Reaction #172379

ord-0bcd3dacdb0645a89ab81d6c03ce4396

Conditions

Detailed conditions
See reaction.notes.procedure_details.

Procedure

Analogously to Example 4A, 1.50 g (4.13 mmol) of the compound from Example 3A are reacted with 704 mg (4.54 mmol) of 2-fluoro-5-nitrotoluene. This gives 570 mg (28% of theory) of the title compound.

Source

DOI: 10.6084/m9.figshare.5104873.v1Patent: US08846934B2uspto-grants-2014_09