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 −135°. If this solution were mixed with 500 mL of a solution containing 8 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

Explanation:

The given data is as follows.

Amount of racemic mixture = 8 g

Amount of pure R compound = 10 g

Therefore, total amount of R compound after missing is calculated as follows.

                 $10 g + \frac{8}{2}$

                 = 14 g

Let us assume that the amount of Si isomer is 4 g.

So, [R] = 11.5 g/L

     [S] = 4 g/L

Now, we will calculate the enantiomeric excess as follows.

          $\frac{\text{amount of pure R compound}}{\text{total amount of R compound}} \times 100$

             = $\frac{10 g}{14 g} \times 100$

             = 71.4%

Therefore, optical purity of the compound is 71.4%.

Now, the specific rotation will be calculated as follows.

          Optical purity = $100 \times \frac{\text{specific rotation}}{-135}$

               71.4% = $0.740 \times \text{specific rotation}$

                         = $-96.4^{o}C$

Therefore, specific rotation of the resulting mixture of the compound is $-96.4^{o}C$.