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How do alkenes undergo addition and haloalkanes undergo substitution?

Describe the addition reactions of alkenes and the substitution reactions of haloalkanes, writing equations and products.

How alkenes undergo addition reactions across the C=C double bond and how haloalkanes undergo substitution, with equations for hydrogenation, halogenation, hydration and hydrolysis.

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  1. What this dot point is asking
  2. Why alkenes add and haloalkanes substitute
  3. Addition reactions of alkenes
  4. Substitution reactions of haloalkanes

What this dot point is asking

You must write equations for the characteristic addition reactions of alkenes and substitution reactions of haloalkanes and name the products.

Why alkenes add and haloalkanes substitute

The C=C double bond of an alkene is a region of high electron density; the relatively weak Ο€ bond breaks easily so that two new single bonds form - an addition reaction. In a haloalkane the C-X bond is polar (the halogen is more electronegative), leaving the carbon slightly positive and open to attack by a nucleophile, which substitutes for the halogen.

Addition reactions of alkenes

Hydrogenation - adding H2\text{H}_2 across the double bond (nickel catalyst) gives an alkane:

CH2=CH2+H2β†’CH3CH3\text{CH}_2{=}\text{CH}_2 + \text{H}_2 \rightarrow \text{CH}_3\text{CH}_3

Halogenation - adding a halogen such as bromine gives a dihaloalkane:

CH2=CH2+Br2β†’CH2BrCH2Br\text{CH}_2{=}\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{BrCH}_2\text{Br}

Hydration - adding water (steam, acid catalyst) gives an alcohol:

CH2=CH2+H2O→CH3CH2OH\text{CH}_2{=}\text{CH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{OH}

Adding a hydrogen halide gives a haloalkane:

CH2=CH2+HBr→CH3CH2Br\text{CH}_2{=}\text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br}

Substitution reactions of haloalkanes

When a haloalkane is warmed with aqueous sodium hydroxide, the hydroxide ion (a nucleophile) replaces the halogen, producing an alcohol. This is a hydrolysis (substitution):

CH3CH2Br+OHβˆ’β†’CH3CH2OH+Brβˆ’\text{CH}_3\text{CH}_2\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{Br}^-