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9701 · 15.1

Halogenoalkanes — practice questions

Practice and worked examples for 9701 Halogenoalkanes. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Draw the mechanism for the reaction between bromoethane (CH3CH2BrCH_3CH_2Br) and a hydroxide ion (OH−OH^−). Include all relevant dipoles, charges, and curly arrows.

Show solution outline

This is an SN2 mechanism as bromoethane is a primary halogenoalkane.

Step 1: Show the polarity of the C-Br bond. The nucleophile, OH−OH^−, is attracted to the δ+ carbon atom.

Step 2: Draw a curly arrow from the lone pair on the oxygen of the OH−OH^− ion to the δ+ carbon atom of bromoethane. This represents the formation of a new C-O bond.

Step 3: Simultaneously, draw a curly arrow from the C-Br bond to the Br atom. This represents the breaking of the C-Br bond, with the bromine leaving as a bromide ion (Br−Br^−).

Mechanism Diagram: OH−+CH3CH2δ+−Brδ−→[HOδ−⋅⋅⋅CH2(CH3)⋅⋅⋅Brδ−]−→HOCH2CH3+Br−OH^− + CH_3CH_2^{\delta+}-Br^{\delta−} \rightarrow [HO^{\delta−} \cdot\cdot\cdot CH_2(CH_3) \cdot\cdot\cdot Br^{\delta−}]^− \rightarrow HOCH_2CH_3 + Br^−

  • The curly arrow from the O in OH−OH^− points to the C of the CH2CH_2 group.
  • A second curly arrow goes from the C-Br bond onto the Br atom.
  • The product, ethanol, shows an inverted configuration compared to the reactant (though not easily visible without stereochemistry).
  • The species in square brackets is the transition state.

Worked example 2

Draw the mechanism for the reaction of 2-bromo-2-methylpropane, (CH3)3CBr(CH_3)_3CBr, with water. Explain why this is an SN1 mechanism.

Show solution outline

This reaction proceeds via an SN1 mechanism because 2-bromo-2-methylpropane is a tertiary halogenoalkane, which forms a stable tertiary carbocation.

Step 1 (slow): The C-Br bond breaks heterolytically to form a tertiary carbocation and a bromide ion. (CH3)3C−Br→(CH3)3C++Br−(CH_3)_3C-Br \rightarrow (CH_3)_3C^+ + Br^−

  • Draw a curly arrow from the C-Br bond to the Br atom.

Step 2 (fast): A water molecule (the nucleophile) attacks the planar carbocation. A lone pair on the oxygen atom forms a bond with the positive carbon. (CH3)3C++H2O→(CH3)3C−OH2+(CH_3)_3C^+ + H_2O \rightarrow (CH_3)_3C-OH_2^+

  • Draw a curly arrow from the lone pair on the oxygen of H2OH_2O to the C+C^+ atom.

Step 3 (fast): The intermediate loses a proton (H+H^+) to another water molecule to form the final product, 2-methylpropan-2-ol. (CH3)3C−OH2++H2O→(CH3)3C−OH+H3O+(CH_3)_3C-OH_2^+ + H_2O \rightarrow (CH_3)_3C-OH + H_3O^+

  • Draw a curly arrow from the O-H bond in the intermediate onto the oxygen atom.

This mechanism is favoured because the intermediate (CH3)3C+(CH_3)_3C^+ is a tertiary carbocation, stabilised by the electron-donating effect of the three methyl groups.

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