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(a) Draw the expected alkene product for the following reaction clearly indicating product stereochemistry Define the stereochemistry of the product as E or Z. (2 ...

Question

(a) Draw the expected alkene product for the following reaction clearly indicating product stereochemistry Define the stereochemistry of the product as E or Z. (2 points)4,GNaQEl EIOH(b) Using Newman projection(s) . draw the mechanism for this reaction that clearly rationalizes the stereochemical oulcome Irom your answer (0 pan points).

(a) Draw the expected alkene product for the following reaction clearly indicating product stereochemistry Define the stereochemistry of the product as E or Z. (2 points) 4,G NaQEl EIOH (b) Using Newman projection(s) . draw the mechanism for this reaction that clearly rationalizes the stereochemical oulcome Irom your answer (0 pan points).



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.

The first part of this problem assets to draw the mechanism of the reaction of tube you tine with one equivalent of beer to. So here's our to view time. So the first thing that will happen is let's make your to like that. So the first thing will happen is one of these pi bonds will come and attack the bro mean? And then the bro me in one of the lone pairs on the Berman will actually come and attack one of the carbons as well. And we'll kick off that second burning. So we get a transition state that looks like this. I only have two bonds there. We used one of the pie bonds to attack. The brewing is attached to both carbons, and it's positive because it has two bonds. We also have this br minus leftover. Um, now the br minus going to come in and attack one side. Um, but because this PR pyramid transition state hurt, not transitions. Intermediate is so big, it has to come in from the opposite side. It cannot attack from the same side that this bromate is on because there's no room. So this bro means on the bottom. This one has to come in from the top and attack one carbon. And then whichever one it didn't attack. Uh, we'll get the other brought me that was already there. So our products will look like this. So if we put this roaming on the bottom, this one has to be on the top now, the more proper way to draw that for an all keen would look like this. So roaming here, rooming here. And then we have our Russell groups on the second part of the question asked us to predict the configuration of the product. So basically, what they're asking is is this year see, because it's an Al Kane s o E means it stands for end gegen, which is the German word that means away from. So it's basically Trans and, uh, Z is what the German work Suzanne man, which means together is cysts. So it's easier for you to think of it like that. Um, this, in this case, the bro means are trans to each other or e. Um, so we know that because if we look at this line right here, this plane that the double bond is on, they're on opposite sides of the plane. Um, so this is and e configuration

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.

So we're giving the following reaction. We have an organic compound that is being reacted with acetic acid and heat were also given that this occurs with anyone Elimination reaction. So we want to go ahead and predict what products we're going to get from this reaction. And of course, the first step we know for anyone elimination reaction is that where are leaving group is first gonna leave, So that would be our bromide. Okay, I'll highlight. So that will leave. And we're gonna get the formation of a carbo Kati on here. In our case, it's going to be a tertiary carbo cat down. And now we can go ahead and do our deep throat nation with our acetic acid. See the gas. It is a pretty weak base, but once we put a lot of heat into it, it's capable of deep resonating. So let's say we're going to d a pro. Nate, this hydrogen here. And so we have our acetic acid right, which is going to come in now a pro. Nate, that and the electrons of this hydrogen are going to get pushed down here, so that is going to give us the following product where we get this formation of this south keen now we can also de protein ate at a different hydrogen as well. So let's say since we have our car broke a tie on intermediate here would say we Oops, make that read here the appropriate at this beta hydrogen. So once again, our acetic acid will come in here and deeper. Nate, this hydrogen, the electrons are going to get push down two to form this new pi bon. And now we get the following Ah all keen product. Well, we also get one more product as well, because we have to think about the hydrogen is on the method group. Right? So let's say you know, make that red. Here there's three hydrogen zahn this method group, but I'm just gonna draw one of them. So what about that hydrogen there? And once again, this is drawn from our tertiary carbo cattle on. So now we can go ahead and perform are deprived nation of that, using our acetic acid again. That's gonna come in Andy Pro Nate, that and ah, we're going to get the electrons from the hydrogen coming down here informing this by bond and now we're going to get the following product are all keen product. And now we want to go ahead and predict which one is going to be the major product. Well, we know that the, um, sidesteps rule the more substance waited, um, or substituted. Sorry, Al Kane is going to be the more, ah, major product. So let's go ahead and count our substitutes for you talking for the 1st 1 we have for constituents. And for the 2nd 1 here looks like we only have three cest issuance. And for the last one here, we only have to serve situates. So since our 1st 1 has forces issuance it, there's going to say Viet sites that's rule, that this is going to be the major product.


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