Topic 11 · page 13 of 21

Primary halogenoalkanes: the SN2 evidence

When the rate of hydrolysis of a primary halogenoalkane depends on both the halogenoalkane and hydroxide ions, both are in the rate-determining step. That points to a one-step mechanism, named SN2.

11 min read11.9

Read first: The rate-determining step and the rate equation

Topic 10 gave you the mechanism of a primary halogenoalkane with hydroxide, but not the evidence. The evidence is a rate equation.

The experiment 11.9

Hydrolysis replaces the halogen with an –OH group. For a primary halogenoalkane such as 1-bromobutane with aqueous alkali:

CHX3CHX2CHX2CHX2Br+OHX−→CHX3CHX2CHX2CHX2OH+BrX− \ce{CH3CH2CH2CH2Br + OH- -> CH3CH2CH2CH2OH + Br-}

To find the orders, you carry out initial-rate experiments. In one set the concentration of the halogenoalkane changes and that of hydroxide is constant, and in another set the reverse. The reaction can be followed by taking quenched samples and titrating the hydroxide, or by measuring conductivity.

Worked exampleOrders for 1-bromobutane

(Illustrative data.)

Experiment [C₄H₉Br] / mol dm⁻³ [OH⁻] / mol dm⁻³ Initial rate / 10⁻⁵ mol dm⁻³ s⁻¹
1 0.10 0.10 3.0
2 0.20 0.10 6.0
3 0.10 0.20 6.0
  1. 1 and 2: [C₄H₉Br] ×2, rate ×2: first order in the halogenoalkane.
  2. 1 and 3: [OH⁻] ×2, rate ×2: first order in hydroxide.
  3. rate = k[C₄H₉Br][OH⁻], second order overall.
  4. k = 3.0 × 10⁻⁵ ÷ (0.10 × 0.10) = 3.0 × 10⁻³ dm³ mol⁻¹ s⁻¹.

What the rate equation shows 11.9

Use page 10. Both the halogenoalkane and the hydroxide ion are in the rate equation, so both are in the rate-determining step. The simplest mechanism that does this has one step: the hydroxide ion attacks the carbon at the same moment that the C–Br bond breaks. That one step is the rate-determining step.

The mechanism of 1-bromobutane with hydroxide in one step, with the first three carbons written as C3H7. On the left, a hydroxide ion with a negative charge and lone pairs on oxygen, and 1-bromobutane with delta plus on the carbon bonded to bromine and delta minus on bromine. A curly arrow goes from a lone pair on the oxygen to the carbon, and a second curly arrow goes from the middle of the C-Br bond to bromine. On the right are butan-1-ol and a bromide ion with a negative charge and four lone pairs. A label above the reaction arrow says one step.
SN2: both curly arrows happen in one step, so both species are in the rate-determining step.

The name carries the evidence in its letters. The "2" does not mean two steps. It counts the particles in the rate-determining step: one halogenoalkane and one hydroxide ion. Primary halogenoalkanes hydrolyse this way, as you drew on page 9 of Topic 10.

Check yourself

  1. 1-bromobutane reacts with aqueous hydroxide with rate = k[C₄H₉Br][OH⁻]. State the order with respect to each reagent, the overall order and the units of k.
  2. Use the data below (illustrative) to find the orders and k, with units.
Experiment [RBr] / mol dm⁻³ [OH⁻] / mol dm⁻³ Initial rate / mol dm⁻³ s⁻¹
1 0.050 0.050 1.5 × 10⁻⁵
2 0.10 0.050 3.0 × 10⁻⁵
3 0.050 0.10 3.0 × 10⁻⁵
  1. Explain how this rate equation provides evidence that the hydrolysis of 1-bromobutane is a one-step reaction.
  2. What do the letters S, N and the number 2 stand for in SN2?
  3. [OH⁻] is doubled and [C₄H₉Br] is halved in the hydrolysis of 1-bromobutane. By what factor does the rate change?
  4. Describe the two curly arrows in the mechanism of 1-chloropropane with aqueous hydroxide.

Answers cover these until you have tried the questions

  1. First order in C₄H₉Br, first order in OH⁻, second order overall. Units of k: dm³ mol⁻¹ s⁻¹.
  2. 1 and 2: [RBr] ×2, rate ×2: first order. 1 and 3: [OH⁻] ×2, rate ×2: first order. rate = k[RBr][OH⁻]. k = 1.5 × 10⁻⁵ ÷ (0.050 × 0.050) = 1.5 × 10⁻⁵ ÷ 2.5 × 10⁻³ = 6.0 × 10⁻³ dm³ mol⁻¹ s⁻¹.
  3. The rate depends on the concentration of both the halogenoalkane and the hydroxide ion, so both are in the rate-determining step. A single step in which hydroxide attacks the carbon as the C–Br bond breaks fits this, so the data are evidence for a one-step mechanism.
  4. Substitution, Nucleophilic, and 2 species (one halogenoalkane and one hydroxide ion) in the rate-determining step.
  5. First order in each, so ×2 × ½ = ×1: the rate is unchanged.
  6. Arrow 1 starts at a lone pair on the oxygen of OH⁻ and points to the δ+ carbon bonded to chlorine. Arrow 2 starts in the middle of the C–Cl bond and points to chlorine, which leaves as Cl⁻. The products are propan-1-ol and Cl⁻.