Transcript Optical Purity

Optical Activity
 Enantiomers have virtually
identical physical properties.
Exactly how do they differ from
each other?
 Solutions of them rotate planepolarized light in opposite
directions.
 They react with enzymes
differently (actually, one usually
reacts and the other does not).
Plane-Polarized Light
 Light with an E (or H) field that
oscillates in a single plane.
Figure 5-10. Schematic diagram of a polarizing filter.
Polarimeter
 A solution of a pure enantiomer
rotates plane-polarized light.
Figure 5-13. Schematic diagram of a polarimeter.
Optical Activity-Designation
 A solution of a pure enantiomer
that rotates plane-polarized light
to the
 left (CCW) is designated (-) or l
(levorotatory)
 right (CW) is designated (+) or d
(dextrorotatory)
 These do NOT correlate with
(R) and (S)!
Specific Rotation
 The specific rotation [α] of a
solution of a pure enantiomer is
[α] = α(observed)
c l
 where c is the concentration of
enantiomer in g/mL and l is the
length of the sample cell in dm.
Specific Rotation
 A solution of 0.5 g of (-)-epinephrine
dissolved in 10 mL of dilute HCl was placed
in a 20-cm polarimeter tube. Using the
sodium D line, the rotation was found to be
-5.0° at 25°C. Determine the specific
rotation of epinephrine.
[α] = α(observed)/(c l)
= -5.0°/[(0.5 g/10 mL)(20 cm x 0.1 dm/1 cm) ]
= -5.0°/[(0.05 g/mL)(2.0 dm)]
= -5.0°/0.10
= -50.°
[ ]
25
D
  50 . 
Racemic Mixture
 A 50/50 mixture of enantiomers.
 Since solutions of pure enantiomers
rotate plane-polarized light in
opposite directions, a 50/50 solution
will NOT rotate plane-polarized light.
 The solution is optically inactive.
 A racemic mixture is also called
 a racemate
 a (d,l) pair
 a (±) pair
Other Mixtures of Enantiomers
% _ optical _ purity  100
( observed _ rotation )
( rotation _ of _ pure _ enantiomer )
% _ enantiomer ic _ excess  100
|d l|
|d l|
 If the mixture contains only the pure enantiomers,
these two equations give the same value. If the
mixture contains other optically active compounds,
op will differ from ee.
Enantiomeric Excess (e.e.)
 Problem 5-12: When optically pure (R)-(-)-2bromobutane is heated with water, 2-butanol is
the product. Twice as much (S)-2-butanol forms
as (R)-2-butanol. Find the e.e. and the observed
rotation of the product. [α]=13.50° for pure (S)-2butanol.
Let x = amount of (R) enantiomer formed
e.e. = 100 |d-l|/|d+l|
= 100 |2x-x|/|2x+x|
= 100 (1/3)
= 33%
Using Optical Purity
Since 2-butanol is the only product, e.e. = o.p.
Since o.p. = 100 (observed rotation)/(rotation of
pure enantiomer)
33 = 100 (observed rotation)/ 13.50°
observed rotation = 33(13.50)/100 = +4.5°
The predominance of the (S)-(+)-2-butanol causes
the specific rotation of the solution to be positive.
Specific Rotation of a Mixture
 Sucrose is a disaccharide of D-glucose and Dfructose. Sucrose has a specific rotation of
+66.5°.
 When sucrose is hydrolyzed by an invertase,
a mixture of glucose (C6H12O6, +52.7°) and
fructose (C6H12O6, -92.4°) results and the
observed rotation of the mixture is (-).
 Because of this, the mixture is called an invert
sugar.
 Can you calculate the specific rotation of invert
sugar?
 (-39.7º, since glucose and fructose have the
same molar mass.)
Chirality of Conformationally
Mobile Systems
 A molecule cannot be optically active if
its chiral conformations are in
equilibrium with their mirror images.
Chirality of Conformationally
Mobile Systems
 Such enantiomers are in equilibrium
with each other through ring flipping.
 One enantiomer cannot be separated from
the other.
Chirality of Conformationally
Mobile Systems
 To determine whether a conformationally mobile
molecule can be optically active, consider tis
most symmetric conformation.
cis-1,2-dibromocyclohexane is achiral…is the trans
isomer achiral?
Chirality of Conformationally
Mobile Systems
 Chiral or achiral?
 Optically active or optically
inactive?
Enantiomers with No Chiral
Carbon Atoms
 Conformers that cannot interconvert
(due to steric hindrance) can be
enantiomers.
Enantiomers with No Chiral
Carbon Atoms
 H2C=C=CH2 is allene.
 A disubstituted allene can be chiral,
because the substituents are in
perpendicular planes.
Why aren’t all the
H’s and Cl’s in
the same plane?