Explain: (i) Presence of carbonyl group in glucose. (ii) Presence of five $-$OH groups attached to different carbon atoms.
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Solution and Explanation
These proofs are based on chemical reactions that are selective for specific functional groups.
Step 2: Analysis (i) Carbonyl group
D-Glucose reacts with hydroxylamine ($NH_2OH$) to form a glucose oxime, and with hydrazines to form glucosazone. It also reduces Tollen's reagent (silver mirror) and Fehling's solution (brick-red precipitate of $Cu_2O$). These are characteristic reactions of an aldehyde group. Furthermore, catalytic hydrogenation of glucose gives sorbitol (a hexahydric alcohol), proving the presence of a $C=O$ group that gets reduced to $-$CHOH.
Step 3: Analysis (ii) Five $-$OH groups
Glucose reacts with excess acetic anhydride to form glucose pentaacetate ($C_6H_7O(OCOCH_3)_5$). Since exactly five acetyl groups are incorporated, there must be exactly five free hydroxyl groups in the glucose molecule. Since glucose has 6 carbons and one is the carbonyl carbon ($-$CHO), the remaining five carbons each bear one $-$OH group.
Final Answer:
(i) Glucose reacts with $NH_2OH$, reduces Tollen's reagent, and undergoes catalytic hydrogenation to a hexahydric alcohol -- all prove the presence of an aldehyde group.
(ii) Glucose forms a pentaacetate with acetic anhydride -- proving five free $-$OH groups, each on a different carbon.
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