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

Explanation

The cyclic hemiacetal form of glucose contains an `-OH` group at `C-1` which gets hydrolysed in aqueous solution of produce open chain aldehydic form which then reacts with `NH_(2)OH`-to form corresponding oxime. Thus, glucose contains an aldehydic group. However, when glucose is reacted with acetic anhydride, the `-OH` group at `C-l` along with the other `-OH` groups of `C-2, C-3, C-4` and `C-6` form a pentaacetate.
Since the penta acetate of 1 glucose does not contain a free `-OH` group at `C-1`, it cannot get hydrolysed in aqueous solution to produce open chain aldehydic form and hence glucose pentaacetage does not react with `NH_(2)OH` to form glucose oxime. The reactions are shown as