Elimination Reactions in Halogenoalkanes

This lesson covers: 

1. How halogenoalkanes undergo elimination reactions
2. The elimination reaction mechanism
3. How changing conditions influences if substitution or elimination occurs

Halogenoalkanes eliminate under alkaline conditions

If a halogenoalkane is heated under reflux with an alkali like potassium hydroxide (KOH) dissolved in ethanol, an elimination reaction occurs to form an alkene.

For example, heating 2-iodopropane with KOH gives propene:

CH₃CHICH₃ + KOH ➔ CH₂=CHCH₃ + H₂O + KI

This is an example of an elimination reaction, where a small group breaks off a larger molecule and is not replaced.

Reaction mechanism

The mechanism of the elimination reaction occurs in three steps:

1. OH^- removes a hydrogen ion (H⁺) from the halogenoalkane, forming water.
2. This leaves the adjacent carbon with a spare electron to form a π bond.
3. The movement of electrons breaks the carbon-halogen bond heterolytically, eliminating the halide ion.

So OH^- acts as a base in elimination reactions by removing a proton.

The mechanism for the elimination of 2-iodopropane with OH^- is:

Anhydrous conditions favour elimination

Halogenoalkanes treated with hydroxide can undergo either substitution or elimination, depending on the choice of solvent.

• Substitution is favoured in aqueous conditions. OH^- acts as a nucleophile.
• Elimination is favoured in ethanolic conditions. OH^- acts as a base.

Using a solvent mixture of water and alcohol allows both reactions to occur, giving a mixture of products.