JEE Mains Chemistry · Biomolecules

Monosaccharides: Classification and Glucose Reactions

Carbohydrates are sorted by how many units hydrolysis gives and by their carbonyl group and chain length; every monosaccharide is a reducing sugar, and each reaction of glucose (HI, NH₂OH, HCN, bromine water, acetic anhydride, nitric acid) proves one feature of its open-chain structure.

All Biomolecules notes JEE Mains Chemistry strategy

Why this matters

Seventeen PYQs, all multiple choice, three from 2026. Eight classify a sugar or name the colour test that picks it out: Seliwanoff's for ketoses, Barfoed's for monosaccharides, Tollens' and Fehling's for any reducing sugar. Nine are on the reactions of glucose and the structure they prove, or on the same reasoning applied to fructose and an unknown sugar; three of those are match-the-list questions pairing a reagent with its product.

Concept 1 of 2: Classes of carbohydrates and their colour tests

Two questions sort any sugar. How many units does it give on hydrolysis: one, a few or hundreds? And is its carbonyl an aldehyde or a ketone, on a chain of how many carbons? The colour tests then tell sugars apart by what they can do: reduce a metal ion, dehydrate fast in acid, or stain with iodine.

Definition

Monosaccharides cannot be hydrolysed further: glucose, fructose, galactose, ribose.
Oligosaccharides give 2 to 10 monosaccharide units. The units need not be the same: sucrose gives glucose and fructose, lactose gives galactose and glucose, maltose gives two glucose.
Polysaccharides give a very large number of units: starch, cellulose, glycogen.
By carbonyl and chain length: glucose is an aldohexose, fructose a ketohexose, ribose an aldopentose, glyceraldehyde an aldotriose.
All monosaccharides, aldose or ketose, are reducing sugars (NCERT). Fructose has no CHO, yet it reduces Tollens' reagent because in base it isomerises through an enediol to glucose and mannose.
In solution a monosaccharide's open-chain and cyclic forms coexist at equilibrium, as for D-(+)-glucose; the small open-chain share is what reduces the reagent.
The named colour tests (Benedict's, Barfoed's, Seliwanoff's, xanthoproteic) come from the practical manual, not from the chapter text of NCERT.

Test	Reagent	Positive for	What you see
Fehling's	Copper(II) sulphate with sodium potassium tartrate in NaOH	Every reducing sugar	Red precipitate of Cu₂O
Benedict's	Copper(II) sulphate with sodium citrate and sodium carbonate	Every reducing sugar	Orange-red precipitate of Cu₂O
Tollens'	$\mathrm{[Ag(NH_3)_2]^+}$ in ammonia	Every reducing sugar, fructose included	Silver mirror
Barfoed's	Copper(II) acetate in dilute acetic acid	Reducing monosaccharides, within about two minutes	Red Cu₂O; reducing disaccharides react only after long heating
Seliwanoff's	Resorcinol in hydrochloric acid	Ketoses quickly; aldoses and pentoses only slowly	Cherry-red colour
Iodine	Iodine in potassium iodide solution	Starch	Blue-black colour
Biuret	Dilute copper(II) sulphate in NaOH	Two or more peptide bonds: proteins, tripeptides, biuret itself	Violet colour
Xanthoproteic	Concentrated nitric acid	Proteins with aromatic side chains	Yellow colour, orange with ammonia

Seliwanoff's and the iodine test use no copper; Fehling's, Benedict's, Barfoed's and the biuret test all do.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q44Moderate

Example 1 · Biomolecules · Monosaccharides: Classification and Glucose Reactions

The incorrect statement from the following with respect to carbohydrates is :

Oligosaccharide units need not be identical

Only maltose gives two identical units. Sucrose gives glucose and fructose, and lactose gives galactose and glucose, so a statement that the units are always the same is false.

Fructose reduces Tollens' reagent without a CHO group

In alkaline solution fructose rearranges through an enediol to glucose and mannose, which carry CHO. So a ketose still gives the silver mirror, and every monosaccharide counts as reducing.

Seliwanoff's test uses no copper

Seliwanoff's reagent is resorcinol in hydrochloric acid; it works by dehydrating the sugar to a furfural that couples with resorcinol. Fehling's, Benedict's, Barfoed's and the biuret test all use copper(II).

Drill 7 more on classes of carbohydrates and their colour tests

Concept 2 of 2: Reactions that prove the open-chain structure of glucose

Each reagent answers one question about glucose. HI strips every oxygen and shows the six carbons are in a straight chain. Hydroxylamine and HCN show a carbonyl group. Bromine water, a mild oxidant, turns it into an acid of the same length, so the carbonyl is an aldehyde. Acetic anhydride counts the OH groups. Nitric acid oxidises both ends, so the other end is a primary alcohol. Read the same way, the reactions of an unknown sugar give its structure.

Definition

HI, prolonged heating → n-hexane: the six carbons form a straight chain. An unknown sugar giving isopentane has a branched chain.
$\mathrm{NH_2OH}$ → oxime; HCN → cyanohydrin: a carbonyl group is present.
Bromine water → gluconic acid, $\mathrm{COOH(CHOH)_4CH_2OH}$ : the carbonyl is an aldehyde, and only the CHO is oxidised.
Acetic anhydride → glucose pentaacetate: five OH groups, on five different carbons, since the compound is stable.
Nitric acid → saccharic (glucaric) acid, $\mathrm{COOH(CHOH)_4COOH}$ : both the CHO and the primary $\mathrm{CH_2OH}$ are oxidised.
$\mathrm{NaHCO_3}$ → no reaction: glucose has no COOH.
Adding HCN and hydrolysing the nitrile adds one carbon: fructose $\mathrm{C_6H_{12}O_6}$ gives the acid $\mathrm{C_7H_{14}O_8}$ . $\mathrm{NaBH_4}$ reduces the carbonyl to a hexitol, $\mathrm{C_6H_{14}O_6}$ , which HI then reduces to n-hexane.
Preparation (NCERT): sucrose boiled with dilute HCl or $\mathrm{H_2SO_4}$ in alcoholic solution gives glucose and fructose; starch boiled with dilute $\mathrm{H_2SO_4}$ at 393 K under 2 to 3 atm gives glucose.
Glucose dissolves in water because of its five OH groups, which hydrogen-bond with water, not because of its aldehyde group.

Glucose with bromine water, nitric acid and acetic anhydride

\mathrm{CHO(CHOH)_4CH_2OH} \xrightarrow{\mathrm{Br_2\ water}} \mathrm{COOH(CHOH)_4CH_2OH} \qquad \mathrm{CHO(CHOH)_4CH_2OH} \xrightarrow{\mathrm{HNO_3}} \mathrm{COOH(CHOH)_4COOH} \qquad M_{\text{acetate}} = M_{\text{sugar}} + 42\,n_{\mathrm{OH}}

Worked example

Glucose (molar mass 180 g mol

^{-1}

) is warmed with excess acetic anhydride. How many acetyl groups add, and what are the molecular formula and molar mass of the product?

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The same idea in a real exam question:

JEE Mains · 2022 · 27 July 2022 · Q137Moderate

Example 2 · Biomolecules · Monosaccharides: Classification and Glucose Reactions

Match List-I with List-II.

List-I	List-II
(A) Glucose $+ HI$	(I) Gluconic acid
(B) Glucose $+ Br_{2}$ water	(II) Glucose pentacetate
(C) Glucose $+$ acetic anhydride	(III) Saccharic acid
(D) Glucose $+ HNO_{3}$	(IV) Hexane

Choose the correct answer from the options given below:

Bromine water stops at one acid group

Bromine water oxidises only the CHO of glucose and gives gluconic acid, a monocarboxylic acid. The dicarboxylic saccharic acid needs nitric acid, which also oxidises the terminal CH₂OH.

Starch is hydrolysed by dilute acid

NCERT boils starch with dilute H₂SO₄ at 393 K under 2 to 3 atm to get glucose. A statement that uses concentrated sulphuric acid is false.

The acetate count is the OH count

A sugar that forms a tetraacetate has four OH groups and a pentaacetate five. Glucose pentaacetate has no free CHO, so it does not react with hydroxylamine or 2,4-DNP.

Drill 8 more on reactions that prove the open-chain structure of glucose

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Reactions that prove the open-chain structure of glucose
Glucose with bromine water, nitric acid and acetic anhydride
$\mathrm{CHO(CHOH)_4CH_2OH} \xrightarrow{\mathrm{Br_2\ water}} \mathrm{COOH(CHOH)_4CH_2OH} \qquad \mathrm{CHO(CHOH)_4CH_2OH} \xrightarrow{\mathrm{HNO_3}} \mathrm{COOH(CHOH)_4COOH} \qquad M_{\text{acetate}} = M_{\text{sugar}} + 42\,n_{\mathrm{OH}}$

Reference tables (1)

Classes of carbohydrates and their colour tests8 rows

Test	Reagent	Positive for	What you see
Fehling's	Copper(II) sulphate with sodium potassium tartrate in NaOH	Every reducing sugar	Red precipitate of Cu₂O
Benedict's	Copper(II) sulphate with sodium citrate and sodium carbonate	Every reducing sugar	Orange-red precipitate of Cu₂O
Tollens'	$\mathrm{[Ag(NH_3)_2]^+}$ in ammonia	Every reducing sugar, fructose included	Silver mirror
Barfoed's	Copper(II) acetate in dilute acetic acid	Reducing monosaccharides, within about two minutes	Red Cu₂O; reducing disaccharides react only after long heating
Seliwanoff's	Resorcinol in hydrochloric acid	Ketoses quickly; aldoses and pentoses only slowly	Cherry-red colour
Iodine	Iodine in potassium iodide solution	Starch	Blue-black colour
Biuret	Dilute copper(II) sulphate in NaOH	Two or more peptide bonds: proteins, tripeptides, biuret itself	Violet colour
Xanthoproteic	Concentrated nitric acid	Proteins with aromatic side chains	Yellow colour, orange with ammonia

Seliwanoff's and the iodine test use no copper; Fehling's, Benedict's, Barfoed's and the biuret test all do.

Watch out for (6)

Oligosaccharide units need not be identical
→ Classes of carbohydrates and their colour tests
Fructose reduces Tollens' reagent without a CHO group
→ Classes of carbohydrates and their colour tests
Seliwanoff's test uses no copper
→ Classes of carbohydrates and their colour tests
Bromine water stops at one acid group
→ Reactions that prove the open-chain structure of glucose
Starch is hydrolysed by dilute acid
→ Reactions that prove the open-chain structure of glucose
The acetate count is the OH count
→ Reactions that prove the open-chain structure of glucose

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