Match List-I with List-II tracking nitrogenous bases structures: List-I (Base) List-II (Chemical Structure Diagram) (A) Adenine (I) (B) Cytosine (II) (C) Thymine (III) (D) Uracil (IV) Choose the correct answer from the options given below :

Let's identify the chemical structures of the nitrogenous bases used in nucleic acids: * **(A) Adenine:** A purine derivative featuring a characteristic fused bicyclic ring system with an amino substituent at position 6 (6-aminopurine) ightarrow **(III)**. * **(B) Cytosine:** A pyrimidine monocyclic derivative with an amino group at position 4 and a carbonyl group at position 2 (4-amino-2-oxo-pyrimidine) ightarrow **(IV)**. * **(C) Thymine:** Found in DNA, this pyrimidine derivative features a methyl substituent at position 5 along with carbonyl groups at positions 2 and 4 (5-methyl-2,4-dioxo-pyrimidine) ightarrow **(II)**. * **(D) Uracil:** Found in RNA, this pyrimidine derivative lacks the methyl group found in thymine, featuring just carbonyl groups at positions 2 and 4 (2,4-dioxo-pyrimidine) ightarrow **(I)**. Matching these structures gives the sequence: (A)-(III), (B)-(IV), (C)-(II), (D)-(I). The biochemical structures correspond to the following configurations: Quick identification keys: - Bicyclic ring = Adenine - Monocyclic ring with a -CH3 group = Thymine - Monocyclic ring without a -CH3 group = Uracil - Monocyclic ring with an amino group (-NH2) = Cytosine Class 12 Chemistry: Biomolecules

Match List-I with List-II tracking nitrogenous bases structures: List-I (Base) List-II (Chemical Structure Diagram) (A) Adenine (I) (B) Cytosine (II) (C) Thymine (III) (D) Uracil (IV) Choose the correct answer from the options given below :

Let's identify the chemical structures of the nitrogenous bases used in nucleic acids: * **(A) Adenine:** A purine derivative featuring a characteristic fused bicyclic ring system with an amino substituent at position 6 (6-aminopurine) ightarrow **(III)**. * **(B) Cytosine:** A pyrimidine monocyclic derivative with an amino group at position 4 and a carbonyl group at position 2 (4-amino-2-oxo-pyrimidine) ightarrow **(IV)**. * **(C) Thymine:** Found in DNA, this pyrimidine derivative features a methyl substituent at position 5 along with carbonyl groups at positions 2 and 4 (5-methyl-2,4-dioxo-pyrimidine) ightarrow **(II)**. * **(D) Uracil:** Found in RNA, this pyrimidine derivative lacks the methyl group found in thymine, featuring just carbonyl groups at positions 2 and 4 (2,4-dioxo-pyrimidine) ightarrow **(I)**. Matching these structures gives the sequence: (A)-(III), (B)-(IV), (C)-(II), (D)-(I). The biochemical structures correspond to the following configurations: Quick identification keys: - Bicyclic ring = Adenine - Monocyclic ring with a -CH3 group = Thymine - Monocyclic ring without a -CH3 group = Uracil - Monocyclic ring with an amino group (-NH2) = Cytosine Class 12 Chemistry: Biomolecules

Match List-I with List-II tracking nitrogenous bases structures: List-I (Base) List-II (Chemical Structure Diagram) (A)…

Year	2026	2025	2024	Total
Questions	9	18	8	35

List-I (Base)	List-II (Chemical Structure Diagram)
(A) Adenine	(I) The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
(B) Cytosine	(II) The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
(C) Thymine	(III) The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
(D) Uracil	(IV) The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.

Q67 jee_main_2026_21_jan_morning Amino Acids

Identify the correct statements. A. Arginine and Tryptophan are essential amino acids. B. Histidine does not contain heterocyclic ring in its structure. C. Proline is a six membered cyclic ring amino acid. D. Glycine does not have chiral centre. E. Cysteine has characteristic feature of side chain as

MeS-CH₂-CH₂-

. Choose the correct answer from the options given below:

A. $C and E Only$ $\text{C and E Only}$
B. $B and E Only$ $\text{B and E Only}$
C. $C and D Only$ $\text{C and D Only}$
D. $A and D Only$ $\text{A and D Only}$

Solution

Core Logic

A. Arginine and Tryptophan are indeed essential amino acids (cannot be synthesized by the human body). (Correct)

B. Histidine contains an imidazole ring, which is a heterocyclic ring. So statement B is incorrect.

C. Proline contains a pyrrolidine ring, which is a five-membered cyclic ring. So statement C is incorrect.

D. Glycine is $NH₂-CH₂-COOH$ $\mathrm{NH_2-CH_2-COOH}$. Because its alpha carbon has two identical hydrogen atoms attached, it lacks a chiral center. (Correct)

E. Cysteine's side chain is a thiol group ( $-CH₂-SH$ $-\mathrm{CH_2-SH}$), not a thioether ( $MeS-CH₂-CH₂-$ $\mathrm{MeS-CH_2-CH_2-}$ which belongs to Methionine). So statement E is incorrect.

Correct statements: A and D.

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Class 12 Chemistry: Biomolecules

Q61 jee_main_2026_21_jan_evening Proteins and Enzymes

The correct statements are: A. Activation energy for enzyme catalysed hydrolysis of sucrose is lower than that of acid catalysed hydrolysis. B. During denaturation, secondary and tertiary structures of a protein are destroyed but primary structure remains intact. C. Nucleotides are joined together by glycosidic linkage between

C₁

and

C₄

carbons of the pentose sugar. D. Quaternary structure of proteins represents overall folding of the polypeptide chain. Choose the correct answer from the options given below:

A. $(1) A, C and D Only$ $(1) \text{ A, C and D Only}$
B. $(2) A, B and D Only$ $(2) \text{ A, B and D Only}$
C. $(3) A and B Only$ $(3) \text{ A and B Only}$
D. $(4) B and C Only$ $(4) \text{ B and C Only}$

Solution

Core Logic

A is correct: Enzymes lower activation energy significantly.
B is correct: Denaturation affects secondary and tertiary structures while primary structure remains intact.
C is incorrect: Nucleotides are joined by phosphodiester linkages, and glycosidic linkages connect bases to sugars.
D is incorrect: Quaternary structure represents aggregation of multiple polypeptide subunits, not single chain folding.

Step 1: Final Conclusion

Only statements A and B are correct, matching option (3).

Pattern Recognition

Sees: Biomolecules structural and enzymatic properties. Trap: Confusing primary structure disruption with denaturation.

Chapter Mix

Class 12 Chemistry: Biomolecules

Q65 jee_main_2026_22_january_morning Carbohydrates

Given below are two statements: Statement I: Sucrose is dextrorotatory. However sucrose upon hydrolysis gives a solution having mixture of products. This solution shows laevorotation. Statement II: Hydrolysis of sucrose gives glucose and fructose. Since the laevorotation of glucose is more than the dextrorotation of fructose the resulting solution becomes laevorotatory. In the light of the above statements, choose the correct answer from the options given below.

A. $Statement I is false but Statement II is true.$
B. $Both Statement I and Statement II are false.$
C. $Both Statement I and Statement II are true.$
D. $Statement I is true but Statement II is false.$

Solution

Related Formula

C₁₂H₂₂O₁₁ + H₂O H^+ D-Glucose + D-Fructose

Core Logic

Statement I: Pure sucrose has a specific rotation $[α]D = +66.5°$ $[\alpha]_D = +66.5^{\circ}$, so it is dextrorotatory. Upon hydrolysis, it forms an equimolar mixture of D-glucose and D-fructose. The net specific rotation of the mixture is negative, resulting in laevorotation. (This is called inversion of sugar). Statement I is true.

Statement II: Hydrolysis gives glucose and fructose. The specific rotation of D-glucose is $+52.5°$ $+52.5^{\circ}$ (dextrorotation), while that of D-fructose is $-92.4°$ $-92.4^{\circ}$ (laevorotation). The statement claims that glucose is laevorotatory and fructose is dextrorotatory, which is strictly backward. Therefore, Statement II is false.

Step 1: Final Conclusion

Statement I is true but Statement II is false.

Pattern Recognition

Invert sugar concepts are very common. Always verify which monomer rotates light in which direction. D-fructose is strongly laevorotatory ( $-92.4°$ $-92.4^{\circ}$).

Chapter Mix

Class 12 Chemistry: Biomolecules

Q60 jee_main_2026_22_january_evening Reactions of Glucose

Match List-I with List-II.

List-I (Reaction of glucose with)	List-II (Product formed)
A. Hydroxylamine	I. Gluconic acid
B. $Br₂$ $Br_2$ water	II. Glucose pentaacetate
C. Excess acetic anhydride	III. Saccharic acid
D. Concentrated $HNO₃$ $HNO_3$	IV. Glucoxime

Choose the correct answer from the options given below:

A. A-I, B-III, C-IV, D-II
B. A-IV, B-I, C-II, D-III
C. A-III, B-I, C-IV, D-II
D. A-IV, B-III, C-II, D-I

Solution

Related Formula

Glucose + NH₂OH arrow Glucoxime

Glucose + Br₂/H₂O arrow Gluconic acid

Glucose + (CH₃CO)₂O arrow Glucose pentaacetate

Glucose + conc. HNO₃ arrow Saccharic acid

Core Logic

Step 1: Match each reagent with its specific reaction product on glucose:

Hydroxylamine (A) reacts with carbonyl to form Glucoxime (IV).
$Br₂$ $\text{Br}_2$ water (B) mildly oxidizes $-CHO$ $-\text{CHO}$ to $-COOH$ $-\text{COOH}$ forming Gluconic acid (I).
Acetic anhydride (C) acetylates 5 hydroxyl groups to form Glucose pentaacetate (II).
Conc. $HNO₃$ $\text{HNO}_3$ (D) oxidizes both $-CHO$ $-\text{CHO}$ and primary $-CH₂OH$ $-\text{CH}_2\text{OH}$ to form Saccharic acid (III).

Step 2: Correct matching pair is A-IV, B-I, C-II, D-III.

Chemical reactions chart for glucose for Q60 - JEE Main 2026 Evening

Pattern Recognition

Sees: Classic NCERT Glucose chemical test reaction matches. Shortcut: Mild oxidant $Br₂/H₂O arrow$ Gluconic acid; Strong oxidant $HNO₃ arrow$ $\text{HNO}_3 \rightarrow$ Saccharic acid.

Chapter Mix

Class 12 Chemistry: Biomolecules

Q64 jee_main_2026_23_january_morning Reducing and Non-Reducing Sugars

From the given following (A to D) cyclic structures, those which will not react with Tollen's reagent are :

A. $B and D$ $\text{B and D}$
B. $A and D$ $\text{A and D}$
C. $A and B$ $\text{A and B}$
D. $B and C$ $\text{B and C}$

Solution

Core Logic

Tollens' reagent oxidizes aldehydes. In cyclic sugars, this requires a free hemiacetal group (an anomeric carbon bonded to an $-OH$ $-OH$ group) to allow mutarotation and open-chain aldehyde formation.

Step 1: Structural Analysis

Structures (A) and (D) possess a free $-OH$ $-OH$ group at the anomeric carbon (hemiacetal linkage). They are reducing sugars and will react with Tollens' reagent. Structures (B) and (C) have an $-OCH₃$ $-OCH_3$ group at the anomeric carbon instead of a free $-OH$ $-OH$. These are full acetals (glycosides) and cannot ring-open. Thus, they are non-reducing sugars.

Pattern Recognition

Hemiacetal ( $-OH$ $-OH$ on anomeric C) = Reducing Sugar. Acetal ( $-OR$ $-OR$ on anomeric C) = Non-Reducing Sugar.

Chapter Mix

Class 12 Chemistry: Biomolecules

Solution & Explanation

Core Logic

Step-by-Step Structural Validation

Pattern Recognition

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