OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2018: Question 7

1 mark · Medium difficulty · Multiple Choice

Identify which reaction for the preparation of ethanol has the lowest atom economy from four given chemical equations.

Practise this question

Question

Multiple choice question 7 asking which reaction has the lowest atom economy for the preparation of ethanol. Four options are given: A, C6H12O6 yields 2C2H5OH plus 2CO2; B, C2H4 plus H2O yields C2H5OH; C, C2H5Br plus H2O yields C2H5OH plus HBr; D, CH3COOC2H5 plus H2O yields C2H5OH plus CH3COOH. There is a box for the answer and a mark allocation of [1] at the bottom right.
Question text

7 Ethanol can be prepared by different reactions.

Which reaction has the lowest atom economy?

A C6H12O6 → 2C2H5OH + 2CO2

B C2H4 + H2O → C2H5OH

C C2H5Br + H2O → C2H5OH + HBr

D CH3COOC2H5 + H2O → C2H5OH + CH3COOH

Your answer

[1]

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer for question 7 is C, with 1 mark awarded.

7 C 1

How to answer it

Atom Economy in Ethanol Preparations

What this question tests

This question assesses your understanding of atom economy calculations in organic synthesis, specifically how to apply the formula using balanced chemical equations without necessarily needing to calculate full numerical molar masses for every option.

Question 7

Determining the Lowest Atom Economy

✅ Correct Answer

C

Reaction C has the lowest atom economy because it produces two separate product molecules (one of which, HBr, is a heavy waste byproduct), whereas addition reactions (like B) have a 100% atom economy.

💡 Key Knowledge

  • Atom Economy Formula:
    (Mr of desired product / Total Mr of all products) × 100%
  • Addition Reactions (Option B): Reactants combine to form a single product. Therefore, atom economy is automatically 100%.
  • Substitution/Elimination/Hydrolysis (Options A, C, D): Produce more than one product, meaning atoms are wasted, lowering the atom economy below 100%.

🧠 Exam Technique

Save valuable time in multiple-choice questions by inspecting stoichiometry first:

  • Spot addition reactions immediately (e.g., alkene + water → alcohol ) — these are always 100% atom economy.
  • Compare reactions with side-products by looking at the molar mass of the wasted byproduct relative to the desired product. In option C, producing HBr (Mr = 80.9) alongside ethanol (Mr = 46.0) results in massive mass wastage compared to smaller byproducts like CO₂ or CH₃COOH.

❌ Common Errors

  • Confusing percentage yield with atom economy: Atom economy is purely theoretical and based strictly on the balanced equation coefficients and Mr values, not practical losses.
  • Ignoring stoichiometric balancing numbers: Forgetting to multiply the Mr of a product by its balancing coefficient (e.g., the 2CO₂ in option A).
Mark Scheme Allocation: 1 mark for selecting letter C.

Topics

Module 2: Foundations in chemistry · Module 4: Core organic chemistry · 2.1 Atoms and reactions · 4.2 Alcohols, haloalkanes and analysis

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