Given below are two statements : Statement I : D-glucose pentaacetate reacts with 2, 4-dinitrophenylhydrazine. Statement II : Starch, on heating with concentrated sulfuric acid at 100^circmathrmC and 2-3 atmosphere pressure produces glucose. In the light of the above statements, choose the correct answer from the options given below

Solution & Explanation

### Core Logic Statement I is false because glucose pentaacetate fixes the cyclic hemiacetal system structure securely into an unreactive ester configuration. As a result, it cannot revert to an open-chain form containing a free aldehyde group, meaning it does not react with carbonyl reagents like 2,4-DNP. Statement II is true because starch, a polysaccharide composed of glucose monomer blocks, undergoes acid-catalyzed hydrolysis to yield glucose when heated under pressure. ### Pattern Recognition Sees: Pentacetate reactivity vs polysaccharide hydrolysis. Shortcut: Acetylation locks the cyclic structure of glucose, preventing reactions that require an open-chain carbonyl group (like 2,4-DNP). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

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Q63 jee_main_2024_30_jan_morning Carbohydrates
Sugar which does not give reddish brown precipitate with Fehling's reagent is:
  • A. textSucrose
  • B. textLactose
  • C. textGlucose
  • D. textMaltose

Solution

### Core Logic Fehling's reagent is reduced by reducing sugars to give a reddish-brown precipitate of 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
Q89 jee_main_2024_31_jan_evening Vitamins and their Classification
From the vitamins A, B_1, B_6, B_12, C, D, E and K, the number vitamins that can be stored in our body is ________
Numerical Answer. Answer: 5 to 5

Solution

### 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, 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. Total number = 5. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules
Q78 jee_main_2024_31_jan_morning Reactions of Glucose
Match List I with List II
LIST-ILIST-II
A. Glucose/NaHCO_3/DeltaI. Gluconic acid
B. Glucose/HNO_3II. No reaction
C. Glucose/HI/DeltaIII. n-hexane
D. Glucose/Bromine waterIV. Saccharic acid
Choose the correct answer from the options given below:
  • A. textA-IV, B-I, C-III, D-II
  • B. textA-II, B-IV, C-III, D-I
  • C. textA-III, B-II, C-I, D-IV
  • D. textA-I, B-IV, C-III, D-II

Solution

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

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