Question

An aqueous solution containing 10 g of an optically pure compound was diluted to 500 mL with water and was found to have a specific rotation of −129°. If this solution were mixed with 500 mL of a solution containing 7 g of a racemic mixture of the compound, what would the specific rotation of the resulting mixture of the compound? What would be its optical purity?

Answer 1

Answer:

Specific rotation = -258°

Optical purity = 200%

Explanation:

Substances that have a chiral carbon ( a carbon which has four different atoms or compounds bond to it) have optical isomers. The different molecules are called enantiomers, and they differ by the side they deflect polarised light. A mixture contained the same number of the enantiomers is not optical and doesn't deflect polarized light. It's called a racemic mixture.

The specific rotation (spec.rot.) can be calculated by the equation:

spec.rot. = α/lc

Where α is the observed rotation of the plane polarised light in degrees, l is the path length in decimeters, and c is the concentration of the solution in g/100 mL.

The initial concentration is 10g/500mL = 2g/100 mL. The racemic mixture doesn't change the specific rotation, so the final concentration is 10g/1000 mL = 1g/100 mL. Then:

$\frac{spec.rot.1}{spec.rot.2} = \frac{\frac{α}{lc1} }{\frac{α}{lc2} }$

with α and l the same:

spec.rot.1/spec.rot.2 = c2/c1

spec.rot.2 = c1xspec.rot.1/c2

spec.rot.2 = 2x(-129°)/1

spec.rot.2 = -258°

If for a -129° the optical purity was 100%:

-129° ---------- 100%

-258° --------- x

By a direct simple three rule:

129x = 25800

x = 200%