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How do halogenoalkanes react and how does their structure affect rate?

Nucleophilic substitution and elimination of halogenoalkanes, the effect of carbon-halogen bond strength on rate, and the environmental impact of CFCs on the ozone layer.

A focused answer to WJEC A-Level Chemistry Unit 2, covering nucleophilic substitution and elimination of halogenoalkanes, how carbon-halogen bond strength controls reactivity, and the role of CFCs in ozone depletion.

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What this dot point is asking

WJEC wants you to describe the nucleophilic substitution and elimination reactions of halogenoalkanes, explain how carbon-halogen bond strength controls the rate of hydrolysis, and discuss the role of CFCs in depleting the ozone layer.

The answer

Nucleophilic substitution

Hydrolysis with aqueous NaOH\text{NaOH} gives an alcohol; reaction with KCN\text{KCN} gives a nitrile (extending the carbon chain); reaction with excess ammonia gives an amine.

Bond strength and rate

Elimination

With hot ethanolic (alcoholic) KOH\text{KOH}, a halogenoalkane undergoes elimination instead, losing HX\text{HX} to form an alkene. The same reagent in aqueous solution favours substitution; in ethanol it favours elimination.

CFCs and ozone

CFCs are very stable, so they reach the stratosphere where UV light breaks a Cβˆ’Cl\text{C}-\text{Cl} bond, releasing chlorine radicals. A chlorine radical catalyses the breakdown of ozone: Clβ‹…+O3β†’ClOβ‹…+O2\text{Cl}\cdot + \text{O}_3 \rightarrow \text{ClO}\cdot + \text{O}_2, then ClOβ‹…+Oβ†’Clβ‹…+O2\text{ClO}\cdot + \text{O} \rightarrow \text{Cl}\cdot + \text{O}_2, regenerating the radical so one atom destroys many ozone molecules.

Useful products from nucleophilic substitution

Each nucleophile gives a different and useful product, so substitution is a versatile synthetic tool. Hydroxide gives an alcohol, opening the route to oxidation products. Cyanide gives a nitrile, lengthening the carbon chain by one and providing a group that can be reduced to an amine or hydrolysed to a carboxylic acid. Excess ammonia gives an amine. Knowing which reagent and conditions deliver each product lets you plan multi-step syntheses, the skill drawn on later in the course.

Hydrolysis rate as practical evidence

The order of Cβˆ’X\text{C}-\text{X} bond strength can be shown experimentally by warming the three halogenoalkanes with aqueous silver nitrate in ethanol and timing the appearance of the silver halide precipitate. The iodoalkane reacts fastest (cream-yellow AgI\text{AgI} appears first), then the bromoalkane (cream AgBr\text{AgBr}), then the chloroalkane (white AgCl\text{AgCl} slowest). This confirms that the weaker Cβˆ’I\text{C}-\text{I} bond breaks most readily, the controlling factor being bond strength rather than bond polarity.

Examples in context

The Montreal Protocol. Recognising that chlorine radicals from CFCs catalytically destroy ozone led to a worldwide phase-out of CFCs, an example of chemistry informing policy. Nitriles in synthesis. Substitution with cyanide adds a carbon and creates a nitrile that can be reduced to an amine or hydrolysed to a carboxylic acid, a key chain-lengthening step in synthesis.

Try this

Q1. Name the organic product when 1-bromopropane reacts with aqueous sodium hydroxide. [1 mark]

  • Cue. Propan-1-ol.

Q2. State the reagent and condition that convert 2-bromobutane into an alkene. [1 mark]

  • Cue. Hot ethanolic potassium hydroxide.

Q3. Write the equation for the chlorine radical reacting with ozone. [1 mark]

  • Cue. Clβ‹…+O3β†’ClOβ‹…+O2\text{Cl}\cdot + \text{O}_3 \rightarrow \text{ClO}\cdot + \text{O}_2.

Q4. Name the product when a halogenoalkane reacts with potassium cyanide. [1 mark]

  • Cue. A nitrile (the chain is lengthened by one carbon).

Q5. State the order of hydrolysis rate for 1-chloro-, 1-bromo- and 1-iodobutane, fastest first. [1 mark]

  • Cue. Iodo, then bromo, then chloro (weakest Cβˆ’X\text{C}-\text{X} bond fastest).

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20194 marksDescribe the mechanism of the reaction between 1-bromobutane and aqueous sodium hydroxide, including a relevant equation and the role of the hydroxide ion.
Show worked answer β†’

The hydroxide ion acts as a nucleophile because it has a lone pair and a negative charge.

The carbon-bromine bond is polar (CΞ΄+βˆ’BrΞ΄βˆ’\text{C}^{\delta+}-\text{Br}^{\delta-}), so the carbon is electron-deficient. The hydroxide lone pair attacks this carbon.

A curly arrow goes from the OHβˆ’\text{OH}^- lone pair to the carbon, and another from the Cβˆ’Br\text{C}-\text{Br} bond to the bromine, which leaves as Brβˆ’\text{Br}^-.

Overall: CH3CH2CH2CH2Br+OHβˆ’β†’CH3CH2CH2CH2OH+Brβˆ’\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{Br}^-.

Markers reward the nucleophile with lone pair, attack on the Ξ΄+\delta+ carbon, the leaving bromide, and the alcohol product.

WJEC 20203 marksExplain why 1-iodobutane is hydrolysed faster than 1-chlorobutane.
Show worked answer β†’

The rate of hydrolysis depends on the strength of the carbon-halogen bond, not on bond polarity.

The Cβˆ’I\text{C}-\text{I} bond is weaker than the Cβˆ’Cl\text{C}-\text{Cl} bond because iodine is larger and the shared electrons are further from the nuclei.

The weaker Cβˆ’I\text{C}-\text{I} bond breaks more easily, so 1-iodobutane reacts faster.

Markers reward identifying bond strength (not polarity) as the controlling factor and the weaker Cβˆ’I\text{C}-\text{I} bond breaking more readily.

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