Q.28 In the following reaction, X is an intermediate and Y is one of the end products.
Which one of the following compounds is the end product Y?

The reaction of acetone with ethylmagnesium iodide (a Grignard reagent), followed by acid hydrolysis, gives a tertiary alcohol: 2‑methylbutan‑2‑ol, which corresponds to option (D).​

Introduction

The Grignard reaction of acetone with ethylmagnesium iodide is a standard CSIR NET and IIT‑level problem used to test understanding of nucleophilic addition to carbonyl compounds and classification of alcohols formed after hydrolysis. In this article, the full mechanism, identification of the intermediate X, the final product Y, and the reasoning behind each multiple‑choice option are discussed in detail for strong exam‑oriented preparation.​

Stepwise mechanism and identification of X and Y

1. Reactants in the question

   • Carbonyl compound: acetone (propanone, CH3COCH3CH3COCH3).​

   • Grignard reagent: ethylmagnesium iodide (C2H5MgIC2H5MgI).​

2. Nucleophilic addition step (formation of X)

   • In a Grignard reagent, the carbon attached to magnesium is nucleophilic (carbanion‑like) and attacks the electrophilic carbonyl carbon of acetone.​

   • The ethyl carbanion equivalent C2H5−C2H5− adds to acetone to give a tetrahedral magnesium alkoxide intermediate:

     CH3COCH3+C2H5MgI→CH3C(OMgI)(CH3)(C2H5)CH3COCH3+C2H5MgI→CH3C(OMgI)(CH3)(C2H5)

     This alkoxide complex is the intermediate X.​

3. Acidic work‑up (formation of Y)

   • Treatment with aqueous acid (H+/H2OH+/H2O) protonates the alkoxide oxygen and breaks the Mg–O bond.​

   • The final product is an alcohol with three carbon substituents on the carbon bearing OH: one ethyl and two methyl groups.​

4. Structure and classification of Y

   • Structure: CH3–C(OH)(CH3)(C2H5)CH3–C(OH)(CH3)(C2H5).​

   • This is 2‑methylbutan‑2‑ol (tert‑amyl alcohol), a tertiary alcohol because the OH‑bearing carbon is attached to three alkyl groups.​

Thus, Y is 2‑methylbutan‑2‑ol, matching the skeletal structure in option (D).​

Why option (D) is correct

• Option (D) shows a central carbon attached to an OH group, two methyl groups and one ethyl group – exactly the arrangement of 2‑methylbutan‑2‑ol formed by adding an ethyl group to acetone and then protonating.​

• The reaction of any ketone with a Grignard reagent followed by hydrolysis produces a tertiary alcohol, and the carbon chain length increases by the length of the Grignard alkyl group (here, ethyl), which is consistent with option (D).​

Detailed analysis of each option

Option (A): Carbonyl compound (incorrect)

• Option (A) represents a ketone (a carbonyl group $\text{C}=O$) rather than an alcohol.

• Since the reaction sequence explicitly includes an acidic work‑up of a Grignard addition product, the carbonyl oxygen must become an OH group; therefore a remaining ketone cannot be the final product, eliminating option (A).​

Option (B): Primary alcohol (incorrect)

• Option (B) shows a primary alcohol, where the OH‑bearing carbon is attached to only one alkyl group and two hydrogens.

• However, Grignard reagents reacting with ketones like acetone always yield tertiary alcohols after hydrolysis, not primary alcohols, so option (B) contradicts the general rule.​

Option (C): Secondary ketone or mis‑matched chain (incorrect)

• Option (C) corresponds to a straight‑chain ketone or alcohol whose connectivity does not match the product obtained by adding an ethyl group to acetone.

• The carbon skeleton produced from acetone ($3$ carbons) plus an ethyl group ($2$ carbons) must contain $5$ carbons with a tertiary alcohol center, which is not depicted in option (C).​

Option (D): 2‑Methylbutan‑2‑ol (correct)

• Option (D) shows a tertiary alcohol with three alkyl groups (two methyl and one ethyl) attached to the carbon bearing OH, which fits the expected product of acetone plus ethylmagnesium iodide followed by hydrolysis.​

• Therefore, option (D) is the correct end product Y.​

Key exam points about this Grignard question

• Grignard reagents act as strong nucleophiles and bases, attacking electrophilic carbonyl carbons to form magnesium alkoxide intermediates.​

• The type of alcohol obtained depends on the carbonyl compound:

  • Formaldehyde → primary alcohol.​

  • Other aldehydes → secondary alcohols.​

  • Ketones (like acetone) → tertiary alcohols, as in this problem.​

These points make the reaction of acetone with ethylmagnesium iodide a high‑yield topic for CSIR NET, GATE and other competitive examinations, and quickly guide to option (D) as the correct structure of Y.​