Problem Set 2 solution (PDF, 345 KB)

PHYSICAL ORGANIC CHEMISTRY
ASSIGNMENT TWO
Question 1
The symmetry of crystals is described using space groups. Space group diagrams are used to
describe the symmetry relationship between moieties (a circle in the diagram is used to
represent a molecule or atom(s) or ion(s) with a + or – sign to indicate whether it is above or
below the plane drawn; a comma within the circle indicates a change of chirality which only
results from certain operators, such as a mirror plane or inversion centre).
(i) Draw the space group diagrams for P-1 (triclinic), P21, P2/m, Cc, and C2/m
(monoclinic). Give the number of general equivalent positions (Z) and list the
symmetry operators that relate them (i.e. (x,y,z)…etc.). In each case write a short
(no more than a few sentences) description of how you chose the origin and
direction of axes, constructed the diagram (cell angle choice for example) and
labelled the general equivalent positions. It is recommended that you check your
diagrams by looking up the space group diagrams in the International Tables for
Crystallography (http://it.iucr.org/Ab/).
These were downloaded from http://img.chem.ucl.ac.uk/sgp/large/sgp.htm.
They’re useful because they contain all the needed information but are in a
nonstandard orientation relative to the international tables. As a consequence it
is best to rotate these pictures by 90° clockwise so that they are consistent with
the standard. You should draw your diagrams in the standard orientation (origin
on top left) as used in the lectures. The conventional diagrams can be found on
http://it.iucr.org/Ab/ which is used as a reference whenever such diagrams are
shown in a book or paper.
P-1
P21
P2/m
Cc
C2/m
(ii) For each of these space groups explain whether or not crystals of that space group
could contain molecules of only one chirality (i.e. only one enantiomer).
P-1
P21
P2/m
Cc
C2/m
Centrosymmetric
(inversion) so R and
S present.
Only a rotation
element present so
it is possible to
crystallize only one
enantiomer (R or S)
in this space group.
2/m generates a
center of inversion
therefore
centrosymmetric. R
and S present.
A
glide
plane
implies
that
a
mirror plane is
present. This space
group can therefore
accommodate both
R and S molecules
and is therefore not
chiral.
However, it does
not contain a center
of inversion and
therefore is also not
centrosymmetric.
It is therefore a
polar space group.
2/m generates a
center of inversion
therefore
centrosymmetric. R
and S present.
Chiral space group
(iii) Can any of the above space groups be classed as being non-centrosymmetric and
also non-chiral? Explain.
See the Cc in the above table.
(iv) In which of the above space groups (more than one in some cases) would
molecules of the following symmetry be able to occupy a special position? In
each case, clearly indicate the position of the special position [a specific
coordinate in the case of an inversion, a line (e.g. 0, y, 0) in the case of a two-fold
axes, a plane (e.g. x, 0, z) in the case of the mirror] on the space group diagram.
a. -1 (inversion)
b. 2
c. m
d. 2/m
P-1
P21
P2/m
Cc
C2/m
Inversion present so
a centrosymmetric
molecule
can
crystallize on any
one of the positions.
Because the two
parts of a screw
axis are translated
by ½ along the b
axis, it is not
possible to have the
two halves of a
molecule
joined
across
this
symmetry (compare
with the other space
groups in the list).
2, m and 2/m
symmetry present.
Because the two
parts of a glide
plane are translated
by ½ along the b
axis, it is not
possible to have the
two halves of a
molecule
joined
across
this
symmetry (compare
with the other space
groups in the list).
2,
m
and
2/m
symmetry present.
One such position
would be the origin
(0,0,0) but others
such as (½, ½, 0)
and (½, ½, ½) are
equally
valid
alternative sites.
Only
molecules
with -1 symmetry
can occupy special
positions in this
space group.
A molecule with 2fold symmetry can
crystallize on a 2fold axis.
Anywhere
along
0,y,0 for example
(symmetry along a
line). Other sites
with 2-fold axes are
also valid.
A molecule with
mirror
symmetry
can
crystallize
anywhere along a
mirror plane.
x,0,z for example.
This molecule is
centrosymmetric
but
it
more
specifically has 2/m
symmetry.
Therefore,
molecules such as
XeF4 and Benzene
will be able to
crystallize on sites
where the 2-fold
axis coincides with
the mirror. One
such site in this
case is the origin
(0,0,0)
Also, CO2 can
crystallize in a 2/m
site but a RS
diastereomer can’t.
A molecule with 2fold symmetry can
crystallize on a 2-fold
axis.
For
example
anywhere along 0,y,0
(symmetry along a
line). Other sites with
2-fold axes are also
valid.
A molecule with
mirror symmetry can
crystallize anywhere
along a mirror plane.
x,0,z for example.
This molecule is
centrosymmetric but it
more specifically has
2/m
symmetry.
Therefore, molecules
such as XeF4 and
Benzene will be able
to crystallize on sites
where the 2-fold axis
coincides with the
mirror. One such site
in this case is the
origin (0,0,0)
It is possible for
molecules with pure
-1
symmetry
to
crystallize in this
space group at say (¼,
¼, 0) but knowing this
requires
detailed
knowledge of the
space group and was
not required.
Also, CO2 can
crystallize in a 2/m
site but a RS
diastereomer can’t.
(v) For each of the space groups above, indicate whether you would expect
systematic absences in the associated diffraction patterns and provide the
expected systematic absence condition (clearly indicating absence or presence –
either interpretation is fine) together with the symmetry related to it.
Only the systematic absences associated with the space group symbol is required unless more
is specifically requested.
P-1
P21
P2/m
Cc
C2/m
No
symmetry
elements
with
translation
components
or
lattice centering so
no
systematic
absence conditions.
Only a 21 screw
axis present.
No
symmetry
elements
with
translation
components
or
lattice centering so
no
systematic
absence conditions.
C centering and
c-glide present in
this space group.
C centering present.
So hkl reflections
present when
h + k = 2n
(i.e. when h + k is
even).
So 0k0 reflections
will be present only
when k = 2n
(i.e. when k is
even).
Systematic absence
condition associated
with
the
C
centering:
hkl
reflections
present when
h + k = 2n
(i.e. when h + k is
even).
For the c-glide, h0l
reflections will be
present only when l
= 2n
(i.e. when l is even).
Though you were
not expected to
include
this
information, if you
look at the space
group diagram a 21
screw axis and an
a-glide
is
also
present.
Therefore
the
following symmetry
absence conditions
also exist for this
space group.
For the 21: 0k0
reflections present
only when k = 2n
For the a-glide: h0l
reflections present
only when h = 2n
Question 2
a) Draw diagrams to show the symmetry of the 4, 41, and 43 axes. What is the
(symmetry) relationship between the 41 and 43 screw axes?
Draw this in the same way that the 31 and 32 axes were drawn in the lectures. The result
should look as shown below. A top view of these would also be acceptable if the translation
component was also included in the diagram as shown below. The 41 is a right handed screw
axis while the 43 is a left handed screw axis.
4
41
43
b) What would the expected the symmetry absence conditions be for the 4, 41, and 43
axes assuming that the rotation axis is parallel to the c axis of the unit cell. [Hint, use
the last page of the symmetry absences lecture notes which lists the absence
conditions for various symmetry operators]
4-fold axis no systematic absence conditions because there is no translation component.
The 41 and 43 screw axes both translate the molecule by ¼ along the screw axis direction.
Assuming that screw axis is acting along the z-axis (standard for tetragonal space groups)
then the condition would be:
00l present only when l = 4n in both cases.
Question 3
(a)
Draw space group diagrams to compare space groups C2 and C21.
The space group C21 looks identical except that the origin starts ¼ of the way along the a
axis as shown on the right. However, it is not so easy to see this if you have inserted objects
into the cell which is why a side view is recommended in (b) below if required.
(b)
From your diagrams, demonstrate that C2 and C21 are in fact the same space group,
(while P2 and P21 are different).
(hints: it helps to redraw the unit cell so that the ab unit cell face is in the plane of
the page, then to extend by translation the C21 diagram followed by looking at the
origins of the unit cell).
The above is from a text book where they used different symbols for the objects but the shift
along the a axis is the same indicating that they are the same space group.
(c)
What are the expected symmetry absence conditions for C2 and C21?
The space group is C centered and has a 21 screw axis.
For the C center: hkl reflections present when h + k = 2n
For the 21 screw axis: 0k0 reflections present when k = 2n
Question 4
The diffraction pattern of a crystal was found to have the following equivalent reflection
conditions and systematic absences.
I(hkl) ≡ I(h-kl) ≡ I(-hkl) ≡ I(hk-l) ≡ I(-h-kl) ≡ I(-hk-l) ≡ I(h-k-l) ≡ I(-h-k-l)
Systematic absences: (hkl): no conditions
Deduce, with reasons the crystal system (triclinic, monoclinic, orthorhombic etc) and lattice
type (P, C, F, I).
hkl no conditions implies P. What would have happened to hkl if you have C, I or F instead?
The equivalent reflections imply that a mirror symmetry is perpendicular to each of the axis
directions and therefore we have an orthorhombic system.
Mirror perpendicular to a axis leads to: I(hkl) ≡ I(-hkl)
Mirror perpendicular to b axis leads to: I(hkl) ≡ I(h-kl)
Mirror perpendicular to c axis leads to: I(hkl) ≡ I(hk-l)
Diffraction patterns are also centrosymmetric (Friedel’s law) and so the inverse of the above
reflections are also equivalent:
I(hkl) ≡ I(-h-k-l) ≡ I(h-k-l) ≡ I(-hk-l) ≡ I(-h-kl)
You can come to the same answer by considering that orthorhombic has mmm symmetry. The
point group for mmm has the following symmetry operators:
x,y,z; -x,-y,z; -x,y,-z; x,-y,-z; -x,-y,-z;
x,y,-z;
x,-y,z; -x,y,z.
Any general diffraction plane (not located on a mirror for example) will be related by the
same symmetries resulting in the above equivalent reflection conditions.

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