Chemistry Forum

C₆H₈ reacts with KMnO₄ in cold aqueous KOH to yield MnO₂ and C₆H₁₂O₄. Write the balanced equation using formulas of real compunds....

I'm confused on if this is a hydroxylation reaction or not. I know when an alkene reacts with KmnO₄, 2 OH groups form but I'm not sure where to stick them.

So far I have :
C₆H₈ + 4KMnO₄ + 8H₂₀ --> 4MnO₂ + 3C₆H₁₂O₄ + 4KOH

but adding water doesn't follow hydroxylation.

Help please?

Where the hydroxyls go depends on which of the various isomers of C₆H₈ you used. From the number of O in the product molecule, one of the two cyclohexadiene isomers looks most likely.

Cyclohexa-1,3-diene will produce cyclohexan-1,2,3,4-tetrol
Cyclohexa-1,4-diene will produce cyclohexan-1,2,4,5-tetrol

3 C6H8 + 4 KMnO4 + 8 H2O -(/cold,alkaline)-> 3 C6H8(OH)4 + 4 KOH + 4 MnO2

Structures (using cyclohexa-1,3-diene to produce cyclohexan-1,2,3,4-tetrol)

Structures (using cyclohexa-1,4-diene to produce cyclohexan-1,2,4,5-tetrol)