The optical rotation for the - -form of a carbohydrate is -107-60- that of the - -form is -52-8- - The solution at equilibrium of these anomers has an optical rotaion of -80-2- The value of the -B is-

Let, nbsp;α form be x. Then nbsp;β form is (1 x). As per given question, 107.6x+52.8(1 x)=80.2 ⇒ 107.6x+52.8 52.8x=80.2 ⇒ 54.8x=27.4 x=0.5 nbsp; ...

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Standard XII

Chemistry

Stereoisomerism

The optical r...

Question

The optical rotation for the $α$ -form of a carbohydrate is $+ {107.6}^{o}$ , that of the $β$ -form is $+ {52.8}^{o}$ . The solution at equilibrium of these anomers has an optical rotaion of $+ {80.2}^{o}$ .
The value of the $\frac{α}{β}$ is:

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Solution

Let, $α$ -form be x.
Then $β$ -form is (1-x).
As per given question,
$107.6 x + 52.8 (1 - x) = 80.2 \Rightarrow 107.6 x + 52.8 - 52.8 x = 80.2$
$\Rightarrow 54.8 x = 27.4 x = 0.5$

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The optical rotation for the α-form of a carbohydrate is +107.6o, that of the β-form is +52.8o. The solution at equilibrium of these anomers has an optical rotaion of +80.2o. The value of the αβ is:

Let, α-form be x. Then β-form is (1-x). As per given question, 107.6x+52.8(1−x)=80.2⇒107.6x+52.8−52.8x=80.2 ⇒54.8x=27.4x=0.5

The optical rotation for the $α$ -form of a carbohydrate is $+ {107.6}^{o}$ , that of the $β$ -form is $+ {52.8}^{o}$ . The solution at equilibrium of these anomers has an optical rotaion of $+ {80.2}^{o}$ .
The value of the $\frac{α}{β}$ is:

Let, $α$ -form be x.
Then $β$ -form is (1-x).
As per given question,
$107.6 x + 52.8 (1 - x) = 80.2 \Rightarrow 107.6 x + 52.8 - 52.8 x = 80.2$
$\Rightarrow 54.8 x = 27.4 x = 0.5$