How do you explain the absence of aldehyde group in the pentaacetate of $D$-glucose?

(N/A) -glucose exists in equilibrium between its open-chain form and cyclic hemiacetal forms. The open-chain form contains a free aldehydic $(-CHO)$ group,which reacts with hydroxylamine $(NH_2OH)$ to form an oxime.
However,when $D$-glucose is treated with acetic anhydride,all five hydroxyl $(-OH)$ groups are acetylated to form $D$-glucose pentaacetate. In this derivative,the hemiacetal hydroxyl group at $C-1$ is also converted into an ester group. Consequently,the pentaacetate cannot undergo ring-opening to form the open-chain aldehyde structure. Due to the absence of a free aldehydic group,$D$-glucose pentaacetate does not react with $NH_2OH$ to form an oxime.