Chapter 92, Problem 1P

Interpretation Introduction

Interpretation : The specific rotation of a substance dissolved in a solvent (0.4 g/mL) and with observed rotation of -10° is to be determined.

Concept Introduction : The specific rotation ( ${\text{[α]}}_{\text{λ}}^{\text{T}}$ ) is a property of chiral compounds. When light passes through a sample of the substance in solution, it is characterized as the shift in the orientation of monochromatic plane-polarized light per unit distance-concentration product.

Answer to Problem 1P

Specific rotation = -50° mL/g.dm

Explanation of Solution

Given information :

Observed rotation, $\text{α}$ = -10°

Length of the sample tube, l = 0.5 dm

Concentration of solvent, C = 0.4 g/mL

:

The formula to calculate the specific rotation ( ${\text{[α]}}_{\text{λ}}^{\text{T}}$ ) of a substance dissolved in a solvent is-

  ${\text{[α]}}_{\text{λ}}^{\text{T}}\text{=}\frac{\text{α}}{\text{C×l}}$

In this formula,

  $\text{α}$ is the observed rotation.

C is the concentration in g/mL

l is the length of the sample tube in unit dm.

T is the temperature which is taken as a constant at 25°C.

? is the wavelength of the light used.

Now, put the given data in this formula and solve for the specific rotation of the given substance.

  $\begin{array}{l}{\text{[α]}}_{\text{λ}}^{\text{T}}\text{=}\frac{{\text{-10}}^{\text{o}}}{\text{(0}\text{.4 g/mL)×(0}\text{.5 dm)}}\\ \therefore {\text{[α]}}_{\text{λ}}^{\text{T}}=-{50}^{\circ }\text{mL/g}\text{.dm}\end{array}$

Conclusion

The specific rotation of the given substance is -50° mL/g.dm.