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Biomolecules appeared 35 times across 3 years — 4.1% of Chemistry. This question is from Hydrolysis of Sucrose.

Year 2026 2025 2024 Total
Questions 9 18 8 35

Given below are two statements: Statement I: D-(+)-glucose + D-(+)-fructose -H₂O sucrose sucrose Hydrolysis D-(+)-glucose + D-(+)-fructose Statement II: Invert sugar is formed during sucrose hydrolysis. In the light of given statements, choose the correct answer from the options given below:

Solution & Explanation

Core Logic

Statement I: Sucrose is formed by condensation of D-(+)-glucose and D-(-)-fructose (levorotatory fructose, not dextrorotatory as claimed). Hydrolysis of sucrose yields D-(+)-glucose and D-(-)-fructose. Thus, Statement I is false.

Statement II: Hydrolysis of dextrorotatory sucrose (+66.5°) yields a mixture of dextrorotatory glucose (+52.5°) and highly levorotatory fructose (-92.4°). Because the overall specific rotation of the mixture becomes levorotatory (-39.9°), the hydrolyzed mixture is called invert sugar. Thus, Statement II is true.

Pattern Recognition

Natural fructose is always levorotatory, D-(-)-fructose. Dextrorotatory fructose mentioned in Statement I is an immediate giveaway that the statement is false.

Chapter Mix

Class 12 Chemistry: Biomolecules

Reference Study Guides

More Biomolecules Previous-Year Questions — Page 7

Q76 jee_main_2024_29_jan_morning Proteins and Amino Acids
Type of amino acids obtained by hydrolysis of proteins is :
  • A. β
  • B. α
  • C. δ
  • D. γ

Solution

Core Logic

Proteins are natural biopolymers. The fundamental building blocks (monomers) of all naturally occurring proteins are α-amino acids. In these molecules, both the amino group (-NH₂) and the carboxyl group (-COOH) are attached to the exact same carbon atom, designated as the α-carbon.

Because they are polymers of α-amino acids linked by peptide bonds, the hydrolysis of proteins (acidic, basic, or enzymatic) will cleave these peptide bonds to yield the constituent α-amino acids.

Chapter Mix

Class 12 Chemistry: Biomolecules

Q84 jee_main_2024_30_january_evening Nucleic Acids (DNA and RNA)
The total number of correct statements, regarding the nucleic acids is A. RNA is regarded as the reserve of genetic information. B. DNA molecule self-duplicates during cell division C. DNA synthesizes proteins in the cell. D. The message for the synthesis of particular proteins is present in DNA E. Identical DNA strands are transferred to daughter cells.
Numerical Answer. Answer: 3 to 3

Solution

Core Logic

A. RNA is regarded as the reserve of genetic information. (False - DNA is the reserve of genetic information). B. DNA molecule self-duplicates during cell division. (True - through replication). C. DNA synthesizes proteins in the cell. (False - RNA molecules synthesize proteins via translation, DNA only provides the code). D. The message for the synthesis of particular proteins is present in DNA. (True - the genetic code lies in the base sequence of DNA). E. Identical DNA strands are transferred to daughter cells. (True - through accurate replication and cell division, genetic continuity is maintained).

Step 1: Final Conclusion

The true statements are B, D, and E. Total number of correct statements is 3.

Chapter Mix

Class 12 Chemistry: Biomolecules

Q63 jee_main_2024_30_jan_morning Carbohydrates
  • A. Sucrose
  • B. Lactose
  • C. Glucose
  • D. Maltose

Solution

Core Logic

Fehling's reagent is reduced by reducing sugars to give a reddish-brown precipitate of Cu₂O. Reducing sugars must have a free aldehyde/ketone group or a hemiacetal linkage that can open to form an aldehyde. Sucrose is a non-reducing sugar because its anomeric carbons (C1 of glucose and C2 of fructose) are tied up in a glycosidic linkage, leaving no free hemiacetal group.

Step 1: Analyzing the options

Lactose, glucose, and maltose are all reducing sugars and will give a positive Fehling's test. Sucrose does not.

Pattern Recognition

Sucrose = non-reducing sugar. Maltose, Lactose = reducing sugars. Monosaccharides (glucose, fructose) = always reducing.

Chapter Mix

Class 12 Chemistry: Biomolecules

Q89 jee_main_2024_31_jan_evening Vitamins and their Classification
From the vitamins A, B₁, B₆, B₁₂, C, D, E and K, the number vitamins that can be stored in our body is ________
Numerical Answer. Answer: 5 to 5

Solution

Core Logic

Vitamins are broadly classified into two groups based on solubility:

  • Fat-soluble vitamins: Vitamins A, D, E, and K. These are stored in the liver and adipose (fat-storing) tissues.
  • Water-soluble vitamins: B group vitamins and Vitamin C. These are readily excreted in urine and cannot be stored in the body (with the exception of Vitamin B₁₂, which can be stored in the liver).
Step 1: Final List

The vitamins that can be stored in the body from the given list are A, D, E, K, and B₁₂. Total number = 5.

Chapter Mix

Class 12 Chemistry: Biomolecules

Q78 jee_main_2024_31_jan_morning Reactions of Glucose
Match List I with List II
LIST-ILIST-II
A. Glucose/NaHCO₃/ΔI. Gluconic acid
B. Glucose/HNO₃II. No reaction
C. Glucose/HI/ΔIII. n-hexane
D. Glucose/Bromine waterIV. Saccharic acid
Choose the correct answer from the options given below:
  • A. A-IV, B-I, C-III, D-II
  • B. A-II, B-IV, C-III, D-I
  • C. A-III, B-II, C-I, D-IV
  • D. A-I, B-IV, C-III, D-II

Solution

Core Logic

Matching the reactions of glucose: (A) Glucose does not react with NaHCO₃, so there is no reaction. (A arrow II) (B) Oxidation of glucose with strong oxidizing agents like HNO₃ yields a dicarboxylic acid called saccharic acid. (B arrow IV) (C) Prolonged heating of glucose with HI forms n-hexane, indicating a straight chain of six carbon atoms. (C arrow III) (D) Oxidation with mild agents like bromine water converts glucose to gluconic acid. (D arrow I)

Chapter Mix

Class 12 Chemistry: Biomolecules

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