A representative example — structures drawn live.
The two-step reaction at a glance
Hydroboration–oxidation converts an alkene into an alcohol using two reagents applied in sequence. First, borane (BH3, usually supplied as the Lewis-base complex BH3·THF, or as its bulkier cousin 9-BBN) adds across the C=C double bond. Second, an oxidation with basic hydrogen peroxide (H2O2, NaOH) replaces the boron with a hydroxyl group.
The net transformation is the addition of water (H and OH) across the double bond. What makes it valuable is that its selectivity is the exact opposite of acid-catalyzed hydration or oxymercuration: you get the anti-Markovnikov alcohol, cleanly and stereospecifically.
Regiochemistry: why boron goes to the less substituted carbon
In the first step, boron and hydrogen add to the two alkene carbons in a single concerted step through a four-center transition state. Two factors push boron onto the less substituted carbon:
- Sterics. Boron carries two more hydrogens (or bulky groups in 9-BBN) and prefers the less hindered, less substituted carbon.
- Electronics. Boron is electron-poor (it acts as the electrophile), so the developing partial positive charge sits better on the more substituted carbon. Hydrogen, delivered as a partial hydride, goes to that more substituted carbon.
Because boron marks the spot where OH will later appear, the OH ends up on the less substituted carbon. That is precisely the anti-Markovnikov outcome — the hydrogen adds to the carbon that already has more hydrogens' more-substituted neighbor, and OH lands where Markovnikov addition would not place it.
Stereochemistry: syn addition and retention
The concerted four-center transition state forces boron and hydrogen to add to the same face of the planar alkene. This is a syn addition. When the alkene is part of a ring or gives a product with two adjacent stereocenters, the H and OH are delivered cis to one another.
The oxidation step then swaps boron for OH with retention of configuration — the oxygen takes the exact position boron occupied, without inverting the carbon. So the stereochemical relationship established in the addition step is preserved in the final alcohol.
No carbocation, no rearrangement
Acid-catalyzed hydration proceeds through a carbocation, which can undergo hydride or methyl shifts to give rearranged (and often unexpected) products. Hydroboration is fundamentally different: the addition is concerted, so no carbocation intermediate ever forms. That means substrates prone to rearrangement — for example, 3-methyl-1-butene — give the straightforward anti-Markovnikov alcohol with no skeletal shuffling.
A worked example
Take 1-methylcyclohexene and treat it with BH3·THF, then H2O2/NaOH. Boron adds to the less substituted ring carbon (C2), and hydrogen adds to the more substituted carbon (C1, which bears the methyl). After oxidation, OH sits on C2. Because H and OH added syn, the H on C1 and the OH on C2 are cis, giving trans-2-methylcyclohexanol as the major product. Compare this with acid-catalyzed hydration, which would put OH on the more substituted C1.
When to reach for it
Choose hydroboration–oxidation whenever a synthesis calls for an anti-Markovnikov alcohol, especially a primary alcohol from a terminal alkene, or when you need predictable syn stereochemistry and want to avoid carbocation rearrangements. It pairs naturally with oxymercuration (the Markovnikov complement) as the two go-to ways of controlling where the OH lands.
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, curved arrows, wedge/dash bonds and a clickable periodic table built in. No account needed.