Question

The specific rotation of a ($\alpha$) and b ($\beta$) forms of a monosaccharide are $+{29}^0$ and $-{17}^0$ respectively. When either form is dissolved in water, the specific rotation of the equilibrium mixture was found to be $+{14}^0$. What is the percentage of b form at equilibrium?

• A. 67.4%
• B. 32. 6%
• C. 14%
• D. 29%

Answer 1

Answer: (B)

32. 6%

Let the percentage of b formed at equilibrium be x%.

The percentage of a form at equilibrium will be (100-x).

$\frac{x\times (-17)+(100-x)\times 29}{x+100-x}=14$

$x=\frac{1500}{46}$

$x=32.6$ .