Ethylmagnesium bromide reacts with methanal, and the adduct formed is hydrolysed with water. Which alcohol is obtained?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Propan-1-ol
A Grignard reagent adds to the carbonyl group of an aldehyde or ketone to form an adduct, and hydrolysis of the adduct gives an alcohol. The alkyl group of the reagent becomes bonded to the carbonyl carbon.
With methanal that carbon ends up as a –CH2OH group, so the product is a primary alcohol with one carbon more than the alkyl group:
\( \displaystyle \ce{HCHO + C2H5MgBr -> CH3CH2CH2OMgBr} \xrightarrow{\ce{H2O}} \ce{CH3CH2CH2OH + Mg(OH)Br} \)
The two carbons of the ethyl group and the carbon of methanal make a chain of three carbons with –OH at the end, which is propan-1-ol.
Why the other options are incorrect
Option (1) Ethanol — Incorrect. Ethanol would be formed from methylmagnesium bromide and methanal. The ethyl group brings two carbons, so the alcohol formed here has three.
Option (2) Propan-2-ol — Incorrect. Propan-2-ol is a secondary alcohol. Grignard reagents give secondary alcohols with aldehydes other than methanal, for example ethanal with methylmagnesium bromide.
Option (4) Butan-2-ol — Incorrect. This secondary alcohol is formed when ethylmagnesium bromide reacts with ethanal, CH3CHO, not with methanal.
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.”