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Organic Chemistry - Some Basic Principles and Techniques appeared 101 times across 3 years — 11.8% of Chemistry. This question is from Sigma and Pi Bond Counting.

Year 2026 2025 2024 Total
Questions 22 49 30 101

Total number of sigma (sigma) and pi(pi) bonds respectively present in hex-1-en-4-yne are:

Solution & Explanation

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 sigma bonds = 2 + 1 + 2 + 3 = 8
  • Number of C-C sigma bonds = 5
  • Total sigma bonds = 8 + 5 = 13.

    Sigma and Pi Bond Counting diagram for Q38 - JEE Main 2025 Evening
    Sigma and Pi Bond Counting diagram for Q38 - JEE Main 2025 Evening

  • Number of pi bonds: 1 from double bond + 2 from triple bond = 3 pi bonds.
Pattern Recognition

Every single bond is 1sigma, every double bond contains 1sigma + 1pi, and every triple bond contains 1sigma + 2pi.

Chapter Mix

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

Reference Study Guides

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

Q67 jee_main_2026_23_january_morning Methods of Purification
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 :
  • A. Statement-I is false but Statement-II is true.
  • B. Statement-I is true but Statement-II is false.
  • C. Both Statement-I and Statement-II are true.
  • D. Both Statement-I and Statement-II are false.

Solution

Core Logic

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)

Step 1: Final Conclusion

Both statements are fundamentally false based on basic thermodynamics and purification principles.

Pattern Recognition

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.

Chapter Mix

Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

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, the central positively charged carbon is attached to three methyl groups. This provides a total of 3 × 3 = 9 α-hydrogens, leading to nine hyperconjugation interactions. This extensively stabilizes the carbocation. Statement I is true.

Statement II: In the methyl carbocation, CH₃, there are zero adjacent carbon atoms, meaning there are ZERO α-hydrogens. Thus, no hyperconjugation interactions are possible in CH₃. The statement incorrectly claims there are three hyperconjugation interactions. Statement II is false.

Pattern Recognition

Always count α-hydrogens strictly from the carbon adjacent to the C center. CH₃ has hydrogens on the C 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%
  • B. 87.65%
  • C. 37.57%
  • D. 34.79%

Solution

Related Formula
% of Halogen = Atomic mass of HalogenMolecular mass of AgX × Mass of AgX formedMass of Organic Compound × 100
Core Logic

Given data: Mass of organic compound (w) = 0.2425 g Mass of AgCl precipitate (w₁) = 0.5253 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) × (0.5253)/(0.2425) × 100 % of Cl = 0.2474 × 2.166 × 100 % of Cl ≈ 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^- V. 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: Strong -M effect. Highly stabilizing. I. -Br: Weak +M effect, but prominent -I effect. Overall stabilizing relative to hydrogen. II. -H: (Plain benzyl anion) Neutral baseline. III. -OCH₃: Strong +M effect. Highly destabilizing. V. -CH₃: +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 is 15 mm).
Numerical Answer. Answer: 15 to 15

Solution

Related Formula
Pdry gas = Ptotal - Aqueous tension

PV = nRT

% N = Mass of N₂Mass of organic compound × 100
Core Logic

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

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

Temperature: T = 300 K

Step 1: Calculate moles and mass of Nitrogen

Using Ideal Gas Law, nN₂ = (PV)/(RT):

nN₂ = (((700)/(760)) × ((70)/(1000)))/(0.0821 × 300)

Mass of N₂ (WN₂) = nN₂ × 28

WN₂ = (700)/(760) × (70/1000)/(0.0821 × 300) × 28 ≈ 0.07324 g
Step 2: Calculate Percentage
% N = (0.07324)/(0.50) × 100 = 14.65 %

Rounding to the nearest integer, it is 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

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

Practice all Organic Chemistry - Some Basic Principles and Techniques previous-year questions →

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