Question

How do you explain the presence of five $-OH$ groups in glucose molecule?

Answer 1

When Glucose is treated with acetic anhydride i.e.$(C{H}_{3}CO{)}_{2}O$, in the presence of $ZnC{l}_2$ or any mineral acid, it undergoes acetylation to form Glucose pentaacetate which confirms the presence of five $-OH$ groups. Since it exists as a stable compound, five $-OH$ groups should be attached to different carbon atoms.