Solve a small-molecule structure

The following is based on the experience of a protein crystallographer who one day obtained a small-molecule dataset and managed to solve and refine it without prior knowledge what the crystallized substance was, and without experience in small-molecule crystallography. It was a very rewarding experience (see the figure at the bottom) which is why it's written up here.

This is just a case study. To understand things, one has to read http://shelx.uni-ac.gwdg.de/SHELX/shelx.pdf .

Reduce the data with your favourite data processing software

I use XDS. The decision about the spacegroup has to be postponed, but it surely helps if the correct Laue group is employed during scaling. In the case considered here, the CORRECT step suggested P222 (XDS really only should suggest "222 point symmetry" because CORRECT does not look at systematic absences at this point).

Determine the spacegroup

There are two ways to determine the spacegroup:

1. use XPREP
2. use CCP4 POINTLESS - latest docs at [1]

These two possibilities also differ in the way how to obtain a file suitable for input to the SHELX program.

If there are different spacegroup possibilities then (downstream, in structure solution and refinement) we need to try all of them in turn, until we hit one that refines really satisfactorily (R-factor below, say, 5%) and gives a structure that makes sense.

use XPREP to find out possible spacegroups

First, convert the reflection file to HKLF 4 format (intensities!). The HKLF 4 format is what the SHELX programs read. I used XDSCONV and the following XDSCONV.INP:

SPACE_GROUP_NUMBER=   1
UNIT_CELL_CONSTANTS=    14.433    28.704     8.488  90.000  90.000  90.000
INPUT_FILE=XDS_ASCII.HKL
OUTPUT_FILE=temp.hkl

It is important that - to preserve the full information about systematic absences, for use in XPREP - XDSCONV runs in spacegroup 1. This does not necessarily mean that CORRECT also has to run in spacegroup 1, because XDS_ASCII.HKL has all observations no matter in which spacegroup the CORRECT step runs. As long as the spacegroup used in the CORRECT step is primitive, this works nicely. But if some re-indexing between CORRECT's spacegroup and P1 is necessary (like in I, F, C, R) then it is probably safest to rather just run CORRECT in P1.

answer the question concerning the cell axes, and then hit <Enter> several (about 6) times until the program suggests a list of spacegroups - this choice is going to be important. It may help to observe whether it's centrosymmetric or not, from the line: Mean |E*E-1| = 0.939 [expected .968 centrosym and .736 non-centrosym]. Fortunately there's only one spacegroup consistent with the data:

SPACE GROUP DETERMINATION


Lattice exceptions:  P      A      B      C      I      F     Obv    Rev    All

N (total) =           0  28832  28824  28788  28823  43222  38376  38344  57564
N (int>3sigma) =      0  17961  18421  18158  17862  27270  24715  24627  36959
Mean intensity =    0.0   22.7   23.7   24.8   23.4   23.7   24.7   24.8   24.8
Mean int/sigma =    0.0    9.6   10.0    9.9    9.6    9.8   10.0   10.0   10.0


Crystal system O and Lattice type P selected

Mean |E*E-1| = 0.939 [expected .968 centrosym and .736 non-centrosym]

Chiral flag NOT set



Systematic absence exceptions:

         b--   c--   n--  21--   -c-   -a-   -n-  -21-   --a   --b   --n  --21 

N      1884  1884  1892     7   988  1014   992    28   545   541   534    72
N I>3s  706   706     0     0   304     0   304     0     0   203   203     0
<I>    25.2  25.2   0.5   0.0  18.2   0.4  18.1   0.4   0.4  25.0  25.4   0.4
<I/s>   7.3   7.3   0.5   0.2   6.6   0.5   6.6   0.5   0.4   7.4   7.6   0.4


Identical indices and Friedel opposites combined before calculating R(sym)

Option  Space Group  No.  Type  Axes  CSD  R(sym) N(eq)  Syst. Abs.   CFOM

[A] Pccn           # 56  centro   3   196  0.023  10123  0.5 /  6.6   2.23

Option [A] chosen

After that, say "c" for "define unit-cell CONTENTS", and input a reasonable number of carbon atoms (I used C20). Get out of this menu with "E". Then, choose "f" for "set up shelxtl FILES". Then, answer the question "XM/SHELXD (M) or XS/SHELXS (S) format [S]:" with "m" since we're going to use shelxd for solving the structure. Answer the question about the name (I used the spacegroup number as I knew I would have to test several possibilities). Finally, "q"uit the program. This writes 56.ins :

TITL 56 in Pccn 
CELL 0.71073  14.4330  28.7040   8.4880  90.000  90.000  90.000
ZERR   11.00   0.0029   0.0057   0.0017   0.000   0.000   0.000
LATT  1
SYMM 0.5-X, 0.5-Y, Z
SYMM -X, 0.5+Y, 0.5-Z
SYMM 0.5+X, -Y, 0.5-Z
SFAC C
UNIT 220
FIND    16
PLOP    22    27    31
MIND 1.0 -0.1
NTRY 1000
HKLF 4
END

use POINTLESS to find the spacegroup

Unless the spacegroup number in XDS_ASCII.HKL already indicates this, pointless needs to be told that the spacegroup may not be restricted to those 65 which occur for crystals from macromolecules:

echo CHIRALITY NONCHIRAL | pointless xdsin XDS_ASCII.HKL

gives

         Zone                Number PeakHeight  SD  Probability  ReflectionCondition

Zones for Laue group P m m m
 1 screw axis 2(1) [a]           11   0.990   0.135   *** 0.972   h00: h=2n
 2 screw axis 2(1) [b]           59   1.000   0.097   *** 0.986   0k0: k=2n
 3 screw axis 2(1) [c]          131   0.997   0.062   *** 0.994   00l: l=2n
 4        glide plane b(a)     3754   0.012   0.050       0.000   0kl: k=2n
 5        glide plane c(a)     3754   0.013   0.050       0.000   0kl: l=2n
 6        glide plane n(a)     3754   0.951   0.061   *** 0.988   0kl: k+l=2n
 7        glide plane a(b)     1961   0.953   0.050   *** 0.990   h0l: h=2n
 8        glide plane c(b)     1961   0.104   0.056       0.004   h0l: l=2n
 9        glide plane n(b)     1961   0.100   0.056       0.004   h0l: h+l=2n
10        glide plane a(c)     1074   0.960   0.058   *** 0.991   hk0: h=2n
11        glide plane b(c)     1074   0.080   0.058       0.003   hk0: k=2n
12        glide plane n(c)     1074   0.072   0.050       0.002   hk0: h+k=2n

<!--SUMMARY_END-->


Possible spacegroups:
--------------------
Indistinguishable space groups are grouped together on successive lines

'Reindex' is the operator to convert from the input hklin frame to the standard spacegroup frame.

'SysAbsProb' is an estimate of the probability of the space group based on
the observed systematic absences.

'Conditions' are the reflection conditions (absences)
'TotProb' is a total probability estimate (unnormalised) including the probability
of the crystal being centrosymmetric from the <|E^2-1|> statistic.
Chiral space groups are marked '*' and centrosymmetric ones 'O'


   Spacegroup         TotProb SysAbsProb     Reindex         Conditions

     <P n a a> ( 56) O  0.823  0.911                         h00: h=2n, 0k0: k=2n, 00l: l=2n, 0kl: k+l=2n, h0l: h=2n, hk0: h=2n (zones 1,2,3,6,7,10)


---------------------------------------------------------------


Selecting space group P n a a as there is a single space group with the highest score

The spacegroup that was used for CORRECT does not matter. The next step then is to generate a HKLF 4 file, using XDSCONV:

SPACE_GROUP_NUMBER=   56
UNIT_CELL_CONSTANTS=    14.433    28.704     8.488  90.000  90.000  90.000
INPUT_FILE=XDS_ASCII.HKL
OUTPUT_FILE=56.hkl

Solve the structure with SHELXD

Just run "shelxd 56". You may interrupt it with Ctrl-C once it has found a good solution, as suggested by

Try 11:20  Peaks 99 92 87 87 87 83 77 73 71 70 68 68 64 64 64 63 62 62 61 60
R = 0.294, Min.fun. = 0.747, <cos> = 0.491, Ra = 0.235
Try    11, CC All/Weak 59.81 / 46.01, best 59.81 / 46.01, best final CC  0.00
Peaklist optimization cycle  1    CC = 77.51 %    BG = 0.322   for   22 atoms
Peaks: 99 90 87 85 82 77 75 74 66 64 64 64 63 63 62 57 39 39 36 36 33 31    
Fragments: 17 5                                                              
Peaklist optimization cycle  2    CC = 88.80 %    BG = 0.225   for   25 atoms
Peaks: 99 95 89 88 87 84 82 79 78 78 77 76 75 75 74 73 73 71 71 69 67 65 40 
Fragments: 25                                                                
Peaklist optimization cycle  3    CC = 88.85 %    BG = 0.223   for   25 atoms
Peaks: 99 96 89 87 86 86 82 79 79 76 76 75 75 75 73 73 72 71 69 69 67 65 63 
Fragments: 25                                                                

The resulting 56.res is:

REM TRY     23   FINAL CC 88.85   TIME       3 SECS
REM Fragments: 25
REM 
TITL 56 in Pccn
CELL 0.71073  14.4330  28.7040   8.4880  90.000  90.000  90.000
ZERR   11.00   0.0029   0.0057   0.0017   0.000   0.000   0.000
LATT  1
SYMM 0.5-X, 0.5-Y, Z
SYMM -X, 0.5+Y, 0.5-Z
SYMM 0.5+X, -Y, 0.5-Z
SFAC C
UNIT 220
C001  1  0.45835  0.41566  0.09083 11.00000 0.1   99.00
C002  1  0.36894  0.55007 -0.58932 11.00000 0.1   95.84
C003  1  0.52129  0.72099 -0.95623 11.00000 0.1   89.35
C004  1  0.67521  0.30725  0.04587 11.00000 0.1   87.55
C005  1  0.40328  0.54911 -0.45947 11.00000 0.1   85.96
...
C021  1  0.60567  0.70055 -0.97749 11.00000 0.1   66.94
C022  1  0.49503  0.62079 -0.48787 11.00000 0.1   64.91
C023  1  0.60066  0.62034 -0.48599 11.00000 0.1   63.62
C024  1  0.63251  0.26331  0.06189 11.00000 0.1   63.01
C025  1  0.47217  0.73227 -1.09548 11.00000 0.1   61.79
HKLF 4
END 

Refine using SHELXL

Copy 56.res to 56.ins. Insert

ACTA
LIST 6
L.S. 10

after the UNIT 220 instruction, and run "shelxl 56". This gives a first refined model, and its electron density map, plus the relevant statistics.

general idea of refining a structure

Starting from a rough guess of the number of atoms, we adjust the model, guided by the refinement results. This is an iterative process, in which we repeatedly edit 56.res to reflect our change of conception of the structure, replace 56.ins with it, and run SHELXL again.

assigning chemical types

Since we know that there's not only carbon atoms, but likely also N, O and H, we modify 56.ins to have

SFAC C N O H
UNIT 200 100 100 40

(the actual numbers after UNIT can be taken from the .lst file of SHELXL, they don't seem to matter much.)

We tell SHELXL the chemical identity by putting a 1 for a C, a 2 for a N, a 3 for an O, and a 4 for a H - the number is just the order of the atom in the SFAC line.

The chemical identity of an atom can be found from geometric parameters, and its electron density. The electron density can be displayed e.g. in coot, by loading the 56.fcf file written by SHELXL. Geometric parameters (in particular distances) are listed in the 56.lst file. Typical bond distances of C-C, C=C, C-O, C=O, C-N and X-H are about 1.54, 1.34, 1.43, 1.24, 1.47 and 1.0 A, respectively.

As a proxy to electron density we can use the refined ADPs. Atoms initially called "C", but with very low U values after refinement, are most likely O or N atoms.

For the H atoms, we just cut-and-paste the atoms from the bottom of the .res file into those lines where the other atoms are, if the distances to existing (heavy) atoms are close to 1 A.

Finishing the structure

Finally we switch to anisotropic refinement by putting an

ANIS

line into 56.ins . More info about refinement options is in the SHELXL article!

Electron density

The figure shows the final electron density (blue), but with an O atom refined as N. This gives strong positive (green) difference electron density.

The difference map also shows distinct bonding electron density on most of the bonds.