AS Chemistry 9701: Substitution vs Elimination of Bromoethane

Bromoethane, $\text{CH}_3\text{CH}_2\text{Br}$ , can be converted into two different organic products depending only on the solvent used with sodium hydroxide.

(a) State the reagent and conditions needed to convert bromoethane into ethanol, and give the balanced equation for this reaction. [2]

(b) State the reagent and conditions needed to convert bromoethane into ethene, and give the balanced equation for this reaction. [2]

(c) Explain, in terms of the role played by the hydroxide ion, why the same ion, $\text{OH}^-$ , can bring about two chemically different types of reaction depending on the solvent. [3]

(d) Describe, in words, the elimination mechanism that converts bromoethane into ethene. [2]

Part (a): Substitution to ethanol

Reagent and conditions: aqueous sodium hydroxide, $\text{NaOH(aq)}$ , heated under reflux.

$\text{CH}_3\text{CH}_2\text{Br} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{NaBr}$

Part (b): Elimination to ethene

Reagent and conditions: sodium hydroxide dissolved in ethanol, $\text{NaOH}$ in $\text{C}_2\text{H}_5\text{OH}$ , heated under reflux.

$\text{CH}_3\text{CH}_2\text{Br} + \text{NaOH} \rightarrow \text{CH}_2=\text{CH}_2 + \text{NaBr} + \text{H}_2\text{O}$

Part (c): Two roles for the same ion

The hydroxide ion has two distinct chemical properties: it is both a good nucleophile (its lone pair can attack an electron-deficient carbon) and a strong base (it can remove a proton, $\text{H}^+$ ).

In water, the polar solvent stabilises the ionic transition state of nucleophilic attack well, and $\text{OH}^-$ predominantly behaves as a nucleophile, attacking the carbon bonded to bromine directly and displacing $\text{Br}^-$ . This is substitution, giving the alcohol.
In ethanol, a less polar solvent, $\text{OH}^-$ is instead more likely to behave as a base, removing a hydrogen ion from a carbon adjacent to the C–Br carbon rather than attacking the C–Br carbon itself. This is elimination, giving the alkene.

So it is the change in solvent (aqueous vs ethanolic), not any change in the identity of the ion itself, that determines whether $\text{OH}^-$ acts as a nucleophile or as a base.

Part (d): The elimination mechanism

$\text{OH}^-$ approaches a hydrogen atom on the carbon adjacent to the carbon bearing the bromine (i.e. the $-\text{CH}_3$ carbon in bromoethane) and removes it as $\text{H}^+$ , forming $\text{H}_2\text{O}$ . As this C–H bond breaks, the pair of electrons that formed it shifts to become a new pi ( $\pi$ ) bond between the two carbon atoms. At the same time, the C–Br bond breaks heterolytically and $\text{Br}^-$ leaves. All of this happens in essentially one step, producing ethene, $\text{H}_2\text{O}$ and $\text{Br}^-$ .

Final answers

(a) $\text{NaOH(aq)}$ , reflux: $\text{CH}_3\text{CH}_2\text{Br} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{NaBr}$
(b) $\text{NaOH}$ in ethanol, reflux: $\text{CH}_3\text{CH}_2\text{Br} + \text{NaOH} \rightarrow \text{CH}_2=\text{CH}_2 + \text{NaBr} + \text{H}_2\text{O}$
(c) In water $\text{OH}^-$ acts as a nucleophile (substitution); in ethanol it acts as a base (elimination).
(d) $\text{OH}^-$ removes an $\text{H}^+$ from the carbon next to C–Br; the released electron pair forms the new C=C $\pi$ bond as $\text{Br}^-$ leaves.