In which pairs, the first ion is more stable than the second?

Stability pair A components for Q30
The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).
Stability pair A components for Q30
The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).
Stability pair A components for Q30
The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).
Stability pair A components for Q30
The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).
Stability pair A components for Q30
The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).
Stability pair A components for Q30
The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).

Solution & Explanation

### Related Formula textStability propto textDelocalization of charge via resonance, mesomeric, and back-bonding effects ### Core Logic Evaluating each specific structural pair: - **Pair (A):** The first carbocation is stabilized by strong +M back-bonding from the methoxy (-OMe) oxygen lone pair, making it significantly more stable than the second. - **Pair (B):** The first carbanion is strongly stabilized by the -M and -I electronic effects of the nitro (-NO_2) group situated at the ortho position, whereas a carbocation in that spot would be destabilized. Thus, the first ion is more stable. - **Pair (C):** The second cation has extended allylic resonance stabilization, meaning the first is less stable. - **Pair (D):** Cation with -OMe backbonding is more stable than tertiary aliphatic carbocation, making the first less stable than the second. Thus, only in **(A) & (B)** is the first ion more stable than the second. ### Pattern Recognition Back-bonding from an adjacent oxygen lone pair always triumphs over standard inductive or hyperconjugative alkyl stability templates. For carbanions, ensure electron-withdrawing groups like -NO_2 match the sign of the charge. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Principles of Organic Chemistry
Carbocation and Carbanion Stability
The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).

More Some Basic Principles of Organic Chemistry Previous-Year Questions — Page 10

Q34 jee_main_2025_24_jan_morning Quantitative Analysis of Organic Compounds
Given below are two statements I and II. Statement I: Dumas method is used for estimation of "Nitrogen" in an organic compound. Statement II: Dumas method involves the formation of ammonium sulphate by heating the organic compound with conc mathrmH_2mathrmSO_4 In the light of the above statements, choose the correct answer from the options given below
  • A. Both Statement I and Statement II are true.
  • B. Statement I is false but Statement II is true
  • C. Both Statement I and Statement II are false.
  • D. Statement I is true but Statement II is false

Solution

### Core Logic Statement I is fully accurate: Dumas method is standardly applied for estimating elemental nitrogen across structural compounds. Statement II is incorrect: The reaction leading to ammonium sulphate generation by intense thermal heating alongside concentrated mathrmH_2mathrmSO_4 describes the **Kjeldahl method**, not the Dumas strategy. The Dumas process instead relies on burning carbonaceous compounds explicitly with copper oxide to convert nitrogen cleanly into free gas (N_2). ### Pattern Recognition Dumas method collects element gas N_2 via volumetric analysis; Kjeldahl maps digestions via standard (mathrmNH_4)_2mathrmSO_4 pathways. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q31 jee_main_2025_28_jan_evening Purification of Organic Compounds
The purification method based on the following physical transformation is : textSolid xrightarrow[(textX)]textHeat textVapour xrightarrow[(textX)]textCool textSolid
  • A. Sublimation
  • B. Distillation
  • C. Crystallization
  • D. Extraction

Solution

### Related Formula Direct phase transition without passing through an intermediate liquid phase defines sublimation: textSolid rightleftharpoons textVapour ### Core Logic The schematic diagram represents a solid turning directly into vapor on heating, which then reverts back to a solid phase upon cooling. This distinct behavior isolates sublimable solids from non-sublimable impurities. ### Step 1: Identification This transformation perfectly defines the laboratory purification process known as **Sublimation**. ### Pattern Recognition Look for the skipping of the liquid state entirely: Solid rightarrow Vapour rightarrow Solid. Common examples include camphor, naphthalene, benzoic acid, and iodine. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q44 jee_main_2025_28_jan_evening Isomerism
Given below are two statements: Statement (I): Oxacyclobutane and prop-2-en-1-ol are isomeric compounds. Statement (II): Propan-1-amine and N-methylethanamine are functional group isomers. In the light of the above statements, choose the correct answer from the options given below :
  • A. Both Statement I and Statement II are false
  • B. Both Statement I and Statement II are true
  • C. Statement I is true but Statement II is false
  • D. Statement I is false but Statement II is true

Solution

### Related Formula Isomers share an identical molecular formula but differ in structural arrangement or functional groups: textSame M_f neq textSame structural layout ### Core Logic Evaluating the structural parameters: - **Statement I**: Oxacyclobutane (a cyclic ether) and prop-2-en-1-ol (an unsaturated alcohol) both possess the molecular formula C_3H_6O. They are functional/ring-chain isomers, so Statement I is true. - **Statement II**: Propan-1-amine (1^circ amine) and N-methylethanamine (2^circ amine) both share the molecular formula C_3H_9N. Because primary, secondary, and tertiary amines contain different functional groups, they act as functional group isomers. Thus, Statement II is true. ### Step 1: Conclusion Match Since both structural statements are valid, both Statement I and Statement II are true.
Skeletal representations for the specified organic isomers
Skeletal representations for the specified organic isomers
### Pattern Recognition Always remember that 1^circ, 2^circ, and 3^circ amines are classified as *different functional groups* in IUPAC nomenclature. Consequently, structural shifts between them with a constant carbon count represent functional group isomerism. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q jee_main_2025_29_jan_morning Nucleophiles and Electrophiles
Total number of nucleophiles from the following is :- mathrm N H _ 3, mathrm P h S H, (mathrm H _ 3 mathrm C) _ 2 mathrm S, mathrm H _ 2 mathrm C = mathrm C H _ 2, stackrel ominus mathrm O mathrm H, mathrm H _ 3 mathrm O ^ oplus, (mathrm C H _ 3) _ 2 mathrm C O, > = mathrm N C H _ 3
  • A. 5
  • B. 4
  • C. 7
  • D. 6

Solution

### Related Formula Nucleophiles are electron-rich species containing lone pairs of electrons or pi-bonds that can donate an electron pair to an electrophilic center. ### Core Logic Let us examine each species: * mathrmNH_3: Contains a lone pair on nitrogen rightarrow Nucleophile * mathrmPhSH: Contains lone pairs on sulfur rightarrow Nucleophile * (mathrmH_3mathrmC)_2mathrmS: Contains lone pairs on sulfur rightarrow Nucleophile * mathrmH_2mathrmC=mathrmCH_2: Contains a nucleophilic pi-bond rightarrow Nucleophile * stackrelominusmathrmOmathrmH: Negatively charged with lone pairs rightarrow Nucleophile * mathrmH_3mathrmO^oplus: Electron deficient, positively charged oxygen cannot donate electrons rightarrow Electrophile * (mathrmCH_3)_2mathrmCO: Carbonyl carbon is electrophilic * >=mathrmNCH_3: Imine carbon is electrophilic Thus, the total number of nucleophiles is 5. ### Pattern Recognition Neutral molecules with lone pairs (N, S) or alkenes/alkynes with available pi-electrons operate as good nucleophiles, along with full anions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q jee_main_2025_29_jan_morning IUPAC Nomenclature of Organic Compounds
Match List-I with List-II. Choose the correct answer from the options given below:
IUPAC Nomenclature of Organic Compounds
IUPAC Nomenclature of Organic Compounds
  • A. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  • B. (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  • C. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  • D. (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

Solution

### 1. RELATED FORMULA IUPAC Rules select the longest principal carbon chain and number it to give substituents the lowest possible locants. ### 2. EXECUTION **CORE LOGIC** Let's systematic decode each name : * (A) Longest chain contains 7 carbons (heptane) with an ethyl group at position 3 and a methyl group at position 5 rightarrow 3-Ethyl-5-methylheptane (II) . * (B) C expanded yields a 7 carbon main chain with two methyl groups at carbon-4 rightarrow 4,4-Dimethylheptane (III) . * (C) 5-carbon diene numbered from the left double bond side rightarrow 2-Methyl-1,3-pentadiene (IV) . * (D) 5-carbon alkene starting from the double bond end rightarrow 4-Methylpent-1-ene (I) . Therefore, matching sequence: (A)-(II), (B)-(III), (C)-(IV), (D)-(I). ### 3. PATTERN RECOGNITION Expanding compressed groupings such as mathrm(C_3H_7)_2 prevents errors regarding parent chain carbon counts. ### 4. EVALUATION RUBRIC / MODEL ANSWER null ### 5. CHAPTER MIX Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)