International
Tables for Crystallography Volume B Reciprocal space Edited by U. Shmueli © International Union of Crystallography 2006 |
International Tables for Crystallography (2006). Vol. B. ch. 2.3, pp. 239-240
Section 2.3.2.3. Finding heavy atoms
aDepartment of Biological Sciences, Purdue University, West Lafayette, Indiana 47907, USA, and bCABM & Rutgers University, 679 Hoes Lane, Piscataway, New Jersey 08854-5638, USA |
The previous two sections have developed some of the useful mechanics for interpreting Pattersons. In this section, we will consider finding heavy-atom positions, in the presence of numerous light atoms, from Patterson maps. The feasibility of structure solution by the heavy-atom method depends on a number of factors which include the relative size of the heavy atom and the extent and quality of the data. A useful rule of thumb is that the ratio should be near unity if the heavy atom is to provide useful starting phase information (Z is the atomic number of an atom). The condition that normally guarantees interpretability of the Patterson function in terms of the heavy-atom positions. This `rule', arising from the work of Luzzati (1953), Woolfson (1956), Sim (1961) and others, is not inviolable; many ambitious determinations have been accomplished via the heavy-atom method for which r was well below 1.0. An outstanding example is vitamin B12 with formula C62H88CoO14P, which gave an for the cobalt atom alone (Hodgkin et al., 1957). One factor contributing to the success of such a determination is that the relative scattering power of Co is enhanced for higher scattering angles. Thus, the ratio, r, provides a conservative estimate. If the value of r is well above 1.0, the initial easier interpretation of the Patterson will come at the expense of poorly defined parameters of the lighter atoms.
A general strategy for determining heavy atoms from the Patterson usually involves the following steps.
Detailed and instructive examples of using Pattersons to find heavy-atom positions are found in almost every textbook on crystal structure analysis [see, for example, Buerger (1959), Lipson & Cochran (1966) and Stout & Jensen (1968)].
The determination of the crystal structure of cholesteryl iodide by Carlisle & Crowfoot (1945) provides an example of using the Patterson function to locate heavy atoms. There were two molecules, each of formula C27H45I, in the unit cell. The ratio is clearly well over the optimal value of unity. The P(x, z) Patterson projection showed one dominant peak at in the asymmetric unit. The equivalent positions for require that an iodine atom at , , generates another at and thus produces a Patterson peak at . The iodine position was therefore determined as 0.217, 0.042. The y coordinate of the iodine is arbitrary for yet the value of is convenient, since an inversion centre in the two-atom iodine structure is then exactly at the origin, making all calculated phases 0 or π. Although the presence of this extra symmetry caused some initial difficulties in the interpretation of the steroid backbone, Carlisle and Crowfoot successfully separated the enantiomorphic images. Owing to the presence of the perhaps too heavy iodine atom, however, the structure of the carbon skeleton could not be defined very precisely. Nevertheless, all critical stereochemical details were adequately illuminated by this determination. In the cholesteryl iodide example, a number of different yet equivalent origins could have been selected. Alternative origin choices include all combinations of and .
A further example of using the Patterson to find heavy atoms will be provided in Section 2.3.5.2 on solving for heavy atoms in the presence of noncrystallographic symmetry.
References
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