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

Step-by-step 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 + (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}} * 100`

=
`(10 g)/(14 g) * 100`

= 71.4%

Therefore, optical purity of the compound is 71.4%.

Now, the specific rotation will be calculated as follows.

Optical purity =
`100 * \frac{\text{specific rotation}}{-135}`

71.4% =
`0.740 * \text{specific rotation}`

=
`-96.4^(o)C`

Therefore, specific rotation of the resulting mixture of the compound is
`-96.4^(o)C`.