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Standard XIII

Chemistry

D/L Nomenclature

The specific ...

Question

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

a-anomer = 36.2% and b-anomer = 63.8%

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a-anomer= 63.8% and a-anomer = 36.2%

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a-anomer =70% and b-anomer =30%

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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,
$a + b = 1 . . . (i)$
$110^{o} a + 19^{o} b = {52.7}^{o} . . . (i i)$
Solving both equations, we get $a = 0.362$ and $b = 0.638$

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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.

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

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

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,
$a + b = 1 . . . (i)$
$110^{o} a + 19^{o} b = {52.7}^{o} . . . (i i)$
Solving both equations, we get $a = 0.362$ and $b = 0.638$