3.2.8 Haloalkanes - Elimination

Specification

Students should:
  • understand concurrent substitution and elimination (including mechanisms) in the reaction of a haloalkane (e.g. 2- bromopropane with potassium hydroxide) and the role of the reagent as both nucleophile and base
  • appreciate the usefulness of this reaction in organic synthesis

Elimination of haloalkanes

Haloalkanes react with a solution of sodium hydroxide in ethanol by eliminating a molecule of water and forming a double bond. They can only do this if there is a hydrogen atom on the carbon atom adjacent to the atom that holds the halogen.

1-chloropropane + sodium hydroxide/ethanol propene + sodium chloride + water

ClCH2CH2CH3   +   NaOH/ethanol CH2=CHCH3   +   NaCl   +   H2O

1-chloropropane structure
chlorodimethylpropane structure
1-chloropropane
(can react)
chlorodimethylpropane
(cannot react)

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Mechanism of elimination reactions

Once again the mechanism depends on the nature of the haloalkane. Primary haloalkanes react via an E2 mechanism and tertiary haloalkanes via an E1 mechanism. Secondary mechanisms are probably a combination of both E1 and E2.

In bimolecular elimination the first stage involves two colliding particles. Firstly, the strong base produced by the solution of sodium hydroxide in ethanol (the ethoxide ion) removes a hydrogen atom from the carbon that is adjacent to the carbon atom holding the halogen.

2-bromopropane + ethoxide ion 2-bromopropane carbanion + ethanol

CH3CHBrCH3  +  EtO CH2CHBrCH3   +  EtOH

Then the carbanion formed makes a double bond, ejecting the halide ion in the process:

2-bromopropane 2-bromopropane carbanion + ethanol

CH2CHBrCH3 CH2=CHCH3  +  Br

Mechanism

Overall: Hydrogen bromide eliminated and alkene formed.

In E1 elimination, the halogen breaks off the tertiary haloakane before the hydrogen ion is collected by the ethoxide ion (base).


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