### Core Logic Let us apply official IUPAC priority indexing rules: 1. **Principal Functional Group**: The hydroxyl group (-textOH$-\text{OH}$) possesses higher naming priority over double bonds and simple alkyl side chains. Thus, the carbon bearing the -textOH$-\text{OH}$ group is assigned position **C-1**. 2. **Numbering Direction**: We must number through the ring towards the double bond to assign it the lowest possible locant. Hence, the alkene carbons are given coordinates **C-2** and **C-3**. 3. **Locating Side Chains**: Proceeding with this direction puts the ethyl group at position **C-4**.

Assembling the structural parts alphabetically: * Substituent: `4-Ethyl` * Parent root: `cyclopent-2-en` * Suffix: `1-ol` Combined IUPAC format: **4-Ethylcyclopent-2-en-1-ol**. ### Pattern Recognition Principal suffix priority hierarchy: -textOH > textDouble bond > textAlkyl side-chain$-\text{OH} > \text{Double bond} > \text{Alkyl side-chain}$. Always fix the highest priority suffix at index 1 and head instantly towards the alkene bond to safely restrict locant numbers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula Percentage of Hydrogen in organic analysis: \%textH = frac218 times fractextMass of H_2OtextMass of Compound times 100$\%\text{H} = \frac{2}{18} \times \frac{\text{Mass of } H_2O}{\text{Mass of Compound}} \times 100$ Percentage of Carbon: \%textC = frac1244 times fractextMass of CO_2textMass of Compound times 100$\%\text{C} = \frac{12}{44} \times \frac{\text{Mass of } CO_2}{\text{Mass of Compound}} \times 100$ Percentage of Oxygen: \%textO = 100 - (\%textC + \%textH)$\%\text{O} = 100 - (\%\text{C} + \%\text{H})$ ### Execution Step 1: Compute the mass percent of Hydrogen: \%textH = frac218 times frac0.1270.210 times 100 = frac0.2543.78 approx 6.72\%$\%\text{H} = \frac{2}{18} \times \frac{0.127}{0.210} \times 100 = \frac{0.254}{3.78} \approx 6.72\%$ Step 2: Compute the mass percent of Carbon: \%textC = frac1244 times frac0.3070.210 times 100 = frac3.6849.24 approx 39.87\%$\%\text{C} = \frac{12}{44} \times \frac{0.307}{0.210} \times 100 = \frac{3.684}{9.24} \approx 39.87\%$ Step 3: Deduce the remaining mass percent of Oxygen: \%textO = 100 - (39.87 + 6.72) = 100 - 46.59 = 53.41\%$\%\text{O} = 100 - (39.87 + 6.72) = 100 - 46.59 = 53.41\%$ Thus, the values of hydrogen and oxygen percentage are 6.72\%$6.72\%$ and 53.41\%$53.41\%$, matches with Option (2). ### Pattern Recognition Always focus on the order requested by the question stem. The query specifies 'hydrogen and oxygen respectively'. Option 2 and Option 4 both show these numbers but reversed—verifying the targeted sequence protects your score line. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### 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