1—2—4酸+(2/3/2)+5/2/1x(-6)-(—0.5)

Document Detail
3,3,6,6-Tetra-methyl-9-[6-(3,3,6,6-tetra-methyl-1,8-dioxo-2,3,4,5,6,7,8,9-octa-hydro-1H-xanthen-9-yl)pyridin-2-yl]-2,3,4,5,6,7,8,9-octa-hydro-1H-xanthene-1,8-dione.
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Abstract/OtherAbstract:
In the title mol-ecule, C(39)H(45)NO(6), the two tetra-methyl-octa-hydroxanthen-1,8-dione substituents are arranged approximately parallel to each other and approximately perpendicular to the plane of the pyridine ring. The six-membered xanthene rings adopt flattened boat conformations with the O and methine C atoms deviating from the plane of the other four atoms.
Antar A A Shaaban Kamel M Mirze A A Atash V G Seik Weng Ng
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Acta crystallographica. Section E, Structure reports online
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Acta Crystallogr Sect E Struct Rep Online
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From MEDLINE(R)/PubMed(R), a database of the U.S. National Library of Medicine
Journal Information
Journal ID (nlm-ta): Acta Crystallogr Sect E Struct Rep Online
Journal ID (publisher-id): Acta Cryst. E
Publisher: International Union of Crystallography
Article Information
A full version of this article is available from Crystallography Journals Online.(C) Abdelhamid et al. 2011
open-access:
Received Day: 21 Month: 2 Year: 2011
Accepted Day: 25 Month: 2 Year: 2011
collection publication date: Day: 01 Month: 4 Year: 2011
Electronic publication date: Day: 05 Month: 3 Year: 2011
pmc-release publication date: Day: 05 Month: 3 Year: 2011
Volume: 67 Issue: Pt 4
First Page: o785 Last Page: o785
ID: 3100053
PubMed Id:
Publisher Id: lh5213
Coden: ACSEBH
Publisher Item Identifier: S7318
3,3,6,6-Tetra-methyl-9-[6-(3,3,6,6-tetra-methyl-1,8-dioxo-2,3,4,5,6,7,8,9-octa-hydro-1H-xanthen-9-yl)pyridin-2-yl]-2,3,4,5,6,7,8,9-octa-hydro-1H-xanthene-1,8-dione
Alternate Title:C39H45NO6
Antar A. Abdelhamid
Shaaban Kamel Mohamed
Mirze A. Allahverdiyev
Atash V. Gurbanov
Seik Weng Ng*
aDepartment of Organic Chemistry, Baku State University, Baku, Azerbaijan
bChemistry & Environmental Science Division, School of Science, Manchester Metropolitan University, UK
cDepartment of Chemistry, University of Malaya, 50603 Kuala Lumpur, Malaysia
Correspondence: Correspondence e-mail: seikweng@um.edu.my
Related literature
For a related structure, see: Mohamed et al. (2011).[Chemical Structure ID: scheme1]
Experimental
Crystal data
Mr = 623.76
Monoclinic,
a = 24.1384 (8) ?
b = 10.0371 (4) ?
c = 14.4408 (5) ?
β = 105.8460 (7)°
Mo Kα radiation
μ = 0.08 mm-1
0.30 × 0.30 × 0.30 mm
Data collection
Bruker APEXII diffractometer
35921 measured reflections
7717 independent reflections
6410 reflections with I & 2σ(I)
Rint = 0.035
Refinement
R[F2 & 2σ(F2)] = 0.041
wR(F2) = 0.111
7717 reflections
415 parameters
H-atom parameters constrained
Δρmax = 0.35 e ?-3
Δρmin = -0.25 e ?-3
Data collection: APEX2 (Bruker, 2005); cell refinement: SAINT (Bruker, 2005); data reduction: SAINT; program(s) used to solve structure: SHELXS97 (Sheldrick, 2008); program(s) used to refine structure: SHELXL97 (Sheldrick, 2008); molecular graphics: X-SEED (Barbour, 2001); software used to prepare material for publication: publCIF (Westrip, 2010).
Supplementary Material
Crystal structure: contains datablocks global, I. DOI: 10./lh5213sup1.cif
Structure factors: contains datablocks I. DOI: 10./lh5213Isup2.hkl
Additional supplementary materials:
crystall 3D checkCIF report
fnu1Supplementary data and figures for this paper are available from the IUCr electronic archives (Reference: LH5213).
We thank Manchester Metropolitan University, Baku State University and the University of Malaya for supporting this study.
supplementary crystallographic
information
Experimental
A mixture of 1,2-ethanediammonium chloride (0.01 mol) and
2,6-pyridinedicarbaldehyde (0.01 mol) in aqueous ethanol (1:1) was heated for
30 minutes. To the mixture was added dimedone (0.04 mol). The mixture was
heated for another 5 fhours. The solvent was evaporated to yield a pale yellow
compound. This was recrystallized from ethanol in 60% m.p. 473 K.
Refinement
Carbon-bound H-atoms were placed in calculated positions [C–H 0.95 to 0.99 ?;
U(H) 1.2 to 1.5U(C)] and were included in the refinement in the
riding model approximation.
Crystal data
Data collection
Refinement
Fractional atomic coordinates and isotropic or equivalent isotropic displacement parameters (?2)
Atomic displacement parameters (?2)
Geometric parameters (?, °)
References
Barbour, L. J. (2001). J. Supramol. Chem.1, 189–191.
Bruker (2005). APEX2 and SAINT Bruker AXS Inc., Madison, Wisconsin, USA.
Mohamed, S. K., Abdelhamid, A. A., Khalilov, A. N., Gurbanov, A. V. & Ng, S. W. (2011). Acta Cryst. E67, o850–o851.
Sheldrick, G. M. (2008). Acta Cryst. A64, 112–122.
Westrip, S. P. (2010). J. Appl. Cryst.43, 920–925.
[Figure ID: Fap1]
Thermal ellipsoid plot (Barbour, 2001) of C39H45NO6 at the 70% hydrogen atoms are drawn as spheres of arbitrary radius.
[Figure ID: Fap2]
View of the molecule along the pyridyl plane. The pyridyl atoms are shown as 70% thermal ellipsoids in their van der Waals surfaces whereas the other atoms are shown in the ball-and-stick style.
[TableWrap ID: d1e135]
F(000) = 1336
Mr = 623.76
Dx = 1.231 Mg m-3
Monoclinic, P21/c
Mo Kα radiation, λ = 0.71073 ?
Hall symbol:
Cell parameters from 9997 reflections
a = 24.1384 (8) ?
θ = 2.2–28.3°
b = 10.0371 (4) ?
u = 0.08 mm-1
c = 14.4408 (5) ?
β = 105.8460 (7)°
Prism, yellow
0.30 × 0.30 × 0.30 mm
[TableWrap ID: d1e262]
Bruker APEXII diffractometer
6410 reflections with I & 2σ(I)
Radiation source: fine-focus sealed tube
Rint = 0.035
θmax = 27.5°, θmin = 0.9°
φ and ω scans
h = -31→31
35921 measured reflections
k = -13→13
7717 independent reflections
l = -18→18
[TableWrap ID: d1e360]
Refinement on F2
Primary atom site location: structure-invariant direct methods
Least-squares matrix: full
Secondary atom site location: difference Fourier map
R[F2 & 2σ(F2)] = 0.041
Hydrogen site location: inferred from neighbouring sites
wR(F2) = 0.111
H-atom parameters constrained
w = 1/[σ2(Fo2) + (0.0516P)2 + 1.4985P]
where P = (Fo2 + 2Fc2)/3
7717 reflections
(Δ/σ)max = 0.001
415 parameters
Δρmax = 0.35 e ?-3
0 restraints
Δρmin = -0.25 e ?-3
[TableWrap ID: d1e519]
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[TableWrap ID: d1e2473]
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C12—O1—C20
118.01 (9)
C18—C21—H21C
C29—O4—C37
117.95 (9)
H21A—C21—H21C
C1—N1—C5
117.26 (10)
H21B—C21—H21C
N1—C1—C2
123.18 (11)
C18—C22—H22A
N1—C1—C6
117.84 (10)
C18—C22—H22B
C2—C1—C6
118.97 (11)
H22A—C22—H22B
C1—C2—C3
118.79 (12)
C18—C22—H22C
C1—C2—H2
H22A—C22—H22C
C3—C2—H2
H22B—C22—H22C
C4—C3—C2
118.78 (11)
C32—C23—C24
108.91 (10)
C4—C3—H3
C32—C23—C5
113.40 (10)
C2—C3—H3
C24—C23—C5
109.91 (10)
C3—C4—C5
119.01 (11)
C32—C23—H23
C3—C4—H4
C24—C23—H23
C5—C4—H4
C5—C23—H23
N1—C5—C4
122.97 (11)
C29—C24—C25
118.76 (11)
N1—C5—C23
118.10 (10)
C29—C24—C23
121.54 (11)
C4—C5—C23
118.87 (11)
C25—C24—C23
119.66 (10)
C15—C6—C7
109.28 (10)
O5—C25—C24
121.22 (12)
C15—C6—C1
112.32 (10)
O5—C25—C26
121.50 (11)
C7—C6—C1
111.35 (10)
C24—C25—C26
117.23 (10)
C15—C6—H6
C25—C26—C27
112.66 (10)
C7—C6—H6
C25—C26—H26A
C1—C6—H6
C27—C26—H26A
C12—C7—C8
118.80 (11)
C25—C26—H26B
C12—C7—C6
122.23 (11)
C27—C26—H26B
C8—C7—C6
118.95 (10)
H26A—C26—H26B
O3—C8—C7
120.99 (11)
C30—C27—C31
108.98 (11)
O3—C8—C9
121.47 (11)
C30—C27—C28
108.65 (11)
C7—C8—C9
117.52 (10)
C31—C27—C28
110.25 (11)
C8—C9—C10
114.25 (10)
C30—C27—C26
110.53 (11)
C8—C9—H9A
C31—C27—C26
110.34 (11)
C10—C9—H9A
C28—C27—C26
108.06 (10)
C8—C9—H9B
C29—C28—C27
112.24 (10)
C10—C9—H9B
C29—C28—H28A
H9A—C9—H9B
C27—C28—H28A
C14—C10—C13
109.05 (11)
C29—C28—H28B
C14—C10—C9
109.82 (11)
C27—C28—H28B
C13—C10—C9
109.49 (11)
H28A—C28—H28B
C14—C10—C11
110.21 (11)
C24—C29—O4
122.55 (11)
C13—C10—C11
108.83 (11)
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Article Categories:Organic Papers
Previous Document:&
Next Document:&& 计算题(1)0.75+34x=6(2)
本题难度:0.60&&题型:解答题
计算题(1)0.75+x=6&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&(2)(3)3.5×2.6+5.4÷18&&&(4)[4.6-1.2×(+15)]÷3&&&&&&&(5)×(1÷-÷1)+1.
来源:2015年7月新人教版小升初数学模拟试卷(73) | 【考点】小数四则混合运算;整数、分数、小数、百分数四则混合运算.
认真审题,细心计算.x:15=13:56&&&&&34:x=54:2&&&&&&x:0.75=81:25x:1=:1.5&&&&&&:=:x&&&&&&=
计算下面各题,能简算的要简算.2.09×0.6÷0.24;0.4×70.6×50;5.75÷0.46-8.2;62.56÷0.8÷34.
计算下面各题,能简算的要简算7.8×10.1=7.28+3.2÷2.5=3.5×3.4+5.5×3.4+3.4=1.25×34.8×80=0.75×18÷0.15=78.5÷4÷2.5=
下面各题,怎样简便就怎样计算.0.75×2.5×0.43.74×8.3+1.7×3.742.16×0.8÷0.27(34.7-17.5)×(0.54÷0.9)2.5×3.2×1.253.74×101-3.74
下面各题,怎样简便就怎样计算.0.75×2.5×0.4&&&&&&&&&&&&&&3.74×8.3+1.7×3.742.16×0.8÷0.27&&&&&&&&&&&&&(34.7-17.5)×(0.54÷0.9)
解析与答案
(揭秘难题真相,上)
习题“计算题(1)0.75+34x=6(2)(3)3.5×2.6+5.4÷18(4)[4.6-1.2×(23+15)]÷3(5)47×(1÷34-34÷1)+156.”的学库宝(/)教师分析与解答如下所示:
【分析】(1)依据等式的性质方程两边同时减去075再同时乘43求解 (2)先算除法再算加法(3)先算乘除法再算加法(4)(5)按四则运算顺序计算:同级运算从左往右依次运算两级运算先算乘除后算加减有括号的先算小括号里面的再算中括号里面的然后算大括号里面的最后算括号外面的.
【解答】解:(1)075+34x=6&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp075+34x-075=6-075&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp34x×43=525×43&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp x=7&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp(2)=15+49=64(3)35×26+54÷18&nbsp&nbsp&nbsp=91+03=94(4)[46-12×(23+15)]÷3&nbsp&nbsp=[46-12×473]÷3&nbsp&nbsp=[46-188]÷3&nbsp&nbsp&nbsp&nbsp=-142÷3=-×(1÷34-34÷1)+156=47×(43-34)+156=47×712+156=13+116=216
【考点】小数四则混合运算;整数、分数、小数、百分数四则混合运算.
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知识点讲解
经过分析,习题“计算题(1)0.75+34x=6(2)”主要考察你对
等考点的理解。
因为篇幅有限,只列出部分考点,详细请访问。
整数、分数、小数、百分数四则混合运算
1.分数与小数运算,一般有三种不同的处理方法。⑴当小数与分数的分母同除以一个数可以化简时, 应先同除以一个数后再计算。⑵当小数与分数的分母同除以一个数不能化简时,或虽然能化简,但计算并不简便,可以把小数化成分数后再计算。⑶当把分数化成小数后能使计算简便时, 可以把分数化成小数后计算。我们要根据不同情况,合理,灵活地选择简便的计算方法。2.百分数可以化成小数进行加减。3.当有小括号的时候注意运算顺序。
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