Oxidise ethanedioic acid with warm acidified manganate(VII)
Use warm acidified KMnO₄ to oxidise ethanedioic acid into carbon dioxide.
Section 33 · Carboxylic acids and derivatives
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 33. Focus on the named compounds, reagents and conditions in each outcome; apply the specified reaction type rather than extending it to unlisted cases. Written practice is self-checked
Before you begin
- Recall carboxylic acids, esters, phenols, alcohols, halogenoalkanes and electrophilic/nucleophilic reaction ideas from earlier topics.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What reagent condition oxidises ethanedioic acid in the syllabus?
What you will learn
- Recall the specified reagent and condition
- Identify carbon dioxide as the product
- Distinguish ethanedioic acid oxidation from methanoic acid oxidation
Oxidise ethanedioic acid
Ethanedioic acid, HOOCCOOH, is oxidised by warm acidified KMnO₄ to carbon dioxide. Both carbon atoms in the dicarboxylic acid end up as CO₂.
The balanced equation is HOOCCOOH + [O] → 2CO₂ + H₂O. In ionic form, the oxidation half-equation is H₂C₂O₄ → 2CO₂ + 2H⁺ + 2e⁻.
Notice the named conditions
The syllabus specifies warm acidified potassium manganate(VII). Do not substitute the hot alkaline condition used for oxidising an alkylbenzene side chain.
The stated product is carbon dioxide; water is also present in the balanced overall equation. If asked for the main organic-chemistry product, identify CO₂.
Pause and explain
How many CO₂ molecules form from one ethanedioic acid molecule on full oxidation?
Worked example
State the conditions and product when ethanedioic acid is oxidised.
Show the worked solution
- Use KMnO₄ in acidic solution.
- Warm the reaction mixture.
- Name CO₂ as the oxidation product.
Answer Warm ethanedioic acid with acidified KMnO₄; carbon dioxide forms.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Complete the overall oxidation equation HOOCCOOH + [O] → …
Give me a hint
Account for both carbon atoms and the two hydrogens.
Compare my reasoning
- Each carboxyl carbon forms CO₂.
- Two carbon atoms give 2CO₂.
- The two hydrogens form H₂O.
HOOCCOOH + [O] → 2CO₂ + H₂O.
Look for these in your work
- Two CO₂
- Water included
- Atoms balanced
Explain why ‘acidified’ matters in the named KMnO₄ condition.
Give me a hint
Recall the reagent as the syllabus specifies it.
Compare my reasoning
- The syllabus names acidified KMnO₄.
- The acid provides the required acidic medium.
- Warm the reaction mixture as well.
Use warm acidified KMnO₄; both the acidic medium and warming are part of the stated condition.
Look for these in your work
- Acidified medium
- Warm condition
- Correct substrate
Compare the products from oxidising one molecule of HCOOH and one molecule of HOOCCOOH under their respective syllabus conditions.
Give me a hint
Track each starting molecule’s carbon atoms.
Compare my reasoning
- Methanoic acid has one carbon and forms one CO₂.
- Ethanedioic acid has two carbons and forms two CO₂.
- Both oxidation outcomes also produce water in a balanced equation.
HCOOH forms CO₂ and H₂O; HOOCCOOH forms 2CO₂ and H₂O.
Look for these in your work
- Methanoic product
- Ethanedioic carbon count
- Water accounted
Common mistakes
- Using alkaline KMnO₄
- Leaving out the warm condition
- Writing one CO₂ for a two-carbon molecule
- Confusing ethanedioic and methanoic acid outcomes
What can you explain now?
What reagent condition oxidises ethanedioic acid in the syllabus?
Compare with the explanation
Warm acidified KMnO₄
Both acid and warmth are specified.
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Make it stick
What exact condition oxidises ethanedioic acid, and how many CO₂ molecules form?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.