One tertiary bromide, two products: substitution versus elimination
A student has 10.0 g of 2-bromo-2-methylpropane (Mr = 137.0; bp 73 °C) and must make two compounds: 2-methylpropan-2-ol (Mr = 74.0; bp 82 °C) for an esterification, and 2-methylpropene (bp −6.9 °C) for a polymer study. The laboratory supplies 2.00 mol dm⁻³ NaOH(aq), 2.00 mol dm⁻³ NaOH in ethanol, ethanol, a fractionating column, and a cold trap. In the first attempt the student warms the bromide with aqueous NaOH at 60 °C for 30 minutes and obtains a liquid that decolourises bromine water. A second student repeats the method with 1-bromobutane and finds the reaction much slower, even though 1-bromobutane is a primary bromide. The student eventually isolates 3.1 g of 2-methylpropan-2-ol.
1. Explain, with a mechanism, why the tertiary bromide reacts readily with aqueous NaOH to give the alcohol, and identify the rate-determining step.
Read the worked answer
Mechanism (SN1) Step 1 (slow, rate-determining): the C–Br bond ionises heterolytically, Br⁻ leaves and a planar tertiary carbocation forms: (CH3)3C–Br → (CH3)3C⁺ + Br⁻ Step 2 (fast): water or OH⁻ attacks the vacant p orbital of the carbocation: (CH3)3C⁺ + OH⁻ → (CH3)3C–OH
Why this is fast and why it is SN1 The tertiary carbocation is stabilised by three electron-donating alkyl groups and by the resulting planar geometry; the C–Br bond is also weakened by the bulk of the three methyls. Consequently ionisation is easy, and the rate depends only on [(CH3)3CBr], not on [OH⁻] - first-order kinetics. The rate equation is rate = k[(CH3)3CBr], and the rate-determining step is the ionisation.
Stereochemistry A planar carbocation could be attacked from either face, but here both faces give the same product because the carbocation has two identical methyl groups and the product is achiral; racemisation would only matter for an unsymmetrical chiral substrate.
Role of the aqueous solvent Water stabilises the developing ions and solvates Br⁻, lowering the activation energy for ionisation. The nucleophile can be either water (giving the alcohol after deprotonation) or OH⁻; both lead to the same product, so a high base concentration mainly affects the competing elimination.
2. Explain why the product decolourises bromine water, and explain how you would change the conditions to make the alkene the major product instead.
Read the worked answer
Why bromine water is decolourised The alkene is 2-methylpropene, (CH3)2C=CH2. Its C=C π electrons attack Br2 to give a dibromo compound, so the orange-brown bromine colour is discharged: (CH3)2C=CH2 + Br2 → (CH3)2C(Br)–CH2Br This is a valid test for the alkene; the saturated alcohol does not react under these conditions.
How the alkene is formed OH⁻ acts as a base and removes a β-hydrogen from the carbocation intermediate (E1), or the base removes H⁺ in a concerted E2 step with Br⁻ departing: (CH3)3C–Br + OH⁻ → (CH3)2C=CH2 + H2O + Br⁻
Conditions to make the alkene major
- Use NaOH in ethanol rather than water. In ethanol, OH⁻ is less strongly solvated and behaves as a stronger base, and the less polar solvent favours the less polar transition state leading to the alkene.
- Heat more strongly (reflux at the boiling point of the mixture). Elimination has a larger activation energy than substitution, so raising the temperature increases the elimination rate proportionally more - a classic case of kinetic versus thermodynamic control.
- Keep the base concentrated and avoid excess water, which would favour solvolysis to the alcohol.
The alkene (bp −6.9 °C) can then be collected in a cold trap; the alcohol (bp 82 °C) remains in the flask.
3. Calculate the percentage yield of 2-methylpropan-2-ol and account for the shortfall, suggesting how the student could obtain purer products.
Read the worked answer
Theoretical yield n(C4H9Br) = 10.0 g ÷ 137.0 g mol⁻¹ = 0.0730 mol 1:1 stoichiometry, so maximum n(alcohol) = 0.0730 mol m(theoretical) = 0.0730 × 74.0 = 5.40 g Percentage yield = (3.1 ÷ 5.40) × 100 = 57%
Where the rest went
- Competing elimination: the alkene is a by-product, and the bromine water test shows it is present. The two pathways share the same carbocation, so the alcohol can never be obtained in 100% yield under these conditions.
- Incomplete reaction: 30 minutes at 60 °C may not be enough for full conversion, especially as the bromide is volatile (bp 73 °C) and can escape from an open flask.
- Work-up losses: the alcohol is water-miscible and volatile (bp 82 °C), so it is lost during washing, drying and distillation; some is also lost as the azeotrope with water.
Improving purity and yield Use a reflux condenser to prevent loss of the volatile bromide and alcohol, use a moderate excess of NaOH(aq) with efficient stirring, and keep the temperature low enough to suppress elimination. After reaction, separate the organic layer, wash with water (and NaHCO3 if acid is present), dry over a drying agent, and distil the alcohol using a fractionating column. For the alkene, use the ethanolic NaOH conditions and collect the product in a cold trap.