Question

Explain: (i) Presence of carbonyl group in glucose. (ii) Presence of five $-$OH groups attached to different carbon atoms.

Hint:

Pentaacetate formation $\rightarrow$ 5 OH groups. Tollen's test/oxime formation $\rightarrow$ aldehyde group. Hydrogenation to sorbitol confirms $-$CHO.

Answer 1

Step 1: Evidence for the carbonyl (aldehyde) group.
Glucose reacts with hydroxylamine ($NH_2OH$) to form a glucose oxime, and with Tollen's reagent gives a silver mirror (confirms aldehyde). Fehling's solution gives a brick-red precipitate of $Cu_2O$. These reactions are specific to aldehyde groups.
Step 2: Further proof of $-CHO$.
Catalytic hydrogenation ($H_2/Ni$) of glucose gives sorbitol (a hexahydric alcohol), proving the $-CHO$ group was present and is reduced to $-CH_2OH$.
Step 3: Evidence for five $-OH$ groups.
Glucose reacts with excess acetic anhydride to form glucose pentaacetate, incorporating exactly five acetyl groups. This proves five free $-OH$ groups. Since glucose has 6 carbons (one is the aldehyde carbon), the remaining five carbons each bear one $-OH$ group.
Step 4: Summary.
(i) Aldehyde group: evidenced by reactions with $NH_2OH$, Tollen's reagent, Fehling's solution, and $H_2$ reduction to sorbitol. (ii) Five $-OH$ groups: evidenced by pentaacetate formation with acetic anhydride.