ChemistryISC Class 12 · Biomolecules

Carbohydrates: Glucose and Disaccharides for ISC Class 12

Updated 15 October 2026 · Fact-checked

Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds that give them on hydrolysis. Glucose is an aldohexose whose straight chain, aldehyde group and five –OH groups are shown by its reactions. Disaccharides join two monosaccharides by a glycosidic linkage; sucrose is non-reducing and hydrolyses to laevorotatory invert sugar.

Understand Carbohydrates: Glucose and Disaccharides

Definition and classification. Many carbohydrates fit $C_{x} (H_{2} O)_{y}$ , but the formula is not a definition (ethanoic acid fits it; rhamnose, a carbohydrate, does not). Chemically, carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds that give these on hydrolysis. By hydrolysis they are: monosaccharides (cannot be hydrolysed further: glucose, fructose, ribose); oligosaccharides (2 to 10 monosaccharide units: disaccharides such as sucrose, maltose, lactose); polysaccharides (many units: starch, cellulose, glycogen; not sweet, called non-sugars). Monosaccharides are aldoses (–CHO) or ketoses (C=O), and are named by carbon number: glucose is an aldohexose, fructose a ketohexose, ribose an aldopentose.

Reducing and non-reducing sugars. Carbohydrates that reduce Fehling's solution and Tollens' reagent are reducing sugars: all monosaccharides (aldoses and ketoses) and disaccharides in which a free aldehyde or ketone group remains (maltose, lactose). In sucrose the reducing groups of both units are used in the glycosidic bond, so it is non-reducing.

Evidence for the open-chain structure of glucose. Molecular formula C₆H₁₂O₆. (i) Prolonged heating with HI gives n-hexane: the six carbons are in a straight chain. (ii) With hydroxylamine it forms an oxime, and with HCN a cyanohydrin: a carbonyl group is present. (iii) With bromine water it is oxidised to gluconic acid (six carbons): the carbonyl group is an aldehyde. (iv) With acetic anhydride it gives a pentaacetate: five –OH groups, on different carbons since the molecule is stable. (v) With nitric acid both glucose and gluconic acid give saccharic (glucaric) acid: a primary alcohol (–CH₂OH) group is present. (vi) With phenylhydrazine it forms an osazone. Together these give OHC–(CHOH)₄–CH₂OH. Some facts (no reaction with Schiff's reagent or NaHSO₃, pentaacetate not reacting with hydroxylamine, two forms α and β) are explained by the cyclic hemiacetal structure, a six-membered (pyranose) ring formed by the C5–OH adding to the C1 aldehyde group; α and β differ only at C1 (anomers).

D and L configuration. These letters do not mean the sign of rotation. They show the relative configuration compared with glyceraldehyde: in the Fischer projection, if the –OH on the lowest asymmetric carbon is on the right, the sugar belongs to the D-series (related to D-(+)-glyceraldehyde); on the left, the L-series. Natural glucose is D-(+)-glucose: D is its configuration, (+) shows it is dextrorotatory. Open-chain glucose has four chiral carbons (C2–C5).

Disaccharides and the glycosidic linkage. Two monosaccharides join by losing water, forming an oxide link C–O–C called a glycosidic linkage. Sucrose: α-D-glucose (C1) linked to β-D-fructose (C2); both anomeric carbons are used, so it is non-reducing. Maltose: two α-D-glucose units linked C1 to C4; the second unit has a free anomeric carbon, so it is reducing. Lactose (milk sugar): β-D-galactose (C1) linked to β-D-glucose (C4); reducing.

Invert sugar. Sucrose is dextrorotatory. On hydrolysis by dilute acid or the enzyme invertase it gives equal amounts of D-(+)-glucose (+52.5°) and D-(–)-fructose (–92.4°): $C_{12} H_{22} O_{11} + H_{2} O \to C_{6} H_{12} O_{6} + C_{6} H_{12} O_{6}$ . Because the laevorotation of fructose is larger than the dextrorotation of glucose, the mixture is laevorotatory. The sign of rotation changes from + to −, so the product is called invert sugar; unlike sucrose, it is reducing.

Importance. Carbohydrates are the main energy source (glucose), stored as starch in plants and glycogen in animals; cellulose forms plant cell walls; ribose and deoxyribose are parts of nucleic acids.

Carbohydrates: Glucose and Disaccharides: formulas and key facts

Open-chain glucose: $O H C - (C H O H)_{4} - C H_{2} O H$; Aldohexose; four chiral carbons (C2–C5).
Hydrolysis of sucrose: $C_{12} H_{22} O_{11} + H_{2} O H^{+} or invertase C_{6} H_{12} O_{6} (glucose) + C_{6} H_{12} O_{6} (fructose)$; Product is laevorotatory invert sugar.
Specific rotations (NCERT): $D -glucose : + 52. 5^{\circ}; D -fructose : - 92. 4^{\circ}$; Equal masses: average rotation is negative.
Glycosidic linkages: $sucrose : α -Glc C1- β -Fru C2; maltose : α C1-C4; lactose : β -Gal C1-Glc C4$; Sucrose non-reducing; maltose and lactose reducing.
Stereoisomer count: $number of stereoisomers = 2^{n} (n chiral carbons)$; Open-chain aldohexose: 2⁴ = 16.

How to solve Carbohydrates: Glucose and Disaccharides questions

ISC carbohydrate questions are classification, structural evidence, reducing behaviour, linkages and short calculations. Tie every answer to the functional groups present.

1Classify first: number of units (mono, di, poly), carbonyl type (aldose/ketose) and carbon number (pentose/hexose).
2For 'what does this reaction show', name the reagent, the product and the group proved (straight chain, carbonyl, aldehyde, five OH, primary alcohol).
3For reducing behaviour, look for a free anomeric carbon (hemiacetal); sucrose has none.
4For linkages, give the two units, the carbons joined and α or β.
5For D/L, look at the –OH on the lowest chiral carbon in the Fischer projection; remember D/L is not the sign of rotation.
6For calculations, use molar masses (glucose 180, sucrose 342, starch unit C₆H₁₀O₅ = 162) and the stoichiometry of hydrolysis.

Carbohydrates: Glucose and Disaccharides in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Acid hydrolysis of starch gives glucose

Suppose a small food unit makes glucose syrup by boiling 81 g of starch with dilute acid until it is completely hydrolysed. Take the starch repeating unit as C₆H₁₀O₅ (M = 162) and glucose as C₆H₁₂O₆ (M = 180). After purification it obtains 72 g of glucose.

Write the equation, find the theoretical mass of glucose and the percentage yield, and say whether the syrup will reduce Fehling's solution.

Try it first, then show the solutionSolution

$(C_{6} H_{10} O_{5})_{n} + n H_{2} O \to n C_{6} H_{12} O_{6}$ : one glucose per repeating unit.
Moles of units $= \frac{81}{162} = 0.50$ mol, so theoretical glucose $= 0.50 \times 180 = 90$ g (the mass rises because water is added).
Percentage yield $= \frac{72}{90} \times 100 = 80%$ .
Glucose has a free aldehyde (hemiacetal) group, so the syrup reduces Fehling's solution; starch itself does not.

Answer: Theoretical 90 g; yield 80%. The glucose syrup is a reducing sugar and gives a red precipitate with Fehling's solution.

Memory hook: Hydrolysis adds water: 162 g of starch units become 180 g of glucose.

Tricks to remember Carbohydrates: Glucose and Disaccharides

Short memory aids. Say each one aloud once, then use it on a question the same day.

Glucose reactions, one clue each
HI → hexane (straight chain); NH₂OH/HCN → carbonyl; Br₂ water → aldehyde; acetic anhydride → five OH; HNO₃ → primary alcohol.
Use it when: Structural evidence questions.
Sucrose holds hands at both ends
Sucrose uses C1 of glucose and C2 of fructose (both 'hands') in the bond, so no free hand is left to reduce Fehling's: non-reducing.
Use it when: Reducing versus non-reducing sugars.
D is not dextro
D/L = position of –OH on the lowest chiral carbon (right = D); (+)/(−) = direction of rotation. D-(−)-fructose shows they are independent.
Use it when: Configuration questions.

Carbohydrates: Glucose and Disaccharides: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Which sugar gives no silver mirror? · 1 mark · MCQBoard level

Which of the following does NOT reduce Tollens' reagent?
(a) glucose
(b) fructose
(c) maltose
(d) sucrose

Try it first, then show the solutionSolution

Monosaccharides and maltose have a free anomeric (reducing) group.
In sucrose both reducing groups are used in the glycosidic bond.

Answer: (d) sucrose.

Example 2: Glucose and HI · 2 marks · Short answerBoard level

What is formed when glucose is heated for a long time with HI, and what does this show about its structure?

Try it first, then show the solutionSolution

Prolonged heating with HI reduces glucose to n-hexane, CH₃(CH₂)₄CH₃ (1 mark).
So the six carbon atoms of glucose are linked in a straight (unbranched) chain (1 mark).

Answer: n-Hexane; the six carbons form a straight chain.

Example 3: Proving the carbonyl group · 3 marks · Short answerHarder board level

Give the products of the reaction of glucose with (a) hydroxylamine and (b) hydrogen cyanide. What do these reactions show? Why does glucose nevertheless not give Schiff's test?

Try it first, then show the solutionSolution

(a) Glucose oxime, HOCH₂(CHOH)₄CH=NOH (1 mark).
(b) Glucose cyanohydrin, HOCH₂(CHOH)₄CH(OH)CN. Both show a carbonyl (C=O) group (1 mark).
In solution glucose exists almost entirely in the cyclic hemiacetal form, with only a tiny amount of free –CHO, too little to give Schiff's test (1 mark).

Answer: Oxime and cyanohydrin, showing C=O; the cyclic form explains the negative Schiff's test.

Example 4: Meaning of D-(+)-glucose · 3 marks · Short answerHarder board level

Explain what 'D' and '(+)' mean in D-(+)-glucose, and how the D configuration is assigned. Why is fructose written D-(−)-fructose?

Try it first, then show the solutionSolution

'D' gives the relative configuration: glucose is related to D-(+)-glyceraldehyde (1 mark).
In the Fischer projection of glucose, the –OH on the lowest asymmetric carbon (C5) is on the right, so it belongs to the D-series. '(+)' means it is dextrorotatory (1 mark).
Fructose has the D configuration (C5–OH on the right) but is laevorotatory, so it is D-(−)-fructose: configuration and sign of rotation are independent (1 mark).

Answer: D = configuration related to D-glyceraldehyde (lowest chiral C–OH on the right); (+) = dextrorotatory; fructose is D but laevorotatory.

Example 5: Why it is called invert sugar · 5 marks · Long answerHOTS / competency

(a) Write the equation for the acid hydrolysis of sucrose and name the products. [1]
(b) Using the specific rotations D-glucose +52.5° and D-fructose −92.4°, estimate the specific rotation of the product mixture, assuming the rotations add in proportion to mass. [2]
(c) Explain the name 'invert sugar'. [1]
(d) Is invert sugar reducing? Explain. [1]

Try it first, then show the solutionSolution

(a) $C_{12} H_{22} O_{11} + H_{2} O \to C_{6} H_{12} O_{6} + C_{6} H_{12} O_{6}$ : D-(+)-glucose and D-(−)-fructose (1 mark).
(b) Equal moles and equal molar masses give equal masses of each (1 mark).
Rotation $\approx \frac{1}{2} (+ 52. 5^{\circ}) + \frac{1}{2} (- 92. 4^{\circ}) = - 20. 0^{\circ}$ (1 mark).
(c) Sucrose is dextrorotatory but the product is laevorotatory: the sign of rotation is inverted (1 mark).
(d) Yes. Glucose and fructose each have a free carbonyl (hemiacetal or hemiketal) group, so they reduce Fehling's and Tollens' reagents (1 mark).

Answer: (a) Glucose + fructose. (b) About −20°. (c) Rotation changes from + to −. (d) Yes, both products are reducing sugars.

Where students lose marks in Carbohydrates: Glucose and Disaccharides

The slips examiners see most often, and what to write instead.

Mistake: Calling sucrose a reducing sugar
ISC examiners reported this error in 2024, along with failure to name glucose and fructose as the hydrolysis products.
Do this: Sucrose is non-reducing; its hydrolysis products, glucose and fructose, are reducing.
Usually costs: 1 mark.
Mistake: Naming a hydrogen bond as the link in disaccharides
Examiners noted 'hydrogen bond' written instead of glycosidic linkage (2025).
Do this: Disaccharides: glycosidic linkage (C–O–C). Proteins: peptide bond.
Usually costs: 1 mark.
Mistake: Confusing D/L with (+)/(−)
Students assume D means dextrorotatory.
Do this: D/L is configuration (relative to glyceraldehyde); (+)/(−) is the measured rotation; D-(−)-fructose proves they are separate.
Usually costs: 1 mark.
Mistake: Stating only the reagent in structural evidence
Students write 'reaction with HI' without the product or the conclusion.
Do this: Give reagent → product → what it proves (e.g. HI → n-hexane → straight chain).
Usually costs: ½ to 1 mark per reaction.

Exam tips

Learn the CISCE list of glucose reactions (HI, hydroxylamine, bromine water, acetic anhydride, nitric acid, phenylhydrazine) with the conclusion of each.
For disaccharides, always give the units, the carbons linked and whether the sugar is reducing.
Use 'invert sugar' with the reason (sign of rotation changes) and the values +52.5° and −92.4°.
In Fischer projections, check the lowest chiral carbon for D/L.
Remember that monosaccharides, maltose and lactose are reducing; sucrose and polysaccharides are not.

Practice questions on Carbohydrates: Glucose and Disaccharides

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The linkage that joins two monosaccharide units in a disaccharide is:
(a) a peptide bond
(b) a hydrogen bond
(c) a glycosidic linkage
(d) a phosphodiester bond
Hint
C–O–C formed with loss of water.
Answer
(c) a glycosidic linkage.
Q2 · 2 marks · Short answerBoard level
Classify carbohydrates on the basis of hydrolysis, with one example each. What is meant by a reducing sugar?
Hint
Three classes.
Answer
Monosaccharides (not hydrolysed further, e.g. glucose), oligosaccharides (2–10 units, e.g. sucrose), polysaccharides (many units, e.g. starch). A reducing sugar reduces Fehling's solution and Tollens' reagent because it has a free aldehyde or ketone group.
Q3 · 2 marks · Short answerHarder board level
Name the monosaccharide units in sucrose, state the carbon atoms joined by the glycosidic linkage, and explain why sucrose is non-reducing.
Hint
Both anomeric carbons are involved.
Answer
α-D-glucose and β-D-fructose, linked between C1 of glucose and C2 of fructose. These are the reducing (anomeric) carbons of both units, so no free aldehyde or ketone group is left: sucrose is non-reducing.
Q4 · 3 marks · Short answerHarder board level
Write the open-chain structure of glucose. How many chiral carbon atoms does it have, and how many stereoisomers are possible for an open-chain aldohexose?
Hint
C1 and C6 are not chiral.
Answer
OHC–CHOH–CHOH–CHOH–CHOH–CH₂OH. C2, C3, C4 and C5 are chiral: 4 chiral carbons. Stereoisomers $= 2^{4} = 16$ (8 D and 8 L).
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Sucrose does not reduce Fehling's solution.
Reason (R): In sucrose, the reducing groups of both glucose and fructose are involved in the glycosidic bond.
(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.
Hint
C1 of glucose and C2 of fructose.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
Classify each as an aldose or a ketose and as a pentose or a hexose: glucose, fructose, ribose, galactose. Which of them is found in RNA?
Hint
Count carbons and find the carbonyl type.
Answer
Glucose: aldohexose. Fructose: ketohexose. Ribose: aldopentose (in RNA). Galactose: aldohexose.
Q7 · 3 marks · Short answerHOTS / competency
1.80 g of glucose (M = 180) reacts completely with Fehling's solution. Assuming 1 mol of glucose gives 1 mol of Cu₂O (M = 143), calculate the mass of precipitate. What group of glucose is responsible?
Hint
Moles of glucose first.
Answer
Glucose $= \frac{1.80}{180} = 0.010$ mol; Cu₂O $= 0.010 \times 143 = 1.43$ g, a reddish-brown precipitate. The aldehyde group (free in the open-chain form) reduces Cu²⁺ to Cu₂O.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a clinic uses Benedict's or Fehling's solution to check for sugar in urine samples. One patient's sample gives a red precipitate; a second sample, which contains only sucrose from a sweet drink spilled into it, gives no precipitate until it is boiled with a little dilute acid and neutralised.
(i) What type of sugar does the first test detect? [1]
(ii) Why does the sucrose sample not react at first? [1]
(iii) Why does it react after boiling with acid? [1]
Hint
Think of reducing groups and hydrolysis.
Answer
(i) A reducing sugar, such as glucose, with a free aldehyde (or hemiacetal) group. (ii) Sucrose is non-reducing: both anomeric carbons are tied up in the glycosidic link. (iii) Acid hydrolyses sucrose into glucose and fructose, both reducing sugars, which then reduce the copper(II) reagent.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What does the reaction of glucose with bromine water show?Show answer
The carbonyl group is an aldehyde (it is oxidised to gluconic acid).
2.Linkage in maltose?Show answer
α-1,4 glycosidic linkage between two α-D-glucose units.
3.Units in lactose?Show answer
β-D-galactose and β-D-glucose (C1 of galactose to C4 of glucose).
4.Why is hydrolysed sucrose called invert sugar?Show answer
The sign of rotation changes from dextro (+) to laevo (−).
5.Define a monosaccharide.Show answer
A carbohydrate that cannot be hydrolysed into simpler polyhydroxy aldehyde or ketone units.
6.What does glucose pentaacetate formation show?Show answer
Glucose has five –OH groups.

Carbohydrates: Glucose and Disaccharides: frequently asked questions

What are reducing sugars?

Carbohydrates that reduce Fehling's solution and Tollens' reagent because they have a free aldehyde or ketone (hemiacetal) group; all monosaccharides, maltose and lactose are reducing.

Why is sucrose a non-reducing sugar?

Its glucose and fructose units are joined through C1 of glucose and C2 of fructose, the carbons that carry the reducing groups.

What is a glycosidic linkage?

The oxide (C–O–C) link formed between two monosaccharide units with loss of a water molecule.

How is the straight-chain structure of glucose proved?

On prolonged heating with HI, glucose gives n-hexane, showing its six carbons form an unbranched chain.

What is invert sugar?

The equimolar mixture of glucose and fructose formed by hydrolysis of sucrose; it is laevorotatory although sucrose is dextrorotatory.