OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2025: Question 6
1 mark · Medium difficulty · Multiple Choice
Identify which reaction producing propene has the lowest atom economy.
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Question text
6 Propene can be prepared using the reactions shown.
Which reaction has the lowest atom economy?
A C3H7Br C3H6 + HBr
B C12H26 2C3H6 + C6H14
C C8H18 C3H6 + C5H12
D C3H7OH C3H6 + H2O
Your answer
[1]
Mark scheme
Show the mark scheme
6 A 1
How to answer it
Determining the Reaction with the Lowest Atom Economy
This question evaluates your understanding of atom economy in organic synthesis reactions (elimination, cracking, dehydration). It tests your ability to identify the desired product versus waste by-products, account for stoichiometric balancing numbers (coefficients), and quickly calculate or deduce percentage atom economy under multiple-choice time constraints.
Question 6: Multiple Choice
Synthesis of Propene (C₃H₆)
✅ Correct Answer
A
Equation: C₃H₇Br → C₃H₆ + HBr
Reaction A produces propene alongside hydrogen bromide (HBr). Because bromine has a high atomic mass (Aᵣ = 79.9), a significant proportion of the reactant's mass ends up in the waste product, yielding the lowest atom economy of approximately 34.2%.
💡 Key Knowledge
- Atom Economy Formula:
Atom Economy = (Mᵣ of desired product / Total Mᵣ of all products) × 100% - Remember that total Mᵣ of all products equals total Mᵣ of all reactants (Law of Conservation of Mass).
- Desired product: Propene, C₃H₆ (Mᵣ = (3 × 12.0) + (6 × 1.0) = 42.0).
- Stoichiometric balancing numbers must always be included (e.g., 2C₃H₆ in reaction B).
📐 Step-by-Step Calculation Breakdown
Calculate the atom economy for each reaction where propene (C₃H₆) is the target product:
| Option | Mass of Desired Product | Total Mass of Reactants / Products | Atom Economy (%) |
|---|---|---|---|
| A: C₃H₇Br → C₃H₆ + HBr | 1 × 42.0 = 42.0 | Mᵣ(C₃H₇Br) = 36.0 + 7.0 + 79.9 = 122.9 | (42.0 / 122.9) × 100 = 34.2% (Lowest) |
| B: C₁₂H₂₆ → 2C₃H₆ + C₆H₁₄ | 2 × 42.0 = 84.0 | Mᵣ(C₁₂H₂₆) = (12 × 12.0) + (26 × 1.0) = 170.0 | (84.0 / 170.0) × 100 = 49.4% |
| C: C₈H₁₈ → C₃H₆ + C₅H₁₂ | 1 × 42.0 = 42.0 | Mᵣ(C₈H₁₈) = (8 × 12.0) + (18 × 1.0) = 114.0 | (42.0 / 114.0) × 100 = 36.8% |
| D: C₃H₇OH → C₃H₆ + H₂O | 1 × 42.0 = 42.0 | Mᵣ(C₃H₇OH) = 36.0 + 8.0 + 16.0 = 60.0 | (42.0 / 60.0) × 100 = 70.0% |
🧠 Exam Technique & Shortcut
- Look at the waste product: In A, C, and D, exactly 1 mole of C₃H₆ is formed. The lowest atom economy must be the reaction with the heaviest waste product:
- Waste in A = HBr (Mᵣ = 1.0 + 79.9 = 80.9)
- Waste in C = C₅H₁₂ (Mᵣ = (5 × 12.0) + 12.0 = 72.0)
- Waste in D = H₂O (Mᵣ = 18.0)
- In B, 84 g of product is made per 86 g of waste (~50%), easily ruling it out. This method lets you solve the question in under 30 seconds without full division!
❌ Common Errors & Pitfalls
- Ignoring the coefficient in Option B: Forgetting that 2 moles of C₃H₆ are formed, calculating 42 / 170 = 24.7% and incorrectly choosing B.
- Confusing Atom Economy with Percentage Yield: Percentage yield depends on practical experimental losses, whereas atom economy is purely theoretical from the stoichiometric equation.
- Misreading "lowest" as "highest": D has the highest atom economy (70.0%); rushing can lead to selecting the wrong extreme.
Topics
Module 2: Foundations in chemistry · Module 4: Core organic chemistry · 2.1 Atoms and reactions · 4.1 Basic concepts and hydrocarbons
Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.