Explain associative interchange reactions? What is the mechanism?

Associative interchange, or `I_a` (as opposed to dissociative interchange, `I_d`), since:

• the departing ligand `X` is assumed to have a fairly weak bond with the metal `M` that the incoming `Y` ligand can easily disrupt. That is, `Y` is a good nucleophile and the `M-X` bond is fairly weak.
• the departing ligand `X` and incoming `Y` basically swap by the time the reaction is over.

However, there are two accepted variations of this mechanism that I know of.

Consider a square planar complex `"MXL"_2"T"`, where `T` is the ligand trans to the departing ligand `X`.

The regular process is described in `(a)`, and a solvent-assisted process is described in `(b)`, in the figure below.

Typically, a 5-coordinate transition state forms, but a 6-coordinate transition state is also known, with assistance from solvent (Inorganic Chemistry, Miessler et al., pg. 458).

A two-term general rate law is frequently used to describe a square-planar substitution process:

`r(t) = overbrace(k_1["Cplx"])^((b)) + overbrace(k_2["Cplx"][Y])^((a))`

where:

• the `k_1` portion describes the pseudo-first-order solvent-assisted process.

• the `k_2` portion describes the regular, second-order association process. The complex referred to pertains to `"MXL"_2"T"`.

Referring to the diagram above:

• In `(a)`, `Y` comes in and forms a 5-coordinate transition state, and displaces `X` with assistance from `T` due to the trans effect. See here for a discussion on the trans effect. As both the complex and `Y` participate in one step, this is second order overall.

• In `(b)`, the solvent `S` comes in and displaces `X` first. This step is presumed slow, while the second step, in which `Y` displaces the solvent `S`, is presumed fast.

Since the solvent is presumed in large excess, the order of `S` is approximately zero, and thus, the solvent-assisted mechanism is pseudo-first-order in the complex.