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Halogenoalkanes

Bond polarity, bond enthalpy and the ease of substitution

  • 3 things to know
  • Components 1 and 2

Overview

Trend in carbon-halogen bond enthalpy down the group and why fluoroalkanes are almost completely unreactive towards nucleophiles.

What you need to know

Answer each one before you reveal it.

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Key term

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

Key term

Explain why the C—F bond is the most polar of the four carbon-halogen bonds.

Key term

Place 1-chlorobutane, 1-bromobutane and 1-iodobutane in order of increasing rate of hydrolysis.

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.

    Model answer

    The order is C—F, then C—Cl, then C—Br, then C—I.

  2. 2

    Explain why the C—F bond is the most polar of the four carbon-halogen bonds.

    Model answer

    Fluorine is the most electronegative element of the group, so it draws the bonding pair furthest away from the carbon.

  3. 3

    State the approximate bond enthalpies of the C—F, C—Cl, C—Br and C—I bonds.

    Model answer

    They are about 467, 346, 290 and 228 kJ mol⁻¹ respectively.

  4. 4

    Explain the trend in carbon-halogen bond enthalpy down the group.

    Model answer

    The halogen atom gets larger down the group so the shared pair is further from its nucleus and the overlap is poorer, which makes the bond weaker.

  5. 5

    Place 1-chlorobutane, 1-bromobutane and 1-iodobutane in order of increasing rate of hydrolysis.

    Model answer

    The order is 1-chlorobutane, then 1-bromobutane, then 1-iodobutane.

  6. 6

    Which factor, bond polarity or bond enthalpy, controls the rate of substitution of a halogenoalkane?

    Model answer

    Bond enthalpy controls the rate, because the carbon-halogen bond has to be broken for the reaction to happen at all.

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