AQA A-Level Chemistry Paper 3, June 2025: Question 18
1 mark · Easy difficulty · Multiple Choice
Identify the correct structural formula for 4-chloro-2-methylpent-2-enoic acid from four multiple choice options.
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Question text
18 What is the correct structural formula for 4-chloro-2-methylpent-2-enoic acid?
[1 mark]
A CH3CCl=CHCH(CH3)COOH
B (CH3)2C=CHCHClCOOH
C CH3CHClCH=C(CH3)COOH
D (CH3)2CHCH=CClCOOH
Mark scheme
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18 C 1 (AO1) CH3CHClCH=C(CH3)COOH
How to answer it
IUPAC Naming & Structural Formulae of Multifunctional Alkenoic Acids
This question evaluates your core understanding of organic nomenclature rules (AO1):
- Identifying the principal functional group to define numbering priority (carboxylic acid C=1).
- Locating unsaturation (alkene C=C double bond position).
- Correctly assigning branch prefixes ( -CH₃ ) and halogeno substituents ( -Cl ) along a 5-carbon parent chain.
- Translating between systematic IUPAC names and condensed structural formulae.
Question 18 Analysis
4-chloro-2-methylpent-2-enoic acid [1 Mark]
✅ Correct Answer
Correct Option: C
CH₃CHClCH=C(CH₃)COOH
💡 Key Knowledge
- Priority Rule: The carboxylic acid carbon atom is automatically designated as C1.
- Parent Chain: "pent" means a 5-carbon continuous chain containing the principal functional group.
- Suffix / Infix: "-2-en-" indicates the C=C double bond starts at carbon 2 and connects to carbon 3.
- Substituents:
- "2-methyl" → a -CH₃ group attached at C2.
- "4-chloro" → a -Cl group attached at C4.
📐 Step-by-Step Chain Construction
Break the systematic IUPAC name down from the principal functional group (C1) through to the end of the chain (C5):
Carboxylic acid carbon: -COOH (Carbon 1).
C2 has a methyl branch and a double bond: =C(CH₃)- . Attached to C1, this forms: =C(CH₃)COOH .
The double bond is between C2 and C3: -CH= . Joined to C2: -CH=C(CH₃)COOH .
C4 is bonded to a hydrogen and a chlorine: -CHCl- . Joined: -CHClCH=C(CH₃)COOH .
Terminal methyl group: CH₃- .
Full condensed formula written from left-to-right (C5 → C1):
CH₃–CHCl–CH=C(CH₃)–COOH, which matches Option C.
🧠 Exam Technique & Option Elimination
- Option A: CH₃CCl=CHCH(CH₃)COOH
Numbering from COOH gives the double bond at C3 and chlorine at C4. This is 4-chloro-2-methylpent-3-enoic acid (wrong alkene position). - Option B: (CH₃)₂C=CHCHClCOOH
Chlorine is at C2 and two methyls are at C4. This is 2-chloro-4-methylpent-3-enoic acid. - Option D: (CH₃)₂CHCH=CClCOOH
Chlorine is at C2 and methyl branch is at C4. This is 2-chloro-4-methylpent-2-enoic acid (substituents inverted).
❌ Common Errors & Traps
- Numbering from the wrong end: Starting numbering from the terminal alkyl group instead of the -COOH carbon leads directly to choosing Option A.
- Confusing the positions of substituents: Selecting D because it has the double bond at position 2, but failing to check that chlorine is at C2 instead of C4.
- Missing the C-count in the main chain: Mistaking condensed brackets such as (CH₃)₂C= in B and D as a simple straight chain rather than a branched terminus.
Topics
Organic Chemistry · 3.3.1 Introduction to Organic Chemistry · 3.3.9 Carboxylic Acids and Derivatives
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.