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A. Draw the three staggered and three eclipsed conformations that result from rotation around the bond labeled in red using Newman projections.b. Label the most sta...

Question

A. Draw the three staggered and three eclipsed conformations that result from rotation around the bond labeled in red using Newman projections.b. Label the most stable and least stable conformation.

a. Draw the three staggered and three eclipsed conformations that result from rotation around the bond labeled in red using Newman projections. b. Label the most stable and least stable conformation.



Answers

Draw the staggered and eclipsed conformations that result from rotation around the C$-$C bond in CH$_3 -$ CH$_2$Br.

Let's drop two more staggered and three eclipsed models of, um, two metal butane. So I've drawn the first one for us to make it a little bit easier. Tio, go forward. So the point here is the front carbon we're looking at and these three groups are coming off of the carbon in the back. So we're literally looking down the bond. I would highly suggest looking at a model if you can't visualize this from the Newman projection alone. So what you can do to get the other staggered confirmations, I would move one only the atoms on one of the carbon. So I only moved the ones in the back. Urban, leave all the groups on the front carbon the same. And I am choosing to move each of these three groups buy one each time. So this next confirmation and it draws gonna have the hydrogen here, this methyl group here in this hydrogen down below. So that gives us this confirmation. So the methyl group is now too the up and left because there's irritation on this bond. So the reason you want to keep the front carbon or the back urban doesn't matter which I'd always choose to keep the front carbon the same. The reason you want to leave that the same is otherwise you're going to be drawing the same con firmer over and over again. So the last confirmation we can draw is moving this methyl group down and the two hydrogen sze up over. So again, leave your front carbon the same. And I literally just did the same rotation I did here. So this method group moved down This hydrogen moved up to be this one in this hydrogen move up to the upper left to give that one. So these are the three staggered confirmations. Now we can draw the eclipsed ones where the atoms were on top of each other. There clips saying each other. So leave your front, Adam the same again. Let's just move. This method grew up so halfway position between these two compounds, so this metal group is going to be right behind the hydrogen. This hydrogen is going to be right behind this method group. This hydrogen is going to move up to be behind that mental group and that'LL give you an eclipse confirmation like that between these two confirmations. This hydrogen is going to be up behind that one. Leave your friend carbon the same. So this hydra jin's up behind that one. So that's where this hydrogen is. This methyl group moves down to be eclipsing this metal group, and this hydrogen moves over to be eclipsed by that front methyl group. Finally, halfway between halfway point. Between this version in this one to the eclipse of these two compounds front carbon still the same. This metal group is going to move over to be eclipsed by the other method group in front of it. This hydrogen is going to be eclipsed by the hydrogen above it in front of it. On this hydrogen is going to move down to be eclipsed by the method so eclipsed compromisers are less stable than staggered. So which of these staggered confirmations is the most stable? Well, this metal meth A LL interaction is a higher energy than a metal hydrogen interaction like this one. So here you have one method all methyl interaction, one methyl hydrogen interaction. Just with this method, that's going to be the difference in all compromise, because all of them have hydrogen metals, has your mental hygiene methane, hydrogen, methane, hydrogen, methane and hydrogen hydrogen. So having this method were between two other method groups is at a higher energy than having the method between, um, Ethel and hydrogen or a metal between a medal and a hydrogen. Therefore the first to come from ER's are the most stable, the least stable it is. Just do hysterics alone. The least stable is going to have to metals that are overlapping because you also have the hydrogen is attached to these metals, which makes it bulkier than a small hydrogen. So these last two confirmations are going to be the least stable because you have two method groups that are overlapping each other that you don't have here. All of these are Hydra Jin's and metals next to each other.

This is the answer to Chapter four. Problem number 19 fromthe Smith Organic Chemistry textbook. And in this problem, we're asked to consider the rotation around the carbon carbon bond in one to die. Claro, ethane. Um, and then using Newman projections were asked to draw all of the staggered and eclipsed confirmations that result from rotation around this bond. And then we're asked also to Graff, uh, energy versus the die. He'd roll angle for rotation around this bond. Um, and so what I've done first year is drawn one to die. Claro Ethane with a red arrow indicating the bond that we're looking at rotation around. I mean, so to start, um, our first Newman projection eyes going to have this molecule in its lowest energy confirmation, which is going to be anti so, uh, one chlorine down and the other chlorine up. So that's Ah, it's going to be an anti confirmation. Obviously, the chlorine CZ wannabe is far apart as possible. That's what makes this of the lowest energy. I mean, So this is gonna be staggered and anti, uh, and we'll call this number one. So then, to get from number one to number two um, our next Newman projection. Remember, it's gonna be a 60 degree rotation. Um, and so we'll just do a clockwise rotation of the back carbons or front carbon is gonna be unchanged here, so we'll leave that cloning down. Um, And those hydrogen zzzz as they are but will rotate our back carbon 60 degrees. So that is going to put, um, these groups here. So this is going to be, uh, eclipsed. I will call this number two. So then you get from 2 to 3 again, we'll we'll do a 60 degree clockwise rotation. Um, and so again, our front groups aren't gonna move. And now, um, our bag herbs are again staggered, but, uh, they're like this. So this is obviously even though this is staggered, this is gonna be ah goche. And so, um, significantly higher energy than one us. Since one is his anti so staggered Goche I can never remember of It's a you Were you a Okay, it is a you Okay? Ah, and we'll call this 13 So from 3 to 4 again will rotate 60 degrees. Um, again, our front groups will be unchanged. So C l H h I mean, So now we are again in an eclipsed confirmation, And this is going to be, um, the highest energy confirmation because our chlorine czar, right on top of one another and remember, they want to be as far apart as possible. Eso This isn't another eclipsed on and we'll call this four. Um, And again, I would expect for to be the highest energy confirmation. Certainly. So another 60 degree rotation is going to take us to five again. Our front groups don't change. S So now we're staggered again. Um, so this time like this. So this. Ah, this is staggered and Goche again. Ah, and comparable and energy to three. So staggered. Goche number five. Ah, and then 60 degree rotation is gonna take us to our sixth and last Newman projection here. Um, again, our front groups remain unchanged. Um, and we have, uh, eclipsing so like this. So this again is eclipsed for six. Right? That's number six. It's eclipsed on. And then one more 60 degree rotation. Well, take us back to number one. So these air all of our staggered and eclipsed, um uh, formations here are confirmations. Um, and these are the Newman projections of each of them. Um and so remember, staggered anti number one is gonna be our lowest energy confirmation. Um, and eclipsed number four here is going to be our highest energy confirmation because the chlorine are sitting right on top of one another. Um, and so Ah, we're asked to represent this graphically. So this is gonna take a new page here. Um, and we're asked to represent this in terms of energy versus die he dro angle. So draw, uh, decent graph here. Okay, Um and so, um, we can just call this e for energy this axis on, then, uh, the the X axis is going to be in terms of, uh, the di hydro angle between two Corinne's. So 180 degrees, 120 degrees, 60 degrees, zero degrees, um, and then back the other way back to 1 80 So 60 1 20 and one 80. Okay, um and so remember, um, our confirmation one, uh, is gonna be when the chlorine Zara's far apart as possible. So when they have 180 degree die, he's your angle. That's gonna be our lowest energy confirmations are most stable. Um, and so from one, we're going to go to two, which is gonna be appear to is eclipsed. Um, and so that's less stable. So then we'll come down somewhat, Uh, 23 Um and so, actually, let me let me do this to just make it more, more parent. So three is gonna be lower energy than two, but still higher than one. So rather than make it ambiguous, let's do this. There we go. So one will be down here. Two is higher than one three, uh, is lower than two. And then four, where there is a zero degree di hydro angle, four is gonna be the highest energy confirmation. Um, and then five is going to be comparable to three. Six is going to be comparable to two. Uh, and then we'll come back. It's a one again. So that's what, uh, the diagram that were asked to draw should look like, um, yeah, so that's a graph of energy versus die. He'd roll angle for rotation around this bond. Um, yeah. And again. Ah, here are the human projections that we need to draw to answer the first part of this question. Um, and that's the answer to Chapter four. Problem number 19

Every job on the bond stick model Teo. Show more of a skeletal structure here and number the carbons. One, two, three and four to the substitute int on carbon one. It's the methyl group. The hydrogen I've drawn is in the equatorial position. Theme. Ethel Group is in the axial position. So, Axl, for that substitute carbon to the hydrogen is going straight up. So it is in the axle position. The Brahman is coming out, so it's in the equatorial position and on carbon for the hydrogen is going straight up. So it's axial and thie. Isopropyl group is equatorial substitue INTs one in two are both going down relative to the hydrogen. So there sits there on the same side of the ring. Substitue INTs two and four are both going down relative to the hydrogen. Since they're on the same side of the rain, they must also be sis and drawing the other chair come from er I find it easiest to flip this carbon up, Miss carbon down. So this is going to be carbon number one, two, three and four. Carbon one still has a method group, and it is still going down relative to the hydrogen, so it's going to flip from the axial position to the equal Torrey Yl. The bromide here is going to still be going down, and the isopropyl group is still going to be going down on carbon for so this would be the way to draw the other chair confirmation. And that puts it, flips all the groups from Axl to Equatorial and vice versa. So the group that was actual is now equatorial in the groups that Rick Material are now in the axial positions. So this is going to be less favored, less stable than the chair compromise that they gave us because now you have two groups in the axial position that's very hysterically hindered, and it's not a stable, so less stable.

So we want to draw the most and least able Newman projections for the indicated bonds. So here this is the method group. The carbon in front with three hydrogen is attached. In the back is the other carbon attached to a propane chain into Hydra Jin's. So this is the bond we're looking down. So this carbon and one directly behind it, we don't actually draw those carbons. This is the most stable confirmation because it's ghoulish. And you don't have quite this Terek interaction that you do with the only other way to draw this, which is thie Eclipse version, where you have hydrogen Hye jin's dark interactions down here in a hydrogen carbon interaction up top. So this is the lowest stable calm from ER. This is the greatest in energy for a con from ER for the other bond. You want to get the alcohol chains as far apart as possible because, yes, you have a clip ste on EST eriks. But you can also have stare IX within gosh interactions. So you want to get the two alcohol chains as far apart as possible in the lowest energy confirmation and the highest energy confirmation the carbon carbon chains will over


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