Rate-determining step and reaction mechanismsEdexcel International A Level Chemistry: Revision notes

A reaction mechanism shows the steps by which reactants become products. The rate-determining step is the slowest step, and it controls the overall rate, like the narrowest part of a pipe. Steps after it are fast and do not affect the rate.

An intermediate is a species formed in one step and used up in a later step. It does not appear in the overall equation. The steps of a mechanism must add up to the overall equation once intermediates are cancelled.

The rate equation shows which species (and how many of each) are in the rate-determining step, or in fast steps before it.

• Each reactant that appears in the rate equation with order 1 is present once in the slow step. Order 2 means two of that species.
• A reactant with zero order is not in the rate-determining step. It reacts in a fast step after it.

Worked example. For NO₂ + CO → NO + CO₂ with rate = k[NO₂]², the slow step is NO₂ + NO₂ → NO₃ + NO and the fast step is NO₃ + CO → NO₂ + CO₂. CO does not appear in the rate equation because it reacts after the slow step.

To predict the rate equation of a proposed mechanism, write the rate equation for the slow step using its reactants.

A proposed mechanism is consistent with the data only if:

• the steps add up to the overall equation, and
• the rate-determining step gives the experimental rate equation.

If the predicted rate equation does not match the experimental one, the mechanism is rejected. A consistent mechanism is possible, not proved, because other mechanisms might also fit.

CH₃COCH₃ + I₂ → CH₃COCH₂I + HI, catalysed by H⁺. Initial-rate experiments show the reaction is first order with respect to propanone, first order with respect to H⁺ and zero order with respect to iodine, so rate = k[CH₃COCH₃][H⁺].

The rate-determining step therefore involves propanone and H⁺ but not iodine. A possible mechanism is:

• Fast: propanone is protonated on the carbonyl oxygen.
• Slow: the protonated propanone loses H⁺ from a CH₃ group to form the enol, CH₂=C(OH)CH₃.
• Fast: the enol reacts with iodine to form CH₃COCH₂I and HI, regenerating H⁺.

Iodine reacts only after the slow step, which explains its zero order.

The rate equation tells us which mechanism operates.

Tertiary halogenoalkane, rate = k[RBr] (first order in the halogenoalkane, zero order in OH⁻): SN1. Slow step: the C–Br bond breaks heterolytically to form a carbocation and Br⁻. Fast step: OH⁻ attacks the carbocation. Only the halogenoalkane is in the rate-determining step.

Primary halogenoalkane, rate = k[RBr][OH⁻] (first order in each): SN2. A single step in which the nucleophile OH⁻ attacks the δ+ carbon as the C–Br bond breaks, through a transition state with both species. There is no intermediate.

Primary carbocations are unstable, so primary halogenoalkanes use SN2. Tertiary carbocations are more stable, so tertiary halogenoalkanes use SN1.