Describe the nucleophilic substitution and elimination reactions of halogenoalkanes, distinguish the SN1 and SN2 mechanisms, relate the mechanism to the class of halogenoalkane, and explain the relative rates of hydrolysis of the halogenoalkanes

What this dot point is asking

SEAB wants you to describe the nucleophilic substitution and elimination reactions of halogenoalkanes, distinguish the SN1 and SN2 mechanisms, relate the mechanism to the class of halogenoalkane (primary, secondary, tertiary), and explain the trend in hydrolysis rates. The SN1/SN2 distinction and the hydrolysis-rate explanation are core organic exam content.

The answer

The reactive site

The carbon-halogen bond is polar ($\text{C}^{\delta+}\text{-X}^{\delta-}$) because the halogen is more electronegative. The slightly positive carbon is open to attack by nucleophiles (electron-pair donors such as $\text{OH}^-$, $\text{CN}^-$, $\text{NH}_3$).

Nucleophilic substitution

A nucleophile replaces the halogen:

• with aqueous $\text{OH}^-$ (warm): gives an alcohol (hydrolysis).
• with $\text{CN}^-$ (in ethanol): gives a nitrile, adding a carbon to the chain (useful in synthesis).
• with excess $\text{NH}_3$ (in ethanol, heat, sealed): gives an amine.

Elimination

With hot ethanolic (not aqueous) $\text{OH}^-$, the halogenoalkane undergoes elimination, losing HX to form an alkene:

Aqueous conditions favour substitution; hot ethanolic conditions favour elimination.

SN2 mechanism (primary halogenoalkanes)

The nucleophile attacks the carbon from the side opposite the halogen in a single concerted step. A single transition state has both the nucleophile and the leaving group partially bonded to carbon. The rate depends on both the halogenoalkane and the nucleophile concentrations (bimolecular):

SN1 mechanism (tertiary halogenoalkanes)

The C-X bond breaks first (slow step) to form a carbocation, which the nucleophile then attacks (fast step). The rate depends only on the halogenoalkane (unimolecular):

Tertiary halogenoalkanes favour SN1 because the tertiary carbocation is stabilised by the inductive effect of three alkyl groups; primary halogenoalkanes favour SN2 because the carbon is open to back-side attack and a primary carbocation is too unstable. Secondary halogenoalkanes can go by either route.

Rates of hydrolysis

The rate of hydrolysis depends mainly on the carbon-halogen bond strength, not its polarity. Down the group the bond gets weaker (longer, electrons further from the nuclei):

So iodoalkanes hydrolyse fastest even though the C-Cl bond is the most polar; the ease of breaking the weaker C-I bond outweighs the polarity.

Examples in context

Example 1. Extending a carbon chain with cyanide. Reacting a halogenoalkane with potassium cyanide in ethanol substitutes the halogen with a nitrile group, adding one carbon to the chain. The nitrile can then be hydrolysed to a carboxylic acid or reduced to an amine, making this a key chain-lengthening step that SEAB tests in multi-step synthesis questions.

Example 2. Kinetic evidence for the mechanism. Measuring how the hydrolysis rate changes with nucleophile concentration distinguishes the mechanisms: a rate independent of $[\text{OH}^-]$ indicates SN1 (tertiary), while a rate proportional to $[\text{OH}^-]$ indicates SN2 (primary). This links the organic mechanism directly to the rate-equation work from Physical Chemistry.

Try this

Q1. State the reagent and conditions to convert bromoethane into (a) ethanol and (b) ethene. [2 marks]

• Cue. (a) Warm aqueous NaOH (substitution). (b) Hot ethanolic NaOH (elimination).

Q2. Explain why 2-bromo-2-methylpropane hydrolyses faster than 1-bromobutane. [3 marks]

• Cue. The tertiary substrate goes by SN1 via a stable tertiary carbocation; the primary substrate goes by slower SN2; the tertiary carbocation forms readily, speeding the reaction.

Q3. Place chloroethane, bromoethane and iodoethane in order of increasing rate of hydrolysis and explain. [2 marks]

• Cue. Chloroethane < bromoethane < iodoethane; the C-X bond weakens down the group, so the weakest C-I bond breaks fastest.