1. Chapter 5 Stereoisomerism and Chirality

2. Donald Cram Source: Michigan State University, Department of Chemistry

3. Chapter 5 Skip p. 388-389 Read p. 475-477 for interest
Don’t memorize names of sugars (p ) or amino acids (p )

4. Chapter 5 Problems 1 a, b, c, e, g, m 2
4 a, b, c, d, f, g, i 5 a, b, e 11, 12
13 b, c, d a, b, d , 59

5. Sect 5.1: Symmetry and Chirality
If a molecule is superimposable on its mirror image, it is achiral (non-chiral).
If a molecule is not superimposable on its mirror image, it is chiral. Enantiomers are isomers that are non-superimposable mirror-images

6. Sect 5.2: Enantiomers
Four different atoms are attached to a chiral carbon atom. Mirror images are non-superimposable.
rotate
this molecule
is chiral
note that the fluorine
and bromine have been
interchanged in the
enantiomer

7. Stereocenters The previous molecule has a stereocenter, and is chiral.
A stereocenter is an atom, or a group of atoms,
that can potentially cause a molecule to be chiral.
stereocenters - can
give rise to chirality

8. An achiral molecule A carbon atom with three identical groups is
achiral! There is a symmetry plane in the plane
of the paper.

9. plane of
symmetry C l C l C l C l B r C l B r C l
side view
edge view

10. Another achiral molecule
This molecule has a plane of symmetry in the plane of the paper.
These two structures are superimposable!

11. Look for carbon atoms with four different groups!
Sect 5.3: identification of stereocenters
Look for carbon atoms with four
different groups! * *

12. * * *

13. Finding stereocenters*
plane
( indicates no stereoisomers )

14. Finding stereocenters* * * * * *
The cis isomer (two methyl groups) - achiral!
Thetrans isomers has two stereocenters!

15. Finding stereocenters* * * * * * * *
n = 8
28 = 256

16. Sect. 5.4 Properties of Enantiomers
Enantiomers interact differently with polarized light.
Enantiomers have equal magnitude, but opposite signs of rotation
Most other properties are identical.
Odor may be different!!

17. Sect Polarimetry
dissolve sample in a solvent and put sample into polarimeter
obtain sign of rotation
value depends on concentration and path length.

18. Optical Activity

19. (+)- d- Dextrorotatory (-)- l- Levorotatory TYPES OF OPTICAL ACTIVITY
new
older
(+)- d-
Dextrorotatory
Rotates the plane of plane-polarized light
to the right.
new
older
(-)- l-
Levorotatory
Rotates the plane of plane-polarized light
to the left.

20. Specific Rotation [a]D
This equation corrects
for differences in cell
length and concentration. a
[a]D = t cl
Specific rotation calculated
in this way is a physical
property of an optically
active substance.
a = observed rotation
You always get the same
c = concentration ( g/mL )
[a]D t
value of
l = length of cell ( dm )
D = yellow light from sodium lamp
t = temperature ( Celsius )

21. ENANTIOMERS HAVE EQUAL VALUES BUT OPPOSITE SIGNS OF ROTATIONSW W
Enantiomers C C Y X X Y Z Z
(+)-numbero
(-)-numbero
dextrorotatory
levorotatory
The numbers are the same, but have opposite signs. All other physical properties are IDENTICAL.

22. Racemic mixture [a]D = 0o an equimolar (50/50) mixture of enantiomers
the effect of each molecule is
cancelled out by its enantiomer

23. Sect Configuration
Arrangement in space of atoms or groups around the stereocenter of the molecule
Enantiomers have different configurations.

24. Sect. 5.7 Specification of Configuration: Cahn-Ingold-Prelog rules

25. Sir Christopher Ingold
Source: Michigan State University, Department of Chemistry

26. Specification of Configuration
SEQUENCE RULE 1: priority depends on the atomic numbers of the 4 atoms attached to the stereocenter; atom with higher atomic number receives the higher priority.
If two atoms are isotopes of the same element, the heavier isotope is assigned the higher priority.

27. R S

28. Specification of Configuration
SEQUENCE RULE 2: If the relative priority of two groups cannot be decided by Rule 1, it shall be determined by a similar comparison of the next atoms in the groups (and so on, if necessary), working outward in ranks from the stereocenter.

29. Specification of Configuration
SEQUENCE RULE 3: A doubly-bonded atom A is treated as if there were two
C-A single bonds Priorities in the expanded representations are assigned on the basis of Rule 2.

30. Remember!
The atoms shown in parentheses (the duplicate representations) do not exist!
They are written only for purposes of assigning priorities.

31. More...
A triply-bonded atom A is treated as if there were three C-A bonds, as in:

32. Corollary of Rule 3
If no other distinction can be made, a real atom outranks a “fictional” atom.
NOTE CAREFULLY: This exception is used only as a last resort! You will rarely see this happen.

33. Sect. 5.8: Compounds with more than one stereocenter: 3-Chloro-2-butanol

34. Fischer formulas (Sect. 5.9)
3- Chloro-2-butanol:
Fischer formulas (Sect. 5.9)
pair of enantiomers-1
pair of enantiomers-2
diastereomers

35. How Many Stereoisomers Are Possible?
maximum number of stereoisomers
= 2n,
where n = number of stereocenters
sometimes fewer
than this number
will exist
For the previous example; two stereocenters = 4

36. Determining the number of possible stereoisomers* *
22 = 4 stereoisomers * * *
23 = 8 stereoisomers

37. 2,3-Dichlorobutane mirror image is identical S R meso diastereomers S
enantiomers

38. 2,3-Dichlorobutane
meso

39. Fischer formulas (Sect. 5.9)
2,3-Dichlorobutane:
Fischer formulas (Sect. 5.9)
Meso! “Pair” becomes one!
pair of enantiomers
diastereomers

40. Tartaric Acid (from fermentation of wine): There are three stereo-isomers
ALSO FOUND
(as a minor component)
[a]D = 0
meso -tartaric acid

41. Sect 5.9: Fischer Formulas

42. EVOLUTION OF THE FISCHER PROJECTION C H O O H H O C H C H O C H O H O
“Sawhorse” Projection
EVOLUTION OF THE
FISCHER PROJECTION C H O O H H O C 2 H
Fischer Projection
Orient the
main chain
vertically with
the most
oxidized group
at the top. C H O C H O H O H H O H C H 2 O C H O H 2
Substituents will
stick out toward
you like prongs
Main chain bends
away from you

43. OPERATIONS WITH FISCHER PROJECTIONS
Mirror images (enantiomers)
are created by switching
substituents to the other side.
enantiomers
diastereomers
All stereocenters must be
switched to get an enantiomer
(the mirror inverts them all).
If you switch only one of the stereocenters,
but not both, you get a diastereomer.

44. . Legal operations with Fischer formulas Rotation by 180o in the plane
of the paper does not change
the molecule.
No other rotation is
allowed.

45. This molecule is meso! There isn’t an enantiomer
The molecule does not have
an enantiomer because it
is not chiral.
It had a plane of symmetry,
rendering it a meso molecule.
plane of symmetry

46. R CHO H H O H OHC C H O H C H O H O H R O H CHO R HOCH2 CHO H O H H C
To determine the configuration (R/S). Make two “switches”. Place the priority 4 group in one of the vertical positions. 2 4
#4 at top position
CHO H 1 H O H
OHC C H O H 4 2 2 3 C H O H O H R 2 3 1
Both H’s are in
back = same
result
alternatively: 1 2 O H R
CHO R 3 2 1
HOCH2
CHO H O H 4 H C H O H 2 4 3
#4 at bottom position

47. Draw all of the stereoisomers!

48. Any more??
NO!

49. There are only four stereoisomers!

50. Sect 5.10: cyclic compounds

51. * stereoisomers (max) : 20 = 1 21 = 2 21 = 2 plane
( indicates no stereoisomers )
stereoisomers (max) :
20 = 1
21 = 2
21 = 2

52. * * * * * * stereoisomers (max) : 22 = 4 22 = 4 22 = 4 plane
( but only if cis )
stereoisomers (max) :
22 = 4
22 = 4
22 = 4

53. 1-Bromo-2-chlorocyclopropane
note that the cis/trans isomers are also diastereomers R S R S
cis
enantiomers
diastereomers R R S S
trans
enantiomers

54. 1,2-Dibromocyclopropane
mirror image identical B r Br Br B r
cis
meso
diastereomers B r B r
trans Br Br
enantiomers

55. 1-Bromo-2-chlorocyclohexane
cyclohexanes may be analyzed using planar rings
cis
enantiomers
diastereomers
trans
enantiomers

56. 1,2-dichlorocyclohexane
mirror image identical
cis
meso
diastereomers
trans
enantiomers

57. Interchanges: See problem 59
ONE STEREOCENTER
CH3
CH3 C C H H Br F F Br
enantiomer
Odd:
…etc interchanges = enantiomer
Even:
etc interchanges = original compound

58. You can use interchanges to answer this question! Next slide.
PROBLEM: ARE THESE IDENTICAL
OR ARE THEY ENANTIOMERS?
CH3 H C C
CH3 H F Br Br F
You can use interchanges to answer
this question! Next slide.

59. C
CH3 Br F H
ENANTIOMER
SAME 1 2 3

60. They are enantiomers! Three switches = enantiomers
Back to the problem: are these identical
or are they enantiomers?
CH3 H C C
CH3 H F Br Br F
They are enantiomers! Three switches = enantiomers

61. MORE THAN ONE STEREOCENTER
To form an enantiomer you must
interchange all of the stereocenters.
enantiomer
two stereocenters
two interchanges

62. If a compound has more than one stereocenter,
and you only interchange one of them,
you willI form a distereomer.
two stereocenters
one interchange
diastereomer

63. Sect 5.11: Stereocenters other than carbon

64. Sect 5.12: Other types of chirality

65. Sect 5.13: Resolution (Skip)

66. Sect : carbohydrates

67. Emil Fischer
Source: Michigan State University, Department of Chemistry

68. Glyceraldehyde (an aldotriose)

69. Notice that the D enantiomer has the R configuration in the R/S sytem.

70. The Aldotetroses Notice that the D enantiomers are (-)!
Notice that the L
Enantiomers are (+)!

71. There is no relationship between
D or L and the sign of rotation in
a polarimeter!
D can either be (+) or (-)
L can either be (+) or (-)

72. Fig. 5-27: the aldopentoses

73. There are four natural pentose sugars. They have the D configuration.
OH points to right! D
OH points
to right! D
OH points to right D
OH points
to right D

74. Fig 5-28 “natural” D-aldohexoses

75. The “unnatural” L-aldohexoses

76. Sect 5.17: natural products and amino acids

77. (R) (S) Amino acids belong to the L-series. Can
You assign the R or S configuration to the
following Fischer structure of Alanine?
Same!
enantiomer
(R)
The original structure is, therefore,
(S)

78. These compounds are enantiomers! They have
identical physical properties, except for odor
and rotation of light in a polarimeter!