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Styrene → HCID 73990325, 화합물

HRID 2416403
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HRID 2416403 반응식
[Si:1]([C:23]1(Br)[CH:28]=[CH:27][CH:26]=[CH:25][CH:24]=1)([C:16]1(Br)[CH:21]=[CH:20][CH:19]=[CH:18][CH:17]=1)([C:9]1(Br)[CH:14]=[CH:13][CH:12]=[CH:11][CH:10]=1)[C:2]1(Br)[CH:7]=[CH:6][CH:5]=[CH:4][CH:3]=1.[CH2:30]=[CH:31][C:32]1[CH:37]=[CH:36][CH:35]=[CH:34][CH:33]=1>>[C:3]1([CH:30]=[CH:31][C:32]2[CH:37]=[CH:36][CH:35]=[CH:34][CH:33]=2)[CH:4]=[CH:5][CH:6]=[CH:7][C:2]=1[Si:1]([C:23]1[CH:28]=[CH:27][CH:26]=[CH:25][C:24]=1[CH:30]=[CH:31][C:32]1[CH:37]=[CH:36][CH:35]=[CH:34][CH:33]=1)([C:16]1[CH:21]=[CH:20][CH:19]=[CH:18][C:17]=1[CH:30]=[CH:31][C:32]1[CH:37]=[CH:36][CH:35]=[CH:34][CH:33]=1)[C:9]1[CH:14]=[CH:13][CH:12]=[CH:11][C:10]=1[CH:30]=[CH:31][C:32]1[CH:37]=[CH:36][CH:35]=[CH:34][CH:33]=1
참가 물질실험 절차후처리출처 및 근거반응 노트

참가 물질

3

반응물 2

HCID 7501StyreneHCID 178419497HCID 178419497, 화합물

생성물 1

HCID 73990325HCID 73990325, 화합물수율 52%

실험 절차

Use of tetrakis(4-bromophenyl)silane (Si(C6H4Br)4) gives access to structural analogs containing a silicon atom at the tetrahedral junction. Thus, reaction of Si(C6H4Br)4 with excess styrene gives tetrastilbenylsilane (Si(STB)4) in 52% yield using Pd(OAc)2 under phase transfer conditions. Similarly, (Si(C6H4Br)4) and 4,4′-tert-butylvinylstilbene yields tetrakis(4-tert-butylstyrylstilbenyl)silane (Si(tBuSSB)4) in 40% yield.

후처리

  1. additioncontaining a silicon atom at the tetrahedral junction

출처 및 근거

ORD 기록
ord-3b433d4d8ff0483ab52a30c5e671a429
출처 데이터셋
uspto-grants-2004_11
DOI
10.6084/m9.figshare.5104873.v1
특허 번호
US06824890B2

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