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:

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

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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

Related Formula

According to Huckel's Rule, a planar, monocyclic, completely conjugated system is aromatic if it contains:

Step 1: Classification

Hence, compounds a, c, d, e, and h follow Huckel's rule and are aromatic, whereas b, f, and g are not aromatic.

Pattern Recognition

Quick check for aromaticity: Count the pairs of localized/delocalized π$\pi$ electrons moving through the continuous loop. Odd number of pairs (1, 3, 5...) means aromatic (2π, 6π, 10π$2\pi, 6\pi, 10\pi$). Even pairs mean anti-aromatic/non-aromatic.

Chapter Mix

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

Related Formula

Free radical stability structural hierarchy sequence:

Core Logic

Let's analyze individual cleavage points across the carbon backbone skeleton:

• Position II yields a propargyl intermediate radical directly adjacent to the alkyne bond. This allows strong resonance stabilization across the π$\pi$ system, making it the most stable radical position.
• Position I places the radical directly on an sp$\mathrm{sp}$-hybridized carbon center. The high electronegativity of sp$\mathrm{sp}$ orbitals tightly holds the unpaired electron, making homolytic cleavage extremely difficult and rendering this intermediate the least stable radical position.

Step 1: Verdict

Therefore, the most stable and least stable positions are II and I, respectively.

Pattern Recognition

Radicals located on sp$\mathrm{sp}$ carbons (vinylic/alkynic) are highly unstable due to poor orbital overlap, while positions next to triple bonds (propargylic) are exceptionally stable due to active resonance delocalization.

Chapter Mix

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

Related Formula

The relative rate of nucleophilic acyl substitution follows the leaving group ability:

Core Logic

Let's analyze the leaving groups across all choices layout-by-row:

• For (p), the leaving group is Cl^-$\mathrm{Cl^-}$ (Very weak base, excellent leaving group).
• For (q), the leaving group is RCOO^-$\mathrm{RCOO^-}$ (Resonance stabilized carboxylate, good leaving group).
• For (r), the leaving group is RO^-$\mathrm{RO^-}$ (Alkoxide, strong base, poor leaving group).
• For (s), the leaving group is NH₂^-$\mathrm{NH_2^-}$ (Extremely strong base, exceptionally poor leaving group due to nitrogen lone pair resonance into the carbonyl).

This structural comparison yields the final sequence: p > q > r > s$\mathrm{p > q > r > s}$.

Pattern Recognition

Acyl chlorides (p) are always the most reactive acid derivatives, while amides (s) are consistently the least reactive due to strong amide resonance stabilizing the carbonyl group.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids