Chirality and Optical Activity - Lecture 2 Notes

Second lecture in the series; prior knowledge from the first lecture is assumed.
Review of assigning R/S configurations to chiral centers.

Step 1: Identify the chiral center: A carbon atom bonded to four different groups.
Step 2: Assign priorities: Based on atomic weight; higher atomic weight gets higher priority.
- O > C > H.
- If the first atom is the same, move outward atom by atom, $CH<em>2CH</em>3$ > $CH_3$ .
Step 3: Orient the molecule: Position the lowest priority group (usually H) pointing away from you.
- Visualize yourself relative to the molecule.
- If H is pointing away, determine the direction (clockwise or counter-clockwise) of the remaining priorities (1->2->3).
Clockwise rotation (1->2->3): R configuration.
Counter-clockwise rotation (1->2->3): S configuration.
In the 2-butanol example, prioritize OH > CH2CH3 > CH_3 > H. If H is pointing away and priorities 1-2-3 are arranged counter-clockwise, it's the S enantiomer.

Example 2: Prioritize OH > C=O > CH_2OH > H. If H is pointing away from you, then 1->2->3 in clockwise directs indicate R enantiomer.
Example 3 (Alanine): Identify the chiral center. Priorities: NH2 > COOH > CH3 > H. With H pointing out, the arrangement appears clockwise, but since H is pointing towards you, the configuration is actually S.
- Recognize that with H oriented towards the viewer, the initially determined configuration will be the opposite of the true one.

Stereocenter: An atom, typically carbon, around which the interchange of two groups gives rise to a stereoisomer.
The number of possible stereoisomers is $2^n$ , where $n$ is the number of stereocenters.
- However, not all stereoisomers are enantiomers of each other.
Example: A molecule with one stereocenter has two possible isomers (enantiomers).
Example: Cholesterol (8 stereocenters) has $2^8 = 256$ possible stereoisomers. However, only one form exists in nature.
Tartaric Acid: Has two stereocenters, but only three stereoisomers (not four). This is due to the presence of a meso compound.

Louis Pasteur's experiment with tartaric acid crystals:
- Observed two types of crystals that were mirror images of each other (chiral crystals).
- Physically separated the two types of crystals.
- Dissolved each type separately and found they rotated polarized light in opposite directions.
- Demonstrated that tartaric acid exists as enantiomers.

Enantiomers: Stereoisomers that are non-superimposable mirror images. If all chiral centers are inverted (R becomes S, and S becomes R), the resulting molecule is the enantiomer.
Diastereomers: Stereoisomers that are not mirror images of each other. If only some chiral centers are inverted, the resulting molecule is a diastereomer.
Meso Compound: A molecule with chiral centers that is achiral due to an internal plane of symmetry. Meso compounds are superimposable on their mirror images.
- Occurs when a molecule with multiple stereocenters has identical substituents on those stereocenters.
- In tartaric acid, the SR and RS configurations are the same due to symmetry, forming a meso compound.

Constitutional Isomers: Same molecular formula, different connectivity.
Enantiomers: Mirror images; all stereocenters inverted.
Diastereomers: Non-mirror images; some (but not all) stereocenters inverted.

Biological systems are chiral; interactions depend on stereochemistry. Biological systems are stereospecific.
Drug activity depends on chirality.
- R and S enantiomers of a drug can have different effects; one may be therapeutic, while the other is toxic.

Observed Rotation: The degree to which a sample rotates polarized light; depends on path length and concentration.
Specific Rotation: Normalizes observed rotation for path length and concentration and temperature, allowing comparison between different experiments.
- $[\alpha] = \frac{\alpha_{observed}}{l \cdot c}$ , where $\,\alpha$ is specific rotation, $l$ is path length in decimeters (dm), and $c$ is concentration in g/mL
Enantiomeric Excess (ee):
- Indicates the excess of one enantiomer over the racemic mixture.
- $ee = \frac{[R] - [S]}{[R] + [S]} \times 100 \%$ , R = concentration of R enantiomer, S = concentration of S enantiomer
- If a mixture contains 75% of one enantiomer and 25% of the other, the enantiomeric excess is 50%.

Given the specific rotation of a pure enantiomer and the observed rotation of a mixture, the enantiomeric excess can be calculated:
- $Enantiomeric \, Excess = (\frac{Observed \, Rotation}{Specific \, Rotation \, of \, Pure \, Enantiomer}) \times 100 \%$ .
The enantiomeric excess can be used to determine molar ratios of enantiomers in a mixture.
- Example: If pure quinine has a specific rotation of -65 degrees, and a solution has an observed rotation of -50 degrees, the enantiomeric excess is approximately 77%. ( $\frac{-50}{-65} \times 100$ )
The ratio of enantiomers is NOT 77:23. The correct calculation is as follows
* 77% of the mixture is ONE enantiomer
* The remaining 23% comprises an equal mixture of both enantiomers. 11.5% each.
* The final ratio is 77 + 11.5 : 11.5, or roughly 88.5 : 11.5
* Another way of calculation
* If the enantiomeric excess is 30%, then ratio of mixture is calculated as follows
$\frac{100 + ee}{2} : \frac{100 - ee}{2}$
$\frac{100 + 30}{2} : \frac{100 - 30}{2}$
$65 : 35$

Recognize and name chiral centers (R/S).
Draw enantiomers (mirror images).
Recognize meso compounds (internal plane of symmetry).
Draw diastereomers (flip one chiral center).
Calculate percent enantiomeric excess and molar ratios from observed optical rotation.