Organic Chemistry · Alkenes: Reactions and Synthesis
Hydration of Alkenes: Addition of H2O by Hydroboration
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In 30 seconds
hydroboration Addition of a B–H bond across a double bond Full entry →–oxidation is a two-step sequence that hydrates an alkene with the opposite regiochemistry from oxymercuration. First, borane (BH3, usually as a tetrahydrofuran complex) adds across the double bond to give a trialkylborane A molecule with three carbon–boron bonds (R3B) Full entry →. Second, oxidation with alkaline hydrogen peroxide (H2O2 / NaOH) replaces the carbon–boron bond with a carbon–oxygen bond. The net result is the anti-Markovnikov addition of water:
alkene + H2O 1. BH3 · THF; 2. H2O2, NaOH⟶ alcohol
The hydroxyl group lands on the less substituted carbon of the double bond. In a terminal alkene, that means the –OH goes to the terminal carbon and a primary alcohol forms. The reaction is also stereospecific in the syn sense: the hydrogen and the hydroxyl add to the same face of the alkene, and no carbocation is involved, so rearrangements never occur.
Why this matters
Hydroboration–oxidation is the standard laboratory route to primary alcohols from terminal alkenes — 1-hexene to 1-hexanol, for example — a transformation that is essentially impossible with Markovnikov methods. Together with oxymercuration–demercuration, it gives you complete regiochemical control over hydration: Markovnikov when you want the –OH on the more substituted carbon, anti-Markovnikov when you want it on the less substituted carbon. The syn stereochemistry also matters in synthesis: cyclic alkenes give cis products, and because the reaction proceeds with retention of configuration A migrating group keeps its 3D arrangement Full entry → at carbon, it is a powerful tool for building specific stereocenters. Industrially, hydroboration chemistry underlies routes to detergents and fatty alcohols from α-olefins.
The college version
Core Concepts
Step 1: Hydroboration (the mechanism)
Borane is an electrophile: boron is electron-poor (it has only six valence electrons) and bears hydrogens that behave as hydride-like. The alkene's π electrons interact with boron in a four-centered transition state, and the B–H unit adds across the double bond in one step — syn addition Both new groups attach from the same face of the alkene Full entry →. The regiochemistry is governed by sterics: boron, being the larger atom, adds to the less hindered (less substituted) carbon, while hydrogen adds to the more substituted carbon:
RCH=CH2 + BH3 ⟶ RCH2CH2BH2
Because each B–H bond can add to an alkene, the sequence continues until a trialkylborane (R3B) forms — three alkenes per borane. The addition is syn and completely regiospecific, and no carbocation forms, so the carbon skeleton never rearranges.
Step 2: Oxidation
Alkaline hydrogen peroxide oxidizes the trialkylborane. The alkyl groups migrate from boron to oxygen with retention of configuration, and hydrolysis then releases the alcohol:
R3B H2O2, NaOH⟶ 3 ROH
The carbon that was attached to boron becomes the carbon bearing the –OH. Because the alkyl group migrates with retention, the stereochemistry set in step 1 is preserved — a feature synthetic chemists rely on.
Net regiochemistry and stereochemistry
For an unsymmetrical alkene, H adds to the more substituted carbon and B (later –OH) to the less substituted carbon. The textbook summary: "boron goes to the less substituted carbon." Since both H and B add from the same face, and B becomes OH without changing its attachment, the overall hydration is syn: H and OH on the same face. For cycloalkenes, this means cis products — for example, 1-methylcyclopentene gives cis-2-methylcyclopentanol.
Practical notes
Borane is used as BH3 · THF or BH3 · SMe2 because free borane is unstable and dangerously pyrophoric — a general laboratory-safety point: handle borane solutions in an inert atmosphere, keep them away from air and moisture, and never let the solution dry out. The oxidation step uses dilute alkaline peroxide, which is exothermic and should be added with cooling and stirring.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| Hydroboration–oxidation (anti-Markovnikov) | Oxymercuration–demercuration (Markovnikov) | The –OH goes to the less substituted carbon in hydroboration; to the more substituted carbon in oxymercuration |
| Syn addition (hydroboration) | Anti addition (halogenation, halohydrins) | Hydroboration adds H and OH to the same face; halogens add to opposite faces |
| Borane as electrophile | Borohydride as nucleophile | BH3 is electron-poor and accepts π electrons; NaBH4 is a hydride donor used in reductions |
| C–B bond → C–OH | C–B bond → C–H | Oxidation with H2O2/NaOH gives the alcohol; other reagents (e.g., acetic acid) can give the alkane instead |
| "Boron goes to the less substituted carbon" | "The –OH goes to the less substituted carbon" | Both are true — boron ends up where the OH will be after oxidation |
| Anti-Markovnikov product = primary alcohol | Anti-Markovnikov always means primary | True for terminal alkenes; internal alkenes give secondary alcohols with the OH on the less substituted internal carbon |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Think of the double bond as two friends holding hands. A boron atom with three hydrogen "helpers" walks up and grabs the friend with fewer friends (the less crowded carbon), while one of its hydrogen helpers holds the other friend's hand — both from the same side. Then peroxide comes along and swaps the boron for an oxygen "hat" without anyone moving their seats. Net result: the –OH hat ends up on the less crowded carbon, on the same side as the hydrogen — the opposite of what Markovnikov's rule would predict.
Worked example
Example 1: 1-Hexene to 1-hexanol
Predict the product of hydroboration–oxidation of 1-hexene (SMILES: C=CCCCC).
Reasoning. The double bond is terminal: C1 is the less substituted (less hindered) carbon, C2 the more substituted. Boron adds to C1, hydrogen to C2 (syn). Oxidation then replaces the C1–B bond with C1–OH.
Answer. The product is 1-hexanol (SMILES: CCCCCCO) — a primary alcohol. The same alkene under oxymercuration–demercuration would give the secondary alcohol 2-hexanol, which shows why the two methods are complementary.
Example 2: Stereochemistry on a ring
Predict the product of hydroboration–oxidation of 1-methylcyclopentene (SMILES: CC1=CCCC1).
Reasoning. The more substituted alkene carbon is C1 (it bears the methyl). Hydrogen adds to C1; boron adds to C2. Both add from the same face (syn). Oxidation replaces the C2–B bond with C2–OH, retaining the geometry set in step 1. Since H (at C1) and B (at C2) arrived on the same face, the methyl (at C1) and the –OH (at C2) end up cis to each other.
Answer. The product is cis-2-methylcyclopentanol (SMILES: C[C@H]1CCC[C@@H]1O, cis relationship). The syn addition is directly visible in this cyclic product.
Example 3: Stoichiometry with dimensional analysis
How many moles of BH3 are required to hydroborate 0.450 mol of 1-hexene completely?
Formula first. One borane delivers three B–H bonds, and each B–H bond adds to one alkene. Therefore:
n(BH3) = n(alkene)3
Substitution.
n(BH3) = 0.450 mol3 = 0.150 mol
Answer. 0.150 mol of BH3 (equivalently, 0.450 mol of B–H bonds) is needed. In the lab you would use a slight excess of borane to ensure complete reaction.
Key takeaways
- Hydroboration–oxidation hydrates alkenes anti-Markovnikov: the –OH goes to the less substituted carbon.
- Reagents: BH3 · THF, then H2O2/NaOH. No carbocation, so no rearrangements.
- Addition is syn: H and –OH land on the same face; cyclic alkenes give cis alcohols.
- Boron adds to the less hindered carbon; the alkyl group migrates to oxygen with retention of configuration.
- One BH3 hydroborates three alkenes (trialkylborane) — the stoichiometry is 1:3.
- Terminal alkenes give primary alcohols — the key synthetic payoff.
- Contrast: oxymercuration–demercuration (Markovnikov, anti addition, no rearrangement) vs. hydroboration–oxidation (anti-Markovnikov, syn addition, no rearrangement).
- General lab-safety principle: borane solutions are pyrophoric; work under inert gas and add alkaline peroxide slowly with cooling.
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
What two steps make up hydroboration–oxidation, and what does each accomplish?
Show answer
(1) Hydroboration: BH3·THF adds B and H across the double bond, forming a trialkylborane (syn, anti-Markovnikov). (2) Oxidation: alkaline H2O2 replaces each C–B bond with C–OH, releasing the alcohol.
Where does the –OH end up relative to the alkene's substitution pattern?
Show answer
The –OH goes to the less substituted carbon (anti-Markovnikov); terminal alkenes give primary alcohols.
Is the addition syn or anti, and what does that mean for cyclic alkenes?
Show answer
Syn addition. Cyclic alkenes give cis products (e.g., cis-2-methylcyclopentanol from 1-methylcyclopentene).
Why does hydroboration–oxidation never show rearrangements?
Show answer
The reaction proceeds through a four-centered transition state and an alkylborane — never a free carbocation — so hydride and methyl shifts cannot occur.
How many moles of BH3 are needed for 0.900 mol of a terminal alkene?
Show answer
0.900 mol / 3 = 0.300 mol of BH3.
What is the product of hydroboration–oxidation of 2-methyl-2-butene, and is it primary, secondary, or tertiary?
Show answer
2-Methyl-2-butene is an internal alkene; H adds to C2 and B (later –OH) to C3, giving 3-methyl-2-butanol, a secondary alcohol (SMILES: CC(O)C(C)C).
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- hydroboration
- Addition of a B–H bond across a double bond
- oxidation (of the borane)
- Replacing the C–B bond with a C–OH bond using H2O2/NaOH
- anti-Markovnikov addition
- The –OH ends up on the less substituted carbon
- syn addition
- Both new groups attach from the same face of the alkene
- trialkylborane
- A molecule with three carbon–boron bonds (R3B)
- retention of configuration
- A migrating group keeps its 3D arrangement
Sources & references
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