Question

The specific rotation of α- and β-forms of a monosaccharide are $\text{ }+{{29}^{0}}\text{ }$ and $\text{ }-{{17}^{0}}\text{ }$ respectively. When either form is dissolved in water the specific rotation of the equilibrium mixture was found to be$\text{ +}{{14}^{0}}\text{ }$. What is the composition of these forms?
(A)$\text{ }\alpha -\text{form}-\text{ 67}\text{.4}{\scriptstyle{}^{0}/{}_{0}}$ , $\beta -\text{form}-\text{ 32}\text{.6}{\scriptstyle{}^{0}/{}_{0}}$
(B)$\text{ }\alpha -\text{form}-\text{ 60}{\scriptstyle{}^{0}/{}_{0}}$ , $\beta -\text{form}-\text{ 40}{\scriptstyle{}^{0}/{}_{0}}$
(C)$\text{ }\alpha -\text{form}-\text{ 70}{\scriptstyle{}^{0}/{}_{0}}$ , $\beta -\text{form}-\text{ 30}{\scriptstyle{}^{0}/{}_{0}}$
(D)$\text{ }\alpha -\text{form}-\text{ 50}{\scriptstyle{}^{0}/{}_{0}}$ , $\beta -\text{form}-\text{ 50}{\scriptstyle{}^{0}/{}_{0}}$

Answer 1

Mutarotation is the deviation from the specific rotation, because of the change in the equilibrium between the two anomeric forms. The specific rotation of the molecule at equilibrium is equal to the product of a specific rotation of $\alpha$ and $\beta$ anomers and its composition in the solution.
${{\text{(Spe}\text{.rotation)}}_{\text{eq}}}=\text{(Comp}\text{.of }\alpha \text{ )spe}\text{.rotation of }\alpha -\text{form + (Comp}\text{.of }\beta \text{ )spe}\text{.rotation of }\beta -\text{form}$

Complete step by step solution:
The mutarotation is defined as the change in the optical rotation of the molecule that is observed when the pure $\alpha$ and $\beta$ anomers are dissolved in the solvent. Mutarotaton is possible only when the anomers are in equilibrium or can interconvert between each other.
We can say that the solution of the molecule contains the $x{\scriptstyle{}^{0}/{}_{0}}$ of $\alpha$ anomers and $y{\scriptstyle{}^{0}/{}_{0}}$of $\beta$ anomers. The entire solution contains $100{\scriptstyle{}^{0}/{}_{0}}$ of the molecule. The fraction of anomers is equal to the 1.
$\alpha -\text{anomer + }\beta \text{-anomer = 1}$
Here, we are given the specific rotation of a molecule. It is given as the 14.
Let’s the fraction of $\beta -\text{form}$ at the equilibrium as ‘x’.
We know that the total optical rotation is equal to the 1.Therefore, the fraction of the $\alpha -\text{form}$will be given as,
$\text{Fraction of }\alpha \text{-form = 1 }-\text{ x}$
The specific rotation of the anomers in the solution is always equal to the sum of the product of the composition of anomers and its specific rotation.
Therefore, we can write the specific rotation of equilibrium in terms of $\alpha -\text{form}$and $\beta -\text{form}$as,

& \begin{matrix} {{\text{(Spe}\text{.rotation)}}_{\text{eq}}} & = & \text{(Comp}\text{.of }\alpha \text{ )spe}\text{.rotation of }\alpha -\text{form} & \+ & \text{(Comp}\text{.of }\beta \text{ )spe}\text{.rotation of }\beta -\text{form} \\\ 14 & = & (1-x)\times \text{spe}\text{.rotation of }\alpha -\text{form} & \+ & (x)\times \text{spe}\text{.rotation of }\beta -\text{form} \\\ \end{matrix} \\\ & \text{ } \\\ & \text{ } \\\ \end{aligned}$$ Let's substitute the values in the above form .we have, $\begin{aligned} & 14\text{ = (1}-x)\times \text{ 29 + (}x)(-17) \\\ & 14=\text{ 29 }-29x\text{ + (}-17x) \\\ & 14\text{ = 29 }-29x-17x \\\ & 14\text{ = 29}-46x \\\ & -15\text{ = }-46x \\\ & \therefore x=\dfrac{15}{46} \\\ \end{aligned}$ Therefore, the value of x is, $$x=\dfrac{15}{46}\text{ }$$ Let's calculate the composition of $\alpha -\text{form}$ at the present in the solution. $(1-x)\text{ = 1}-\text{ }\dfrac{15}{46}\text{ = }\dfrac{46-15}{46}\text{ = }\dfrac{31}{46}\text{ = 0}\text{.674 }\times \text{ 100}{\scriptstyle{}^{0}/{}_{0\text{ }}}=67.4{\scriptstyle{}^{0}/{}_{0\text{ }}}$ Now let us calculate the composition of $\beta -\text{form}$ the present in the solution. $(x)\text{ = }\dfrac{15}{46}\text{ = 0}\text{.326 }\times \text{ 100}{\scriptstyle{}^{0}/{}_{0\text{ }}}=32.6{\scriptstyle{}^{0}/{}_{0\text{ }}}$ So, the composition of $\alpha -\text{form}$ the solution is $67.4{\scriptstyle{}^{0}/{}_{0\text{ }}}$and $\beta -\text{form}$is$32.6{\scriptstyle{}^{0}/{}_{0\text{ }}}$. **Hence, (A) is the correct option.** **Note:** Specific rotation is the optical rotation of selective anomers in the solution. However, the mutarotation is the specific rotation of the molecule at the equilibrium. The sugar is in equilibrium with its straight-chain or linear form with the ring form. So, even though we start with the $100{\scriptstyle{}^{0}/{}_{0}}$ sample of pure $\alpha -\text{form}$or$\beta -\text{form}$, once it is dissolved in the water it attains the equilibrium between the straight-chain forms to the other anomers. In short, the anomer obeys the zeroth law.