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40 points Which cydlic Ostcr (4 cord) first undcrEocs hydrolyscs toyicld a ring-opcned beta-keto acid and akconol which then undcrgoes decarboxyIation to givc COzan...

Question

40 points Which cydlic Ostcr (4 cord) first undcrEocs hydrolyscs toyicld a ring-opcned beta-keto acid and akconol which then undcrgoes decarboxyIation to givc COzand thc product shown Hint: a carborylic acid and akcoholcanreact to yicldan&ster One answer+C0zwater heat

40 points Which cydlic Ostcr (4 cord) first undcrEocs hydrolyscs toyicld a ring-opcned beta-keto acid and akconol which then undcrgoes decarboxyIation to givc COzand thc product shown Hint: a carborylic acid and akcoholcanreact to yicldan&ster One answer +C0z water heat



Answers

Predict which acid of each pair is the stronger acid. Briefly explain how you arrived at your answer. (a) $\mathrm{H}_{3} \mathrm{PO}_{4}$ or $\mathrm{H}_{3} \mathrm{~A}_{5} \mathrm{O}_{4} ;$ (b) $\mathrm{H}_{3} \mathrm{As} \mathrm{O}_{3}$ or $\mathrm{H}_{3} \mathrm{As} \mathrm{O}_{+}$. (Hint: Review Chap- ter 10 and Appendix $\mathrm{F}$.)

Okay. This problem is asking us to predict the products of these directions if we're interact each of these with age to us before and heat. So the purpose of specifying that their seat in the directions is because in reality, if we didn't have heat, then we could have a mixture of substitution and elimination products. But by specifying heat, we know that we're going to prefer to perform the elimination product. So that is going to be our major product in each of these. Okay, So knowing that I'm going to erect this alcohol with age towards a four and that's a sight, a signed note is that this is a tertiary alcohol and it's a Chirchir alcohol because the alcohol is right there. Here's my carbon that that alcohol's attached to, and that carbon is connected to 123 other carbons. So for each of these, I would recommend identifying, um, what alcohol you have, whether it's primary, secondary or tertiary, and then proceeding with the reaction because that you can kind of figure out whether you can do we one or you to. So because I do have an E, a tertiary alcohol in this one I know I'm gonna prefer to dio e one. And that's because otherwise I would have too much Sterk endurance to proceed with an e two reaction. Okay, so I know that I have aged to us before, which is a super strong acid. I'm going to protein ate my most basic compound, our most basic Adam on this molecule, which is my auction. So it's gonna form always to just water connected to my carbon. And because this is an a one reaction that that means I'm gonna have a you nima left transition state. So that means I'm gonna have to, um, sequential steps not simultaneous. Just two different steps. So I'm gonna result in the explosion of water. I'm gonna have this as my intermediate, in which I have basically the same exact compound, the only difference being that I have the boss of water and a positive charge. Now on this carbon. Okay, so any to reaction is where a longer believes. Hence, this water and then a base such as water, because I just had that leave. A base comes in and deep protein. It's a hydrogen to eventually make an Okay, okay, so that, um, That water has a couple of different options it wants to relieve. They want the whole purpose of making it okay is to leave this carbon of its positive charge. So in order to relieve that carbon of its posits charge, I must de protein ate a hydrogen, which is adjacent to this carbon. So what I mean by that is this 100. If r two d promenade that it would make enough can right there. If r two d probate this 100 it would make it okay right there. And if I were to Deep Throat Nate, this 100 it would make an RPG right there, all representing the, um the relief of this positive charge. So I can't depreciate this 100 or this one or this one, because they're just too far away would have nothing to do with relieving that carbon of its posit. Charge. Okay, So knowing all that information I'm going to go ahead and do this step by step are very methodically. So first things first, I'm going to dip, rotate that 100 resulting in the loss of a hydrogen, and I'm gonna move the electrons from that carbon 100 bond onto the signal bond to create an OK, OK, and when I do that, it'll make an arcane right here. Okay, so this is considered to be a die substitute. Okay, we know to die substitute O'Kane because if we're looking at what's only connected to my Hokkien, we have those two carbons and we have to identify what is connected to those two carbons. So we have connected to this carbon. We have three to hundreds, which aren't considered in the naming of the substance on my okay, but we do consider carbons. So I have this carbon and this carbon immediately connected to that carbon. So that means that I have a die substitute lk So dai subsidy tokens are relatively stable, but obviously we would want it to be as stable as possible, which would be a catcher substance, cocaine, in which we have four connections. So I would have to connection to this carbon and then two connections to this carpenter said you have Okay, so no one all that information, I'm gonna go on to the next one because this one isn't particularly favorable, right? I want to be very favorable to make a die substitute. Okay. Okay. The next thing I'm gonna de protein ate the left hydrogen, which is this one. If I d printed that 100 I'm going to result in a knocking right there. Okay, so that is considered to be a try. Substitute Joaquin. And it's a try subsea joking, Because here's my, um, component of the arcane, my two carbons, and connected to this carbon. I have one carbon and then connected to this carbon. I have 12 carbons that's two plus one equals three. That's a try. Some state. Okay. Okay, that's pretty stable. But again, we want to make sure that we can, um we have at least we exhausted. Our eggs are our options. So knowing that I have a tricep see that Okay, in there, I'm gonna move onto the next one, okay? And that would be the loss of this proton via H 20 If I lose that proton, I'm gonna form the, um, the double bond right there. So making elkin right there. And will this be this one would be a tetra subsidy. Okay, Because connected this carbon, I have 12 carbons in connected to this carbon. I have 12 carbons. So that is a tetra substitute. Okay, that is considered to be my most stable. Okay, someone perform. This is my major product. So see, saves three. And then this is a double bond. No. And then this carbon is connected to not a hydrogen anymore, because I got ready. That one is connected to C H three and see if three Okay, So this is my most stable, um, product. And it's my major product. Okay, next up, we'll do this one. So this is an alcohol and the same thing as always. We're going to go ahead and protein ate that oxygen. So it's gonna result in O h two or the positive charge all connected to that carbon. Okay, So again, this is either going to behave in an e one or e T. Reaction. We're going to do the one because with secondary of a choice between a one and a two, it's gonna preferentially Dewey one. I'm gonna result in the loss of that, um, water. I'm gonna have this as my product released my intermediate. So have that volcano unaffected. And I have that positive carbon got on right there. Okay, So now I have a choice. I have to relieve that carbon of its posit charge. So I have to De Protein ate one of these two hundreds, right, Because those are connected to the carbons which are immediately connected to that carbon. So the hydrogen right there, they had you right there. They're gonna get deeper donated by water. But only one of them is. The question is, which one? So if our two deponent that one old results in an AL pain right there, so this would be the product of that one. So that and then if r two d protein eat this one, do you want on the bottom of the results in and Elcano right there? So I would get this product. Okay, So which one is considered to be the major product? It's going to be this one, and that simply because in this one, we form a conjugated system, which is not necessarily important for this particular problem, but conjugated system is considered to be more stable than not having a cardigan system. Okay, so that's just an basically alternating signal bonds and pi bonds, which is pretty stable. Okay, so next up we have this one. So this one, this is Oops. I have to wonder 100. Ok, so this one is a secondary alcohol, and that's because the carbon in which my alcohol is attached to is connected to 12 other carbons. Okay, so that's a secondary alcohol. So I know I'm gonna be focusing primarily on an E one type of reaction. Okay, So first things first, I'm gonna go ahead and Protein ate that alcohol forming. Oh, h two or the positive charge. Okay. And again, I'm gonna make this alcohol slash water leave, resulting in the formation of this product or intermediate to duplicate that. Move it down, and I'm gonna have a positive charge on that carbon that my water just left off. Okay, so there's my positive charge. And in the previous ones, we had our posit charges, and then we just immediately had our corresponding bases depot, Nate, the neighboring hydrogen. So we conform in all cane. But this one, we have a secondary carbon iodine, and that's that's not bad, But we have the potential to be to do even better. So similar to some previous problems in which we moved our carbon carbon cotton around so that we can form a more stable um carro cada an intermediate. We're going to the same exact thing here, except we're gonna do it in an E one type of reaction. So in the previous ones, we did it with s and one. Now we're just doing the same thing. Nothing except with an e one reaction. Okay, so I have my secondary carver cut on. It has the potential to be even better. And I'm gonna do that by moving this muscle group onto that carpet. So by doing that, I will leave this carbon of its posit charge, and I'll move my posit charge onto this carbon and that carbon because it's connected to 123 other carbons. That's gonna be a tertiary carbon iodine, as opposed to a second date card cut out. Okay, so let's go ahead and do that. Gonna move this car metal group onto that carbon resulting in this So ch three ch two c h, which is not connected to a metal group. And then I have the connection to that Carmen, which is connected to C H three and C h three. Okay. So again, this is gonna have a positive charge now, because I moved electrons away from that carbon and onto this one. So again, this is going to behave in an E one type of reaction. So I need to focus on the hydrogen is which are surrounding this carbon. Okay, In this case, it's going to be either this one, this one or this one. And if I were to dip, rotate this one, it'll be the same dynasty permitted that one. So I basically have two options. Okay, So if r two d prone eight, this 100 results in the formation of a now keen right here, okay, that would be considered to be a die subsea joking. That's because if I'm only focusing on what's my talking? I have the connection to 12 of the carbons on this carbon and no carbon with that one. So that's a dissected talking that's not particularly stable. We have the potential to be even better than that. And we're gonna do that. We're gonna do that by deep resonating in this region. If I d Pernetti that hydrogen and I will result in an Al cane right here, and that Al Kane is going to be considered to be tetra substantive Because I have this carbon, my 12 carbons and connect to this carbon. I have my 12 apartments. That is a Tetris subsidy. Cocaine that's gonna be considered to be the most stable. Okay, so seize three. Stage two. See, it doesn't have 100 any longer because it has double once said. And then I have the connection to my C C H three and C C history. Okay, so that is my major product. My Tetris substituted. Okay, OK, moving on to this one. So this one is a primary alcohol, and the reason we know that is because here's my alcohol. Carbon in which my alcohol is attached to, is connected to only the one other carbon. Okay, so this is going to proceed in an E two type of reaction. So again we have The choice is between accent one s and two e one and e two. We eliminated the S and one s and two because we have the presence of heat. So we know we're only gonna have elimination products, and then we eliminate at e one because he won and focuses primarily on tertiary and sometimes secondary alcohols in this case, but primary is exclusive. T two. Okay, so knowing that I'm going to go ahead and Protein ate my which resulting in a one to minus O h two with a positive charge and that will result in the eventual loss of that as water. Okay, so when to draw out the bond so I can show the so I can demonstrate this more easily. So if I have always to that is a better leading group, so I can make that leave eventually. But again, I can't make that leave without forming a primary carbon iodine, and we know that that isn't very stable. So what I'm gonna dio instead is I'm going to go ahead and alter this in a way. So if I were to move this around, it would result in this compound. In fact, I'll just go ahead and you draw this so I'll get this a primary carpet cut. I'm which we know is not very stable. So what I'm gonna dio is I'm going to utilize this hydrogen to relieve this carbon of its posit charge and move the primary carbon cotton around to eventually make a secondary carpet. So I'm going to go do a hydride shift so connected to this carbon, I have a hydrogen, that hydrogen is gonna go and basically attack that carbon forming ch three on that carbon and then only ch on the other one. It's like this series three. I see it too. Ch two c h and then series three. So, of course, this one is gonna have a positive charge because I got rid of electrons. Okay, so the purpose of doing that was to show that I can't have a primary carbon iodine unless I have the potential to be even better than that. Okay, so in this case, I had the potential to make a secondary car because I'm so I utilized that. Okay, so I moved the primary carbon got on to my ch, So no have a secondary carbon iodine. And now is where I can form my out keen. So I got rid of water. That water can come in after the base and D protein ate a hedge into get rid of that positive charge on my carbon so I can either focus on this carbon with 100 or this carbon with their had reduced. So if r two d Protonix these hundreds uniform in l cane right here. So this would be considered to be a model subsidy cocaine, because here, my carbons involved connected to this carbon. I don't have any carbons and then connected to this carbon. I have just that carpet. So that is gonna be amount of subsidy looking. Whereas if I were to deeper tonight, the other one So if I were to deprogram it, this one than our form an arcane right here, and this one would be a So here's my Here's a component of my okay right there. This would be a die subsidy. Okey. Because I have a connection to this carbon for this carbon and this carbon for this carbon. So that would be a die subsidy looking, which is considered to be much better than a mono subsidy. Okay, Okay, so this is going to be my product. I'll just go ahead and read all this. So Series Three it's used to see Sorry. I think this is going to be ch and then C h ch three. Okay, so that's that moving onto the next one and this one. I have my primary alcohol, several similarly, and I'm gonna go ahead and Protein ate that O H two or the positive charge. So if I were to make that leave would result in a primary carbon cut on which I'll drive right now. So ch two double bonded to ch ch two and then ch two with a positive charge upon that Carter. So again, I don't like primary carpet patterns because they're not considered to be stable. If I have the potential, make a secondary or even tertiary car because I'm going to do that. So what I'm gonna do is utilize similar, similar to the last one where moved this 100 over. I wouldn't do that same thing. So that's gonna go ahead and make this intermediate. Soc is too ch ch. And then that's gonna be a CH three over here with a positive charge upon this one instead. So that is going to go ahead and be stabilized by the presence of this arcane so that all can can move electrons around to make that more stable. So now I'm going to utilize the water so water is gonna come in and Deep resonate Hodgins to relieve this of its positive charge. So if I were to deep rotate this one, then I would result in an AL keen right here, Okay? And it all came right there isn't particularly stable because that would form and SP hybridized carbon, which isn't the most stable thing in the world. OK, instead, I'm going to D protein eight, um, this 100 year, resulting in the formation of this product, two C H two connected to ch connected to ch connected to C H two. Okay, so this is considered to be a, um I just made a price mono subsidy looking, but it stabilized because I formed a conjugated system. Okay, so, Ahmad a substance tokens aren't the most able thing, but if they're part of a conjugated system, they're stabilized. Okay? And that's basically just a set of pi bonds separated by a single bond separated by a pie bond. Okay, so that's that Next up is this one. So this is a primary alcohol, and that's we know that because the carbon in which my calls attached to is only connected to one of the credit. Okay, so that's my second start, My primary alcohol that's going to go ahead and get pregnant by my H two s 04 So we're gonna have a positive charge right there and again. I'm gonna go ahead and make that water leave, resulting in the formation of this compound. So see H with a positive charge on that carbon. Okay, so again, we want to form the most stable possible alky. So what I'm gonna do is utilize water. Water is going to come in, and d protein 81 of the judges to make a knocking somewhere. But what I'm gonna do instead is I'm going to utilize, um, basically shifting stuff around. I'm gonna move this 100 onto that carbon. And in the process of doing that, I will move my car, but cut on around. So go ahead and make my series three and a four minute positive charge right there. So I went from a primary carbon time to instead a tertiary carbon cotton, which is much, much more, much more stable. So now I can go ahead and utilize my water so my water is gonna come in and attack a proton. Um, let's see what happened if I departed this hydrogen. If I d printed that 101 former Nelkin right there, which is considered to be a subsidy docking. If I were to D protein ate this 100 you that I would form a try subsidy. Okay. And try substitute is much more stable than a di substituted. Okay, so that's what I want. That's the one that we're gonna you place. So again, I'm deep resonating because hydrogen and moving the electrons onto that bond. So that's gonna form this product. That is a try. Subside, joking. That's pretty stable. So these are all the answers.

This is the answer to Chapter 12. Problem number 47 Fromthe Smith Organic chemistry Textbook in this problem asks us to draw a stepwise mechanism for the reduction of a pox side, a toe alcohol be using lithium aluminum hydride. And then we're also asked what product would be formed if lithium aluminum Ah, do Tara Reid were used as a re agent. Ah, and were asked to indicate the stereo chemistry of all stereo genic centers in the product using wedges and dashes. Okay, um and so, um, the entire point of this problem is, too, to hammer home the idea that lithium aluminum hydride does not deliver a hydrogen to the oxygen in this situation on that holds true for the reduction of key tones and Alba hides as well. So it delivers a high dried to it carbon never to an oxygen. Um and so with that in mind, here's the stepwise mechanism. It is fortunately a very short mechanism, so that's always pleasant. But so the hydride will be delivered to this carbon. Um, and now, in terms of stereo chemistry, since the pox side is on dashes and so is facing away from us the hydra. I just gonna be delivered on our side of this molecule so it's gonna end up on a wedge. Um and so you don't necessarily need to indicate the stereo chemistry there because it's not a stereo genic center anymore. Now that it has two hydrogen is on it. There's no Kat reality there. Remember, a requirement of Cairo ality is four different bonds. But waken indicated here, um, you know, just just to do it. So there's there's hydride that we just delivered. I'm in. So here on a dash is the alc oxide that now results from from that high drug delivery. So this oxygen has a negative charge. Three lone pairs. Um, it would like to make another bond. I mean, so here is where the water work up comes in. So whenever we use lithium aluminum hydride, we always tend to finish with a water work up. And this is the reason why So this oxygen one of the loan Paris Congrats. One of these hydrogen sze. And so now we've pro nated our Alcock side. Ah, and we have a neutral molecule. And indeed this is the final product that we were asked to draw the mechanism for. So here's our final product. Um, metal group put this hydrogen in here against for consistency. Um, and now we have our alcohol there. So, um, so that's part A. As I said, this is a very simple, straightforward mechanism. Um, and so if we were going to use lithium aluminum to tear eyed instead, eso part be here on the product would be exactly the same. Except where we delivered that hydride we would deliver. Um, e yes. It would be called a Dou te ride. Uh, so ah, were we delivered that hydride? We would instead have, uh, deuterium there. And so nothing else about the molecule changes. Our final product is the same. Sterile chemistries are the same, but here we now have a deuterium. And so again, the whole point here is to just hammer home the idea that lithium aluminum hydride delivers its hydride never to the oxygen, always to a carbon that is bound to the oxygen. Whether it's the reduction of a pox side like this reaction or the reduction of an alga Hideaki tone, it always works that way. And so that's the answer to Chapter 12. Problem number 47

In this problem, I can write the idiots Annette at sea people burn CH In pageants of one HDD and so forth, 20% As two or so four. It will change into see http. See it too. This is a and it is in position so see http. Mg. X. As to it will Tending to CH three c h o H CH three. And when it is oxidized in pageants of an oxidizing event, it will change into CS Teresi. You see http so option B so option B is correct answer for this problem.

Question is which The products opened via Oxy Mercury Ation of Built one. I'll be So build one is not this product, and when it is react with oxygen circulation circulation. Then, with the help of Marcon Kickoff Edition, which will add on the the stable cargo goddamn carbon without rearrangement. That's why. Which will lead on this carbon? So CS three CH 2 c. Well, that's a double bonus yesterday, But this is unacceptable because of it. Can open your eyes and give your debts. He has. He has to CST. Thank you. So, here's the very problem.


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