Question

Which reagent oxidizes glucose to saccharic acid?

• A. $\text{dil. HNO}_3$
• B. $\text{NH}_2\text{OH}$
• C. HCN
• D. $\text{Br}_2\ \text{water}$

Hint:

Remember the oxidation gradient for glucose:
1. Mild oxidizing agent ($\text{Br}_2\ \text{water}$) $\rightarrow$ Mono-carboxylic acid (Gluconic acid).
2. Strong oxidizing agent ($\text{dil. HNO}_3$) $\rightarrow$ Di-carboxylic acid (Saccharic acid).

Answer 1

The Correct Option is A

Step 1: Understand the target product.
We want to turn glucose into saccharic acid. Saccharic acid has a $-\text{COOH}$ group at both ends of the carbon chain. So we need a reagent that can do this.
Step 2: Look at the two ends of glucose.
Glucose has an aldehyde group $-\text{CHO}$ at one end and a primary alcohol group $-\text{CH}_2\text{OH}$ at the other end. To get saccharic acid, BOTH of these must change into $-\text{COOH}$.
Step 3: Know that this needs a strong oxidiser.
Turning an alcohol all the way to an acid is hard. A weak reagent will not do it. We need a strong oxidising agent that attacks both ends.
Step 4: Check the weak and non-oxidising options.
Bromine water is mild. It only changes the $-\text{CHO}$ end into $-\text{COOH}$ (giving gluconic acid) and leaves the alcohol end alone. $\text{NH}_2\text{OH}$ and HCN are not oxidisers at all; they just add on to the carbonyl. So these three are ruled out.
Step 5: Pick the strong oxidiser.
Dilute nitric acid ($\text{dil. HNO}_3$) is a strong oxidising agent. It oxidises both the aldehyde end and the alcohol end together: \[ \text{CHO}-(\text{CHOH})_4-\text{CH}_2\text{OH} \xrightarrow{\text{dil. HNO}_3} \text{COOH}-(\text{CHOH})_4-\text{COOH} \]
Step 6: State the product.
This product with acid groups at both ends is saccharic acid. So the right reagent is dilute nitric acid. \[ \boxed{\text{dil. HNO}_3} \]