Q.56 Chemistry ⭐ 360° Option Elimination Active

NEET 2023 Chemistry Question 56 Solution

Topic: Alcohols, Phenols and Ethers · Official NTA Booklet Code G5

Question

Consider the following reaction and identify the product (P). CH₃−CH(CH₃)−CH(OH)−CH₃ (3-Methylbutan-2-ol) → (HBr) Product (P)

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Verified Against NTA Final Answer Key

Option (2) : 2-Bromo-2-methylbutane

📚 Alcohols, Phenols and Ethers Medium

📋 Given Parameters & Problem Setup

Substrate 3-Methylbutan-2-ol, (CH₃)₂CH–CH(OH)–CH₃

Reagent HBr

Mechanism S_N1 via carbocation intermediate

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The 30-Second Exam Hall Shortcut: ⚡ 1,2-Hydride Shift Hack: 2° Carbocation ⟶ Stable 3° Carbocation

⏱️ Save ~45 seconds

Loss of H₂O generates 2° carbocation: (CH₃)₂CH–C⁺H–CH₃. 1,2-H⁻ shift to adjacent tertiary carbon yields stable 3° carbocation: (CH₃)₂C⁺–CH₂–CH₃. Br⁻ attacks 3° carbon to give 2-bromo-2-methylbutane in 15 seconds!

1 Protonate alcohol -OH to form good leaving group -OH₂⁺.

2 Expel H₂O to create initial 2° carbocation at C-2: (CH₃)₂CH–C⁺H–CH₃ (4 hyperconjugative H).

3 Adjacent C-3 is a tertiary carbon holding a hydrogen.

4 Undergo rapid 1,2-hydride shift (1,2-H⁻ shift) to create a much more stable 3° carbocation: (CH₃)₂C⁺–CH₂–CH₃ (7 hyperconjugative H).

5 Bromide nucleophile (Br⁻) attacks the 3° carbocation.

6 Product is 2-bromo-2-methylbutane: CH₃–C(Br)(CH₃)–CH₂–CH₃.

7 Select Option (2) directly!

🛡️

Why You Can 100% Trust This Shortcut: Organic Reaction Mechanisms · Carbocation Rearrangements (Carbocations undergo 1,2-hydride or 1,2-alkyl shifts whenever a more stable tertiary intermediate is accessible).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Protonation of the Secondary Hydroxyl Group

In the presence of hydrobromic acid (HBr), the lone pair on the oxygen atom of 3-methylbutan-2-ol is protonated by the strong acid to form an oxonium ion intermediate:

(\text{CH}_3)_2\text{CH}-\text{CH}(\text{OH})-\text{CH}_3 + \text{H}^+ \rightleftharpoons (\text{CH}_3)_2\text{CH}-\text{CH}(\text{O}^+\text{H}_2)-\text{CH}_3

Step 2

Heterolytic Cleavage to Form Initial Secondary Carbocation

Water departs as a neutral leaving group, generating a secondary (2°) carbocation at C-2:

(\text{CH}_3)_2\text{CH}-\text{CH}(\text{O}^+\text{H}_2)-\text{CH}_3 \xrightarrow{-\text{H}_2\text{O}} (\text{CH}_3)_2\text{CH}-\overset{+}{\text{C}}\text{H}-\text{CH}_3 \quad (2^\circ\text{ Carbocation})

Step 3

Thermodynamically Favored 1,2-Hydride Shift

The adjacent C-3 carbon has a tertiary C-H bond. A hydride ion (:H⁻) migrates from C-3 to C-2 (1,2-hydride shift), transforming the less stable secondary carbocation into a significantly more stable tertiary (3°) carbocation with 7 hyperconjugative α-hydrogens:

(\text{CH}_3)_2\overset{\text{H}}{\text{C}}-\overset{+}{\text{C}}\text{H}-\text{CH}_3 \xrightarrow{1,2-\text{H}^-\text{ shift}} (\text{CH}_3)_2\overset{+}{\text{C}}-\text{CH}_2-\text{CH}_3 \quad (3^\circ\text{ Carbocation})

Step 4

Nucleophilic Capture by Bromide Ion

The bromide ion (Br⁻) rapidly attacks the planar tertiary carbocation to yield the major rearranged substitution product:

(\text{CH}_3)_2\overset{+}{\text{C}}-\text{CH}_2-\text{CH}_3 + \text{Br}^- \longrightarrow \text{CH}_3-\text{C}(\text{Br})(\text{CH}_3)-\text{CH}_2-\text{CH}_3 \quad (\text{2-Bromo-2-methylbutane})

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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 (1)

✗ Incorrect

Verdict & Scientific Reason:

Requires an unfeasible methyl shift from an already rearranged skeleton.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate misapplied Wagner-Meerwein shift rules.

2 (2)

✓ 100% Correct

Verdict & Scientific Reason:

Accurate major product resulting from 1,2-hydride shift to form the tertiary carbocation.

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Why this is airtight: None. Candidate correctly applied carbocation rearrangement principles.

3 (3)

✗ Incorrect

Verdict & Scientific Reason:

Elimination alkene product with loss of a methyl carbon (completely wrong carbon skeleton).

⚠️

Student Fallacy (Why candidates pick this trap): Candidate confused acid substitution with an elimination reaction and truncated the chain.

4 (4)

✗ Incorrect

Verdict & Scientific Reason:

Minor unrearranged product formed without hydride shift.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate forgot that secondary carbocations adjacent to tertiary carbons ALWAYS rearrange via hydride shift.

⚠️

NTA Trap Alert: Direct Replacement Without Carbocation Rearrangement (Picking Option 4)

Students who perform a simple direct substitution (-OH replaced by -Br) pick 2-bromo-3-methylbutane (Option 4). Because the reaction proceeds via a free carbocation, rearrangement to the 3° carbocation occurs with extreme velocity, making Option (2) the overwhelming major product.

🛡️ How to avoid the -1 mark: Whenever you see a secondary alcohol adjacent to a branch (tertiary carbon), ALWAYS check for a 1,2-hydride shift!

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Core NEET Syllabus Standard

National Eligibility cum Entrance Test (NEET UG) · Alcohols, Phenols and Ethers Standard Benchmark

Core Scientific Principle Tested:

"Secondary alcohols react with hydrogen halides through carbocation intermediates, undergoing 1,2-hydride shift to give more stable tertiary products."

Syllabus Focus: Reactions of Alcohols – Cleavage of C–O Bond

Preserved in Rationalized Curriculum Syllabus & Academic Standards ↗

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NEET 2023 Chemistry Question 56 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ 1,2-Hydride Shift Hack: 2° Carbocation ⟶ Stable 3° Carbocation

Step-by-Step Rigorous Workflow & Mathematical Proof

Protonation of the Secondary Hydroxyl Group

Heterolytic Cleavage to Form Initial Secondary Carbocation

Thermodynamically Favored 1,2-Hydride Shift

Nucleophilic Capture by Bromide Ion

360° Option Elimination Engine Key Feature

NTA Trap Alert: Direct Replacement Without Carbocation Rearrangement (Picking Option 4)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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