Given below are two statements: Statement I: (CH₃)₃C$(CH_{3})_{3}C^{\oplus}$ is more stable than CH₃$CH_{3}^{\oplus}$ as nine hyperconjugation interactions are possible in (CH₃)₃C$(CH_{3})_{3}C^{\oplus}$. Statement II: CH₃$CH_{3}^{\oplus}$ is less stable than (CH₃)₃C$(CH_{3})_{3}C^{\oplus}$ as only three hyperconjugation interactions are possible in CH₃$CH_{3}^{\oplus}$. In the light of the above statements, choose the correct answer from the options given below.

Core Logic

Count the number of alpha hydrogens (α-H$\alpha-H$) adjacent to the carbocation in each structure. (A)

Step 1: Final Conclusion

Both cations (A) and (C) have 6 α$\alpha$-hydrogens, meaning they share the same number of hyperconjugative structures.

Pattern Recognition

Hyperconjugation count corresponds strictly to the number of C-H$C-H$ bonds on the carbon atoms immediately adjacent (sp³$sp^3$) to the positive center.

Chapter Mix

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

Related Formula

Core Logic

Mass of organic compound = 0.314 g$0.314\text{ g}$ Mass of BaSO₄$BaSO_4$ formed = 0.4813 g$0.4813\text{ g}$ Substituting the values:

Step 1: Final Conclusion

The percentage of sulphur is approximately 21.05%$21.05\%$.

Pattern Recognition

Straight application of Carius method for Sulphur estimation formula.

Chapter Mix

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

Related Formula

Core Logic

Let's first draw the structural formula of 3,3-dimethylhex-1-ene-4-yne:

Step 1: Identify Hybridization of Each Carbon

We count the sigma (σ$\sigma$) bonds or pi (π$\pi$) bonds on each carbon:

• C₁$\mathrm{C}_1$: Involved in a double bond (CH₂ =$\mathrm{CH_2 =}$) sp²$\implies \mathrm{sp^2}$
• C₂$\mathrm{C}_2$: Involved in a double bond (=CH-$=\mathrm{CH-}$) sp²$\implies \mathrm{sp^2}$
• C₃$\mathrm{C}_3$: Single bonds only (bonded to C₂$\mathrm{C}_2$, C₄$\mathrm{C}_4$, and two methyl carbons) sp³$\implies \mathrm{sp^3}$
• Two methyl carbons attached to C₃$\mathrm{C}_3$: Single bonds only 2 × sp³$\implies 2 \times \mathrm{sp^3}$
• C₄$\mathrm{C}_4$: Involved in a triple bond (-C ≡$-\mathrm{C} \equiv$) sp$\implies \mathrm{sp}$
• C₅$\mathrm{C}_5$: Involved in a triple bond (≡ C-$\equiv \mathrm{C}-$) sp$\implies \mathrm{sp}$
• C₆$\mathrm{C}_6$: Single bonds only (-CH₃$-\mathrm{CH_3}$) sp³$\implies \mathrm{sp^3}$

Step 2: Calculate the Count

Summing the hybridization counts:

• sp³$\mathrm{sp^3}$ carbons: C₃$\mathrm{C}_3$, C₆$\mathrm{C}_6$, and 2 × CH₃$2 \times \mathrm{CH_3}$ on C₃$\mathrm{C}_3$ = 4$= 4$ carbon atoms
• sp²$\mathrm{sp^2}$ carbons: C₁$\mathrm{C}_1$ and C₂$\mathrm{C}_2$ = 2$= 2$ carbon atoms
• sp$\mathrm{sp}$ carbons: C₄$\mathrm{C}_4$ and C₅$\mathrm{C}_5$ = 2$= 2$ carbon atoms

Thus, the number of sp³$\mathrm{sp^3}$, sp²$\mathrm{sp^2}$ and sp$\mathrm{sp}$ hybridized carbons is 4, 2, 2$4, 2, 2$ respectively.

Pattern Recognition

Quick Tip: Always draw side substituents (like methyl groups) explicitly. A common mistake is to skip counting the methyl substituent carbons as sp³$\mathrm{sp^3}$.

Chapter Mix

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

Related Formula

Core Logic

Let's align each purification technique to its designated mixtures based on NCERT guidelines:

• (A) Simple distillation: Used for liquids having a significant difference in their boiling points (>30~K$>30~\mathrm{K}$ or 30^$30^\circ\mathrm{C}$). Chloroform (b.p. 334~K$334~\mathrm{K}$) and aniline (b.p. 457~K$457~\mathrm{K}$) are separated easily using simple distillation arrow$\rightarrow$ (III).
• (B) Fractional distillation: Used if boiling point differences of the components are very close (less than 25~K$25~\mathrm{K}$). Separation of petrochemical fractions such as diesel and petrol uses this technique arrow$\rightarrow$ (I).
• (C) Distillation under reduced pressure: Used for liquids that tend to decompose at or below their normal boiling points. Glycerol is separated from spent-lye in soap manufacturing industry using this vacuum method to prevent glycerol decomposition arrow$\rightarrow$ (IV).
• (D) Steam distillation: Applied to substances which are steam-volatile and completely immiscible in water. Aniline and water are separated using this technique arrow$\rightarrow$ (II).

Step 1: Conclusion

Thus, the correct match is: (A)-(III), (B)-(I), (C)-(IV), (D)-(II) This corresponds perfectly to option (4).

Pattern Recognition

Glycerol from spent-lye is a highly tested practical chemistry concept. Remember that vacuum distillation lowers the boiling point, permitting evaporation without decomposition.

Chapter Mix

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

Related Formula

Core Logic

Dumas' method estimates nitrogen by collecting dry nitrogen gas (N₂$N_2$). We must subtract the aqueous tension (vapor pressure of water) to find the pressure exerted solely by the dry nitrogen gas.

Step 1: Calculate Pressure of Dry Nitrogen

Converting pressure to atmospheres:

Step 2: Calculate Moles of Nitrogen Gas

Using the ideal gas law with R = 0.0821~ L~atm~mol⁻¹~K⁻¹$R = 0.0821~\mathrm{L~atm~mol^{-1}~K^{-1}}$, T = 300~K$T = 300~\mathrm{K}$, and V = 60~mL = 60 × 10⁻³~L$V = 60~\mathrm{mL} = 60 \times 10^{-3}~\mathrm{L}$:

Step 3: Calculate Mass and Percentage of Nitrogen

The molar mass of N₂$\mathrm{N_2}$ is 28~ g~mol⁻¹$28~\mathrm{g~mol^{-1}}$:

Now find the percentage in 0.5~g$0.5~\mathrm{g}$ of organic compound:

Pattern Recognition

Watch out! Always subtract the aqueous tension from the wet gas pressure first to find the dry gas pressure. Forgetting this step is the most common source of error in Dumas calculations.

Chapter Mix

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