AQA · A-Level · Chemistry

    Course version: 7405 · Version from Sept 2016

    The concurrent substitution and elimination reactions of a halogenoalkane…

    Full specification wording

    Elimination: The concurrent substitution and elimination reactions of a halogenoalkane (eg 2-bromopropane with potassium hydroxide).

    Hydroxide ion can behave in two ways towards 2-bromopropane, and both happen at the same time.

    Checked against the AQA 7405 specification (opens in a new tab) on .

    • Point 1 of 5
    • 3 key terms

    Study notes

    From the checked teaching notes for this specification point.

    The idea

    Hydroxide ion can behave in two ways towards 2-bromopropane, and both happen at the same time. As a nucleophile it attacks the δ⁺ carbon and displaces bromide, giving propan-2-ol by nucleophilic substitution. As a base it removes a hydrogen from a carbon next to the C–Br carbon; those bonding electrons swing in to form a C=C bond while bromide leaves, giving propene by elimination. Conditions shift the balance rather than switching one route off: warm aqueous potassium hydroxide favours substitution, while hot potassium hydroxide dissolved in ethanol favours elimination. Exam answers must name both mechanisms correctly and say which role the hydroxide is playing in each, since the same reagent is doing two different jobs.

    What earns the marks

    • Naming elimination as the route to the alkene and nucleophilic substitution as the route to the alcohol.
    • Stating that hydroxide acts as a base in one reaction and as a nucleophile in the other.
    • The conditions that favour each route, aqueous and warm against ethanolic and hot.
    • The three curly arrows in the elimination, including the arrow from the C–H bond into the C–C bond.

    Common mistakes and how to put them right

    • Assuming two different reagents are used, when the same potassium hydroxide does both jobs.

    • Naming the alkene route "nucleophilic elimination", which contradicts the base role of the hydroxide.

    • Drawing the hydroxide arrow to the carbon instead of to the hydrogen in the elimination.

    • Writing that changing the solvent stops one reaction entirely rather than changing the proportions.

    Exam tips

    • Name the solvent, not just the reagent: "KOH in ethanol" and "KOH(aq)" are what distinguish the two answers.
    • Label which hydrogen you are removing; from 2-bromopropane it must come from a CH₃ group.
    • If a question says "concurrent", it expects both products, so give both and say which conditions favour which.

    Try the lesson

    The start of the Elimination: The concurrent substitution and elimination reactions of a halogenoalkane (eg 2-bromopropane with potassium hydroxide). lesson, made from checked teaching notes for this specification point.

    Have a go first

    When 2-bromopropane reacts with potassium hydroxide dissolved in ethanol, two organic products are formed. Which pair of products is formed?

    Listen

    Hydroxide ion can behave in two ways towards 2-bromopropane, and both happen at the same time. As a nucleophile it attacks the δ⁺ carbon and displaces bromide, giving propan-2-ol by nucleophilic substitution. As a base it removes a hydrogen from a carbon next to the C–Br carbon; those bonding electrons swing in to form a C=C bond while bromide leaves, giving propene by elimination. Conditions shift the balance rather than switching one route off: warm aqueous potassium hydroxide favours substitution, while hot potassium hydroxide dissolved in ethanol favours elimination. Exam answers must name both mechanisms correctly and say which role the hydroxide is playing in each, since the same reagent is doing two different jobs. A strong answer earns marks for these points. naming elimination as the route to the alkene and nucleophilic substitution as the route to the alcohol. stating that hydroxide acts as a base in one reaction and as a nucleophile in the other. the conditions that favour each route, aqueous and warm against ethanolic and hot. the three curly arrows in the elimination, including the arrow from the C–H bond into the C–C bond. Exam tip: Name the solvent, not just the reagent: "KOH in ethanol" and "KOH(aq)" are what distinguish the two answers.

    See it

    A top-down flowchart. At the top, one node reads '2-bromopropane with potassium hydroxide'. Two arrows lead from it to two nodes: 'Hydroxide acts as a nucleophile' and 'Hydroxide acts as a base'. From the nucleophile node, an arrow leads to 'Attacks the δ⁺ carbon and displaces bromide', which leads to 'Propan-2-ol by nucleophilic substitution', which leads to 'Warm aqueous potassium hydroxide favours substitution'. From the base node, an arrow leads to 'Removes a hydrogen from a carbon next to the C–Br carbon', which leads to 'Bonding electrons form a C=C bond while bromide leaves', which leads to 'Propene by elimination', which leads to 'Hot potassium hydroxide in ethanol favours elimination'.
    It shows the same hydroxide reagent acting as a nucleophile to give propan-2-ol by substitution and as a base to give propene by elimination, with the conditions that favour each route.

    Key words

    Say what each word means before you turn the card.

    competing reactions

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    role of solvent

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    elimination to an alkene

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    See every definition
    competing reactions
    The concurrent substitution and elimination reactions of a halogenoalkane, such as 2-bromopropane with potassium hydroxide, both happening at the same time.
    role of solvent
    Conditions shift the balance rather than switching one route off: warm aqueous potassium hydroxide favours substitution, while hot potassium hydroxide dissolved in ethanol favours elimination.
    elimination to an alkene
    As a base the hydroxide removes a hydrogen from a carbon next to the C–Br carbon, and those bonding electrons swing in to form a C=C bond while bromide leaves, giving propene.

    Label the diagram

    Some boxes have lost their names. Pick the right name for each number.

    The diagram again, with 3 boxes numbered instead of named. A top-down flowchart. At the top, one node reads '2-bromopropane with potassium hydroxide'. Two arrows lead from it to two nodes: 'Hydroxide acts as a nucleophile' and 'Hydroxide acts as a base'. From the nucleophile node, an arrow leads to 'Attacks the δ⁺ carbon and displaces bromide', which leads to 'Propan-2-ol by nucleophilic substitution', which leads to 'Warm aqueous potassium hydroxide favours substitution'. From the base node, an arrow leads to 'Removes a hydrogen from a carbon next to the C–Br carbon', which leads to 'Bonding electrons form a C=C bond while bromide leaves', which leads to 'Propene by elimination', which leads to 'Hot potassium hydroxide in ethanol favours elimination'.

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