Trimethyl borate → 3-Fluoro-5-(4-fluorophenoxy)phenylboronic acid
Reaction equation
[F:1][C:2]1[CH:3]=[C:4](Br)[CH:5]=[C:6]([O:8][C:9]2[CH:14]=[CH:13][C:12]([F:15])=[CH:11][CH:10]=2)[CH:7]=1.C([Li])CCC.[B:22](OC)([O:25]C)[O:23]C>O1CCCC1>[F:1][C:2]1[CH:3]=[C:4]([B:22]([OH:25])[OH:23])[CH:5]=[C:6]([O:8][C:9]2[CH:14]=[CH:13][C:12]([F:15])=[CH:11][CH:10]=2)[CH:7]=1Participants
5
Reactant 3
Product 1
Reagents, catalysts & solvents 1
Procedure
This compound is prepared in a manner analogous to that of Step B of Example 6, using 9.1 grams (0.032 mole) of 3-fluoro-5-(4-fluorophenoxy)phenyl bromide, 14 mL (0.035 mole) of n-butyllithium (2.5M in hexanes), and 10.4 mL (0.095 mole) of trimethyl borate in 100 mL of tetrahydrofuran, yielding 3-fluoro-5-(4-fluorophenoxy)phenylboronic acid.