Question

Which of the following compound is obtained when glucose is treated with dilute nitric acid?

• A. Glucose oxime
• B. Gluconic acid
• C. Saccharic acid
• D. Glucose cyanohydrin

Hint:

To remember the oxidation products of glucose easily, keep this summary in mind:

• Glucose + Bromine water ($\text{Br}_2/\text{H}_2\text{O}$) $\rightarrow$ Gluconic acid (Monocarboxylic acid)

• Glucose + Nitric acid ($\text{HNO}_3$) $\rightarrow$ Saccharic acid (Dicarboxylic acid)

Answer 1

The Correct Option is C

Step 1: Identify the reactive ends of glucose.
Glucose, $\text{CHO}-(\text{CHOH})_4-\text{CH}_2\text{OH}$, has an aldehyde at C1 and a primary alcohol at C6.
Step 2: Recall the strength of the oxidant.
Dilute nitric acid is a strong enough oxidising agent to attack both terminal groups, unlike mild bromine water which touches only the aldehyde.
Step 3: Oxidise the aldehyde end.
The C1 aldehyde $-\text{CHO}$ is oxidised to a carboxylic acid $-\text{COOH}$.
Step 4: Oxidise the alcohol end.
At the same time the C6 primary alcohol $-\text{CH}_2\text{OH}$ is oxidised to a carboxylic acid $-\text{COOH}$.
Step 5: Write the product.
Both ends now carry $-\text{COOH}$, giving $\text{COOH}-(\text{CHOH})_4-\text{COOH}$, a six-carbon dicarboxylic acid.
Step 6: Name it.
This dicarboxylic acid is saccharic acid (glucaric acid). The other options come from milder or different reagents.
\[ \boxed{\text{Saccharic acid, option (3)}} \]