Rates of Hydrolysis and Bond Enthalpy
A concise revision guide to comparing the rates of hydrolysis of halogenoalkanes with aqueous silver nitrate in ethanol: the precipitate colours and times, the chloro-, bromo- and iodo- trend explained by bond enthalpy, and, as an extension, the primary, secondary and tertiary trend.
- 3.3.3.1iv
What these spec points say
- 3.3.3.1iv explain why the carbon-halogen bond enthalpy influences the rate of reaction
GCSE Recap: The Halide Test
Type a short answer to each question on the GCSE halide test.
Comparing Rates with Silver Nitrate
The rate at which different halogenoalkanes are hydrolysed is compared with aqueous silver nitrate in ethanol.
Equal amounts of each halogenoalkane are placed in test tubes with ethanol and warmed in a water bath at about 50 °C.
Aqueous silver nitrate at the same temperature is added to each at the same moment, and the time taken for a precipitate to appear is recorded.
Water is the nucleophile: R–X + H₂O → R–OH + H⁺ + X⁻. The halide ion released reacts with silver ions, Ag⁺ + X⁻ → AgX, and the sooner the precipitate forms, the faster the hydrolysis.
The ethanol is a co-solvent that lets the water-insoluble halogenoalkane mix with the aqueous silver nitrate.
The water bath keeps every tube at the same temperature so that the comparison is fair.
The colour of the precipitate confirms which halogen has left: white silver chloride, cream silver bromide, yellow silver iodide.
The class practical comparing rates of hydrolysis follows exactly this method; control the volumes, concentrations and temperature and start the clock when the silver nitrate is added.
Three halogenoalkanes in aqueous silver nitrate and ethanol in a water bath: the precipitate colours, the times, and the bond enthalpies that explain them. The Structure row is an extension beyond the AQA specification.
Key idea: Faster precipitate means faster hydrolysis. Chloro slowest, iodo fastest.
Quick Check: The Silver Nitrate Experiment
Answer the questions on the method and the first precipitate.
Chloro, Bromo and Iodo: the Bond Enthalpy Trend
For primary compounds with the same carbon chain, the order of reactivity is iodoalkane > bromoalkane > chloroalkane.
1-iodobutane gives a yellow precipitate within seconds, 1-bromobutane a cream precipitate in a minute or two, and 1-chlorobutane only a faint white cloudiness after many minutes. Fluoroalkanes give no precipitate at all.
The explanation is the bond enthalpy of the carbon–halogen bond: C–Cl 346, C–Br 290, C–I 228 kJ mol⁻¹.
The rate-determining step involves breaking the C–X bond, so the weaker the bond, the lower the activation energy and the faster the reaction.
Bond polarity would predict the opposite order, because C–Cl is the most polar of the three.
Exam focus: The exam answer must be about bond strength: the C–I bond is the longest and weakest and breaks most easily.
| Compound | C–X bond enthalpy / kJ mol⁻¹ | Precipitate | Relative rate |
|---|---|---|---|
| 1-chlorobutane | 346 | white AgCl, very slow | slowest |
| 1-bromobutane | 290 | cream AgBr, a few minutes | middle |
| 1-iodobutane | 228 | yellow AgI, seconds | fastest |
Exam answer model: “The C–I bond has the lowest bond enthalpy, so less energy is needed to break it, the activation energy is lower and hydrolysis is fastest. Bond polarity does not explain the trend.”
Quick Check: Explaining the Halogen Trend
Write a short explanation of the trend using the bond enthalpies.
Fair Tests: One Factor at a Time
Questions on this experiment often give times for several compounds and ask whether the comparison is fair.
To compare the halogens, use primary compounds with the same carbon chain, such as 1-chlorobutane, 1-bromobutane and 1-iodobutane, so that the C–X bond is the only thing that changes.
Keep the volume and concentration of silver nitrate, the amount of halogenoalkane, the solvent and the temperature the same in every tube.
Exam tip: Change one thing at a time. If two compounds differ in both the halogen and the carbon skeleton, the difference in time cannot be put down to the C–X bond alone.
Exam focus: State the trend, state the evidence (time for the precipitate), then give the reason (the C–X bond enthalpy).
Extension: beyond the AQA specification
For the same halogen, the order of reactivity is tertiary > secondary > primary. 2-bromo-2-methylpropane gives a precipitate almost at once at room temperature, 2-bromopropane takes longer, and 1-bromopropane is the slowest.
A tertiary compound can lose its halide ion first to form a carbocation, which is stabilised by the three alkyl groups around the positive carbon, and water then attacks the carbocation rapidly.
A primary compound cannot form a stable carbocation, so it must wait for the water molecule to attack the crowded δ+ carbon directly, which is slower.
To test the structure fairly, use three bromo compounds with the same number of carbons. You may meet this in wider reading; AQA does not examine it.
For the same halogen a tertiary compound is hydrolysed fastest and a primary compound slowest; a fair test changes only the halogen or only the structure.
Quick Check: Fair Tests
Answer each quick question on fair testing.
Quick Check: Halogen Trend and Fair Testing
Pick the one accurate statement in each round to check the whole page.
Common Exam Points
Say
“Aqueous silver nitrate in ethanol, water bath at the same temperature, time the precipitate.” “Cream precipitate of silver bromide.” “C–I weakest, so fastest.”
Do not say
“Silver nitrate is the nucleophile” (water is). “Iodoalkanes are fastest because iodine is the most reactive halogen.”
Watch for
Fair-test questions: same volume and concentration of silver nitrate, same amount of halogenoalkane, same temperature, same solvent.
FAQs
Use these quick answers to check the hydrolysis experiment.
Why is ethanol added to the test tubes?
The halogenoalkane does not dissolve in water and the silver nitrate does not dissolve in the halogenoalkane. Ethanol mixes with both, so the reactants meet in one solution.
What is the nucleophile in this experiment?
Water. The silver ions do not take part in the hydrolysis; they react with the halide ion after it has been released to form the precipitate.
Why is a water bath used?
To hold every tube at the same temperature, so that the rates can be compared fairly and the reaction is fast enough to time.
Why does the C–I bond break most easily if it is the least polar?
Bond enthalpy falls as the halogen atom gets larger and the bond longer: C–I is only 228 kJ mol⁻¹ against 346 for C–Cl. Less energy is needed to break it, so the activation energy is lower.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
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