Organic Chemistry - Some Basic Principles and Techniques Previous Year Questions — JEE Main Chemistry
101 past-year Organic Chemistry - Some Basic Principles and Techniques questions from JEE Main (Chemistry) — page 19.
Q57 (2026)
Given below are two statements:
Statement I: $(CH_{3})_{3}C^{\oplus}$ is more stable than $CH_{3}^{\oplus}$ as nine hyperconjugation interactions are possible in $(CH_{3})_{3}C^{\oplus}$.
Statement II: $CH_{3}^{\oplus}$ is less stable than $(CH_{3})_{3}C^{\oplus}$ as only three hyperconjugation interactions are possible in $CH_{3}^{\oplus}$.
In the light of the above statements, choose the correct answer from the options given below.
- $\text{Statement I is true but Statement II is false}$
- $\text{Both Statement I and Statement II are true}$
- $\text{Both Statement I and Statement II are false}$
- $\text{Statement I is false but Statement II is true}$
### Related Formula
$\text{Number of hyperconjugation structures} = \text{Number of } \alpha\text{-hydrogens}$
### Core Logic
Statement I: In the tert-butyl carbocation, $(CH_{3})_{3}C^{\oplus}$, the central positively charged carbon is attached to three methyl groups. This provides a total of $3 \times 3 = 9$ $\alpha$-hydrogens, leading to nine hyperconjugation interactions. This extensively stabilizes the carbocation. Statement I is true.
Statement II: In the methyl carbocation, $CH_{3}^{\oplus}$, there are zero adjacent carbon atoms, meaning there are ZERO $\alpha$-hydrogens. Thus, no hyperconjugation interactions are possible in $CH_{3}^{\oplus}$. The statement incorrectly claims there are three hyperconjugation interactions. Statement II is false.
### Pattern Recognition
Always count $\alpha$-hydrogens strictly from the carbon adjacent to the $C^{\oplus}$ center. $CH_3^{\oplus}$ has hydrogens on the $C^{\oplus}$ itself, not on an adjacent alpha carbon, hence 0 hyperconjugative structures.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: General Organic Chemistry
Q65 (2026)
In Carius method $0.2425 \text{ g}$ of an organic compounds gave $0.5253 \text{ g}$ silver chloride. The percentage of chlorine in the organic compound is
- $53.58\%$
- $87.65\%$
- $37.57\%$
- $34.79\%$
### Related Formula
$$\% \text{ of Halogen} = \frac{\text{Atomic mass of Halogen}}{\text{Molecular mass of AgX}} \times \frac{\text{Mass of AgX formed}}{\text{Mass of Organic Compound}} \times 100$$
### Core Logic
Given data:
Mass of organic compound ($w$) = $0.2425 \text{ g}$
Mass of AgCl precipitate ($w_1$) = $0.5253 \text{ g}$
Molar mass of AgCl = $108 (\text{Ag}) + 35.5 (\text{Cl}) = 143.5 \text{ g/mol}$
### Step 1: Perform Calculation
Substitute the values into the Carius formula:
$$\% \text{ of Cl} = \frac{35.5}{143.5} \times \frac{0.5253}{0.2425} \times 100$$
$$\% \text{ of Cl} = 0.2474 \times 2.166 \times 100$$
$$\% \text{ of Cl} \approx 53.58\%$$
### Pattern Recognition
Carius estimation is a direct plug-and-play stoichiometric ratio calculation. 1 mole of AgCl contains exactly 1 mole of Cl atoms.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: General Organic Chemistry
Q69 (2026)
Arrange the following carbanions in the decreasing order of stability
I. $p-\text{Br}-\text{C}_6\text{H}_4-\text{CH}_2^-$
II. $\text{C}_6\text{H}_5-\text{CH}_2^-$
III. $p-\text{CH}_3\text{O}-\text{C}_6\text{H}_4-\text{CH}_2^-$
IV. $p-\text{CHO}-\text{C}_6\text{H}_4-\text{CH}_2^-$
V. $p-\text{CH}_3-\text{C}_6\text{H}_4-\text{CH}_2^-$
Choose the correct answer from the options given below :
- $\text{I} > \text{II} > \text{IV} > \text{V} > \text{III}$
- $\text{I} > \text{IV} > \text{II} > \text{V} > \text{III}$
- $\text{IV} > \text{I} > \text{II} > \text{V} > \text{III}$
- $\text{IV} > \text{II} > \text{I} > \text{III} > \text{V}$
### 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. $-\text{CHO}$: Strong -M effect. Highly stabilizing.
I. $-\text{Br}$: Weak +M effect, but prominent -I effect. Overall stabilizing relative to hydrogen.
II. $-\text{H}$: (Plain benzyl anion) Neutral baseline.
III. $-\text{OCH}_3$: Strong +M effect. Highly destabilizing.
V. $-\text{CH}_3$: +I and +H (hyperconjugation) effects. Destabilizing, but less so than strong +M.
### Step 1: Final Conclusion
Based on the effects, the stability order is:
$-\text{CHO} (-M) > -\text{Br} (-I) > -\text{H} > -\text{CH}_3 (+I, +H) > -\text{OCH}_3 (+M)$
Therefore: $\text{IV} > \text{I} > \text{II} > \text{V} > \text{III}$.
### Pattern Recognition
For carbanions, think: "EWG stabilizes, EDG destabilizes". This is the exact opposite of carbocation stability rules.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q74 (2026)
In Dumas method for estimation of nitrogen, $0.50 \text{ g}$ of an organic compound gave $70 \text{ mL}$ of nitrogen collected at $300 \text{ K}$ and $715 \text{ mm}$ pressure. The percentage of nitrogen in the organic compound is ____%
(Aqueous tension at $300 \text{ K}$ is $15 \text{ mm}$).
### Related Formula
$$P_{\text{dry gas}} = P_{\text{total}} - \text{Aqueous tension}$$
$PV = nRT$
$$\% N = \frac{\text{Mass of } N_2}{\text{Mass of organic compound}} \times 100$$
### Core Logic
Pressure of dry $N_2$ gas:
$P_{N_2} = (715 - 15) \text{ mm} = 700 \text{ mm Hg} = \frac{700}{760} \text{ atm}$
Volume of $N_2$ gas:
$V_{N_2} = 70 \text{ mL} = \frac{70}{1000} \text{ L}$
Temperature:
$T = 300 \text{ K}$
### Step 1: Calculate moles and mass of Nitrogen
Using Ideal Gas Law, $n_{N_2} = \frac{PV}{RT}$:
$$n_{N_2} = \frac{\left(\frac{700}{760}\right) \times \left(\frac{70}{1000}\right)}{0.0821 \times 300}$$
Mass of $N_2$ ($W_{N_2}$) = $n_{N_2} \times 28$
$$W_{N_2} = \frac{700}{760} \times \frac{70/1000}{0.0821 \times 300} \times 28 \approx 0.07324 \text{ g}$$
### Step 2: Calculate Percentage
$$\% N = \frac{0.07324}{0.50} \times 100 = 14.65 \%$$
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 $(P_1V_1)/T_1 = (P_{STP}V_{STP})/T_{STP}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q60 (2026)
Find out the statements which are not true.
A. Resonating structure with more number of covalent bonds and lesser charge separation are more stable.
B. In electromagnetic effect, an unsaturated system shows + E effect with nucleophile and -E effect with electrophile.
C. Inductive effect is responsible for high melting point, boiling point and dipole moment of polar compounds.
D. The greater the number of alkyl groups attached to the doubly bonded carbon atoms, higher is the heat of hydrogenation.
E. Stability of carbanion increases with the increase in s-character of the carbon carrying the negative charge.
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Choose the correct answer from the options given below.
- A, D & E only
- B, D & E only
- A, C & D only
- B & D only
### Core Logic
Let's analyze the statements:
Statement A: Resonating structure with more covalent bonds and lesser charge separation are indeed more stable. (True)
Statement B: In the electromeric effect, when the pi electrons shift towards the attacking reagent (electrophile), it's +E. When they shift away from the attacking reagent (nucleophile), it's -E. The statement says +E with nucleophile and -E with electrophile, which is reversed. (False)
Statement C: Inductive effect is a permanent polarization that contributes to the dipole moment and intermolecular forces, thereby influencing boiling/melting points. (True)
Statement D: The greater the number of alkyl groups attached to the double bond, the more stable the alkene is (due to hyperconjugation). More stable alkenes release LESS energy upon hydrogenation, meaning they have a LOWER heat of hydrogenation. (False)
Statement E: Stability of a carbanion increases with the s-character of the carbon atom because an orbital with more s-character is closer to the nucleus, stabilizing the negative charge better ($sp > sp^2 > sp^3$). (True)
### Step 1: Conclusion
Statements B and D are NOT true.
### Pattern Recognition
Heat of Hydrogenation (HOH) is inversely proportional to alkene stability. More substituted alkene = more stable = lower HOH.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
