Propanone (acetone) is treated with an excess of methylmagnesium bromide, and the adduct formed is hydrolysed with water. What mass of alcohol is obtained from 11.6 g of propanone, assuming complete conversion? (Atomic masses: H = 1, C = 12, O = 16)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 14.8 g
The Grignard reagent adds to the carbonyl group to form an adduct, and hydrolysis of the adduct gives an alcohol. With a ketone the product is a tertiary alcohol. Propanone and CH3MgBr give 2-methylpropan-2-ol, one mole for each mole of propanone:
\( \displaystyle \ce{(CH3)2C=O + CH3MgBr -> (CH3)3C-OMgBr} \)
\( \displaystyle \ce{(CH3)3C-OMgBr + H2O -> (CH3)3C-OH + Mg(OH)Br} \)
\( \displaystyle n(\ce{CH3COCH3}) = \frac{11.6\ \mathrm{g}}{58\ \mathrm{g\ mol^{-1}}} = 0.200\ \mathrm{mol} \)
\( \displaystyle M(\ce{(CH3)3COH}) = 4 \times 12 + 10 \times 1 + 16 = 74\ \mathrm{g\ mol^{-1}} \)
\( \displaystyle m(\ce{(CH3)3COH}) = 0.200\ \mathrm{mol} \times 74\ \mathrm{g\ mol^{-1}} = 14.8\ \mathrm{g} \)
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.”