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CH. 4
1
Chapter 4 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.
Stereoisomers are isomers that have the same connectivity but different arrangements of the atoms in space. An example of stereoisomers that we have already seen the cis and trans isomers.
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 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. 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.
C2H6O CH3 O CH3diethyl ether
CH3CH2 OHethanol
C CCH3
H
CH3
H
cis-2-butene
C CCH3
H
H
CH3trans-2-butene
CH. 4
2
For an object to have the property of chirality that 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 Non-chiral Hands socks gloves spoon Feet piece of paper Shoes 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.
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 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. 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.
HC
BrCl
F
carbon that has 4 different sustituents
HC
BrCl
F
enantiomer of the molecule on the leftenantiomer of the
molecule on the right
CH. 4
3
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.
HC
BrCl
F
carbon that has 4 different sustituents
HC
BrCl
F
enantiomer of the molecule on the left
enantiomer of the molecule on the right
A B
rotate 180°HC
BrFCl
The Cl and F are swiched in B; no amount of rotation can make the molecules the same.
B
mirror
HC
BrClCl
HC
BrCl
ClC
rotate 180°HC
BrClClD
Here, C and D are mirror images but they are superimposable, which indicates that C and D are the same molecule. It is achiral and does not have an enantiomer.
D
CO
CH
CH3
H
Hchirality center
C
CH3
H CH2
CH3
chirality center
CH. 4
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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.
CH
CH3
BrCl
plane that cuts the molecule in two; here the two sides are not the same; the molecule therefore is chiral.
chiral
CH
CH3
ClCl
plane that cuts the molecule in two; here the two sides are the same; the molecule therefore is achiral.
achiral
C
CH3
H CH2
CH3
plane that cuts the molecule in two; here the two sides are not the same; the molecule therefore is chiral.
chiral
C
CH3
H CH2
CH3achiral
plane that cuts the molecule in two; here the two sides are the same; the molecule therefore is achiral.
Normal light made up of the superposition of many electro-magnetic waves travelling in all possible planes.
Normal light as seen from the direction of the propagation of the EM waves. Vectors are in all possible orientations.
CH. 4
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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.
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.
Polaroid filter with tiny slit that allows light travelling only in one plane to pass through.
Plane polarized light.
Plane polarized light.
CH3
CCH2CH3OH
H
angle of rotation = - α
Plane polarized light.
CH3
CCH2CH3OH
H
angle of rotation = - α
CH. 4
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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.
A racemic mixture is a equal, 50:50 mixture of each enantiomer. A racemic mixture is optically inactive. 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.
Plane polarized light.
CH3
CCH2CH3OH
H
angle of rotation = - α
CH3
CCH2CH3H
OH
angle of rotation = +αangle of rotation = - α net rotation =
0
specific rotation [α] = 100 α
cll = length of polarimeter in dmc= concentration in grams per 100 mL of solution
α = observed rotation
CH3
CCH2CH3OH
H
CH3
CCH2CH3H
OH
(+)-rotator or (-)-rotator?
CH. 4
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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 C-OH bond.
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). Rules for assigning Priority 1. The higher atomic number takes priority: I > Br > Cl > F > O > N > C > H 2. When two atoms directly attached to the chirality center are the same, compare the atoms attached to them, again based on atomic number. List the atoms in order from heaviest to lowest. Precedence is determined at the first point of difference.
For:
CH3 CHOH
CH CH2
[α] = +33.2°
H2, Pd CH3 CHOH
CH2 CH3
[α] = +13.5
CH3
CH2CH3Cl
H
The clorine is the heaviest. Then compare the other two substitutents. The ethyl group outranks the methyl group: list the substituents in order from highest to lowest. Look for the first point of difference.
1 2
3
-C(C,H,H Vs.C(H,H,H,H
carbon is heavier than H at the first point of differenceSo, this molecule has the S configuration.
CH. 4
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If there is a tie, go on to the next substituent, comparing these atom by atom, looking for the first point of difference.
If there is a double bond, count that atom twice; if there is a triple bond, count that atom three times.
The correct, full name of the following molecule is (S)-2-butanol. The mirror image of this molecule will be its enantiomer and will have the R-configuration.
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
CH2
CH2CH3Cl
H
OH
1
2
3
Here, the CH2OH group outranks the ethyl group:-C(O,H,H) vs. -C(C,H,H,H
heavier
So this molecule is R.
CH3
CH2CH2HC
HCl
CH3CH3
3
1 2
The isoproyl group is heavier than the ethyl chloro group and the methyl group is the lightest. Compare the substituents attached to the carbon atoms attached to the chirality center. Note that the chlorine atom is not counted because the rule says to look at only the first point of difference.
-C(C,C,H)isopropyl group chloroethyl group
-C(C,H,H)
methyl group-C(H,H,H)
1 2 3This molecule is S.
CH3
CHHC
H
CH3CH3
CH2
-C(H,H,H)
-C(C,C,H)
-C(C,C,H)
1. Compare the first atoms attached to the chirality center. They are all carbons, so order the substituents as before.In this example we find that two of the carbons Marked by #) attached to the chirality center have the same priority.2. Move out along the chain, comparing the next substituents, in this case carbons marked by*
#
#**
-C(C,H,H)
-C(H,H,H)Higher
lower
12
3
Therefore, this molecule has the R configuartion.
CH3
CCH2CH3HO
H
CH3
CCH3CH2
HOH
mirror
12
3 Moving from hisghest to lowest is clockwise, so this is the R configuraton as expected.
(S)-2-butanol (R)-2-butanol
CH. 4
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the assignment when the smallest group is not pointing away and you get R for the configuration, the correct answer is then S.
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.
It does not matter which substituents you rotate. You will get the correct answer for you assignment.
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
CCH2CH3H
Cl
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.
1
2
3In 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.
CH3
CCH2CH3H
ClCH3
CClCH3CH2
H
Hold the methyl group the same, and rotate the other three so that the hydrogen points away.
12
3
Clockwise = R
CH3
CCH2CH3H
Cl
ClC
CH2CH3CH3H
1
23
Clockwise = R
CH3 H
HH
CCH2CH2
CH3 H
CH
CCCC
HH
1' 2'12
C1' and C1 are the same so go on to compare C2' and C2.
C2'(C,C,H)
C2(C,H,H)
1
2
3
larger/heavier substitutent
Clockwise = R
CH. 4
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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.
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).
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
H
F
Br Cl
going back into the page
coming out of page
C
H
F
ClBr
CH3
CH2CH3
HO H C
CH3
CH2CH3
HOHrotate 180°around vertical axis
H OHCH3
CH2CH3
1
2
3
Clockwise = R
CH. 4
11
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.
The alkene is flat and the peroxy acid can attack from the top side to form one enantiomer or from the bottom side to form the other enantiomer. Both reactions are equally likely.
Another example:
CH3CH CH2 + CH3 CO
OOHperoxy acetic acid
CH2 CO H
CH3chiralachiral
achiral
C CHH
HCH3
CH3 CO
OO
H
C CO HH
HCH3top attack C C
HH
HCH3
O
O
!
2
3
Clockwise = R enantiomer
C CHH
HCH3
CH3 CO
OO
H
bottomattack
C CHH
HCH3 O
C CHH
HCH3 O
O
1
23Counterclockwise = S enantiomer
CH3CH2CH CH2
1-butene, achiral
H3O+
H2OCH3CH2CH CH3
OH
2-butanol, chiral, racemic mixture
CH. 4
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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.
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.
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.
C CHH
HCH3CH2
H OH
H
H2O C CH2
H
CH3CH2
HOH
H
OH
H
C CH3
H
CH3CH2
OH H
top
OH
H
C CH3
H
CH3CH2
OH
bottom
C CH3
H
CH3CH2 OHH
OH
H
C CH3
H
CH3CH2 OH
2
1
3
S-2-butanol, 50%
1
23
R-2-butanol, 50%
C CHHO
CHOH
CH31 2 3 4* *
OHO
C2
C3
H
OHC
OHCH3
H
O
HO
A
C2
C3
H
CHO
H
CH3
O
OH
HO
B2S, 3S 2R, 3R
CH. 4
13
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.
For chirality center C3:
Now, assign the priorities for C2 and C3 of the enantiomer B. C2 should be R and C3 should be R as well.
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.
C2
C3
H
OHC
OHCH3
H
O
HO
1
C(O,C,H)
C(O,O,O)
larger
smallest
2
3
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 highest to lowest, so the C2 center is S.
C2
C3
H
OHC
OHCH3
H
O
HO
Treat as one of the four substitutents attached to C3. It has priority over the CH3 group.
1
2
3C(H,H,H)
C(O,C,H)Movement from highest to lowest priority is counterclockwise, so the C3 center is S.
C2
C3
H
CHO
H
CH3
O
OH
HO
1
3
2Movement from highest to lowest is clockwise, so C2 has the R configuration.
C2
C3
H
CHO
H
CH3
O
OH
HO1
2
3
Movement from highest to lowest at C3 is clockwise, so C3 has the R oinfiguration.
CH. 4
14
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.
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?
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
C2
C3
H
CHO
OH
H
OOH
CH3
C
C2
C3
H
OHC
CH3
H
HO
OHO
D2S, 3R 2R, 3S
C2
C3
H
CHO
OH
H
OOH
CH3
C2
C3
H
OHC
OHCH3
H
O
HO
2S, 3S
A C
2R, 3S
C2
C3
H
OHC
OHCH3
H
O
HO
2S, 3S
A
C2
C3
H
OHC
CH3
H
HO
OHO
D
C2
C3
H
CHO
H
CH3
O
OH
HO
B2R, 3R
C2
C3
H
CHO
OH
H
OOH
CH3
C
2R, 3S
C2
C3
H
CHO
H
CH3
O
OH
HO
B2R, 3R
C2
C3
H
OHC
CH3
H
HO
OHO
D
2S, 3R
2S, 3R
diastereomers diastereomers
diastereomersdiastereomers
CH. 4
15
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. 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.
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.
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.
CH3 CHHO
CHOH
CH31 2 3 4
* * two chirality centers,; expect 22 stereoisomers
C2
C3
H
CH3HO
OHCH3
HA
C2
C3
H
OHCH3
H
CH3HO
B2S, 3S 2R, 3R
CH. 4
16
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.
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.
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.
C2
C3
H
OHCH3
OH
HC
CH3
2R, 3S
C2
C3
H
CH3HO
CH3
H
HO
2S, 3RD
rotate 180° C3
C2
H
OHCH3
OH
H
CH3
same as C
in the plane of the page
CH3CH3
HH plane of symmetry
achiral
HCH3
CH3H
HCH3
CH3H
3-dimensional drawing
flat drawing
cis-1,2-dimethylcyclopropane
CH3CH3
HH
trans-1,2-dimethylcyclopropane
No plane of symmetry;chiral
CH. 4
17
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.
ClCl
HH
cis-1,2-dichlorocyclobutane
Cl
Cl
H
Hplane of symmetry; achiral
HCl
ClH Cl
H
Cl
HNo plane of symmetry; chiral
trans-1,2-dichlorocyclobutane
