Elimination Reactions: Nucleophiles vs. Bases Revision Notes | A Level AQA Chemistry

1. The Same Reagent, Two Different Outcomes

Here is a puzzle worth thinking about: you add potassium hydroxide to 2-bromopropane and you can get either propan-2-ol or propene — depending entirely on the conditions you choose. Same reagent, same substrate, two completely different products.

The reason is that OH⁻ has a dual identity. It carries a lone pair, which makes it a nucleophile. It is also a strong Brønsted–Lowry base, meaning it can accept a proton. Which role it plays is decided by the reaction conditions — and that is exactly what this lesson unpacks.

2. The Dual Role of the Hydroxide Ion

Think back to what defines a nucleophile: an electron-rich species that donates a lone pair to an electron-deficient carbon. The hydroxide ion (OH⁻) fits perfectly — it has three lone pairs and a full negative charge, making it strongly attracted to the $\delta^+$ carbon bonded to the halogen.

Now think about what defines a base: a species that accepts a proton. OH⁻ does this too. When it approaches a hydrogen on the carbon adjacent to the C–X carbon (rather than attacking the C–X carbon itself), it pulls that proton away. The C–H bond breaks, the C–X bond breaks, and a double bond forms between those two carbons.

This competition is not a flaw in the chemistry — it is a feature you can exploit. By choosing the right conditions, you can push the reaction strongly towards one product or the other. The key variables are solvent and concentration of KOH.

3. Nucleophilic Substitution: The Mechanism

In nucleophilic substitution, OH⁻ approaches the back of the C–Br bond in 2-bromopropane. The carbon bearing the bromine is electron-deficient ($\delta^+$) because bromine is more electronegative and withdraws electron density through the C–Br bond.

The OH⁻ donates its lone pair to that carbon in a concerted, single-step process. As the new O–C bond forms, the C–Br bond breaks heterolytically, and Br⁻ departs as the leaving group. This is an $S_N2$ mechanism — bimolecular nucleophilic substitution — because both the halogenoalkane and the nucleophile are involved in the rate-determining step.

The overall equation for this substitution is: $\text{CH}_3\text{CHBrCH}_3\text{(l)} + \text{OH}^-\text{(aq)} \rightarrow \text{CH}_3\text{CH(OH)CH}_3\text{(l)} + \text{Br}^-\text{(aq)}$

The primary condition favouring substitution is aqueous KOH (water as solvent). Lower temperature is a secondary factor; the solvent choice is the more important variable for AQA purposes. Water stabilises the ionic species and the polar transition state of the Sₙ2 mechanism, making nucleophilic substitution more competitive than elimination.

4. Elimination: The Mechanism

Now picture the same 2-bromopropane molecule, but this time OH⁻ does not head for the C–Br carbon. Instead, it approaches one of the hydrogen atoms on the carbon next to the C–Br carbon — the adjacent carbon. This is the base pathway.

In a concerted step, OH⁻ abstracts the proton (H⁺) from the adjacent carbon. The electrons from the C–H bond shift to form a new $\pi$ bond between the two carbons. Simultaneously, the C–Br bond breaks and Br⁻ departs. The result is propene, water, and bromide ion.

The overall equation for elimination is: $\text{CH}_3\text{CHBrCH}_3\text{(l)} + \text{OH}^-\text{(aq)} \rightarrow \text{CH}_3\text{CH=CH}_2\text{(g)} + \text{H}_2\text{O(l)} + \text{Br}^-\text{(aq)}$

The conditions that favour elimination are ethanolic KOH (ethanol as solvent) and a higher temperature. Ethanol is a less polar solvent than water; it does not stabilise the ionic transition state of substitution as effectively, so the base pathway becomes relatively more favourable. Higher temperature also provides the additional energy needed to break two bonds (C–H and C–Br) in the same step.

5. Concurrent Reactions: Substitution and Elimination Together

A critical point for AQA exams: substitution and elimination do not take turns — they happen at the same time. Whenever you mix a halogenoalkane with KOH, both pathways are active. The conditions you choose shift the ratio of products, not switch one pathway completely off.

With aqueous KOH at room temperature, the substitution product (propan-2-ol) predominates. With ethanolic KOH at higher temperature, the elimination product (propene) predominates. In both cases, a small amount of the other product is also present.

The solvent effect is worth understanding at a deeper level. Water stabilises the ionic transition state and the charged species involved in the S_N2 mechanism, making substitution more favourable relative to elimination under these conditions. Ethanol is less polar, so it provides less stabilisation for the substitution pathway, making the elimination pathway relatively more competitive.

6. Worked Example: Identifying Products and Mechanisms

Exam questions on this topic typically ask you to identify the product, state the mechanism type, draw curly arrows, or explain the role of the reagent. The step-by-step walkthrough below models how to approach a multi-part question on 2-bromopropane.

Worked example: Predicting Products and Mechanisms for 2-Bromopropane with KOH

2-Bromopropane is reacted with KOH under two different sets of conditions: (i) aqueous KOH at room temperature; (ii) ethanolic KOH under reflux. For each condition, identify the organic product, name the mechanism, and state the role of OH⁻.

1. 2-Bromopropane: $\text{CH}_3\text{CHBrCH}_3$. The C–Br bond is polarised $\delta^+$ on C and $\delta^-$ on Br because Br is more electronegative. — Recognising the electrophilic carbon (C–Br) and the adjacent C–H bonds is the starting point for deciding which pathway operates.
2. OH⁻ (nucleophile) attacks the $\delta^+$ carbon bearing Br from the back. C–O bond forms, C–Br bond breaks, Br⁻ leaves. Product: propan-2-ol, $\text{CH}_3\text{CH(OH)CH}_3$. — Water stabilises the polar transition state, favouring the nucleophilic substitution ($S_N2$) pathway. OH⁻ acts as a nucleophile here.
3. OH⁻ (base) abstracts H from the carbon adjacent to C–Br. C–H bond breaks, C=C forms, C–Br bond breaks, Br⁻ leaves. Product: propene, $\text{CH}_3\text{CH=CH}_2$. — Ethanol is less polar than water, so it does not stabilise the substitution transition state as well. Higher temperature provides energy for the concerted two-bond-breaking elimination step. OH⁻ acts as a base here.
4. Condition (i): nucleophilic substitution ($S_N2$). Condition (ii): elimination (E2). — AQA requires you to name the mechanism type. Both are bimolecular and concerted — the nucleophile/base and the substrate both feature in the rate-determining step.
5. Both reactions occur simultaneously under either set of conditions. The conditions shift the ratio of products, not switch one pathway off entirely. — Exam questions may ask you to explain why a mixture of products is obtained even when conditions favour one pathway.

Answer: (i) Aqueous KOH, room temperature: propan-2-ol via nucleophilic substitution; OH⁻ acts as nucleophile. (ii) Ethanolic KOH, reflux: propene via elimination; OH⁻ acts as base. Both reactions occur concurrently.

7. Curly Arrow Rules for These Mechanisms

Curly arrows are the language of organic mechanisms, and AQA mark schemes are precise about where they start and end. A full curly arrow (double-headed, showing an electron pair) must always start from an electron-rich site — either a lone pair or an existing bond — and point towards an electron-poor site.

• In the substitution mechanism: draw one arrow from the lone pair on O (in OH⁻) to the C bearing Br; draw a second arrow from the C–Br bond to the Br atom.
• In the elimination mechanism: draw one arrow from the lone pair on O (in OH⁻) to the H on the adjacent carbon; draw a second arrow from the C–H bond to the C–C bond (forming C=C); draw a third arrow from the C–Br bond to the Br atom.
• Never start a curly arrow from an atom label alone — it must originate from a lone pair or a bond.
• Half-headed (fish-hook) arrows show single-electron movement in radical mechanisms; do not use them here.

A common place to lose marks is drawing the substitution arrow from the wrong face of the carbon. In an $S_N2$ mechanism, attack is from the back of the C–X bond (180° to the leaving group). While AQA does not always require you to show stereochemistry explicitly at this level, the back-side attack description is expected in written explanations.

8. Exam Technique and Common Mistakes

Questions on this topic appear regularly in AQA Paper 1 and Paper 2, often as 4–6 mark mechanism questions or as 'explain' questions about conditions. The marks are very specific, so the language you use matters as much as the chemistry.