For the compound 1-ethyl-3-isopropylcyclopentane

Question

Structural representation of 1-ethyl-3-isopropylcyclopentane showing cis and trans isomers.

(c) What is the stereochemical relationship between trans isomers?

Accepted Answer

Hi, everybody. Welcome back. Let's look at our next problem. It says, consider the isomers of one bromo two chocy butane identify the relationship between the trans isomers and our molecules drawn without any orientation in space, we have the cyclo butane ringing which is a square. And on the carbon on the left, we have a chlorine and the carbon on the right top, we have a bromine. So they're adjacent to each other on the top of the square. It's one bromo two Chloro because that's the convention of naming that we name by alphabetical order, our answer choices for the relationship between the trans isomers. A identical B dimers C and anti or D conformers. So to identify the relationship, we first need to draw the trans isomers. Now, we might be a little surprised at first thinking we've only got single bonds. How can we have a trans isomer? But recall that when they're confined in a ring structure, even a, an alkane. In this case, a cycle alkane can have cyst and trans isomers because there is no free rotation of the single bonds in that ring. So cyst and trans, there it refers to how they're oriented with regard to the plane of the ring. So let's draw, we know we have more than one trans isomer because we need the relationship between them. So we'll draw two squares. There's our cyclone butane ring. So we have the chlorine and the bromine for trans isomer, we'd want them one coming out above the plane of the ring, one going down below the plane of the ring. In this case, we'll call the plane of the ring, the plane of the screen. So one should predict out towards us and one behind. So let's start with the chlorine projecting out front. So we'll draw that as a solid wedge to represent coming out towards us, there's chlorine and of course, the other bond on that carbon it's implied is a hydrogen. So then we'll draw a, one of those dashed wedges to represent going back behind the screen and there's the hydrogen and on the carbon with a bromine, same thing, we have an implied hydrogen if we want the trans arrangement while the chlorine is coming out towards us. So the bromine must be projecting out backwards. So we'll draw those little, a little dashed wedge for the bromine and a little solid wedge for the hydrogen. So again, this is a trans isomer because the two halogens are pointing opposite directions, one out behind the plane and one out in front of the plane. However, we could flip them and have another trans isomer. So the other trans isomer would involve the chlorine with a dashed wedge going back behind the plane of the screen and the hydrogen with its solid wedge coming out forward and then on the carbon, on the top right, the bromine projecting up towards us and the hydrogen with a dashed wedge projecting out behind. So again, both of those are the trans arrangements as the bromine and chlorine are opposite sides of the plane, but they have them arranged slightly differently. So we've got our two trans isomers. But how do we determine the relationships? Well, that requires labeling the stereo centers. We need to identify the stereo centers but identify stereo centers and label them R or S according to the R and S system. And then we can tell from whether each one is R or S what kind of isomers they are. So if all stereo centers in the molecule are opposite or excuse me, not stereo isomers if all stereo centers in the molecule. So in the one isomer ers of the other are opposite each other in terms of their RRS status, then the molecules are in Antium. And that makes sense when you look at a mirror image, everything in that mirror image is opposite. If all of the stereo centers in the two isomers are the same, you have two identical molecules makes sense. The arrangement in space around each atom is identical in both isomers. And then if the stereo centers are mixed, some are the same, some are opposite. Then you have ditherers which makes sense as dimers are those configuration isomers that differ in spatial arrangement but are not mirror images of each other. Whereas the an an tumors are non superimposable mirror images. Finally, we have one other choice. We see identical dimers and anti we have conformers as a choice. But conformers are stereo isomers that differ by only by rotation around a single bond, no bonds need to be broken to turn them into each other. They involve the free rotation around a single bond. And again, we know that our molecule doesn't have freely rotating single bonds because they're constrained in a ring. So choice D conformers is not going to apply to this situation. So let's get on with the business of labeling these stereo centers. So we have two stereo centers in each of our molecules. And that would be the two carbons with the chlorine and bromine on them as they each have four different groups bonded to them. Since our molecule is not symmetrical, whereas the two bottom carbons on the square, so that would be carbons, three and four each have two hydrogens. So they are not stereo centers. So we can kind of put a little asterisk by our two stereo centers. So we need to compare the stereo centers to each other. So that gets sort of a, a complicated or maybe not complicated. I should say multi step process that we need to keep really straight. It is easy to mix up. So step one of that process is label the priority of the four different groups on our stereo centers. So let's look, I know bromine is car is carbon one by naming convention. But I think it feels more logical to go left to right. So we'll start with our carbon with a chlorine on it as that's on the left here, our top priority substituent will be chlorine as that has the highest atomic number, the lowest rate obviously will be hydrogen. And since the middle two are going to be those two carbons, so I'll label chlorine number one hydrogen number four. Now we have two carbon atoms. So we need to look at the next atom in line sort of like in alphabetical order, you look at the second letter if the first is the same. So our carbon on the top or carbon that has the bond across the top has a roaming atom attached to it. So that would be carbon one, whereas carbon three down on the bottom has another carbon attached to it. So priority group number two will be carbon one prior to group number three will be carbon three. So now the next thing we look at is is our lowest priority substituent in the back or in the front. So in this case, our hydrogen is in the back. So I'm going to label this corner and I will say low for lowest equals back. Now, why do we need to do that? Because we're next going to be looking at are our priority groups? One through three, are they in a clockwise relationship or counterclockwise relationship? And with our lowest priority group in the back, that's sort of our default position. So we can think of that if you think of the fact that maybe the last priority to board the airplane is last in line, that's kind of the normal order. Whereas if your lowest priority in the front, that would be the opposite order. So our lowest priority in the back sort of status quo. Now, we need to look at if we follow it around from highest priority in terms of 123. Are we going in a clockwise or counterclockwise direction? Well, from 1 to 2 to 3, in this case, we are turning in a clockwise direction since we're in our default lowest party in the back position, turning clockwise is turning to the right. So I'm going to write clockwise. So that's going to be the R arrangement again, really easy to mix this up. So kind of think through this very carefully step by step. Now let's move on to stereo center number two with the bromine. In this case, our highest priority will be bromine hydrogen again, will be number four. Number two will be that carbon two with the chlorine bound to it. Whereas number three will be carbon four going down, which has a second carbon bound to it. So in this case, our lowest priority substituent the hydrogen top of low equals front. The hydrogen is on that solid wedge coming out to the front. So again, the lowest part in the front is the opposite situation. And so let's look at which direction our other uh groups are going. So from 1 to 2 to 3, we are going in the counter clockwise direction. Again, I'm going to write counter again. It's I like to write it down because it's so easy to get mixed up thinking through this. We're in that sort of opposite situation with the lowest in front. So counterclockwise is going to be the R arrangement. Again, think about the fact that going counterclockwise is kind of like turning left. Usually you would think of that as s left sinister, but again, we have lowest parity in front. So we're in the opposite situation. You can see why that's easy to mix up. So we have our two stereo centers each in the R arrangement. So now we need to look at our other trans isomer. So looking at that carbon with the chlorine attached to it, let's label our priorities easy since we're essentially doing the same molecule as the other one. So chlorine number one, hydrogen number four, carbon, number one is priority two, carbon, number three is priority three. So our lowest priority hydrogen is in the front. This time to go from 1 to 2 to 3, we are moving clockwise, all right, clockwise. Now clockwise is turning to the right, but we're in the opposite situation with the lowest priority in the front. So that will be the s arrangement. So now let's look at that carbon on the right with the roaming attached to it. Bromine will be number one. Hydrogen. Number four, the carbon two is number two and carbon four is number three. So let's think about which direction we need to go from 1 to 2 to 3. In this case, is counterclockwise. So we will jot down that. Our lowest priority is in the back. In this case, hydrogen is has that dashed wedge, we turn counter clockwise, all right, counter we're in the, you know, normal standard position, lowest priority in the back. So in that case, counter clock white, which is just turning left is the S position. So here we go, we've worked all the way through. We see our first trans isomer is RR for the two stereo centers. Our second trans isomer is SS for the two stereo centers, stereo centers. So these situations, all stereo centers in this case are opposite. So what was that situation? Again, that means they are an anti non superimposable mirror images. So our answer choice will be C and again, choice A will be incorrect because our R and S status is opposite and choice B dias termers will be incorrect because again, it's not mixed, both of them are opposite each other. So once again, the is the trans isomers of one bromo two choc cyclo butane, these two are choice C and an Timor. See you in the next video.

For the compound 1-ethyl-3-isopropylcyclopentane

(c) What is the stereochemical relationship between trans isomers?

Key Concepts

Stereochemistry

Trans Isomers

Stereochemical Relationships

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