Given below are two statements: Statement I: textD-(+)-glucose + textD-(+)-fructose xrightarrow-mathrmH_2mathrmO textsucrose qquad textsucrose xrightarrowtextHydrolysis textD-(+)-glucose + textD-(+)-fructose$\text{D-(+)-glucose} + \text{D-(+)-fructose} \xrightarrow{-\mathrm{H}_2\mathrm{O}} \text{sucrose} \qquad \text{sucrose} \xrightarrow{\text{Hydrolysis}} \text{D-(+)-glucose} + \text{D-(+)-fructose}$ Statement II: Invert sugar is formed during sucrose hydrolysis. In the light of given statements, choose the correct answer from the options given below:

### Core Logic Fehling's reagent is reduced by reducing sugars to give a reddish-brown precipitate of Cu_2O$Cu_2O$. Reducing sugars must have a free aldehyde/ketone group or a hemiacetal linkage that can open to form an aldehyde. Sucrose is a non-reducing sugar because its anomeric carbons (C1 of glucose and C2 of fructose) are tied up in a glycosidic linkage, leaving no free hemiacetal group. ### Step 1: Analyzing the options Lactose, glucose, and maltose are all reducing sugars and will give a positive Fehling's test. Sucrose does not. ### Pattern Recognition Sucrose = non-reducing sugar. Maltose, Lactose = reducing sugars. Monosaccharides (glucose, fructose) = always reducing. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

### Core Logic Vitamins are broadly classified into two groups based on solubility: 1) Fat-soluble vitamins: Vitamins A, D, E, and K. These are stored in the liver and adipose (fat-storing) tissues. 2) Water-soluble vitamins: B group vitamins and Vitamin C. These are readily excreted in urine and cannot be stored in the body (with the exception of Vitamin B_12$B_{12}$, which can be stored in the liver). ### Step 1: Final List The vitamins that can be stored in the body from the given list are A, D, E, K, and B_12$B_{12}$. Total number = 5$5$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

### Core Logic Matching the reactions of glucose: (A) Glucose does not react with NaHCO_3$NaHCO_3$, so there is no reaction. (A rightarrow$\rightarrow$ II) (B) Oxidation of glucose with strong oxidizing agents like HNO_3$HNO_3$ yields a dicarboxylic acid called saccharic acid. (B rightarrow$\rightarrow$ IV) (C) Prolonged heating of glucose with HI$HI$ forms n-hexane, indicating a straight chain of six carbon atoms. (C rightarrow$\rightarrow$ III) (D) Oxidation with mild agents like bromine water converts glucose to gluconic acid. (D rightarrow$\rightarrow$ I) ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules