Q.78 Chemistry ⭐ 360° Option Elimination Active

NEET 2025 Chemistry Question 78 Solution

Topic: Aldehydes, Ketones and Carboxylic Acids & Amines · Official NTA Booklet Code 45

Question

The major product of the following reaction is: 4-oxo-4-phenylbutanenitrile (C₆H₅−CO−CH₂−CH₂−CN) → (i) CH₃MgBr (excess) (ii) H₃O⁺

✓

Verified Against NTA Final Answer Key

Option (2) : Ph-C(CH₃)(OH)-CH₂-CH₂-CO-CH₃ (Tertiary alcohol at C4, methyl ketone at C1)

📚 Aldehydes, Ketones and Carboxylic Acids & Amines Medium

📋 Given Parameters & Problem Setup

Substrate 4-Oxo-4-phenylbutanenitrile (Ph-CO-CH₂-CH₂-CN)

Step (i) Reagents CH₃MgBr (excess)

Step (ii) Reagent H₃O⁺ (Acidic hydrolysis workup)

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The 30-Second Exam Hall Shortcut: ⚡ Grignard on Ketone + Nitrile Fingerprint

⏱️ Save ~45 seconds

CH₃MgBr adds to Ketone (Ph-CO-) ⟹ Tertiary Alcohol Ph-C(CH₃)(OH)-! CH₃MgBr adds to Nitrile (-CN) to form imine salt, which on H₃O⁺ hydrolyzes to Methyl Ketone -CO-CH₃! Combining both gives Ph-C(CH₃)(OH)-CH₂-CH₂-CO-CH₃ (Option 2)!

1 Identify functional groups: Ph-CO- (ketone) and -CN (nitrile).

2 Reaction of Ketone with CH₃MgBr: Nucleophilic attack of methyl carbanion gives magnesium alkoxide, which upon H₃O⁺ workup yields a tertiary alcohol: Ph-C(CH₃)(OH)-.

3 Reaction of Nitrile with CH₃MgBr: Nucleophilic addition yields magnesium ketimine salt R-C(CH₃)=N-MgBr.

4 Crucial point: The imine magnesium salt does NOT add a second Grignard reagent under standard conditions.

5 Upon acidic aqueous workup (H₃O⁺), the ketimine hydrolyzes to a methyl ketone: -C(=O)-CH₃.

6 The final molecule has a tertiary alcohol on the phenyl side and a methyl ketone on the aliphatic chain.

7 Match Option (2) in under 20 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Grignard Addition to Nitriles and Carbonyls (NCERT Chemistry · Aldehydes, Ketones and Carboxylic Acids).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Grignard Addition to the Ketone Carbonyl

The methyl Grignard reagent (CH₃MgBr) acts as a strong nucleophile, attacking the electrophilic carbonyl carbon of the ketone:

\text{Ph}-\text{C}(=\text{O})-\text{CH}_2- + \text{CH}_3\text{MgBr} \longrightarrow \text{Ph}-\text{C}(\text{CH}_3)(\text{O}^-\text{MgBr}^+)-\text{CH}_2-

• Subsequent protonation during acidic workup (H₃O⁺) affords a tertiary alcohol: Ph-C(CH₃)(OH)-.

Step 2

Grignard Addition to the Nitrile Group and Imine Hydrolysis

Nitriles react with Grignard reagents by nucleophilic attack on the carbon-nitrogen triple bond to form an imine intermediate salt:

-\text{CH}_2-\text{C}\equiv\text{N} + \text{CH}_3\text{MgBr} \longrightarrow -\text{CH}_2-\text{C}(\text{CH}_3)=\text{N}-\text{MgBr}

• The negative charge on the imine nitrogen prevents a second Grignard addition.
• Upon acidic hydrolysis (H₃O⁺), the imine bond is cleaved to liberate a methyl ketone group:
• -\text{CH}_2-\text{C}(\text{CH}_3)=\text{N-MgBr} \xrightarrow{\text{H}_3\text{O}^+} -\text{CH}_2-\text{C}(=\text{O})-\text{CH}_3 + \text{NH}_4^+ + \text{Mg}^{2+} + \text{Br}^-.

Step 3

Consolidate the Major Product Structure

Combine the modified functionalities at both ends of the molecule:

\text{Major Product} = \text{Ph}-\text{C}(\text{CH}_3)(\text{OH})-\text{CH}_2-\text{CH}_2-\text{C}(=\text{O})-\text{CH}_3

• This structure corresponds exactly to Option (2).

⭐

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

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Leaves the nitrile group completely unreacted despite excess Grignard reagent.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming Grignard reagent does not react with nitriles.

2 B

✓ 100% Correct

Verdict & Scientific Reason:

100% correct. Ketone becomes 3° alcohol; nitrile becomes methyl ketone on hydrolysis.

💡

Why this is airtight: None; rigorous multi-functional organic synthesis.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Falsely assumes a second addition of Grignard to the nitrile to form a diol.

⚠️

Student Fallacy (Why candidates pick this trap): Treating nitrile like an ester and adding two equivalents of Grignard.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Leaves ketone unreduced and only reacts nitrile.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming nitrile is more reactive than ketone towards Grignard reagent.

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NTA Trap Alert: Treating Nitriles Like Esters (The Diol Trap)

Esters react with TWO equivalents of Grignard reagent to form tertiary alcohols (Option 3). But nitriles react with ONLY ONE equivalent because the imine magnesium salt cannot undergo elimination to regenerate a carbonyl in situ! Hydrolysis gives a KETONE, not a tertiary alcohol!

🛡️ How to avoid the -1 mark: Remember: Ester + excess Grignard = 3° Alcohol; Nitrile + Grignard + H₃O⁺ = KETONE!

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Core NEET Syllabus Standard

NEET Chemistry Core Syllabus · Aldehydes, Ketones and Carboxylic Acids Standard Benchmark

Core Scientific Principle Tested:

"R-CN + R'MgBr followed by H₃O⁺ yields ketone R-CO-R'."

Syllabus Focus: Preparation of Ketones from Nitriles & Reactions of Carbonyls with Grignard Reagent

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Ketone becomes 3° Alcohol; Nitrile stops at Ketone ⟹ Ph-C(OH)(Me)...CO-Me (Option 2)!

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NEET 2025 Chemistry Question 78 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Grignard on Ketone + Nitrile Fingerprint

Step-by-Step Rigorous Workflow & Mathematical Proof

Grignard Addition to the Ketone Carbonyl

Grignard Addition to the Nitrile Group and Imine Hydrolysis

Consolidate the Major Product Structure

360° Option Elimination Engine Key Feature

NTA Trap Alert: Treating Nitriles Like Esters (The Diol Trap)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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