Q.60 Chemistry ⭐ 360° Option Elimination Active

NEET 2021 Chemistry Question 60 Solution

Topic: Alcohols, Phenols and Ethers · Official NTA Booklet Code M1

Question

What is the IUPAC name of the organic compound formed in the following chemical reaction? Acetone → (i) C₂H₅MgBr, dry Ether (ii) H₂O, H⁺ → Product

✓

Verified Against NTA Final Answer Key

Option (4) : 2-methyl butan-2-ol

📚 Alcohols, Phenols and Ethers Easy

📋 Given Parameters & Problem Setup

Substrate Acetone (CH3-CO-CH3)

Grignard Reagent C2H5MgBr in dry ether

Hydrolysis Acidic aqueous workup (H2O, H+)

⚡

The 30-Second Exam Hall Shortcut: ⚡ Carbon Count: 3 (Acetone) + 2 (Ethyl) = 5 Carbons; Ketone $\implies$ 3° Alcohol

⏱️ Save ~20 seconds

Acetone ( $3$ carbons) $+$ ethylmagnesium bromide ( $2$ carbons) yields a $5$ -carbon tertiary alcohol. The ethyl group attacks the carbonyl carbon: $(CH_3)_2C(OH)-CH_2CH_3$ . Longest chain is 4 carbons (butane) with methyl and -OH at C-2: 2-methylbutan-2-ol! Match Option (4)!

1 Acetone structure:

CH_3 - C(=O) - CH_3

2 Nucleophilic ethyl carbanion (

\bar{C}_2H_5

) from Grignard reagent attacks carbonyl carbon:

(CH_3)_2C(OMgBr)-CH_2CH_3

3 Acidic hydrolysis protonates the alkoxide:

(CH_3)_2C(OH)-CH_2CH_3

4 Number the longest continuous carbon chain: 4 carbons (butane) containing the -OH group at position 2 and a methyl branch at position 2.

5 IUPAC name: 2-methylbutan-2-ol.

6 Match Option (4).

🛡️

Why You Can 100% Trust This Shortcut: Canonical Grignard addition to ketones yielding tertiary alcohols.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Nucleophilic Addition of Grignard Reagent to Acetone

Grignard reagents ( $RMgX$ ) act as powerful carbon nucleophiles. The polar organomagnesium bond provides a nucleophilic ethyl group ( $C_2H_5^-$ ) that attacks the electrophilic carbonyl carbon of acetone (propan-2-one):

CH_3-\overset{\overset{O}{\parallel}}{C}-CH_3 + C_2H_5MgBr \xrightarrow{\text{dry ether}} CH_3-\overset{\overset{OMgBr}{\vert}}{\underset{\underset{C_2H_5}{\vert}}{C}}-CH_3

Step 2

Hydrolysis of the Magnesium Halide Alkoxide Complex

Acidic aqueous hydrolysis of the magnesium alkoxide intermediate decomposes it into the corresponding tertiary alcohol and basic magnesium bromide:

(CH_3)_2C(OMgBr)CH_2CH_3 + H_2O \xrightarrow{H^+} (CH_3)_2C(OH)CH_2CH_3 + Mg(OH)Br

Step 3

IUPAC Nomenclature of the Product Alcohol

Identify the principal functional group and the longest continuous carbon chain: - Longest carbon chain containing the $-OH$ group has $4$ carbons (butane). - Numbering starts from the right to give the principal $-OH$ group the lowest locant ( $2$ ): - Carbon 1: $-CH_3$ - Carbon 2: $-C(CH_3)(OH)-$ - Carbon 3: $-CH_2-$ - Carbon 4: $-CH_3$ - The substituents are a methyl group at position 2 and an -ol group at position 2.

\mathbf{\text{IUPAC Name: 2-methylbutan-2-ol}}

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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:

This 4-carbon alcohol ( $tert$ -butanol) would be formed if methylmagnesium bromide ( $CH_3MgBr$ ) were used, not ethylmagnesium bromide.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming a methyl Grignard reagent was added.

2 2

✗ Incorrect

Verdict & Scientific Reason:

This is a secondary alcohol formed from pentanal or propanal with ethyl Grignard, not from acetone.

⚠️

Student Fallacy (Why candidates pick this trap): Treating the product as a straight 5-carbon chain secondary alcohol.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Secondary alcohol formed from propionaldehyde and ethyl Grignard.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming symmetrical secondary alcohol formation.

4 4

✓ 100% Correct

Verdict & Scientific Reason:

Exact tertiary alcohol product formed by nucleophilic addition of ethyl group to acetone.

💡

Why this is airtight: None; textbook Grignard synthesis.

⚠️

NTA Trap Alert: Longest Carbon Chain Selection Trap

Students who count $(CH_3)_2C(OH)C_2H_5$ sometimes overlook the ethyl group as two carbons, calling it a derivative of pentan-2-ol (Option 2) or forgetting the chain length. The longest straight chain through ethyl is 4 carbons (butane), not 5.

🛡️ How to avoid the -1 mark: Expand ethyl:

-CH_2-CH_3

. The chain is

CH_3-C(CH_3)(OH)-CH_2-CH_3 \implies

4 carbons

\implies

2-methylbutan-2-ol.

📖

Core NEET Syllabus Standard

NCERT Chemistry · Alcohols, Phenols and Ethers Standard Benchmark

Core Scientific Principle Tested:

"Addition of a Grignard reagent to a ketone yields a tertiary alcohol following acidic hydrolysis."

Syllabus Focus: Preparation of Alcohols from Grignard Reagents

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Chemistry Question 60 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Carbon Count: 3 (Acetone) + 2 (Ethyl) = 5 Carbons; Ketone ⟹ \implies⟹ 3° Alcohol

Step-by-Step Rigorous Workflow & Mathematical Proof

Nucleophilic Addition of Grignard Reagent to Acetone

Hydrolysis of the Magnesium Halide Alkoxide Complex

IUPAC Nomenclature of the Product Alcohol

360° Option Elimination Engine Key Feature

NTA Trap Alert: Longest Carbon Chain Selection Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Alcohols, Phenols and Ethers

Total Number of Cyclic Ether Isomers (Structural and Stereoisomers) of C₄H₈O

Lucas Test and Reactivity Order of Alcohols

Intramolecular versus Intermolecular Hydrogen Bonding in Nitrophenols

Reaction of 3-Methylbutan-2-ol with HBr via 1,2-Hydride Shift

Solve 4 NEET Questions a Day — One per Subject

The 30-Second Exam Hall Shortcut: ⚡ Carbon Count: 3 (Acetone) + 2 (Ethyl) = 5 Carbons; Ketone $\implies$ 3° Alcohol