To develop the knowledge

1. ORGANIC CHEMISTRY
STEREOCHEMISTRY
ANSWERS:
What is Stereoisomerism ? Give an Example.
Stereoisomers (also called configurational isomers)
differ in the way their atoms are arranged in space.
Stereoisomers are different compounds that do not readily
interconvert. Therefore, they can be kind separated. There
are two kinds of stereoisomers: cis–trans isomers and isomers
that contain chirality (ky-RAL-i-tee) centers.
Define Optical isomerismwith examples.
l isomerism
ptical isomers?
isomers are named like this because of their effect on plane polarised light.
substances which show optical isomerism exist as two isomers known as enantiomers.
A solution of one enantiomer rotates the plane of polarisation in a clockwise direction. This enantiomer is known as
he (+) form.
For example, one of the optical isomers (enantiomers) of the amino acid alanine is known as (+)alanine.
A solution of the other enantiomer rotates the plane of polarisation in an anti-clockwise direction. This enantiomer is k

2. as the (-) form. So the other enantiomer of alanine is known as or (-)alanine.
the solutions are equally concentrated the amount of rotation caused by the two isomers is exactly the same - but in
opposite directions.
When optically active substances are made in the lab, they often occur as a 50/50 mixture of the two enantiomers. Th
known as a racemic mixture or racemate. It has no effect on plane polarised light.
Note: One of the worrying things about optical isomerism is the number of obscure words that
get thrown at you. Bear with it - things are soon going to get more visual!
There is an alternative way of describing the (+) and (-) forms which is potentially very confusing
involves the use of the lowercase letters d- and l-, standing for dextrorotatory and laevorotatory
respectively. Unfortunately, there is another different use of the capital letters D- and L- in this to
is totally confusing! Stick with (+) and (-).
ptical isomers arise
amples of organic optical isomers required at A' level all contain a carbon atom joined to four different groups. These
each have the same groups joined to the central carbon atom, but still manage to be different:
sly as they are drawn, the orange and blue groups aren't aligned the same way. Could you get them to align by rotati
molecules? The next diagram shows what happens if you rotate molecule B.
ll aren't the same - and there is no way that you can rotate them so that they look exactly the same. These are isome
her.
e described as being non-superimposable in the sense that (if you imagine molecule B being turned into a ghostly
you couldn't slide one molecule exactly over the other one. Something would always be pointing in the wrong direct
appens if two of the groups attached to the central carbon atom are the same? The next diagram shows this possibili

3. models are aligned exactly as before, but the orange group has been replaced by another pink one.
g molecule B this time shows that it is exactly the same as molecule A. You only get optical isomers if all four groups
d to the central carbon are different.
and achiral molecules
sential difference between the two examples we've looked at lies in the symmetry of the molecules.
are two groups the same attached to the central carbon atom, the molecule has a plane of symmetry. If you imagine
the molecule, the left-hand side is an exact reflection of the right-hand side.
there are four groups attached, there is no symmetry anywhere in the molecule.
cule which has no plane of symmetry is described as chiral. The carbon atom with the four different groups attached
this lack of symmetry is described as a chiral centre or as an asymmetric carbon atom.
lecule on the left above (with a plane of symmetry) is described as achiral.
iral molecules have optical isomers.
ationship between the enantiomers
the enantiomers is simply a non-superimposable mirror image of the other one.
words, if one isomer looked in a mirror, what it would see is the other one. The two isomers (the original one and its
have a different spatial arrangement, and so can't be superimposed on each other.

4. hiral molecule (one with a plane of symmetry) looked in a mirror, you would always find that by rotating the image in
ld make the two look identical. It would be possible to superimpose the original molecule and its mirror image.
eal examples of optical isomers
2-ol
ymmetric carbon atom in a compound (the one with four different groups attached) is often shown by a star.
emely important to draw the isomers correctly. Draw one of them using standard bond notation to show the 3-dimens
ment around the asymmetric carbon atom. Then draw the mirror to show the examiner that you know what you are d
n the mirror image.
Help! If you don't understand this bond notation, follow this link to drawing organic molecules b
you go on with this page.
hat you don't literally draw the mirror images of all the letters and numbers! It is, however, quite useful to reverse larg
- look, for example, at the ethyl group at the top of the diagram.
n't matter in the least in what order you draw the four groups around the central carbon. As long as your mirror image
accurately, you will automatically have drawn the two isomers.
h of these two isomers is (+)butan-2-ol and which is (-)butan-2-ol? There is no simple way of telling that. For A'level
es, you can just ignore that problem - all you need to be able to do is to draw the two isomers correctly.
oxypropanoic acid (lactic acid)

5. gain the chiral centre is shown by a star.
enantiomers are:
ortant this time to draw the COOH group backwards in the mirror image. If you don't there is a good chance of you jo
e central carbon wrongly.
raw it like this in an exam, you won't get the mark for that isomer even if you have drawn everything else perfectly.
opropanoic acid (alanine)
ypical of naturally-occurring amino acids. Structurally, it is just like the last example, except that the -OH group is rep
enantiomers are:
e of these isomers occurs naturally: the (+) form. You can't tell just by looking at the structures which this is.
however, been possible to work out which of these structures is which. Naturally occurring alanine is the right-hand
e, and the way the groups are arranged around the central carbon atom is known as an L- configuration. Notice the

6. ital L. The other configuration is known as D-.
may well find alanine described as L-(+)alanine.
eans that it has this particular structure and rotates the plane of polarisation clockwise.
you know that a different compound has an arrangement of groups similar to alanine, you still can't say which way it
he plane of polarisation.
er amino acids, for example, have the same arrangement of groups as alanine does (all that changes is the CH3 gro
re (+) forms and others are (-) forms.
e common for natural systems to only work with one of the enantiomers of an optically active substance. It isn't too di
why that might be. Because the molecules have different spatial arrangements of their various groups, only one of the
fit properly into the active sites on the enzymes they work with.
b, it is quite common to produce equal amounts of both forms of a compound when it is synthesised. This happens j
and you tend to get racemic mixtures.
Note: For a detailed discussion of this, you could have a look at the page on the addition of HC
aldehydes

116. Note: I am being deliberately unkind here! Normally when a molecule like cholesterol is discuss
this context, extra detail is often added to the skeletal structure. For example, important hydroge
or methyl groups are drawn in. It is good for you to have to do it the hard way!
how many chiral centres did you find? In fact, there are 8 chiral centres out of the total of 9 carbons marked. If you di

117. eight, go back and have another look before you read any further. It might help to sketch the structure on a piece of p
w in any missing hydrogens attached to the numbered carbons, and write in the methyl groups at the end of the bran
done for you below, but it would be a lot better if you did it yourself and then checked your sketch afterwards.
with the easy one - it is obvious that carbon 9 has two methyl groups attached. It doesn't have 4 different groups, an
chiral.
ake a general look at the rest, it is fairly clear that none of them has a plane of symmetry through the numbered carbo
re they are all likely to be chiral centres. But it's worth checking to see what is attached to each of them.
1 has a hydrogen, an -OH and two different hydrocarbon chains (actually bits of rings) attached. Check clockwise an
kwise, and you will see that the arrangement isn't identical in each direction. Four different groups means a chiral ce
2 has a methyl and three other different hydrocarbon groups. If you check along all three bits of rings , they are all d
er chiral centre. This is also true of carbon 6.
s 3, 4, 5 and 7 are all basically the same. Each is attached to a hydrogen and three different bits of rings. All of these
entres.
carbon 8 has a hydrogen, a methyl group, and two different hydrocarbon groups attached. Again, this is a chiral cent
looks difficult at first glance, but it isn't. You do, however, have to take a great deal of care in working through it - it is
gly easy to miss one out.
3.Geometrical isomerism
Geometric Isomers are compounds with different spatial arrangements of groups attached to
the carbons of a double bond. In alkenes, the carbon-carbon double bond is rigidly fixed. Even
though the attachment of atoms is the same, the geometry (the way the atoms "see" each other)
is different.
When looking for geometric isomers, a guiding principle is that there MUST BE TWO
DIFFERENT "GROUPS" ON EACH CARBON OF THE DOUBLE BOND. A "group" can be
hydrogen, alkyls, halogens, etc.

118. Identical compounds may appear to have different arrangements as written, but closer
examination by rotation or turning will result in the molecules being superimposed. If they are
super impossible or if they have identical names, then the two compounds are in fact identical.
Isomers of compounds have a different arrangement of the atoms. Isomer compounds will differ
from identical compounds by the arrangement of the atoms. See example below.
Both compounds have the same number of atoms, C5H12. They are isomers because in the left
molecule the root is 4 carbons with one branch. In the right molecule, the root is 3 carbons with
2 branches. They are isomers because they have the same number of atoms but different
arrangements of those atoms.
Completely different compounds: If the number of each element is different, the two
compounds are merely completely different. A simple count of the atoms will reveal them as
different.
1,2-dichlorethene
In the example on the left, the chlorine atoms can be opposite or across from each other in
which case it is called the "trans" isomer. If the the chlorine atoms are next to or adjacent each
other, the isomer is called "cis".
If one carbon of the double bond has two identical groups such as 2 H's or 2 Cl's or 2 CH3 etc.
there cannot be any geometric isomers.
2-butene
Consider the longest chain containing the double bond: If two groups (attached to the carbons
of the double bond) are on the same side of the double bond, the isomer is a cis alkene. If the
two groups lie on opposite sides of the double bond, the isomer is a trans alkene. One or more
of the "groups" may or may not be part of the longest chain. In the case on the left, the "group"
is a methyl - but is actually part of the longest chain.
A common mistake is to name this compound as 1,2-dimethylethene. Look at all carbons for the
longest continuous chain - the root is 4 carbons - butene
4. Fischer Projection.
The wedge and dash representations of stereochemistry can often become
cumbersome, especially for large molecules which contain a number of
stereocenters. An alternative way to represent stereochemistry is the Fischer
Projection, which was first used by the German chemist Emil Fischer. The

119. Fischer Projection represents every stereocenter as a cross. The horizontal
line represents bonds extending out of the plane of the page, whereas the
vertical line represents bonds extending into the plane of the page.
Figure %: Drawing Fischer Projections Manipulations of Fischer
Projections
When working with Fischer Projections, keep in mind the following rules:
• Because the "up" and "down" aspects of the bonds don't change, a Fischer
projection may be rotated by 180 degrees without changing its meaning.
• A Fischer projection may not be rotated by 90 degrees. Such a rotation
typically changes the configuration to the enantiomer.
• To find the enantiomer of a molecule drawn as a Fischer projection, simply
exchange the right and left horizontal bonds.
• To determine whether the molecule in Fischer projection is a meso
compound, draw a horizontal line through the centre of the molecule and
determine whether the molecule is symmetric about that line.

120. Figure %: Operations on Fischer projection
5.Projections;
Wedge-hash diagrams
Wedge-hash (or wedge-dash) diagrams are the most common representation
used to show 3D shape as they are ideally suited to showing the structure of
sp3
hybridised (tetrahedral atoms). They are rarely needed for
sp2
(e.g. alkenes) or sp systems (e.g. alkynes).
Wedge-hash diagrams are usually drawn with two bonds in the plane of the
page, one infront of the plane, and one behind the plane. This gives the
molecule 3D perspective: we envisage the bold lines being closer to us and
the hashes fading away in the background.

121. When drawing wedge-hash it is a good idea to visualise the tetrahedral
arrangement (or the appropriate geometry) of the groups and try to make the
diagram look like this. It is worth looking at the drawing and asking yourself
does it make geometric sense ? Can you see the shape you are trying to
depict ? If the answer to that question is "no", then the diagram is inadequate
and should be redrawn.
As a suggestion, they seem to be most effective when the "similar" pairs of
bonds (2-in-plane, 2-out-of-plane) are next to each other, as shown in the left
box above.
Here are some other examples to review:
Remember that diagrams are being used to present the required information
efficiently. Organic chemists use line diagrams to represent structures as part
of the symbolic code because they are quicker and easier to draw as we can
just leave out the C atoms and the H attached to those C atoms because we

122. know to just assume that they are there. This idea also carries over into
wedge-hash diagrams. A common scenario is shown below where the bond to
an H has been omitted and it is assumed that we know that it is there.
Sawhorse
Sawhorse diagrams are similar to wedge-dash diagrams, but without trying to
use "shading" to denote the perspective. The representation of propane
shown below has been drawn so that we are looking at the molecule which is
below us and to our left.
Newman Projections
Newman projections are drawn by looking directly along a
particular bond in the system (here a C-C bond) and arranging the
substituents so that they are equally spaced around the atoms at each end of
that bond. The protocol requires that the atoms within the central bond are
shown as a dot and circle as defined below. Think of them as an end on view
of a particular bond and the showing the arrangement of the groups around
that bond.
In order to draw a Newman projection from a wedge-dash diagram, it is useful
to imagine putting your "eye" in line with the central bond in order to look
along it.

123. Let's work through an example, consider drawing a Newman projection by
looking at the following wedge-dash diagram of propane from the left hand
side.
• First draw the dot and circle to represent the front and back C
respectively
• Since the front carbon atom has an H atom in the plane of the page
pointing up we can add that first
• The back carbon atom has an H atom in the plane of the page pointing
down
• Now add the other bonds to each C so that it is symmetrical
• The groups / bonds (blue) that were forward of the plane of the page in
the original wedge-dash diagram are now to our right
• Those behind (green) the plane are now to our left

124. Colour code
wedged groups
Colour code
hashed groups
Reset colours
Rotate to view
from
RIGHT (unselect
to undo the
rotation)
Rotate to view
from
LEFT (unselect
to undo the
rotation)
• Now you try the same thing, but looking from the right to generate the
other Newman projection.
• Fischer Projections
• Fischer projections are most commonly used to represent biomolecules such as
amino acids and carbohydrates (sugars) as the provide a quick way of
representing one or more multiple stereocenters. They tend to be less commonly
used by organic chemists because they represent the molecules in an
unfavourable conformation, i.e. an "unnatural position".
• A Fischer projection is the view of a molecule oriented with the carbon chain
oriented vertically and all the horizontal bonds oriented towards the observer (like
arms coming out the hug you).
•
• Note that Fischer projections of carbohydrates are typically drawn with the
longest chain oriented vertically and with the more highly oxidised C (the
carbonyl group) at the top. Here we see the Fischer projections of the simplest
carbohydrate, glyceraldehyde in its (S)-(-)- and (R)-(+)- forms:

125. S-(-)-glyceraldehyde R-(+)-glyceraldehyde
• Here is an example of a Fischer diagram with the stereochemistry at 2 centres.
It's the same idea, carbon chain vertical and the horizontal bonds towards
you. One way to "decode" a Fischer projection, is to draw in wedges and hashes
to "reveal" the orientation of the bonds. This can help visualise the shape.
•ROTATION OF MOLECULES
CAHN-INGOLD-PRELOG
RULES. Priority Rules
The purpose of the Cahn-In gold-Prelog priority rules (CIP priority rules) is so that
chemists can correctly and unambiguously name a specific stereoisomer of a
molecule. Stereoisomers are compounds that have the same chemical formula, the
same atom connectivity, but differ in their three-dimensional orientation.
By using the CIP priority rules, it's possible to correctly prioritize each atom or group of
atoms bonded to a specific carbon atom (stereocenter) within a molecule so that
chemists can assign a designator to that specific carbon. Something that's important for
us to realize, however, is that in order for a carbon to be a stereocenter within a
compound it has to have four different atoms or groups of atoms bonded to it.

126. Using the CIP Priority Rules
Luckily for us as long as we can count and use the periodic table of the elements, the
CIP priority rules are quite easy to navigate by following a few simple steps:
1. Look up the atomic numbers of each of the atoms bonded to the stereocenter carbon
of interest. The atom with the highest number always receives the top priority number (1
in most cases) and each remaining atom is ordered accordingly based on their atomic
numbers (priority numbers 2-4).
2. If there is a tie between two or more atoms, we must consider atoms at a two-bond
distance from the stereocenter to break the tie.
3. If there is still a tie, then we must consider atoms at a three-bond distance from the
stereocenter until the tie is broken.
Examples of Using the CIP Priority Rules
Consider as an example a carbon atom that is bonded to a fluorine atom, a hydrogen
atom, a bromine atom, and a hydroxyl (-OH) group. First and foremost notice that this
carbon would be classified as a stereocenter since it's bonded to four different groups of
atoms. Let's see if we can use the CIP priority rules to assign a priority number to each
of the substituents.
6A stereocenter carbon atom bonded to four different
substituents.
Let's start by giving the atomic number for each atom directly bonded to the carbon:

127. • Fluorine: 9
• Bromine: 35
• Hydrogen: 1
• Oxygen: 8
According to the CIP priority rules atoms with a higher atomic number receive highest
priority and we rank others accordingly. With that being said, the bromine atom would
be priority group 1, the fluorine atom would be priority group 2, the oxygen (of the
hydroxyl group) would be priority group 3, and finally the hydrogen would be priority
group 4. Easy enough right?
Let's look
at a slightly more complicated case. Notice that the carbon atom bearing the chlorine in
2-chlorobutane is the stereocenter in this compound since it's the only carbon that's
bonded to four different substituents. If we step out one atom at a time from the
stereocenter we hit a chlorine, a hydrogen, and two carbon atoms. The chlorine atom
will receive the top priority (number 1) since the atomic number is greater than both
hydrogen and carbon.

128. 6 R.S.NOTATION OF
OPTICAL ISOMERS
WITH ONE ASYMMETRIC
CARBON ATOMS Cahn-In gold-Prelog R/S
Notation
The Cahn-In gold-Prelog R/S rules are used for naming enantiomers and diastereomers
NB: The term chirality has superceded the term stereogenic or chiral centre.
• Identify the chirality centres (most commonly an sp3
C with 4 different
groups attached)
• Assign the priority to each group (high = 1, low = 4) based on atomic
number of the atom attached to the chirality centre (remember the first
point of difference rule)
• Position the lowest priority group away from you as if you were looking
along the C-(4) s bond. If you are using a model, grasp the group in
your fist.
• For the other 3 groups, determine the direction of high to low priority (1
to 3)
• If this is clockwise, then the centre is R (Latin: rectus = right)
• If this is counter clockwise, then it is S (Latin: sinister = left)
Example: bromofluoroiodomethane
The chirality centre is easy to spot, and the four attached

129. Highlight
chirality centre
groups are I (purple), Br (maroon), F(green) and H (white),
listed in priority order, highest to lowest.
Rotate the JSMOL image on the left so the you are looking
along the C-H bond and the H is away from you, then
determine the direction of high to low priority.
It decreases clockwise (I to Br to F), so this is
the Renantiomer.
Sub rules:
• Isotopes: H vs D ? Since isotopes have identical atomic numbers, the
mass number is used to discriminate them so D > H
• Same atom attached ? By moving out one unit at a time, locate the first
point of difference and apply rules there.
Can you convey this on a piece of paper?
With one asymmetric carbon
atoms. If a compound has more than one asymmetric carbon, the steps used to
determine whether an asymmetric carbon has the R or the S configuration must be applied to
each of the asymmetric carbons individually. As an example, let’s name one of the
stereoisomers of 3-bromo-2-butanol.
First, we will determine the configuration at C-2. The OH group has the highest priority, the C-3

130. carbon (the C attached to Br, C, H) has the next highest priority, CH3 is next, and H has the
lowest priority. Because the group with the lowest priority is bonded by a hatched wedge, we
can immediately draw an arrow from the group with the highest priority to the group with the
next highest priority. Because that arrow points counterclockwise, the configuration at C-2 is S.
Now we need to determine the configuration at C-3. Because the group with the lowest priority
(H) is not bonded by a hatched wedge, we must put it there by temporarily switching two
groups.
The arrow going from the highest priority group (Br) to the next highest priority group (the C
attached to O, C, H) points counterclockwise, suggesting it has the S configuration. However,
because we switched two groups before we drew the arrow, C-3 has the opposite
configuration—it has the R configuration. Thus, the isomer is named (2S,3R)-3-bromo-2-
butanol.
Fischer projections with two asymmetric carbons can be named in a similar manner. Just apply
the steps to each asymmetric carbon that you learned for a Fischer projection with one

131. asymmetric carbon. For C-2, the arrow from the group with the highest priority to the group
with the next highest priority points clockwise, suggesting it has the R configuration. But
because the group with the lowest priority is on a horizontal bond, we can conclude that C-2
has the S configuration (Section 5.6).
By repeating these steps for C-3, you will find that it has the R configuration. Thus, the isomer is
named (2S,3R)-3-bromo-2-butanol.
The four stereoisomers of 3-bromo-2-butanol are named as shown here. Take a few minutes to
verify their names.
i.,e

132. Notice that enantiomers have the opposite configuration at both asymmetric carbons, whereas
diastereomers have the same configuration at one asymmetric carbon and the opposite
configuration at the other. Tartaric acid has three stereoisomers because each of its two
asymmetric carbons has the same set of four substituents. The meso compound and the pair of
enantiomers are named as shown.
The physical properties of the three stereoisomers of tartaric acid are listed in Table 5.1. The
meso compound and either one of the enantiomers are diastereomers. Notice that the physical
properties of the enantiomers are the same, whereas the physical properties of the
diastereomers are different. Also notice that the physical properties of the racemic mixture
differ from the physical properties of the enantiomers.

133. Two Asymmetric
Carbon atoms:
A compound with two asymmetric carbons that
has only three stereoisomers is 2,3-
dibromobutane.The “missing “isomer is the mirror
image of 1because 1and its mirror image are the
same molecule. This can be seen more clearly if you
look either at the perspective for-mulas drawn in
their eclipsed conformations or at the Fischer
projection.

134. Hi
It is obvious that 1and its mirror image are identical
when looking at the perspective formula in the
eclipsed conformation. To convince yourself that

135. the Fischer projection of 1and its mirror image are
identical, rotate the mirror image by 180°.
(Remember, you can move Fischer projections only
by rotating them 180°in the plane of the paper.
Stereoisomer 1is called a meso compound. Even
though a meso(mee-zo)compound has asymmetric
carbons, it is an achiral molecule because it is
superim-posable on its mirror image. Mesosis the
Greek word for “middle. A meso com-pound is
achiral—when polarized light is passed through a
solution of a mesocompound,the plane of
bipolarization is not rotated. A meso compound can
be recognized by the fact that it has two or more
asymmetric carbons and a plane of symme-try. If a
compound has a plane of symmetry, it will not be
optically active even though it has asymmetric
carbons. A plane of symmetry cuts the molecule in
half, and one-half is the mirror image of the other
half. Stereoisomer 1has a plane of symmetry, which
means that it does not have a nonsuperimposable
mirror image—it does not have an enantiomer.

136. 1 2
3
OPTICAL ACTIVITIES IN
COMPOUNDS
NOT CONTAINING ASYMMETRIC CARBON
ATOMS :
ALLENES:

137. 1. Allenes are optically active as long as they have different
groups on each end of the cumulative double bonds. They are
constrained in rotation, so they can be resolved. There are some
naturally occurring allenes, some being optically active. Allene
itself has D2d symmetry, and is not chiral.
Let us take 1,2-prop-diene, the prototype of allenes, and add two
halogen atoms to it, one fluorine and the other a chlorine atom.
This molecule has C1 symmetry, and is definitely optically
active! Generally, any molecule that has C1 symmetry only is
optically active or can have mirror images that do not
superimpose over each other. Try holding up a hand mirror to the
allene molecule picture, to see for yourself.
2Allenes are the compounds having commulative double bonds. The allenes exhibit optical
isomerism providedthe two groups attached to each terminal carbonatom are different.

138. Optical isomerism in allenes was first of all predicted b Van’t Hoff
in 1875 on the basis of the inhibition of free rotation of the groups
attached to ethylenic carbon atoms and the tetrahedral nature of
these carbon atoms. Now a days, optical activity of allenes can
better be explained in terms of molecular orbital theory.
In allenes the two terminal carbon atoms are in sp² hybridised
state and the central carbon atom sp hybridised. The central
carbon atom is linked to adjacent carbon atoms by two collinear
sigma bonds and two pi bonds. The two pi bonds are in planes at
right angles to each other. One pi bond is present on plane of
paper and other pi bond is perpendicular to that plane.
As well known that in trigonal state the pi bond is at right angle to
plane of the sigma bonds, the terminal groups attached to trigonal
carbon atom lie in planes perpendicular to each other. The

139. molecule ,thus, possesses neither plane nor a centre of symmetry
and forms a non-superimposable mirror image, hence the allene
molecules are asymmetric and optically active.
BIPHENYL
Chirality Without a Stereocenter -Biphenyls is a small group, the
single bond connecting easy rotation and result in racemization the two
phenyl rings would Chirality resulting from restricted rotation about a
single bond is called undergo If Xis
Atropisomerism under.
SPIRANES
Oneorbothofthedoublebondsinallenes maybe
replacedby rings. Whenbotharerings,thecompounds

140. areknownasspirans.Suchcompoundsarealsoaxially
asymmetric.
CONFORMATIONAL ANALYSIS: