Geometrical Isomerism

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Take two compounds with exactly the same formula: platinum, with two ammonia molecules and two chloride ions attached. Same metal, same ligands. Yet only one of them is used as an anticancer drug. So what could be different? Hold that thought, because by the end of this video you will know. Questions on these isomers come up often in JEE, so let's begin.

0:31Cis and trans in a square plane

So, look at the first two drawings. The black dot is the metal, M, with four ligands at the corners of a square, grey for ligand A and orange for ligand B, two different ligands. So the formula is M A two B two, where the small two below a letter means two of that ligand. Okay? In the first drawing, the two orange B ligands sit at adjacent corners, next to each other. We call this cis. In the second, they sit opposite each other, across the metal, and we call this trans. Same formula, different arrangement in space, and these are geometrical isomers. Clear? Let me ask you this. With four corners, two A ligands and two B ligands, can you place them in a third way, neither cis nor trans? The answer is no, you cannot. Any other placement is just one of these two, turned around. Two B ligands are either next to each other or opposite. So square planar M A two B two has exactly two geometrical isomers. Got it? Now a question many students get wrong. Suppose the same four ligands sit at the corners of a tetrahedron instead. Do we again get cis and trans forms? Oh no, careful, the answer is no! In a tetrahedron, every position is adjacent to every other one, with no opposite corner at all. So the two B ligands are always next to each other, and tetrahedral complexes show no geometrical isomerism. Right?

2:04Octahedral: cis, trans, fac and mer

Now, six ligands at the corners of an octahedron, in the middle two drawings, labelled cis and trans M A four B two. The six positions form three opposite pairs along three axes, so each has one opposite position and four neighbours. So the two B ligands can be cis, adjacent, or trans, opposite. Two isomers again. With me? Now let me make it harder, with three A ligands and three B ligands, M A three B three. Look at the last two drawings, labelled fac and mer. In each one, where do the three orange B ligands sit? Is any B opposite another B? Let's see. In fac, the three B ligands sit on one face of the octahedron, all next to each other, with no B opposite another. In mer, they lie on a meridian, like a line of longitude, so two of them are opposite. There is no third way, because either no two B ligands are opposite, or exactly two are. So M A three B three has two isomers, fac and mer. Clear? And for M A five B, with only one different ligand, all six positions are equivalent, so B gives the same complex wherever it goes. Just one form, so no geometrical isomerism. Right?

3:20Counting strategy: three steps

Alright, for harder formulas we need a method, not guesswork. Step one: fix the ligand that occurs least often. Okay? Step two: for each pair, decide whether the two ligands are cis or trans, next to each other or opposite. Right? And step three, the one students forget: check that no two arrangements are related by rotation. If you can turn one into the other, it is the same isomer. Got it? Let us try it. Square planar MABCD, with four different ligands. How many geometrical isomers does it have? Very good if you said three! Fix ligand A first. Now only one thing matters, which ligand sits opposite to it, and it can be B, C or D, so three isomers. Clear?

4:10Bidentate ligands: AA and AB

Now, bidentate ligands, which bind through two donor atoms. We write a symmetric one as A A, like e n, ethane one two diamine, and an unsymmetric one as A B, like g l y, glycinate, with a nitrogen end and an oxygen end. Both ends must span cis positions, never trans. Okay? So, think about square planar M, with one A A ligand in round brackets, and B two. Can the two B ligands be trans? The answer is no, they cannot. The A A ligand takes two adjacent corners, so the two B ligands get the other two, also adjacent, giving just one form. But with two A B ligands, the two A ends can be cis or trans, so two isomers. Right? In the octahedron, with two A A ligands, the two B ligands can still be cis or trans, so two isomers. With three A A ligands, all six positions are taken in pairs, leaving one geometrical form. With me? Finally, three A B ligands. Their three A ends can sit on one face, or on a meridian, so again fac and mer, two isomers. Got it?

5:19Example 1: cis and trans Pt(NH3)2Cl2, which is cisplatin

Alright, Example one, and it is your turn. Draw the geometrical isomers of open square bracket, P t, N H three in round brackets with a small two, C l two, close square bracket. That is platinum with two ammonia and two chloride ligands. Which one is the anticancer drug? Pause the video and try it first. Done? Okay, let's check. This platinum complex is square planar, with two ammonia and two chlorides, so it is of type M A two B two. Right? So the two chlorides sit either at adjacent corners, the cis form, or at opposite corners, the trans form, just like the first two drawings. Clear? So there are two geometrical isomers. And that solves our puzzle: the drug is the cis form, called cis platin. Same formula, but only the cis arrangement is used as the drug. Okay?

6:14Example 2: five geometrical isomers of MA2B2C2

Now a harder one, Example two. Show that M A two B two C two has five geometrical isomers. The example on screen is open square bracket, P t, N H three in round brackets with a small two, C l two, p y in round brackets with a small two, close square bracket, with a raised two plus outside. Here p y is pyridine. Pause the video and try it first. Done? Okay, let's check. We have three pairs: two A ligands, two B ligands and two C ligands. Each pair is either cis or trans, so we count case by case. With me? Case one: all three pairs trans, with the A pair on one axis, the B pair on the second and the C pair on the third. That can be done in only one way, so one isomer. Right? Now, what about exactly two pairs trans, and the third pair cis? Can that happen? Oh no, careful, it cannot happen! Two trans pairs fill two axes completely, so the last pair gets the third axis and is forced to be trans too. Clear? So the next case: one pair trans, and the other two pairs cis. The trans pair can be the A pair, the B pair or the C pair, so three isomers. Got it? And the last case: all three pairs cis. There is only one such arrangement, but it is chiral, which means it cannot be superimposed on its mirror image. With me? So the total is one, plus three, plus one, which gives five geometrical isomers. And the all cis form also has a mirror image partner, so there are six stereoisomers in all. Okay?

7:52Quick revision: cis-trans, fac-mer, counting in 3 steps

So, quick revision. Square planar complexes can show cis and trans, but tetrahedral ones show no geometrical isomerism. In an octahedron, M A four B two gives cis and trans, and M A three B three gives fac and mer. To count, fix the rarest ligand, decide cis or trans, and check rotations. Now go and try the practice problems on this page. All the best!