Oxidation of hydroquinone to quinone: practice and mechanism
Hydroquinone (1,4-dihydroxybenzene) is oxidised to 1,4-benzoquinone: both OH become C=O and two C=C double bonds remain in the ring.
Example
Phenols with two OH groups in para position are easily oxidised: each OH loses an H, the ring loses its aromaticity and 1,4-benzoquinone with two C=O groups forms. The reaction is reversible (redox pair).
Mechanism step by step
- 1
The oxidant removes an electron and a proton (from one OH group) from the hydroquinone.
- 2
The resulting semiquinone radical is resonance-stabilised and gives up a second electron and proton.
- 3
1,4-Benzoquinone forms: two C=O groups and two C=C bonds, no longer aromatic.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- The quinone keeps two C=C double bonds. Oxidation only removes two H atoms (from the OH groups) and two electrons; nothing is hydrogenated.
- Both OH groups are oxidised, not just one. A quinone has two C=O groups para to each other.
Often confused with: Strong oxidation (chromic acid)
Frequently asked questions
Why does the ring give up aromaticity so easily?
Two C=O bonds and a fully conjugated system partly compensate for the loss. The reaction is also easily reversible (quinone/hydroquinone redox couple).
Where does this occur in nature?
Ubiquinone (coenzyme Q) in the respiratory chain and plastoquinone in photosynthesis transfer electrons using exactly this redox couple.