Part 3: What rate equations reveal about mechanisms

Test proposed steps against both the overall equation and experimental rate law, including acid-catalysed iodination and SN1/SN2 hydrolysis.

A mechanism must satisfy two tests

A mechanism is a sequence of elementary steps whose sum gives the overall chemical equation. An intermediate is formed in one step and consumed later, so it cancels from the overall equation. A catalyst is used and then regenerated, so it also cancels overall but can appear in the rate law.

The rate-determining step is the step controlling the overall rate in the proposed mechanism, often the slowest step. For a simple elementary step, its rate depends on the species that collide in that step. Thus a proposed slow A + B → intermediate is consistent with rate proportional to [A][B].

If a slow step uses an intermediate made in a preceding fast equilibrium, relate the intermediate concentration to that equilibrium before comparing with the measured rate law. It is not generally valid to insert arbitrary earlier reactants into the rate equation without this reasoning.

A matching rate law supports a mechanism; it does not prove it is unique. Different mechanisms can lead to the same overall equation and rate dependence. Orders are measured facts, while elementary stages are a model that must also conserve atoms and charge.

Worked example: a zero-order reactant can react later

Consider the overall reaction 2NO₂ + F₂ → 2NO₂F, with an illustrative measured law rate = k[NO₂][F₂]. A plausible two-step model is NO₂ + F₂ → NO₂F + F• (slow), followed by F• + NO₂ → NO₂F (fast). Adding the steps cancels the fluorine atom intermediate and gives the overall equation.

The slow first step contains one NO₂ and one F₂, consistent with first order in each despite the coefficient two for NO₂ overall. A one-step model requiring two NO₂ molecules and one F₂ molecule would imply a different concentration dependence and is not supported by that law.

For a hypothetical overall A + B → P with rate = k[A], the model A → X (slow) followed by X + B → P (fast) is consistent within its working concentration range. B is consumed but changing it does not affect the controlling step. This does not imply reaction can continue indefinitely with no B present.

Iodination: the slow process happens before iodine reacts

The experimental law rate = k[CH₃COCH₃][H⁺] is first order in propanone and acid and zero order in iodine under the usual conditions. It suggests that the rate-controlling formation of a reactive form of propanone depends on propanone and acid, while iodine reacts in a later rapid stage.

One chemically plausible model begins with fast reversible protonation of the carbonyl oxygen. The protonated species then loses an adjacent proton slowly, producing an enol. The enol reacts rapidly with iodine and returns to the carbonyl form. The simplified equations show formal proton bookkeeping; actual proton transfers are mediated by solvent.

For the first equilibrium, the amount of protonated propanone is proportional to [propanone][H⁺] when the relevant equilibrium conditions apply. If its conversion to enol is slow, the resulting rate law is proportional to those two concentrations. Thus both species can appear in the measured law even though the slow elementary event is not a direct collision of neutral propanone with H⁺.

Iodine being zero order indicates its involvement after the controlling stage in this model, not that iodine is a catalyst. The overall reaction consumes I₂ and produces H⁺ and I⁻. The acid is still a catalyst for the enol-forming route; in CP13a its large initial excess makes the extra H⁺ formed comparatively unimportant.

Hydrolysis rate laws distinguish plausible substitution routes

For a primary halogenoalkane reacting with hydroxide, rate = k[RCH₂X][OH⁻] is consistent with an SN2 route. The nucleophile approaches the δ⁺ carbon and C–O bond formation occurs as the C–X bond breaks in one concerted step. Both reacting species affect that step, explaining second order overall. Primary carbon is relatively accessible and a free primary carbocation would be poorly stabilised.

For a tertiary halogenoalkane, rate = k[R₃CX] is consistent with SN1. Slow heterolytic cleavage forms R₃C⁺ and X⁻; rapid nucleophilic attack then forms the alcohol, directly with OH⁻ or through protonated alcohol with water followed by deprotonation. Only the halogenoalkane controls the first-step rate in this model.

Alkyl groups stabilise a tertiary carbocation through electron donation, while the crowded carbon makes direct backside SN2 attack difficult. This connects kinetic evidence to structure. A primary SN2 and tertiary SN1 classification is a useful pattern under suitable conditions, not a claim that solvent, nucleophile, temperature or elimination pathways never matter.

A constant solvent concentration can be hidden in an observed rate constant. When interpreting an unfamiliar data set, ask which concentrations were actually varied before concluding that a species has no mechanistic role. A rate law alone gives evidence for a route, not a photograph of a transition state.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources