Sulfuric Acid → I2-Ionone
Reaction equation
[CH3:1][C:2]([CH3:14])=[CH:3][CH2:4][CH2:5]/[C:6](/[CH3:13])=[CH:7]/[CH:8]=[CH:9]/[C:10]([CH3:12])=[O:11].S(=O)(=O)(O)O>C(Cl)Cl>[CH3:13][C:6]1[CH2:5][CH2:4][CH2:3][C:2]([CH3:14])([CH3:1])[C:7]=1/[CH:8]=[CH:9]/[C:10]([CH3:12])=[O:11]Participants
4
Reactant 2
Product 1
Reagents, catalysts & solvents 1
Procedure
The procedure was carried out exactly as described in Example 9, except that the solution of 600 g of a 96% pure pseudoionone in 1,200 ml of methylene chloride, and the 1,000 ml of sulfuric acid, were introduced simultaneously into the reactor in the course of 5 minutes instead of 10 minutes. The distillation gave 490 g of a 98% pure β-ionone, corresponding to a yield of β-ionone of 83.4% of theory.
Workup
- distillationThe distillation
Sources & evidence
- ORD record
- ord-e2be22b826b449e9841d51ff860caf31
- Source dataset
- uspto-grants-1986_01
- Patent number
- US04565894
Reaction notes
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