Hydroboration–Oxidation of Alkenes

Quick answer Hydroboration–oxidation adds H and OH across a C=C with the OH on the less substituted carbon (anti-Markovnikov) and both on the same face (syn). No carbocation forms, so nothing rearranges.

⚙ Mechanism · Hydroboration–Oxidation3 steps

Step 1 — B–H adds across the double bond in one step.

Boron and hydrogen add to the same face at the same time — syn addition, no carbocation anywhere. Boron is the bulky end and also the electron-poor end, so it lands on the less substituted carbon. That is where anti-Markovnikov comes from.

Step 2 — hydroperoxide adds to the empty orbital on boron.

Boron has only six electrons, so it is a Lewis acid and HOO⁻ simply adds to it. Nothing has happened to the carbon yet.

Step 3 — the alkyl group migrates from boron to oxygen.

The carbon slides from boron onto the neighbouring oxygen as the weak O–O bond breaks, then hydrolysis frees the alcohol. Crucially the carbon never becomes a free centre, so it keeps its configuration — OH ends up exactly where the boron was.

The anti-Markovnikov, syn route to alcohols — the complement to acid hydration and oxymercuration.

Propene → 1-propanol: the OH lands on the terminal (less substituted) carbon. Structures drawn live.

1. Hydroboration–Oxidation Adds Water Across an Alkene in Two Separate Steps

Step 1, borane (BH₃·THF) adds across the C=C to a trialkylborane; step 2, basic peroxide (H₂O₂, NaOH) swaps boron for OH.

1-Butene (alkene in)

1-Butanol (alcohol out)

2. The OH Ends Up on the Less Substituted Carbon — Anti-Markovnikov Selectivity

The OH lands on the less substituted carbon, so a terminal alkene gives the primary alcohol — propene yields 1-propanol, not 2-propanol.

1-Propanol — anti-Markovnikov (hydroboration)

2-Propanol — Markovnikov (acid hydration)

3. Boron Adds to the Less Hindered Carbon Because of Both Sterics and Electronics

In the concerted four-center transition state two effects agree that boron adds to the less substituted carbon:

Sterics. Bulky boron fits better on the less hindered carbon.
Electronics. Electron-poor boron leaves partial positive charge on the more substituted carbon, which takes the H.

Methylenecyclohexane

Cyclohexylmethanol (primary alcohol)

4. The Addition Is Syn — H and OH Add to the Same Face of the Alkene

Boron and hydrogen are delivered together, so they add to the same face (syn), and oxidation replaces boron with OH with retention — with 1-methylcyclohexene that gives trans-2-methylcyclohexanol.

1-Methylcyclohexene

trans-2-Methylcyclohexanol (syn product)

5. No Carbocation Forms, So the Skeleton Never Rearranges

The concerted addition forms no carbocation, so substrates that rearrange under acid — 3-methyl-1-butene is the classic case — give the clean anti-Markovnikov alcohol with the skeleton intact.

3-Methyl-1-butene → 3-methyl-1-butanol: no hydride shift, no rearranged product.

6. Worked Example: 1-Methylcyclohexene → trans-2-Methylcyclohexanol

Boron adds to C2 (OH anti-Markovnikov) and H to the methyl-bearing C1, and syn addition places methyl and OH trans — trans-2-methylcyclohexanol, whereas acid hydration would put OH on C1.

The full anti-Markovnikov, syn outcome on a trisubstituted ring alkene.

7. Summary

Anti-Markovnikov (OH on the less substituted carbon) · syn (H and OH same face) · oxidation with retention · no carbocation, no rearrangement · the anti-Markovnikov partner to oxymercuration.

Worked example

Problem. Product of propene with 1) BH₃ 2) H₂O₂, NaOH?

Boron adds to the less-substituted carbon, H to the other — anti-Markovnikov.
Addition is syn and concerted: no carbocation, so no rearrangement.
Oxidation swaps C–B for C–OH at the same position, with retention.

Answer. Propan-1-ol — OH on the terminal carbon (anti-Markovnikov), opposite to acid-catalysed hydration.

How each reagent works — the arrow pushing

Electron flow only. Follow the arrows; the structures do the talking.

BH₃; H₂O₂, NaOH — hydroboration–oxidation2 steps

Why it works · Boron has only six valence electrons — it is electron-deficient (δ+), a Lewis acid. B and H add across the alkene in one concerted, syn, 4-center step (no carbocation → no rearrangement). Sterics put the bulky boron on the less-substituted carbon and H on the other, so oxidation (H₂O₂/NaOH, which swaps C–B for C–OH with retention) gives the anti-Markovnikov, syn alcohol.

Quiz yourself

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1-Butanol (the primary alcohol). Boron and therefore OH add to the terminal, less substituted carbon — the anti-Markovnikov position — so the OH is on C1, not C2.

Boron and hydrogen are delivered together in a single concerted four-center transition state, so they must add to the same face of the planar alkene. The oxidation then replaces boron with OH with retention, preserving that syn relationship.

Hydroboration is concerted and forms no carbocation. Rearrangements (hydride/methyl shifts) require a cationic intermediate, so without one the carbon skeleton stays intact and you get 3-methyl-1-butanol cleanly.

They are complementary. Oxymercuration gives the Markovnikov alcohol (OH on the more substituted carbon) with no rearrangement; hydroboration gives the anti-Markovnikov alcohol (OH on the less substituted carbon) with syn stereochemistry. Together they let you choose which carbon bears the OH.

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