In each case of List-I, the carbonyl compound is treated with the Grignard reagent and the adduct formed is hydrolysed. Match List-I (reactants) with List-II (alcohol formed).
| List-I | List-II |
|---|---|
| (A) \( \ce{HCHO + CH3CH2CH2MgBr} \) | (I) 2-Methylpropan-1-ol |
| (B) \( \ce{CH3CHO + CH3CH2MgBr} \) | (II) Butan-2-ol |
| (C) \( \ce{CH3COCH3 + CH3MgBr} \) | (III) 2-Methylpropan-2-ol |
| (D) \( \ce{HCHO + (CH3)2CHMgBr} \) | (IV) Butan-1-ol |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-II, C-III, D-I
A Grignard reagent first adds to the C=O group: the alkyl group of RMgX goes to the carbonyl carbon and an adduct is formed. Hydrolysis of the adduct gives the alcohol.
The class of the alcohol depends on the carbonyl compound: methanal gives a primary alcohol, other aldehydes give secondary alcohols and ketones give tertiary alcohols.
All four products here are C4H10O isomers, so this rule decides each pair.
(A) \( \ce{HCHO + CH3CH2CH2MgBr} \) — This matches (IV) Butan-1-ol. Methanal gives a primary alcohol; the propyl group joins the CH2 that carries the –OH group, giving a straight four-carbon chain.
\( \displaystyle \ce{HCHO} \xrightarrow[\text{(ii) } \ce{H2O}]{\text{(i) } \ce{CH3CH2CH2MgBr}} \ce{CH3CH2CH2CH2OH} \)
(B) \( \ce{CH3CHO + CH3CH2MgBr} \) — This matches (II) Butan-2-ol. Ethanal is an aldehyde other than methanal, so a secondary alcohol is formed.
\( \displaystyle \ce{CH3CHO} \xrightarrow[\text{(ii) } \ce{H2O}]{\text{(i) } \ce{CH3CH2MgBr}} \ce{CH3CH(OH)CH2CH3} \)
(C) \( \ce{CH3COCH3 + CH3MgBr} \) — This matches (III) 2-Methylpropan-2-ol. Propanone is a ketone, so a tertiary alcohol is formed; the carbon carrying –OH now has three methyl groups.
\( \displaystyle \ce{CH3COCH3} \xrightarrow[\text{(ii) } \ce{H2O}]{\text{(i) } \ce{CH3MgBr}} \ce{(CH3)3C-OH} \)
(D) \( \ce{HCHO + (CH3)2CHMgBr} \) — This matches (I) 2-Methylpropan-1-ol. Methanal again gives a primary alcohol, but the branched isopropyl group gives a branched chain.
\( \displaystyle \ce{HCHO} \xrightarrow[\text{(ii) } \ce{H2O}]{\text{(i) } \ce{(CH3)2CHMgBr}} \ce{(CH3)2CHCH2OH} \)
NCERT (2026-27 edition):
7.4.1 Preparation of Alcohols – “3. 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.”