What is hydroboration oxidation in alkynes?

1 Answer
Truong-Son N.
Jul 18, 2015

It's similar to for alkenes, but besides creating #[B(OH)_4]^(-)#, #H_2O#, 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 #R-OH# donates its proton to the base (#OH^(-)# from #M^(+)OH^(-)#) to form an enolate. (#H_2O# forms, now)
  • Then, the enolate's oxygen moves its electrons down to form a #pi# bond (double bond = 1 #sigma# + 1 #pi#) and the #pi# bond down at the bottom donates its pi electrons to the resultant #H_2O# 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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