Write the reactions of D-Glucose with the following: (3 × 1)
(a) HI
(b) Br₂ water
(c) Conc. HNO₃
Write the reactions of D-Glucose with the following: (3 × 1)
(a) HI
(b) Br₂ water
(c) Conc. HNO₃
Reactions of D-Glucose (open-chain, CH₂OH(CHOH)₄CHO):
(a) With HI (heated with red phosphorus):
Glucose is completely reduced to n-hexane, which proves that its six carbon atoms are joined in a straight (unbranched) chain.
- CH₂OH(CHOH)₄CHO + HI → CH₃(CH₂)₄CH₃ (n-hexane)
(b) With Br₂ water (a mild oxidising agent):
Only the terminal –CHO group is oxidised to –COOH, giving D-gluconic acid.
- CH₂OH(CHOH)₄CHO + Br₂ + H₂O → CH₂OH(CHOH)₄COOH + 2HBr
(c) With conc. HNO₃ (a strong oxidising agent):
Both the –CHO group and the terminal –CH₂OH group are oxidised to –COOH, giving saccharic acid (D-glucaric acid).
- CH₂OH(CHOH)₄CHO + [O] → HOOC(CHOH)₄COOH
Marking Scheme
- 11 mark: reaction with HI giving n-hexane (accept mention of red phosphorus / prolonged heating).
- 21 mark: reaction with Br₂ water giving gluconic acid (–CHO oxidised to –COOH only).
- 31 mark: reaction with conc. HNO₃ giving saccharic acid / glucaric acid (both –CHO and –CH₂OH oxidised to –COOH).
- 4Full/half credit for correct product name even if the exact condensed formula is not written, provided the transformation is correct.
Hint
Reagent strength decides how much gets oxidised: HI reduces (→ n-hexane); Br₂ water is mild (aldehyde → mono-acid); conc. HNO₃ is strong (both ends → di-acid).
Quick Oral Answer
HI reduces glucose to n-hexane proving a straight six-carbon chain; mild Br₂ water oxidises only the aldehyde to give gluconic acid; strong conc. HNO₃ oxidises both the aldehyde and the terminal alcohol to give the dicarboxylic saccharic acid.
Analysis & Explanation
This question tests whether you can distinguish the behaviour of the aldehyde, the primary alcohol, and the whole carbon chain of glucose under three different reagents.
Concept — strength of the reagent decides the product:
- HI is a reducing agent, not an oxidant. On prolonged heating with red P it strips every –OH and the –CHO down to –CH, converting all six carbons into n-hexane. This classic result was key historical evidence for the straight six-carbon chain.
- Br₂ water is mild — it can only touch the most easily oxidised group, the aldehyde. So –CHO → –COOH and we stop at the mono-carboxylic gluconic acid.
- Conc. HNO₃ is strong — it oxidises both ends (the –CHO and the far –CH₂OH) to –COOH, giving the di-acid saccharic (glucaric) acid.
Exam trap: Students often write gluconic acid for conc. HNO₃ as well. Remember: Br₂ water = one end oxidised (mono-acid); conc. HNO₃ = both ends oxidised (di-acid).
Real-world link: Because ketoses like fructose lack a free –CHO in the reactive form, Br₂ water does not oxidise fructose — this reaction is used to chemically distinguish an aldose (glucose) from a ketose (fructose).
Common Mistakes
- 1Writing saccharic (di-)acid as the product of Br₂ water — Br₂ water is mild and oxidises only the –CHO end to give gluconic acid.
- 2Writing gluconic acid for conc. HNO₃ — the strong acid oxidises BOTH the –CHO and the terminal –CH₂OH, giving the dicarboxylic saccharic acid.
- 3Saying HI oxidises glucose — HI is a reducing agent and gives n-hexane, not an acid.
Interesting Facts
The reduction of glucose to n-hexane with HI was one of the pieces of evidence Emil Fischer used to establish the straight-chain structure of sugars; Fischer received the 1902 Nobel Prize in Chemistry for this work.
Bromine water oxidises aldoses (glucose) but not ketoses (fructose), so it is a simple bench test to tell an aldose from a ketose.
Gluconic acid and its salts (e.g. calcium gluconate, used as a calcium supplement and antidote for hydrofluoric-acid burns) are manufactured industrially by oxidising glucose.
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Frequently Asked Questions
Why does bromine water give gluconic acid but conc. HNO₃ give saccharic acid?
Bromine water is a mild oxidising agent that can only oxidise the reactive –CHO group to –COOH, so it stops at the mono-carboxylic gluconic acid. Conc. HNO₃ is a strong oxidising agent that oxidises both the –CHO group and the far-end primary –CH₂OH group to –COOH, producing the dicarboxylic saccharic (glucaric) acid.
What does the reaction of glucose with HI prove about its structure?
On heating with HI (and red phosphorus), glucose is reduced all the way to n-hexane. Since all six carbons end up in a single straight-chain alkane, this shows that the six carbon atoms of glucose are linked together in an unbranched (straight) chain.