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Halogenoalkanes

Reactivity of the carbon–halogen bond and the silver nitrate test

  • 3 things to know
  • AS Unit 2

Overview

28 recall questions on reactivity of the carbon–halogen bond and the silver nitrate test.

What you need to know

Answer each one before you reveal it.

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Key term

Fluoroalkanes are in fact the least reactive halogenoalkanes towards hydroxide ions. Explain this apparent contradiction.

Key term

Why is ethanol added to the mixture in the silver nitrate test?

Key term

Why must sodium hydroxide not be used in place of water in the silver nitrate test?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Place the C–F, C–Cl, C–Br and C–I bonds in order of decreasing polarity, and explain the order.

    Model answer

    The order is C–F, then C–Cl, then C–Br, then C–I, because electronegativity falls down the group from fluorine at 4.0 through chlorine at 3.0 and bromine at 2.8 to iodine at 2.5, so the difference from carbon at 2.5 gets smaller and the C–I bond is essentially non-polar.

  2. 2

    Place the C–F, C–Cl, C–Br and C–I bonds in order of decreasing bond enthalpy.

    Model answer

    The order is C–F at about 467 kJ mol⁻¹, then C–Cl at about 346 kJ mol⁻¹, then C–Br at about 290 kJ mol⁻¹, then C–I at about 228 kJ mol⁻¹, since the halogen atom gets larger down the group so the shared pair is further from its nucleus and the overlap is poorer.

  3. 3

    If bond polarity alone decided the rate of hydrolysis, which halogenoalkane would react fastest?

    Model answer

    A fluoroalkane would react fastest, because the C–F bond is by far the most polar and so has the most strongly δ+ carbon for a nucleophile to attack.

  4. 4

    Fluoroalkanes are in fact the least reactive halogenoalkanes towards hydroxide ions. Explain this apparent contradiction.

    Model answer

    Bond polarity governs how strongly the carbon attracts the nucleophile, but the slow step of the reaction requires the carbon–halogen bond to break, and the C–F bond is by far the strongest of the four at about 467 kJ mol⁻¹, so bond enthalpy rather than polarity is the factor that controls the rate.

  5. 5

    Arrange 1-chlorobutane, 1-bromobutane and 1-iodobutane in order of increasing rate of hydrolysis and justify the order.

    Model answer

    1-Chlorobutane is slowest, 1-bromobutane is intermediate and 1-iodobutane is fastest, because the carbon–halogen bond enthalpy falls from about 346 kJ mol⁻¹ for C–Cl to about 290 kJ mol⁻¹ for C–Br and about 228 kJ mol⁻¹ for C–I, so less energy is needed to break the bond and the activation energy is lower.

  6. 6

    Why does the carbon–halogen bond get weaker as you go down the halogen group?

    Model answer

    The halogen atom becomes larger, so its bonding electrons lie further from the nucleus and are shielded by more inner shells; the orbitals overlapping with carbon are also more diffuse, giving poorer overlap and a longer, weaker bond.

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