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

Year	2026	2025	2024	Total
Questions	22	49	30	101

Year

2026

2025

2024

Total

Questions

101

Solution & Explanation

Related Formula

Moles of Compound X = Moles of N₂

VN₂ (at STP) = nN₂ \times 22400~mL

Core Logic

From the molecular structure of compound X, its molecular formula is $C₅H₁₀N₂$ $\mathrm{C_5H_{10}N_2}$.

Molar Mass of X $= (5 \times 12) + (10 \times 1) + (2 \times 14) = 60 + 10 + 28 = 86 ~g~mol⁻¹$ .

Step 1: Calculation

Moles of compound X in $0.42~g$ $0.42\mathrm{~g}$:

nX = (0.42)/(86)~mol

By Conservation of Atomic Mass (POAC) on Nitrogen atoms:

2 \times nX = 2 \times nN₂ nN₂ = nX = (0.42)/(86)~mol

Volume of $N₂$ $\mathrm{N_2}$ liberated at STP:

VN₂ = (0.42)/(86) \times 22400~mL \approx 109.395~mL

Using $22.7~L$ $22.7\mathrm{~L}$ standard STP molar volume: $(0.42)/(86) \times 22700 \approx 110.88~mL \approx 111~mL$ .

Pattern Recognition

Dumas method volume calculation: $V = mMcompound \times 22.4 L/mol \times 1000 mL = 111 mL$ .

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

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

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Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Match with the Question Layout

Pattern Recognition

Chapter Mix

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Core Logic

Pattern Recognition

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Core Logic

Pattern Recognition

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More Organic Chemistry - Some Basic Principles and Techniques Questions — jee_main_2025_03_april_morning

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