Equilibrium Mixture of Monosaccharides from Alkaline Treatment of D Glucose

When D-glucose is treated with aqueous sodium hydroxide ($\text{NaOH}$), it undergoes a base-catalyzed keto-enol tautomerization known as the Lobry de Bruyn-Alberda van Ekenstein rearrangement.

Reaction Mechanism

1. Formation of Enolate/Enediol Intermediate: The base ($\text{OH}^-$) abstracts the acidic $\alpha$-hydrogen atom from $\text{C-2}$ of D-glucose to form a resonance-stabilized enolate anion, which picks up a proton from water to form a 1,2-enediol intermediate:

$\text{D-Glucose} \underset{-\text{H}_2\text{O}}{\xrightarrow{\text{OH}^-}} \text{Enolate Ion} \underset{+\text{H}_2\text{O}}{\xrightarrow{\quad}} \text{1,2-Enediol}$

The structure of the 1,2-enediol intermediate is: $\text{HO-CH=C(OH)-(CHOH)}_3\text{-CH}_2\text{OH}$

1. Tautomerization to D-Mannose: Reprotonation at the $\text{C-2}$ position from the opposite face of the enediol double bond converts the intermediate into the $\text{C-2}$ epimer of D-glucose, which is D-mannose.

2. Tautomerization to D-Fructose: Proton transfer at the $\text{C-1}$ oxygen followed by ketonization converts the enediol into the ketose form, which is D-fructose.

Thus, the reaction produces an equilibrium mixture containing:

1. D-glucose
2. D-fructose
3. D-mannose

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Fischer Projections of the Monosaccharides

1. D-Glucose:

$$\begin{array}{c} \phantom{\text{HO}}\text{CHO} \\ | \\ \phantom{\text{O}}\text{H} - \text{C} - \text{OH} \\ | \\ \text{HO} - \text{C} - \text{H}\phantom{\text{O}} \\ | \\ \phantom{\text{O}}\text{H} - \text{C} - \text{OH} \\ | \\ \phantom{\text{O}}\text{H} - \text{C} - \text{OH} \\ | \\ \phantom{\text{H}_2\text{OH}}\text{CH}_2\text{OH} \end{array}$$

2. D-Fructose:

$$\begin{array}{c} \phantom{\text{H}_2\text{OH}}\text{CH}_2\text{OH} \\ | \\ \phantom{= \text{O}}\text{C} = \text{O} \\ | \\ \text{HO} - \text{C} - \text{H}\phantom{\text{O}} \\ | \\ \phantom{\text{O}}\text{H} - \text{C} - \text{OH} \\ | \\ \phantom{\text{O}}\text{H} - \text{C} - \text{OH} \\ | \\ \phantom{\text{H}_2\text{OH}}\text{CH}_2\text{OH} \end{array}$$

3. D-Mannose:

$$\begin{array}{c} \phantom{\text{HO}}\text{CHO} \\ | \\ \text{HO} - \text{C} - \text{H}\phantom{\text{O}} \\ | \\ \text{HO} - \text{C} - \text{H}\phantom{\text{O}} \\ | \\ \phantom{\text{O}}\text{H} - \text{C} - \text{OH} \\ | \\ \phantom{\text{O}}\text{H} - \text{C} - \text{OH} \\ | \\ \phantom{\text{H}_2\text{OH}}\text{CH}_2\text{OH} \end{array}$$

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Conclusion

Comparing these Fischer projections with the given options, Option (C) correctly depicts the three monosaccharides present in the equilibrium mixture.

Correct Option: (C)