Robinson Annulation - Video Tutorials & Practice Problems
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1,5-dicarbonyl compounds generated from the Michael Reaction (enone + enolate) have the ability to undergo intramolecular self-condensation into 6-membered enones. Are you impressed yet?
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concept
Robinson Annulation
Video duration:
4m
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Are you guys ready to have some serious fun? Good, Because this topic is gonna get nuts. I'm serious. So anyway, we've learned from prior topics guys that won five die carbon eels have the ability to undergo intra molecular self condensation and form six members in owns. Okay, remember, we drew these before. Okay, Now, what's different is that now we've also learned a reaction that makes die carbon compounds. And that's a Michael reaction. Right? Well, when the 15 die carbon Neil is generated from a Michael reaction, that means that you had an unknown and in a late come together to make a die Carbonell. And when that molecule cycle eyes is, that's called a Robinson annual ation. So, essentially, this is an AL doll. Times three. Okay, how does that? Well, because think about it. You had to use one Aldo to make the original Eno right. Write the alphabet. It's actually that's one Al doll. Then use the second algal. The Anna late toe. Attack it via Congregate edition. We needed the congee edition. Now you made the 15 die carbon Neil. But now we're saying is that Now that 15 die, Carbonell is going to psych lies on itself and make a new six member e known. That's the third album, That's all Italy mediated. So I know it's It's, like too much fun, right? Not allowed to have this much fun in or go. So anyway, guys, here's an example of a Michael product, right? A Michael product, because I've got a 15 die carbon eel. And again, I would be faced with the choice. Where do I put the family? So I put it in the blue or in the red position. Well, you're gonna put it in the position that's gonna give you the sixth numbered ring. So if I use the red one to attack this Carbondale, do I get a six or a ring? I get four. 1234 However, if I use the blue one 45 Yes, I get what I'm looking for, guys. So this is called the Robinson annual Asian. So what I'm gonna do first isn't gonna help you guys draw the product and then we'll do one from scratch. Okay, so for this, this molecules really easy to draw Once you have everything numbered out, you can draw your six member bring. And I'm going to continue to keep those positions the way that I had him. Where have Okay, now I just have to figure out what am I missing on each of these? Adams. Okay, so too is gonna be a key. Tone six is gonna be What? Well, six is gonna be that alcohol, right? And it's also gonna have a metal. So that's the product. But we know that these like to dehydrate, right? So then I guess I shouldn't have drawn it in the box. My bad. But the final product of this would be metha here dope on here. And that is your cyclic in on. And we've done this before, guys. But now, specifically, this is called the Robinson annual ation. Because it started off with a 15 die carbon deal that was made through a Michael reaction. Okay, So similar to what we've already done, but just connected to Congregate Edition. So what I want you guys to do for this product is noticed that it was starting off from what is it called from the Michael reaction in this one. I'm starting. Start from Michael. Right, Because what I have isn't a known. And what is going to be in easily? Okay, So I want you guys to react those together, get your Michael product, and then do the Robinson annual ation for that, try your hardest, and then I will come in and save the day eventually.
2
example
Predict the Major Product
Video duration:
5m
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Okay, so let's start off with the Michael product. That part's pretty easy to draw guys because I told you that for Michael Reactions, I always just like Thio draw the, you know, in the same way it was before and then just put the family on top so I would just put my family on the top like this. Okay, so that's not bad, because I know that my English is gonna form there. Fair. Okay, Cool. Alright, guys. So now I have to figure out where to form an email late so that it can Psychlos into a six for a ring I have got, I guess, four possibilities. I have got my black I've got my red, I've got my blue and I've got my green Now notice I've got a bunch of different possibilities here. We were wondering is which one is gonna make the best six member ripped? Well, there's two that I can count out immediately, and that's red and blue. The reason is because red is only four away from this carbon eel blue is only four away from this one, so I can cross those out. I'm just gonna across across okay, So those are gonna work now. It's between black and green. Okay, Black is six carbons away, so this looks like it's a possibility. Green is one, 23456 carbons away as well. But we haven't issue, which is that three of the carbons within on three of the carbons within this chain are also in a ring. So if I use the green one, I'm gonna wind up getting a bys. I click product. Okay. So, ideally, I'm trying to avoid using multiple atoms in my chain that are also in a ring. Okay, Notice that if I use look, look at the difference. If I use black, how many of my atoms are in the ring just to I have one. Let me start. I have too many numbers going on now. Okay. I just hope I hope you guys we're learning about this. Okay? So if I start from black, I have 123456 So notice that only two of the atoms are within the ring. Now, if I start from green 123 the first three atoms of that chain or in the ring, so I'm gonna wind up. Getting a bridge compound is gonna be kind of difficult to draw. So let's draw the other one instead. Okay? And the other one is gonna be more stable. Okay, so now for this compound because I have a ring involved, I would actually recommend against doing my system where you draw cycle Hexen on the number it I would try to rearrange this so it looks more like the one that it's gonna make soon. I'm gonna rotate this bond. If you rotate this bond, it's gonna look more like what it will eventually look like. So I'm gonna draw it like this now. So I've got this molecule here. It's all 10 but now I've got my metha here and my carbon, you'll hear. Okay, so notice that all I did was I rotated it. Okay, Now, this is much better, guys, because if my negative is here and my positive is here, I see that I could just do this, okay? And that's going to a six member bring. So what I'm gonna get is a six member green that looks like this. A new six mirroring that has another six member ring on it. Okay. This is the fun part about using Michael reactions. Okay, so then what are we missing on this thing? Well, this key tone still exists. Nothing happened to that. Now, here, I should actually have an own negative, but that's gonna get protein ated toe in alcohol, and then eventually this is going to be dehydrated. And the direction it gets dehydrated is towards the Alfa Beta. So that means that my n on is gonna form right here. And that's gonna be the final product of this reaction, my cyclic in on. Okay, guys. So as you can see, this is one of the hardest reactions. Uh, inorganic chemistry. It's difficult to visualize. It's kind of bulky, clunky difficulty, even conceptualize. Sometimes I hope that I've made it a little bit more clear. I'm really trying to do this in a step wise fashion. So that every step of the way you're like E guess that makes sense. You're not completely lost. But in the end of the day, Robinson emulation is a product that you'll have to practice drawing because you have to get good at it. Alright, guys. So that's it for this topic. Let's move on.
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Problem
Problem
Provide the product for the following Robinson Annulation Reaction.
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B
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D
If you noticed a methyl group that vanished around 3:50 in the above video, you are not alone! Place a methyl there in your final answer and you will be good:)
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Problem
Problem
How would you prepare the following compound using a Robinson annulation reaction between a di-ketone and an alpha, beta unsaturated ketone?
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C
D
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Problem
Problem
How would you prepare the following compound using a Robinson annulation reaction between a ketone and an alpha, beta unsaturated ketone?
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B
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D
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