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Question #44952
The Question
Consider the behavior of D-glucopyranose () and the nucleoside uridine () in an aqueous buffer at and . While exhibits a time-dependent change in specific rotation (mutarotation), maintains a constant specific rotation. Which of the following statements correctly explains the fundamental stereoelectronic reason for the inability of to undergo mutarotation?
Options
- A.
The N-glycosidic linkage in is an N,O-acetal where the uracil moiety exists predominantly in the diketo (lactam) form. The N1 lone pair is delocalized via amide resonance, drastically reducing its basicity and leaving group ability, thereby preventing ring-opening to an acyclic intermediate at pH 7.0.
- B.
The uracil ring in exists predominantly in the enol (lactim) form, which provides a strong effect that thermodynamically destabilizes the oxocarbenium ion required for mutarotation.
- C.
Mutarotation in is kinetically feasible via an open-chain intermediate, but the equilibrium is completely shifted to the -anomer due to severe 1,3-diaxial interactions present in the -anomer.
- D.
The N-glycosidic bond in undergoes rapid reversible cleavage to an oxocarbenium ion, but the keto-enol tautomerization of the uracil base directs nucleophilic attack exclusively to the -face via neighboring group participation.
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Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Exception_Exposure
Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.
Concepts Behind This Question
D-Glucose Anomeric Equilibrium|Nitrogenous Base Keto-Enol Tautomerism|Nucleoside vs Nucleotide
Why This Question Is Tricky
Common Mistakes
This question tests the student's ability to differentiate between the reactivity of hemiacetals (which undergo mutarotation) and N,O-acetals (which are locked).
The 'Hidden Variable' is the keto-enol (lactam-lactim) tautomerism of the nitrogenous base.
Students often memorize that nucleosides don't mutarotate but fail to understand the stereoelectronic reason: the predominant lactam form involves the N1 lone pair in amide resonance, rendering it a terrible leaving group and preventing the ring-opening mechanism entirely at neutral pH.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Biomolecules - Class 12 |
| Concept | D-Glucose Anomeric Equilibrium |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |