Rate-determining steps and mechanisms

The rate-determining step (spec 11.7)

The rate-determining step (RDS) is the slowest step in a mechanism. Species that appear in the rate equation take part in the RDS, or in a fast step before it. The order with respect to each species equals the number of particles of it involved up to and including the RDS. Species that react only after the RDS are zero order.

Proposing a mechanism (spec 11.8)

A valid mechanism must:

1. add up to the overall balanced equation (intermediates cancel)

2. have an RDS containing exactly the species and numbers shown by the rate equation

An intermediate is formed in one step and used up in a later one; a catalyst is used in one step and regenerated later.

Evidence for SN1 and SN2 (spec 11.9)

Hydrolysis of halogenoalkanes by OH-:

• Primary, such as 1-bromobutane: rate = k[RBr][OH-]. Both species are in the one step, so the mechanism is SN2 (substitution, nucleophilic, 2 species in the RDS). OH- attacks as Br- leaves.
• Tertiary, such as 2-bromo-2-methylpropane: rate = k[RBr]. The slow step is the C-Br bond breaking to give a carbocation, (CH3)3C+, which then reacts quickly with OH-. This is SN1 (1 species in the RDS).

Tertiary carbocations are stabilised by the three electron-releasing alkyl groups; three bulky groups also block attack by OH- from behind, so SN2 is hindered. Secondary halogenoalkanes can react by both routes.

Worked example

2NO2(g) + F2(g) -> 2NO2F(g) has rate = k[NO2][F2]. Suggest a two-step mechanism.

1. The RDS must contain one NO2 and one F2.
2. Step 1 (slow): NO2 + F2 -> NO2F + F
3. Step 2 (fast): NO2 + F -> NO2F
4. The F atom is an intermediate; the steps add up to the overall equation.

Answer: Slow NO2 + F2 -> NO2F + F, then fast NO2 + F -> NO2F

Check yourself

1. What is the rate equation for the hydrolysis of 2-bromo-2-methylpropane by hydroxide ions?
   • rate = k[(CH3)3CBr][OH-]
   • rate = k[OH-]
   • rate = k[(CH3)3CBr]
   • rate = k[(CH3)3CBr]²
   Show answer

   rate = k[(CH3)3CBr]. Tertiary halogenoalkanes react by SN1; only the halogenoalkane is in the slow step.

2. The hydrolysis of 1-bromobutane by OH- has rate = k[C4H9Br][OH-]. Which mechanism does this support?
   • SN2
   • SN1
   • Electrophilic addition
   • Free-radical substitution
   Show answer

   SN2. Two species in the rate-determining step means SN2.

3. A mechanism is: step 1 (slow) A + B -> X; step 2 (fast) X + B -> C. What is the rate equation?
   • rate = k[A][B]²
   • rate = k[X][B]
   • rate = k[A][B]
   • rate = k[A]
   Show answer

   rate = k[A][B]. Only the species in the slow step appear: one A and one B.

4. In the mechanism step 1 NO2 + F2 -> NO2F + F, step 2 NO2 + F -> NO2F, which species is an intermediate?
   • NO2
   • F
   • F2
   • NO2F
   Show answer

   F. F is made in step 1 and used in step 2, so it is not in the overall equation.

5. H2 + 2ICl -> I2 + 2HCl has rate = k[H2][ICl]. Which slow step fits?
   • 2ICl -> I2 + Cl2
   • H2 + 2ICl -> I2 + 2HCl
   • HI + ICl -> I2 + HCl
   • H2 + ICl -> HI + HCl
   Show answer

   H2 + ICl -> HI + HCl. The slow step must contain one H2 and one ICl.

6. What is the order with respect to OH- in the SN1 hydrolysis of a tertiary halogenoalkane?
   Show answer

   0. OH- reacts with the carbocation only after the slow step, so it is zero order.

7. Why do tertiary halogenoalkanes react by SN1 rather than SN2?
   • Tertiary halogenoalkanes have no C-X bond polarity
   • OH- cannot act as a nucleophile
   • Tertiary halogenoalkanes are gases
   • The tertiary carbocation is stabilised by alkyl groups, and bulky groups hinder attack by the nucleophile
   Show answer

   The tertiary carbocation is stabilised by alkyl groups, and bulky groups hinder attack by the nucleophile. Electron-releasing alkyl groups stabilise the carbocation, and the crowded carbon blocks backside attack.

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Sources

• Pearson Edexcel IAL Chemistry (2018) specification
• Pearson Edexcel IAL Chemistry (2018) scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.