Properties and Reactions of Sugars and Invert Sugar

To determine the correct statement(s), we analyze each option individually:

1. Evaluation of Option (A):

• When glucose is treated with concentrated nitric acid ($\text{HNO}_3$), both the terminal aldehyde group ($-\text{CHO}$) and primary alcohol group ($-\text{CH}_2\text{OH}$) are oxidized to carboxylic acid groups ($-\text{COOH}$), yielding saccharic acid (glucaric acid): $\text{CHO}-(\text{CHOH})_4-\text{CH}_2\text{OH} \xrightarrow{\text{HNO}_3} \text{HOOC}-(\text{CHOH})_4-\text{COOH}$
• Similarly, gluconic acid ($\text{HOOC}-(\text{CHOH})_4-\text{CH}_2\text{OH}$) is also oxidized to saccharic acid by $\text{HNO}_3$.
• Therefore, the statement claiming that "glucose is not oxidized to saccharic acid" is incorrect.

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2. Evaluation of Option (B):

• Fructose is a ketohexose containing an $\alpha$-hydroxy ketone functional group.
• In an alkaline medium (such as Fehling's reagent), fructose undergoes the Lobry de Bruyn-Alberda van Ekenstein transformation.
• Via a 1,2-enediol intermediate, D-fructose isomerizes into D-glucose and D-mannose (another aldohexose): $\text{D-fructose} \underset{\text{OH}^-}{\rightleftharpoons} \text{1,2-enediol intermediate} \underset{\text{OH}^-}{\rightleftharpoons} \text{D-glucose} + \text{D-mannose}$
• These generated aldohexoses reduce Fehling's reagent to give a red precipitate of cuprous oxide ($\text{Cu}_2\text{O}$), resulting in a positive Fehling's test.
• Therefore, Statement (B) is correct.

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3. Evaluation of Option (C):

• Hydrolysis of sucrose (a dextrorotatory disaccharide) in the presence of dilute acid or the enzyme invertase yields an equimolar mixture of $\text{D-(+)-glucose}$ and $\text{D-(-)-fructose}$: $\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \underset{\text{D-(+)-glucose}}{\text{C}_6\text{H}_{12}\text{O}_6} + \underset{\text{D-(-)-fructose}}{\text{C}_6\text{H}_{12}\text{O}_6}$
• This resulting equimolar mixture is defined as invert sugar.
• Therefore, Statement (C) is correct.

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4. Evaluation of Option (D):

• An enantiomer pair has equal magnitude but opposite sign for its specific rotation:
  • Since the specific rotation of $\text{L-(-)-glucose}$ is $-52.5^\circ$, the specific rotation of $\text{D-(+)-glucose}$ is: $[\alpha]_{\text{D-glucose}} = +52.5^\circ$
  • Since the specific rotation of $\text{L-(+)-fructose}$ is $+92.5^\circ$, the specific rotation of $\text{D-(-)-fructose}$ is: $[\alpha]_{\text{D-fructose}} = -92.5^\circ$
• Invert sugar is an equimolar (1:1) mixture of $\text{D-glucose}$ and $\text{D-fructose}$. Its specific rotation $[\alpha]_{\text{invert sugar}}$ is the average of the two specific rotations: $[\alpha]_{\text{invert sugar}} = \frac{[\alpha]_{\text{D-glucose}} + [\alpha]_{\text{D-fructose}}}{2} = \frac{(+52.5^\circ) + (-92.5^\circ)}{2} = \frac{-40^\circ}{2} = -20^\circ$
• The option states $-40^\circ$, which is incorrect.
• Therefore, Statement (D) is incorrect.

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Conclusion:

The correct statements are (B) and (C).