Design a controlled comparison, explain why a precipitate appears and use timed observations to compare C–X bonds and halogenoalkane classes.
The visible event follows the hydrolysis
Hydrolysis of a halogenoalkane substitutes OH for the halogen through reaction with water. In this experiment water is the nucleophile. Ethanol helps the organic reactant mix with the aqueous solution; silver nitrate supplies Ag⁺, which detects halide ions released after C–X breaks. Silver ions are not simply replacing Br in an intact bromoalkane.
The alcohol formation and silver halide precipitation are separate events. Ionic chloride already in solution can precipitate immediately with Ag⁺, whereas a covalently bonded chlorine in a chloroalkane must first be released. Record cloudiness/precipitation as the observation and hydrolysis as the inference.
| Halide released | Precipitate | Observation |
|---|---|---|
| Cl⁻ | AgCl | White precipitate |
| Br⁻ | AgBr | Cream precipitate |
| I⁻ | AgI | Yellow precipitate |
Control the mixture and the temperature
The Pearson CP4 method compares 1-chloro-, 1-bromo- and 1-iodobutane, then compares 1-bromobutane, 2-bromobutane and 2-bromo-2-methylpropane. In each set, change only the structural feature under investigation. Comparing a tertiary iodoalkane with a primary chloroalkane confounds halogen and carbon skeleton.
Prepare labelled tubes with equal specified volumes of ethanol and equal measured quantities of halogenoalkane. Place them in a water bath around the specified temperature, typically 50 °C. Warm equal volumes of the same silver nitrate solution separately so mixing does not introduce a different starting temperature. After equilibration, mix one pair, start timing immediately and use a consistent mixing procedure.
Stop timing at the first reproducible appearance of cloudiness, using the same lighting and background. Keep each tube in the bath while timing. Repeat for each reactant and collect repeat readings. The published worksheet uses drops as a convenient quantity; calibrated small-volume pipettes or equal-concentration stock solutions can improve control because different liquids may make different-sized drops.
Use eye protection, suitable ventilation and a hot-water bath away from flames because ethanol and halogenoalkanes are flammable; halogenoalkanes are harmful and silver nitrate is corrosive at the stated concentration. Collect silver-containing and organic waste as directed by the laboratory. These notes explain the practical; safe technique and competence must still be demonstrated under supervision.
Changing the halogen: bond strength controls the simple trend
For comparable primary compounds under these conditions, iodoalkanes hydrolyse fastest, then bromoalkanes, then chloroalkanes. C–I has the lowest bond enthalpy and is easiest to break; C–Cl is stronger. The C–Cl bond is more polar than C–I, but a polarity-only argument predicts the wrong observed order. Use the factor relevant to breaking the bond.
Typical average bond enthalpies are about 238, 276 and 338 kJ mol⁻¹ for C–I, C–Br and C–Cl respectively; exact values depend on the data source and molecule. Their role here is the trend. Bond enthalpy is not numerically identical to the activation energy of a solution reaction.
The chloroalkane may give no visible precipitate in the chosen observation window. Record “no precipitate within 600 s”, for example, rather than “does not hydrolyse”. Absence of a detectable event within a limited time is a limit on the measurement, not proof of zero reaction.
Changing the carbon skeleton: use the experimental context
For the aqueous ethanol hydrolysis comparison with the same halogen, the usual order is tertiary fastest, then secondary, then primary. Tertiary compounds can react by a pathway involving a relatively stable tertiary carbocation; three alkyl groups donate electron density and stabilise its positive centre. A primary carbocation is much less stable, so primary substitution normally proceeds by a different, concerted pathway.
This contextual explanation does not make tertiary compounds fastest for every nucleophile in every solvent. In a direct one-step substitution, crowding can hinder attack at a tertiary carbon. Keep the CP4 observation separate from a universal reactivity claim. Drawing a detailed tertiary substitution mechanism is explanatory extension; 6.36 requires the primary hydroxide/ammonia mechanisms.
Use reciprocal time as a comparative measure
If approximately the same small amount of product must form to make cloudiness visible, a shorter time suggests a faster reaction, and 1/t can be used as a relative rate measure with units s⁻¹. It is not an absolute molar reaction rate, because the threshold amount was not measured and may differ between precipitates.
Worked illustrative repeats for one bromoalkane are 42 s, 45 s and 43 s. Mean time = (42 + 45 + 43)/3 = 43.3 s. Comparative rate = 1/43.3 = 0.0231 s⁻¹. A comparable sample reaching the same endpoint in 130 s has 1/t = 0.00769 s⁻¹, so the first is about 3.00 times as fast by this measure.
If endpoint uncertainty is estimated as ±2 s, the approximate percentage uncertainty for 43.3 s is (2/43.3) × 100 = 4.62%. Faster reactions have a larger percentage timing uncertainty for the same absolute error. This helps justify a lower bath temperature or a sensor for an excessively rapid reaction; changing the stopwatch alone may not solve subjective cloudiness detection.
| Specific limitation | Likely effect | Targeted improvement |
|---|---|---|
| Unequal starting temperatures | Warmer tubes can appear intrinsically more reactive | Equilibrate both solutions in a thermostated bath |
| Different drop sizes | Different reactant amounts alter the endpoint time | Use measured volumes or equal-concentration stock solutions |
| Subjective first cloudiness | Changes when the timer is stopped | Use a common visual standard or a light sensor threshold |
| Delayed start after mixing | Recorded time is too short; 1/t too large | Start on mixing with one consistent operator/procedure |
| Different silver-halide appearance/solubility | Visible endpoint is not precisely the same conversion | Treat 1/t as comparative; avoid claiming an absolute rate constant |
Write method, observation and inference as one argument
A strong explanation says that ethanol helps reactants share a phase, the bath controls temperature, hydrolysis releases halide ions and Ag⁺ precipitates those ions. Each statement has a chemical purpose. “Use the same conditions for a fair test” leaves the actual controls unspecified.
Pearson 8CH0/02 June 2023 Q4(a)(i–ii) assessed these links. The examiner report highlighted failure to explain the solvent and water’s role. When asked to explain why cloudiness appears, connect water attack, halide release and precipitation; when asked to compare rates, add the observed order and the relevant C–X bond-strength reasoning.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.
Q1. Why are both ethanol and water present in CP4?Show answer
Ethanol helps the halogenoalkane mix with the aqueous phase. Water acts as the nucleophile in hydrolysis. Treating ethanol as the hydrolysing reagent misses the role of water.
Q2. Why does AgBr appear only after bromoalkane hydrolysis?Show answer
Bromine is initially covalently bonded to carbon. Hydrolysis releases Br⁻, which reacts with Ag⁺ to form insoluble AgBr. Ag⁺ does not simply precipitate a bromine atom while it is still bonded to carbon.
Q3. A comparable chloroalkane takes longer than the iodoalkane. Explain why the more polar bond does not make it faster.Show answer
C–Cl has higher bond enthalpy than C–I and is harder to break. In this comparison bond strength gives the observed trend and outweighs a simple polarity argument.
Q4. Times are 58, 62 and 60 s. Calculate the mean time and its reciprocal.Show answer
Mean = (58 + 62 + 60)/3 = 60.0 s. 1/t = 1/60.0 = 0.0167 s⁻¹. This is a comparative measure, not mol dm⁻³ s⁻¹.
Q5. A student starts timing 3 s after mixing and concludes that the reaction is especially fast. Predict the bias and improve the method.Show answer
The recorded time is too short, so its reciprocal overestimates comparative rate. Start timing at mixing, apply a fixed mixing procedure and repeat; a sensor can improve endpoint detection but does not by itself correct a late start.
Sources
Sources and examiner guidance (reviewed 9 October 2026)
- Pearson Edexcel 9CH0 specification — Issue 3 — Topic 6, 6.33–6.35 and Core Practical 4; printed pages 17–20. Reviewed 9 October 2026.
- Pearson Edexcel 8CH0 specification — Issue 3 — Topic 6, printed pages 15–18; AS scope checked against the A-Level outcomes.
- Chemrevise — Edexcel Organic Chemistry I — Pages 19–21; explanatory coverage reference. Teaching and practice here are original.
- Pearson Core Practical 4 — teacher and student sheets — Teacher printed pages 1–2; method, comparisons and practical controls. Reviewed 9 October 2026.
- Pearson 8CH0/02 — June 2023 mark scheme — Q4(a–b), PDF pages 17–19; solvent, temperature, halide release and class comparison.
- Pearson 8CH0/02 — June 2023 examiner report — Q4(a), PDF pages 4–5; solvent and water-role misconceptions.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.
