0.5~g$0.5\mathrm{~g}$ of an organic compound on combustion gave 1.46~g$1.46\mathrm{~g}$ of CO₂$\mathrm{CO}_2$ and 0.9~g$0.9\mathrm{~g}$ of H₂O$\mathrm{H}_2\mathrm{O}$. The percentage of carbon in the compound is _____. (Nearest integer) [Given: Molar mass (in g~mol⁻¹$\mathrm{g~mol^{-1}}$) C: 12, H: 1, O: 16$\mathrm{C: 12, H: 1, O: 16}$]

Related Formula

Metamers: Different alkyl chains attached to the same polyvalent functional group.

Functional Isomers: Same molecular formula, different functional groups.

Core Logic

Statement A: Pentan-3-one and pentan-2-one have different alkyl groups attached to the carbonyl carbon (diethyl vs methyl-propyl), so they are metamers. (Correct)

Statement B: Cyanides (-CN$-\mathrm{CN}$) and isocyanides (-NC$-\mathrm{NC}$) have different functional groups, making them functional isomers. (Correct)

Statement C: Ethanol and methanol differ by a -CH₂-$-\mathrm{CH_2}-$ unit; they are homologues, not position isomers. (Incorrect)

Statement D: Methylamine and ethylamine are homologues, but among the available options, option 3 (A & B only) represents the correct combination.

Step 1: Final Selection

Statements A and B are unambiguously correct.

Pattern Recognition

Pentan-2-one vs Pentan-3-one = Metamers. Cyanide vs Isocyanide = Functional isomers.

Chapter Mix

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

Related Formula

Core Logic

Order of acidity among functional groups:

• Benzenesulfonic acid (Ph-SO₃H$\mathrm{Ph-SO_3H}$) arrow$\rightarrow$ Conjugate base stabilized by 3 equivalent resonance structures. (Most acidic)
• Benzoic acid (Ph-COOH$\mathrm{Ph-COOH}$) arrow$\rightarrow$ Conjugate base stabilized by 2 equivalent carboxylate resonance structures.
• Phenol (Ph-OH$\mathrm{Ph-OH}$) arrow$\rightarrow$ Phenoxide ion resonance-stabilized over the aromatic ring.
• Terminal alkyne (EtO₂C-C ≡ CH$\mathrm{EtO_2C-C \equiv CH}$) arrow$\rightarrow$ Proton lost from sp$sp$-hybridized carbon. Though electron-withdrawing ester group increases C-H$C-H$ acidity, terminal alkynes are significantly less acidic than phenols, carboxylic acids, and sulfonic acids (pKₐ ≈ 18--20$pK_a \approx 18\text{--}20$ vs pKa ≈ 10$pKa \approx 10$ for phenol).

Step 1: Final Conclusion

Compound (A) is the least acidic among the given choices.

Pattern Recognition

Acidity order: -SO₃H > -COOH > -OH (phenol) > -C ≡ CH (terminal alkyne)$-\mathrm{SO_3H} > -\mathrm{COOH} > -\mathrm{OH} (phenol) > -\mathrm{C \equiv CH} (terminal alkyne)$.

Chapter Mix

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

Related Formula

Core Logic

From the molecular structure of compound X, its molecular formula is C₅H₁₀N₂$\mathrm{C_5H_{10}N_2}$.

Molar Mass of X = (5 × 12) + (10 × 1) + (2 × 14) = 60 + 10 + 28 = 86 ~g~mol⁻¹$= (5 \times 12) + (10 \times 1) + (2 \times 14) = 60 + 10 + 28 = 86\mathrm{~g~mol^{-1}}$.

Step 1: Calculation

Moles of compound X in 0.42~g$0.42\mathrm{~g}$:

By Conservation of Atomic Mass (POAC) on Nitrogen atoms:

Volume of N₂$\mathrm{N_2}$ liberated at STP:

Using 22.7~L$22.7\mathrm{~L}$ standard STP molar volume: (0.42)/(86) × 22700 ≈ 110.88~mL ≈ 111~mL$\frac{0.42}{86} \times 22700 \approx 110.88\mathrm{~mL} \approx 111\mathrm{~mL}$.

Pattern Recognition

Dumas method volume calculation: V = mMcompound × 22.4 L/mol × 1000 mL = 111 mL$V = \frac{m}{M_{\text{compound}}} \times 22.4\text{ L/mol} \times 1000\text{ mL} = 111\text{ mL}$.

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