OCR A-Level Chemistry AS Depth in chemistry (02), June 2022: Question 5

13 marks · Medium difficulty · Structured Questions

Analyze the preparation of 2-bromobutane through radical substitution, electrophilic addition, and nucleophilic substitution reactions involving atom economy, percentage yield calculations, and purification procedures.

Practise this question

Question

A structured chemistry exam question about the preparation of 2-bromobutane via three different reactions (Reaction 5.1, Reaction 5.2, and Reaction 5.3). Part (a) covers free radical substitution of butane, asking for conditions, propagation equations, and reasons for low yield. Part (b) covers atom economy comparison between reactions 5.1 and 5.2. Part (c) involves a percentage yield calculation for preparing 2-bromobutane from butan-2-ol, and describing the practical purification procedure using a separating funnel, drying agent, and distillation.
Question text

5 2-Bromobutane, CH3CH2CHBrCH3, can be prepared by three different methods.

The relative molecular mass, Mr, of 2-bromobutane is 136.9.

(a) 2-Bromobutane can be prepared by reacting butane with bromine (Reaction 5.1).

CH3CH2CH2CH3 + Br2 CH3CH2CHBrCH3 + HBr Reaction 5.1

The reaction is initiated by the formation of bromine radicals from bromine.

(i) State the conditions for the formation of bromine radicals from bromine.

… [1]

(ii) Write two equations for the propagation steps in the mechanism for Reaction 5.1.

Use structural formulae for organic species and dots (•) for unpaired electrons on

radicals.

CH3CH2CH2CH3 + … + …

… + … + …

[2]

(iii) The yield of CH3CH2CHBrCH3 is only 30%.

Suggest two reasons why the yield of CH3CH2CHBrCH3 is so low.

1 …

2 …

[2]

(b) 2-Bromobutane can also be prepared by reacting but-2-ene, CH3CH=CHCH3, with hydrogen

bromide, HBr (Reaction 5.2).

CH3CH=CHCH3 + HBr CH3CH2CHBrCH3 Reaction 5.2

Explain, in terms of atom economy, why Reaction 5.2 is more sustainable than

Reaction 5.1.

Include calculations to justify your answer.

… 13 [2]

(c) 2-Bromobutane can also be prepared by reacting butan-2-ol, CH3CH2CHOHCH3, with

sodium bromide and sulfuric acid (Reaction 5.3).

CH CH CHOHCH + H+ + Br– CH CH CHBrCH + H O Reaction 5.3

32 3 3 2 3 2

2-Bromobutane is a liquid with a boiling point of 91 °C and does not mix with water.

(i) A student plans to prepare 10.0 g of 2-bromobutane using Reaction 5.3.

The percentage yield is 67.0%.

Calculate the mass of CH3CH2CHOHCH3 needed for this preparation.

Give your answer to 3 significant figures.

mass = … g [3]

(ii) The student mixes butan-2-ol, sodium bromide and sulfuric acid in a pear-shaped flask,

and refluxes the mixture.

After 1 hour, the mixture in the flask has separated into two layers: an aqueous layer

and an organic layer.

Describe the procedures the student would need to carry out to obtain a pure, dry

sample of 2-bromobutane from this mixture.

… [3]

Mark scheme

Show the mark scheme The official mark scheme providing answers and guidance for all parts of question 5, including UV conditions, radical propagation equations, low yield explanations involving further substitution, atom economy percentage calculations, stoichiometry calculations for mass of butan-2-ol (8.07 g), and purification steps using a separating funnel, drying agent, and distillation.

AO

Question Answer Marks Guidance

element

5 (a) (i) UV OR ultraviolet 1 AO1.1 ALLOW Sunlight

IGNORE Temperature

(ii) 2 AO2.5 ALLOW Displayed or Skeletal formulae

CH3CH2CH2CH3 + Br • → CH3CH2CHCH3 + HBr ×2 ALLOW 1 mark if BOTH equations are

• ‘correct’ using molecular formulae, i.e.

CH3CH2CH2CH3 + Br • → C4H9 • + HBr

CH3CH2CHCH3 + Br2 → CH3CH2CHBrCH3 + Br •

• C4H9 • + Br2 → C4H9Br + Br •

IGNORE position of • within CH3CH2CHCH3 •

ALLOW 1 mark if incorrect structure of

intermediate radical is used, e.g.

CH3CH2CH2CH2 • for CH3CH2CHCH3 •

(iii) Further substitution 2 AO3.2 ALLOW multisubstitutation, including

OR ×2 examples

formation of di/ tri / etc. bromobutanes ALLOW an example of a different termination

OR product

produces different termination products ALLOW more than one hydrogen (atom) can

OR be replaced

more than one termination step ALLOW radicals react with each other to form

other products

Formation of 1-bromobutane

OR (Br) subsitution in a different position

AO

Question Answer 20 Marks Guidance

element

(b) % atom economy for butane and bromine (5.1) 2 AO2.2 Calculator: 62.85583104

136.9

= 217.8 × 100 = 62.9%

AO1.2 ALLOW calculation for 5.2

atom economy for but-2-ene and HBr (5.2) is 100%

ALLOW Calculations not expressed as a %

i.e. 0.629 and 1.

(c) (i) FIRST CHECK THE ANSWER ON ANSWER LINE 3 AO2.4 ALLOW ECF throughout

If answer = 8.07 g award 3 marks ×3

CARE: Intermediate rounding may give 8.06 g which is IGNORE trailing zeroes in intermediate

acceptable for 3 marks working, e.g. 0.073 for 0.0730

--------------------------------------------------------------------

ALLOW 3 SF or more, correctly rounded

n(2-bromobutane)

10.0 Calculator: 0.7304601899

= = 0.073(0)…. (mol)

136.9

n(CH3CH2CHOHCH3) Calculator: 0.1089552239

= 0.0730…. × 67.0 = 0.109 (mol) ALLOW alternative method mass

• Theoretical mass of 2-bromobutane

mass CH3CH2CHOHCH3 10.0

= 100 × = 14.9….. (g)

= 0.109 × 74.0 = 8.07 g 67.0

3 SF required Calculator: 14.925373

• Theoretical n(CH3CH2CHBrCH3)

14.923373

= = 0.1902 (mol)

136.9

• Mass of CH3CH2CHOHCH3

= 0.109 × 74.0 = 8.07 g

Common Errors for 2 marks

5.41 g (no % yield)

3.62 g (inverted yield)

AO

element

(ii) Separating funnel (to separate aqueous and organic layers) 3 AO3.3

×3

Dry organic layer with anhydrous salt ALLOW Use a drying agent

ALLOW appropriate example of an

anhydrous salt e.g. MgSO4, CaCl2

Distil and collect fraction at 91oC

Total 13

How to answer it

Preparation and Synthetic Routes of 2-Bromobutane

What this question tests

This question evaluates your core organic chemistry synthesis and mechanism knowledge, including free radical substitution mechanisms, atom economy comparisons, multi-step reacting mole calculations involving percentage yield, and practical purification techniques such as separation, drying, and distillation.

Question 5 (a)

Free Radical Substitution of Butane

✅ Correct Answers

  • (i) Conditions: UV light (or ultraviolet / sunlight).
  • (ii) Propagation Step 1: CH₃CH₂CH₂CH₃ + Br• → CH₃CH₂CH•CH₃ + HBr
  • (ii) Propagation Step 2: CH₃CH₂CH•CH₃ + Br₂ → CH₃CH₂CHBrCH₃ + Br•
  • (iii) Reasons for low yield: Further substitution (multi-substitution / di-, tri-bromobutanes), and formation of 1-bromobutane (substitution at a different position).

💡 Key Knowledge

  • Free radical substitution requires UV light to homolytically split bromine molecules into radicals.
  • Propagation equations must feature a radical on both sides of each equation and use dots ( • ) for unpaired electrons.

🧠 Exam Technique

When writing propagation steps, check your balancing carefully. A radical must react to form a new radical in every propagation sequence. Do not confuse initiation with propagation steps.

❌ Common Errors

Placing the radical dot incorrectly on carbon atoms or omitting the dot entirely will cost you marks. Students also frequently lose marks in part (iii) by giving vague answers like "side reactions" instead of naming further substitution or isomeric substitution products.

Question 5 (b)

Atom Economy Comparison

✅ Correct Answers

Reaction 5.1 atom economy = 62.9% (or 63%).

Reaction 5.2 atom economy = 100%.

Conclusion: Reaction 5.2 is more sustainable because it is an addition reaction with only one product, resulting in zero waste.

📐 Calculation Breakdown

  1. Molar mass of 2-bromobutane: 136.9 g mol⁻¹
  2. Molar mass of butane + bromine (Rxn 5.1): 58.0 + 159.8 = 217.8 g mol⁻¹
  3. Atom Economy: (136.9 / 217.8) × 100 = 62.855...% = 62.9%
  4. Reaction 5.2: Addition reaction combining all atoms into a single product = 100%.
Question 5 (c) (i)

Percentage Yield and Stoichiometry Calculation

✅ Correct Answers

Final Mass of Butan-2-ol needed = 8.07 g (3 significant figures required).

Mark scheme accepts answers resulting from intermediate rounding giving 8.06 g. ECF applies throughout.

📐 Step-by-Step Calculation

  1. Moles of 2-bromobutane desired: n = 10.0 / 136.9 = 0.07304 mol
  2. Account for percentage yield (67.0%): Because yield is less than 100%, you need more starting material. Theoretical moles of 2-bromobutane = 0.07304 × (100 / 67.0) = 0.10895 mol
  3. Moles of butan-2-ol needed (1:1 ratio): 0.10895 mol
  4. Mass of butan-2-ol: mass = moles × Mr = 0.10895 × 74.0 = 8.07 g (to 3 SF).

❌ Common Calculation Traps

Inverted Percentage Yield: Multiplying by (67.0 / 100) instead of dividing gives 5.41 g (common error). Always ask yourself: Do I need more or less starting material to get my final product? Since yield is partial, you always need a larger mass of reactant.

Question 5 (c) (ii)

Practical Purification Techniques

✅ Correct Answers

  • Step 1: Use a separating funnel to separate the aqueous and organic layers.
  • Step 2: Add an anhydrous salt (e.g., anhydrous MgSO₄ or CaCl₂) to dry the organic layer.
  • Step 3: Distill and collect the fraction boiling at 91 °C.

🧠 Exam Technique & Terminology

Examiners are very strict with practical terminology. You must specify an anhydrous salt for drying—stating just "salt" will lose the mark. Ensure you mention collecting the specific fraction at the specified boiling point (91 °C) during distillation.

Topics

Module 2: Foundations in chemistry · Module 4: Core organic chemistry · Practical Activity Groups · PAG 5: Synthesis of an organic liquid · 2.1 Atoms and reactions · 4.1 Basic concepts and hydrocarbons · 4.2 Alcohols, haloalkanes and analysis

Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.