Elimination

01Elimination

Elimination reactions

This section covers elimination of a hydrogen halide from a halogenoalkane using ethanolic hydroxide to form an alkene.

From halogenoalkane to alkene

In an elimination reaction, atoms are removed from an organic molecule and a multiple bond forms. For a halogenoalkane, a hydrogen and a halogen are removed from neighbouring carbon atoms, corresponding to the loss of a hydrogen halide.

Functional-group change Halogenoalkane → alkene
Reagent Potassium hydroxide or sodium hydroxide
Solvent Ethanol
Conditions Heat under reflux
Role of OHX\ce{OH^-} Base
Organic product Alkene

For example, heating 1-bromopropane with ethanolic potassium hydroxide produces propene:

CHX3CHX2CHX2Br+KOHCHX3CH=CHX2+KBr+HX2O\ce{CH3CH2CH2Br + KOH -> CH3CH=CH2 + KBr + H2O}

1-bromopropane

The molecule contains the C–Br bond and a hydrogen on the neighbouring carbon.

Propene

Removal of H and Br allows a carbon–carbon double bond to form.

Exam Tip: For elimination of a halogenoalkane, state ethanolic hydroxide and heat. Do not give aqueous hydroxide if the required organic product is an alkene.

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