3.3.3.2 Elimination

Question 1

What happens during the elimination reaction of halogenoalkanes?

Answer 1

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

Question 2

How do you draw an elimination reaction in halogenoalkanes plus mechanisms? (2-bromoalkane)

Answer 2

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

Question 3

How does the OH- ion react with a halogenoalkane?

Answer 3

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

Question 4

What are the steps of bromoethane reacting with potassium/sodium hydroxide in heat?

Answer 4

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

Question 5

What is the difference between a substitution and elimination reaction?

Answer 5

Substitution: cold OH ions in water, forms an alcohol
Elimination: hot OH ions in ethanol, forms an alkene

Question 6

How do you draw an elimination reaction in halogenoalkanes plus mechanisms? (1-bromoalkane)

Answer 6

See card