Ammonia → HCID 23147641, compound

HRID 495288Notes

Reaction equation

HRID 495288 reaction equation
C(Cl)(=O)OC.C(N(CC)CC)C.[CH2:13]([C:15]1[C:20]([O:21]C(OC)=O)=[CH:19][C:18]([O:26]C(OC)=O)=[C:17]([C:31]2[CH:36]=[CH:35][CH:34]=[C:33]([C:37]3[CH:42]=[CH:41][CH:40]=[CH:39][CH:38]=3)[CH:32]=2)[C:16]=1[CH2:43][CH2:44][O:45][CH3:46])[CH3:14].[BH4-].[Na+].N>CO.O1CCCC1.O>[CH2:13]([C:15]1[C:20]([OH:21])=[CH:19][C:18]([OH:26])=[C:17]([C:31]2[CH:36]=[CH:35][CH:34]=[C:33]([C:37]3[CH:42]=[CH:41][CH:40]=[CH:39][CH:38]=3)[CH:32]=2)[C:16]=1[CH2:43][CH2:44][O:45][CH3:46])[CH3:14]

Participants

9

Reactant 5

HCID 222AmmoniaHCID 6586Methyl chloroformateHCID 8471TriethylamineHCID 4311764Sodium borohydrideHCID 23147532HCID 23147532, compound

Product 1

HCID 23147641HCID 23147641, compoundYield 52%

Reagents, catalysts & solvents 3

HCID 887MethanolSolventHCID 962WaterSolventHCID 8028TetrahydrofuranSolvent2×

Procedure

In the same manner as in Example 118, 1-[4,6-bis(methoxycarbonyloxy)-2-(2-methoxyethyl)-3-(3-phenylphenyl)phenyl]ethanone was obtained from Compound 120 (65 mg, 0.18 mmol) obtained in Example 120, using methyl chlorocarbonate (0.03 mL, 0.39 mmol), triethylamine (0.06 mL, 0.43 mmol) and tetrahydrofuran (3.0 mL); and from the resulting compound, obtained was 4-ethyl-1,5-bis(methoxycarbonyloxy)-3-(2-methoxyethyl)-2-(3-phenylphenyl)benzene, using sodium borohydride (30 mg, 0.79 mmol), water (2.0 mL) and tetrahydrofuran (2.0 mL). Then, the resulting compound was treated with methanol solution (5.0 mL) of 7 mmol/L ammonia to obtain Compound 132 (33 mg, 52%).

Workup

  1. customand from the resulting compound, obtained

Sources & evidence

ORD record
ord-7240105bec00493c9fddbb6bb4826302
Source dataset
uspto-grants-2009_05
Patent number
US07538241B2

Reaction notes

Loading…