Lithium Hydroxide → 9-Bromo-10-fluoro-3-formyl-2,5-diazatetracyclo[11.1.1.0[2,6].0[7,12]]pentadeca-3,5,7(12),8,10-pentaene-4-carboxylic acid

HRID 575135Notes

Reaction equation

HRID 575135 reaction equation
[Br:1][C:2]1[C:15]([F:16])=[CH:14][C:13]2[CH:12]3[CH2:17][CH:10]([CH2:11]3)[N:9]3[C:5](=[N:6][C:7]([C:20]([O:22]C)=[O:21])=[C:8]3[CH:18]=[O:19])[C:4]=2[CH:3]=1.[Li+].[OH-]>O.CO>[Br:1][C:2]1[C:15]([F:16])=[CH:14][C:13]2[CH:12]3[CH2:11][CH:10]([CH2:17]3)[N:9]3[C:5](=[N:6][C:7]([C:20]([OH:22])=[O:21])=[C:8]3[CH:18]=[O:19])[C:4]=2[CH:3]=1

Participants

4

Reactant 2

HCID 3939Lithium HydroxideHCID 86693159Methyl 9-bromo-10-fluoro-3-formyl-2,5-diazatetracyclo[11.1.1.0[2,6].0[7,12]]pentadeca-3,5,7(12),8,10-pentaene-4-carboxylate

Product 1

HCID 866931609-Bromo-10-fluoro-3-formyl-2,5-diazatetracyclo[11.1.1.0[2,6].0[7,12]]pentadeca-3,5,7(12),8,10-pentaene-4-carboxylic acidYield 74.8%

Reagents, catalysts & solvents 1

HCID 22242641Methanol--water (1/1)Solvent

Procedure

Methyl 9-bromo-10-fluoro-3-formyl-2,5-diazatetracyclo[11.1.1.0[2,6].0[7,12]]pentadeca-3,5,7(12),8,10-pentaene-4-carboxylate (125 mg) was saponified with LiOH in H2O/MeOH to give 90 mg of 9-bromo-10-fluoro-3-formyl-2,5-diazatetracyclo[11.1.1.0[2,6].0[7,12]]pentadeca-3,5,7(12),8,10-pentaene-4-carboxylic acid.

Sources & evidence

ORD record
ord-ee4796d266014e13891969d3ecf09527
Source dataset
uspto-grants-2015_05
Patent number
US09034866B2

Reaction notes

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