What is hydroboration oxidation in alkynes?

Question

What is hydroboration oxidation in alkynes?

1 Answer

Impact of this question

3970 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-18T00:55:46

It's similar to for alkenes, but besides creating ${[B {(O H)}_{4}]}_{4}^{-}$ $[B(OH)_4]^(-)$ , $H_{2} O$ $H_2O$ , and $O O H^{-}$ $OOH^(-)$ (hydrogen peroxide's conjugate base), instead of getting the three $m o l s$ $mols$ 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 $M^{+} O H^{-}$ $M^(+)OH^(-)$ available within the reaction vessel.

An example of this is:

First, the enol's $R - O H$ $R-OH$ donates its proton to the base ( $O H^{-}$ $OH^(-)$ from $M^{+} O H^{-}$ $M^(+)OH^(-)$ ) to form an enolate. ( $H_{2} O$ $H_2O$ forms, now)
Then, the enolate's oxygen moves its electrons down to form a $π$ $pi$ bond (double bond = 1 $σ$ $sigma$ + 1 $π$ $pi$ ) and the $π$ $pi$ bond down at the bottom donates its pi electrons to the resultant $H_{2} O$ $H_2O$ that just formed, grabbing a proton off and reforming the base ( $O H^{-}$ $OH^(-)$ from $M^{+} O H^{-}$ $M^(+)OH^(-)$ ).

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

Science

Math

Humanities

... and beyond

What is hydroboration oxidation in alkynes?

1 Answer

Related questions

Impact of this question