Cobalt(III) trifluoroacetylacetonate, Co(tfac)₃, is a sixc oordinate, octahedral metal chelate in which three planar, bidentate tfac ligands are attached to a central Co atom:

(a) Draw all possible diastereoisomers and enantiomers of Co(tfac)₃.

Video transcript

Hi, everybody. Let's look at our next problem. This problem starts out with a long chemical name because we've got some ligands or a ligand that has a very long name. It's abbreviated D TB P I. So with that in mind, we start with tris open parentheses, two, open parentheses, die church beetle phosphene, close parentheses, one pinol, close parentheses, Rhodium two and its molecular formula is we have in brackets with an overall charge of positive two Rh. And then in parentheses, D TB P I subscript three is an octahedral complex with three bate ligands surrounding the central metal atom. And then we have sort of a shorthand structure here with a note that D TB P I is represented as we have N and P with a little arc. So the arc represents the rest of the ligand structure. So we know our bi 10th 8 ligand has two different binding atoms nitrogen and phosphorus. So our sort of shorthand diagram shows our Rh with a bond to the N A bond to the P 90 degrees apart and then the arc between them brackets around the whole thing. And a subscript of three. What are the structures of all the dimers of the complex, which isomers are in Antium. So we recall ditherers have different arrangements of the atoms around various stereo centers, but they are non identical and non mirror images. While the an anim are those mirror images that are not superimposable, so not identical. So with that in mind, let's think about the possible arrangements of this molecule. We have the N and the P that always have to be adjacent to each other. So let's draw our octahedron here. So I have drawn the rhodium with its eight bonds arranged around it in that octahedral arrangement. Now, let's note that we can draw this. So let's draw it as just alternating N and P around the bonds like in a clockwise fashion. And then we have to draw the arcs between each NP pair. So when we look at this, we drew it sort of drawing going around in a circle because we have a two dimensional drawing. But this is a three dimensional structure. And when we look at it, notice that there's three nitrogen atoms and three phosphorus atoms. So this creates a situation that's comparable to the fac and mirr isomers. You see when you have three of a single kind of isomer or ligand, excuse me, notice I say comparable because the three nitrogens are not three separate ligands, they're part of a single ligand where they're lumped in there with phosphorus. So it's not the same thing, but we can think about it to help us visualize the possibilities of different stereo isomers. So in this case, we have a nitrogen on the top and two nitrogens projecting out of the screen towards us. So they're all on the same face of the octahedron, whereas the three phosphorus, the two projecting back and the one down are also on the same face. So this is analogous to a f isomer, put that in parentheses. So just keeping in mind because then we'll have this mirror possibility to help us think of another sting ice swimmer. So now let's look at the mirror image here and see if these are identical or if they're a pair of an ante MERS. Oh And I need to draw brackets around my whole structure to remember that it has a two plus there, a positive two charge on the whole thing. So put all those brackets. Now I need to put my rodi again with its eight bonds. The easiest way to start on these mirror images is to start with the closest to. So we have phosphorus on top nitrogen on the bottom, on the left side, my top and bottom atoms, nitrogen, nitrogen and phosphorus of course, will be the same. And then let's draw the outer phosphorus and nitrogen. And then we just have to be careful when we fill in the um links between the nitrogen and phosphorus. The remainder of the ligand to make sure I'm connecting them in the right direction. So I note that within the square plane, I have a phosphorus and nitrogen linked that was on the left side, it's now on the right side, then my top nitrogen is linked to the phosphorus in the back. So that's now on the left side, my bottom phosphorus is linked to the nitrogen in the front, which was on the right side and now it's on the left side. So we've drawn the mirror image. Can we superimpose them? Well, let's note that, imagine we tried to flip over my square plane, sort of flip it over 100 80 degrees. So that the square planar phosphorus and nitrogen, the ligand that's in a square plane is in the same place. Well, what would happen there is that then my phosphorus on the bottom would be on the top and the nitrogen that was on the top would be on the bottom. So let me just draw that real quick. We're going to have to scroll up to make a little room here. I'll draw it in red. So I'm imagining flipping the whole square plane over. So the phosphorus and nitrogen that are linked within the same plane, well, they match my mirror image, but now my nitrogen and phosphorus have been inverted. So nitrogen is now on the bottom. And then the uh in in the original, the nitrogen phosphorus that were on the right, they've swung around and been flipped over as well. And now that nitrogen in the bottom is linked to the phosphorus in the back. The phosphorus that's now on the top is linked to the nitrogen in the front. And when we look at our Anant tumor, this just does not match the various propellers of the blade are in the wrong location. They're in the wrong plane. So these are not super imposable mirror images. So this is a pair of Anan tumors. So we'll highlight that. So we've got two dimers that happen to exist as a pair of an anti emer. Now, let's think about our other arrangements. So it will be comparable to a mirr isomer. So let's put brackets here and put our rhodium with its eight bonds. Well, it could be tricky to try and think of putting the nitrogens on and figuring out where all the rest of the ligand is a way to generate. This is if we think of the fact that when we drew our first structure, we went around clockwise NPNPNP. Well, if we just invert one of them, so for the first one, write it as PN and then NPNP and then link them together. When we look at this, observe that the nitrogens are now all within the square plane, all three nitrogens in the same plane. And then the phosphorus atoms are in the plane that goes through top and bottom and then projects out through the back phosphorus. So we have generated a structure that is analogous to the mur isomer. So we know from the fact that F and M are two different Diasa MS. So are these, they're kind of a little shortcut, maybe to know what's possible. And then let's draw the mirror image. So I've got my rhodium with its bonds. I have two nitrogens on the right side that I reflect here on the left. Now, my phosphorus top and bottom and then a phosphorus and nitrogen that are now on the right side and draw on the bonds. So first the bond within the square plane and then we'll do the bond from the top phosphorus. Now to the back nitrogen bond from the bottom phosphorus. Now to the front nitrogen, we can be pretty certain that these will be an anti ierse since this is analogous to the other structure. But just to be thorough, we can go through and draw that mirror image. So my rhodium with its eight bonds and now let's look at the mirror image if I try and again, flip the whole thing over. So the square plane has been flipped over by 180 degrees. I do line up my phosphorus and nitrogen with the li in between them. On the right side, I, in this case, I still have phosphorus top and bottom because I have the same atom. So I haven't changed that. But then I have on the left side, now I have two nitrogens. So on a superficial level, they might look super imposable. But when I draw in the connection of the ligands, then I realize that it's not. So my top phosphorus was the bottom one. And if we imagine flipping this whole thing over, it's now linked to the nitrogen in the front. Whereas my bottom phosphorus is now linked to the one in the back. It would be a little hard to see that. But you can see that that's not going to match up. It matches on the right side, the bottom to the goes into the back. Whereas on our an ante or the bottom phosphorus is linked to the front nitrogen and the top phosphorus is linked to the back nitrogen. So if we didn't see the rest of the ligand, if they were just individual phosphorus and nitrogens, it wouldn't be and, and it would be, you would be able to superimpose it, they would be identical. But because of these linkages to the ligands, these are indeed a pair of an anti imer. So an anti immerse, now we can't generate any more. You can kind of, again, you can kind of tell that because we have this analogy to F and Mr. But let's just imagine, we know we generated that structure, we just drew by flipping one of our PN orientations. The other way could we flip two of them? Well, that would just give us the identical molecule as our mirror image because when we look at our mirror images, we would have PNPN and then NP so we have all our possibilities here, we have a total of four stereo isomers. We have structures of them and they exist in two pairs of an anti emmers. See you in the next video.