Question

The specific optical rotations of pure $a-$ and $b-D$ mannopyranose are $+ 29.3^{o}$ and$ -17.0$, respectively. When either form is dissolved in water, specific optical rotation of the equilibrium mixture is found to be $+ 14.2^{o}$. Calculate the percentage of a anomer at equilibrium.

Answer 1

Let fraction of a-anomer $ = x$
$\therefore $ Fraction of b-anomer $ = \left( 1 - x\right)$
Hence $x\left(+ 29.3^{o}\right)+\left(1 -x\right)\left(-17.0^{o}\right) = +14.2^{o}$
$29.3^{o} x +\left(-17^{o} + 17^{o}x\right) = + 14.2^{o}$
$29.3^{o}x + 17^{o}x = 14.2^{o} + 17^{o}$
$46.3^{o}x = 31.2^{o}$
$x = 0.67$
$\therefore $ Fraction of a-anomer$= 0.67$ or $67\%$