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首页> 外文期刊>Journal of chemical crystallography >Synthesis and crystal structures of α-phenyl- and α-trifluoromethyl-α-(2-pyridyl-N-oxide)ethanols and α-phenyl-α-(2-pyridyl-N-oxide)ethylene
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Synthesis and crystal structures of α-phenyl- and α-trifluoromethyl-α-(2-pyridyl-N-oxide)ethanols and α-phenyl-α-(2-pyridyl-N-oxide)ethylene

机译:α-苯基-和α-三氟甲基-α-(2-吡啶基-N-氧化物)乙醇和α-苯基-α-(2-吡啶基-N-氧化物)乙烯的合成和晶体结构

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The X-ray structures of two α-substituted α-(2-pyridyl-N-oxide) ethanols, the α-phenyl- (3a) and α-trifluoromethyl- (3b) derivatives, were determined. On dehydration of 3a, compound 4a was obtained. This study brought a clear proof that the α-(2-pyridyl-N-oxide)ethanols are formed following oxidation with peroxides of their corresponding pyridines. A different molecular packing was observed for the α-trifluoromethyl derivative due to an additional hydrogen bonding. Compound 3a crystallized in the monoclinic space group P 2_1 with unit cell parameters a = 5.7416(1) ?, b = 14.3841(4) ?, c = 13.2821(3) ?, β = 94.918(2)°, V = 1092.90(4) ?~3, Z = 4, D = 1.308 Mg/m ~3. Compound 3b crystallized in the triclinic space group P -1 with unit cell parameters a = 6.1209(2) ?, b = 8.1938(4) ?, c = 9.4675(4) ?, α = 73.363(3)°, β = 73.166(3)°, γ = 71.659(3)°, V = 421.32(3) ?~3, Z = 2, D = 1.633 Mg/m ~3. Compound 4a crystallized in the monoclinic space group P 2 _1 with unit cell parameters a = 9.3579(3) ?, b = 12.7340(3) ?, c = 9.8579(3) ?, β = 117.3249(15)°, V = 1043.63(5) ?~3, Z = 4, D = 1.255 Mg/m~3.
机译:确定了两种α-取代的α-(2-吡啶基-N-氧化物)乙醇的X射线结构,即α-苯基-(3a)和α-三氟甲基-(3b)衍生物。在3a脱水后,获得化合物4a。这项研究提供了明确的证据,证明α-(2-吡啶基-N-氧化物)乙醇是用相应吡啶的过氧化物氧化后形成的。由于另外的氢键作用,观察到α-三氟甲基衍生物的分子堆积不同。在单斜晶空间群P 2_1 / n中结晶的化合物3a,晶胞参数a = 5.7416(1)α,b = 14.3841(4)α,c = 13.2821(3)β,β= 94.918(2)°,V = 1092.90(4)α〜3,Z = 4,D = 1.308 Mg / m〜3。在晶胞空间群P -1中结晶的化合物3b,晶胞参数a = 6.1209(2)α,b = 8.1938(4)β,c = 9.4675(4)α,α= 73.363(3)°,β= 73.166 (3)°,γ= 71.659(3)°,V = 421.32(3)α〜3,Z = 2,D = 1.633Mg / m〜3。在单斜晶空间群P 2 _1 / n中结晶的化合物4a,晶胞参数为a = 9.3579(3)α,b = 12.7340(3)α,c = 9.8579(3)α,β= 117.3249(15)°,V = 1043.63(5)= 3,Z = 4,D = 1.255Mg / m〜3。

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