Substitution vs Elimination

So far we have treated nucleophilic substitution as a clean, one-product reaction. In reality, whenever a haloalkane meets a base there are two competing pathways:

• Substitution — the base attacks the δ+ carbon, replacing the halogen with a new group.
• Elimination — the base removes a β-hydrogen, creating a new C=C double bond and kicking out the halide.

Both happen in the same flask with the same reagents; which dominates depends on conditions, particularly the solvent, temperature, and nature of the base. Being able to predict and control the outcome is a cornerstone of A-Level synthesis.

This lesson covers the OCR A-Level Chemistry A (H432) specification point 4.2.2 (e): comparison of substitution and elimination reactions of haloalkanes; the effect of reaction conditions on the product formed.

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1. The Two Reactions Side by Side

Consider 2-bromopropane, (CH₃)₂CHBr, reacted with sodium hydroxide.

Substitution (in aqueous NaOH):

$CH_3\text{-}CHBr\text{-}CH_3 + OH^- \xrightarrow{\text{aq, warm}} CH_3\text{-}CHOH\text{-}CH_3 + Br^-$CH3​-CHBr-CH3​+OH−aq, warm​CH3​-CHOH-CH3​+Br−

The OH⁻ attacks the C bonded to Br, displacing Br⁻. Product: propan-2-ol.

Elimination (in ethanolic NaOH, hot):

$CH_3\text{-}CHBr\text{-}CH_3 + OH^- \xrightarrow{\text{ethanolic, hot}} CH_3\text{-}CH\text{=}CH_2 + Br^- + H_2O$CH3​-CHBr-CH3​+OH−ethanolic, hot​CH3​-CH=CH2​+Br−+H2​O

The OH⁻ acts as a base, taking a β-hydrogen from the CH₃ next door. The electrons of the C–H bond fold into a new C=C bond and Br⁻ leaves. Product: propene (and water).

Same reagents, different solvents, completely different products.

graph TD
  A[2-Bromopropane + NaOH] --> B{Conditions?}
  B -->|Aqueous NaOH<br/>warm| C[SUBSTITUTION<br/>Propan-2-ol + NaBr]
  B -->|Ethanolic NaOH<br/>hot| D[ELIMINATION<br/>Propene + NaBr + H2O]

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2. The Mechanism of Elimination (E2)

At A-Level you are not required to draw the curly-arrow mechanism of elimination for OCR H432, but understanding it helps you decide when it will happen.

The elimination of an H and a halogen from adjacent carbons, under the action of a base, is called E2 — elimination, bimolecular. Like SN2, it is concerted:

• The base removes an H from the β-carbon (the carbon next to the one with the halogen).
• The electrons from the C–H bond become the π-bond of the new C=C.
• The halide leaves at the same time.

graph LR
  A["Base: + H-C-C-X"] -->|concerted<br/>3 arrows| B["Base-H + C=C + X-"]

Three curly arrows:

1. From the lone pair on the base to the β-H.
2. From the C–H bond into the C–C bond (forming the π-bond).
3. From the C–X bond to the X.

Key Structural Point: The β-Carbon

You can only have an elimination if there is a β-hydrogen — a hydrogen on a carbon adjacent to the C–X carbon. If there is no β-H (for example, in halomethane or a neopentyl halide) you can only do substitution.

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3. What Favours Which Pathway?

OCR tests a simple but reliable set of rules. Learn this table:

Factor	Favours Substitution	Favours Elimination
Solvent	Water (aqueous)	Ethanol (ethanolic)
Temperature	Warm (~40–60 °C)	Hot (reflux, ~70 °C+)
Nature of base/nucleophile	Small, high charge density (OH⁻, CN⁻)	Bulky, less nucleophilic (e.g. t-butoxide)
Class of haloalkane	Primary	Tertiary
Concentration of base	Dilute	Concentrated

3.1 Solvent Effect — Why Ethanol vs Water?

In water, OH⁻ is strongly solvated — surrounded by water molecules hydrogen bonding to it. This makes it a reasonably good nucleophile (it still has a lone pair pointing outwards) but its basicity is tempered because the proton is hard to accept into a solvent shell. So it substitutes.

In ethanol, the OH⁻ is less well solvated (ethanol is a poorer H-bond donor) and exists "more free". It behaves more as a base, ripping a proton off the β-carbon. So it eliminates.

3.2 Temperature Effect

Elimination generates more disorder (one molecule → three particles) than substitution (one + one → one + one). Higher temperature favours the higher-entropy pathway by the Gibbs equation ΔG = ΔH − TΔS. Hence hot conditions favour elimination.

3.3 Class of Haloalkane

• Primary haloalkanes strongly prefer substitution (SN2 attack at C is easy because there is no steric hindrance).
• Tertiary haloalkanes, even in water, often give a lot of elimination because the back of the C is blocked by three R groups and the base "can't get to" the C, so it takes the β-H instead.
• Secondary haloalkanes sit between the two and can give mixtures unless you pick the conditions carefully.