Match the reaction in List-I with its main product in List-II.
| List-I (Reaction) | List-II (Main product) |
|---|---|
| (A) Propene + H2O, in the presence of an acid catalyst | (I) Butan-1-ol |
| (B) Propene, (i) B2H6 (ii) H2O2 / aqueous NaOH | (II) Propan-1-ol |
| (C) Butan-2-one + NaBH4 | (III) Butan-2-ol |
| (D) Butanoic acid, (i) LiAlH4 (ii) H2O | (IV) Propan-2-ol |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-IV, B-II, C-III, D-I
From one alkene, the two addition methods give different alcohols: acid catalysed hydration follows Markovnikov’s rule, while hydroboration–oxidation gives the product that looks like an anti-Markovnikov addition of water. In reductions, the –OH group appears on the carbon that carried the C=O group.
(A) Propene + H2O, in the presence of an acid catalyst — This matches (IV) Propan-2-ol. H+ adds to the CH2 end, the more stable secondary carbocation forms on C-2, and water attacks there.
\( \displaystyle \ce{CH3CH=CH2 + H2O} \xrightarrow{\ce{H+}} \ce{CH3CH(OH)CH3} \)
(B) Propene, (i) B2H6 (ii) H2O2 / aqueous NaOH — This matches (II) Propan-1-ol. Boron attaches to the sp2 carbon carrying more hydrogen atoms (the CH2 end), and oxidation puts the –OH group there.
\( \displaystyle \ce{CH3CH=CH2} \xrightarrow[\text{(ii) }\ce{H2O2}/\ce{OH^-}]{\text{(i) }\ce{B2H6}} \ce{CH3CH2CH2OH} \)
(C) Butan-2-one + NaBH4 — This matches (III) Butan-2-ol. A ketone is reduced to a secondary alcohol.
\( \displaystyle \ce{CH3COCH2CH3} \xrightarrow{\ce{NaBH4}} \ce{CH3CH(OH)CH2CH3} \)
(D) Butanoic acid, (i) LiAlH4 (ii) H2O — This matches (I) Butan-1-ol. LiAlH4, a strong reducing agent, reduces a carboxylic acid to a primary alcohol with the same carbon chain.
\( \displaystyle \ce{CH3CH2CH2COOH} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4}} \ce{CH3CH2CH2CH2OH} \)
NCERT (2026-27 edition):
7.4.1 Preparation of Alcohols – “(i) By acid catalysed hydration: Alkenes react with water in the presence of acid as catalyst to form alcohols. In case of unsymmetrical alkenes, the addition reaction takes place in accordance with Markovnikov’s rule.”
“The addition of borane to the double bond takes place in such a manner that the boron atom gets attached to the sp2 carbon carrying greater number of hydrogen atoms. The alcohol so formed looks as if it has been formed by the addition of water to the alkene in a way opposite to the Markovnikov’s rule.”
“It is also prepared by treating aldehydes and ketones with sodium borohydride (NaBH4) or lithium aluminium hydride (LiAlH4). Aldehydes yield primary alcohols whereas ketones give secondary alcohols.”
“(ii) By reduction of carboxylic acids and esters: Carboxylic acids are reduced to primary alcohols in excellent yields by lithium aluminium hydride, a strong reducing agent.”