule, it remains untouched. So now we do have that lone pair, their negative one formal charge, but we have this ketone going on. So now we can go ahead and bring in our second reagent. So we'll just use this on top then. So what's gonna happen is that this lone pair here is gonna go ahead attack right over here at this carbon. Then this pipe bomb is gonna go ahead and break and form a pi bond here. And then this pipe bomb is gonna go ahead and break and form a lone pair on the oxygen. So I'm moving forward. What's going to happen again is basically we're attaching our new molecule in six member ring remains untouched with alternating pi bonds. Alright. So we have keto no more pie ponds in our original molecule and then were extended in our new addition. Here, we have a pi bond here, single bond oxygen, three lone pairs negative one formal charge. So now we have again sort of like a resonance structure. So what's gonna happen now is the long, personally want to reform that carbon Neil pi bond. Then this pipe is going to go ahead and form a long pair on this carbon. See here, we have a long pair and a negative one formal charge and our ketone is reformed. Alrighty. Now scrolling down here for more space. What happens now is that again, our H20 that we have sort of just floating around comes in. What's gonna happen is that the lone pair on the carbon is gonna go ahead pony itself. Of course, the bond between oxygen and hydrogen will break forming a lone pair on the oxygen. So we're going down here. So what's happening again? The spending ring is gonna remain as is alright. So after we have pollinated that carbon, what's going to happen is again, our hydroxide is going to come in three lone pairs on the hydroxide negative one charge on the hydroxide lone pairs on the oxygen of our hydroxide is going to go ahead and prone eight itself. We're gonna go ahead and break the bond to form another pi bond near the carbon carbon. And this pi bond between the carbon oxygen is going to go and break and form a long pair on the oxygen. Again, I'm just following the arrow pushing that we have The six member ring on the left remains untouched. Alright, so we have the single bond oxygen, three lone pairs negative one formal charge. And then this new pi bonds like so where we know that the carbon steel pipe on likes to reform? So that's exactly what's gonna come down and do. And then this pipe on here that we're just gonna come out here on the terminal carbon and become a long pair sort of like resonance structures. Then we'll go ahead and add that bracket in resonance arrows following air pushing again, we have a six member ring alternating pi bonds, another six member bring here with a key tone on the right. So now what's gonna happen is we have this here did not register at let's go ahead and try that. Alright, double bound oxygen compare on the terminal carbon negative one charge on the terminal carbon. All right. Now what's going to happen is a inter molecular reaction? So this lone pair on the terminal carbon is going to go ahead and check this carbon Neil carbon. The pipeline will go ahead and break forming a lone pair on the oxygen. So go ahead and draw the product of that. We'll have three rings. Then so again, aromatic ring here, alternating pi bonds, another six member ring, we have this auction just sticking out your single bond. Then another six member ring like so and a ketone group, Singapore, an auction has three lone pairs negative one formal charge re bringing our H 20. What's gonna happen is that this negatively charged auction wants to go ahead and promenade itself breaking the hydrogen oxygen single bond to form a lone pair. Then we're gonna go ahead and go down here for more space following the air pushing again, We have to go ahead and draw our 36 remembered rings like so ketone roommates in tact. Now we have an O H group adding in our alpha hydrogen here. What's going to happen? It's kind of repetitive. At this point. We have our hydroxide group coming in again, three lone pairs negative one from our charge lone pair on the hydroxide is gonna go ahead and pollinate itself. The bond will go ahead and break, forming a pi bond. This carbon needs to be relieved of stress. So the pi bond between the carbon and oxygen will break and form a lone pair on the oxygen six member ring here on the left of the molecule remains intact still. All right. So what's gonna happen now is that we have a new pipe bond right over here, single bound oxygen, three lone pairs, negative one formal charge and the alcohol group remains as is. So what happens now serve like again resonance structure? So adding those arrows and the brackets. So what's gonna happen is again, of course, our carbonell reforming so well for my pipe on there and then this, we'll go ahead and form a lone pair on that carbon right over here, right? So new lone pair here, negative one formal charge, we have a ketone going on. Let's end our resonance. So now what's gonna happen is the lone pair on this carbon is going to go ahead and form a pi bond is going to go ahead and inject this hydroxy or alcohol group because we're in basic condition. So we can do this otherwise we would have wanted to eject water instead in acidic conditions. Alright. So now moving downwards here, once we have ejected the hydroxide or the alcohol group, and this pi bond forms here. Now everything will be neutral and that will be our final product. So let's go ahead and draw in those three rings again, we have a ketone group here and new pi bond here. So the reaction starts with the formation of the late, let's scroll up. Alright. So Russian starts with the formation of the late of the cycle Hickson derivative followed by congregate edition of the late with metal vinyl ketone to give a delta die ketone delta die ketone then undergoes a subsequent inter molecular algo con sensation to produce the conjugated cyclo hexane own product right over here. And that's going to be my final answer for this problem.

Propose a mechanism for the following reaction.

Key Concepts

Robinson Annulation

Michael Addition

Aldol Condensation

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