91Ó°ÊÓ

A specific rotation constitutes a property of the chiral molecule. It can be identified by measuring the energy and temperature of a plane polarized light.The formula used for evaluating the specific rotation of a mixture is:

$\begin{array}{c}\mathrm{Specific}\mathrm{rotation}\mathrm{of}\mathrm{the}\mathrm{mixture}=\mathrm{Specific}\mathrm{rotation}\mathrm{of}A×\mathrm{fraction}\mathrm{of}D-\mathrm{glucose}\mathrm{in}\mathrm{the}\mathrm{Î\pm ´Ú´Ç°ù³¾}+\\ \mathrm{Specific}\mathrm{rotation}\mathrm{of}B×\mathrm{fraction}\mathrm{of}D-\mathrm{glucose}\mathrm{in}\mathrm{the}\mathrm{β´Ú´Ç°ù³¾}\end{array}$

Let A denote the fraction of D-glucose in the $\mathrm{Î\pm }$from and B denote the fraction of D-glucose in the $\mathrm{β}$form.

$\begin{array}{c}A+B=1\\ B=1−A\end{array}$

The specific rotation of A is 112.2.

The specific rotation of B is 18.7.

The specific rotation of the equilibrium mixture is 52.7.

The specific rotation of the mixture can be found as:

$\begin{array}{c}\text{Specificrotationofthemixture=SpecificrotationofA×fractionofD-glucoseintheα´Ú´Ç°ù³¾+}\\ \text{SpecificrotationofB×fractionofD-glucoseintheβ´Ú´Ç°ù³¾}\\ \text{52.7=112.2A+}(1−A)18.7\\ 52.7=112.2A+18.7−18.7A\\ 34.0=93.5A\end{array}$

The value of A can be found as:

$\begin{array}{c}A=\frac{34.0}{93.5}\\ =0.36\end{array}$

The value of B can be found as:

$\begin{array}{c}B=1−A\\ =1−0.36\\ =0.64\end{array}$

This calculation indicates that 36% is in the data-custom-editor="chemistry" $\mathrm{Î\pm }$-form and 64% is the data-custom-editor="chemistry" $\mathrm{β}$-form. This agrees with the values in Section 21.10.