Topic 11 · page 12 of 21

Deducing a mechanism

Write the slow step from the rate equation, then add fast steps until the steps add up to the overall equation. Check both conditions, and remember that the rate equation can rule a mechanism out but cannot prove it.

14 min read11.811.6(iii)

Read first: The rate-determining step and the rate equation, Iodination of propanone: data and predictions

A method that works 11.8

You are given the overall (stoichiometric) equation and the rate equation. You are asked for a mechanism that fits both. The rate equation fixes the slow step and the overall equation fixes the total.

MethodDeducing a mechanism

  1. From the rate equation, list the species in the rate-determining step and how many of each (the order).
  2. Write the slow step with those species as the reactants. Choose products, including any intermediate, so that atoms and charges balance.
  3. Add fast steps that use the intermediates and the remaining reactants to make the final products.
  4. Check 1: add the steps and cancel the intermediates. The result must be the overall equation.
  5. Check 2: the slow step must contain exactly the species in the rate equation.

Worked exampleNitrogen dioxide and fluorine

The reaction 2 NOX2+FX2→2 NOX2F\ce{2NO2 + F2 -> 2NO2F} has rate = k[NO₂][F₂]. Suggest a mechanism.

  1. The slow step contains one NO₂ and one F₂.
  2. Step 1 (slow): NOX2+FX2→NOX2F+F\ce{NO2 + F2 -> NO2F + F}. The product F is an intermediate.
  3. The second NO₂ in the overall equation has not yet reacted, so step 2 (fast): NOX2+F→NOX2F\ce{NO2 + F -> NO2F}.
  4. Check 1: add the steps: NO₂ + F₂ + NO₂ + F → NO₂F + F + NO₂F. Cancel F: 2NO₂ + F₂ → 2NO₂F.
  5. Check 2: the slow step has one NO₂ and one F₂. ✓

Worked exampleHydrogen and iodine monochloride

The reaction HX2+2 ICl→IX2+2 HCl\ce{H2 + 2ICl -> I2 + 2HCl} has rate = k[H₂][ICl]. Suggest a mechanism.

  1. The slow step contains one H₂ and one ICl.
  2. Step 1 (slow): HX2+ICl→HI+HCl\ce{H2 + ICl -> HI + HCl}.
  3. One ICl is still unused, and HI is an intermediate. Step 2 (fast): HI+ICl→IX2+HCl\ce{HI + ICl -> I2 + HCl}.
  4. Check 1: H₂ + ICl + HI + ICl → HI + HCl + I₂ + HCl. Cancel HI: H₂ + 2ICl → I₂ + 2HCl.
  5. Check 2: the slow step has one H₂ and one ICl. ✓

If the stoichiometric equation has a species that is not in the rate equation, it reacts in a fast step after the slow one. Here the second ICl is such a species.

The iodination of propanone 11.6(iii)11.8

For the iodination of propanone (page 11), rate = k[CH₃COCH₃][H⁺]. Iodine is zero order. The mechanism is deduced in the same way.

Worked exampleA possible mechanism for iodination

CHX3COCHX3+IX2→CHX3COCHX2I+HI \ce{CH3COCH3 + I2 -> CH3COCH2I + HI}
  1. The slow step contains one propanone and one H⁺, and no iodine.
  2. Step 1 (slow): CHX3COCHX3+HX+→[CHX3C(OH)CHX3]X+\ce{CH3COCH3 + H+ -> [CH3C(OH)CH3]+}. This is protonated propanone, an intermediate.
  3. Step 2 (fast): [CHX3C(OH)CHX3]X+→CHX2=C(OH)CHX3+HX+\ce{[CH3C(OH)CH3]+ -> CH2=C(OH)CH3 + H+}. This forms an enol, a second intermediate, and the H⁺ is given back.
  4. Step 3 (fast): CHX2=C(OH)CHX3+IX2→CHX3COCHX2I+HI\ce{CH2=C(OH)CH3 + I2 -> CH3COCH2I + HI}. Iodine reacts only here, after the slow step.
  5. Check 1: add the steps. The intermediates cancel and so does H⁺, leaving CH₃COCH₃ + I₂ → CH₃COCH₂I + HI. ✓
  6. Check 2: the slow step has one propanone and one H⁺, as in the rate equation. ✓

Three features answer the specification points. H⁺ is a catalyst: it is used in step 1 and made again in step 2, so it is not in the overall equation. It appears in the rate equation because it is in the slow step. Iodine is zero order because it enters in step 3. The mechanism is only a possible mechanism: it fits the data, and that is what is asked for.

Check yourself

  1. The decomposition 2NO₂Cl → 2NO₂ + Cl₂ has rate = k[NO₂Cl]. Suggest a two-step mechanism, with a slow step and a fast step.
  2. A reaction 2A + B → C has rate = k[A]². Suggest a two-step mechanism using an intermediate X.
  3. A student suggests that the slow step in the iodination of propanone is CH₃COCH₃ + I₂ → CH₃COCH₂I + HI. Explain why this does not fit the rate equation.
  4. A mechanism is: step 1 (slow) A + B → C + D; step 2 (fast) C + B → E. Write the overall equation and the rate equation.
  5. Why is a mechanism described as "possible" rather than "proved" when it has been deduced from a rate equation?
  6. Write the three steps of the mechanism for the iodination of propanone, name the two intermediates and say which species is the catalyst.

Answers cover these until you have tried the questions

  1. Step 1 (slow): NO₂Cl → NO₂ + Cl. Step 2 (fast): NO₂Cl + Cl → NO₂ + Cl₂. Check: NO₂Cl + NO₂Cl + Cl → NO₂ + Cl + NO₂ + Cl₂. Cancel Cl: 2NO₂Cl → 2NO₂ + Cl₂ ✓. The slow step contains one NO₂Cl, as in the rate equation ✓.
  2. The slow step contains two A. Step 1 (slow): A + A → X. Step 2 (fast): X + B → C. Check: 2A + X + B → X + C, so 2A + B → C ✓.
  3. That step contains iodine and no H⁺, so it would give rate = k[CH₃COCH₃][I₂]. The measured rate equation has no iodine (zero order) and does have H⁺, so the rate-determining step cannot contain I₂ and must contain H⁺.
  4. Add the steps: A + B + C + B → C + D + E. Cancel C: A + 2B → D + E. The slow step is step 1, which has one A and one B, so rate = k[A][B].
  5. Several different mechanisms can give the same rate equation. The rate equation shows which species are in the slow step but not what else happens, so the data can rule a mechanism out but cannot prove that one is correct.
  6. Step 1 (slow): CH₃COCH₃ + H⁺ → [CH₃C(OH)CH₃]⁺. Step 2 (fast): [CH₃C(OH)CH₃]⁺ → CH₂=C(OH)CH₃ + H⁺. Step 3 (fast): CH₂=C(OH)CH₃ + I₂ → CH₃COCH₂I + HI. The intermediates are [CH₃C(OH)CH₃]⁺ and CH₂=C(OH)CH₃. The catalyst is H⁺.