In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass 32text g mol^-1). Molar mass of barium sulphate is 233text g mol^-1.

Solution & Explanation

### 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 ### Core Logic Molar mass of BaSO_4 = 233 g/mol. Mass of Sulfur (S) in 1 mole of BaSO_4 = 32 g. Mass of BaSO_4 formed = 1.2 g. Mass of organic compound = 0.75 g. \% mathrmS = frac32233 times frac1.20.75 times 100 \% mathrmS = frac32 times 1.2 times 100233 times 0.75 = frac3840174.75 approx 21.97\% ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

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Organic Chemistry Some Basic Principles And Techniques

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 2

Q jee_main_2025_02_april_evening Quantitative Estimation (Dumas Method)
In Dumas' method for estimation of nitrogen, 0.5 gram of an organic compound gave 60~mathrmmL of nitrogen collected at 300mathrmK temperature and 715~mathrmmmHg pressure. The percentage composition of nitrogen in the compound (Aqueous tension at 300mathrmK = 15~mathrmmmHg) is
  • A. 1.257
  • B. 20.87
  • C. 18.67
  • D. 12.57

Solution

### Related Formula p_mathrmN_2 = p_texttotal - p_textaq n_mathrmN_2 = fracp_mathrmN_2 VR T \% mathrmN = fractextMass of nitrogentextMass of organic compound times 100 ### Core Logic Dumas' method estimates nitrogen by collecting dry nitrogen gas (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 Converting pressure to atmospheres: p_mathrmN_2 = frac700760~mathrmatm ### Step 2: Calculate Moles of Nitrogen Gas Using the ideal gas law with R = 0.0821~mathrmL~atm~mol^-1~K^-1, T = 300~mathrmK, and V = 60~mathrmmL = 60 times 10^-3~mathrmL: n_mathrmN_2 = fracleft(frac700760right) times 60 times 10^-30.0821 times 300 n_mathrmN_2 = frac0.92105 times 0.06024.63 approx 2.244 times 10^-3~mathrmmol ### Step 3: Calculate Mass and Percentage of Nitrogen The molar mass of mathrmN_2 is 28~mathrmg~mol^-1: textMass of mathrmN_2 = n_mathrmN_2 times 28 = 2.244 times 10^-3 times 28 approx 0.06283~mathrmg Now find the percentage in 0.5~mathrmg of organic compound: \% mathrmN = frac0.06283~mathrmg0.5~mathrmg 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
Q jee_main_2025_02_april_morning Aromaticity and Huckel's Rule
Designate whether each of the following compounds is aromatic or not aromatic:
Aromaticity and Huckel's Rule diagram for Q26 - JEE Main 2025 Morning
The diagram displays eight different cyclic conjugated hydrocarbon compounds labeled (a) through (h) to evaluate for aromatic character.
Choose the correct answer from the options given below:
  • A. text(1) e, g aromatic and a, b, c, d, f, h not aromatic
  • B. text(2) b, e, f, g aromatic and a, c, d, h not aromatic
  • C. text(3) a, b, c, d aromatic and e, f, g, h not aromatic
  • D. text(4) a, c, d, e, h aromatic and b, f, g not aromatic

Solution

### 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)
Aromaticity analysis solutions diagram for Q26
The diagram displays eight different cyclic conjugated hydrocarbon compounds labeled (a) through (h) to evaluate for aromatic character.
Aromaticity analysis solutions diagram for Q26
The diagram displays eight different cyclic conjugated hydrocarbon compounds labeled (a) through (h) to evaluate for aromatic character.
### 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 electrons moving through the continuous loop. Odd number of pairs (1, 3, 5...) means aromatic (2pi, 6pi, 10pi). 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
Q jee_main_2025_02_april_morning Free Radical Stability
Consider the following compound (X) beginarrayc mathrm I \\ mathrm H - mathrm C equiv mathrm C - mathrm C H _ 2 - mathrm C H - mathrm C H _ 3 \\ mathrm I \\ mathrm C H _ 3 endarray The most stable and least stable carbon radicals, respectively, produced by homolytic cleavage of corresponding mathrmC - H bond are :
  • A. (1)\ textII, IV
  • B. (2)\ textIII, II
  • C. (3)\ textI, IV
  • D. (4)\ textII, I

Solution

### Related Formula Free radical stability structural hierarchy sequence: textResonance Stabilized (Propargyl/Allyl) > 3^circ > 2^circ > 1^circ > textVinylic/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 system, making it the most stable radical position. * **Position I** places the radical directly on an mathrmsp-hybridized carbon center. The high electronegativity of mathrmsp orbitals tightly holds the unpaired electron, making homolytic cleavage extremely difficult and rendering this intermediate the least stable radical position.
Free Radical Stability
Free Radical Stability
### Step 1: Verdict Therefore, the most stable and least stable positions are II and I, respectively. ### Pattern Recognition Radicals located on mathrmsp 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
Q jee_main_2025_02_april_morning Nucleophilic Acyl Substitution and Hydrolysis
Consider the following molecules :
Nucleophilic Acyl Substitution and Hydrolysis
Nucleophilic Acyl Substitution and Hydrolysis
The correct order of rate of hydrolysis is :
  • A. (1)\ r > q > p > s
  • B. (2)\ q > p > r > s
  • C. (3)\ p > r > q > s
  • D. (4)\ p > q > r > s

Solution

### 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
Nucleophilic Acyl Substitution and Hydrolysis
Nucleophilic Acyl Substitution and Hydrolysis
### Core Logic Let's analyze the leaving groups across all choices layout-by-row: * For **(p)**, the leaving group is mathrmCl^- (Very weak base, excellent leaving group). * For **(q)**, the leaving group is mathrmRCOO^- (Resonance stabilized carboxylate, good leaving group). * For **(r)**, the leaving group is mathrmRO^- (Alkoxide, strong base, poor leaving group). * For **(s)**, the leaving group is mathrmNH_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. ### 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
Q42 jee_main_2025_02_april_morning Empirical Formula Derivation
On complete combustion 1.0mathrm~g of an organic compound (X) gave 1.46mathrm~g of mathrmCO_2 and 0.567mathrm~g of mathrmH_2mathrmO. The empirical formula mass of compound (X) is ________ g. Given molar mass in mathrmg cdot mol^-1\ C:12,\ H:1,\ O:16
  • A. (1)\ 30
  • B. (2)\ 45
  • C. (3)\ 60
  • D. (4)\ 15

Solution

### Related Formula Elemental content calculation system equations: textMoles of C = fractextMass of mathrmCO_244 textMoles of H = 2 times fractextMass of mathrmH_2O18 ### Core Logic Let's perform the stoichiometry layout step-by-step: * Moles of mathrmC inside sample system: mathrmn_C = frac1.4644 = 0.033mathrm~mol textMass of C = 0.033 times 12 = 0.396mathrm~g * Moles of mathrmH inside sample system: mathrmn_H = 2 times frac0.56718 = 0.063mathrm~mol textMass of H = 0.063 times 1 = 0.063mathrm~g * Determine Oxygen mass by subtracting values from total starting mass: textMass of O = 1.0 - (0.396 + 0.063) = 0.541mathrm~g mathrmn_O = frac0.54116 = 0.033mathrm~mol * Find atomic whole-number ratio profile: mathrmC : H : O = 0.033 : 0.063 : 0.033 approx 1 : 2 : 1. * This gives an empirical configuration of mathrmCH_2O. ### Step 1: Evaluation Calculating formula mass: textEmpirical Mass = 12 + (2 times 1) + 16 = 30mathrm~g ### Pattern Recognition When calculated mole properties output identical numbers for two elements (0.033 for both C and O), their structural subscript ratio is exactly 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

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