Q61 / 180 · AIIMS 2009 · Chemistry

The equilibrium constant for mutarotation $\alpha$-D Glucose $\rightleftharpoons \beta$-D Glucose is 1.8 . What percentage of $\alpha$ form remains at equilibrium?

  • Correct answer
    A
    $35.7$
  • B
    $64.3$
  • C
    $55.6$
  • D
    $44.4$

Solution

$\alpha$-D glucose $\rightleftharpoons \beta$-D glucose $\begin{array}{lcc}\text { Initial } & 1 & 0 \\ \text { At equilibrium } & 1-\alpha & \alpha\end{array}$ $K=\frac{\alpha}{1-\alpha}=1.8$ Solving $\alpha=0.642$ $1-\alpha=0.358$ Percent of $\alpha$-D glucose remaining at equilibrium $=35.8 \%$.

Hence, the correct answer is option: 'A'.

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