Which compound reacts with propene to form tripropylborane?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Diborane
In hydroboration, diborane, \( \ce{(BH3)2} \) or \( \ce{B2H6} \), adds its B–H bonds across the C=C of an alkene. Boron goes to the sp2 carbon carrying more hydrogen atoms (the end carbon of propene), and each BH3 unit adds to three alkene molecules to give a trialkylborane:
\( \displaystyle \ce{6CH3-CH=CH2 + B2H6 -> 2(CH3CH2CH2)3B} \)
Oxidation of tripropylborane by \( \ce{H2O2} \) in aqueous NaOH then gives propan-1-ol (hydroboration–oxidation).
Why the other options are incorrect
Option (1) Boron trifluoride — Incorrect. \( \ce{BF3} \) has no B–H bond, so it has no hydrogen to add across the double bond.
Option (3) Boric acid — Incorrect. \( \ce{B(OH)3} \) also has no B–H bond; its three B–OH groups cannot add a hydrogen and a boron across the C=C.
Option (4) Sodium borohydride — Incorrect. \( \ce{NaBH4} \) is used to reduce aldehydes and ketones to alcohols; it is not the hydroboration reagent for alkenes.
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. This is oxidised to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.”