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Halogenoalkanes

Elimination reactions forming alkenes

  • 3 things to know
  • Components 1 and 2

Overview

Recall questions on the reagent and conditions for elimination, the hydroxide ion acting as a base rather than a nucleophile, the equations and the products besides the alkene, why some halogenoalkanes give two alkenes, which class gives the most elimination, and how the solvent decides the outcome.

What you need to know

Answer each one before you reveal it.

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Key term

What besides the alkene is formed in an elimination, how does the solvent decide the outcome, and which class gives most alkene?

Key term

Describe the three curly arrows in the elimination of hydrogen bromide from 1-bromopropane.

Key term

Name the class of organic product formed when a halogenoalkane undergoes elimination.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Name the class of organic product formed when a halogenoalkane undergoes elimination.

    Model answer

    An alkene is formed.

  2. 2

    State the reagent and conditions needed for the elimination reaction of a halogenoalkane.

    Model answer

    Potassium hydroxide dissolved in ethanol is used, and the mixture is heated under reflux.

  3. 3

    Write the equation for the elimination of hydrogen bromide from 1-bromopropane by ethanolic potassium hydroxide.

    Model answer

    CH₃CH₂CH₂Br + KOH → CH₃CH=CH₂ + KBr + H₂O.

  4. 4

    In an elimination reaction, is the hydroxide ion acting as a nucleophile or as a base?

    Model answer

    It acts as a base, because it removes a hydrogen ion rather than bonding to a carbon atom.

  5. 5

    From which carbon atom is the hydrogen removed during an elimination reaction?

    Model answer

    It is removed from a carbon atom next to the one carrying the halogen.

  6. 6

    Describe the three curly arrows in the elimination of hydrogen bromide from 1-bromopropane.

    Model answer

    One arrow runs from a lone pair on the hydroxide ion to the hydrogen on the carbon next to the C—Br, a second runs from that C—H bond into the gap between the two carbons to form the double bond, and a third runs from the C—Br bond to the bromine atom.

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