What is hydroboration oxidation in alkynes?

Question

Santiago Alfonso · Accepted Answer

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.

Ella Allen · Answer

undefined Hydroboration-oxidation in alkynes is a chemical reaction used to convert alkynes (carbon-carbon triple bonds) into aldehydes or ketones. In this reaction, the alkyne reacts with borane (BH3) in the presence of a suitable solvent, such as tetrahydrofuran (THF), to form a boron-containing intermediate called an alkylborane. This intermediate is then oxidized with hydrogen peroxide (H2O2) and a base, such as sodium hydroxide (NaOH) or potassium hydroxide (KOH), to yield the corresponding aldehyde or ketone. The hydroboration-oxidation reaction proceeds through anti-Markovnikov addition, meaning that the boron atom adds to the less substituted carbon atom of the triple bond, leading to the formation of a less substituted alkylborane intermediate.