Question

How do you explain the following ? (a) Presence of a carbonyl group in glucose (b) Presence of five --OH groups in glucose

Hint:

Oxime and cyanohydrin formation indicate the presence of a carbonyl group, while formation of glucose pentaacetate confirms the presence of five hydroxyl groups in glucose.

Answer 1

Step 1: Evidence for carbonyl group in glucose.
Glucose reacts with $\text{NH}_2\text{OH}$ to give glucose oxime, and with HCN to give cyanohydrin. Both are characteristic reactions of carbonyl ($\text{C=O}$) compounds. Reduction with HI and red P gives n-hexane, confirming an unbranched 6-carbon aldehyde chain.
Step 2: Evidence for five -OH groups in glucose.
Glucose reacts with excess acetic anhydride: \[\text{Glucose} + 5(\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{Glucose pentaacetate}\] Each -OH group reacts with one mole of acetic anhydride. Formation of a pentaacetate proves exactly five hydroxyl groups.
Step 3: Structural summary.
Together these results establish glucose as an aldohexose ($\text{C}_6\text{H}_{12}\text{O}_6$): one C=O (aldehyde) at C-1 and one -OH each at C-2, C-3, C-4, C-5, and C-6, consistent with all the above reactions.