Reactions of Glucose with HI, Bromine Water and Nitric Acid
Chemistry · Biomolecules · NEET
Three reagents give three different products from glucose. Long heating with HI gives n-hexane (a straight 6-carbon chain). Mild bromine water gives gluconic acid (a mono-carboxylic acid, from the -CHO group). Strong nitric acid (HNO3) gives saccharic acid (a di-carboxylic acid, from both the -CHO and the end -CH2OH). Memory hook: "HI Hexane, Bromine one-acid, Nitric two-acid" — the stronger the oxidiser, the more acid groups you get.
Glucose gives n-hexane with HI (proves straight chain), gluconic acid with mild bromine water (one -COOH, proves aldehyde), and saccharic acid with strong HNO3 (two -COOH, proves terminal -CH2OH).
Your doubts, answered
Why does glucose give n-hexane with HI, and what does it prove?
On long (prolonged) heating with HI (hydroiodic acid, a strong reducing agent), every oxygen in glucose is stripped off and all six carbons stay joined. The product is n-hexane, a straight 6-carbon alkane. This proves that the six carbon atoms of glucose are in an unbranched, straight chain. Remember: 'HI reduces glucose to n-Hexane.'
What is the difference between bromine water and nitric acid for glucose?
Bromine water is a MILD oxidiser: it only oxidises the aldehyde (-CHO) group to -COOH, giving gluconic acid (ONE -COOH, five carbons still have -OH/CH2OH). Nitric acid (HNO3) is a STRONG oxidiser: it oxidises BOTH the aldehyde end (-CHO) and the far -CH2OH end to -COOH, giving saccharic acid (TWO -COOH groups). Stronger oxidiser = more -COOH groups.
Why does glucose give a carboxylic acid (gluconic acid) with bromine water?
Bromine water gently oxidises the carbonyl group of glucose. Because the product is a carboxylic acid (gluconic acid), it means the carbonyl was an ALDEHYDE (-CHO), not a ketone. Only aldehydes are oxidised to acids this easily. So this reaction proves glucose carries an aldehyde group (it is an aldose).
What does the formation of saccharic acid with HNO3 prove?
Saccharic acid is a di-carboxylic acid (a -COOH at BOTH ends). One -COOH comes from the aldehyde (-CHO). The other -COOH forms only if the last carbon was a primary alcohol (-CH2OH), because HNO3 oxidises -CH2OH to -COOH. So saccharic acid proves glucose has a terminal primary alcoholic (-CH2OH) group.
Does gluconic acid also give saccharic acid with HNO3?
Yes. NCERT states that BOTH glucose and gluconic acid give saccharic acid on oxidation with nitric acid. Gluconic acid already has one -COOH (from -CHO); HNO3 then oxidises its -CH2OH end to a second -COOH, giving saccharic acid. This is a common NEET trap statement.
⚠️ The NEET trap ✗ Bromine water oxidises glucose all the way to saccharic acid (a dicarboxylic acid). ✓ Bromine water is MILD — it stops at gluconic acid (mono-carboxylic acid). Only strong HNO3 goes further to saccharic acid (dicarboxylic acid). 🧠 Match the strength to the number of -COOH: mild bromine = ONE acid group, strong nitric = TWO acid groups.
Real NEET questions
NEET 2024
The reagents with which glucose does NOT react to give the corresponding test/product are: A. Tollen's reagent B. Schiff's reagent C. HCN D. NH2OH E. NaHSO3. Choose the correct option.
A · A and D
B · B and E ✓
C · E and D
D · B and C
Solution: In water, glucose stays mostly in the cyclic hemiacetal (pyranose) form, so a truly free -CHO is scarce. Schiff's test (B) and the NaHSO3 bisulphite addition (E) both need a free aldehyde, so glucose fails both. Tollen's (A), HCN (C) and NH2OH (D) work through the tiny open-chain amount. This connects to bromine water: bromine water succeeds in oxidising that same aldehyde group to gluconic acid. Non-reacting reagents = B and E, so option B.
NEET 2016 Phase 1
Which one given below is a non-reducing sugar?
A · Maltose
B · Lactose
C · Glucose
D · Sucrose ✓
Solution: Glucose has a free -CHO group, which is exactly why bromine water oxidises it to gluconic acid — so glucose is a REDUCING sugar. Maltose and lactose keep one free anomeric carbon and are also reducing. Sucrose locks both anomeric carbons in its glycosidic bond, so it has no free -CHO and is NON-reducing. Answer: (D) Sucrose.
Solved Biomolecules NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What are the products of glucose with HI, bromine water and HNO3?
HI gives n-hexane, bromine water gives gluconic acid, and nitric acid (HNO3) gives saccharic acid.
Is gluconic acid a monocarboxylic or dicarboxylic acid?
Gluconic acid is a MONO-carboxylic acid — it has just one -COOH group (from the aldehyde). Saccharic acid is a DI-carboxylic acid with two -COOH groups.
Which reagent proves the straight chain of glucose?
HI. Prolonged heating with HI gives n-hexane, a straight 6-carbon alkane, proving all six carbons of glucose are in an unbranched chain.
Which reagent proves the primary -CH2OH group in glucose?
Nitric acid (HNO3). It oxidises the terminal -CH2OH to a second -COOH, forming saccharic acid (dicarboxylic), which proves a primary alcohol group is present.
Why does bromine water not oxidise ketones like fructose to acids?
Bromine water only oxidises aldehyde (-CHO) groups to -COOH. Fructose has a keto group, not a free aldehyde, so it does not give an acid with bromine water. This is a useful test to tell aldoses from ketoses.
