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Question

# The specific rotation of two glucose anomers are α=+110o and β=+19o and for the constant equilibrium mixture it is +52.7o. Calculate the per cent composition of anomers in the equilibrium mixture.

A
a-anomer = 36.2% and b-anomer = 63.8%
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B
a-anomer= 63.8% and a-anomer = 36.2%
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C
a-anomer =70% and b-anomer =30%
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D
a-anomer = 30% and b-anomer = 70%
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Solution

## The correct option is A a-anomer = 36.2% and b-anomer = 63.8%Let a and b be the mole fractions of α and β− anomers in equilibrium mixture. Thus, a+b=1 ...(i) 110oa+19ob=52.7o ...(ii) Solving both equations, we get a=0.362 and b=0.638

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