Chemistry · Lesson

Regioselectivity, stereoselectivity and kinetic or thermodynamic control

Compare product site and configuration preferences with conversion and yield. Follow one reversible model from early kinetic preference to late equilibrium at a fixed temperature, using consistent forward and reverse barriers.

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By the end, you can

Distinguish regioselectivity, stereoselectivity and stereospecificity.
Normalize parallel-channel rates under explicit assumptions.
Connect early and late product ratios within one reversible network.
Keep activation-energy inference, product stability, conversion and yield separate.

A selectivity claim must name the alternatives

Regioselectivity concerns preferred bond-making or bond-breaking sites that lead to different constitutional relationships. Stereoselectivity concerns preferential formation of stereoisomeric products; enantioselectivity compares enantiomers, and diastereoselectivity compares diastereomers. State the actual alternatives and the conditions rather than using selective as a free-standing property.

Stereospecificity describes a relationship between the starting configurations and the product configurations: different stereoisomeric starting materials follow different stereochemical outcomes in the stated pathway. A high product ratio from one starting material is not enough to establish that relationship. A familiar mechanistic name or flat drawing is likewise insufficient to supply an experimental ratio.

A constant branch ratio can coexist with changing yield

For an invented closed irreversible S→A and S→B model, take k_A=0.0200 s⁻¹, k_B=0.00500 s⁻¹, fixed medium and initial normalized amounts (1,0,0). Then S=exp(−0.0250t), A=0.800(1−S) and B=0.200(1−S). For t>0, A/B=4.00; at t=0 the product ratio is undefined because both product amounts are zero.

At 40.0 s, S≈ 0.367879, A≈ 0.505696 and B≈ 0.126424. The 80.0% A fraction among formed products differs from its 50.5696% fraction of the initial amount at that time. Different reactant dependence, additional pathways, loss or product interconversion would require a different balance.

One consistent network can favour different products early and late

Now use a separate invented ideal dilute isomer model S⇌A and S⇌B at one fixed temperature, 298 K. All three states have the same molecularity and share a standard-state convention. Assign molar standard Gibbs levels G_S=0, G_A=−5.00 and G_B=−8.00 kJ mol⁻¹, with transition levels 75.0 and 80.0 kJ mol⁻¹ relative to S for the A and B connections. These are illustrative levels, not measured compound energies.

On a first reading, compare the displayed early and late amounts and check that S+A+B stays one. Return to calculating these rate coefficients after revisiting reaction rates and the thermodynamic treatment of activation barriers. Take first-order transition-state coefficients k=(k_B T/h) exp(−ΔG‡/(RT)), with transmission coefficient one for every direction and an equal prefactor. Here k_B is Boltzmann’s constant, h Planck’s constant and R the molar gas constant. Forward barriers are 75.0 and 80.0 kJ mol⁻¹; the same transition levels give reverse barriers 80.0 and 88.0 kJ mol⁻¹. They must be measured from the respective starting states.

The resulting rate coefficients in s⁻¹ are k_SA≈ 0.443607, k_AS≈ 0.0589651, k_SB≈ 0.0589651 and k_BS≈ 0.00233534. Starting from S alone, the initial formation-flux ratio A/B approaches k_SA/k_SB≈ 7.52321 as t→0⁺. The amount ratio at t=0 itself remains undefined. This early preference comes from competing activation barriers, not the final product Gibbs levels.

Thermodynamic control requires available equilibration

For the same reversible model, equilibrium requires k_SA S=k_AS A and k_SB S=k_BS B. Its dimensionless ratios are A/S≈ 7.52321 and B/S≈ 25.2490. Normalize S+A+B=1 to obtain S≈ 0.0296101, A≈ 0.222763 and B≈ 0.747627, so equilibrium A/B≈ 0.297960. B dominates late even though A forms preferentially early.

At 10 s the exact model amounts are approximately S=0.0944267, A=0.757308 and B=0.148265; A/B≈ 5.10781. At 1000 s they are approximately 0.0296255, 0.222898 and 0.747476, with A/B≈ 0.298201. A and B exchange through S, and the plotted amount changes follow the full reversible balance. Product stability alone cannot give the time needed to approach equilibrium.

All those comparisons use the same fixed 298 K. High temperature by itself does not prove thermodynamic control, and low temperature by itself does not prove kinetic control. Available interconversion paths, barriers, concentrations, time and measurement context matter. The model provides no temperature extrapolation and no measured real product distribution.

Early and late product preferences in the same reversible network

Normalized amount fraction

u = log₁₀(t / s)

Starting state S
Product A
Product B
— Exact reversible product A

Underlying data

u = log₁₀(t / s)	Normalized amount fraction
-2	0.9949881988274505 · Starting state S
-1	0.9511117130519005 · Starting state S
0	0.6142675830123691 · Starting state S
1	0.09442670421736138 · Starting state S
2	0.05847588125604808 · Starting state S
3	0.02962553452347072 · Starting state S
4	0.029610113139059147 · Starting state S
-2	0.004423635852337501 · Product A
-1	0.04313792042347497 · Product A
0	0.33924441772090486 · Product A
1	0.7573083779456354 · Product A
2	0.4758608918471685 · Product A
3	0.22289823209205145 · Product A
4	0.22276301588636807 · Product A
-2	0.000588165320211987 · Product B
-1	0.005750366524624576 · Product B
0	0.04648799926672609 · Product B
1	0.14826491783700319 · Product B
2	0.4656632268967834 · Product B
3	0.7474762333844779 · Product B
4	0.7476268709745728 · Product B

Original exact S⇌A, S⇌B first-order isomer toy at fixed 298 K, common standard states, equal transition-state prefactor and transmission coefficient one. Initial normalized amounts are (1,0,0). Markers show three state amounts whose sum is conserved; the dashed curve shows only A. The horizontal coordinate is u=log₁₀(t/s), covering positive times 0.01–10000 s: one unit of u multiplies time by ten. No point at t=0 or log of zero is drawn. Stipulated Gibbs levels and kinetics are not experimental data.

A product ratio gives a barrier difference only within a model

For a separate invented product pair containing only two enantiomers in fractions 0.900 and 0.100, their ratio is 9:1 and enantiomeric excess is 80.0%. That says nothing by itself about optical-rotation sign or an R/S assignment.

Only if those amounts reflect two irreversible competing routes from a common reactant, with identical molecularity, the same prefactor and transmission coefficient, does the stipulated transition-state model give ΔG‡_minor−ΔG‡_major=RT ln 9≈ 5.44409 kJ mol⁻¹ at 298 K. This is an activation difference, not the equilibrium Gibbs difference between the products. Unequal prefactors, reversibility or different pathways prevent that simple inversion from giving a unique barrier difference.

Conversion, selectivity and yield answer different questions

In an invented one-to-one amount ledger, 0.00500 mol initial substrate leaves 0.00100 mol unreacted, with 0.00300 mol desired product and 0.00100 mol alternative product. With only these outcomes and no losses, conversion is (0.00500−0.00100)/0.00500=0.800. Desired-product selectivity among converted material is 0.00300/0.00400=0.750. Desired-product amount yield relative to the initial substrate is 0.00300/0.00500=0.600.

Each number uses a named denominator. Isolated yield additionally requires recovery and purity evidence; this simple amount balance supplies no isolation measurement. A reported preference among products cannot replace conversion, and conversion cannot establish how much desired material is available.

Worked examples

Separate product composition from amount yield

Use the irreversible k_A=0.0200 and k_B=0.00500 s⁻¹ model at t=40.0 s.

S = exp(−0.0250 × 40.0) ≈ 0.367879.
A = 0.800 × (1−S) ≈ 0.505696; B ≈ 0.126424.
A/(A+B) = 0.800 while A/initial amount ≈ 0.505696.

80.0% product selectivity and approximately 50.5696% desired amount yield in this ideal balance at that time.

Normalize both equilibrium ratios

For the reversible 298 K toy, use A/S=7.523207 and B/S=25.249038.

Normalization denominator is 1 + 7.523207 + 25.249038 ≈ 33.772245.
S ≈ 0.0296101, A ≈ 0.222763, B ≈ 0.747627.
A/B ≈ 0.297960, whereas the initial formation-flux ratio tends to 7.52321.

The same reversible network prefers A in early formation and B at equilibrium. Time and reverse paths connect those limits.

Keep excess and activation inference separate

For the separate invented 90:10 enantiomer pair, calculate ratio, excess and the conditional barrier difference at 298 K.

Enantiomer ratio = 0.900 / 0.100 = 9.00.
Excess = (0.900−0.100)/(0.900+0.100) = 0.800.
Under the stated irreversible equal-prefactor model, R × 298 × ln 9 ≈ 5444.09 J mol⁻¹.

9:1 and 80.0% ee; the conditional minor-minus-major activation difference is 5.44409 kJ mol⁻¹. No optical sign or equilibrium energy was inferred.

Use three explicit denominators

For initial, desired, alternative and unreacted amounts 0.00500, 0.00300, 0.00100 and 0.00100 mol, calculate the selected metrics.

Conversion = 0.00400 / 0.00500 = 0.800.
Desired selectivity = 0.00300 / 0.00400 = 0.750.
Desired amount yield = 0.00300 / 0.00500 = 0.600.

80.0% conversion, 75.0% desired selectivity and 60.0% desired amount yield in the stated one-to-one closed balance.

Try it yourself

Does a 99:1 stereoisomer ratio from one starting material by itself establish stereospecificity?
Show the answer and reasoning
No.
Stereospecificity concerns the relationship between different starting configurations and their outcomes in the pathway.
What is A/B at t=0 in the models starting with S alone?
Show the answer and reasoning
Undefined.
Both product amounts are zero; an initial flux limit is a different quantity.
Does high temperature alone demonstrate that the products have equilibrated?
Show the answer and reasoning
No.
Interconversion paths, rates, time and conditions must permit equilibration.
Is RT ln 9 inferred from the stipulated irreversible 9:1 route model a product equilibrium Gibbs difference?
Show the answer and reasoning
No.
Under its stated assumptions it is the difference between activation barriers.

What to carry forward

Name the sites or configurations being compared.
Keep stereoselectivity distinct from stereospecificity.
State the assumptions behind a kinetic branch ratio.
Require available equilibration before claiming thermodynamic control.
Keep activation energies separate from product Gibbs levels.
Give conversion, selectivity and yield their own denominators.

References

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