Lesson 3.3.3.2

3.3.3.2 Halogenoalkanes: Elimination Quiz: AQA Chemistry, Unit 3

20 questions

Written by Revision Ninja · Last reviewed 11 October 2026

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Lesson 3.3.3.2, Halogenoalkanes: Elimination: 20 multiple choice questions for the AQA Chemistry (7405), Unit 3: Organic chemistry, written with Revision Ninja.

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The 20 questions

  1. What type of reaction occurs when 2-bromopropane is heated with ethanolic potassium hydroxide?

    • Elimination
    • Addition
    • Oxidation only
    • Free-radical substitution

  2. What is the organic product of elimination from 2-bromopropane?

    • Propan-1-ol
    • Propane
    • Propan-2-ol
    • Propene

  3. What is the role of the hydroxide ion in the elimination of 2-bromopropane?

    • It acts as a base by removing a proton
    • It acts as a radical initiator
    • It acts as an electrophile by accepting a pair of electrons
    • It acts as a catalyst that is not involved in the mechanism

  4. What is the role of the hydroxide ion in the substitution of 2-bromopropane with aqueous OH-?

    • It acts as an electrophile
    • It is a catalyst with no part in the product
    • It acts as a nucleophile, substituting the bromine
    • It is a radical that breaks the C-Br bond

  5. Why does ethanolic KOH favour elimination over substitution?

    • The reagent is strongly basic and the solvent reduces the nucleophilicity of OH-
    • Ethanol makes the hydroxide ion neutral with no lone pair
    • Ethanolic KOH contains no hydroxide ions
    • Elimination needs no reagent at all

  6. Which product is formed by substitution of 2-bromopropane with aqueous KOH?

    • Propene
    • Propan-1-ol
    • Propan-2-ol
    • 2-bromopropanol

  7. Which feature of 2-bromopropane leads to concurrent substitution and elimination?

    • The C-Br bond is adjacent to carbons with hydrogens that can be removed
    • The molecule is fully saturated with no polar bonds
    • The molecule is an alkene with a C=C bond
    • The bromine atom is bonded to a hydrogen only

  8. In the elimination of HBr from 2-bromopropane, which bond is formed?

    • A C-Br double bond
    • A C-OH single bond
    • A C=C double bond
    • A C-K ionic bond

  9. Which description of the reagent in the reaction with 2-bromopropane is correct?

    • OH- acts as both a nucleophile and a base
    • OH- acts only as an acid
    • OH- acts only as a free radical
    • OH- acts only as an oxidising agent

  10. What is the mechanism of elimination of HBr from a halogenoalkane by a strong base such as OH- in ethanol?

    • Electrophilic addition to a C=C bond
    • SN1 with a carbocation only
    • Free-radical chain reaction
    • E2: the base removes a proton as the C-Br bond breaks and a C=C bond forms

  11. Which of these is the product of elimination of 2-bromobutane?

    • Butane
    • But-2-ene
    • But-1-ene only
    • Butan-2-ol

  12. Which product is formed from 1-bromopropane with ethanolic KOH by elimination?

    • Propan-2-ol
    • Propene
    • Propan-1-ol
    • Propanal

  13. What is the molecular formula of the elimination product of 2-bromopropane?

    • C3H5Br
    • C3H8
    • C3H6
    • C3H7OH

  14. Which reagent gives a mixture of alcohol and alkene from 2-bromopropane?

    • Dilute HCl only
    • Concentrated sulfuric acid only
    • Aqueous KOH at room temperature, with some ethanolic KOH
    • Chlorine gas in UV light

  15. Why is the nucleophilic substitution product more likely in aqueous solution than in ethanol?

    • Ethanol is a stronger nucleophile than water
    • Water removes all the bromine from the molecule
    • Water is a more polar solvent that favours substitution, reducing the basicity effect
    • Ethanol is an oxidising agent

  16. Why must 2-bromopropane be heated with KOH in the elimination reaction?

    • Heating removes the alkene product immediately
    • The reaction needs energy to break the C-Br bond and to remove HBr
    • The reaction is spontaneous at 0 C only
    • Heating converts KOH into a radical

  17. A student heats 2-bromopropane with ethanolic KOH. Which product is most likely to be the main organic product?

    • 2-bromopropanol
    • Propene
    • Propanone
    • Propan-2-ol

  18. Explain why 2-bromopropane gives a mixture of products when heated with aqueous KOH.

    • Hydroxide acts as both a nucleophile and a base, so substitution and elimination occur at the same time
    • Aqueous KOH is a radical initiator producing two products
    • The reaction is reversible so both products form at equilibrium
    • Potassium ions react to form two products

  19. Explain why a secondary halogenoalkane is more likely to undergo elimination than a primary halogenoalkane with a strong base.

    • Elimination is only possible for tertiary halogenoalkanes
    • Secondary halogenoalkanes contain no C-H bonds
    • Primary halogenoalkanes cannot be attacked by any nucleophile
    • Secondary halogenoalkanes have more adjacent hydrogens to remove, and the steric hindrance at the carbon reduces substitution

  20. Explain the role of the solvent in the competing reactions of 2-bromopropane with hydroxide ions.

    • The solvent is the reagent that forms the alkene
    • Ethanol reduces the nucleophilicity and basicity balance towards elimination, while water favours substitution
    • The solvent is a catalyst that is consumed completely
    • The solvent converts OH- into H+

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