In the given structure, number of sp and sp^2 hybridized carbon atoms present respectively are :
Hybridization of Carbon diagram for Q30 - JEE Main 2025 Evening
The skeletal molecular diagram presents a chain containing carbonyl, double bonds, a triple bond, and a nitrile terminal functional grouping.

Solution & Explanation

### Core Logic To determine carbon atom hybridization within skeletal networks, evaluate the count of steric components (sigma-bonds attached to each carbon): * 4 text sigmatext-bonds ightarrow sp^3 * 3 text sigmatext-bonds ightarrow sp^2 (typically carbons forming one double bond like mathrmC=O or mathrmC=C) * 2 text sigmatext-bonds ightarrow sp (typically carbons forming a triple bond like -mathrmCequiv C- or -mathrmCequiv N) ### Step 1: Specific Atom Assignment Let's perform an audit across the skeletal sequence:
Hybridization of Carbon solution diagram for Q30 - JEE Main 2025 Evening
The skeletal molecular diagram presents a chain containing carbonyl, double bonds, a triple bond, and a nitrile terminal functional grouping.
* sp^2 Carbons: 1. Carbonyl carbon (mathrmC=O) 2. Carbons sharing the first alkene motif (2 atoms) 3. Carbons sharing the second alkene motif (2 atoms) Total sp^2 carbons = 1 + 2 + 2 = 5 * sp Carbons: 1. Carbons bound in the central alkyne group (-mathrmCequiv C-) (2 atoms) 2. Terminal nitrile carbon (-mathrmCequiv N) (1 atom) Total sp carbons = 2 + 1 = 3 ### Pattern Recognition Count all carbons involved in triple bonds (alkynes, nitriles) to find sp items. Count all double-bonded carbons (alkenes, ketones) to quickly isolate the sp^2 population. ### 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 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

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