Consider the following ethylenediamine complexes.

(a) Which complexes are chiral, and which are achiral?

(b) Draw the enantiomer of each chiral complex.

(c) Which, if any, of the chiral complexes are enantiomers of one another?

Video transcript

Hi, everyone. Welcome back. Let's look at our next question. The structures of oxalate complexes are shown below. And then we have three tasks. Number one, classify each isomer as either chiral or a chiral. Number two, sketch the anant tumor of each chiral complex. And number three identify the pair of chiral complexes that are an Antium. And we're given four different oxalate complexes with differing numbers of oxalate and water ligands. Our models labeled A through D all have a central atom represented as a blue ball which is scandium. They're an octra hadal arrangement. They have water ligands represented as gray balls and oxalate ligands. C 204 in which the bonding oxygens are represented as two red balls linked by a little arc that represents sort of shorthand for the rest of the molecule that just helps us focus on how the two are linked together. So let's start with choice. A, the first thing to note, we have two water ligands and they're in a trans arrangement. Whenever we see a trans arrangement like this, it's far more likely to be an uh an a chiral molecule than if they're in a cyst arrangement. So just kind of a way to, especially if you're in a hurry, um think through this and then we have two oxalate ligands. So let's look at this and identify is it Carval or a chiral? Well, the first thing we want to do is look for a plane of symmetry. Is there a plane of symmetry? Well, in this case, there is, although we have to be very careful with these um ligands like oxalate where we have this kind of propeller blade structure that can deceive us into looking at the two dimensional structure and thinking it's symmetrical. But in this case, we can make a plane and I've sort of sketched in this plane running through our two oxalate ligands with the water molecules sticking out behind and in front of the plane I've drawn. So again, a little tricky to visualize, but we can see that in this plane, the oxalate ligands that are in the square plane of our octahedral complex are 180 degrees apart. And then the top and bottom oxygen atoms that are linked to them are also 180 degrees apart. They are make this sort of plane arrangement again with the water sticking out in front and behind. So since this has a plane of symmetry molecule, A is a chiral, therefore, it will not have an an an temor. So we don't need to sketch anything for molecule A. Now let's take a look at molecule B molecule B also has two water ligands and two oxalate ligands. But the water ligands are insists positions to each other. They are 90 degrees apart. Then the oxalate ligands fill the other war spots. Now, it can be tempting to look at this molecule and think there's a plane of symmetry. We might imagine it running right down the middle of this molecule slicing it in half down the square plane. However, when we look at that, we need to keep in mind the linkages of the oxalate ions. If we cut the molecule in half like that, we would have our top oxygen atom bound with the linkage going behind our plane. Whereas our bottom oxygen atom and the other oxalate ligand is bound to an oxygen atom projecting in front of that plane. So it's not a symmetrical arrangement after all, again, always have to be careful of that 3D arrangement in space. But again, we're kind of alerted to that because we have this cyst arrangement. So we would definitely want to draw our mirror image and we can check if it's super imposable. So I will sketch complex B and its mirror image. So I have complex B on the left and its mirror image on the right. So let's see if there's a way to rotate B to make its mirror image super imposable. So as we look at our complex, we see, we've got water ligands on the right of the first molecule. And on the left of the second molecule, and if we rotated the whole molecule 180 degrees, using the top and bottom as the axis of rotation, we could end up with waters on the left and two Os from oxalates on the right linked to top and bottom. But again, we have to pay attention to that arrangement of those propeller blades. So let's draw our rotated molecule down below. I've sketched in just the atoms which are in the same arrangement as the mirror image. But the trick comes when we sketch in the rest of the oxalate molecule and that can be a little tricky to keep track of. So I'm going to use colored highlighting in my original B structure. I'll highlight my top oxygen atom from oxalate in yellow and it's linked to the back left. But when I rotate around 180 degrees, my top oxygen still in the same place as that's the axis of rotation. So I'll highlight in that in yellow, but that back left oxygen in the square plane has swung 180 degrees to become front, right? In addition, and now I'll use blue for my other oxalate. The bottom oxalate was linked to the front left. Now our bottom oxalate same place that front left oxy oxygen has become the back, right. So when we look at the arrangement of the rest of the molecule, the linkages are now different with the top one being linked to the front molecule and the bottom one being linked to the back molecule. Whereas when we look at our mirror image, they're linked in the opposite way. So this is a non superimposable mirror image. So molecule B must be chiral and we have drawn it's an anti Amer here, it's mirror image. So I'm just going to put a little wiggle around these two. So we'll say that the one on the right is the anant humor of B. Now let's move on to complex C complex C also has a trans arrangement of water molecules. They are 180 degrees apart on top and bottom with the two oxalates within the square plane. Well, let's analyze this molecule and of course, our oxalate molecules are within one plane, the square plane. It's just easier to see on this molecule. It is definitely a chiral with Plaintiff symmetry. But if we look carefully, we can see this is actually the same molecule as a, if we rotated molecule a 90 degrees. So put a little arrow with the two water ligands moving up 90 degrees to become top and bottom. We see that we would end up with complex C. Actually, that was a little confusing. Let me redraw the rotation I was indicating here so that the axis of rotation is the two oxygen atoms of the oxalates that are in the back, left and front, right, we spin that 90 degrees. So that the waters are top and bottom. Our oxalate ligands are in the same plane, each link to its adjacent oxygen, each oxygen atom linked to its adjacent oxygen atom. Sorry, that's a very wordy way. It's hard to verbally describe the rotations here but A and C are the same molecule, both a chiral, obviously, since they are the same. Finally, let's look at d in this case, we have just one oxalate ligand and four water ligands. Well, here we are going to have an a chiral molecule because we definitely have a plane of symmetry, we only have the one oxalate ligand. So it's a little easier but similar to molecule A the two water ligands in the front, left and back, right are sticking out on either side of our plane. This one's definitely a little simpler since we just have the one oxalate to deal with. So we have identified each isomer, we have molecule A as being a chiral molecule B as being chiral and we drew it's an an tumor molecule C is a chiral because it's the same molecules A and molecule D is also a chiral. However, number three said identify the pair of chiral complexes that are in Antium but neither of the, none of these excuse me are an Antium of each other. We just have A and C that are the same molecule um but not non superposes mirror images. See you in the next video