Which of the following are the starting materials for the Grignard sy…

The correct answer is Option (1) → CH₃MgBr + CH₃COCH₃

A Grignard reagent adds to the carbonyl group, and hydrolysis of the adduct gives an alcohol: a ketone gives a tertiary alcohol. tert-Butyl alcohol, \( \ce{(CH3)3C-OH} \), has three methyl groups on the carbinol carbon, so it needs acetone (two CH₃) and methylmagnesium bromide (the third CH₃):

    \( \displaystyle \ce{(CH3)2C=O + CH3MgBr} \xrightarrow{\text{dry ether}} \ce{(CH3)3C-OMgBr} \)

    \( \displaystyle \ce{(CH3)3C-OMgBr + H2O -> (CH3)3C-OH + Mg(OH)Br} \)

Why the other options are incorrect

Option (2) CH₃CH₂MgBr + CH₃COCH₃ — Incorrect. The ethyl group joins the carbonyl carbon, giving \( \ce{(CH3)2C(OH)CH2CH3} \), 2-methylbutan-2-ol, not tert-butyl alcohol.

Option (3) CH₃MgBr + CH₃CH(OH)CH₃ — Incorrect. An alcohol has no carbonyl group. Its –OH hydrogen destroys the Grignard reagent, giving methane: \( \ce{CH3MgBr + (CH3)2CHOH -> CH4 + (CH3)2CHOMgBr} \).

Option (4) (CH₃)₂CHMgBr + HCHO — Incorrect. Methanal gives a primary alcohol: here \( \ce{(CH3)2CHCH2OH} \), 2-methylpropan-1-ol.

NCERT (2026-27 edition):
7.4.1 Preparation of Alcohols – “The first step of the reaction is the nucleophilic addition of Grignard reagent to the carbonyl group to form an adduct. Hydrolysis of the adduct yields an alcohol.”

“You will notice that the reaction produces a primary alcohol with methanal, a secondary alcohol with other aldehydes and tertiary alcohol with ketones.”