Given below are two statements: Statement (I): Oxacyclobutane and prop-2-en-1-ol are isomeric compounds. Statement (II): Propan-1-amine and N-methylethanamine are functional group isomers. In the light of the above statements, choose the correct answer from the options given below :

Isomers share an identical molecular formula but differ in structural arrangement or functional groups: Evaluating the structural parameters: - **Statement I**: Oxacyclobutane (a cyclic ether) and prop-2-en-1-ol (an unsaturated alcohol) both possess the molecular formula C3H6O. They are functional/ring-chain isomers, so Statement I is true. - **Statement II**: Propan-1-amine (1^ amine) and N-methylethanamine (2^ amine) both share the molecular formula C3H9N. Because primary, secondary, and tertiary amines contain different functional groups, they act as functional group isomers. Thus, Statement II is true. Since both structural statements are valid, both Statement I and Statement II are true. Always remember that 1^, 2^, and 3^ amines are classified as *different functional groups* in IUPAC nomenclature. Consequently, structural shifts between them with a constant carbon count represent functional group isomerism. Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Given below are two statements: Statement (I): Oxacyclobutane and prop-2-en-1-ol are isomeric compounds. Statement (II): Propan-1-amine and N-methylethanamine are functional group isomers. In the ligh

Isomers share an identical molecular formula but differ in structural arrangement or functional groups: Evaluating the structural parameters: - **Statement I**: Oxacyclobutane (a cyclic ether) and prop-2-en-1-ol (an unsaturated alcohol) both possess the molecular formula C3H6O. They are functional/ring-chain isomers, so Statement I is true. - **Statement II**: Propan-1-amine (1^ amine) and N-methylethanamine (2^ amine) both share the molecular formula C3H9N. Because primary, secondary, and terti

Given below are two statements: Statement (I): Oxacyclobutane and prop-2-en-1-ol are isomeric compounds. Statement (II): Propan-1-amine and N-methylethanamine are functional group isomers. In the light of the above statements, choose the correct answer from the options given below :

Isomers share an identical molecular formula but differ in structural arrangement or functional groups: Evaluating the structural parameters: - **Statement I**: Oxacyclobutane (a cyclic ether) and prop-2-en-1-ol (an unsaturated alcohol) both possess the molecular formula C3H6O. They are functional/ring-chain isomers, so Statement I is true. - **Statement II**: Propan-1-amine (1^ amine) and N-methylethanamine (2^ amine) both share the molecular formula C3H9N. Because primary, secondary, and tertiary amines contain different functional groups, they act as functional group isomers. Thus, Statement II is true. Since both structural statements are valid, both Statement I and Statement II are true. Always remember that 1^, 2^, and 3^ amines are classified as *different functional groups* in IUPAC nomenclature. Consequently, structural shifts between them with a constant carbon count represent functional group isomerism. Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Given below are two statements: Statement (I): Oxacyclobutane and prop-2-en-1-ol are isomeric compounds. Statement (II): Propan-1-amine and N-methylethanamine are functional group isomers. In the ligh

Isomers share an identical molecular formula but differ in structural arrangement or functional groups: Evaluating the structural parameters: - **Statement I**: Oxacyclobutane (a cyclic ether) and prop-2-en-1-ol (an unsaturated alcohol) both possess the molecular formula C3H6O. They are functional/ring-chain isomers, so Statement I is true. - **Statement II**: Propan-1-amine (1^ amine) and N-methylethanamine (2^ amine) both share the molecular formula C3H9N. Because primary, secondary, and terti

Given below are two statements: Statement (I): Oxacyclobutane and prop-2-en-1-ol are isomeric compounds. Statement…

Year	2026	2025	2024	Total
Questions	22	49	30	101

Year

2026

2025

2024

Total

Questions

101

Solution & Explanation

Related Formula

Isomers share an identical molecular formula but differ in structural arrangement or functional groups:

Same Mf \neq Same structural layout

Core Logic

Evaluating the structural parameters:

Statement I: Oxacyclobutane (a cyclic ether) and prop-2-en-1-ol (an unsaturated alcohol) both possess the molecular formula $C₃H₆O$ $C_3H_6O$. They are functional/ring-chain isomers, so Statement I is true.
Statement II: Propan-1-amine ( $1^°$ $1^\circ$ amine) and N-methylethanamine ( $2^°$ $2^\circ$ amine) both share the molecular formula $C₃H₉N$ $C_3H_9N$. Because primary, secondary, and tertiary amines contain different functional groups, they act as functional group isomers. Thus, Statement II is true.

Step 1: Conclusion Match

Since both structural statements are valid, both Statement I and Statement II are true.

Skeletal representations for the specified organic isomers

Pattern Recognition

Always remember that $1^°$ $1^\circ$, $2^°$ $2^\circ$, and $3^°$ $3^\circ$ amines are classified as different functional groups in IUPAC nomenclature. Consequently, structural shifts between them with a constant carbon count represent functional group isomerism.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 9

Q28 jee_main_2025_29_jan_evening Chromatographic Techniques

Given below are two statements: Statement (I): In partition chromatography, stationary phase is thin film of liquid present in the inert support. Statement (II): In paper chromatography, the material of paper acts as a stationary phase. In the light of the above statements, choose the correct answer from the options given below:

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

Solution

Core Logic

Statement I is true: In partition chromatography, the stationary phase is indeed a thin film of liquid held on the surface of an inert solid support. Statement II is false: In paper chromatography, the water molecules trapped inside the cellulose network of the paper act as the stationary phase, not the paper material itself.

Pattern Recognition

Remember that paper chromatography is a type of partition chromatography where moisture content (water) adsorbed on the paper serves as the stationary liquid phase.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q38 jee_main_2025_29_jan_evening Sigma and Pi Bond Counting

Total number of sigma

(sigma)

and

pi(pi)

bonds respectively present in hex-1-en-4-yne are:

A. 13 and 3
B. 11 and 3
C. 3 and 13
D. 14 and 3

Solution

Core Logic

The structural formula of hex-1-en-4-yne is given by:

CH₂ = CH - CH₂ - C equiv C - CH₃

Let's count the chemical bonds chronologically:

Number of $C-H$ $C-H$ $sigma$ $sigma$ bonds = $2 + 1 + 2 + 3 = 8$ $2 + 1 + 2 + 3 = 8$
Number of $C-C$ $C-C$ $sigma$ $sigma$ bonds = $5$ $5$

Total $sigma$ $sigma$ bonds = $8 + 5 = 13$ $8 + 5 = 13$.

Sigma and Pi Bond Counting diagram for Q38 - JEE Main 2025 Evening

Number of $pi$ $pi$ bonds: $1$ $1$ from double bond + $2$ $2$ from triple bond = $3$ $3$ $pi$ $pi$ bonds.

Pattern Recognition

Every single bond is $1sigma$ $1sigma$, every double bond contains $1sigma + 1pi$ $1sigma + 1pi$, and every triple bond contains $1sigma + 2pi$ $1sigma + 2pi$.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q49 jee_main_2025_29_jan_evening Quantitative Estimation of Sulphur

In the sulphur estimation,

0.20 g

of a pure organic compound gave

0.40 g

of barium sulphate. The percentage of sulphur in the compound is

x \times 10⁻¹%

, where

x

= ________. (Molar mass:

O=16

S=32

Ba=137 in g mol⁻¹

)

Numerical Answer. Answer: 275 to 275

Solution

Related Formula

%S = (32)/(233) \times Mass of BaSO₄Mass of organic compound \times 100

Core Logic

Let's substitute the given values into the formula:

Mass of BaSO₄ = 0.40 g

Mass of organic compound = 0.20 g

Molar mass of BaSO₄ = 137 + 32 + (4 \times 16) = 233 g/mol

%S = (32)/(233) \times (0.40)/(0.20) \times 100 = (32 \times 2 \times 100)/(233) approx 27.468%

Step 1: Match with the Question Layout

Rounding to the standard value given in the official key:

%S = 27.5% = 275 \times 10⁻¹% implies x = 275

Pattern Recognition

Carius method calculations depend heavily on standard conversion factors. The constant factor for sulphur gravimetry is $(32)/(233)$ $\frac{32}{233}$.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q42 jee_main_2025_28_jan_morning Carbocation Stability

The correct order of stability of following carbocations is :

Carbocation structures for Q42 - JEE Main 2025 Morning — The images show different structural models labeled A, B, C, and D for evaluating stability variations.

A. $A > B > C > D$
B. $B > C > A > D$
C. $C > B > A > D$
D. $C > A > B > D$

Solution

Core Logic

To evaluate carbocation stability, apply the priority rules: Aromaticity > Resonance > Hyperconjugation.

C: Represents a cyclopropenyl cation derivative which achieves full aromatic stabilization due to its planar cyclic conjugated system satisfying Huckel's rule ( $2π$ $2\pi$ electrons). This makes it the most stable.
A: Stabilized by extended resonance from multiple phenyl groups.
B: Contains fewer phenyl rings participating in active cross-conjugation relative to A.
D: Stabilized solely by simple aliphatic hyperconjugation, making it the least stable.

Visual alignment chart:

Stability ranking structural chart for Q42 - JEE Main 2025 Morning — The images show different structural models labeled A, B, C, and D for evaluating stability variations.

Hence, the correct stability hierarchy is:

C > A > B > D

Pattern Recognition

Sees: Mixed aromatic, benzylic, and aliphatic carbocations. Shortcut: Isolate the cyclopropenyl system as an aromatic champion to confidently lead the sequence.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q43 jee_main_2025_28_jan_morning Acidity of Organic Compounds

The compounds that produce

CO₂

with aqueous

NaHCO₃

solution are: A.

The prompt lists five structures labeled A through E evaluating structural acidities.

Choose the correct answer from the options given below:

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

Solution

Core Logic

Organic compounds react with sodium bicarbonate ( $NaHCO₃$ $\mathrm{NaHCO}_3$) to liberate $CO₂$ $\mathrm{CO}_2$ gas if they are stronger acids than carbonic acid ( $H₂CO₃$ $\mathrm{H}_2\mathrm{CO}_3$). Evaluating the structures:

A: Benzoic acid, which is significantly more acidic than carbonic acid.
C: Picric acid (2,4,6-trinitrophenol). Due to three strong electron-withdrawing nitro groups, its acidity exceeds typical carboxylic acids and $H₂CO₃$ $\mathrm{H}_2\mathrm{CO}_3$.
D: Benzenesulfonic acid, a highly strong mineral-like organic acid.
B & E: Standard phenols or weakly substituted phenols, which are less acidic than carbonic acid and do not liberate $CO₂$ $\mathrm{CO}_2$.

Therefore, structures A, C, and D give a positive test result.

Pattern Recognition

Sees: Sodium bicarbonate test for organic systems. Shortcut: Only carboxylic acids, sulfonic acids, and highly nitrated phenols like picric acid possess sufficient proton acidity to displace $CO₂$ $\mathrm{CO}_2$ from bicarbonate ions.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Given below are two statements: Statement (I): Oxacyclobutane and prop-2-en-1-ol are isomeric compounds. Statement (II): Propan-1-amine and N-methylethanamine are functional group isomers. In the light of the above statements, choose the correct answer from the options given below :

Solution & Explanation

Related Formula

Core Logic

Step 1: Conclusion Match

Pattern Recognition

Chapter Mix

Reference Study Guides

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 9

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Match with the Question Layout

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

More Organic Chemistry - Some Basic Principles and Techniques Questions — jee_main_2025_28_jan_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back