Read the passage given below to answer questions:
Most of haloalkanes and haloarenes react with certain metals to give compound containing carbon metal bonds called organometallic compounds. Grignard reagent is such organomagnesium compound. Grignard reagent can be prepared from halogen derivative of alkanes/arenes with magnesium metal in dry ether. These are highly reactive compounds and react with source of proton to give hydrocarbon. Grignard reagents undergo addition with carbonyl compounds to give corresponding alcohols.
The product (P) formed in the following reaction is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → CH3CH(OH)CH3 (propan-2-ol)
The methyl group of \( \ce{CH3MgBr} \) adds to the carbonyl carbon of ethanal (acetaldehyde); hydrolysis of the adduct gives the alcohol:
\( \ce{CH3CHO + CH3MgBr} \xrightarrow{\text{dry ether}} \ce{(CH3)2CH-OMgBr} \xrightarrow{\ce{H3O+}} \ce{(CH3)2CH-OH + Mg(OH)Br} \)
An aldehyde other than formaldehyde gives a secondary alcohol; here it is propan-2-ol, drawn in option (2) as \( \ce{H3C-CH(OH)-CH3} \).
Why the other options are incorrect
Option (1) — Incorrect. 2-methylpropan-2-ol, a tertiary alcohol, would need a ketone (acetone) as the carbonyl compound.
Option (3) — Incorrect. An ether; the oxygen ends up as –OH, not –\( \ce{OCH3} \).
Option (4) — Incorrect. Propanoic acid is formed from a Grignard reagent and \( \ce{CO2} \), not from an aldehyde.