Draw the structures of all possible diastereoisomers of an octahedral complex with the formula MA₂B₂C₂. Which of the diastereoisomers, if any, can exist as enantiomers?

Video transcript

Hi, everyone. Welcome back. Let's take a look at our next problem. It says for the octahedral complex with a generic formula of MX two Y two Z two, draw the structures of all possible dimers and identify which can exist as an anti emer. So let's recall that distinction between ditherers and an anti ditherers. Of course, being those molecules that have different arrangements of the atoms in space though the same bonding but do not exist as mirror images. Whereas in anti MERS of course, are pairs of non super imposable mirror images. So we have our generic formula here, we have a metal and it has an octahedral arrangement of bonds. So one top and bottom and our square planar arrangement and we note that we have six ligands total that are in three pairs. So with that arrangement where there all, all the ligands aren't even numbers we can expect, especially with two of each that there won't be that many an tear. A lot of these arrangements are going to have planes of symmetry through the molecule. So just keep that in mind as you're looking. So we know that when we have a pair of ligands that they can exist in a cis or trans arrangement. So let's work our way through in an orderly fashion. We'll start with trans arrangements because they're slightly more likely to have a plane of symmetry. So let's start by having all of our ligands be in a trans arrangement. So we'll draw our two xs top and bottom and then our Ys 180 degrees apart and our Z is 180 degrees apart in the square plane. Well, we can tell right away that there's a very obvious plane of symmetry right through the middle of this molecule. And we can double check by looking at the mirror image and seeing if it has a super imposable mirror image. If we rotated our first molecules, we have our mirror image. Now often we look, we look for mirror image superimposes, we look at a 180 degree rotation. But in this case, I'm just going to rotate 90 degrees so that each of our ligands in the square plane moves over by one position. And you can see that if I do that, of course, my exes stay in the same place, the Z at the front left moves over to be in the front, right. And everybody else just moves around one spot. And if I do that, you can see that that is identical with the mirror image. So they are super imposable. And therefore this does not exist as a pair of an anti ier. So I'm going to go ahead and cross out that mirror image is being super imposable and I'll just erase my drawing of my rotation that I did there to get some more space. So here is one of my ditherers and now let's look at an arrangement where just one of our liens is trans and the other are sis. So let's start with X. So I'll put my X's top and bottom 180 degrees apart. And I'll put the Ys on the left side of my square plane and the Zs on the right side. So let's look at our mirror image here. In this case, I'm not even really going to bother to draw, to draw the rotation. It's pretty straightforward to see that if I rotate my original molecule 180 degrees, the Ys will flip over and be on the right side, the Zs will flip over and be on the left side and this will be identical to the mirror image molecule. So this also is not a pair of an anti imer. So highlight our trans X arrangement as another dimer but not a pair of an anti. So now by analogy, we can see that any one of the ligands, if we did trans for that particular ligand and cyst for the others, we would also have a different ditherer but no an anti imer because it will have the same plane of symmetry and rotation will always make it super imposable on its mirror image. So let's go ahead and draw those dite ier and maybe this time we'll put them trans uh across the square plane. So we can see that the principle is the same. So let's imagine a scenario where we put Y 101 180 degrees apart in the square plane and then our other ligand cis. So we have X on the front left on the bottom and Z on the top and the back, right, would this make a super imposable mirror image? I've drawn its mirror image next to it. And again, importantly, we can see that we do have a plane of symmetry running through the original molecule. So as we expect our molecules a chiral and we can make a super imposable mirror image if we rotate our original image 90 degrees so that everybody in our square plane moves over to the right. You see that the X on the bottom left will now be in the front, right? And everybody will move over and it will be the same as this mirror image. So as we'd expect from the analogy to the trans X situation trans Y also not a pair of an anti MERS. So I'm going to erase the mirror image and all my notations just to focus on that ditherer. So you can see it does not matter where on the molecule, the two trans lids are located, if just, if one of them is trans and the other two are cysts, it's not a pair of an ante. So there's another dias Starmer with the Y ligands being trans. And so it's quite logical to see that the Z ligands being trans is yet another dimer that is not a pair of an anti immerse. So I've drawn that trans Z molecule. So we're up to four ditherers. So far, we've done the arrangement with all of them being trans and then with each ligand at a time being trans and the others being cyst. So the last one we have left, we might for a moment think we need to look at two of them being trans and one being cis. But when you think about it, there's only six spots. If two of them are trans, the last one has to be trans and we have our initial situation. So our last dia where that comes up is when all of our ligands are ciss. So let's draw that situation. So we'll just fill in starting at the top, going around clockwise xxyyzz. And now let's look at our mirror image but drawn in the mirror image. Do we have a plane of symmetry in the original molecule? Well, if we drew it through the middle of the two XS where they have two Z's on one side, but there's no symmetry on the other. And again, if we drew through the middle of the square plane, the top and bottom Arent symmetrical. So we don't have a pale of symmetry in our molecule. And just to check, let's draw our try and rotate our original molecule around all drawn in red and kind of tiny to squeeze in here. When I look at this molecule X and Y at the top and bottom, of course, will remain the same. I'd want to rotate it around 180 degrees to try and match up those two Zs move them over to the right side. So let's put Xyzz, they all match. But what happens to the X and Y on the right? Well, when I rotate it around X ends up in the front and Y in the back. And as you can see that is not super imposable with my mirror image. So this one is our one that exists as a pair of an anti emer. And that makes sense because the cyst all cyst arrangement, we'd expect to be less symmetrical and more likely to be an anti emer. So we've got one last pair of dimers that existed as a paro of anor. See you in the next video.