Question

A natural product was isolated in the laboratory, and its observed rotation was +10° when measured in a 1 dm sample tube containing 1.0 g of compound in 10 mL of water. What is the specific rotation of this compound?

Short answer

The specific rotation is $\text{+100°}$.

Step-by-step solution

Specific rotation

Specific rotation comes from the interaction of the chiral compound with monochromatic light.

Every chiral compound has its tendency for plane-polarized light’s rotation, and it depends upon the concentration of the sample, length of the corresponding tube, and the observed rotation.

Formula to calculate the specific rotation    

The formula to calculate the specific rotation is as follows:

$\left[\text{α}\right]\text{=}\frac{\text{α}}{\text{l×c}}$

Where,$\left[\text{α}\right]$is the specific rotation,

$\text{α}$is the observed rotation,

l is the length of the tube,

c is the concentration of the sample.

Explanation

The given data is:

$\text{α}$is $\text{+10°}$.

l is 1 dm.

c is the $\frac{\text{1g}}{\text{10mL}}$.

Therefore, the specific rotation is calculated as follows:

$$\begin{gathered} \left[\text{α}\right]\text{=}\frac{10°}{\text{1×}{\displaystyle \frac{\text{1g}}{\text{10mL}}}} \\ \text{=+100°} \end{gathered}$$

Therefore, the specific rotation is $\text{+100°}$.