Thermodynamic vs Kinetic Control — With the Numbers Worked Out

By Aniket Bhardwaj · 12 September 2026 · Advanced Chemistry

When one starting material can give two products, two completely different questions decide which one you isolate. Which transition state is lower? decides the answer when the reaction cannot go backwards. Which product is more stable? decides the answer when it can. Most students can recite this. Far fewer can put numbers on it — and putting numbers on it is what turns a two-mark answer into a full-mark one at CSIR-NET, GATE and MSc level. This article gives you both equations, four fully computed examples, the classic laboratory cases, and the Curtin–Hammett principle that catches people out.

The energy landscape, described precisely

Since we cannot draw a diagram here, read the profile as a list of four energies measured from a common starting material S:

Quantity	Pathway to product K (kinetic)	Pathway to product T (thermodynamic)
Free energy of activation, ΔG^‡	Lower — the smaller hill	Higher — the bigger hill
Free energy of the product, ΔG°	Higher — the shallower well	Lower — the deeper well
Formed faster?	Yes	No
More stable?	No	Yes
Barrier for the reverse reaction (product → S)	Small — escapes easily	Large — trapped

Everything else follows from that last row. Product K sits in a shallow well, so if it can go back to S it will; the system then drains, one molecule at a time, into the deeper well of product T. Whether that drainage happens is entirely a matter of whether the reverse barrier is climbable at your temperature and in your time.

Condition	You get the KINETIC product	You get the THERMODYNAMIC product
Temperature	Low (often −78 °C, dry ice/acetone)	High
Time	Short; quench immediately	Long; allow equilibration
Reversibility	Step is effectively irreversible	Step is reversible
Reagent stoichiometry	Strong base in excess, no starting material left to shuttle protons	Catalytic base, or excess substrate, so equilibration can occur
What decides the ratio	ΔΔG^‡ (difference in barrier heights)	ΔΔG° (difference in product stabilities)

The two equations

Kinetic control — the ratio is fixed by the barriers:
[K] / [T] = k_K / k_T = exp( −ΔΔG^‡ / RT ), where ΔΔG^‡ = ΔG^‡_K − ΔG^‡_T

Thermodynamic control — the ratio is the equilibrium constant:
[T] / [K] = K_eq = exp( −ΔG° / RT ) and ΔG° = −RT ln K_eq = ΔH° − TΔS°

Both use R = 8.314 J K⁻¹ mol⁻¹. The pre-exponential factors are assumed similar for two closely related pathways, which is the standard approximation and worth stating in an answer.

Worked example 1 — how much selectivity does a small barrier difference buy?
Two competing pathways differ by ΔΔG^‡ = 8.0 kJ mol⁻¹.

At −78 °C (T = 195 K):
RT = 8.314 × 195 = 1621.2 J mol⁻¹
ΔΔG^‡/RT = 8000 ⁄ 1621.2 = 4.9345
ratio = e^4.9345 = 139 → 139 ⁄ 140 = 99.3% of the kinetic product.

At 25 °C (T = 298 K):
RT = 8.314 × 298 = 2477.6 J mol⁻¹
ΔΔG^‡/RT = 8000 ⁄ 2477.6 = 3.2290
ratio = e^3.2290 = 25.3 → 25.3 ⁄ 26.3 = 96.2%.

The same 8 kJ mol⁻¹ gives 139:1 in a dry-ice bath and only 25:1 at room temperature. This is the whole reason synthetic chemists reach for −78 °C when they want kinetic selectivity: nothing about the molecule changes, only the value of RT in the exponent.

Worked example 2 — what barrier difference do I need for 95:5?
Required ratio = 95 ⁄ 5 = 19, at 298 K.
ΔΔG^‡ = RT ln 19 = 2477.6 × 2.9444 = 7295 J mol⁻¹ = 7.3 kJ mol⁻¹.
Cross-check by the forward route: exp(7295 ⁄ 2477.6) = e^2.9444 = 19.00 ✓, and 19 ⁄ 20 = 95.0% ✓.
Fewer than 8 kJ mol⁻¹ — about the strength of one modest hydrogen bond — is enough to give 95:5 selectivity. That is why apparently tiny steric or electronic differences produce clean synthetic outcomes.

Worked example 3 — the equilibrium ratio under thermodynamic control.
Product T is more stable than product K by ΔΔG° = 12.0 kJ mol⁻¹ at 298 K.
K_eq = exp(12 000 ⁄ 2477.6) = e^4.8434 = 127
→ 127 ⁄ 128 = 99.2% of the thermodynamic product once equilibrium is reached.

Cross-check by the reverse route (this is the check to write out in an exam):
ΔG° = −RT ln K_eq = −2477.6 × ln 127 = −2477.6 × 4.8442 = −12 000 J mol⁻¹ = −12.0 kJ mol⁻¹ ✓
The two routes agree, so the arithmetic is sound.

Worked example 4 — the same reaction, two answers.
Suppose product K has the lower barrier by 8.0 kJ mol⁻¹ while product T is the more stable by 12.0 kJ mol⁻¹.
• Run the reaction at 195 K and quench after minutes: the reverse reaction is far too slow to matter, so you isolate 99.3% K (Example 1).
• Warm the same flask, allow hours, and let both products revert to starting material: you now isolate 99.2% T (Example 3).
Same substrate, same reagents, opposite products. Nothing changed except whether the system was given the chance to go backwards.

Four cases you are expected to know

1. Enolate formation from 2-methylcyclohexanone. The ring has two different α positions. Lithium diisopropylamide (LDA) in THF at −78 °C, with the base in slight excess, removes a proton from the less hindered C6 — a faster deprotonation, because the base is bulky and the proton is more accessible. That is the kinetic enolate, and it is the less substituted one. Using a smaller or weaker base, at higher temperature, with unreacted ketone present so that protons can shuttle between molecules, lets the two enolates equilibrate; the system then settles on the more substituted enolate, because a more substituted C=C is more stable. Trapping either one with an alkyl halide or with trimethylsilyl chloride is the standard way to prove which formed.

2. Sulfonation of naphthalene. Around 80 °C the electrophile attacks the more reactive α (1-) position and you obtain naphthalene-1-sulfonic acid — the kinetic product, reached over the lower barrier. Around 160 °C the sulfonation becomes reversible, and the system moves to naphthalene-2-sulfonic acid, in which the bulky −SO₃H group escapes the peri crowding with the hydrogen at C8. Heat the 1-isomer with sulfuric acid and it isomerises to the 2-isomer, which is the direct experimental proof that the first product is under kinetic and the second under thermodynamic control.

3. Addition of HBr to buta-1,3-diene. The allylic cation formed on protonation can be captured at either end. Capture at the nearer carbon gives the 1,2-adduct (3-bromobut-1-ene) and is faster, because the bromide is already beside the larger positive charge; capture at the far carbon gives the 1,4-adduct (1-bromobut-2-ene), which has the more substituted, more stable internal double bond. Run the reaction cold and the 1,2-product dominates; warm the mixture and it converts, through the same allylic cation, into the 1,4-product. Reported product ratios vary between textbooks with solvent, temperature and concentration, so quote the direction of the change confidently and treat any specific percentages as the conditions of that particular experiment.

4. endo vs exo Diels–Alder adducts. Cyclopentadiene and maleic anhydride give the endo adduct on mixing at low temperature — the kinetic product, favoured by secondary orbital interactions in the transition state. The exo adduct is usually the more stable, being less sterically congested, and reactions that can reverse (furan with a dienophile is the standard case, because the retro-Diels–Alder is easy) drift to exo on heating.

The Curtin–Hammett principle — the trap

A very common exam scenario: a substrate exists as two rapidly interconverting conformers (or two rapidly equilibrating intermediates), A and B, which react to give different products. The instinctive answer — "the major product comes from the major conformer" — is wrong.

If A ⇌ B is much faster than either reaction, then
[product from A] / [product from B] = exp( −ΔΔG^‡ / RT )
where ΔΔG^‡ is the difference between the two transition state energies measured from a common reference — the populations of A and B do not appear.

So a conformer present at only 1% can be responsible for essentially all of the product, if its transition state is the lower one. The classic illustration is the reaction of a substituted cyclohexane through the minor axial conformer. Whenever a question says "rapidly interconverting", check whether Curtin–Hammett applies before you reach for populations.

A related rule worth pairing with this

The Hammond postulate tells you what the transition state looks like: for a strongly exothermic step the transition state resembles the reactants (early), and for an endothermic step it resembles the products (late). That is what lets you argue that a more stable intermediate is also reached over a lower barrier — an argument you may use only for endothermic, product-like transition states. Using Hammond to justify a kinetic preference in an early, reactant-like transition state is a standard error.

Mistakes that cost marks

Assuming the more stable product always forms. It forms only if the reaction can reverse. If the first step is irreversible, product stability is simply irrelevant.
Assuming the kinetic product is by definition less stable. The definition is about barriers. The two often differ, which is why the question is interesting — but if the same product had both the lower barrier and the lower energy there would be no competition at all.
Using ΔH instead of ΔG. Entropy frequently decides these competitions, especially when one route forms a ring or expels a small molecule. Both equations above take free energies.
Forgetting to convert kJ to J. ΔΔG^‡/RT with R in J K⁻¹ mol⁻¹ and ΔΔG^‡ in kJ mol⁻¹ gives a number a thousand times too small, and an answer of "1.003:1". If your ratio comes out very close to 1, check the units first.
Getting the sign wrong. Write the ratio the way round you want it and then decide the sign, rather than memorising it. A useful check: the lower barrier must produce the larger number.
Quoting the Curtin–Hammett result as conformer populations. Under fast pre-equilibrium the populations cancel out. Only the transition-state energies survive.
Saying "low temperature favours the kinetic product" without saying why. Two separate effects operate: the exponential in ΔΔG^‡/RT becomes larger, and the reverse reaction that would allow equilibration becomes too slow. Both are worth a sentence.
Claiming a fixed percentage for a textbook case. Reported ratios for HBr addition to butadiene differ between books because the conditions differ. Give the direction and the reason; attach numbers only to the conditions they were measured under.

Where this appears in the exam

Exam	Typical demand
CSIR-NET Chemical Sciences	Identify which product forms under stated conditions; calculate a product ratio from ΔΔG^‡; apply Curtin–Hammett; explain enolate regiochemistry
GATE Chemistry	Numerical: ratio from a barrier difference or ΔG° from an equilibrium constant; sketch and label the two-pathway energy profile
IIT-JAM / CUET-PG	1,2 vs 1,4 addition to conjugated dienes; naphthalene sulfonation; effect of temperature on product distribution
MSc coursework	Hammond postulate reasoning; kinetic vs thermodynamic enolates in synthesis planning; reversible Diels–Alder chemistry

These questions are won on the exponential. Every ratio in this article came from ΔG° = −RT ln K or from exp(−ΔΔG^‡/RT), and the marks go to whoever gets the units and the sign right. The Gibbs Free Energy calculator does ΔG = ΔH − TΔS and the ΔG°/K conversion in one place, so you can check a barrier difference against a product ratio in seconds.

Open the Gibbs Free Energy Calculator →

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