In Dumas' method 292text mg$292\text{ mg}$ of an organic compound released 50text mL$50\text{ mL}$ of nitrogen gas (textN_2)$(\text{N}_2)$ at 300text K$300\text{ K}$ temperature and 715text mm Hg$715\text{ mm Hg}$ pressure. [cite: 435, 451] The percentage composition of 'N' in the organic compound is dots$\dots$ % (Nearest integer) [cite: 452, 455] (Aqueous tension at 300text K = 15text mm Hg$300\text{ K} = 15\text{ mm Hg}$)

### Core Logic Let us review the chemical basis for each qualitative test: * **A. Sodium bicarbonate (textNaHCO_3$\text{NaHCO}_3$) solution**: Carboxylic acids are sufficiently acidic to decompose textNaHCO_3$\text{NaHCO}_3$, liberating carbon dioxide gas observed as vigorous effervescence. Therefore, textA rightarrow textII$\text{A} \rightarrow \text{II}$. * **B. Neutral ferric chloride (textFeCl_3$\text{FeCl}_3$)**: Phenols react with neutral textFeCl_3$\text{FeCl}_3$ solution to form characteristic deeply colored violet coordination complexes. Therefore, textB rightarrow textIII$\text{B} \rightarrow \text{III}$. * **C. Ceric ammonium nitrate (CAN)**: Alcohols react with CAN reagent to cause a distinct color shift to deep dark red due to complexation. Therefore, textC rightarrow textIV$\text{C} \rightarrow \text{IV}$. * **D. Alkaline textKMnO_4$\text{KMnO}_4$ (Baeyer's Reagent)**: Reacts readily via syn-hydroxylation across carbon-carbon double/triple bonds, resulting in decolored solutions alongside brown textMnO_2$\text{MnO}_2$ precipitates. This detects unsaturation. Therefore, textD rightarrow textI$\text{D} \rightarrow \text{I}$. ### Step 1: Assembly Combining the validated relationships gives: textA-II, B-III, C-IV, D-I$$\text{A-II, B-III, C-IV, D-I}$$ This maps perfectly to Option (1). ### Pattern Recognition Baeyer's test (alkaline textKMnO_4$\text{KMnO}_4$) always tests for alkenes/alkynes. textNaHCO_3$\text{NaHCO}_3$ is unique for acidic groups like carboxylic acids. Matching these two reliable pairs isolates the correct option without needing to review the entire table. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Alcohols, Phenols and Ethers

### Core Logic Statement I is true: In partition chromatography, the stationary phase is indeed a thin film of liquid held on the surface of an inert solid support. Statement II is false: In paper chromatography, the water molecules trapped inside the cellulose network of the paper act as the stationary phase, not the paper material itself. ### Pattern Recognition Remember that paper chromatography is a type of partition chromatography where moisture content (water) adsorbed on the paper serves as the stationary liquid phase. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic The structural formula of hex-1-en-4-yne is given by: CH_2 = CH - CH_2 - C equiv C - CH_3$$CH_2 = CH - CH_2 - C equiv C - CH_3$$ Let's count the chemical bonds chronologically: * Number of C-H$C-H$ sigma$sigma$ bonds = 2 + 1 + 2 + 3 = 8$2 + 1 + 2 + 3 = 8$ * Number of C-C$C-C$ sigma$sigma$ bonds = 5$5$ Total sigma$sigma$ bonds = 8 + 5 = 13$8 + 5 = 13$.

* Number of pi$pi$ bonds: 1$1$ from double bond + 2$2$ from triple bond = 3$3$ pi$pi$ bonds. ### Pattern Recognition Every single bond is 1sigma$1sigma$, every double bond contains 1sigma + 1pi$1sigma + 1pi$, and every triple bond contains 1sigma + 2pi$1sigma + 2pi$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula %S = frac32233 times fractextMass of BaSO_4textMass of organic compound times 100$$%S = \frac{32}{233} \times \frac{\text{Mass of } BaSO_4}{\text{Mass of organic compound}} \times 100$$ ### Core Logic Let's substitute the given values into the formula: textMass of BaSO_4 = 0.40text g$$\text{Mass of } BaSO_4 = 0.40\text{ g}$$ textMass of organic compound = 0.20text g$$\text{Mass of organic compound} = 0.20\text{ g}$$ textMolar mass of BaSO_4 = 137 + 32 + (4 times 16) = 233text g/mol$$\text{Molar mass of } BaSO_4 = 137 + 32 + (4 \times 16) = 233\text{ g/mol}$$ %S = frac32233 times frac0.400.20 times 100 = frac32 times 2 times 100233 approx 27.468%$$%S = \frac{32}{233} \times \frac{0.40}{0.20} \times 100 = \frac{32 \times 2 \times 100}{233} approx 27.468%$$ ### Step 1: Match with the Question Layout Rounding to the standard value given in the official key: %S = 27.5% = 275 times 10^-1% implies x = 275$$%S = 27.5% = 275 \times 10^{-1}% implies x = 275$$ ### Pattern Recognition Carius method calculations depend heavily on standard conversion factors. The constant factor for sulphur gravimetry is frac32233$\frac{32}{233}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### 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 (2pi$2\pi$ 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: mathrmC > mathrmA > mathrmB > mathrmD$$\mathrm{C} > \mathrm{A} > \mathrm{B} > \mathrm{D}$$ ### Pattern Recognition Sees: Mixed aromatic, benzylic, and aliphatic carbocations. Shortcut: Isolate the cyclopropenyl system as an aromatic champion to confidently lead the sequence. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques