How to identify meso compounds.
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- Would the meso compound be a diastereomer of the enatiomers? If so, would that give a total number of 4 stereoisomers?(12 votes)
- Yes, a meso compound is a diastereomer of the enantiomers.
No, there are only three stereoisomers: the two enantiomers and the one meso compound.(16 votes)
- How to differentiate in an exam with no model(11 votes)
- One way is you have to very carefully and methodically draw the molecule one step at a time to try and get it to look like the other molecule. So first you might rotate it 180 degrees in the plane of the paper and then you might flip the molecule in a horizontal mirror plane. Once the two molecules you are comparing look more similar you may be able to see that they are actually the same molecule (eg meso compounds) or are actually enantiomers.
So for the example above (diol example) you can rotate molecule 4 by 180 degrees in the plane of the page and see that it's actually identical to moIecule 3.
It's tricky but with practice you can get good at it. Hope this helps.(9 votes)
- Can a Meso compund have only one chirality center?(6 votes)
- No, it has to have at least two chirality centres.(17 votes)
- at9:03, doesn't rotating the bond create a whole new stereoisomer? i know he was looking for a plane of symmetry to define the molecule as a meso compound but i'm having trouble understanding why you can rotate the bond and still consider it the same(6 votes)
- As long as you are breaking no bonds, you still have the same molecule.
Rotation about a single bond gives only a different conformation of the same molecule.(17 votes)
- the meso compound is optically inactive due to its plan of symmetry ,so how i can consider it is an optical isomer to the enantiomers (the optically active) ?
& why an optically inactive meso compounds cannot be separated into optically active components ?(1 vote)
- Optical isomers are molecules with chiral centres that differ in the three-dimensional arrangement of their bonds.
The definition says nothing about their optical activities.
Hence, a meso compound is a stereoisomer of its optically active isomers.
When you draw the two mirror image isomers of a meso compound, you are just looking at different views of the same compound.
A meso compound is a single compound. You can't separate it into two different compounds.(6 votes)
- how to identify mess compound? is it by the plane of symmetry.(1 vote)
- Yes. If a compound in a group of stereoisomers has a plane of symmetry, it is achiral and thus meso. :)(5 votes)
- Are molecules 2 and 3 diasteroisomers? From my point of view, if I rotate the molecule 180 degrees, I get one matching OH... so, would that make them diasteroisomers?(2 votes)
- So looking at the first example we decided that 3 and 4 were enantiomers and 1 and 2 were a meso compound. Would that then mean that diastereomers would include: 2/3, 2/4, 1/3, and 1/4?(2 votes)
- Would you ever compare drawing one with drawing four or drawing two with drawing three? How come drawings one and two always get compared with each other as well as drawing three and four?(2 votes)
- It was a random pick. He could have chosen the one and three, it's completely random, there's absolutly no order to do the check of the stereoisomerization.(1 vote)
- Does compound 1 and 2 could be Meso compound ? Because when rotate compound 2 (180 degrees) will look exactly like compound 1 so it will be superimposable and they identical to each other.
Thank you(1 vote)
- I'm not sure whether you are referring to the cyclopropane derivative or the diol molecule? For both examples the possible isomers are labelled I and II.
The conclusions in the video are correct. In the first case, I and II are superimposable and therefore identical molecules, while in the second example, I and II are enantiomers.(2 votes)
- [Instructor] A meso compound is a compound that has chirality centers but is achiral. We're gonna come back to this definition in a few minutes. Right now, let's focus in on this drawing. And our goal is to draw all possible stereoisomers for this dot structure. So we know from earlier videos that this carbon is a chirality center, and so is this one. And we would expect two to the n stereoisomers, where n is the number of chirality centers. And since we have two chiral centers here, we would expect two to the second power, or four stereoisomers. So this is really just a maximum number, so I'm gonna put a question mark right here. So do we get four stereoisomers? Let's draw out the four possibilities. Our first stereoisomer could have both bromines coming out at us in space. So let me go ahead and draw that in. So we could have both bromines coming out at us. For the second possibility, we might have both bromines going away from us in space. So I'll draw that in there. For the third possibility, we could have one bromine up and one bromine down. So I'll put those in. And for the fourth, just reverse them. Have the top bromine down and the bottom bromine up. like that. Well, all right, let's examine the relationships between our stereoisomers. And let's start with stereoisomer possibility three and four. So let's compare these and let's figure out the relationship. On the left is stereoisomer three. We can see there's a bromine coming at us and a bromine going away from us. On the right is stereoisomer four. Now we have a bromine going away from us at the top carbon and a bromine coming out at us here. So let's compare our two stereoisomers. If I rotate the one on the right, we can see these are mirror images of each other. And if I try to superimpose one on top of the other, here we get one pair of bromines to line up, but the other pair doesn't match. If we try to get the other pair of bromines to line up, now the first pair doesn't match. So these are non-superimposable mirror images of each other. These are enantiomers. So three and four are enantiomers. They are non-superimposable mirror images. And we could have guessed that by looking at the drawing, because at this carbon we have bromine coming out at us in space, and then now we have bromine going away from us in space for the other stereoisomer. So that's an opposite configuration at that carbon. And at this carbon, we go from a dash to a wedge. So that's an opposite configuration at this one too. So since we have opposite configurations at all chirality centers, we would expect these two to be enantiomers of each other. What about the relationship between one and two? Well, at first we might say oh, those are enantiomers because here we have a wedge, and then over here we have a dash and here we have a wedge, and here we have a dash. So that should be the opposite configuration at both chirality centers. So those might be enantiomers. But let's go to the video to see if that's true. On the left is a model of drawing one, with the two bromines coming out at us in space. On the right is a model of drawing two, with the two bromines going away from us in space. And if I rotate the model on the right, we can see that these are mirror images of each other. But they are superimposable mirror images. So if I put that one on top of the other, you'll see that they are superimposable. So these actually are two models of the same molecule. This is a meso compound. It's a compound that has chirality centers, but it is achiral, the mirror image is superimposable. So one and two really represent the same molecule. This is a meso compound, a compound that has chirality centers but is achiral. The mirror image is superimposable on itself. So we thought we would have four stereoisomers, but really we only have three. We have a pair of enantiomers and we have one meso compound. So to look for a meso compound, one thing you could do is what we did in the video. We had the mirror image and we were able to superimpose the mirror image on itself. Another way to look for a meso compound is to look for a plane of symmetry. So if I draw a line here, think about this as being a plane, and look for symmetry on either side. So you can see, it's symmetrical. I drew in the plane of symmetry with a dashed line here, but it's hard to visualize it. So up here is a better picture. Here you can see the plane dividing the molecule in half. And on the left side, we have our bonds here and then we have our bromine going up and our hydrogen going down. The right side is symmetrical with the left side. So look for symmetry on both sides of the plane. Let's do another example. This one's a little bit harder than the last one. We know that we have two chiral centers. So that's a chiral center and so is this one. So we would expect two to the second stereoisomer. So that's, of course, four. So I'll put a question mark here again because we're not sure if we actually will get four stereoisomers. That's a maximum number. Down here I have the four possibilities. So I've drawn them out just to save some time. So we have one, two, three and four. And let's examine the relationship between one and two first. On the left is stereoisomer one, and I've left the hydrogens off the methyl groups and the OH just so we can see the models better. So here's our carbon chain, and we have both OHs coming out at us in space. And then for stereoisomer two, here's the carbon chain, and we have both OHs going away from us. So I'm gonna hold them in the way they are in the drawing and I'm gonna rotate the one on the right. And when I do that, we can see that these are mirror images of each other. But if I try to superimpose one on the other, so I'll just rotate this one back here and flip it over, you can see they don't match up. So the atoms don't line up here. So they're non-superimposable. Doesn't matter how you do it. I'll rotate it again, and we can see we can't superimpose our atoms. So these are non-superimposable mirror images of each other. These are enantiomers. And we could check for a plane of symmetry, so I could take one of these and I can rotate it so we have our hydrogens going away from us in space. And I look for a plane of symmetry, but I don't see one. So this is not a meso compound. So one and two are enantiomers. They're non-superimposable mirror images of each other. And we could've guessed that because if we look at our chiral centers here, so this one has an OH coming out at us, and then that one has it going away from us, this one has an OH coming out at us and this one has it going away from us, so we have opposite configurations at both chirality centers. Let's look at three and four next. So what is the relationship between these two? Well, first, we might think these could be enantiomers because at this carbon, we have OH on a wedge, and then here we have OH on a dash. And then here we have OH on a dash, and here we have it on a wedge. So that might be your first guess. But let's look at the video and let's look at the model sets to help us out. Remember, I'm leaving the hydrogens off the methyl groups and the hydrogens off the oxygens in the video just to help us see the molecule more clearly. On the left, we have a model of drawing three. So here's our carbon chain with an OH going away from us in space, and an OH coming out at us in space. On the right is a model of drawing four. Here's our carbon chain with an OH coming out at us in space and an OH going away from us in space. So I'll hold the two models and we'll compare them. First let's see if they are mirror images of each other. So I'll take the one on the right and I'll rotate it and I'll hold it up next to the one on the left. And now we can see that these two are mirror images of each other. So next, let's see if one is superimposable on the other. So I'll go back to the starting point and I'll rotate it around like that, and let's see if we can superimpose the one on the right, the mirror image, on the molecule on the left. And notice that we can. All of the atoms line up. So all of the hydrogens, carbons and oxygens are in the same place. So this is a compound that has chirality centers, but it is achiral, the mirror image is superimposable on itself. So we should be able to find a plane of symmetry. So I'll just pick one of these models, doesn't matter which one because they represent the same compound, and I'll rotate, I'll rotate it around so we can see a plane of symmetry. So right there is our plane of symmetry. This is a meso compound. So three and four actually represent the same compound. So this is one meso compound. So these two are the same, and we have one meso compound. It's not really obvious looking at these bond line structures that these represent the same molecule. So definitely get a model set and try this out for yourself. So we thought there might be four stereoisomers, but actually there are only three. We have a pair of enantiomers, and we have one meso compound.