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 12

Q62 jee_main_2024_27_jan_morning Isomerism
  • A. Structure A
  • B. Structure B
  • C. Structure C
  • D. Structure D

Solution

### 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.
Enol conversion pathway diagram for Q62 - JEE Main 2024 Morning
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
Q64 jee_main_2024_27_jan_morning Basic Strength
Which of the following is strongest Bronsted base?
  • A. Option 1
  • B. Option 2
  • C. Option 3
  • D. Option 4

Solution

### Core Logic Option (4) is a cyclic secondary aliphatic amine (piperidine derivative) where the nitrogen atom is textsp^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.
Lone pair localization logic diagram for Q64 - JEE Main 2024 Morning
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
Q66 jee_main_2024_27_jan_morning Acidic Strength
Which of the following has highly acidic hydrogen?
  • A. Structure 1
  • B. Structure 2
  • C. Structure 3
  • D. Structure 4

Solution

### 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- carbon produces a conjugate base that is strongly stabilized via extensive delocalization across both oxygen atoms.
Conjugate base stabilization resonance scheme for Q66 - JEE Main 2024 Morning
Four carbonyl organic structures presented as options.
### Pattern Recognition Look for hydrogens between two -M / -I carbonyl complexes rightarrow Active methylene effect. ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q74 jee_main_2024_27_jan_morning Classification of Organic Compounds
Cyclohexene is textquadquad 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
Cyclohexene chemical ring diagram for Q74 - JEE Main 2024 Morning
### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Q77 jee_main_2024_27_jan_morning IUPAC Nomenclature
IUPAC name of following compound (P) is:
Substituted ring framework structural layout for Q77 - JEE Main 2024 Morning
Substituted cyclohexane molecule labeled as compound P.
Substituted ring framework structural layout for Q77 - JEE Main 2024 Morning
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), whereas setting it at the ethyl-bearing carbon gives (1,3,3). The set (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.
Locant indexing direction chart for Q77 - JEE Main 2024 Morning
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
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%)