Under specific reaction conditions, halogenoalkanes undergo an elimination reaction instead of nucleophilic substitution. In an elimination reaction, a hydrogen halide (such as \(\text{HBr}\) or \(\text{HCl}\)) is removed from the molecule, producing an alkene.
🔑 Key Principle
An elimination reaction involves the removal of a small molecule from a larger starting material. In halogenoalkanes, a hydroxide ion acts as a base to remove a proton from a carbon atom adjacent to the C-X carbon (the \(\beta\)-carbon). The bonding pair of electrons of this C-H bond then forms a C=C double bond, displacing the halide leaving group.
Key Definitions
A chemical reaction in which a small molecule is removed from a single reactant, leaving behind a saturated bond that is converted into an unsaturated bond (like a C=C double bond).
A proton (\(\text{H}^+\)) acceptor. In elimination reactions, the hydroxide ion acts as a base by using its lone pair of electrons to accept a proton from a carbon atom.
Hydroxide: Nucleophile vs Base
The hydroxide ion is a versatile species. Depending on the solvent, concentration, and temperature, it can behave as either a nucleophile or a base. This competition determines whether a halogenoalkane undergoes substitution or elimination:
| Reaction Parameter | Favours Substitution | Favours Elimination |
|---|---|---|
| Role of OH⁻ | Nucleophile (electron pair donor to C) | Base (proton acceptor from adjacent C) |
| Solvent | Water (aqueous sodium/potassium hydroxide) | Ethanol (ethanolic sodium/potassium hydroxide) |
| Temperature | Warm / moderate temperatures | Hot / high reflux temperatures |
| Halogenoalkane | Primary halogenoalkanes favour substitution | Tertiary halogenoalkanes favour elimination |
Remember this rule of thumb: Water for substitution, ethanol for elimination. If you are asked to state the conditions to produce an alkene from a halogenoalkane, you must specify ethanolic sodium hydroxide (or potassium hydroxide) and heating under reflux.
The Elimination Mechanism
The elimination reaction is represented by a single-step mechanism containing three curly arrows. Below is the mechanism for the reaction of 2-bromopropane with ethanolic sodium hydroxide to form propene:
Elimination in Asymmetric Halogenoalkanes
If the halogenoalkane is asymmetric (e.g. 2-bromobutane), elimination can occur on either side of the carbon atom bonded to the halogen. This results in the formation of a mixture of isomeric alkene products.
Let's look at the elimination of 2-bromobutane:
\[ \text{CH}_3\text{CH(Br)CH}_2\text{CH}_3 + \text{OH}^- \rightarrow \text{Alkenes} + \text{H}_2\text{O} + \text{Br}^- \]
The bromine is bonded to Carbon 2. Elimination can target the hydrogens on either Carbon 1 or Carbon 3:
- Removal of hydrogen from Carbon 1: Produces but-1-ene.
- Removal of hydrogen from Carbon 3: Produces but-2-ene.
Because but-2-ene exhibits stereoisomerism due to restricted rotation about the C=C double bond, it exists as a mixture of two geometric isomers:
- E-but-2-ene (trans isomer, with methyl groups on opposite sides)
- Z-but-2-ene (cis isomer, with methyl groups on the same side)
Therefore, the elimination of 2-bromobutane yields a mixture of three alkene products: but-1-ene, E-but-2-ene, and Z-but-2-ene.
Worked Examples
Mechanism steps:
- Draw the structure of 2-bromopropane, showing the C-H bond on one of the methyl groups (C1 or C3) and the C-Br bond.
- Draw a hydroxide ion (\(\text{OH}^-\)) with its lone pair.
- Draw Arrow 1 starting from the lone pair of the hydroxide ion pointing to the hydrogen atom on Carbon 1.
- Draw Arrow 2 starting from the middle of the C-H bond on Carbon 1 and pointing to the single bond between Carbon 1 and Carbon 2.
- Draw Arrow 3 starting from the C-Br bond and pointing to the Bromine atom.
- Draw the products: propene (\(\text{CH}_3\text{CH=CH}_2\)), water (\(\text{H}_2\text{O}\)), and a bromide ion (\(\text{Br}^-\)).
Answer:
In 2-chlorobutane, the chlorine is on Carbon 2. Adjacent carbons containing hydrogens are Carbon 1 (\(-\text{CH}_3\)) and Carbon 3 (\(-\text{CH}_2-\)).
- Eliminating hydrogen from Carbon 1 forms but-1-ene: \(\text{CH}_2\text{=CH-CH}_2\text{-CH}_3\). This has no geometric isomers because Carbon 1 is bonded to two identical hydrogen atoms.
- Eliminating hydrogen from Carbon 3 forms but-2-ene: \(\text{CH}_3\text{-CH=CH-CH}_3\). Since both carbons in the double bond are attached to two different groups (H and methyl), but-2-ene displays stereoisomerism.
The products are: but-1-ene, E-but-2-ene, and Z-but-2-ene.
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