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(2S,3R)-2-Bromo-3-phenylbutane undergoes an E2 elimination when treated wth sodium ethoxide_ Draw all possible Newman projections for the bond relevant for the elim...

Question

(2S,3R)-2-Bromo-3-phenylbutane undergoes an E2 elimination when treated wth sodium ethoxide_ Draw all possible Newman projections for the bond relevant for the eliminaton reacton; and use those Newman projections to explain the stereochemical outcome of the reaction. Draw the fnal product and provde its IUPAC name_

(2S,3R)-2-Bromo-3-phenylbutane undergoes an E2 elimination when treated wth sodium ethoxide_ Draw all possible Newman projections for the bond relevant for the eliminaton reacton; and use those Newman projections to explain the stereochemical outcome of the reaction. Draw the fnal product and provde its IUPAC name_



Answers

What stereochemistry do you expect for the alkene obtained by E2 elimination of $(1 R, 2 R)$ 1,2-dibromo-1,2 diphenylethane? Draw a Newman projection of the reacting conformation.

So we're giving the following reaction and we want to try to understand why e to elimination would result in the Z I summer rather than the EI summer. And we want to explain this using Newman projection. So I'm going to go ahead and quickly convert our reactant into ah bond line structure. So this is going to more easily help us draw Ah Newman protection. So our bahn my structure are reacting is going to look like the following. I'm gonna abbreviate our final group here is just Ph. Okay, so, um, now we're gonna go ahead and convert this into our ah Newman projection so it's more easily Ah, were easily able to do this. Our number projections going to look like the following. We have our tosel it group for the front carbon are tossed like group is on the top, right? Our hydrogen is top left, and our ah ch three group are Method Group is facing down. And for our back carbon, we have our ch three groups facing up and our general group facing right in our hydrogen group facing left all right, and quickly here, we can actually see that. You know, eat to elimination. Reaction we know requires anti ah anti elimination between our beta hydrogen and are leaving group, which in this case is tossed lit. And that's exactly what we see here. So for a front carbon we have are tossed like here and 180 degrees from that we have our beta hydrogen, so we know that this is going to undergo e to elimination. So we go inshore air pushing mechanism for that real quick. So we're going to have our base come in and the appropriate that hydrogen, beta hydrogen and these electrons are gonna get shared between the front and carbon. I'm sorry. The front and back carbon and are tossed like group is going to leave, right? And so we're going to get the following Newman projection. No, I'm sorry, but that and what we're going to see here is that for our front carbon, we're going to have our looks like r H on the left ch three group on the right, and for our back carbon, we're going to have our ch three group on the left and our final group on the right. All right, so when we draw that out and bond line structure. Um, that's gonna look like the following. So we see we have our as a group up top here on the left, carbon, hydrogen on the bottom and for our right carbon fennel group on the top ch three group on the bottom. And that will be in quickly assigned priorities. We see that for the left carbonara Method group as higher priority over the hydrogen and our final group for the right carbon has priority over the method group. So we see that in fact, we do have the ah, see, I summer.

This is problem 15 of Chapter 27 looking at Before, plus to Psychlo edition of the reaction that drawn here. Since this is a four plus to Psychlo edition, we know it will involve the four Pi electrons of our dying, which we see here, and the two pi electrons of a Dana file. And we'll use one of the double bonds on the Benzo Queen own. And it's a thermal for plus to Psychlo addition, since it's run under heat, as indicated here, which means it is a deal's older reaction are dying has shown here in the S. Trans confirmation s refers to this single bond and we could see the two double bonds are trans to each other across a single bond. But for a deal's older reaction, the dying must be in the S sis confirmation. You can check this for yourself If you try a deal's older mechanism with the S Trans dying, it will lead to a cyclo taxi with a trans double bonds which is impossible to form. So let's redraw are dying in the S sis confirmation and now we can draw our mechanism arrows to see what our product will look like to answer part A of this problem. If we follow our arrows when we can draw our product, then we see we've formed a new six members ring, which is what we expect from a four plus to Psychlo edition. By forming this ring, we've also formed three new stereo centers, which all mark here, here and here part. They also asks to indicate the stereo chemistry at these new stereo centers and to determine that let's move on and look at part B part Be asked us to draw the P orbital's of the bonds we used in this reaction. So let's draw this a little differently to try to visualize how the new bonds are being formed. So here we have our Diana file and are dying so we can imagine that the dying is being introduced from below the Diana file. And now let's draw the P orbital's on this diagram. Here are the pure petals on the data file in the P Orbital's on the dying Ah, thermal Psychlo edition happens between the ground state Homo of one of the reactant CE and the grown ST Louis Mo of the other reactive. Let's draw in the ground. ST Louis, Mo. Orbital's of the Diana file and the ground state Homo orbital's of the dying. Now we could also consider the loom O of the Dying and the homo of the Diana file. You can try it for yourself, and you'll see the result will be the same. So as we can see the lobes of the dying in the Diana file match up perfectly, allowing the two new bonds to be formed. And since we're imagining the dying to come in from below, we can imagine that thes hydra Jin's on the day, you know file will be simultaneously pushed up and out of the way. And similarly, we can imagine the Esther on the Dying will be pushed down and out of the way of the newly forming bonds. So now we can indicate the stereo chemistry of our new stereo centers. The two hydrogen sze were both pushed in the same direction. So there, sis to each other, and since we said they were being pushed up, we'll show them up on our product, and we said the Esther was pushed down, so we'll show the Esther pointing down on our product now, we could also consider the dying to be introduced from above the Diana file, in which case we would draw. The hydrogen is pointing down and the Esther pointing up. This would give us the same relative stereo chemistry, and in this case it would be identical to the product I've drawn here. You can prove this to yourself by building a model of those two products. This completes problem 15 of Chapter 27 for Part A. We drew the product of the four plus to Psychlo edition and indicated the new stereo chemistry. And for Part B, we drew the P Orbital's and explained how the orbital overlap led to the new stereo chemistry.

This is problem 15 of Chapter 27 looking at Before, plus to Psychlo edition of this given reaction. Since this is a four plus to Psychlo addition, we know it will involve the or pie electrons of our dying, which we see here, and the two pi electrons of a dying A file. And we'll use one of the double bonds on the Benzo Quinn known. And it's a thermal four plus two psycho addition. Since it's run under Heat has indicated here, which means it is a deal's older reaction. Are dying is shown here in the S. Trans confirmation s refers to this single bond, and we could see the double bonds Airtran's to each other across the single bond. But for a deal's older reaction, the dying must be in the S cysts confirmation because then s trans dying would lead to a cycle heck scene with a trans double bond which is impossible to form. So let's redraw are dying in the S sis confirmation. And now we can draw our mechanism arrows to see what a product will look like for the answer to part a of this problem. And if we follow our arrows. Then we can draw our product and we see we formed a new six member ring, which is what we expect from a four plus to Psychlo edition. By forming this ring, we've also formed three new stereo centers, which I'll indicate here, here and here part. They also asks to indicate the stereo chemistry at these new stereo centers and to determine that, let's move on and look at part B part be asked us to draw the P orbital's of the bonds we used in this reaction. So let's draw this a little differently to try to visualize how the new bonds air being formed. So here's our Dana file, and here is our dying so we can imagine the dying is being introduced from Be low, the dining A file And now let's draw the P orbital's. On this diagram. You're the P orbital's for the data file and the pee or petals on the dying. A thermal Psychlo addition happens between the ground state homo of one of the reactors and the ground, ST Louis, Mo. Of the other reactive. Let's dry in the ground. ST Louis, Mo. Orbital's of the Diana file and the ground state Homo Orbital's of the dying. We could also consider the loom over the dying and the home of the Diana file. You could try that for yourself to see. The result will be the same so we can see the lobes of the dying and the Dana file match up perfectly, allowing the two new bonds to be formed. Since we're imagining the dying to come in from below, we can imagine that the hydrogen sze on the data file are being pushed up and out of the way. And similarly, the Esther on the Dying is being pushed down and out of the way of the newly forming bonds so we can now indicate the stereo chemistry of our new stereo centers. The two hydrogen sze were both pushed in the same direction. So there, sis to each other and we said they were being pointed out, are pushed up. So let's show them pointing up on our product. And we said the Esther was being pushed down. So let's show it pointing down on our product now. We could also consider the dying to be introduced from above Dana file, in which case we would draw the hydrogen is pointing down and the Esther pointing up. This would give us the same relative stereo chemistry, and in this case it would be the in Antium er of the product I've drawn here. You can prove this to yourself by building a model of the two products. This completes problem. 15 of Chapter 27 Her part A. We drew the product of the four plus to Psychlo Addition and indicated the new stereo chemistry. And for Part B, we drew the P Orbital's and explained how the orbital overlap led to the new stereo chemistry.

Mmm. This is the answer to Chapter 11. Problem number 53 from the Smith Organic Chemistry textbook. In this problem, it says, from what you have learned about finals and the hydration of al kinds, predict what product is formed by the acid catalyzed hydration of this molecule that it spells out there and draw a stepwise mechanism that illustrates how it is formed. Okay. Ah. And so, uh, the methyl ester is the product that's formed. I'm not draw that at the end. Um, and so let's jump right into the stepwise mechanism. And so this is an acid catalyzed hydration. Ah, And so the first step is gonna involve the acid s o the al Kaine. The electrons of of one of the bonds of the triple bond will pick up a proton from the acid and become pro donated. And that will get us to this product. Okay, so we have formed a carbo cat eye on here at this carbon. Um, and so as since this is a hydration, the next thing that's gonna happen is the actual hydration. The addition of water, and so water will come. Uh, and the one of the lone pairs in the oxygen will act as nuclear file and attacked this carbon on. And so we get the addition of water to that carbo cat eye on, uh, and so that gets us to this product. All right. Suppose i should say this intermediate. So there's our water. That's added. It has three bonds and only one lone pair, and so it's gonna have a positive charge on it now, um, and, um, another molecule of water can come along. Oops. Uh, can come along and grab one of these protons. And so, uh, now that's been de protein ated on, so it's just going to be alcohol. Okay, Uh, so too in pairs on it now, um and so what can happen here? Is that one of these lone pairs? Oh, no. You know what? I apologize. Eso before one of those loan Paris can attack, um, we're going to have to protein ate that double bond on DSO. We made a molecule of acid in the previous step, and so we can just, uh, use that molecule of acid to accomplish this. So h 30 plus the prototypical acid. So lone pair. Positive charge. Uh, the um, electrons of this double bond can grab one of these protons. Uh, those electrons will revert to that oxygen. Okay. And so now we've pro nated this double bond. Uh, and so we will have this. Okay, um, and there's ah, carbo cat eye on on this carbon new. And there's two lone pairs on this alcohol oxygen, and one of those lone pairs I can ad in here. And when that happens, we will get this product. So now we're starting to resemble I the metal Lester here. Um, and in fact, the only thing that has to happen is a deep throat nation. And so this oxygen is only gonna have one lone pair. It's gonna have a positive charge on it. Um, but we have, ah, water floating around. Obviously, we've used water several times so far in this reaction. So, uh, we have water with its tool in pairs. Can you grab that proton? Those electrons can revert to that oxygen. Ah, and that is going thio for more carbon. You'll and so will get to our final product here, which, as I said, is the metal Esther. For the sake of completeness, we should put all of our lone pairs in. So there they are. I suppose I omitted them on this other oxygen throughout the entire mechanism. So let me go back and correct that. Uh okay. Yeah. And so, uh, this is the stepwise mechanism for this reaction. Um, it's going thio proceed. Just like an acid catalyzed hydration. Um, uh, however, um, the presence of ah, that other oxygen eyes going to mean that, um, instead of forming ah kee tone as we otherwise would, we're going form the the methyl ester here. And so that's what we do. That's the product on this is the stepwise mechanism for it. And that's the answer to chapter 11.


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