Edexcel A-Level Chemistry Paper 3, June 2019: Question 8
15 marks · Hard difficulty · Practical Techniques and Data Analysis
Describe and explain the preparation, purification, and calculation of yield/molecules for 1-bromobutane from butan-1-ol.
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Question text
8 1-bromobutane can be prepared from butan-1-ol and hydrogen bromide.
CH3CH2CH2CH2OH + HBr o CH3CH2CH2CH2Br + H2O
Hydrogen bromide can be made from sodium bromide and 50% concentrated sulfuric acid.
(a) The steps for the preparation of impure 1-bromobutane are summarised.
Step 1 Dissolve the sodium bromide in distilled water in a pear-shaped flask and
then add 20.0 cm3 of butan-1-ol.
Step 2 Surround the flask with an ice bath to cool the mixture, before adding
concentrated sulfuric acid drop by drop.
Step 3 Remove the flask from the ice bath and add a few anti-bumping granules
to the reaction mixture.
Step 4 Set up the apparatus for heating under reflux. Heat the mixture in the
flask for 30 minutes and then allow the apparatus to cool.
Step 5 Rearrange the apparatus for distillation and heat the mixture until no
more 1-bromobutane distils over.
(i) Parts of the method are given in bold type in Steps 2, 3 and 4.
Give a reason why each of these parts is necessary.
(3)
(ii) A student drew a diagram of the apparatus used for heating under reflux in*P58308A02032*
Step 4. There are three errors in the apparatus shown in the diagram.
Assume the apparatus is suitably clamped.
Identify the three errors, including the effect of each error.
(3)
(iii) The student corrected the errors.*P58308A02132*
While the mixture was heating under reflux, the student noticed a small
amount of a brown vapour was formed.
Explain why the brown vapour forms.
(2)
(b) The distillate collected in Step 5 is a mixture consisting of two layers.
There is an aqueous layer and a layer containing impure 1-bromobutane.
Data
Densities:
water 1.00 g cm–3
butan-1-ol 0.81 g cm–3
1-bromobutane 1.27 g cm–3
Boiling temperature of 1-bromobutane = 102 °C
The steps for the purification of the 1-bromobutane are summarised.
Step 6 Transfer the mixture from Step 5 to a separating funnel and remove the
aqueous layer.
Step 7 Wash the impure 1-bromobutane with concentrated hydrochloric acid in
the separating funnel. Remove the aqueous layer.
Step 8 Add aqueous sodium hydrogencarbonate to the impure 1-bromobutane
in the separating funnel.
Step 9 Shake the mixture in the separating funnel and, from time to time, invert
the funnel and open the tap.
Step 10 Collect the 1-bromobutane layer from Step 9 in a small conical flask.
Add anhydrous sodium sulfate and swirl the flask until the liquid
becomes clear.
Step 11 Decant the 1-bromobutane into a clean pear-shaped flask and redistil it.
Measure the volume of 1-bromobutane produced.
(i) State the position of the aqueous layer in the separating funnel at the start of
Step 6. Justify your answer.
(1)
… *P58308A02232*
(ii) Concentrated hydrochloric acid is used to remove any unreacted butan-1-ol in
the mixture in Step 7.
Give the reasons for carrying out Steps 8, 9 and 10.
(3)
(iii) Give a suitable temperature range over which to collect the pure
1-bromobutane in the redistillation in Step 11.
(1)
(iv) The volume of 1-bromobutane collected was 12.0 cm3.
Calculate the number of molecules of 1-bromobutane produced in this experiment.
Give your answer to an appropriate number of significant figures.
(2)
(Total for Question 8 = 15 marks)
Mark scheme
Show the mark scheme
How to answer it
Preparation and Purification of 1-Bromobutane
What this question tests
This question assesses your practical chemistry knowledge regarding the preparation of a halogenoalkane via nucleophilic substitution/haloform reactions. It tests your understanding of reflux apparatus setup and safety, identification of chemical errors in diagrams, side-reactions involving concentrated sulfuric acid, practical purification steps using a separating funnel, acid-base neutralization, drying agents, distillation temperature ranges, and stoichiometry calculations involving density, moles, and Avogadro's constant.
Functions of Practical Steps
Explain the necessity of ice cooling, anti-bumping granules, and heating under reflux.
✅ Correct Answers (Mark Scheme)
- Ice bath (Step 2): Cools the mixture because the reaction with concentrated sulfuric acid is very exothermic / releases a lot of heat.
- Anti-bumping granules (Step 3): Prevents violent/sudden/localised boiling or superheating/large bubbles.
- Reflux (Step 4): Prevents the loss of any volatile substances/reactants/products/organic compounds by condensing vapours back down.
❌ Common Errors
- Saying reflux "prevents gas escaping" without specifying organic vapours/reactants/products.
- Stating anti-bumping granules "prevent explosions" (not accepted; must refer to superheating or smooth boiling).
- Confusing exothermic cooling with quenching reactions.
Identifying Reflux Apparatus Errors
Identify three errors in the provided reflux diagram and state the effect of each.
✅ Correct Answers & Effects
- Error 1: Gap between the flask and condenser. Effect: Vapour/gas/reactants/products will escape.
- Error 2: Water entering at the top and leaving at the bottom of the condenser. Effect: Condenser won't fill completely / inefficient cooling / airlock.
- Error 3: Stopper placed on top of the condenser. Effect: Build-up of pressure leading to an explosion/glass shattering.
🧠 Exam Technique
Always examine laboratory diagrams systematically from bottom to top: check joint seals, water jacket flow direction (always bottom-in, top-out), and whether any closed systems (stoppers) have been incorrectly introduced to heating setups.
Side Reactions and Brown Vapours
Explain why a small amount of brown vapour forms during reflux.
💡 Key Knowledge
- The brown vapour is bromine ( Br₂ ).
- It is formed because bromide ions ( Br⁻ ) or HBr are oxidized by the concentrated sulfuric acid. Concentrated H₂SO₄ acts as an oxidizing agent here.
❌ Common Errors
Students often lose marks by writing just Br instead of Br₂ , or by naming sodium bromide instead of bromide ions as the species being oxidized.
Separating Funnel Layers
State the position of the aqueous layer in the separating funnel and justify using density data.
✅ Correct Answer
The aqueous layer is on the top.
Justification: Water has a lower density ( 1.00 g cm⁻³ ) compared to 1-bromobutane ( 1.27 g cm⁻³ ).
❌ Common Errors
Writing that water is "lighter" (examiners prefer explicit comparison of density values) or confusing the densities of butan-1-ol with 1-bromobutane.
Purification Steps
Give the reasons for carrying out Steps 8, 9, and 10.
💡 Key Knowledge
- Step 8 ( NaHCO₃ added): Reacts with / neutralizes / removes remaining acidic impurities (hydrochloric acid / H⁺ ions).
- Step 9 (Invert funnel and open tap): Releases carbon dioxide gas formed during neutralization to prevent a dangerous build-up of pressure.
- Step 10 (Anhydrous Na₂SO₄ added): Acts as a drying agent to remove/absorb residual water from the organic product.
❌ Common Errors
Stating that sodium sulfate "dehydrates" the product or reacts chemically with water to form hydrates in a way that destroys the product. It simply absorbs moisture as a drying agent.
Redistillation Temperature Range
Give a suitable temperature range over which to collect pure 1-bromobutane.
✅ Correct Answer
Starting temperature: 99 °C, 100 °C, or 101 °C
Final temperature: 103 °C, 104 °C, or 105 °C
(Based on the given boiling point of 102 °C , a standard ±2 °C collection range is accepted).
❌ Common Errors
Giving a single temperature value (e.g., exactly 102 °C ) instead of a functional collection range.
Calculation of Number of Molecules
Calculate the number of molecules of 1-bromobutane produced from 12.0 cm³ . Give your answer to an appropriate number of significant figures.
📐 Step-by-Step Calculation
- Find mass of 1-bromobutane:
Mass = Volume × Density = 12.0 cm³ × 1.27 g cm⁻³ = 15.24 g - Calculate Molar Mass ( Mᵣ ) of 1-bromobutane ( C₄H₉Br ):
Mᵣ = (4 × 12.0) + (9 × 1.0) + 79.9 = 136.9 g mol⁻¹ - Calculate moles of 1-bromobutane:
Moles = Mass / Mᵣ = 15.24 / 136.9 = 0.11132 mol - Calculate number of molecules using Avogadro's constant ( L = 6.02 × 10²³ ):
Number of molecules = 0.11132 × 6.02 × 10²³ = 6.70 × 10²²
Final Answer: 6.7 × 10²² or 6.70 × 10²² (to 2 or 3 significant figures).
Topics
Organic Chemistry · Core Practicals · Physical Chemistry · Core Practical 7: Identify unknown organic liquids and inorganic solids · Core Practical 15: Analyse organic and inorganic unknowns · Core Practical 16: Synthesise aspirin from 2-hydroxybenzoic acid · Topic 6: Organic Chemistry I · Topic 18: Organic Chemistry III · Topic 5: Formulae, Equations and Amounts of Substance
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.