Hydrochloric Acid → Catechol
Reaction equation
[C:1]1([OH:7])[CH:6]=[CH:5][CH:4]=[CH:3][CH:2]=1.Cl.[H][H].[O:11]=O>CO>[CH:6]1[C:1]([OH:7])=[CH:2][CH:3]=[C:4]([OH:11])[CH:5]=1.[C:4]1(=[O:11])[CH:5]=[CH:6][C:1](=[O:7])[CH:2]=[CH:3]1.[C:1]1([C:6](=[CH:5][CH:4]=[CH:3][CH:2]=1)[OH:11])[OH:7]Participants
8
Reactant 4
Product 3
Reagents, catalysts & solvents 1
Procedure
A phenol oxidation reaction test was carried out in the following way by use of the catalysts obtained in Example 3 and Comparative Example 4. Each catalyst of 0.1 g, 0.01-normal hydrochloric acid of 10 ml, methanol of 10 ml and phenol of 1.0 g were put in a glass reactor vessel. While the resultant mixture was agitated at 70° C., hydrogen and oxygen were supplied to the reactor vessel at their respective flow rates of 40 ml/min and 40 ml/min. The amounts of p-hydroquinone (HQ), of benzoquinone (BQ), and of catechol (CAT), produced in the reaction solution at 12 hours after the start of the reaction, were obtained. The proportions of these were as follows.
Sources & evidence
- ORD record
- ord-fd6624cb41554035a0d59e6ea4cce22e
- Source dataset
- uspto-grants-2012_07
- Patent number
- US08222173B2
Reaction notes
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