No, there’s no free carbocation.

Thanks for this comment. The problem is not regioselectivity but that 1,1-dibora products form, which is why bulky boranes must be used. See JACS 1967 vol. 89 p. 291. https://pubs.acs.org/doi/10.1021/ja00978a022

it will be a ketone

Is the regioselectivity of BH3 any good? Seems like hydroboration via 9-BBN or some other bulky R2BH reagent greatly enhances the anti-Markovnikov product via sterics. While the electronegativity difference is real, you’re hoping for a collision that leads to the π bond attacking boron for the syn transfer of the H to the more substituted side of the alkene (or alkyne)–is there any way something as small as BH3 can reliably select for the anti-M product?

Hydration of a terminal alkyne will first form an enol following Markovnikov’s rule. The enol will then undergo Keto-Enol Tautomerization and rearrange to a methyl ketone.

For H2so4 reactions its always a ketone for Bh3 reaction it can be ketone or aldehyde