AQA · A-Level · Chemistry
Course version: 7405 · Version from Sept 2016
Halogenoalkanes
Master the reactions and mechanisms of Halogenoalkanes! This guide covers everything from nucleophilic substitution and elimination to the environmental impact of CFCs on the ozone layer.
- 5 min read
- 3 worked examples
- 5 practice questions
- 4 key terms
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The start of the Halogenoalkanes lesson, made from checked teaching notes for this specification point.
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Ammonia reacts with bromoethane in a nucleophilic substitution. Which product is formed in the first substitution step?
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The halogen is more electronegative than carbon, so the carbon it is bonded to carries a δ⁺ charge and attracts a lone pair. Ammonia attacks that carbon, the carbon-halogen bond breaks heterolytically and the halide ion leaves, giving an alkylammonium salt. A second ammonia molecule then removes a proton, freeing the primary amine. But that amine still has a lone pair, so it attacks another halogenoalkane, and the sequence runs on through the secondary amine and the tertiary amine to the quaternary ammonium salt, which has four alkyl groups, no N-H bond and so no proton left to lose. The product ratio is set by the conditions: excess ammonia favours the primary amine, excess halogenoalkane the quaternary salt. A strong answer earns marks for these points. the carbon being δ⁺ because the halogen is more electronegative. the halide ion identified as the leaving group. a second molecule of ammonia or amine removing H⁺ to free the amine. the quaternary ammonium salt having four alkyl groups and no N-H bond. Exam tip: Draw all four organic products in order when the question says 'a mixture of products is formed'; the marks are usually one per species.
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Key words
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leaving group
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further substitution
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quaternary ammonium salts
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controlling the product
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- leaving group
- The halide ion, which leaves when the carbon-halogen bond breaks heterolytically as ammonia attacks the carbon.
- further substitution
- The sequence runs on because the amine still has a lone pair, so it attacks another halogenoalkane, through the secondary and tertiary amine to the quaternary ammonium salt.
- quaternary ammonium salts
- A salt with four alkyl groups and no N-H bond, so there is no proton left to lose.
- controlling the product
- The product ratio is set by the conditions: excess ammonia favours the primary amine, excess halogenoalkane the quaternary salt.
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Study Notes
Overview
Halogenoalkanes are organic compounds where one or more hydrogen atoms in an alkane have been replaced by a halogen atom (fluorine, chlorine, bromine, or iodine). They are incredibly important in organic synthesis because they are much more reactive than alkanes.
This reactivity stems from the polar carbon-halogen bond. Because halogens are more electronegative than carbon, the carbon atom becomes electron-deficient (\delta+), making it a prime target for nucleophiles. Understanding these mechanisms is crucial, as they form the foundation for many synthetic pathways in chemistry.
In exams, you will frequently be asked to draw mechanisms, explain reactivity trends based on bond enthalpies, and compare substitution versus elimination reactions. You'll also need to know the environmental impact of chlorofluorocarbons (CFCs).
Key Concepts
Concept 1: Nucleophilic Substitution ($S_N2$)
Because the carbon attached to the halogen is \delta+, it is susceptible to attack by nucleophiles (electron pair donors). The nucleophile donates a lone pair to the carbon, forming a new bond, while the carbon-halogen bond breaks heterolytically, releasing the halide ion (the leaving group).
Key Reactions:
- With aqueous NaOH/KOH: Forms alcohols. The OH^- ion acts as the nucleophile. Conditions: warm, aqueous solvent.
- With KCN in ethanol: Forms nitriles. The CN^- ion is the nucleophile. Conditions: warm, ethanolic solvent. Examiner Tip: This reaction is vital because it extends the carbon chain by one carbon atom!
- With excess NH_3: Forms primary amines. The NH_3 molecule is the nucleophile. Conditions: heated in a sealed tube.
Concept 2: Elimination Reactions
Under different conditions, the hydroxide ion (OH^-) can act as a base (proton acceptor) rather than a nucleophile. It removes a hydrogen ion (H^+) from a carbon atom adjacent to the C-X bond. The electrons from the C-H bond form a carbon-carbon double bond, and the halide ion leaves.
Key Conditions:
- Reagent: NaOH or KOH
- Solvent: Ethanol (no water present)
- Conditions: Hot (reflux)
- Product: Alkene
Examiner Tip: The solvent is the critical difference! Aqueous = Substitution (Alcohol). Ethanolic = Elimination (Alkene).
Concept 3: Bond Enthalpy and Reactivity
The rate of reaction of halogenoalkanes depends on the strength of the carbon-halogen bond (bond enthalpy), NOT the polarity of the bond.
- C-F bond: Strongest bond (highest bond enthalpy). Fluoroalkanes are the least reactive.
- C-I bond: Weakest bond (lowest bond enthalpy). Iodoalkanes are the most reactive.
Reactivity Trend: Iodoalkanes > Bromoalkanes > Chloroalkanes > Fluoroalkanes.
Concept 4: Ozone Depletion by CFCs
Chlorofluorocarbons (CFCs) were widely used as refrigerants and propellants because they are unreactive under normal conditions. However, in the upper atmosphere, UV radiation provides enough energy to break the C-Cl bond homolytically, forming chlorine radicals (Cl\bullet).
These chlorine radicals act as catalysts in the breakdown of ozone (O_3) into oxygen (O_2).
The Catalytic Cycle:
- Cl\bullet + O_3 \rightarrow ClO\bullet + O_2
- ClO\bullet + O_3 \rightarrow 2O_2 + Cl\bullet
Overall equation: 2O_3 \rightarrow 3O_2
The chlorine radical is regenerated, meaning a single CFC molecule can destroy thousands of ozone molecules.
Practical Applications
**Testing the Rate of Hydrolysis (Required Practical Context)**You can compare the reactivity of different halogenoalkanes by hydrolysing them in the presence of silver nitrate solution (AgNO_3(aq)). The water acts as a weak nucleophile.
RX + H_2O \rightarrow ROH + H^+ + X^-
As the halide ions (X^-) are produced, they react with the silver ions (Ag^+) to form a precipitate of silver halide (AgX).
- Iodoalkane: Yellow precipitate (AgI) forms rapidly.
- Bromoalkane: Cream precipitate (AgBr) forms slower.
- Chloroalkane: White precipitate (AgCl) forms very slowly.
This proves experimentally that the C-I bond breaks most easily.
Podcast Episode
Listen to our 10-minute deep dive on Halogenoalkanes for expert tips and a quick-fire recall quiz!
Worked Examples
3 worked examples — open one to explore the question and available guidance.
Practice Questions
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State the reagent and conditions required to convert 1-bromobutane into butanenitrile.
Hint: Think about how to add a carbon atom to the chain.
Explain why chlorofluorocarbons (CFCs) are damaging to the ozone layer. Include relevant equations in your answer.
Hint: What does UV light do to the C-Cl bond? How does the resulting species react with ozone?
A student reacts 2-bromo-2-methylpropane with hot ethanolic potassium hydroxide. Name the mechanism, state the role of the hydroxide ion, and draw the structure of the organic product.
Hint: Look closely at the solvent. What type of reaction happens in ethanol?
Compare the rates of hydrolysis of 1-chlorobutane, 1-bromobutane, and 1-iodobutane. Explain your answer.
Hint: Which bond is the strongest? Which is the weakest?
When 1-bromopropane reacts with ammonia, the product is propan-1-amine. Why must excess ammonia be used?
Hint: What happens to the HBr produced in the reaction?
Exam technique
How to turn what you know into marks
How to approach the question
Always check the solvent conditions first! If you see 'NaOH', immediately look for 'aq' or 'ethanol' to determine if it's substitution or elimination.
How to structure your answer
For mechanism questions: Draw the organic molecule first. Add $\delta+$ and $\delta-$ charges. Draw the attacking species with its lone pair. Then draw the curly arrows.
Timing
Approximately 1 minute per mark, so a 6-mark mechanism and explanation question should take about 6-7 minutes.
Common pitfalls
- Forgetting to draw the lone pair on the nucleophile before drawing the curly arrow.
- Starting a curly arrow from a positive charge (arrows show movement of negative electrons!).
- Confusing the role of $OH^-$: it's a nucleophile in aqueous conditions, but a base in ethanolic conditions.
What the command words want
- Compare
- Identify similarities AND differences - use comparative language.
- Explain
- Give the reason why - e.g., 'Because the C-I bond has a lower bond enthalpy'.
- Describe
- Say what you would observe - e.g., 'A yellow precipitate forms'.
- Evaluate
- Consider both sides, use evidence, make a judgement.
- Calculate
- Show working, use correct formula, include units.
- State/Give
- Brief factual answer - e.g., 'Nucleophile'. No explanation needed.
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