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: Understand the clue given.
Both glucose and gluconic acid, when oxidised by dilute nitric acid, give the same product, saccharic acid. We must say what this proves about glucose.

Step 2: Know what dilute nitric acid does.
Dilute $\text{HNO}_3$ is a mild oxidiser. It turns an aldehyde group (-CHO) into -COOH, and it also turns a primary alcohol group ($\text{-CH}_2\text{OH}$) into -COOH. It does not touch the secondary alcohol groups in the middle.

Step 3: Look at glucose.
Glucose has a -CHO at one end (carbon 1) and a $\text{-CH}_2\text{OH}$ at the other end (carbon 6), with -OH groups in between.
\[ \text{HOCH}_2\text{-(CHOH)}_4\text{-CHO} \rightarrow \text{HOOC-(CHOH)}_4\text{-COOH} \]
So both ends become -COOH, giving saccharic acid.

Step 4: Look at gluconic acid.
Gluconic acid already has -COOH at carbon 1 but still has the $\text{-CH}_2\text{OH}$ at carbon 6.
\[ \text{HOCH}_2\text{-(CHOH)}_4\text{-COOH} \rightarrow \text{HOOC-(CHOH)}_4\text{-COOH} \]

Step 5: Compare the two results.
Gluconic acid gives the same saccharic acid only because its leftover $\text{-CH}_2\text{OH}$ at the far end got oxidised to -COOH. This shows glucose has exactly one such primary alcohol group at the end.

Step 6: Choose the answer.
The experiment proves glucose contains one primary alcoholic group, which is option 3.
\[ \boxed{\text{one primary alcoholic group}} \]