تفاعل #69610

معادلة التفاعل

12

2-Phenoxy-3-tert-butyl-5-methyl-N-[(1E)-2-pyridylmethylene]aniline

nBu2O

phenyl lithium

→

C1

المردود 62.0%

2-Phenoxy-5-methyl-N-[phenyl(2-pyridyl)methyl]aniline

المردود 62.0%

المذيبات

THF

ظروف التفاعل

المعالجة

1. 1
   أخرىthe reaction was quenched with water (1 mL)
2. 2
   workup.ADDITIONMgSO4 was added
3. 3
   ترشيحthe mixture filtered
4. 4
   أخرىto give a yellow solution
5. 5
   أخرىRemoval of the solvent under vacuum

الإجراء التجريبي

To a solution of 12 (300 mg, 1.12 mmol) in THF was added 1.24 mL of a 1.8M nBu2O solution of phenyl lithium at −40° C. After stirring for 15 min, the reaction was quenched with water (1 mL). MgSO4 was added and the mixture filtered to give a yellow solution. Removal of the solvent under vacuum gave C1 as a yellow oil (240 mg, 62% yield).

المصدر