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.
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.
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.
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.
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
- 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.
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.
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.