Question

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

Answer 1

D-glucose reacts with hydroxylamine (NH₂OH) to form an oxime because of the presence of aldehydic (−CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with NH₂OH to give an oxime.

But pentaacetate of D-glucose does not react with NH₂OH. This is because pentaacetate does not form an open chain structure.

Answer 2

The cyclic hemiacetal form of glucose contains an -OH group at C-l which gets hydrolysed in aqueous solution to produce open chain aldehydic form which then reacts with NH₂OH -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 at C-2, C-3, C-4 and C-6 form a pentaacetate.
Since the penta acetate of1 glucose does not contain a free -OH group at C-l, it cannot get hydrolysed in aqueous solution to produce open chain aldehydic form and hence glucose pentaacetate does not react with NH₂OH to form glucose oxime. The reactions are shown as: