Chapter 5 Stereochemistry - University of Northern … 5 Stereochemistry. ... Racemic mixtures ......
Transcript of Chapter 5 Stereochemistry - University of Northern … 5 Stereochemistry. ... Racemic mixtures ......
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Chapter 5Stereochemistry
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ÚThe study of the 3-dimensional structure of molecules
ÚStereoisomers: have the same bonding sequence, but differ in the orientation of their atoms in space
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Chirality and Enantiomers (5-2)ÚChiral objects: are those that have
right-handed and left-handed formsThe chirality of an object can be determined by looking at its mirror image
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image
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ÚA chiral object has a mirror image that is different from the original object (it is non-superimposable)
Figure 5-1
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Chiral Achiral: object thatis not chiral
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Molecules can either be chiral or achiral
Figure 5-6Chiral compound
Figure 5-7 Achiral compound
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In B: the original molecule and its mirror image are non-
A BFigure 5-3
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In B: the original molecule and its mirror image are non-superimposable (no plane of symmetry)
Chiral
In A: the original molecule and its mirror image are superimposable (plane of symmetry)
Achiral
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ÚTwo molecules are said to be superimposable if they can be placed on top of each other and the 3-D position of each atom on one molecule coincides with the equivalent atom on the other molecule
ÚMolecules that are non-superimposable
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ÚMolecules that are non-superimposable mirror images are called:
Enantiomers
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Nomenclature of Asymmetric Carbon Atoms(5-3)
ÚFor a carbon atom to be chiral, it must have 4 different substituents. In this case the carbon atom is called:
Figure 5-6
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– Chiral carbon– Chiral centre– Asymmetric carbon– Stereocentre
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ÚAsymmetric carbons are marked with a *
Figure 5-4
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ÚGeneralizations
– If a compound has no chiral carbon, it is usually achiral
– If a compound has just one chiral carbon, it is always chiral
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carbon, it is always chiral
– If a compound has more than one chiral carbon, it may or may not be chiral
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Rule of thumb:
Any molecule that has an internal plane of symmetry cannot be chiral, even though it may contain chiral carbon atoms
Figure 5-3
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ÚBecause enantiomers are actually 2 distinct molecules with different properties, a notation (nomenclature) system for naming configurations of chiral carbon atoms was proposed by:
Cahn-Ingold-Prelog
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Cahn-Ingold-Prelog
– It assigns a letter (R) or (S) to the chiral carbon
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Procedure to assign “R” and “S”
– Assign a priority to each group attached to the chiral centre
– 1) highest priority2) next highest3) next highest
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3) next highest4) lowest priority
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Úhigher atomic # (or atomic mass) gets the highest priority (this is for the atom directly attached to the chiral carbon)
– Question:Assign the priorities for all the groups in this molecule.
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molecule.
C
CH3
FH
NH2
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Ú In case of a tie, use the next atom along the chain as tie breaker– Question:
Assign the priorities for all the groups in this molecule.
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ÚTreat double and triple bond as if each were bonds to a separate atoms
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– Question:Assign the priorities for all the groups in this molecule.
H
OHO
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CH3CH2NH2
H
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Ú Using a molecular model or a 3-D drawing, place the group with the lowest priority in the back (pointing away from you)Look at the molecule along the bond from the chiral centre to the lowest priority group
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Ú Draw an arrow going from the first priority group to the second and third
Ú If the arrow follows a clockwise rotation: notation “R”
Ú If the arrow follows a counterclockwise rotation: “S”
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ÚWhen the lowest priority group is not located in the back, a rotation may be needed in order to facilitate the determination of the configuration.
OH1
C
CH2CH33
rotate
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CCH3CH2CH2
H
CH2CH32
3
4
CCH3CH2CH2
OH
H
1
2 4rotate
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ÚQuestion:Predict the configuration. C
CH3HO
CH2CH3
H
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Ú Tricks:It is often difficult to rotate an entire molecule. In this case, use either of the two tricks below.(A) if you look through the 4 to C priority bond (ie the opposite to what you should do), simply reverse your answer since you are looking at the mirror image of the original molecule.
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C
CH3HO
CH2CH3
H
12
3
4
Looking through 4-C, a S configurationis obtained….therefore, the correctanswer is the mirror image of this one,ie: R
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(B) Or you can use your right (R) or left (S) hand to determine the configuration. Use your thumb to point through the C-4 bond, the tip of your other fingers are now the arrow head. If you use your right hand = R configuration, left hand = S configuration.
CH3
3 Using your left hand, your fingertips Go from 1 to 3 (wrong hand)
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C
CH3HO
CH2CH3
H
12
4
Go from 1 to 3 (wrong hand)
With your right hand, the fingertips Go from 1-2-3 (correct).
R configuration
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Practice QuestionsÚWhat is the configuration of the chiral
centers in the following compounds?
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H
Br COOH
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Ú What is the configuration of the chiral centers in the following molecules?
O OH
Br
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Br
OH
CO2Me
CO2Me
SCH3CH3O
OCH3
CH3S
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Optical Activity (5-4)
Ú Chiral compounds have a physical property that other compounds do not have:
Ú Optical activity: it is the ability of chiral molecules to rotate the plane of polarized light.
Ú Molecules that can rotate the plane of polarized
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Ú Molecules that can rotate the plane of polarized light are:
Optically activeInstrument used is: Polarimeter
Figure 5-13
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ÚRotations of the plane that are clockwise are:
Dextrorotatory and (+)ÚRotations of the plane that are
counterclockwise are:Levorotatory and (-)
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NOTE: These signs and terms have nothing to do with “R” and
“S”notations
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ÚSpecific Rotationsymbolized by (α): measure of optical activity. This is a physical property.
α = rotation observedc = concentration of the sample (in g/mL)l = length of the cell (in decimeter)T = temperature
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ÚDepends on: – concentration of sample– Path length of the cell used– Temperature– Wavelength of light source
T = temperatureλ = wavelength of light source
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Practice QuestionÚThe observed rotation of 2.0g of a compound in
50mL of solution in a polarimeter tube of 20cm long is +13.4o. What is the specific rotation of the compound?
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ÚWhat would be the optical rotation of the enantiomer of this compound?
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Racemic mixtures (5-6)
Suppose we have an equal amount of – (+)-2-butanol [α]D = 13.5o
– (-)-2-butanol [α]D = - 13.5o
Ú The observed optical rotation for this mixture will be zero
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be zero
Ú An equal mixture of 2 enantiomers is called:
“racemic mixture” or “racemate” and it is optically inactive
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Enantiomeric excess (ee) (5-7)This is the measure of optical purity
ÚIf a mixture is neither a racemate nor an optically pure compound, the optical purity must be determined
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optical purity must be determined
optical purity = ee = [experimental rotation]
[rotation of pure enantiomer]x 100
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ÚPractice QuestionA mixture of (+)-2-butanol and (-)-2-butanol gives an optical rotation of 9.54o. What is the optical purity of the mixture?
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Ú If you know the amounts of each compounds present in the mixture, it is not necessary to know the optical rotation of the pure enantiomer in order to calculate the ee.
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ee = R - SR + S
=(+) + (-)
(+) - (-)
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ÚExampleA mixture is composed of 6g of (+)-2-butanol and 4g of (-)-2-butanol. What is the ee of the mixture?
ee = 6 - 46 + 4
= 20%
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6 + 4ÚExample 2
What is the ee of a mixture made of R/S = 5/1?
ee = 5 - 15 + 1
= 67%
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Practice QuestionÚ (+)-Mandelic acid has a specific rotation of
+158o. What would be the observed specific rotation of the following mixtures?(a) 25% (-)-isomer and 75% (+)-isomer(b) 50% (-)-isomer and 50% (+)-isomer
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(b) 50% (-)-isomer and 50% (+)-isomer
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Fisher Projections (5-10)looks like a cross with the horizontal bonds projecting out towards the viewer and the vertical bonds projecting away.In Fisher projections, only the central carbon atom is in the plane
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Rules for drawing Fisher Projections
– Longest carbon chain is always vertical– Most oxidized carbon is on top– Rotation of 180o do not change the
molecule (must be kept in the plane)
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molecule (must be kept in the plane)
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ÚPractice QuestionDetermine the configurations of the chiral centers in these compounds.
H
CH3Br
Br H
CHO
H
CH OH
OHCH2CH3
H Br
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Br H
CH3CH2OH CH3
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ÚDraw the Fisher projections of the two enantiomers of :
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Diastereomers (5-11 to 5-13)
Summary of Isomerism
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ÚDiastereomers:Configurational diastereomer with chiral centers must have at least 2 chiral carbons
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ÚNOTE diastereomers have at least 2 chiral carbons, one of which has the same configuration
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ÚUsually, the number of possible stereoisomers for a given molecule is equal to: # stereoisomers = 2n
where n = # of chiral carbons
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ÚPractice QuestionHow many stereoisomers? Draw all possibilities.
Br
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ÚUsing the answer from the previous examples, pair the enantiomers and the diastereomers.
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ÚHow many stereoisomers are expected? Can you draw them all? Can you pair them as enantiomers and diastereomers?
OH
COOMe
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COOMe
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ÚExceptionSometimes this general rule does not give the exact number of stereoisomers
Example: 2,3-dibromobutane
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ÚResult: only 3 stereoisomers2 enantiomers (optically active if pure)1 diastereomer (always optically inactive since it is achiral)
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ÚThis achiral diastereomer is called:Meso diastereomer (compound)
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ÚMeso compoundsAre achiral compounds with chiral carbons. These compounds are also optically inactive.
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Practice QuestionÚ Which of the following compounds has a stereoisomer
that is a meso compound?(a) 2,4-dibromohexane(b) 2,4-dibromopentane(c) 2,4-dimethylpentane(d) 1,3-dichlorocyclohexane
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(d) 1,3-dichlorocyclohexane(e) 1,4-dichlorocyclohexane(f) 1,2-dichlorocyclobutane
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Study ProblemsÚ 5-26 For each structure:
Star (*) the asymmetric atomsassign a configuration (R or S) label structure as chiral, achiral and mesoCH2Br
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CH2BrBrH
CH2BrBrH
Br
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Ú 5-32Calculate the specific rotation of the following sample taken at 25oC using the sodium D line.1.0 g sample is dissolved in 20.0 mL of ethanol. Then 5.0 mL of this solution is placed in a 20.0 cm polarimeter tube. The observed rotation is 1.25o
counterclockwise.
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Ú 5-36Draw all the stereoisomers of 1,2,3-trimethylcyclopentane and give the relationships between them.
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Ú 5-33(+)-Tartaric acid has a specific rotation of +12.0o. Calculate the specific rotation of a mixture of 68% (+)-tartaric acid and 32% (-)-tartaric acid.
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Ú 5-29Convert the following Fisher Projections to perspective formulas.
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