We just learned about oxidative addition, so now let’s learn about what is essentially the reverse
process, reductive elimination. To see what this reaction accomplishes we can truly just take the
generalized equation for oxidative addition and reverse it, so that the complex with ligands A and
B and the metal in the n+2 oxidation state is now the reactant, and this will reduce to the metal
in the n oxidation state, while also yielding A-B. Just like with oxidative addition, complexes with
stable Mn and Mn+2 oxidation states will be good candidates, and in particular those that seek to
eliminate a formal positive charge, or those that have electron withdrawing substituents.
Let’s take for example this iridium complex with a polyhapto ligand, a trimethyl phosphine,
a hydride, and a methanide. If this were to undergo reductive elimination, what would result?
Well we just learned that oxidative addition allows molecular hydrogen and alkanes to split
apart and coordinate to the metal, and reductive elimination is the reverse process, so it makes
the most sense for hydride and methanide to combine to form methane. Once again, iridium
will have its oxidation state reduced by two as a result of this process. Complexes that undergo
reductive elimination are primed for oxidative addition as they now have lower oxidation states
and coordination sites available, so a molecule like benzene could then react,
to produce this complex, as it is the case that even aromatic C-H bonds can undergo
oxidative addition. Indeed oxidative additions and reductive eliminations
are basic steps in some important catalytic reactions we will see later in the series.
Reductive elimination is quite straightforward mechanistically, and it is extremely common,
often taking place in order to allow for another reaction to occur. Take for example this tungsten
complex with two Cp ligands, a methanide, and a hydride. Let’s say this is to react with benzene
in some manner. What is the count on this complex? Well each of the four ligands has a 1- charge,
and the complex is neutral, so tungsten is 4+, meaning it has only two valence electrons.
Plus 12 from the cyclic ligands and another four from the other two, that brings us to 18.
This means the complex is coordinatively saturated, and also quite crowded sterically.
Benzene will not be able to coordinate. But this complex is capable of reductive elimination.
The methanide and hydride ligands can leave as methane, and then there will be open sites for
benzene to undergo oxidative addition, leaving the phenyl ligand and a new hydride ligand.
So again, we will often see reductive elimination in conjunction with other reactions.
As we would expect, just in the way that oxidative addition leaves two new ligands cis to
one another, the ligands that leave via reductive elimination must usually be cis to one another in
order for the new bond to form. Here is an iridium complex (text p147) with two triphenyl phosphines,
two hydrides, a chlorine, and a carbonyl. The metal is 3+ and the electron count is 18.
These two hydrides are cis to one another, and can therefore eliminate to form H2,
so the resulting complex will have a 1+ charge and an electron count of 16. The
relevant intermediate is difficult to observe but it is thought that the H2 molecule will
remain coordinated to the metal in this manner for some time prior to dissociating. So we can
see that similar to stepwise oxidative addition, stepwise reductive elimination is also possible.
We must remember that since oxidative addition and reductive elimination are reverse processes, the
products of one can serve as the reactants of the other, and these reactions can both be reversible.
Usually one side will be thermodynamically favored, which can be utilized in synthesis. And
as we mentioned, sometimes reductive elimination followed by oxidative addition is a good way to
substitute one set of ligands for another, if the reactions required for such a transformation
are favorable. However, we will see later that in some catalytic cycles, steps can occur that are
not thermodynamically favorable, as long as the change in Gibbs free energy is negative overall.
But for now, let’s move forward and check out some more organometallic reactions.