Question

Glucose and gluconic acid on oxidation with dilute nitric acid forms saccharic acid. This reaction confirms that glucose contains

• A. four primary alcoholic groups.
• B. two primary alcoholic groups.
• C. one primary alcoholic group.
• D. five hydroxyl groups.

Hint:

Mild oxidizing agents like bromine water ($\text{Br}_2/\text{H}_2\text{O}$) target only the aldehyde group to make gluconic acid. Stronger agents like dilute $\text{HNO}_3$ target both the aldehyde and the single primary alcohol group to make saccharic acid. Comparing these reactions isolates the presence of that single primary alcohol.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks what specific functional group feature inside a glucose molecule is validated when both glucose and its mono-carboxylic derivative (gluconic acid) yield the exact same dicarboxylic acid (saccharic acid) upon oxidation with dilute $\text{HNO}_3$.

Step 2: Detailed Explanation:
Dilute nitric acid ($\text{HNO}_3$) behaves as a moderately powerful oxidizing agent capable of converting both aldehyde groups (-CHO) and primary alcoholic groups ($\text{-CH}_2\text{OH}$) into carboxylic acid functional units (-COOH). It is, however, incapable of breaking carbon-carbon bonds or oxidizing secondary alcoholic groups (-CH(OH)-) into acids under these conditions. Let's evaluate the structures and chemical changes:
1.

Glucose possesses one terminal aldehyde group (-CHO) at $\text{C}_1$, four secondary alcoholic groups, and one primary alcoholic group ($\text{-CH}_2\text{OH}$) located at the opposite terminal end ($\text{C}_6$).
2. When glucose is treated with dilute $\text{HNO}_3$, both the terminal -CHO group and the terminal primary $\text{-CH}_2\text{OH}$ group undergo complete oxidation to form a dicarboxylic acid called saccharic acid (also known as glucaric acid):
$$ \text{HOCH}_2-(\text{CHOH})_4-\text{CHO} \xrightarrow{\text{dil. HNO}_3} \text{HOOC}-(\text{CHOH})_4-\text{COOH} $$
3.

Gluconic acid is a monocarboxylic derivative where the $\text{C}_1$ position is already completely oxidized to a carboxylic acid (-COOH), leaving the $\text{C}_6$ primary alcohol intact:
$$ \text{HOCH}_2-(\text{CHOH})_4-\text{COOH} \xrightarrow{\text{dil. HNO}_3} \text{HOOC}-(\text{CHOH})_4-\text{COOH} $$ Since the oxidation of gluconic acid yields the exact same dicarboxylic architecture (saccharic acid), it proves that there was exactly one primary alcoholic group ($\text{-CH}_2\text{OH}$) ready at the terminal end to be oxidized.

Step 3: Final Answer: This chemical process explicitly confirms the presence of one primary alcoholic group, pointing directly to option (C).