Chemistry · Lesson

Ligand design, selectivity and catalyst limitations

Connect ligand interactions to competing pathways while separating selectivity, yield and persistence. Evaluate invented barrier, steric-volume and deactivation models within explicit limits.

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By the end, you can

Relate a selected barrier difference to a conditional parallel-pathway ratio.
Separate enantiomeric excess from total and major-product yield.
Calculate an invented buried-volume descriptor with its geometry convention.
Integrate a stipulated deactivation model and test it against a finite substrate capacity.

Ligands affect several parts of a reacting system

Ligand donation, acceptor interactions, shape and flexibility can influence geometry, substrate association, barriers and competing catalyst states. A modification can favour one product pathway while changing overall rate or lifetime. There is no single steric or electronic number that universally predicts the outcome of every catalytic system.

Selectivity describes a specified competition with its denominator. Regioselectivity, chemoselectivity and stereoselectivity compare different product distinctions. Evidence for preferential formation is not automatically stereospecificity, proof of one transition state or a measured equilibrium ratio.

A rate ratio derived from barriers requires a shared reference and matching assumptions

On a first reading, retain that competing pathways can differ in speed and that the ratio depends on the model. The following transition-state-theory calculation is an optional deeper example. At stipulated 298.15 K, choose a common starting state, two irreversible first-order channels with activation free energies 60000 and 65000 J mol⁻¹, unit transmission coefficients κ_A=κ_B=1 and the same TST prefactor.

Use Avogadro constant N_A=6.02214076×10²³ mol⁻¹ and Boltzmann constant k_B=1.380649×10⁻²³ J K⁻¹ to form molar gas constant R=N_Ak_B=8.31446261815324 J mol⁻¹ K⁻¹. Denote the two channel coefficients k_path,A and k_path,B; their ratio is k_path,A/k_path,B=exp[(65000−60000)/(RT)]≈ 7.515573. The product fractions are ratio/(1+ratio)≈ 0.882568 and 1/(1+ratio)≈ 0.117432 under these parallel-channel assumptions. With the selected TST prefactor k_BT/h, exact Boltzmann constant k_B=1.380649× 10⁻²³ J K⁻¹ and Planck constant h=6.62607015× 10⁻³⁴ J s, the model coefficients are approximately 191.301 and 25.4539 s⁻¹.

If A and B specifically form a pair of enantiomers and have no later interconversion, the selected fractions imply 76.5136% enantiomeric excess. If they are other product classes, that ee label does not apply. Multiple resting states, reversible steps, unequal prefactors or subsequent equilibration can change the relationship. These invented barriers are not measured catalyst predictions.

Enantiomeric excess does not give the chemical yield

For a stipulated two-enantiomer pool containing 95 mmol R and 5 mmol S, ee=(95−5)/(95+5)=0.900, or 90%. The enantiomeric ratio is 19:1 and major fraction 0.950 within this pair. The label R does not determine positive optical rotation.

If the initial substrate amount was 125 mmol with one product molecule per counted event, total pair yield is 100/125=0.800 and R-product yield 95/125=0.760. Neither is 95% merely because the major enantiomer constitutes 95% of the selected pair. Chemically different impurities, other products and sample purity have their own inventories.

A buried volume is defined by a chosen geometry boundary

Percent buried volume compares ligand-occupied space within a specifically positioned sphere with that sphere’s total volume. Actual calculations require a geometry, sphere radius, atom radii, treatment of H atoms and a numerical method; overlapping occupied regions must not be counted twice. A consistent convention makes comparisons interpretable.

In an invented descriptor toy, radius 3.50 Å gives sphere volume 4πr³/3≈ 179.594 Å³. Suppose 400 of 1000 equal-volume sample cells are classified as buried. The fraction is 0.400, giving 40.0% buried, about 71.8378 Å³ buried and 107.757 Å³ free under that selected sampling model. These are not coordinates or a computed map for any real ligand.

Two differently arranged ligands can share a scalar 40% while leaving different access paths or asymmetric regions. A scalar descriptor does not uniquely specify an activation barrier or ee. A real comparison requires the same geometry conventions and evidence connecting the descriptor to a defined reaction.

A returned state can still lose activity over time

In an invented constant-reactant-activity model, initially active catalyst fraction is one and decays by first-order loss coefficient k_d=0.0100 s⁻¹ with zero residual activity. Initial loaded-normalized TOF is 0.200 s⁻¹. Active fraction is exp(−k_dt), instantaneous loaded TOF=0.200exp(−k_dt), and cumulative TON is(0.200/0.0100)[1−exp(−0.0100t)].

The model activity half-life is ln 2/0.0100≈ 69.3147 s and its maintained-reservoir lifetime TON limit is 20. At 60.0 s, active fraction≈ 0.548812, instantaneous loaded TOF≈ 0.109762 s⁻¹ and cumulative TON≈ 9.02377. At 300 s, cumulative TON≈ 19.0043. These are invented model outputs, not actual catalyst longevity or proof of a specific molecular deactivation pathway.

A separate closed batch with 0.0100 mol available substrate and 0.00200 mol loaded catalyst can produce at most TON=5 under one-product-per-substrate stoichiometry. The maintained-reservoir prediction 9.02377 at 60 s would exceed that finite capacity, so it cannot apply to that closed batch. Substrate depletion and changing activity need an additional balance; a formal long-time catalyst model does not override a finite feed.

Some real systems show residual activity or several loss processes. A fitted exponential can summarize observations within its conditions without uniquely identifying the molecular route. Compare lifetime, activity and product selectivity separately, with the catalyst denominator and material boundary visible.

Cumulative turnover under an invented maintained-reservoir deactivation model

Cumulative TON (loaded basis)

Model elapsed time / s

Selected maintained-reservoir times
— Restricted zero-residual deactivation integral

Underlying data

Model elapsed time / s	Cumulative TON (loaded basis)
0	0 · Selected maintained-reservoir times
30	5.183635586365643 · Selected maintained-reservoir times
60	9.02376727811947 · Selected maintained-reservoir times
120	13.976115761755958 · Selected maintained-reservoir times
300	19.00425863264272 · Selected maintained-reservoir times

Original fixed-reactant-activity model with k_d=0.0100 s⁻¹, initial loaded-normalized TOF=0.200 s⁻¹ and zero residual activity. Markers select five times; the dashed curve is TON=20[1−exp(−0.0100t)]. Its limit 20 assumes a maintained substrate reservoir. It cannot be transferred to the separate finite batch whose stoichiometric capacity is TON=5. No measured catalyst lifetime or molecular deactivation mechanism is supplied.

Worked examples

Normalize the two channels after their rate ratio

Use the stipulated common-state 65000−60000 J mol⁻¹ difference at 298.15 K.

Ratio exp[5000/(RT)]≈ 7.515573.
A fraction≈ 0.882568 and B≈ 0.117432.
Only if A/B are the selected enantiomer pair does 100×their difference give 76.5136% ee.

A conditional parallel-channel calculation, without measured real-catalyst selectivity.

Keep the pair and substrate denominators separate

Use 95 mmol R, 5 mmol S and 125 mmol initial substrate.

ee90% and ratio 19:1 use the 100 mmol pair.
Total pair yield 100/125=0.800.
R yield 95/125=0.760.

Major fraction 0.950, total yield 0.800 and R yield 0.760 are distinct measures.

Use the same sample cells and sphere boundary

Use invented radius 3.50 Å and 400 buried of 1000 equal-volume cells.

Sphere volume≈ 179.594 Å³.
Buried fraction 0.400 gives 40.0%.
Buried/free volumes≈ 71.8378/107.757 Å³.

A restricted descriptor toy, without a real ligand map or a unique selectivity prediction.

Check the integrated model against finite material capacity

Use the maintained-reservoir model at 60.0 s and the separate finite batch.

Active fraction exp(−0.6)≈ 0.548812.
Cumulative TON=20(1−exp(−0.6))≈ 9.02377.
Finite batch maximum 0.0100/0.00200=5.
The reservoir result exceeds that capacity and cannot describe that closed batch.

An explicit model-scope failure caught by the substrate balance, rather than a physical production prediction.

Try it yourself

A separate two-enantiomer pool has 80 mmol R and 20 mmol S. Find its ee and major fraction.
Show the answer and reasoning
60% ee; major fraction 0.800.
Use the 100 mmol pair denominator: difference 60/100 and major 80/100. Chemical yield still needs initial-substrate information.
At one activity half-life in the maintained-reservoir model, find active fraction, loaded TOF and cumulative TON.
Show the answer and reasoning
0.500;0.100 s⁻¹; TON=10.0.
exp(−k_dt)=1/2 halves initial 0.200 TOF and gives 20(1−1/2), retaining the reservoir assumptions.
A closed one-product-per-substrate batch contains 0.0200 mol substrate and 0.00400 mol loaded catalyst. What stoichiometric TON ceiling follows?
Show the answer and reasoning
5.00.
0.0200/0.00400=5.00; a kinetic lifetime integral cannot exceed the available substrate capacity.
Why can two ligands with the same scalar buried-volume percentage have different catalytic selectivity?
Show the answer and reasoning
The scalar omits the detailed arrangement of accessible and occupied regions and the electronic/pathway effects.
Geometry conventions and reaction evidence are needed; one percentage cannot uniquely identify every barrier.

What to carry forward

Define the selectivity comparison and its denominator.
Rate ratios derived from barrier differences require matching pathway assumptions.
ee, major fraction and chemical yield are separate quantities.
A steric descriptor needs a consistent geometry boundary.
Integrate activity loss only within its material-balance scope.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.