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Question

The specific rotations of freshly prepared aqueous solutions of I and II are +112 and +18.7, respectively. On standing, the optical rotation of the aqueous solution of I slowly decreases to give a final value of +52.7 due to equilibration with II. Under this state of equilibrium, what is the ratio of concentration of II:I?

A
0.57
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B
1.00
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C
1.75
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D
5.9
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Solution

The correct option is C 1.75
Specific rotation of α-glucose(I)=+112
β-glucose (II)=+18.7

At equilibrium,
[I]=x,[II]=(1x)
(112×x)+(18.7×(1x))=52.7
112x+18.718.7x=52.7
93.3x=34
x=3493.3=0.36=α
(1x)=(10.36)=0.63=β
βα=0.630.36=1.75

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