Catalytic cycles and turnover

A returned catalyst state closes a proposed cycle

A catalyst participates in the reacting sequence and is regenerated in the selected net cycle. Its state can change at each step, and activation, side reactions and deactivation may lie outside a simple drawing. Catalysis changes a reaction pathway and rate without changing the overall standard reaction Gibbs energy for the same net and conditions.

Choose a class-level mononuclear Pd model retaining two neutral monodentate donors L. PdL₂ has formal Pd0 and fourteen electrons. Oxidative addition of a selected R–X gives PdL₂(R)(X), assigned Pd+2 and sixteen. A symbolic transfer replaces X by R′, giving PdL₂(R)(R′) with the same metal state and count. Reductive elimination releases R–R′ and returns PdL₂ to fourteen. R/R′ are assigned organyl −1 and X halide −1 in this selected ledger.

This 14→16→16→14 cycle organizes the chosen connectivity changes. It omits actual precatalyst activation, bases, solvent ligation, aggregation, off-cycle pools and alternative oxidation-state pathways. A circular diagram is a hypothesis or bounded schematic, rather than independent proof that every arrow occurs under an actual set of conditions.

A net must include the transfer partner and coproducts

For one original formula-balance illustration, write C₆H₅Br+C₆H₇BO₂+NaOH→C₁₂H₁₀+NaBr+B(OH)₃. The selected reactants represent phenyl bromide, phenylboronic acid and sodium hydroxide; the product ledger includes the coupled organic skeleton and inorganic coproducts. Both sides have C₁₂H₁₃BO₃NaBr and net charge zero.

This is formal net stoichiometry, with no amounts, conditions, procedure, yield or claim of one universal base/transmetalation mechanism. Omitting the transfer partner or coproducts would make the apparent cycle’s carbon, boron or halide balance incomplete. The metal-state ledger and a fully balanced written net address different constraints.

State what is counted and over what interval

Here cumulative turnover number TON is amount of specified product divided by the stated catalyst amount, dimensionless for one product per counted event. Average turnover frequency over a stated interval is TON divided by that elapsed time. Instantaneous TOF instead uses the product formation rate at that time divided by the defined catalyst or active-site amount, giving inverse-time units. Terminology varies in older sources, so inspect the actual equation.

For an invented batch, load 0.00200 mol catalyst and 0.500 mol substrate. After 20.0 min, suppose 0.450 mol substrate has been consumed and 0.400 mol desired product formed, with one product molecule per counted substrate event. Loaded-catalyst TON is 200 and full-interval average TOF=200/1200≈ 0.166667 s⁻¹, or 600 h⁻¹. Catalyst loading is 0.400 mol%.

Conversion is 0.900, desired-product selectivity 0.400/0.450≈ 0.888889 and desired yield 0.800. They refer to different denominators. The measured-window TON is not automatically an ultimate lifetime turnover number or an initial TOF, and neither establishes every loaded catalyst entity as active.

A known active fraction changes the normalization

In a separate stipulated fixed-volume comparison, product formation rate is 1.00× 10⁻⁵ mol s⁻¹, loaded catalyst 0.00200 mol and independently known active catalyst 0.000500 mol, with one site per active entity. Loaded normalization gives 0.00500 s⁻¹, while active-site normalization gives 0.0200 s⁻¹. Their fourfold difference follows the active fraction 0.250.

If the active-site count is unknown, a loaded normalization can still be reported honestly, but should not be compared as though it were the same as a separately established site-normalized value. Product event definitions, changing volume, concentration versus amount rates and accessible-site counting also need consistency. A larger reported number can reflect a different denominator rather than a faster actual process.

A cycle can spend time in several states

For an optional deeper model, choose irreversible serial states A→B→C→A with maintained reactant activities, independent exponential waiting steps and rates 1.00, 0.500, 0.100 s⁻¹. Count one product per completed cycle. Mean waiting times 1, 2, 10 s sum to 13 s, so stationary turnover frequency is 1/13≈ 0.0769231 s⁻¹.

The stationary state fractions are proportional to waiting times:1/13, 2/13, 10/13. Multiplying each fraction by its outgoing coefficient gives the same cycle flux 1/13. Making the first step tenfold faster changes the total time to 0.1+2+10=12.1 s and TOF≈ 0.0826446 s⁻¹. Making the last tenfold faster gives 1+2+1=4 s and TOF=0.250 s⁻¹.

This invented fixed-activity model explains why a long waiting step can strongly affect throughput without making every other step irrelevant. It has no reversible paths, substrate depletion, activation or catalyst loss. Real cycles can have several resting states and conditions-dependent rate control; one prominent intermediate or a drawn slow arrow is not a complete kinetic proof.

Worked examples

Try it yourself

1. Under the same product-event convention, 0.150 mol product forms from 0.00100 mol loaded catalyst over 600 s. Find TON and interval-average TOF.

   Show the answer and reasoning

   TON=150; average TOF=0.250 s⁻¹.

   Divide product by loaded catalyst, then by the full elapsed time; neither is an independently known active-site normalization.

2. In a separate three-step fixed-activity serial toy, rates are 1, 1, 0.5 s⁻¹. Find the mean cycle time and stationary TOF.

   Show the answer and reasoning

   4 s;0.25 s⁻¹.

   Independent mean waiting times 1+1+2=4 s; count one product per completed cycle.

3. Why must inorganic coproducts appear in the full phenyl-coupling net?

   Show the answer and reasoning

   They carry atoms absent from the coupled organic product.

   Boron, halide, sodium and the remaining H/O inventory cannot disappear when the catalyst is regenerated.

4. Does a drawn closed cycle establish the actual active catalyst and the rate-controlling state?

   Show the answer and reasoning

   No.

   Activation, speciation and kinetic evidence are separate tests of the schematic.

What to carry forward

• Track the returned catalyst and the complete net.
• Treat a class cycle as a bounded model.
• Define TON, TOF, event and time window.
• State loaded versus active-site normalization.
• Several waiting steps can contribute to cycle throughput.