Edexcel A-Level Chemistry AS Paper 2, June 2025: Question 2

1 mark · Medium difficulty · Multiple Choice

Identify which alkene must be named using the E/Z naming system rather than the cis/trans system.

Practise this question

Question

Multiple-choice question asking: 'Which compound must be named using the E/Z system and could not be named using the cis/trans system?'. Option A: HOOC—CH2—CH2—COOH. Option B: HOOC—CH2—CBr=CH—COOH. Option C: HOOC—CH2—CBr=CBr2. Option D: HOOC—(CH2)2—CH=CH—(CH2)2—COOH. Total for Question 2 is 1 mark.

Mark scheme

Show the mark scheme Mark scheme table for Question 2 showing that the only correct answer is B (HOOC—CH2—CBr=CH—COOH). It explains: A is incorrect because this molecule is not unsaturated, i.e. it has no C=C so it cannot be named using either system; C is incorrect because this molecule has two identical groups on one of the C=C carbons so it cannot be named using either system; D is incorrect because this molecule has two identical groups on both C=C carbons so it can be named using either system. Mark is 1.

How to answer it

Naming Geometric Isomers: E/Z vs cis/trans Nomenclature

📋 What this question tests

This question assesses your understanding of stereoisomerism in alkenes, specifically:

  • The conditions required for geometric (geometric/stereoisomerism) to occur around a C=C double bond.
  • The crucial limitation of cis/trans notation (requires two identical groups across the double bond).
  • The universal application of the Cahn-Ingold-Prelog (CIP) E/Z priority system.

Question 2 Analysis

Multiple Choice (1 Mark)

✅ Correct Answer

B: HOOC—CH₂—CBr=CH—COOH

This molecule satisfies the conditions for geometric isomerism because each carbon of the C=C double bond is bonded to two different groups:

  • C2: bonded to —H and —COOH
  • C3: bonded to —Br and —CH₂COOH

Because all four groups attached to the double-bond carbons are completely different, there is no shared reference group to compare as cis or trans. Hence, it must be designated using the CIP E/Z priority system.

💡 Key Knowledge

  • Requirement for geometric isomerism: Restricted rotation around the C=C bond AND two different groups attached to each individual carbon atom of the double bond.
  • When to use cis/trans: Only valid when there is a common, identical group on both carbon atoms of the double bond (e.g., both carbons have an —H attached).
  • When E/Z is required: Applies to any alkene showing stereoisomerism using atomic number priority (CIP rules). If all four substituents are distinct, cis/trans breaks down completely and E/Z must be used.

📐 Step-by-Step Option Elimination

  1. Check Option A (HOOC—CH₂—CH₂—COOH):
    Contains only single C—C bonds (saturated chain). Free rotation exists, so no geometric isomerism is possible at all.
  2. Check Option C (HOOC—CH₂—CBr=CBr₂):
    The right-hand carbon is bonded to two identical bromine atoms (—Br and —Br). Because it has two identical groups attached to the same double-bonded carbon, it cannot exhibit stereoisomerism under either system.
  3. Check Option D (HOOC—(CH₂)₂—CH=CH—(CH₂)₂—COOH):
    Both carbons have an —H and an —(CH₂)₂COOH group. Because identical groups (—H or —(CH₂)₂COOH) exist on both sides, this molecule can be named using either cis/trans (specifically cis or trans depending on orientation) or E/Z. The question asks for the one that could not be named with cis/trans.
  4. Conclusion: Molecule B has geometric isomerism (two different groups on each carbon) but four completely distinct ligands across the double bond (—H, —COOH, —Br, —CH₂COOH). Therefore, it must be named using E/Z and cannot use cis/trans.

🧠 Exam Technique

  • Draw out the C=C center: When given condensed structural formulas, sketch the two carbon atoms of the C=C double bond and show their four separate bonds branching off at 120° angles.
  • Scan for identical groups on the same carbon first: Look at each alkene carbon individually. If either carbon holds two identical atoms/groups (like =CBr₂ in option C), eliminate immediately!
  • Remember the rule for cis/trans: You need a matching pair across the bond (one on each carbon). No matching pair means cis/trans is impossible.

❌ Common Errors & Examiner Traps

  • Confusing Option B and D: Students often pick D because it is a symmetrical alkene that clearly shows E/Z, forgetting that D can also be named cis/trans. Pay close attention to the negative constraint: "could NOT be named using cis/trans".
  • Assuming all alkenes have geometric isomers: Missing the =CBr₂ in Option C and assuming all unsaturated molecules show E/Z isomerism.
  • Misreading the saturated chain: Glancing at Option A and missing that it has no double bond at all.
Mark Scheme Breakdown:
• B is the only correct answer [1 mark].
• A is incorrect: saturated (no C=C double bond).
• C is incorrect: right-hand carbon has two identical —Br atoms.
• D is incorrect: both C=C carbons share identical groups (—H and —(CH₂)₂COOH), allowing cis/trans.

Topics

Organic Chemistry · Topic 6: Organic Chemistry I

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.