Vanadium in the 5 oxidation state oxidises 1,4 dihydroxybenzene to quinone fairly easily, and also the 1,2 dihydroxy isomer. The reaction for 1,3 dihydroxybenzene (resorcinol) is more difficult as the resultant molecule is not conjugated over the whole structure as is quinone.
The colour change shows that the vanadium V is being reduced to a lower oxidations state as the resorcinol is being oxidised.
The question is the identity of the oxidation product...
There cannot be conjugation with two carbonyl groups at positions 1 and 3 on the ring (which is now not aromatic) and consequently there is complete degradation to CO2 and H2O.
As to how to mix them together without a reaction.... no idea - but why would you want to?
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