What is the enantiomeric composition of a mixture that has a specific rotation of + 75° and has an enantiomeric excess of 60%?

Question

| Organic Chemistry 1

Accepted Answer

Let's say, we have x % of (+) isomer and y % of (-) isomer ⇒ x + y = 100

ee = $\frac{# moles of 1 enantio . - # moles of the other enantio .}{# moles of both enantiomers} \times 100$ = $\frac{x - y}{100} \times 100$ ⇒ x - y = 60

x + y = 100, x - y = 60 ⇒ 2x = 160 ⇒ x = 80, y = 20.

80% of (+) isomer, 20% of (-) isomer.

Enantiomeric excess | Stereoisomers

Organic Chemistry 1 - Enantiomeric excess