What is the product of the given reaction?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1-Pentanol
The reagents (i) \( \ce{BH3} \)/THF and (ii) \( \ce{H2O2} \)/\( \ce{OH^{-}} \) carry out hydroboration–oxidation. Borane adds to the double bond with the boron going to the less substituted carbon (the terminal \( \ce{CH2} \)); alkaline hydrogen peroxide then replaces boron by –OH. The net result is addition of water against Markovnikov’s rule.
\( \ce{CH3CH2CH2CH=CH2} \xrightarrow[\ce{(ii) H2O2/OH-}]{\ce{(i) BH3/THF}} \ce{CH3CH2CH2CH2CH2OH} \)
Pent-1-ene gives pentan-1-ol (1-pentanol).
NCERT (2026-27 edition):
From Alkenes – “… (ii) By hydroboration–oxidation: Diborane (BH3)2 reacts with alkenes to give trialkyl boranes as addition product. This is oxidised to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.”