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NCERT Exemplar · Q34

Q.Consider compound (A): a cyclic (pyranose) form of glucose, drawn as a Fischer-type projection closed into a ring by an oxygen that joins the top carbon (the former carbonyl / anomeric carbon) to the ring carbon bearing the terminal -CH2OAc group. The key feature is that the anomeric (top) carbon carries an acetylated oxygen, -OAc, rather than a free hemiacetal -OH; the terminal group is -CH2OAc and the remaining ring carbons carry -OAc or -OH substituents. Why does compound (A) not react with hydroxylamine (NH2OH) to form an oxime?

Uttarakhand UbseShort· 2mImportance★★★★★est
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An oxime forms only from a free carbonyl group reacting with NH2OH. In compound (A) the anomeric hydroxyl is acetylated (-OAc), which locks the cyclic acetal so it cannot open into the open-chain aldehyde. With no free -CHO available, no oxime is produced.

Concept

Glucose exists mainly in the cyclic hemiacetal (pyranose) form but is in equilibrium with a small amount of the open-chain aldehyde. It is that open-chain -CHO that condenses with hydroxylamine to give an oxime:

-CHO + NH2OH -> -CH=N-OH + H2O

The ring can open to the aldehyde only because the anomeric carbon carries a free hemiacetal -OH.

Why compound (A) is inert to NH2OH …

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