Question 9.SSE.7: How do you explain the absence of aldehyde group in the pent......
How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
The cyclic hemiacetal form of glucose contains an OH group at C-l which gets hydrolysed in the aqueous solution to produce the open chain aldehydic form which then with NH _2 OH to form the
\begin{matrix} \overset{5\left( CH _3 CO _2\right)_2 O}{\underset{-5 CH _3 COOH}{\xrightarrow {\hspace{2 cm}} }} \\\\\\\\\\\\\\ \end{matrix} 
\begin{matrix} \alpha-\text {Glucose Pentaacetate} \\ + \\ \beta -\text {Glucose Pentaacetate} \\ \downarrow NH_2OH \\ \text {No oxime} \end{matrix}
Corresponding oxime. Thus, glucose contains an aldehydic group. In contrast, when glucose is reacted with acetic anhydride, the OH group at C-l, 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-l it cannot get hydrolysed in aqueous solution to produce the open chain aldehydic form and hence glucose pentaacetate does not react with NH_2OH to form glucose form glucose oxime.
Thus, Glucose pentaacetate does not contain the aldehyde group.