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

Reference Study Guides

More Organic Chemistry - Some Basic Principles and Techniques Previous-Year Questions — Page 13

Q85 jee_main_2024_27_jan_morning Stereoisomerism
3-Methylhex-2-ene on reaction with textHBr in presence of peroxide forms an addition product (A). The number of possible stereoisomers for 'A' is textquadquad.
Numerical Answer. Answer: 4 to 4

Solution

### Core Logic In the presence of peroxide, textHBr undergoes an anti-Markovnikov free-radical addition mechanism across the alkene bond of 3-methylhex-2-ene. Bromine appends to carbon-2 while hydrogen attaches at carbon-3. The generated dynamic addition product (A) corresponds structurally to 2-bromo-3-methylhexane.
Addition product chiral identification map for Q85 - JEE Main 2024 Morning
Addition product chiral identification map for Q85 - JEE Main 2024 Morning
### Step 1: Check chiral centers count Inspecting the structure of 2-bromo-3-methylhexane reveals two asymmetric chiral centers: - Carbon-2 carrying text-H, -CH_3text, -Br, and -CH(CH_3text)CH_2textCH_2textCH_3 - Carbon-3 carrying text-H, -CH_3text, -CH(Br)CH_3text, and -CH_2textCH_2textCH_3 Since both stereocenters are unsymmetrical (n=2): textTotal Stereoisomers = 2^n = 2^2 = 4 ### Pattern Recognition Anti-Markovnikov hydrobromination often generates multiple stereocenters. Count unsymmetrical C^* elements systematically. ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 11 Chemistry: Hydrocarbons
Q86 jee_main_2024_27_jan_morning Aromaticity
Among the given organic compounds, the total number of aromatic compounds is textquadquad.
Structures mapping A, B, C, D molecular properties for Q86 - JEE Main 2024 Morning
Four distinctive hydrocarbon ring structures depicted as options A, B, C, D.
Structures mapping A, B, C, D molecular properties for Q86 - JEE Main 2024 Morning
Four distinctive hydrocarbon ring structures depicted as options A, B, C, D.
Numerical Answer. Answer: 3 to 3

Solution

### Core Logic To satisfy Huckel's rules for aromaticity, a system must be planar, cyclic, completely conjugated, and possess (4n+2)pi delocalized electrons. Evaluating structures B, C, and D against these criteria shows they follow the (4n+2)pi electron counting rules successfully. Thus, three structures are fully aromatic. ### Pattern Recognition Verify ring conjugation continuity and apply Huckel's electron count constraints (2, 6, 10, 14dots). ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 11 Chemistry: Hydrocarbons
Q87 jee_main_2024_27_jan_morning Electrophilic Aromatic Substitution
Among the following, total number of meta directing functional groups is (Integer based): -textOCH_3, -textNO_2, -textCN, -textCH_3, -textNHCOCH_3, -textCOR, -textOH, -textCOOH, -textCl
Numerical Answer. Answer: 4 to 4

Solution

### Core Logic Meta-directing functional groups are electron-withdrawing groups via resonance or induction effects (-M or -I). Let's audit the list: - -textOCH_3 rightarrow textortho/para (+M) - -textNO_2 rightarrow textmeta (-M, -I) - -textCN rightarrow textmeta (-M, -I) - -textCH_3 rightarrow textortho/para (+I, hyperconjugation) - -textNHCOCH_3 rightarrow textortho/para (+M) - -textCOR rightarrow textmeta (-M, -I) - -textOH rightarrow textortho/para (+M) - -textCOOH rightarrow textmeta (-M, -I) - -textCl rightarrow textortho/para (+M counteracted by strong -I) The meta-directing groups are -textNO_2, -textCN, -textCOR, and -textCOOH. ### Pattern Recognition Groups whose attachment atom carries a double or triple bond to an electronegative element typically function as meta-directors. ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Haloalkanes and Haloarenes
Q64 jee_main_2024_29_jan_morning Resonance and Resonance Energy
The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is
  • A. textelectromagnetic energy
  • B. textresonance energy
  • C. textionization energy
  • D. texthyperconjugation energy

Solution

### Core Logic By definition, a resonance hybrid (the actual structure) is always more stable than any of its contributing canonical structures (resonance structures). The difference in potential energy between the most stable contributing resonance structure and the actual resonance hybrid is called the resonance energy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques
Q69 jee_main_2024_29_jan_morning Qualitative Analysis of Organic Compounds
Appearance of blood red colour, on treatment of the sodium fusion extract of an organic compound with FeSO_4 in presence of concentrated H_2SO_4 indicates the presence of element/s
  • A. textBr
  • B. textN
  • C. textN and S
  • D. textS

Solution

### Core Logic In the Lassaigne's test for elemental analysis, if an organic compound contains both Nitrogen and Sulphur together, the sodium fusion extract contains sodium thiocyanate (NaSCN) instead of sodium cyanide (NaCN) and sodium sulphide (Na_2S). Na + C + N + S rightarrow NaSCN When this extract is treated with Iron (II) sulphate (FeSO_4) and acidified with concentrated H_2SO_4 (which oxidizes some Fe^2+ to Fe^3+), the Fe^3+ ions react with the thiocyanate ions to form a blood-red colored complex. ### Step 1: The Reaction Fe^2+ xrightarrow[textConc. H_2SO_4, H^+ Fe^3+ Fe^3+ + 3SCN^- rightarrow Fe(SCN)_3 quad text(blood red colour) The appearance of the blood red colour specifically confirms the simultaneous presence of both Nitrogen and Sulphur. ### Pattern Recognition N alone rightarrow Prussian Blue (Fe_4[Fe(CN)_6]_3). S alone rightarrow Purple colour with sodium nitroprusside or black ppt of PbS. N + S together rightarrow Blood red colour (Fe(SCN)_3). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

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