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Hydrogen iodide → 4-Iodo-6-(4-ethoxyphenyl)pyrimidine

HRID 1752004
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반응식

HRID 1752004 반응식
F[C:2]1[CH:3]=[C:4](B(O)O)[CH:5]=[CH:6][C:7]=1[O:8][CH2:9][CH3:10].Cl[C:15]1[CH:20]=[C:19](Cl)[N:18]=[CH:17][N:16]=1.[IH:22]>>[I:22][C:15]1[CH:20]=[C:19]([C:4]2[CH:5]=[CH:6][C:7]([O:8][CH2:9][CH3:10])=[CH:2][CH:3]=2)[N:18]=[CH:17][N:16]=1
참가 물질실험 절차후처리출처 및 근거반응 노트

참가 물질

4

반응물 3

HCID 24841Hydrogen iodideHCID 709434,6-DichloropyrimidineHCID 2782821(4-Ethoxy-3-fluorophenyl)boronic acid

생성물 1

HCID 513555714-Iodo-6-(4-ethoxyphenyl)pyrimidine

실험 절차

The compound was prepared according to Example 1 using 3-fluoro-4-ethoxyphenylboronic acid and 4,6-dichloropyrimidine. The resultant chloro compound was converted to iodo with hydroiodic acid as described in the general procedure.

후처리

  1. customThe compound was prepared

출처 및 근거

ORD 기록
ord-06a354a8cc4c4b6b8452ec38e1ce6158
출처 데이터셋
uspto-grants-2015_10
DOI
10.6084/m9.figshare.5104873.v1
특허 번호
US09162987B2

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