What is the product of the given reaction?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Pentan-1-ol
The reactant is pent-1-ene, and the two reagents carry out hydroboration–oxidation: it adds H and OH across the double bond in the anti-Markovnikov way.
(i) \( \ce{BH3} \)/THF: borane adds to the C=C bond with boron on the end carbon (the \( \ce{=CH2} \), which carries more hydrogen atoms) and H on C-2, giving a trialkyl borane.
\( \displaystyle \ce{3CH3CH2CH2CH=CH2 + BH3} \xrightarrow{\ce{THF}} \ce{(CH3CH2CH2CH2CH2)3B} \)
(ii) \( \ce{H2O2} \)/\( \ce{OH^-} \): hydrogen peroxide in aqueous alkali replaces each boron–carbon bond by a C–OH bond, so the OH ends up on the end carbon.
\( \displaystyle \ce{(CH3CH2CH2CH2CH2)3B + 3H2O2} \xrightarrow{\ce{OH^-}} \ce{3CH3CH2CH2CH2CH2OH + B(OH)3} \)
The product, \( \ce{CH3CH2CH2CH2CH2OH} \), is the primary alcohol pentan-1-ol.
Why the other options are incorrect
Option (2) Pentan-2-ol — Incorrect. This is the Markovnikov product, formed by acid-catalysed hydration of pent-1-ene; hydroboration–oxidation puts the OH on the other carbon.
Option (3) Pentanal — Incorrect. The reaction gives an alcohol, not a carbonyl compound; the alkaline \( \ce{H2O2} \) only converts the C–B bond into a C–OH bond.
Option (4) Pentane-1,2-diol — Incorrect. Only one OH group is added: boron (later OH) goes to one carbon and hydrogen to the other.
NCERT (2026-27 edition):
7.4.1 Preparation of Alcohols – “(ii) By hydroboration–oxidation: Diborane (BH3)2 reacts with alkenes to give trialkyl boranes as addition product. … 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.”