### Related Formula P_textdry textN_2 = P_texttotal - textAqueous Tension $P_{\text{dry } \text{N}_2} = P_{\text{total}} - \text{Aqueous Tension} $ PV = nRT implies n = fracPVRT $PV = nRT \implies n = \frac{PV}{RT} $ \%textN = fractextMass of NitrogentextMass of organic compound times 100 $\%\text{N} = \frac{\text{Mass of Nitrogen}}{\text{Mass of organic compound}} \times 100 $ ### Core Logic First, isolate the pressure contribution of the dry nitrogen gas: P_textdry textN2 = 715 - 15 = 700text mm Hg = frac700760text atm $P_{\text{dry } \text{N}2} = 715 - 15 = 700\text{ mm Hg} = \frac{700}{760}\text{ atm} $ Using the ideal gas parameters: - V = 50text mL = 0.050text L$V = 50\text{ mL} = 0.050\text{ L}$ [cite: 1073, 1076] - T = 300text K$T = 300\text{ K}$ - R = 0.0821text L atm mol^-1textK^-1$R = 0.0821\text{ L atm mol}^{-1}\text{K}^{-1}$ ### Step 1: Compute Moles and Mass Calculate total moles of textN_2$\text{N}_2$ molecules collected: ntextN2 = fracleft(frac700760 ight) times 0.0500.0821 times 300 approx 1.868 times 10^-3text mol $n{\text{N}2} = \frac{\left(\frac{700}{760} ight) \times 0.050}{0.0821 \times 300} \approx 1.868 \times 10^{-3}\text{ mol} $ Compute corresponding mass of atomic Nitrogen elements (2 times 14 = 28text g/mol$2 \times 14 = 28\text{ g/mol}$): [cite: 1016, 1017] textMass of N = ntextN2 times 28 = 1.868 times 10^-3 times 28 approx 0.0523text g = 52.3text mg$\text{Mass of N} = n{\text{N}2} \times 28 = 1.868 \times 10^{-3} \times 28 \approx 0.0523\text{ g} = 52.3\text{ mg}$ ### Step 2: Calculate Percentage Composition Applying the fraction formulation against total sample mass: [cite: 1021, 1022] \%textN = frac52.3text mg292text mg times 100 approx 17.91\% approx 18\% $\%\text{N} = \frac{52.3\text{ mg}}{292\text{ mg}} \times 100 \approx 17.91\% \approx 18\% $ ### Pattern Recognition Dumas analysis safety check: Always strip away the vapor pressure of water (aqueous tension) from the measured barometric value before computing the chemical molar counts. For standard conditions shortcuts, remember that 1text mole = 22400text mL$1\text{ mole} = 22400\text{ mL}$ at STP can act as an alternate path if values are normalized. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic To determine carbon atom hybridization within skeletal networks, evaluate the count of steric components (sigma$\sigma$-bonds attached to each carbon): * 4 text sigmatext-bonds ightarrow sp^3$4 \text{ } \sigma\text{-bonds} ightarrow sp^3$ * 3 text sigmatext-bonds ightarrow sp^2$3 \text{ } \sigma\text{-bonds} ightarrow sp^2$ (typically carbons forming one double bond like mathrmC=O$\mathrm{C=O}$ or mathrmC=C$\mathrm{C=C}$) * 2 text sigmatext-bonds ightarrow sp$2 \text{ } \sigma\text{-bonds} ightarrow sp$ (typically carbons forming a triple bond like -mathrmCequiv C-$-\mathrm{C\equiv C}-$ or -mathrmCequiv N$-\mathrm{C\equiv N}$) ### Step 1: Specific Atom Assignment Let's perform an audit across the skeletal sequence:

* sp^2$sp^2$ Carbons: 1. Carbonyl carbon (mathrmC=O$\mathrm{C=O}$) 2. Carbons sharing the first alkene motif (2 atoms) 3. Carbons sharing the second alkene motif (2 atoms) Total sp^2$sp^2$ carbons = 1 + 2 + 2 = 5$1 + 2 + 2 = 5$ * sp$sp$ Carbons: 1. Carbons bound in the central alkyne group (-mathrmCequiv C-$-\mathrm{C\equiv C}-$) (2 atoms) 2. Terminal nitrile carbon (-mathrmCequiv N$-\mathrm{C\equiv N}$) (1 atom) Total sp$sp$ carbons = 2 + 1 = 3$2 + 1 = 3$ ### Pattern Recognition Count all carbons involved in triple bonds (alkynes, nitriles) to find sp$sp$ items. Count all double-bonded carbons (alkenes, ketones) to quickly isolate the sp^2$sp^2$ population. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic Let's evaluate each statement based on electrophilic aromatic substitution guidelines: * Statement (A): Both -mathrmOCH_3$-\mathrm{OCH}_{3}$ and -mathrmNHCOCH_3$-\mathrm{NHCOCH}_{3}$ have lone pairs on the atom directly attached to the benzene ring. These lone pairs undergo resonance delocalization into the ring, increasing electron density and activating it toward substitution. This statement is true. * Statement (B): While -mathrmCN$-\mathrm{CN}$ is a deactivating meta-directing group, -mathrmOH$-\mathrm{OH}$ is a strongly activating ortho/para-directing group due to resonance. This statement is false. * Statement (C): Both -mathrmCN$-\mathrm{CN}$ and -mathrmSO_3mathrmH$-\mathrm{SO}_{3}\mathrm{H}$ withdraw electron density via inductive and resonance effects (-I, -M$-I, -M$). This deactivates the ring and directs substitution to the meta position. This statement is true. * Statement (D): Activating groups increase electron density primarily at the ortho and para positions via resonance, directing incoming electrophiles to those sites. This statement is true. * Statement (E): Halogens are an exception: they are deactivating due to their strong inductive effect (-I$-I$), but ortho/para-directing due to resonance (+M$+M$). This statement is false. Therefore, statements (A), (C), and (D) are correct, which matches Option (1). ### Pattern Recognition Groups that activate the aromatic ring by donating electrons via resonance always direct incoming electrophiles to the ortho and para positions. Halogens are a unique exception: they deactivate the ring but still direct to the ortho and para positions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes

### Related Formula For an unsymmetrical molecule with n$n$ distinct stereogenic units (chiral centers or stereogenic double bonds): textTotal Stereoisomers = 2^n$\text{Total Stereoisomers} = 2^n$ ### Core Logic Let's examine the structure of 5-phenylpent-4-en-2-ol: mathrmPh-CH=CH-CH_2-CH(OH)-CH_3 $\mathrm{Ph-CH=CH-CH_2-CH(OH)-CH_3} $ Identify the stereogenic units: 1. **Double Bond (mathrm-CH=CH-$\mathrm{-CH=CH-}$):** Positioned between carbons 4 and 5, this alkene group can exist in 2 distinct geometric configurations: *cis* (Z$Z$) or *trans* (E$E$). 2. **Chiral Carbon Center (*mathrmC$*\mathrm{C}$):** Carbon-2 is attached to four distinct groups: -mathrmH$-\mathrm{H}$, -mathrmOH$-\mathrm{OH}$, -mathrmCH_3$-\mathrm{CH}_3$, and -mathrmCH_2-CH=CH-Ph$-\mathrm{CH_2-CH=CH-Ph}$. This asymmetric carbon center can exist in 2 distinct optical configurations: (R$R$) or (S$S$). Since the molecule is unsymmetrical, the two stereogenic units behave independently (n = 2$n = 2$): textTotal Stereoisomers = 2^2 = 4$\text{Total Stereoisomers} = 2^2 = 4$ ### Visual Mapping The structural tracking confirms the presence of these stereogenic sites:

### Pattern Recognition Always break the molecule down to count chiral centers and stereogenic double bonds independently. Since the molecule has asymmetric ends, you can safely use the simplified 2^n$2^n$ formula without worrying about meso configurations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula For a hydrocarbon with the molecular formula mathrmC_x mathrmH_y$\mathrm{C}_x \mathrm{H}_y$, the Double Bond Equivalent (DBE) or Degree of Unsaturation (DU) is given by: textDU = x + 1 - fracy2$\text{DU} = x + 1 - \frac{y}{2}$ ### Core Logic 1. Calculate the mass of carbon in 1 mole of the hydrocarbon: textMass of Carbon = 80text g cdot 90\% = 72text g$\text{Mass of Carbon} = 80\text{ g} \cdot 90\% = 72\text{ g}$ 2. Find the number of carbon atoms (x$x$): x = frac7212 = 6$x = \frac{72}{12} = 6$ 3. Find the mass and number of hydrogen atoms (y$y$): textMass of Hydrogen = 80text g - 72text g = 8text g$\text{Mass of Hydrogen} = 80\text{ g} - 72\text{ g} = 8\text{ g}$ y = frac81 = 8$y = \frac{8}{1} = 8$ Thus, the molecular formula of hydrocarbon (X) is mathrmC_6mathrmH_8$\mathrm{C}_{6}\mathrm{H}_{8}$. 4. Calculate the degree of unsaturation: textDU = 6 + 1 - frac82 = 7 - 4 = 3$\text{DU} = 6 + 1 - \frac{8}{2} = 7 - 4 = 3$ ### Pattern Recognition First, use the percentage composition and total molar mass to determine the exact number of carbon and hydrogen atoms. Once you have the molecular formula, plug it into the standard textDU = x + 1 - fracy2$\text{DU} = x + 1 - \frac{y}{2}$ equation to find the total number of rings and/or pi bonds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques