Identify the correct statements from the following: A. textCH_3textCH_2textCOCH_2textCH_3$\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$ and textCH_3textCOCH_2textCH_2textCH_3$\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3$ are metamers B. textCH_3textCH_2textCH_2textCN$\text{CH}_3\text{CH}_2\text{CH}_2\text{CN}$ and textCH_3textCH_2textCH_2textNC$\text{CH}_3\text{CH}_2\text{CH}_2\text{NC}$ are functional isomers C. 2-methylphenol and 3-methylphenol are position isomers D. textCH_3textCH_2textNH_2$\text{CH}_3\text{CH}_2\text{NH}_2$ and textCH_3textCH_2textCH_2textNH_2$\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$ are homologous Choose the correct answer from the options given below.

### Related Formula p_mathrmN_2 = p_texttotal - p_textaq$$p_{\mathrm{N_2}} = p_{\text{total}} - p_{\text{aq}}$$ n_mathrmN_2 = fracp_mathrmN_2 VR T$$n_{\mathrm{N_2}} = \frac{p_{\mathrm{N_2}} V}{R T}$$ \% mathrmN = fractextMass of nitrogentextMass of organic compound times 100$$\% \mathrm{N} = \frac{\text{Mass of nitrogen}}{\text{Mass of organic compound}} \times 100$$ ### Core Logic Dumas' method estimates nitrogen by collecting dry nitrogen gas (N_2$N_2$). We must subtract the aqueous tension (vapor pressure of water) to find the pressure exerted solely by the dry nitrogen gas. ### Step 1: Calculate Pressure of Dry Nitrogen p_mathrmN_2 = 715~mathrmmmHg - 15~mathrmmmHg = 700~mathrmmmHg$$p_{\mathrm{N_2}} = 715~\mathrm{mmHg} - 15~\mathrm{mmHg} = 700~\mathrm{mmHg}$$ Converting pressure to atmospheres: p_mathrmN_2 = frac700760~mathrmatm$$p_{\mathrm{N_2}} = \frac{700}{760}~\mathrm{atm}$$ ### Step 2: Calculate Moles of Nitrogen Gas Using the ideal gas law with R = 0.0821~mathrmL~atm~mol^-1~K^-1$R = 0.0821~\mathrm{L~atm~mol^{-1}~K^{-1}}$, T = 300~mathrmK$T = 300~\mathrm{K}$, and V = 60~mathrmmL = 60 times 10^-3~mathrmL$V = 60~\mathrm{mL} = 60 \times 10^{-3}~\mathrm{L}$: n_mathrmN_2 = fracleft(frac700760right) times 60 times 10^-30.0821 times 300$$n_{\mathrm{N_2}} = \frac{\left(\frac{700}{760}\right) \times 60 \times 10^{-3}}{0.0821 \times 300}$$ n_mathrmN_2 = frac0.92105 times 0.06024.63 approx 2.244 times 10^-3~mathrmmol$$n_{\mathrm{N_2}} = \frac{0.92105 \times 0.060}{24.63} \approx 2.244 \times 10^{-3}~\mathrm{mol}$$ ### Step 3: Calculate Mass and Percentage of Nitrogen The molar mass of mathrmN_2$\mathrm{N_2}$ is 28~mathrmg~mol^-1$28~\mathrm{g~mol^{-1}}$: textMass of mathrmN_2 = n_mathrmN_2 times 28 = 2.244 times 10^-3 times 28 approx 0.06283~mathrmg$$\text{Mass of } \mathrm{N_2} = n_{\mathrm{N_2}} \times 28 = 2.244 \times 10^{-3} \times 28 \approx 0.06283~\mathrm{g}$$ Now find the percentage in 0.5~mathrmg$0.5~\mathrm{g}$ of organic compound: \% mathrmN = frac0.06283~mathrmg0.5~mathrmg times 100 = 12.566\% approx 12.57\%$$\% \mathrm{N} = \frac{0.06283~\mathrm{g}}{0.5~\mathrm{g}} \times 100 = 12.566\% \approx 12.57\%$$ ### Pattern Recognition Watch out! Always subtract the aqueous tension from the wet gas pressure first to find the dry gas pressure. Forgetting this step is the most common source of error in Dumas calculations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula According to Huckel's Rule, a planar, monocyclic, completely conjugated system is aromatic if it contains: (4n + 2)pi quad textelectrons (where n = 0, 1, 2, dots)$$(4n + 2)\pi \quad \text{electrons (where } n = 0, 1, 2, \dots)$$

### Step 1: Classification Hence, compounds a, c, d, e, and h follow Huckel's rule and are aromatic, whereas b, f, and g are not aromatic. ### Pattern Recognition Quick check for aromaticity: Count the pairs of localized/delocalized pi$\pi$ electrons moving through the continuous loop. Odd number of pairs (1, 3, 5...) means aromatic (2pi, 6pi, 10pi$2\pi, 6\pi, 10\pi$). Even pairs mean anti-aromatic/non-aromatic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 11 Chemistry: Hydrocarbons

### Related Formula Free radical stability structural hierarchy sequence: textResonance Stabilized (Propargyl/Allyl) > 3^circ > 2^circ > 1^circ > textVinylic/Alkyne Center$$\text{Resonance Stabilized (Propargyl/Allyl)} > 3^\circ > 2^\circ > 1^\circ > \text{Vinylic/Alkyne Center}$$ ### Core Logic Let's analyze individual cleavage points across the carbon backbone skeleton: * **Position II** yields a propargyl intermediate radical directly adjacent to the alkyne bond. This allows strong resonance stabilization across the pi$\pi$ system, making it the most stable radical position. * **Position I** places the radical directly on an mathrmsp$\mathrm{sp}$-hybridized carbon center. The high electronegativity of mathrmsp$\mathrm{sp}$ orbitals tightly holds the unpaired electron, making homolytic cleavage extremely difficult and rendering this intermediate the least stable radical position.

### Step 1: Verdict Therefore, the most stable and least stable positions are II and I, respectively. ### Pattern Recognition Radicals located on mathrmsp$\mathrm{sp}$ carbons (vinylic/alkynic) are highly unstable due to poor orbital overlap, while positions next to triple bonds (propargylic) are exceptionally stable due to active resonance delocalization. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula The relative rate of nucleophilic acyl substitution follows the leaving group ability: textRate of Hydrolysis propto textLeaving Group Ability propto frac1textBasic Strength of Leaving Group$$\text{Rate of Hydrolysis} \propto \text{Leaving Group Ability} \propto \frac{1}{\text{Basic Strength of Leaving Group}}$$

### Core Logic Let's analyze the leaving groups across all choices layout-by-row: * For **(p)**, the leaving group is mathrmCl^-$\mathrm{Cl^-}$ (Very weak base, excellent leaving group). * For **(q)**, the leaving group is mathrmRCOO^-$\mathrm{RCOO^-}$ (Resonance stabilized carboxylate, good leaving group). * For **(r)**, the leaving group is mathrmRO^-$\mathrm{RO^-}$ (Alkoxide, strong base, poor leaving group). * For **(s)**, the leaving group is mathrmNH_2^-$\mathrm{NH_2^-}$ (Extremely strong base, exceptionally poor leaving group due to nitrogen lone pair resonance into the carbonyl). This structural comparison yields the final sequence: mathrmp > q > r > s$\mathrm{p > q > r > s}$. ### Pattern Recognition Acyl chlorides (p) are always the most reactive acid derivatives, while amides (s) are consistently the least reactive due to strong amide resonance stabilizing the carbonyl group. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Related Formula Elemental content calculation system equations: textMoles of C = fractextMass of mathrmCO_244$$\text{Moles of C} = \frac{\text{Mass of } \mathrm{CO_2}}{44}$$ textMoles of H = 2 times fractextMass of mathrmH_2O18$$\text{Moles of H} = 2 \times \frac{\text{Mass of } \mathrm{H_2O}}{18}$$ ### Core Logic Let's perform the stoichiometry layout step-by-step: * Moles of mathrmC$\mathrm{C}$ inside sample system: mathrmn_C = frac1.4644 = 0.033mathrm~mol$$\mathrm{n_C} = \frac{1.46}{44} = 0.033\mathrm{~mol}$$ textMass of C = 0.033 times 12 = 0.396mathrm~g$$\text{Mass of C} = 0.033 \times 12 = 0.396\mathrm{~g}$$ * Moles of mathrmH$\mathrm{H}$ inside sample system: mathrmn_H = 2 times frac0.56718 = 0.063mathrm~mol$$\mathrm{n_H} = 2 \times \frac{0.567}{18} = 0.063\mathrm{~mol}$$ textMass of H = 0.063 times 1 = 0.063mathrm~g$$\text{Mass of H} = 0.063 \times 1 = 0.063\mathrm{~g}$$ * Determine Oxygen mass by subtracting values from total starting mass: textMass of O = 1.0 - (0.396 + 0.063) = 0.541mathrm~g$$\text{Mass of O} = 1.0 - (0.396 + 0.063) = 0.541\mathrm{~g}$$ mathrmn_O = frac0.54116 = 0.033mathrm~mol$$\mathrm{n_O} = \frac{0.541}{16} = 0.033\mathrm{~mol}$$ * Find atomic whole-number ratio profile: mathrmC : H : O = 0.033 : 0.063 : 0.033 approx 1 : 2 : 1$\mathrm{C : H : O} = 0.033 : 0.063 : 0.033 \approx 1 : 2 : 1$. * This gives an empirical configuration of mathrmCH_2O$\mathrm{CH_2O}$. ### Step 1: Evaluation Calculating formula mass: textEmpirical Mass = 12 + (2 times 1) + 16 = 30mathrm~g$$\text{Empirical Mass} = 12 + (2 \times 1) + 16 = 30\mathrm{~g}$$ ### Pattern Recognition When calculated mole properties output identical numbers for two elements (0.033$0.033$ for both C and O), their structural subscript ratio is exactly 1:1$1:1$. This pattern significantly speeds up empirical calculations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques