Q.76 Chemistry ⭐ 360° Option Elimination Active

NEET 2025 Chemistry Question 76 Solution

Topic: Hydrocarbons · Official NTA Booklet Code 45

Question

Among the given compounds I-III, the correct order of bond dissociation energy of C–H bond marked with * is: I. benzene (aromatic C–H) II. phenylacetylene (terminal alkyne C–H) III. cyclopropene (ring CH₂ C–H)

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Verified Against NTA Final Answer Key

Option (1) : II > I > III

📚 Hydrocarbons Medium

📋 Given Parameters & Problem Setup

Compound I Benzene (sp² C-H bond)

Compound II Phenylacetylene (sp C-H bond on terminal alkyne)

Compound III Cyclopropene (sp³ allylic C-H bond)

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The 30-Second Exam Hall Shortcut: ⚡ s-Character and Radical Stability Shortcut

⏱️ Save ~40 seconds

Bond Dissociation Energy (BDE) ∝ s-character of C ∝ 1 / (Radical Stability)! Terminal alkyne (sp, 50% s) = STRONGEST bond (II, highest BDE). Benzene (sp², 33% s) = intermediate (I). Cyclopropene allylic C-H gives resonance-stabilized allylic radical = WEAKEST bond (III, lowest BDE)! ⟹ II > I > III!

1 Recall relationship: Higher s-character brings bonding electrons closer to nucleus, resulting in a shorter, stronger C-H bond with higher BDE.

2 Alternatively: BDE is inversely proportional to the stability of the free radical formed upon homolysis.

3 Compound II (Phenylacetylene): C-H is sp hybridized (50% s-character). It has the highest electronegativity and forms an unstable sp-radical ⟹ HIGHEST BDE (II).

4 Compound I (Benzene): C-H is sp² hybridized (33.3% s-character). The phenyl radical cannot delocalize into the aromatic π-system ⟹ intermediate BDE (I).

5 Compound III (Cyclopropene): Homolytic cleavage of the allylic C-H bond generates a resonance-stabilized allylic cyclopropenyl radical ⟹ LOWEST BDE (III).

6 Decreasing order of BDE: II > I > III.

7 Select Option (1) in under 15 seconds!

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Why You Can 100% Trust This Shortcut: Hybridization, s-Character, and Bond Strengths (NCERT Chemistry · Chemical Bonding and Hydrocarbons).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Fundamental Factors Governing C-H Bond Dissociation Energy

Bond dissociation energy (BDE) of a C-H bond is defined by homolytic cleavage: R-H ⟶ R• + H•. Thermodynamically, BDE reflects both orbital hybridization (s-character) and the stability of the resulting carbon radical:

\text{BDE} \propto \text{s-character of C-H bond} \propto \frac{1}{\text{Stability of } R^\bullet}

• Greater s-character holds the bonding pair closer to carbon, increasing bond strength.
• Forming an unstable radical requires significantly higher energy input.

Step 2

Evaluate Hybridization and BDE for Compounds II and I

Compare the terminal alkyne and aromatic C-H bonds:

\text{II (Phenylacetylene): } sp\text{-hybridized C-H } (50\%\text{ s-character}) \implies \text{BDE} \approx 545-555\text{ kJ/mol}

• An sp-hybridized C-H bond is the strongest among all hydrocarbon C-H bonds.
• I (Benzene): sp²-hybridized C-H (33.3% s-character) has BDE ≈ 470 kJ/mol. The phenyl radical resides in an sp² orbital orthogonal to the aromatic π-electron cloud and cannot be resonance-stabilized.
• Thus BDE(II) > BDE(I).

Step 3

Evaluate Compound III (Allylic Cyclopropenyl C-H)

Homolysis of the saturated C-H bond in cyclopropene yields an allylic free radical:

\text{III (Cyclopropene): } \text{Cleavage produces an allylically delocalized radical} \implies \text{BDE} \approx 360-375\text{ kJ/mol}

• The unpaired electron is resonance-delocalized over the adjacent double bond, substantially lowering the energy needed to break the bond.
• Therefore, Compound III has the lowest BDE.
• Final decreasing order of BDE: II > I > III, matching Option (1).

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360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 A

✓ 100% Correct

Verdict & Scientific Reason:

100% correct. sp alkyne (II) > sp² phenyl (I) > allylic cyclopropenyl (III).

💡

Why this is airtight: None; flawless physical organic chemistry analysis.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Falsely places sp² benzene above sp alkyne.

⚠️

Student Fallacy (Why candidates pick this trap): Underestimating the immense strength of an sp C-H bond.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Inverts the entire order, confusing stability with bond energy.

⚠️

Student Fallacy (Why candidates pick this trap): Inverting the relationship between radical stability and BDE.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Places allylic C-H ahead of benzene C-H.

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Student Fallacy (Why candidates pick this trap): Thinking ring strain in cyclopropene raises BDE above benzene.

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NTA Trap Alert: Confusing Ring Strain with C-H Bond Strength

Some candidates assume the ring strain in cyclopropene makes its C-H bond harder to break. In fact, generating the allylic radical relieves angle strain and provides allylic resonance, making this C-H bond the EASIEST to cleave (lowest BDE)!

🛡️ How to avoid the -1 mark: BDE follows hybridization: sp (alkyne) > sp² (benzene) > allylic (cyclopropene)!

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Core NEET Syllabus Standard

NEET Chemistry Core Syllabus · Hydrocarbons Standard Benchmark

Core Scientific Principle Tested:

"C-H bond strength increases with s-character: sp > sp² > sp³."

Syllabus Focus: Alkynes and Aromatic Hydrocarbons: C-H Bond Energies

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

s-character rules: Triple (sp, II) > Double/Aromatic (sp², I) > Allylic (III) ⟹ II > I > III!

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NEET 2025 Chemistry Question 76 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ s-Character and Radical Stability Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Fundamental Factors Governing C-H Bond Dissociation Energy

Evaluate Hybridization and BDE for Compounds II and I

Evaluate Compound III (Allylic Cyclopropenyl C-H)

360° Option Elimination Engine Key Feature

NTA Trap Alert: Confusing Ring Strain with C-H Bond Strength

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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