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

14.3 

D-glucose reacts with hydroxylamine (NH₂OH) to form oxime due to the presence of the aldehyde functional group (-CHO). This is due to the cyclic structure of glucose which forms an open chain structure in an aqueous medium, which then reacts to give an oxime.

But in case of pentaacetate of D-glucose, it does not form open chain structure in an aqueous medium so it does not react with NH₂OH.

<p><strong>14.3&nbsp;</strong></p><p>D-glucose reacts with hydroxylamine (NH₂OH) to form oxime due to the presence of the aldehyde functional group (-CHO). This is due to the cyclic structure of glucose which forms an open chain structure in an aqueous medium, which then reacts to give an oxime.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1686812289phpcp01fV_480x360.jpeg" media=" (max-width: 500px)"></picture></div><div><picture><img src="https://images.shiksha.com/mediadata/images/articles/1686812289phpcp01fV.jpeg" alt="" width="362" height="154"></picture></div><div><p>But in case of pentaacetate of D-glucose, it does not form open chain structure in an aqueous medium so it does not react with NH₂OH.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1686812308phpxR1rqr_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1686812308phpxR1rqr.jpeg" alt="" width="362" height="249"></picture></div></div></div></div>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment