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Question

Glucose reacts with excess acetic anhydride to form:

A
pentaacetate
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B
hexaacetate
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C
tetraacetate
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D
octaacetate
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Solution

The correct option is A pentaacetate
Acetylation of glucose with ​5 mole of acetic anhydride gives glucose pentaacetate which confirms the​ presence of five –OH groups.​
Since it exists as a stable compound, five –OH groups should be attached to different carbon atoms.​ Hence option (a) is correct.

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