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

Answer

The cyclic hemiacetal form of glucose contains an `OH` group at `C-1` which gets hydrolysed in aqueous solution to produce the open chain aldehydic form which then reacts with `NH_(2)OH` to form the corresponding oxime. Thus, glucose contains an aldehydic group. In contrast, when glucose is reacted with acetic anhydride, the `OH` group at `C-1`, along with the four other `OH` groups at `C-2,C-3,C-4` and `C-6` form a pentaacetate. Since the pentaacetate of glucose does not contain a free `OH` group at `C-1`, it cannot get hydrolysed in aqueous solution to produce the open chain aldehydic form and hence glucose pentaacetate does not react with `NH_(2) OH` to form glucose oxime. This proves that glucose pentaacetate does not contain the aldehyde group.
`(##RES_ORG_CHM_V02_XII_C04_E01_015_A01##)`.