Topic 5 · page 7 of 15

Oxidation to a diol and the bromine test

Acidified potassium manganate(VII) oxidises a C=C to a diol and turns from purple to colourless. Bromine water is decolourised by a C=C, which makes it the test for an alkene.

13 min read5.45.4(v)5.5

Read first: Adding hydrogen and halogens

Oxidation by acidified manganate(VII) 5.45.4(v)

Acidified potassium manganate(VII) is a strong oxidising agent. Shaken with an alkene at room temperature, it oxidises the C=C. The π bond breaks, and an –OH group is added to each carbon of the C=C. The product is a diol: a compound with two –OH groups.

PracticalOxidising an alkene

  1. Put a few drops of the alkene (or a small volume of its solution) in a test tube.
  2. Add a few drops of dilute potassium manganate(VII) solution, acidified with dilute sulfuric acid.
  3. Shake at room temperature and watch the colour.
CHX2=CHX2+[O]+HX2O→HOCHX2CHX2OH \ce{CH2=CH2 + [O] + H2O -> HOCH2CH2OH}

[O] stands for oxygen supplied by the oxidising agent. The product from ethene is ethane-1,2-diol. In general, an –OH group is added to each of the two carbons that were joined by the C=C.

Two before-and-after pairs of test tubes. Acidified potassium manganate(VII) is purple. After shaking with cyclohexene, an alkene, the liquid is colourless. After shaking with cyclohexane, an alkane, it is still purple.
Purple to colourless with an alkene. No change with an alkane. Schematic colours.

Observation: the purple solution turns colourless. Explanation: manganate(VII) oxidises the C=C to a diol, and in doing so it is itself reduced, so its purple colour disappears.

Worked exampleBalancing with [O]

Write a balanced equation for the oxidation of propene to propane-1,2-diol.

  1. Propene: CH₃CH=CH₂. Add one O atom from [O] and one H₂O.
  2. Atoms on the left: C 3, H 6 + 2 = 8, O 1 + 1 = 2.
  3. Propane-1,2-diol, CH₂(OH)CH(OH)CH₃, is C₃H₈O₂: C 3, H 8, O 2. It balances.
  4. CH₃CH=CH₂ + [O] + H₂O → CH₂(OH)CH(OH)CH₃

The test for a C=C: bromine water 5.5

To test whether a compound contains a C=C double bond, add bromine water. This is a test you must be able to describe.

PracticalTesting for a C=C double bond

  1. Put about 2 cm³ of the liquid to be tested in a test tube.
  2. Add a few drops of bromine water. (Bromine water is harmful: use a dilute solution and small volumes, and wear eye protection.)
  3. Shake the tube and observe the colour.
Sample Observation
alkene (contains C=C) bromine water turns from orange to colourless at once
alkane (no C=C) bromine water stays orange
Two before-and-after pairs of test tubes. Bromine water is orange. After shaking with cyclohexene, an alkene, the liquid is colourless. After shaking with cyclohexane, an alkane, it is still orange.
Orange to colourless with an alkene. No change with an alkane. Schematic colours.

Explanation: the bromine adds across the C=C (see Adding hydrogen and halogens). The products are colourless, so the orange colour of bromine disappears. In bromine water some of the product is a bromo-alcohol, as you will see in Ethene and bromine; the colour change is the same.

Bromine water Acidified manganate(VII)
Alkene orange to colourless purple to colourless
Alkane stays orange stays purple
Reaction with the alkene electrophilic addition of bromine oxidation of the C=C to a diol

Check yourself

  1. Describe a test to tell cyclohexane from cyclohexene, and give the observation with each.
  2. Ethene is shaken with acidified potassium manganate(VII). Give the colour change and the name of the organic product.
  3. Write an equation, using [O], for the oxidation of but-2-ene to a diol, and name the diol.
  4. Explain why the bromine water test can distinguish an alkene from an alkane.
  5. A liquid has molecular formula C₆H₁₂ and does not decolourise bromine water. Suggest what it could be, and say why the formula alone could not tell you.
  6. Name the reagents and the conditions for the oxidation of ethene to ethane-1,2-diol.

Answers cover these until you have tried the questions

  1. Shake each with bromine water. Cyclohexene turns the bromine water from orange to colourless. Cyclohexane leaves it orange.
  2. Purple to colourless. The product is ethane-1,2-diol.
  3. CH₃CH=CHCH₃ + [O] + H₂O → CH₃CH(OH)CH(OH)CH₃. Balance check: left C 4, H 8 + 2 = 10, O 2; right C₄H₁₀O₂: C 4, H 10, O 2. The diol is butane-2,3-diol.
  4. Bromine adds across the C=C of an alkene, using up the bromine, and the product is colourless, so the orange colour disappears. An alkane has no C=C to add to, so there is no reaction and the colour stays.
  5. It could be cyclohexane, C₆H₁₂, which has no C=C. The formula C₆H₁₂ also fits an alkene (hexene), so the formula alone cannot decide; the test (no decolourisation) shows there is no C=C.
  6. Acidified potassium manganate(VII) (dilute sulfuric acid) at room temperature.