The correct order of stability of following carbocations is :

A

B

C

D

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

Related Formula

The percentage of carbon via combustion analysis is given by:

Core Logic

Let us substitute the given parameters:

• Mass of organic compound = 0.5 g$= 0.5\text{ g}$
• Mass of CO₂$\text{CO}_2$ collected = 1.46 g$= 1.46\text{ g}$

Step 1: Numerical Calculation

Rounding to the nearest integer gives 80.

Pattern Recognition

Shortcut: (12)/(44) ≈ 0.2727$\frac{12}{44} \approx 0.2727$. Multiply 0.2727 × 1.46$0.2727 \times 1.46$ to find the mass of carbon (≈ 0.398 g$\approx 0.398\text{ g}$). Since 0.398 g$0.398\text{ g}$ out of 0.5 g$0.5\text{ g}$ is roughly (4)/(5)$\frac{4}{5}$, the value is right around 80%$80\%$.

Evaluation Rubric / Model Answer

80

Chapter Mix

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

Related Formula

Core Logic

Analyzing the molecules:

• In molecule (R), according to Bredt's rule, the bridgehead nitrogen has a localized lone pair which cannot participate in resonance. Thus, it is highly available and most basic.
• In molecule (Q), the nitrogen lone pair is involved in cross-conjugation with the carbonyl group, reducing its availability.
• In molecule (P), the lone pair on nitrogen is directly conjugated with the carbonyl group (amide resonance), making it the least available.

Therefore, the correct basicity order is: R > Q > P$R > Q > P$

Step 1: Final Identification

Comparing availability, structure R has localized electrons, Q has cross-conjugation, and P has standard amide resonance. Hence, option (4) is correct.

Pattern Recognition

Look for localized vs delocalized lone pairs on nitrogen. Bridgehead nitrogen lone pairs that violate Bredt's rule for double bond formation remain strictly localized, drastically increasing basicity compared to conjugated amides.

Chapter Mix

Class 11 Chemistry: Some Basic Principles of Organic Chemistry