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3B3ADraw the most stable isomer of 1,3-dimethylcyclohexaneDraw the most stable conformation of cis- -1-chloro-2-isopropylcyclohexane3C3DDraw the ring-flipped confor...

Question

3B3ADraw the most stable isomer of 1,3-dimethylcyclohexaneDraw the most stable conformation of cis- -1-chloro-2-isopropylcyclohexane3C3DDraw the ring-flipped conformation:Draw the stereoisomer of 2-bromo-1,3- dimethylcyclohexane which is LEAST susceptible to E2 reactionsCH3Br

3B 3A Draw the most stable isomer of 1,3-dimethylcyclohexane Draw the most stable conformation of cis- -1-chloro-2-isopropylcyclohexane 3C 3D Draw the ring-flipped conformation: Draw the stereoisomer of 2-bromo-1,3- dimethylcyclohexane which is LEAST susceptible to E2 reactions CH3 Br



Answers

cis-1,2-Dimethylcyclobutane is less stable than its trans isomer, but cis-1, 3-dimethylcyclobutane is more stable than its trans isomer. Draw the most stable conformations of both, and explain.

This is the answer to Chapter four. Problem number 27 Fromthe Smith Organic chemistry. Textbook on. And there are several parts to this problem. So we are asked to consider one to dimethyl Psychlo. Heck sane our ask to draw structures for the system. Trans. I simmers using a hexagon. I'm instead of the chair confirmation. So then and be were asked to draw two possible chair confirmations for assists for the sis I simmer on and were asked which is more stable than were asked to draw two possible chair confirmations from Trans I simmer and which is more stable. I'm and then we're asked overall, which I summer cysts or trans is more stable and why? Um Okay, so, uh, to start, um, start with a So we're asked to draw a hexagon, and I should slow down, draw better hexagons. So where has to draw the hexagon version of for each cyst and trans? And so this is 12 dime Ethel. Uh, so the cysts, it's gonna look like this with both methyl groups. They could be both dashed or both wedged. Um, I always just default toe wedged. Ah, but both dashed would be acceptable to Justus, long as they are both the same. Um, and then for the trans, um, again, it doesn't really matter which is which of one needs to be wedged and the other needs to be dashed. S o. This is insists when they're on the same side and trans when they're opposites. One another. So then four part B of this problem we're asked to draw to chair confirmations for the cysts. I simmer. Um, so that is going to look like this. Okay, so there's our chair, um, and four assists. We're going to want to put them both on the same side. So there's a c h three and here would be a ch three, so both up will put them on the same side. So there is both of them up. Um, so then we're asked to draw both possible chair confirmations, so we'll have to flip our chair here. So that's gonna look like this when the chair flips. So now they were both up, So, um, will keep them up. Um, but of course, uh, one was actually own, one was equatorial and one will still be axial and the other equatorial, so As I said, we'll keep them both up. So here's our axial warn. And here's our equatorial one. Um And then is there more to port be which, uh, which if either is more stable and the answer is neither is more stable. They each have one axial and one equatorial method, and so neither one of them has an advantage. Um, so then we can look at sea and C says to do the same thing. But for the Trans I summers, or rather, the trans I simmer and so we can do the same thing. Draw a chair. So there's our chair, Um, And then let's see. Uh, so we can keep this math a group as it waas There's a method group. And then, of course, the difference is going to be this method group. Uh, now the equatorial or no, I will now be accident. Right? So let me Actually, it's sort of unambiguous. Um, the way that I drew it was ambiguous. It was not not equatorial or actual. And I can tell you from lots of experience grading exams, professors and T A's that grade hate to see that because it looks like you're hedging your bets if you don't know whether or not you're trying to say it's actual or equatorial, Um and that is a good way to get no credit for a question. So even if you're not 100% I would always say pick one and make sure it's decidedly axial or equatorial. Um, okay, so the flipped chair version of this is going to look like this, right? And so, um, this time what we can definitively say eyes here, both of these are axial. But in this flip chair, the second chair, both of these are equatorial. And so, since both of these are equatorial, uh, this structure on the right is going to be more stable than the structure on the left. Um And so that leads us into part D of this problem. Uh, which says Which of these is more stable? And the answer is going to be the trans I simmers more stable because there's a chair confirmation. Ah, And again, it's the one that we just drew on the right, right above this chair con for mation with both methyl groups equatorial. And remember that, uh, having large substitutions and equatorial positions is going to minimize their Starik interactions. And so it's going to lead to more stable, lower energy structures. Um, right, so not Equatoria Hills. Right? So the answer is Trans because there's a chair confirmation where both metal groups are advertorial. Um, And again, um, I'll circle it here and read. So it's this confirmation. Um, yeah. And so, uh, that's the way to approach this problem. We just need to remember what assistant trans mean. We need to remember how axiron equatorial cyst issuance work on the chair structure on. And then we just need to consider which are going to, you know, have interactions and which structures were going to minimize Starik interactions. Um, and we always want the fewest. The most minimal hysteric interactions possible. Uh, and that's the answer to Chapter four. Problem number 27

This is the answer to Chapter four. Problem number 57 from the Smith Organic Chemistry Textbook. And in this problem, Ah, we are given three molecules, three cyclone vaccine derivatives. Um, and we're asked to draw the system Trans. I simmers using a hexagon for the six member dring. Uh, we're then as to draw two possible chairs for the cysts. Two possible chairs for the trends. Talk about the stability of each of those on and then talk about whether the cyst or trans itself is more stable. And so, um Okay, so to start, um, well, so I've drawn of drawing each of these on its own page because there's kind of a lot of drawing in this answer. So rather than crowd a couple of women on one page, I figured that was easier. So ah, for a here, um, our sister Isom Er, remember, sis is going to mean that both of these substitutes are on the same side of the ring and so you can do both dashed or both wedged. I myself always just stick with wedges. Um, if if I'm not, you know, from given the choice. So here's a s. Oh, this is gonna be assists. Uh, the Trans is going to be with the two groups on opposite sides of the ring. And so, in order to do that, um, all that we have to do is make one of these dashed and the other wedged so we could do it like that. We could do the opposite of this. It doesn't matter as long as one is one, and the other is the other. So that's a Furby. We're asked to draw these, um, in their chair confirmations. Um, okay, so we're asked to draw, uh, the two chairs for the cysts. So the first chair, it's gonna look like this. So these substitutes have a 13 relationship, so we need to put them on the same side. I'm so probably the easiest way to do that is going to be to put them both actually, here to start in a 13 relationship. So there they are. They're both axial. Um, and they're cysts. They're on the same side. Um, the other possible chair confirmation. Four assists, eyes going to be flipped. And so that's gonna look like this. So here is our flip chair. Um, and So now they are each going to the equatorial, so that is going to look like this. So there we go. Um, so that's what it would look like with both of them in equatorial positions again. They're both their their sister one another. They're on the same side. Um, so then for part c were asked to do the same thing for the trans version of this molecule. So that's gonna look like this. Oh, you know what? I did forget something here, didn't I, um so were also asked to talk about er which of these chair confirmations is gonna be more stable. Um, and so, uh, looking at looking back, it be, um So here in the first chair for B, they're both axial. Ah, in the second chair. Both of these substitutes are equatorial. So, uh, this one is going to be more stable, right? Because substitue INTs in equatorial positions make for a more stable molecule. So we have a version here where they're both equatorial versus a version where they're both axial. So obviously the double equatorial is going to be other more stable molecule here. Um, I don't like the way that I've drawn this chair, so I'm gonna erase it on redraw it. Okay, so for C. There we go. That's better. So, for See, um, we could do the same thing. I'll put them at the same positions I'm and so it doesn't really matter. Uh, which way that we do this? We could do it either way, as long as we draw ah, one actual here and the other equatorial so that they're on different sides of the ring. So that's gonna look like that. Um, and of course, the flip chair version of this is gonna look like this. Ah, And so when we flipped them, the one that was axial will now be equatorial and the one that was equatorial, I will now be axial. And in terms of which of these two is more stable. Um, so the, uh, the isopropyl group is a larger group than the math group. And so when we have the ice of Pro Bowl group in an equatorial, uh, confirmation. So the larger drink equitorial s. So that's gonna be the more stable confirmation here. Okay. And so that's part C from number one. Um and then So we're asked essentially to decide which is the most stable of these four chair structures that that we drew. And so we can say that this sis confirmation with both of these substantial ins equatorial is gonna be the most stable of it. And so that's so That's the answer, Thio Part D from number one. So this confirmation eyes the most stable. Okay? And so, um, for molecules to and three, we're just going to do exactly the same thing. I'm gonna try to move a little faster here. Um, so a is gonna look like this So, sis, we can put them both wedged or both dashed. As I said, it doesn't matter. This assists, uh, trains is gonna look like this. Leave this one wedge and we'll make this one dashed. Okay? And so that's a B is gonna look like this. Um, so for these to be sis, uh, let's see. Let's put the ethel group axial. Ah, And then on the next carbon, let's put, um, the metal group equatorial. Okay. Um and so the flipped chair version of that he's gonna look like this. So when they fled, um, Ethel group is now equatorial Metal group. Is now, axial. I'm in a ce faras. Which one of these is more stable? Um, again, Estelle was bigger than metal. So, uh, the SL group in an equatorial position is going thio confer more stability than having the effort group in an axial position. So then, see, we can do exactly the same way. Mmm two tilted. So see, to make these trains. Um, so we went actual equatorial for a one to relationship to be cysts. Last time in part. Be so here in part. C, we're gonna go axial axial to start, and that will give these a trans relationship. There we go. Trans relationship there. Um, and of course, the flipped chair. It's gonna look like this that doesn't look in drawing chairs. Is is somewhat difficult, even with, you know, lots and lots of practice. So I encourage you to practice as much as possible. Okay. There's a passable chair, I guess. Um and so we went axial axios. So this time we'll go equatorial Equatorial, So that's gonna look like that's not really Oh, yeah, it is. That's gonna look like this. So equatorial equatorial here. Um, and obviously with both substitue INTs in an equatorial position that's gonna be more stable versus both substitutes in an actual position. Ah, and then overall for Port de, um, this structure where both substitue INTs are equatorial eyes going to be the most stable structure. So Ah, right. So there we go. So the Trans is actually the more stable there. Um okay. And then see our three Rather is is gonna go exactly the same way. Um and I'm just gonna draw. Ah, Try not to talk as much. Oops. Okay. Ah, And so it looks like I did forget again. I'm so here we have a Turk beauty group Equatorial versus Turd Beetle group axial. I mean, obviously the equatorial torpedo group is going to be the more stable Her beautiful is Ah, very large group, Okay. And so that's part. Be part. See, um, to have these Trans is gonna look like this. So here they're both axial. Um And so of course, when we flip it, they're both gonna be equitorial. Um, And by now, you should know that if they're both equatorial, that's gonna be the more stable confirmation here. So equatorial and equatorial, so more stable. And then for a part three D. Here, Uh, this one will be most stable. And again, that's going to be because they are both equatorial You. Okay, so that was a pretty long answer. I do apologize. I don't know how I could shorten that up. Um, but, uh yeah, so that's the way to approach these problems. Are this problem? We just need to remember. Ah, what system? Trans mean? How to draw that, Um in a hexagon. Ah, representation. How to draw a traer chair, structures, flip chair structures. And then remember that Ah, larger substitutes. Being equatorial is Maur stable. And if there are more than one substitute a int a confirmation where they're both equatorial is going to be the most stable on. That's the answer to Chapter four. Problem number 57.

So in this problem we're looking at 123456 Hex aclara cyclo vaccine. So let's draw out the stereo customers first. We can first have all the chlorine pointing up. Then we can flip one of them. Then we can flip two of them. And there's several ways we can flip two of them. Okay? We can do one and 2. We can do hoops, we can do one in three. Okay? Or we can do one in 4. And we can also flip three of these. We can flip 1, 2 and three. We can flip one skip one foot too. We can also alternate. And and finally, there's one more which we can get by the shifting the previous ones over. Yeah. Okay. So if we zoom out here are all the stereo customers. Okay. Now, which two here are mirror images of each other. It's going to be these two. All right. And you can look at them directly. These are going to be mirror images. Now, which one is going to be the most stable I summer. It's going to be the one where all of their chorines our trans to each other. And so we can draw that confirmation here. One chlorine here, one chlorine here, one chlorine here, one more in here, one chlorine here, one chlorine here, I'm afraid. And this confirmation is the most stable because they're all in the equatorial position um and they're all trans to each other. So it gives each of the it gives each of the chlorine atoms the most room. So has the best Eriks. So it's the most stable. And that's the problem.

So here we have nine stereo Wiseman's off. 123456 Hex Claure, Cyclo Hexen. So they're all listed below, where you can see that we have that cyclo hexane core. And then we have six substitue INTs, one on each carbon off a chlorine atom on. Now what's changing between all of the's stereo ISMs is the confirmation about the chlorine atom. So, for example, in some structures they're all wedged like this one. However, in some structures like this one, we have some wedged in, some dashed. So the structure above that I have labeled one are a pair of a non Chalmers, This one and this one that are both labeled one on the structure above that is labeled two is the most stable. So this is because we can place the chlorine atom inthe e equatorial position on every single carbon. When this cyclo hexane is in the chair, confirmation on DSO we'll avoid those 13 die axial interactions


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