Q.What happens when D-glucose is treated with the following reagents?
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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 (C12H22O11) is a disaccharide composed of one molecule of D-galactose and one molecule of D-glucose linked by a β(1→4) glycosidic bond. Upon hydrolysis (reaction with water), this bond is cleaved, yielding the two monosaccharides:
Lactose+H2Olactase or acidD-Galactose+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 (C6H12O6) but differ in the arrangement of the hydroxyl group on carbon 4 (they are C-4 epimers).
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: β(1→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 β(1→4) glucose
- The bond is between:
- Carbon-1 of galactose (in β configuration)
- Carbon-4 of glucose
So the structural formula is:
Galactose−O−Glucose
2. What does hydrolysis do?
Hydrolysis means "splitting with water." The reaction is:
Lactose+H2Olactase or acidGalactose+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:
CX12HX22OX11+HX2OCX6HX12OX6+CX6HX12OX6
- Lactose: CX12HX22OX11
- Water: HX2O
- Products: two molecules of CX6HX12OX6 (one galactose, one glucose)
Why the same molecular formula?
Both galactose and glucose are aldohexoses — they have the same molecular formula CX6HX12OX6 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. …
The key idea is that each reagent targets a specific functional group in glucose, producing characteristic oxidation or reduction products.
Step 1 — Reaction with HI: Hydroiodic acid is a strong reducing agent. It reduces the aldehyde group (−CHO) to a methyl group (−CH3), and also reduces all the hydroxyl groups (−OH) to hydrogen atoms, yielding n-hexane as the final product.
Step 2 — Reaction with bromine water: Bromine water (Br2/H2O) is a mild oxidising agent that selectively oxidises the aldehyde group to a carboxylic acid group, without affecting the alcohol groups. This gives gluconic acid. …
D-glucose reacts differently with each reagent: HI reduces it to n-hexane (cleaving all C–OH bonds), bromine water oxidises only the aldehyde group to give gluconic acid, and HNO₃ oxidises both ends to yield a dicarboxylic acid (glucaric acid). The key is recognising which functional groups each reagent attacks.
Let’s understand why each reagent does what it does. D-glucose is an aldohexose — it has an aldehyde group at C1 and hydroxyl groups on every other carbon. The behaviour of these reagents depends on their oxidising or reducing power and their selectivity.
1. Reaction with HI (hydroiodic acid)
HI is a strong reducing agent. In hot, concentrated HI, all the –OH groups in glucose are replaced by –I, and then the C–I bonds are reduced to C–H bonds. This is a reductive deoxygenation — every hydroxyl group gets removed, and the aldehyde group also gets reduced to a methyl group. The entire carbon chain survives intact, but all oxygen is stripped off.
The product is n-hexane (CH₃–CH₂–CH₂–CH₂–CH₂–CH₃).
A common mistake is to think HI only reduces the aldehyde. In fact, HI under these conditions reduces every C–OH bond, not just the carbonyl.
2. Reaction with bromine water
Bromine water (Br₂ in H₂O) is a mild oxidising agent. It selectively oxidises the aldehyde group (–CHO) to a carboxylic acid (–COOH) without touching the alcohol groups. This is because bromine water specifically targets aldehydes (and not ketones or alcohols) under neutral or slightly acidic conditions.
So D-glucose gives gluconic acid (a monocarboxylic acid where C1 is –COOH, and the rest of the chain remains unchanged).
Bromine water is the classic test for an aldehyde group in sugars. It won’t oxidise secondary alcohols, so it’s perfect for distinguishing aldoses from ketoses. …
Method: Reagent-Specific Functional Group Analysis
This method uses the functional group reactivity of D-glucose (an aldohexose) to predict products with each reagent.
Steps
Step 1: Identify the functional groups in D-glucose
- One aldehyde group (–CHO) at C1
- Four secondary alcohol groups (–OH) at C2, C3, C4, C5
- One primary alcohol group (–CH₂OH) at C6
Step 2: Apply each reagent based on its known reaction with these groups
(i) With HI (hydroiodic acid) — Reductive cleavage
- HI is a strong reducing agent that cleaves C–O bonds and reduces all oxygen-containing groups.
- All –OH groups are replaced by –H, and the aldehyde is reduced to –CH₃.
- Product: n-Hexane (CH3–CH2–CH2–CH2–CH2–CH3)
(ii) With bromine water (Br2/H2O) — Oxidation of aldehyde only
- Bromine water is a mild oxidizing agent that selectively oxidizes the aldehyde group to a carboxylic acid (–COOH).
- Alcohol groups remain unchanged.
- Product: Gluconic acid (a six-carbon aldonic acid)
(iii) With HNO3 (nitric acid) — Oxidation of both ends
- HNO3 is a strong oxidizing agent that oxidizes both: …
Here are the common mistakes students make when answering questions about the reactions of D-glucose with HI, Bromine water, and HNO₃, along with how to avoid each.
Mistake 1: Confusing the reaction with HI as simple reduction
The error: Students often write that HI simply reduces the aldehyde group (−CHO) to an alcohol (−CH2OH), treating it like a mild reducing agent.
Why it’s wrong: HI is a strong reducing agent under heat. It does not stop at the aldehyde. It reduces all the hydroxyl groups (−OH) to hydrogen atoms (−H), converting the sugar into a straight-chain hydrocarbon.
Correct outcome:
D-glucose + excess HI (heat) → n-Hexane (CH3(CH2)4CH3).
The entire carbon chain is reduced, and all oxygen atoms are removed.
How to avoid:
- Remember: HI + heat = complete deoxygenation of sugars.
- Think of it as “stripping” every −OH group, leaving only a saturated alkane.
Mistake 2: Thinking bromine water oxidizes all parts of glucose
The error: Students assume bromine water (Br2/H2O) is a strong enough oxidant to break the glucose chain or oxidize all carbon atoms.
Why it’s wrong: Bromine water is a mild, selective oxidant. It only oxidizes the aldehyde group (−CHO) to a carboxylic acid group (−COOH). It does not affect the alcohol groups (−OH) on the rest of the chain.
Correct outcome:
D-glucose + Bromine water → Gluconic acid (a monocarboxylic acid).
The aldehyde at C1 becomes −COOH, while all other −OH groups remain unchanged.
How to avoid:
- Memorize: Bromine water = specific test for aldehydes in sugars.
- Do not confuse it with strong oxidants like HNO3 (see next mistake).
Mistake 3: Confusing the action of HNO3 with bromine water
The error: Students write that HNO3 also gives gluconic acid, or that it only oxidizes the aldehyde group.
Why it’s wrong: HNO3 is a strong oxidizing agent. It oxidizes both ends of the glucose chain — the aldehyde group (−CHO) at C1 and the primary alcohol group (−CH2OH) at C6 — into carboxylic acid groups (−COOH).
Correct outcome: …
- AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQQ.Two statements are given below Statement I: Cane sugar is disaccharide of α−D−glucose and β−D−fructose Statement II: Milk sugar is disaccharide of β−D−galactose and β−D−glucose Correct answer is (A) Statements I and II both are correct (B) Statements I and II both are not correct (C) Statement I is correct, but statement II is not correct (D) statement I is not correct but statement II is correct
›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 α-D-glucose and β-D-fructose. Lactose, by contrast, is a reducing sugar because glucose's anomeric carbon is left free; the fixed unit is β-D-galactose joined via β-1,4 linkage to D-glucose, and standard descriptions state it as β-D-galactose and β-D-glucose.
Step-by-Step Solution
- Statement I: Sucrose = α-D-glucopyranose + β-D-fructofuranose, linked C1(glucose)→C2(fructose). This is the textbook description. Correct. …
- AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQQ.Consider the following Statement-I : Lactose is composed of α-D-glucose and β-D-glucose. Statement-II : Lactose is a reducing sugar. The correct answer is (A) Both statement-I and statement-II are not correct (B) Both statement-I and statement-II are correct (C) Statement-I is correct, but statement-II is not correct (D) Statement-I is not correct, but statement-II is correct
›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 = β-D-galactose + D-glucose, joined β(1→4) between galactose C1 and glucose C4.
- Statement-I claims lactose is "α-D-glucose + β-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. …
- AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQQ.Consider the following Statement-I: Cane sugar is a disaccharide of α-D-glucose and β-D-fructose Statement-II: Milk sugar is a diasaccharide of α-D-glucose and β-D-galactose The correct answer is (A) Both statement-I and statement-II are correct (B) Both statement-I and statement-II are not correct (C) Statement-I is correct, but statement-II is not correct (D) Statement-I is not correct, but statement-II is correct
›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 (α or β) 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 α-D-glucose and C2 of β-D-fructose. Statement-I matches this exactly — correct.
- Lactose (milk sugar): formed by a glycosidic bond between C1 of β-D-galactose and C4 of β-D-glucose (glucose unit here is in β form as it provides the free anomeric carbon, though the ring can open to α/β equilibrium — the standard textbook description names β-D-galactose and glucose, not α-D-glucose). …
- AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQQ.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: [α]D=+66.5∘ (dextrorotatory)
- Glucose: [α]D=+52.7∘ (dextrorotatory)
- Fructose: [α]D=−92.4∘ (laevorotatory)
-
Hydrolysis reaction
Sucrose+H2O→Glucose+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:
Net rotation=2(+52.7∘)+(−92.4∘)=2−39.7∘=−19.85∘
This is negative (laevorotatory). So the mixture is laevorotatory, even though glucose alone is dextrorotatory.
- Evaluate Assertion (A) …
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- AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQQ.Type of Glycosidic bonds in cellulose and starch respectively ___________________ (A) β-1-4, β-1-6 and α-1-4, α-1-6 glycosidic bonds (B) β-1-4 and α-1-4 glycosidic bonds only (C) β-1-6 and α-1-4, α-1-6 glycosidic bonds (D) β-1-4, α-1-4 and α-1-6 glycosidic bonds
›Reveal solutionSolution
Cellulose has only β-1,4 glycosidic bonds; starch (via amylopectin's branching) has both α-1,4 and α-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 β-D-glucose units connected end to end by β-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 α linkages).
Starch, by contrast, is made of α-D-glucose units. Its two components are amylose (a straight chain held together by α-1,4 bonds) and amylopectin (a branched molecule with an α-1,4 backbone plus α-1,6 bonds at branch points, roughly every 24–30 residues). Because starch as a storage polysaccharide is really this combination, both α-1,4 and α-1,6 bonds are correctly attributed to it.
Step-by-Step Solution
- Identify cellulose's bond type: only β-1,4-glycosidic bonds (no branching, no α bonds). …
- AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQQ.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 α-D-glucopyranose and β-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, [α]D=+66.5∘.
- Acid hydrolysis (or the enzyme invertase) breaks the glycosidic bond: Sucrose+H2O→Glucose+Fructose.
- Free D-glucose is dextrorotatory, [α]D=+52.5∘.
- Free D-fructose is strongly laevorotatory, [α]D=−92∘ (fructose's rotation is large and negative — it is sometimes called laevulose for this reason). …
- AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQQ.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 α(1→4) 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 α(1→4) glycosidic linkage). Enzyme maltase hydrolyses this bond ⇒ 2 glucose molecules only.
- Sucrose = glucose + fructose, hydrolysed by sucrase/invertase ⇒ gives glucose and fructose, not glucose alone.
- Lactose = glucose + galactose, hydrolysed by lactase ⇒ gives glucose and galactose, not glucose alone. …
- AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQQ.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 C12H22O11.
- Acid hydrolysis reaction: C12H22O11+H2OH+C6H12O6(glucose)+C6H12O6(fructose). …
- AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQQ.Identify the product of the following reaction. (C6H10O5)n+nH2OH+, 393K2-3 atm ? (Starch) (A) Fructose (B) Glucose (C) Lactose (D) Maltose
›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 (α-1,4 and α-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 (C6H10O5)n.
- Complete hydrolysis adds one water molecule per glycosidic bond broken: (C6H10O5)n+nH2OH+,393K2-3 atmnC6H12O6.
- The product, C6H12O6, is glucose — the single repeating monosaccharide unit of starch. …
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