A dipeptide, "x" on complete hydrolysis gives "y" and "z". "y" on treatment with aq. HNO_2 produces lactic acid. On the other hand "z" on heating gives the following cyclic molecule. Based on the information given, the dipeptide X is:

Cyclic molecule from dipeptide residue heating for Q29 - JEE Main 2025 Evening
The image shows a heterocyclic six-membered cyclic molecule containing amide groups, formed by heating amino acid residue z.
Cyclic molecule from dipeptide residue heating for Q29 - JEE Main 2025 Evening
The image shows a heterocyclic six-membered cyclic molecule containing amide groups, formed by heating amino acid residue z.

Solution & Explanation

### Related Formula textDipeptide X xrightarrowtextHydrolysis textAmino Acid y + textAmino Acid z ### Core Logic - Since **y** reacts with nitrous acid (HNO_2) to give lactic acid (CH_3-CH(OH)-COOH), **y** must be alanine (CH_3-CH(NH_2)-COOH). - When glycine (NH_2-CH_2-COOH) is heated, two molecules undergo intermolecular cyclization to produce a six-membered diketopiperazine ring as shown in the problem diagram. Therefore, **z** is glycine. Hence, combining residue **y** (alanine) and **z** (glycine), the dipeptide X is **alanine-glycine**. ### Step 1: Stepwise Degradation Overview Reaction scheme: 1. Alanine-Glycine linkage rightarrow Alanine + Glycine 2. textAlanine + HNO_2 rightarrow textLactic acid + N_2uparrow + H_2O 3. 2 times textGlycine xrightarrowDelta textCyclic diketopiperazine + 2H_2O ### Pattern Recognition Lactic acid generation from alpha-amino acids via nitrous acid deamination is a definitive chemical fingerprint for alanine. The unsubstituted cyclic diketopiperazine product confirms glycine as the second component. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules
Amino Acids and Peptides
The image shows a heterocyclic six-membered cyclic molecule containing amide groups, formed by heating amino acid residue z.

More Biomolecules Previous-Year Questions — Page 6

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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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)