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Biomolecules appeared 36 times across 3 years — 4.2% of Chemistry. This question is from Reducing and Non-Reducing Sugars.

Year 2026 2025 2024 Total
Questions 9 19 8 36

From the given following (A to D) cyclic structures, those which will not react with Tollen's reagent are :
Reducing and Non-Reducing Sugars
Reducing and Non-Reducing Sugars
Reducing and Non-Reducing Sugars
Reducing and Non-Reducing Sugars
Reducing and Non-Reducing Sugars
Reducing and Non-Reducing Sugars
Reducing and Non-Reducing Sugars
Reducing and Non-Reducing Sugars

Solution & Explanation

Core Logic

Tollens' reagent oxidizes aldehydes. In cyclic sugars, this requires a free hemiacetal group (an anomeric carbon bonded to an -OH group) to allow mutarotation and open-chain aldehyde formation.

Step 1: Structural Analysis

Structures (A) and (D) possess a free -OH group at the anomeric carbon (hemiacetal linkage). They are reducing sugars and will react with Tollens' reagent. Structures (B) and (C) have an -OCH₃ group at the anomeric carbon instead of a free -OH. These are full acetals (glycosides) and cannot ring-open. Thus, they are non-reducing sugars.

Pattern Recognition

Hemiacetal (-OH on anomeric C) = Reducing Sugar. Acetal (-OR on anomeric C) = Non-Reducing Sugar.

Chapter Mix

Class 12 Chemistry: Biomolecules

Reference Study Guides

More Biomolecules Previous-Year Questions — Page 8

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

More Biomolecules Questions — jee_main_2026_23_january_morning

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)