(D) $1$. Glucose reacts with acetic anhydride to form glucose pentaacetate.$2$. This reaction indicates that there are $5$ hydroxyl $(-OH)$ groups pre

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

presence of five hydroxyl groups

(D) $1$. Glucose reacts with acetic anhydride to form glucose pentaacetate.$2$. This reaction indicates that there are $5$ hydroxyl $(-OH)$ groups present in the glucose molecule.$3$. The reaction is: $C_6H_{12}O_6 + 5(CH_3CO)_2O \rightarrow C_6H_7O(OCOCH_3)_5 + 5CH_3COOH$.