How do you explain the presence of five $-OH$ groups in a glucose molecule?
(N/A) Glucose reacts with acetic anhydride to form glucose pentaacetate.
Since acetic anhydride reacts with $-OH$ groups to form esters (acetylation),the formation of a penta-derivative indicates the presence of five $-OH$ groups in the glucose molecule.
The reaction is:
$CHO-(CHOH)_4-CH_2OH + 5(CH_3CO)_2O \rightarrow CHO-(CHOCOCH_3)_4-CH_2OCOCH_3 + 5CH_3COOH$.
Since acetic anhydride reacts with $-OH$ groups to form esters (acetylation),the formation of a penta-derivative indicates the presence of five $-OH$ groups in the glucose molecule.
The reaction is:
$CHO-(CHOH)_4-CH_2OH + 5(CH_3CO)_2O \rightarrow CHO-(CHOCOCH_3)_4-CH_2OCOCH_3 + 5CH_3COOH$.
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