Glucose is soluble in benzene. If α-D-glucose is dissolved in benzene, essentially no β-D-glucose is formed. However, if small amounts of phenol and pyridine are added, then the equillibrium amount of β-D-glucose is formed but slowly. If, to a solution of α-D-glucose in benzene, a small amount of 2-hydroxy-pyridine is added, the equilibrium amount of β-D-glucose is formed much more rapidly than when the small amount of phenol and pyridine were added. If a small amount of 4-hydroxypyridine is added to a solution of α-D-glucose , the rate of the formation of β-D-glucose isn't any different from the rate of reaction when a small amount of phenol and pyridine are added. Fully explain all these observations.
The mutarotation of alpha-D-glucose into beta-D-glucose takes place rapidly in a polar solvent, such as water. The reason that mutarotation takes place in a polar solvent is because the ring form of alpha-D-glucose must open up to form the open-chain, and then cyclicize again to form the hemiacetal beta-D-glucose isomer.
Benzene is a nonpolar solvent; and therefore, we would not expect to see any mutarotation of glucose in a benzene solvent.
But, phenol is a weak acid and pyridine is a weak base. Therefore, when they are added in small quantities to benzene, they can react together in a typical acid/base reaction. The hydrogen of the -OH on phenol gets picked off by ...
This solution explains several observations, addressing glucose, benzene, 2-hydroxypyridine and D-glucose