(Solved): Optical Rotation A solution of one stereoisomer of a given monosaccharide rotates plane-polarized l ...

Optical Rotation A solution of one stereoisomer of a given monosaccharide rotates plane-polarized light to the left (counterclockwise) and is called the leverotary isomer, designated (-); the other stereoisomer rotates plane-polarized light to the same extent but to the right (clockwise) and is called the dextrorotary isomer, designated $(+)$. An equimolar mixture of the $(+)$ and $(?)$ forms does not rotate plane polarized light. The optical activity of a stereoisomer is expressed quantitatively by its optical rotation, the number of degrees of freedom by which plane-polarized light is rotated on passage through a given path length of solution of a compound at a given concentration. The specific rotation of an optically active compound is defined as: $[?{]}_D=\frac{\text{ Optical rotation ( })}{(\text{ Path length }(dm))(\text{ Concentration }(g/ml))}$ The temperature and wavelength of the light employed (usually the sodium D-line) must be specified in the definition. 31 point A freshly prepared solution of the $?$-D-glucose shows a specific rotation of $+112^{?}$. Over time the rotation of the solution gradually decreases and reaches an equilibrium value corresponding to +52.5 . In contrast, a freshly prepared solution of $?$-Dglucose has a specific rotation of $+19^{?}$. The rotation of this solution increases with time to the same equilibrium value as the $?$ -anomer. Based on the data above, what percentage of the solution does the $??\mathrm{D}$-glucose anomer represent?