Carbohydrates are classified on the basis of their behaviour on hydrolysis and also as reducing or non-reducin

NCERT Exemplar · Q54

Q. Carbohydrates are classified on the basis of their behaviour on hydrolysis and also as reducing or non-reducing sugar. Sucrose is a _______. (Note: one or more of the following options may be correct.)

(i) monosaccharide

(ii) disaccharide

(iii) reducing sugar

(iv) non-reducing sugar

Andhra Pradesh BieapMCQ· 1mImportance★★★★★

79% · 87/110 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 →

Sucrose is a disaccharide composed of glucose and fructose linked via their anomeric carbons, which blocks the free aldehyde/ketone group — so it is a non-reducing sugar. The correct options are (B) disaccharide and (D) non-reducing sugar.

Sucrose is common table sugar. To classify it, we need two ideas: how many sugar units it breaks into on hydrolysis, and whether it can reduce Tollens’ or Fehling’s reagent.

Number of monosaccharide units

Hydrolysis of sucrose gives one molecule of D-glucose and one molecule of D-fructose. That’s exactly two monosaccharide units. By definition, a carbohydrate that yields two monosaccharides on hydrolysis is a disaccharide.

So option (B) is correct; (A) monosaccharide is wrong.
Reducing vs non-reducing behaviour

A sugar is reducing if it has a free aldehyde group (in an aldose) or a free ketone group that can tautomerise to an aldehyde (in a ketose). In sucrose, the glycosidic bond is formed between the anomeric carbon of glucose (C1) and the anomeric carbon of fructose (C2). Both anomeric carbons are involved in the bond — neither is free.

Without a free anomeric carbon, the sugar cannot open to an aldehyde form and therefore cannot reduce Cu²⁺ or Ag⁺ ions. Hence sucrose is a non-reducing sugar.

Option (C) is wrong; (D) is correct. …

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

Here are the common mistakes students make on this question, along with how to avoid each.

Mistake 1: Confusing "Reducing" with "Non-Reducing" based on taste or common use

The Mistake: Many students assume that because sucrose (table sugar) is sweet and commonly used in cooking, it must be a "reducing sugar." They associate "sugar" with the ability to reduce reagents like Fehling’s or Benedict’s solution.
Why it’s wrong: The reducing property depends on the presence of a free aldehyde ( $-CHO$ ) or free ketone ( $-CO-$ ) group (specifically a hemiacetal or hemiketal group). In sucrose, the glucose and fructose units are linked via their anomeric carbons (C1 of glucose and C2 of fructose). This bond blocks both potential reducing ends.
How to avoid: Always check the glycosidic linkage. If the anomeric carbon of both monosaccharides is involved in the bond (as in sucrose: $\alpha$ ‑1,2‑ $\beta$ ), the sugar is non-reducing. If at least one anomeric carbon is free, it is reducing.

Mistake 2: Forgetting that Sucrose is a Disaccharide

The Mistake: Students sometimes tick only (D) "non-reducing sugar" and forget to tick (B) "disaccharide," thinking the question asks for only one property.
Why it’s wrong: The question explicitly states "one or more of the following options may be correct." Sucrose is both a disaccharide and a non-reducing sugar.
How to avoid: Read the instruction carefully. When a question says "one or more," treat it as a multiple-select question. List all correct classifications before marking.

Mistake 3: Confusing Hydrolysis Products with the Sugar Itself

The Mistake: Students recall that sucrose hydrolyses to glucose and fructose (both reducing sugars) and then incorrectly label sucrose itself as a reducing sugar.
Why it’s wrong: Hydrolysis changes the molecule. The properties of the original sucrose are determined by its intact structure, not its breakdown products.
How to avoid: Separate the original molecule from its products. Sucrose is non-reducing; its hydrolysis products (glucose + fructose) are reducing. Always ask: "Does the intact molecule have a free anomeric carbon?"

Mistake 4: Misclassifying Sucrose as a Monosaccharide

The Mistake: Some students confuse "sucrose" with "glucose" or "fructose" and mark (A) monosaccharide. …

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.

AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$\alpha-D-glucose$
›Reveal solutionSolution
Both statements accurately describe the standard disaccharide compositions of sucrose and lactose.

Concept and Intuition

Disaccharides are named by which two monosaccharide units (and in which anomeric form) are joined by a glycosidic bond. Sucrose's non-reducing character comes specifically from the fact that BOTH anomeric carbons (C1 of glucose and C2 of fructose) are involved in the glycosidic bond, locking the ring forms as $\alpha$ -D-glucose and $\beta$ -D-fructose. Lactose, by contrast, is a reducing sugar because glucose's anomeric carbon is left free; the fixed unit is $\beta$ -D-galactose joined via $\beta$ -1,4 linkage to D-glucose, and standard descriptions state it as $\beta$ -D-galactose and $\beta$ -D-glucose.

Step-by-Step Solution

Statement I: Sucrose = $\alpha$ -D-glucopyranose + $\beta$ -D-fructofuranose, linked $C1(glucose) \to C2(fructose)$ . This is the textbook description. Correct. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
$\alpha$
›Reveal solutionSolution
Tests whether you remember lactose's actual monomer composition and why it is still classed as a reducing sugar.

Concept and Intuition

Disaccharides are named by which two monosaccharides are linked and through which carbons. Sucrose (glucose + fructose, both anomeric carbons involved) is the classic non-reducing sugar because neither free anomeric OH survives the glycosidic bond. Lactose and maltose are the classic reducing sugars because one anomeric carbon is left free.

Step-by-Step Solution

Lactose = $\beta$ -D-galactose $+$ D-glucose, joined $\beta(1\to4)$ between galactose C1 and glucose C4.

Statement-I claims lactose is " $\alpha$ -D-glucose $+$ $\beta$ -D-glucose" — this describes maltose's/only-glucose composition, not lactose's actual galactose+glucose composition. False.

Because the glycosidic bond uses galactose's C1 and glucose's C4, glucose's own C1 (anomeric carbon) stays free with a free -OH. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$\alpha$
›Reveal solutionSolution
This tests the exact monosaccharide composition and anomeric forms in sucrose vs lactose. The answer is (C).

Concept and Intuition

Disaccharides are formed by a glycosidic linkage between two monosaccharide units, and the specific anomeric form ( $\alpha$ or $\beta$ ) of each unit is a precise structural fact that must be remembered correctly, since sucrose and lactose have different compositions and linkages.

Step-by-Step Solution

Sucrose (cane sugar): formed by a glycosidic bond between C1 of $\alpha$ -D-glucose and C2 of $\beta$ -D-fructose. Statement-I matches this exactly — correct.

Lactose (milk sugar): formed by a glycosidic bond between C1 of $\beta$ -D-galactose and C4 of $\beta$ -D-glucose (glucose unit here is in $\beta$ form as it provides the free anomeric carbon, though the ring can open to $\alpha$ / $\beta$ equilibrium — the standard textbook description names $\beta$ -D-galactose and glucose, not $\alpha$ -D-glucose). …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
$Q. Given below are two statements Assertion (A): Hydrolysis of sucrose results in change in the optical rotation from dextro (+) to laevo (-) Reason (R): Both the products from the hydrolysis are leavorotatory The correct answer is (A) Both A and R are correct and R in the correct explanation of A (B) Both A and R are correct but R is not the correct explanation of A (C) A is correct but R is incorrect (D) A is incorrect but R is correct$
›Reveal solutionSolution
The assertion is correct: sucrose hydrolysis inverts optical rotation from dextro to laevo. The reason is incorrect because one product (glucose) is dextrorotatory, not both laevorotatory. So the correct choice is (C).

Concept & Intuition

Optical rotation measures how a substance rotates plane-polarized light. Sucrose is a disaccharide made of glucose and fructose. When hydrolyzed, it breaks into these two monosaccharides. The key is that sucrose itself is dextrorotatory (rotates light to the right, +), but the mixture of glucose and fructose after hydrolysis is laevorotatory (rotates light to the left, –). This phenomenon is called inversion of sucrose, and the product mixture is called invert sugar. The reason given claims both products are laevorotatory — that’s the trap. In reality, glucose is dextrorotatory, fructose is strongly laevorotatory, and the net effect is laevorotatory because fructose’s leftward rotation outweighs glucose’s rightward rotation.

Step-by-step reasoning

Identify the specific rotations

Sucrose: $[ \alpha ]_D = +66.5^\circ$ (dextrorotatory)

Glucose: $[ \alpha ]_D = +52.7^\circ$ (dextrorotatory)

Fructose: $[ \alpha ]_D = -92.4^\circ$ (laevorotatory)

Hydrolysis reaction

$\text{Sucrose} + \text{H}_2\text{O} \rightarrow \text{Glucose} + \text{Fructose}$
One molecule of sucrose yields one molecule each of glucose and fructose.

Net rotation after hydrolysis The observed rotation of the mixture is the weighted average of the rotations of the products. Since both are produced in equal molar amounts:

$\text{Net rotation} = \frac{(+52.7^\circ) + (-92.4^\circ)}{2} = \frac{-39.7^\circ}{2} = -19.85^\circ$
This is negative (laevorotatory). So the mixture is laevorotatory, even though glucose alone is dextrorotatory.

Evaluate Assertion (A) …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQ
$\beta$
›Reveal solutionSolution
Cellulose has only $\beta$ -1,4 glycosidic bonds; starch (via amylopectin's branching) has both $\alpha$ -1,4 and $\alpha$ -1,6 bonds.

Concept and Intuition

Both cellulose and starch are glucose polymers, but the way the glucose units are joined determines their structure and digestibility. Cellulose is built entirely of $\beta$ -D-glucose units connected end to end by $\beta$ -1,4-glycosidic bonds. This linkage lets the chains lie flat and hydrogen-bond into rigid, fibrous microfibrils — ideal for cell walls, but not digestible by human enzymes (which only cleave $\alpha$ linkages).

Starch, by contrast, is made of $\alpha$ -D-glucose units. Its two components are amylose (a straight chain held together by $\alpha$ -1,4 bonds) and amylopectin (a branched molecule with an $\alpha$ -1,4 backbone plus $\alpha$ -1,6 bonds at branch points, roughly every 24–30 residues). Because starch as a storage polysaccharide is really this combination, both $\alpha$ -1,4 and $\alpha$ -1,6 bonds are correctly attributed to it.

Step-by-Step Solution

Identify cellulose's bond type: only $\beta$ -1,4-glycosidic bonds (no branching, no $\alpha$ bonds). …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQ
$Q. Hydrolysis of sucrose gives (A) Dextrorotatory glucose & Laevorotatory fructose (B) Dextrorotatory fructose & Laevorotatory glucose (C) Dextrorotatory glucose & Dextrorotatory fructose (D) Laevorotatory glucose & Laevorotatory fructose$
›Reveal solutionSolution
Hydrolysis of sucrose ("inversion") gives dextrorotatory glucose and laevorotatory fructose — the sign flip (net + to net −) is why it's called inversion of sugar.

Concept and Intuition

Sucrose is built from $\alpha$ -D-glucopyranose and $\beta$ -D-fructofuranose joined C1–C2 through their anomeric carbons, which locks both anomeric centres and makes sucrose a non-reducing sugar with no free aldehyde/ketone. Hydrolysing this glycosidic bond liberates both monosaccharides in their free, mutarotating forms, each with its own intrinsic optical rotation.

Step-by-Step Solution

Sucrose itself is dextrorotatory, $[\alpha]_D = +66.5^\circ$ .

Acid hydrolysis (or the enzyme invertase) breaks the glycosidic bond: $\text{Sucrose} + H_2O \rightarrow \text{Glucose} + \text{Fructose}$ .

Free D-glucose is dextrorotatory, $[\alpha]_D = +52.5^\circ$ .

Free D-fructose is strongly laevorotatory, $[\alpha]_D = -92^\circ$ (fructose's rotation is large and negative — it is sometimes called laevulose for this reason). …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
$Q. Hydrolysis of which disaccharide in presence of enzyme maltase give glucose only? (A) Sucrose (B) Cellulose (C) Lactose (D) Maltose$
›Reveal solutionSolution
Maltose is a disaccharide built from two glucose units, so its enzymatic hydrolysis by maltase produces glucose exclusively — unlike sucrose or lactose, which each yield two different monosaccharides.

Concept and Intuition

Disaccharides hydrolyse into their two constituent monosaccharides, and the specific enzyme named must match the specific glycosidic bond being cleaved. Maltase is the enzyme that hydrolyses the $\alpha(1\to4)$ bond in maltose; since maltose's two building blocks are both glucose, hydrolysis gives only glucose as product.

Step-by-Step Solution

Maltose = glucose + glucose (joined by an $\alpha(1\to4)$ glycosidic linkage). Enzyme maltase hydrolyses this bond $\Rightarrow$ 2 glucose molecules only.

Sucrose = glucose + fructose, hydrolysed by sucrase/invertase $\Rightarrow$ gives glucose and fructose, not glucose alone.

Lactose = glucose + galactose, hydrolysed by lactase $\Rightarrow$ gives glucose and galactose, not glucose alone. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$Q. If sucrose is boiled with dilute. HCl in alcoholic solution the ratio in which glucose and fructose are formed is (A) 1:1 (B) 1:2 (C) 2:1 (D) 4:1$
›Reveal solutionSolution
Acid hydrolysis of sucrose (inversion) cleaves its single glycosidic linkage to give exactly one glucose and one fructose molecule per sucrose molecule — a 1:1 ratio, option (A).

Concept and Intuition

Sucrose is a disaccharide formed by the condensation of one molecule of alpha-D-glucose and one molecule of beta-D-fructose, joined through a glycosidic linkage between C1 of glucose and C2 of fructose. Because this glycosidic bond is the only bond joining the two monosaccharide units, hydrolyzing it (by boiling with dilute acid, a reaction historically called 'inversion' because the optical rotation changes sign) breaks sucrose into exactly one glucose unit and one fructose unit — there is no possibility of an unequal split, since each sucrose molecule contains precisely one of each monosaccharide.

Step-by-Step Solution

Recall the structure of sucrose: glucose + fructose joined by one glycosidic bond (1→2 linkage), with the molecular formula $C_{12}H_{22}O_{11}$ .

Acid hydrolysis reaction: $C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6\,(\text{glucose}) + C_6H_{12}O_6\,(\text{fructose})$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
$(C_6H_{10}O_5)_n + nH_2O \xrightarrow[2\text{-}3\ atm]{H^+,\ 393K}$
›Reveal solutionSolution
Complete acid hydrolysis of starch under heat and pressure breaks the glycosidic bonds all the way down to its monosaccharide unit, glucose.

Concept and Intuition

Starch is a polysaccharide made of many glucose units linked by glycosidic bonds ( $\alpha$ -1,4 and $\alpha$ -1,6 linkages in amylose/amylopectin). Acid-catalyzed hydrolysis under heat and elevated pressure cleaves all these glycosidic bonds completely, releasing the individual glucose monomer units — this is the industrial process used to make glucose syrup from starch.

Step-by-Step Solution

Starch's repeating unit formula is $(C_6H_{10}O_5)_n$ .

Complete hydrolysis adds one water molecule per glycosidic bond broken: $(C_6H_{10}O_5)_n + nH_2O \xrightarrow[2\text{-}3\ atm]{H^+, 393K} nC_6H_{12}O_6$ .

The product, $C_6H_{12}O_6$ , is glucose — the single repeating monosaccharide unit of starch. …
Start your 14-day free trial to see the full solution →

←Previous questionQ53 — Write the structures of fragments produced on complete hydrolysis of D…→Next questionQ55 — Proteins can be classified into two types on the basis of their molecu…

Q.Carbohydrates are classified on the basis of their behaviour on hydrolysis and also as reducing or non-reducing sugar. Sucrose is a _______. (Note: one or more of the following options may be correct.)

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 Carbohydrates

Method: Structural Analysis of Sucrose

Steps:

Mistake 1: Confusing "Reducing" with "Non-Reducing" based on taste or common use

Mistake 2: Forgetting that Sucrose is a Disaccharide

Mistake 3: Confusing Hydrolysis Products with the Sugar Itself

Mistake 4: Misclassifying Sucrose as a Monosaccharide

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.Carbohydrates are classified on the basis of their behaviour on hydrolysis and also as reducing or non-reducing sugar. Sucrose is a _______. (Note: one or more of the following options may be correct.)

🎓Unlock everything free for 14 days

Unlock everything free for 14 days