Hydration via Hydroboration-Oxidation

Key Questions

  • Answer:

    Hydroboration-oxidation is a method of making alcohols from alkenes.

    Explanation:

    It involves the addition of BH3 to an alkene, followed by oxidation with alkaline hydrogen peroxide to form an alcohol.

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    The reaction is a Markovnikov addition of BH3 to the alkene

    On oxidation of the boron intermediate, the OH group ends up on the less substituted carbon.

    This is opposite to the position of the OH group in the acid-catalyzed Markovnikov addition of water to an alkene, so the reaction is often called the anti-Markovnikov addition of water to an alkene.

  • Answer:

    The borane-THF complex (BTHF) is used for hydroboration for reasons of safety and convenience.

    Explanation:

    The active ingredient is borane, BH3, but borane is a highly toxic gas.

    Borane exists naturally as the dimer B2H6 (diborane), but diborane mixes easily with air and forms explosive mixtures.

    Also, it ignites spontaneously in moist air at room temperature.

    In a solution in THF, borane exists as a loose Lewis acid-base complex. This allows boron to have an octet and makes the reagent more stable.

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    The solution is commercially available in a 1 mol/L concentration in volumes from 25 to 800 mL.

    It is much more convenient to work with the solution than with a gas. Even so, the solution must be stored at 2 to 8 °C, and it must have a stabilizer added.

    Borane forms a more stable and more soluble Lewis acid-base complex with dimethyl sulfide:

    H3-B-+S(CH3)2

    It is available in concentrations of 2, 5, and 10 mol/L and in volumes from 25 mL to 18 L.

    That should make it a more convenient reagent than the BTHF complex.

    There is only one problem: It has the highly disagreeable smell of rotten cabbage!

  • It's similar to for alkenes, but besides creating [B(OH)4], H2O, and OOH (hydrogen peroxide's conjugate base), instead of getting the three mols of an alcohol, you have an enol. This enol can undergo keto-enol tautomerization.

    In this case it is in basic (base-ic) conditions, with M+OH available within the reaction vessel.

    An example of this is:

    • First, the enol's ROH donates its proton to the base (OH from M+OH) to form an enolate. (H2O forms, now)
    • Then, the enolate's oxygen moves its electrons down to form a π bond (double bond = 1 σ + 1 π) and the π bond down at the bottom donates its pi electrons to the resultant H2O that just formed, grabbing a proton off and reforming the base (OH from M+OH).

    You then form a ketone or aldehyde, depending on the location of the triple bond. Also, remember that hydroboration is anti-Markovnikov.

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