Q147 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Chemistry
Identify the product formed when glucose is treated with dil. $\text{HNO}_3$.
- Correct answerASaccharic acid
- BOxalic acid
- CGluconic acid
- DMalonic acid
Solution
Glucose on oxidation with dilute nitric acid ($\text{HNO}_3$) yields a dicarboxylic acid known as saccharic acid.
Dilute $\text{HNO}_3$ is a strong oxidizing agent that oxidizes both the terminal aldehyde group ($-\text{CHO}$) and the terminal primary alcohol group ($-\text{CH}_2\text{OH}$) of glucose into carboxylic acid groups ($-\text{COOH}$).
$\text{CHO}-(\text{CHOH})_4-\text{CH}_2\text{OH} \xrightarrow{\text{dil. HNO}_3} \text{COOH}-(\text{CHOH})_4-\text{COOH}$
Thus, the product formed is saccharic acid.
Answer: Saccharic acid
Dilute $\text{HNO}_3$ is a strong oxidizing agent that oxidizes both the terminal aldehyde group ($-\text{CHO}$) and the terminal primary alcohol group ($-\text{CH}_2\text{OH}$) of glucose into carboxylic acid groups ($-\text{COOH}$).
$\text{CHO}-(\text{CHOH})_4-\text{CH}_2\text{OH} \xrightarrow{\text{dil. HNO}_3} \text{COOH}-(\text{CHOH})_4-\text{COOH}$
Thus, the product formed is saccharic acid.
Answer: Saccharic acid
Hence, the correct answer is option: 'A'.