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

To evaluate carbocation stability, apply the priority rules: Aromaticity > Resonance > Hyperconjugation.

• C: Represents a cyclopropenyl cation derivative which achieves full aromatic stabilization due to its planar cyclic conjugated system satisfying Huckel's rule (2π$2\pi$ electrons). This makes it the most stable.
• A: Stabilized by extended resonance from multiple phenyl groups.
• B: Contains fewer phenyl rings participating in active cross-conjugation relative to A.
• D: Stabilized solely by simple aliphatic hyperconjugation, making it the least stable.

Visual alignment chart:



Hence, the correct stability hierarchy is:

Pattern Recognition

Sees: Mixed aromatic, benzylic, and aliphatic carbocations. Shortcut: Isolate the cyclopropenyl system as an aromatic champion to confidently lead the sequence.

Chapter Mix

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

Core Logic

Organic compounds react with sodium bicarbonate (NaHCO₃$\mathrm{NaHCO}_3$) to liberate CO₂$\mathrm{CO}_2$ gas if they are stronger acids than carbonic acid (H₂CO₃$\mathrm{H}_2\mathrm{CO}_3$). Evaluating the structures:

• A: Benzoic acid, which is significantly more acidic than carbonic acid.
• C: Picric acid (2,4,6-trinitrophenol). Due to three strong electron-withdrawing nitro groups, its acidity exceeds typical carboxylic acids and H₂CO₃$\mathrm{H}_2\mathrm{CO}_3$.
• D: Benzenesulfonic acid, a highly strong mineral-like organic acid.
• B & E: Standard phenols or weakly substituted phenols, which are less acidic than carbonic acid and do not liberate CO₂$\mathrm{CO}_2$.

Therefore, structures A, C, and D give a positive test result.

Pattern Recognition

Sees: Sodium bicarbonate test for organic systems. Shortcut: Only carboxylic acids, sulfonic acids, and highly nitrated phenols like picric acid possess sufficient proton acidity to displace CO₂$\mathrm{CO}_2$ from bicarbonate ions.

Chapter Mix

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

Core Logic

Let us check the statements step-by-step:

• Statement A: Pentan-3-one and pentan-2-one have different alkyl groups attached on either side of the divalent polyfunctional carbonyl group (-CO-$-\text{CO}-$). Hence, they are metamers.

  

• Statement B: Cyanides (-CN$-\text{CN}$) and Isocyanides (-NC$-\text{NC}$) contain distinct functional groups, so they are functional isomers.

  

• Statement C: Phenol structures containing a methyl substituent at positions 2 and 3 are structural position isomers.
• Statement D: The given structures represent members of a homologous series because they differ sequentially by a -CH₂-$-\text{CH}_2-$ unit.

Step 1: Verification

Evaluating according to standard multi-choice options, statements A and B are perfectly validated.

Pattern Recognition

Shortcut: Metamers require variable alkyl distribution across a polyvalent heteroatom group. Functional isomers require changes like -CN$-\text{CN}$ vs -NC$-\text{NC}$.

Chapter Mix

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

Core Logic

Let us check the conjugate bases formed upon losing a proton:

• Compounds (A), (B), and (C) generate conjugate bases stabilized by resonance through the aromatic ring or strong electron-withdrawing groups.
• Compound (D) represents an ethynyl group in a terminal alkyne structure (EtO₂C-C$\text{EtO}_2\text{C}-\text{C}\equiv\text{CH}$). Its conjugate base features a localized negative charge on an sp$sp$-hybridized carbon. Because there is no resonance stabilization present for this anion, it is significantly less stable than the conjugate bases of the other functional groups.

Step 1: Conclusion

Since a less stable conjugate base implies a weaker parent acid, the terminal alkyne compound (D) is the least acidic.

Pattern Recognition

Shortcut: A resonance-stabilized anion is always more stable than a localized one. Look for the alkyne carbon (sp-C-H$sp\text{-C}-\text{H}$) versus resonance-delocalized oxygen or active methylene centers.

Evaluation Rubric / Model Answer

Option (A)

Chapter Mix

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

Related Formula

Using the Principle of Atom Conservation (POAC) for Nitrogen:

Core Logic

The molecular weight of the given heterocyclic amine organic structure X$X$ (piperazine, C₄H₁₀N₂$\text{C}_4\text{H}_{10}\text{N}_2$) is calculated as:

Given mass of compound = 0.42 g$= 0.42\text{ g}$:

Step 1: Calculating STP Volume

Since each molecule contains 2$2$ nitrogen atoms, 1 mol$1\text{ mol}$ of compound produces 1 mol$1\text{ mol}$ of N₂$\text{N}_2$ gas:

Using standard molar volume at STP (22700 mL/mol$22700\text{ mL/mol}$ per IUPAC convention, or 22400 mL/mol$22400\text{ mL/mol}$ in traditional calculations):

(Note: If calculated using 22400 mL/mol$22400\text{ mL/mol}$, Volume = (0.42)/(86) × 22400 ≈ 109.4 mL ≈ 109 mL$\text{Volume} = \frac{0.42}{86} \times 22400 \approx 109.4\text{ mL} \approx 109\text{ mL}$.)

Rounding to the nearest integer gives 111 (official accepted range: 109 to 111).

Pattern Recognition

Shortcut: Determine the molar mass (M = 86 g/mol$M = 86\text{ g/mol}$) and nitrogen atom count (2 N atoms 1 mol N₂ per mol of compound$2\text{ N atoms} \implies 1\text{ mol } \text{N}_2\text{ per mol of compound}$). Multiply moles directly by molar volume at STP to find the liberated gas volume.

Evaluation Rubric / Model Answer

111

Chapter Mix

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