Competing Reaction Channels

Introduction

A single reactant can reach several products through different saddle points. The lowest product energy does not automatically identify the fastest formation route, and the lowest isolated saddle does not always control the measured mixture. Product branching depends on barriers, populations, reversibility and sometimes trajectories after a shared saddle. A reaction profile becomes predictive only when its alternative channels are placed on a common kinetic and thermodynamic footing.

Core explanation

Suppose reactant R forms P1 with first-order rate k1 and P2 with rate k2 through independent elementary channels. If products do not interconvert or react further, the instantaneous fluxes are k1[R] and k2[R]. The kinetic branching fractions are f1 = k1/(k1+k2) and f2 = k2/(k1+k2). Under conventional transition-state theory, each rate depends strongly on its activation Gibbs energy, so a small difference in barriers can yield a sizable selectivity. Yet each barrier must use the same reactant reference, temperature and standard state.

For a flexible reactant, each channel may have several conformer-specific saddles. Then rates to P1 and P2 are sums of population-weighted contributions, not single barriers chosen by visual inspection. For a gas-phase activated intermediate, branching can depend on internal energy and collisional energy transfer; a single thermal rate expression may not suffice. For a catalytic cycle, product flux can depend on the entire network and catalyst resting-state populations. The two-channel formula is a clear starting model, not a universal law.

Kinetic control of product composition means the ratio is governed by relative rates of parallel forward reactions under the relevant conditions. Thermodynamic control means product composition reflects equilibrium constants because products or precursors can interconvert on the experimental timescale. A less stable product can dominate early or under low-temperature conditions if it forms faster and cannot equilibrate. On warming or waiting, reversible chemistry may shift composition toward a more stable product. Temperature alone does not guarantee a switch; reversibility and actual rates are the decisive factors.

Energy diagrams should therefore show both the forward activation free energies and product free energies. From one common R, P1 can have a lower barrier but higher final G, while P2 has a higher barrier and lower final G. Calling P1 “the kinetic product” and P2 “the thermodynamic product” is justified only if their paths and interconversion behavior match the experimental conditions. If P1 and P2 do not equilibrate, their relative final energies do not determine their observed ratio.

There is another important exception: a post-transition-state bifurcation can send trajectories from one saddle into multiple products without a separate conventional saddle for each branch. In such a case, comparing two transition-state free energies cannot predict branching. Dynamical simulations or other evidence of how trajectories leave the shared saddle are needed. Similarly, spin crossings or solvent motions can create channels absent from a single static surface.

An experimentally observed selectivity may also be altered by detection or downstream chemistry. If P1 decomposes faster than P2, a late measurement need not reflect initial branching. Mechanistic interpretation should compare time-resolved yields, isotope effects or spectroscopic intermediates where available, not merely an isolated computational product-energy table.

Step-by-step reasoning

Identify all plausible elementary routes and verify each saddle's endpoints. Place every reactant, saddle and product on a common corrected free-energy scale. Determine whether starting conformers equilibrate and whether products interconvert. For simple irreversible parallel routes, estimate rate constants and calculate f1 and f2. If the network is reversible or sequential, write its kinetic equations. Check whether one saddle bifurcates before assigning separate barrier-controlled fractions.

Visual explanation

Draw R at left with two arrows rising to separate transition states and descending to P1 and P2. Make P1's peak lower but its final valley higher, and P2's peak higher but final valley deeper. Add a second diagram in which both products descend from one shared saddle to illustrate why the two-barrier formula fails there. Annotate early flux versus long-time equilibrium composition.

Real-world analogy

Two doors can lead from a crowded room to different destinations. The easier door may receive more traffic immediately, even if the place beyond the other door is more attractive. If people can later move freely between destinations, the final distribution can change. Molecules differ because “ease” is a free-energy and dynamical rate problem, not simply door width.

Real-world example

An allylic intermediate may be trapped at two positions to form regioisomeric products. A lower barrier toward one product can dominate early trapping, while a lower-energy alternative may accumulate if reversible formation permits equilibration. The actual outcome depends on solvent, temperature, reaction time and subsequent transformations. Computational profiles should therefore be tied to measured rates and product ratios rather than interpreted as universal labels.

Why?

Why compare activation Gibbs energies from one reference? The rate ratio depends on access from the same reactant pool. Why can a higher-energy product appear first? Product stability concerns the final state, while formation rate concerns the path to it. Why check for reversibility? Equilibrium control requires interconversion. Why mention bifurcation? One saddle can sometimes feed more than one product, defeating a simple count of saddles.

Common misconception

The most stable product is not necessarily the major product in a finite-time reaction. Also, “lower barrier” is incomplete unless both barriers include comparable corrections and the same reference state. A final misconception is that every distinct product requires its own separate transition state; dynamical branching can violate that picture.

Worked example

Question: Parallel irreversible routes from R have k1 = 3.0 s⁻¹ and k2 = 1.0 s⁻¹. P2 is 15 kJ mol⁻¹ more stable than P1. What are the initial branching fractions, and does product stability change them in this model?

Reasoning: The total disappearance rate is (k1+k2)[R] = 4[R] s⁻¹. Channel fractions are 3/4 and 1/4. The products are assumed not to interconvert or reverse, so their relative thermodynamic stability does not enter the initial flux equations. If the assumption changed, the longer-time composition would require reversible kinetics.

Answer: Initial fractions are 0.75 to P1 and 0.25 to P2; P2's greater stability does not alter this irreversible-model branching.

Quick check

1. What additional condition is needed before product energies can determine an equilibrium product ratio? Answer: Products or their relevant precursors must interconvert sufficiently to reach equilibrium under the stated conditions.

Exam focus

Write branching fractions from rates only for simple parallel irreversible channels. Distinguish formation flux from equilibrium composition and check common reference states. Explain that multiple conformers, reversible networks and post-saddle bifurcations require extensions beyond a two-barrier sketch.

Advanced insight

For two TST channels with similar prefactors and a common reactant state, the selectivity often depends approximately on exp[−(ΔG1‡−ΔG2‡)/RT]. A calculation error of only a few kilojoules per mole can therefore change a predicted product ratio substantially. At a bifurcation, phase-space momentum and transverse motion affect outcomes even if a minimum-energy path points toward one valley. Robust predictions should test methodological uncertainty and compare observed branching over temperature, pressure and time.

Summary

Competing channels divide reactant flux among products. Simple irreversible branching uses relative rate constants, whereas thermodynamic control requires product or precursor equilibration. Conformer populations and kinetic networks complicate the comparison. A common-saddle bifurcation can make static barrier ranking inadequate. Always connect each proposed route, state the kinetic assumptions and compare predictions with actual product data.

Practice questions

1. If two irreversible channels have rates kA and kB, what is the fraction forming B? Answer: kB/(kA+kB), assuming the same reactant pool and no downstream conversion.

2. Why is a product's low Gibbs energy insufficient to predict its early yield? Answer: Early yield is set by formation rates and accessibility, not only final-state stability.

3. What changes if P1 and P2 interconvert rapidly? Answer: Their long-time ratio can approach an equilibrium ratio determined by relative free energies.

4. Why may one saddle fail to define one product channel? Answer: Trajectories can bifurcate after the saddle and reach different products without separate upstream saddles.

Sources: IUPAC Gold Book, kinetic control; IUPAC Gold Book, thermodynamic control; Journal of Physical Chemistry A, selectivity on bifurcating surfaces.