Hydroboration of alkynes using R2BH or BH3 to give aldehydes
Description: When alkynes are treated with BH3 and subsequently treated with H2O2, they are converted into aldehydes.
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Could this reaction also occur with reagents; 1.) BH3 2.) H2O2, OH, H2O?
In many textbooks, yes – in practice, not so much.
With BH3, hydroboration happens twice. You still get the aldehyde after oxidation, but because BH3 is so small, the selectivity isn’t as good for giving the aldehyde vs the ketone.
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.
” 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!
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!