Ethyl alcohol can be prepared from the following set of reactants and subsequent hydrolysis of adduct:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → HCHO + CH3MgX
A Grignard reagent adds to the carbonyl group of an aldehyde or ketone; hydrolysis of the adduct gives an alcohol. The alkyl group of the reagent joins the carbonyl carbon. With methanal the product is a primary alcohol with one carbon more than the Grignard reagent; other aldehydes give secondary alcohols and ketones give tertiary alcohols.
Ethyl alcohol (\( \ce{CH3CH2OH} \)) is a primary alcohol with two carbons, so it needs methanal (1 C) + methylmagnesium halide (1 C):
\( \ce{HCHO + CH3MgX -> CH3CH2OMgX} \xrightarrow{\ce{H2O}} \ce{CH3CH2OH + Mg(OH)X} \)
Why the other options are incorrect
Option (1) HCHO + CH3CH2MgX — Incorrect. Methanal with an ethyl group gives a three-carbon primary alcohol, propan-1-ol (\( \ce{CH3CH2CH2OH} \)).
Option (2) CH3CHO + CH3CH2MgX — Incorrect. Ethanal with an ethyl group gives a secondary alcohol, butan-2-ol (\( \ce{CH3CH(OH)CH2CH3} \)).
Option (4) CH3CHO + CH3MgX — Incorrect. Ethanal with a methyl group gives a secondary alcohol, propan-2-ol (\( \ce{(CH3)2CHOH} \)).
NCERT (2026-27 edition):
From Grignard Reagents – “Alcohols are produced by the reaction of Grignard reagents (Unit 6, Class XII) with aldehydes and ketones. 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.”