Correct Answer - C
An open- chain structure of glucose has 4 chiral centres. The open chain structure exists in equilibrium with two cyclic forms.
The cyclic forms of `D-(+)-` glucose are hemiacetals formed by an intramoleculer reaction of the `-OH` group of `C5` with the aldehyde group. Cyclization creates a new stereocenter at `C1` and thus there are 5 chiral centres in the cyclic hemiacetal form of glucose. This new stereocentre explains how two cyclic forms are possible.
