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Halogenoalkanes: Question 3

Syllabus 31.1

Multiple choice A2 1 mark

The table gives approximate C–Cl bond data for two compounds:

Compound C–Cl bond length / pm C–Cl bond enthalpy / kJ mol⁻¹
Chloromethane, CH3Cl\text{CH}_3\text{Cl} 178 338
Chlorobenzene, C6H5Cl\text{C}_6\text{H}_5\text{Cl} 169 399

1-Chlorooctane and chlorobenzene are separately heated under reflux with aqueous sodium hydroxide. 1-Chlorooctane is readily hydrolysed under these conditions, but chlorobenzene shows no detectable reaction.

Which statement correctly uses the bond data to explain this difference in reactivity?

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

Interpreting the bond data

Chlorobenzene’s C–Cl bond is both shorter (169 pm vs 178 pm) and stronger (399 kJ mol⁻¹ vs 338 kJ mol⁻¹) than the C–Cl bond in chloromethane (a good model for the bond in any simple chloroalkane, such as 1-chlorooctane). A shorter, stronger bond needs more energy to break, not less, so this data alone already rules out chlorobenzene reacting faster.

Why the bond is shorter and stronger: delocalisation

In chlorobenzene, the chlorine atom’s carbon is part of the aromatic ring, and one of chlorine’s lone pairs of electrons occupies a p-orbital that can overlap with the ring’s π\pi system. This lone pair partially delocalises into the ring, giving the C–Cl bond some double-bond character. Two consequences follow:

  • The bond becomes shorter and stronger (matching the higher bond enthalpy and shorter bond length in the table), so it is harder to break homolytically or heterolytically.
  • The delocalisation also reduces the electron density that would otherwise be concentrated at the carbon atom bonded to chlorine, so there is less positive character on that carbon for a nucleophile such as OH−\text{OH}^- to be attracted to and attack.

By contrast, in 1-chlorooctane the chlorine’s lone pairs cannot delocalise into anything (there is no adjacent π\pi system), so the C–Cl bond remains a normal, weaker single bond, and the carbon retains enough partial positive character to be attacked readily by OH−\text{OH}^-, which is exactly why 1-chlorooctane is hydrolysed under these conditions while chlorobenzene is not.

Why the other options are wrong

  • B gets the energetics backwards: a stronger bond requires more, not less, energy to break.
  • C is factually wrong. Chlorine remains considerably more electronegative than carbon even in chlorobenzene, so the C–Cl bond is still polar; delocalisation reduces the electron deficiency at the ring carbon, but it does not make the bond non-polar.
  • D misidentifies the reason as steric: the accepted explanation is electronic (delocalisation strengthening the bond and lowering the electrophilicity of the carbon), not the ring being “too bulky” to approach. The ring carbon bonded to chlorine is no more sterically hindered than many other reactive carbons.

Final answer

A, delocalisation of a chlorine lone pair into the aromatic ring shortens and strengthens the C–Cl bond and lowers the electron deficiency at that carbon, so chlorobenzene resists nucleophilic substitution under conditions that readily hydrolyse a chloroalkane such as 1-chlorooctane.