Transcript No Slide Title

Chapter 5
Stereoisomerism
and Chirality
Donald Cram
Source: Michigan State University, Department of Chemistry
http://www.chemistry.msu.edu/Portraits/PortraitsHH_collection.shtml
Chapter 5
• Skip p. 388-389
• Read p. 475-477 for interest
• Don’t memorize names of sugars (p.
460-465) or amino acids (p. 472-474)
Chapter 5 Problems
•
•
•
•
•
•
•
1 a, b, c, e, g, m 2
4 a, b, c, d, f, g, i 5 a, b, e
13 b, c, d
14 - 18
23 - 28
30
38
40, 41
45
48
59
11, 12
20 a, b, d
32 - 36
43
52
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
W
W
C
C
X
Y
Z
Y
X
Z
Sect 5.2: Enantiomers
Four different atoms are attached to a chiral carbon
atom. Mirror images are non-superimposable.
Cl
Cl
rotate
H
F
Br
this molecule
is chiral
Cl
H
F
Br
Br
F
H
note that the fluorine
and bromine have been
interchanged in the
enantiomer
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
An achiral molecule
A carbon atom with three identical groups is
achiral! There is a symmetry plane in the plane
of the paper.
Cl
Br
Cl
Cl
Cl
Cl
Cl
Br
plane of
symmetry
Cl
Br
Cl
Cl
Cl
Cl
Br
Cl
side view
edge view
Another achiral molecule
This molecule has a plane of
symmetry in the plane of the paper.
F
Br
F
Cl
Cl
Cl
Cl
These two structures are superimposable!
Br
Sect 5.3: identification of stereocenters
Look for carbon atoms with four
different groups!
C H 2O H
CH3
C
*
CH 2 C H
CH3
CH3
OH
CH
*
CH3
CH 3
*
**
CH 3
OH
CH
CH 3
Finding stereocenters
O
O
O
C H3
*
*
C H3
plane
( indicates no stereoisomers )
C H3
Finding stereocenters
O
O
O
C H3
*
*
C H3
C H3
*
*
C H3
C H3
*
*
C H3
The cis isomer (two
methyl groups) - achiral!
Thetrans isomers has
two stereocenters!
Finding stereocenters
C H3
*
C H3
*
C H3
*
HO
*
*
C H3
*
*
C H3
*
H
n=8
28 = 256
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!!
Sect. 5.5 Polarimetry
• dissolve sample in a solvent and put
sample into polarimeter
• obtain sign of rotation
• value depends on concentration and
path length.
Optical Activity
TYPES OF OPTICAL ACTIVITY
Dextrorotatory
new
older
(+)-
d-
Rotates the plane of plane-polarized light
to the right.
Levorotatory
new
older
(-)-
l-
Rotates the plane of plane-polarized light
to the left.
Specific Rotation [a]D
t
[a ]D =
a
a
cl
= observed rotation
c = concentration ( g/mL )
l = length of cell ( dm )
This equation corrects
for differences in cell
length and concentration.
Specific rotation calculated
in this way is a physical
property of an optically
active substance.
You always get the same
value of [a]D
D = yellow light from sodium lamp
t = temperature ( Celsius )
t
ENANTIOMERS HAVE EQUAL VALUES
BUT OPPOSITE SIGNS OF ROTATIONS
W
C
X
Z
W
Enantiomers
Y
(+)-numbero
C
Y
Z
X
(-)-numbero
dextrorotatory
levorotatory
The numbers are the same, but have opposite signs. All other
physical properties are IDENTICAL.
Racemic mixture
an equimolar (50/50) mixture of enantiomers
[a]D = 0o
the effect of each molecule is
cancelled out by its enantiomer
Sect. 5.6 Configuration
• Arrangement in space of atoms or
groups around the stereocenter of the
molecule
• Enantiomers have different
configurations.
Sect. 5.7 Specification of
Configuration: Cahn-Ingold-Prelog
rules
1
2
2
C
4
1
C
4
3
R
3
S
Sir Christopher Ingold
Source: Michigan State University, Department of Chemistry
http://www.chemistry.msu.edu/Portraits/PortraitsHH_collection.shtml
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.
F
I
I
C
C
F
Br
Cl
R
Cl
Br
S
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.
Specification of Configuration
SEQUENCE RULE 3: A doubly-bonded
atom A is treated as if there were two
C-A single bonds.
A
C
A
(C )
C
(A )
Priorities in the expanded
representations are assigned on the
basis of Rule 2.
Remember!
The atoms shown in parentheses (the
duplicate representations) do not exist!
They are written only for purposes of
assigning priorities.
More...
A triply-bonded atom A is treated as if
there were three C-A bonds, as in:
C
A
(A )
(C )
C
A
(A )
(C )
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.
Sect. 5.8: Compounds with
more than one stereocenter:
3-Chloro-2-butanol
S
C H3
R
H
C
OH
H
C
Cl
HO
Cl
R
C H3
C H3
R
C H3
S
C H3
C
H
HO
C
H
H
C
OH
C
H
H
C
Cl
Cl
C
H
S
R
C H3
e n a n tio m e rs
S
C H3
e n a n tio m e rs
d ia ste re o m e rs
C H3
3- Chloro-2-butanol:
Fischer formulas (Sect. 5.9)
pair of
enantiomers-1
C H3
C H3
OH
HO
Cl
Cl
C H3
pair of
enantiomers-2
C H3
C H3
OH
HO
Cl
C H3
C H3
diastereomers
Cl
C H3
How Many Stereoisomers
Are Possible?
maximum number of stereoisomers
sometimes fewer
= 2n,
than this number
will exist
where n = number of stereocenters
For the previous example; two stereocenters = 4
Determining the number of possible
stereoisomers
CH3
C H 2O H
OH
C
CH
*H
CH 2 C
CH3
CH3
*
CH3
22 = 4 stereoisomers
CH 3
*
*
CH 3
*
OH
CH
CH 3
23 = 8 stereoisomers
C H 3C H C H C H 3
2,3-Dichlorobutane
Cl Cl
Cl Cl
S
Cl Cl
R
C H3
C H3
H
H
diastereomers
mirror image
C H 3 is identical C H 3
meso
H
Cl Cl
Cl Cl
S
S
H
H
C H3
C H3
H
R
enantiomers
H
C H3
R
C H3
H
2,3-Dichlorobutane
C H3
C H3
C H3
C H3
H
C
Cl
Cl
C
H
Cl
C
H
H
C
Cl
H
C
Cl
Cl
C
H
H
C
Cl
Cl
C
H
C H3
C H3
C H3
e n a n tio m e rs
meso
C H3
2,3-Dichlorobutane:
Fischer formulas (Sect. 5.9)
Meso! “Pair”
becomes one!
CH
CH
pair of
enantiomers
CH
3
Cl
Cl
Cl
Cl
3
CH
3
CH
3
Cl
Cl
Cl
CH
3
diastereomers
3
CH
3
Cl
CH
3
O H Acid
OH
Tartaric
(from fermentation
meso
OC
C O O H are three stereoofH Owine):
There
H
H
isomers
OH OH
HO O C
OH OH
H
COOH
H
(+)-tartaric acid
enantiomers
OH OH
HO O C
meso
COOH
H
H
meso -tartaric acid
OH OH
H
HOOC
COOH
H
(-)-tartaric acid
ALSO FOUND
(as a minor component)
[a]D = 0
OH OH
Sect 5.9: Fischer Formulas
“Sawhorse” Projection
CHO
HOCH2
Orient the
main chain
vertically with
the most
oxidized group
at the top.
EVOLUTION OF THE
FISCHER PROJECTION
OH
H
Fischer Projection
CHO
H
OH
CH2OH
Substituents will
stick out toward
you like prongs
CHO
H
OH
CH2OH
Main chain bends
away from you
OPERATIONS WITH FISCHER PROJECTIONS
C HO
C HO
OH
HO
OH
HO
C H 2O H
Mirror images (enantiomers)
are created by switching
substituents to the other side.
C H 2O H
C H 2O H
enantiomers
OH
HO
diastereomers
C HO
OH
C HO
C H 2O H
OH
HO
C H 2O H
HO
C H3
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.
Legal operations with Fischer formulas
C HO
.
C H 2O H
OH
HO
OH
HO
C H 2 O H 180o
Rotation by 180o in the plane
of the paper does not change
the molecule.
C HO
No other rotation is
allowed.
This molecule is meso! There isn’t
an enantiomer
CH3
Br
Br
CH3
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
To determine the configuration (R/S). Make two
“switches”. Place the priority 4 group in one of the
vertical positions.
2
4
CHO
4
H
OH
1
CH2OH
OHC
CH2OH
2
OH R 3
3
alternatively:
1
1
2
CHO
4
H
OH
3
1
CH2OH
3
#4 at top position
H
R OH
2
R
CHO
HOCH2
4
Both H’s are in
back = same
result
H
#4 at bottom position
Draw all of the stereoisomers!
CH 3
CH
CH
CH
Cl
OH
Cl
CH 3
CH
CH
3
Cl
Cl
OH
HO
Cl
CH
3
Cl
CH
3
m eso
3
m eso
Any more??
CH
CH
3
Cl
Cl
HO
OH
Cl
CH
3
CH
3
Cl
CH
3
Cl
HO
NO!
Cl
OH
Cl
CH
3
3
Cl
CH
3
CH
3
There are only four stereoisomers!
Sect 5.10: cyclic
compounds
O
O
O
C H3
*
*
C H3
C H3
plane
( indicates no stereoisomers )
stereoisomers (max) :
20 = 1
21 = 2
21 = 2
O
O
O
C H3
*
*
C H3
*
*
C H3
stereoisomers (max) :
22 = 4
22 = 4
C H3
C H3
*
*
C H3
plane
( but only if cis )
22 = 4
1-Bromo-2-chlorocyclopropane
note that the cis/trans isomers are also diastereomers
Br
R
S
Cl
Cl
R
S
Br
cis
enantiomers
diastereomers
Br
R
R
S
S
Br
trans
Cl
Cl
enantiomers
1,2-Dibromocyclopropane
mirror image identical
Br
Br
Br
Br
cis
diastereomers
meso
Br
Br
trans
Br
Br
enantiomers
1-Bromo-2-chlorocyclohexane
cyclohexanes may be analyzed using planar rings
Br
Cl
Cl
Br
cis
enantiomers
diastereomers
Br
Br
Cl
Cl
enantiomers
trans
1,2-dichlorocyclohexane
mirror image identical
Cl
Cl
Cl
Cl
cis
meso
diastereomers
Cl
Cl
Cl
Cl
enantiomers
trans
Interchanges: See problem 59
ONE STEREOCENTER
Br
CH3
CH3
C
C
F
H
F
Br
H
enantiomer
Odd:
Even:
1..3..5…etc interchanges = enantiomer
2..4..6...etc interchanges = original compound
PROBLEM: ARE THESE IDENTICAL
OR ARE THEY ENANTIOMERS?
Br
CH3
H
C
C
F
H
CH3
Br
F
You can use interchanges to answer
this question! Next slide.
CH3
Br
2
1
C
Br
F
CH3
F
C
C
C
H
H
H
H
ENANTIOMER
CH3
SAME
F
Br
3
CH3
F
Br
ENANTIOMER
Back to the problem: are these identical
or are they enantiomers?
Br
CH3
H
C
C
F
H
CH3
Br
F
They are enantiomers! Three
switches = enantiomers
Cl Cl
MORE THAN ONE STEREOCENTER
Cl Cl
Cl Cl
To form an enantiomer you must
interchange all of the stereocenters.
C H3
C H3
C H3
C H3
H
H
H
H
Cl Cl
C H3
H
H
C H3
two stereocenters
two interchanges
Cl Cl
H
C H3
C H3
H
enantiomer
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
Sect 5.11: Stereocenters
other than carbon
Sect 5.12: Other types of
chirality
Sect 5.13: Resolution
(Skip)
Sect 5.14-5.16:
carbohydrates
Emil Fischer
Source: Michigan State University, Department of Chemistry
http://www.chemistry.msu.edu/Portraits/PortraitsHH_collection.shtml
Glyceraldehyde
(an aldotriose)
H
H
O
C
C
H
C
OH
C H2 O H
D (+ )
O
HO
C
H
C H2 O H
L (-)
H
H
O
C
C
H
C
O
OH
HO
C
H
C H2 O H
C H2 O H
(R )
(S )
Notice that the D enantiomer
has the R configuration in the
R/S sytem.
The Aldotetroses
H
H
O
O
C
C
OH
H
HO
H
E ryth ro se
OH
H
HO
H
C H2 O H
C H2 O H
D (-)
L
(+ )
H
H
O
O
C
C
HO
H
OH
H
T h re o se
OH
H
Notice that the D
enantiomers are (-)!
HO
H
C H2 O H
C H2 O H
D (-)
L
(+ )
Notice that the L
Enantiomers are (+)!
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 (-)
Fig. 5-27: the aldopentoses
CHO
CHO
CHO
CHO
H
OH
HO
H
HO
H
OH
HO
H
H
OH
HO
H
H
OH
HO
H
H
OH
HO
H
C H2 O H
C H2 O H
D (-)
L (+ )
H
H
C H2 O H
C H2 O H
D (-)
L (+ )
A ra b in o se
R ib o se
CHO
H
CHO
OH
HO
HO
H
H
H
OH
HO
C H2 O H
H
HO
H
H
OH
OH
HO
H
H
OH
H
L (-)
X ylo se
CHO
CHO
C H2 O H
D (+ )
OH
H
OH
HO
H
C H2 O H
C H2 O H
D (-)
L (+ )
L yxo se
There are four natural pentose sugars. They
have the D configuration.
CHO
CHO
H
OH
H
OH
H
OH
OH points
to right!
D
HO
H
OH
H
OH
A ra b in o se
R ib o se
CHO
CHO
HO
H
OH points
to right!
D
C H2 O H
C H2 O H
H
H
OH
H
OH
C H2 O H
X ylo se
OH points
to right
D
HO
H
HO
H
H
OH
C H2 O H
L yxo se
OH points
to right
D
Fig 5-28 “natural” D-aldohexoses
CHO
CHO
CHO
H
C HO
OH
HO
H
H
HO
H
H
OH
HO
H
OH
H
OH
HO
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
C H2 O H
(+ )-A llo se
(+ )-A ltro se
C H2 O H
(+ )-G lu co se
(+ )-M a n n o se
CHO
CHO
H
OH
HO
H
OH
H
H
C H2 O H
C H2 O H
CHO
HO
H
H
OH
C H2 O H
(-)-G u lo se
HO
H
H
H
CHO
OH
HO
H
OH
HO
H
HO
H
H
HO
H
HO
H
OH
C H2 O H
(-)-Id o se
H
OH
C H2 O H
(+ )-G a la cto se
H
OH
C H2 O H
(+ )-T a lo se
The “unnatural” L-aldohexoses
CHO
CHO
CHO
HO
C HO
H
H
OH
OH
H
OH
HO
H
H
HO
H
HO
H
H
HO
H
HO
H
HO
H
HO
H
HO
H
HO
H
HO
H
HO
H
C H2 O H
(-)-A llo se
(-)-A ltro se
C H2 O H
(-)-G lu co se
(-)-M a n n o se
CHO
CHO
HO
H
H
HO
H
HO
HO
C H2 O H
C H2 O H
CHO
H
OH
OH
H
C H2 O H
(+ )-G u lo se
H
HO
OH
HO
CHO
H
H
OH
H
H
OH
H
OH
OH
H
OH
H
OH
H
C H2 O H
(+ )-Id o se
HO
H
C H2 O H
(-)-G a la cto se
HO
H
C H2 O H
(-)-T a lo se
Sect 5.17: natural products
and amino acids
Amino acids belong to the L-series. Can
You assign the R or S configuration to the
following Fischer structure of Alanine?
O
O
O
4
O
H
H
H
sw itch
H 2N
H 2N
CH
H
2
1
O
H 2N
O
3
CH
3
Same!
The original structure is, therefore,
(S)
CH
3
3
enantiomer
(R)
H
CH
CH
3
O
3
O
2
3
1
H 2C
H
CH
H
3
CH
2
H 3C
(S ) ca ra w a y o il
(R ) S p e a rm in t o il
M irro r p la n e
These compounds are enantiomers! They have
identical physical properties, except for odor
and rotation of light in a polarimeter!