Question

The specific rotation of a pure substance is 1.68°. What is the specific rotation of a mixture containing 75% of this isomer and 25% of the (-) isomer?
A) +1.68°
B) 0°
C) +1.26°
D) +0.84°
E) +.042°

Answer 1

Answer:

Specific rotation of mixture = D. 0.84 °

Explanation:

Enantiomeric excess = % of one enantiomer - % of the other enantiomer = 75-25 % = 50 %

Also,

Enantiomeric excess = $\frac {optical\ rotation\ of\ mixture}{optical\ rotation\ of\ pure\ enantiomer}\times 100$

Optical rotation of pure enantiomer = 1.68  °

Applying the values as:-

$50=\frac {optical\ rotation\ of\ mixture}{1.68^0}\times 100$

$Optical\ rotation\ of\ mixture=\frac {1.68^0}{100}\times 50$

Specific rotation of mixture = 0.84 °