Ethanethiol → HCID 13882884, compound
Reaction equation
[CH2:1]([SH:3])[CH3:2].[F:4][C:5]([F:23])([F:22])[C:6]1[C:11]([C:12]([O:14][CH2:15][CH3:16])=[O:13])=[C:10](Cl)[CH:9]=[C:8]([C:18]([F:21])([F:20])[F:19])[N:7]=1.[O-:24]CC.[Na+].[Na].C1C=C(Cl)C=C(C(OO)=O)C=1>C(O)C>[F:4][C:5]([F:23])([F:22])[C:6]1[C:11]([C:12]([O:14][CH2:15][CH3:16])=[O:13])=[C:10]([S:3]([CH2:1][CH3:2])=[O:24])[CH:9]=[C:8]([C:18]([F:21])([F:20])[F:19])[N:7]=1Participants
Conditions
- Temperature
- 5 °C
Procedure
Ethanethiol (5.33 ml, 0.072 mol) followed by 14.72 g (0.046 mol) of product of Example 19 was added to a solution of sodium ethoxide prepared from 1.66 g (0.072 mol) of sodium in 100 ml of ethanol and stirred at reflux overnight. The reaction mixture was filtered, poured into 200 ml of water, extracted with ether, washed with 10% NaOH, dried (MgSO4) and concentrated in vacuo to 9.45 g of yellow oil. This oil in 150 ml of CH2Cl2 was cooled to 5° C. and 3.73 g (0.022 mol) of MCPBA was added. Reaction was monitored by TLC. The reaction mixture was filtered, washed with NaHCO3, dried (MgSO4) and concentrated in vacuo to give 5.06 g of yellow oil. This was purified by HPLC using 10% ethyl acetate/cyclohexane as eluting solvent to give 3.81 g (65.6%) of an oil. A portion (1.71 g) of this oil was kugelhohr distilled at 22 Pa, pot temperature 95° C., affording 1.45 g of product as a clear oil, nD25 1.4575.
Workup
- temperatureat reflux overnight
- filtrationThe reaction mixture was filtered
- additionpoured into 200 ml of water
- extractionextracted with ether
- washwashed with 10% NaOH
- dry with materialdried (MgSO4)
- concentrationconcentrated in vacuo to 9.45 g of yellow oil
- customReaction
- filtrationThe reaction mixture was filtered
- washwashed with NaHCO3
- dry with materialdried (MgSO4)
- concentrationconcentrated in vacuo
Sources & evidence
- ORD record
- ord-2123c353fe994b17b7cf1c94f41ea3c5
- Source dataset
- uspto-grants-1987_04
- Patent number
- US04655816
Reaction notes
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