Using the Fischer projection of glucose shown above, show what happens to glucose when it is...
Using the Fischer projection of glucose shown above, show what happens to glucose when it is placed in a basic solution (specific names of products are not necessary but Fischer projections of those structures are, mechanistic arrows may help your explanation, draw Fischer projections of any intermediates that help your explanation)
Homework Answers
Key Reaction: Enediol Rearrangement (Lobry de Bruyn–van Ekenstein)
In basic solution, glucose undergoes tautomerization via an enediol intermediate, leading to isomerization and epimerization. Below is the stepwise process using Fischer projections:
1. Starting Material: D-Glucose (Fischer Projection)
CHO │ H │ OH │ H │ OH │ H │ CH₂OH
2. Deprotonation at C1 (Base Attack)
OH⁻ abstracts the acidic α-hydrogen (C1-H), forming a enediolate intermediate:
C(OH)= │ H │ OH │ H │ OH │ H │ CH₂OH
3. Enediol Intermediate (Resonance Stabilization)
The enediolate tautomerizes to a neutral enediol:
CH(OH)- │ C(OH)= │ H │ OH │ H │ CH₂OH
4. Reprotonation (Forms Isomers)
Reprotonation at C2 yields two major products:
D-Fructose (ketose formed if C2 is protonated):
CH₂OH │ C=O │ H │ OH │ H │ CH₂OH
D-Mannose (epimer of glucose at C2, if C1 is reprotonated):
CHO │ H │ OH │ OH │ H │ CH₂OH
Summary of Products:
D-Fructose (ketose form, C2=O).
D-Mannose (C2 epimer of glucose).
Key Notes:
Mechanistic Drivers: Base-catalyzed enolization leads to reversible C1-C2 bond reshuffling.
No Oxidation: Unlike acidic conditions, base mediates isomerization, not degradation.
Visual Clue: The enediol intermediate is planar (sp² hybridized at C1 and C2).
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PLEASE ANSWER ALL A-G, will
give thumbs up!! FOR A, just need explanation why reaction occurs
at C2 position rather than C3, already have mechanism drawn...
PLEASE EXPLAIN OR AT LEAST
JUST ANSWER ALL ASAP, BEFORE TOMORROW!!
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