JEE Main · Chemistry ↓ Falling

Organic Chemistry - Some Basic Principles and Techniques appeared 101 times across 3 years — 11.8% of Chemistry. This question is from Acidic Strength of Organic Compounds.

Year	2026	2025	2024	Total
Questions	22	49	30	101

Practice all Organic Chemistry - Some Basic Principles and Techniques questions → View full JEE Main weightage & trends →

The least acidic compound, among the following is:

Acidic Strength of Organic Compounds

A. Compound (D)
Acidic Strength of Organic Compounds
B. Compound (A)
Acidic Strength of Organic Compounds
C. Compound (B)
Acidic Strength of Organic Compounds
D. Compound (C)

Solution & Explanation

Core Logic

Let us check the conjugate bases formed upon losing a proton:

Compounds (A), (B), and (C) generate conjugate bases stabilized by resonance through the aromatic ring or strong electron-withdrawing groups.
Compound (D) represents an ethynyl group in a terminal alkyne structure ( $EtO₂C-C$ ). Its conjugate base features a localized negative charge on an $sp$ $sp$-hybridized carbon. Because there is no resonance stabilization present for this anion, it is significantly less stable than the conjugate bases of the other functional groups.

Step 1: Conclusion

Since a less stable conjugate base implies a weaker parent acid, the terminal alkyne compound (D) is the least acidic.

Pattern Recognition

Shortcut: A resonance-stabilized anion is always more stable than a localized one. Look for the alkyne carbon ( $sp-C-H$ $sp\text{-C}-\text{H}$) versus resonance-delocalized oxygen or active methylene centers.

Evaluation Rubric / Model Answer

Option (A)

Chapter Mix

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

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 10

Q jee_main_2025_03_april_morning Structural Isomerism

Identify the correct statements from the following: Choose the correct answer from the options given below.

Structural Isomerism

A. C & D only
B. B & C only
C. A & B only
D. A, B & C only

Solution

Core Logic

Let us check the statements step-by-step:

Statement A: Pentan-3-one and pentan-2-one have different alkyl groups attached on either side of the divalent polyfunctional carbonyl group ( $-CO-$ $-\text{CO}-$). Hence, they are metamers.
Metamerism illustration for Q32 - JEE Main 2025 Morning
Statement B: Cyanides ( $-CN$ $-\text{CN}$) and Isocyanides ( $-NC$ $-\text{NC}$) contain distinct functional groups, so they are functional isomers.
Metamerism illustration for Q32 - JEE Main 2025 Morning
Statement C: Phenol structures containing a methyl substituent at positions 2 and 3 are structural position isomers.
Statement D: The given structures represent members of a homologous series because they differ sequentially by a $-CH₂-$ $-\text{CH}_2-$ unit.

Step 1: Verification

Evaluating according to standard multi-choice options, statements A and B are perfectly validated.

Pattern Recognition

Shortcut: Metamers require variable alkyl distribution across a polyvalent heteroatom group. Functional isomers require changes like $-CN$ $-\text{CN}$ vs $-NC$ $-\text{NC}$.

Chapter Mix

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

Q jee_main_2025_03_april_morning Quantitative Analysis - Dumas Method

During estimation of nitrogen by Dumas' method of compound X (0.42 g):

The image shows the molecular skeletal architecture of compound X with structural parameters revealing a formula corresponding to a molecular mass of 86 g/mol.

mL of

N2

gas will be liberated at STP. (nearest integer) (Given molar mass in g mol: C: 12, H: 1, N: 14)

The image shows the molecular skeletal architecture of compound X with structural parameters revealing a formula corresponding to a molecular mass of 86 g/mol.

Numerical Answer. Answer: 111 to 111

Solution

Related Formula

Using the Principle of Atom Conservation (POAC) for Nitrogen:

ncompound \times (atoms of N per molecule) = 2 \times nN₂

Core Logic

The molecular weight of the given heterocyclic amine organic structure $X$ $X$ (piperazine, $C₄H₁₀N₂$ $\text{C}_4\text{H}_{10}\text{N}_2$) is calculated as:

Molar Mass = (4 \times 12) + (10 \times 1) + (2 \times 14) = 86 g/mol

Stoichiometric parsing matrix step for Q46 — The image shows the molecular skeletal architecture of compound X with structural parameters revealing a formula corresponding to a molecular mass of 86 g/mol.

Given mass of compound $= 0.42 g$ $= 0.42\text{ g}$:

Moles of compound X = (0.42)/(86) mol

Step 1: Calculating STP Volume

Since each molecule contains $2$ $2$ nitrogen atoms, $1 mol$ $1\text{ mol}$ of compound produces $1 mol$ $1\text{ mol}$ of $N₂$ $\text{N}_2$ gas:

nN₂ = ncompound = (0.42)/(86) mol

Using standard molar volume at STP ( $22700 mL/mol$ $22700\text{ mL/mol}$ per IUPAC convention, or $22400 mL/mol$ $22400\text{ mL/mol}$ in traditional calculations):

Volume of N₂ at STP = (0.42)/(86) \times 22700 mL \approx 110.86 mL \approx 111 mL

(Note: If calculated using $22400 mL/mol$ $22400\text{ mL/mol}$, $Volume = (0.42)/(86) \times 22400 \approx 109.4 mL \approx 109 mL$ .)

Rounding to the nearest integer gives 111 (official accepted range: 109 to 111).

Pattern Recognition

Shortcut: Determine the molar mass ( $M = 86 g/mol$ $M = 86\text{ g/mol}$) and nitrogen atom count ( $2 N atoms 1 mol N₂ per mol of compound$ ). Multiply moles directly by molar volume at STP to find the liberated gas volume.

Evaluation Rubric / Model Answer

111

Chapter Mix

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

Q jee_main_2025_03_april_morning Quantitative Analysis - Estimation of Carbon

0.5 g of an organic compound on combustion gave 1.46 g of

CO₂

and 0.9 g of

H₂O

. The percentage of carbon in the compound is _____. (Nearest integer) [Given: Molar mass (in

g mol⁻¹

) C: 12, H: 1, O: 16]

Numerical Answer. Answer: 80 to 80

Solution

Related Formula

The percentage of carbon via combustion analysis is given by:

% C = (12)/(44) \times Mass of CO₂Mass of organic compound \times 100

Core Logic

Let us substitute the given parameters:

Mass of organic compound $= 0.5 g$ $= 0.5\text{ g}$
Mass of $CO₂$ $\text{CO}_2$ collected $= 1.46 g$ $= 1.46\text{ g}$

Step 1: Numerical Calculation

% C = (12)/(44) \times (1.46)/(0.5) \times 100

% C = (12 \times 1.46)/(22) \times 100 \approx 79.64%

Rounding to the nearest integer gives 80.

Pattern Recognition

Shortcut: $(12)/(44) \approx 0.2727$ $\frac{12}{44} \approx 0.2727$. Multiply $0.2727 \times 1.46$ $0.2727 \times 1.46$ to find the mass of carbon ( $\approx 0.398 g$ $\approx 0.398\text{ g}$). Since $0.398 g$ $0.398\text{ g}$ out of $0.5 g$ $0.5\text{ g}$ is roughly $(4)/(5)$ $\frac{4}{5}$, the value is right around $80%$ $80\%$.

Evaluation Rubric / Model Answer

80

Chapter Mix

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

Q jee_main_2025_04_april_evening Basicity of Organic Bases

The correct order of basicity for the following molecules is:

Basicity of Organic Bases diagram for Q26 - JEE Main 2025 Evening — The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.

Basicity of Organic Bases diagram for Q26 - JEE Main 2025 Evening — The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.

Basicity of Organic Bases diagram for Q26 - JEE Main 2025 Evening — The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.

Basicity of Organic Bases diagram for Q26 - JEE Main 2025 Evening — The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.

A. $P > Q > R$ $P > Q > R$
B. $R > P > Q$ $R > P > Q$
C. $Q > P > R$ $Q > P > R$
D. $R > Q > P$ $R > Q > P$

Solution

Related Formula

Basicity \propto Availability of lone pair of electrons on Nitrogen

Core Logic

Analyzing the molecules:

In molecule (R), according to Bredt's rule, the bridgehead nitrogen has a localized lone pair which cannot participate in resonance. Thus, it is highly available and most basic.
In molecule (Q), the nitrogen lone pair is involved in cross-conjugation with the carbonyl group, reducing its availability.
In molecule (P), the lone pair on nitrogen is directly conjugated with the carbonyl group (amide resonance), making it the least available.

Therefore, the correct basicity order is: $R > Q > P$ $R > Q > P$

Step 1: Final Identification

Basicity breakdown explanation diagram for Q26 — The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.

Basicity breakdown explanation diagram for Q26 — The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.

Basicity breakdown explanation diagram for Q26 — The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.

Comparing availability, structure R has localized electrons, Q has cross-conjugation, and P has standard amide resonance. Hence, option (4) is correct.

Pattern Recognition

Look for localized vs delocalized lone pairs on nitrogen. Bridgehead nitrogen lone pairs that violate Bredt's rule for double bond formation remain strictly localized, drastically increasing basicity compared to conjugated amides.

Chapter Mix

Class 11 Chemistry: Some Basic Principles of Organic Chemistry

SYSTEM

Syllabus Analysis & Trend Mapping

Dashboard Chapter Trends Formula Recognition Questions Feed Weightage

Rankbit System

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%)