Hydroboration of alkynes using BH3 to give aldehydes

by James

Description: When alkynes are treated with BH3 and subsequently treated with H2O2, they are converted into aldehydes.

Notes: Note that only terminal alkynes (i.e. alkynes with an H on one end) are converted into aldehydes. The reaction is anti-Markovnikoff selective. The addition occurs syn but the stereochemistry is lost at the end.


Notes: Note that in example 3 a ketone is formed since this is not a terminal alkyne. And in example 4, both sides of the alkyne are equally substituted, so according to Markovnikoff’s rule neither of the carbons are favored.

Mechanism: This is a loooong mechanism. The first step is hydroboration, where boron adds to the less substituted carbon (anti-Markovnikoff) (Step 1, arrows A and B). When hydroboration is complete, addition of NaOH and H2O2 begins the oxidation step, which starts with deprotonation of hydrogen peroxide (Step 2, arrows C and D). The deprotonated hydrogen peroxide is a better nucleophile than HOOH, and it now attacks the boron to give a charged boron species (Step 3, arrow E).  Then comes the weird step! The carbon-boron bond breaks, and the electrons from this bond migrate to the oxygen, which breaks the weak oxygen-oxygen bond (Step 4, arrows F and G). Then, cleavage of O–BH2 (the “boronic ester”) with NaOH (Step 5, arrows H and I) gives the free oxygen (an “enolate”) which is protonated (Step 6, arrows J and K) to give the enol. Finally (second weird step!) the enol rearranges into its more stable constitutional isomer, the aldehyde. This is called “tautomerization” (Step 7, arrows L, M, N, O. P).

Notes: It’s probably not necessary to actually show Step 2 if you just show HOO(-) attacking boron in Step 3. It’s also reasonable to switch the order of the protonation and tautomerization (i.e. do step 7 before step 6) or to show an acid other than water in Step 6.


{ 5 comments… read them below or add one }


This is SOO helpful! Thank you. I’m only in OChem 1 and often they tell us not to worry about knowing these longer mechanisms, but just seeing them and having them explained once or twice helps me understand everything else!
Thanks again for breaking it down into bite-sized pieces!



” Then comes the weird step! The carbon-boron bond breaks, and the examples from this bond migrate to the oxygen”
Do you mean to say examples here or do you mean something else (electrons?)



Oops. Yes, thank you!


Payam Zohdi

Can example 3 also form two products equally like example 4 because the double bond being on the adjacent carbon would still result it being on a carbon that is equally substituted correct? I am having a hard time seeing the difference between example 4 and 3 both look like the two options are both on secondary carbons



In example 3, no matter what side of the alkyne is added to, the product is still 2-butanone. remember that there is no such thing as 3-butanone since we always number the molecule in such a way as to keep the number the lowest.


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