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

Answer: 67

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 $67\%$