### 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

### 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