Explain associative interchange reactions? What is the mechanism?

1 Answer
Aug 14, 2017

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.

Inorganic Chemistry, Miessler et al., pg. 458Inorganic Chemistry, Miessler et al., pg. 458

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.