乙醇 → 对碘苯甲酸乙酯

HRID 1222

反应方程式

HRID 1222 反应方程式
[I:1][C:2]1[CH:10]=[CH:9][C:5]([C:6]([OH:8])=[O:7])=[CH:4][CH:3]=1.S(Cl)(Cl)=O.[CH2:15](O)[CH3:16]>>[CH2:15]([O:7][C:6](=[O:8])[C:5]1[CH:9]=[CH:10][C:2]([I:1])=[CH:3][CH:4]=1)[CH3:16]

参与物

4

反应物 3

HCID 702乙醇HCID 120854-碘苯甲酸HCID 24386氯化亚砜

产物 1

HCID 142891对碘苯甲酸乙酯

实验过程

To a suspension of 10 g (40.32 mmol) of 4-iodobenzoic acid in 100 ml absolute ethanol was added 2 ml thionyl chloride and the mixture was then heated at reflux for 3 hours. Solvent was removed in vacuo and the residue was dissolved in 100 ml ether. The ether solution was washed with saturated NaHCO3 and saturated NaCl solutions and dried (MgSO4). Solvent was then removed in vacuo and the residue kugelrohr distilled (100 degrees C.; 0.55 mm) to give the title compound as a colorless oil, PMR (CDCl3): δ1.42 (3H, t, J~7 Hz), 4.4 (2H, q, J~7 Hz), 7.8 (4H).

后处理

  1. temperaturethe mixture was then heated
  2. temperatureat reflux for 3 hours
  3. customSolvent was removed in vacuo
  4. dissolutionthe residue was dissolved in 100 ml ether
  5. washThe ether solution was washed with saturated NaHCO3 and saturated NaCl solutions
  6. dry with materialdried (MgSO4)
  7. customSolvent was then removed in vacuo
  8. distillationthe residue kugelrohr distilled (100 degrees C.; 0.55 mm)

来源与证据

ORD记录
ord-6724f220dc7c4e89ac64b79ac6a33d74
来源数据集
uspto-grants-1998_03
专利号
US05723620

关联笔记

加载中…