Lesson 15 of 34
Hot oxidation reveals what was attached to each double-bond carbon
Cut C=C into two carbonyl-containing fragments, then account for further oxidation.
A-Level 9476 (2026-2027)
I can describe alkene chemistry using ethene and analogous structures.Syllabus
Syllabus 9476, 11.4(b). Describe alkene chemistry using ethene and analogous structures.
Hot oxidation reveals what was attached to each double-bond carbon
Cut C=C into two carbonyl-containing fragments, then account for further oxidation.
| Original alkene carbon | Final oxidation fragment | Why |
|---|---|---|
| =CH2 | CO2, with water in the overall balance. | The terminal carbon is oxidised beyond methanal/methanoic acid under the vigorous conditions. |
| =CH-R | RCOOH. | An initially aldehyde-like fragment is further oxidised to a carboxylic acid. |
| =C(R)R' | RCOR', a ketone. | No H is attached to this carbon, so the corresponding ketone is formed. |
Worked exampleDeduce an alkene from its cleavage products
A single acyclic alkene gives propanone and ethanoic acid on heating with acidified KMnO4. Suggest its structure.
Try it first. Then show the worked stepsHide the worked steps
- The propanone carbonyl carbon was originally attached to two methyl groups.
- The ethanoic-acid carboxyl carbon was originally attached to one methyl group and one H.
- Replace the two C=O groups by a C=C joining those original carbons, restoring the H on the acid-derived carbon.
(CH3)2C=CHCH3, 2-methylbut-2-ene. Its overall oxidation can be written C5H10 + 3[O] → CH3COCH3 + CH3COOH.
A cyclic alkene opens into one molecule with two oxygen-containing ends, not two separate molecules. Count the carbon atoms in every product, including CO2, before you suggest the starting structure. Cold alkaline and hot acidified manganate(VII) give different results. Cold alkaline gives a diol. Hot acidified breaks the double bond.
Check your understanding
But-2-ene, CH3CH=CHCH3, is heated with acidified KMnO4. What forms?
Show the answer
Two molecules of ethanoic acid, CH3COOH
The C=C breaks. Each =CH-CH3 fragment first gives an aldehyde, which is oxidised further to a carboxylic acid under these vigorous conditions.