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Standard XII

Chemistry

Stereoisomerism

The specific ...

Question

The specific rotation of $α$ - and $β$ -forms of a monosaccharide are $+ 29^{o}$ and $- 17^{o}$ respectively. When either form is dissolved in water the specific rotation of the equilibrium mixture was found to be $+ 14^{o}$ . What is the composition of these forms?

α

-form-

67.4 %

β

-form-

32.6 %

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α

-form-

60 %

β

-form-

40 %

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α

-form-

70 %

β

-form-

30 %

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α

-form-

50 %

β

-form-

50 %

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Solution

The correct option is A $α$ -form- $67.4 %$ , $β$ -form- $32.6 %$
Suppose that the fraction of $β$ -form $=$ x then, fraction of $α$ -form $= 1 - x$
$x \times (- 17) + (1 - x) 29 = 14$
or, $- 17 x + 29 - 29 x = 14$
or, $- 46 x + 29 = 14$
or, $15 = 46 x$
or, $x = \frac{15}{46} = 0.326 = 32.6 % = β$ -form
$1 - 0.326 = 0.674 = 67.4 % = α$ -form.

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The specific rotation of α- and β-forms of a monosaccharide are +29o and −17o respectively. When either form is dissolved in water the specific rotation of the equilibrium mixture was found to be +14o. What is the composition of these forms?

The correct option is A α-form- 67.4%, β-form- 32.6%Suppose that the fraction of β-form =x then, fraction of α-form =1−xx×(−17)+(1−x)29=14or, −17x+29−29x=14or, −46x+29=14or, 15=46xor, x=1546=0.326=32.6%=β-form1−0.326=0.674=67.4%=α-form.

The specific rotation of $α$ - and $β$ -forms of a monosaccharide are $+ 29^{o}$ and $- 17^{o}$ respectively. When either form is dissolved in water the specific rotation of the equilibrium mixture was found to be $+ 14^{o}$ . What is the composition of these forms?