Given below are two statements: Statement-I : Sublimation is used for the separation and purification of compounds with low melting point. Statement-II : The boiling point of a liquid increases as the external pressure is reduced. In the light of the above statements, choose the correct answer from the options given below :

Assess theoretical principles of purification processes. Statement-I: Sublimation is a process used for separating sublimable compounds from non-sublimable impurities. It does not strictly depend on a 'low melting point'. Sublimable solids bypass the liquid phase altogether when heated. (False) Statement-II: The boiling point of a liquid is the temperature at which its vapor pressure equals the external atmospheric pressure. If external pressure is reduced, the liquid needs less vapor pressure (and thus lower temperature) to boil. Hence, boiling point decreases with reduced external pressure. (False) Both statements are fundamentally false based on basic thermodynamics and purification principles. Lower pressure = Lower boiling point (used in vacuum distillation). Sublimation relies on vapor pressure of solid overcoming external pressure without melting, independent of specifically 'low' melting points. Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

Given below are two statements: Statement-I : Sublimation is used for the separation and purification of compounds with low melting point. Statement-II : The boiling point of a liquid increases as the

Assess theoretical principles of purification processes. Statement-I: Sublimation is a process used for separating sublimable compounds from non-sublimable impurities. It does not strictly depend on a 'low melting point'. Sublimable solids bypass the liquid phase altogether when heated. (False) Statement-II: The boiling point of a liquid is the temperature at which its vapor pressure equals the external atmospheric pressure. If external pressure is reduced, the liquid needs less vapor pressure (

Given below are two statements: Statement-I : Sublimation is used for the separation and purification of compounds with low melting point. Statement-II : The boiling point of a liquid increases as the external pressure is reduced. In the light of the above statements, choose the correct answer from the options given below :

Assess theoretical principles of purification processes. Statement-I: Sublimation is a process used for separating sublimable compounds from non-sublimable impurities. It does not strictly depend on a 'low melting point'. Sublimable solids bypass the liquid phase altogether when heated. (False) Statement-II: The boiling point of a liquid is the temperature at which its vapor pressure equals the external atmospheric pressure. If external pressure is reduced, the liquid needs less vapor pressure (and thus lower temperature) to boil. Hence, boiling point decreases with reduced external pressure. (False) Both statements are fundamentally false based on basic thermodynamics and purification principles. Lower pressure = Lower boiling point (used in vacuum distillation). Sublimation relies on vapor pressure of solid overcoming external pressure without melting, independent of specifically 'low' melting points. Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

Given below are two statements: Statement-I : Sublimation is used for the separation and purification of compounds with low melting point. Statement-II : The boiling point of a liquid increases as the

Assess theoretical principles of purification processes. Statement-I: Sublimation is a process used for separating sublimable compounds from non-sublimable impurities. It does not strictly depend on a 'low melting point'. Sublimable solids bypass the liquid phase altogether when heated. (False) Statement-II: The boiling point of a liquid is the temperature at which its vapor pressure equals the external atmospheric pressure. If external pressure is reduced, the liquid needs less vapor pressure (

Given below are two statements: Statement-I : Sublimation is used for the separation and purification of compounds with…

Year	2026	2025	2024	Total
Questions	22	49	30	101

More Organic Chemistry Some Basic Principles and Techniques Previous-Year Questions — Page 3

Q57 jee_main_2026_23_january_evening Carbocation Stability and Hyperconjugation

Given below are two statements: Statement I:

(CH₃)₃C

is more stable than

CH₃

as nine hyperconjugation interactions are possible in

(CH₃)₃C

. Statement II:

CH₃

is less stable than

(CH₃)₃C

as only three hyperconjugation interactions are possible in

CH₃

. In the light of the above statements, choose the correct answer from the options given below.

A. $Statement I is true but Statement II is false$
B. $Both Statement I and Statement II are true$
C. $Both Statement I and Statement II are false$
D. $Statement I is false but Statement II is true$

Solution

Related Formula

$Number of hyperconjugation structures = Number of α-hydrogens$

Core Logic

Statement I: In the tert-butyl carbocation, $(CH₃)₃C$ $(CH_{3})_{3}C^{\oplus}$, the central positively charged carbon is attached to three methyl groups. This provides a total of $3 \times 3 = 9$ $3 \times 3 = 9$ $α$ $\alpha$-hydrogens, leading to nine hyperconjugation interactions. This extensively stabilizes the carbocation. Statement I is true.

Statement II: In the methyl carbocation, $CH₃$ $CH_{3}^{\oplus}$, there are zero adjacent carbon atoms, meaning there are ZERO $α$ $\alpha$-hydrogens. Thus, no hyperconjugation interactions are possible in $CH₃$ $CH_{3}^{\oplus}$. The statement incorrectly claims there are three hyperconjugation interactions. Statement II is false.

Pattern Recognition

Always count $α$ $\alpha$-hydrogens strictly from the carbon adjacent to the $C$ $C^{\oplus}$ center. $CH₃$ $CH_3^{\oplus}$ has hydrogens on the $C$ $C^{\oplus}$ itself, not on an adjacent alpha carbon, hence 0 hyperconjugative structures.

Chapter Mix

Class 11 Chemistry: General Organic Chemistry

Q65 jee_main_2026_23_january_evening Quantitative Analysis

In Carius method

0.2425 g

of an organic compounds gave

0.5253 g

silver chloride. The percentage of chlorine in the organic compound is

A. $53.58%$ $53.58\%$
B. $87.65%$ $87.65\%$
C. $37.57%$ $37.57\%$
D. $34.79%$ $34.79\%$

Solution

Related Formula

% of Halogen = Atomic mass of HalogenMolecular mass of AgX \times Mass of AgX formedMass of Organic Compound \times 100

Core Logic

Given data: Mass of organic compound ( $w$ $w$) = $0.2425 g$ $0.2425 \text{ g}$ Mass of AgCl precipitate ( $w₁$ $w_1$) = $0.5253 g$ $0.5253 \text{ g}$ Molar mass of AgCl = $108 (Ag) + 35.5 (Cl) = 143.5 g/mol$

Step 1: Perform Calculation

Substitute the values into the Carius formula:

% of Cl = (35.5)/(143.5) \times (0.5253)/(0.2425) \times 100

% of Cl = 0.2474 \times 2.166 \times 100

% of Cl \approx 53.58%

Pattern Recognition

Carius estimation is a direct plug-and-play stoichiometric ratio calculation. 1 mole of AgCl contains exactly 1 mole of Cl atoms.

Chapter Mix

Class 11 Chemistry: General Organic Chemistry

Q69 jee_main_2026_24_january_morning Stability of Carbanions

Arrange the following carbanions in the decreasing order of stability I.

p-Br-C₆H₄-CH₂^-

II.

C₆H₅-CH₂^-

III.

p-CH₃O-C₆H₄-CH₂^-

IV.

p-CHO-C₆H₄-CH₂^-

p-CH₃-C₆H₄-CH₂^-

Choose the correct answer from the options given below :

A. $I > II > IV > V > III$
B. $I > IV > II > V > III$
C. $IV > I > II > V > III$
D. $IV > II > I > III > V$

Solution

Core Logic

The stability of a carbanion increases when electron-withdrawing groups (EWG) are present, as they help disperse the negative charge through -I or -M effects. Electron-donating groups (EDG) decrease stability by intensifying the negative charge through +I or +M effects.

Evaluating the para-substituents: IV. $-CHO$ $-\text{CHO}$: Strong -M effect. Highly stabilizing. I. $-Br$ $-\text{Br}$: Weak +M effect, but prominent -I effect. Overall stabilizing relative to hydrogen. II. $-H$ $-\text{H}$: (Plain benzyl anion) Neutral baseline. III. $-OCH₃$ $-\text{OCH}_3$: Strong +M effect. Highly destabilizing. V. $-CH₃$ $-\text{CH}_3$: +I and +H (hyperconjugation) effects. Destabilizing, but less so than strong +M.

Step 1: Final Conclusion

Based on the effects, the stability order is: $-CHO (-M) > -Br (-I) > -H > -CH₃ (+I, +H) > -OCH₃ (+M)$ Therefore: $IV > I > II > V > III$ .

Pattern Recognition

For carbanions, think: "EWG stabilizes, EDG destabilizes". This is the exact opposite of carbocation stability rules.

Chapter Mix

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

Q74 jee_main_2026_24_january_morning Quantitative Analysis Dumas Method

In Dumas method for estimation of nitrogen,

0.50 g

of an organic compound gave

70 mL

of nitrogen collected at

300 K

and

715 mm

pressure. The percentage of nitrogen in the organic compound is ____% (Aqueous tension at

300 K

15 mm

Numerical Answer. Answer: 15 to 15

Solution

Related Formula

Pdry gas = Ptotal - Aqueous tension

$PV = nRT$ $PV = nRT$

% N = Mass of N₂Mass of organic compound \times 100

Core Logic

Pressure of dry $N₂$ $N_2$ gas: $PN₂ = (715 - 15) mm = 700 mm Hg = (700)/(760) atm$

Volume of $N₂$ $N_2$ gas: $VN₂ = 70 mL = (70)/(1000) L$

Temperature: $T = 300 K$ $T = 300 \text{ K}$

Step 1: Calculate moles and mass of Nitrogen

Using Ideal Gas Law, $nN₂ = (PV)/(RT)$ $n_{N_2} = \frac{PV}{RT}$:

nN₂ = (((700)/(760)) \times ((70)/(1000)))/(0.0821 \times 300)

Mass of $N₂$ $N_2$ ( $WN₂$ $W_{N_2}$) = $nN₂ \times 28$ $n_{N_2} \times 28$

WN₂ = (700)/(760) \times (70/1000)/(0.0821 \times 300) \times 28 \approx 0.07324 g

Step 2: Calculate Percentage

% N = (0.07324)/(0.50) \times 100 = 14.65 %

Rounding to the nearest integer, it is $15 %$ $15 \%$.

Pattern Recognition

Always subtract aqueous tension from total pressure before plugging into the ideal gas law. Alternatively, convert volume to STP directly using $(P₁V₁)/T₁ = (PSTPVSTP)/TSTP$ .

Chapter Mix

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

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Organic Chemistry Some Basic Principles and Techniques Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Perform Calculation

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate moles and mass of Nitrogen

Step 2: Calculate Percentage

Pattern Recognition

Chapter Mix

More Organic Chemistry Some Basic Principles and Techniques Questions — jee_main_2026_23_january_morning

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