Given below are two statements: Statement (I): In partition chromatography, stationary phase is thin film of liquid…

Year	2026	2025	2024	Total
Questions	19	49	25	93

Year

2026

2025

2024

Total

Questions

Solution & Explanation

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

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 8

Q38 jee_main_2025_08_april_evening Quantitative Elemental Analysis

On complete combustion,

0.210 g

of an organic compound containing C, H, and O yielded

0.127 g

H₂O

and

0.307 g

CO₂

. The mass percentages of hydrogen and oxygen in the given organic compound respectively are:

A. 53.41, 39.6
B. 6.72, 53.41
C. 7.55, 43.85
D. 6.72, 39.87

Solution

Related Formula

Percentage of Hydrogen in organic analysis:

%H = (2)/(18) \times Mass of H₂OMass of Compound \times 100

Percentage of Carbon:

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

Percentage of Oxygen:

%O = 100 - (%C + %H)

Execution

Step 1: Compute the mass percent of Hydrogen:

%H = (2)/(18) \times (0.127)/(0.210) \times 100 = (0.254)/(3.78) \approx 6.72%

Step 2: Compute the mass percent of Carbon:

%C = (12)/(44) \times (0.307)/(0.210) \times 100 = (3.684)/(9.24) \approx 39.87%

Step 3: Deduce the remaining mass percent of Oxygen:

%O = 100 - (39.87 + 6.72) = 100 - 46.59 = 53.41%

Thus, the values of hydrogen and oxygen percentage are $6.72%$ $6.72\%$ and $53.41%$ $53.41\%$, matches with Option (2).

Pattern Recognition

Always focus on the order requested by the question stem. The query specifies 'hydrogen and oxygen respectively'. Option 2 and Option 4 both show these numbers but reversed—verifying the targeted sequence protects your score line.

Chapter Mix

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

Q42 jee_main_2025_08_april_evening Qualitative Analysis of Functional Groups

Match the reagents in LIST-I with the corresponding chemical functional groups they detect in LIST-II:

LIST-I (Reagent)	LIST-II (Functional Group detected)
A. Sodium bicarbonate solution	I. double bond / unsaturation
B. Neutral ferric chloride	II. carboxylic acid
C. Ceric ammonium nitrate	III. phenolic - OH
D. Alkaline $KMnO₄$ $\text{KMnO}_4$	IV. alcoholic - OH

Choose the correct answer from the options given below:

A. $A-II, B-III, C-IV, D-I$ $\text{A-II, B-III, C-IV, D-I}$
B. $A-II, B-III, C-I, D-IV$ $\text{A-II, B-III, C-I, D-IV}$
C. $A-III, B-II, C-IV, D-I$ $\text{A-III, B-II, C-IV, D-I}$
D. $A-II, B-IV, C-III, D-I$ $\text{A-II, B-IV, C-III, D-I}$

Solution

Core Logic

Let us review the chemical basis for each qualitative test:

A. Sodium bicarbonate ( $NaHCO₃$ $\text{NaHCO}_3$) solution: Carboxylic acids are sufficiently acidic to decompose $NaHCO₃$ $\text{NaHCO}_3$, liberating carbon dioxide gas observed as vigorous effervescence. Therefore, $A arrow II$ $\text{A} \rightarrow \text{II}$.
B. Neutral ferric chloride ( $FeCl₃$ $\text{FeCl}_3$): Phenols react with neutral $FeCl₃$ $\text{FeCl}_3$ solution to form characteristic deeply colored violet coordination complexes. Therefore, $B arrow III$ $\text{B} \rightarrow \text{III}$.
C. Ceric ammonium nitrate (CAN): Alcohols react with CAN reagent to cause a distinct color shift to deep dark red due to complexation. Therefore, $C arrow IV$ $\text{C} \rightarrow \text{IV}$.
D. Alkaline $KMnO₄$ $\text{KMnO}_4$ (Baeyer's Reagent): Reacts readily via syn-hydroxylation across carbon-carbon double/triple bonds, resulting in decolored solutions alongside brown $MnO₂$ $\text{MnO}_2$ precipitates. This detects unsaturation. Therefore, $D arrow I$ $\text{D} \rightarrow \text{I}$.

Step 1: Assembly

Combining the validated relationships gives:

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

This maps perfectly to Option (1).

Pattern Recognition

Baeyer's test (alkaline $KMnO₄$ $\text{KMnO}_4$) always tests for alkenes/alkynes. $NaHCO₃$ $\text{NaHCO}_3$ is unique for acidic groups like carboxylic acids. Matching these two reliable pairs isolates the correct option without needing to review the entire table.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Alcohols, Phenols and Ethers

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

Solution & Explanation

Core Logic

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Chapter Mix

Reference Study Guides

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 8

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Step 1: Assembly

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Core Logic

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Related Formula

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More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 8

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