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:

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. Remember that paper chromatography is a type of partition chromatography where moisture content (water) adsorbed on the paper serves as the stationary liquid phase. Class 11 Chemistry: Organic Chemistry - Some Basic Principles and 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

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. Remember that paper chromatography is a type of partition chromatography where moisture content (water) adsorbed on the paper serves as the stationary liquid phase. Class 1

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:

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. Remember that paper chromatography is a type of partition chromatography where moisture content (water) adsorbed on the paper serves as the stationary liquid phase. Class 11 Chemistry: Organic Chemistry - Some Basic Principles and 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

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. Remember that paper chromatography is a type of partition chromatography where moisture content (water) adsorbed on the paper serves as the stationary liquid phase. Class 1

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

Q42 jee_main_2025_02_april_morning Empirical Formula Derivation

On complete combustion

1.0~g

of an organic compound (X) gave

1.46~g

of

CO₂

and

0.567~g

of

H₂O

. The empirical formula mass of compound (X) is ________ g. Given molar mass in

g \cdot mol⁻¹ C:12, H:1, O:16

A. $(1) 30$ $(1)\ 30$
B. $(2) 45$ $(2)\ 45$
C. $(3) 60$ $(3)\ 60$
D. $(4) 15$ $(4)\ 15$

Solution

Related Formula

Elemental content calculation system equations:

Moles of C = Mass of CO₂44

Moles of H = 2 \times Mass of H₂O18

Core Logic

Let's perform the stoichiometry layout step-by-step:

Moles of $C$ $\mathrm{C}$ inside sample system:

nC = (1.46)/(44) = 0.033~mol

Mass of C = 0.033 \times 12 = 0.396~g

Moles of $H$ $\mathrm{H}$ inside sample system:

nH = 2 \times (0.567)/(18) = 0.063~mol

Mass of H = 0.063 \times 1 = 0.063~g

Determine Oxygen mass by subtracting values from total starting mass:

Mass of O = 1.0 - (0.396 + 0.063) = 0.541~g

nO = (0.541)/(16) = 0.033~mol

Find atomic whole-number ratio profile: $C : H : O = 0.033 : 0.063 : 0.033 \approx 1 : 2 : 1$ .
This gives an empirical configuration of $CH₂O$ $\mathrm{CH_2O}$.

Step 1: Evaluation

Calculating formula mass:

Empirical Mass = 12 + (2 \times 1) + 16 = 30~g

Pattern Recognition

When calculated mole properties output identical numbers for two elements ( $0.033$ $0.033$ for both C and O), their structural subscript ratio is exactly $1:1$ $1:1$. This pattern significantly speeds up empirical calculations.

Chapter Mix

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

Q jee_main_2025_03_april_evening IUPAC Nomenclature of Multi-substituted Benzenes

What is the correct IUPAC name of the compound given below?

Chemical structure for Q44 - JEE Main 2025 Evening — Substituted benzene derivative with carboxyl, hydroxyl, bromo, and nitro substituents.

A. 3-Bromo-2-hydroxy-5-nitrobenzoic acid
B. 3-Bromo-4-hydroxy-1-nitrobenzoic acid
C. 2-Hydroxy-3-bromo-5-nitrobenzoic acid
D. 5-Nitro-3-bromo-2-hydroxybenzoic acid

Solution

Related Formula

According to IUPAC rules for nomenclature of aromatic compounds:

Principal functional group has highest priority:

-COOH > -OH

The principal functional group carbon is designated as Carbon-1, and numbering is directed to give substituents the lowest possible locants.

Core Logic

Assign priority and number the ring:

Carbon-1: $-COOH$ $-\mathrm{COOH}$ (Carboxyl carbon, parent name 'benzoic acid')
Carbon-2: $-OH$ $-\mathrm{OH}$ (Hydroxyl substituent)
Carbon-3: $-Br$ $-\mathrm{Br}$ (Bromo substituent)
Carbon-5: $-NO₂$ $-\mathrm{NO}_2$ (Nitro substituent)

This numbering yields substituent locants at positions 2, 3, and 5.

Step 1: Arrange alphabetically

List the substituents alphabetically with locants:

3-Bromo
2-Hydroxy
5-Nitro

Combining these names:

3-Bromo-2-hydroxy-5-nitrobenzoic acid

This matches Option (1).

Pattern Recognition

Carboxylic acid always dictates position 1 in ring numbering over alcohol. Numbering clockwise gives 2-hydroxy, 3-bromo, and 5-nitro, whereas counterclockwise numbering would yield much higher locants (2-nitro, 4-bromo, 5-hydroxy) which violates the lowest-locant rule.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Q jee_main_2025_03_april_evening Stoichiometry of Nitration

X~g

of nitrobenzene on nitration gave

4.2~g

of m-dinitrobenzene. The value of

X

is ________

g

. (nearest integer) [Given: molar mass (in

g~mol⁻¹

) C: 12, H: 1, O: 16, N: 14]

Numerical Answer. Answer: 3 to 3

Solution

Related Formula

Balanced reaction for nitration of nitrobenzene:

C₆H₅NO₂ + HNO₃ arrow C₆H₄(NO₂)₂ + H₂O

Moles = MassMolar Mass

Core Logic

From the balanced stoichiometry:

1 mole of nitrobenzene yields 1 mole of m-dinitrobenzene.

Step 1: Determine molar masses

Molar mass of Nitrobenzene ( $C₆H₅NO₂$ $\mathrm{C_6H_5NO_2}$):

M₁ = 6(12) + 5(1) + 14 + 2(16) = 72 + 5 + 14 + 32 = 123~g/mol

Molar mass of m-Dinitrobenzene ( $C₆H₄(NO₂)₂$ $\mathrm{C_6H_4(NO_2)_2}$):

M₂ = 6(12) + 4(1) + 2(14) + 4(16) = 72 + 4 + 28 + 64 = 168~g/mol

Stoichiometry of Nitration Step 2: Calculate moles and find X Moles of m-dinitrobenzene produced:

n = 4.2~g168~g/mol = 0.025~mol

Since stoichiometry is$ Since stoichiometry is $1:1, the moles of nitrobenzene required is also$, the moles of nitrobenzene required is also $0.025\mathrm{~mol}:

Mass of nitrobenzene X = 0.025~mol \times 123~g/mol = 3.075~g

Rounding to the nearest integer gives $ Rounding to the nearest integer gives $3$. Pattern Recognition Electrophilic aromatic substitution stoichiometry is straightforward: each aromatic precursor ring converts to exactly one product ring. Finding moles from the heavier substituted product and converting back using the reactant's molecular weight quickly yields the answer. Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Amines

Q jee_main_2025_03_april_evening Isomerism in Benzene Derivatives

The total number of structural isomers possible for the substituted benzene derivatives with the molecular formula

C₉H₁₂

is ________.

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

Degrees of Unsaturation (Double Bond Equivalents, DBE):

DBE = C + 1 - (H)/(2) + (N)/(2)

For formula $C₉H₁₂$ $\mathrm{C_9H_{12}}$:

DBE = 9 + 1 - (12)/(2) = 4

These 4 degrees of unsaturation match a benzene ring exactly (one ring + three double bonds).

Core Logic

Since the question specifies 'substituted benzene derivatives', we must keep the benzene core ( $C₆H₅-$ $\mathrm{C_6H_5-}$ or similar) intact. This leaves 3 carbon atoms to be distributed as alkyl substituents.

Step 1: Categorize by substitution patterns

Mono-substituted benzene (one propyl group containing 3 carbons):
n-Propylbenzene: $C₆H₅-CH₂-CH₂-CH₃$ $\mathrm{C_6H_5-CH_2-CH_2-CH_3}$ (Isomer 1)
Isopropylbenzene (Cumene): $C₆H₅-CH(CH₃)₂$ $\mathrm{C_6H_5-CH(CH_3)_2}$ (Isomer 2)
Di-substituted benzene (one ethyl group and one methyl group):
1-Ethyl-2-methylbenzene (ortho-ethylmethylbenzene) (Isomer 3)
1-Ethyl-3-methylbenzene (meta-ethylmethylbenzene) (Isomer 4)
1-Ethyl-4-methylbenzene (para-ethylmethylbenzene) (Isomer 5)

Step 2: Tri-substituted benzenes

Tri-substituted benzene (three methyl groups):
1,2,3-Trimethylbenzene (Hemimellitene) (Isomer 6)
1,2,4-Trimethylbenzene (Pseudocumene) (Isomer 7)
1,3,5-Trimethylbenzene (Mesitylene) (Isomer 8)

Step 3: Total Count

Summing all options:

Total structural isomers = 2 + 3 + 3 = 8

Pattern Recognition

For alkyl benzenes with $N$ $N$ extra carbons, systematically group them as single chain substituents down to multiple methyl substituents. This hierarchical sorting prevents duplicates or missing patterns.

Chapter Mix

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

Q36 jee_main_2025_03_april_evening Dumas' Method for Nitrogen Estimation

In Dumas' method for estimation of nitrogen

0.4~g

of an organic compound gave

60~mL

of nitrogen collected at

300~K

temperature and

715~mm~Hg

pressure. The percentage composition of nitrogen in the compound is : (Given: Aqueous tension at

300~K = 15~mm~Hg

)

A. 15.71%
B. 20.95%
C. 17.46%
D. 7.85%

Solution

Related Formula

Pressure of dry nitrogen gas:

PN₂ = Ptotal - Aqueous tension

Using Ideal Gas Law:

nN₂ = PN₂ VR T

%N = Mass of nitrogenMass of organic compound \times 100

Core Logic

Given parameters:

Mass of compound $m = 0.4~g$ $m = 0.4\mathrm{~g}$
Volume of nitrogen $V = 60~mL = 0.060~L$ $V = 60\mathrm{~mL} = 0.060\mathrm{~L}$
Total pressure $Ptotal = 715~mm~Hg$ $P_{\text{total}} = 715\mathrm{~mm~Hg}$
Temperature $T = 300~K$ $T = 300\mathrm{~K}$
Aqueous tension $= 15~mm~Hg$ $= 15\mathrm{~mm~Hg}$

Step 1: Calculate dry nitrogen pressure

PN₂ = 715~mm~Hg - 15~mm~Hg = 700~mm~Hg

PN₂ = (700)/(760)~atm \approx 0.921~atm

Step 2: Calculate moles of nitrogen gas Using$ Step 2: Calculate moles of nitrogen gas Using $R = 0.0821\mathrm{~L\cdot atm\cdot K^{-1}\cdot mol^{-1}}:

nN₂ = (((700)/(760)) \times 0.060)/(0.0821 \times 300) = (0.05526)/(24.63) \approx 2.2436 \times 10⁻³~mol

Mass of$ Mass of $\mathrm{N}_2gas:

Mass = 2.2436 \times 10⁻³ \times 28~g \approx 0.06282~g

Step 3: Calculate percentage of Nitrogen

\%\mathrm{N} = \frac{0.06282\mathrm{~g}}{0.4\mathrm{~g}} \times 100 \approx 15.71\%$$

This matches Option (1).

Pattern Recognition

In Dumas' method calculations, always subtract the aqueous tension to obtain the pressure of dry nitrogen gas. Do not use the raw moist gas pressure, as doing so will overestimate the nitrogen content.

Chapter Mix

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

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Step 1: Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Arrange alphabetically

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine molar masses

Step 2: Calculate moles and find X

Step 2: Calculate moles and find X

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Categorize by substitution patterns

Step 2: Tri-substituted benzenes

Step 3: Total Count

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate dry nitrogen pressure

Step 2: Calculate moles of nitrogen gas

Step 2: Calculate moles of nitrogen gas

Step 3: Calculate percentage of Nitrogen

Step 3: Calculate percentage of Nitrogen

Pattern Recognition

Chapter Mix

More Organic Chemistry - Some Basic Principles and Techniques Questions — jee_main_2025_29_jan_evening

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