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What starting material is needed to synthesize each compound by a ring-closing metathesis reaction?...

Question

What starting material is needed to synthesize each compound by a ring-closing metathesis reaction?

What starting material is needed to synthesize each compound by a ring-closing metathesis reaction?



Answers

What starting material is needed to synthesize each compound by a ring-closing metathesis reaction?

This is the answer to Chapter 26. Problem number 19 Fromthe Smith Organic chemistry textbook. Ah, this problem asks us what product is formed by rain closing metastases of each compound. Okay. And so for a we have, um, a nine Adam chain here s 07 carbons and two oxygen's. Ah. And so we're gonna end up making ah seven member dring with two oxygen's in it, and so that is gonna look like this. Okay, so there's the ring that will make by by the ring clothing closing metastases ce Ah, and then we just need to draw the rest. The molecule. Okay, s So that's the answer to part A For part B. We're starting with, um, with an eight carbon dying chain. And so that is going to close up to make a six member ring. Ah, and so I'm just gonna put this over here to the right. Just there's no space issues. So here is our six member during, and actually, um, let me go back for part. A. I forgot to include, um, the al Keen. So they're at the top of the ring. Would be a double bond. Eso for part B here. Um, are all keen is gonna be here, So we've made a six member drink. Now keenan it. There will be a metal group here. Ah, and then we have our two Ethel Esther's, uh, coming off of the same carbon up top here. C o to e T c 02 e t. Okay, s so there we go. That is part B. Ah, And again, these air just to ring closing metastases products. Um, so I for these generally just count the carbon Jane that we will be cycle izing on. Remember that we're gonna lose two carbons. Eso es. We start with 10 atoms of seven carbons to oxygen's. We lose two carbons. So we make a seven member bring with two oxygen's in it on then for B. We're starting from any carbon chain, so we make a six member drink. Ah, and that's the answer to Chapter 26. Problem number 19

Yeah, This is the answer to Chapter 27 problem number 26 from the Smith Organic Chemistry Textbook. And so this problem is asking us about electro cyclic ring openings. Andi, it's asking us to consider the different products that'll be formed under the two different types of conditions that we've talked about all throughout this chapter. Thermal conditions versus electro photo electric conditions. And so, for each of these electricity flickering openings remember, it's just the same type of electro cyclic reaction we've been talking about. So electrons are going to move all at once and all around on dso. Since we're starting from a ring, we do need to consider the number of bonds, the number of double bonds and our product a supposed to the number of double bonds and are starting materials on DSO for a well, look at thermal first on DH. So when we do this under thermal conditions, the ring will open Rome and so we can draw this backbone where the the ring has opened on DH. We have three double bonds now. 112 and three. There they are. Our method groups over there on the left side are going to be unaffected. And then we need to think about whether this is going to be a con Rhoda Torrey or destroy auditory on DH. So, like I said, we need to consider the number of pie bonds and our products on DSO here. Our product has three pie bots on DSO three pie Vons Odd number on. And so this is going to be a disrobe hitori rotation that yet And so these both start coming towards us. And so they're going to rotate in opposite ways. So this one will end up going up, This one will end up going down. Um, and so again, like I said, this will be destro Dettori s. So we know that since thermal in this instance is destroyed hitori um, we know that photo electric is gonna have to be the opposite. So Conroe auditory and again the product is largely going to look the same carbon sort of backbone of the molecule will be exactly the same. So there we go. Put our double bonds in 12123 Um, and then now what we need to consider is, uh, the stereo chemistry here. So again, they both start over you are coming out of the page if you wantto want to call it that on DH, so they're going to rotate in the same way. So up and up. And so the top bond here will be will be Trans, and the bottom bond is going to end up being A system will bond. And so that's That's our con road. It's worry rotation, and then So when we look ATT be everything's going to be exactly the same. Except that the two method groups that we're looking at are starting trans to one another. So everything's going to be just exactly opposite of how it was in part A. And so when we do the thermal electric sick, lingering opening, we'll get this same sort of backbone that we've been considering. 23 double bonds, those two method groups on DH. Then we'll actually get the the same exact product as we got for the photo chemical ring opening from a So there's the thermal product from B. Like I said, it's exactly the same as the the photo chemical product from a and so again thermal, it's going to be destroyed, Hitori Ah, But since he's method groups were on opposite sides of the molecule on, and they're going to rotate in opposite ways now. They're both in the same orientation. And so conversely, the photochemical, um, conditions are going to give us the same product as the thermal conditions gave us in, eh? So and draw the backbone of our molecule. Put our three double bonds in, put our on affected metal groups and on DH, Then again, these air starting Trans. But they are rotating the same way. And so they're going to also have opposite orientations in the product on DH. So they do. And that's Ah, the answer for Be on. That is the answer to Chapter 27. Problem number 26.

This is the answer to Chapter 27. Problem number 40 from the Smith Organic Chemistry textbook. Ah, And in this problem, we're asked basically to work backwards and to look at each of these three products. Ah, and determined from what? Starting materials. We could make these products by making use of a thermal four plus to Psychlo addition. Eso deals all the reaction. Um And so, uh, probably the best way to do this is to start, uh, by looking at at each of these and thinking. Okay, so So what? Bonds are probably breaking here, Um, and thinking about how that would come about. So when we look at a, um, the bonds that are going to break our this bond in this bond, um, and then if we think Okay, um, so what would those bonds breaking look like? We can draw the movement of electrons because, remember, all of these reactions are at least theoretically reversible. All Paris cyclic reactions, Um, and so that would look like this. So the electrons of that bond would move to there. The electrons of this double bond in the product would move to there, and the electrons of this bond would move, uh, that way. And so when we're looking at this, we see that we actually have an al kind is one of our starting materials, and that's okay for, ah, for a cyclo addition for deals older. Um, that's perfectly fine to have in our kind as a starting material. And so now that we know the bonds that are breaking in the way that electrons are moving, we can go ahead and start to fill in. What are our starting materials? We're going to look like So we're going to have, um, a backbone that looks like that on. Then we will have, Ah, backbone. That looks like this with the al kind are the triple bond. Um and so let's put the season. Why not? So that's, uh, that's fairly straightforward. Um, and then since since we know that we have that in the AL kind, we can go ahead and fill in the rest of the AL kind, these two ch three and C o to see h three. So there are all kind for for a and then we do just need to consider, um, what the stereo chemistry is saying about this other starting material. And so remember, uh, these thermal four plus two cycle auditions are superficial, and so we're going to need to put the same starter chemistry on, uh, these two, um, substitutes over here. Yeah, right. Okay. Um, so there we go like that. Ah. And so that is going to be our answer for part a, um because these two starting materials, when, uh, put together to do ah, four plus two thermal cycler addition will give us the product that were given for a So then, for B, we can go ahead and do the exact same thing so we can look at what bonds are breaking in forming here. So the bombs that were gonna want a break are again bonds there and there. Um, we can just a CZ we did for a weaken. Say Okay. So in order to make that happen, how are you electrons gonna move? So, just like this in a Paris cyclic fashion s. So now we see that unlike in a where we had an Al Qaeda and a dying, we're going to have a dying and an alky so we can do exactly what we did again, we can I start with at least the dying's backbone there, Uh, and then we can look at our Al Keen, and then we need to think about the stereo chemistry on these various subsidiary. It's so we see that the two co two ch three substitue INTs our sister one another. They have the same the same stereo chemistry. Ah. And so since they're cysts one another in the product there Ah, we're going to make them sister one another in the starting material as well. So Theo, too ch three go to ch three, okay. And so that's the AL keen piece of our, um or the Al Keen reacting of the two reactions were asked to draw. Uh, there is that piece and then we just need to look at how the substitue INTs on the dying are going to look. And so, for a they were sister one another. But here and be, they're trans to one another then, So we need to account for that so we can draw the tops of situ int exactly the same way that we did in a has the same stereo chemistry it's dashed on then to get the Trans Ethel Group on the other side to get the wedged at the group. We're just going to have to, uh, change the stereo chemistry from how we had it. Annette. So for a we had, uh, both of the double bonds had, um, Trans Uh uh e chemistry. So we're just going to change them? You are changed this one at least, and make it make it easy. So it'll be it'll be cysts. So there we go. Um, that's how that will look on. When these two reactions were put together, we will get the product that we were asked to start from s So then for C, we're gonna follow the exact same pattern that we followed for and be so there are the bonds that we want to break Here is the movement of the electrons that we need to facilitate that. So again, we're gonna end up with a dying and an AL Keen so we can go ahead and, uh, draw at least the pieces that we know to get started. So dying Al Keen. So now, um, we see that the two co two ch three groups are trans to one another here. So in order to make them trans to one another in our product, we will go ahead juicy h three and make them trans to one another in our starting material. So, um so that's the AL Keen piece handled. And then for the other piece we see that are too Ethel groups are once again trains to another and wedged her Pardon me and dashed. And so we can go ahead and, ah, draw them the same way that we have been one. And there's two. So that matches how they were drawn in a endeavors from how they were drawn and be. And of course, that makes sense because that is the stereo. Chemistry of those positions correlates with different ways that we are drawing starting materials. And so this problem is a good overview on how the stereo chemistry, at each possible position of the dying and the Al Keen or Al Qaeda, goes on to effect your stereo chemistry in the product. And that is the answer to Chapter 27. Problem number 40

Okay, So for an algal reaction, we have, say, some nuclear file attacking some all the hide the nuclear Filic Alfa Carbon retains the car venal nature. But the alga hide is transformed into an alcohol. No. So keeping this in mind, we can easily go backwards from this beta hydroxy key tone we think of the nuclear file is being essentially the exact same with the alcohol came from Thie Aldo Hide. Same thing can be said for these other products. Here we have the beta hydroxy key turn acid 10 will be our nuclear file. On this other side will be her Elektra Filic. I'll go hide. Finally we have something that looks slightly different. We have this Alfa Beta unsaturated ketone on this case. You just have to remember that if we take our beta hydroxy key tone when I go through a dehydration reaction Well, actually lose water at this side. The form, the alphabet on Saturday and system so really at this position here was were are hydroxy group originally was. Which means that's where are I'll go. Hide was originally so this alf bid on saturated system really came from this nuclear file reacting with this Aldo hide


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