The compound formed on reaction of propanone with methylmagnesium bromide followed by hydrolysis is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2-methylpropan-2-ol
A Grignard reagent (RMgX) adds to the carbonyl group of an aldehyde or ketone by nucleophilic addition: the alkyl group R goes to the carbonyl carbon and MgX to the oxygen. Hydrolysis of this adduct gives an alcohol with one more C–C bond:
methanal gives a primary alcohol,
other aldehydes give secondary alcohols,
ketones give tertiary alcohols.
Propanone is a ketone, so with methylmagnesium bromide it gives a tertiary alcohol:
\( \displaystyle \ce{(CH3)2C=O + CH3MgBr -> (CH3)3C-OMgBr} \)
\( \displaystyle \ce{(CH3)3C-OMgBr + H2O -> (CH3)3C-OH + Mg(OH)Br} \)
So the product is 2-methylpropan-2-ol (tert-butyl alcohol), the structure in Option (2).
Why the other options are incorrect
Option (1) propan-2-ol — Incorrect. This is the reduction product of propanone (\( \ce{H2} \)/Ni or \( \ce{NaBH4} \)); it has no new carbon atom.
Option (3) \(CH_3CH_2CH_2OH\) — Incorrect. Propan-1-ol is a primary alcohol; it is made from methanal and ethylmagnesium bromide, not from a ketone.
Option (4) \(CH_3CH_2CH_2CH_2OH\) — Incorrect. Butan-1-ol is also a primary alcohol (from methanal and propylmagnesium bromide); a ketone can never give a primary alcohol.
NCERT (2026-27 edition):
From Grignard Reagents
