The optical rotation of the alpha-form of a p | vedclass

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Question

(A) Let the fraction of the $\alpha$-form be $X$ and the fraction of the $\beta$-form be $(1 - X)$.The equilibrium rotation is given by the weighted a

Vedclass · Accepted Answer

$28$

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(A) Let the fraction of the $\alpha$-form be $X$ and the fraction of the $\beta$-form be $(1 - X)$.The equilibrium rotation is given by the weighted average: $150.7(X) + 52.8(1 - X) = 80.2$.Expanding the equation: $150.7X + 52.8 - 52.8X = 80.2$.Simplifying the terms: $97.9X = 80.2 - 52.8$.$97.9X = 27.4$.$X = \frac{27.4}{97.9} \approx 0.2799$.Converting to percentage: $0.2799 	imes 100 \approx 28 \%$.Thus,the percentage of the $\alpha$-form is $28 \%$.

List-$I$ (Reaction of glucose with)	List-$II$ (Product formed)
$A$. Hydroxylamine	$I$. Gluconic acid
$B$. $Br_2$ water	$II$. Glucose pentaacetate
$C$. Excess acetic anhydride	$III$. Saccharic acid
$D$. Concentrated $HNO_3$	$IV$. Glucoxime

The optical rotation of the $\alpha$-form of a pyranose is $+150.7^{\circ}$,that of the $\beta$-form is $+52.8^{\circ}$. In solution,an equilibrium mixture of these anomers has an optical rotation of $+80.2^{\circ}$. The percentage of the $\alpha$-form in the equilibrium mixture is $....... \%$.

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