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Module II: Stereochemistry (6 hrs)
Conformations: Conformations of ethane, cyclohexane and
methylcyclohexane - Explanation of
stability.
Geometrical Isomerism: Definition – Condition – Geometrical
isomerism in but-2-ene and but-2-ene-1,4-dioic acid - Methods of
distinguishing geometrical isomers using melting point and
dipole
moment.
Optical Isomerism: Optical activity – Chirality – Enantiomers -
Meso compounds - Diastereoisomers
–Optical isomerism in lactic acid and tartaric acid -
Racemisation and resolution (elementary idea).
Conformations of ethane
Conformations-Different spatial arrangements (structures) of a
molecule which are generated by the
relatively easy rotation around a single bond (usually a C-C
single bond).
Ethane, which is comprised of two methyl groups attached to each
other, has properties very similar
to those of methane. However, the complete 3-dimensional shape
of ethane cannot be specified by
these bond lengths and bond angles alone because ethane can
internally rotate about its C-C bond. Ethane has this extra degree
of freedom is due to the possibility of free rotation about the
cylindrically symmetrical C-C sigma bonds. The sigma bond can
maintain a full degree of overlap
while its two ends rotate. Hence, the energetic barrier to
rotation about sigma bonds is generally very
low. Unlike pi bonds in alkenes, the C-C sigma bond does not
hold the two methyl groups in fixed positions relative to one
another. The different spatial arrangements formed by rotations
about a single
bond are called conformations or conformers.
Although there are seven sigma bonds in the ethane molecule,
rotation about the six carbon-hydrogen
bonds does not result in any change in the shape of the molecule
because the hydrogen atoms are
essentially spherical.
A Newman projection can be used to specify the conformation of a
particular bond with clarity and
detail. A Newman projection represents the head-on look down the
bond of interest. A Newman
projection visualizes the conformation of a chemical bond from
front to back, with the front atom represented by a dot and the
back carbon as a circle. The front carbon atom is called
proximal,
while the back atom is called distal. This type of
representation clearly illustrates the specific
dihedral angle between the proximal and distal atoms. The six
carbon-hydrogen bonds are shown as solid lines protruding from the
two carbons at 120°angles, which is what the actual tetrahedral
geometry looks like when viewed from this perspective and
flattened into two dimensions.
Dihedral angle is the angle formed between bonds on the front
atom and bonds on the rear atom.
While there are an infinite number of conformations about any
sigma bond, in ethane two particular conformers are noteworthy and
have special names. In the eclipsed conformation, the C-H bonds
on
the front and back carbons are aligned with each other with
dihedral angles of 0 degrees. If we now
rotate the front CH3 group 60° clockwise, the molecule which was
in the highest energy ‘eclipsed'
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conformation, where the hydrogens on the front carbon are as
close as possible to the hydrogens on the back carbon will change
to a new conformation named staggered .In the staggered
conformation,
the C-H bonds on the rear carbon lie between those on the front
carbon with dihedral angles of 60
degrees. The process of rotation around cylindrically
symmetrical C-C sigma bond can be continued all around the
360°circle, with three eclipsed conformations and three staggered
conformations, in
addition to an infinite number of variations in between.
The eclipsed conformation of ethane is less stable than the
staggered conformation by 3 kcal/mol.
(Note that this figure is very small compared to the energy
required to rotate around double bonds,
which is 60 kcal/mol (the bond energy of a C-C pi bond) The
staggered conformation is the most stable of all possible
conformations of ethane, since the angles between C-H bonds on the
front
and rear carbons are maximized at 60 degrees. In the eclipsed
form, the electron densities on the
C-H bonds are closer together than they are in the staggered
form. When two C-H bonds are brought into a dihedral angle of zero
degrees, their electron clouds experience repulsion, which raises
the
energy of the molecule. This additional instability factor,
arises due to this repulsion is called
torsional strain. The eclipsed conformation of ethane has three
such C-H eclipsing interactions, so we can infer that each eclipsed
C-H "costs" roughly 1 kcal/mol. At room temperature, ethane
molecules have enough energy to be in a constant state of rotation.
Because of this rapid rotation, it is
impossible to isolate any particular conformation
Energy diagram for rotation about the C-C bond in ethane
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Conformations of cyclohexane
Cyclopropane is necessarily planar (flat), with the carbon atoms
at the corners of an equilateral triangle. The 60º bond angles are
much smaller than the optimum 109.5º angles of a normal
tetrahedral carbon atom, and the resulting angle strain
dramatically influences its chemical behaviour.
Cyclopropane also suffers substantial eclipsing strain or
torsional strain, since all the carbon-carbon
bonds are fully eclipsed. Cyclobutane reduces some
bond-eclipsing strain by folding (the out-of-plane dihedral angle
is about 25º), but the total eclipsing and angle strain remains
high. Cyclopentane has
very little angle strain (the angles of a pentagon are 108º),
but its eclipsing strain would be large
(about 10 kcal/mol) if it remained planar. Consequently, the
five-membered ring adopts non-planar puckered conformations
whenever possible.Rings larger than cyclopentane would have angle
strain if
they were planar. However, this strain, together with the
eclipsing strain inherent in a planar structure,
can be relieved by puckering the ring.
Cyclohexane is a good example of a carbocyclic system that
virtually eliminates eclipsing and angle
strain by adopting non-planar conformations. A planar structure
for cyclohexane is clearly improbable. The bond angles would
necessarily be 120º, 10.5º larger than the ideal tetrahedral
angle.
Also, every carbon-carbon bond in such a structure would be
eclipsed. The resulting angle and
eclipsing strains would severely destabilize this structure. If
two carbon atoms on opposite sides of the
six-membered ring are lifted out of the plane of the ring, much
of the angle strain can be eliminated.
This boat structure, which is almost free of angle strain, still
has two eclipsed bonds and severe steric
crowding of two hydrogen atoms on the "bow" and "stern" of the
boat. This steric crowding is often
called known as the flagpole interaction Also it has torsional
strain associated with eclipsed bonds at
the four of the C atoms that form the side of the boat.
A twist relieves some of the torsional strain of the boat and
moves the flagpole H further apart reducing the steric strain.
Consequently the twist boat or skew-boat is slightly more stable
than the
boat, but the twist-boat conformer still retains some of the
strains that characterize the boat conformer.
Finally, by lifting one carbon above the ring plane and the
other below the plane, a relatively strain-
free 'chair' conformer is formed. This is the predominant
structure adopted by molecules of
cyclohexane.
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In chair conformation of cyclohexane, we find that the twelve
hydrogens are not structurally equivalent. Six of them are located
about the periphery of the carbon ring, and are termed
equatorial.
The other six are oriented above and below the approximate plane
of the ring (three in each location),
and are termed axial.
Conformational rotation of cyclohexane interconverts the
conformations as there are two equivalent
chair conformations of cyclohexane in rapid equilibrium and all
the twelve hydrogens have 50% equatorial and 50% axial character.
This proceeds from one chair to twist boat to boat to twist boat
to
the other chair conformation. This process is often referred to
as "ring flipping".
Conformations methylcyclohexane
A substituent group in a substituted cyclohexane, such as the
methyl group in methyl cyclohexane,
can be in either an equatorial or an axial position.
The "equatoriality principle" provides that any substituent
prefers to occupy the less sterically hindered equatorial position,
if at all possible.in the case of di- or poly-substituted
cyclohexanes not
all substituents can occupy all equatorial positions in every
isomer, but that isomer will be the most
stable in which all of the substituents occupy equatorial
positions.
All axial bonds in any chair conformation are necessarily
parallel to each other, substituents larger
than hydrogen generally suffer greater steric crowding when they
are oriented axial rather than equatorial. Consequently,
substituted cyclohexanes will preferentially adopt conformations in
which
the larger substituents assume equatorial orientation. When the
ring flip occurs, however, it
converts to axial methylcyclohexane. These two conformations are
in rapid equilibrium at room
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temperature, but can be frozen out as distinct compounds at
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an axial position it suffers steric crowding by the two axial
hydrogens
located on the same side of the ring. This steric interaction,
called, 1,3-Diaxial interactions results in
a repulsive destabilizing effect between an axial substituent
located on carbon atom 1 of a
cyclohexane ring and the hydrogen atoms (or other substituents)
located on carbon atoms 3 and 5.It
results in the axial isomer to have about 1.8 kcal/mol of steric
strain.
Geometric (cis/trans) isomerism
Isomers are molecules that have the same molecular formula, but
have a different arrangement of the atoms in space. That excludes
any different arrangements which are simply due to the molecule
rotating as a whole, or rotating about particular bonds. Where
the atoms making up the various
isomers are joined up in a different order, this is known as
structural isomerism. In stereoisomerism, the atoms making up the
isomers are joined up in the same order, but still manage to have a
different
spatial arrangement. Geometric isomerism is one form of
stereoisomerism. These isomers occur where you have restricted
rotation somewhere in a molecule,examples usually just involve the
carbon-
carbon double bond.
These two molecules are not the same. The carbon-carbon double
bond won't rotate and so you would
have to take the models to pieces in order to convert one
structure into the other one. That is a simple test for isomers. If
you have to take a model to pieces to convert it into another one,
then you've got
isomers. If you merely have to twist it a bit, then you
haven't!
In one, the two chlorine atoms are locked on opposite sides of
the double bond. This is known as the
trans isomer. (trans: from latin meaning "across" - as in
transatlantic). In the other, the two chlorine
atoms are locked on the same side of the double bond. This is
known as the cis isomer. (cis: from latin
meaning "on this side")
There are problems as compounds get more complicated. For
example, could you name these two
isomers using cis and trans?
Because everything attached to the carbon-carbon double bond is
different, there aren't any obvious things which you can think of
as being "cis" or "trans" to each other. The E-Z system gets around
this
problem completely.
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To get geometric isomers you must have:
restricted rotation (often involving a carbon-carbon double bond
for introductory purposes);
two different groups on the left-hand end of the bond and two
different groups on the right-
hand end. It doesn't matter whether the left-hand groups are the
same as the right-hand ones or not.
The E-Z system
In the example below, at the left-hand end of the bond, it turns
out that bromine has a higher priority
than fluorine. And on the right-hand end, it turns out that
chlorine has a higher priority than hydrogen.
If the two groups with the higher priorities are on the same
side of the double bond, that is described
as the (Z)- isomer. So you would write it as (Z)-name of
compound. The symbol Z comes from a
German word (zusammen) which means together.
Rules for determining priorities (Cahn-Ingold-Prelog (CIP)
rules)
Among first at the atoms attached directly to the carbon atoms
at each end of the double bond, the
atom which has the higher atomic number is given the higher
priority.
Just consider the first isomer - and look separately at the
left-hand and then the right-hand carbon
atom. Compare the atomic numbers of the attached atoms to work
out the various priorities.
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Notice that the atoms with the higher priorities are both on the
same side of the double bond. That counts as the (Z)- isomer.The
second isomer obviously still has the same atoms at each end, but
this
time the higher priority atoms are on opposite sides of the
double bond. That's the (E)- isomer.
Consider the following isomers
Chlorine has a higher atomic number than hydrogen, and so has
the higher priority. That, of course, is equally true of all the
other carbon atoms in these two isomers. In the first isomer, the
higher priority
groups are on opposite sides of the bond. That must be the (E)-
isomer. The other one, with the higher
priority groups on the same side, is the (Z)- isomer.
And now but-2-ene
This adds the slight complication that you haven't got a single
atom attached to the double bond, but a
group of atoms. Consider the atom directly attached to the
double bond - in this case the carbon in
the CH3 group. For this simple case, you can ignore the hydrogen
atoms in the CH3 group entirely.
Here is one of the isomers of but-2-ene:
The CH3 group has the higher priority because its carbon atom
has an atomic number of 6 compared
with an atomic number of 1 for the hydrogen also attached to the
carbon-carbon double bond.The
isomer drawn above has the two higher priority groups on
opposite sides of the double bond. The
compound is (E)-but-2-ene.
Rule to allow for isotopes
For example, consider the case of hydrogen; deuterium is an
isotope of hydrogen having a relative
atomic mass of 2. It still has only 1 proton, and so still has
an atomic number of 1. However, it isn't
the same as an atom of "ordinary" hydrogen, and so these two
compounds are geometric isomers:
The hydrogen and deuterium have the same atomic number - so on
that basis, they would have the
same priority. In a case like that, the one with the higher
relative atomic mass has the higher priority.
So in these isomers, the deuterium and chlorine are the higher
priority groups on each end of the
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double bond. That means that the left-hand isomer in the last
diagram is the (E)- form, and the right-
hand one the (Z)-.
Rules for more complicated molecules
Let's illustrate this by taking the following molecule, and
seeing how easy it is to find out whether it is
a (Z)- or (E)- isomer by applying an extra rule.
Focus on the left-hand end of the molecule. What is attached
directly to the carbon-carbon double
bond? In both of the attached groups, a carbon atom is attached
directly to the bond. Those two atoms
obviously have the same atomic number and therefore the same
priority. So that doesn't help.
In this sort of case, you now look at what is attached directly
to those two carbons (but without
counting the carbon of the double bond) and compare the
priorities of this next lot of atoms.
In the CH3 group: The atoms attached to the carbon are H H
H.
In the CH3CH2 group: The atoms attached directly to the carbon
of the CH2 group are C H H.
In the second list, the C is written first because it has the
highest atomic number. Now compare the
two lists atom by atom. The first atom in each list is an H in
the CH3 group and a C in the CH3CH2
group. The carbon has the higher priority because it has the
higher atomic number. So that gives the
CH3CH2 group a higher priority than the CH3 group.
Now look at the other end of the double bond. The extra thing
that this illustrates is that if you have a
double bond, you count the attached atom twice. Here is the
structure again.
So, again, the atoms attached directly to the carbon-carbon
double bond are both carbons. We
therefore need to look at what is attached to those carbons.
In the CH2OH group: The atoms attached directly to the carbon
are O H H.
In the CHO group: The atoms attached directly to the carbon are
O O H.
Remember that the oxygen is counted twice because of the
carbon-oxygen double bond.
In both lists, the oxygens are written first because they have a
higher atomic number than hydrogen.
So, what is the priority of the two groups? The first atom in
both lists is an oxygen - that doesn't help. Look at the next atom
in both lists. In the CH2OH group, that's a hydrogen; in the CHO
list, it's an
oxygen.
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The oxygen has the higher priority - and that gives the CHO
group a higher priority than the CH2OH group.The isomer is
therefore a (Z)- form, because the two higher priority groups (the
CH3CH2 group
and the CHO group) are both on the same side of the bond.
One more example to make a couple of additional minor points
Look first at the left-hand groups.
In both the top and bottom groups, you have a CH2 group attached
directly to the carbon-carbon
double bond, and the carbon in that CH2 group is also attached
to another carbon atom. In each case, the list will read C H H.
There is no difference between the priorities of those groups, so
what are you
going to do about it? The answer is to move out along the chain
to the next group. And if
necessary, continue to do this until you have found a
difference.
Next along the chain at the top left of the molecule is another
CH2 group attached to a further carbon
atom. The list for this group is again C H H.
But the next group along the chain at the bottom left is a CH
group attached to two more carbon
atoms. Its list is therefore C C H.
Comparing these lists atom by atom, leads you to the fact that
the bottom group has the higher
priority.
Now look at the right-hand groups. Here is the molecule
again:
The top right group has C H H attached to the first carbon in
the chain.
The bottom right one has Cl H H.
The chlorine has a higher atomic number than carbon, and so the
bottom right group has the higher priority of these two groups.
This molecule is a (Z)- isomer because the higher priority groups
at each
end are on the same side of the double bond.
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Can you easily translate cis- and trans- into (Z)- and (E)-?
You might think that for simple cases, cis- will just convert
into (Z)- and trans- into (E)-.Look for
example at the 1,2-dichloroethene and but-2-ene cases.
But it doesn't always work! Think about this relatively
uncomplicated molecule.
This is clearly a cis- isomer. It has two CH3 groups on the same
side of the double bond. But work out
the priorities on the right-hand end of the double bond. The two
directly attached atoms are carbon and bromine. Bromine has the
higher atomic number and so has the higher priority on that end.
At
the other end, the CH3 group has the higher priority. That means
that the two higher priority groups
are on opposite sides of the double bond, and so this is an (E)-
isomer - NOT a (Z)-.
cis- and trans-isomers have different properties depending on
the influence of the substituted
group in the molecule
Geometrical isomerism of but-2-ene-1,4-dioic acid and difference
in melting points
The two geometric isomers of butenedioic acid (maleic and
fumaric acid) demonstrate different
physical properties. The melting points are very different as
the close proximity of the two -COOH groups allows for the
formation of intramolecular hydrogen bonds. This decreases the
possibility of
intermolecular hydrogen bonding and reduces the melting
point.
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cis-butenedioic acid - melting point = 135ºC (intramolecular
hydrogen bond reduces melting point)
trans-butenedioic acid - melting point = 287ºC
Geometrical isomerism of 1,2-dichloroethene and difference in
melting/boiling points
The polarity strongly influences the relative boiling point as
it determines the strength of the intermolecular forces. For
example, cis-1,2-dichloroethene has a net dipole moment and
dipole-dipole
attractions between its molecules in addition to the van der
Waals' forces, whereas trans-1,2-
dichloroethene which is non-polar has only van der Waals'
forces. The boiling point of the cis-isomer
is therefore higher. Melting point is generally more influenced
by the symmetry of the molecules as this affects the packing in the
solid state. The trans-isomers are able to pack more closely due to
their
greater symmetry, so the intermolecular forces are more
effective than in the cis-isomer. The melting
point of the trans-isomer is therefore higher.
Optical Isomerism
Optical isomerism is a form of stereoisomerism. In
stereoisomerism, the atoms making up the isomers
are joined up in the same order, but still manage to have a
different spatial arrangement. Optical
isomers are named like this because of their effect on plane
polarised light.
Simple 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 the (+)
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
known as the (-) form. So the other enantiomer of alanine is known
as or (-)alanine.
If both enatiomers are presentin 1:1ratio, this is known as a
racemic mixture or racemate. It has no effect on plane polarised
light.
In 1874 Jacobus van't Hoff and Joseph Le Bel recognized that a
compound that contains a single tetrahedral carbon atom with four
different substituents could exist in two forms that were
mirror
images of each other. Consider the CHFClBr molecule, for
example, which contains four different
substituents on a tetrahedral carbon atom. The figure below
shows one possible arrangement of these
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substituents and the mirror image of this structure. By
convention, solid lines are used to represent bonds that lie in the
plane of the paper. Wedges are used for bonds that come out of the
plane of the
paper toward the viewer; dashed lines describe bonds that go
behind the paper.
If we rotate the molecule on the right by 180 around the CH bond
we get the structure shown on the
right in the figure below.
These structures are different because they cannot be
superimposed on each other, as shown in the
figure below.
CHFClBr is therefore a chiral molecule that exists in the form
of a pair of stereoisomers that are mirror images of each other. As
a rule, any tetrahedral atom that carries four different
substituents is a
stereocenter, or a stereogenic atom. However, the only criterion
for chirality is the non-
superimposable nature of the object. A test for achirality is
the presence of a mirror plane within
the molecule. If a molecule has a plane within it that will cut
it into two symmetrical halves, then it is achiral. Therefore, lack
of such a plane indicates a molecule is chiral. Compounds that
contain a
single stereo-center are always chiral.
In 1813 Jean Baptiste Biot noticed that plane-polarized light
was rotated either to the right or the left
when it passed through single crystals of quartz or aqueous
solutions of tartaric acid or sugar. Because they interact with
light, substances that can rotate plane-polarized light are said to
be optically active.
Those that rotate the plane clockwise (to the right) are said to
be dextrorotatory (from the Latin
dexter, "right"). Those that rotate the plane counterclockwise
(to the left) are called levorotatory
(from the Latin laevus, "left").
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In 1848 Louis Pasteur noted that sodium ammonium tartrate forms
two different kinds of crystals that are mirror images of each
other, much as the right hand is a mirror image of the left hand.
By
separating one type of crystal from the other with a pair of
tweezers he was able to prepare two
samples of this compound. One was dextrorotatory when dissolved
in aqueous solution, the other was
levorotatory. Since the optical activity remained after the
compound had been dissolved in water, it could not be the result of
macroscopic properties of the crystals. Pasteur therefore concluded
that
there must be some asymmetry in the structure of this compound
that allowed it to exist in two forms.
Once techniques were developed to determine the
three-dimensional structure of a molecule, the
source of the optical activity of a substance was recognized:
Compounds that are optically active
contain molecules that are chiral. Chirality is a property of a
molecule that results from its
structure. Compounds, such as CHFClBr, that contain a single
stereocenter are the simplest to
understand. One enantiomer of these chiral compounds is
dextrorotatory; the other is levorotatory. To decide whether a
compound should be optically active, we look for evidence that the
molecules are
chiral. The instrument with which optically active compounds are
studied is a polarimeter, shown in
the figure below.
If an achiral molecule (one with a plane of symmetry) looked in
a mirror, you would always find that
by rotating the image in space, you could make the two look
identical. It would be possible to
superimpose the original molecule and its mirror image.
Examples of optical isomers
Butan-2-ol: The asymmetric carbon atom in a compound (the one
with four different groups attached)
is often shown by a star.
It's extremely important to draw the isomers correctly. Draw one
of them using standard bond
notation to show the 3-dimensional arrangement around the
asymmetric carbon atom. Then draw the
mirror to show the examiner that you know what you are doing,
and then the mirror image.
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2-hydroxypropanoic acid (lactic acid): Once again the chiral
centre is shown by a star.
The two enantiomers are:
Absolute Configuration: R-S Sequence Rules
The Cahn-Ingold-Prelog priority rules are used for naming chiral
centers and geometric isomers (e.g.
E- or Z-alkenes). These rules are used to establish the priority
of the groups attached to the chiral
center and are based on atomic number, and the first point of
difference.
Rule 1
First, examine at the atoms directly attached to the
stereocenter of the compound. A substituent with a
higher atomic number takes precedence over a substituent with a
lower atomic number. Hydrogen is
the lowest possible priority substituent, because it has the
lowest atomic number.
1. When dealing with isotopes, the atom with the higher atomic
mass receives higher priority.
2. When visualizing the molecule, the lowest priority
substituent should always point away from the viewer (a dashed line
indicates this). To understand how this works or looks, imagine
that a clock and a pole. Attach the pole to the back of the clock,
so that when when looking at
the face of the clock the pole points away from the viewer in
the same way the lowest priority substituent should point away.
3. Then, draw an arrow from the highest priority atom to the 2nd
highest priority atom to the 3rd highest priority atom. Because the
4th highest priority atom is placed in the back, the arrow should
appear like it is going across the face of a clock. If it is going
clockwise, then it is an R-enantiomer; If it is going
counterclockwise, it is an S-enantiomer.
When looking at a problem with wedges and dashes, if the lowest
priority atom is not on the dashed line pointing away, the molecule
must be rotated.
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Wedges indicate coming towards the viewer.
Dashes indicate pointing away from the viewer.
Rule 2
If there are two substituents with equal rank, proceed along the
two substituent chains until there is a
point of difference. First, determine which of the chains has
the first connection to an atom with the
highest priority (the highest atomic number). That chain has the
higher priority.
If the chains are similar, proceed down the chain, until a point
of difference.
For example: an ethyl substituent takes priority over a methyl
substituent. At the connectivity of the
stereocenter, both have a carbon atom, which are equal in rank.
Going down the chains, a methyl has
only has hydrogen atoms attached to it, whereas the ethyl has
another carbon atom. The carbon atom
on the ethyl is the first point of difference and has a higher
atomic number than hydrogen; therefore
the ethyl takes priority over the methyl.
Rule 3
If a chain is connected to the same kind of atom twice or three
times, check to see if the atom it is
connected to has a greater atomic number than any of the atoms
that the competing chain is connected
to.
If none of the atoms connected to the competing chain(s) at the
same point has a greater atomic number: the chain bonded to the
same atom multiple times has the greater priority
If however, one of the atoms connected to the competing chain
has a higher atomic number: that chain has the higher priority.
Determining R/S when The #4 Substituent Is In Front (i.e. on the
“solid Wedge”): When the #4 priority is on a wedge you can just
reverse the rules. So now we have two sets of rules:
If the #4 priority is on a hollow wedge: Clockwise = R
anticlockwise = S
If the #4 priority is on a solid wedge, reverse the typical
rules: Clockwise = S anticlockwise = R
If The #4 Group Is In The Plane Of The Page:
Take the #4 substituent (in the plane of the page) and swap it
with the substituent in the back [If
the configuration is R, this will switch it to S. If the
configuration is S, this will flip it to R. We’ll
need to account for this in step #3].
Redraw the chiral centre, and determine R/S on the new chiral
centre which now has group #4 in
the back.
Whatever result you got, flip it to its opposite to account for
the fact that you did a single swap in
step #1.
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Meso Compounds: Meso compounds are achiral compounds that have
multiple chiral centers. It is
superimposed on its mirror image and is optically inactive
despite its stereocenters. In general, a meso
compound should contain two or more identical substituted
stereocenters. Also, it has an internal symmetry plane that divides
the compound in half. These two halves reflect each other by the
internal
mirror. The stereochemistry of stereocenters should "cancel
out". What it means here is that when we
have an internal plane that splits the compound into two
symmetrical sides, the stereochemistry of
both left and right side should be opposite to each other, and
therefore, result in optically inactive.
Identification
If A is a meso compound, it should have two or more
stereocenters, an internal plane, and the stereochemistry should be
R and S.
Look for an internal plane, or internal mirror, that lies in
between the compound. The stereochemistry (e.g. R or S) is very
crucial in determining whether it is a meso compound or not.
As mentioned above, a meso compound is optically inactive, so
their stereochemistry should cancel
out. For instance, R cancels S out in a meso compound with two
stereocenters.
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An interesting thing about single bonds or sp3-orbitals is that
we can rotate the substituted groups that
attached to a stereocenter around to recognize the internal
plane. As the molecule is rotated, its
stereochemistry does not change. For example:
The following cyclic molecule has a plane of symmetry (the
vertical plane going through the red broken line perpendicular to
the plane of the ring) and, therefore, is achiral, but has has two
chiral
centers. Thus, its is a meso compound.
Diastereomers
We have seen that enantiomers are stereoisomers that are
non-superimposable mirror images of each
other. Thus far we have only dealt with compounds that contain
only a single stereogenic center. For
these compounds, we can produce the enantiomer by changing the
configuration at that stereocenter; that is, the enantiomers differ
only in their spatial arrangements at the stereocenter. What
happens
when we consider molecules with more than one stereocenter? It
turns out that the mirror image of
such a molecule has all its stereocenters inverted. Hence the
enantiomer of a molecule has precisely the opposite configuration
at every stereocenter. Every (R) configuration becomes (S), and
vice versa.
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The interesting case, however, occurs when only some of the
stereocenters are inverted. Stereoisomers that differ at some
stereocenters but not at others are not mirror images, so they are
not
enantiomers. Instead, they are diastereomers. A diastereomer is
simply any stereoisomer that is not an
enantiomer.
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The optical isomers of 2-hydroxypropanoic acid (lactic acid) are
a simple example of optical isomerism. The (+) enantiomer of lactic
acid is found in muscle. Sour milk contains a racemic mixture
of the two enantiomers.
2,3-dihydroxypropanal or glyceraldehyde is used as a standard by
which other chiral molecules are
compared. There are two enantiomers of glyceraldehyde, depending
on the position of the –OH (hydroxyl) group on the molecule. These
are known as D-glyceraldehyde and L-glyceraldehyde. The
little capital letters D and L are deliberate. The positions of
the hydroxyl groups on other chiral
molecules can be compared with glyceraldehyde to see if they are
the D-enantiomer or the L-enantiomer. This is very common in
Biology (natural sugars are D-enantiomers and amino acids are
L-enantiomers). However, knowing whether a molecule is the D or
L-enantiomer does not tell us
whether it is the (+) or (–) enantiomer
Tartaric acid has four basic isomers
1. d-(-)-tartaric acid* 2. l-(+)-tartaric acid* 3. racemic acid
4. meso-tartaric acid.
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Racemisation and resolution
Racemization is the conversion of an enantiomerically pure
mixture (one where only one enantiomer is present) into a mixture
where more than one of the enantiomers are present. If the
racemization
results in a mixture where the D and L enantiomers are present
in equal quantities, the resulting
sample is described as a racemic mixture or a racemate.
Chiral molecules have two forms which differ in their optical
characteristics: The levorotatory form
and the dextrorotatory which are non-superimposable when rotated
in 3-dimensional space, are said to be enantiomers. Racemization
occurs when one pure form of an enantiomer is converted into
equal
proportion of both enantiomers, forming a racemate. When there
are both equal numbers of
dextrorotating and levorotating molecules, the net optical
rotation of a racemate is zero.
Racemate may have different physical properties from either of
the pure enantiomers because of the differential intermolecular
interactions. The change from a pure enantiomer to a racemate can
change
its density, melting point, solubility, heat of fusion,
refractive index, and its various spectra.
Crystallization of a racemate can result in separate (+) and (−)
forms, or a single racemic compound.
Separation of racemates into their component enantiomers is a
process called resolution. In 1848, Louis Pasteur was the first to
perform the stereochemical resolution of a racemic mixture.
While
studying crystals of sodium ammonium tartrate under a
microscope, he observed that there were two
different crystalline forms that were non-superimposable mirror
images of each other. He painstakingly physically separated these
two forms and discovered that the two samples rotated plane-
polarized light in opposite directions. This led to his
deduction that the crystals were composed of
molecules that were also non-superimposable mirror images
(enantiomers). Pasteur was fortunate in
that he was studying a system in which the enantiomers
crystallized separately (called spontaneous resolution), which is
rare among organic compounds. However, physically separating
crystals by hand
is not feasible when working with large quantities of
material.
Since enantiomers have identical physical properties, such as
solubility and melting point, resolution
is extremely difficult. Diastereomers, on the other hand, have
different physical properties, and this fact is used to achieve
resolution of racemates. Reaction of a racemate with an
enantiomerically pure
chiral reagent gives a mixture of diastereomers, which can be
separated. For example, if a racemic
mixture of a chiral alcohol is reacted with a enantiomerically
pure carboxylic acid, the result is a mixture of diastereomers: in
this case, because the pure (R) entantiomer of the acid was used,
the
product is a mixture of (R-R) and (R-S) diastereomeric esters,
which can, in theory, be separated by
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their different physical properties. Subsequent hydrolysis of
each separated ester will yield the 'resolved' (enantiomerically
pure) alcohols.
Rule 1Rule 2Rule 3