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$32\text{ g mol}^{-1}$). Molar mass of barium sulphate is 233text g mol^-1$233\text{ g mol}^{-1}$.
A.4.55\%$4.55\%$
B.10.30\%$10.30\%$
C.21.97\%$21.97\%$
D.16.48\%$16.48\%$
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$$\text{Percentage of Sulphur} = \frac{\text{Mass of } S \text{ in } BaSO_4}{\text{Molar mass of } BaSO_4} \times \frac{\text{Mass of } BaSO_4 \text{ formed}}{\text{Mass of organic compound}} \times 100$$
### 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
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### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
### Core Logic
The enol form of structure (2) produces a fully conjugated, aromatic ring system (phenol derivative) which provides immense resonance stabilization. Therefore, the equilibrium lies heavily toward the enol form. Four choices depicting structure inputs for keto compounds tautomerizing to enols.
### Pattern Recognition
Look for enol forms that attain aromaticity. Aromatic stabilization overrides typical keto-preference factors.
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
### Core Logic
Option (4) is a cyclic secondary aliphatic amine (piperidine derivative) where the nitrogen atom is textsp^3$\text{sp}^3$ hybridized and its lone pair is entirely localized, making it highly available to accept a proton. In contrast, options (1), (2), and (3) have lone pairs involved in resonance with aromatic systems or unsaturated structures. Four different amine ring structures listed as options.
### Pattern Recognition
Localized aliphatic amines are consistently stronger Bronsted bases than aromatic or delocalized analogs.
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
### Core Logic
Option (4) features an active methylene group flanked directly between two electron-withdrawing carbonyl groupings. The removal of a proton from this central -textCH_2-$-\text{CH}_2-$ carbon produces a conjugate base that is strongly stabilized via extensive delocalization across both oxygen atoms. Four carbonyl organic structures presented as options.
### Pattern Recognition
Look for hydrogens between two -M$-M$ / -I$-I$ carbonyl complexes rightarrow$\rightarrow$ Active methylene effect.
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q74jee_main_2024_27_jan_morningClassification of Organic Compounds
Cyclohexene is textquadquad$\text{\quad\quad}$ type of an organic compound.
A. Benzenoid aromatic
B. Benzenoid non-aromatic
C. Acyclic
D. Alicyclic
Solution
### Core Logic
Cyclohexene features an aliphatic carbon ring framework containing an unsaturated double bond but lacks an aromatic sextet ring structure. It belongs to the alicyclic category (aliphatic + cyclic compounds). Cyclohexene chemical ring diagram for Q74 - JEE Main 2024 Morning
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q77jee_main_2024_27_jan_morningIUPAC Nomenclature
IUPAC name of following compound (P) is:
Substituted cyclohexane molecule labeled as compound P.Substituted cyclohexane molecule labeled as compound P.
A. 1-Ethyl-5, 5-dimethylcyclohexane
B. 3-Ethyl-1,1-dimethylcyclohexane
C. 1-Ethyl-3, 3-dimethylcyclohexane
D. 1,1-Dimethyl-3-ethylcyclohexane
Solution
### Core Logic
Number the ring to give substituents the lowest possible locants. Setting locant 1 at the carbon carrying the two methyl groups provides a locant list of (1,1,3)$(1,1,3)$, whereas setting it at the ethyl-bearing carbon gives (1,3,3)$(1,3,3)$. The set (1,1,3)$(1,1,3)$ wins by the lowest locant rule. Then arrange alphabetically: 3-ethyl precedes 1,1-dimethyl. Hence, the correct systematic tag is 3-Ethyl-1,1-dimethylcyclohexane. Substituted cyclohexane molecule labeled as compound P.
### Pattern Recognition
Lowest locant grouping set takes ultimate priority before checking alphabetical organization rules.
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
More Organic Chemistry - Some Basic Principles and Techniques Questions — jee_main_2026_21_jan_morning
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