OCR A-Level Chemistry AS Depth in chemistry (02), June 2018: Question 5

10 marks · Medium difficulty · Practical Questions

Describe the preparation, purification, and percentage yield calculation of 1-bromobutane from butan-1-ol, and determine the rate of reaction from a concentration-time graph.

Practise this question

Question

Question 5 shows three parts about 1-bromobutane. Part a(i) asks to draw a labelled diagram to set up apparatus for preparation and outline purification from a reaction mixture containing butan-1-ol, sulfuric acid, and sodium bromide. Part a(ii) provides moles and percentage yield to calculate the mass of 1-bromobutane. Part b shows a graph of hydroxide ion concentration against time for the reaction of 1-bromobutane with aqueous hydroxide ions, and asks to calculate the rate of reaction at 30 minutes using a tangent.
Question text

5 (a) 1-Bromobutane is an organic liquid with a boiling point of 102 °C.

A student prepares 1-bromobutane by reacting butan-1-ol with sulfuric acid and sodium

bromide. The student boils the mixture for one hour.

The equation is shown below.

CH CH CH CH OH + H+ + Br – CH CH CH CH Br + H O

32 2 2 3 2 2 2 2

The student obtains a reaction mixture containing an organic layer (density = 1.27 g cm–3)

and an aqueous layer (density = 1.00 g cm–3).

(i)* Draw a labelled diagram to show how you would safely set up apparatus for the

preparation. Outline a method to obtain a pure sample of 1-bromobutane from the

reaction mixture.

… [6]

(ii) The student used 0.150 mol of butan-1-ol. The student obtained a 61.4% percentage

yield of 1-bromobutane.

Calculate the mass of 1-bromobutane obtained.

Give your answer to three significant figures.

mass = … g [2]

(b) A student investigates the rate of reaction of 1-bromobutane with aqueous hydroxide ions.

The graph shows how the hydroxide ion concentration, [OH–(aq)], changes during the

reaction.

0.45

0.40

0.35

0.30

[OH–(aq)] 0.25

/ mol dm–3

0.20

0.15

0.10

0.05

0.00

0 50 100 150 200

time/min

Using the graph, calculate the rate of reaction, in mol dm–3 min–1, at 30 minutes.

Show your working on the graph and in the space below.

rate of reaction = … mol dm–3 min–1 [2]

Mark scheme

Show the mark scheme Mark scheme for question 5 detailing level of response criteria for reflux and purification in part a(i), calculation steps for mass in part a(ii) awarding 2 marks for 12.6g, and tangent drawing and gradient calculation for part b awarding 2 marks.

Question Answer Marks Guidance

5 (a) (i)* Please refer to the marking instructions on page 5 of this mark 6 Indicative scientific points may include:

scheme for guidance on how to mark this question.

Apparatus set up for reflux:

Level 3 (5–6 marks)

round-bottom/pear shaped flask

Correctly labelled diagram of reflux apparatus that works, with

no safety problems heat source

AND condenser

An appreciation of most of the purification steps required to Detail: water flow in condenser bottom to

gain a pure sample top; open system.

Purification

There is a well-developed line of reasoning which is clear and Use of a separating funnel to separate

logically structured. The information presented is relevant and organic and aqueous layers

substantiated. Detail: Collect lower organic layer

density greater

Level 2 (3–4 marks) Drying with an anhydrous salt,

Labelled diagram of apparatus (either reflux or distillation) but Detail: e.g. MgSO4, CaCl2, etc.

with safety/procedural problems OR clear diagram of reflux Redistillation

apparatus without labelling Detail: Collect fraction distilling at 102ºC.

AND

Some details of further purification steps

There is a line of reasoning presented with some structure. The

information presented is relevant and supported by some evidence.

Level 1 (1–2 marks)

Diagram of apparatus (reflux OR separation OR distillation)

drawn with no labelling OR labelled diagram with significant

safety/procedural

AND / OR

Few or imprecise details about further purification stages

There is an attempt at a logical structure with a line of reasoning. The

information is in the most part relevant.

0 marks No response or no response worthy of credit.

H032/02 Mark scheme June 2018

5 (a) (ii) FIRST, CHECK THE ANSWER ON ANSWER LINE 17 2 Common errors:

IF answer = 12.6 (g) award 2 marks 33.4 (0.150 x 100/61.4 = 0.244 x 136.9)

1 mark

61.4

n(1-bromobutane) = 0.150 = 0.0921 (mol) ALLOW ECF for incorrect moles or incorrect Mr of

1-bromobutane (provided answer is to 3 SF)

DO NOT ALLOW 6.82 (using Mr of butan-1-ol)

Mass 1-bromobutane = 0.0921 136.9 = 12.6 (g)

3 SF required ALLOW calculation using masses, e.g.

Theoretical = 0.150 136.9 = 20.535 (g)

(ALLOW 20.535 rounded back to 20.5)

61.4

Actual mass = 20.535 = 12.6 (g)

(20.5 also gives 12.6)

(b) Tangent on graph 2 DO NOT ALLOW interpolation (taking a direct

drawn at approximately t = 30 min (±10 mins) reading from graph), answer must be derived from

taking a gradient

Calculation of rate

= Gradient (y/x) of tangent drawn

0.19 –3 -3 -1 ALLOW ecf from incorrectly drawn tangent

e.g. 72 = 2.64 10 / 0.00264 (mol dm min )

Tolerance:

Readings from y axis should be ± 0.01 mol dm-3

(i.e. within 1 square)

Readings from x axis should be ± 5 minutes (i.e.

within 0.5 of a square)

IGNORE units

IGNORE sign

Total 10

How to answer it

Preparation and Reaction Kinetics of 1-Bromobutane

What this question tests

This question assesses your practical organic chemistry knowledge regarding the synthesis, isolation, and purification of halogenoalkanes via reflux and separation techniques. It also tests stoichiometric percentage yield calculations and the graphical determination of rates of reaction by drawing tangents.

Question 5 (a) (i)

Reflux Setup and Organic Purification

💡 Key Knowledge: Reflux & Separation

  • Reflux: Heat reaction mixture without losing volatile components. Requires a round-bottom/pear-shaped flask, a heating source, and a vertical condenser with water flowing bottom-to-top. Must be an open system (no stopper!).
  • Separating Funnel: Used to separate immiscible layers based on density. Lower layer = aqueous layer (density = 1.00 g cm⁻³), Upper layer = organic layer containing 1-bromobutane (density = 1.27 g cm⁻³). Wait, check the densities carefully! Since 1-bromobutane has a density of 1.27 g cm⁻³, it is the lower layer.

🧠 Exam Technique (Level of Response)

Marked out of 6 using levels of response. To secure Level 3 (5–6 marks), you must describe:

  1. A working reflux apparatus diagram setup (vertically aligned condenser, flask, heat).
  2. Use of a separating funnel to discard the aqueous layer and retain the organic layer (knowing which is which using density values).
  3. Drying the organic layer using an anhydrous salt (e.g., anhydrous MgSO₄ or CaCl₂).
  4. Redistillation, collecting the fraction distilling at 102 °C.

❌ Common Errors & Examiner Pitfalls

  • Sealed apparatus: Putting a bung or stopper in the top of the condenser creates a closed system, which can explode under heating!
  • Wrong layer retained: Assuming the organic layer is always on top. Always check the density values given in the stem.
  • Missing purification steps: Forgetting to dry the product with an anhydrous salt before final distillation.
Question 5 (a) (ii)

Percentage Yield Calculation

📐 Step-by-Step Calculation

Step 1: Find the molar mass (Mᵣ) of 1-bromobutane (C₄H₉Br)

Mᵣ = (4 × 12.0) + (9 × 1.0) + 79.9 = 136.9 g mol⁻¹

Step 2: Calculate actual moles obtained using percentage yield

Moles of 1-bromobutane = 0.150 × (61.4 / 100) = 0.0921 mol

Step 3: Convert moles to mass

Mass = moles × Mᵣ = 0.0921 × 136.9 = 12.607... g

Final Answer (3 SF): 12.6 g

✅ Alternative Method

You can also calculate the theoretical maximum mass first:

Theoretical mass = 0.150 mol × 136.9 g mol⁻¹ = 20.535 g

Actual mass = 20.535 × (61.4 / 100) = 12.6 g

Award 2 marks for correct final answer on the answer line (12.6 g). ECF allowed if moles/Mᵣ were wrong provided final value is given to 3 SF.

❌ Common Calculation Traps

  • Using the Mᵣ of butan-1-ol (74.0) instead of 1-bromobutane in your final mass calculation (leads to 4.54g or 6.82g errors).
  • Failing to round to the requested three significant figures.
Question 5 (b)

Graphical Rate of Reaction Determination

🧠 Exam Technique: Drawing Tangents

  1. Locate t = 30 min on the x-axis.
  2. Go straight up to the curve and mark the point.
  3. Carefully draw a straight line (tangent) that touches the curve only at t = 30 min, ensuring equal steepness of angle on both sides of the touch point.
  4. Choose large, easy-to-read points on your tangent line to calculate the gradient ( Δy / Δx ).

📐 Working Out the Gradient

Using points from a typical accurate student tangent:

Gradient = Δy / Δx = (0.19 - 0.00) / (72 - 0) = 2.64 × 10⁻³ mol dm⁻³ min⁻¹

Mark Scheme Tolerance:
• Tangent drawn at t = 30 min (±10 mins tolerance for angle).
• Rate value accepted within range based on your drawn tangent gradient.
• Unit required implicitly by question context: mol dm⁻³ min⁻¹ .

❌ Common Errors

  • Interpolation: Taking a single point reading off the curve (e.g., reading y at 30 min) instead of finding the gradient of a tangent. A curve requires a gradient to find instantaneous rate!
  • Drawing a short, sloppy tangent that makes calculating Δy/Δx inaccurate. Always make your gradient triangle as large as possible.

Topics

Module 1: Development of practical skills in chemistry · Module 2: Foundations in chemistry · Module 4: Core organic chemistry · Module 5: Physical chemistry and transition elements · Practical Activity Groups · PAG 5: Synthesis of an organic liquid · PAG 9: Rates of reaction – continuous monitoring method · 2.1 Atoms and reactions · 4.2 Alcohols, haloalkanes and analysis · 5.1 Rates, equilibrium and pH

Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.