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“Molecular structure analysis using single crystal (and powder) X-ray diffraction”

Chemical crystallography

Chemical Crystallography

« The great advantage of X-ray analysis as a methodof chemical structure analysis is its power to show totally unexpected and surprising structure with, at thesame time, complete certainty ».

Nobel lecture, 1964, Prof. Dorothy Crowfoot Hodgkin


The determination of the 3D structure of moleculesusing the technique of X-ray diffractioncan be applied to both crystals and in some cases powders

Knowledge of this technique is essential for chemiststoday. For example, the knowledge of accurate molecular structure isessential for structure based functional studies to aidin the development of effective therapeutic agents anddrugs.

Reference book: Crystal Structure Analysis: A Primer. J. P. Glusker & K. N. Trueblood. Oxford University Press, ISBN 0-19-503543-7 (pbk.)


The development of this technique has revolutionized chemistry as shown by the many Nobel Prizes awarded in the field of Crystallography, for example:

von Laue in 1914, the Braggs, father and son 1915,

Pauling 1954, Perutz and Kendrew 1962,

Crick, Watson and Wilkens 1962, Hodgkin 1964,

Barton and Hassel 1969, Lipscomb 1976, Klug 1982 and

Hauptmann & Karle, 1985….

To view an enlarged image of an object (dimensions mm)we can use an optical microscope

Light

object

Lens

Image

Scattered rays (T)

Refocused (T-1)

(Fourier transform = T; inverse Fourier transform = T-1)


Diffraction of light by an object can be described mathematically by a Fourier Transform. If we have an object given by the function f(x) theFourier transform will be given by

F = T[f(x)]

For any Fourier transform there exists an inverse Fourier transform by which we can recuperate the inital object

f(x) = T-1 (F)

f(x) = T-1 {T[f(x)]}

So the Fourier transform of the Fourier transform of an object gives us an image of the original object; exactly as in the optical microscope.

Crystalsize < 0.5 mm

X-rays

o oo o

oo

ooo oooo o

computer

Ocrystallographer

mathematics

Image plate detector (IPD)

To view atoms in a moleculeseparated by 1-2 Å1 Angstrom (Å) = 0.00000001 cm = 10-8cm = 10-10 m we will need to use X-rays, for example, with a wavelength of0.71073 Å (MoKα )or 1.5418 Å (CuKα)

No lens exists which iscapable of refocusing X-rays,so the lens is replaced bya computer, a crystallographerand a lot of mathematics

Molecules

3D electron density map

Ref: Crystal Structure Analysis: A Primer. J. P. Glusker & K. N. Trueblood. Oxford University PressISBN 0-19-503543-7 (pbk.)

(T)

(T-1)



Diffraction of X-rays by a crystal can be described mathematically by a Fourier Transform. If we have a crystal given by the function f(x) theFourier transform will be given by

F = T[f(x)]

For any Fourier transform there exists an inverse Fourier transform by which we can recuperate the inital object

f(x) = T-1 (F)

f(x) = T-1 {T[f(x)]}

So the Fourier transform of the Fourier transform of an object gives us an image of the original object; the electron density.


A single crystal is the convolution of the basic unit (the unit cell) and a

3D-lattice. The unit cell can contain 1 or more molecules.

Single crystal = unit cell * 3D lattice

If we place the crystal in the X-ray beam we will be able to register the

diffraction image, which is the Fourier transform (T) of the crystal

or the Fourier transform of the convolution.


So the diffraction image is the Fourier transform of the single crystal

T(single crystal)

= T(unit cell * 3D lattice)

which by the theory of convolution is

= T(unit cell) . T(3D lattice)

The T(3D lattice) is equal to the reciprocal lattice,

so what we will see is the Fourier transform of the unit cell contents

distributed over the lattice points of the reciprocal lattice.

StoeStoe IPDS IPDS Image Plate Diffraction Image Plate Diffraction SystemSystem

single crystalsize < 0.5 mm

Chemical crystallography – single crystal analysis

Diffraction image Diffraction image fromfrom a single a single crystalcrystalStoeStoe IPDS Image Plate Diffraction IPDS Image Plate Diffraction SystemSystem


Chemical CrystallographyAs we cannot use a lens to perform the inverse Fourier transform (T-1) itmust be done mathematically. But to do this we need all the information describing the diffracted waves, that is

Fhkl = |Fhkl| exp(iαhkl)

where |Fhkl| is the amplitude of the wave and αhkl is the phase.(hkl are the Miller indices used to characterize the diffacted wave)

The electron density at any point (x,y,z) in the unit cell is given by

The problem resides in the fact that when we record the diffraction image, we measure the intensity of the various diffracted waves (Ihkl ~ |Fhkl| 2)but we loose the information concerning the phase of each wave (αhkl).

Hence the expression, the « phase problem »

0

1( , , ) cos 2 ( )hkl hkl

h k lx y z F hx ky lz

Vρ π α

≥

= | | + + −∑ ∑∑


It can be shown that

the phase αhkl is related to the positions of the atoms in the unit cell.

This of course is just the information we want to describe the structure ofthe molecule – so we have a problem!

sin 2 ( )tan

cos 2 ( )

j j j j

hkl

j j j j

j

j

f hx ky lz

f hx ky lz

πα

π

+ +=

+ +

∑∑


It was Lindo Patterson in 1934 who realised that if you take the Fourier transform of the Intensity, Ihkl [ ~ |Fhkl|2)]

rather than Fhkl [ = |Fhkl| exp(iαhkl)]

the resulting electron density map would look like a radial distribution function.

That is, it would be possible to determine the distances between the atomsin the unit cell. Vector (u,v,w) = [(x1,y1, z1) – (x2, y2, z2)].

0

1( , , ) cos 2 ( )hkl

h k lP u v w I hu kv lw

Vπ

≥

= + +∑ ∑∑

f1

fn

Fhkl

00

900

|Fhkl|

Σfjsinαj

Σfjcosαj

α1

f1sin2π(hx1 + kx1 + lx1)tanαhkl ~ tanα1 = -------------------------------

f1cos2π(hx1 + kx1 + lx1)

Now with the knowledge of the heavy atom

position we can substitute the values of x1, y1, z1 of

the heavy atom in the expression for tan αhkl

If one of the atoms was heavy ( 16 or more electrons) then it will be possible to locate the« heavy atom-heavy atom » vectors andhence, determine the position of this atom, (1), in the unit cell: tanα1 ~ tanαhkl

Hence the expression, the « heavy atom method »


Using this preliminary value of α one can calculate an electron density mapand so we will be able to locate the other atoms in our molecule.

0

1( , , ) cos 2 ( )hkl hkl

h k lx y z F hx ky lz

Vρ π α

≥

= | | + + −∑ ∑∑


Example: TZN001 Prof. Rafael Lopez Garzon

Reaction: TREN + 2ZnCl2 [(TREN)Zn2Cl4]

N

N

NO

HN NH2

CH3

O

N

NH3++H3N

. H2O + 2ZnCl2

N

N

N

HN NH2

CH3

O

N

NH2H2N

Zn Cl

Cl

O

Zn Cl

Cl

. 2H2O

-2HCl

+ H2O

[C11H26N8O4Zn2Cl4 ]

Cl- Cl-

Chemical Crystallography Example: TZN001 Prof. Rafael Lopez Garzon

STOE XRED32 1.04 19-Jan-2003 12:37Input File(s) : E:\Data-HSTE\TZN001\TZN001.x

Cell Parameters : 13.8365(6), 20.6708(7), 16.2932(7), 90.00, 111.885(3), 90.00Cell Volume : 4324.2(3)Spacegroup : P 21/a (centrosymmetric)Laue Symmetry : Monoclinic 2/m (b)

Number of reflections read : 57356 Rejected : 226Minimum and maximum 2Theta : 3.2, 59.1Minimum and maximum H,K,L : -19,-27,-19 19,28,22Avg. I/Sigma for all data : 24.08Systematic absent reflections : 901 Average I/Sigma = 0.23Number of unique reflections : 12044

Completeness of data set [%] : 99.4 Redundancy : 4.67R(int) = 0.0638 ( 56153 contributors ) R(Sigma) = 0.0310

Average I/Sigma vs. 2Theta2Theta 3.2 26.0 33.1 37.8 42.0 45.5 49.1 52.0 55.0 57.3 59.1Resol. 12.84 1.58 1.25 1.10 0.99 0.92 0.86 0.81 0.77 0.74 0.72N(Refl) 5789 5836 5531 5792 5817 6505 6018 6496 5636 3710I/Sigma 64.64 42.33 34.36 28.12 22.06 15.43 10.94 7.93 6.57 5.03Completeness 99.6 100.0 100.0 100.0 100.0 100.0 100.0 100.0 100.0 92.8Redundancy 4.9 4.9 4.9 4.8 4.7 4.7 4.6 4.6 4.5 4.0R(int) 0.055 0.055 0.057 0.062 0.067 0.077 0.092 0.118 0.152 0.184I/Sigma < 0 2 4 8 16 32 64 128 256 512N(Refl) 3151 14432 20382 27501 35684 43943 50872 55378 57028 57130Percent 5.5 25.2 35.5 47.9 62.2 76.6 88.7 96.6 99.4 99.6


Spacegroup Determination Crystal System : Monoclinic_BLattice N(+) N(>4s) Avg I/S Max N(-) N(>4s) Avg I/S Max

A 28593 18695 24.5 396.0 28537 18053 23.7 429.1B 28559 18279 23.7 310.7 28571 18469 24.5 429.1C 28594 18116 23.9 396.0 28536 18632 24.2 429.1I 28530 18458 24.2 429.1 28600 18290 24.0 396.0F 42873 27545 24.0 396.0 14257 9203 24.2 429.1

R(obv) 38090 24437 23.9 429.1 19040 12311 24.4 369.6R(rev) 38061 24508 24.1 429.1 19069 12240 24.0 396.0

Symm.Element-> -a- 848 1 0.2 4.3 792 657 50.9 369.6

-c- 845 350 23.8 296.2 795 308 25.6 369.6-n- 825 349 24.4 296.2 815 309 25.0 369.6

-> -21- 25 0 0.4 1.9 27 25 61.0 153.4

Possible spacegroup(s) : P 21/a (c,-b,a) -> P 21/cE - Statistics

Average |Z-1| (Z-1)^2 (Z-1)^3 |Z-1|^3 Number

all data 0.968 1.958 7.380 8.070 57130H,K,L 0.974 2.005 7.721 8.393 51567

0,K,L 0.951 1.980 8.059 8.767 1917H,0,L 0.943 1.234 1.276 2.573 1640H,K,0 0.850 1.346 3.111 3.723 1859

theoreticalacentric 0.736 1.000 2.000 2.415

centrosym. 0.968 2.000 8.000 8.691


Space group: P21/a (No. 14, unique axis b, cell choice 3)

In general Z = 4 (4 identical molecules related by a 2-fold screw axis, a glide plane and a center of symmetry)

Positions (1) x, y, z (2) –x + 0.5, y + 0.5, -z (3) –x, -y, -z (4) x + 0.5, -y + 0.5, z

Refection condiditions: h0l : h = 2n a glide plane0k0 : k = 2n 2-fold screw axis

(h00 : h = 2n) (a glide plane)

Patterson vectors: 1) – 2) 2x – 0.5, -0.5, 2z1) – 3) 2x, 2y, 2z1) – 4) -0.5, 2y – 0.5, 0


0

1( , , ) cos 2 ( )hkl

h k lP u v w I hu kv lw

Vπ

≥

= + +∑ ∑∑ Super-sharp Patterson for TZN001

Patterson vectors:U V W

1) – 2) 2x – 0.5, +

Maximum = 999.10, minimum = -68.60 highest memory used = 9220 /1497940.4 seconds CPU time

Rms Patterson density excluding points close to the origin or an equivalentlattice point is 6.03

U V W Weight Peak Sigma Length1 0.0000 0.0000 0.0000 4. 999. 165.73 0.00 2 0.5000 0.2949 0.0000 2. 98. 16.27 9.223 0.5000 0.2320 0.0000 2. 93. 15.49 8.424 0.5000 0.1514 0.0000 2. 86. 14.32 7.595 0.7156 0.2228 0.2015 1. 82. 13.65 7.566 0.6721 0.2380 0.3374 1. 80. 13.22 9.677 0.4135 0.4296 0.0501 1. 80. 13.20 10.438 0.9140 0.2779 0.0516 1. 78. 12.93 5.999 0.2133 0.0709 0.2037 1. 78. 12.91 3.8210 0.7994 0.5000 0.1508 2. 78. 12.86 11.2111 0.2589 0.1929 0.2863 1. 76. 12.57 6.1712 0.9710 0.0328 0.4883 1. 75. 12.50 8.1413 0.1719 0.4698 0.3373 1. 75. 12.48 10.9714 0.6114 0.4604 0.4616 1. 75. 12.41 14.3615 0.6281 0.5000 0.2537 2. 72. 11.93 12.8916 0.8619 0.5000 0.1760 2. 72. 11.91 11.0817 0.4734 0.2615 0.4868 1. 70. 11.66 9.8118 0.5000 0.0616 0.0000 2. 70. 11.65 7.0319 0.0799 0.0000 0.1633 2. 70. 11.62 2.4720 0.7572 0.0402 0.2857 1. 70. 11.59 6.73

0.5, 2z1) – 3) 2x, 2y, 2z1) – 4) +0.5, 2y – 0.5, 0

Result of the interpretation of the Patterson synthesis.

Zn


Σ||Fobshkl| - |Fcalc

hkl||R1 = -----------------------------

Σ|Fobshkl|

R factor

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ SHELXL-97 - CRYSTAL STRUCTURE REFINEMENT - HUGE W95/98/NT/2000 VERSION ++ Copyright(C) George M. Sheldrick 1993-2001 Release 97-2 ++ d:\solvias-conference-may-2005\tzn started at 11:06:57 on 23-May-2005 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

TITL TEST STRUCTURE D:\SOLVIAS-CONFERENCE-MAY-2005\TZN001\TZN001CELL 0.71073 13.83600 20.67100 16.29300 90.00000 111.88500 90.00000ZERR 4 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000LATT 1SYMM -X+ 0.50000, Y+ 0.50000, -ZSFAC C H N O CL ZNUNIT 120 120 16 20 16 8

V = 4324.04 F(000) = 1624.0 Mu = 1.39 mm-1 Cell Wt = 3196.48 Rho = 1.228

PLAN 30FMAP 2L.S. 2BONDCONFACTAFVAR 0.27601WGHT 0.10000ZN1 6 0.350590 0.603610 0.424190 11.000000 0.040000ZN2 6 0.322760 0.634260 0.913810 11.000000 0.040000ZN3 6 0.566000 0.674330 0.628250 11.000000 0.040000CL4 5 0.502180 0.718310 1.125960 11.000000 0.040000CL5 5 0.433410 0.595130 0.586760 11.000000 0.040000CL6 5 0.531800 0.781610 1.021290 11.000000 0.040000CL7 5 0.333400 0.679800 1.055500 11.000000 0.040000CL8 5 0.631390 0.692200 0.518750 11.000000 0.040000CL9 5 0.680770 0.638450 0.756780 11.000000 0.040000CL10 5 0.484050 0.765890 0.636780 11.000000 0.040000HKLF 4

Peaks located from the Patterson synthesis are introduce here used to calulate phase values, and then a difference Fourier synthesis is calculated to locate new atomic positions.

TEST STRUCTURE D:\SOLVIAS-CONFERENCE-MAY-2005\TZN001\TZN001

ATOM x y z sof U11 – thermal parameter

Zn1 0.34979 0.60361 0.42447 1.00000 0.02303

Zn2 0.32230 0.63415 0.91325 1.00000 0.02610

Zn3 0.56517 0.67366 0.62713 1.00000 0.02499

Cl4 0.50079 0.71872 1.12476 1.00000 0.01071

Cl5 0.43605 0.59507 0.58754 1.00000 0.02539

Cl6 0.53152 0.78099 1.02133 1.00000 0.03389

Cl7 0.33512 0.67971 1.05574 1.00000 0.03679

Cl8 0.63058 0.69156 0.51903 1.00000 0.03041

Cl9 0.68142 0.63834 0.75698 1.00000 0.03395

Cl10 0.48403 0.76623 0.63666 1.00000 0.03609

Final Structure Factor Calculation for TEST STRUCTURE D:\SOLVIAS-CONFERENCE-MAY-2005\TZN001\TZN001

Total number of l.s. parameters = 41 Maximum vector length = 511 Memory required = 871 / 25039

wR2 = 0.7556 before cycle 3 for 12044 data and 0 / 41 parameters

GooF = S = 7.705; Restrained GooF = 7.705 for 0 restraints

Weight = 1 / [ sigma^2(Fo^2) + ( 0.1000 * P )^2 + 0.00 * P ] where P = ( Max ( Fo^2, 0 ) + 2 * Fc^2 ) / 3

R1 = 0.3728 for 10125 Fo > 4sig(Fo) and 0.3952 for all 12044 datawR2 = 0.7556, GooF = S = 7.705, Restrained GooF = 7.705 for all data

Electron density synthesis with coefficients Fo-Fc = Difference Fourier synthesis coefficients Fobshkl – Fcalc

hklHighest peak 21.81 at 0.6087 0.6330 0.1700 [ 2.26 A from CL4 ]Deepest hole -3.08 at 0.3535 0.3634 0.2827 [ 0.65 A from CL9 ]

Mean = 0.00, Rms deviation from mean = 1.00 e/A^3, Highest memory used = 2128 /159269

Fourier peaks appended to .res file

x y z sof U Peak Distances to nearest atoms (including symmetry equivalents)

Q1 1 0.6087 0.6330 1.1700 1.00000 0.05 21.81 2.26 CL4 3.67 CL7 3.80 CL6 4.57 ZN2Q2 1 0.5031 0.7793 1.2388 1.00000 0.05 19.81 2.23 CL4 3.66 CL7 3.71 CL6 4.61 CL8Q3 1 0.1130 0.6931 0.8402 1.00000 0.05 11.11 2.95 ZN2 3.24 CL10 3.57 CL6 3.72 CL7Q4 1 0.1858 0.6248 0.4337 1.00000 0.05 10.21 2.37 ZN1 3.49 CL5 4.21 CL8 5.12 ZN3Q5 1 0.2941 0.6111 0.2783 1.00000 0.05 10.07 2.22 ZN1 4.13 CL7 4.69 CL5 4.94 CL6Q6 1 0.3226 0.4604 0.4215 1.00000 0.05 9.72 2.98 ZN1 3.28 CL8 3.41 ZN3 3.54 CL9Q7 1 0.2628 0.5353 0.9532 1.00000 0.05 9.69 2.38 ZN2 3.38 CL7 5.12 CL4 5.34 CL6Q8 1 0.3081 0.5942 0.7788 1.00000 0.05 9.65 2.28 ZN2 4.12 CL5 4.73 CL7 5.10 CL10Q9 1 0.7272 0.6237 0.3814 1.00000 0.05 9.50 3.32 CL8 4.63 CL4 4.91 CL10 5.16 CL5Q10 1 0.4796 0.5630 0.4088 1.00000 0.05 8.99 2.08 ZN1 3.25 CL5 3.44 CL8 3.46 CL5Q11 1 0.6313 0.6699 0.9318 1.00000 0.05 8.92 3.24 CL9 3.29 CL6 4.18 CL7 4.24 ZN2Q12 1 0.3535 0.8965 0.6035 1.00000 0.05 8.90 3.17 CL10 3.40 CL8 4.11 CL9 4.39 ZN3Q13 1 0.3564 0.7017 0.4189 1.00000 0.05 8.90 2.03 ZN1 3.37 CL5 3.53 CL8 3.58 CL10Q14 1 0.3235 0.7212 0.8506 1.00000 0.05 8.80 2.07 ZN2 3.40 CL7 3.40 CL6 3.52 CL9Q15 1 0.1240 0.4734 0.6137 1.00000 0.05 8.76 5.15 CL5 5.69 ZN2 5.72 CL10 5.77 CL8Q16 1 0.9331 0.6338 0.8105 1.00000 0.05 8.74 3.26 CL9 3.64 CL6 3.78 CL10 4.92 ZN3Q17 1 0.2734 0.5129 0.4202 1.00000 0.05 8.65 2.14 ZN1 3.29 CL5 4.43 CL8 4.45 CL9Q18 1 0.4911 0.9619 0.6066 1.00000 0.05 8.58 4.08 CL10 5.41 ZN1 5.63 CL8 5.73 ZN2Q19 1 0.4663 0.5912 0.9491 1.00000 0.05 8.51 2.06 ZN2 3.47 CL7 3.79 CL4 4.10 CL6Q20 1 0.2268 0.5442 0.7400 1.00000 0.05 8.44 3.22 ZN2 4.58 CL5 5.03 CL10 5.55 CL7Q21 1 0.1558 0.6407 0.8717 1.00000 0.05 8.32 2.15 ZN2 3.20 CL7 3.83 CL6 4.15 CL10Q22 1 0.9453 0.5301 0.8622 1.00000 0.05 8.32 4.07 CL9 4.59 CL6 5.41 ZN2 5.46 CL6Q23 1 0.1802 0.5105 0.4189 1.00000 0.05 8.27 3.01 ZN1 3.99 CL5 4.83 CL8 5.04 CL9Q24 1 0.0359 0.5766 0.4207 1.00000 0.05 8.27 4.36 ZN1 5.03 CL10 5.07 CL8 5.19 CL5Q25 1 0.5393 0.9272 1.0895 1.00000 0.05 8.25 3.21 CL6 4.41 CL4 4.86 CL7 5.58 ZN1Q26 1 0.1159 0.4822 0.9368 1.00000 0.05 8.09 4.36 ZN2 4.79 CL6 5.02 CL7 5.33 CL6Q27 1 -0.0265 0.5207 0.3981 1.00000 0.05 7.94 5.34 ZN1 5.35 CL10 5.84 CL10 6.10 ZN2Q28 1 0.1975 0.5293 0.6432 1.00000 0.05 7.89 3.97 CL5 4.62 ZN2 5.01 ZN1 5.13 CL10Q29 1 0.1836 0.5899 0.2306 1.00000 0.05 7.80 3.15 ZN1 4.22 CL6 4.52 CL7 4.67 CL4Q30 1 0.3656 0.5709 0.2519 1.00000 0.05 7.62 2.98 ZN1 3.80 CL7 4.37 CL9 4.47 CL4

Begin by adding 10-20 new peaks, calculate new values of tanαhkl and repeat cycle untilall the atoms have been locate and no new peaks can be found.

Chemical CrystallographyExample: TZN001 Prof. Rafael Lopez Garzon

Final R1 value after full-matrix least-squares refinement = 0. 0321


Structure solution by « Direct Methods »

These are based on statistics and were first developed in the late 1940’s.

By « direct » we mean that we can obtain the phases, αhkl, directly withouthaving to locate an atomic position(s) as we do for the « heavy atom method ».

Phase determination is based on the premise that; the electron-density map should have no negative areas, the peaks are only at discrete locations (atomic positions).

Phase relationships between groups of three reflections(« triplets ») are developed to give relative phase angles for which an electron density map may be calculated.

Only certain values for the phases are consistent with the above conditions.

Ehkl = Fhkl / (ε Σfj2)1/2


« Direct Methods »

For a centrosymmetric structure the phase values αhkl are either 0° or 180°; Cosine(αhkl) will be either +1 or -1; or the sign is either + or -

Using triplets of very strong/intense reflections,

By a triplet of Bragg reflections: H, K, and (H -K)

where H = h,k,l and H = K + (H-K)

For eg. if K = -2, 4, 7 and H – K = 3, -1, 1

Then reflection H must be 1, 3, 8

Sayre found thatEH = const ΣEKEH-K

Chemical Crystallography « Direct Methods »

For 3 strong reflections

sign(H) ~ sign(K) x sign(H - K) Σ2 relationship

Probability relationships – Karle & Hauptmann (1950), Nobel prize 1985

sign(H) x sign(K) x sign(H - K) ~ +1

This relation is called the « Triple-product sign relationship ».

If you know the sign of K and H-K then you can deduce theprobable sign of reflection H.

[Note: For a centro-symmetric structure:If the sign is + then αhkl is equal to 0°If the sign is – then αhkl is equal to 180°

For a non-centrosymmetric structure the situation is more complicatedas αhkl can take any value between 0 - 360°]

Chose 3 strong reflections (large E values) with manyΣ2 relations to define the origin.H, K, H-K1, 2, 3 Fix their signs as +Symbolic additionThe combination of the indices ofreflections 1 & 2 gives those of a

new reflection number 41 & 2 gives 4, whose sign will be = + . + = +1 & 3 gives 52 & 3 gives 6 4 & 3 gives 74 & 5 gives 8Etc……. If the phase probability falls below say 0.98, we must stop and introduce a symbol, for eg. reflection « a » (a strong reflection) whose phase value is 0 or 180º,that is, sign + or -, and symbolic addition continues1 & a gives 32 (= + . a)2 & a gives 33 (= + . a)Etc…


The absorption coefficient of an atom for X-rays shows discontinuities when plottedas a function of the incident X-radiation.Absorption edge – near which the energy ofthe X-radiation is sufficient either to excite an electron in the strongly absorbing atomto a higher quantum state or to eject theelectron completely from the atom.

Anomalous dispersion and absolute configurationThis has an effect on the phase change on scattering. The scattering factor for an atom, fj, becomes « complex »and is replaced by

fj + Δfj’ + iΔfj’’

Δfj ’’

Δfj ’’

I hkl≠ I –h-k-l and αhkl≠ -α-h-k-l

|Fhkl|

|F-h-k-l|

αhkl

α-h-k-l

Δfj’

Δfj’


Sodium Rubidium (+)-tartrate:comparison of Ihkl and I-h-k-l

h k l observed Ihkl/I-h-k-l calculated Ihkl/I-h-k-l

1 5 1 ? 1.081 6 1 > 1.301 7 1 < 0.831 8 1 > 1.251 9 1 > 1.411 10 1 > 1.191 11 1 < 0.662 6 1 > 1.102 7 1 > 3.002 8 1 > 1.072 9 1 ? 1.022 10 1 < 0.84


Example: unknown

N

O O

O

SOO

C20 H29 N O5 S Mw 395.50

STOE XRED32 1.26 01-Apr-2005 11:17

Input File(s) : E:unknown.x

Cell Parameters : 6.9087(3), 9.9225(5), 29.2554(20), 90.00, 90.00, 90.00Cell Volume : 2005.51(20)

Spacegroup : P 21 21 21 (acentric)Laue Symmetry : Orthorhombic m m m

Number of reflections read : 16603 Rejected : 50Minimum and maximum 2Theta : 2.8, 51.3Minimum and maximum H,K,L : -8,-12,-35 7,11,35Avg. I/Sigma for all data : 22.88

Systematic absent reflections : 68 Average I/Sigma = 0.29

Number of unique reflections : 3787 Friedel pairs have not been mergedCompleteness of data set [%] : 99.6 Redundancy : 4.35

R(int) = 0.0518 ( 16460 contributors ) R(Sigma) = 0.0347

Average I/Sigma vs. 2Theta2Theta 2.8 22.6 28.7 33.3 36.9 40.0 43.1 45.6 48.2 50.3 51.3Resol. 14.63 1.82 1.43 1.24 1.12 1.04 0.97 0.92 0.87 0.84 0.82N(Refl) 1589 1631 1801 1640 1769 1895 1829 1863 1722 814I/Sigma 61.43 38.38 33.02 26.11 22.85 13.39 11.49 7.72 6.35 5.09Completeness 99.7 100.0 100.0 100.0 100.0 100.0 100.0 100.0 100.0 93.0Redundancy 4.5 4.5 4.5 4.4 4.4 4.4 4.3 4.2 4.2 4.1R(int) 0.046 0.044 0.044 0.043 0.047 0.056 0.058 0.079 0.167 0.222

I/Sigma < 0 2 4 8 16 32 64 128 256 512N(Refl) 336 2322 3861 6364 9601 12635 15157 16340 16546 16553Percent 2.0 14.0 23.3 38.3 57.8 76.1 91.3 98.4 99.7 99.7

********************************************************************************General extinction analysis for unknown********************************************************************************Class Rule N(+) N(-) Q(+) Q(-) Symbol Direction________________________________________________________________________________

hkl none 16553 16553 22.88 22.88 --> P h+k+l=2n 8261 8292 22.66 23.11 I h+k=2n 8292 8261 22.68 23.08 C k+l=2n 8286 8267 22.23 23.53 A h+l=2n 8267 8286 23.30 22.46 B h+k,k+l,h+l=2n 4146 12407 22.45 23.03 F -h+k+l=3n 5515 11038 22.62 23.01 R, obverse h-k+l=3n 5531 11022 22.59 23.03 R, reverse

________________________________________________________________________________

hk0 h=2n 117 131 22.07 19.48 a || (001) k=2n 112 136 16.86 23.87 b || (001) h+k=2n 127 121 16.16 25.47 n || (001) h+k=4n 63 185 17.25 21.88 d || (001)

________________________________________________________________________________

0kl k=2n 745 750 25.80 28.18 b || (100) l=2n 749 746 27.10 26.88 c || (100) k+l=2n 745 750 27.83 26.16 n || (100) k+l=4n 373 1122 25.38 27.53 d || (100)

________________________________________________________________________________

h0l h=2n 531 603 26.78 31.73 a || (010) l=2n 563 571 24.95 33.81 c || (010) h+l=2n 574 560 32.04 26.71 n || (010) h+l=4n 284 850 30.55 29.03 d || (010)

________________________________________________________________________________

00l l=2n 46 46 54.38 0.29 --> 21,42,63 || c l=3n 30 62 11.23 35.13 31,32,62,64 || c l=4n 21 71 69.62 14.83 41,43 || c l=6n 14 78 23.35 28.05 61,65 || c

________________________________________________________________________________

h00 h=2n 11 10 24.82 0.37 --> 21,42,63 || a h=3n 5 16 1.22 16.92 31,32,62,64 || a h=4n 6 15 19.15 10.79 41,43 || a h=6n 3 18 1.65 15.10 61,65 || a

________________________________________________________________________________

0k0 k=2n 12 12 42.68 0.25 --> 21,42,63 || b k=3n 8 16 27.07 18.66 31,32,62,64 || b k=4n 6 18 57.41 9.48 41,43 || b k=6n 4 20 53.97 14.96 61,65 || b

________________________________________________________________________________

Three 2-fold screw axes mutually perpendicuar to one another, space group P212121 , No. 19, International Tables for Crystallography, Vol A.

Example: unknown Direct methods: « non-centrosymmetric structure »

SIR97 : Phase routine Release 97.01

test structure D:\unknown\unknown++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

*** converge / diverge section ***

Origin fixing reflexion(s)

phase assignedcode h k l E restriction phase

9 0 8 15 2.62 0,180 3606 0 7 16 2.77 90,270 90 18 1 0 8 2.41 0,180 360

Selected symbols

phase code h k l E restriction

174 2 1 0 1.60 0,180 16 0 1 27 2.42 90,270 92 5 2 11 1.80 any101 5 3 10 1.77 any7 1 8 7 2.68 any

Enantiomorph fixing reflexion

code h k l E

39 2 0 27 2.08

Solution from direct methods program SIR97 -

Solution from direct methods program SIR97; R1 = 0.0758, after refining atomic positions and isotropic thermal parametersContinue by identifying correctly the N and O atoms.

O

N

Final R factor is R1 = 0. 0241, after adding hydrogen atoms and refineingall atoms anisotopically.The absolute structure parameter (Flack X factor) = -0.01(5), hence the atomic coordinates correspond to the absolute structure of the molecule in the crystal.

STADI-P Stoe Powder diffractometer

powder samplein glass capillary

Chemical crystallography – powder analysis

powder

Diffraction image Diffraction image fromfrom a a powderpowder

Coun

ts

1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0

.0

1.0

2-Theta, deg x 10

Cou

nts

x 10

5

powder

Chemical crystallography – powder analysis

N

N

O

O

OCH3

OCH3

green micro-crystaline powder

Organic ligand + metal

+ copper dichloride (CuCl2)

Chemical analysis indicated :38.65% of carbon, 2.40% of hydrogen, 13.28% nitrogen powder X-ray diffractogramCompound is stable to 240ºC2 different C=O groups present (IR spectroscopy)

X-ray analysis of a powder samplePossible structure consists of

two molecules of organic ligand

and one metal atom

Observed profile (+) ; calculated profile (-) for [Cu(C7H5N2O4)2]n

The difference curve is presented below on the same scale.Structure confirmed by the Reliability factors, eg. Rp 0.027

X-ray analysis of a powder sample: [Cu(C7H5N2O4)2]n

2 organic ligands and 1 metal atom

N

N

O

O

OO

N

N

O

O

OO

Cu

CH3

H3C


Example:

N

N

NH2

NH2H2N

H2N

O

OO

O

C8H8N6O4 (Pz-amide)

Crystal system : Triclinic

space group : P -1

Unit cell : a = 4.995 b = 7.143 c = 7.196 Åα = 78.88° β = 86.11° γ = 73.80 °

N

N

NH2

NH2H2N

H2N

O

OO

O

Pz-amide : Solution from powder diffraction data

RF = 0.048

N

N

NH2

NH2H2N

H2N

O

OO

O


« The great advantage of X-ray analysis as a methodof chemical structure analysis is its power to show totally unexpected and surprising structure with, at thesame time, complete ’’chemical’’ certainty ».

Nobel lecture, 1964, Prof. Dorothy Crowfoot Hodgkin


Reference book used for this lecture: Crystal Structure Analysis: A Primer. J. P. Glusker & K. N. Trueblood. Oxford University Press, ISBN 0-19-503543-7 (pbk.)

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