OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2019: Question 4

1 mark · Medium difficulty · Multiple Choice

Identify the correct E-Z stereoisomer term for the given alkene structure

Practise this question

Question

Multiple choice question 4 showing the chemical structure of a stereoisomer with a central carbon-carbon double bond, substituted by H3C and CH3CH2 on the left carbon, and CH(CH3)2 and CH2CH2CH3 on the right carbon. Below the structure, four multiple-choice options are given: A (cis-), B (trans-), C (E-), and D (Z-), along with an answer box.
Question text

4 The structure of a stereoisomer is shown below.

H3C CH(CH3)2

C C

CH3CH2 CH2CH2CH3

Which term correctly describes this stereoisomer?

A cis-

B trans-

C E-

D Z-

Your answer

[1]

Mark scheme

Show the mark scheme Mark scheme for question 4 indicating that the correct answer is C, awarding 1 mark.

4 C 1 AO1.2 ALLOW E (This is the correct term)

How to answer it

Determining E/Z Stereoisomerism

What this question tests

This question assesses your ability to apply Cahn-Ingold-Prelog (CIP) priority rules to assign E/Z stereoisomerism to an alkene with four different groups attached to the C=C double bond, testing AO1.2 knowledge and application.

Question 4 Analysis

Marks: 1

✅ Correct Answer

The correct option is C (E-).

Mark Scheme: C (ALLOW E) — [1 mark]

💡 Key Knowledge

  • C=C Double Bond Restriction: Restricted rotation around the double bond creates stereoisomers.
  • CIP Priority Rules: Priority is assigned based on atomic number (higher atomic number = higher priority).
  • E vs Z Definitions: E (entgegen) means higher priority groups are on opposite sides of the double bond. Z (zusammen) means they are on the same side.

🧠 Exam Technique

Split the C=C bond down the middle vertically into a left carbon and a right carbon. Evaluate the two attached groups on the left carbon independently from the two on the right carbon using CIP rules before comparing their relative positions.

❌ Common Errors

Students frequently default to using cis- and trans- terminology. These terms are only appropriate when each carbon of the C=C bond has at least one identical substituent (e.g., two hydrogens). When all four groups are different, you must use E/Z nomenclature.

Step-by-Step Breakdown

📐 Applying CIP Priority Rules

  1. Left-hand carbon of C=C: Attached to -CH₃ and -CH₂CH₃ (ethyl). Compare the atoms directly attached to the C=C carbon: both are carbon atoms. Move to the next atoms along: -CH₃ has H atoms attached, whereas -CH₂CH₃ has a carbon attached (higher atomic number than H). Therefore, -CH₂CH₃ has higher priority on the left carbon.
  2. Right-hand carbon of C=C: Attached to -CH(CH₃)₂ (isopropyl) and -CH₂CH₂CH₃ (propyl). Both are bonded via a carbon atom. Looking further down the chains, the isopropyl group branches into two carbons, giving it higher priority than the linear propyl chain. Therefore, -CH(CH₃)₂ has higher priority on the right carbon.
  3. Compare positions: The higher priority ethyl group on the left points downwards, while the higher priority isopropyl group on the right points upwards. Because the higher priority groups are on opposite sides, the configuration is E.

Topics

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

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.