George Hardgrove’s Program for 2D Patterson Maps using MathCad.

(Color code: Red is used to highlight action steps for you.)

Open the Hardgrove folder on Kastner Public Space and copy the Patterson folder down to your desktop.

Open (double click) the file im1run.xmcd ... it should open in MathCad.

Move through the file by scrolling using the arrow or mouse: this triggers the calculations. Only use the scroll bar to search between calculations.

We will look at three structures:

im1 - Bromobenzoic Acid C7H6O2Br	The crystal structure of the compound has been reported by G. Ferguson and G. A. Sim in Acta Crystallographica 15, page 346, (1962). The ultimate objective here will be to determine which isomer is present and to describe the hydrogen bonding - chains or dimers.
im2 - Bromo-1,4-napthoquinone C10H5O2Br	This molecule is a derivative of 1,4-naptho-quinone shown at the left. The ultimate objective of this exercise is to determine to location of the Br atom in the molecule. The structure was solved by J. Gaultier and C. Hauw, Acta Cryst. 18, page. 604 (1965)
im3 - Bromo-amino-1,4-napthoquinone C10H6O2NBr	This molecule is a derivative of 1,4-naptho- quinone shown at the left. The ultimate objective of this exercise is to determine to location of the Br and NH2 groups in the molecule. The structure was solved by J. Gaultier and C. Hauw, Acta Cryst. 20, page. 620 (1966)

The Patterson Maps for these structures are given below. Word.doc version.

CRYSTAL STRUCTURE IMAGES AND THE PHASE PROBLEM

One way to think about the image of a crystal unit cell is to picture an electron density function. It may be calculated at many points X,Y,Z in the unit cell. At the centers of atoms it will have large values, while in the empty spaces between molecules it will have values near zero. The formula for the lectron density is:

where Fo is the amplitude of the reflection hkl and V the volume of the unit cell. If we could calculate this function directly we would be able to locate the peaks and determine the positions of the atoms. It is not so easy. What we measure in the laboratory are values of Fo² or the intensities of the diffracted X-rays. When we take the square root of Fo² a sign ambiguity arises - the value of Fo can be + or -. This sign problem (or in general phase angle problem) is called the phase problem in X-ray crystallography. A great deal of time and effort has been expended to find the signs or relative phases of X-ray reflections. One route to solving the phase problem was developed by A. L. Patterson which uses values of Fo² rather than Fo for the calculation.

The Patterson function is not the same as the electron density, but it is an interatomic vector map. This function describes the simultaneous probability of finding an atom at coordinates X,Y,Z and another atom at coordinates X+U,Y+V,Z+W. Thus for every pair of atoms in the structure there is a vector U,V,W from the first to the second and another vector
-U,-V,-W from the second to the first. The heights of peaks in the Patterson function at these two positions are proportional to the product of the number of electrons of the two atoms. For a structure with n atoms in the unit cell there can be n² peaks (actually n(n-1) if we exclude vectors from each atom to itself) in the resulting Patterson function.

The Patterson function is most useful when one of the atoms in a molecule has a much larger atomic number than the rest. Bromobenzoic acid is a good test case. The peaks representing the probability of finding a Br atom at X,Y,Z and another Br atom at X+U,Y+V,Z+W will be much larger than Br-C. C-O, C-C or other combinations. These very large peaks can be used to locate the Br atoms.

The crystal symmetry often dictates the relationship of the coordinates of one heavy atom with another. For example, if two molecules such as bromobenzoic acid are related by a screw axis through the origin extending in the y direction, then one Br atom will have coordinates X,Y,Z and the other Br will have coordinates -X, 1/2+Y,-Z. The Patterson function will have large peaks corresponding to the vectors representing the differences in these coordinates. We shall construct a table displaying the differences between the coordinates across the top and those down the side.

[Example: (x, y, z) - (x, y, z) = 0, 0, 0 and (x, y, z) - (-x, y + 1/2, -z) = 2x, -1/2, 2z.]

	x, y, z	-x, y + 1/2, -z
x, y, z	0, 0, 0	2x, -1/2, 2z
-x, y + 1/2, -z	2x, -1/2, 2z	0, 0, 0

For example, consider a Patterson map of a structure with this symmetry with peaks at

U,V,W = 0.58, 0.50, 0.28 and 0.42 ,0.50, 0.72. Note that -0.58, -0.50, -0.28 is identical to 0.42 ,0.50, 0.72. That is, 0.42 is equal to 1 - 0.58, so 0.42 is equal to -0.58 by translation. (Add and subtract one at whim, says MEK). If you replace variables x, y, z in the table above with the solution the Patterson (the X values of the two heavy atoms are +0.21 and -0.21 (or +0.79) and the z values are +0.36 and -0.36 (or +0.64).

	0.79, y, 0.64	0.21, y + 1/2, 0.36
0.79, y, 0.64	0, 0, 0	-0.58, -0.5, -0.28
0.21, y + 1/2, 0.36	0.58, 0.50, 0.28	0, 0, 0

The value of Y cannot be determined in an absolute sense, but in this case one Y value could be arbitrarily set at 0 and the other is at y=0.50.

COMPUTATION OF THE PATTERSON FUNCTION

For the P2 symmetry of bromobenzoic acid the Patterson function
for the U, W projection using the h0l reflections becomes:

Using this formula it is possible to calculate the function summing over all of the Fo² values for each of several thousand points. This takes a great amount of computer time. Using the formula cos(a+b)=cos(a)cos(b)-sin(a)sin(b) the expression becomes

where the sum over h*U can be performed first and the sums over l*W performed later with a great saving of computer time.

Don't worry about all the details of how the calculation, but the information is there if you are curious.

INTERPRETATION OF THE PATTERSON FUNCTION

Just to review. There are two molecules in the unit cell of this X, Z projection. These molecules are related by the symmetry group P2 which requires that there are two-fold axes perpendicular to the X, Z plane at the origin 0, 0. Because of translational symmetry there are two-fold axes at 0, 1/2; 1/2, 0; and 1/2.1/2. The equivalent positions are X, Z and -X, -Z. This means that U = 2X and W = 2Z.

Here is the Patterson Maps for the bromobenzoic acid as given on page 5 of the MathCad file. Find the values for U and W, the Br-Br vectors, and calculate X and Z, the coordinates of the bromine atom for the bromobenzoic acid. Note that the axes are not always in the same orientation.

You have now located the two heaviest atoms in the unit cell. Set natoms=1 (on the bottom of page 8 of the MathCad program) and enter x, 0, z for the Ele = 5 for Br in the matrix (page 9). The 5 tells the program to use the scattering factor for bromine.

We set y=0 since we are working in projection down the y axis. Press F9 to get the program to calculate the electron density.

STRUCTURE FACTORS FOR BROMOBENZOIC ACID

Our long-range objective is to visualize the structure using the electron density function.

To calculate this function we need the values of Fo with their algebraic signs. From the X-ray measurements we can only obtain the absolute magnitudes of the Fo values When we know the positions of the atoms (or at least some of them) we can calculate values of Fcalc with their signs. Then we attach the signs of F_calc to Fo, and calculate the electron density function. The formula for F_calc for the mth reflection is as follows:

where f_n is the scattering factor for the n^th atom and x_n, y_n, z_n are the fractional coordinates of the n^th atom in the unit cell. The summation is carried out over the atoms in the unit cell. The f values are the scattering powers of the atoms for X-rays. We shall use the tables of the f's published in the International Tables for Crystallography. They are read in and plotted in the computational document. This graph shows dramatically how Br dominates the scattering relative to the other atoms. The thermal motion of the atoms will be accounted for by use of an overall temperature factor B.

For this crystal structure with P2 symmetry and including only h0l reflections

Using this formula we need only include atoms from one molecule. The scattering from symmetrically located atoms is included automatically.

Use the matrix display feature (Fobs=, then F9) to calculate arrays Fobs, Fcalc, and Fcoef. Scroll to page 12 to see the arrays. In what way is Fobs like Fcoef? In what way is Fcalc like Fcoef? Why? Watch this change after each calculation following addition of more atomic fractional crystal coordinates to the array on page 9.

Also watch the change in the R factor on page 10 after each addition.

The electron density map (page 12) should make increasing chemical sense as you add atomic coordinates and the calculation of the Fcalc's is based on this growing information..

Resize the map on page 12, push the "print screen" button --- third from the right at the top of your keyboard. Open Programs, accessories, paint and paste (control V). Select the map and paste it into Word. Then print just that one page.

Can you spot the other atoms in the molecule? Mark the atoms on your hard-copy and read off the coordinates of each atom and make a table of Ele, X, and Z for each atom (back on page 9). Be careful to check the labelling of the axes. Also be sure to add the number of new atoms to the ( natoms = ) on the bottom of page 8.

Repeat the instructions in the previous four paragraphs until you have found all non-hydrogen atoms.

For benzoic acid can you find evidence of hydrogen bonding?

Has the R factor improved?

Has the map improved?

Exit the program.

Open im3run.xmcd. Repeat to solve this structure.

Homework:  Repeat the above with im2run.xmcd.  Hand in a copy of the final electron density map (can be a group activity if you wish).