3.3.3.2 Elimination
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What happens during the elimination reaction of halogenoalkanes?
Halogenoalkane + sodium/potassium hydroxide + heat OH- ion acts as a base A hydrogen halide is eliminated from the molecule, leaving a double bond in its place so that an alkene is formed
How do you draw an elimination reaction in halogenoalkanes plus mechanisms? (2-bromoalkane)
See card 1st arrow: OH ion pointing to a hydrogen atom 2nd arrow: from the C - H bond of the hydrogen atom being attacked towards the C - C bond next to it 3rd arrow: from the C - Halogen bond to the halogen Produces an alkene, water and a halogen ion
How does the OH- ion react with a halogenoalkane?
Cold aqueous OH: nucleophile and its lone pair attacks a halogenoalkane at the positive carbon to form an alcohol Hot OH in ethanol: acts as a base, removes a hydrogen ion from the molecule to form an alkene and a hydrogen halide
What are the steps of bromoethane reacting with potassium/sodium hydroxide in heat?
The OH ion uses its lone pair to form a bond with one of the hydrogen atoms on the carbon next to the C-Br bond These hydrogen atoms are very slightly positive The electron pair from the C-H bond now becomes part of a carbon carbon double bond The bromine takes the pair of electrons in the C-Br bond and leave as a bromide ion
What is the difference between a substitution and elimination reaction?
Substitution: cold OH ions in water, forms an alcohol Elimination: hot OH ions in ethanol, forms an alkene
How do you draw an elimination reaction in halogenoalkanes plus mechanisms? (1-bromoalkane)
See card