Associative and Dissociative Substitution

Introduction

Knowing that a ligand must leave or arrive is only half the story; the order of events matters. Does the old ligand leave before the new one arrives, or does the newcomer push in first? These two limiting pathways — dissociative and associative substitution — give different rate laws, different activation parameters and, crucially, different electron counts in their intermediates. Reading those signatures tells you how a catalyst opens its sites.

Core explanation

Dissociative (D). The leaving ligand departs first:

L nM–X → L nM + X (slow) L nM + Y → L nM–Y (fast)

The rate depends only on the complex: rate = k[L nM–X]. The intermediate has one fewer ligand and two fewer electrons. For an 18-electron complex this gives a reasonable 16-electron species, which is why saturated complexes such as Ni(CO)₄ and Cr(CO)₆ substitute dissociatively.

Associative (A). The entering ligand binds first:

L nM–X + Y → L nM(X)(Y) (slow) L nM(X)(Y) → L nM–Y + X (fast)

Rate = k[L nM–X][Y]. The intermediate has an extra ligand and two more electrons. This is favourable only when the starting complex has 16 or fewer electrons, as in square-planar d⁸ complexes of Pt(II), Pd(II), Rh(I) and Ir(I), which pass through five-coordinate 18-electron intermediates.

Interchange (I). Many real reactions lie between the extremes. In an Iₐ mechanism bond making dominates; in I d bond breaking dominates. No intermediate is detected.

Kinetic signatures.

Feature D A --- --- --- Rate law first order in complex, zero order in Y first order in complex and in Y ΔS‡ positive (more particles) negative (fewer particles) ΔV‡ positive negative Sensitivity to Y small large Steric bulk on metal speeds reaction slows reaction

Solvent pathways. Square-planar complexes often show a two-term rate law, rate = (k₁ + k₂[Y])[complex]. The k₂ term is direct associative attack; the k₁ term is associative attack by solvent, followed by fast replacement of solvent by Y.

Ring slippage. Complexes such as (η⁵-C₅H₅)Rh(CO)₂ are 18-electron yet substitute associatively. The Cp ring slips to η³, freeing two electrons so the incoming ligand can bind without exceeding 18. Indenyl ligands slip even more easily — the "indenyl effect" — giving rate enhancements of up to about 10⁸.

Step-by-step reasoning

1. Count the electrons of the starting complex. 2. If 18-electron, expect D (or ring slippage). 3. If 16-electron square planar, expect A. 4. Test by varying [Y]: a rate independent of [Y] points to D. 5. Measure ΔS‡ and ΔV‡: positive values support D; negative values support A.

Visual explanation

Draw two energy profiles. For D, the first hump leads to a trough labelled "16 e⁻, lower coordination". For A, the trough is labelled "18 e⁻, trigonal bipyramid". Placing the intermediates side by side makes the electron-count argument obvious.

Real-world analogy

Changing a player in a football match can happen two ways. In one, a player walks off and the team briefly plays one short before the substitute runs on (dissociative). In the other, the substitute steps onto the pitch first and the outgoing player then leaves (associative). The rules of the game — here, the 18-electron limit — decide which is allowed.

Real-world example

The anticancer drug cisplatin, cis-[PtCl₂(NH₃)₂], is square-planar Pt(II). Inside cells, chloride is replaced by water through an associative, solvent-assisted pathway, and the aqua complex then binds to guanine in DNA. The slow, associative substitution helps the drug survive in blood, where chloride concentration is high.

Why?

Why does steric bulk speed dissociative but slow associative substitution? In D, crowding is relieved as a ligand leaves, lowering the barrier. In A, a sixth or fifth ligand must squeeze in, so crowding raises the barrier.

Common misconception

"A first-order rate law proves a dissociative mechanism." A solvent-assisted associative pathway also appears first order in complex and zero order in Y, because solvent concentration is constant. Activation parameters and solvent effects are needed to decide.

Worked example

Question: For substitution of CO in Ni(CO)₄ by PPh₃, the rate is independent of [PPh₃] and ΔS‡ is positive. Assign the mechanism and explain using electron counts.

Reasoning: Ni(0) is d¹⁰; four CO ligands add 8 electrons, so Ni(CO)₄ is 18-electron. Associative attack would need a 20-electron intermediate. Loss of CO gives 16-electron Ni(CO)₃, consistent with the zero order in PPh₃ and positive ΔS‡.

Answer: Dissociative (D) substitution.

Quick check

1. Which mechanism is expected for ligand substitution in square-planar [PtCl₄]²⁻, and why? Answer: Associative, because the 16-electron d⁸ complex can accept a fifth ligand to form an 18-electron intermediate.

Exam focus

Link each mechanism to its rate law, the sign of ΔS‡ and ΔV‡, and the electron count of its intermediate. Be able to explain ring slippage as the way an 18-electron complex can react associatively.

Advanced insight

Activation volumes measured under high pressure are among the most reliable mechanistic probes: values near +10 cm³ mol⁻¹ indicate dissociative character, and values near −10 cm³ mol⁻¹ associative character. Combined with linear free-energy relationships for different entering nucleophiles, they place a reaction on the continuum from A through I to D.

Summary

Dissociative substitution loses a ligand first, gives a lower-electron intermediate, is zero order in the incoming ligand and has positive ΔS‡; it suits 18-electron complexes. Associative substitution binds the newcomer first, is first order in it, has negative ΔS‡ and suits 16-electron complexes. Interchange mechanisms lie between, and ring slippage lets saturated complexes react associatively.

Practice questions

1. Write the rate law for a purely associative substitution. Answer: Rate = k[complex][Y], first order in both the complex and the entering ligand. 2. What sign of ΔS‡ is expected for dissociative substitution, and why? Answer: Positive, because one molecule splits into two in the transition state, increasing disorder. 3. Why does an indenyl complex substitute faster than its cyclopentadienyl analogue? Answer: The indenyl ring slips from η⁵ to η³ more easily because the benzo ring regains aromaticity, opening a site for associative attack. 4. Explain the two-term rate law for substitution in square-planar complexes. Answer: One term is direct associative attack by Y; the other is associative attack by solvent followed by fast replacement of the solvent by Y.