8.5 Hydration of Alkenes: Addition of H2O by Hydroboration

Hydroboration-Oxidation Reaction

Hydroboration-oxidation converts alkenes into alcohols through a two-step sequence. What makes it stand out from other hydration methods is its anti-Markovnikov regiochemistry and syn stereochemistry, giving you precise control over where the $-OH$ group ends up and how it's oriented in 3D space.

Reaction Sequence

The overall transformation requires two distinct steps:

Step 1: Hydroboration Borane ($BH_3$) adds across the alkene double bond, forming a trialkylborane intermediate. The boron atom attaches to the less substituted carbon because it's less sterically crowded. Since $BH_3$ has three B–H bonds, one molecule of borane can react with up to three equivalents of alkene. In practice, $BH_3$ is often used as a complex with tetrahydrofuran ($BH_3 \cdot THF$) for easier handling.

Step 2: Oxidation The trialkylborane is then treated with hydrogen peroxide ($H_2O_2$) in aqueous base ($NaOH$). This replaces each carbon–boron bond with a carbon–oxygen bond, yielding the alcohol. The oxygen ends up on the same carbon that was bonded to boron.

Regiochemistry and Stereochemistry

These two features are what make hydroboration-oxidation distinct from acid-catalyzed hydration and oxymercuration-reduction.

Regiochemistry: Anti-Markovnikov

The $-OH$ group ends up on the less substituted carbon of the original double bond. This is the opposite of Markovnikov's rule. The reason traces back to the hydroboration step: boron, being a relatively large atom, preferentially bonds to the less hindered carbon of the alkene.

• Acid-catalyzed hydration → Markovnikov ($-OH$ on more substituted carbon)
• Oxymercuration-reduction → Markovnikov
• Hydroboration-oxidation → anti-Markovnikov

If an exam question asks you to place an $-OH$ on the less substituted carbon of an alkene, hydroboration-oxidation is your go-to reaction.

Stereochemistry: Syn Addition

Both the boron and the hydrogen add to the same face of the double bond during the hydroboration step. Because oxidation replaces boron with $-OH$ while retaining the configuration at that carbon, the overall result is syn addition of $H$ and $-OH$.

• Contrast this with halogenation ($Br_2$), which proceeds through anti addition.
• Contrast this with acid-catalyzed hydration, which is not stereospecific (the carbocation intermediate allows rotation).

Predicting Products

To figure out the product of a hydroboration-oxidation, follow these steps:

1. Identify the double bond and determine which carbon is less substituted.
2. Place the $-OH$ on the less substituted carbon (anti-Markovnikov).
3. Assign stereochemistry using syn addition. The $H$ and $-OH$ both add from the same face of the double bond.

Here's how this plays out with different alkene types:

• Monosubstituted (e.g., 1-butene): $-OH$ goes on the terminal carbon, giving a primary alcohol (1-butanol).
• Disubstituted, unsymmetrical: $-OH$ goes on the less substituted carbon. If both carbons have the same degree of substitution, steric effects from nearby groups determine the major product.
• Trisubstituted: $-OH$ goes on the one carbon that bears fewer substituents.
• Cyclic alkenes: Same rules apply. For cyclohexene, $-OH$ and $H$ add syn to the ring, which matters when substituents are already present on the ring.

For stereochemical outcomes with defined alkene geometry:

• An (E)-alkene produces the product where $-OH$ and the groups that were trans end up on the same face (the overall relationship can appear "anti" between the original substituents).
• A (Z)-alkene produces the product where $-OH$ and those groups end up with a syn relationship.

The key point: the addition itself is always syn. The apparent relationship between substituents in the product depends on the starting alkene geometry.

Mechanism and Intermediates

Hydroboration Step:

1. $BH_3$ approaches the alkene. Borane is electron-deficient (empty p orbital on boron), so it acts as the electrophile. The alkene's $\pi$ electrons act as the nucleophile.
2. A four-centered transition state forms, in which the C–C $\pi$ bond, the B–H bond, and the new C–B and C–H bonds are all partially formed or broken simultaneously.
3. Both the B and H deliver to the same face of the double bond in a single concerted step. No carbocation intermediate forms, which is why there are no rearrangements.

Oxidation Step:

1. Hydroxide ion ($OH^-$) attacks the electrophilic boron atom of the trialkylborane.
2. An alkyl group migrates from boron to the adjacent oxygen of the peroxide ($-OOH$), with loss of hydroxide. This migration happens with retention of configuration at carbon.
3. The process repeats until all three alkyl groups have migrated.
4. Hydrolysis of the resulting borate ester releases three equivalents of the alcohol product.