Q.What happens when D-glucose is treated with the following reagents?

The key idea is that each reagent targets a specific functional group in glucose, producing characteristic oxidation or reduction products.

Step 1 — Reaction with $HI$: Hydroiodic acid is a strong reducing agent. It reduces the aldehyde group ($-CHO$) to a methyl group ($-CH_3$), and also reduces all the hydroxyl groups ($-OH$) to hydrogen atoms, yielding n-hexane as the final product.

Step 2 — Reaction with bromine water: Bromine water ($Br_2/H_2O$) is a mild oxidising agent that selectively oxidises the aldehyde group to a carboxylic acid group, without affecting the alcohol groups. This gives gluconic acid. …

Let’s understand why each reagent does what it does. D-glucose is an aldohexose — it has an aldehyde group at C1 and hydroxyl groups on every other carbon. The behaviour of these reagents depends on their oxidising or reducing power and their selectivity.

1. Reaction with HI (hydroiodic acid)

HI is a strong reducing agent. In hot, concentrated HI, all the –OH groups in glucose are replaced by –I, and then the C–I bonds are reduced to C–H bonds. This is a reductive deoxygenation — every hydroxyl group gets removed, and the aldehyde group also gets reduced to a methyl group. The entire carbon chain survives intact, but all oxygen is stripped off.

The product is n-hexane (CH₃–CH₂–CH₂–CH₂–CH₂–CH₃).

2. Reaction with bromine water

Bromine water (Br₂ in H₂O) is a mild oxidising agent. It selectively oxidises the aldehyde group (–CHO) to a carboxylic acid (–COOH) without touching the alcohol groups. This is because bromine water specifically targets aldehydes (and not ketones or alcohols) under neutral or slightly acidic conditions.

So D-glucose gives gluconic acid (a monocarboxylic acid where C1 is –COOH, and the rest of the chain remains unchanged).

Method: Reagent-Specific Functional Group Analysis

This method uses the functional group reactivity of D-glucose (an aldohexose) to predict products with each reagent.

Steps

Step 1: Identify the functional groups in D-glucose

• One aldehyde group (–CHO) at C1
• Four secondary alcohol groups (–OH) at C2, C3, C4, C5
• One primary alcohol group (–CH₂OH) at C6

Step 2: Apply each reagent based on its known reaction with these groups

(i) With $HI$ (hydroiodic acid) — Reductive cleavage

• $HI$ is a strong reducing agent that cleaves C–O bonds and reduces all oxygen-containing groups.
• All –OH groups are replaced by –H, and the aldehyde is reduced to –CH₃.
• Product: n-Hexane ($\text{CH}_3\text{–CH}_2\text{–CH}_2\text{–CH}_2\text{–CH}_2\text{–CH}_3$)

(ii) With bromine water ($\text{Br}_2/\text{H}_2\text{O}$) — Oxidation of aldehyde only

• Bromine water is a mild oxidizing agent that selectively oxidizes the aldehyde group to a carboxylic acid (–COOH).
• Alcohol groups remain unchanged.
• Product: Gluconic acid (a six-carbon aldonic acid)

(iii) With $HNO_3$ (nitric acid) — Oxidation of both ends

• $HNO_3$ is a strong oxidizing agent that oxidizes both: …

Here are the common mistakes students make when answering questions about the reactions of D-glucose with HI, Bromine water, and HNO₃, along with how to avoid each.

Mistake 1: Confusing the reaction with HI as simple reduction

The error: Students often write that HI simply reduces the aldehyde group ($-CHO$) to an alcohol ($-CH_2OH$), treating it like a mild reducing agent.

Why it’s wrong: HI is a strong reducing agent under heat. It does not stop at the aldehyde. It reduces all the hydroxyl groups ($-OH$) to hydrogen atoms ($-H$), converting the sugar into a straight-chain hydrocarbon.

Correct outcome:

D-glucose + excess HI (heat) → n-Hexane ($CH_3(CH_2)_4CH_3$).

The entire carbon chain is reduced, and all oxygen atoms are removed.

How to avoid:

• Remember: HI + heat = complete deoxygenation of sugars.
• Think of it as “stripping” every $-OH$ group, leaving only a saturated alkane.

Mistake 2: Thinking bromine water oxidizes all parts of glucose

The error: Students assume bromine water ($Br_2/H_2O$) is a strong enough oxidant to break the glucose chain or oxidize all carbon atoms.

Why it’s wrong: Bromine water is a mild, selective oxidant. It only oxidizes the aldehyde group ($-CHO$) to a carboxylic acid group ($-COOH$). It does not affect the alcohol groups ($-OH$) on the rest of the chain.

Correct outcome:

D-glucose + Bromine water → Gluconic acid (a monocarboxylic acid).

The aldehyde at C1 becomes $-COOH$, while all other $-OH$ groups remain unchanged.

How to avoid:

• Memorize: Bromine water = specific test for aldehydes in sugars.
• Do not confuse it with strong oxidants like $HNO_3$ (see next mistake).

Mistake 3: Confusing the action of $HNO_3$ with bromine water

The error: Students write that $HNO_3$ also gives gluconic acid, or that it only oxidizes the aldehyde group.

Why it’s wrong: $HNO_3$ is a strong oxidizing agent. It oxidizes both ends of the glucose chain — the aldehyde group ($-CHO$) at C1 and the primary alcohol group ($-CH_2OH$) at C6 — into carboxylic acid groups ($-COOH$).

Correct outcome: …