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

Specific rotation of mixture = D. 0.84 °

Step-by-step 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}* 100`

Optical rotation of pure enantiomer = 1.68 °

Applying the values as:-

`50=\frac {optical\ rotation\ of\ mixture}{1.68^0}* 100`

`Optical\ rotation\ of\ mixture=\frac {1.68^0}{100}* 50`

Specific rotation of mixture = 0.84 °