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$300\text{ K} = 15\text{ mm Hg}$)

### Related Formula textPercentage of Sulphur = fractextMass of S text in BaSO_4textMolar mass of BaSO_4 times fractextMass of BaSO_4 text formedtextMass of organic compound times 100$$\text{Percentage of Sulphur} = \frac{\text{Mass of } S \text{ in } BaSO_4}{\text{Molar mass of } BaSO_4} \times \frac{\text{Mass of } BaSO_4 \text{ formed}}{\text{Mass of organic compound}} \times 100$$ ### Core Logic Molar mass of BaSO_4$BaSO_4$ = 233 g/mol. Mass of Sulfur (S) in 1 mole of BaSO_4$BaSO_4$ = 32 g. Mass of BaSO_4$BaSO_4$ formed = 1.2 g. Mass of organic compound = 0.75 g. \% mathrmS = frac32233 times frac1.20.75 times 100$$\% \mathrm{S} = \frac{32}{233} \times \frac{1.2}{0.75} \times 100$$ \% mathrmS = frac32 times 1.2 times 100233 times 0.75 = frac3840174.75 approx 21.97\%$$\% \mathrm{S} = \frac{32 \times 1.2 \times 100}{233 \times 0.75} = \frac{3840}{174.75} \approx 21.97\%$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic In structure 3, there are positive formal charges on two adjacent atoms (Oxygen and the Carbon adjacent to it). Like charges on adjacent atoms cause extreme electrostatic repulsion, making the structure highly unstable.

### Pattern Recognition Rules of resonance stability: 1. Complete octets are more stable. 2. More covalent bonds = more stable. 3. Least charge separation is more stable. 4. Like charges on adjacent atoms create massive destabilization (Least stable scenario). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic A. Propanal (CH_3CH_2CHO$CH_3CH_2CHO$) and propanone (CH_3COCH_3$CH_3COCH_3$) have different functional groups (aldehyde vs ketone) but the same molecular formula. They are functional isomers. (Correct) B. Ethoxyethane (C_2H_5-O-C_2H_5$C_2H_5-O-C_2H_5$) and methoxypropane (CH_3-O-C_3H_7$CH_3-O-C_3H_7$) differ in the alkyl chains attached to the polyvalent oxygen atom. They are metamers. (Correct) C. But-2-ene shows geometrical isomerism (cis-trans), but no optical isomerism as it lacks a chiral center. (Incorrect) D. But-1-ene and but-2-ene differ in the position of the double bond. They are position isomers, not functional isomers. (Incorrect) E. Pentane (straight chain) and 2,2-dimethyl propane (branched chain) have the same formula C_5H_12$C_5H_{12}$ but different carbon skeletons. They are chain isomers. (Correct) Correct statements: A, B, and E. ### Pattern Recognition Metamerism arises when there is a difference in the alkyl groups attached to a polyvalent functional group (e.g., -O-, -S-, -NH-, -CO-). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula textHybridization of Carbon = begincases mathrmsp^3 & 4~sigmatext-bonds \\ mathrmsp^2 & 3~sigmatext-bonds, 1~pitext-bond \\ mathrmsp & 2~sigmatext-bonds, 2~pitext-bonds endcases$$\text{Hybridization of Carbon} = \begin{cases} \mathrm{sp^3} & 4~\sigma\text{-bonds} \\ \mathrm{sp^2} & 3~\sigma\text{-bonds}, 1~\pi\text{-bond} \\ \mathrm{sp} & 2~\sigma\text{-bonds}, 2~\pi\text{-bonds} \end{cases}$$ ### Core Logic Let's first draw the structural formula of **3,3-dimethylhex-1-ene-4-yne**:

overset6mathrmCmathrmH_3 - overset5mathrmC equiv overset4mathrmC - overset3mathrmC(mathrmCH_3)_2 - overset2mathrmCmathrmH = overset1mathrmCmathrmH_2$$\overset{6}{\mathrm{C}}\mathrm{H}_3 - \overset{5}{\mathrm{C}} \equiv \overset{4}{\mathrm{C}} - \overset{3}{\mathrm{C}}(\mathrm{CH}_3)_2 - \overset{2}{\mathrm{C}}\mathrm{H} = \overset{1}{\mathrm{C}}\mathrm{H}_2$$ ### Step 1: Identify Hybridization of Each Carbon We count the sigma (sigma$\sigma$) bonds or pi (pi$\pi$) bonds on each carbon: - **mathrmC_1$\mathrm{C}_1$**: Involved in a double bond (mathrmCH_2 =$\mathrm{CH_2 =}$) implies mathrmsp^2$\implies \mathrm{sp^2}$ - **mathrmC_2$\mathrm{C}_2$**: Involved in a double bond (=mathrmCH-$=\mathrm{CH-}$) implies mathrmsp^2$\implies \mathrm{sp^2}$ - **mathrmC_3$\mathrm{C}_3$**: Single bonds only (bonded to mathrmC_2$\mathrm{C}_2$, mathrmC_4$\mathrm{C}_4$, and two methyl carbons) implies mathrmsp^3$\implies \mathrm{sp^3}$ - **Two methyl carbons** attached to mathrmC_3$\mathrm{C}_3$: Single bonds only implies 2 times mathrmsp^3$\implies 2 \times \mathrm{sp^3}$ - **mathrmC_4$\mathrm{C}_4$**: Involved in a triple bond (-mathrmC equiv$-\mathrm{C} \equiv$) implies mathrmsp$\implies \mathrm{sp}$ - **mathrmC_5$\mathrm{C}_5$**: Involved in a triple bond (equiv mathrmC-$\equiv \mathrm{C}-$) implies mathrmsp$\implies \mathrm{sp}$ - **mathrmC_6$\mathrm{C}_6$**: Single bonds only (-mathrmCH_3$-\mathrm{CH_3}$) implies mathrmsp^3$\implies \mathrm{sp^3}$ ### Step 2: Calculate the Count Summing the hybridization counts: - mathrmsp^3$\mathrm{sp^3}$ carbons: mathrmC_3$\mathrm{C}_3$, mathrmC_6$\mathrm{C}_6$, and 2 times mathrmCH_3$2 \times \mathrm{CH_3}$ on mathrmC_3$\mathrm{C}_3$ = 4$= 4$ carbon atoms - mathrmsp^2$\mathrm{sp^2}$ carbons: mathrmC_1$\mathrm{C}_1$ and mathrmC_2$\mathrm{C}_2$ = 2$= 2$ carbon atoms - mathrmsp$\mathrm{sp}$ carbons: mathrmC_4$\mathrm{C}_4$ and mathrmC_5$\mathrm{C}_5$ = 2$= 2$ carbon atoms Thus, the number of mathrmsp^3$\mathrm{sp^3}$, mathrmsp^2$\mathrm{sp^2}$ and mathrmsp$\mathrm{sp}$ hybridized carbons is 4, 2, 2$4, 2, 2$ respectively. ### Pattern Recognition Quick Tip: Always draw side substituents (like methyl groups) explicitly. A common mistake is to skip counting the methyl substituent carbons as mathrmsp^3$\mathrm{sp^3}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula textMethod Selection = fleft( Delta T_mathrmb, textdecomposition threshold, textvolatility with steam right)$$\text{Method Selection} = f\left( \Delta T_{\mathrm{b}}, \text{decomposition threshold}, \text{volatility with steam} \right)$$ ### Core Logic Let's align each purification technique to its designated mixtures based on NCERT guidelines: - **(A) Simple distillation**: Used for liquids having a significant difference in their boiling points (>30~mathrmK$>30~\mathrm{K}$ or 30^circmathrmC$30^\circ\mathrm{C}$). Chloroform (b.p. 334~mathrmK$334~\mathrm{K}$) and aniline (b.p. 457~mathrmK$457~\mathrm{K}$) are separated easily using simple distillation rightarrow$\rightarrow$ **(III)**. - **(B) Fractional distillation**: Used if boiling point differences of the components are very close (less than 25~mathrmK$25~\mathrm{K}$). Separation of petrochemical fractions such as diesel and petrol uses this technique rightarrow$\rightarrow$ **(I)**. - **(C) Distillation under reduced pressure**: Used for liquids that tend to decompose at or below their normal boiling points. Glycerol is separated from spent-lye in soap manufacturing industry using this vacuum method to prevent glycerol decomposition rightarrow$\rightarrow$ **(IV)**. - **(D) Steam distillation**: Applied to substances which are steam-volatile and completely immiscible in water. Aniline and water are separated using this technique rightarrow$\rightarrow$ **(II)**. ### Step 1: Conclusion Thus, the correct match is: **(A)-(III), (B)-(I), (C)-(IV), (D)-(II)** This corresponds perfectly to option (4). ### Pattern Recognition Glycerol from spent-lye is a highly tested practical chemistry concept. Remember that vacuum distillation lowers the boiling point, permitting evaporation without decomposition. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques