JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
Mathematics · Physics · Chemistry
This MCQ module is based on: Carbohydrates Monosaccharides
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Carbohydrates Monosaccharides
🎓 Class 12
Chemistry
CBSE
Theory
Ch 10 – Biomolecules
⏱ ~14 min
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Carbohydrates — Structure of Glucose and Fructose
A living system grows, sustains and reproduces itself. The most amazing thing about it is that it is composed of non-living atoms and molecules. The pursuit of what goes on chemically within a living system is the domain of biochemistry. Living systems are built from complex biomolecules — carbohydrates, proteins, nucleic acids and lipids — which interact to constitute the molecular logic of life.
10.1 Carbohydrates
Carbohydrates are primarily produced by plants and form a very large group of naturally occurring organic compounds. Common examples are cane sugar, glucose and starch. Most have the general formula Cx(H₂O)y and were once considered hydrates of carbon — hence the name.
Why the old definition fails. Glucose (C₆H₁₂O₆) fits C₆(H₂O)₆ perfectly. But acetic acid CH₃COOH also fits C₂(H₂O)₂ and is not a carbohydrate, while rhamnose C₆H₁₂O₅ is a carbohydrate but does not fit the formula at all. The formula is a historical accident, not a definition.
The modern chemical definition. Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds which produce such units on hydrolysis.
Some carbohydrates, being sweet in taste, are also called sugars. The most common sugar used in our homes is sucrose, while the sugar present in milk is lactose. Carbohydrates are also called saccharides (Greek sakcharon, sugar).
10.1.1 Classification of carbohydrates
Carbohydrates are classified on the basis of their behaviour on hydrolysis into three broad groups.
| Class | Behaviour on hydrolysis | Examples |
|---|---|---|
| Monosaccharides | cannot be hydrolysed further to a simpler polyhydroxy aldehyde or ketone | glucose, fructose, ribose |
| Oligosaccharides | yield two to ten monosaccharide units | sucrose, maltose, lactose (all disaccharides) |
| Polysaccharides | yield a large number of monosaccharide units | starch, cellulose, glycogen, gums |
About 20 monosaccharides are known to occur in nature. Oligosaccharides are further classified as disaccharides, trisaccharides, tetrasaccharides and so on depending on the number of monosaccharides they provide on hydrolysis; the most common are the disaccharides. The two units obtained may be the same or different — one molecule of sucrose gives one glucose and one fructose, whereas maltose gives two molecules of glucose only. Polysaccharides are not sweet in taste, hence they are also called non-sugars.
Reducing and non-reducing sugars. All carbohydrates which reduce Fehling's solution and Tollens' reagent are called reducing sugars. All monosaccharides, whether aldose or ketose, are reducing sugars.
10.1.2 Monosaccharides
Monosaccharides are further classified on the basis of the number of carbon atoms and the functional group present. A monosaccharide containing an aldehyde group is an aldose; one containing a keto group is a ketose. The number of carbon atoms is built into the name.
Table 10.1 — Different types of monosaccharides
| Carbon atoms | General term | Aldehyde | Ketone |
|---|---|---|---|
| 3 | Triose | Aldotriose | Ketotriose |
| 4 | Tetrose | Aldotetrose | Ketotetrose |
| 5 | Pentose | Aldopentose | Ketopentose |
| 6 | Hexose | Aldohexose | Ketohexose |
| 7 | Heptose | Aldoheptose | Ketoheptose |
10.1.2.1 Glucose
Glucose occurs freely in nature as well as in the combined form. It is present in sweet fruits and honey, and ripe grapes contain it in large amounts.
Preparation of glucose
1. From sucrose (cane sugar). If sucrose is boiled with dilute HCl or H₂SO₄ in alcoholic solution, glucose and fructose are obtained in equal amounts.
C₁₂H₂₂O₁₁ + H₂O —[H⁺]→ C₆H₁₂O₆ + C₆H₁₂O₆
sucrose glucose fructose
sucrose glucose fructose
2. From starch. Commercially glucose is obtained by hydrolysis of starch by boiling it with dilute H₂SO₄ at 393 K under 2–3 atm pressure.
(C₆H₁₀O₅)n + nH₂O —[H⁺, 393 K, 2–3 atm]→ nC₆H₁₂O₆
Structure of glucose — the six evidences
Glucose is an aldohexose and is also known as dextrose. It is the monomer of many larger carbohydrates, namely starch and cellulose, and is probably the most abundant organic compound on earth. Its open-chain structure was assigned on the basis of the following evidences.
| # | Observation | Structural conclusion |
|---|---|---|
| 1 | Molecular formula found to be C₆H₁₂O₆ | six carbons, six oxygens |
| 2 | On prolonged heating with HI it forms n-hexane | all six carbon atoms are linked in a straight chain |
| 3 | Forms an oxime with hydroxylamine and adds HCN to give a cyanohydrin | a carbonyl group (>C=O) is present |
| 4 | Oxidised by mild bromine water to a six-carbon acid, gluconic acid | the carbonyl group is present as an aldehydic group |
| 5 | Acetylation with acetic anhydride gives glucose pentaacetate | five –OH groups, each on a different carbon (since the compound is stable) |
| 6 | On oxidation with HNO₃, both glucose and gluconic acid give the dicarboxylic saccharic acid | a primary alcoholic –OH group is present |
CHO–(CHOH)₄–CH₂OH —[Br₂ water]→ COOH–(CHOH)₄–CH₂OH (gluconic acid)
CHO–(CHOH)₄–CH₂OH —[HNO₃]→ COOH–(CHOH)₄–COOH (saccharic acid)
CHO–(CHOH)₄–CH₂OH —[HNO₃]→ COOH–(CHOH)₄–COOH (saccharic acid)
The exact spatial arrangement of the different –OH groups was given by Fischer after studying many other properties.
D and L notation
Glucose is correctly named D(+)-glucose. Here ‘D’ represents the configuration whereas ‘(+)’ represents the dextrorotatory nature of the molecule.
The distinction students most often blur. ‘D’ and ‘L’ have no relation with the optical activity of the compound, and they are not related to the small letters ‘d’ and ‘l’. D tells you a configuration; (+) or (−) tells you the direction of rotation, which must be measured.
The letters D or L before the name indicate the relative configuration of a stereoisomer with respect to a compound whose configuration is known. In carbohydrates this reference is glyceraldehyde, which contains one asymmetric carbon and exists in two enantiomeric forms.
The (+) isomer of glyceraldehyde has the D configuration, meaning the –OH group lies on the right-hand side in its Fischer projection. All compounds chemically correlated to D(+)-glyceraldehyde have D configuration; those correlated to L(−)-glyceraldehyde have L configuration, with the –OH on the left.
The rule for assigning D or L to a sugar. Look only at the lowest asymmetric carbon atom — the one furthest from the carbonyl group. In (+)-glucose the –OH on this carbon is on the right, comparable to (+)-glyceraldehyde, so (+)-glucose is assigned the D configuration. The other asymmetric carbons are not considered. The structure is always written with the most oxidised carbon (–CHO) at the top.
Cyclic structure of glucose
The open-chain structure explained most properties of glucose, but three facts could not be explained.
- Despite having an aldehyde group, glucose does not give Schiff's test and does not form the hydrogensulphite addition product with NaHSO₃.
- The pentaacetate of glucose does not react with hydroxylamine, indicating the absence of a free –CHO group.
- Glucose exists in two different crystalline forms, α and β. The α-form (m.p. 419 K) is obtained by crystallisation from a concentrated solution at 303 K, while the β-form (m.p. 423 K) is obtained by crystallisation from a hot saturated aqueous solution at 371 K.
It was therefore proposed that one of the –OH groups adds to the –CHO group to form a cyclic hemiacetal structure. It was found that glucose forms a six-membered ring in which the –OH at C-5 is involved in ring formation. This explains both the absence of a free –CHO group and the existence of two forms. The two cyclic forms exist in equilibrium with the open-chain structure.
Anomers and the pyranose name. The two cyclic hemiacetal forms differ only in the configuration of the hydroxyl group at C-1, called the anomeric carbon (the aldehyde carbon before cyclisation). Such isomers, the α-form and the β-form, are called anomers. The six-membered cyclic structure is called the pyranose structure, in analogy with pyran — a cyclic compound with one oxygen and five carbon atoms in the ring. The cyclic structure is more correctly represented by the Haworth structure.
10.1.2.2 Fructose
Fructose is an important ketohexose. It is obtained along with glucose by the hydrolysis of the disaccharide sucrose. It is a natural monosaccharide found in fruits, honey and vegetables, and in its pure form is used as a sweetener.
Fructose also has the molecular formula C₆H₁₂O₆. On the basis of its reactions it was found to contain a ketonic functional group at carbon number 2 and six carbons in a straight chain, as in glucose. It belongs to the D-series and is laevorotatory, so it is appropriately written as D-(−)-fructose.
D and (−) together in one name. Fructose is the clearest proof that the D/L label and the sign of rotation are independent. D-(−)-fructose has the D configuration yet rotates plane-polarised light to the left.
Fructose also exists in two cyclic forms, obtained by addition of the –OH at C-5 to the keto group. The ring thus formed is five-membered and is named furanose, by analogy with furan — a five-membered cyclic compound with one oxygen and four carbon atoms. The cyclic structures of the two anomers are represented by Haworth structures.
| Feature | Glucose | Fructose |
|---|---|---|
| Molecular formula | C₆H₁₂O₆ | C₆H₁₂O₆ |
| Functional group | aldehyde at C-1 → aldohexose | keto at C-2 → ketohexose |
| Ring size | six-membered pyranose | five-membered furanose |
| –OH involved in ring | C-5 | C-5 |
| Optical rotation | dextrorotatory, D-(+) | laevorotatory, D-(−) |
| Other name | dextrose | fruit sugar / laevulose |
🧪 Activity 10.1 — Building the case against the open-chain structureL5 Evaluate
Scientists did not simply decide glucose was cyclic. They were forced into it by three experiments the open-chain structure could not explain. This activity reconstructs that reasoning.
Predict: Glucose reacts with hydroxylamine to form an oxime, which proves a carbonyl group is present. So why does it fail Schiff's test, which also detects aldehydes? Write your idea first.
- List the three anomalies: no Schiff's test, no NaHSO₃ addition product; pentaacetate unreactive towards hydroxylamine; two crystalline forms with different melting points.
- For each anomaly, state what the open-chain structure predicts and what is actually observed.
- Propose one structural change that accounts for all three at once.
- Check your proposal: does it also explain why glucose still forms an oxime at all?
The single change that explains everything: the –OH at C-5 adds across the C-1 aldehyde to form a six-membered cyclic hemiacetal.
Anomaly 1. Open chain predicts a positive Schiff's test and a bisulphite adduct. Observed: neither. Explanation — in the cyclic hemiacetal there is no free –CHO group; only a trace exists in the open form at equilibrium, too little for these tests.
Anomaly 2. Open chain predicts the pentaacetate still has its –CHO and should form an oxime. Observed: it does not. Explanation — acetylation locks the ring shut by capping the C-1 hydroxyl, so the molecule can no longer open to the aldehyde at all.
Anomaly 3. Open chain predicts one substance, one melting point. Observed: α-form m.p. 419 K and β-form m.p. 423 K. Explanation — cyclisation creates a new asymmetric centre at C-1, the anomeric carbon, giving two diastereomers called anomers.
The check in step 4 is the subtle part. Glucose does still form an oxime because the cyclic and open forms are in equilibrium. Hydroxylamine reacts irreversibly with the small amount of open-chain aldehyde present; as it is consumed, the equilibrium shifts to replace it, and eventually all the glucose is converted. Schiff's reagent and NaHSO₃ form reversible adducts too weak to pull that equilibrium, so they give no visible result. The difference between these tests is therefore about thermodynamics, not about whether an aldehyde exists — an excellent illustration of how a mechanism can be probed by choosing reagents of different reversibility.
Intext questions
Intext 10.1 — Why glucose and sucrose dissolve in water but cyclohexane and benzene do not
Glucose and sucrose carry many –OH groups which form hydrogen bonds with water molecules, and the energy released on hydration more than compensates for breaking the crystal lattice. Cyclohexane and benzene are non-polar hydrocarbons with no –OH groups; they cannot hydrogen bond with water, so they are insoluble. Being six-membered rings is irrelevant — what matters is the functional groups on the ring.
Intext 10.3 — Absence of the aldehyde group in glucose pentaacetate
In aqueous solution glucose exists mainly in the cyclic hemiacetal form, with only a trace of the open-chain aldehyde in equilibrium. On acetylation, the five –OH groups including the anomeric –OH at C-1 are converted to acetyl esters. Capping C-1 prevents the ring from opening, so the open-chain aldehyde can no longer form. Hence glucose pentaacetate does not react with hydroxylamine and shows no aldehyde behaviour.
Competency-Based Questions
A food technologist is analysing three white crystalline powders labelled X, Y and Z. All three have molecular formula C₆H₁₂O₆ or are built from such units. Tests show: X reduces Tollens' reagent and forms an oxime, but gives no colour with Schiff's reagent. Y also reduces Tollens' reagent and on treatment with bromine water gives no acid, while its ketonic nature is confirmed separately. Z gives no reaction with Tollens' reagent until it has been boiled with dilute acid.
1. Explain why X forms an oxime but fails Schiff's test. L4 Analyse
X exists predominantly as a cyclic hemiacetal, in equilibrium with a very small amount of open-chain aldehyde. Hydroxylamine reacts irreversibly with that trace of aldehyde, and as it is consumed the equilibrium shifts to supply more, so an oxime is eventually obtained from all the sugar. Schiff's reagent forms only a weak reversible adduct that cannot pull the equilibrium, so no colour develops. X is glucose.
2. Y is a reducing sugar yet is a ketose. Reconcile these two statements. L4 Analyse
There is nothing to reconcile — NCERT states explicitly that all monosaccharides, whether aldose or ketose, are reducing sugars. The bromine-water result is the discriminating test: bromine water is a mild oxidising agent that oxidises only an aldehydic group to a carboxylic acid. Y gives no acid with bromine water, which confirms it has a keto group and not an aldehyde. Y is fructose, a ketohexose.
3. What does the behaviour of Z tell you about its class, and what are the likely products of boiling it with dilute acid? L3 Apply
Z is non-reducing until hydrolysed, so it is not a monosaccharide — it is an oligosaccharide (a disaccharide) in which the reducing groups of both units are tied up in the linkage. The classic example is sucrose, which on boiling with dilute HCl or H₂SO₄ gives equal amounts of glucose and fructose: C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆. Both products are reducing sugars, which is why Tollens' reagent responds after hydrolysis.
4. Glucose is oxidised by bromine water to gluconic acid and by nitric acid to saccharic acid. What does each result establish? L2 Understand
Bromine water is mild and attacks only the aldehyde, giving the mono-carboxylic gluconic acid — this establishes that the carbonyl group present is an aldehydic one, not ketonic. Nitric acid is stronger and oxidises both ends, giving the dicarboxylic saccharic acid — and since gluconic acid also gives saccharic acid on the same treatment, the other end that gets oxidised must have been a primary alcoholic –OH group.
5. A student writes: "D-fructose must be dextrorotatory because it is a D-sugar." Evaluate this statement. L5 Evaluate
The statement is wrong, and fructose is the standard counter-example. The prefix D refers only to configuration — specifically, to the fact that the –OH on the lowest asymmetric carbon lies on the right in the Fischer projection, matching D-(+)-glyceraldehyde. The sign of rotation is an experimentally measured property of the whole molecule and is determined by every chiral centre acting together, not by the reference centre alone. Fructose is correctly written D-(−)-fructose: it has the D configuration yet is laevorotatory. NCERT states the point directly — D and L have no relation with the optical activity of the compound, and they are also not related to the small letters d and l.
Assertion–Reason Questions
For each pair choose: (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
Assertion (A): Glucose does not give the Schiff's test although it contains an aldehyde group.
Reason (R): Glucose exists predominantly in a cyclic hemiacetal form in which the aldehyde group is not free.
Answer: A. The –OH at C-5 adds to the C-1 aldehyde to form the six-membered pyranose ring, so almost no free –CHO is present and the weakly binding Schiff's reagent gives no colour.
Assertion (A): Glucose on prolonged heating with HI gives n-hexane.
Reason (R): All six carbon atoms of glucose are linked in a straight chain.
Answer: A. Complete reduction of every oxygen-bearing carbon leaves only the carbon skeleton, and obtaining the straight-chain n-hexane rather than a branched isomer is precisely the evidence that the skeleton is unbranched.
Assertion (A): The α and β forms of glucose are enantiomers.
Reason (R): They differ in the configuration of the hydroxyl group at C-1.
Answer: D. The assertion is false — they are anomers, a special kind of diastereomer, not enantiomers. Enantiomers must be non-superimposable mirror images differing at every chiral centre, whereas α- and β-glucose differ at only one centre, C-1. The reason is a correct statement and is in fact the definition of anomers.
Coming next. Part 2 takes up Sections 10.1.3 to 10.1.5 — the disaccharides sucrose, maltose and lactose, glycosidic linkage, invert sugar, the polysaccharides starch, cellulose and glycogen, and the importance of carbohydrates.
Frequently Asked Questions
How are carbohydrates defined chemically?
Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds which produce such units on hydrolysis. The older description as hydrates of carbon with formula Cx(H₂O)y is unreliable, because acetic acid fits that formula yet is not a carbohydrate, while rhamnose (C₆H₁₂O₅) is a carbohydrate but does not fit it.
What is the difference between a reducing and a non-reducing sugar?
Reducing sugars are those which reduce Fehling's solution and Tollens' reagent. All monosaccharides, whether aldose or ketose, are reducing sugars. A disaccharide is non-reducing if the reducing groups of both monosaccharide units are locked into the glycosidic linkage, as in sucrose; if a free aldehyde or ketone group can be regenerated, as in maltose and lactose, the sugar is reducing.
Why does glucose not give Schiff's test or form a bisulphite addition product?
Because glucose exists predominantly as a six-membered cyclic hemiacetal, formed when the –OH at C-5 adds across the C-1 aldehyde. Only a trace of the open-chain aldehyde is present at equilibrium. Schiff's reagent and NaHSO₃ form weak reversible adducts that cannot pull that equilibrium, so no reaction is seen. Hydroxylamine reacts irreversibly with the trace of aldehyde, which is why an oxime can still be obtained.
What are anomers and what is the anomeric carbon?
When glucose cyclises, C-1 — the carbon that was the aldehyde — becomes a new asymmetric centre called the anomeric carbon. The two cyclic forms that differ only in the configuration of the hydroxyl group at this carbon are called anomers, designated α and β. The α-form melts at 419 K and is crystallised from concentrated solution at 303 K; the β-form melts at 423 K and is crystallised from hot saturated solution at 371 K.
What do the prefixes D and L mean in sugar names?
They indicate relative configuration with respect to glyceraldehyde, not the direction of optical rotation. A sugar is assigned D if the –OH on its lowest asymmetric carbon lies on the right in the Fischer projection, matching D-(+)-glyceraldehyde, and L if it lies on the left. D and L have no relation with optical activity and are not related to the small letters d and l — which is why D-(−)-fructose is laevorotatory despite being a D-sugar.
What is the difference between the pyranose and furanose structures?
Pyranose describes the six-membered cyclic form, named by analogy with pyran, a ring of one oxygen and five carbon atoms; glucose adopts this form when the –OH at C-5 adds to the C-1 aldehyde. Furanose describes the five-membered cyclic form, named by analogy with furan, a ring of one oxygen and four carbon atoms; fructose adopts this form when the –OH at C-5 adds to the C-2 keto group.
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