Halogenoalkane substitution and elimination

What you’ll learn

• Classify halogenoalkanes and predict reaction types.
• Describe electron-pair movement in primary substitution.
• Explain hydrolysis-rate trends using bond strength.

Understand the idea

A halogenoalkane contains a halogen bonded to a saturated carbon. Classify it as primary, secondary or tertiary by the number of carbon groups attached to that carbon, not by the total carbons in the molecule. The polar carbon–halogen bond makes carbon partially positive and susceptible to a nucleophile, an electron-pair donor.

For substitution of a primary halogenoalkane by hydroxide, a curly arrow starts at an oxygen lone pair and ends at the carbon bearing the halogen; another moves the C–X bond pair onto X. Hydroxide in aqueous conditions acts as a nucleophile, replacing halide with an OH group. With ammonia, nitrogen donates a lone pair to carbon; loss of a proton from the initially formed alkylammonium ion then gives an amine.

When hydroxide acts as a base in ethanolic conditions, elimination can remove a hydrogen from a neighbouring carbon while halide leaves, forming an alkene. The syllabus also includes carbon-chain extension through substitution by cyanide to form a nitrile; this is a conceptual transformation involving highly toxic reagents and is not a home experiment.

Hydrolysis releases halide ions that can be detected by silver-halide precipitation. For comparable primary compounds, iodoalkanes usually hydrolyse faster than bromoalkanes, which are faster than chloroalkanes, because C–I is weaker than C–Br and C–Cl. Under comparable aqueous-ethanolic conditions, tertiary compounds often react faster than secondary and primary isomers. Keep temperature, concentration and solvent conditions controlled when interpreting a comparison.

Key formulae

$$\mathrm{CH_3CH_2Br+OH^-\rightarrow CH_3CH_2OH+Br^-}$$

$$\mathrm{CH_3CH_2Br+2NH_3\rightarrow CH_3CH_2NH_2+NH_4^++Br^-}$$

$$\mathrm{CH_3CH_2Br+OH^-\rightarrow CH_2{=}CH_2+H_2O+Br^-}$$

$$\mathrm{R-X+CN^-\rightarrow R-CN+X^-}$$

Worked example

Explain the electron-pair movements when hydroxide substitutes bromide in bromoethane.

1. The C–Br bond is polar, with a partially positive carbon.
2. A full-headed curly arrow starts at an oxygen lone pair of hydroxide and points to that carbon, forming the C–O bond.
3. A second arrow goes from the C–Br bond to bromine, showing bromide leaving with the pair.

Answer: Ethanol and bromide ions form by nucleophilic substitution.