OCR A-Level Chemistry AS Breadth in chemistry (01), June 2018: Question 19

1 mark · Medium difficulty · Multiple Choice

Identify the correct E/Z isomerism for two double bonds in a given branched alkene molecule from multiple-choice options

Practise this question

Question

Multiple choice question 19 shows a skeletal formula of an alkene with two double bonds, labelled 1 and 2. Below it, a table lists four options (A, B, C, D) with combinations of E or Z stereochemistry for double bond 1 and double bond 2.
Question text

19 The molecule below has two double bonds, labelled 1 and 2.

The arrangement around each double bond can be identified as E or Z.

Which row in the table is correct for double bond 1 and double bond 2?

Double bond 1 Double bond 2

A E Z

B Z E

C E E

D Z Z

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme for question 19 indicating the correct answer is C with 1 mark.

19 C 1

How to answer it

Determining E/Z Isomerism in Polyunsaturated Molecules

Question 19 • Multiple Choice • 1 Mark

What this question tests

This question assesses your ability to apply Cahn-Ingold-Prelog (CIP) priority rules to assign E and Z stereoisomers to carbon-carbon double bonds in organic molecules containing multiple sites of unsaturation.

Exam Question Breakdown: Part 19

Evaluating Double Bond 1 and Double Bond 2

✅ Correct Answer

Row C: Double Bond 1 = E, Double Bond 2 = E

Awarded 1 mark for selecting option C.

💡 Key Knowledge

  • Z (zusammen): Higher priority groups are on the same side of the double bond.
  • E (entgegen): Higher priority groups are on opposite sides of the double bond.
  • CIP Priority: Based on atomic number (higher atomic number = higher priority).

🧠 Exam Technique

  • Split each double bond down the middle vertically into left and right carbons.
  • Identify the two groups attached to each carbon atom of the double bond.
  • Compare atomic numbers of the atoms directly attached to each C of the double bond to assign priorities (1 and 2).

❌ Common Errors

  • Confusing E / Z with cis / trans , especially when carbon chains are complex or symmetrical.
  • Failing to consider hidden hydrogen atoms attached to the carbon-carbon double bond carbons.
  • Comparing entire chain masses instead of strictly looking at the atomic number of the first atom directly attached to the double bond carbon.

📐 Step-by-Step Analysis

  1. Analyze Double Bond 1 (Left):
    • Left carbon has: an ethyl group ( -CH₂CH₃ ) and an implicit -H . Priority: ethyl (C > H).
    • Right carbon has: a methyl group ( -CH₃ ) and a carbon chain continuing to the right. Priority: the continuing carbon chain (C vs C—tie-break resolves further along, but clearly higher priority than H/methyl).
    • Position check: The higher priority ethyl group on the left and the higher priority chain on the right are on opposite sides across the double bond axis. Therefore, it is E.
  2. Analyze Double Bond 2 (Right):
    • Left carbon has: the rest of the chain/double bond 1 system and an implicit -H . Priority: the carbon chain.
    • Right carbon has: an ethyl group ( -CH₂CH₃ ) and an implicit -H . Priority: ethyl.
    • Position check: The higher priority carbon chain on the left and the higher priority ethyl group on the right are on opposite sides across the double bond axis. Therefore, it is also E.
  3. Conclusion: Both double bonds are E , matching option C.

Topics

Module 4: Core organic chemistry · 4.1 Basic concepts and hydrocarbons

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.