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

Q55 jee_main_2026_28_january_morning Separation Techniques

Method used for separation of mixture of products (B and C) obtained in the following reaction is:

Reaction yielding mixture B and C — Sequence of reactions on benzene yielding ortho and para substituted isomers.

A. Simple distillation
B. sublimation
C. steam distillation
D. fractional distillation

Solution

Core Logic

The reaction of Benzene with $Br₂ / FeBr₃$ $Br_2 / FeBr_3$ yields Bromobenzene (A). Nitration of Bromobenzene ( $conc. HNO₃ / conc. H₂SO₄$ $conc.\; HNO_3 / conc.\; H_2SO_4$) yields a mixture of ortho-bromonitrobenzene (B) and para-bromonitrobenzene (C).

Separation of Ortho and Para isomers — Sequence of reactions on benzene yielding ortho and para substituted isomers.

Step 1: Justification of Separation Technique

The ortho and para isomers of bromonitrobenzene have differing boiling points but not sufficiently far apart for simple distillation. Thus, Fractional Distillation is the appropriate method to separate these positional isomers accurately based on slight differences in boiling points.

Pattern Recognition

Mixtures of isomeric organic liquids (like ortho/para derivatives) that differ mildly in boiling points are classically separated by fractional distillation.

Chapter Mix

Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques Class 12 Chemistry: Haloalkanes and Haloarenes

Q74 jee_main_2026_28_january_morning Quantitative Analysis of Organic Compounds

0.53~g

of an organic compound (x) when heated with excess of nitric acid (concentrated) and then with silver nitrate gave

0.75~g

of silver bromide precipitate.

1.0~g

of (x) gave

1.32~g

of

CO₂

gas on combustion. The percentage of hydrogen in the compound (x) is _____ %. [Nearest Integer] [Given : Molar mass in

g~mol⁻¹

H : 1, C : 12, Br : 80, Ag : 108, O : 16; Compound (x) :

CₓHyBrz

]

Numerical Answer. Answer: 4 to 4

Solution

Step 1: Calculate Percentage of Carbon

$1.0~g$ $1.0\mathrm{~g}$ of compound gives $1.32~g$ $1.32\mathrm{~g}$ of $CO₂$ $\mathrm{CO}_2$.\nMoles of $CO₂ = (1.32)/(44) = 0.03~mol$ .\nMass of Carbon $= 0.03 \times 12 = 0.36~g$ $= 0.03 \times 12 = 0.36\mathrm{~g}$.\n $Percentage of C = ((0.36)/(1.0)) \times 100 = 36%$ .

Step 2: Calculate Percentage of Bromine

$0.53~g$ $0.53\mathrm{~g}$ of compound gives $0.75~g$ $0.75\mathrm{~g}$ of $AgBr$ $\mathrm{AgBr}$ (Molar mass = $108 + 80 = 188~g/mol$ $108 + 80 = 188\mathrm{~g/mol}$).\n $Mass of Br = ((80)/(188)) \times 0.75 \approx 0.319~g$ .\n $Percentage of Br = ((0.319)/(0.53)) \times 100 = 60.2%$ .

Step 3: Calculate Percentage of Hydrogen

The compound strictly consists of C, H, and Br (as formula is given as $CₓHyBrz$ $\mathrm{C}_x\mathrm{H}_y\mathrm{Br}_z$).\n $Percentage of H = 100 - (%C + %Br)$ \n $Percentage of H = 100 - (36 + 60.2) = 100 - 96.2 = 3.8%$ .\nRounding off to the nearest integer, we get $4%$ $4\%$.

Pattern Recognition

For basic gravimetric elemental analysis: always map precipitate mass back to elemental mass using strict molar mass ratios, then find mass fractions, ensuring everything sums to 100%.

Chapter Mix

Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

Q60 jee_main_2026_28_january_evening Hyperconjugation

The cyclic cations having the same number of hyperconjugation are : A.

B.

C.

D.

Choose the correct answer from the options given below :

A. $(1) A and C Only$ $(1)\text{ A and C Only}$
B. $(2) B and C Only$ $(2)\text{ B and C Only}$
C. $(3) A and B Only$ $(3)\text{ A and B Only}$
D. $(4) A, C and D only$ $(4)\text{ A, C and D only}$

Solution

Core Logic

Count the number of alpha hydrogens ( $α-H$ $\alpha-H$) adjacent to the carbocation in each structure. (A)

α-H = 6

(B)

α-H = 7

(C)

α-H = 6

(D)

α-H = 5

Step 1: Final Conclusion

Both cations (A) and (C) have 6 $α$ $\alpha$-hydrogens, meaning they share the same number of hyperconjugative structures.

Pattern Recognition

Hyperconjugation count corresponds strictly to the number of $C-H$ $C-H$ bonds on the carbon atoms immediately adjacent ( $sp³$ $sp^3$) to the positive center.

Chapter Mix

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

Q70 jee_main_2026_28_january_evening Quantitative Analysis Of Elements

A student has been given

0.314 g

of an organic compound and asked to estimate Sulphur. During the experiment, the student has obtained

0.4813 g

of barium sulphate. The percentage of sulphur present in the compound is (Given Molar mass in

g mol⁻¹

S:32

,

BaSO₄:233

)

A. $(1) 42.10%$ $(1)\text{ 42.10\%}$
B. $(2) 63.15%$ $(2)\text{ 63.15\%}$
C. $(3) 21.05%$ $(3)\text{ 21.05\%}$
D. $(4) 48.24%$ $(4)\text{ 48.24\%}$

Solution

Related Formula

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

Core Logic

Mass of organic compound = $0.314 g$ $0.314\text{ g}$ Mass of $BaSO₄$ $BaSO_4$ formed = $0.4813 g$ $0.4813\text{ g}$ Substituting the values:

% S = (32)/(233) \times (0.4813)/(0.314) \times 100

% S = 0.1373 \times 1.5328 \times 100 \approx 21.052%

Step 1: Final Conclusion

The percentage of sulphur is approximately $21.05%$ $21.05\%$.

Pattern Recognition

Straight application of Carius method for Sulphur estimation formula.

Chapter Mix

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

Q jee_main_2025_02_april_evening Hybridization in Organic Compounds

In 3, 3-dimethylhex-1-ene-4-yne, there are

sp³

,

sp²

and

sp

hybridised carbon atoms respectively:

A. $4, 2, 2$ $4, 2, 2$
B. $3, 3, 2$ $3, 3, 2$
C. $2, 4, 2$ $2, 4, 2$
D. $2, 2, 4$ $2, 2, 4$

Solution

Related Formula

Hybridization of Carbon = cases sp³ & 4~σ-bonds sp² & 3~σ-bonds, 1~π-bond sp & 2~σ-bonds, 2~π-bonds cases

Core Logic

Let's first draw the structural formula of 3,3-dimethylhex-1-ene-4-yne:

6CH₃ - 5C \equiv 4C - 3C(CH₃)₂ - 2CH = 1CH₂

Step 1: Identify Hybridization of Each Carbon

We count the sigma ( $σ$ $\sigma$) bonds or pi ( $π$ $\pi$) bonds on each carbon:

$C₁$ $\mathrm{C}_1$: Involved in a double bond ( $CH₂ =$ $\mathrm{CH_2 =}$) $sp²$ $\implies \mathrm{sp^2}$
$C₂$ $\mathrm{C}_2$: Involved in a double bond ( $=CH-$ $=\mathrm{CH-}$) $sp²$ $\implies \mathrm{sp^2}$
$C₃$ $\mathrm{C}_3$: Single bonds only (bonded to $C₂$ $\mathrm{C}_2$, $C₄$ $\mathrm{C}_4$, and two methyl carbons) $sp³$ $\implies \mathrm{sp^3}$
Two methyl carbons attached to $C₃$ $\mathrm{C}_3$: Single bonds only $2 \times sp³$ $\implies 2 \times \mathrm{sp^3}$
$C₄$ $\mathrm{C}_4$: Involved in a triple bond ( $-C \equiv$ $-\mathrm{C} \equiv$) $sp$ $\implies \mathrm{sp}$
$C₅$ $\mathrm{C}_5$: Involved in a triple bond ( $\equiv C-$ $\equiv \mathrm{C}-$) $sp$ $\implies \mathrm{sp}$
$C₆$ $\mathrm{C}_6$: Single bonds only ( $-CH₃$ $-\mathrm{CH_3}$) $sp³$ $\implies \mathrm{sp^3}$

Step 2: Calculate the Count

Summing the hybridization counts:

$sp³$ $\mathrm{sp^3}$ carbons: $C₃$ $\mathrm{C}_3$, $C₆$ $\mathrm{C}_6$, and $2 \times CH₃$ $2 \times \mathrm{CH_3}$ on $C₃$ $\mathrm{C}_3$ $= 4$ $= 4$ carbon atoms
$sp²$ $\mathrm{sp^2}$ carbons: $C₁$ $\mathrm{C}_1$ and $C₂$ $\mathrm{C}_2$ $= 2$ $= 2$ carbon atoms
$sp$ $\mathrm{sp}$ carbons: $C₄$ $\mathrm{C}_4$ and $C₅$ $\mathrm{C}_5$ $= 2$ $= 2$ carbon atoms

Thus, the number of $sp³$ $\mathrm{sp^3}$, $sp²$ $\mathrm{sp^2}$ and $sp$ $\mathrm{sp}$ hybridized carbons is $4, 2, 2$ $4, 2, 2$ respectively.

Pattern Recognition

Quick Tip: Always draw side substituents (like methyl groups) explicitly. A common mistake is to skip counting the methyl substituent carbons as $sp³$ $\mathrm{sp^3}$.

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 4

Solution

Core Logic

Step 1: Justification of Separation Technique

Pattern Recognition

Chapter Mix

Solution

Step 1: Calculate Percentage of Carbon

Step 2: Calculate Percentage of Bromine

Step 3: Calculate Percentage of Hydrogen

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Identify Hybridization of Each Carbon

Step 2: Calculate the Count

Pattern Recognition

Chapter Mix

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

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