OCR A-Level Chemistry AS Breadth in chemistry (01), June 2022: Question 17

1 mark · Medium difficulty · Multiple Choice

Identify which alkene is an E stereoisomer from four given structural formulae.

Practise this question

Question

Multiple choice question 17 asking 'Which alkene is an E stereoisomer?'. Four options A, B, C, and D show different alkene structures containing a central carbon-carbon double bond with various alkyl groups and hydrogen atoms attached. An answer box is provided at the bottom left.
Question text

17 Which alkene is an E stereoisomer?

CH3CH2 CH3

A C C

H H

H3C CH2CH3

B C C

H CH3

H CH3

C C C

CH3CH2 C(CH3)3

H3C CH3

D C C

H CH(CH3)2

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme shows the correct answer for question 17 is D.

17 D 1 AO1.2

How to answer it

Identifying E Stereoisomers

What this question tests

This question assesses your understanding of E/Z stereoisomerism in alkenes (specifically AO1.2 knowledge application). You need to apply Cahn-Ingold-Prelog (CIP) priority rules to carbon atoms attached to a C=C double bond and determine whether higher priority groups are on opposite sides (E) or the same side (Z).

Question 17

Full Mark Breakdown & Solution Guide

✅ Correct Answer

D

Mark: 1 / 1

💡 Key Knowledge (CIP Rules)

  • Assign priorities to the two groups attached to each carbon of the C=C double bond based on atomic number (higher atomic number = higher priority).
  • E (Entgegen / Opposite): Higher priority groups are on opposite sides of the double bond.
  • Z (Zusammen / Together): Higher priority groups are on the same side of the double bond.

🧠 Exam Technique

  • Draw a vertical dashed line through the C=C double bond to clearly split the left and right carbons.
  • Examine the left carbon first: compare the two attached groups and label their priority.
  • Examine the right carbon independently: compare its two attached groups and label their priority.
  • Check the spatial arrangement of the two high-priority groups to assign E or Z.

❌ Common Errors

  • Assuming 'E' simply means looking at hydrogen positions (e.g., looking for hydrogens on opposite sides, which fails when four different groups are attached).
  • Confusing the Cahn-Ingold-Prelog priority rules with cis/trans nomenclature.
  • Incorrectly evaluating complex alkyl branching (e.g., comparing -CH(CH₃)₂ vs -CH₃ ).

🔍 Step-by-Step Analysis of Option D

  1. Left carbon of C=C: Attached to -CH₃ and -H . Carbon (atomic number 6) outranks Hydrogen (atomic number 1). Therefore, -CH₃ is the higher priority group on the left. It is pointing up.
  2. Right carbon of C=C: Attached to -CH₃ and -CH(CH₃)₂ (isopropyl group). Both start with carbon, so look at atoms bonded further down: isopropyl has carbons attached to the next carbon, whereas the methyl carbon only bonds to hydrogens. Therefore, -CH(CH₃)₂ is the higher priority group on the right. It is pointing down.
  3. Final E/Z Assignment: The higher priority groups ( -CH₃ pointing up on the left, and -CH(CH₃)₂ pointing down on the right) are on opposite sides of the double bond. This makes it the E stereoisomer.

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 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.