the phase problem or the problem of crystallography what is the problem? how was the problem solved?

The Phase Problemor

The Problem of Crystallography

What is the problem?How was the problem solved?

In the context of a single crystal x-ray study

• A Structure Query (What did I make?)• Needed: A Single Crystal (0.2 mm)• Collection of X-Ray Diffraction Data• Solution of the Phase Problem

(To get a Preliminary Model)• Structure Model Parameter Refinement• Interpretation of the Result/Geometry Analysis• Validation of the Analysis • Report and Publication

Example of an Oil-mounted Crystal

Diffraction Pattern of a crystal rotated over 1 degree

H,K,L, I

6

Glusker and Trueblood

Microscope versus X-ray diffraction

same principle, no lenses

Bragg’s Law

n l = 2dsinq

8

Summation of waves

A wave:wavelength, speed, amplitide (F), phase (j)

The result of a two wavessummation depends ontheir amplitudes and (relative) phase

10

Diffraction from any object

• X-rays will scatter on each atom of the object:• scatter predominantly on electron shells, not nuclei• elastic (=same energy)• in all directions

• The intensity of diffracted radiation in a particular direction will depend on the interference (=sum) of scattered waves from every atom of the object

The structure factor is a mathematical function describing the amplitude and phase of a wave diffracted from crystal lattice planes characterized by Miller indices h,k,l.

The structure factor may be expressed as

where the sum is over all atoms in the unit cell, xj,yj,zj are the

positional coordinates of the jth atom, fj is the scattering factor of

the jth atom, and αhkl is the phase of the diffracted beam.

Fourier Theory

• Diffraction pattern related to object• Mathematical operation called

Fourier Transform• Can be inverted to give pattern

of electron density• Requires amplitude and phase

of diffracted waves

The Phase Problem

F T

Fourier Transform of a molecule

Fourier Transform of a crystal

16

Diffraction as Fourier transformReal space (x,y,z):

electron density r(x,y,z)

‘Reciprocal space’ (h,k,l):

diffracted waves F(h,k,l), j(h,k,l)

Physics tells us that the diffracted waves are Fourier transforms of the electron density:

xyz

lzkyhxilkhi dxdydzezyxelkhF )(2),,( ),,(),,(

Moreover, a backward transform (synthesis) should bring us from wavesback to the electron density:

dhdkdlelkhFconstzyxhkl

lkhilzkyhxi ),,()(2),,(),,(

I.e. once we know the amplitudes and phases of diffracted waves we can calculate the electron density!

Summary. Structure Factor F(hkl)

How to get to an electron density map from the intensities of reflections?

• Intrinsic amplitude of scattered X-ray from each type of atom depends on its electron density r(xyz) (described by atomic scattering factor f)

• f is used to describe scattering vector for each atom for one reflection (with Miller indices h,k,l)

• Sum of atomic scattering vectors is molecular scattering factor F, also called structure factor (can be calculated for a given structure):

F(hkl) = r(xyz) e-2pi(hx+ky+lz) Fourier Transformation

The Phase Problem

Continued: How to get to an electron density map from the intensities of reflections?

Fourier Transformation (other way around)

r(xyz) = F(hkl) e2pi(hx+ky+lz)

Magnitude of F is proportional to square root of intensity: |F(hkl)| ~ √ I(hkl)

Thus: r(xyz) = 1/NV ∑ ∑ ∑ √I(hkl) e-i(hkl) e2pi(hx+ky+lz)

h k l

Electron Density Intensity Phase

Missing Information to get electron density!

What is the Phase Problem?

• From diffraction experiment; only measure the intensities (amplitude2)

• Phase information is lost!• ..hence the ‘phase problem’• Can we survive without the

lost phase information?.....

20

Phases are more important than amplitudes!

Phase Problem

• What we measure is not F directly, but the intensity I where

• But F = (amplitude)ei(phase) and

so that we lose all the phase information and cannot do the inverse FT exactly

2 2I E F

2 2( )F amplitude

Phase Problem

• What we need:

• What we have:

• What we can get:• What we miss:

• Phase and Amplitude of diffracted waves

• Number of X-ray photons in each spot

• Intensity = Amplitude2

• Relative phase angles for different spotsPhase

has been lost

What next?

• We can measure:– Diffraction spot intensities– Unit cell dimensions– Symmetry group

• We want to find out:– Location of each atom in unit cell– Type of atom and expected F– Phase associated with each atom

Solution of the Phase Problem

• Early Method: Trial & Error (Salts such as NaCl, Silicates etc.)

• Patterson Methods (Heavy Atom)• Direct Methods (SHELXS, DIRDIF, SIR)• Isomorphous Replacement• New: Charge Flipping (Ab-initio)

• Phase Problem Solved! Given reasonable data.

Recovering phases…the Patterson function

• Invented by W. L. Patterson for small molecules• Patterson map is calculated with the square of structure

factor amplitude and a phase of zero• This is an interatomic vector map• Each peak corresponds to a vector between atoms in the

crystal• Peak intensity is the product of electron densities of each

atom

Patterson Function

Number of spots = N2 where N = # atoms - gets huge!

Once phases have been determined as accurately as possible, an e-density map (the Fourier transform of the structure factors with phases) is prepared and interpreted.

Use observed structure amplitudes |Fobs(hkl)| with best phases, best(hkl) to give experimentally derived structure factors, Fexp(hkl):

Fexp(hkl) = |Fobs(hkl)| exp[ i best(hkl)]

Fourier transform creates e-density exp(xyz):

From model building to developing a structureFourier synthesis

Iterative Process: Model is subtly changed and structure factors are calculated (Fcalc) and compared with measured Fobs to see how well a model fits the data.

Direct Methods

• Derive phases directly from the magnitudes• Positivity and atomicity• Equal atom structure

' ''

' ''

1 sq sq sq

sq

F F f f F F FV

fF F F

f V

h h h h h h h hh

h h h hh

2 ( )

1

2

1

( , , )

( ) j

N i h k lyx z jj jj

j

N i

jj

F h k l f

F f

e

e

h r

h

* * *

jjj j

h k l

yx z

h a b c

r a b c

Structure Factors

Sh Sh’ Sh - h’ or S-hSh’ Sh - h’ +1

' ''

sq

fF F F

f V h h h hh

Sayre Equation

Sign relationship. An important outcome of the Sayre Equation

1947 D. Harker and J. Kasper

1952 D. Sayre

1950’s J. Karle and H. Hauptman

1964 I. L. Karle and J. Karle

1970’s M. M. Woolfson

1985 Nobel Prize awarded to H. Hauptman and J. Karle

Isomorphous replacement

• Used early 1900’s for small molecules by Groth (1908); Beevers and Lipson (1934) ; Robertson (1935)

• Perutz (1956) and Kendrew (1958) used it on proteins• Use of heavy atom substitution in a crystal • “Isomorphous” – same shape• “Replacement” – heavy atom might be replacing light salts or

solvent molecules• Why heavy atoms? large atomic numbers; Contribute

disproportionately to the intensities• Principle: change the crystal (with heavy atoms); perturb the

structure factors; conclude phases from how the structure factors are perturbed

• Collect two (or more) datasets: ‘native’ and ‘derivative’ data

The unit cell contains two centrosymmetric molecules. The metal atom and the four surrounding isoindole nitrogen atoms must all lie strictly in one plane

The centrosymmetric space group was confirmed by the absence of a piezoelectric and pyroelectric effect, which is seen in resorcinol

If a reflection from the free compound corresponds in phase to a peak at the centre of symmetry, then the same reflection from the metal compound will be of greater intensity owing to the addition of the peak due to the metal atom alone, and the phase constants of both these reflections will be 0 (or positive sign). If, however, the reflection from the free compound corresponds in phase to a trough at the centre of symmetry, then the addition of the peak due to the metal atom alone will tend to fill up this trough, and the resultant reflection from the metal compound will usually be of smaller intensity, the phase constants of both reflections in this case being x (or negative sign). If the reflection from the free compound corresponds to a very small trough (weak intensity), the phase may be reversed by the addition of the metal atom. When the structure amplitudes are expressed in absolute measure there is practically no ambiguity in the principal (h01) z one, and the phase constants of these reflections, numbering about 300, from phthalocyanine and its nickel derivative have been determined with certainty.

The usual difficulty of the unknown phase constant, has been overcome by comparing absolute measurements of corresponding measurements from nickel phthalocyanine and the metal-free compound which leads to a direct determination of all the significant phase constants in the (h0l) zones of the two compounds, numbering about 300. A Fourier analysis of these results determines two co-ordinates of each carbon and nitrogen atom in the structure and the regularity of the projection shows beyond doubt that the molecule is planar.

Isomorphous replacement contd…

• Calculate the Isomorphous difference Fiso

• Use Fiso in direct or Patterson method to deduce position of heavy atoms

• Then deduce possible values for (protein) phase angles

Completing the structure

• Given the Diffraction Data and (Approximate) Phases a 3D Electron Density Map can be Calculated.

rx,y,z = 1/V Shkl Fhkl exp{-2pi(hx + ky + lz)}

Fhkl = |Fhkl|exp(fhkl)

• Following is a section through such a map

Interpretation in Terms of Atoms

• Position of highest density => Position x,y,z• Deviation of the density shape from the ideal atomic

electron density => Thermal motion parameters:• Isotropic: U(iso) or• Anisotropic: U11,U12,U13,U22,U23,U33

(Displacement Parameters) => ORTEP Note: ORTEP does NOT represent the electron

distribution

Structural Refinement

Interpretation in Terms of Bonds

• Bonds between atoms of type A and B are assigned on the basis of atomic covalent radii with: d < R(A) + R(B) + 0.4

• ‘Crystallographic Bonds’ are not necessarily Chemical Bonds.

• Van der Waals Radii are used to detect isolated molecular species or short contacts.

Display Options

• Ball-and-Stick Simple but may hide problems with a structure.

• ORTEP Often preferred because it visualizes most model parameters and possible problems.

• CPK Spacefilling PLOT illustrating the shape etc

A pretty picture, but what about the numbers …

Summary. Central Formulae

• Diffraction Spots: 2dhkl sinqhkl = nl

• Solve the Phase problem. Get the model• Electron Density Map (3D Fourier Map)

rx,y,z = 1/V Shkl Fhkl exp{-2pi(hx + ky + lz)}• Structure Factor (Improve the model) Fhkl(calc) = Sj fj Tj,hkl exp{2pi(hxj+kyj+lzj)}• Least Squares Model Refinement

Minimize: Shkl [whkl(|Fhkl(obs)|2-|Fhkl(calc)|2)]2

• Convergence Criteria: R1, wR2, S

End of First Lecture