Monosaccharides contain carbonyl group hence are classified as aldose or ketose. The number of carbon atoms pr

NCERT Exemplar · Q25

Q. Monosaccharides contain carbonyl group hence are classified as aldose or ketose. The number of carbon atoms present in the monosaccharide molecule are also considered for classification. In which class of monosaccharide will you place fructose?

Karnataka PUCShort· 2mImportance★★★★★

53% · 58/110 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Lactose Hydrolysis Products

Lactose Hydrolysis Products – From Intuition to Precision

Imagine you have a glass of milk. That slightly sweet taste comes from a sugar called lactose. But lactose is a disaccharide – it's actually two smaller sugar units joined together. If you could "unstick" those two units, you'd get two simpler sugars. That unsticking process is hydrolysis (water + breaking), and the two simpler sugars you get are the hydrolysis products.

The Intuition: Breaking a Sugar Chain

Think of lactose as a train with exactly two carriages. The coupling between them is a chemical bond. When you add water and the right conditions (like an enzyme called lactase, or an acid), that bond snaps. The train splits into two separate carriages. Each carriage is now a free, smaller sugar molecule.

So the hydrolysis products are simply the two individual sugar units that were originally linked to form lactose.

The Precise Statement

Lactose ( $\text{C}_{12}\text{H}_{22}\text{O}_{11}$ ) is a disaccharide composed of one molecule of D-galactose and one molecule of D-glucose linked by a $\beta(1 \rightarrow 4)$ glycosidic bond. Upon hydrolysis (reaction with water), this bond is cleaved, yielding the two monosaccharides:

\text{Lactose} + \text{H}_2\text{O} \xrightarrow{\text{lactase or acid}} \text{D-Galactose} + \text{D-Glucose}

The two hydrolysis products are:

D-Galactose – a monosaccharide (aldohexose, C6H12O6)
D-Glucose – a monosaccharide (aldohexose, C6H12O6)

Both are reducing sugars, and both have the same molecular formula ( $\text{C}_6\text{H}_{12}\text{O}_6$ ) but differ in the arrangement of the hydroxyl group on carbon 4 (they are C-4 epimers).

Note

In the body, the enzyme lactase (present in the small intestine) performs this hydrolysis so that the resulting glucose and galactose can be absorbed into the bloodstream. Lactose intolerance occurs when lactase activity is low, leaving lactose undigested.

Why This Matters for Exams

Always name both products: galactose and glucose. Never just "sugars" or "monosaccharides."
Know the bond: $\beta(1 \rightarrow 4)$ glycosidic linkage. Hydrolysis breaks this specific bond. …

Why this formula?

Lactose Hydrolysis Products — Understanding the Why

Lactose is a disaccharide composed of two monosaccharides linked by a glycosidic bond. When it undergoes hydrolysis, the bond is broken, yielding specific products. Let's build the reasoning step by step.

1. What is lactose chemically?

Lactose = galactose $\beta(1 \to 4)$ glucose
The bond is between:
- Carbon-1 of galactose (in $\beta$ configuration)
- Carbon-4 of glucose

So the structural formula is:

\text{Galactose} - O - \text{Glucose}

2. What does hydrolysis do?

Hydrolysis means "splitting with water." The reaction is:

\text{Lactose} + \text{H}_2\text{O} \xrightarrow{\text{lactase or acid}} \text{Galactose} + \text{Glucose}

The water molecule adds across the glycosidic bond:

The H from water attaches to the oxygen of the galactose (forming a free –OH on galactose)
The OH from water attaches to the carbon-1 of glucose (forming a free –OH on glucose)

3. Why are the products exactly galactose and glucose?

Because the glycosidic bond is between specific carbons:

Galactose contributes its anomeric carbon (C1)
Glucose contributes its C4

When the bond breaks, each sugar regains its free anomeric carbon (in the case of galactose) or free hydroxyl at C4 (in the case of glucose). No rearrangement occurs — the monosaccharides are released as they were originally linked.

4. Key formula — the hydrolysis equation

The balanced chemical equation:

\ce{C12H22O11 + H2O -> C6H12O6 + C6H12O6}

Lactose: $\ce{C12H22O11}$
Water: $\ce{H2O}$
Products: two molecules of $\ce{C6H12O6}$ (one galactose, one glucose)

Why the same molecular formula?

Both galactose and glucose are aldohexoses — they have the same molecular formula $\ce{C6H12O6}$ but differ in the arrangement of –OH groups (epimers at C4).

5. The "why" behind the formula

Mass conservation: The total number of C, H, O atoms before and after must match. …

Start your 14-day free trial to unlock the full solution →

Fructose is a ketohexose — it has a ketone functional group (making it a ketose) and six carbon atoms (making it a hexose). This dual classification places it in the ketohexose category.

The classification of monosaccharides is a two‑step process: first, identify the carbonyl group (aldehyde or ketone), and second, count the number of carbon atoms. Let’s see how this applies to fructose.

Identify the carbonyl group.

Fructose has a ketone ( $\ce{C=O}$ ) group at the second carbon, not an aldehyde ( $\ce{-CHO}$ ) at the end. Therefore, it is a ketose.
Count the carbon atoms.

The molecular formula of fructose is $\ce{C6H12O6}$ , which means it contains six carbon atoms. Monosaccharides with six carbons are called hexoses.
Combine the two classifications.

Since fructose is both a ketose and a hexose, its full classification is ketohexose. …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2025Set 2025-M1 markMCQ
$\alpha-D-$
›Reveal solutionSolution
Maltose is a reducing disaccharide from starch hydrolysis, but its hydrolysis yields two glucose units, not glucose and fructose; thus Assertion is correct, Reason is wrong, so option (C) is correct.

Concept & Intuition

Maltose is a disaccharide composed of two glucose molecules linked by an α(1→4) glycosidic bond. Because one of the glucose units retains a free anomeric carbon (the hemiacetal group), maltose can reduce Cu²⁺ ions (e.g., in Benedict’s test) — that’s what makes it a reducing sugar. The assertion correctly states that maltose is obtained by partial hydrolysis of starch using the enzyme diastase. However, the reason claims that hydrolysis of maltose gives one glucose and one fructose — that’s actually the hydrolysis product of sucrose, not maltose. The classic pitfall here is confusing the hydrolysis products of common disaccharides.

Step-by-step reasoning

Check the Assertion (A)

Maltose is indeed a reducing sugar because its structure has a free anomeric carbon on the non-reducing end? Actually, careful: In maltose, the glycosidic bond involves C1 of one glucose and C4 of the other. The glucose unit that provides C1 has its anomeric carbon tied up in the bond, so it cannot open to a free aldehyde. But the other glucose unit (the one with the free C1) retains a hemiacetal group, which can open to an aldehyde form. Thus maltose reduces Tollens’ or Benedict’s reagent.

Partial hydrolysis of starch (a polymer of glucose) with the enzyme diastase (an amylase) does yield maltose as a major product.

Therefore, Assertion (A) is correct.

Check the Reason (R)

Hydrolysis of one mole of maltose (C₁₂H₂₂O₁₁) with water yields two moles of D-glucose.

The reaction: $\text{Maltose} + \text{H}_2\text{O} \xrightarrow{\text{acid or enzyme}} 2\, \text{D-glucose}$ …
Start your 14-day free trial to see the full solution →
KCET 2024Set B-41 markMCQ
$Q. Stanley Miller simulated the conditions of pre-biotic earth using spark-discharge apparatus. Which organic compounds were observed by him on analysing the end product of his experiment? (A) Pigments (B) Fats (C) Nitrogen bases (D) Amino acids$
›Reveal solutionSolution
Miller's spark-discharge experiment recovered simple amino acids, demonstrating that organic monomers can form abiotically from a reducing atmosphere.

Step 1 — What the experiment set out to test.

Oparin (Russia) and Haldane (England) had proposed that the first form of life arose from pre-existing non-living organic molecules (e.g. RNA, proteins) — chemical evolution preceding biological evolution. The claim needed an experimental test: can organic molecules form from purely inorganic precursors under primitive-Earth conditions?

Step 2 — The apparatus and the conditions simulated.

Stanley L. Miller, a student of Harold Urey, built a closed flask (1953) and created conditions like those of the primitive, reducing (oxygen-free) atmosphere:

Gases: methane ( $\mathrm{CH_4}$ ), ammonia ( $\mathrm{NH_3}$ ) and hydrogen ( $\mathrm{H_2}$ ), in a $2:1:2$ ratio.

Water vapour supplied by boiling water in a connected flask.

Temperature about $800^\circ\text{C}$ .

Electric discharge between electrodes, simulating lightning — the energy source.

The vapour was condensed and collected in a U-trap, and the accumulated liquid analysed after a week.

Step 3 — The result. …
Start your 14-day free trial to see the full solution →
KCET 2023Set D-21 markMCQ
$Q. Sucrose is dextrorotatory but after hydrolysis the mixture show laevorotation, this is because of (A) Laevorotation of glucose is more than dextrorotation of fructose. (B) Sucrose is a non-reducing sugar. (C) Racemic mixture is formed. (D) Laevorotation of fructose is more than dextrorotation of glucose.$
›Reveal solutionSolution

Add the specific rotations of the two hydrolysis products: fructose's large negative rotation outweighs glucose's smaller positive one, so the sign of the mixture inverts.

Step 1 — The reaction
$\underset{\text{sucrose}}{\mathrm{C_{12}H_{22}O_{11}}} + \mathrm{H_2O} \xrightarrow{\;\mathrm{H^+/invertase}\;} \underset{\text{glucose}}{\mathrm{C_6H_{12}O_6}} + \underset{\text{fructose}}{\mathrm{C_6H_{12}O_6}}$
Sucrose is a 1→2 glycosidic disaccharide of $\alpha$ -D-glucose and $\beta$ -D-fructose. Hydrolysis cleaves it into equimolar glucose and fructose.

Step 2 — The specific rotations

Species Specific rotation $[\alpha]_D$
Sucrose $+66.5^\circ$ (dextrorotatory)
D-(+)-Glucose $+52.5^\circ$
D-(−)-Fructose $-92.4^\circ$

Step 3 — Net rotation of the product mixture

Optical rotations of components in a mixture are additive. For a 1:1 mixture the net rotation goes as
$[\alpha]_{\text{mix}} \propto \tfrac12\big(+52.5^\circ\big) + \tfrac12\big(-92.4^\circ\big) = \tfrac12\big(-39.9^\circ\big) < 0 .$
The magnitude of fructose's laevorotation ( $92.4^\circ$ ) exceeds glucose's dextrorotation ( $52.5^\circ$ ), so the sum is negative — the mixture rotates plane-polarised light to the left.

Step 4 — Why this is called "inversion" …
Start your 14-day free trial to see the full solution →
COMEDK 2022Set 20221 markMCQ
$\alpha$
›Reveal solutionSolution

So the monosaccharide units of maltose are alpha-D-glucose and alpha-D-glucose.

Concept: structure of common disaccharides.

Maltose = two units of alpha-D-GLUCOSE joined by an alpha(1->4) glycosidic linkage (C1 of the first glucose to C4 of the second). It is a reducing sugar since the second glucose retains a free anomeric -OH. …
Start your 14-day free trial to see the full solution →
COMEDK 2021Set 20211 markMCQ
$\alpha$
›Reveal solutionSolution
Hence the correct pair is alpha-D-glucopyranose and beta-D-fructofuranose.

Concept: Structure of the disaccharide sucrose.

Sucrose is a non-reducing disaccharide in which the anomeric C1 of glucose and the anomeric C2 of fructose are joined head-to-head through a glycosidic linkage, so no free anomeric -OH (hemiacetal) remains - that is why it is non-reducing.

The two units are:

alpha-D-glucopyranose (six-membered pyranose ring, alpha at C1)

beta-D-fructofuranose (five-membered furanose ring, beta at C2) …
Start your 14-day free trial to see the full solution →

Species	Specific rotation $[\alpha]_D$
Sucrose	$+66.5^\circ$ (dextrorotatory)
D-(+)-Glucose	$+52.5^\circ$
D-(−)-Fructose	$-92.4^\circ$

←Previous questionQ24 — Under what conditions glucose is converted to gluconic and saccharic a…→Next questionQ26 — The prefixes 'D' and 'L' before a sugar's name give the configuration …

Q.Monosaccharides contain carbonyl group hence are classified as aldose or ketose. The number of carbon atoms present in the monosaccharide molecule are also considered for classification. In which class of monosaccharide will you place fructose?

Concept understanding — Lactose Hydrolysis Products

Lactose Hydrolysis Products – From Intuition to Precision

The Intuition: Breaking a Sugar Chain

The Precise Statement

Why This Matters for Exams

Why this formula?

Lactose Hydrolysis Products — Understanding the Why

1. What is lactose chemically?

2. What does hydrolysis do?

3. Why are the products exactly galactose and glucose?

4. Key formula — the hydrolysis equation

5. The "why" behind the formula

Concept: Classification of Monosaccharides

Method: Functional Group + Carbon Chain Classification

Steps:

The Concept in One Line

Common Mistakes & How to Avoid Them

✗ Mistake 1: Calling fructose an aldose

✗ Mistake 2: Saying fructose is a pentose

✗ Mistake 3: Writing “ketohexose” but then giving wrong carbon number in explanation

✗ Mistake 4: Forgetting that “ketose” includes the position of the carbonyl …

Step 1 — The reaction

Step 2 — The specific rotations

Step 3 — Net rotation of the product mixture

Step 4 — Why this is called "inversion" …

Q.Monosaccharides contain carbonyl group hence are classified as aldose or ketose. The number of carbon atoms present in the monosaccharide molecule are also considered for classification. In which class of monosaccharide will you place fructose?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days