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Add water to propene with dilute acid and you get propan-2-ol. Many questions want propan-1-ol instead, with the OH on the end carbon. Hydroboration-oxidation does that. This post explains how the two steps work, compares it with acid hydration on three alkenes, sets out the mole ratios that trip students up and lists the common mistakes.
Where is this in the syllabus?
The NCERT Class 12 chapter Alcohols, Phenols and Ethers (2026-27 reprint) lists hydroboration-oxidation as a preparation of alcohols from alkenes. The CBSE 2026-27 Chemistry curriculum includes this unit. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list "general methods of preparation" for organic compounds containing oxygen. The JEE (Advanced) 2026 brochure names acid-catalysed hydration of alkenes but does not name hydroboration, so treat it there as NCERT background. Check the current year's documents.
The method is linked to Herbert C. Brown, who shared the 1979 Nobel Prize in Chemistry with Georg Wittig "for their development of the use of boron- and phosphorus-containing compounds, respectively, into important reagents in organic synthesis". An early report was Brown and Rao's 1957 note on a remarkably fast room-temperature addition of diborane to olefins.
How do the two steps work?
NCERT's description: diborane, (BH₃)₂, reacts with alkenes to give trialkylboranes, which hydrogen peroxide in aqueous sodium hydroxide oxidises to alcohols. Boron attaches to the sp² carbon carrying more hydrogen atoms, so the alcohol looks as if water added opposite to Markovnikov's rule, in excellent yield.
OpenStax Organic Chemistry adds the mechanism:
Hydroboration: the C–H and C–B bonds form at the same time, from the same face of the double bond, in one step with no carbocation. Boron goes to the less crowded carbon.
Each BH₃ can add three times, so one BH₃ and three alkenes give a trialkylborane.
Oxidation: H₂O₂ and OH⁻ replace each C–B bond with C–OH, keeping the same stereochemistry.
Worked example 1: two routes from 2-methylpropene
Dilute H₂SO₄ and water: through the tertiary carbocation, giving 2-methylpropan-2-ol, a tertiary alcohol.
B₂H₆, then H₂O₂/NaOH: boron on the CH₂ end, giving 2-methylpropan-1-ol, a primary alcohol.
Both are C₄H₁₀O (M = 74.1 g/mol), so only a test can tell them apart. The Lucas test, covered in our post on alcohols, phenols and ethers, turns cloudy at once for the tertiary alcohol.
Now the mole ratio. Take 16.8 g of 2-methylpropene (M = 56.1 g/mol), which is 0.300 mol.
If all three B–H bonds are used, it needs 0.300 ÷ 3 = 0.100 mol of BH₃, which is 0.0500 mol of B₂H₆ (about 1.38 g).
It can give at most 0.300 mol of 2-methylpropan-1-ol, which is 0.300 × 74.1 ≈ 22.2 g.
So one B₂H₆ can serve up to six alkene molecules and give up to six alcohol molecules.
Worked example 2: no rearrangement
3,3-Dimethylbut-1-ene, CH₂=CH–C(CH₃)₃, makes the point. With HCl, the secondary carbocation can undergo a methyl shift, and OpenStax reports an equal mixture of rearranged and unrearranged chlorides. Hydroboration-oxidation forms no carbocation, so the skeleton survives and the product is 3,3-dimethylbutan-1-ol. Our post on carbocation rearrangements explains the shift.
Worked example 3: which face?
This is beyond the JEE Main syllabus but appears in JEE Advanced-style questions. OpenStax shows 1-methylcyclopentene giving trans-2-methylcyclopentanol. H and B add to the same face; OH then takes boron's place, so the methyl group and the OH end up on opposite faces. The six-membered ring behaves the same way: 1-methylcyclohexene gives trans-2-methylcyclohexan-1-ol.
Which mistakes cost marks?
Confusing it with the peroxide effect. That applies to HBr only; see our post on Markovnikov's rule and the peroxide effect.
Putting OH on the more substituted carbon.
Drawing a carbocation. There is none, which is why there is no rearrangement.
Using a 1 : 1 ratio of B₂H₆ to alkene. Up to six alkenes react per B₂H₆.
Leaving out the second step. The trialkylborane is not an alcohol until H₂O₂/NaOH acts on it.
A practice routine
For propene, but-1-ene and 2-methylbut-2-ene, write the products of acid hydration and of hydroboration-oxidation side by side.
Name each alcohol and classify it as 1°, 2° or 3°.
Redo the mole ratio for 8.4 g of propene (M = 42.1 g/mol). You should get about 0.20 mol and a maximum of about 12.0 g of propan-1-ol.
For 3-methylbut-1-ene, explain why acid hydration can rearrange and hydroboration cannot.
When you mix up which route gives which alcohol, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, draw propene going both ways on one page and mark where the OH lands in each.
