CH. 7
Chapter 7
Stereochemistry
In this chapter we will examine stereochemistry in more detail.
There are several kinds of isomers. We have already seen constitutional isomers. These
are molecules that have the same formula but different connectivities of the atoms.
C2H6O
CH3
O CH3
CH3CH2 OH
ethanol
diethyl ether
Stereoisomers are isomers that have the same connectivity but different arrangements of
the atoms in space. Examples of stereoisomers that we have already seen are the cis- and
trans-isomers of disubstituted alkenes.
CH3
CH3
C C
H
H
cis-2-butene
H
CH3
C C
H
CH3
trans-2-butene
There are two kinds of stereoisomers. We will distinguish them by their definitions.
(1) Enantiomers (en-an-tee-oh-mur): these are stereoisomers that are non-superimposable
on their mirror images.
(2) Diastereomers (die-uh-stare-ee-oh-mur): these are stereoisomers that are not mirror
images and are non-superimposable. An example of two isomeric molecules that are
diastereomers is cis- and trans-2-butene.
Enantiomers:
Molecules are three-dimensional objects. Many three-dimensional objects, including many
molecules, have the property of chirality. A chiral object is non-superimposable on its
mirror image. Every object, of course, has a mirror image (except vampires!) but many
objects are superimposable on the mirror image. In other words, many objects are exactly
the same as their mirror image. If two objects are superimposable they are the same in
every way and they are two copies of the same thing. But other objects are not
superimposable on their mirror image. For example, your two hands are nonsuperimposable. Your hands are mirror images of each other but if you turn them and try
to superimpose them but putting the palm of one hand on the back of the other, you find
that your thumbs are on opposite sides. Your hands are not the same and have the property
of chirality and the relationship between them is that they are enantiomers.
1
CH. 7
For an object to have the property of chirality there must be no plane of symmetry in the
object. If there is a plane of symmetry the object will be non-chiral and superimposable on
its mirror image. It will be the same as its mirror image and will not exist as enantiomers.
Chiral objects
Hands
Feet
Shoes
Piece of paper
Non-chiral
gloves
spoon
socks
book with front and back covers
A molecule that has a carbon that is attached to four different groups or atoms is a chiral
molecule and will exist as two different stereoisomers. The relation between these
stereoisomers is that they are enantiomers.
enantiomer of the
molecule on the right
Cl
F
H
H
C
C Cl
Br
F
Br
enantiomer of the
molecule on the left
carbon that has 4
different sustituents
A carbon that has four different substituents is called a chirality center.
Look at molecules A and B. We use the dotted line between them to symbolize a mirror.
Recall that a mirror transposes the object. What is on the left in the object, is on the right
in the mirror image and vice versa. Look at the front of an ambulance. The word
“ambulance” is written backwards. Now look at an ambulance in your rear view mirror as
you are driving and you will be able to read the word clearly when you see it behind you in
the mirror. The mirror transposes the word “ambulance” so that it reads normally from left
to right.
In the example below, no amount of rotation will transform molecule A into molecule B.
They are not they same molecule. They are non-superimposable mirror images and are
therefore enantiomers.
Note that by exchanging any two substituents, one enantiomer is converted into its
enantiomer. It does not matter which substituents are switched.
2
CH. 7
H
H
enantiomer of the
molecule on the right
Cl
C
F
A
Br
carbon that has 4
different sustituents
H
rotate 180°
C Cl
F C
Br
F
Cl
Br
B
B
enantiomer of the
The Cl and F are swiched in
molecule on the left B; no amount of rotation can
make the molecules the same.
mirror
But if two of the substituents are the same, the molecule will be superimposable on its
mirror image. The molecule will be achiral and will be the same molecule as its mirror
image.
One criterion for chirality then for a molecule with one carbon atom is that there are four
different substituents attached to that carbon. If there are three or fewer different
substituents, the molecule cannot be chiral.
Carbons in rings can be chirality centers.
Another way to determine whether a molecule is chiral is whether or not it has a plane of
symmetry. If it has a plane of symmetry, then it cannot be chiral.
3
CH. 7
Optical Activity
Molecules that are enantiomers are the same in every physical property. Enantiomers have
the same melting points, the same boiling points, the same densities, the same polarities,
etc. It is in fact very difficult to separate two enantiomers by physical means. Since all
drugs sold today that are chiral must be sold as one pure enantiomer, this creates real
problems in the pharmaceutical industry.
One property, however, in which enantiomers are different is in their behavior toward
plane polarized light. Normal light consists of elector-magnetic waves that are oriented in
all possible planes. Using a polaroid filter, which is simply a very thin slit, all of the
planes of the electro-magnetic waves can be filtered out, except for one. Therefore, the
light passing through a polaroid filter is oriented in one plane.
4
CH. 7
When this light is passed through a solution that contains only one enantiomer of a chiral
molecule, the plane of this light is rotated from its original orientation and the amount of
this rotation, called the specific rotation α, is a characteristic physical property
of the particular compound, just like the melting point or boiling point.
The other enantiomer will rotate the light in exactly the same amount but in the opposite
direction.
5
CH. 7
angle of rotation = - α
CH3
H C CH CH
2
3
OH
Plane polarized light.
When a compound consists of one pure enantiomer and it rotates plane-polarized light, it is
said to be optically active. A solution that consists of a 50:50 mixture of both enantiomers
of a chiral compound is optically inactive. It will not rotate plane-polarized light. The
light will be rotated in the +α direction when it strikes one enantiomer and then it will be
rotated back in the -α direction when it strikes the other enantiomer and the net rotation
will be zero.
angle of rotation = - α
CH3
CH3
H C CH CH
2
3
OH
Plane polarized light.
OH C CH CH
2
3
H
angle of rotation = - α
angle of rotation = +α
net rotation =
0
In order for a solution to be optically active it must contain an excess of one enantiomer.
The optical purity is a measure of the percent of the excess of one enantiomer over the
other and is defined as percent of one enantiomer minus percent of the other. For example,
a material that is 50 optically pure contains 75% of one enantiomer and 25% of the other.
The device for measuring optical rotation is called a polarimeter. It is a long, closed tube
with a light source at one end. Typically a sodium lamp is used. It emits light of a single
wavelength (589 nm). This light then passes through a polaroid filter and then through a
smaller tube that contains a solution of the chiral molecule whose optical rotation is being
measured. The observed rotation depends on the number of molecules that the light
encounters. We define the specific rotation, [α], as a function of the measured or observed
rotation α that is dependent on the concentration of the solution and the length of the
smaller tube inside the polarimeter that contains the sample solution.
specific rotation [α] =
100 α
α = observed rotation
l = length of polarimeter in dm
cl
c= concentration in grams
per 100 mL of solution
A racemic mixture is an equal, 50:50 mixture of each enantiomer. A racemic mixture is
optically inactive.
6
CH. 7
Absolute and Relative Configuration
The absolute configuration is the exact arrangement of the atoms in space. The absolute
configuration of any molecule was not known until 1951 when the configuration of (+)tartaric acid was determined. For example, it was known before 1951 that one of the
enantiomers of 2-butanol rotated light in the (+) direction and the other rotated it in the (-)
direction but it was not known which enantiomer was the (+) one and which was the (-)
one.
CH3
CH3
H C CH CH
2
3
OH
OH C CH CH
2
3
H
(+)-rotator or (-)-rotator?
But chemists had determined the relative configuration, relating one molecule to another.
For example, 3-buten-2-ol and 2-butanol must have the same configuration at the C-OH
bond, since reduction of 3-buten-2-ol changes it into 2-butanol but does not affect the COH bond.
CH3
CH
CH CH2
H2, Pd
CH3
CH
CH2 CH3
OH
[α] = +13.5
OH
[α] = +33.2°
Assigning the Absolute Configuration
Rules for Assigning Absolute Configuration:
(1) Rank the substituents by atomic number from highest to lowest.
(2) Orient the molecule so that the lowest ranked substituent is pointing away. Use the
first point of difference rule if two atoms have the same atomic number.
(3) Ignore the group that is pointing away (this should be the smallest group.) If you
move in a clockwise direction going from highest to lowest, this is R (from the Latin
rectus, meaning right).
If you move in a counterclockwise direction going from highest to lowest, this is S (from
the Latin sinister, meaning left).
3
C(H,H,H)
CH3
(1) The smallest group, the hydrogen, is pointing away.
(2) Oxygen is biggest, then there are two carbons so look at
4H C
CH2CH3
the substituents on each carbon.
HO 2
C(C,H,H) (3) Ignore the smallet group. Moving from highest to lowest substituent, this is
1
counterclockwise. This is the S configuratiion.
7
CH. 7
So the correct, full name of this molecule is (S)-2-butanol. The mirror image of this
molecule will be its enantiomer and will have the R-configuration.
3 CH
CH3
C
H C CH CH
2
3
HO
(S)-2-butanol
CH3CH2
2
mirror
Moving from hisghest to lowest is clockwise, so this is the R
configuraton as expected.
3
H
OH
1
(R)-2-butanol
The most common mistake that students make in assigning the absolute configuration is
not rotating the molecule so that the smallest group points away (dashed line going back
into the paper). If the smallest group is not pointing away, you will get the wrong answer.
Actually, you can assign the absolute configuration when the smallest group is not pointing
away but then you must change your answer to the correct one. For example, if you make
the assignment when the smallest group is not pointing away and you get R for the
configuration, the correct answer is then S.
Here the small group is not pointing away. Ignore the group that is ointing away and
rank the other three in terms of priority from highest to lowest as before.
CH3 2
Cl
C
H
3
CH2CH3
1
In this case, the movement from highest to lowest is in the counterclockwise direction,
giving the S configuration. But this is the wrong answer. The correct answer is then R
for this molecule.
Rules for rotating the smallest group away: Any molecule can be changed into a different
view of the same molecule by rotating any three of the substituents in either the clockwise
or counterclockwise direction while holding fourth substituent in the same position.
3
CH3
CH3
C
Cl
H
CH2CH3
Hold the methyl group the
same, and rotate the other
three so that the hydrogen
points away.
H C Cl
CH3CH2
1
2
Clockwise = R
It does not matter which substituents you rotate. You will get the correct answer for you
assignment.
CH3
C
Cl
H
CH2CH3
Cl
1
Clockwise = R
H C CH CH
2
3
CH3
2
3
8
CH. 7
But note: you must rotate three of the substituents. If you simply switch two of them, then
you have changed the molecule from one enantiomer to the other enantiomer.
For cyclic molecules, break the ring into two substituents at the first point of difference
and assign priority as described above.
CH3 H
1
CH3 H
H
C 1'
1
CH2
CH2 C 2'
2
C
2
C
C C H
H
Clockwise = R
3
H
H
C1' and C1 are the same so go on to compare C2' and C2.
C2'(C,C,H)
larger/heavier substitutent
C2(C,H,H)
Fischer Projections
There is another convention for drawing molecules with chirality centers. This is called
the Fischer convention after the great carbohydrate chemist, Emil Fischer, who invented it.
It is a shorthand method and is very useful for quickly drawing sugar molecules
(carbohydrates) that have multiple chirality centers. Glucose, for example, the most
common sugar, has four adjacent chirality centers.
To use the Fischer projection, make a flat representation. The vertical lines are going back
into the paper and the horizontal lines are coming out of the paper. It is customary to have
the carbon backbone oriented in the vertical direction.
H
Br
going back into the page
H
Cl
F
Br
C
Cl
F
coming out of page
To assign R and S, rotate the molecule so that the small group is pointing away. To do
this, rotate the molecule 180 degrees around the vertical axis. The substituents that were
up will down be down and the substituents that were down will now be pointing up. The
OH and H will switch places but the substituents on the vertical axis will be in the same
positions (except that they will now be pointing out of the page).
9
CH. 7
CH3
HO
3
CH3
CH3
OH
H
C
H
rotate 180°
around vertical axis
CH2CH3
CH2CH3
H
OH
1
Clockwise = R
CH2CH3
2
Physical Properties of Enantiomers
As mentioned, enantiomers are the same in all physical properties – melting point, boiling
point, polarity, refractive index, etc. – except for the rotation of plane-polarized light. It is
therefore very difficult to separate enantiomers. They cannot be separated using normal
techniques like fractional distillation or recrystallization since the boiling points and
polarities are the same for each enantiomers. Special techniques must be used. We will
not discuss the details of these techniques.
Reactions that Create a Chirality Center
If a chiral product is created from achiral reactants, the product must be a racemic mixture
and the reaction mixture is optically inactive.
For example, when achiral propene reacts with achiral peroxy acetic acid the product forms
an epoxide with a chirality center. The solution will consist of an equal, 50:50 mixture of
both enantiomers.
O
O
CH3CH CH2 +
achiral
CH3
CH2 C
C OOH
peroxy acetic acid
H
CH3
chiral
achiral
The alkene is flat and the peroxy acid can attack from the topside to form one enantiomer
or from the bottom side to form the other enantiomer. Both reactions are equally likely.
10
CH. 7
O
CH3
C O
H
H
H
H
top attack
C C
H
CH3
H
H
CH3
C O
H
O
C
C
3
H 2
CH3 O
H
C C
attack
H
H
H
bottom
C C
CH3
!
H
H O
C C 2
CH3
O H
3
O H
C C
H
CH3
O
CH3
H
O
O
Clockwise = R enantiomer
Counterclockwise = S enantiomer
1
H
O
Another example:
OH
CH3CH2CH CH2
H3O+
1-butene, achiral
H2O
CH3CH2CH CH3
2-butanol, chiral, racemic mixture
H
H
H
H
H O H
H
C C
CH3CH2
H
H
O H
H
C CH2
H2O
CH3CH2
O H
H
H
O H
H
H
1
H O
C CH3
CH3CH2
3
2
S-2-butanol, 50%
top
H O
C CH3
CH3CH2
bottom
H
C CH3
CH3CH2 O
H
H
H
O H
3
H
2
C CH3
CH3CH2
O
H
1
R-2-butanol, 50%
Chiral Molecules with More Than One Chirality Center
As a general rule, if a molecule has n chirality centers, it will have up to 2n stereoisomers.
For example, a molecule with two chirality centers will have up to 22 = 4 stereoisomers.
For example, look at the following diol. There are two chirality centers and there may be
up to 4 stereoisomers. The chirality centers are marked with an asterisk.
O HO
HO C * CH
1
2
OH
CH*
3
CH3
4
11
CH. 7
To find all of the stereoisomers, first draw one enantiomer and then make its mirror image.
It is arbitrary which enantiomer you draw first but it is useful to orient the molecule so that
the smallest group is pointing away at each chirality center. This allows you to easily
assign the absolute configuration. You are also encouraged to draw the molecules as
shown here.
HO
O
H
C
C2
OH
C3
OH
CH3
A
H
O
HO
C2
C
HO
C3
CH3
H
2S, 3S
OH
B
H
2R, 3R
Molecules A and B are mirror images and they are non-superimposable; therefore, they are
enantiomers. This can be confirmed by assigning the absolute configuration to each
chirality center. The absolute configuration for each center will be the opposite
configuration in each enantiomer (i.e. R will be S and S will be R in the enantiomer).
Bother centers will be changed.
To assign R or S consider each chirality center separately. For chirality center C2, look at
the substituents attached to C2. There is an OH, an H, a CH(OH)CH3 and a CO2H. The
OH is heaviest, then the CO2H, then the CH(OH)CH3. Note that the small group is
pointing away, so the assignment can be made without any rotations.
smallest
2
C(O,O,O)
HO
larger
O
H
C
C2
OH
C3
OH
CH3
H
1
To assign R/S, consider the four substituents attached to
C2. The H is smallest, then the OH, then the CO2H, then
the CH(OH)CH3. Rotation is counterclockwise from
C(O,C,H) highest to lowest, so the C2 center is S.
3
For chirality center C3:
HO
C(H,H,H) CH3
3
Treat as one of the four substitutents attached to C3.
It has priority over the CH3 group.
O
H
C
C2
OH 2 C(O,C,H)
C3
OH
1
Movement from highest to lowest priority is
counterclockwise, so the C3 center is S.
H
12
CH. 7
Now, assign the priorities for C2 and C3 of the enantiomer B. C2 should be R and C3
should be R as well.
H
O
HO
C2
C
HO
C3
CH3
H
3
H
O
HO
C2
C
HO
1
C3
CH3
1
H
2
OH
2
OH
Movement from highest to lowest is clockwise, so C2 has the R configuration.
Movement from highest to lowest at C3 is clockwise, so C3 has the R oinfiguration.
3
To find the other two isomers, change one of the chirality centers and keep the other the
same for either isomer A or B. It doesn’t matter which isomer you choose or which
chirality center you pick, either C2 or C3. You will identify the two remaining isomers
either way.
Let’s change the chirality center at C2 for isomer A by switching two of the substituents.
Again, it doesn’t matter which two substituents you pick, but for convenience sake, pick
two that keep the small group pointing away and the carbon skeleton on the vertical axis.
Then, to find the fourth isomer, make the mirror image of C. Now make the absolute
configuration assignments as before. Clearly, the two new stereoisomers, C and D, are
enantiomers. Both chirality centers have changed configuration.
H
O
HO
C2
C OH
CH3
C3
OH
C
H
2S, 3R
O
HO C
H
C2
HO C3
D
OH
CH3
H
2R, 3S
And also, A and B are enantiomers but what is the relationship between A and C, between
A and D, between B and C, between B and D?
13
CH. 7
diastereomers
diastereomers
HO
O
H
C
C2
OH
HO
C2
C OH
C3
OH
CH3
C3
OH
CH3
A
H
O
HO
C
C2
OH
C3
OH
A
H
2R, 3S
2S, 3S
H
CH3
C
H
O
H
O
HO
C2
C
HO
C3
H
2R, 3R
C2
HO C3
OH
CH3
H
D
2S, 3R
2S, 3S
diastereomers
H
OH
HO C
H
diastereomers
H
O
O
H
O
O
HO
C2
C OH
HO
C2
C
CH3
CH3
C3
OH
HO
C3
CH3
B
C
H
2R, 3S
H
2R, 3R
OH
B
HO C
H
C2
HO C3
D
OH
CH3
H
2S, 3R
A and C are not mirror images and they are not superimposable. Therefore they must be
diastereomers. The same is true for A and D, B and C, B and D. There are four sets of
diastereomers.
Note that in A and C and in all of the other diastereomeric pairs, one of the chirality
centers has the same configuration in each, but the other chirality center has a different
configuration. (i.e. in A, C2 has the S configuration but in C it has the R configuration
and the configuration at C3 in both is S)
This is a very common type of diastereomerism. It will occur when you have two or more
chirality centers in a molecule.
Note also again, in enantiomers both chirality centers are different: in A C2 is S and C3 is
S and in B C2 is R and C3 is R. This is true for all enantiomeric pairs.
One very important point is that diastereomers, unlike enantiomers, have different
physical properties and so can be separated by normal physical means such as fractional
distillation and recrystallization. Diastereomers have different boiling points, different
melting points, different polarities, etc.
Meso Compounds
There are certain cases in which a molecule with n chirality centers will have fewer than
the 2n stereoisomers. This is due to a plane of symmetry in the molecule.
14
CH. 7
For example, consider 2,3-butanediol. This molecule has n = 2 chirality centers and so we
would expect to find 22 = 4 stereoisomers but in fact we find only three.
HO
OH
CH3 * CH CH* CH3
1
3
2
two chirality centers,; expect 22 stereoisomers
4
We find them in the same way that we did for 2,3-pentanediol. A and B are mirror
images and non-superimposable. Therefore they are enantiomers.
H
HO
CH3
A
H
C2
CH3
CH3
C2
OH
C3
OH
HO
C3
CH3
H
2S, 3S
B
H
2R, 3R
For the other two stereoisomers, take A, keep one of the chirality centers the same, switch
the other one. Again, it does not matter which center you switch or which molecule you
pick, A or B, you will end up with the correct result.
We get two more stereoisomers, C and D. Rotate D 180° in the plane of the paper and we
see that D is really the same as C. C and D are really the same molecule. They are mirror
images but they are superimposable, indicating that they are the same molecule.
So there are only three stereoisomers for 2,3-butanediol: A, B, and C.
We can see that C (or D) has a plane of symmetry and is therefore achiral even though it
does have two chirality centers. A molecule with two or more chirality centers but that
has a plane of symmetry is called a meso compound and it will have fewer than the 2n
stereoisomers.
H
H
CH3
CH3
C
C2
OH
C3
OH
H
2R, 3S
HO
H
C2
CH3
HO C3
CH3
D
rotate 180°
in the plane of
the page
H
2S, 3R
15
CH3
CH3
C3
OH
C2
OH
H
same as C
CH. 7
Disubstituted Cyclic Alkanes
To determine whether or not a disubstituted cyclic alkane is chiral, look for a plane of
symmetry in the molecule. With cyclohexanes in the chair conformation, it can be helpful
to make a flat drawing. If the molecule has a plane of symmetry, it is achiral.
Look at cis- and trans-1,2-dimethylcyclopropane. The cis-isomer is achiral because it has
a plane of symmetry but the trans-isomer is chiral because there is no plane of symmetry.
trans-1,2-dimethylcyclopropane
cis-1,2-dimethylcyclopropane
CH3 CH3
CH3
H
H
3-dimensional
drawing
H
plane of symmetry
achiral
CH3 H
CH3
CH3
H
flat drawing
H
CH3
H
H
CH3
No plane of symmetry;
chiral
For disubstituted cyclobutanes, the 1,2-cis-isomer is achiral. The flat drawing makes it
easier to see the plane of symmetry. And the 1,2-transisomer is chiral, since there is no
plane of symmetry.
cis-1,2-dichlorocyclobutane
Cl
Cl
Cl
H
H
trans-1,2-dichlorocyclobutane
H
H
Cl
Cl
Cl
Cl
H
No plane of symmetry; chiral
H
Cl
H
plane of symmetry; achiral
H
For the 1,3-derivative, however, both the cis- and trans-isomers are achiral, since there is
a plane of symmetry for each of these isomers. In fact the, 1,3-cis-isomer has two planes
of symmetry. The plane of symmetry is more difficult to see in the 1,3-trans-isomer in
either the bent or flat pictures, but it cuts the molecule into two equal portions.
Remember, the plane is an imaginary line. Picture the Cl and the H as each being sliced
in half at both the 1- and 3-positions so that the molecule is divided into two equal
portions.
16
CH. 7
For disubstituted cyclopentanes and cyclohexanes the same principles hold: the 1,2-cis and
1,3-cis derivatives (where the substituents are the same) are achiral but the 1,2-trans and
1,3-trans-derivatives are chiral. Both the 1,4-cis- and 1,4-trans-derivatives are achiral,
regardless of whether the two substituents are the same or different. As with the smaller
rings, making a flat picture can be helpful in seeing the plane of symmetry.
Do not try to memorize these relationships but learn to look for a plane of symmetry. With
the cyclohexanes, the 1,2-cis-derivative must first undergo a ring flip in order to be
superimposable on its mirror image but since this occurs rapidly at room temperature,
these molecules are considered to be achiral under standard conditions.
Again, strictly speaking the chair conformation of 1,2-disubstituted cyclohexanes is chiral
but since both enantiomers are present at room temperature, the molecule is considered to
be achiral (as evident in the flat picture). One enantiomer ring flips into its mirror image.
So, A and B of the 1,2-cis-derivatives are called conformational enantiomers.
17
CH. 7
And, as mentioned, for the 1,2- and 1,3-cis-cases the two substituents must be the same;
otherwise the molecule is achiral.
Molecules that have chirality centers and double bonds will exist as diastereomers.
CH3
H
C C
H
C CH3
HO H
A
(2R,3E)-3-penten-2-ol
H
CH3
C C
CH3
C
HO
H
H
B
(2S,3E)-3-penten-2-ol
18
A and B are enantiomers.
CH. 7
CH3
H
C C
H
C CH3
HO H
A
H
H
A and C are diastereomers. They are nonsuperimposable and they are not mirror mages.
C C
C CH3
HO H
CH3
C
B and C are also diastereomers.
(2R,3Z)-3-penten-2-ol
(2R,3E)-3-penten-2-ol
Reactions that Produce Diastereomers
Look at he addition of bromine to 2-butene.
CH3CH CHCH3 +
Br2
CH3
*
*
CH CH CH3
Br
Br
Two chirality centers are produced. Four stereoisomers are possible but this is a meso
compound, so we see only three.
Recall that this reaction is stereospecific: we see only anti-addition. So cis-2-butene gives
a different product(s) than trans-2-butene.
Br
H
C C
CH3
CH3 Br
H
Br CH
3
+
C C
CH3
H
Br
H
H
CH3
C C
CH3 Br H
+
Br
trans-2-butene
Br
Br
CH3
H C C
H
Br
CH3
CH3
CH3 Br
C C
H
H
Br
H C C
H
Br
These molecules are the same due
to the plane of symmetry. This is the
meso form.
CH3
Br
CH3
CH3
C C
H Br H
CH3 Br CH
3
+
C C
H
H
Br
CH3
+
Br
cis-2-butene
Br
Br
CH3
CH3
C C H
Br
2R, 3R
H
Br
CH3
CH3
C C H
Br
2S, 3S
H
19
These molecules are enantiomers.
CH. 7
Diastereomers have different physical properties, as we mentioned, as can react at different
rates in chemical reactions since one side of the molecule or one angle of approach may be
more hindered or less hindered in one diastereomer than in the other one. And, as in the
example below, one diastereomer may be formed faster than another.
H
H
H
CH3
CH3
H2, Pt
CH2
H
CH3
+
H
CH3
CH3
The 1,2-cis-dimethylcyclohexane is formed
faster because approach of the hydrogen is
faster from the less hindered side away from
the methyl group.
32%
68%
Stereospecific reactions
Stereospecific reactions are those that when carried out with stereoisomeric starting
materials gives a product from one reactant that is a stereoisomer of the product from the
other reactant. The example that we just saw, addition of bromine to alkenes is
stereospecific. Cis-2-butene gives one set of products while its stereoisomer, trans-2butene gives a different set of products and these two sets of products are stereoisomers of
each other. Reactions that involve syn-addition, such as hydroboration, or anti-addition,
such as bromination or bromohydrin formation, are stereospecific.
Stereoselective reactions
Stereoselective reactions are those in which one starting material gives more of one isomer
when two or more are possible. The hydrogenation reaction we just saw above is an
example of a stereoselective reactive reaction. Reactions are stereoselective when the
approach of the reagent is less hindered from one side of the substrate.
Resolution of Enantiomers
The enantiomers of a racemic mixture can be separated in a variety of ways. (Note,
racemic mixture comes from the Greek racemus = bunch of grapes since the original
studies on enantiomers was done on tartaric acid, a naturally chiral molecule that is very
cheap and readily available since it is a by-product of wine making from grapes.)
One very important strategy for separating the two enantiomers of a racemic mixture is to
convert the enantiomers temporarily to diastereomers. Diastereomers have different
physical and chemical properties and so can be separated by the normal means such as
fractional crystallization. Once the two diastereomers have been separated, they can then
be converted back into their pure enantiomers. This works particularly well with chiral
carboxylic acids and chiral amines, which will form diastereomeric salts on simple mixing.
An example involving a racemic mixture of 1-phenylethylamine and (S)-malic acid is
20
CH. 7
shown below. The racemic mixture is mixed in solution with enantiomerically pure (S)malic acid. A mixture of diastereomers is formed which can then be separated by
fractional crystallization. Treatment of the separated diastereomers with base will
deprotonate the amine and allow the carboxylate salts to be extracted into the aqueous
layer, leaving the pure, resolved, enantiomerically pure amine in the organic layer.
O
C OH
O
+
2 HO C
H CH2 C OH
+
H C
H2N
CH3
(S)-malic acid
+
NH2 C
H
CH3
NH3 C
H
CH3
H C
NH3
CH3
racemic mixture of 1-phenylethylamine
O
O
C O
C O
O
HO C
H CH2 C OH
O
HO C
H CH2 C OH
mixture of diastereomeric salts
separate by fractional
crystallization
1.treat with NaOH
H C
H2N
CH3
2. extract amine
into organic
solvent
pure (R)-1phenylethylamine
(S)-malic acid stays in
aqueous layer.
H C
NH3
CH3
and
O
NH3 C
H
CH3
O
C O
C O
O
HO C
H CH2 C OH
O
HO C
H CH2 C OH
1.treat with NaOH
2. extract amine
into organic
solvent
NH2 C
H
CH3
(S)-malic acid stays in
aqueous layer.
pure (S)-1phenylethylamine
Stereoregular Polymers
In many polymerization reactions, a new chirality center is created. For example, in the
polymerization of propene, every third carbon has a chirality center. With the
development of the Zeigler-Natta catalyst, it was possible to synthesize stereoregular
polymers. In the case of polypropylene there are two possible stereoregular polymers,
21
CH. 7
isotactic and syndiotactic. In the isotactic polymer, all the CH3 groups are oriented in the
same direction (i.e. all cis) and in the syndiotactic polymer, the CH3 groups alternate front
and back (i.e. all trans to each other). In atactic polypropylene that is no specific
orientation of the CH3 groups. They are oriented randomly.
CH3 HCH3 H CH3 HCH3 H
isotactic
CH3 H HCH3CH3 H H CH
3
syndiotactic
H CH3H CH3H
H
CH3
H
CH3 CH3 CH3H
atactic
The stereoregularity of the polymer has a great effect on its properties. Stereoregular
polymers allow for efficient packing. Atactic polymers do not associate strongly and have
lower densities and lower melting points than stereoregular polymers.
Chirality Centers Other Than Carbon
Molecules with atoms other than carbon can form chirality centers. Nitrogen with three
substituents has the potential for being chiral since we can consider the lone pair as a
fourth substituent. Nitrogen chirality centers, however, under rapid inversion at room
temperature and are therefore achiral.
H
N
CH3
H
r.t.
N
CH3CH2
CH2CH3
CH3
Phosphorus compound undergo much slower inversion. Tri-coordinate sulfur compounds
do not invert easily and are chiral.
O
CH3CH2
S
chiral
CH3
22