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.

Choose the correct answer from the options given below.

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) arrow$\rightarrow$ Highest electronegativity, most stable. CH₂=CH⁻$\bullet \; CH_{2}=CH^{-}$ (sp²$^{2}$ hybridized, 33.3%$33.3\%$ s-character) arrow$\rightarrow$ Intermediate electronegativity. CH₃-CH₂⁻$\bullet \; CH_{3}-CH_{2}^{-}$ (sp³$^{3}$ hybridized, 25%$25\%$ s-character) arrow$\rightarrow$ 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

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).

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

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$\mathrm{CO}_2 = \frac{1.32}{44} = 0.03\mathrm{~mol}$.\nMass of Carbon = 0.03 × 12 = 0.36~g$= 0.03 \times 12 = 0.36\mathrm{~g}$.\nPercentage of C = ((0.36)/(1.0)) × 100 = 36%$\text{Percentage of } C = \left(\frac{0.36}{1.0}\right) \times 100 = 36\%$.

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$108 + 80 = 188\mathrm{~g/mol}$).\nMass of Br = ((80)/(188)) × 0.75 ≈ 0.319~g$\text{Mass of } Br = \left(\frac{80}{188}\right) \times 0.75 \approx 0.319\mathrm{~g}$.\nPercentage of Br = ((0.319)/(0.53)) × 100 = 60.2%$\text{Percentage of } Br = \left(\frac{0.319}{0.53}\right) \times 100 = 60.2\%$.

Step 3: Calculate Percentage of Hydrogen

The compound strictly consists of C, H, and Br (as formula is given as CₓHyBrz$\mathrm{C}_x\mathrm{H}_y\mathrm{Br}_z$).\nPercentage of H = 100 - (%C + %Br)$\text{Percentage of } H = 100 - (\%\text{C} + \%\text{Br})$\nPercentage of H = 100 - (36 + 60.2) = 100 - 96.2 = 3.8%$\text{Percentage of } H = 100 - (36 + 60.2) = 100 - 96.2 = 3.8\%$.\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