Question:medium

Write the reactions of D-Glucose with the following:

(a) HI
(b) \(Br_2\) water
(c) Conc. \(HNO_3\)

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Glucose has an open-chain aldohexose structure with one aldehyde group, five carbons bearing hydroxyl groups (one primary, four secondary) and a straight carbon chain.
Updated On: Jun 16, 2026
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Solution and Explanation

Step 1: Look at the structure of glucose.
Open-chain glucose is an aldohexose: it has one aldehyde \((-CHO)\) group at the top, four secondary \(-OH\) groups, one primary \(-OH\) group at the bottom, and a straight six-carbon chain. Each reagent attacks a different feature.

Step 2: Reaction with HI.
Hot HI is a strong reducing agent. It removes all the oxygen functions and reduces the whole chain to a straight-chain alkane, n-hexane. \[ \text{Glucose} \xrightarrow{\text{HI, heat}} CH_3(CH_2)_4CH_3\ (\text{n-hexane}) \]

Step 3: Reaction with bromine water.
\(Br_2\) water is a mild oxidising agent. It oxidises only the aldehyde group to a carboxylic acid, giving gluconic acid (a six-carbon monocarboxylic acid). \[ \text{Glucose} \xrightarrow{Br_2/H_2O} \text{Gluconic acid (CHO} \to \text{COOH)} \]

Step 4: Reaction with concentrated nitric acid.
Conc. \(HNO_3\) is a strong oxidising agent. It oxidises both the top aldehyde group and the bottom primary \(-OH\) group to \(-COOH\), giving saccharic acid (glucaric acid), a dicarboxylic acid. \[ \text{Glucose} \xrightarrow{\text{conc. } HNO_3} \text{Saccharic acid} \]

Step 5: Summarise.
HI gives n-hexane, \(Br_2\) water gives gluconic acid, and conc. \(HNO_3\) gives saccharic acid.

Answer: (a) HI gives n-hexane. (b) \(Br_2\) water gives gluconic acid. (c) Conc. \(HNO_3\) gives saccharic (glucaric) acid.
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