Racemic Mixtures, Optical Purity and Configuration

Racemic mixtures, optical purity, and what reactions do to a chiral centre: retention, inversion, racemisation, resolution, asymmetric synthesis and the Walden cycle.

Builds on: Part 10 · Lactic Acid, Tartaric Acid and Counting Stereoisomers.

Optical purity

Enantiomeric excess

5 g (+) and 3 g (−): 3 g of each cancel as a racemate and the extra 2 g rotates the light. ee = (5 − 3)/8 × 100 = 25 %, so [α]obs = 0.25 × 13.5° = +3.375°.

Resolution

Through diastereomeric salts

React the racemic acid with one enantiomer of a chiral base such as (−)-strychnine. The two salts are diastereomers with different solubilities: crystallise them apart, then add mineral acid to free each enantiomer.

Walden inversion

A cycle that inverts

(+)-Malic acid → PCl₅ → (−)-chlorosuccinic acid → AgOH → (−)-malic acid, and back by the same steps from (−)-malic acid. PCl₅ and KOH cause inversion; moist Ag₂O does not.

Walden cycle diagram. — The Walden cycle.

Summary

Key formulas

% ee = \frac{n _{1} - n _{2}}{n _{1} + n _{2}} \times 100

Enantiomeric excess and optical purity

% ee = \frac{[ α ] _{obs}}{[ α ] _{pure}} \times 100

Enantiomeric excess and optical purity

[α]_{obs} = \frac{% ee}{100} \times [α]_{pure}

Enantiomeric excess and optical purity

Worked examples

From the video

1. 5 g (+)-2-butanol + 3 g (−)-2-butanol; pure (+) has [α] = +13.5°. Find ee and [α]obs.

ee = 25 %; [α]obs = +3.375°.

JEE-style question

Your turn

[α]obs = +6.75°, pure (+)-2-butanol = +13.5°. % of (+) form?

(a)50%

(b)75%

(c)25%

(d)100%

Show the answer and the traps

ee = 6.75/13.5 = 50%. That excess is (+); the other 50% is racemic, half of it (+). So (+) = 50 + 25 = 75%. Option (b).

50% is the ee itself. 25% is the amount of (−). 100% would rotate the full +13.5°.

Watch out

Common mistakes

Assuming a change of sign means a change of configuration.(−)-2-methylbutan-1-ol → (+)-1-chloro-2-methylbutane keeps its configuration.

Reporting the ee as the percentage of the major enantiomer.ee 50 % means 75 % of the major form.

Practice

Try these

1. A mixture is 80 % (+) and 20 % (−). Find the ee.

60 %.

2. Pure (−)-enantiomer has [α] = −20°. What does a sample with 40 % ee of (−) show?

−8°.

3. Why can diastereomeric salts be separated by crystallisation but enantiomers can't?

Diastereomers have different solubilities; enantiomers have identical physical properties.