Question

# Pyranose ring structure of glucose is due to hemiacetal formation between :a.) ${C_1}$ and ${C_5}$b.) ${C_1}$ and ${C_4}$c.) ${C_1}$ and ${C_3}$d.) ${C_2}$ and ${C_4}$

In the above diagram, we see the structure of Glucose in linear chain and its conversion to glucopyranose and now by counting the carbon number; we can find out that hemiacetal linkage occurs at ${C_1}$ and ${C_5}$ positions.