CUET UG Chemistry Booster Test - 2 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

1 / 20

QUESTION 1 OF 20

Consider the following statements regarding the carbohydrate formula:
1. The general formula is Cx(H₂O)y, and glucose accurately fits it as C₆(H₂O)₆.
2. Rhamnose (C₆H₁₂O₅) is a carbohydrate but does not fit the hydrate formula.
3. Acetic acid [C₂(H₂O)₂] fits the formula and is therefore classified as a carbohydrate.
Which of the above are correct?

Answer Review

1 Consider the following statements regarding the carbohydrate formula:
1. The general formula is Cx(H₂O)y, and glucose accurately fits it as C₆(H₂O)₆.
2. Rhamnose (C₆H₁₂O₅) is a carbohydrate but does not fit the hydrate formula.
3. Acetic acid [C₂(H₂O)₂] fits the formula and is therefore classified as a carbohydrate.
Which of the above are correct?

A 1 and 2

B 2 and 3

C 1 and 3

D 1, 2, and 3

💡 Short Explanation

�� Glucose fits the general hydrate formula. �� Rhamnose is an exception among carbohydrates. �� Acetic acid is not a carbohydrate despite fitting the formula.

✅ Correct Answer Explanation

Statement 1 is correct because glucose can be represented as C₆(H₂O)₆. Statement 2 is correct because rhamnose is a carbohydrate but does not satisfy the general hydrate formula. Statement 3 is incorrect because acetic acid is a carboxylic acid, not a carbohydrate, even though its formula can be written as C₂(H₂O)₂. Therefore, Statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Statement 3 is incorrect.
�� Option C → Statement 2 is also correct.
�� Option D → Includes incorrect Statement 3.

🎯 Strategy

Used

Exception-Based Elimination

Application:

Identify compounds that fit the formula but are not carbohydrates.

Final Logic:

Statements 1 and 2 are true; Statement 3 is false.

🧠 Memory Trick

Formula Fits ≠ Carbohydrate

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Carbohydrates as Hydrates of Carbon

2 The chemical formation of n-hexane upon prolonged heating of glucose with HI signifies that:

A Glucose contains an active aldehyde group.

B Glucose lacks any branched carbon skeleton.

C The hydroxyl groups are present on five adjacent carbons.

D Glucose fundamentally has a cyclic pyranose structure.

💡 Short Explanation

�� HI reduces glucose to n-hexane. �� n-Hexane possesses a straight carbon chain. �� This proves glucose has an unbranched skeleton.

✅ Correct Answer Explanation

On prolonged heating with HI, glucose is reduced to n-hexane. Since n-hexane contains six carbon atoms arranged in a straight chain without branching, this proves that the carbon skeleton of glucose is also unbranched. Thus, glucose possesses a straight-chain carbon framework.

❓ Why Other Options Are Incorrect

�� Option A → Aldehyde presence is confirmed by oxidation reactions.
�� Option C → HI reduction does not directly prove hydroxyl positions.
�� Option D → HI reduction supports the carbon chain structure, not pyranose formation.

🎯 Strategy

Used

Reaction Interpretation

Application:

Identify the structural information obtained from HI reduction.

Final Logic:

n-Hexane formation confirms an unbranched six-carbon chain.

🧠 Memory Trick

HI → Hexane → Straight Chain

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

3 Which of the following reactions specifically confirms the presence of a carbonyl group (>C=O) in glucose but does not inherently distinguish whether it is an aldehyde or ketone?

A Oxidation with nitric acid

B B. Acetylation with acetic anhydride

C Oxidation with mild bromine water

D Reaction with hydroxylamine to form an oxime

💡 Short Explanation

�� Aldehydes and ketones both form oximes. �� Oxime formation confirms a carbonyl group. �� It does not distinguish aldehyde from ketone.

✅ Correct Answer Explanation

Hydroxylamine reacts with carbonyl compounds to form oximes. Since both aldehydes and ketones undergo this reaction, oxime formation confirms the presence of a carbonyl group but does not identify whether the carbonyl is aldehydic or ketonic. Therefore, oxime formation is evidence for a carbonyl group.

❓ Why Other Options Are Incorrect

�� Option A → Gives oxidation information.
�� Option C → Confirms hydroxyl groups.

Option D → Specifically confirms an aldehyde group.

🎯 Strategy

Used

Functional Group Analysis

Application:

Identify the reaction common to aldehydes and ketones.

Final Logic:

Oxime formation proves carbonyl presence.

🧠 Memory Trick

Oxime = Carbonyl Proof

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Carbonyl Group in Glucose

4 What is the specific stable compound formed when glucose is treated with acetic anhydride, validating the presence of five separate –OH groups?

A Glucose pentanitrate

B Glucose pentaacetate

C Gluconic acid

D Saccharic acid

💡 Short Explanation

�� Acetic anhydride acetylates hydroxyl groups. �� Five hydroxyl groups are acetylated. �� The product formed is glucose pentaacetate.

✅ Correct Answer Explanation

Glucose reacts with acetic anhydride to produce glucose pentaacetate. The formation of a pentaacetate derivative confirms the presence of five hydroxyl groups in the glucose molecule. Thus, glucose pentaacetate provides direct evidence for five –OH groups.

❓ Why Other Options Are Incorrect

�� Option A → Not formed during acetylation.
�� Option C → Produced by oxidation with bromine water.
�� Option D → Produced by oxidation with nitric acid.

🎯 Strategy

Used

Reaction Identification

Application:

Identify the product formed during acetylation.

Final Logic:

Pentaacetate confirms five hydroxyl groups.

🧠 Memory Trick

Pentaacetate = 5 OH Groups

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

📄 Passage

Glucose gets oxidised to a six-carbon carboxylic acid (gluconic acid) on reaction with a mild oxidising agent like bromine water. This indicates that the carbonyl group is present as an aldehydic group. Furthermore, on oxidation with nitric acid, both glucose and gluconic acid yield a dicarboxylic acid, saccharic acid. This highlights a specific functional group located elsewhere in the carbon chain.

Question: The conversion of glucose to gluconic acid by bromine water primarily confirms:

A The presence of a primary alcohol.

B The straight-chain nature of glucose.

C The presence of a terminal aldehydic group.

D The existence of a cyclic hemiacetal linkage.

💡 Short Explanation

�� Bromine water is a mild oxidising agent. �� It oxidises aldehydes to carboxylic acids. �� Gluconic acid formation proves an aldehyde group.

✅ Correct Answer Explanation

Bromine water oxidises the aldehydic group of glucose to form gluconic acid. Since ketones do not undergo this reaction under similar conditions, the reaction confirms that glucose contains a terminal aldehyde group. Therefore, glucose is an aldohexose.

❓ Why Other Options Are Incorrect

�� Option A → Not demonstrated by bromine water oxidation.
�� Option B → Established through HI reduction.
�� Option D → Not proven by oxidation.

🎯 Strategy

Used

Passage-Based Analysis

Application:

Determine which functional group is oxidised.

Final Logic:

Gluconic acid formation proves an aldehydic group.

🧠 Memory Trick

Bromine Water → Aldehyde Proof

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

📄 Passage

Question: Why does the formation of saccharic acid from gluconic acid using nitric acid uniquely indicate the presence of a primary alcoholic group in glucose?

A Nitric acid selectively oxidises only secondary alcohols.

B Only a primary alcohol at the terminal end can be oxidised to a carboxyl group, converting the monocarboxylic acid to a dicarboxylic acid.

C Saccharic acid is generated via the cleavage of the central carbon chain.

D Gluconic acid completely lacks an aldehyde group for oxidation.

💡 Short Explanation

�� Gluconic acid already contains one carboxyl group. �� Nitric acid oxidises the terminal alcohol. �� A second carboxyl group is formed.

✅ Correct Answer Explanation

Gluconic acid already possesses one carboxyl group at C-1. When nitric acid oxidises the primary alcohol present at C-6, another carboxyl group is formed, producing saccharic acid. This conversion proves the presence of a terminal primary alcohol group in glucose.

❓ Why Other Options Are Incorrect

�� Option A → Nitric acid oxidises primary alcohols as well.
�� Option C → No carbon-chain cleavage occurs.
�� Option D → The important evidence comes from oxidation of the alcohol group.

🎯 Strategy

Used

Passage-Based Analysis

Application:

Track the structural change from monoacid to diacid.

Final Logic:

Second carboxyl group arises from oxidation of a primary alcohol.

🧠 Memory Trick

Monoacid → Diacid = Primary Alcohol Present

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

7 In the notation D-(+)-glucose, the (+) symbol specifically quantifies which of the following parameters?

A The molecule contains purely alpha anomers.

B The position of the –OH group on the lowest asymmetric carbon is on the right.

C The relative temperature of crystallisation is positive.

D The compound rotates plane-polarised light to the right (dextrorotatory).

💡 Short Explanation

�� (+) indicates optical rotation. �� Positive rotation means rightward rotation. �� It does not indicate configuration.

✅ Correct Answer Explanation

The (+) sign denotes that glucose rotates plane-polarised light to the right and is therefore dextrorotatory. This notation is related to optical activity and is independent of D/L configuration. Thus, (+) signifies positive optical rotation.

❓ Why Other Options Are Incorrect

��Option A → Does not describe anomeric composition.

Option B → Refers to D-configuration, not optical rotation.

�� Option C → Unrelated.

🎯 Strategy

Used

Concept Clarification

Application:

Differentiate configuration from optical activity.

Final Logic:

(+) means dextrorotatory.

🧠 Memory Trick

Plus = Right Rotation

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

8 Match the stereochemical terms with their correct implications for glucose.

List I	List II
1. — D-configuration	a. — Dextrorotatory nature
2. — L-configuration	b. — –OH on the lowest asymmetric carbon is on the left
3. — (+) rotation	c. — Laevorotatory nature
4. — (–) rotation	d. — Correlates to (+)-glyceraldehyde

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

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

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

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

💡 Short Explanation

�� D/L refers to configuration. �� (+)/(–) refers to optical rotation. �� The two concepts are independent.

✅ Correct Answer Explanation

Matching: • D-configuration → Correlates with (+)-glyceraldehyde • L-configuration → OH group on lowest asymmetric carbon is on the left • (+) rotation → Dextrorotatory • (–) rotation → Laevorotatory Hence: 1-d, 2-b, 3-a, 4-c

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Option Grouping

Application:

Match configuration terms with their meanings.

Final Logic:

Only Option A provides all correct matches.

🧠 Memory Trick

D/L = Configuration; +/– = Rotation

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

9 Arrange the following physical parameters of glucose anomers in decreasing numerical order (highest K to lowest K):
A. Melting point of β-form
B. Melting point of α-form
C. Crystallisation temperature of β-form from hot saturated solution
D. Crystallisation temperature of α-form from concentrated solution

A A > B > C > D

B A > C > B > D

C A > B > D > C

D C > A > B > D

💡 Short Explanation

β-form melting point = 423 K.
α-form melting point = 419 K.
Crystallisation temperatures are 371 K and 303 K.

✅ Correct Answer Explanation

The values are:

β-form melting point = 423 K
α-form melting point = 419 K
β-form crystallisation temperature = 371 K
α-form crystallisation temperature = 303 K

Therefore:

423 K > 419 K > 371 K > 303 K

Hence the order is A > B > C > D, which corresponds to Option 1.

❓ Why Other Options Are Incorrect

Option 2 incorrectly arranges the numerical values.
Option 3 incorrectly arranges the numerical values.
Option 4 incorrectly arranges the numerical values.

🎯 Strategy

Ordering

Application:

Arrange the given temperatures from highest to lowest.

Final Logic:

423 > 419 > 371 > 303.

🧠 Memory Trick

423 > 419 > 371 > 303

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Alpha and Beta Forms of Glucose.

10 The pyranose structure of glucose involves a cyclic hemiacetal. Which carbon's hydroxyl group is specifically involved in the ring formation with the aldehyde group?

A C-3

B C-4

C C-5

D C-6

💡 Short Explanation

�� The aldehyde group is present at C-1. �� The hydroxyl group at C-5 reacts intramolecularly. �� A six-membered pyranose ring is formed.

✅ Correct Answer Explanation

In glucose, the hydroxyl group attached to C-5 reacts with the aldehyde group at C-1 to form a cyclic hemiacetal. This intramolecular reaction produces the six-membered pyranose ring structure. Therefore, the hydroxyl group involved belongs to C-5.

❓ Why Other Options Are Incorrect

�� Option A → Does not form the stable pyranose ring.
�� Option B → Not involved in the major ring formation.
�� Option D → Does not participate in hemiacetal formation.

🎯 Strategy

Used

Structural Analysis

Application:

Identify the hydroxyl group participating in cyclisation.

Final Logic:

C-5 OH forms the pyranose ring.

🧠 Memory Trick

C1 + C5 = Pyranose Ring

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

11 Regarding the α-form of glucose, which of the following statements is true?
1. It differs from the β-form strictly in the configuration at C-1.
2. It is obtained by crystallisation from a concentrated glucose solution at 371 K.
3. It possesses a higher melting point than the β-form.

A Only 1

B 1 and 2

C 1 and 3

D Only 3

💡 Short Explanation

�� α- and β-glucose differ only at C-1. �� α-Glucose crystallises at 303 K, not 371 K. �� β-Glucose has the higher melting point.

✅ Correct Answer Explanation

Statement 1 is correct because α- and β-glucose differ only in the configuration of the hydroxyl group attached to the anomeric carbon (C-1). Statement 2 is incorrect because α-glucose is obtained by crystallisation from a concentrated solution at 303 K, whereas 371 K is associated with β-glucose. Statement 3 is incorrect because β-glucose has a melting point of 423 K, which is higher than the melting point of α-glucose (419 K). Therefore, only Statement 1 is correct.

❓ Why Other Options Are Incorrect

�� Option B → Statement 2 is incorrect.
�� Option C → Statement 3 is incorrect.
�� Option D → Statement 3 is incorrect.

🎯 Strategy

Used

Statement Verification

Application:

Evaluate each statement using known properties of α- and β-glucose.

Final Logic:

Only Statement 1 is true.

🧠 Memory Trick

Alpha = 303 K, Beta = Higher MP

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Alpha and Beta Forms of Glucose

12 The uniquely identifiable C1 carbon in the cyclic hemiacetal structure of glucose, which determines whether the molecule is the α-form or β-form, is named the:

A Epimeric carbon

B Asymmetric central carbon

C Anomeric carbon

D Chiral polar centre

💡 Short Explanation

�� C-1 becomes a new chiral centre after cyclisation. �� α and β forms differ at this carbon. �� It is called the anomeric carbon.

✅ Correct Answer Explanation

During cyclisation of glucose, the aldehyde carbon (C-1) becomes a new stereogenic centre. The orientation of the hydroxyl group attached to this carbon determines whether the molecule exists as α-glucose or β-glucose. This carbon is known as the anomeric carbon.

❓ Why Other Options Are Incorrect

�� Option A → Epimers differ at any one chiral carbon, not specifically C-1.
�� Option B → Not a recognized term for C-1 in glucose.
�� Option D → Not used in carbohydrate nomenclature.

🎯 Strategy

Used

Keyword Identification

Application:

Identify the carbon responsible for anomerism.

Final Logic:

C-1 in cyclic glucose is the anomeric carbon.

🧠 Memory Trick

Anomer = Anomeric Carbon

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Anomeric Carbon

13 Unlike the six-membered pyranose ring of glucose, fructose forms a five-membered furanose ring. This relies on the addition of the –OH group at which carbon to the ketonic group?

A C-4

B C-5

C C-6

D C-3

💡 Short Explanation

�� Fructose contains a ketonic group at C-2. �� The hydroxyl group at C-5 participates in ring formation. �� This produces a five-membered furanose ring.

✅ Correct Answer Explanation

In fructose, the hydroxyl group attached to C-5 reacts with the ketonic carbon at C-2 to form a cyclic hemiketal. This intramolecular reaction generates a five-membered furanose ring structure. Therefore, the hydroxyl group involved belongs to C-5.

❓ Why Other Options Are Incorrect

�� Option A → Does not produce the common furanose ring.
�� Option C → Not involved in standard furanose formation.
�� Option D → Does not participate in the major cyclic structure.

🎯 Strategy

Used

Structural Analysis

Application:

Identify the hydroxyl group participating in cyclisation.

Final Logic:

C-5 hydroxyl reacts with the C-2 ketonic group.

🧠 Memory Trick

Fructose: C2 Keto + C5 OH

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

14 Despite being a ketose, fructose exhibits certain optical reaction phenomena. Which structural feature explicitly dictates why it is appropriately written as D-(–)-fructose?

A It belongs to the D-series and rotates light clockwise.

B It correlates to (+)-glyceraldehyde's lowest asymmetric carbon but rotates light to the left.

C It contains an aldehyde group hidden in its open chain.

D It naturally forms a massive pyranose ring structure.

💡 Short Explanation

�� D/L refers to configuration. �� (+)/(–) refers to optical rotation. �� D-fructose rotates light to the left.

✅ Correct Answer Explanation

The D-notation of fructose is assigned by comparing the configuration of its lowest asymmetric carbon with (+)-glyceraldehyde. However, its optical rotation is negative, meaning it rotates plane-polarised light to the left. Therefore, fructose is correctly represented as D-(–)-fructose.

❓ Why Other Options Are Incorrect

�� Option A → D-fructose is laevorotatory, not dextrorotatory.
�� Option C → Fructose is a ketose.
�� Option D → Fructose predominantly forms a furanose ring.

🎯 Strategy

Used

Concept Clarification

Application:

Separate configuration from optical activity.

Final Logic:

D and (–) describe different properties.

🧠 Memory Trick

D ≠ Dextrorotatory

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

15 When a robust glycosidic linkage structurally forms between two monosaccharides, which molecule is eliminated in the process?

A Hydrogen gas

B Carbon dioxide

C A water molecule

D An oxygen atom

💡 Short Explanation

�� Glycosidic bonds form through condensation. �� Condensation removes water. �� The resulting linkage joins monosaccharides.

✅ Correct Answer Explanation

Formation of a glycosidic bond occurs through a condensation reaction between two monosaccharides. During this process, one molecule of water is eliminated. Thus, water is the molecule removed during glycosidic bond formation.

❓ Why Other Options Are Incorrect

�� Option A → Hydrogen gas is not released.
�� Option B → Carbon dioxide is not produced.
�� Option D → Oxygen is not removed independently.

🎯 Strategy

Used

Reaction Analysis

Application:

Recall the mechanism of glycosidic bond formation.

Final Logic:

Condensation releases water.

🧠 Memory Trick

Glycosidic Bond = Loss of H₂O

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

16 Match the disaccharide with its correct chemical hydrolysis products.

List I	List II
1. — Sucrose	a. — Two molecules of α-D-glucose
2. — Maltose	b. — Equimolar mixture of D-(+)-glucose and D-(–)-fructose
3. — Lactose	c. — β-D-galactose and β-D-glucose

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

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

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

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

💡 Short Explanation

�� Sucrose → Glucose + Fructose �� Maltose → Glucose + Glucose �� Lactose → Galactose + Glucose

✅ Correct Answer Explanation

Matching: • Sucrose → Equimolar mixture of D-(+)-glucose and D-(–)-fructose • Maltose → Two molecules of α-D-glucose • Lactose → β-D-galactose and β-D-glucose Thus: 1-b, 2-a, 3-c

❓ Why Other Options Are Incorrect

�� Options B, C and D contain incorrect hydrolysis products.

🎯 Strategy

Used

Option Grouping

Application:

Match each disaccharide with its hydrolysis products.

Final Logic:

Only Option A provides all correct matches.

🧠 Memory Trick

Sucrose GF, Maltose GG, Lactose Gal-Glu

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

17 Why is sucrose termed a non-reducing sugar while maltose is a reducing sugar?

A Sucrose consists of two identical glucose units that neutralise each other.

B .The aldehydic and ketonic reducing groups of glucose and fructose are both bound in the glycosidic bond in sucrose.

C .Sucrose yields a laevorotatory mixture upon hydrolysis.

D .Sucrose actively forms a non-reactive furanose ring.

💡 Short Explanation

Sucrose has no free reducing group.
Both anomeric carbons participate in bonding.
Therefore sucrose is non-reducing.

✅ Correct Answer Explanation

In sucrose, the anomeric carbon of glucose and the anomeric carbon of fructose are both involved in glycosidic bond formation. Since no free aldehydic or ketonic group can be generated, sucrose cannot act as a reducing sugar.

Hence, sucrose is non-reducing whereas maltose possesses a free reducing end.

❓ Why Other Options Are Incorrect

Option 1 → Sucrose contains glucose and fructose, not two glucose units.
Option 3 → Hydrolysis product does not explain non-reducing behavior.
Option 4 → Not the reason for non-reducing nature.

🎯 Strategy

Functional Group Analysis

Application:

Determine whether a free reducing group exists.

Final Logic:

No free reducing group = Non-reducing sugar.

🧠 Memory Trick

Sucrose = Locked Ends

📚 Class XII – Chemistry Part II – Chapter: Biomolecules – Topic: Non-Reducing Nature of Sucrose.

18 Maltose possesses reducing properties primarily because:

A Both anomeric carbons are permanently locked in the glycosidic linkage.

B It can be rapidly hydrolysed by mild acidic catalysts.

C A free reactive aldehyde group can be transiently produced at C1 of the second glucose unit in aqueous solution.

D It contains a uniquely exposed ketonic functional group.

💡 Short Explanation

�� Maltose contains a free anomeric carbon. �� This carbon can generate an aldehyde group. �� Hence maltose behaves as a reducing sugar.

✅ Correct Answer Explanation

The glycosidic bond in maltose involves only one anomeric carbon. The second glucose unit retains a free anomeric carbon that can open into the aldehydic form in solution. Therefore, maltose can reduce Fehling's solution and Tollens' reagent.

❓ Why Other Options Are Incorrect

�� Option A → Would make maltose non-reducing.
�� Option B → Hydrolysis is not the reason for reducing behavior.
�� Option D → Maltose contains no ketonic group.

🎯 Strategy

Used

Functional Group Analysis

Application:

Identify the source of reducing behavior.

Final Logic:

Free anomeric carbon = Reducing sugar.

🧠 Memory Trick

Maltose = One Free End

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

19 The exact sequence of the glycosidic linkage in lactose occurs between:

A C1 of α-D-glucose and C2 of β-D-fructose

B C1 of β-D-galactose and C4 of β-D-glucose

C C1 of α-D-glucose and C4 of α-D-glucose

D C1 of β-D-glucose and C4 of β-D-galactose

💡 Short Explanation

�� Lactose contains galactose and glucose. �� The linkage is β(1→4). �� Galactose C1 joins glucose C4.

✅ Correct Answer Explanation

Lactose consists of β-D-galactose and β-D-glucose joined through a β(1→4) glycosidic linkage. Specifically, C1 of galactose is linked to C4 of glucose. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents sucrose linkage.
�� Option C → Represents maltose linkage.
�� Option D → Reverses the actual linkage sequence.

🎯 Strategy

Used

Direct Recall

Application:

Recall the linkage pattern of lactose.

Final Logic:

Lactose = β(1→4) Galactose–Glucose.

🧠 Memory Trick

Lactose = Gal(1→4)Glu

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

20 Which specific common name is assigned to lactose due to the environment of its primary natural occurrence?

A Cane sugar

B Invert sugar

C Milk sugar

D Malt sugar

💡 Short Explanation

�� Lactose occurs naturally in milk. �� It is the principal sugar present in milk. �� Therefore it is called milk sugar.

✅ Correct Answer Explanation

Lactose is the naturally occurring disaccharide found in milk and dairy products. Because of this characteristic occurrence, lactose is commonly known as milk sugar.

❓ Why Other Options Are Incorrect

�� Option A → Cane sugar refers to sucrose.
�� Option B → Invert sugar is obtained from sucrose hydrolysis.
�� Option D → Malt sugar refers to maltose.

🎯 Strategy

Used

Direct Recall

Application:

Recall the common names of disaccharides.

Final Logic:

Lactose = Milk sugar.

🧠 Memory Trick

Lactose → Lactation → Milk

📚 Bottom of Form

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

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

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