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Halogenoalkanes

Relative rates of hydrolysis and carbon–halogen bond enthalpy

  • 3 things to know
  • Unit AS 2

Overview

Recall questions on the silver nitrate experiment comparing three halogenoalkanes, why ethanol is the solvent, the order of hydrolysis rates and its explanation by bond enthalpy rather than polarity, the four carbon–halogen bond enthalpies, the precipitate colours, and the variables that must be controlled.

What you need to know

Answer each one before you reveal it.

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Key term

Explain the order of the rates of hydrolysis of chloro-, bromo- and iodoalkanes in terms of bond enthalpy

Key term

Describe how the relative rates of hydrolysis of three halogenoalkanes are compared in the laboratory.

Key term

Why is ethanol used as the solvent when rates of hydrolysis are compared with aqueous silver nitrate?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Describe how the relative rates of hydrolysis of three halogenoalkanes are compared in the laboratory.

    Model answer

    Equal amounts of 1-chlorobutane, 1-bromobutane and 1-iodobutane are added to separate tubes containing ethanol and aqueous silver nitrate at the same temperature, and the time taken for a precipitate to appear in each tube is measured.

  2. 2

    Why is ethanol used as the solvent in this experiment?

    Model answer

    Halogenoalkanes do not dissolve in water, and ethanol dissolves both the halogenoalkane and the aqueous silver nitrate so that the two can mix and react.

  3. 3

    Which of the three halogenoalkanes gives a precipitate first?

    Model answer

    1-iodobutane gives its precipitate first, because it is hydrolysed the most rapidly of the three.

  4. 4

    State the order of the rate of hydrolysis of primary halogenoalkanes with different halogen atoms.

    Model answer

    The order of decreasing rate is the iodo compound, then the bromo compound, then the chloro compound.

  5. 5

    Explain the order of the rates of hydrolysis in terms of bond enthalpy.

    Model answer

    The carbon–iodine bond has the lowest bond enthalpy of the three and is broken most easily, while the carbon–chlorine bond has the highest and is the hardest to break, so the rate follows the ease with which the carbon–halogen bond breaks.

  6. 6

    Give approximate values for the bond enthalpies of the four carbon–halogen bonds.

    Model answer

    They are about 467 kJ mol⁻¹ for C–F, about 346 kJ mol⁻¹ for C–Cl, about 290 kJ mol⁻¹ for C–Br and about 228 kJ mol⁻¹ for C–I.

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