Reaction of methacrylic acid with BH3

If methacrylic acid (2-methylprop-2-enoic acid) is reacted with first $\ce{BH\_3/THF}$ and second $\ce{H\_2O\_2/HO^{-}}$ then $\ce{OH}$ group is attached at less subtituted carbon. 


But if it is treated with $\ce{CH\_3COOH}$ in the second step, what is the end product?


![methacrylic acid](https://i.stack.imgur.com/12d9g.jpg)




It's simply a protonolysis:
[![enter image description here](https://i.stack.imgur.com/WH6Xd.png)](https://i.stack.imgur.com/WH6Xd.png)


Except in the picture D is used instead of H. Hydroboration based reactions always proceed via this boron-ate complex (negatively charged) then migration and loss of leaving group. Search google for carbonylation, amination, cyanidation using boron.


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The acid would be reduced by the borane to the primary alcohol first. Subsequent hydroboration of the product allylic alcohol would probably give you the diol.


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