AQA A-Level Chemistry AS Paper 2, June 2025: Question 13
1 mark · Medium difficulty · Multiple Choice
Determine the molecular formula of the main organic product formed when 2-methylpentane-1,2,4-triol is refluxed with excess acidified potassium dichromate(VI).
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Question text
13 2-Methylpentane-1,2,4-triol is refluxed with an excess of
acidified potassium dichromate(VI).
What is the molecular formula of the main organic product?
[1 mark]
A C6H8O4
B C6H10O3
C C6H10O4
D C6H12O3
Mark scheme
Show the mark scheme
13 C 1 (AO3) C6H10O4
How to answer it
Oxidation of 2-Methylpentane-1,2,4-triol
This question tests your ability to deduce the structural formula of a polyfunctional alcohol from its systematic IUPAC name, classify each alcohol group as primary (1°), secondary (2°), or tertiary (3°), predict the products of complete oxidation under reflux with excess acidified potassium dichromate(VI), and calculate the final molecular formula.
Identifying the Oxidation Product
AQA A-Level Chemistry • Organic Chemistry
✅ Correct Answer
C: C₆H₁₀O₄
💡 Key Knowledge: Alcohol Oxidation
- Primary (1°) Alcohols: Oxidised to aldehydes, and then fully oxidised to carboxylic acids (-COOH) under reflux with excess oxidising agent.
- Secondary (2°) Alcohols: Oxidised to ketones (-C=O).
- Tertiary (3°) Alcohols: Not oxidised by acidified potassium dichromate(VI) because there is no hydrogen atom on the carbon attached to the -OH group.
📐 Step-by-Step Structural & Molecular Deduction
- Draw and identify the functional groups of the reactant:
Reactant: 2-methylpentane-1,2,4-triol (Formula: C₆H₁₄O₃)
Carbon chain numbered 1 to 5:- Carbon 1: -CH₂OH → Primary (1°) alcohol
- Carbon 2: -C(CH₃)(OH)- → Tertiary (3°) alcohol
- Carbon 3: -CH₂- → Alkane bridge
- Carbon 4: -CH(OH)- → Secondary (2°) alcohol
- Carbon 5: -CH₃ → Methyl end group
- Apply oxidation rules under reflux with excess K₂Cr₂O₇ / H⁺:
- C1 (1° alcohol): Completely oxidised to a carboxylic acid: -CH₂OH + 2[O] → -COOH + H₂O
- C2 (3° alcohol): Unchanged (no oxidation): remains -C(OH)(CH₃)-
- C4 (2° alcohol): Oxidised to a ketone: -CH(OH)- + [O] → -C(=O)- + H₂O
- Determine the structure of the product:
IUPAC Name: 2-hydroxy-2-methyl-4-oxopentanoic acid
Condensed structure: HOOC-C(CH₃)(OH)-CH₂-CO-CH₃ - Count atoms to find the molecular formula:
- Carbon (C): 5 (chain) + 1 (methyl branch) = 6 carbons
- Hydrogen (H): 1 (in COOH) + 3 (in CH₃ at C2) + 1 (in OH at C2) + 2 (at C3) + 0 (at C4) + 3 (in CH₃ at C5) = 10 hydrogens
- Oxygen (O): 2 (in COOH) + 1 (in OH at C2) + 1 (in C=O at C4) = 4 oxygens
🧠 Exam Technique: Quick Elimination Strategy
- Oxygen count check:
• 1° alcohol oxidises to -COOH (gains 1 O).
• 2° alcohol oxidises to ketone (no change in O count, loses 2 H).
• 3° alcohol does not react (no change).
Starting molecule has 3 oxygens. Product must have 3 + 1 = 4 oxygens. This instantly eliminates options B and D! - Hydrogen count check:
Starting with C₆H₁₄O₃:
• 1° alcohol to -COOH: loses 2 H.
• 2° alcohol to ketone: loses 2 H.
Net loss = 4 H atoms → 14 - 4 = 10 H.
This leads directly to C₆H₁₀O₄ without drawing full structures!
❌ Common Student Traps
- Stopping at the aldehyde (Option B): Thinking the 1° alcohol only oxidises to an aldehyde (-CHO). The question states reflux and excess oxidising agent, which guarantees complete oxidation to -COOH.
- Oxidising the tertiary alcohol (Option A): Incorrectly assuming all -OH groups oxidise, leading to double-bond formation or cleavage (C₆H₈O₄). Tertiary alcohols cannot undergo oxidation under these conditions.
- Miscounting chain carbons: Forgetting the methyl branch at carbon-2 and assuming 5 total carbons instead of 6.
Topics
Organic Chemistry · 3.3.1 Introduction to Organic Chemistry · 3.3.5 Alcohols
Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.