OCR A-Level Chemistry Unified chemistry (03), June 2022: Question 3

10 marks · Hard difficulty · Extended Response

Describe the preparation of 1-bromobutane by refluxing butan-1-ol with sodium bromide and sulfuric acid, including a reflux diagram, percentage yield calculation, purification steps, and the two-step mechanism.

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Question text

3 Information about 1-bromobutane and butan-1-ol is shown in the table.

Compound Melting point / °C Boiling point / °C Density / g cm–3

1-bromobutane –113 102 1.268

butan-1-ol –90 118 0.810

A student prepares a sample of 1-bromobutane by refluxing 9.25 g of butan-1-ol with sodium

bromide and sulfuric acid.

After reflux, the reaction mixture is purified.

The student obtains 6.10 cm3 of pure 1-bromobutane.

(a)* Draw a diagram to show how the student would have carried out the reflux and calculate the

percentage yield of 1-bromobutane that the student obtains.

Describe how the student could have obtained pure 1-bromobutane from the reaction

mixture obtained after reflux. [6]

Additional answer space if required.

(b) Butan-1-ol reacts with sodium bromide and sulfuric acid to form 1-bromobutane by

nucleophilic substitution.

The mechanism for this reaction takes place by two steps.

Step 1 The oxygen atom of the alcohol group accepts a proton to form a positively-

charged intermediate.

Step 2 Bromide ions react with the intermediate from Step 1 by nucleophilic substitution

to form 1-bromobutane.

Show both steps in this mechanism.

[4]

Mark scheme

Show the mark scheme

Question Answer Marks AO Guidance

element

3 (a)* Refer to marking instructions on page 4 of mark scheme for 6 Indicative scientific points may include:

guidance on marking this question. AO2.8 Diagram

×2 Diagram draw with condenser above flask

Level 3 (5-6 marks) Labels including

Diagram showing reflux with most labels

AND AO3.3 • condenser

A CORRECT calculation of the % yield of 1-bromobutane ×4 • water in at bottom and out at top

AND • pear-shaped or round-bottom flask

A detailed description of most purification steps.

Calculation of % yield of 1-bromobutane

There is a well-developed line of reasoning which is clear and

logically structured. The information presented is relevant and 9.25

• n(butan-1-ol) = 74.0 = 0.125 (mol)

substantiated.

• mass 1-bromobutane = 6.10 × 1.268 = 7.7348 g

Level 2 (3-4 marks) 7.7348

• n(1-bromobutane) = 136.9 = 0.0565 (mol)

Diagram showing reflux with some labels

AND 0.0565

• % yield = 0.125 × 100 = 45.2%

Calculates the % yield of 1-bromobutane with some errors

OR ALLOW 45.2 ± 0.2 for small slip/rounding

Diagram showing reflux with most labels -------------

AND NOTE Use of 6.1 g (omission of density)

describes some purification steps, with some detail 6.10

• n(1-bromobutane) = 136.9 = 0.044558… (mol)

Calculates the % yield of 1-bromobutane with some errors 0.044558…

• % yield = 0.125 × 100 = 35.6%

AND

describes some purification steps, with some detail

Purification

There is a line of reasoning presented with some structure. The • In separating funnel, organic layer is on

information presented is relevant and supported by some bottom

evidence. • Drying with an anhydrous salt by

formula or name,

e.g. MgSO4, Na2SO4, CaCl2

• Redistil at 102ºC

Examples of detail in bold (NOT INCLUSIVE)

NOTE: ‘Use a separating funnel’, dry, and ‘redistil’

on their own are NOT detailed descriptions

Question Answer 18 Marks AO Guidance

element

Level 1 (1-2 marks)

Diagram showing reflux

Attempts to calculate the % yield of 1-bromobutane

Describes few purification steps.

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.

element

(b) ALLOW any combination of skeletal OR

19 4 AO3.2 structural OR displayed formula as long as

×4 unambiguous

Step 1 The oxygen atom of the alcohol group accepts a For CH3CH2CH2, ALLOW CH3(CH2)2, C3H7

proton to form a positively-charged intermediate.

2 marks IGNORE dipoles

Curly arrow ------------------------------------------------

ALLOW curly arrow to

H of H–O–SO3H OR H–Br

IGNORE absence of curly arrow from H–O or

from H–Br

Intermediate + charge MUST be on O of intermediate

Step 2 Bromide ions react with the intermediate by

nucleophilic substitution to form 1-bromobutane.

2 marks

2 possible routes: Curly arrow must

EITHER • start from, OR be traced back to any point

Curly across width of lone pair on :Br– OR :OH

arrow OR start from – charge on Br–

(Lone pair NOT needed if curly arrow shown

Curly –

from – charge on Br )

arrow

IGNORE final products:

OR 1-bromobutane and H2O

Curly

arrow

IF C H CH –O+H is not shown,

37 2 2

ALLOW intermediate mark for carbocation:

Curly C3H7CH2+

arrow

20 element

ALLOW 2 marks max for mechanism without

positively charge intermediate, i.e.

Curly

arrow

Curly

arrow

If in doubt, contact Team Leader

How to answer it

Preparation, Purification, and Mechanism of 1-bromobutane

What this question tests

This multi-step synoptic question assesses practical chemistry skills (drawing reflux setups, organic liquid separation, and purification techniques), stoichiometry (percentage yield calculations involving density and molar mass), and organic reaction mechanisms (nucleophilic substitution of alcohols using a protonated intermediate).

Question Part (a) - Reflux, Yield Calculation & Purification

Practical Techniques & Percentage Yield

✅ Expected Responses (Level 3 Criteria)

Reflux Diagram: Pear-shaped/round-bottom flask connected to a vertical condenser with water-in at the bottom and out at the top. No airtight seals (must be open to the air).
Yield Calculation: Correctly calculates 45.2% (or 35.6% if density error made).
Purification Steps: Use separating funnel, run off lower aqueous layer, wash with an acid/alkali/water, dry with an anhydrous salt (e.g., anhydrous MgSO₄ or Na₂SO₄ ), and redistill collecting the fraction boiling at 102°C.

💡 Key Knowledge

Reflux allows heating organic reaction mixtures for prolonged periods without losing volatile reactants or products, as vapours condense and drip back down.
Density is required to convert the volume of product obtained ( 6.10 cm³ ) into mass ( mass = volume × density ).
Anhydrous salts must be insoluble in the organic product and chemically unreactive with it.

🧠 Exam Technique (Levels of Response)

This is a 6-mark Level of Response question. To hit Level 3, you must successfully include all three elements: a labelled diagram, a correct numerical yield, and detailed purification steps.
Do not write vague statements like "wash it" or "distill it"—specify *how* using a separating funnel, name an appropriate drying agent, and state the exact boiling point collection temperature.

❌ Common Errors

Forgetting to use density ( 1.268 g cm⁻³ ) and dividing the volume ( 6.10 ) directly by molar mass instead of calculating mass first.
Drawing a closed reflux apparatus (trapped systems can explode!).
Failing to identify which layer is the organic layer in the separating funnel (1-bromobutane has a higher density than water, so it forms the bottom layer).

📐 Step-by-Step Calculation Guide

Moles of reactant (butan-1-ol): Mass = 9.25 g . Molar mass of C₄H₉OH = 74.0 g mol⁻¹ .
Moles = 9.25 / 74.0 = 0.125 mol .
Mass and moles of actual product (1-bromobutane): Volume = 6.10 cm³ , Density = 1.268 g cm⁻³ .
Mass = 6.10 × 1.268 = 7.7348 g .
Molar mass of C₄H₉Br = 136.9 g mol⁻¹ .
Moles = 7.7348 / 136.9 = 0.0565 mol .
Percentage Yield: ( 0.0565 / 0.125 ) × 100 = 45.2% .

Mark Scheme Note: Level 3 (5–6 marks) requires all 3 components well described. Level 3 (3–4 marks) for 2 components, Level 1 (1–2 marks) for 1 component.

Question Part (b) - Organic Mechanism

Nucleophilic Substitution Mechanism (2-Step)

✅ Correct Answer Structure

Step 1: Protonation of the alcohol oxygen by H⁺ forming a positively charged intermediate ( C₃H₇—CH₂—OH₂⁺ ). Curly arrow from oxygen lone pair to H⁺ .
Step 2: Bromide ion ( Br⁻ ) acts as a nucleophile, using a lone pair to attack the CH₂ carbon, releasing a water molecule. Curly arrow from Br⁻ lone pair to carbon, and C-O bond arrow breaking towards the oxygen.

💡 Key Knowledge

Alcohols are poor substrates for direct nucleophilic substitution because the hydroxide ion ( OH⁻ ) is a poor leaving group.
Adding an acid ( H₂SO₄ + NaBr in situ) protonates the —OH group, turning it into —OH₂⁺ (water), which is an excellent leaving group.

🧠 Exam Technique

Curly arrows must start precisely at a bond or a lone pair and point towards the destination atom or bond.
Ensure the positive charge ( ⁺ ) is explicitly shown on the oxygen atom of the intermediate —OH₂⁺ .
Accepts either structural formulas ( C₃H₇ ) or full displayed formulas for the alkyl chain.

❌ Common Errors

Starting curly arrows in mid-air instead of directly from a lone pair or bond.
Omitting the positive charge on the intermediate oxygen, which loses an immediate marking point.
Forgetting the second curly arrow in Step 2 that shows the departure of the water molecule.

Mark Scheme Note: 4 marks total. 2 marks for Step 1 (protonation + intermediate structure with correct charges), 2 marks for Step 2 (nucleophilic attack of Br⁻ with correct curly arrows forming 1-bromobutane and water).

Topics

Module 4: Core organic chemistry · Module 6: Organic chemistry and analysis · Practical Activity Groups · 4.2 Alcohols, haloalkanes and analysis · PAG 5: Synthesis of an organic liquid · PAG 7: Qualitative analysis of organic functional groups · 6.2 Nitrogen compounds, polymers and synthesis

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.