6-Methoxy-1,2,3,4-tetrahydroisoquinoline hydrochloride → 2-[[6-Fluoro-2-(4-fluorophenyl)quinazolin-4-yl]carbonyl]-6-methoxy-1,2,3,4-tetrahydroisoquinoline

HRID 717472

반응식

HRID 717472 반응식
[F:1][C:2]1[CH:3]=[C:4]2[C:9](=[CH:10][CH:11]=1)[N:8]=[C:7]([C:12]1[CH:17]=[CH:16][C:15]([F:18])=[CH:14][CH:13]=1)[N:6]=[C:5]2[C:19](O)=[O:20].Cl.[CH3:23][O:24][C:25]1[CH:26]=[C:27]2[C:32](=[CH:33][CH:34]=1)[CH2:31][NH:30][CH2:29][CH2:28]2>>[F:1][C:2]1[CH:3]=[C:4]2[C:9](=[CH:10][CH:11]=1)[N:8]=[C:7]([C:12]1[CH:13]=[CH:14][C:15]([F:18])=[CH:16][CH:17]=1)[N:6]=[C:5]2[C:19]([N:30]1[CH2:29][CH2:28][C:27]2[C:32](=[CH:33][CH:34]=[C:25]([O:24][CH3:23])[CH:26]=2)[CH2:31]1)=[O:20]

참가 물질

3

반응물 2

HCID 29203356-Methoxy-1,2,3,4-tetrahydroisoquinoline hydrochlorideHCID 904107136-Fluoro-2-(4-fluorophenyl)quinazoline-4-carboxylic acid

생성물 1

HCID 862677832-[[6-Fluoro-2-(4-fluorophenyl)quinazolin-4-yl]carbonyl]-6-methoxy-1,2,3,4-tetrahydroisoquinoline수율 50.6%

실험 절차

Reaction of 6-fluoro-2-(4-fluorophenyl)quinazoline-4-carboxylic acid with 6-methoxy-1,2,3,4-tetrahydroisoquinoline hydrochloride gave compound 99 (50.6% yield). 1H NMR (400 MHz, DMSO-d6) δ 2.80 and 3.03 (2t, 2H), 3.54 and 4.02 (2t, 2H), 3.72 and 3.74 (2s, 3H), 4.47 and 4.92 (2s, 2H), 6.65-7.27 (m, 3H), 7.40-7.48 (m, 1H), 7.62-7.82 (m, 2H), 8.05-8.10 (m, 1H), 8.20-8.39 (m, 3H); MS (ESI) m/z 432 ([M+H]+).

출처 및 근거

ORD 기록
ord-357ccab4ac54465c96c852c671547cbd
출처 데이터셋
uspto-grants-2015_11
특허 번호
US09193707B2

연결된 노트

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