Topic 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)

Hot oxidation reveals what was attached to each double-bond carbon

Cut C=C into two carbonyl-containing fragments, then account for further oxidation.

Hot acidified manganate(VII): fate of each alkene carbon
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 example

Deduce an alkene from its cleavage products

A single acyclic alkene gives propanone and ethanoic acid on heating with acidified KMnO₄. Suggest its structure.

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.

Answer

(CH₃)₂C=CHCH₃, 2-methylbut-2-ene. Its overall oxidation can be written C₅H₁₀ + 3[O] → CH₃COCH₃ + CH₃COOH.

A cyclic alkene can open into one molecule containing two oxygenated ends rather than two separate molecules. Count the carbon atoms in every product, including CO₂, before proposing the original structure. Cold alkaline and hot acidified manganate(VII) are not interchangeable conditions: one gives a diol, the other cleaves the double bond.