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.