4-Ethynyloxan-4-ol → HCID 18366119, compound

HRID 366709Notes

Reaction equation

HRID 366709 reaction equation
[C:1]([C:3]1([OH:9])[CH2:8][CH2:7][O:6][CH2:5][CH2:4]1)#[CH:2].[F:10][C:11]1[C:12](I)=[C:13]2[C:17](=[CH:18][CH:19]=1)[NH:16][C:15](=[O:20])/[C:14]/2=[CH:21]\[C:22]1[NH:23][CH:24]=[CH:25][C:26]=1[O:27][CH3:28]>C1C=CC([P]([Pd]([P](C2C=CC=CC=2)(C2C=CC=CC=2)C2C=CC=CC=2)([P](C2C=CC=CC=2)(C2C=CC=CC=2)C2C=CC=CC=2)[P](C2C=CC=CC=2)(C2C=CC=CC=2)C2C=CC=CC=2)(C2C=CC=CC=2)C2C=CC=CC=2)=CC=1.CN(C=O)C.CCN(CC)CC>[F:10][C:11]1[C:12]([C:2]#[C:1][C:3]2([OH:9])[CH2:8][CH2:7][O:6][CH2:5][CH2:4]2)=[C:13]2[C:17](=[CH:18][CH:19]=1)[NH:16][C:15](=[O:20])/[C:14]/2=[CH:21]\[C:22]1[NH:23][CH:24]=[CH:25][C:26]=1[O:27][CH3:28]

Participants

6

Procedure

Using Method C above, 4-ethynyl-tetrahydro-pyran-4-ol (41.1 mg, 0.33 mmol) (Example 85A) was coupled with (Z)-1,3-dihydro-5-fluoro-4-iodo-3-[(3-methoxy-1H-pyrrol-2-yl)methylene]-2H-indol-2-one (50 mg, 0.13 mmol) (Starting Material 6) using (Ph3P)4Pd (15 mg) and Cul (2.5 mg) as catalyst in DMF (4 mL) and Et3N (4 mL) as solvent at 85° C. for 7 h to yield (Z)-1,3-Dihydro-5-fluoro-4-[(4-hydroxy-tetrahydro-pyran-4-yl)ethynyl]-3-[(3-methoxy-1H-pyrrol-2-yl)methylene]-2H-indol-2-one. (Yield 36 mg, 75%).

Sources & evidence

ORD record
ord-6d4d62c21fc14baa8c311c82a47990d4
Source dataset
uspto-grants-2000_10
Patent number
US06130239

Reaction notes

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