JEE Advanced levelSingle correct optionDifficulty 3 of 3bm038
Pure alpha-D-glucopyranose has specific rotation +112.2∘, the beta anomer +18.7∘ and the equilibrium mixture +52.7∘. Ignoring the open-chain form, the mole percent of the alpha anomer at equilibrium is
A36.4%
B52.7%
C47.3%
D63.6%
Answer: A
Let the alpha fraction be x: 112.2x+18.7(1−x)=52.7. Then x=(52.7−18.7)/(112.2−18.7)=34.0/93.5=0.364, i.e. 36.4%. The 63.6% option is the beta fraction.