Sodium Hydroxide → HCID 66932435, compound
Reaction equation
C[O:2][C:3](=[O:29])/[CH:4]=[CH:5]/[C:6]1[CH:7]=[C:8]2[C:25](=[CH:26][CH:27]=1)[O:24][C:11]1([CH2:16][CH2:15][N:14]([C:17](OC(C)(C)C)=O)[CH2:13][CH2:12]1)[CH2:10][C:9]2=[O:28].[F:30][C:31]1[CH:32]=[C:33]2[C:37](=[CH:38][CH:39]=1)[NH:36][CH:35]=[CH:34]2.[OH-].[Na+]>>[F:30][C:31]1[CH:32]=[C:33]2[C:37](=[CH:38][CH:39]=1)[NH:36][CH:35]=[C:34]2[CH2:17][N:14]1[CH2:15][CH2:16][C:11]2([CH2:10][C:9](=[O:28])[C:8]3[C:25](=[CH:26][CH:27]=[C:6](/[CH:5]=[CH:4]/[C:3]([OH:2])=[O:29])[CH:7]=3)[O:24]2)[CH2:12][CH2:13]1Participants
Procedure
(E)-3-{1′-(5-Fluoro-1H-indol-3-ylmethyl)-4-oxo-spiro[chromane-2,4′-piperidine]-6-yl}-acrylic acid methyl ester was obtained starting from (E)-3-{4-oxo-spiro[chromane-2,4′-piperidine]-6-yl}-acrylic acid methyl ester (135 mg, 0.45 mmol, Intermediate 1, free base) and 5-fluoro-1H-indole, according to the procedure described in Example 50, Step A (170 mg, 85%). The methyl ester was hydrolyzed with 4 M NaOH following the procedure described in Example 40, Step B, giving (E)-3-{1′-(5-fluoro-1H-indol-3-ylmethyl)-4-oxo-spiro[chromane-2,4′-piperidine]-6-yl}-acrylic acid (100 mg, 63%). The resulting product was treated with NH2OTHP according to the procedure described in Example 40, Step C, giving (E)-3-{1′-(5-fluoro-1H-indol-3-ylmethyl)-4-oxo-spiro[chromane-2,4′-piperidine]-6-yl}-N-(tetrahydro-pyran-2-yloxy)-acrylamide. Finally, removal of the THP protecting group following the procedure described in Example 40, Step C gave (E)-3-{1′-(5-fluoro-1H-indol-3-ylmethyl)-4-oxo-spiro[chromane-2,4′-piperidine]-6-yl}-N-hydroxy-acrylamide hydrochloride (76 mg) as a white solid.
Sources & evidence
- ORD record
- ord-1a9c04c73ec546128d8a32adee32db79
- Source dataset
- uspto-grants-2015_08
- Patent number
- US09115148B2
Reaction notes
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