Treatment of D-glucose with aqueous NaOH results in a mixture of monosaccharides, which are
(A) D-ma

Q: Why doesn't the base attack the aldehyde carbon directly?

While hydroxide can attack the carbonyl carbon to form a hydrate, this is a reversible and non-productive step. The abstraction of the acidic $\alpha$-hydrogen leads to the stable enediol intermediate, which drives the isomerization process.

Q: Why do we get D-mannose and not some other epimer?

The enediol intermediate only involves carbons 1 and 2. The stereocenters at carbons 3, 4, and 5 remain completely untouched. Therefore, when tautomerization happens back to an aldose, only the stereochemistry at C-2 can change, yielding the C-2 epimer, D-mannose.

Q: Is this reaction reversible?

Yes, absolutely! The Lobry de Bruyn-van Ekenstein transformation is an equilibrium process. If you start with pure D-fructose or pure D-mannose in a basic medium, you will eventually get the exact same mixture of all three sugars.

Q: Why doesn't the reaction continue down the chain to C-3?

Prolonged exposure to strong base can indeed cause further isomerization and even degradation of the sugar chain. However, the initial and most rapid equilibrium established is between glucose, mannose, and fructose via the 1,2-enediol.

Q: How do I quickly identify the products in an exam?

Just remember the triad: an aldose in base will equilibrate with its C-2 epimer and its corresponding ketose. For D-glucose, the C-2 epimer is D-mannose, and the ketose is D-fructose.

Question

Treatment of D-glucose with aqueous NaOH results in a mixture of monosaccharides, which are

Sigma Percentile · Accepted Answer

### The Magic of Basic Medium

Let's start by looking at our reactant, D-glucose. Notice the aldehyde group at the top, carbon-1, and the specific arrangement of hydroxyl groups on carbons 2, 3, 4, and 5. This exact stereochemistry is what makes it D-glucose.

Now, what happens when we treat this with aqueous sodium hydroxide? The base acts as a catalyst. This triggers a very famous reaction in carbohydrate chemistry called the **Lobry de Bruyn-van Ekenstein transformation**. It is a mouthful, but the concept is beautiful and highly logical.

### The Enediol Intermediate: A Chemical Crossroads

Look closely at the mechanism. The hydroxide ion from the base targets the slightly acidic $\alpha$-hydrogen on carbon-2. It pulls it off, creating a double bond between carbon-1 and carbon-2, and pushing the electrons up to the oxygen.

After protonation, we get a crucial intermediate: the **enediol**. Notice the double bond and the two hydroxyl groups. This intermediate is the crossroads of our reaction.

### The Tautomerization Split

There is a catch here. This enediol is not stable. It desperately wants to tautomerize back to a more stable carbonyl form. But when the double bond breaks and carbon-2 regains a hydrogen, the hydrogen can attack from either face of the planar double bond!

If it attacks from the opposite side compared to the original glucose structure, the $-OH$ group ends up on the left side in the Fischer projection. This gives us a completely new sugar, **D-mannose**, which is the C-2 epimer of glucose.

But wait, there is another possibility! What if the tautomerization happens at carbon-1 instead? The double bond shifts, and the oxygen on carbon-2 becomes a carbonyl group. Visualize this: carbon-2 is now a ketone! This transforms our aldose into a ketose, specifically **D-fructose**.

### The Final Equilibrium

So, starting from just D-glucose, the basic medium creates a dynamic equilibrium. We end up with a mixture of D-glucose, its epimer D-mannose, and the ketose D-fructose.

Matching this with our given options, we see that Option (C) perfectly lists these three monosaccharides. It is a very simple question if you remember the mechanism. Always look for the $\alpha$-hydrogen when dealing with sugars in a basic medium!

Animated Solution for Chemistry - Organic Chemistry: Treatment of D-glucose with aqueous NaOH results in a mixture of monosaccharides, which are

Select Answer:

Visualized Solution

$D -Glucose Structure$

$Base-Catalyzed Isomerization$

$Formation of Enediol Intermediate$

$Tautomerization to D-Mannose$

$Tautomerization to D-Fructose$

$The Equilibrium Mixture$

$Epimers and Ketoses$

The Sigma Insight: Biomolecules

The Magic of Basic Medium

The Enediol Intermediate

The Tautomerization Split

The Final Equilibrium

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