Appeared in: 2025 (2 sets)
How do you explain the following ?
(a) Presence of an aldehydic group in glucose.
(b) Presence of five $-\text{OH}$ groups in glucose.
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Official Evaluation Steps & EL10 Insight
(a) Presence of an aldehydic group in glucose:
- Step 1 (Governing Reaction): [0.5 Mark] Glucose reacts with hydroxylamine ($\text{NH}_2\text{OH}$) to form an oxime and adds a molecule of hydrogen cyanide ($\text{HCN}$) to give cyanohydrin.
- Step 2 (Chemical Equation): [0.5 Mark] $$ \text{CHO}(\text{CHOH})_4\text{CH}_2\text{OH} + \text{NH}_2\text{OH} \rightarrow \text{CH}=\text{NOH}(\text{CHOH})_4\text{CH}_2\text{OH} + \text{H}_2\text{O} $$ $$ \text{CHO}(\text{CHOH})_4\text{CH}_2\text{OH} + \text{HCN} \rightarrow \text{CH}(\text{OH})(\text{CN})(\text{CHOH})_4\text{CH}_2\text{OH} $$
(b) Presence of five -OH groups in glucose:
- Step 1 (Governing Reaction): [0.5 Mark] Acetylation of glucose with acetic anhydride ($\text{(CH}_3\text{CO)}_2\text{O}$) indicates the presence of five hydroxyl groups by forming glucose pentaacetate.
- Step 2 (Chemical Equation): [0.5 Mark] $$ \text{CHO}(\text{CHOH})_4\text{CH}_2\text{OH} + 5(\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{CHO}(\text{CHOCOCH}_3)_4\text{CH}_2\text{OCOCH}_3 + 5\text{CH}_3\text{COOH} $$
💡 EL10 Concept Insight: Think of these tests like a "chemical detective" kit: the aldehyde group acts like a specific lock that only opens with keys like HCN, while the five -OH groups act like five identical hooks that all get "capped" by the acetic anhydride to prove they are there.