Hydroboration of alkynes: practice and mechanism
The bulky (Sia)₂BH adds anti-Markovnikov to the end of the alkyne. Oxidation gives an enol that immediately tautomerises to the aldehyde (with Hg²⁺/H₂O: ketone).
Example
The bulky borane adds anti-Markovnikov: boron goes to the terminal carbon. After oxidation an enol with the OH at the chain end forms, which tautomerises to the aldehyde. HgSO₄/H₂SO₄ would give the methyl ketone instead.
Mechanism step by step
- 1
The bulky disiamylborane adds to the triple bond: B goes to the terminal, less hindered carbon (anti-Markovnikov, syn addition).
- 2
H₂O₂/NaOH replaces the boron by OH: an enol forms.
- 3
The enol tautomerises at once to the keto form; because the OH was on the terminal carbon, an aldehyde forms.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
Often confused with: Hydration of alkynes · Hydroboration–oxidation
Frequently asked questions
Why (Sia)₂BH instead of BH₃?
BH₃ would add twice to the alkyne. The bulky dialkylborane adds only once and strengthens the regioselectivity.
How do I get the methyl ketone instead?
With H₂O, H₂SO₄ and HgSO₄: Markovnikov hydration, the OH goes to the inner carbon.