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:

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)$-\text{CHO} (-M) > -\text{Br} (-I) > -\text{H} > -\text{CH}_3 (+I, +H) > -\text{OCH}_3 (+M)$ Therefore: IV > I > II > V > III$\text{IV} > \text{I} > \text{II} > \text{V} > \text{III}$.

Pattern Recognition

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

Chapter Mix

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

Related Formula

PV = nRT$PV = nRT$

Core Logic

Pressure of dry N₂$N_2$ gas: PN₂ = (715 - 15) mm = 700 mm Hg = (700)/(760) atm$P_{N_2} = (715 - 15) \text{ mm} = 700 \text{ mm Hg} = \frac{700}{760} \text{ atm}$

Volume of N₂$N_2$ gas: VN₂ = 70 mL = (70)/(1000) L$V_{N_2} = 70 \text{ mL} = \frac{70}{1000} \text{ L}$

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

Step 1: Calculate moles and mass of Nitrogen

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

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

Step 2: Calculate Percentage

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$(P_1V_1)/T_1 = (P_{STP}V_{STP})/T_{STP}$.

Chapter Mix

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

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 > sp² > sp³$sp > sp^2 > sp^3$). (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.

Chapter Mix

Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

Related Formula

Core Logic

Combustion equation: CₓHyOz + O₂ arrow CO₂ + H₂O$\mathrm{C_xH_yO_z + O_2 \rightarrow CO_2 + H_2O}$

Potash (KOH) tube absorbs CO₂$CO_2$. So, mass of CO₂$CO_2$ produced = 0.1837 g (approximated as 0.18 g in solution data for simplicity, but strictly 0.1837 based on prompt. The solution explicitly uses 0.18 for C calculation, let's trace: Mass of 'C' = (0.18)/(44) × 12$= \frac{0.18}{44} \times 12$). CaCl₂$CaCl_2$ tube absorbs H₂O$H_2O$. So, mass of H₂O$H_2O$ produced = 0.15 g.

Mass of Carbon (C) = (12)/(44) × 0.18 0.049 0.05 gm$= \frac{12}{44} \times 0.18 \simeq 0.049 \simeq 0.05 \text{ gm}$ Mass of Hydrogen (H) = (2)/(18) × 0.15 = 0.0166 0.017 gm$= \frac{2}{18} \times 0.15 = 0.0166 \simeq 0.017 \text{ gm}$

Step 1: Calculate Mass of Oxygen

Since the total mass of compound A is 0.25 gm: Mass of Oxygen (O) = 0.25 - (Mass of C + Mass of H)$= 0.25 - (\text{Mass of C} + \text{Mass of H})$ Mass of 'O' = 0.25 - 0.05 - 0.017 = 0.183 gm$= 0.25 - 0.05 - 0.017 = 0.183 \text{ gm}$

Step 2: Calculate Percentage

Mass % of 'O' = (0.1833)/(0.25) × 100 = 73.32%$= \frac{0.1833}{0.25} \times 100 = 73.32\%$ Rounding to the nearest integer gives 73.

Pattern Recognition

In Liebig's combustion method, the CaCl₂$CaCl_2$ U-tube maps strictly to H₂O$H_2O$ mass, and the Potash bulb maps strictly to CO₂$CO_2$ mass. Find carbon and hydrogen masses, subtract from total sample mass to find the third element (Oxygen).

Chapter Mix

Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

Core Logic

The stability of a carbanion is directly proportional to the electronegativity of the carbon atom bearing the negative charge. The electronegativity of carbon increases with the increase in s-character of its hybridization state.

Step 1: Hybridization Analysis

CH≡ C⁻$\bullet \; CH\equiv C^{-}$ (sp hybridized, 50%$50\%$ s-character) -> Highest electronegativity, most stable.\nCH₂=CH⁻$\bullet \; CH_{2}=CH^{-}$ (sp²$^{2}$ hybridized, 33.3%$33.3\%$ s-character) -> Intermediate electronegativity.\nCH₃-CH₂⁻$\bullet \; CH_{3}-CH_{2}^{-}$ (sp³$^{3}$ hybridized, 25%$25\%$ s-character) -> Lowest electronegativity, least stable.

Final Conclusion

Order of stability: CH≡ C⁻ > CH₂=CH⁻ > CH₃-CH₂⁻$CH\equiv C^{-} > CH_{2}=CH^{-} > CH_{3}-CH_{2}^{-}$

Pattern Recognition

Stability of carbanion ∝ % s-character$\propto \% \; s\text{-character}$. More s-character pulls the electron pair closer to the nucleus, stabilizing the negative charge.

Chapter Mix

Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques