CP 5 Hydrolysis of Halogenoalkanes
In Core Practical 5 you compare how quickly halogenoalkanes are hydrolysed by timing the appearance of a silver halide precipitate after aqueous silver nitrate is added.
Relative rate is 1/t, so a shorter time means a faster reaction. Part 1 changes the halogen; Part 2 changes the structure around the carbon bonded to the halogen.
Hydrolysis Timing Bench
Time how long each halogenoalkane takes to give a precipitate with aqueous silver nitrate at 50 °C, then compare rates using 1/t.
© Dr. Mohammed Al-Fatah – onlinelearningsystem.net
GCSE Recap: Testing for Halide Ions
Three quick questions on the silver nitrate test you met at GCSE: the colours, the reason for the acid and what a precipitate is.
The Five Halogenoalkanes Used
Skeletal formulae of the five halogenoalkanes with the carbon bonded to the halogen circled. Three are primary, one is secondary and one is tertiary.
Before you interpret the results you must be able to classify each halogenoalkane.
Find the carbon atom bonded to the halogen and count the carbon atoms bonded to that carbon: one gives a primary halogenoalkane, two a secondary and three a tertiary.
Remember: The length of the chain is irrelevant; only the environment of the C–X carbon counts.
| Halogenoalkane | Classification | Where it is used | Comparison being made |
|---|---|---|---|
| 1-chlorobutane | Primary | Part 1 | Effect of the halogen |
| 1-bromobutane | Primary | Parts 1 and 2 | Control compound in both parts |
| 1-iodobutane | Primary | Part 1 | Effect of the halogen |
| 2-bromobutane | Secondary | Part 2 | Effect of structure |
| 2-bromo-2-methylpropane | Tertiary | Part 2 | Effect of structure |
1-Bromobutane appears in both parts on purpose. In Part 1 it is the middle member of the halogen series; in Part 2 it is the primary compound against which the secondary and tertiary bromoalkanes are judged.
The tertiary compound, 2-bromo-2-methylpropane, is the bromo analogue of the product you make in Core Practical 6 (chlorination of 2-methylpropan-2-ol).
Exam wording: “The carbon bonded to the bromine is attached to three other carbon atoms, so 2-bromo-2-methylpropane is a tertiary halogenoalkane.” Name the carbon, count its neighbours, state the class.
Check: Classifying by the C–X Carbon
Classify six halogenoalkanes that are not on this page as primary, secondary or tertiary.
Method: Step by Step
The timing set-up: bath at 50 °C with a thermometer, three reagent tubes and the warmed silver nitrate, a pipette delivering 1 cm³, a stop clock and a white tile behind the tubes.
The reaction runs in a water bath at about 50 °C. At room temperature 1-chlorobutane takes far too long to give a visible precipitate (the sample time below is already 10 minutes at 50 °C).
A controlled temperature is essential because rate depends on temperature. Every step below carries its reason: the reason is what the mark scheme rewards.
The control variables an examiner expects you to list are: the temperature of the bath, the volume of ethanol, the volume and concentration of the silver nitrate solution, and the amount of halogenoalkane.
Two more are the same observer judging the endpoint against the same background, and the tubes staying in the bath throughout. The only variable that changes is the halogenoalkane.
Evaluation point: Using the same number of drops of each liquid gives approximately, not exactly, equal amounts of substance: the halogenoalkanes have different densities and molar masses. A fairer method measures equal moles by mass or with a graduated pipette.
Check: Fair Test
Three things a student did wrong while comparing three different compounds; decide which variable is no longer controlled and which way the time moves.
What Happens During Hydrolysis?
Water attacks the δ+ carbon of a primary halogenoalkane in one step as the C–Br bond breaks; the protonated alcohol then loses H⁺.
Hydrolysis of a halogenoalkane is a nucleophilic substitution. The halogen is more electronegative than carbon, so the C–X bond is polar and the carbon carries a partial positive charge, δ+.
Water is a nucleophile because its oxygen atom has two lone pairs of electrons.
In the primary halogenoalkanes of Part 1 the oxygen lone pair attacks the δ+ carbon while the C–X bond breaks, the halide ion leaves and the protonated alcohol then loses H⁺.
R–X + H₂O → R–OH + H⁺ + X⁻
CH₃CH₂CH₂CH₂Br(l) + H₂O(l) → CH₃CH₂CH₂CH₂OH(aq) + H⁺(aq) + Br⁻(aq)
Water is a weak nucleophile, much weaker than hydroxide ions, which is why the reaction is slow enough to time with a stop clock. Direct attack on the δ+ carbon is not the only route, and this matters for Part 2.
The two routes are drawn out as full mechanisms later in the course. For this practical you need the idea behind them.
A primary halogenoalkane reacts when water attacks the δ+ carbon directly.
A tertiary halogenoalkane loses its halide first to give a tertiary carbocation, (CH₃)₃C⁺, which is stabilised by the positive inductive effect of three alkyl groups pushing electron density towards the positive carbon.
Because that carbocation forms readily, 2-bromo-2-methylpropane hydrolyses fastest, and a secondary halogenoalkane sits between the two.
Key idea: Whatever the route, the C–X bond must break. In Part 1 the strength of that bond controls the rate; in Part 2 the halogen is the same, so the stability of the carbocation that can form controls the rate.
How the Silver Halide Precipitate Shows the Rate
Silver halide precipitate colours: uniform cloudy suspensions of white silver chloride, cream silver bromide and pale yellow silver iodide, with the ionic equations.
The precipitate can only form once hydrolysis has released halide ions, so the time to the first cloudiness measures how quickly the halogenoalkane has hydrolysed.
Because Ag⁺(aq) and X⁻(aq) react instantly, the precipitation step adds nothing to the time.
The solid forms as a fine suspension throughout the liquid, which is why the endpoint is described as cloudiness rather than a lump at the bottom of the tube.
| Halide ion released | Ionic equation | Colour of AgX(s) | What it tells you |
|---|---|---|---|
| Cl⁻ | Ag⁺(aq) + Cl⁻(aq) → AgCl(s) | White | A chloroalkane has hydrolysed |
| Br⁻ | Ag⁺(aq) + Br⁻(aq) → AgBr(s) | Cream | A bromoalkane has hydrolysed |
| I⁻ | Ag⁺(aq) + I⁻(aq) → AgI(s) | Yellow | An iodoalkane has hydrolysed |
White and cream are hard to tell apart in a cloudy tube, and the colour is not what the practical measures: the compounds are known, so the colour is a check, not an identification.
If a question asks how you would confirm which silver halide has formed, use ammonia solution: dilute NH₃(aq) dissolves AgCl, concentrated NH₃(aq) dissolves AgBr, and AgI dissolves in neither.
The full test is covered in Core Practical 8 (analysis of inorganic and organic unknowns).
Key idea: Yellow AgI means iodide ions were released by hydrolysis, so the original compound was an iodo compound. It does not mean iodine molecules, I₂, are present.
Part 2: Comparing Primary, Secondary and Tertiary Bromoalkanes
Part 2 at 50 °C: 1-bromobutane, 2-bromobutane and 2-bromo-2-methylpropane with their times and 1/t values, and why the tertiary compound is fastest.
In Part 2 the halogen is kept the same, bromine, so the only variable is the structure around the C–Br carbon.
Part 2 was a separate run, which is why 1-bromobutane appears again as the primary control: times are only comparable within one run, so each part carries its own 1-bromobutane reading (87 s in the Part 1 run, 59 s in this one).
Remember: Never compare a Part 1 time with a Part 2 time.
| Halogenoalkane | Classification | Time to first cloudiness / s | 1/t / s⁻¹ (2 s.f.) | Relative rate |
|---|---|---|---|---|
| 2-bromo-2-methylpropane | Tertiary | 3 | 0.33 | Fastest |
| 2-bromobutane | Secondary | 34 | 0.029 | Intermediate |
| 1-bromobutane | Primary | 59 | 0.017 | Slowest |
The order is tertiary > secondary > primary, and it needs an explanation because bond enthalpy cannot give one: every compound here has a C–Br bond of about 290 kJ mol⁻¹.
If water simply attacked the δ+ carbon in every case, the crowded tertiary carbon should be the slowest. It is the fastest because it reacts by a different route.
The C–Br bond breaks first to give the tertiary carbocation (CH₃)₃C⁺, which is stabilised by the positive inductive effect of its three alkyl groups, and water then attacks the carbocation.
Key idea: The more alkyl groups on the C–Br carbon, the more stable the carbocation, the lower the activation energy and the faster the hydrolysis.
Evaluation point: A time of 3 s is too short to measure reliably: reaction time alone is about ±0.2 s at the start and at the stop.
To improve Part 2, run it at a lower temperature (room temperature is enough for the tertiary compound) or use more dilute reagents so that every time is at least 30 s.
Check: Why the Tertiary Compound Is Fastest
A student explains the tertiary result with a bond enthalpy argument; pick the statement that would earn the marks.
What This Practical Is Testing
In Core Practical 5 you compare how quickly different halogenoalkanes are hydrolysed.
Hydrolysis is a reaction with water: the water molecule acts as a nucleophile, the C–X bond breaks and the products are an alcohol, a hydrogen ion and a halide ion.
The halide ion is detected by aqueous silver nitrate, which gives a precipitate of the silver halide. The rates are compared by timing how long the precipitate takes to appear.
Key idea: relative rate ∝ 1/t, so a shorter time means a faster hydrolysis. The experiment gives relative rates, not a rate in mol dm⁻³ s⁻¹.
The experiment has two linked comparisons. In Part 1 three primary halogenoalkanes with different halogens are compared: 1-chlorobutane, 1-bromobutane and 1-iodobutane.
In Part 2 the halogen is kept the same and the structure is changed: 1-bromobutane (primary), 2-bromobutane (secondary) and 2-bromo-2-methylpropane (tertiary).
Questions on the Unit 3 practical skills paper (WCH13), which asks for methods, apparatus choices, data processing, uncertainty and evaluation ask you to describe the method, explain the role of each reagent, process the times and evaluate the comparison.
| Stage | What happens | What you see |
|---|---|---|
| Hydrolyse | Water attacks the halogenoalkane; the C–X bond breaks and X⁻ is released. | Nothing yet: the solution stays clear. |
| Detect | Ag⁺(aq) from the silver nitrate reacts with X⁻(aq) to give AgX(s). | The liquid turns cloudy throughout. |
| Compare | The time to the first cloudiness is recorded and converted to 1/t. | The shortest time is the fastest hydrolysis. |
Key idea: Silver nitrate does not cause the hydrolysis. The water it is dissolved in is the nucleophile, and the Ag⁺ ions detect the halide ion as it is released.
This is why no precipitate can form before the aqueous silver nitrate is added, and why timing starts at that moment.
Safety and Apparatus
The reagents are flammable, volatile and, in the case of silver nitrate, staining. A hazard question expects you to name the hazard and the matching precaution, not just “wear goggles”.
Hazards and precautions: Ethanol and the halogenoalkanes are highly flammable: no naked flames, so the water bath is prepared from a kettle.
The halogenoalkanes are harmful and volatile: work in a well ventilated laboratory or fume cupboard and keep the tubes stoppered until the silver nitrate is added.
Silver nitrate stains skin and is an irritant: wear gloves and wash spills off at once.
Kettle water scalds: mix it with cold water to reach 50 °C before filling the beaker. Eye protection throughout.
Why Ethanol, Water and Silver Nitrate Are Used
Each reagent has one job, and students lose marks by swapping them round: the commonest error is calling ethanol the nucleophile.
Why not use hydroxide ions? Aqueous sodium hydroxide would hydrolyse the halogenoalkane faster, but OH⁻ would give a brown precipitate of silver oxide, Ag₂O, the moment silver nitrate was added: 2Ag⁺(aq) + 2OH⁻(aq) → Ag₂O(s) + H₂O(l).
That masks the halide test. OH⁻ is also a much stronger nucleophile, so the reaction would be over too quickly to time.
Exam wording: “Water, from the aqueous silver nitrate, is the nucleophile. Ethanol is the solvent that allows the halogenoalkane and the aqueous solution to mix.” Never write “ethanol is the nucleophile” or “silver nitrate hydrolyses the halogenoalkane”.
Check: Roles of the Reagents
Fill the gaps for the hydrolysis of a compound not used on this page: nucleophile, solvent, detector and the colour you would see.
Part 1: Comparing Chloro-, Bromo- and Iodoalkanes
In Part 1 all three halogenoalkanes are primary, so the only variable is the halogen. The sample results below are for one run at 50 °C.
The column that examiners look for is 1/t: relative rate is proportional to 1/t, so the largest value is the fastest reaction.
| Halogenoalkane | Bond broken | Time to first cloudiness / s | 1/t / s⁻¹ (2 s.f.) | Mean bond enthalpy / kJ mol⁻¹ | Relative rate |
|---|---|---|---|---|---|
| 1-iodobutane | C–I | 52 | 0.019 | 228 | Fastest |
| 1-bromobutane | C–Br | 87 | 0.011 | 290 | Intermediate |
| 1-chlorobutane | C–Cl | 606 | 0.0017 | 346 | Slowest |
The order is iodo > bromo > chloro. The mean bond enthalpy falls down Group 7: C–Cl 346 kJ mol⁻¹, C–Br 290 kJ mol⁻¹, C–I 228 kJ mol⁻¹ (values from the Edexcel data booklet).
The C–X bond is broken in the rate-determining step, so the weakest bond, C–I, needs the lowest activation energy and 1-iodobutane hydrolyses fastest.
The ratio of the 1/t values shows the size of the effect: 1-iodobutane reacts about 0.019 / 0.0017 ≈ 11 times faster than 1-chlorobutane at 50 °C.
Bond polarity predicts the opposite. Chlorine is the most electronegative of the three halogens, so the C–Cl bond is the most polar and its carbon the most δ+; if polarity controlled the rate the chloro compound would be fastest.
Key idea: The data show it is slowest, so bond enthalpy, not polarity, controls the rate.
Exam wording: “The C–I bond has the lowest bond enthalpy (228 kJ mol⁻¹), so it breaks most easily in the rate-determining step and 1-iodobutane hydrolyses fastest.” Quote the values, say which step the bond breaks in, and link weak bond to low activation energy.
Check: New Data at 50 °C
Times for three propyl compounds you have not met on this page: convert to 1/t, rank, explain and predict the effect of a cooler bath.
Exam-Style Data Interpretation
A common exam version gives times for equal moles (equal amounts of substance) of 2-chloropropane, 2-bromopropane and 2-iodopropane in the same total volume at 50 °C, together with electronegativities and bond enthalpies, and asks you to explain the trend.
Equal moles in the same total volume gives the same concentration of each halogenoalkane, so concentration cannot be the cause of any difference.
| Halogenoalkane | Time to first cloudiness / s | 1/t / s⁻¹ (2 s.f.) | Electronegativity of the halogen | Mean C–X bond enthalpy / kJ mol⁻¹ |
|---|---|---|---|---|
| 2-iodopropane | 31 | 0.032 | 2.7 | 228 |
| 2-bromopropane | 74 | 0.014 | 3.0 | 290 |
| 2-chloropropane | 241 | 0.0041 | 3.2 | 346 |
Why not electronegativity? Chlorine is the most electronegative, so the C–Cl carbon is the most δ+ and should attract the nucleophile most strongly. If that controlled the rate, 2-chloropropane would be fastest. It is slowest, so the polarity argument fails.
Why does bond enthalpy explain the trend? The C–X bond is broken in the rate-determining step. A stronger bond needs more energy to break, so the activation energy is higher and a smaller fraction of collisions succeed.
C–Cl (346 kJ mol⁻¹) is the strongest and 2-chloropropane is slowest; C–I (228 kJ mol⁻¹) is the weakest and 2-iodopropane is fastest.
Working with 1/t. 1/31 = 0.032 s⁻¹, 1/74 = 0.014 s⁻¹, 1/241 = 0.0041 s⁻¹. The iodo compound reacts about 0.032 / 0.0041 ≈ 8 times faster than the chloro compound. Quote 1/t to 2 significant figures, because the times themselves are only known to the nearest second.
Exam wording: “Equal moles in the same volume gives equal concentrations, so the difference in rate is due to the halogenoalkane.
The C–I bond has the lowest bond enthalpy, so it breaks most easily and 2-iodopropane hydrolyses fastest, even though iodine is the least electronegative halogen.”
Check: Describe and Explain
Write a six-mark answer describing how you would compare three pentyl compounds and explaining the expected order; the key terms are marked automatically.
Errors, Uncertainty and Improvements
The stop clock reads to 0.01 s, but the real uncertainty is your reaction time (about ±0.2 s at the start and again at the stop, so ±0.4 s in total).
Far larger is the judgement of “first cloudiness”, which can differ by ±2 s between observers. Percentage uncertainty = (uncertainty / reading) × 100.
Reaction time on 606 s: (0.4 / 606) × 100 = 0.066 %. Reaction time on 52 s: (0.4 / 52) × 100 = 0.77 %. Reaction time on 3 s: (0.4 / 3) × 100 = 13 %. Judgement of the endpoint (±2 s) on 3 s: (2 / 3) × 100 = 67 %. Thermometer (±0.5 °C) on 50 °C: (0.5 / 50) × 100 = 1.0 %. Graduated pipette (±0.05 cm³) on 1 cm³ of silver nitrate: (0.05 / 1) × 100 = 5.0 %. The short Part 2 times, not the apparatus, dominate the uncertainty.
Key idea: The apparatus errors are small (1 % to 5 %); the endpoint judgement and reaction time on very short times are large. An improvement that lengthens the times, or replaces the eye with a sensor, does more than a better pipette.
Common Mistakes
- “Silver nitrate causes the hydrolysis.” Water hydrolyses the halogenoalkane; silver ions only detect the halide ion released.
- “Ethanol is the nucleophile.” Ethanol is the solvent. The nucleophile is water from the aqueous silver nitrate.
- “A yellow precipitate shows iodine is present.” Yellow AgI shows iodide ions, released by hydrolysis of an iodo compound.
- “Chlorine is most electronegative, so the chloro compound is fastest.” The data show the opposite: bond enthalpy controls the rate, and C–Cl is the strongest bond.
- “The tertiary compound is fastest because its C–Br bond is weaker.” The bond is the same; the tertiary carbocation is more stable.
- Taking the tube out of the bath to look at it. The temperature drops and the comparison is no longer fair; hold the tube in the bath against a white background.
- Comparing a Part 1 time with a Part 2 time. They are separate runs; compare only within a run.
- Quoting a rate in mol dm⁻³ s⁻¹. The experiment gives relative rates from 1/t, nothing more.
Check: Rates of Hydrolysis Summary
Six rounds; in each pick the one accurate statement about reagents, endpoint, 1/t, bond enthalpy, carbocations and precipitate colours.
Common Exam Points
Say
“Water from the aqueous silver nitrate is the nucleophile; ethanol is the solvent.”
“Relative rate is proportional to 1/t, so the shortest time is the fastest.”
“The C–I bond has the lowest mean bond enthalpy (228 kJ mol⁻¹), so it breaks most easily in the rate-determining step.”
“The tertiary carbocation is stabilised by three alkyl groups, so the C–Br bond breaks first and the tertiary halogenoalkane hydrolyses fastest.”
Do not say
“Silver nitrate hydrolyses the halogenoalkane.” “Ethanol is the nucleophile.” “The chloro compound is fastest because C–Cl is the most polar bond.” “The yellow precipitate is iodine.” “The rate was 52 s.”
Watch for
Every equation needs state symbols: Ag⁺(aq) + Br⁻(aq) → AgBr(s). Give 1/t to 2 significant figures with the unit s⁻¹.
In an evaluation, name the variable that was not controlled and say which way the time moves (too long or too short), then give the improvement.
FAQs
Quick answers to the questions students ask most about Core Practical 5 and the hydrolysis of halogenoalkanes.
How is the rate measured in this practical?
It is not measured directly. The time from adding the aqueous silver nitrate to the first cloudiness is recorded, and relative rate is taken as 1/t.
A shorter time means a faster hydrolysis, and the 1/t values can be compared within one run at one temperature.
Why does no precipitate form before the silver nitrate is added?
Because there is no water in the tube. Ethanol and the halogenoalkane do not react; the water that acts as the nucleophile arrives with the aqueous silver nitrate, so hydrolysis and the release of halide ions start at that moment.
Why does 1-iodobutane react faster than 1-chlorobutane?
The C–I bond has a lower mean bond enthalpy (228 kJ mol⁻¹) than C–Cl (346 kJ mol⁻¹), so it breaks more easily in the rate-determining step and the activation energy is lower. Electronegativity would predict the opposite order, so it is not the controlling factor.
Why is 2-bromo-2-methylpropane so much faster than 1-bromobutane?
Both have a C–Br bond, so bond enthalpy cannot explain it.
The tertiary compound loses its bromide first to form the carbocation (CH₃)₃C⁺, stabilised by the positive inductive effect of three alkyl groups. That stable carbocation forms readily, so the reaction is fast.
The two routes are drawn out as mechanisms later in the course.
Why is it hard to tell silver chloride from silver bromide?
White and cream look alike in a cloudy tube. Add ammonia solution: dilute ammonia dissolves silver chloride, concentrated ammonia dissolves silver bromide and silver iodide dissolves in neither.
Why must there be no naked flames?
Ethanol and the halogenoalkanes are highly flammable and volatile, so the water bath is filled from a kettle and never heated with a Bunsen burner. Bung the tubes while they warm so that the vapour is not lost.
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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