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Question #44963
The Question
An optically active \ce{D}-ketohexose gives a positive Tollens' test. In the alkaline medium of the reagent, undergoes the Lobry de Bruyn–van Ekenstein transformation via an intermediate to form an equilibrium mixture containing and two aldohexoses and . Oxidation of and with warm \ce{HNO3} yields aldaric acids and , respectively. It is observed that is optically inactive (meso) while is optically active. Which of the following correctly identifies , the intermediate , and the stereochemical relationship between and ?
Options
- A.
= \ce{D}-Tagatose; = 1,2-enediol; and are C-2 epimers
- B.
= \ce{D}-Fructose; = 1,2-enediol; and are C-2 epimers
- C.
= \ce{D}-Sorbose; = 1,2-enediol; and are C-2 epimers
- D.
= \ce{D}-Tagatose; = 2,3-enediol; and are C-3 epimers
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Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Mechanism_Trap
Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.
Concepts Behind This Question
Reducing Sugar Criteria|D-Glucose Anomeric Equilibrium|Epimerization vs Anomerization
Why This Question Is Tricky
Common Mistakes
This question tests the hidden mechanism of the Lobry de Bruyn–van Ekenstein transformation and stereochemical analysis of aldaric acids.
Students often blindly associate a positive Tollens' test for ketoses exclusively with D-Fructose.
The trap is failing to evaluate the stereochemistry of the resulting aldaric acids.
D-Fructose yields two optically active aldaric acids, whereas D-Tagatose yields one meso (galactaric acid) and one optically active (talaric acid) product, correctly satisfying the conditions.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Biomolecules - Class 12 |
| Concept | Reducing Sugar Criteria |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |