CUET UG Chemistry Booster Test - 3 Structure and Reactions of Carbohydrates -

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CUET UG Chemistry Booster Test - 2 Structure and Reactions of Carbohydrates

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Chemistry Booster Test - 2 Structure and Reactions of Carbohydrates

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QUESTION 1 OF 20

While the molecular formula of glucose is C₆H₁₂O₆, not all compounds fitting the general carbohydrate formula Cₓ(H₂O)y are carbohydrates. Which of the following defines the modern chemical nature of carbohydrates beyond just the general molecular formula?

Answer Review

1 While the molecular formula of glucose is C₆H₁₂O₆, not all compounds fitting the general carbohydrate formula Cₓ(H₂O)y are carbohydrates. Which of the following defines the modern chemical nature of carbohydrates beyond just the general molecular formula?

A They must invariably be sweet-tasting crystalline solids.

B They must be optically active polyhydroxy aldehydes or ketones, or compounds yielding such units.

C They must exist exclusively in a thermodynamically stable cyclic hemiacetal form.

D They must yield only D-(+)-glucose upon complete enzymatic hydrolysis.

💡 Short Explanation

�� The hydrate formula is not universally applicable. �� Some carbohydrates do not fit the formula. �� The functional-group-based definition is accepted today.

✅ Correct Answer Explanation

The modern definition classifies carbohydrates as optically active polyhydroxy aldehydes or ketones, or substances that yield such compounds upon hydrolysis. This definition accommodates compounds like rhamnose that do not fit the general hydrate formula.

❓ Why Other Options Are Incorrect

�� Option A → Not all carbohydrates are sweet.
�� Option C → Not all carbohydrates exist exclusively in cyclic form.
�� Option D → Different carbohydrates yield different hydrolysis products.

🎯 Strategy

Used

Concept Clarification

Application:

Differentiate the modern definition from the older hydrate-of-carbon concept.

Final Logic:

Carbohydrates are defined by their functional groups rather than only their formula.

🧠 Memory Trick

Carbohydrates = Polyhydroxy Aldehydes/Ketones

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Modern Definition of Carbohydrates

2 Analyze the following analytical statements about the structural elucidation of glucose:
1. Prolonged heating with HI entirely cleaves all carbon-carbon bonds.
2. The formation of n-hexane with HI definitively confirms an unbranched, straight carbon skeleton.
3. The open-chain structure fully explains all observed chemical reactions of glucose, including Schiff\\\'s test.
Which of the statements is/are conceptually accurate?

A Only 1

B Only 2

C 1 and 3

D 2 and 3

💡 Short Explanation

�� HI reduction produces n-hexane. �� This confirms a straight-chain carbon skeleton. �� The open-chain structure fails to explain some observations.

✅ Correct Answer Explanation

Statement 1 is incorrect because HI reduction does not involve complete cleavage of all carbon-carbon bonds. Statement 2 is correct because formation of n-hexane establishes that glucose contains six carbon atoms arranged in an unbranched chain. Statement 3 is incorrect because the open-chain structure cannot explain why glucose fails to give Schiffs test and why its pentaacetate does not react with hydroxylamine. Therefore, only Statement 2 is correct.

❓ Why Other Options Are Incorrect

�� Option A → Statement 1 is false.
�� Option C → Statements 1 and 3 are false.
�� Option D → Statement 3 is false.

🎯 Strategy

Used

Statement Verification

Application:

Evaluate each statement using experimental evidence.

Final Logic:

Only Statement 2 is supported.

🧠 Memory Trick

HI → Hexane → Straight Chain

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Structural Elucidation of Glucose

3 Match the structural evidence discovered in glucose with the chemical reagent directly used to confirm it.

List I	List II
1. — Straight chain of six carbons	a. — Bromine water
2. — Presence of a carbonyl group	b. — Acetic anhydride
3. — Presence of five hydroxyl groups	c. — Hydrogen cyanide
4. — Terminal aldehydic functional group	d. — Hydrogen iodide

A 1-d, 2-c, 3-b, 4-a

B 1-c, 2-d, 3-b, 4-a

C 1-d, 2-a, 3-c, 4-b

D 1-b, 2-c, 3-d, 4-a

💡 Short Explanation

�� HI confirms a straight chain. �� HCN confirms a carbonyl group. �� Acetic anhydride confirms hydroxyl groups. �� Bromine water confirms an aldehyde.

✅ Correct Answer Explanation

Matching: 1. Straight chain → Hydrogen iodide 2. Carbonyl group → Hydrogen cyanide 3. Five hydroxyl groups → Acetic anhydride 4. Aldehydic group → Bromine water Thus: 1-d, 2-c, 3-b, 4-a

❓ Why Other Options Are Incorrect

�� Options B, C and D contain incorrect reagent assignments.

🎯 Strategy

Used

Option Grouping

Application:

Match each structural feature with the experiment that proves it.

Final Logic:

Only Option A contains all correct matches.

🧠 Memory Trick

HI–Chain, HCN–Carbonyl, Acetic–OH, Bromine–CHO

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Structural Evidence for Glucose

4 The acetylation of glucose with acetic anhydride yields a stable glucose pentaacetate. Conceptually, this specific reaction proves that the five –OH groups are attached to different carbon atoms because:

A Chemical compounds containing multiple –OH groups on the same carbon atom are notoriously unstable.

B Acetic anhydride completely cleaves carbon-carbon bonds during substitution.

C Acetylation inherently converts the terminal aldehyde group directly into an alcohol.

D The molecule is instantaneously converted into a furanose ring preventing multiple acetylations.

💡 Short Explanation

�� Glucose forms a stable pentaacetate. �� Five hydroxyl groups are independently acetylated. �� This indicates they are attached to different carbon atoms.

✅ Correct Answer Explanation

The existence of stable glucose pentaacetate indicates that glucose contains five distinct hydroxyl groups. If several hydroxyl groups were attached to the same carbon atom, the structure would be unstable. Therefore, the hydroxyl groups must be distributed across different carbon atoms.

❓ Why Other Options Are Incorrect

�� Option B → Acetylation does not break carbon-carbon bonds.
�� Option C → Acetylation does not reduce aldehydes.
�� Option D → Furanose formation is unrelated to this conclusion.

🎯 Strategy

Used

Conceptual Reasoning

Application:

Interpret the significance of pentaacetate formation.

Final Logic:

Five stable acetylations require five separate hydroxyl-bearing carbons.

🧠 Memory Trick

Pentaacetate = Five Separate OH Groups

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Hydroxyl Groups in Glucose

5 During the mild oxidation of glucose with bromine water, the product formed is gluconic acid. This specific chemical conversion firmly establishes that the carbonyl group at C-1 is:

A A ketone that strongly resists further oxidation.

B A primary alcohol capable of bypassing the aldehyde phase.

C An aldehyde that is easily oxidisable compared to a ketone.

D Actively involved in forming a complex peptide linkage.

💡 Short Explanation

�� Bromine water oxidises aldehydes. �� Glucose forms gluconic acid. �� Therefore, glucose contains an aldehyde group.

✅ Correct Answer Explanation

The conversion of glucose into gluconic acid by bromine water demonstrates that the carbonyl group behaves as an aldehyde. Aldehydes are readily oxidised by mild oxidising agents, whereas ketones generally resist such oxidation. Thus, glucose contains an aldehydic carbonyl group.

❓ Why Other Options Are Incorrect

�� Option A → Ketones are not readily oxidised by bromine water.
�� Option B → The reaction involves aldehyde oxidation.
�� Option D → Peptide linkages are unrelated.

🎯 Strategy

Used

Reaction Analysis

Application:

Identify the functional group oxidised by bromine water.

Final Logic:

Gluconic acid formation confirms an aldehyde.

🧠 Memory Trick

Bromine Water → Aldehyde Proof

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Oxidation of Glucose

6 The distinct dicarboxylic acid obtained when either glucose or gluconic acid undergoes harsh oxidation with nitric acid is scientifically named:

A Acetic acid

B Saccharic acid

C Ascorbic acid

D Pyruvic acid

💡 Short Explanation

�� Nitric acid is a strong oxidising agent. �� Both terminal groups become carboxylic acids. �� The resulting product is saccharic acid.

✅ Correct Answer Explanation

Nitric acid oxidises both the aldehyde group and the terminal primary alcohol group of glucose, producing a dicarboxylic acid known as saccharic acid. The same product is obtained when gluconic acid is further oxidised.

❓ Why Other Options Are Incorrect

�� Option A → Simple monocarboxylic acid.
�� Option C → Vitamin C.
�� Option D → Keto acid unrelated to glucose oxidation.

🎯 Strategy

Used

Direct Recall

Application:

Recall the oxidation product of glucose with nitric acid.

Final Logic:

Strong oxidation gives saccharic acid.

🧠 Memory Trick

Nitric Acid → Saccharic Acid

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Saccharic Acid Formation

7 Arrange the following steps sequentially to properly deduce whether a monosaccharide is assigned a relative D-configuration:
A. Locate the lowest asymmetric carbon atom in the structural projection.
B. Ensure the most oxidised carbon (e.g., –CHO) is placed cleanly at the top.
C. Check if the –OH group situated on this specific lowest carbon is oriented on the right side.
D. Establish correlation with the (+) isomer of the glyceraldehyde reference standard.

A 1. B, A, C, D

B . A, B, D, C

C 3. D, B, A, C

D 4. C, A, B, D

💡 Short Explanation

Proper Fischer projection is required first.
Lowest chiral carbon is then identified.
Position of OH determines D or L configuration.

✅ Correct Answer Explanation

The correct procedure is:

1.Place the most oxidised carbon at the top.

2.Locate the lowest asymmetric carbon.

3.Observe the position of the OH group on that carbon.

4.Compare with glyceraldehyde to assign D or L configuration.

Therefore, the sequence is B → A → C → D.

❓ Why Other Options Are Incorrect

3. Why Other Options Are Incorrect

Options B, C and D do not follow the accepted stereochemical procedure.

🎯 Strategy

4. Strategy Used:

Sequential Reasoning

Application:

Arrange the steps needed for configuration assignment.

Final Logic:

Proper projection precedes stereochemical comparison.

🧠 Memory Trick

Top → Lowest Chiral → OH → D/L

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: D and L Configuration

8 How many asymmetric (chiral) carbon atoms are actively present in the reference compound, glyceraldehyde, used to map the D/L configuration of complex sugars?

A Zero

B One

C Two

D Three

💡 Short Explanation

�� Glyceraldehyde is the simplest chiral carbohydrate. �� It contains only one asymmetric carbon atom. �� It serves as the D/L reference standard.

✅ Correct Answer Explanation

Glyceraldehyde contains a single carbon atom attached to four different groups, making it chiral. Because it has only one stereogenic centre, it is used as the standard reference compound for assigning D and L configurations.

❓ Why Other Options Are Incorrect

�� Option A → Glyceraldehyde is chiral.
�� Option C → Contains only one chiral carbon.
�� Option D → Too many chiral centres.

🎯 Strategy

Used

Direct Recall

Application:

Recall the stereochemical structure of glyceraldehyde.

Final Logic:

Glyceraldehyde has one chiral carbon.

🧠 Memory Trick

Glyceraldehyde = One Chiral Centre

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Glyceraldehyde Reference Standard

📄 Passage

The open-chain structure of glucose could not explain several of its observed properties. For instance, despite ostensibly having an aldehyde group, glucose does not give Schiff's test, and its stable pentaacetate form does not react with hydroxylamine. To resolve this, it was proposed that one of the –OH groups adds to the internal –CHO group to form a cyclic hemiacetal. It was found that glucose naturally forms a six-membered pyranose ring dynamically involving the –OH at C-5.

Question: The analytical fact that glucose pentaacetate does not react with hydroxylamine proves the absence of:

A A free primary alcohol group in the molecule.

B A cyclic hemiacetal equilibrium.

C A free reactive –CHO group in the pentaacetate derivative.

D Anomeric carbons across the structure.

💡 Short Explanation

�� Hydroxylamine reacts with free carbonyl groups. �� Glucose pentaacetate fails to react. �� Therefore, no free aldehyde group is present.

✅ Correct Answer Explanation

Since hydroxylamine reacts readily with aldehydes and ketones, the inability of glucose pentaacetate to react with hydroxylamine indicates the absence of a free carbonyl (–CHO) group in the derivative. This observation supports the cyclic structure of glucose.

❓ Why Other Options Are Incorrect

�� Option A → Not established by this experiment.
�� Option B → Cyclic equilibrium is actually supported.
�� Option D → Anomeric carbons are present.

🎯 Strategy

Used

Passage-Based Analysis

Application:

Determine what hydroxylamine normally reacts with.

Final Logic:

No reaction means no free carbonyl group.

🧠 Memory Trick

No Oxime → No Free CHO

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Cyclic Structure of Glucose

📄 Passage

Question: The cyclic structure of glucose is named pyranose in direct structural analogy with pyran. Pyran is a fundamental cyclic organic compound precisely consisting of:

A Five carbon atoms and one oxygen atom in the ring.

B Four carbon atoms and one oxygen atom in the ring.

C Six carbon atoms continuously bound in the ring.

D Five carbon atoms and two distinct oxygen atoms in the ring.

💡 Short Explanation

�� Pyran is a six-membered heterocyclic ring. �� The ring contains one oxygen atom. �� The remaining five atoms are carbon atoms.

✅ Correct Answer Explanation

Pyran is a six-membered heterocyclic compound consisting of five carbon atoms and one oxygen atom in the ring. Because the cyclic structure of glucose resembles pyran, it is known as a pyranose structure.

❓ Why Other Options Are Incorrect

�� Option B → Represents a five-membered ring.
�� Option C → Contains no oxygen atom.
�� Option D → Pyran contains only one oxygen atom.

🎯 Strategy

Used

Structural Comparison

Application:

Compare the ring structure of glucose with pyran.

Final Logic:

Pyran = 5 carbons + 1 oxygen.

🧠 Memory Trick

Pyranose = Pyran-Like Ring

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Pyranose Structure of Glucose

11 The fundamental structural difference defining the separation between the α and β anomers of cyclic D-glucose lies strictly in the spatial arrangement (configuration) around:

A C-5, the carbon anchoring the ring formation.

B C-1, the hemiacetal anomeric carbon.

C C-6, the external primary alcohol carbon.

D C-2, the carbon adjacent to the anomeric carbon.

💡 Short Explanation

�� α- and β-glucose differ at only one carbon atom. �� This carbon is created during cyclisation. �� It is called the anomeric carbon.

✅ Correct Answer Explanation

The α and β forms of glucose differ only in the orientation of the hydroxyl group attached to C-1 after ring formation. This carbon originates from the aldehyde carbon of open-chain glucose and becomes the anomeric carbon in the cyclic structure. Therefore, the distinction between α and β anomers is based on the configuration at C-1.

❓ Why Other Options Are Incorrect

�� Option A → C-5 participates in ring formation but does not determine α/β forms.
�� Option C → C-6 remains unchanged in both anomers.
�� Option D → C-2 is unrelated to anomerism.

🎯 Strategy

Used

Concept Identification

Application:

Identify the carbon responsible for anomerism.

Final Logic:

α and β forms differ only at C-1.

🧠 Memory Trick

Anomer = Difference at Anomeric Carbon

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Anomers of Glucose

12 What is the experimentally determined melting point of the β-form of glucose, distinguishing it physically from the lower-melting α-form?

A 419 K

B 303 K

C 371 K

D 423 K

💡 Short Explanation

�� α-Glucose melts at 419 K. �� β-Glucose melts at a higher temperature. �� The value is 423 K.

✅ Correct Answer Explanation

Experimental studies show that β-glucose has a melting point of 423 K, whereas α-glucose melts at 419 K. This difference provides a physical method for distinguishing the two crystalline forms.

❓ Why Other Options Are Incorrect

�� Option A → Melting point of α-glucose.
�� Option B → Crystallisation temperature of α-glucose.
�� Option C → Crystallisation temperature of β-glucose.

🎯 Strategy

Used

Direct Recall

Application:

Recall the melting point of β-glucose.

Final Logic:

β-Glucose melts at 423 K.

🧠 Memory Trick

Beta = Bigger Melting Point = 423 K

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Physical Properties of Glucose Anomers

13 Evaluate the core structural characteristics defining fructose:
1. It inherently possesses a ketonic functional group specifically at carbon 2.
2. It forms a six-membered pyranose ring by the addition of –OH at C-6 to the central keto group.
3. The cyclic structure is named furanose in strict analogy with the compound furan.
Which statements are factually correct?

A 1 and 2

B 1 and 3

C 2 and 3

D 1, 2, and 3

💡 Short Explanation

�� Fructose is a ketohexose. �� The keto group is located at C-2. �� Its common cyclic form is furanose.

✅ Correct Answer Explanation

Statement 1 is correct because fructose contains a ketonic group at C-2. Statement 2 is incorrect because fructose commonly forms a five-membered furanose ring rather than a six-membered pyranose ring. Statement 3 is correct because the cyclic structure resembles furan and is therefore called a furanose. Thus, Statements 1 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Statement 2 is incorrect.
�� Option C → Statement 2 is incorrect.
�� Option D → Includes incorrect Statement 2.

🎯 Strategy

Used

Statement Verification

Application:

Evaluate each statement separately.

Final Logic:

Statements 1 and 3 are true.

🧠 Memory Trick

Fructose = Keto at C2 + Furanose Ring

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Structure of Fructose

14 Fructose is chemically written as D-(–)-fructose. What physical experimental phenomenon fundamentally dictates the "(–)" sign embedded in its notation?

A The direction it actively rotates the plane of polarised light (laevorotatory).

B Its strong reduction kinetics when exposed to sodium amalgam.

C The spatial configuration of its lowest asymmetric carbon matching (–)-glyceraldehyde.

D Its spontaneous reaction speed with Tollens\\\' reagent.

💡 Short Explanation

�� The (–) sign indicates optical activity. �� Fructose rotates plane-polarised light to the left. �� Therefore it is laevorotatory.

✅ Correct Answer Explanation

The symbol (–) refers to the direction in which a compound rotates plane-polarised light. Fructose rotates light to the left and is therefore described as laevorotatory. The D designation refers to configuration and is unrelated to the sign of optical rotation.

❓ Why Other Options Are Incorrect

�� Option B → Unrelated to optical notation.
�� Option C → D/L configuration is independent of (+)/(–).
�� Option D → Not connected to optical activity.

🎯 Strategy

Used

Concept Clarification

Application:

Distinguish optical rotation from stereochemical configuration.

Final Logic:

(–) indicates leftward optical rotation.

🧠 Memory Trick

Minus = Left Rotation

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Optical Activity of Fructose

15 The structural connection formed between two monosaccharides by the elimination of a water molecule leaves a solitary oxygen atom bridging the two units. What is the specific biochemical terminology assigned to this stable bond?

A Peptide bond

B Hydrogen bond

C Phosphodiester bond

D Glycosidic linkage

💡 Short Explanation

�� Monosaccharides combine through condensation. �� Water is eliminated. �� The resulting bridge is called a glycosidic linkage.

✅ Correct Answer Explanation

When two monosaccharides join together, a water molecule is removed and an oxygen bridge remains between them. This linkage is known as a glycosidic linkage and is responsible for the formation of disaccharides and polysaccharides.

❓ Why Other Options Are Incorrect

�� Option A → Found in proteins.
�� Option B → Weak intermolecular interaction.
�� Option C → Found in nucleic acids.

🎯 Strategy

Used

Definition Recall

Application:

Identify the bond characteristic of carbohydrates.

Final Logic:

Sugar units are connected by glycosidic linkages.

🧠 Memory Trick

Glyco = Sugar Bond

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Glycosidic Linkage

16 When a complex non-reducing sugar like sucrose is hydrolysed by dilute acids or selective enzymes, the resulting chemical product is:

A A homopolymer comprised of pure α-D-glucose units.

B An equimolar mixture of structurally distinct reducing monosaccharides.

C A pair of entirely identical non-reducing monosaccharides.

D Exclusively dextrorotatory unbound fructose.

💡 Short Explanation

�� Sucrose hydrolysis produces glucose and fructose. �� The products are reducing sugars. �� They are formed in equal amounts.

✅ Correct Answer Explanation

Hydrolysis of sucrose yields one molecule of D-(+)-glucose and one molecule of D-(–)-fructose. These monosaccharides are reducing sugars and are produced in equimolar amounts. Hence, the product is an equimolar mixture of structurally different reducing sugars.

❓ Why Other Options Are Incorrect

�� Option A → No polymer is formed.
�� Option C → Products are not identical.
�� Option D → Glucose is also formed.

🎯 Strategy

Used

Reaction Analysis

Application:

Identify the products of sucrose hydrolysis.

Final Logic:

Sucrose → Glucose + Fructose.

🧠 Memory Trick

Sucrose = G + F

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Hydrolysis of Sucrose

17 Match the characteristic chemical feature with the appropriate sugar property/molecule.

List I	List II
1. — Glycosidic linkage bridging C1(α) and C2(β)	a. — D-(+)-glucose
2. — Strongly Laevorotatory (–92.4°)	b. — Invert sugar
3. — Moderately Dextrorotatory (+52.5°)	c. — D-(–)-fructose
4. — Resulting rotational mixture after sucrose hydrolysis	d. — Sucrose

A 1-d, 2-c, 3-a, 4-b

B 1-a, 2-b, 3-c, 4-d

C 1-d, 2-a, 3-c, 4-b

D 1-b, 2-d, 3-a, 4-c

💡 Short Explanation

�� Sucrose contains the α(1→2)β linkage. �� Fructose is strongly laevorotatory. �� Glucose is dextrorotatory.

✅ Correct Answer Explanation

Matching: 1. C1(α)-C2(β) linkage → Sucrose 2. Strongly laevorotatory → D-(–)-fructose 3. Moderately dextrorotatory → D-(+)-glucose 4. Product after sucrose hydrolysis → Invert sugar Therefore: 1-d, 2-c, 3-a, 4-b

❓ Why Other Options Are Incorrect

�� Options B, C and D contain incorrect associations.

🎯 Strategy

Used

Option Grouping

Application:

Match each property with the appropriate sugar.

Final Logic:

Only Option A gives all correct pairings.

🧠 Memory Trick

Sucrose Linkage, Fructose Left, Glucose Right

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Sucrose, Glucose and Fructose

18 Consider the chemical reasoning dictating maltose as a reducing sugar:
1. It consists strictly of two α-D-glucose units.
2. The core linkage spans between C1 of one glucose unit and C4 of another.
3. A free reactive aldehyde group can spontaneously be produced at C1 of the second glucose unit in solution.
Which is the primary functional reason for its reducing nature?

A Only 1

B Only 2

C Only 3

D 1 and 2

💡 Short Explanation

�� Reducing sugars require a free anomeric carbon. �� Maltose can generate a free aldehyde group. �� This causes its reducing behavior.

✅ Correct Answer Explanation

Although Statements 1 and 2 describe the structure of maltose, the actual reason for its reducing nature is that the second glucose unit possesses a free anomeric carbon capable of opening into an aldehydic form in solution. Therefore, Statement 3 provides the primary explanation.

❓ Why Other Options Are Incorrect

�� Option A → Structural fact, not the cause.
�� Option B → Structural description only.
�� Option D → Neither explains reducing behavior directly.

🎯 Strategy

Used

Cause-and-Effect Analysis

Application:

Identify the feature responsible for reducing properties.

Final Logic:

Free aldehyde generation causes reduction.

🧠 Memory Trick

Free C1 = Reducing Sugar

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Reducing Nature of Maltose

19 Arrange the structural components involved in the formation of lactose sequentially, starting from the C1 linking unit directly to the receiving unit:

A A → B → C → D

B D → B → C → A

C A → C → B → D

D B → A → D → C

💡 Short Explanation

�� Lactose contains galactose linked to glucose. �� The linkage is β(1→4). �� Galactose donates C1 to glucose C4.

✅ Correct Answer Explanation

In lactose, C1 of β-D-galactose is connected through a glycosidic oxygen atom to C4 of β-D-glucose. Therefore, the structural sequence proceeds from the donor monosaccharide through the oxygen bridge to the receiving carbon and finally the receiving sugar. Hence: A → B → C → D

❓ Why Other Options Are Incorrect

�� Options B, C and D disrupt the actual linkage sequence.

🎯 Strategy

Used

Structural Sequencing

Application:

Trace the lactose linkage from donor to acceptor.

Final Logic:

Galactose → Oxygen → C4 → Glucose.

🧠 Memory Trick

Gal(1→4)Glu

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Structure of Lactose

20 Lactose actively reduces Fehling's solution. Which dynamic structural mechanism allows this chemical reaction to persistently occur despite being a bonded disaccharide?

A The C1 anomeric carbon of the galactose unit remains chemically free.

B The C1 of the bound glucose unit can undergo rapid ring opening to form a free, reactive aldehyde group.

C The robust glycosidic linkage breaks rapidly in the basic environment of Fehling\\\'s solution.

D Lactose inherently contains a transient ketonic group that isomerises to an aldehyde.

💡 Short Explanation

�� Lactose contains a free anomeric carbon. �� The glucose unit can generate an aldehyde group. �� This gives lactose reducing properties.

✅ Correct Answer Explanation

In lactose, the anomeric carbon of the glucose unit remains free. It can undergo ring opening in solution to generate a reactive aldehyde group capable of reducing Fehling\\\'s solution and Tollens\\\' reagent. Therefore, lactose behaves as a reducing sugar.

❓ Why Other Options Are Incorrect

�� Option A → Galactose C1 participates in the linkage.
�� Option C → The glycosidic bond does not rapidly break.
�� Option D → Lactose does not contain a ketonic group.

🎯 Strategy

Used

Functional Group Analysis

Application:

Identify the source of reducing behavior.

Final Logic:

Free glucose anomeric carbon enables aldehyde formation.

🧠 Memory Trick

Free Glucose C1 = Reducing Lactose

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Reducing Nature of Lactose

CUET UG Chemistry Booster Test – 3 Structure and Reactions of Carbohydrates

CUET UG Chemistry Booster Test - 2 Structure and Reactions of Carbohydrates

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