Animated Solution for Chemistry - Organic Chemistry: The correct statement(s) regarding sugars is (are) Given : Specific rotations of L-(-)-glucose and L-(+)-fructose are -52.5° and +92.5°, respectively.

Carbohydrates are not just a source of energy; they are a playground of fascinating chemical reactions and stereochemical puzzles. This question is a beautiful amalgamation of oxidation reactions, tautomerization, hydrolysis, and optical activity. Let's dissect each statement to uncover the hidden truths of these sweet molecules.

The Oxidizing Power of Nitric Acid

When we treat sugars with oxidizing agents, the products depend heavily on the strength of the reagent. Nitric acid (${\text{HNO}}_3$) is a strong oxidizing agent.

When D-glucose is treated with ${\text{HNO}}_3$, it doesn't just stop at oxidizing the aldehyde group at C1. It goes a step further and oxidizes the primary alcohol group at C6 as well. The result is a dicarboxylic acid known as saccharic acid (or glucaric acid).

Similarly, if we start with gluconic acid (where C1 is already a carboxylic acid), ${\text{HNO}}_3$ will oxidize the C6 primary alcohol, again yielding saccharic acid. Therefore, the statement claiming that glucose is not oxidized to saccharic acid is fundamentally incorrect.

The Magic of Alkaline Medium

Fructose is a ketose. A standard rule of thumb in organic chemistry is that ketones do not reduce mild oxidizing agents like Fehling's or Tollen's reagents. So, why does fructose give a positive Fehling's test?

The secret lies in the medium. Fehling's solution is strongly basic (contains ${\text{OH}}^{-}$ ions). In an alkaline environment, fructose undergoes a fascinating rearrangement known as the Lobry de Bruyn-van Ekenstein transformation.

The base abstracts an acidic alpha-hydrogen, causing the double bond to shift and forming an unstable enediol intermediate. This intermediate can then rearrange not just back to fructose, but also into D-glucose and D-mannose. Because both glucose and mannose are aldohexoses, they readily reduce Fehling's solution. Thus, fructose gives a positive test by proxy. Statement B is absolutely correct.

The Sweetness of Invert Sugar

Sucrose, common table sugar, is a disaccharide composed of an $\alpha$-D-glucose unit and a $\beta$-D-fructose unit linked by a glycosidic bond.

When sucrose undergoes acidic hydrolysis, the glycosidic bond breaks, releasing its constituent monosaccharides:

Because one molecule of sucrose yields exactly one molecule of glucose and one molecule of fructose, the resulting product is a 1:1 equimolar mixture. This specific mixture is famously known as invert sugar. Statement C is perfectly accurate.

The Optical Rotation Trap

This is where the examiners set a brilliant trap. The question provides the specific rotations for the L-isomers: $[\alpha {]}_{\text{L-glucose}}=-52.5^{\circ }$ $[\alpha {]}_{\text{L-fructose}}=+92.5^{\circ }$

However, naturally occurring sucrose hydrolyzes into D-isomers. Since enantiomers have equal but opposite specific rotations, we must first invert the signs: $[\alpha {]}_{\text{D-glucose}}=+52.5^{\circ }$ $[\alpha {]}_{\text{D-fructose}}=-92.5^{\circ }$

Now, to find the specific rotation of the invert sugar mixture, we must remember that it is an equimolar mixture. The specific rotation of such a mixture is the arithmetic mean of the specific rotations of its components:

The option states the rotation is $-40^{\circ }$, which is merely the sum, not the average. Therefore, statement D is incorrect.

By carefully navigating through oxidation rules, base-catalyzed tautomerization, and stereochemical calculations, we confidently arrive at the conclusion that only statements (B) and (C) are correct.