What are the key steps involved when a halogenoalkane undergoes an elimination reaction in the presence of a strong base like sodium hydroxide (NaOH)?

  1. OH- abstracts an H+ from the carbon adjacent to the C-X bond, forming H2O.
  2. The C-X bond breaks, eliminating the halide ion (X-).
  3. A C=C double bond forms between adjacent carbons.

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What are the products formed when a halogenoalkane undergoes an elimination reaction with hydroxide ions?

The products are an alkene, a hydrogen halide and water.

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Explain the role of OH- in the elimination reaction mechanism of a halogenoalkane.

The OH- ion acts as a base by removing a proton (H+ ion) from the halogenoalkane.

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Does the carbon-halogen bond in a halogenoalkane undergo heterolytic or homolytic fission during an elimination reaction?

Heterolytic fission

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What is the mechanism for the elimination reaction above?

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What factor determines whether a halogenoalkane will undergo substitution or elimination when reacting with a hydroxide ion?

The type of solvent (aqueous or ethanolic).

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How does the choice of solvent influence the tendency of a halogenoalkane to undergo substitution or elimination?

Aqueous conditions (water as the solvent) favour substitution, whereas ethanolic conditions (ethanol as the solvent) favour elimination.

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Name the 3 possible alkene products formed during the elimination of 2-chlorobutane.

  1. But-1-ene
  2. cis-but-2-ene
  3. trans-but-2-ene

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