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5 $ ~stubolecinlg ; influence has When multiple omijbiced t0 the Newman others. Draw conformations the structure such - mat onc for the 1 conformation - favored Ne...

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5 $ ~stubolecinlg ; influence has When multiple omijbiced t0 the Newman others. Draw conformations the structure such - mat onc for the 1 conformation - favored Newman much more fonooabletion

5 $ ~stubolecinlg ; influence has When multiple omijbiced t0 the Newman others. Draw conformations the structure such - mat onc for the 1 conformation - favored Newman much more fonooabletion



Answers

(a) Using Newman projections, draw all staggered and eclipsed conformations that result from rotation around the indicated bond in each molecule; (b) draw a graph of energy versus dihedral angle for rotation around this bond.

This is the answer to Chapter four, problem number 49 from the Smith Organic Chemistry Textbook. And in this problem, we are given three molecules and were asked to convert these skeletal structures to Newman projections. Um, and so we are given, uh, while a carbon carbon bond, and each is indicated as the bond, uh, along which to look to make the Newman projection. Um, And so for a, um, are Newman projection is gonna look like this. So remember, we always start with a circle. I'll go a little bigger here. Um, so we always start with the circle. Um, And then if we are looking, uh, along the first carbon here, um, we're gonna have one group up, one down into the right and one down into the left. Um and so, uh, the group that is up is going to be a methyl group. The group that is down to the right is going to be a booming now, and to the left will be a hydrogen. So then we need to think about the substitute prints on the back carbon. And so our two up groups are going to be hydrogen, so hydrogen hydrogen Uh, and our bottom group is going to be another mental group. Okay. And so there's a So then, for B, we're going to do the same thing. Um, we always start with a circle for Newman projection. So there's our circle. So, uh, looking along the indicated carbon, we are going to have, um, a hydrogen up, and then we're gonna have to halogen down to the sides. So we're going to have a broom in to the left and a chlorine to the right. Uh, so B is a little different than a the way that B is drawn. Um, it looks like, uh, these two carbons having eclipsed confirmation. So we need to draw this molecule in an eclipsed confirmation. And so we can do that like this. Mhm. Um, And when we do that, we're gonna have a chlorine up a hydrogen to the left, Uh, and a broom aiming to the right. Okay. And so there we go. Um, so that's b so then looking at sea, um, we'll start the same way again. So there's our circle. Um, c is going to be staggered like a was so unlike B was, um, So our front Carbon is gonna have groups like this are back. Carbon will have groups like this, and then we just need to fill the information in. So our top group on the front carbon is going to be an ethyl group, so we can put ch two ch three for Ethel. Then we're gonna have a chlorine to the right and a hydrogen to the left. Okay, um, and then on our back carbon, our bottom group here will be an ethyl group. So ch two ch three. Um, And then it looks like we are also going to have a chlorine to the right there and a hydrogen to the left there. And so let's see. And that's the way to approach this problem. We just need to remember how to draw a human projection. Um, and so there's an indicated bond in each of these molecules. We need to just pretend that we're looking at that bond. End on, uh, so that we only see the front carbon. Um, and then we just need to redraw it from that perspective. And that's the answer to Chapter four. Problem number 49

Let's draw the Newman projections for the standard and Eclipse confirmations for two metal panting for rotation about the carbon to and carbon three bond. So let's take a look at to muffle Pantene and identify carbons two and three So we'll have that h three c c h ch three ch two ch two ch three. And if we identify, this is the carbon to carbon three. So we're gonna identify the rotation about the carbon, too. Carbon three bond. So let's start by drawing the eclipsed confirmations. We have ch three each three each. One, 23 and the eclipse configuration. Yeah, a little bit better here with one to three coming across this way. Here each H c h two c h three. This will be ch three h three c h. They're equipped with be each he ch two ch three and the third eclipsed confirmation will be each age ch two c h three are staggered confirmations way will have th 38 h three c and then they're staggered. We will have sure h h ch two ch three h three h d c c This confirmation here would be h each ch two ch three and ch 38 age three C that each each c two c h three. So there would be our eclipsed and staggered confirmations for two methyl painting about the C two C three bond rotation. Which confirmation is lowest and energy. So the staggered confirmation in which one methyl group is opposite the folk group. Yeah. Mm hmm. Um, em mother group is nothing. Nothing. Um, in both, um ghost and And he there's two of them are the most stable and the lowest energy confirmations. And those two staggered confirmations that would be the most stable would be these two here.

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.

Okay. This problem is having a strong Newman projections of the following compounds. So for this first order, we have three methyl plantain. So a three month old paint in the first thing I'm gonna do is draw the actual skeletal structure. So I have a pen Tate, and then on the third carbon I have that methyl group. So for analyzing human projections, we have to know what bonds were analyzing the Newman projection on. So in this case, we're analyzing it between carbon number two and carbon number three. So I'm gonna write my hypothetical disc along this bond case without circle that I just drew. That corresponds to the circle of my Newman projection. Okay, so this problem is asking us to draw it in the most stable conform. Er So what I'm gonna do is I'm just gonna isolate one of the sides. So the dark side right here, that's what a correspond to decide. And then the light side, which obviously we can't see is gonna be on the back of decide. So on this side, draw my circle. We have to hard agents. Okay, So these two Hardin's Court 1 200 right there on carbon number two. Basically 200 units. And then right here we have my method group. So my method groups ch three. Okay, so on the dark side of the disc, which is on the other side of this, we have what's rest the rest of it. So right there. Okay, so on that other side, we have a hydrogen. So I'm gonna write that, um, somewhere we don't know yet. But basically, we have to know what are my other two substitutes? Okay, so we have this metal group and we have this effort group. So knowing that I have to make this into the most stable conform, er, I know that my Ethel group is gonna be considered the most bulky. Okay, So between my metal, my hydrogen and my ethel group, my ethel is considered the most bulky. And when I'm comparing it to what I have in front here, the most Eric Kendricks would be if our tohave an eclipse two version of the CH three right here and then if our to write that Ethel group on the other side. Okay, so I don't want that at all. I want in the exact opposite of that position. So that means I'm gonna have in a staggered position my Stage two c three. My little group here. I want that on the opposite side of my method group, Which on this side is my my boat. Most bulky compound. Okay, so if I have my whole group there, then that just means I have either my hydrogen here or my 100 in there and then my method group on the corresponding other side. So because it's symmetrical, because I'm comparing between these 200 it doesn't matter which way I draw, So I'm just gonna write down my ch through here and my hydrogen. So this ch three right there that correspondents to this one, my ethel group, corresponds to this and that. My hydrogen correspondents to this hydrogen good. So that is that one has so much they will conform of that one. And then here we have three methyl hexi. So again, I'm gonna drive the compound. So I have I have to say on the third carbon, I have it right there. Okay, so it's the third carbon and then this one is asking us to draw the new in projection along carbon three and four. Okay, so here's my Here's my disk. And on the, um, let's see, I'll drop my dark side corresponding to this side. Okay? So that means that my light side, which I'll draw here, is corresponding to this side. So what I'm drawing are now is the light side slash front side of my Newman projection. And on that name and projection, I can basically do it in any way that I want because I'm just gonna have the backside be corresponding to whatever I have in the front. So in the front, I'm gonna have my effort group. We'll just write that ch two, 63 So this Essel group right there that corresponds to that. Okay. And then, of course, I have my method group ch three. And then last but not least, I have my hydrogen. So this hydrogen is corresponding to the hydrogen coming off of this carpet. Okay, okay. And then next up, I have, um, carbon number four. So this is carbon over three, and that's carbon cover number four carbon over three. Okay, so a lot of carbon number four I have to hundreds, and I also have this Ethel group. Okay, so the ethnic group is the one that I care about because I want to put that as far away as possible from my most bulky compound on the front side of my new and projection. So the most bulky compound are substituted on my carbon number three. Is this Ethel Group? Okay, so that means that I want this death grip on the backside of monument protection to be in the exact opposite of wherever this Ethel group is. So that means I'm gonna have my ethel group right there. Ch two ch three. So this Ethel group corresponds to this one. And of course, this group corresponds to this one. So even though even though this group is right next to the surgery, it's still a little bit Sterkel hindered. But at least it's as far away as possible from this one, which is the one we mostly care about. OK, so that's that. And then, of course, I have my other two hydrogen which can go on either side. Those don't matter as much. Okay, so those two hundreds was just threw out. Those correspond to the hydrogen is coming off a number four. Okay, so that's the most able conform er of three month Alexei. And now we have 1233 dimethyl Hexi. So something has always draw the compound in a skeletal structure. That's vaccine. And then 33 dimethyl have uncover number three. I have two methyl groups. OK, so this one is also asking us for the Newman projection between carbon number three and four. So when your own projections gonna go right there and the dark side is going to correspond to this side Okay, So my dark side over here is the underside of this circle and the light side correspondent to decide. Okay, so on carbon number three, I have a methyl group se three. Actually, I just start like this Series three. I have a Series three, and then I also have my ethel Group C H two and then 63 So this effort group correspondents to this one and then these two method groups correspond to these two. Okay, So that's that so far. And now let's go ahead and do my backside. So the backside, which is attached coming in before I have two things attached to carbon there. Sorry, three things. It's hatched. Carbon number four. I have my 200 I also have this Ethel group. So getting this Ethel group in comparison to these two hundreds is the most bulky. Is the most book a substitute off a number carbon off of carbon number four. Okay, so I have toe consider where I put my bulky compound. In this case, my effort group in relation to whatever is the book is compound on this front side of my Newman projection. Okay, So the most boki compound on this Newman production on Karma number three on the front side is this ethic group. So that means that I want this Ethel group complete opposite to that I wanted in a staggered position. It's a stage two ch three. Okay. And then I could just put my head, regions wherever I went. In this case, it's gonna also be staggered. Okay, So this would be considered the most stable conform er 33 dimethyl hexi


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