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**Hint:**The equilibrium constant describes the relation between the reactant which is taking part in the reaction and the product which is formed at equilibrium. For the general reaction

$A + B \rightleftharpoons C + D$

The equilibrium constant is given as shown below,

$K = \dfrac{{[C][D]}}{{[A][B]}}$

**Complete answer:**

Mutarotation is defined as the deviation from the specific rotation because of the equilibrium change between the two anomeric forms that are $\alpha$-form and $\beta$-form in the aqueous solution.

Given,

The percentage of $\beta$-D glucose is 63.6 %.

The conversion of $\alpha$-D glucose into $\beta$-D glucose is shown below.

$\alpha - D - glu\cos e \rightleftharpoons \beta - D - glu\cos e$

Assume that the initial percentage of $\alpha$-D glucose is 100% and the initial percentage of $\beta$-D glucose is 0.

Then, the equilibrium percentage of $\alpha$-D glucose is calculated as shown below.

$\alpha - D - Glucose\% = 100\% - 63.6\%$

$\Rightarrow \alpha - D - Glucose\% = 36.4\%$

The equilibrium constant of the mutarotation is shown below.

${K_c} = \dfrac{{\beta - D - glu\cos e}}{{\alpha - D - glu\cos e}}$

Substitute the values of equilibrium percentage of $\alpha$-D glucose and $\beta$-D glucose as shown below.

${K_c} = \dfrac{{63.6}}{{36.4}}$

${K_c} = 1.747$

Thus, the value of equilibrium constant for mutarotation is 1.747.

**Therefore, the correct option is B.**

**Additional Information:**The carbohydrate fructose and glucose can undergo mutarotation but sucrose and cellulose does not undergo mutarotation because they do not contain hydroxyl functional groups at the anomeric position.

**Note:**When the value of equilibrium constant is greater than 1, then the equilibrium of the reaction favors the product. When the value of equilibrium constant is less than 1, then the equilibrium of the reaction favors the reactant.

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