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 :

Solution & Explanation

### 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

More Organic Chemistry - Some Basic Principles and Techniques 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
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