Q.48 Chemistry ⭐ 360° Option Elimination Active

NEET 2026 Chemistry Question 48 Solution

Topic: Organic Chemistry: Some Basic Principles and Techniques · Official NTA Booklet Code 50

Question

The following carbocation is stabilized by the interaction of the empty p orbital with: C₆H₅−C⁺H−CH₃ (1-phenylethyl cation)

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Verified Against NTA Final Answer Key

Option (1) : filled σ and filled π orbitals

📚 Organic Chemistry: Some Basic Principles and Techniques Medium

📋 Given Parameters & Problem Setup

Species 1-phenylethyl cation [PhCH(⁺)CH₃]

Carbocation Carbon sp² hybridized with a vacant, unhybridized 2p orbital

Adjacent Groups Phenyl ring (π system) and methyl group (C-H σ bonds)

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The 30-Second Exam Hall Shortcut: ⚡ Electron Donor Rules: Empty Accepts from Filled!

⏱️ Save ~60 seconds

Stabilization requires electron density to flow INTO the vacant p orbital from FILLED orbitals!

1 The carbocation carbon has an EMPTY p-orbital (electron-deficient).

2 To stabilize this deficiency, neighboring groups must DONATE electron density.

3 Donation comes from: (i) the FILLED π-orbitals of the aromatic ring (resonance), and (ii) the FILLED σ-orbitals of the adjacent C-H bonds (hyperconjugation).

4 Empty cannot donate to empty! Both donating orbitals must be FILLED (filled σ and filled π orbitals). Option (1) is instantly deduced in 10 seconds!

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Why You Can 100% Trust This Shortcut: Fundamental molecular orbital theory: Chemical stabilization via delocalization always involves overlap between a filled donor orbital (HOMO/bonding σ or π) and an empty acceptor orbital (LUMO/vacant p).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Analyze the Electronic Structure of the Carbocation

In the 1-phenylethyl carbocation, the positively charged carbon is sp² hybridized. It forms three σ bonds in a trigonal planar geometry, leaving one unhybridized 2p orbital that is completely empty (contains 0 electrons).

\text{Carbocation C: } [\text{sp}^2]^3 \quad (\text{empty } 2p_z \text{ orbital, LUMO})

• Hybridization of C⁺: sp².
• Acceptor orbital: vacant unhybridized 2p orbital.

Step 2

Resonance Stabilization with the Aromatic Benzene Ring

The empty 2p orbital on the benzylic carbon aligns parallel to the adjacent p-orbitals of the aromatic ring. This enables delocalization of the filled aromatic π-bonding electrons into the empty p-orbital, dispersing the positive charge across the ortho and para positions of the benzene ring.

\pi_{\text{filled}}(\text{ring}) \longrightarrow p_{\text{empty}}(\text{C}^+) \quad (\text{Resonance / Mesomeric effect})

• Donor: Filled π-bonding molecular orbitals of the benzene ring.
• Effect: Extensive benzylic resonance stabilization.

Step 3

Hyperconjugative Stabilization with the Adjacent Methyl Group

The empty 2p orbital also interacts with the adjacent sp³-hybridized C-H σ-bonding orbitals of the methyl group. This no-bond resonance (hyperconjugation) involves the overlap of the filled C-H σ orbital with the empty p orbital.

\sigma_{\text{C-H, filled}} \longrightarrow p_{\text{empty}}(\text{C}^+) \quad (\text{Hyperconjugation})

• Donor: Filled σ orbitals of the adjacent C-H bonds.
• Overall Interaction: Empty p orbital interacts with filled σ and filled π orbitals.

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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:

Exact physical description. Filled C-H σ-orbitals donate via hyperconjugation, and filled aromatic π-orbitals donate via resonance into the vacant p-orbital.

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Why this is airtight: None; rigorous application of frontier molecular orbital theory.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Overlap between two empty orbitals provides zero electron density and zero electronic stabilization.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming orbitals of the same occupancy interact to stabilize.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. The empty p-orbital does not get stabilized by empty σ* antibonding orbitals; it is stabilized by filled bonding σ orbitals.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing hyperconjugation with back-bonding into antibonding orbitals.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Empty antibonding orbitals cannot donate electrons to a positively charged empty orbital.

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Student Fallacy (Why candidates pick this trap): Selecting antibonding orbitals without considering electron occupancy.

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NTA Trap Alert: Confusing Donor and Acceptor Roles in Hyperconjugation

Students frequently confuse whether hyperconjugation involves $\sigma \to p$ or $\sigma \to \pi^*$ . In carbocations, the acceptor is the vacant p-orbital, and the donors are the FILLED bonding σ and π orbitals.

🛡️ How to avoid the -1 mark: Remember: Stabilization of a cation requires electron donation. Donors must be FILLED; acceptors are EMPTY.

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Core NEET Syllabus Standard

NEET Chemistry Core Syllabus · General Organic Chemistry Standard Benchmark

Core Scientific Principle Tested:

"Carbocations are stabilized by overlap of the vacant p orbital with filled adjacent σ and π orbitals."

Syllabus Focus: Carbocation stability, Resonance, and Hyperconjugation

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Empty p needs electron feast: Filled σ and Filled π feed the beast!

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NEET 2026 Chemistry Question 48 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Electron Donor Rules: Empty Accepts from Filled!

Step-by-Step Rigorous Workflow & Mathematical Proof

Analyze the Electronic Structure of the Carbocation

Resonance Stabilization with the Aromatic Benzene Ring

Hyperconjugative Stabilization with the Adjacent Methyl Group

360° Option Elimination Engine Key Feature

NTA Trap Alert: Confusing Donor and Acceptor Roles in Hyperconjugation

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Organic Chemistry: Some Basic Principles and Techniques

Matching Organic Mixtures with Specific Distillation Techniques

Distinction Between Lassaigne’s Sodium Fusion Test and Copper Oxide Test for Carbon

Purification of Solids by Sublimation

Stability Order of Alkyl Carbocations (Hyperconjugation and Inductive Effect)

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