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'

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

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.

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

temperature, but can be frozen out as distinct compounds at -78degrees. When the methyl group in the structure above occupies 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.

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.

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

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.

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.

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.

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

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").

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.

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.

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.

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.

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.

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.

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.

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

their different physical properties. Subsequent hydrolysis of each separated ester will yield the 'resolved' (enantiomerically pure) alcohols.

Rule 1Rule 2Rule 3