OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2025: Question 10
1 mark · Medium difficulty · Multiple Choice
Determine the number of stereoisomers for the compound CH3CH(Br)CH=CHCH(Cl)CH3.
Practise this questionQuestion
Question text
10 How many stereoisomers does the compound CH3CH(Br)CH=CHCH(Cl)CH3 have?
A 2
B 4
C 6
D 8
Your answer
[1]
Mark scheme
Show the mark scheme
10 D 1 ALLOW 8
How to answer it
Stereoisomerism in Multifunctional Alkenes
This multiple-choice question assesses your ability to identify different forms of stereoisomerism within a single organic molecule and determine the total number of stereoisomers:
- Identifying chiral centres (optical isomerism) in a carbon chain.
- Identifying carbon–carbon double bonds ( C=C ) exhibiting E/Z isomerism (geometric isomerism).
- Recognising molecular asymmetry (distinct functional ends prevent meso-forms).
- Applying the mathematical rule 2ⁿ to calculate the total number of stereoisomers.
Question 10 Breakdown
Compound: CH₃CH(Br)CH=CHCH(Cl)CH₃
✅ Correct Answer
Option D: 8 stereoisomers
📐 Step-by-Step Analysis
- Chiral centre 1 (C2):
Attached to -H , -CH₃ , -Br , and -CH=CHCH(Cl)CH₃ (4 different groups → 2 optical isomers). - Alkene unit (C3=C4):
Both C3 and C4 have two distinct groups attached ( -H and an alkyl halide group) → 2 geometric isomers (E and Z). - Chiral centre 2 (C5):
Attached to -H , -CH₃ , -Cl , and -CH=CHCH(Br)CH₃ (4 different groups → 2 optical isomers). - Symmetry check:
One terminal group has bromine ( -Br ) and the other has chlorine ( -Cl ). The molecule is unsymmetrical, meaning no meso forms exist. - Total stereoisomers:
2 × 2 × 2 = 2³ = 8 .
💡 Key Knowledge
- Stereoisomers have the same structural formula but a different arrangement of atoms in 3D space.
- They include both optical isomers (enantiomers around chiral carbons) and geometric isomers (E/Z isomers across restricted rotation double bonds).
- When an unsymmetrical molecule contains n independent stereocentres/stereogenic units, the maximum number of stereoisomers is given by: Total Isomers = 2ⁿHere, n = 3 (two chiral carbons + one E/Z double bond).
🧠 Exam Technique
- Draw out the skeletal or displayed formula: Condensed formulas hide geometric isomerism. Drawing the C=C double bond flat with trigonal planar angles (120°) immediately reveals whether E/Z isomerism is possible.
- Check both ends of the chain: If both ends had the same halogen (e.g., both -Br ), meso stereoisomers would reduce the total count. Because one is -Br and the other is -Cl , all 8 combinations are unique.
- Tree Diagram Approach:
• (E) alkene → 4 optical combinations (RR, RS, SR, SS)
• (Z) alkene → 4 optical combinations (RR, RS, SR, SS)
Total = 4 + 4 = 8.
❌ Common Errors
- Forgetting the double bond (choosing B: 4): Many students spot the two chiral carbons and calculate 2² = 4 , completely forgetting that "stereoisomers" includes E/Z isomerism.
- Only counting optical or geometric isomerism: Assuming stereoisomerism only means E/Z (choosing A: 2).
- Incorrectly assuming internal symmetry: Assuming the two chiral centres are identical and cancelling out enantiomers as meso forms without checking the halogen substituents.
Topics
Module 4: Core organic chemistry · Module 6: Organic chemistry and analysis · 4.1 Basic concepts and hydrocarbons · 6.2 Nitrogen compounds, polymers and synthesis
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.