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C. Elimination vs. Substitution

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    This page discusses the factors that decide whether halogenoalkanes undergo elimination reactions or nucleophilic substitution when they react with hydroxide ions from, say, sodium hydroxide or potassium hydroxide.

    The reactions

    Both reactions involve heating the halogenoalkane under reflux with sodium or potassium hydroxide solution.

    Nucleophilic substitution

    The hydroxide ions present are good nucleophiles, and one possibility is a replacement of the halogen atom by an -OH group to give an alcohol via a nucleophilic substitution reaction.


    In the example, 2-bromopropane is converted into propan-2-ol.


    Halogenoalkanes also undergo elimination reactions in the presence of sodium or potassium hydroxide.


    The 2-bromopropane has reacted to give an alkene - propene.

    Notice that a hydrogen atom has been removed from one of the end carbon atoms together with the bromine from the centre one. In all simple elimination reactions the things being removed are on adjacent carbon atoms, and a double bond is set up between those carbons.

    What decides whether you get substitution or elimination?

    The reagents you are using are the same for both substitution or elimination - the halogenoalkane and either sodium or potassium hydroxide solution. In all cases, you will get a mixture of both reactions happening - some substitution and some elimination. What you get most of depends on a number of factors.

    The type of halogenoalkane

    This is the most important factor.

    type of halogenoalkane substitution or elimination?
    primary mainly substitution
    secondary both substitution and elimination
    tertiary mainly elimination

    For example, whatever you do with tertiary halogenoalkanes, you will tend to get mainly the elimination reaction, whereas with primary ones you will tend to get mainly substitution. However, you can influence things to some extent by changing the conditions.

    The solvent

    The proportion of water to ethanol in the solvent matters.

    • Water encourages substitution.
    • Ethanol encourages elimination.

    The temperature

    Higher temperatures encourage elimination.

    Concentration of the sodium or potassium hydroxide solution

    Higher concentrations favor elimination.

    In summary

    For a given halogenoalkane, to favour elimination rather than substitution, use:

    • heat
    • a concentrated solution of sodium or potassium hydroxide
    • pure ethanol as the solvent

    The role of the hydroxide ions

    The role of the hydroxide ion in a substitution reaction

    In the substitution reaction between a halogenoalkane and OH- ions, the hydroxide ions are acting as nucleophiles. For example, one of the lone pairs on the oxygen can attack the slightly positive carbon. This leads on to the loss of the bromine as a bromide ion, and the -OH group becoming attached in its place.


    The role of the hydroxide ion in an elimination reaction

    Hydroxide ions have a very strong tendency to combine with hydrogen ions to make water - in other words, the OH- ion is a very strong base. In an elimination reaction, the hydroxide ion hits one of the hydrogen atoms in the CH3 group and pulls it off. This leads to a cascade of electron pair movements resulting in the formation of a carbon-carbon double bond, and the loss of the bromine as Br-.



    Jim Clark (

    This page titled C. Elimination vs. Substitution is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jim Clark.

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