A mixture of two enantiomers with a composition of 65.0 R has an observed rotation of 25.3 in a polarimeter tube If the mixture has a concentration of 2.038 g mL at 25 C what is the predicted of an optically pure sample of the S enantiomer

- 41.3°

Explanation:

Observed rotation for the mixture can be converted to specific rotation for the mixture through

[α] = α/(lc)

[α] = specific rotation = ?

when written as [α]²⁵D, 25 refers to the temperature at which the optical exercise was carried out, and D indicates that a particular wavelength of Sodium light was used.

α = observed rotation = 25.3°

l = cell length or length of the polarimeter tube in dm = 1 dm (Because 10cm = 1 dm)

c = concentration of the mixture in g/mL = 2.038 g/mL

[α] = 25.3/(1×2.038)

[α] = 12.4°

The mixture contains 65% R enantiomer and automatically 35% S enantiomer.

That is,

Fraction of R = 0.65

Fraction of S = 0.35

Let the predicted specific rotation of the pure S enantiomer be +x

The predicted specific rotation of the pure R enantiomer will automatically be -x

Rotation (R/S Mixture) = [Fraction(S) × Rotation (S)] + [Fraction(R) × Rotation (R)]

12.4° = [0.35x] + [0.65(-x)]

12.4 = 0.35x - 0.65x

12.4 = -0.30x

x = -(12.4/0.3)

x = -41.3°

The predicted specific rotation of the pure S enantiomer = x = - 41.3°.

Hope this Helps!

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