Oxidise an alkene gently to a diol
Describe cold, dilute, acidified potassium manganate(VII) addition across a double bond.
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Apply the stated reagents and conditions, balance equations, and keep mechanistic claims within the named examples. Written practice is self-checked
Before you begin
- Use displayed and structural formulas, covalent bond polarity, balanced equations, and the organic reaction terms from Topic 13.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the main organic product class from cold dilute acidified KMnO₄ and an alkene?
What you will learn
- Recall the cold dilute acidified reagent
- Identify a diol product
- Distinguish mild oxidation from bond-cleaving conditions
Mild oxidation adds two hydroxy groups
Cold, dilute, acidified potassium manganate(VII), KMnO₄, oxidises an alkene to a diol by adding oxygen-containing groups across C=C.
For propene the product is propane-1,2-diol. The carbon–carbon bond becomes a single bond; each former alkene carbon gains an –OH group.
Conditions control the product
Hot, concentrated acidified manganate(VII) can cleave the C=C bond. That is a different oxidation outcome and is used to infer where the double bond was located.
Do not substitute the hot concentrated condition when the question asks for the diol. State cold, dilute and acidified for the diol pathway.
Pause and explain
Which condition set gives the diol pathway?
Worked example
Predict the product class when propene is treated with cold, dilute, acidified KMnO₄.
Show the worked solution
- Identify the C=C bond.
- Add –OH to each carbon that formed the double bond.
- Name the product as a two-hydroxy propane.
Answer Propane-1,2-diol, CH₃CH(OH)CH₂OH.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
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.
Give the organic product from but-1-ene under the cold dilute acidified manganate(VII) conditions.
Give me a hint
Place an –OH group on each former alkene carbon.
Compare my reasoning
- Keep the four-carbon skeleton.
- C1 and C2 each receive –OH.
- Name the vicinal diol.
Butane-1,2-diol, HOCH₂CH(OH)CH₂CH₃.
Look for these in your work
- C₄ skeleton retained
- Both OH groups added
- Product named
Explain why cold dilute and hot concentrated KMnO₄ should not be treated as identical conditions.
Give me a hint
Compare addition at C=C with cleavage of C=C.
Compare my reasoning
- Cold dilute conditions give a diol.
- Hot concentrated conditions can rupture the C=C bond.
- Different products answer different syllabus tasks.
Cold dilute reagent forms a diol; hot concentrated reagent cleaves C=C and forms products used to locate the original linkage.
Look for these in your work
- Cold outcome stated
- Hot outcome stated
- Purpose distinguished
A student gives propanone as the product from cold dilute KMnO₄ and propene. Correct the answer.
Give me a hint
Check whether the specified conditions are mild or bond-cleaving.
Compare my reasoning
- Cold dilute conditions add two –OH groups across C=C.
- The carbon skeleton remains intact.
- The product is propane-1,2-diol, not a cleavage carbonyl.
The expected diol is propane-1,2-diol; propanone is not the intact-skeleton product of this mild pathway.
Look for these in your work
- Conditions applied
- Skeleton retained
- Product corrected
Common mistakes
- Producing a cleavage product under cold dilute conditions
- Adding only one –OH group
- Omitting temperature, concentration or acidification
What can you explain now?
What is the main organic product class from cold dilute acidified KMnO₄ and an alkene?
Compare with the explanation
A diol
Two hydroxy groups are added across the double bond.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
How do the products of cold dilute and hot concentrated manganate(VII) oxidation differ?
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.