Catalytic Cycles and Turnover

Introduction

Organometallic catalysts perform the same chemical task thousands or millions of times. The standard way to describe this is a catalytic cycle : a loop of elementary steps — ligand loss, oxidative addition, insertion, elimination — that returns the catalyst to where it started. Reading a cycle correctly, finding its net equation and measuring how many times it turns are core skills for understanding any homogeneous catalytic process.

Core explanation

Anatomy of a cycle. A cycle is drawn as a loop of metal complexes joined by arrows. Arrows entering the loop show substrates consumed; arrows leaving show products released. Each arrow inside the loop is one elementary step that you should be able to name and electron-count.

The net reaction. Add every species entering and leaving the cycle and cancel any that appear on both sides. All catalyst species cancel because the loop is closed. For alkene hydrogenation, H₂ and the alkene enter, the alkane leaves, and the net reaction is simply alkene + H₂ → alkane.

Precatalyst and activation. The compound weighed out is often not in the cycle. A precatalyst such as Pd(OAc)₂ or [Rh(COD)(diphosphine)]⁺ must first be reduced, lose a throwaway ligand or react with a base to enter the loop. The activation steps are drawn outside the cycle, feeding into it.

Resting state and turnover-limiting step. Most of the catalyst sits in one species, the resting state , waiting for the slow step. The step with the largest energy span controls the overall rate. These two ideas are distinct: the resting state is a population, the turnover-limiting step is a barrier.

Turnover number and frequency.

TON = moles of product ÷ moles of catalyst

TOF = TON ÷ time

TON measures productivity over the catalyst's life; TOF measures speed. A catalyst can be very fast (high TOF) but die quickly (low TON), or slow but long-lived. Industrial homogeneous processes typically need TON above about 10⁴–10⁶ to be economic, especially for precious metals like rhodium.

Catalyst loading. Loadings are often given in mol%. A loading of 0.1 mol% means one mole of catalyst per 1000 moles of substrate, so full conversion corresponds to a TON of 1000.

Off-cycle species. Some species sit outside the loop, such as dimers or complexes with excess ligand. They are in equilibrium with cycle intermediates and reduce the amount of active catalyst without being destroyed.

Formulae

TON = n(product) ÷ n(catalyst). TOF = TON ÷ t, in h⁻¹ or s⁻¹. Catalyst loading (mol%) = 100 × n(catalyst) ÷ n(substrate).

Step-by-step reasoning

To analyse a catalytic cycle:

1. Identify the species entering and leaving the loop. 2. Write the net reaction and check it balances. 3. Name each step and track oxidation state and electron count around the loop. 4. Confirm the final species is the same as the first. 5. Use the amounts given to compute TON and TOF.

Visual explanation

The SIM-CAT-001 cycle shows a clock-face loop with the active catalyst at twelve o'clock. Substrate arrows curve in at three and six o'clock, and the product arrow curves out at nine. Each complex is labelled with its oxidation state and electron count, so the redox rhythm of the cycle can be read at a glance.

Real-world analogy

A catalytic cycle is like a bicycle wheel. Each spoke is an elementary step; one full revolution moves the bicycle forward by one product molecule. TOF is how fast the wheel spins, while TON is how far the bicycle travels before a tyre punctures.

Real-world example

The Monsanto acetic acid process uses a rhodium iodide catalyst at very low concentration. Its high TON and TOF allow each rhodium atom to make a very large number of acetic acid molecules, which is essential given the cost of rhodium. Its successor, the iridium-based Cativa process, improved rates further and reduced by-products.

Why?

Why do catalyst species cancel when summing a cycle? Every catalyst species appears once as a product of one step and once as a reactant of the next. Because the loop closes, each is formed and consumed exactly once per turnover, leaving only substrates and products in the net equation.

Common misconception

"The resting state is the species that does the key chemistry." The resting state is merely the most abundant species. It sits just before the turnover-limiting step and is often the least reactive member of the cycle.

Worked example

Question: 0.050 mmol of a palladium catalyst converts 40.0 mmol of aryl bromide into product in 2.0 hours before becoming inactive. Calculate the loading, TON and TOF.

Reasoning: Loading = 100 × 0.050 ÷ 40.0 = 0.125 mol%. TON = 40.0 ÷ 0.050 = 800. TOF = 800 ÷ 2.0 = 400 h⁻¹.

Answer: 0.125 mol%; TON = 800; TOF = 400 h⁻¹ (averaged over the run).

Quick check

1. A catalyst at 0.01 mol% gives complete conversion. What is the minimum TON achieved? Answer: 10 000, because each mole of catalyst converted 10 000 moles of substrate.

Exam focus

Be able to extract the net reaction, distinguish precatalyst from active catalyst, and calculate loading, TON and TOF with correct units. State clearly that TOF measured at the start (initial TOF) usually differs from an average over the whole reaction.

Advanced insight

The energetic span model (Kozuch and Shaik) shows that TOF is controlled not by the single highest barrier but by the energy difference between a TOF-determining intermediate and a TOF-determining transition state, which may be several steps apart. This explains why stabilising an intermediate too much can slow a catalyst, even if every individual barrier looks small.

Summary

A catalytic cycle is a closed loop of elementary steps; summing it gives the net reaction with all catalyst species cancelling. Precatalysts enter via activation steps, and off-cycle species hold back catalyst reversibly. The resting state is the most abundant species; the turnover-limiting step sets the rate. TON measures lifetime productivity and TOF measures speed.

Practice questions

1. Define turnover frequency and give a typical unit. Answer: The number of turnovers per catalyst per unit time, usually expressed in h⁻¹ or s⁻¹. 2. Why is Pd(OAc)₂ described as a precatalyst in many cross-couplings? Answer: It is Pd(II) and must be reduced to Pd(0) before it can enter the cycle by oxidative addition. 3. A rhodium catalyst gives TON = 5000 in 10 hours. What is its average TOF? Answer: 500 h⁻¹. 4. Explain the difference between an off-cycle species and a decomposition product. Answer: An off-cycle species is in reversible equilibrium with the cycle and can re-enter it; a decomposition product is irreversibly inactive.