Biomolecules

Concept:Sugars are carbohydrates that exhibit specific reactions and optical properties.Explanation:Option (A):Glucose has both an aldehyde group ($-CHO$) and a primary alcohol group ($-CH_2OH$).Treatment with $HNO_3$ oxidizes both ends to $COOH$, forming saccharic acid.Gluconic acid has the aldehyde already oxidized to $COOH$, so $HNO_3$ can still oxidize the $CH_2OH$ end to $COOH$, yielding saccharic acid.Thus, both glucose and gluconic acid can be oxidized to saccharic acid by $HNO_3$.The statement says glucose is NOT oxidized to saccharic acid, which is false.Hence, Option (A) is incorrect.Option (B):Fehling’s reagent contains a weak base (dilute $NaOH$).In basic medium, fructose undergoes tautomerization to form glucose and mannose (aldohexoses).Glucose and mannose are reducing sugars and give a positive Fehling’s test.Therefore, fructose also gives a positive Fehling’s test due to isomerization.This statement is correct.Option (C):Sucrose is a disaccharide of $D$-glucose and $D$-fructose.Upon acid-catalyzed hydrolysis, sucrose breaks into an equimolar mixture of $D$-glucose and $D$-fructose.This mixture is called invert sugar.Thus, Option (C) is correct.Option (D):Specific rotation of $D$-glucose: $[ \alpha ] = +52.5^\circ$.Specific rotation of $D$-fructose: $[ \alpha ] = -92.5^\circ$.For an equimolar mixture (invert sugar), the net specific rotation is the average:$$[ \alpha ]_{\text{invert}} = \frac{+52.5^\circ + (-92.5^\circ)}{2} = \frac{-40^\circ}{2} = -20^\circ.$$The given value ($-40^\circ$) is incorrect.Hence, Option (D) is false.Answer:The correct statements are (B) and (C).